[
{
"id": 20001,
"subject": "Mathematics (Olympiad)",
"question": "Let $M$ be a subset of $\\{1, 2, \\dots, 2006\\}$ such that the sum of any two distinct elements in $M$ is greater than the largest element in $M$. What is the largest possible size of $M$?",
"options": [],
"answer": "See solution",
"solution": "The answer is $1004$.\n\nWhen $M = \\{1003, 1004, \\dots, 2006\\}$, the sum of any two elements in $M$ is at least $2007$, which is larger than the largest element in $M$. Therefore, $x + y > z$ for any $x, y, z \\in M$. This gives a possible case for $|M| = 1004$.\n\nNext, consider any subset $M$ satisfying the constraint. Suppose the largest element of $M$ is $a$. Then $M$ contains at most one element in each of the pairs\n\n$$\n(1, a-1), (2, a-2), \\dots, \\left( \\left\\lfloor \\frac{a}{2} \\right\\rfloor, \\left\\lceil \\frac{a}{2} \\right\\rceil \\right)\n$$\n\n(note that the last pair may contain the same number, which does not affect the validity of our claim). Therefore, $M$ contains at most $\\frac{a}{2}$ elements smaller than $a$. Since $a$ is the largest element, $M$ cannot contain elements larger than $a$. Therefore,\n\n$$\n|M| \\leq \\frac{a}{2} + 1 \\leq 1003 + 1 = 1004.\n$$\n\nHence, the largest possible size of $M$ is $1004$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20002,
"subject": "Mathematics (Olympiad)",
"question": "Find all triples $(a, b, c)$ of real numbers such that\n\n$$\n\\cos(ax) + \\cos(bx) = 2 \\cos(cx)\n$$\n\nholds for all $x \\in \\mathbb{R}$.",
"options": [],
"answer": "See solution",
"solution": "The triples we are looking for have the forms $(t, t, t)$, $(-t, t, t)$, $(t, -t, t)$, $(t, t, -t)$ where $t \\in \\mathbb{R}$.\n\nIf $c = 0$, then $\\cos(ax) + \\cos(bx) = 2$ for all $x \\in \\mathbb{R}$. Since $\\cos(ax) \\leq 1$ and $\\cos(bx) \\leq 1$, we must have $\\cos(ax) = 1$ and $\\cos(bx) = 1$ for any $x$. Thus $a = b = 0$, which clearly works.\n\nNow, suppose that $c \\neq 0$. Plugging in $x = \\frac{2\\pi}{c}$, we find $\\cos\\left(2\\pi \\cdot \\frac{a}{c}\\right) + \\cos\\left(2\\pi \\cdot \\frac{b}{c}\\right) = 2$. Therefore $\\cos\\left(2\\pi \\cdot \\frac{a}{c}\\right) = 1$ and $\\cos\\left(2\\pi \\cdot \\frac{b}{c}\\right) = 1$. This means that $2\\pi \\frac{a}{c} = 2k\\pi$ and $2\\pi \\frac{b}{c} = 2l\\pi$ for some integers $k$ and $l$, i.e., $a = ck$ and $b = cl$.\n\nNote that $k \\neq 0$. Indeed, otherwise the left-hand side is always $\\geq 0$, but the right-hand side can take negative values. For a similar reason, $l \\neq 0$. We shall prove that $|k| = |l| = 1$. Suppose otherwise. Assume without loss of generality that $|k| \\geq |l|$. Plug in $x = \\frac{\\pi}{a}$. We obtain $-1 + \\cos\\left(\\frac{l}{k}\\pi\\right) = 2\\cos\\left(\\frac{1}{k}\\pi\\right)$. Note that the left-hand side $< -1 + 1 = 0$. Moreover, if $|k| \\geq 2$ then the right-hand side is nonnegative, yielding a contradiction. Therefore we have $|k| = 1$. This leads to $|l| = 1$. This means $|a| = |b| = |c|$. Clearly, such triples work. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20003,
"subject": "Mathematics (Olympiad)",
"question": "Let the radius of the circumcircle of triangle $AXY$ be $\\sqrt{2014}$.\n\nIn triangle $AXY$, the angle opposite side $XY$ is $60^\\circ$ because triangle $ABC$ is equilateral.\n\nIf $R$ is the radius of the circumcircle of triangle $AXY$, then\n\n$$\n|XY| = 2R \\sin 60^\\circ = \\sqrt{2014} \\cdot \\sqrt{3}.\n$$\n\nLet $|AX| = m$, $|AY| = n$, where $m$ and $n$ are positive integers.\n\n\n\nApply the cosine rule to triangle $AXY$:\n\n$$\nm^2 + n^2 - 2mn \\cos 60^\\circ = |XY|^2,\n$$\nwhich leads to\n$$\nm^2 + n^2 - mn = 2014 \\cdot 3.\n$$\n\nIs it possible for $R = \\sqrt{2014}$?",
"options": [],
"answer": "See solution",
"solution": "Notice that $2014 \\cdot 3$ is even. If $m$ and $n$ have different parity or are both odd, then $m^2 + n^2 - mn$ is odd, which is impossible.\n\nIf $m$ and $n$ are both even, then $m^2 + n^2 - mn$ is divisible by $4$, but $2014 \\cdot 3$ is not, so this is also impossible.\n\nTherefore, the initial assumption is wrong, and $R$ cannot be equal to $\\sqrt{2014}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20004,
"subject": "Mathematics (Olympiad)",
"question": "A polynomial $p(x)$ with real coefficients is called a square if and only if it is not a constant and there exists a polynomial $q(x)$ with real coefficients such that $p(x) = q(x)^2$.\n\nSuppose that $f(x)$ and $g(x)$ are non-constant polynomials with real coefficients such that neither of them is a square, but $f(g(x))$ is. Show that $g(f(x))$ is not a square.",
"options": [],
"answer": "See solution",
"solution": "We can easily extend the definition of a square polynomial to polynomials with complex coefficients. In all the arguments below, we consider polynomials with complex coefficients.\n\n**Lemma:** If $p(x)$ is a square and $\\alpha$ is a nonzero complex number, then $p(x) - \\alpha$ is not a square.\n\n**Proof of Lemma:** Suppose $p(x) = q(x)^2$ and $p(x) - \\alpha = r(x)^2$, with both $q(x)$ and $r(x)$ being non-constant polynomials. Then $\\alpha = (q(x) - r(x))(q(x) + r(x))$. Clearly, either $q(x) - r(x)$ or $q(x) + r(x)$ is not a constant polynomial, and hence a contradiction. This proves the lemma.\n\n**Continuation of the solution:**\n\nWe can write $f(x)$ as $f_1(x)^2(x - \\alpha_1)(x - \\alpha_2) \\cdots (x - \\alpha_k)$, where $f_1(x)$ is a polynomial and $\\alpha_1, \\alpha_2, \\ldots, \\alpha_k$ are distinct complex numbers. Then\n$$\nf(g(x)) = f_1(g(x))^2 (g(x) - \\alpha_1) \\cdots (g(x) - \\alpha_k)\n$$\nis a square. It follows that $(g(x) - \\alpha_1)(g(x) - \\alpha_2) \\cdots (g(x) - \\alpha_k) = h(x)^2$ for some polynomial $h(x)$.\n\nLet $\\beta$ be such that $g(\\beta) = \\alpha_1$. Then $h(\\beta) = 0$ and hence $h(x) = (x - \\beta)h_1(x)$. Note that $g(\\beta) - \\alpha_i = 0$ for $i = 1$. Therefore, it follows that $(x - \\beta)^2$ divides $g(x) - \\alpha_1$. We can then conclude that $g(x) - \\alpha_1$ is a square. Similarly, $g(x) - \\alpha_i$ is a square for $i = 1, 2, \\ldots, k$. By the above lemma, it follows that $k = 1$, so $f(x) = f_1(x)^2(x - \\alpha)$ and $g(x) = g_1(x)^2 + \\alpha$ for some nonzero complex number $\\alpha$.\n\nTherefore, $g(f(x)) = g_1(f(x))^2 + \\alpha$ and hence by the above lemma it follows that $g(f(x))$ is not a square. This completes the proof.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20005,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for any real $x, y$, the following equality holds:\n\n$$\nf(x + y f(x + y)) = f(y^2) + x f(y) + f(x).\n$$",
"options": [],
"answer": "See solution",
"solution": "Let us analyze the given functional equation:\n\n$$\nf(x + y f(x + y)) = f(y^2) + x f(y) + f(x). \\tag{1}\n$$\n\n**Step 1: Substitute $x = y = 0$**\n\n$$\nf(0) = 2f(0) \\implies f(0) = 0.\n$$\n\n**Step 2: Substitute $x = 0$**\n\n$$\nf(y f(y)) = f(y^2). \\tag{2}\n$$\n\n**Step 3: Substitute $y = -x$**\n\n$$\nf(x) = f(x^2) + x f(-x) + f(x) \\implies f(x^2) = -x f(-x). \\tag{3}\n$$\n\n**Step 4: Substitute $x = -x$ in (3)**\n\n$$\nf(x^2) = x f(x) \\implies -f(x) = f(-x). \\tag{4}\n$$\n\n**Step 5: Substitute $y = -y$ in (1)**\n\n$$\nf(x - y f(x - y)) = f(y^2) + x f(-y) + f(x).\n$$\n\nBy symmetry, we get:\n\n$$\nf(x + y f(y - x)) = f(y^2) - x f(y) + f(x). \\tag{5}\n$$\n\n**Step 6: Subtract (5) from (1):**\n\n$$\nf(x + y f(x + y)) - f(x + y f(y - x)) = 2x f(y). \\tag{6}\n$$\n\n**Step 7: Zero solution**\n\nSuppose $f(x_0) = 0$ for some $x_0 \\neq 0$. Substitute $y = x_0 - x$ in (1):\n\n$$\nf(x) = f((x_0 - x)^2) + x f(x_0 - x) + f(x) \\implies f((x_0 - x)^2) = -x f(x_0 - x).\n$$\n\nBut from (3):\n\n$$\nf((x_0 - x)^2) = -(x_0 - x) f(x - x_0) = (x_0 - x) f(x_0 - x).\n$$\n\nEquating:\n\n$$\n-x f(x_0 - x) = (x_0 - x) f(x_0 - x) \\implies f(x_0 - x) = 0, \\forall x \\in \\mathbb{R}.\n$$\n\nThus, $f(x) = 0$ for all $x$ is a solution.\n\n**Step 8: Injectivity**\n\nSuppose $f(x_1) = f(x_2)$ for $x_1 \\neq x_2$. Substitute $x = \\frac{x_1 - x_2}{2}$, $y = \\frac{x_1 + x_2}{2}$ in (6):\n\n$$\nf\\left(\\frac{x_1 - x_2}{2} + \\frac{x_1 + x_2}{2} f(x_1)\\right) - f\\left(\\frac{x_1 - x_2}{2} + \\frac{x_1 + x_2}{2} f(x_2)\\right) = (x_1 - x_2) f\\left(\\frac{x_1 + x_2}{2}\\right).\n$$\n\nSince $f(x_1) = f(x_2)$, the left side is zero, so $f\\left(\\frac{x_1 + x_2}{2}\\right) = 0$. But then $x_1 + x_2 = 0$, so $x_1 = -x_2$.\n\nFrom (4):\n\n$$\nf(x_2) = f(-x_1) = -f(x_1) = -f(x_2) \\implies f(x_2) = 0 \\implies x_2 = 0 \\implies x_1 = 0.\n$$\n\nContradiction unless $x_1 = x_2 = 0$. Thus, $f$ is injective.\n\n**Step 9: Find all solutions**\n\nFrom (2) and injectivity:\n\n$$\nf(y f(y)) = f(y^2) \\implies y f(y) = y^2 \\implies f(y) = y.\n$$\n\nThus, the only solutions are:\n\n$$\nf(x) = 0 \\quad \\text{and} \\quad f(x) = x.\n$$\n\nBoth satisfy the original equation.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20006,
"subject": "Mathematics (Olympiad)",
"question": "Let $P$ be an internal point of triangle $ABC$. The line through $P$ parallel to $AB$ meets $BC$ at $L$, the line through $P$ parallel to $BC$ meets $CA$ at $M$, and the line through $P$ parallel to $CA$ meets $AB$ at $N$. Prove that\n\n$$\n\\frac{BL}{LC} \\cdot \\frac{CM}{MA} \\cdot \\frac{AN}{NB} \\le \\frac{1}{8}\n$$\n\nand locate the position of $P$ in triangle $ABC$ when equality holds.",
"options": [],
"answer": "See solution",
"solution": "We denote the area of the triangle $EGH$ by $S(EGH)$. Let $AP$, $BP$ and $CP$ cut $AB$, $BC$ and $CA$ at $X$, $Y$ and $Z$ respectively.\n\n\n\nWe have\n\n$$\n\\begin{aligned}\n\\frac{BL}{LC} &= \\frac{ZP}{PC} && (\\text{as } PLC \\sim ZBC) \\\\\n&= \\frac{S(BPZ)}{S(BPC)} && (\\text{triangles with same height}) \\\\\n&= \\frac{S(APZ)}{S(APC)}\n\\end{aligned}\n$$\n\nand therefore\n\n$$\n\\frac{BL}{LC} = \\frac{S(BPZ) + S(APZ)}{S(BPC) + S(APC)} = \\frac{S(APB)}{S(BPC) + S(CPA)}\n$$\n\nSimilarly,\n\n$$\n\\begin{aligned}\n\\frac{CM}{MA} &= \\frac{S(BPC)}{S(CPA) + S(APB)} \\\\\n\\frac{AN}{NB} &= \\frac{S(CPA)}{S(APB) + S(BPC)}\n\\end{aligned}\n$$\n\nLet $S(BPC) = x$, $S(CPA) = y$ and $S(APB) = z$. Then we are required to show that\n\n$$\n\\frac{z}{x+y} \\times \\frac{x}{y+z} \\times \\frac{y}{z+x} \\le \\frac{1}{8}\n$$\n\nBy the AM-GM inequality,\n\n$$\nx + y \\ge 2\\sqrt{xy}\n$$\n\n$$\ny + z \\ge 2\\sqrt{yz}\n$$\n\n$$\nz + x \\ge 2\\sqrt{zx}\n$$\n\nand these inequalities can be multiplied together to obtain\n\n$$\n(x + y)(y + z)(z + x) \\ge 8xyz.\n$$\n\nThis clearly implies the desired inequality.\n\nEquality holds if and only if $x = y = z$, which occurs if and only if $P$ is the centroid of triangle $ABC$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20007,
"subject": "Mathematics (Olympiad)",
"question": "A slider on the scrolling bar of Paul's mail client shows the proportion of emails which are preceding the email which is currently open. Paul noticed that before deleting some emails the slider was on 10%. Paul deleted some consecutive emails, starting from the one which was currently open. After that, the slider was on 50%. What is the percentage of the emails that Paul deleted?",
"options": [],
"answer": "See solution",
"solution": "Let $n$ be the number of emails before deletion. Since the slider showed 10% before deletion, there were $0.1n$ emails preceding the currently open email. After deletion, these same emails accounted for 50% of the remaining emails, so $0.1n$ is 50% of the new total. Thus, the number of remaining emails is $0.2n$. Therefore, Paul deleted $n - 0.2n = 0.8n$ emails, which is 80% of the original emails.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20008,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. For which $n$ is $n^8 - n^2$ divisible by $72$?",
"options": [],
"answer": "See solution",
"solution": "Suppose $n$ is of the form $n = 4k - 2$ where $k \\in \\mathbb{Z}^+$. Then\n\n$$\nn^8 - n^2 = n^2(n^6 - 1) = 4(2k - 1)^2(n^6 - 1).\n$$\n\nSince both $2k-1$ and $n^6-1$ are odd, this number is not divisible by $8$, and hence not divisible by $72$.\n\nFor any other $n$, $n^8 - n^2$ is divisible by both $8$ and $9$:\n\n- If $n$ is even, $n$ must be a multiple of $4$. In this case $n^8 - n^2$ is divisible by $n^2$, and hence by $8$.\n- If $n$ is odd, it is well-known that $8 \\mid n^2 - 1$. Thus,\n\n$$\nn^8 - n^2 = n^2(n^2 - 1)(n^4 + n^2 + 1)\n$$\n\nis divisible by $8$.\n\nFor divisibility by $9$:\n- If $3 \\mid n$, then $n^8 - n^2$ is divisible by $n^2$, and hence by $9$.\n- If $3 \\nmid n$, then $3 \\mid n^2 - 1$. By the lifting the exponent lemma,\n\n$$\n3^2 \\mid (n^2)^3 - 1^3 = n^6 - 1,\n$$\n\nso $9 \\mid n^2(n^6 - 1) = n^8 - n^2$.\n\nTherefore, $72 \\mid n^8 - n^2$ if and only if $n \\not\\equiv 2 \\pmod{4}$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20009,
"subject": "Mathematics (Olympiad)",
"question": "Let $x, y$ be two positive real numbers and $\\ell, k$ with $\\ell \\ge k$ be natural numbers. Let $x_i, y_i$ for $i = 1, 2, \\dots, \\ell$ satisfy:\n\n$$\nx_i \\ge 0, \\quad y_i \\ge 0, \\quad x_i + y_i \\le \\frac{x+y}{k}, \\quad \\text{for } i = 1, 2, \\dots, \\ell;\n$$\n\n$$\n\\sum_{i=1}^{\\ell} x_i = x, \\quad \\sum_{i=1}^{\\ell} y_i = y.\n$$\n\nProve that\n\n$$\n\\sum_{i=1}^{\\ell} x_i y_i \\le \\frac{xy}{k}.\n$$\n\nEquality holds only if $x_i = x/k, \\ y_i = y/k$ for $i = 1, 2, \\dots, k$ and $x_i = y_i = 0$ for $i > k$.",
"options": [],
"answer": "See solution",
"solution": "Let $f(x, y) := \\sum_{i=1}^{\\ell} x_i y_i$, where $x = (x_1, \\dots, x_\\ell)$, $y = (y_1, \\dots, y_\\ell)$. The conditions determine a compact set, so $f$ attains its maximum at some $(x'_i, y'_i)$. Assume $x'_1 \\ge x'_2 \\ge \\dots \\ge x'_\\ell$. We show $y'_i$ are also in decreasing order. If not, suppose $y'_i < y'_{i+1}$. Set\n\n$$\nx_i = x_{i+1} := \\frac{x'_i + x'_{i+1}}{2}, \\quad y_i = y_{i+1} := \\frac{y'_i + y'_{i+1}}{2}.\n$$\n\nThen (by Chebyshev's inequality):\n\n$$\nx_i y_i + x_{i+1} y_{i+1} > x'_i y'_i + x'_{i+1} y'_{i+1},\n$$\n\ncontradicting maximality. Next, if $x'_1 + y'_1 < (x+y)/k$, we can increase $x'_1, y'_1$ slightly and decrease $x'_2, y'_2$ to get a larger $f$, so $x'_1 + y'_1 = (x+y)/k$.\n\nLet $k'$ be the largest index with $x_{k'} > 0$ and $y_{k'} > 0$. Similarly, $x'_i + y'_i = (x+y)/k$ for $i = 1, \\dots, k'$, so $k' \\le k$. If $k' < k$, we can redistribute to get a larger $f$, contradiction. Thus $k' = k$.\n\nSuppose $x'_j \\ne x/k$ for some $j \\le k$. If $x'_j < x/k$, then for some $i > j$, $x'_i > x/k$, so the sequence is not decreasing, contradiction. Thus $x'_i = x/k$, $y'_i = y/k$ for $i = 1, \\dots, k$.\n\nTherefore, $f(x, y) = k \\cdot (x/k)(y/k) = xy/k$, as required. Equality holds only in this case. $\\Box$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20010,
"subject": "Mathematics (Olympiad)",
"question": "An integer is written in each cell of a $9 \\times 9$ square table. For every $k$ numbers in the same row (or column), their sum is in the same row (or column). Find the smallest possible number of zeros in the table if:\n\na) $k = 5$;\n\nb) $k = 8$.",
"options": [],
"answer": "See solution",
"solution": "a) Example: Number the rows and columns from $1$ to $9$. Write $1$ in the fields $(i, i)$ for $i = 1, \\dots, 9$; $-1$ in field $(1, 9)$ and in fields $(i, i-1)$ for $i = 2, \\dots, 9$; $0$ in other fields. Possible sums are $1$, $0$, and $-1$.\n\nEvaluation: Suppose there are at least $19$ non-zero numbers. By the pigeonhole principle, there will be at least three non-zero numbers in some row, and therefore at least two non-zero numbers with the same sign, say positive (the negative case is analogous). Arrange the numbers in order: $a_1 \\leq a_2 \\leq \\dots \\leq a_9$, where $a_9 \\geq a_8 > 0$. If $a_5 \\geq 0$, then $a_5 + a_6 + a_7 + a_8 + a_9 \\geq a_8 + a_9 > a_9$ must be in the same row, a contradiction. If $a_5 < 0$, then $a_1 + a_2 + a_3 + a_4 + a_5 < a_1$ must be in the same row, a contradiction.\n\nb) A possible example without zeros is as follows (works because $5 \\cdot 3 + 3 \\cdot (-4) = 3$ and $4 \\cdot 3 + 4 \\cdot (-4) = -4$):\n\n\n\n$\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20011,
"subject": "Mathematics (Olympiad)",
"question": "Considere todas las sucesiones de 2004 números reales $$(x_0, x_1, x_2, \\ldots, x_{2003})$$ tales que\n\n$$\n\\begin{array}{l}\nx_0 = 1, \\\\\n0 \\leq x_1 \\leq 2x_0, \\\\\n0 \\leq x_2 \\leq 2x_1, \\\\\n\\vdots \\\\\n0 \\leq x_{2003} \\leq 2x_{2002}.\n\\end{array}\n$$\n\nEntre todas estas sucesiones, determine aquella para la cual la siguiente expresión toma su mayor valor:\n\n$$S = \\pm x_1 \\pm x_2 \\pm \\cdots \\pm x_{2002}$$\n\ndonde el último término, $x_{2002}$, tiene coeficiente $+1$, y los anteriores tienen coeficiente $+1$ o $-1$. Demuestre que, aunque no se conozca exactamente la expresión de $S$, se puede determinar con certeza la sucesión que la maximiza.",
"options": [],
"answer": "See solution",
"solution": "La clave está en que, sin importar la elección de los signos, el término $x_{2002}$ siempre tiene coeficiente $+1$. Para maximizar $S$, se debe maximizar $x_{2002}$, ya que los otros términos pueden sumarse o restarse arbitrariamente. Dado que cada $x_{k}$ está acotado por $0 \\leq x_k \\leq 2x_{k-1}$ y $x_0 = 1$, la sucesión que maximiza $x_{2002}$ es aquella donde $x_k = 2x_{k-1}$ para todo $k$, es decir, $x_k = 2^k$. Así, la sucesión óptima es $x_k = 2^k$ para $k = 0, 1, \\ldots, 2003$, independientemente de los signos en $S$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20012,
"subject": "Mathematics (Olympiad)",
"question": "Solve in positive integers the equation\n\n$$\nm^{\\frac{1}{n}} + n^{\\frac{1}{m}} = 2 + \\frac{2}{mn(m+n)^{\\frac{1}{m}+\\frac{1}{n}}}.\n$$",
"options": [],
"answer": "See solution",
"solution": "There are no solutions in positive integers. Without loss of generality, assume that $m \\ge n$. If $n \\ge 3$, then\n\n$$\n\\left(1 + \\frac{1}{m}\\right)^m = \\sum_{k=0}^{m} \\binom{m}{k} \\frac{1}{m^k} < \\sum_{k=0}^{m} \\frac{1}{k!} < \\sum_{k=0}^{\\infty} \\frac{1}{k!} = e < n \\implies n^{\\frac{1}{m}} > 1 + \\frac{1}{m}\n$$\n\nSimilarly, $m^{\\frac{1}{n}} > 1 + \\frac{1}{n}$, whence\n\n$$\nm^{\\frac{1}{n}} + n^{\\frac{1}{m}} > 2 + \\frac{1}{m} + \\frac{1}{n} \\ge 2 + \\frac{2}{mn} > 2 + \\frac{2}{mn(m+n)^{\\frac{1}{m}+\\frac{1}{n}}}.\n$$\n\nThe cases $n = 1, 2$ can be solved directly.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20013,
"subject": "Mathematics (Olympiad)",
"question": "Find the greatest real number $a$ such that for every positive real numbers $x, y, z$ we have\n\n$$\n\\frac{x+1}{y} + \\frac{2y+1}{z} + \\frac{3z+1}{x} > a.\n$$",
"options": [],
"answer": "See solution",
"solution": "For any positive real numbers $x, y, z$ we have\n\n$$\n\\begin{aligned}\n\\frac{x+1}{y} + \\frac{2y+1}{z} + \\frac{3z+1}{x} &= \\frac{x}{y} + \\frac{2y}{z} + \\frac{3z}{x} + \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} \\\\\n&\\geq 3\\sqrt{6} + \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} > 3\\sqrt{6}.\n\\end{aligned}\n$$\n\nHence $a \\geq 3\\sqrt{6}$.\n\nConsider the triples $(x, y, z)$ satisfying the equality case in the previous AM-GM inequality. We have\n\n$$\n\\frac{x}{y} = \\frac{2y}{z} = \\frac{3z}{x} = \\sqrt{6}.\n$$\n\nIt follows that $x = \\frac{3}{\\sqrt{6}}z$, $y = \\frac{\\sqrt{6}}{2}z$. So all the desired triples are given by $\\left(\\frac{3}{\\sqrt{6}}z, \\frac{\\sqrt{6}}{2}z, z\\right)$, and we obtain\n\n$$\n3\\sqrt{6} + \\left( \\frac{\\sqrt{6}}{3} + \\frac{2}{\\sqrt{6}} + 1 \\right) \\cdot \\frac{1}{z} > a,\n$$\n\nFor $z \\to +\\infty$, we deduce that $3\\sqrt{6} \\geq a$.\n\nThus, the answer is $a = 3\\sqrt{6}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20014,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $n$ such that $n^n + 1$ is divisible by $n+1$.",
"options": [],
"answer": "See solution",
"solution": "For odd integers $n$, we can factor:\n\n$$\nn^n + 1 = (n+1)\\left(n^{n-1} - n^{n-2} + n^{n-3} - \\dots + 1\\right).\n$$\n\nThus, $n+1$ divides $n^n + 1$ for all odd $n$.\n\nSuppose there exists an even $n = 2k$ such that $2k + 1$ divides $(2k)^{2k} + 1$. Then $2k + 1$ also divides $2kK = (2k)^{2k+1} + 2k = \\left((2k)^{2k+1} + 1\\right) + (2k - 1)$. Since $(2k)^{2k+1} + 1 = (2k+1)\\left((2k)^{2k} - (2k)^{2k-1} + \\dots + 1\\right)$, $2k - 1$ must also be divisible by $2k + 1$, which is impossible. Therefore, no even $n$ satisfies the condition.\n\n*Answer*: All odd positive integers $n$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20015,
"subject": "Mathematics (Olympiad)",
"question": "We say that a sequence $a_1, a_2, \\dots, a_k$ is *m-fold 1-spaced* if $1$ can be written in at least *m* different ways as the difference of two elements $a_i$ and $a_j$ from the sequence.\n\nShow that if there exists an $m$-fold 1-spaced sequence of length $k$, then $k \\ge 2m$.\n\nAdditionally, construct an example of an $m$-fold 1-spaced sequence of length $2m$.",
"options": [],
"answer": "See solution",
"solution": "Let $(a_1, a_2, \\dots, a_k)$ be an $m$-fold 1-spaced sequence. Define $b_i = a_{i+1} - a_i$. Each way to write $1$ as a difference $a_{j+t} - a_j = 1$ corresponds to a sum of consecutive $b$'s: $b_j + b_{j+1} + \\dots + b_{j+t-1} = 1$ with $1 \\le j$, $1 \\le t$, and $j+t \\le k$.\n\nSuppose $1 = a_{j_1+t_1} - a_{j_1} = \\dots = a_{j_m+t_m} - a_{j_m}$ are $m$ different ways, with $1 \\le j_1 < j_2 < \\dots < j_m$ and $1 \\le t_1 < t_2 < \\dots < t_m$. If $t_i \\ge t_{i+1}$ for some $i$, then\n\n$$\n\\begin{align*}\n1 &= a_{j_{i+1}+t_{i+1}} - a_{j_{i+1}} = b_{j_{i+1}} + \\dots + b_{j_{i+1}+t_{i+1}-1} \\\\\n&\\le b_{j_i} + \\dots + b_{j_i+t_{i+1}-1} \\\\\n&\\le b_{j_i} + \\dots + b_{j_i+t_i-1} = a_{j_i+t_i} - a_{j_i} = 1.\n\\end{align*}\n$$\n\nEquality can only occur if $j_i = j_{i+1}$ and $t_i = t_{i+1}$, contradicting the choices being different. Similarly, if $j_i = j_{i+1}$, then $a_{j_{i+1}+t_{i+1}} = a_{j_i+t_i}$, again a contradiction. Thus, $j_1 < j_2 < \\dots < j_m$ and $t_1 < t_2 < \\dots < t_m$, so $k \\ge j_m + t_m \\ge m + m = 2m$.\n\nTo construct an $m$-fold 1-spaced sequence of length $2m$, define $b_1 = 1$. Choose $0 < \\epsilon_1 < \\frac{1}{6}$, set $b_2 = \\frac{1}{2} + \\epsilon_1$, $b_3 = \\frac{1}{2} - \\epsilon_1$. Recursively, for $2 \\le i \\le m-1$, choose $0 < \\epsilon_i < \\frac{b_i-b_{i+1}}{2}$ and $\\epsilon_i < \\frac{b_{i-1}-b_i}{2} - \\epsilon_{i-1}$, then set $b_{2i} = \\frac{b_i}{2} + \\epsilon_i$, $b_{2i+1} = \\frac{b_i}{2} - \\epsilon_i$. This ensures $b_1, b_2, \\dots, b_{2i+1}$ is decreasing and $b_{2i} + b_{2i+1} = b_i$.\n\nBy induction, for $1 \\le j \\le m$:\n\n$$\nb_j + b_{j+1} + \\dots + b_{2j-1} = 1.\n$$\n\nDefine $a_0 = 0$, $a_{i+1} = a_i + b_i$. Then $a_{2j} - a_j = 1$ for all $j \\ge 1$. Thus, the first $2m$ terms of $a$ form an $m$-fold 1-spaced sequence of length $2m$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20016,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be an odd integer greater than $1$ and let $c_1, c_2, \\dots, c_n$ be integers. For each permutation $a = (a_1, a_2, \\dots, a_n)$ of $\\{1, 2, \\dots, n\\}$, define $S(a) = \\sum_{i=1}^{n} c_i a_i$. Prove that there exist permutations $b$ and $c$, $b \\neq c$, such that $n!$ divides $S(b) - S(c)$.",
"options": [],
"answer": "See solution",
"solution": "Let $\\sum_a$ denote the sum over all $n!$ permutations $a = (a_1, a_2, \\dots, a_n)$. We compute $\\sum_a S(a) \\bmod n!$ in two ways, one of which assumes the desired conclusion is false, and reach a contradiction.\n\nSuppose, for the sake of contradiction, that the claim is false. Then each $S(a)$ must have a different remainder $\\bmod n!$. Since there are exactly $n!$ such permutations $a$, there exists exactly one permutation $a$ such that $S(a) \\equiv s \\pmod{n!}$ for each $s = 1, 2, \\dots, n!$. Since $n > 1$, $n!$ is even and $n! + 1$ is odd. Hence,\n\n$$\n\\sum_a S(a) \\equiv \\sum_{s=1}^{n!} s \\equiv \\frac{n!}{2} (n! + 1) \\pmod{n!},\n$$\n\nor\n\n$$\n\\sum_a S(a) \\equiv \\frac{n!}{2} \\pmod{n!}. \\qquad (1)\n$$\n\nOn the other hand, for $i, k \\in \\{1, \\dots, n\\}$, we have $a_i = k$ in exactly $(n-1)!$ permutations $a$. Thus, for $1 \\le i \\le n$,\n\n$$\n\\sum_a a_i = (n-1)! (1 + 2 + \\dots + n) = n! \\cdot \\frac{n+1}{2}.\n$$\n\nHence,\n\n$$\n\\sum_a S(a) = \\sum_a \\sum_{i=1}^n c_i a_i = \\sum_{i=1}^n \\left( c_i \\sum_a a_i \\right).\n$$\n\nBecause $n+1$ is even, $n!$ divides $\\sum_a a_i = n! \\cdot \\frac{n+1}{2}$ for each $i$. It follows that $n!$ divides $\\sum_a S(a)$, contradicting (1). Therefore, the initial assumption was false, and there do exist distinct permutations $b$ and $c$ such that $n!$ divides $S(b) - S(c)$.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20017,
"subject": "Mathematics (Olympiad)",
"question": "Consider a $55 \\times 55$ grid of unit squares. Initially, all squares are white. In one operation, you may select any rectangle formed by the grid lines and color all the squares inside it either black or white (overwriting previous colors). What is the minimum number of such operations required to reach a configuration in which:\n\n1. Every $1 \\times 1$ square is black or white.\n2. For every pair of adjacent $1 \\times 1$ squares, their colors are different.\n3. For every $2 \\times 2$ square, not all four $1 \\times 1$ squares are the same color?\n\nFind the minimum number of operations needed.",
"options": [],
"answer": "See solution",
"solution": "First, we show that the answer we seek is at least $784$.\n\nLet us call the vertices of the squares of the grid lattice points. There are $56^2$ lattice points. For each lattice point, call the set of all the squares of the grid having this lattice point as a vertex its neighborhood. We will first show that in order to reach the configuration satisfying the conditions of the problem, it is necessary to include in the sequence of operations, for every lattice point, an operation involving a rectangle for which it is a vertex.\n\nNote that if a configuration satisfies the conditions of the problem, then for every lattice point there must be at least one black square in its neighborhood. For a lattice point $P$, let us say that an operation is an operation containing $P$ if it chooses a rectangle having $P$ either in its interior or on its sides. If for some lattice point $P$, no operation containing $P$ is performed throughout the process, then all the squares in the neighborhood of $P$ remain white, so the process does not attain the configuration desired. Therefore, at least one operation containing $P$ must be performed.\n\nSo, suppose for a lattice point $P$ we consider the last operation containing $P$ that has been performed. If this operation chooses a rectangle containing $P$ in the interior (hence $P$ is not a vertex of the rectangle), then some pair of adjacent squares in the neighborhood of $P$ must have the same color. Thus this process cannot reach the desired configuration, and this implies that the last operation containing $P$ must pick a rectangle having $P$ as one of the vertices.\n\nSince there are $4$ vertices for a rectangle chosen for an operation, and since there are $56^2$ lattice points, we see that the number of operations in a process necessary to achieve the desired configuration is at least $\\frac{56^2}{4} = 784$.\n\nNext, we show that there is a process consisting of exactly $784$ operations to reach the desired configuration.\n\nFor this purpose, let us consider $28$ rectangles consisting of all the squares lying on $1$st, $3$rd, $\\ldots$, $55$th row (i.e., odd-numbered rows) from the top and go through $28$ operations of coloring all the squares in the rectangle black starting with the top row. Next, take $27$ rectangles consisting of all the squares lying on $2$nd, $4$th, $\\ldots$, $54$th column (i.e., even-numbered columns) from the left and go through $27$ operations of coloring all the squares in the rectangle white starting with the left-most row. After these operations, all the squares located at the intersection of an odd-numbered row and an odd-numbered column are colored black and all other squares in the grid are colored white. Finally, for each of $27^2$ squares located at the intersection of an even-numbered row and an even-numbered column, we go through the operation of picking one of these squares in some order and coloring it black till we exhaust all of these squares. Then, we see that we end up with the configuration for which all the three conditions of the problem are satisfied. The number of operations necessary to go through this process is $28 + 27 + 27^2 = 784$, and therefore, the answer we seek is $784$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20018,
"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 such four numbers after several moves?\n\na) $2015$, $2016$, $2017$, $2018$\n\nb) $2016$, $2017$, $2019$, $2022$",
"options": [],
"answer": "See solution",
"solution": "**Answer:** a), b) that is not possible.\n\n**Solution.**\n\na) Let us look at the numbers modulo $3$. Obviously, the amount of numbers divisible by $3$ cannot decrease. If both $a, b$ 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 $3$. At the beginning, we had only one number divisible by $3$, and in the end only one, thus such a situation is impossible.\n\nb) Now, consider the situation when exactly three numbers are divisible by $3$. Thus, those numbers equal $0, 0, 0, k$, where $k \\in \\{1, 2\\}$ modulo $3$. These four numbers will never change. Let's check when the amount of numbers divisible by $3$ can increase. Then $a, b$ should be $1, 2$ modulo $3$. However, then numbers $0, 2$ appear. Thus, in the situation where exactly three numbers are divisible by $3$, they should equal $0, 0, 0, 2$, and four numbers from the condition equal $0, 0, 0, 1$. Thus, we get a contradiction.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20019,
"subject": "Mathematics (Olympiad)",
"question": "Let $M(n)$ be the minimum number of cells that must be coloured in the $6 \\times n$ square grid to ensure that every $2 \\times 3$ rectangle (in any orientation) contains a coloured cell. Is it true that $M(n) = p_n + k_n^3$ for all $n \\ge 2$ where $p_n$ is a prime and $k_n$ is a nonnegative integer?",
"options": [],
"answer": "See solution",
"solution": "No, it is not true. First, note that $M(n)$ is a nondecreasing function and $M(n) \\ge n$ for all $n \\ge 2$, since a $6 \\times n$ rectangle can be cut into $n$ rectangles of size $2 \\times 3$, each of which must contain a coloured cell.\n\nNext, we show that $M(n+1) \\leq M(n) + 2$ for all $n \\geq 2$. Assume we have the minimal colouring for the $6 \\times n$ grid (blue rectangle in the following figure) and we add one more column to the right.\n\n\n\nAt least one of the red or green cells is coloured. If a red cell is coloured, then colouring cells marked A and C will complete the minimal colouring for the $6 \\times (n+1)$ grid. If a green cell is coloured, then colouring B and C is enough.\n\nNeither $35$ nor $36$ can be expressed in the form $p + k^3$, where $p$ is a prime and $k$ is a nonnegative integer (just check $k \\in \\{0, 1, 2, 3\\}$). As $M(n)$ is an unbounded function which increases at each step by at most two, we will have that either $M(n) = 35$ or $M(n) = 36$ for some $n$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20020,
"subject": "Mathematics (Olympiad)",
"question": "Solve the following system of equations in the set of rational numbers:\n\n$$\n(x^2 + 1)^3 = y + 1\n$$\n$$\n(y^2 + 1)^3 = z + 1\n$$\n$$\n(z^2 + 1)^3 = x + 1.\n$$",
"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 integer values of $p$ 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": 20021,
"subject": "Mathematics (Olympiad)",
"question": "Given a point $A$ not on a line $\\ell$ in the plane, describe the locus of points $Q$ such that:\n\n1. $AQ = PQ$ for some point $P$ on $\\ell$,\n2. The angle $\\angle PAQ = \\alpha$ is constant.\n\n*Note:* If $A \\in \\ell$, the locus consists of the two lines through $A$ making an angle $\\alpha$ with $\\ell$.",
"options": [],
"answer": "See solution",
"solution": "Let $A(0, 2\\mu)$ and $\\ell$ be the $x$-axis. For $P(2\\lambda, 0)$, the midpoint $M(\\lambda, \\mu)$. For $Q(x, y)$, the conditions are:\n\n1. $MQ^2 + AM^2 = AQ^2$\n2. $MQ = AM \\tan \\alpha$\n3. $AQ^2 = PQ^2$\n\nFrom these, we derive:\n\n$$\n(y + (t^2 - 1)\\mu)^2 = (t x)^2\n$$\nwhere $t = \\tan \\alpha$.\n\nThus, the locus is the union of two lines:\n$$\ny = \\pm t x + (1 - t^2)\\mu\n$$\nwith slopes $\\pm \\tan \\alpha$ and $y$-intercept $(1 - t^2)\\mu$.\n\nAlternatively, vector or trigonometric approaches yield the same result.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20022,
"subject": "Mathematics (Olympiad)",
"question": "Non-isosceles triangle $ABC$ is given, in which $2AC = AB + BC$. Let $I$ be the incenter of the inscribed circle in $ABC$, $K$ be the middle of the sector $ABC$ of the circumscribed circle. Let $T$ be such point on the line $AC$ that $\\angle TIB = 90^\\circ$. Prove that line $TB$ is tangent to the circumcircle of $\\triangle KBI$.\n\n\n\n**Fig. 11**\n\n\n\n**Fig. 12**",
"options": [],
"answer": "See solution",
"solution": "Without loss of generality, suppose $AB < BC$. Let the angular bisector of $\\angle ABC$ intersect the circumcircle of the triangle a second time at point $W$, and let point $M$ be the midpoint of $AC$. Let us prove that $BI = IW$. From $I$ and $W$, draw perpendiculars $II_1$ and $WW_1$ to line $BC$ (see Fig. 14). It is known that\n\n$$\nBW_1 = \\frac{1}{2}(AB + BC) = AC, \\quad \\text{and}\n$$\n\n$$\nBI_1 = \\frac{1}{2}(AB + BC - AC) = \\frac{1}{2}AC, \\quad \\text{hence, } BI_1 = I_1W_1 \\quad \\text{and}\n$$\n\n$$\nBI = WI.\n$$\n\nNotice also that quadrilateral $TIMW$ is cyclic with diameter $TW$.\n\nLet us show that $\\angle BKI = \\angle IMA$. From the trillium theorem, $WA = WC = WI$, i.e.\n\n$$WI^2 = WM \\cdot WK. \\quad \\text{Then,}$$\n$$\\triangle WIM \\sim \\triangle KIW, \\quad \\text{which means}$$\n$$\\angle WMI = \\angle WIK \\Rightarrow \\angle BKI = \\angle IMA.$$\n\nNow, in $\\triangle BTW$, $TI$ is both altitude and median, which makes $\\triangle BTW$ isosceles, and so\n\n$$\\angle TBI = \\angle TWI = \\angle TMI = \\angle BKI,$$\nwhich yields the statement of the problem.\n\n\n\n**Fig. 13**",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20023,
"subject": "Mathematics (Olympiad)",
"question": "Нека $x$, $y$ и $z$ се позитивни реални броеви така што $x^4 + y^4 + z^4 = 3$. Докажи дека\n\n$$\n\\frac{9}{x^2 + y^2 + z^2} + \\frac{9}{x^4 + y^4 + z^4} + \\frac{9}{x^6 + y^6 + z^6} \\le x^6 + y^6 + z^6 + 6.\n$$\n\nКога важи равенство?",
"options": [],
"answer": "See solution",
"solution": "Ако го искористиме неравенството на Коши–Буњаковски–Шварц за позитивните броеви $(x, y^2, z^3)$ и $(x^3, y^2, z)$ добиваме\n\n$$\n(x^4 + y^4 + z^4)^2 \\leq (x^2 + y^4 + z^6)(x^6 + y^4 + z^2), \\text{ т.е.}\n$$\n$$\n\\frac{1}{x^2 + y^4 + z^6} \\leq \\frac{x^6 + y^4 + z^2}{9} \\tag{1}\n$$\n\nАналогно, со користење на неравенство на Коши–Буњаковски–Шварц за позитивните броеви $(x^2, y^3, z)$ и $(x, y^3, z^3)$, како и за позитивните броеви $(x^3, y, z^2)$ и $(x, y^3, z^2)$, добиваме\n\n$$\n\\frac{1}{x^4 + y^6 + z^2} \\le \\frac{x^4 + y^2 + z^6}{9}, \\tag{2}\n$$\n\nОДНОСНО\n\n$$\n\\frac{1}{x^6 + y^2 + z^4} \\le \\frac{x^2 + y^6 + z^4}{9}. \\tag{3}\n$$\n\nСега, со собирање на неравенствата (1), (2) и (3) го добиваме неравенството\n\n$$\n\\frac{1}{x^2 + y^4 + z^6} + \\frac{1}{x^4 + y^6 + z^2} + \\frac{1}{x^6 + y^2 + z^4} \\le \\frac{x^6 + y^6 + z^6 + x^4 + y^4 + z^4 + x^2 + y^2 + z^2}{9}. \\tag{4}\n$$\n\nОд неравенство меѓу аритметичка и квадратна средина за позитивните броеви $x^2$, $y^2$ и $z^2$ добиваме дека $x^2 + y^2 + z^2 \\le 3\\sqrt{\\frac{x^4 + y^4 + z^4}{3}} = 3$, па ако замениме во (4) го добиваме бараното неравенство. Равенство во (1) важи ако и само ако $\\frac{x}{x^3} = \\frac{y^2}{y^2} = \\frac{z^3}{z}$, т.е. $x = z = 1$ и од $x^4 + y^4 + z^4 = 3$ следува дека $y = 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20024,
"subject": "Mathematics (Olympiad)",
"question": "Each employee in a company is assigned a weekly schedule, represented as a 7-tuple of 0's and 1's, where each coordinate corresponds to a day of the week (1 for 'working day', 0 for 'non-working day'). For every pair of employees, their weekly schedules must differ in at least 3 days (i.e., their tuples differ in at least 3 coordinates).\n\nWhat is the maximum possible number of employees the company can have under this condition?",
"options": [],
"answer": "See solution",
"solution": "We will show that the maximum possible number of employees is $n = 16$.\n\nFirst, we prove that $n \\leq 16$. Each possible weekly schedule is a 7-tuple of 0's and 1's. For each employee's schedule $v_i$, there are 7 neighbors (tuples differing in exactly one coordinate). For any $i \\neq j$, $v_j$ is not a neighbor of $v_i$, and $v_i$ and $v_j$ do not share neighbors, since otherwise they would differ in at most 2 coordinates, contradicting the condition.\n\nThus, the $n$ schedules and their $7n$ neighbors are all distinct, so $8n \\leq 2^7 = 128$, giving $n \\leq 16$.\n\nTo show that $n = 16$ is possible, here is an explicit example of 16 such schedules:\n\n$$\n\\begin{align*}\nv_1 &= (0, 0, 0, 0, 0, 0, 0), & v_2 &= (1, 1, 1, 0, 0, 0, 0), & v_3 &= (1, 0, 0, 1, 1, 0, 0), & v_4 &= (1, 0, 0, 0, 0, 1, 0), \\\\\nv_5 &= (0, 1, 0, 1, 0, 0, 1), & v_6 &= (0, 1, 0, 0, 1, 1, 0), & v_7 &= (0, 0, 1, 1, 0, 1, 0), & v_8 &= (0, 0, 1, 0, 1, 0, 1), \\\\\nv_9 &= (1, 1, 1, 1, 1, 1, 1), & v_{10} &= (0, 0, 0, 1, 1, 1, 1), & v_{11} &= (0, 1, 1, 0, 0, 1, 1), & v_{12} &= (0, 1, 1, 1, 1, 0, 0), \\\\\nv_{13} &= (1, 0, 1, 0, 1, 1, 0), & v_{14} &= (1, 0, 1, 1, 0, 0, 1), & v_{15} &= (1, 1, 0, 0, 1, 0, 1), & v_{16} &= (1, 1, 0, 1, 0, 1, 0)\n\\end{align*}\n$$\n\nFor all $i \\neq j$, the tuples $v_i$ and $v_j$ differ in at least 3 coordinates, as required.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20025,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying the following functional equation:\n\n$$\nf(x + f(y)) = f(f(y)) + 2x^2y^2 + f(y^2)\n$$\nfor all real numbers $x, y$.",
"options": [],
"answer": "See solution",
"solution": "Let us label the equation given in the problem by $(\\star)$. We will show that the functions $f$ satisfying the equation $(\\star)$ are given by the following:\n\n$$\nf(x) = 0, \\quad x^2, \\quad -x^2, \\quad x^2 - 1, \\quad 1 - x^2.\n$$\n\nIt is easy to check that all of these functions satisfy the equation $(\\star)$.\n\nWe will show in the sequel that there are no other functions which satisfy the equation $(\\star)$.\n\nSo, let a function $f$ satisfy the equation $(\\star)$ and suppose for some $t$, $f(t) \\neq 0$ is true. Suppose for some pair of real numbers $a, b$ the condition $f(a) = f(b)$ is satisfied. By comparing the results obtained by substituting into $(\\star)$ $(x, y) = (a, t)$ and $(x, y) = (b, t)$, we see that $2a^2f(t) = 2b^2f(t)$ must hold, from which we get $a = \\pm b$.\n\nLet $f(0) = k$. Substituting $(x, y) = (0, 0)$ into the equation $(\\star)$, we see that $f(k) = 0$ must be satisfied. Also, if we substitute $(x, y) = (0, k)$ into $(\\star)$ and use the fact $f(k) = 0$, we get $f(k^2) = 0$. Thus, we see that we must have $k^2 = \\pm k$, and therefore $k = 0, \\pm 1$. Finally, substituting $(x, y) = (0, y)$ into $(\\star)$, we get $f(k - f(y)) = f(y^2)$, from which we can conclude that $k - f(y) = \\pm y^2$ for any $y$, i.e., for any $y$, $f(y) \\in \\{k + y^2, k - y^2\\}$ must hold.\n\n**(1) Let us consider the case $k=0$.**\n\nSuppose there exist non-zero real numbers $z, w$ for which $f(z) = z^2$, $f(w) = -w^2$ are satisfied. By substituting $(x, y) = (w, w)$ into the equation $(\\star)$, we obtain $0 = f(f(w)) + 2w^4 + f(w^2)$. Since $k = 0$, we have $f(f(w)) = \\pm w^4$ and $f(w^2) = \\pm w^4$, and we can check that $f(w^2) = -w^4$ must hold, since $w \\neq 0$. Finally, if we substitute $(x, y) = (z, w)$ into the equation $(\\star)$, we get\n\n$$\n\\pm(z^2 + w^2)^2 = \\pm z^4 + 2z^2w^2 - w^4.\n$$\n\nBy simplifying each of the 4 cases arising from the combinations of $\\pm$ signs, we can check easily that there are no $(z, w)$ with $z^2 > 0$, $w^2 > 0$ which satisfy the last equation. We can thus conclude that if $k = 0$, then either $f(x) = x^2$ for all $x$ or $f(x) = -x^2$ for all $x$.\n\n**(2) Consider the cases when $k = \\pm 1$.**\n\nBy substituting $(x, y) = (x, k)$ into the equation $(\\star)$, we get\n\n$$\nf(f(x) - k) = f(f(x)) - 2kx^2 + k.\n$$\n\nFrom $f(x) - k = \\pm x^2$, we have $f(f(x) - k) = k \\pm x^4$, and since $f(f(x)) = k \\pm f(x)^2$ we get\n\n$$\n(\\star\\star) \\qquad \\pm f(x)^2 = \\pm x^4 + 2kx^2 - k.\n$$\n\n- Suppose $k = 1$ holds.\n\nIf there exists a $c \\neq 0$ for which $f(c) = 1 + c^2$, then by substituting $x = c$ into the equation $(\\star\\star)$ above, we get\n\n$$\n\\pm(1 + c^2)^2 = \\pm c^4 + 2c^2 = 1.\n$$\n\nBut we can check easily that this is impossible for any combination of $\\pm$ signs, since $c^2 > 0$. Hence, if $k = 1$ we must have $f(x) = 1 - x^2$ for all $x$.\n\n- Suppose $k = -1$ holds.\n\nIf there exists a $c \\neq 0$ for which $f(c) = 1 - c^2$, we get from the equation $(\\star\\star)$ that\n\n$$\n\\pm(-1 - c^2)^2 = \\pm c^4 - 2c^2 + 1.\n$$\n\nBut as for the case of $k = 1$, this identity (for any combination of $\\pm$ signs) does not hold since $c^2 > 0$. Thus we must have $f(x) = -1 + x^2$ for all $x$.\n\nPutting together the arguments made above, we conclude that there are only 5 possibilities: $0, x^2, -x^2, x^2 - 1, 1 - x^2$ for $f$ satisfying the equation $(\\star)$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20026,
"subject": "Mathematics (Olympiad)",
"question": "設三角形 $ABC$ 的內心為 $I$，$X \\neq A$ 在外接圓 $\\Omega$ 上使得 $AI = XI$。內切圓分別和 $AC, AB$ 切於 $E, F$，$M_a, M_b, M_c$ 分別為三邊 $BC, CA, AB$ 的中點。再設 $T$ 為 $M_bF$ 和 $M_cE$ 的交點，$S$ 為 $AT$ 和 $\\Omega$ 的另一個交點。\n\n試證：$X, M_a, S, T$ 共圓。\n\n",
"options": [],
"answer": "See solution",
"solution": "設 $O$ 為 $\\triangle ABC$ 的外心，$J$ 為 $EF$ 和 $OI$ 的交點，$L$ 為 $AJ$ 和 $\\Omega$ 異於 $A$ 的交點。\n\n**Claim.** $O, M_a, X, J, L$ 五點共圓。\n\n*Proof.* 由於 $JO$ 為 $\\angle XJA$ 的角平分線且 $XO = LO$，因此 $J, L, O, X$ 共圓。另一方面，設 $M$ 在 $OI$ 上為 $AX$ 中點，$Y$ 在 $\\Omega$ 上使得 $XY \\parallel BC$，那麼由 $A, F, I, M, E$ 共圓且 $AFIE$ 為調和四邊形，可知\n\n$$\n-1 = (F, E; A, I) \\stackrel{M}{\\cong} (F, E; MA \\cap EF, J) \\stackrel{A}{\\cong} (B, C; X, L) \\stackrel{Y}{\\cong} (B, C; \\infty_{BC}, BC \\cup YL)\n$$\n\n因此 $Y, M_a, L$ 共線，即 $OM_a$ 為 $\\angle KM_aX$ 的角平分線，代表 $M_a, X, J, O$ 共圓，從而 $O, M_a, X, J, L$ 五點共圓。\n\n設 $N$ 是弧 $BC$ 中點（不包含 $A$ 的），由 Claim：\n\n$$\n\\angle XJM_a = \\angle XOM_a = 2\\angle XAN = 2\\angle XAI = 2\\angle (OI, ED),\n$$\n\n且 $AJ$ 和 $XJ$ 關於 $OI$ 對稱，我們有 $AJ$ 和 $M_aJ$ 關於 $EF$ 對稱。注意到線段 $AT, M_bM_c, EF$ 的中點們共線（完全四線形 $\\{CA, AB, EM_c, FM_b\\}$ 的牛頓線），將其關於 $A$ 位似 2 倍可得 $TM_a$ 過 $A$ 關於 $EF$ 的對稱點，故 $M_a, J, T$ 共線，再由 Claim：\n\n$$\n\\angle TM_aX = \\angle JM_aX = \\angle JLX = \\angle ALX = \\angle ASX = \\angle TSX,\n$$\n\n即 $X, M_a, S, T$ 共圓。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20027,
"subject": "Mathematics (Olympiad)",
"question": "Prove that\n$$\n(ABC) = \\frac{a^2}{2}(\\cot B + \\cot C)\n$$\nwhere $(ABC)$ denotes the area of triangle $ABC$ with side $a$ opposite angle $A$.\n\nAlso, show that for all acute-angled triangles $ABC$,\n$$\n\\cos A \\cos B \\cos C \\le \\frac{1}{8}.\n$$",
"options": [],
"answer": "See solution",
"solution": "We recall that $\\sin(B + C) = \\sin(A)$ because $\\angle A + \\angle B + \\angle C = 180^\\circ$ and obtain\n\n$$\n\\begin{aligned}\n\\cot B + \\cot C &= \\frac{\\cos B \\sin C + \\cos C \\sin B}{\\sin B \\sin C} = \\frac{\\sin(B + C)}{\\sin B \\sin C} \\\\\n&= \\frac{\\sin A}{\\sin B \\sin C} = \\frac{a}{b \\sin C} \\quad \\text{(Sine Rule)} \\\\\n&= \\frac{a^2}{ab \\sin C} = \\frac{a^2}{2(ABC)},\n\\end{aligned}\n$$\n\nthe required result.\n\nHere is an alternative way to prove this formula. Let $D$ be the foot of the altitude from $A$, $x = |AD|$, and $y = |CD|$ where $x$ is taken negative if $\\angle B$ is obtuse and $y$ is taken negative if $\\angle C$ is obtuse. We then have $a = x + y$, $\\cot B = x/h$ and $\\cot C = y/h$, hence $\\cot B + \\cot C = a/h$. Using $2(ABC) = ah$ this turns into the desired formula.\n\nTo show that $\\cos A \\cos B \\cos C \\le \\frac{1}{8}$ for all acute-angled triangles $ABC$, we use the area formula shown above. Since $B, C$ are acute angles, $\\cot B$ and $\\cot C$ are positive and we can use the AM-GM inequality to obtain\n\n$$\n\\sqrt{\\cot B \\cot C} \\le \\frac{\\cot B + \\cot C}{2} = \\frac{a^2}{4(ABC)},\n$$\n\nwith equality iff $\\angle B = \\angle C$. Whence\n\n$$\n\\begin{aligned}\n(ABC) &\\le \\frac{a^2}{4} \\sqrt{\\tan B \\tan C}, \\\\\n&= \\frac{a}{2} \\frac{\\sqrt{\\left(\\frac{1}{2}ab \\sin C\\right) \\left(\\frac{1}{2}ac \\sin B\\right)}}{\\sqrt{bc \\cos B \\cos C}} = \\frac{a}{2} \\frac{(ABC)}{\\sqrt{bc \\cos B \\cos C}},\n\\end{aligned}\n$$\n\nand so\n\n$$\n\\cos B \\cos C \\le \\frac{a^2}{4bc},\n$$\n\nwith equality iff $\\angle B = \\angle C$. Similarly,\n\n$$\n\\cos C \\cos A \\le \\frac{b^2}{4ca}, \\quad \\text{with equality iff } \\angle C = \\angle A\n$$\n\nand\n\n$$\n\\cos A \\cos B \\le \\frac{c^2}{4ab}, \\quad \\text{with equality iff } \\angle A = \\angle B.\n$$\n\nHence\n\n$$\n(\\cos A \\cos B \\cos C)^2 \\le \\frac{1}{64},\n$$\n\nwith equality iff $\\angle A = \\angle B = \\angle C$, whence the desired inequality follows.\n\nAn alternative proof of inequality above, not using the area formula, may use the Cosine Rule as follows:\n\n$$\n\\begin{aligned}\n\\cos B \\cos C &= \\frac{(a^2 - b^2 + c^2)}{2ac} \\cdot \\frac{(a^2 + b^2 - c^2)}{2ab} \\\\\n&= \\frac{a^4 - (b^2 - c^2)^2}{4a^2bc} \\le \\frac{a^4}{4a^2bc} = \\frac{a^2}{4bc},\n\\end{aligned}\n$$\n\nwith equality iff $b = c$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20028,
"subject": "Mathematics (Olympiad)",
"question": "Let $G$ be a graph with $n$ vertices. Prove that $G$ is a permutation graph if and only if both $G$ and its complement $\\bar{G}$ are good graphs (i.e., they can be oriented to be acyclic and transitively closed).",
"options": [],
"answer": "See solution",
"solution": "We proceed in several steps:\n\nFirst, we show by induction on $n$ that every good graph with $n$ vertices is a divisibility graph.\n\n**Base case:** For $n = 1$, the claim is obvious.\n\n**Inductive step:** Assume the claim holds for $n = k-1$. Consider a good graph $G$ with $n = k$ vertices. Since a good graph cannot have a cycle (otherwise, a cycle $v_1 \\to v_2 \\to \\dots \\to v_t \\to v_1$ would force $v_1 \\to v_t$ to be an edge, contradicting simplicity), $G$ is acyclic. Thus, there exists a vertex $v$ with in-degree zero. The graph $G - \\{v\\}$ is a good graph with $k-1$ vertices, so by induction, it is a divisibility graph. Assign $n(v) = p$ (a prime not dividing any $n(u)$ for $u \\in G - \\{v\\}$). For any $u \\neq v$ with $v \\to u$, set $n'(u) = p \\cdot n(u)$. Then $G$ is a divisibility graph.\n\nNow, to prove the main claim:\n\nSuppose $G$ is a permutation graph. Label the vertices by $i$ and let $\\pi$ be the permutation. For each edge $(v, u)$, direct it from $v$ to $u$ if $i < j$ and $\\pi(j) < \\pi(i)$. This orientation is acyclic and transitively closed, so $G$ is good. For the complement $\\bar{G}$, direct edges from $v$ to $u$ if $i < j$ and $\\pi(i) < \\pi(j)$. This orientation is also acyclic and transitively closed, so $\\bar{G}$ is good.\n\nFor the converse, assume both $G$ and $\\bar{G}$ are good graphs. We prove the following lemma:\n\n**Lemma:** If both $G$ and $\\bar{G}$ are good, then there is an ordering of the vertices $v_1, \\dots, v_n$ such that all edges in both $G$ and $\\bar{G}$ are directed from $v_i$ to $v_j$ only if $i < j$.\n\n*Proof of Lemma:* Since $G$ is acyclic, such an ordering exists. Suppose, for contradiction, that in $\\bar{G}$, some edge $v_k \\to v_t$ is directed backward ($k > t$) and the distance $k-t$ is minimal. If $v_k$ and $v_t$ are not adjacent, there exists $v_s$ between them. Considering possible edge directions and the goodness of $G$ and $\\bar{G}$, we reach a contradiction. Thus, $v_k$ and $v_t$ must be adjacent, and swapping their positions reduces the number of backward edges, contradicting minimality. Thus, all edges are forward.\n\nNow, define a permutation $\\pi$ by $\\pi(i) = i - 1 + d^+(v_i) - d^-(v_i)$, where $d^+(v_i)$ and $d^-(v_i)$ are the out-degree and in-degree of $v_i$ in $G$. For $i < j$, we show $v_i \\to v_j$ if and only if $\\pi(i) > \\pi(j)$, by considering the sets of vertices with respect to $v_i$ and $v_j$ and analyzing degree differences. The calculations show that $\\pi(i) - \\pi(j) > 0$ if $v_i \\to v_j$, and $< 0$ otherwise.\n\nTherefore, $G$ is a permutation graph.\n\n*■*",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20029,
"subject": "Mathematics (Olympiad)",
"question": "Determine if there are polynomials $p(x)$ and $q(x)$ with real coefficients such that\n\n$$\n\\frac{p(n)}{q(n)} = 1 + \\frac{1}{2!} + \\frac{1}{3!} + \\dots + \\frac{1}{n!}\n$$\n\nfor every positive integer $n$.",
"options": [],
"answer": "See solution",
"solution": "Assume that there are polynomials $p, q \\in \\mathbb{R}[X]$ such that\n\n$$\n\\frac{p(n)}{q(n)} = 1 + \\frac{1}{2!} + \\dots + \\frac{1}{n!}, \\quad n \\ge 1.\n$$\n\nThen\n\n$$\n\\frac{p(n+1)}{q(n+1)} - \\frac{p(n)}{q(n)} = \\frac{1}{(n+1)!}, \\quad n \\ge 1,\n$$\n\nso\n\n$$\n\\frac{p(n+1)q(n) - p(n)q(n+1)}{q(n)q(n+1)} = \\frac{1}{(n+1)!}, \\quad n \\ge 1.\n$$\n\nDefine the polynomials $u, v \\in \\mathbb{R}[X]$ by\n\n$$\nu(x) = p(x+1)q(x) - p(x)q(x+1), \\quad v(x) = q(x)q(x+1).\n$$\n\nFrom above it follows that $u$ is not the zero polynomial and we have\n\n$$\n\\frac{u(n)}{v(n)} = \\frac{1}{(n+1)!}, \\quad n \\ge 1\n$$\n\nand\n\n$$\n\\frac{u(n+1)}{v(n+1)} = \\frac{1}{(n+2)!}, \\quad n \\ge 1.\n$$\n\nIt follows\n\n$$\n\\frac{u(n+1)}{u(n)} \\cdot \\frac{v(n)}{v(n+1)} = \\frac{1}{n+2}, \\quad n \\ge 1.\n$$\n\nWe have\n\n$$\n\\lim_{n \\to \\infty} \\frac{u(n+1)}{u(n)} = \\lim_{n \\to \\infty} \\frac{v(n+1)}{v(n)} = 1,\n$$\n\nand from above we obtain the contradiction $1 = 0$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20030,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a non-negative integer with binary representation $n = (a_r a_{r-1} \\cdots a_0)_2 = \\sum_{j=0}^{r} a_j 2^j$, where $a_j = 0$ or $1$ for $j = 0, 1, \\dots, r$. Define $s(n) = \\sum_{j=0}^{r} a_j$ as the number of ones in the binary representation of $n$.\n\nThe power $m$ of the factor $2^m$ in the factorization of $n!$ can be expressed as:\n\n$$\n\\lfloor \\frac{n}{2} \\rfloor + \\lfloor \\frac{n}{4} \\rfloor + \\cdots + \\lfloor \\frac{n}{2^m} \\rfloor + \\cdots = n - s(n).\n$$\n\nConsider the binomial coefficient $\\binom{2012}{k} = \\frac{2012!}{k! (2012-k)!}$.\n\n- If $\\binom{2012}{k}$ is odd, then $s(k + m) = s(k) + s(m)$, where $m = 2012 - k$; that is, the binary addition of $k + m = 2012$ has no carrying.\n- If $\\binom{2012}{k}$ is even but not a multiple of 4, then $s(k + m) = s(k) + s(m) - 1$; that is, there is exactly one carrying in the binary addition.\n\nGiven $2012 = (11111011100)_2$, which has eight 1's and three 0's, determine how many $k$ in $0 \\leq k \\leq 2012$ make $\\binom{2012}{k}$ a multiple of 2012.",
"options": [],
"answer": "See solution",
"solution": "First, count the number of $k$ such that $\\binom{2012}{k}$ is odd. Since $2012 = (11111011100)_2$ has eight 1's, there are $2^8 = 256$ such $k$.\n\nNext, count the number of $k$ such that $\\binom{2012}{k}$ is even but not a multiple of 4. There are $2^7 = 128$ such $k$ for each of the two possible carrying positions, totaling $256$.\n\nThus, the number of $k$ such that $\\binom{2012}{k}$ is a multiple of 4 is $2013 - 256 - 256 = 1501$.\n\nHowever, for $k = 0, p, 2p, 3p, 4p$ (with $p = 503$), $\\binom{4p}{k}$ is not a multiple of $p$. Specifically, $\\binom{4p}{0}$, $\\binom{4p}{p}$, $\\binom{4p}{2p}$, $\\binom{4p}{3p}$, and $\\binom{4p}{4p}$ are not multiples of $p$, but only three of these are multiples of 4 but not multiples of $p$.\n\nTherefore, the number of $k$ such that $\\binom{2012}{k}$ is a multiple of 2012 is $1501 - 3 = 1498$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20031,
"subject": "Mathematics (Olympiad)",
"question": "Let $x$, $y$, $n$ be positive integers such that\n\n$$\nn = \\frac{x^2 - 1}{2} = \\frac{y^2 - 1}{3}.\n$$\n\nProve that $n = y^2 - x^2$ and $20$ divides $n$.",
"options": [],
"answer": "See solution",
"solution": "a) Since $x^2 = 2n + 1$ and $y^2 = 3n + 1$, we have\n$$\ny^2 - x^2 = (3n + 1) - (2n + 1) = n.\n$$\n\nb) Since $x$ is odd, both $x-1$ and $x+1$ are even, so $4$ divides $x^2 - 1 = 2n$, implying $n$ is even. Then $y^2 = 3n + 1$ is odd, so $y$ is odd. Again, $4$ divides $y^2 - 1$, so $n = (y^2 - 1) - (x^2 - 1)$ is divisible by $4$; in fact, by $8$.\n\nA perfect square modulo $5$ gives remainders $0$, $1$, or $4$. The equality $x^2 + y^2 = 5n + 2$ implies both $x^2$ and $y^2$ leave remainder $1$ modulo $5$, so $5$ divides $y^2 - x^2$. Thus, $20$ divides $n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20032,
"subject": "Mathematics (Olympiad)",
"question": "Suppose there are 1999 cups, each either facing upward or downward. In each move, you may select any 100 cups and flip them (change their orientation). Is it possible to make all 1999 cups face downward using such moves? What about if there are 1998 cups?",
"options": [],
"answer": "See solution",
"solution": "No, it is impossible to make all 1999 cups face downward. In each move, if you choose 100 cups where $k$ of them are facing downward and $100-k$ are facing upward, the number of cups facing downward changes by $$(100-k) - k = 100 - 2k.$$ Since the initial number of cups facing downward is even, and each move preserves the parity, it is always even. Thus, with 1999 cups, it is impossible to reach all cups facing downward (which would be odd).\n\nHowever, with 1998 cups, it is possible. For example, if you flip cups $1, 2, \\ldots, 100$ and then cups $2, 3, \\ldots, 101$, only cups $1$ and $101$ are flipped twice, showing you can flip any two cups in two moves. Since $2$ divides $1998$, you can repeat this process to turn all cups downward.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20033,
"subject": "Mathematics (Olympiad)",
"question": "Determine the values of the parameter $\\alpha \\in \\mathbb{R}$ for which the equation\n\n$$\nx^2 + (\\alpha - 2)x - (\\alpha - 1)(2\\alpha - 3) = 0\n$$\n\nhas two roots such that one of them is equal to the square of the other.",
"options": [],
"answer": "See solution",
"solution": "The discriminant is\n$$\nD = (\\alpha - 2)^2 + 4(\\alpha - 1)(2\\alpha - 3) = (3\\alpha - 4)^2,\n$$\nso the equation has roots $x_1 = \\alpha - 1$ and $x_2 = -2\\alpha + 3$.\n\nWe seek $\\alpha$ such that\n$$\nx_1 = x_2^2 \\quad \\text{or} \\quad x_2 = x_1^2.\n$$\nThat is,\n$$\n\\begin{align*}\n\\alpha - 1 &= (-2\\alpha + 3)^2 \\\\\n-2\\alpha + 3 &= (\\alpha - 1)^2\n\\end{align*}\n$$\nSolving these:\n\nFirst equation:\n$$\n\\alpha - 1 = 4\\alpha^2 - 12\\alpha + 9 \\\\\n4\\alpha^2 - 13\\alpha + 10 = 0 \\\\\n\\alpha = 2 \\quad \\text{or} \\quad \\alpha = \\frac{5}{4}\n$$\n\nSecond equation:\n$$\n-2\\alpha + 3 = \\alpha^2 - 2\\alpha + 1 \\\\\n\\alpha^2 = 2 \\\\\n\\alpha = \\sqrt{2} \\quad \\text{or} \\quad \\alpha = -\\sqrt{2}\n$$\n\nThus, the possible values are $\\boxed{2,\\ \\frac{5}{4},\\ \\sqrt{2},\\ -\\sqrt{2}}$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20034,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a fixed acute-angled triangle. Let $E$ and $F$ be points on sides $AC$ and $AB$, respectively, and let $M$ be the midpoint of $EF$. Let the perpendicular bisector of $EF$ intersect line $BC$ at $K$, and let the perpendicular bisector of $MK$ intersect lines $AC$ and $AB$ at $S$ and $T$, respectively. If the quadrilateral $KSAT$ is cyclic, prove that $\\angle KEF = \\angle KFE = \\angle A$.",
"options": [],
"answer": "See solution",
"solution": "Let the circumcircle of quadrilateral $KSAT$ be $\\omega_1$. Let the line $AM$ intersect $ST$ at $N$, and let $AM$ intersect $\\omega_1$ again at $L$, as shown below.\n\n\n\nSince $EF \\parallel TS$ and $M$ is the midpoint of $EF$, $N$ is also the midpoint of $ST$. Moreover, since $K$ and $M$ are symmetric with respect to $ST$, we have $\\angle KNS = \\angle MNS = \\angle LNT$. Thus, $K$ and $L$ are symmetric with respect to the perpendicular bisector of $ST$, so $KL \\parallel ST$.\n\nLet $G$ be the reflection of $K$ over $N$. Then $G$ lies on line $EF$. Without loss of generality, suppose $G$ lies on the ray $MF$. Then\n\n$$\n\\angle KGE = \\angle KNS = \\angle SNM = \\angle KLA = 180^\\circ - \\angle KSA\n$$\n\n(When $K = L$, $\\angle KLA$ refers to the angle between $AL$ and the tangent to $\\omega$ at $L$.) Therefore, $K, G, E, S$ are concyclic. Since $KSGT$ is a parallelogram, $\\angle KEF = \\angle KSG = 180^\\circ - \\angle TKS = \\angle A$. Also, since $KE = KF$, by symmetry $\\angle KFE = \\angle KEF = \\angle A$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20035,
"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\na) Prove that the center $J$ of the circumcircle of triangle $SMN$ lies on a fixed line.\n\nb) 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.\n\n",
"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\nb) 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\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$$\n\\triangle SAD \\sim \\triangle SKL \\sim \\triangle SBC.\n$$\n\nFrom the above pairs of similar triangles, we have the ratio transformation\n\n$$\n\\frac{UB}{UK} = \\frac{AQ}{AD} = \\frac{BP}{BC} = \\frac{AQ + BP}{AD + BC}.\n$$\n\nFinally, since $AQ + BP = 2LK$, according to the property of the midline of the trapezoid and Ptolemy's theorem,\n\n$$\nKA(JA + JB) = JK \\cdot AB\n$$\n\nso we get the following\n\n$$\n\\begin{aligned}\n\\frac{TI}{LK} &= \\frac{AQ + BP}{2LK} \\\\\n&= \\frac{AD + BC}{2LK} \\cdot \\frac{UB}{UK} \\\\\n&= \\frac{JA + JB}{2JK} \\cdot \\frac{UB}{UK} \\\\\n&= \\frac{AB}{2KA} \\cdot \\frac{UB}{UK} \\\\\n&= \\frac{UB}{UK} \\cdot \\sin GKA \\\\\n&= \\frac{GI}{GK}.\n\\end{aligned}\n$$\n\nSo the equality is proved and $GT$ passes through $L$. Therefore, the median from vertex $G$ in triangle $GPQ$ passes through the midpoint of arc $CD$ which does not contain $S$ of the circumcircle of triangle $SCD$, which is a fixed point. $\\square$\n\n",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20036,
"subject": "Mathematics (Olympiad)",
"question": "Given a semicircle $(c)$ with diameter $AB$ and center $O$. On $(c)$, take point $C$ such that the tangent at $C$ intersects the line $AB$ at point $E$. The perpendicular from $C$ to $AB$ meets $AB$ at $D$. On $(c)$, let points $H$ and $Z$ satisfy $CD = CH = CZ$. The line $HZ$ meets $CO$, $CD$, and $AB$ at points $S$, $I$, and $K$ respectively. The line through $I$ parallel to $AB$ meets $CO$ and $CK$ at $L$ and $M$ respectively. Consider the circumcircle $(k)$ of triangle $LMD$, which meets $AB$ and $CK$ again at $P$ and $U$ respectively. Let $(e_1)$, $(e_2)$, $(e_3)$ be the tangents to $(k)$ at $L$, $M$, $P$ respectively, and define $R = (e_1) \\cap (e_2)$, $X = (e_2) \\cap (e_3)$, $T = (e_1) \\cap (e_3)$. Prove that if $Q$ is the center of $(k)$, the lines $RD$, $TU$, $XS$ are concurrent at a point lying on the line $IQ$.",
"options": [],
"answer": "See solution",
"solution": "Since $CH = CZ$, we have $OC \\perp HZ$. So from the cyclic quadrilateral $SODI$ we get\n\n$$\nCS \\cdot CO = CI \\cdot CD. \\qquad (1)\n$$\n\n\n\nDraw the perpendicular $(v)$ to $HC$ at $H$. Let $J$ be the intersection of $(v)$ and $CO$. Then $CJ$ is a diameter of the circle $(O, OA)$ and\n\n$$\nCJ = 2CO. \\qquad (2)\n$$\n\nFrom right triangle $JHC$,\n\n$$\nHC^2 = CS \\cdot CJ. \\qquad (3)\n$$\n\nTherefore, from (1), (2), and (3),\n\n$$\nCS \\cdot \\frac{1}{2}CJ = CI \\cdot CD \\quad \\text{or} \\quad HC^2 = 2CI \\cdot CD. \\qquad (3)\n$$\n\nHowever, $HC = CD$ and thus $CD = 2CI$. Thus, $I$ is the midpoint of $CD$. Also, $LM \\parallel OK$, so $L$ and $M$ are the midpoints of $CO$ and $CK$ respectively. Therefore, the circumcircle $(k)$ of triangle $LMD$ is the *Euler circle* of $COK$ and thus passes through $S$.\n\nWe have $QS = QU$ and from right triangles $OSK$, $OUK$ we get $PS = PU = \\frac{OK}{2}$.\n\nTherefore, $P$ and $Q$ lie on the perpendicular bisector of $SU$. Now, $SU \\parallel TX$ because $QP \\perp (e_3)$. Similarly, $DU \\parallel RT$ and $SD \\parallel RX$.\n\nSince triangles $SUD$ and $XTR$ are homothetic, the lines $RD$, $TU$, $XS$ are concurrent at the center $M$ of homothety.\n\nThe points $I$ and $Q$ are the incenters of homothetic triangles $SUD$ and $XTR$, respectively. Thus, the line $IQ$ passes through $M$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20037,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $x$, $y$, and $z$ such that\n\n$$\n1 + 2^x 3^y = z^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "It is easily seen that for $z = 1, 2, 3$ the given equation has no solution. Let $z \\geq 4$. Then $2^x 3^y = (z-1)(z+1)$.\n\nBoth $z-1$ and $z+1$ cannot be divisible by $3$ (since if they were, $3 \\mid (z+1)-(z-1) = 2$, which is impossible). From $2 \\mid (z-1)(z+1)$, it follows that $z-1$ and $z+1$ are both even, and only one can be divisible by $4$.\n\nHence, we have two cases:\n\n**Case a)** $z+1 = 2 \\cdot 3^y$, $z-1 = 2^{x-1}$\n\nSubtracting, $z+1 - (z-1) = 2 \\cdot 3^y - 2^{x-1} = 2$, so $3^y - 2^{x-2} = 1$.\n\n- For $x=2$, $3^y - 1 = 1 \\implies 3^y = 2$ (no solution).\n- For $x=3$, $3^y - 2 = 1 \\implies 3^y = 3 \\implies y=1$, $z=5$. So $(x, y, z) = (3, 1, 5)$.\n- For $x \\geq 4$, $3^y \\equiv 1 \\pmod{4}$, so $y$ is even. Let $y = 2y_1$.\n\nSubstitute: $3^{2y_1} - 2^{x-2} = 1 \\implies (3^{y_1} - 1)(3^{y_1} + 1) = 2^{x-2}$.\n\nThis implies $3^{y_1} - 1 = 2$, $3^{y_1} + 1 = 2^{x-3}$, so $y_1 = 1$, $y = 2$, $x = 5$, $z = 17$. So $(x, y, z) = (5, 2, 17)$.\n\n**Case b)** $z+1 = 2^{x-1}$, $z-1 = 2 \\cdot 3^y$\n\nSubtracting: $2^{x-1} - 2 \\cdot 3^y = 2 \\implies 2^{x-2} - 3^y = 1$.\n\n- For $y=1$, $2^{x-2} - 3 = 1 \\implies 2^{x-2} = 4 \\implies x=4$, $z=7$. So $(x, y, z) = (4, 1, 7)$.\n- For $y \\geq 2$, $2^{x-2} \\equiv 1 \\pmod{3}$, so $x-2$ is even, $x-2 = 2x_1$.\n\nSubstitute: $3^y = 2^{2x_1} - 1 = (2^{x_1} - 1)(2^{x_1} + 1)$.\n\nThis implies $2^{x_1} - 1 = 1$ or $2^{x_1} - 1 = 3$.\n- If $2^{x_1} - 1 = 1$, $x_1 = 1$, $x = 4$ (already found).\n- If $2^{x_1} - 1 = 3$, $x_1 = 2$, $x = 6$, $3^y = 15$ (no solution).\n\n**Final solutions:**\n\n$$\n(x, y, z) = (3, 1, 5),\\ (5, 2, 17),\\ (4, 1, 7).\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20038,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $k$ and all non-negative integers $m$ for which the equation\n$$\nn_1^2 + n_2^2 + \\dots + n_k^2 = 5^{k+m}\n$$\nhas positive integer solutions $(n_1, n_2, \\dots, n_k)$.",
"options": [],
"answer": "See solution",
"solution": "Let $k = 2$. If $n_1 = 5^l$ and $n_2 = 2 \\cdot 5^l$, then $n_1^2 + n_2^2 = 5^{2l+1}$. If $n_1 = 3 \\cdot 5^l$ and $n_2 = 4 \\cdot 5^l$, then $n_1^2 + n_2^2 = 5^{2l+2}$. In both cases, $l$ is an arbitrary non-negative integer. Hence, the equation $n_1^2 + n_2^2 = 5^l$ has solutions for all positive integers $t$.\n\nNext, consider the case $k = 3$. Let $n_1 = 3a_1$, $n_2 = 4a_1$, and $n_3 = 5^2 a_2$. Then $n_1^2 + n_2^2 + n_3^2 = 5^2(a_1^2 + a_2^2)$. For every non-negative integer $m$, there exists a solution to the equation $a_1^2 + a_2^2 = 5^{1+m}$, so the equation $n_1^2 + n_2^2 + n_3^2 = 5^{3+m}$ has solutions for all non-negative integers $m$.\n\nFor $k \\ge 3$, proceed by induction. Assume that the equation $a_1^2 + a_2^2 + \\dots + a_l^2 = 5^{l+m}$ has positive integer solutions for all non-negative integers $m$ and for all positive integers $l$ such that $2 \\le l \\le k$. We wish to show that the equation $n_1^2 + n_2^2 + \\dots + n_k^2 + n_{k+1}^2 = 5^{k+1+m}$ also has positive integer solutions for all non-negative integers $m$.\n\nSince $5^{k+1+m} = 5^{k+m} + 4 \\cdot 5^{k+m}$ and $k > k-1 \\ge 2$, the induction hypothesis implies that for all non-negative integers $i$, there exist positive integers $a_1, a_2, \\dots, a_{k-1}$ such that\n$$\na_1^2 + a_2^2 + \\dots + a_{k-1}^2 = 5^{k-1+i}\n$$\nIn particular, for $i = m + 1$, there exist positive integers $a_1, a_2, \\dots, a_{k-1}$ such that\n$$\na_1^2 + a_2^2 + \\dots + a_{k-1}^2 = 5^{k+m}\n$$\nAs shown earlier, there exist positive integers $b_1$ and $b_2$ such that $b_1^2 + b_2^2 = 5^{k+m}$. So,\n$$\n\\begin{aligned}\n5^{k+1+m} &= 5^{k+m} + 4 \\cdot 5^{k+m} \\\\\n&= a_1^2 + a_2^2 + \\dots + a_{k-1}^2 + 4(b_1^2 + b_2^2) \\\\\n&= a_1^2 + a_2^2 + \\dots + a_{k-1}^2 + (2b_1)^2 + (2b_2)^2.\n\\end{aligned}\n$$\nSince $m$ was arbitrary, this concludes the induction.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20039,
"subject": "Mathematics (Olympiad)",
"question": "Prove that\n\n$$\n\\sum_{\\text{cyc}} (x + y) \\sqrt{(z + x)(z + y)} \\geq 4(xy + yz + zx)\n$$\n\nfor all positive real numbers $x, y, z$.",
"options": [],
"answer": "See solution",
"solution": "We will obtain the inequality by adding the inequalities\n\n$$\n(x + y) \\sqrt{(z + x)(z + y)} \\geq 2xy + yz + zx\n$$\n\nfor cyclic permutations of $x, y, z$.\n\nSquaring both sides of this inequality, we obtain\n\n$$\n(x + y)^2 (z + x)(z + y) \\geq 4x^2y^2 + y^2z^2 + z^2x^2 + 4xy^2z + 4x^2yz + 2xyz^2\n$$\n\nwhich is equivalent to\n\n$$\nx^3 y + x y^3 + z(x^3 + y^3) \\geq 2x^2y^2 + xyz(x + y)\n$$\n\nwhich can be rearranged to\n\n$$\n(xy + yz + zx)(x - y)^2 \\geq 0,\n$$\n\nwhich is clearly true.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20040,
"subject": "Mathematics (Olympiad)",
"question": "For an integer $n > 1$, let $gpf(n)$ denote the greatest prime factor of $n$. A *strange pair* is an unordered pair of distinct primes $p$ and $q$ such that $\\{p, q\\} = \\{gpf(n), gpf(n + 1)\\}$ for no integer $n > 1$. Prove that there exist infinitely many strange pairs.",
"options": [],
"answer": "See solution",
"solution": "We show that there are infinitely many strange pairs of the form $\\{2, q\\}$ where $q$ is an odd prime.\n\n**Lemma.** If some primes $2 < q_1 < q_2$ satisfy $ord_{q_1}(2) = ord_{q_2}(2)$, then $\\{2, q_1\\}$ is a strange pair.\n\n*Proof.* Suppose, for contradiction, that $2 = \\text{gpf}(n)$ and $q_1 = \\text{gpf}(n+1)$. Then $n = 2^k$ for some $k$, and $q_1 \\mid 2^k + 1$. Thus $q_1 \\mid 2^{2k} - 1$, so $ord_{q_2}(2) = ord_{q_1}(2) \\mid 2k$. Therefore, $q_2 \\mid 2^{2k} - 1 = (2^k - 1)(2^k + 1)$, but $q_2 \\nmid 2^k - 1$, so $q_2 \\mid 2^k + 1$. Thus $\\text{gpf}(n+1) \\geq q_2$, a contradiction.\n\nSimilarly, if $2 = \\text{gpf}(n+1)$ and $q_1 = \\text{gpf}(n)$, then $n+1 = 2^k$, so $ord_{q_2}(2) = ord_{q_1}(2) \\mid k$ and $q_2 \\mid 2^k - 1$. Thus $\\text{gpf}(n+1) \\geq q_2$, a contradiction. $\\square$\n\nIt remains to show that there exist infinitely many disjoint pairs of primes $q_1 < q_2$ satisfying the Lemma's conditions.\n\nLet $p = 2r - 1 > 5$ be a prime, and let $N = 2^{2p} + 1$. We prove:\n\n1. $N$ has at least two distinct prime factors greater than $5$.\n2. $ord_q(2) = 4p$ for every prime factor $q > 5$ of $N$.\n\nThus, every prime $p > 5$ provides a pair of odd primes satisfying the Lemma. Moreover, (2) shows that distinct primes $p > 5$ provide disjoint such pairs, whence the conclusion.\n\nTo prove (1), note $3 \\nmid N$, and write $N = (4+1)(4^{p-1} - 4^{p-2} + \\cdots + 1) \\equiv 5p \\pmod{25}$, so $25 \\nmid N$.\n\nNext, $N = (2^p + 1)^2 - 2^{p+1} = (2^p - 2^r + 1)(2^p + 2^r + 1)$. The two factors are coprime (since they are odd and their difference is $2^{r+1}$), and each is larger than $5$, so each has a prime factor greater than $5$. This proves (1).\n\nTo prove (2), let $q > 5$ be a prime factor of $N$. Then $ord_q(2) \\mid 4p$, since $q \\mid N \\mid 2^{4p} - 1$. If $ord_q(2) < 4p$, then either $ord_q(2) \\mid 2p$ or $ord_q(2) \\mid 4$. The former is impossible since $2^{2p} - 1 = N - 2 \\equiv -2 \\pmod{q}$, the latter since $q \\nmid 15 = 2^4 - 1$. This proves (2) and completes the proof.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20041,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be a triangle. Let $D$ be the foot of the internal bisector of angle $A$. The perpendicular from $D$ to the tangent $AT$ (where $T$ belongs to $BC$) to the circumscribed circle of $ABC$ intersects the altitude $AH_a$ at the point $I$ (where $H_a$ belongs to $BC$).\n\nIf $P$ is the midpoint of $AB$ and $O$ is the center of the circumcircle, $TI$ intersects $AB$ at $M$ and $PT$ intersects $AD$ at $F$. Prove that $MF$ is perpendicular to $AO$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $Q$ be the midpoint of $AC$ and $N$ the intersection of $AD$ and $PQ$. Then $N$ is the midpoint of $AD$. As $DE$ is perpendicular to $AT$, with $E$ the intersection point of $DI$ and $AT$, and as $OA$ is perpendicular to $AT$, we get that $DE$ is parallel to $OA$, so the angles $OAN$ and $ADE$ are equal. As a consequence, triangles $ADE$ and $DAH_a$ are congruent.\n\nIn particular, angle $DAT$ equals angle $H_aAD$, that is, $ATD$ is isosceles and point $I$ is the orthocenter of $ABC$.\n\nSo, $TI$ is perpendicular to $AD$, and the intersection point of $TI$ and $AD$ is the midpoint of $AD$ ($N$). The four points $M, N, I, T$ are collinear.\n\nWe apply Ceva's theorem in triangle $APT$ with cevians $PN$, $AD$, and $TM$:\n\n$$\n\\frac{FP}{FT} \\cdot \\frac{MA}{PM} = 1 \\quad \\Leftrightarrow \\quad \\frac{PF}{TF} = \\frac{MP}{MA}.\n$$\n\n(Observe that $NP$ cuts $AT$ at its midpoint.)\n\nSo, $MF$ is parallel to $AT$, and thus $MF$ is perpendicular to $AO$, as claimed.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20042,
"subject": "Mathematics (Olympiad)",
"question": "Let $H$ denote the intersection of the altitudes $AD$ and $CE$ of an acute triangle $ABC$. Let $M$ and $N$ denote the midpoints of the sides $AB$ and $BC$ respectively. The rays $MH$ and $NH$ intersect the circumcircle $\\omega$ of $ABC$ at points $K$ and $L$ respectively. If the circumcircles of the triangles $EHK$ and $DHL$ intersect $\\omega$ again at $P$ and $Q$, show that the points $D, E, P, Q$ lie on a common circle.",
"options": [],
"answer": "See solution",
"solution": "Let $O$ denote the center of the circumcircle $\\omega$ and denote $\\alpha := \\angle BAC$, $\\beta := \\angle ABC$, and $\\gamma := \\angle BCA$. Let $A' := (AO) \\cap \\omega$. First we show that $N \\in A'H$.\n\n\n\nIndeed, let $N' := AH \\cap BC$. Since $O$ is the midpoint of $AA'$, the centroid $G$ of the triangle $AHA'$ is the point on $OH$ satisfying $HG = 2GO$. By Euler's theorem, $G$ is also the centroid of the triangle $ABC$. It follows that $HBA'C$ is a parallelogram and thus $N' = N$. Now using the fact that $PEHK$, $BDHE$ and $PKCA$ are circumscribed, we compute\n\n$$\n\\begin{align*}\n\\angle PED &= \\angle PEH - \\angle DEH \\\\\n&= (180^\\circ - \\angle PKH) - \\angle DBH \\\\\n&= (180^\\circ - (\\angle PKC - 90^\\circ)) - (90^\\circ - \\gamma) \\\\\n&= 180^\\circ - \\angle PKC + \\gamma \\\\\n&= \\angle PAC + \\gamma \\\\\n&= \\angle PAB + \\alpha + \\gamma.\n\\end{align*}\n$$\n\nSimilarly, using the fact that $PQAL$, $QDHL$ and $LEHA$ are circumscribed, we compute\n\n$$\n\\begin{align*}\n\\angle PQD &= \\angle PQL + \\angle LQD \\\\\n&= \\angle PAL + (180^\\circ - \\angle LHD) \\\\\n&= \\angle PAL + (180^\\circ - (\\angle LHE + \\angle EHD)) \\\\\n&= \\angle PAL - \\angle LHE + \\beta \\\\\n&= \\angle PAL - \\angle LAE + \\beta \\\\\n&= \\beta - \\angle PAB.\n\\end{align*}\n$$\n\nIt follows that $\\angle PED + \\angle PQD = \\alpha + \\beta + \\gamma = 180^\\circ$, hence $DEPQ$ is circumscribed.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20043,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for all $x, y \\in \\mathbb{R}$,\n\n$$\nf(f(x)f(y)) + f(x+y) = f(xy).\n$$",
"options": [],
"answer": "See solution",
"solution": "The solutions are $f(x) = 0$, $f(x) = x - 1$, and $f(x) = 1 - x$. It is straightforward to verify that these satisfy the given functional equation.\n\nLet us analyze the equation:\n\n$$\nf(f(x)f(y)) + f(x+y) = f(xy) \\quad \\text{for all } x, y \\in \\mathbb{R}. \\qquad (1)\n$$\n\n**Case 1:** There exists $a \\neq 1$ such that $f(a) = 0$.\n\nSet $x = a$ in (1):\n\n$$\nf(0) + f(a + y) = f(ay).\n$$\n\nSince $a \\neq 1$, we can solve $a + y = ay$ for $y$, which allows us to deduce $f(0) = 0$. Then, setting $x = 0$ in (1) gives $f(y) = 0$ for all $y$.\n\n**Case 2:** $f(1) = 0$ and $f(a) \\neq 0$ for all $a \\neq 1$.\n\nFrom $f(f(0)^2) = 0$, we get $f(0)^2 = 1$. Note that if $f(x)$ is a solution, so is $-f(x)$. So, consider $f(0) = 1$.\n\nSet $y = 0$ in (1):\n\n$$\nf(f(x)) + f(x) = 1 \\quad \\text{for all } x. \\qquad (2)\n$$\n\nReplace $x$ with $f(x)$ in (2):\n\n$$\nf(f(f(x))) + f(f(x)) = 1. \\qquad (3)\n$$\n\nSubtract (2) from (3):\n\n$$\nf(f(f(x))) = f(x). \\qquad (4)\n$$\n\nIf $f$ is injective, (4) implies $f(f(x)) = x$. Plugging into (2) gives $f(x) = 1 - x$.\n\nSimilarly, if $f(0) = -1$, a similar argument gives $f(x) = x - 1$.\n\nIt remains to prove injectivity. Suppose $f(a) = f(b)$. From $f(x + 1) = f(x) + 1$ (by setting $y = 1$ in (1)), induction gives $f(x + n) = f(x) + n$ for all $n \\in \\mathbb{N}^+$. Then,\n\n$$\nf(a + n + 1) = f(b + n) + 1. \\qquad (6)\n$$\n\nNow, consider the system:\n\n$$\nx + y = a + n + 1 \\tag{8a}\n$$\n$$\nxy = b + n \\tag{8b}\n$$\n\nBy Vieta's formulas, $x$ and $y$ are roots of $z^2 - (a + n + 1)z + b + n = 0$, which has real solutions for large $n$.\n\nChoose such $n$ and plug $x, y$ into (1):\n\n$$\nf(f(x)f(y)) + f(a + n + 1) = f(b + n).\n$$\n\nFrom (6), $f(a + n + 1) = f(b + n) + 1$, so\n\n$$\nf(f(x)f(y)) + 1 = 0 \\implies f(f(x)f(y)) = -1.\n$$\n\nBut in this case, $f(x)f(y) = 0$, so $f(x) = 0$ or $f(y) = 0$. Since we are in case 2, $x = 1$. Then (8b) gives $y = b + n$, and (8a) gives $a = b$. Thus, $f$ is injective.\n\nTherefore, the only solutions are $f(x) = 0$, $f(x) = x - 1$, and $f(x) = 1 - x$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20044,
"subject": "Mathematics (Olympiad)",
"question": "In the plane, fix two points $A$ and $B$ ($A \\neq B$). Consider a point $C$ moving in the plane such that $\\angle ACB = \\alpha$, where $\\alpha$ is a given angle ($0^\\circ < \\alpha < 180^\\circ$). The circle with center $I$, inscribed in triangle $ABC$, touches the sides $AB$, $BC$, and $CA$ at points $D$, $E$, and $F$, respectively. The lines $AI$ and $BI$ intersect $EF$ at $M$ and $N$, respectively. Show that:\n\n1. The segment $MN$ has constant length.\n2. The circumcircle of triangle $DMN$ passes through a fixed point.\n\n",
"options": [],
"answer": "See solution",
"solution": "1. Consider triangle $AFM$:\n\n$$\n\\begin{align*}\n\\angle AMF &= 180^\\circ - (\\angle MFA + \\angle FAM) = 90^\\circ - (\\angle EFI + \\angle FAM) = 90^\\circ - (\\angle ECI + \\angle FAM) \\\\\n&= 90^\\circ - \\left(\\frac{C}{2} + \\frac{A}{2}\\right) = \\frac{B}{2} = \\angle IBA.\n\\end{align*}\n$$\n\nWe also have $\\angle NIM = \\angle AIB$.\n\nHence, $\\triangle IMN \\sim \\triangle IBA$. (*)\n\nDraw $IH \\perp MN$, then:\n\n$$\n\\frac{MN}{BA} = \\frac{IH}{ID} = \\frac{IH}{IF} = \\sin \\angle EFI = \\sin \\frac{\\alpha}{2}.\n$$\n\nConsequently, $MN = BA \\cdot \\sin \\frac{\\alpha}{2}$, which is constant.\n\n2. The points $F$ and $D$ are symmetric with respect to line $AM$. By (*), $\\angle IMD = \\angle IBD$, so quadrilateral $IMBD$ is cyclic.\n\nThus, $\\triangle BMA$ is right at $M$. Let $P$ be the midpoint of $AB$, then $\\angle BPM = 2\\angle BAM = \\angle BAC$.\n\nSince $E$ and $D$ are symmetric with respect to $BN$, by (*)\n\n$$\n\\angle MND = 2\\angle INM = 2\\angle IAB = \\angle BAC.\n$$\n\nThus, $\\angle BPM = \\angle MND$, so $M$, $N$, $D$, $P$ are concyclic.\n\nTherefore, the circumcircle of $\\triangle DMN$ passes through the fixed point $P$, the midpoint of $AB$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20045,
"subject": "Mathematics (Olympiad)",
"question": "Let $s, t \\neq 0$, and define a pair $(s, t)$ as *good* if, for any prime divisor $p$ of $s^2 + t^2$, either $p$ divides both $s$ and $t$, or there exists $k \\in \\mathbb{Z}^+$ such that $\\gcd(s + k t, t - k s) \\geq p$ after $k$ moves, where a move transforms $(s, t)$ to $(s + k t, t - k s)$.\n\n(a) Show that $(s, t)$ is a good pair.",
"options": [],
"answer": "See solution",
"solution": "Yes. Since $s, t \\neq 0$, $s^2 + t^2 \\geq 2$. Let $p$ be any prime divisor of $s^2 + t^2$.\n\nIf $p \\mid t$, then $p \\mid s$, so $\\gcd(s, t) \\geq p$. Thus, $(s, t)$ is good.\n\nIf $p \\nmid t$, then there exists $k \\in \\mathbb{Z}^+$ such that $t k \\equiv -s \\pmod{p}$. Then $p \\mid s + k t$.\n\nAlso,\n$$\nt - k s = t - (-t^{-1} s) s = t^{-1}(t^2 + s^2) \\equiv 0 \\pmod{p}.\n$$\nThus, $\\gcd(s + k t, t - k s) \\geq p$. This shows $(s, t)$ is a good pair, as we can apply $k$ moves to $(s, t)$ to obtain $(s + k t, t - k s)$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20046,
"subject": "Mathematics (Olympiad)",
"question": "Show that $2010$ cannot be written as the difference of two squares.",
"options": [],
"answer": "See solution",
"solution": "Assume there are integers $x$ and $y$ such that\n\n$$\n2010 = x^2 - y^2 = (x - y)(x + y).\n$$\n\nThe factors $x - y$ and $x + y$ have the same parity.\n\n- If both are even, their product would be divisible by $4$, but $2010$ is not.\n- If both are odd, their product would be odd, but $2010$ is even.\n\nTherefore, there are no such integers.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20047,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle with $AB \\neq AC$. Let the angle bisector of $\\angle BAC$ intersect $BC$ at $P$, and intersect the perpendicular bisector of segment $BC$ at $Q$. Prove that\n$$\n\\frac{PQ}{AQ} = \\left(\\frac{BC}{AB+AC}\\right)^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "We first show that $Q$ lies on the circumcircle of triangle $ABC$.\n\n\n\nLet the perpendicular bisector of $BC$ intersect the circumcircle of $ABC$ at $Q'$. Since $Q'$ bisects the arc $BC$, we have $\\angle BAQ' = \\angle CAQ'$. Thus $Q'$ lies on the angle bisector of $\\angle BAC$. Hence $Q = Q'$.\n\nLet $BC = a$, $AC = b$, and $AB = c$. From the Angle Bisector Theorem we have\n$$\nBP = \\frac{ac}{b+c} \\quad \\text{and} \\quad PC = \\frac{ab}{b+c}. \\qquad (1)\n$$\nThe triangles $ABP$ and $AQC$ are similar, thus\n$$\n\\frac{c}{AP} = \\frac{AQ}{b}. \\qquad (2)\n$$\nOn the other hand, computing the power of point $P$ with respect to the circumcircle of $ABC$ we get\n$$\nBP \\cdot PC = AP \\cdot PQ. \\qquad (3)\n$$\nCombining (1), (2), and (3) yields\n$$\n\\frac{PQ}{AQ} = \\frac{BP \\cdot PC}{bc} = \\frac{ac}{b+c} \\cdot \\frac{ab}{b+c} \\cdot \\frac{1}{bc} = \\left(\\frac{a}{b+c}\\right)^2 = \\left(\\frac{BC}{AB+AC}\\right)^2\n$$\nas required.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20048,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs $\\big(p, q\\big)$ of prime numbers such that\n\n$$\np(p^2 - p - 1) = q(2q + 3).\n$$",
"options": [],
"answer": "See solution",
"solution": "We show that the only solution is $(p, q) = (13, 31)$.\n\nFirst, suppose that $p = q$. Then $p^2 - p - 1 = 2q + 3 = 2p + 3$, so $(p-4)(p+1) = 0$. As neither $4$ nor $-1$ are prime numbers, there are no solutions $(p, q)$ with $p = q$.\n\nHence $p \\neq q$. From the equation, it follows that $p \\mid 2q+3$ and $q \\mid p^2-p-1$. As $2q+3$ and $p^2-p-1$ are positive, it follows that $p \\leq 2q+3$ and $q \\leq p^2-p-1$. To find a better lower bound for $p$, we multiply these relations:\n\n$$\n\\begin{aligned}\npq &\\mid (2q + 3)(p^2 - p - 1) \\\\\n&= 2qp^2 - 2qp - 2q + 3(p^2 - p - 1) \\\\\n&= 3(p^2 - p - 1) - 2q.\n\\end{aligned}\n$$\n\nNote that $3(p^2 - p - 1) - 2q \\geq 3q - 2q = q > 0$. Therefore, the relation above yields\n\n$$\n\\begin{aligned}\npq &\\leq 3(p^2 - p - 1) - 2q \\\\\n&= 3p^2 - 3p - (2q + 3) \\\\\n&\\leq 3p^2 - 3p - p \\\\\n&= 3p^2 - 4p.\n\\end{aligned}\n$$\n\nAdding $4p$ to both sides, then dividing both sides by $p$, we find that $q + 4 \\leq 3p$, so\n\n$$\n\\frac{2q+3}{6} < \\frac{q+4}{3} \\leq p.\n$$\n\nAs $p$ is a divisor of $2q+3$, we deduce that $2q+3 = kp$ with $k \\in \\{1, 2, 3, 4, 5\\}$.\n\n- If $k = 1$, then $2q + 3 = p$ and therefore also $q = p^2 - p - 1$. This implies that $p = 2q + 3 = 2(p^2 - p - 1) + 3 = 2p^2 - 2p + 1$. This in turn implies that $(2p - 1)(p - 1) = 0$, but this equation does not have prime solutions.\n\n- Note that $k = 2$ and $k = 4$ are not possible either, since then $kp$ would be even, while $2q + 3$ is odd.\n\n- If $k = 3$, then it follows from $2q + 3 = 3p$ that $3 \\mid q$, and therefore $q = 3$. Hence $p = \\frac{2q+3}{3} = 3$. However, this does not give a solution of the given equation.\n\nTherefore $k = 5$. Then we have $5p = 2q + 3$, and therefore $5q = p^2 - p - 1$ as well. Substituting this gives $25p = 5(2q + 3) = 2(p^2 - p - 1) + 15 = 2p^2 - 2p + 13$, and therefore $(p - 13)(2p - 1) = 0$. As $p$ is prime, it follows that $p = 13$, and that $q = \\frac{5p-3}{2} = 31$. Note that $(p, q) = (13, 31)$ is indeed a solution of the given equation, so it is the only solution of the given equation. $\\square$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20049,
"subject": "Mathematics (Olympiad)",
"question": "Suppose the numbers $\\{1, 2, \\dots, 25\\}$ are arranged in some order in a $5 \\times 5$ array, with each number appearing exactly once. Find the maximal positive integer $k$ such that, for any arrangement, there is always a $2 \\times 2$ subarray whose numbers have a sum not less than $k$.",
"options": [],
"answer": "See solution",
"solution": "We will prove that $k_{\\text{max}} = 45$.\n\nFirst, consider all possible $2 \\times 2$ subarrays in the $5 \\times 5$ array. The total sum of all numbers is $1 + 2 + \\dots + 25 = 325$.\n\nBy an averaging argument, for any arrangement, there must exist a $2 \\times 2$ subarray whose sum is at least $45$.\n\nTo show that $45$ is the best possible, consider the following arrangement:\n\n\n\nIn this array, every $2 \\times 2$ subarray has a sum less than or equal to $45$. Thus, $k_{\\text{max}} = 45$.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 20050,
"subject": "Mathematics (Olympiad)",
"question": "Given two positive integers $m$ and $n$, determine the minimum number of distinct roots the polynomial $$\\prod_{k=1}^{m} (f + k)$$ may have, as $f$ runs through the set of polynomials of degree $n$ with complex coefficients.",
"options": [],
"answer": "See solution",
"solution": "The required minimum is $n(m - 1) + 1$ and is achieved by any of the polynomials $X^n - k$, for $k = 1, \\dots, m$.\n\nWe now proceed to prove that $n(m - 1) + 1$ is a global lower bound in the setting under consideration.\n\nFor any $f \\in \\mathbb{C}[X]$, $f \\neq 0$, and any $z \\in \\mathbb{C}$, let $\\text{ord}_z f = \\text{ord}_{X-z} f$ be the highest power of $X-z$ dividing $f$. Clearly, $\\text{ord}_z f = 0$ for all but finitely many $z$, $Z(f) = \\{z : z \\in \\mathbb{C}, \\text{ord}_z f \\neq 0\\}$ is precisely the set of distinct roots of $f$, and (with the customary convention that empty sums are zero)\n\n$$\n\\sum_{z \\in \\mathbb{C}} \\mathrm{ord}_z f = \\sum_{z \\in Z(f)} \\mathrm{ord}_z f = \\mathrm{deg} f.\n$$\n\nRewrite the latter as\n\n$$\n|Z(f)| + \\sum_{z \\in Z(f)} (\\mathrm{ord}_z f - 1) = \\mathrm{deg} f,\n$$\n\nand notice that\n\n$$\n\\sum_{z \\in Z(f)} (\\mathrm{ord}_z f - 1) = \\sum_{z \\in \\mathbb{C}} \\mathrm{ord}_z (f, f'),\n$$\n\nwhere $f'$ is the derivative of $f$, and $(f, f')$ is the highest common factor of $f$ and $f'$, to get\n\n$$\n|Z(f)| + \\sum_{z \\in \\mathbb{C}} \\mathrm{ord}_z (f, f') = \\mathrm{deg} f. \\quad (*)\n$$\n\nGiven $f \\in \\mathbb{C}[X]$, $f \\neq 0$, let $g = \\prod_{k=1}^{m} (f + a_k)$, where $m$ is a positive integer and the $a_k$ are pairwise distinct complex numbers, and write $(*)$ for $g$:\n\n$$\n|Z(g)| + \\sum_{z \\in \\mathbb{C}} \\mathrm{ord}_z (g, g') = \\mathrm{deg} g = m \\mathrm{deg} f.\n$$\n\nSince $g' = f' \\sum_{k=1}^{m} \\prod_{j \\neq k} (f + a_j)$, and the polynomials $f + a_k$ are pairwise coprime if $\\mathrm{deg} f \\ge 1$, it follows that $(g, g')$ divides $f'$, so\n\n$$\n\\sum_{z \\in \\mathbb{C}} \\mathrm{ord}_z (g, g') \\le \\sum_{z \\in \\mathbb{C}} \\mathrm{ord}_z f' = \\mathrm{deg} f' = \\mathrm{deg} f - 1;\n$$\n\nnotice that this would make no sense if $\\mathrm{deg} f = 0$. Consequently,\n\n$$\n|Z(g)| \\ge (m - 1) \\mathrm{deg} f + 1,\n$$\n\nand the conclusion follows.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20051,
"subject": "Mathematics (Olympiad)",
"question": "Determine all positive integers $n$ for which\n$$\n\\frac{n^2 + 1}{\\lfloor \\sqrt{n} \\rfloor^2 + 2}\n$$\nis an integer.\n(Here $\\lfloor r \\rfloor$ denotes the greatest integer less than or equal to $r$.)",
"options": [],
"answer": "See solution",
"solution": "Let $m = \\lfloor \\sqrt{n} \\rfloor$ ($m \\in \\mathbb{N}^+$).\n\nTherefore $m \\leq \\sqrt{n} < m + 1$, so $m^2 \\leq n < m^2 + 2m + 1$. Hence $n = m^2 + k$ where $0 \\leq k \\leq 2m$.\n\nWe require $m^2 + 2 \\mid (m^2 + k)^2 + 1$. Thus,\n$$\n\\begin{aligned}\n(m^2 + k)^2 + 1 &\\equiv 0 \\pmod{m^2 + 2} \\\\\n(k - 2)^2 + 1 &\\equiv 0 \\pmod{m^2 + 2}\n\\end{aligned}\n$$\nSo $(k - 2)^2 + 1 = l(m^2 + 2)$ for some integer $l > 0$.\n\nAlso, since $k \\leq 2m$,\n$$\n\\begin{aligned}\nl(m^2 + 2) &= (k - 2)^2 + 1 \\\\\n&< k^2 + 5 \\\\\n&\\leq 4m^2 + 5\n\\end{aligned}\n$$\nThus $l < 4$. We consider cases:\n\n**Case 1:** $l = 1$. Then $(k - 2)^2 + 1 = m^2 + 2$.\nThis rearranges to $(k - 2 + m)(k - 2 - m) = 1$, so $m = 0$, a contradiction.\n\n**Case 2:** $l = 2$. Then $(k - 2)^2 = 2m^2 + 3$.\nIf $3 \\mid m$, then $3 \\mid k - 2$, so $(k - 2)^2 \\equiv 0 \\pmod{9}$, but $2m^2 + 3 \\equiv 3 \\pmod{9}$, contradiction.\nIf $3 \\nmid m$, then $m^2 \\equiv 1 \\pmod{3}$, so $(k - 2)^2 \\equiv 2 \\pmod{3}$, impossible.\n\n**Case 3:** $l = 3$. Then $(k - 2)^2 = 3m^2 + 5$. Thus $(k - 2)^2 \\equiv 2 \\pmod{3}$, impossible.\n\nTherefore, there is no positive integer $n$ satisfying the given condition.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20052,
"subject": "Mathematics (Olympiad)",
"question": "Let $m_1 < m_2 < \\dots < m_n$ be the number of points of each colour. We call $m_1, m_2, \\dots, m_n$ the colour distribution. Then $m_1 + \\dots + m_n = 2012$ and the number of multi-coloured sets is $M = m_1 m_2 \\dots m_n$.\n\nFind the colour distribution that maximizes $M$.",
"options": [],
"answer": "See solution",
"solution": "We analyze the properties of the colour distribution:\n\n1. $m_1 > 1$. If $m_1 = 1$, then $M = m_1 m_2 \\dots m_n < m_2 m_3 \\dots m_{n-1} (1 + m_n)$, so using $n-1$ colours with distribution $m_2, m_3, \\dots, m_{n-1}, (1+m_n)$ gives a larger $M$.\n\n2. $m_{i+1} - m_i \\le 2$ for all $i$. If $m_{k+1} - m_k \\ge 3$, replacing $m_k, m_{k+1}$ by $m_k + 1, m_{k+1} - 1$ increases $M$.\n\n3. $m_{i+1} - m_i = 2$ for at most one $i$. If there are two such gaps, adjusting as above increases $M$.\n\n4. $m_{i+1} - m_i = 2$ for exactly one $i$. If all gaps are $1$, then $m_1 + \\dots + m_n = n m_1 + \\frac{n(n-1)}{2} = 2012$. Solving, $n(2m_1 - 1 + n) = 8 \\cdot 503$. Since $503$ is prime, $n=8$ and $m_1=248$, but splitting $m_1$ into $2, 246$ (with $n+1$ colours) gives a larger $M$.\n\n5. $m_1 = 2$. If $m_n - m_{n-1} = 2$, then $m_1 + \\dots + m_n = n m_1 + \\frac{n(n-1)}{2} + 1 = 2012$. Solving, $n(2m_1 - 1 + n) = 2 \\cdot 2011$, and since $2011$ is prime, $n=2$ and $m_1=1005$, which leads to a contradiction as above. Thus $m_n - m_{n-1} = 1$ and $m_{i+1} - m_i = 2$ for some $1 \\le i \\le n-2$. If $m_1 \\ge 3$, splitting $m_{i+2}$ into $2, m'$ with $m' = m_{i+2} - 2$ increases $M$, so $m_1 = 2$.\n\nTherefore, the maximum $M$ is achieved for $n=61$ colours with the distribution $2, 4, 5, 6, \\dots, 63$ (i.e., $m_1=2$, $m_2=4$, $m_3=5$, ..., $m_{61}=63$).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20053,
"subject": "Mathematics (Olympiad)",
"question": "In the diagram, $A$, $B$, $C$, $D$, and $E$ must each be replaced by one of five consecutive positive integers, not necessarily in that order. The numbers inside the triangle add up to $29$. The numbers inside the circle add up to $47$. The numbers inside the square add up to $30$. All five numbers add up to $75$. What is the value of $C$?\n\n",
"options": [],
"answer": "See solution",
"solution": "To have a sum of $75$, the five consecutive numbers must be $13$, $14$, $15$, $16$, and $17$. Since the numbers in the triangle total $29$, the two possibilities are $13$ and $16$ or $14$ and $15$. Since the numbers in the square total $30$, the two possibilities are $14$ and $16$ or $13$ and $17$. Since the numbers in the circle total $47$, the only possibility is $14$, $16$, and $17$. The only possible letter that could represent $15$ is thus $A$, from which it follows that $B = 14$, $E = 13$, $D = 17$, and finally $C = 16$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20054,
"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$$\n\\begin{array}{l} (bg)(bg)(gb) \\dots (..) \\underbrace{gg \\dots g}_{t \\ge 2} b \\dots, \\\\ (bg)(bg)(gb) \\dots (..) \\underbrace{bb \\dots b}_{t \\ge 2} g \\dots \\end{array}\n$$\n\nThen, it is easy to check that, by moving one girl or one boy as shown in the arrangements above, 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.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20055,
"subject": "Mathematics (Olympiad)",
"question": "a) Find the smallest positive integer which, when multiplied by $2520$, gives a square of a positive integer.\n\nb) Prove that the sum of two consecutive odd integers is divisible by $4$.",
"options": [],
"answer": "See solution",
"solution": "a) For $2520$ we have $2520 = 2^3 \\cdot 3^2 \\cdot 5 \\cdot 7$. To obtain a perfect square, we need to multiply by $2 \\cdot 5 \\cdot 7 = 70$. Thus, $2520 \\cdot 70 = 420^2$, so the desired number is $70$.\n\nb) Let the two consecutive odd integers be $2k-1$ and $2k+1$, where $k \\in \\mathbb{Z}$. Their sum is:\n$$\n(2k-1) + (2k+1) = 4k\n$$\nwhich is divisible by $4$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20056,
"subject": "Mathematics (Olympiad)",
"question": "Да се определат сите природни броеви $x$, $y$ и $z$ за кои\n$$\n1 + 2^x 3^y = z^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "Лесно се проверува дека за $z=1,2,3$ дадената равенка нема решение.\n\nНека $z \\ge 4$. Имаме $2^x 3^y = (z-1)(z+1)$. Најмногу еден од $z-1$ и $z+1$ се дели со 3, бидејќи ако $3 \\mid z-1$ и $3 \\mid z+1$ следува дека $3 \\mid (z+1)-(z-1)=2$, што не е можно. Исто така, бидејќи $2 \\mid (z-1)(z+1)$, добиваме дека и $z-1$ и $z+1$ се делат со 2 и само еден од нив може да се дели со 4, бидејќи ако $4 \\mid z-1$ и $4 \\mid z+1$ следува дека $4 \\mid (z+1)-(z-1)=2$, што не е можно.\n\nПоради ова ги имаме следниве случаи:\n\n$$\n1^{\\circ} \\quad z+1=2 \\cdot 3^{y}, \\; z-1=2^{x-1} \\qquad 2^{\\circ} \\quad z+1=2^{x-1}, \\; z-1=2 \\cdot 3^{y}.\n$$\n\n**Случај 1:** Нека $z+1=2 \\cdot 3^{y}$, $z-1=2^{x-1}$.\n\nСо одземање на дадените равенки имаме\n$$\n2 \\cdot 3^y - 2^{x-1} = 2 \\implies 3^y - 2^{x-2} = 1.\n$$\nЗа $x=2$ лесно се проверува дека дадената равенка нема решение.\n\nЗа $x=3$ имаме $3^y = 1+2 = 3$, т.е. $y=1$ и лесно се добива $z=5$.\n\nЗначи едно решение е $(x, y, z) = (3, 1, 5)$.\n\nАко $x \\ge 4$, имаме $3^y \\equiv 1 \\pmod{4}$, па следува дека $y$ е парен, т.е. $y=2y_1$, $y_1 \\in \\mathbb{N}$.\n\nСега имаме\n$$\n3^{2y_1} - 1 = 2^{x-2} \\implies (3^{y_1} - 1)(3^{y_1} + 1) = 2^{x-2},\n$$\nод каде лесно се добива дека $3^{y_1} - 1 = 2$ и $3^{y_1} + 1 = 2^{x-3}$, т.е. $y_1 = 1$ односно $y=2$, сега лесно се добива дека $x=5$ и $z=17$.\n\nЗначи решение е $(x, y, z) = (5, 2, 17)$.\n\n**Случај 2:** Нека $z+1=2^{x-1}$, $z-1=2 \\cdot 3^{y}$, тогаш $2^{x-1}-2 \\cdot 3^{y}=2$, т.е. $2^{x-2}-3^{y}=1$.\n\nЗа $y=1$ имаме $x=4$ и $z=7$, па решение е $(x, y, z) = (4, 1, 7)$.\n\nАко $y \\ge 2$ тогаш имаме $2^{x-2} \\equiv 1 \\pmod{3}$, т.е. $x-2$ мора да е парен број, па нека $x-2=2x_1$. Па имаме $3^y = 2^{2x_1} - 1 = (2^{x_1} - 1)(2^{x_1} + 1)$, од каде следува дека $2^{x_1} - 1 = 1$ или $3$. Ако $2^{x_1} - 1 = 1$ добиваме $x_1 = 1$, т.е. $x=4$, па го добиваме решението $(4,1,7)$.\n\nАко $2^{x_1} - 1 = 3$ следува $x_1 = 2$, т.е. $x=6$ и добиваме $3^y = 15$, што не е можно.\n\nЗначи решенија на дадената равенка се $(x, y, z) = (3, 1, 5),\\; (5, 2, 17),\\; (4, 1, 7)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20057,
"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 condiciones:\n\n1. $f(x)f(y) \\geq f(xy)$ para todos $x, y \\in Q_0$;\n2. $f(x + y) \\geq f(x) + f(y)$ 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 varias propiedades que debe cumplir cualquier función $f$ que satisfaga las condiciones del enunciado:\n\n1. $f(1) \\geq 1$. Tomando $x = 1$, $y = a$ en (1) y usando (3), se obtiene $a = f(a) \\leq f(1)f(a) = af(1)$, así que $f(1) \\geq 1$.\n\n2. Para todo entero positivo $n$ y todo racional positivo $x$, se tiene $f(nx) \\geq n f(x)$, 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 $n$, $f(n) \\geq n f(1)$, con igualdad si y sólo si $f(k) = k f(1)$ para $k = 1, 2, \\dots, n$. Esto se prueba por inducción usando (2).\n\n3. Para todo entero positivo $u$ y todo racional positivo $x$, se tiene $f(x^u) \\leq (f(x))^u$, con igualdad si y sólo si $f(x^k) = (f(x))^k$ para $k = 1, 2, \\dots, u$. Esto se prueba por inducción usando (1).\n\n4. $f$ toma sólo valores positivos. Para enteros positivos $m, n$, se tiene $f\\left(\\frac{m}{n}\\right) \\geq \\frac{m f(1)}{f(n)} > 0$. Esto se deduce de (2) y (1).\n\n5. $f$ es estrictamente creciente. Si $z > x$ son racionales positivos, existe $y = z - x > 0$ y por (2), $f(z) = f(x + y) \\geq f(x) + f(y) > f(x)$, ya que $f(y) > 0$ por (4).\n\nSupongamos que existe un entero positivo $N$ tal que $f(N) > N$. Entonces existe $M$ tal que $M(f(N) - N) \\geq 1$. Por (2), $f(MN) > M f(N) \\geq MN + 1$. Para $n \\geq MN$, $f(n) = f(MN + (n - MN)) \\geq f(MN) + f(n - MN) \\geq MN + 1 + (n - MN) f(1) \\geq n + 1$.\n\nConsideremos la sucesión $a, a^2, a^3, \\dots$ (con $a > 1$). Sea $u$ tal que $a^u \\geq MN$ y $k$ la parte entera de $a^u$. Entonces $a^u \\geq k > a^u - 1$ y, usando (3), (3) y (5), $k + 1 > a^u = (f(a))^u \\geq f(a^u) \\geq f(k) \\geq k + 1$, contradicción. Por 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 lo anterior, $f\\left(\\frac{1}{n}\\right) \\geq \\frac{1}{n}$ para todo $n$. Usando (2), para todo racional $\\frac{m}{n}$, $f\\left(\\frac{m}{n}\\right) \\geq m f\\left(\\frac{1}{n}\\right) \\geq \\frac{m}{n}$.\n\nSupongamos que existe $q = \\frac{m}{n}$ (con $m, n$ coprimos) tal que $f(q) \\neq q$. Entonces $f(q) > q$ y por (2), $m = f(m) = f(nq) \\geq n f(q) > n q = m$, contradicción. Por tanto, $f(q) = q$ para todo racional positivo $q$, como queríamos demostrar.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20058,
"subject": "Mathematics (Olympiad)",
"question": "a) Assume that $N$ is the orthocenter of $\\triangle ABC$. Show that the respective reflections of $m_a, m_b, m_c$ through each bisector of angles $\\angle BNC, \\angle CNA, \\angle ANB$ are the same line.\n\nb) Assume that $N$ is the nine-point center of $\\triangle ABC$. Show that the respective reflections of $m_a, m_b, m_c$ through $BC, CA, AB$ concur.",
"options": [],
"answer": "See solution",
"solution": "First, we present two following lemmas. Let $ABC$ be a triangle inscribed in $(O)$ and $N$ is the nine-point center of $\\triangle ABC$.\n\n**Lemma 1.** Let $K$ be the center of $(BOC)$, then $AN, AK$ are isogonal with respect to $\\angle BAC$.\n\n**Lemma 2.** Let $B', C'$ be the reflections of $B, C$ through $AC, AB$, respectively, then $AK$ is perpendicular to $B'C'$.\n\n\n\n1. Let $X$ be the reflection of $O$ through $BC$. Let $H$ be the orthocenter of $\\triangle ABC$. We have $AH = OX$, and $AH$ is parallel to $OX$. Hence, $AX$ cuts $OH$ at the midpoint of $OH$, which is $N$, and thus $N$ is the midpoint of $AX$.\n\nWe have $\\angle OXB = \\angle KOB = \\angle KBO$, so $\\triangle OKB \\sim \\triangle OBX$, implying that $\\frac{OK}{OB} = \\frac{OB}{OX}$, or\n\n$$\nOX \\cdot OK = OB^2 = OA^2.\n$$\n\nHence, we have $\\frac{OK}{OA} = \\frac{OA}{OX}$, implying that $\\triangle OKA \\sim \\triangle OAX$. Therefore, $\\angle OKA = \\angle XAO = \\angle DAK$.\n\nSince $AH, AO$ are isogonal with respect to $\\angle BAC$, we have $\\angle BAK = \\angle NAC$, implying that $AN, AK$ are isogonal with respect to $\\angle BAC$.\n\n2. Let $Y, Z$ be the reflection of $O$ through $CA, AB$. Similar to the argument above, we have $N$ is the midpoint of $BY, CZ$. Let $N_b, N_c$ be the reflection of $N$ through $CA, AB$. Since $OC'$ is the reflection of $ZC$ through $AB$, and $N$ is the midpoint of $ZC$, then $N_c$ is the midpoint of $OC'$. Similarly, $N_b$ is the midpoint of $OB'$. Hence, $N_b N_c$ is parallel to $B'C'$. We have\n\n$$\n\\angle KAN_b = \\angle N_b AC + \\angle KAC = \\angle NAC + \\angle NAB = \\angle BAC,\n$$\n\nand similarly, $\\angle KAN_c = \\angle BAC$. On the other hand, we have $AN_b = AN = AN_c$, and since $\\angle KAN_b = \\angle KAN_c$, we must have $AK$ perpendicular to $N_b N_c$. So $AK \\perp B'C'$, as desired.\n\nBack to the main problem, let $O_a, O_b, O_c$ be circumcenters of $\\triangle NBC, \\triangle NCA, \\triangle NAB$ and $K_a, K_b, K_c$ be circumcenters of $\\triangle O_a BC, \\triangle O_b CA, \\triangle O_c AB$. By Lemma 2, $m_a, m_b, m_c$ are $NK_a, NK_b, NK_c$, respectively.\n\na) If $N$ is the orthocenter $H$, let $m'_a, m'_b, m'_c$ be reflections of $m_a, m_b, m_c$ through the bisectors of angles $BHC, CHA, AHB$. Let $J$ be the nine-point center of $\\triangle ABC$. Note that $(J)$ is also the nine-point circle of $\\triangle BHC$. Then, by Lemma 1, we have $HK_a, KJ$ are isogonal with respect to $\\angle BHC$, so $m'_a$ goes through $J$. Similarly, $m'_b, m'_c$ also go through $J$, and thus $m'_a \\equiv m'_b \\equiv m'_c \\equiv HJ$.\n\n\n\nb) Assume that $N$ is the nine-point center of $\\triangle ABC$. Let $N_a, N_b, N_c$ be the reflections of $N$ through $BC, CA, AB$, and let $(X), (Y), (Z)$ be the nine-point circles of $\\triangle NBC, \\triangle NCA, \\triangle NAB$. We will show that the reflections of $m_a, m_b, m_c$ through $BC, CA, AB$ concur on $(N_a N_b N_c)$. $\\square$\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20059,
"subject": "Mathematics (Olympiad)",
"question": "A square has side length $a$. An octagon is inscribed in the square such that its vertices touch the sides of the square. What is the side length $x$ of the inscribed octagon?\n\n",
"options": [],
"answer": "See solution",
"solution": "Denote by $x$ the side length of the inscribed octagon. The four triangles formed at the vertices of the square are isosceles right-angled triangles. Since their hypotenuse is of length $x$, their legs are of length $\\frac{x}{\\sqrt{2}}$. Thus,\n$$\na = x + 2\\frac{x}{\\sqrt{2}} = x + x\\sqrt{2} = x(\\sqrt{2} + 1).\n$$\nFrom this we deduce\n$$\nx = \\frac{a}{\\sqrt{2} + 1} = \\frac{a(\\sqrt{2} - 1)}{2 - 1} = a(\\sqrt{2} - 1).\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20060,
"subject": "Mathematics (Olympiad)",
"question": "Let $S(\\text{AMC})$ and $S(\\text{BMD})$ denote the areas of triangles $AMC$ and $BMD$, respectively. Let $AM$, $MC$, $BM$, $MD$ be the lengths of the respective segments, and $\\angle AMC$, $\\angle BMD$ the respective angles. Prove that if\n\n$$\n\\frac{S(\\text{AMC})}{\\tan \\angle AMC} = \\frac{S(\\text{BMD})}{\\tan \\angle BMD},\n$$\n\nthen\n\n$$\nAM^2 + MC^2 - 4 \\frac{S(\\text{AMC})}{\\tan \\angle AMC} = BM^2 + MD^2 - 4 \\frac{S(\\text{BMD})}{\\tan \\angle BMD}.\n$$\n\n",
"options": [],
"answer": "See solution",
"solution": "Since\n\n$$\nS(\\text{AMC}) = 0.5 \\cdot AM \\cdot MC \\sin \\angle AMC, \\quad S(\\text{BMD}) = 0.5 \\cdot BM \\cdot MD \\sin \\angle BMD,\n$$\n\nwe have\n\n$$\nAM \\cdot MC = \\frac{2S(\\text{AMC})}{\\sin \\angle AMC}, \\quad BM \\cdot MD = \\frac{2S(\\text{BMD})}{\\sin \\angle BMD}. \\tag{1}\n$$\n\nBy the law of cosines,\n\n$$\nAC^2 = AM^2 + MC^2 - 2AM \\cdot MC \\cos \\angle AMC,\n$$\n\n$$\nBD^2 = BM^2 + MD^2 - 2BM \\cdot MD \\cos \\angle BMD.\n$$\n\nFrom (1) it follows\n\n$$\nAC^2 = AM^2 + MC^2 - 4S(\\text{AMC}) \\frac{\\cos \\angle AMC}{\\sin \\angle AMC} = AM^2 + MC^2 - \\frac{4S(\\text{AMC})}{\\tan \\angle AMC}, \\tag{2}\n$$\n\n$$\nBD^2 = BM^2 + MD^2 - 4S(\\text{BMD}) \\frac{\\cos \\angle BMD}{\\sin \\angle BMD} = BM^2 + MD^2 - \\frac{4S(\\text{BMD})}{\\tan \\angle BMD}. \\tag{3}\n$$\n\nBy the given condition,\n\n$$\n\\frac{S(\\text{AMC})}{\\tan \\angle AMC} = \\frac{S(\\text{BMD})}{\\tan \\angle BMD},\n$$\n\nso (2) and (3) give the required equality.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20061,
"subject": "Mathematics (Olympiad)",
"question": "Let $x$, $y$, and $z$ be positive real numbers such that $xy + yz + zx = 3xyz$. Prove that\n$$\nx^2 y + y^2 z + z^2 x \\geq 2(x + y + z) - 3\n$$",
"options": [],
"answer": "See solution",
"solution": "The given condition can be rearranged to $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = 3$. Using this, we obtain:\n$$\n\\begin{aligned}\nx^2 y + y^2 z + z^2 x - 2(x + y + z) - 3 &= y \\left(x - \\frac{1}{y}\\right)^2 + z \\left(y - \\frac{1}{z}\\right)^2 + x \\left(z - \\frac{1}{x}\\right)^2 \\\\\n&\\geq 0\n\\end{aligned}\n$$\n\nEquality holds if and only if $x = y = z = 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20062,
"subject": "Mathematics (Olympiad)",
"question": "For integers $n = 0, \\dots, 2021$, let $a \\leq b \\leq c$ be three numbers on the blackboard after a certain procedure is done $n$ times. Let $p_n = c - b$ and $q_n = b - a$. Here, $n = 0$ means the initial status. When the procedure is done once, the three numbers $a \\leq b \\leq c$ are replaced with $\\frac{a+b}{2} \\leq \\frac{a+c}{2} \\leq \\frac{b+c}{2}$. All three numbers after $2021$ procedures are positive integers. What is the minimum possible value of the sum of the initial three numbers?",
"options": [],
"answer": "See solution",
"solution": "Let $p_n = c - b$ and $q_n = b - a$ after $n$ steps. After one step, $p_{n+1} = \\frac{q_n}{2}$ and $q_{n+1} = \\frac{p_n}{2}$. Thus, after $2021$ steps, $p_{2021} = \\frac{q_0}{2^{2021}}$ and $q_{2021} = \\frac{p_0}{2^{2021}}$. Since the initial numbers are distinct, $p_0, q_0 \\neq 0$, so $p_{2021}, q_{2021} \\neq 0$. All three numbers after $2021$ steps are positive integers, so $p_{2021} \\geq 1$, $q_{2021} \\geq 1$, which gives $p_0 \\geq 2^{2021}$, $q_0 \\geq 2^{2021}$. The minimum initial number is at least $1$, so the sum is at least $1 + (1 + 2^{2021}) + (1 + 2 \\cdot 2^{2021}) = 3 \\cdot 2^{2021} + 3$.\n\nNow, let the initial numbers be $1$, $1 + 2^{2021}$, and $1 + 2 \\cdot 2^{2021}$. Then $p_0 = q_0 = 2^{2021}$, so $p_{2021} = q_{2021} = 1$. The sum remains $3 \\cdot 2^{2021} + 3$ throughout the process. After $2021$ steps, the numbers are $2^{2021}$, $2^{2021} + 1$, and $2^{2021} + 2$, all integers. Thus, the minimum possible sum is $3 \\cdot 2^{2021} + 3$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20063,
"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 $k$. 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": "Therefore, the answer is $$2^5 \\cdot 3 \\cdot 17 = 1632.$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 20064,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an isosceles triangle with $|AB| = |AC|$. The bisector of angle $ABC$ meets the side $AC$ at the point $D$.\n\na) Is the triangle $ABD$ isosceles whenever the triangle $BCD$ is isosceles?\n\nb) Is the triangle $BCD$ isosceles whenever the triangle $ABD$ is isosceles?",
"options": [],
"answer": "See solution",
"solution": "a) Denote $\\angle BAC = \\alpha$ and $\\angle ABC = \\angle ACB = \\beta$. Assume that the triangle $BCD$ is isosceles. If $|CB| = |CD|$, then the angles $\\angle CBD$, $\\angle CDB$, and $\\angle BCD$ would be $\\frac{\\beta}{2}$, $\\frac{\\beta}{2}$, and $\\beta$ respectively, which implies $2 \\cdot \\frac{\\beta}{2} + \\beta = 180^\\circ$, giving $\\beta = 90^\\circ$. This is impossible, since the triangle $ABC$ has two angles equal to $\\beta$. If $|DB| = |DC|$, then we would get $\\frac{\\beta}{2} = \\beta$, which is also impossible. This leaves the only option $|BC| = |BD|$. Then, the triangles $ABC$ and $BCD$ are similar, since all corresponding angles are the same. Therefore $\\angle DBA = \\angle DBC = \\angle BAC = \\angle BAD$, showing that the triangle $ABD$ is isosceles with $|DA| = |DB|$.\n\n\n\nb) If the angles of the triangle $ABC$ are $\\frac{3}{7} \\cdot 180^\\circ$, $\\frac{2}{7} \\cdot 180^\\circ$, $\\frac{2}{7} \\cdot 180^\\circ$, then $\\angle ADB = 180^\\circ - \\angle BAD - \\angle ABD = 180^\\circ - \\frac{3}{7} \\cdot 180^\\circ - \\frac{1}{7} \\cdot 180^\\circ = \\frac{3}{7} \\cdot 180^\\circ = \\angle BAD$, which shows that the triangle $ABD$ is isosceles with $|BA| = |BD|$. At the same time, the angles in the triangle $BCD$ are $\\frac{1}{7} \\cdot 180^\\circ$, $\\frac{2}{7} \\cdot 180^\\circ$, $\\frac{4}{7} \\cdot 180^\\circ$, which are pairwise different, so the triangle $BCD$ is not isosceles.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20065,
"subject": "Mathematics (Olympiad)",
"question": "From an unlimited supply of 1-cent coins, 10-cent coins, and 25-cent coins, Silas wants to find a collection of coins that has a total value of $N$ cents, where $N$ is a positive integer. He uses the so-called _greedy algorithm_, successively choosing the coin of greatest value that does not cause the value of his collection to exceed $N$. For example, to get 42 cents, Silas will choose a 25-cent coin, then a 10-cent coin, then 7 1-cent coins. However, this collection of 9 coins uses more coins than necessary to get a total of 42 cents; indeed, choosing 4 10-cent coins and 2 1-cent coins achieves the same total value with only 6 coins.\n\nIn general, the greedy algorithm _succeeds_ for a given $N$ if no other collection of 1-cent, 10-cent, and 25-cent coins gives a total value of $N$ cents using strictly fewer coins than the collection given by the greedy algorithm. Find the number of values of $N$ between 1 and 1000 inclusive for which the greedy algorithm succeeds.",
"options": [],
"answer": "See solution",
"solution": "For easier exposition, call the coins by their American names: 1-cent pennies, 10-cent dimes, and 25-cent quarters. Also change the range to $0 \\leq N \\leq 999$, which will not change the answer because the greedy algorithm succeeds for both $N = 0$ and $N = 1000$.\n\nIf $N$ is not a multiple of 5, then both the collection obtained by the greedy algorithm and a least-sized collection will use 1, 2, 3, or 4 additional pennies as necessary to account for the remainder when $N$ is divided by 5. Therefore the greedy algorithm will succeed for $N$ if and only if it produces a smallest collection for the greatest multiple of 5 less than or equal to $N$. It follows that it suffices to solve the problem for multiples of 5 and then multiply by 5 to obtain the requested number of values.\n\nSuppose that $N$ is one of the 200 multiples of 5 in the range $[0, 999]$. There are five cases, depending on the remainder when $N$ is divided by 25. In three of these cases, when $N$ gives remainder 0, 10, or 20 when divided by 25, the greedy algorithm will use as many quarters as possible and fill the rest with dimes, which is optimal.\n\n- If $N \\equiv 5 \\pmod{25}$, then except for $N = 5$, the greedy algorithm is not optimal, because that algorithm will choose $\\left\\lfloor \\frac{N}{25} \\right\\rfloor$ quarters and 5 pennies, but replacing 1 quarter and 5 pennies with 3 dimes results in a smaller collection with the same total. The greedy algorithm succeeds for $N = 5$; it will choose 5 pennies.\n- If $N \\equiv 15 \\pmod{25}$, then except for $N = 15$, the greedy algorithm is not optimal, because that algorithm will choose $\\left\\lfloor \\frac{N}{25} \\right\\rfloor$ quarters, 1 dime, and 5 pennies, but replacing 1 quarter and 5 pennies with 3 dimes results in a smaller collection with the same total. The greedy algorithm succeeds for $N = 15$; it will choose 1 dime and 5 pennies.\n\nThus the greedy algorithm will produce a smallest collection in $\\frac{3}{5}$ of the 200 cases, together with the 2 exceptional cases $N = 5$ and $N = 15$, a total of $\\frac{3}{5} \\cdot 200 + 2 = 122$ cases. The requested number of values is therefore $122 \\cdot 5 = 610$.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20066,
"subject": "Mathematics (Olympiad)",
"question": "令 $x$ 和 $y$ 是正整數。試證：若對所有的正整數 $n$，$2^n y + 1$ 整除 $x^{2n} - 1$，則 $x = 1$。",
"options": [],
"answer": "See solution",
"solution": "首先，我們證明對於所有正整數 $y$，存在無限多個質數 $p \\equiv 3 \\pmod{4}$ 使得 $p$ 整除某些形如 $2^n y + 1$ 的數。\n\n顯然只需考慮 $y$ 為奇數的情況。令\n\n$$\n2y + 1 = p_1^{e_1} \\cdots p_r^{e_r}\n$$\n\n為 $2y+1$ 的質因數分解。假設存在有限多個質數 $p_{r+1}, \\ldots, p_{r+s} \\equiv 3 \\pmod{4}$ 整除某些形如 $2^n y + 1$ 的數，但不整除 $2y + 1$。\n\n我們想找到一個 $n$ 使得 $p_i^{e_i} \\mid 2^n y + 1$ 對所有 $1 \\leq i \\leq r$ 成立，且 $p_i \\nmid 2^n y + 1$ 對所有 $r+1 \\leq i \\leq r+s$ 成立。對此，只需取\n\n$$\nn = 1 + \\varphi(p_1^{e_1+1} \\cdots p_r^{e_r+1} p_{r+1} \\cdots p_{r+s}),\n$$\n\n則有\n\n$$\n2^n y + 1 \\equiv 2y + 1 \\pmod{p_1^{e_1+1} \\cdots p_r^{e_r+1} p_{r+1} \\cdots p_{r+s}}.\n$$\n\n因此 $2^n y + 1$ 的質因數分解包含 $p_1^{e_1}, \\ldots, p_r^{e_r}$ 和質數的冪次方同餘 $1 \\pmod{4}$。因為 $y$ 是奇數，有\n\n$$\n2^n y + 1 \\equiv p_1^{e_1} \\cdots p_r^{e_r} \\equiv 2y + 1 \\equiv 3 \\pmod{4}.\n$$\n\n但 $n > 1$，這導致矛盾。因此 $2^n y + 1 \\equiv 1 \\pmod{4}$。\n\n若 $p$ 是 $2^n y + 1$ 的質因數，則對 $d = 2^n$，有 $x^d \\equiv 1 \\pmod{p}$。由費馬小定理，當 $d = p-1$ 時同餘式成立，因此也成立於 $d = \\gcd(2^n, p-1)$。對於 $p \\equiv 3 \\pmod{4}$，有 $\\gcd(2^n, p-1) = 2$，因此 $x^2 \\equiv 1 \\pmod{p}$。\n\n總結：所有 $p \\equiv 3 \\pmod{4}$ 且整除某些 $2^n y + 1$ 的質數也整除 $x^2 - 1$。這只可能發生在 $x = 1$，否則正整數 $x^2 - 1$ 會有無窮多個質因數，矛盾。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20067,
"subject": "Mathematics (Olympiad)",
"question": "The parabola with equation $y = x^2 - 4$ is rotated $60\\degree$ counterclockwise around the origin. The unique point in the fourth quadrant where the original parabola and its image intersect has $y$-coordinate $\\frac{a-\\sqrt{b}}{c}$, where $a$, $b$, and $c$ are positive integers, and $a$ and $c$ are relatively prime. Find $a + b + c$.",
"options": [],
"answer": "See solution",
"solution": "Because the graph of $y = x^2 - 4$ is symmetric about the $y$-axis, the line $\\ell$ through the origin with negative slope making a $30\\degree$ angle with the $y$-axis is a line of symmetry for the figure consisting of the original parabola and its image. The desired intersection of the two curves lies on this line, as can be seen in the figure. (Another way to see this is to note that the given rotation is the composition of a reflection across the $y$-axis—which takes the parabola to itself—followed by a reflection across line $\\ell$. The composition of two reflections is a rotation through twice the angle between the two reflecting lines.)\n\n\n\nBecause the line forms a $60\\degree$ angle with the positive $x$-axis, and the legs of a $30$-$60$-$90\\degree$ right triangle are in the ratio of $\\sqrt{3} : 1$, the equation of the line is $y = -\\sqrt{3}x$. Substituting for $x$ in the equation $y = x^2 - 4$ yields $y^2 - 3y - 12 = 0$, and the Quadratic Formula gives negative solution $y = \\frac{3-\\sqrt{57}}{2}$. The requested sum is $3 + 57 + 2 = 62$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20068,
"subject": "Mathematics (Olympiad)",
"question": "Prove that there are infinitely many positive integers $n$ such that $2^{2^n+1} + 1$ is divisible by $n$, but $2^n + 1$ is not.",
"options": [],
"answer": "See solution",
"solution": "Throughout the solution, $n$ stands for a positive integer.\n\nBy Euler's theorem, $$(2^{3^n} + 1)(2^{3^n} - 1) = 2^{2 \\cdot 3^n} - 1 \\equiv 0 \\pmod{3^{n+1}}.$$ Since $2^{3^n} - 1 \\equiv 1 \\pmod{3}$, it follows that $2^{3^n} + 1$ is divisible by $3^{n+1}$.\n\nThe number $$\\frac{2^{3^{n+1}} + 1}{2^{3^n} + 1} = 2^{2 \\cdot 3^n} - 2^{3^n} + 1$$ is greater than $3$ and congruent to $3$ modulo $9$, so it has a prime factor $p_n > 3$ that does not divide $2^{3^n} + 1$ (otherwise, $2^{3^n} \\equiv -1 \\pmod{p_n}$, so $2^{2 \\cdot 3^n} - 2^{3^n} + 1 \\equiv 3 \\pmod{p_n}$, contradicting the fact that $p_n$ is a factor greater than $3$ of $2^{2 \\cdot 3^n} - 2^{3^n} + 1$).\n\nWe now show that $a_n = 3^n p_n$ satisfies the conditions in the statement. Since $2^{a_n} + 1 \\equiv 2^{3^n} + 1 \\not\\equiv 0 \\pmod{p_n}$, it follows that $a_n$ does not divide $2^{a_n} + 1$.\n\nOn the other hand, $3^{n+1}$ divides $2^{3^n} + 1$ which in turn divides $2^{a_n} + 1$, so $2^{3^{n+1}} + 1$ divides $2^{2^{a_n}+1} + 1$. Finally, both $3^n$ and $p_n$ divide $2^{3^{n+1}} + 1$, so $a_n$ divides $2^{2^{a_n}+1} + 1$.\n\nAs $n$ runs through the positive integers, the $a_n$ are clearly pairwise distinct and the conclusion follows.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20069,
"subject": "Mathematics (Olympiad)",
"question": "Given real numbers $a, b, c$ such that $abc = 1$, prove that for all integers $k \\ge 2$,\n\n$$\n\\frac{a^k}{a+b} + \\frac{b^k}{b+c} + \\frac{c^k}{c+a} \\ge \\frac{3}{2}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Since\n\n$$\n\\frac{a^k}{a+b} + \\frac{1}{4}(a+b) + \\underbrace{\\frac{1}{2} + \\frac{1}{2} + \\dots + \\frac{1}{2}}_{k-2} \\ge k \\cdot \\sqrt[\\underline{k}]{\\frac{a^k}{2^k}} = \\frac{k}{2}a,\n$$\nthen\n\n$$\n\\frac{a^k}{a+b} \\ge \\frac{k}{2}a - \\frac{1}{4}(a+b) - \\frac{k-2}{2}.\n$$\n\nSimilarly,\n\n$$\n\\frac{b^k}{b+c} \\ge \\frac{k}{2}b - \\frac{1}{4}(b+c) - \\frac{k-2}{2},\n$$\n\n$$\n\\frac{c^k}{c+a} \\ge \\frac{k}{2}c - \\frac{1}{4}(c+a) - \\frac{k-2}{2}.\n$$\n\nAdding the three inequalities above, we obtain\n\n$$\n\\begin{aligned}\n& \\frac{a^k}{a+b} + \\frac{b^k}{b+c} + \\frac{c^k}{c+a} \\\\\n\\ge & \\frac{k}{2}(a+b+c) - \\frac{1}{2}(a+b+c) - \\frac{3}{2}(k-2) \\\\\n= & \\frac{k-1}{2}(a+b+c) - \\frac{3}{2}(k-2) \\\\\n\\ge & \\frac{3}{2}(k-1) - \\frac{3}{2}(k-2) \\\\\n= & \\frac{3}{2},\n\\end{aligned}\n$$\n\nas desired.\n\nRemark: The problem could also be proved by the Cauchy inequality or the Chebyshev inequality.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20070,
"subject": "Mathematics (Olympiad)",
"question": "Let $m$ and $n$ be positive integers with $m \\ge n \\ge 2022$. Prove that for arbitrary real numbers $a_1, a_2, \\dots, a_n$ and $b_1, b_2, \\dots, b_n$, the number of ordered pairs $(i, j)$ ($1 \\le i, j \\le n$) satisfying $|a_i + b_j - ij| \\le m$ does not exceed $3n\\sqrt{m \\ln n}$.",
"options": [],
"answer": "See solution",
"solution": "**Solution**\n\nColor all $(i, j)$ satisfying $|a_i + b_j - ij| \\le m$ red.\n\n**Lemma**: If $(i_1, j_1)$, $(i_1, j_2)$, $(i_2, j_1)$, $(i_2, j_2)$ are all red, then\n\n$$\n|(i_2 - i_1)(j_2 - j_1)| \\le 4m.\n$$\n\n**Proof of lemma**: By definition of red points,\n\n$$\n|a_{i_1} + b_{j_1} - i_1 j_1| \\le m, \\quad |a_{i_1} + b_{j_2} - i_1 j_2| \\le m,\n$$\n$$\n|a_{i_2} + b_{j_1} - i_2 j_1| \\le m, \\quad |a_{i_2} + b_{j_2} - i_2 j_2| \\le m.\n$$\n\nTaking two differences to eliminate $a_{i_1}$, $a_{i_2}$, $b_{j_1}$, and $b_{j_2}$, we obtain\n$|i_1j_1 - i_1j_2 - i_2j_1 + i_2j_2| \\le 4m$, which gives the lemma. \\text{\\textopenbullet}\n\nReturn to the original problem. For given $1 \\le i_1 < i_2 \\le n$, we need to find the number of $j$ such that both $(i_1, j)$ and $(i_2, j)$ are red. Let $d = i_2 - i_1$. Any two such $j$ differ by at most $\\frac{4m}{d}$, so there are at most $\\frac{4m}{d} + 1$ such $j$.\n\nIf $d$ is fixed, there are $n-d$ ways to choose $i_1$ and $i_2$. Thus, the number of red pairs $(i_1, j)$ and $(i_2, j)$ with $1 \\le i_1 < i_2 \\le n$, $1 \\le j \\le n$ does not exceed\n\n$$\n\\begin{aligned}\n\\sum_{d=1}^{n-1} (n-d) \\left( \\frac{4m}{d} + 1 \\right) &= \\sum_{d=1}^{n-1} (n-d) + 4m \\sum_{d=1}^{n-1} \\frac{n-d}{d} \\\\\n&< \\frac{n(n-1)}{2} + 4mn \\ln n.\n\\end{aligned}\n$$\n\nFor $1 \\le j \\le n$, let the number of red points in $(1, j), (2, j), \\dots, (n, j)$ be $x_j$. Then\n\n$$\n\\sum_{j=1}^{n} \\binom{x_j}{2} < \\frac{n(n-1)}{2} + 4mn \\ln n,\n$$\n\nso\n\n$$\n\\sum_{j=1}^{n} \\left( x_j - \\frac{1}{2} \\right)^2 < n^2 - \\frac{1}{4}n + 8mn \\ln n.\n$$\n\nBy Cauchy-Schwarz,\n\n$$\n\\sum_{j=1}^{n} \\left( x_j - \\frac{1}{2} \\right)^2 \\le \\sqrt{n \\left( n^2 - \\frac{1}{4}n + 8mn \\ln n \\right)},\n$$\n\nso\n\n$$\n\\sum_{j=1}^{n} x_j \\le \\frac{n}{2} + n\\sqrt{n - \\frac{1}{4} + 8m \\ln n}.\n$$\n\nWhen $m \\ge n \\ge 2022$, a simple estimate shows the right-hand side is $\\le 3n\\sqrt{m \\ln n}$. \\text{\\textopenbullet}\n\n**Remark**: This result comes from new methods in harmonic analysis, with the following variation: there exists a constant $C > 0$ such that for any positive integers $m \\ge n^2$ and any functions $f, g, h: \\{1, 2, \\dots, n\\} \\to \\mathbb{R}$, the number of ordered pairs $(x, y)$ ($x, y \\in \\{1, 2, \\dots, n\\}$) satisfying\n\n$$\n|f(x) + y g(x) + h(y) - x y^2| \\le m\n$$\n\ncannot exceed\n\n$$\nC m^{1/3} n (\\ln n + 1)^{2/3}.\n$$\n\nIn this statement, the leading order term $m^{1/3} n$ is optimal (consider $f(x) + y g(x) + h(y) - x y^2 = \\frac{1}{3}(y-x)^3$). For expressions like $f(x) + g(y) + y h(x) + x l(y) - x^2 y^2$, even with large $m$, whether a bound of leading order $m^{1/4} n^{1+\\delta}$ exists is still open.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20071,
"subject": "Mathematics (Olympiad)",
"question": "How many 4-digit palindromes are divisible by 9?",
"options": [],
"answer": "See solution",
"solution": "A 4-digit palindrome has the form $abba$, where $a$ and $b$ are digits. The sum of the digits is $2a + 2b = 2(a + b)$. For divisibility by 9, $a + b$ must be 9 or 18. The possible pairs $(a, b)$ are $\\{(1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1), (9,0), (9,9)\\}$, giving the palindromes 1881, 2772, 3663, 4554, 5445, 6336, 7227, 8118, 9009, 9999. Thus, there are **10** such palindromes.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20072,
"subject": "Mathematics (Olympiad)",
"question": "A $9 \\times 12$ rectangle is partitioned into unit squares. The centers of all the unit squares, except for the four corner squares and eight squares sharing a common side with one of them, are coloured red.\n\nIs it possible to label these red centres $C_1, C_2, \\dots, C_{96}$ in such a way that the following two conditions are both fulfilled:\n\n1. The distances $C_1C_2, \\dots, C_{95}C_{96}, C_{96}C_1$ are all equal to $\\sqrt{13}$.\n2. The closed broken line $C_1C_2\\dots C_{96}C_1$ has a centre of symmetry?",
"options": [],
"answer": "See solution",
"solution": "Assume the broken line exists and denote its center of symmetry as $O$.\n\nFirst, $O$ must also be the center of symmetry of the whole rectangle. Under symmetry about $O$, each segment of the broken line maps to another segment of the broken line, so each red center maps to another red center. Thus, the 96 red centers form 48 symmetric pairs about $O$, making $O$ the centroid of all red centers and the rectangle's center.\n\nNow, color all red centers black and white so that any two centers at distance 1 have different colors (a checkerboard coloring). Each red center and its mate (its symmetric pair about $O$) have different colors. Also, any two centers at distance $\\sqrt{13}$ have different colors, so each segment of the broken line connects centers of different colors.\n\nIf $XY$ is a segment of the broken line, then the segment connecting $X$'s mate and $Y$'s mate is also part of the broken line. To see this, consider a point $Z$ on $XY$ not at any intersection. Its symmetric image $Z_1$ (about $O$) is also not at any intersection and must lie on a segment of the broken line, namely the segment connecting $X$'s mate and $Y$'s mate.\n\nArrange the 96 centers on a circle in the order they appear on the broken line, forming a regular 96-gon. The segment connecting any point and its mate is a diameter, so there are 47 points between $A$ and its mate. Since adjacent centers are of different colors, two centers with an odd number of points between them must be the same color. But $A$ and its mate are of different colors—a contradiction.\n\nTherefore, such a labeling is not possible.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20073,
"subject": "Mathematics (Olympiad)",
"question": "Let $p$ be an odd prime and let $N = \\frac{1}{4}(p^3 - p) - 1$. The numbers $1, 2, \\dots, N$ are painted arbitrarily in two colors, red and blue. For any positive integer $n \\le N$, denote by $r(n)$ the fraction of integers in $\\{1, 2, \\dots, n\\}$ that are red (number of red numbers divided by $n$). Prove that there exists a positive integer $a \\in \\{1, 2, \\dots, p-1\\}$ such that $r(n) \\ne \\frac{a}{p}$ for all $n = 1, 2, \\dots, N$.",
"options": [],
"answer": "See solution",
"solution": "Denote by $R(n)$ the number of red numbers in $\\{1, 2, \\dots, n\\}$, i.e., $R(n) = n r(n)$. Similarly, denote by $B(n)$ and $b(n) = B(n)/n$ the number and proportion of blue numbers in $\\{1, 2, \\dots, n\\}$, respectively. Notice that $B(n) + R(n) = n$ and $b(n) + r(n) = 1$. Therefore, the statement of the problem does not change after swapping the colors.\n\nArguing indirectly, for every $a \\in \\{1, 2, \\dots, p-1\\}$ choose some positive integer $n_a$ such that $r(n_a) = a/p$ and, hence, $R(n_a) = a n_a / p$. Clearly, $p \\mid n_a$, so that $n_a = p m_a$ for some positive integer $m_a$, and $R(n_a) = a m_a$. Without loss of generality, we assume that $m_1 < m_{p-1}$, as otherwise one may swap the colors. Notice that\n\n$$\nm_a \\le \\frac{N}{p} < \\frac{p^2-1}{4} \\quad \\text{for all } a=1,2,\\dots,p-1.\n$$\n\nThe solution is based on a repeated application of the following simple observation.\n\n*Claim.* Assume that $m_a < m_b$ for some $a, b \\in \\{1, 2, \\dots, p-1\\}$. Then\n\n$$\nm_b \\ge \\frac{a}{b} m_a \\quad \\text{and} \\quad m_b \\ge \\frac{p-a}{p-b} m_a.\n$$\n\n*Proof.* The first inequality follows from $b m_b = R(n_b) \\ge R(n_a) = a m_a$. The second inequality is obtained by swapping colors. Let $q = (p-1)/2$. We distinguish two cases.\n\n**Case 1:** All $q$ numbers $m_1, m_2, \\dots, m_q$ are smaller than $m_{p-1}$.\n\nLet $m_a$ be the maximal number among $m_1, m_2, \\dots, m_q$; then $m_a \\ge q \\ge a$. Applying the Claim, we get\n\n$$\nm_{p-1} \\ge \\frac{p-a}{p-(p-1)} m_a \\ge (p-q)q = \\frac{p^2-1}{4}\n$$\n\nwhich contradicts (1).\n\n**Case 2:** There exists $k \\le q$ such that $m_k > m_{p-1}$. Choose $k$ to be the smallest index satisfying $m_k > m_{p-1}$; by our assumptions, we have $1 < k \\le q < p-1$.\n\nLet $m_a$ be the maximal number among $m_1, m_2, \\dots, m_{k-1}$; then $a \\le k-1 \\le m_a < m_{p-1}$. Applying the Claim, we get\n\n$$\n\\begin{align*}\nm_k &\\ge \\frac{p-1}{k} m_{p-1} \\\\\n&\\ge \\frac{p-1}{k} \\cdot \\frac{p-a}{p-(p-1)} m_a \\\\\n&\\ge \\frac{p-1}{k} \\cdot (p-k+1)(k-1) \\\\\n&\\ge \\frac{k-1}{k} \\cdot (p-1)(p-q) \\\\\n&\\ge \\frac{1}{2} \\cdot \\frac{p^2-1}{2}\n\\end{align*}\n$$\n\nwhich contradicts (1) again. $\\square$\n\n",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20074,
"subject": "Mathematics (Olympiad)",
"question": "An integer is written in each square of the shown table, so that the sum of the numbers in the white squares is 23 and the sum of the numbers in the squares from the odd-numbered columns is 40. We then replace the numbers in the white squares with their opposites. What is now the sum of the numbers in the squares from the odd-numbered lines?\n\n",
"options": [],
"answer": "See solution",
"solution": "Denote $w_e$ as the sum of the numbers in the white squares from the even-numbered lines and $w_o$ as the sum of the numbers in the white squares from the odd-numbered lines. Then $w_e + w_o = 23$.\n\nThe set of black squares from the odd-numbered columns is the same as the set of black squares from the odd-numbered lines, and the set of white squares from the odd-numbered columns is the same as the set of white squares from the even-numbered lines.\n\nLet $n$ be the sum of the numbers in the black squares from the even-numbered columns. From the above and the problem statement, we know that $n + w_e = 40$ and we have to find $n - w_o$. We get:\n\n$$n - w_o = (n + w_e) - (w_e + w_o) = 40 - 23 = 17$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20075,
"subject": "Mathematics (Olympiad)",
"question": "Let $p_1, p_2, \\dots, p_{2009} \\in \\mathbb{R}^+$. Prove the inequality\n\n$$\n\\sum_{i=1}^{2009} \\frac{p_i}{p_{i+1} + p_{i+2} + \\dots + p_{i+1004}} \\ge \\frac{2009}{1004},\n$$\n\nwhere the indexes are taken mod $2009$.",
"options": [],
"answer": "See solution",
"solution": "Denote\n\n$$\nL = \\frac{p_1}{p_2 + p_3 + \\dots + p_{1005}} + \\frac{p_2}{p_3 + p_4 + \\dots + p_{1006}} + \\dots + \\frac{p_{2009}}{p_1 + p_2 + \\dots + p_{1004}}\n$$\n\nWe will use the Cauchy-Schwarz inequality in the form:\n\nFor every $a_1, a_2, \\dots, a_n \\in \\mathbb{R}$, $b_1, b_2, \\dots, b_n \\in \\mathbb{R}^+$ it holds\n\n$$\n\\frac{a_1^2}{b_1} + \\frac{a_2^2}{b_2} + \\dots + \\frac{a_n^2}{b_n} \\ge \\frac{(a_1 + a_2 + \\dots + a_n)^2}{b_1 + b_2 + \\dots + b_n}\n$$\n\nwith equality iff\n\n$$\n\\frac{a_1}{b_1} = \\frac{a_2}{b_2} = \\dots = \\frac{a_n}{b_n}.\n$$\n\nThen\n\n$$\nL = \\frac{p_1^2}{p_1(p_2 + p_3 + \\dots + p_{1005})} + \\frac{p_2^2}{p_2(p_3 + p_4 + \\dots + p_{1006})} + \\dots + \\frac{p_{2009}^2}{p_{2009}(p_1 + p_2 + \\dots + p_{1004})}\n$$\n\nSo\n\n$$\nL \\ge \\frac{(p_1 + p_2 + \\dots + p_{2009})^2}{\\sum_{i=1}^{2009} p_i (p_{i+1} + \\dots + p_{i+1004})}.\n$$\n\nNotice that the denominator of the last fraction is $\\sum_{i,j: i < j} p_i p_j$. Let\n\n$$\n\\alpha = \\frac{(p_1 + p_2 + \\dots + p_{2009})^2}{\\sum_{i=1}^{2009} p_i (p_{i+1} + \\dots + p_{i+1004})}.\n$$\n\nThen $(\\sum p_i)^2 = \\frac{\\alpha}{2}((\\sum p_i)^2 - (\\sum p_i^2))$, so we have\n\n$$\n\\frac{\\alpha}{2} \\sum p_i^2 = \\left(\\frac{\\alpha}{2} - 1\\right) \\left(\\sum p_i\\right)^2 \\quad \\text{i.e.} \\quad \\frac{\\alpha}{\\alpha-2} \\left(\\sum p_i^2\\right) = \\left(\\sum p_i\\right)^2.\n$$\n\nSo $0 < \\frac{\\alpha}{\\alpha-2} \\le 2009$ (we get the second inequality from Cauchy-Schwarz or the inequality between the quadratic and arithmetic mean). This inequality is actually equivalent to $\\alpha \\ge \\frac{2009}{1004}$. Thus, $L \\ge \\frac{2009}{1004}$ and equality holds iff $p_1 = p_2 = \\dots = p_{2009}$.",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 20076,
"subject": "Mathematics (Olympiad)",
"question": "Given circles $\\omega_1$ and $\\omega_2$ intersecting at points $X$ and $Y$, let $l_1$ be a line through the center of $\\omega_1$ intersecting $\\omega_2$ at points $P$ and $Q$, and let $l_2$ be a line through the center of $\\omega_2$ intersecting $\\omega_1$ at points $R$ and $S$. Prove that if $P, Q, R,$ and $S$ lie on a circle, then the center of this circle lies on line $XY$.",
"options": [],
"answer": "See solution",
"solution": "Let $\\omega$ denote the circumcircle of $P$, $Q$, $R$, $S$ and let $O$ denote the center of $\\omega$. Line $XY$ is the radical axis of circles $\\omega_1$ and $\\omega_2$. It suffices to show that $O$ has equal power to the two circles; that is, to show that\n\n$$\nOO_1^2 - O_1S^2 = OO_2^2 - O_2Q^2 \\quad \\text{or} \\quad OO_1^2 + O_2Q^2 = OO_2^2 + O_1S^2.\n$$\n\nLet $M$ and $N$ be the intersections of lines $O_2O$, $l_1$ and $O_1O$, $l_2$. Because circles $\\omega$ and $\\omega_2$ intersect at points $P$ and $Q$, we have $PQ \\perp OO_2$ (or $l_1 \\perp OO_2$). Hence\n\n$$\nOO_1^2 - OQ^2 = (OM^2 + MO_1^2) - (OM^2 + MQ^2) = (O_2M^2 + MO_1^2) - (O_2M^2 + MQ^2) = O_2O_1^2 - O_2Q^2\n$$\n\nor\n\n$$\nO_2O_1^2 + OQ^2 = OO_1^2 + O_2Q^2.\n$$\n\nLikewise, we have $O_2O_1^2 + OS^2 = OO_2^2 + O_1S^2$. Because $OS = OQ$, we obtain that $OO_1^2 + O_2Q^2 = OO_2^2 + O_1S^2$, which is what was to be proved.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20077,
"subject": "Mathematics (Olympiad)",
"question": "Find all ordered triples of positive integers $(a, b, c)$ such that each of the numbers $ab + c$, $ac + b$, and $bc + a$ is a power of $2$.",
"options": [],
"answer": "See solution",
"solution": "We will prove that there are only two families of solutions:\n\n$$\n(1, 1, 2^x - 1), \\quad (1, 2^x - 1, 2^x + 1),\n$$\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\n$$\na \\equiv b \\equiv c \\pmod{2}.\n$$\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\n$$\nbc + a = 2^d,\n$$\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\n$$\nab + c, \\quad ac + b, \\quad bc + a\n$$\n\nare powers of $2$. Since $a \\le b \\le c$, we have\n\n$$\nab + c \\le ac + b \\le bc + a.\n$$\n\n*Step II:* $a = 1$. Suppose, by contradiction, that $1 < a$. Then $1 < a < b < c$, and\n\n$$\nab + c = 2^k, \\quad ac + b = 2^m, \\quad bc + a = 2^n,\n$$\n\nwith $3 \\le k < m < n$ (since $2^k \\ge 3 + 5 = 8$).\n\nFrom the above equalities, we get the congruences:\n\n$$\nab \\equiv -c \\pmod{2^k}, \\quad ac \\equiv -b \\pmod{2^k}, \\quad bc \\equiv -a \\pmod{2^k}.\n$$\n\nMultiplying these congruences and noting that $a, b, c$ are odd, we obtain\n\n$$\nabc \\equiv -1 \\pmod{2^k}\n$$\n\nand\n\n$$\na^2 \\equiv b^2 \\equiv c^2 \\equiv 1 \\pmod{2^k}\n$$\n\n(by multiplying the first congruence by $c$, the second by $b$, and the third by $a$).\n\nSince for any odd natural number $x$,\n\n$$\n\\gcd(x - 1, x + 1) = 2,\n$$\n\nwe deduce that\n\n$$\na \\equiv \\pm 1 \\pmod{2^{k-1}}, \\quad b \\equiv \\pm 1 \\pmod{2^{k-1}}, \\quad c \\equiv \\pm 1 \\pmod{2^{k-1}}.\n$$\n\nFrom these congruences, it follows that\n\n$$\na, b, c \\ge 2^{k-1} - 1.\n$$\n\nThis inequality implies\n\n$$\n2^k = ab + c \\ge (2^{k-1} - 1)^2 + 2^{k-1} - 1 = 2^{2k-2} - 2^{k-1},\n$$\n\nwhich means $2 \\ge 2^{k-1} - 1 \\ge 3$, a contradiction since $k > 3$. Therefore, the assumption is false and $a = 1$.\n\n**Step III: the solutions.** If $b = 1$, then clearly $c = 2^x - 1$. Otherwise, assume $a = 1$ and $1 < b < c$. From Step II, we know that $c > b \\ge 2^{k-1} - 1$. Since $b + c = 2^k$ and $b, c$ are odd, the only possibility is $b = 2^{k-1} - 1$, $c = 2^{k-1} + 1$.\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": 20078,
"subject": "Mathematics (Olympiad)",
"question": "On a plane, there are given points $A_0$, $B_0$, $C_0$ (not necessarily distinct) such that $A_0B_0 + B_0C_0 + C_0A_0 = 1$. Points $A_1$, $B_1$, $C_1$ (not necessarily distinct) are chosen so that $A_1B_1 = A_0B_0$ and $B_1C_1 = B_0C_0$. Points $A_2$, $B_2$, $C_2$ are chosen as a permutation of $A_1$, $B_1$, $C_1$. Finally, points $A_3$, $B_3$, $C_3$ (not necessarily distinct) are chosen so that $A_3B_3 = A_2B_2$ and $B_3C_3 = B_2C_2$. Find the least and greatest possible value of $A_3B_3 + B_3C_3 + C_3A_3$.",
"options": [],
"answer": "See solution",
"solution": "The least and greatest possible values are $\\frac{1}{3}$ and $3$.\n\nDenote the lengths $A_0B_0$, $B_0C_0$, $C_0A_0$ by $x$, $y$, $z$ in non-increasing order. Similarly, denote $A_1B_1$, $B_1C_1$, $C_1A_1$ by $x'$, $y'$, $z'$ in non-increasing order, and $A_3B_3$, $B_3C_3$, $C_3A_3$ by $x''$, $y''$, $z''$ in non-increasing order. (Permuting the points does not change the distances, so we do not need a separate vector for $A_2B_2$, $B_2C_2$, $C_2A_2$.)\n\nWe have $x + y + z = 1$, $y + z \\ge x$, $y' + z' \\ge x'$, $y'' + z'' \\ge x''$. By construction, triples $(x, y, z)$ and $(x', y', z')$ have two values in common (not necessarily at corresponding places), and similarly for $(x', y', z')$ and $(x'', y'', z'')$.\n\nUsing these observations:\n\n$$\n\\begin{aligned}\nx'' + y'' + z'' &\\le 2(y'' + z'') \\le 2(x' + y') \\le 2(y' + y' + z') \\\\\n&\\le 2(x + x + y) \\le 6x \\le 3(x + y + z) = 3.\n\\end{aligned}\n$$\n\nWe can achieve the value $3$ as follows. Let $A_0B_0 = \\frac{1}{2}$ and $C_0 = A_0$. Let $A_1 = A_0$, $B_1 = B_0$ and $\\overrightarrow{B_1C_1} = -\\overrightarrow{B_0C_0}$. Let $A_2 = A_1$, $B_2 = C_1$, $C_2 = B_1$. Finally, let $A_3 = A_2$, $B_3 = B_2$ and $\\overrightarrow{B_3C_3} = -\\overrightarrow{B_2C_2}$. By construction, $A_3B_3 = 1$, $B_3C_3 = \\frac{1}{2}$, $C_3A_3 = \\frac{3}{2}$, so $A_3B_3 + B_3C_3 + C_3A_3 = 3$.\n\nThis establishes the upper bound. For the lower bound, note that all steps are reversible and the 3-step process itself is symmetric. By scaling, we can also make the initial configuration satisfy the conditions. Hence, all processes satisfying the conditions and achieving a final value $t$ correspond to processes achieving $\\frac{1}{t}$. Thus, the lower bound is $\\frac{1}{3}$.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20079,
"subject": "Mathematics (Olympiad)",
"question": "AC is the hypotenuse of right triangle $ABC$, and $BH$ is its altitude. Points $M$ and $N$ are the midpoints of segments $AH$ and $CH$, respectively. Lines $BM$ and $BN$ intersect the circumscribed circle of triangle $ABC$ a second time at points $P$ and $Q$, respectively. Segments $AQ$ and $CP$ intersect at point $R$. Prove that line $BR$ passes through the midpoint of segment $MN$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $K$ be the midpoint of segment $BH$, $S$ be the intersection point of $AK$ and $BP$, and $T$ be the intersection point of $CK$ and $BQ$. Then $SK$ and $KT$ are one third of the corresponding medians, and $ST$ is parallel to $AC$.\n\nThe triangles $ABH$ and $BCH$ are similar. From this similarity and properties of inscribed angles, we have\n\n$$\n\\angle KAB = \\angle NBC = \\angle QAC.\n$$\n\nHence $\\angle BAC = \\angle KAR$. But also $\\angle BAC = \\angle BPC = \\angle BPR$ as inscribed angles. Therefore, quadrilateral $SAPR$ is cyclic and $\\angle ARS = \\angle APS = \\angle APB = \\angle AQB$. So, $RS$ is parallel to $BT$. By analogous reasoning, $RT$ is parallel to $BS$. Hence $BSRT$ is a parallelogram.\n\nDiagonal $BR$ of this parallelogram splits diagonal $ST$ into two equal parts; therefore, it also splits the segment $MN$, which is parallel to $ST$, into two equal parts. QED.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20080,
"subject": "Mathematics (Olympiad)",
"question": "Define the *length* of a binary sequence to be the number of digits it contains.\n\nFor $n = 1, 2, 3$, consider the following observations:\n\n1. In year $2^n$, the plant pattern is $11101110\\ldots1110111\\ldots01110111$ and has length $2^{n+1} - 1$.\n2. In year $2^n + 1$, the plant pattern is $100\\ldots0\\ldots001$ and has length $2^{n+1} + 1$.\n\nProve by induction that these statements are true for all $n \\geq 1$.\n\nAdditionally, given that $128 = 2^7$, determine the number of plants in year 128.",
"options": [],
"answer": "See solution",
"solution": "We use induction to prove the statements for all $n \\geq 1$.\n\nAssume statements (1) and (2) are true for $n = k$.\n\n- In year $2^k$, the length of the plant pattern on either side of the center 0 is $2^k - 1$.\n- In year $2^k + 1$, there are $2^k - 1$ vacant positions between each plant and the center vacant position.\n\nBy assumption, in year $2^{k+1} = 2^k + 2^k$, the plant pattern is two copies of the pattern for year $2^k$ separated by one 0. Thus, in year $2^{k+1}$, the pattern has length $2(2^{k+1} - 1) + 1 = 2^{k+2} - 1$.\n\nIn year $2^{k+1} + 1$, the pattern is $100\\ldots0\\ldots001$ and has length $2^{k+1} + 1$.\n\nTherefore, the statements hold for $n = k + 1$, and by induction, for all $n \\geq 1$.\n\nFor $128 = 2^7$, the plant pattern for year 128 has length $2^8 - 1 = 255$. Placing a 0 at the right end gives $2^8 = 256$ digits, of which $2^8/4 = 64$ are 0. Thus, the number of plants in year 128 is $2^8 - 2^8/4 = 256 - 64 = 192$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20081,
"subject": "Mathematics (Olympiad)",
"question": "Find the value of $y$ so that $\\sqrt{y^2 + 2y + 1}$, $\\frac{y^2 + 3y - 1}{3}$, and $y-1$ are consecutive terms in an arithmetic progression.",
"options": [],
"answer": "See solution",
"solution": "Let $p, q, r$ be consecutive terms in an arithmetic progression, so $p + r = 2q$.\n\nThus,\n$$\n\\sqrt{y^2 + 2y + 1} + (y - 1) = 2 \\cdot \\frac{y^2 + 3y - 1}{3}\n$$\nSimplify $\\sqrt{y^2 + 2y + 1} = |y+1|$:\n$$\n|y+1| + y - 1 = \\frac{2}{3}(y^2 + 3y - 1)\n$$\nMultiply both sides by $3$:\n$$\n3(|y+1| + y - 1) = 2(y^2 + 3y - 1)\n$$\n$$\n3|y+1| + 3y - 3 = 2y^2 + 6y - 2\n$$\n$$\n3|y+1| = 2y^2 + 3y + 1\n$$\nNow consider two cases:\n\n**Case 1:** $y + 1 \\ge 0$ (so $|y+1| = y+1$)\n\n$$\n3(y+1) = 2y^2 + 3y + 1\n$$\n$$\n2y^2 + 3y + 1 - 3y - 3 = 0\n$$\n$$\n2y^2 - 2 = 0 \\implies y^2 = 1 \\implies y = 1 \\text{ or } y = -1\n$$\nBut $y + 1 \\ge 0$ means $y \\ge -1$, so $y = 1$ or $y = -1$.\n\n**Case 2:** $y + 1 < 0$ (so $|y+1| = -(y+1)$)\n\n$$\n3(-y-1) = 2y^2 + 3y + 1\n$$\n$$\n-3y - 3 = 2y^2 + 3y + 1\n$$\n$$\n2y^2 + 6y + 4 = 0\n$$\n$$\n(y + 2)(y + 2) = 0 \\implies y = -2\n$$\nCheck $y + 1 < 0 \\implies y < -1$, so $y = -2$ is valid.\n\n**Final answer:**\n$$\ny \\in \\{-2, -1, 1\\}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20082,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a cyclic quadrilateral. The internal angle bisectors of $\\angle ABC$ and $\\angle ADC$ intersect at a point lying on diagonal $AC$. Let $M$ be the midpoint of $AC$. The line parallel to $BC$ passing through $D$ intersects $BM$ at $E$ and the circumcircle of $ABCD$ at $F$ (with $F \\neq D$). Prove that $BCEF$ is a parallelogram.",
"options": [],
"answer": "See solution",
"solution": "We prove the statement in reverse, as this approach is more natural for the problem.\n\nIf $BCEF$ is a parallelogram, its diagonals bisect each other, so the intersection point $G = BE \\cap CF$ must be the midpoint of $CE$.\n\nIf $G$ is the midpoint of $CE$, then $\\triangle GBC$ and $\\triangle GEF$ are congruent: $CG = GF$, and $FE \\parallel BC$ implies $\\angle GEF = \\angle GBC$ and $\\angle GFE = \\angle GCB$. Thus, $BG = GE$, so $BCEF$ is a parallelogram since its diagonals bisect each other. Therefore, $G$ being the midpoint of $CF$ is equivalent to the problem statement.\n\nSince $M$ is the midpoint of $AC$, by the midline theorem in $\\triangle ACF$, $G$ is the midpoint of $CF$ if and only if $MG \\parallel AF$. Thus, we need to prove $BM \\parallel AF$.\n\nGiven $FD \\parallel BC$, this is equivalent to $\\angle AFD = \\angle MBC$.\n\nAlso, $\\angle AFD = \\angle ABD$ (angles subtended by the same chord), so our claim is equivalent to $\\angle ABD = \\angle MBC$.\n\nDepending on the position of $F$, we may have $\\pi - \\angle AFD = \\angle MBC$, but then $\\pi - \\angle AFD = \\angle ABD$, so the conclusion still holds.\n\nWe know $\\angle BDA = \\angle BCM$ (angles subtended by the same chord), so our claim is equivalent to $\\triangle BCM \\sim \\triangle BDA$.\n\nThis angle equality implies $\\frac{BC}{CM} = \\frac{AD}{BD}$. Since $M$ is the midpoint of $AC$, $CM = \\frac{AC}{2}$, so our claim is equivalent to $2AD \\cdot BC = BD \\cdot AC$.\n\nBy the angle bisector theorem in $\\triangle ABC$ and $\\triangle CDA$:\n\n$$\n\\frac{AB}{BC} = \\frac{AI}{CI} = \\frac{AD}{CD}.\n$$\n\nThus, our claim is equivalent to $AB \\cdot CD + AD \\cdot BC = BD \\cdot AC$, which is Ptolemy's theorem for cyclic quadrilaterals.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20083,
"subject": "Mathematics (Olympiad)",
"question": "Iulia and Ștefan shared the 52 playing cards from a deck so each got 26 cards. The cards from 2 to 10 are assigned their own value, the ace is worth 11 points, the jack 12 points, the queen 13 points and the king 14 points. Ștefan noticed that he had no ace in his stack, no 2 and no four cards of the same value. Iulia noticed that from her cards above 11 points, she doesn't have more than two of the same value. Iulia adds up all her points. What is the lowest value she can get? What is the highest?\n\nA deck has 52 cards: four with the value 2, four with the value 3, ..., four with the value 10, 4 aces, 4 jacks, 4 queens and 4 kings.",
"options": [],
"answer": "See solution",
"solution": "If we denote by $x_i$ the number of cards with the value $i$ that Iulia has in her stack, then $x_i \\geq 1$, $x_{11} = x_4 = 4$, $1 \\leq x_{12} \\leq 2$, $1 \\leq x_{13} \\leq 2$, $1 \\leq x_{14} \\leq 2$.\n\nIf $M$ and $m$ represent the highest and the lowest value that Iulia's stack can have then:\n\n$$\n\\begin{align*}\nM &= 2 \\cdot 14 + 2 \\cdot 13 + 2 \\cdot 12 + 4 \\cdot 11 + 4 \\cdot 10 + 2 \\cdot 9 + 1 \\cdot 8 + 1 \\cdot 7 + 1 \\cdot 6 + 1 \\cdot 5 + 1 \\cdot 4 + 1 \\cdot 3 + 4 \\cdot 2 = 221, \\\\\nm &= 1 \\cdot 14 + 1 \\cdot 13 + 1 \\cdot 12 + 4 \\cdot 11 + 1 \\cdot 10 + 1 \\cdot 9 + 1 \\cdot 8 + 1 \\cdot 7 + 1 \\cdot 6 + 2 \\cdot 5 + 4 \\cdot 4 + 4 \\cdot 3 + 4 \\cdot 2 = 169.\n\\end{align*}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20084,
"subject": "Mathematics (Olympiad)",
"question": "Let $E$ be a set with $n$ elements. Suppose $A_1, A_2, \\dots, A_k$ are $k$ distinct non-empty subsets of $E$, with the property that for any $1 \\leq i < j \\leq k$, either $A_i \\cap A_j = \\emptyset$ or one includes the other (i.e., $A_i \\subset A_j$ or $A_j \\subset A_i$). Find the maximum value of $k$.\n\n",
"options": [],
"answer": "See solution",
"solution": "We claim that the maximum value of $k$ is $2n-1$.\n\nTo prove this, we first give an example that achieves $k = 2n-1$. Let $E = \\{1, 2, \\dots, n\\}$, and define\n\n$$\nA_i = \\begin{cases} \\{i\\} & \\text{for } 1 \\leq i \\leq n, \\\\ \\{1, 2, \\dots, i-n+1\\} & \\text{for } n+1 \\leq i \\leq 2n-1. \\end{cases}\n$$\n\nIt is easy to see that these sets satisfy the given property.\n\nNow we prove that $k \\leq 2n - 1$ by induction.\n\nWhen $n = 1$, the statement is clearly true.\n\nAssume it holds for $n \\leq m-1$. For $n = m$, consider the set among $A_1, \\dots, A_k$ (excluding $E$) with the most elements; suppose it is $A_1$ with $t$ ($\\leq m-1$) elements. Partition the sets into three categories:\n\n1. The set $E$.\n2. Sets included in $A_1$.\n3. Sets disjoint from $A_1$.\n\n(Categories 1 and 2 may be empty.)\n\n- The number of sets in category 1 is at most 1.\n- By induction, the number in category 2 is at most $2t-1$.\n- The union of sets in category 3 contains at most $m-t$ elements, so by induction, there are at most $2(m-t)-1$ such sets.\n\nThus,\n\n$$\nk \\leq 1 + (2t-1) + [2(m-t)-1] = 2m-1.\n$$\n\nTherefore, the proposition holds for $n = m$.\n\nBy induction, for any set $E$ of $n$ elements, $k \\leq 2n-1$. This completes the proof. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20085,
"subject": "Mathematics (Olympiad)",
"question": "Нека $k$ е опишаната кружница околу рамнокракиот триаголник $\\triangle ABC$ со основа $BC$. Нека $E$ е пресечната точка на симетралите на внатрешните агли во темињата $B$ и $C$. Нека $D$ и $F$ се пресечните точки на симетралите на аглите во темињата $B$ и $C$ со $k$ соодветно. Докажи дека $EDAF$ е ромб.",
"options": [],
"answer": "See solution",
"solution": "Нека $\\alpha = \\angle ABC = \\angle ACB$ и нека $\\gamma = \\angle CAB$. Тогаш, бидејќи $BD$ е симетрала на $\\angle CBA$ и $CF$ е симетрала на $\\angle BCA$, следува дека:\n\n$$\n\\angle ABE = \\angle CBE = \\frac{\\alpha}{2} = \\angle BCE = \\angle ACE\n$$\n\nПа $\\angle CED = \\alpha$. Па оттука и $\\angle BEF = \\alpha$. $\\angle CFA = \\angle CBA = \\alpha$ како агли над ист кружен лак, па $\\angle CFA = \\alpha = \\angle CED$. Од каде $FA \\parallel ED$. Аналогно $EF \\parallel AD$.\n\nСега $\\angle CAD = \\angle CBD = \\frac{\\alpha}{2}$ и $\\angle BCF = \\angle BAF = \\frac{\\alpha}{2}$, од каде добиваме\n\n$$\n\\angle FAC = \\angle FAB + \\gamma = \\frac{\\alpha}{2} + \\gamma\n$$\nи\n$$\n\\angle DAB = \\angle DAC + \\gamma = \\frac{\\alpha}{2} + \\gamma\n$$\nЗначи $\\angle FAC = \\angle DAB$.\n\nСега, бидејќи и $\\angle ABD = \\frac{\\alpha}{2} = \\angle ACF$, $\\overline{AC} = \\overline{AB}$, следува дека $\\triangle ACF \\cong \\triangle ABD$, па $\\overline{AF} = \\overline{AD}$. Па $EDAF$ е ромб.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20086,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $n$ such that there exist positive integers $a, b, c$ satisfying $\\gcd(a, b, c) = 1$ and $a + b + c = \\gcd(ab + c, ac - b) = n$.",
"options": [],
"answer": "See solution",
"solution": "All positive integers $n$ whose prime divisors are all of the form $4k + 1$.\n\nWe use the following lemma:\n\n**Lemma.** Let $m$ be an odd positive integer. The equation $x^2 \\equiv -1 \\pmod m$ has an integer solution if and only if all prime divisors of $m$ are of the form $4k + 1$.\n\nFirst, all prime divisors of $n$ must be of the form $4k + 1$.\n\n**Claim 1.** $\\gcd(n, abc) = 1$.\n\n*Proof.* Assume the contrary. Let $p$ be a prime such that $p \\mid \\gcd(n, abc)$. Then $p \\mid n$, $p \\mid a$, $p \\mid b$, $p \\mid c$, contradicting $\\gcd(a, b, c) = 1$. $\\square$\n\n**Claim 2.** $n$ is odd.\n\n*Proof.* If $n$ is even, at least one of $a, b, c$ is even since $n = a + b + c$, contradicting Claim 1. $\\square$\n\n**Claim 3.** $a^2 \\equiv -1 \\pmod n$.\n\n*Proof.* Since $\\gcd(ab + c, ac - b) = n$, we have $ab \\equiv -c \\pmod n$ and $ac \\equiv b \\pmod n$. Multiplying gives $a^2bc \\equiv -bc \\pmod n$, so $n | (a^2 + 1)bc$. By Claim 1, $n$ and $bc$ are coprime, so $n | a^2 + 1$, i.e., $a^2 \\equiv -1 \\pmod n$. $\\square$\n\nBy the lemma, all prime divisors of $n$ are of the form $4k + 1$.\n\nConversely, for any $n$ with all prime divisors of the form $4k + 1$, there exists $x$ such that $x^2 \\equiv -1 \\pmod n$ and $1 < x < n-1$. Since $n$ is odd, one of $x$ or $n-x$ is even; assume $x$ is even. Set $a = x$, $b = \\frac{n-x+1}{2}$, $c = \\frac{n-x-1}{2}$.\n\nThen $a + b + c = n$, $\\gcd(a, b, c) = 1$ (since $b - c = 1$), and\n\n$$\nab + c = \\frac{n(x + 1) - x^2 - 1}{2}$$\n$$ac - b = \\frac{n(x - 1) - x^2 - 1}{2}$$\n\nSince $n$ is odd and $n | x^2 + 1$, $n | ab + c$ and $n | ac - b$. Also, $(ab + c) - (ac - b) = n$, so $\\gcd(ab + c, ac - b) = n$.\n\nThus, all requirements are satisfied.\n\n**Remark.** For any such triplet, $b = c + 1$ and $a$ is even.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20087,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an isosceles triangle with $AB = AC$. Let $D$ be a point on the segment $BC$ such that $BD = 2DC$. Let $P$ be a point on the segment $AD$ such that $\\angle BAC = \\angle BPD$. Prove that $\\angle BAC = 2\\angle DPC$.",
"options": [],
"answer": "See solution",
"solution": "Extend $AD$ to $E$ such that $PE = PB$. Join $EB$ and $EC$.\n\n\n\n$$\n\\angle BPE = \\angle BAC \\text{ and } \\frac{PB}{PE} = \\frac{AB}{AC} = 1.\n$$\n\nHence it follows that $CAB$ is similar to $EFB$. Thus $\\angle PEB$. This shows that $A$, $B$, $E$, $C$ are concyclic. In turn we obtain\n\n$$\n\\angle AEC = \\angle ABC = \\angle ACB = \\angle AEB.\n$$\n\nWe conclude that $AE$ bisects $\\angle BEC$.\n\nLet $M$ be the midpoint of $BE$. Join $PM$ and $PC$. Since $EA$ bisects $\\angle BEC$, we have\n\n$$\n\\frac{CE}{EB} = \\frac{CD}{DB} = \\frac{CD}{2CD} = \\frac{1}{2}.\n$$\n\nThus $CE = EB/2 = EM$. We also observe that $\\angle MEP = \\angle CEP$. This shows that $CEP$ is congruent to $MEP$. This implies that\n\n$$\n\\angle BAC = \\angle BPE = 2\\angle MPE = 2\\angle DPC.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20088,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. Find the number of odd coefficients of the polynomial $(x^2 - x + 1)^n$.",
"options": [],
"answer": "See solution",
"solution": "Let $P(x)$ and $Q(x)$ be polynomials with integer coefficients. If $P(x) - Q(x)$ has all coefficients even, we say $P(x) \\sim Q(x)$. In this case, $P(x)$ and $Q(x)$ have the same number of odd coefficients, denoted $\\beta(P)$. Clearly,\n\n$$\n(x^2 - x + 1)^n \\sim (x^2 + x + 1)^n.\n$$\n\nWe now discuss the number of odd coefficients of $P_n(x) = (x^2 + x + 1)^n$.\n\nBy induction, when $n = 2^q$ ($q$ a positive integer), $P_n(x) \\sim x^{2n} + x^n + 1$. Thus, we can convert $n$ to binary. Consider the case $n = 2^m - 1$, $m$ a positive integer.\n\nLet $m = 2k + 1$, $k \\ge 0$. Then\n\n$$\nn = 2^{2k+1} - 1 \\equiv 1 \\pmod{3}.\n$$\n\nConsider the polynomial\n\n$$\nR(x) = (x+1) \\left( \\sum_{k=0}^{\\frac{n-1}{3}} x^{n+3k} + \\sum_{k=0}^{\\frac{n-4}{3}} x^{3k} \\right) + x^{n-1}.\n$$\n\nThen $\\beta(R) = \\frac{2^{m+2}-1}{3}$.\n\n$$\n\\begin{aligned}\nR(x)(x^2+x+1) &\\sim (x+1) \\left( \\sum_{k=0}^{n+1} x^{n+k} + \\sum_{k=0}^{n-2} x^k \\right) + x^{n-1}(x^2+x+1) \\\\\n&\\sim x^{2n+2} + x^{n+1} + 1,\n\\end{aligned}\n$$\n\nand $P_n(x)(x^2+x+1) \\sim x^{2n+2} + x^{n+1} + 1$.\n\nThus $\\beta(P_n) = \\beta(R) = \\frac{2^{m+2}+1}{3}$.\n\nNow let $m = 2k + 1$, $k$ a positive integer. Then\n\n$$\nn = 2^{2k} - 1 \\equiv 0 \\pmod{3}.\n$$\n\nConsider the polynomial\n\n$$\nQ(x) = (x+1) \\sum_{k=0}^{\\frac{n-3}{3}} (x^{n+2+3k} + x^{3k}) + x^n.\n$$\n\nSimilarly, $\\beta(P_n) = \\beta(Q) = \\frac{2^{m+2}-1}{3}$.\n\nTherefore,\n\n$$\n\\beta(P_{2^m-1}(x)) = \\frac{2^{m+2} + (-1)^{m+1}}{3}.\n$$\n\nFor the general case, write $n$ in binary:\n\n$$\nn = (\\underbrace{11\\cdots1}_{a_k \\text{ digits}} \\underbrace{00\\cdots0}_{b_k \\text{ digits}} \\underbrace{11\\cdots1}_{a_{k-1} \\text{ digits}} \\underbrace{00\\cdots0}_{b_{k-1} \\text{ digits}} \\cdots \\underbrace{11\\cdots1}_{a_1 \\text{ digits}} \\underbrace{00\\cdots0}_{b_1 \\text{ digits}})_2,\n$$\n\nwhere $a_i, b_i$ are positive integers, $b_1 \\ge 0$.\n\nLet $S_i = \\sum_{j=1}^{i-1} (b_j + a_j) + b_i$ for $i = 1, 2, \\dots, k$.\n\nThen\n\n$$\nn = \\sum_{i=1}^{k} 2^{S_i}(2^{a_i} - 1),\n$$\n\n$$\nP_n(x) = \\prod_{i=1}^{k} (x^2 + x + 1)^{2^{S_i} (2^{a_i} - 1)}\n$$\n\n$$\n\\sim \\prod_{i=1}^{k} (x^{2^{S_i+1}} + x^{2^{S_i}} + 1)^{2^{a_i}-1}.\n$$\n\nTherefore,\n\n$$\n\\beta(P_n) = \\prod_{i=1}^{k} \\frac{2^{a_i+2} - (-1)^{a_i}}{3}.\n$$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20089,
"subject": "Mathematics (Olympiad)",
"question": "Solve in the set of positive integers the equation\n\n$$\n3^{x} - 5^{y} = z^{2}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Working modulo 3 on both sides of the equation, we get\n$z^2 \\equiv -(-1)^y \\pmod{3}$.\n\nWhen $y$ is even, $z^2 \\equiv -1 \\pmod{3}$, which is impossible. Hence,\n\n$$\n\\boxed{y = 2y_1 + 1}, \\quad y_1 = 0, 1, 2, \\dots\n$$\n\nWorking modulo 4, we get $z^2 \\equiv (-1)^x - 1^y \\pmod{4}$.\n\nIf $x$ is odd, $z^2 \\equiv 2 \\pmod{4}$, which is impossible. Thus,\n\n$$\n\\boxed{x = 2x_1}, \\quad x_1 = 1, 2, \\dots\n$$\n\nSubstituting into the original equation:\n\n$$\n3^{2x_1} - 5^{2y_1+1} = z^2 \\implies 5^{2y_1+1} = (3^{x_1} - z)(3^{x_1} + z)\n$$\n\nSince $(3, z) = 1$ (otherwise $3 \\mid z$, which is absurd),\n\n$$\n(3^{x_1} - z, 3^{x_1} + z) \\mid 2 \\cdot (3^{x_1}, z) = 2\n$$\n\nThus, $(3^{x_1} - z, 3^{x_1} + z) = 1$ or $2$. Since $z$ is even, we have\n\n$$\n(3^{x_1} - z, 3^{x_1} + z) = 1\n$$\n\nFrom above, $3^{x_1} - z > 0$ and $3^{x_1} + z > 1$, so\n\n$$\n\\begin{gathered}\n3^{x_1} - z = 1, \\quad 3^{x_1} + z = 5^{2y_1+1} \\\\\n\\implies 2 \\cdot 3^{x_1} = 5^{2y_1+1} + 1\n\\end{gathered}\n$$\n\nConsider cases:\n\nIf $x_1 = 2x_2$, then $2 \\cdot 9^{x_2} = 5 \\cdot 25^{y_1} + 1$. Modulo 24:\n\n$$\n2 \\cdot 9^{x_2} \\equiv 5 \\cdot 1^{y_1} + 1 \\pmod{24} \\implies 9^{x_2} \\equiv 3 \\pmod{12}\n$$\n\nThis is impossible since $9^{x_2} \\equiv 9 \\pmod{12}$ always.\n\nIf $x_1 = 2x_2 + 1$, then\n\n$$\n2 \\cdot 3 \\cdot 9^{x_2} = 5 \\cdot 25^{y_1} + 1\n$$\n\nFor $x_2 = 0$, $y_1 = 0$, so $y = 1$, $x = 2$, $z = 2$.\n\nIf $x_2 \\geq 1$, modulo 9:\n\n$5 \\cdot 7^{y_1} \\equiv -1 \\pmod{9}$, which is valid only when $y_1 \\equiv 1 \\pmod{3}$, i.e., $y_1 = 3y_2 + 1$.\n\nIf $y_2 = 0$, no solution. For $y_2 \\geq 1$, substitute $y_1$ into above and consider modulo 7:\n\n$$\n\\begin{align*}\n6 \\cdot 2^{x_2} &\\equiv 5 \\cdot 4^{3y_2+1} + 1 \\pmod{7} \\\\\n&\\equiv (-1) \\cdot 1^{y_2} + 1 \\pmod{7} \\\\\n&\\equiv 0 \\pmod{7}\n\\end{align*}\n$$\n\nThis is impossible.\n\nHence, the unique solution is $(x, y, z) = (2, 1, 2)$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20090,
"subject": "Mathematics (Olympiad)",
"question": "Find all integers $x$ such that for every positive integer $n$, if $p$ is any prime dividing $10^n + n$, then $p$ also divides $x^n + n$.",
"options": [],
"answer": "See solution",
"solution": "Assume $x \\neq 10$ and let $p$ be a prime that does not divide $x - 10$. By the Chinese Remainder Theorem, there exists a positive integer $n$ such that\n\n$$\n\\begin{cases}\nn \\equiv 1 \\pmod{p-1} \\\\\nn \\equiv -10 \\pmod{p}\n\\end{cases}\n$$\n\nBy Fermat's Little Theorem, $10^n \\equiv 10 \\pmod{p}$, so\n\n$$\n10^n + n \\equiv 10 + n \\equiv 10 - 10 = 0 \\pmod{p}.\n$$\n\nBut $x^n + n \\equiv x + n \\equiv x - 10 \\not\\equiv 0 \\pmod{p}$, since $p$ does not divide $x - 10$. Thus, $p$ divides $10^n + n$ but not $x^n + n$, contradicting the condition. Therefore, $x = 10$ is the only solution.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20091,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\omega$ be the circumscribed circle of $\\triangle ABC$. Chord $AD$ is the angle bisector of $\\triangle ABC$ and meets $BC$ at $L$. Chord $DK$ is perpendicular to $AC$ and meets it at $M$. Find the ratio $\\frac{AM}{MC}$, given that $\\frac{BL}{LC} = \\frac{1}{2}$.",
"options": [],
"answer": "See solution",
"solution": "Using the property of the angle bisector, we have $\\frac{BL}{AB} = \\frac{LC}{AC}$, so $\\frac{AC}{AB} = \\frac{LC}{BL} = 2$, hence $AC = 2AB$.\n\n$$\n\\angle BAD = \\angle DAC \\quad (AD \\text{ is a bisector of } \\angle BAC), \\text{ thus } BD = DC.\n$$\n\nConsider $S$ as the midpoint of $AC$. Then $AS = SC = AB$. Thus, $\\triangle BAD = \\triangle SAD$ (they have two equal sides and the angle between them). This implies $BD = DS$ and $DC = BD = DS$.\n\nHence, $\\triangle SDC$ is isosceles with base $SC$. Its altitude $DM$ is also a median. Thus, $MC = \\frac{1}{2}SC = \\frac{1}{2} \\cdot \\frac{1}{2}AC = \\frac{1}{4}AC$. We have $AM = AC - MC = AC - \\frac{1}{4}AC = \\frac{3}{4}AC$. Therefore,\n\n$$\n\\frac{AM}{MC} = \\frac{\\frac{3}{4}AC}{\\frac{1}{4}AC} = 3.\n$$\n\nNote: In the acute-angled triangle $ABC$ with $AC = 2AB$, $\\angle ACB$ does not exceed $30^\\circ$. This implies that the perpendicular from point $D$ to line $AC$ meets segment $AC$.\n\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20092,
"subject": "Mathematics (Olympiad)",
"question": "We consider a group of trees in a nature reserve, all of which have a positive integral age. The average age is $41$ years. After destruction of a tree with an age of $2010$ years by lightning, the average age of the remaining trees is $40$ years.\n\nDetermine the original number of trees in the group. What is the maximal number of trees of an age of $2010$ years in the original group?",
"options": [],
"answer": "See solution",
"solution": "Let the original number of trees be $n$ and the sum of their ages be $s$.\n\nThen:\n$$\ns = 41n.\n$$\nAfter removing a tree of age $2010$:\n$$\ns - 2010 = 40(n - 1).\n$$\nSolving these equations:\n$$\n41n - 2010 = 40(n - 1) \\\\\n41n - 2010 = 40n - 40 \\\\\nn = 1970\n$$\nSo the sum of ages is:\n$$\ns = 41 \\times 1970 = 80770.\n$$\nTo maximize the number of trees aged $2010$, let $k$ be their number:\n$$\n2010k + \\text{(sum of other ages)} = 80770\n$$\nSince all ages are positive integers and $2010k \\leq 80770$, the maximal $k$ is:\n$$\nk = \\left\\lfloor \\frac{80770}{2010} \\right\\rfloor = 40\n$$\nBut if $k = 40$, the remaining ages sum to $80770 - 2010 \\times 40 = 370$, which must be distributed among $1970 - 40 = 1930$ trees, each with positive integer age. This is not possible, as $370 < 1930$.\n\nTry $k = 39$:\n$$\n80770 - 2010 \\times 39 = 2380\n$$\nNow $2380$ can be distributed among $1931$ trees, which is possible (e.g., $1930$ trees of age $1$ and one tree of age $450$).\n\nThus, the maximal number of trees aged $2010$ is $39$.\n\n*Example*: $1930$ trees of age $1$, $1$ tree of age $450$, and $39$ trees of age $2010$.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 20093,
"subject": "Mathematics (Olympiad)",
"question": "Determine all triples of positive integers $(p, a, b)$ such that $p$ is a prime and $p^a + p^b$ is a perfect cube.",
"options": [],
"answer": "See solution",
"solution": "Case 1: $a = b$.\n\nThe problem then is to determine all pairs $(p, a)$ of positive integers such that $p$ is a prime and $2p^a$ is a perfect cube. But since every even perfect cube is divisible by $8$, this is exactly the case when $p = 2$ and $a \\equiv 2 \\pmod{3}$.\n\nCase 2: $a < b$.\n\nWe have $p^a + p^b = p^a(p^s + 1)$, where $s = b - a > 0$.\n\nSuppose $p^a(p^s + 1)$ is a perfect cube. Since $p^a$ and $p^s + 1$ are relatively prime, $p^a$ and $p^s + 1$ must both be perfect cubes. Hence $a = 3t$ for some positive integer $t$.\n\nWe write $p^s + 1 = u^3$. Then $p^s = u^3 - 1 = (u-1)(u^2 + u + 1)$.\n\nHence both $u-1$ and $u^2 + u + 1$ are non-negative powers of $p$. Since $u-1 < u^2 + u + 1$, it follows that $u-1 \\mid u^2 + u + 1$.\n\nBut $u^2 + u + 1 = (u-1)(u+2) + 3$. So $u-1 \\mid 3$. Thus $u = 2$ or $u = 4$.\n\nCase 2a: $u = 2$.\n\nThen $p^s = 7$, implying $p = 7$ and $s = 1$, so that $b = 3t + 1$.\n\nCase 2b: $u = 4$.\n\nThen $p^s = 63$, which is not a prime power.\n\nHence the only possible triples that satisfy the conditions are:\n\n$$\n(2, 3t-1, 3t-1) \\text{ and } (7, 3t, 3t+1), \\quad t = 1, 2, 3, \\dots\n$$\n\nBy inspection, one finds that all these triples satisfy the given conditions.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20094,
"subject": "Mathematics (Olympiad)",
"question": "An acute scalene triangle $ABC$ is inscribed in a circle $k$. The bisector of angle $\\angle ABC$ meets side $BC$ at point $D$. Let $I$ be an arbitrary point on the segment $AD$, and let $H$ be the orthogonal projection of $I$ onto $BC$. Circle $\\omega$ is centered at $I$ and passes through $H$, and $U$ is the internal homothety center of the circles $k$ and $\\omega$. Prove that $U$ lies on a fixed line as $I$ moves on $AD$, and that $AI$ bisects the angle $HAU$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $M$, $N$ be the midpoints of the minor and major arcs $BC$ of circle $(O)$, respectively, then points $A$, $D$, $M$ are collinear. Let $T$ be the intersection of the two tangent lines at $B$ and $C$ of $(O)$, and $J$ the midpoint of $BC$. First, by angle chasing, we have $\\angle TBM = \\angle BAM = \\angle MBC$, so $BM$ is the internal bisector of $\\angle CBT$. Thus, $M$ is the center of the incircle of triangle $BTC$, denoted by $(M)$. Then it is easy to see that $D$ is the internal homothety center of $(\\omega)$ and $(M)$; and $T$ is the external homothety center of $(M)$ with $(O)$.\n\nApplying Monge D'Alembert's theorem to three circles $(I)$, $(M)$, $(O)$, we see that $U$, $D$, $T$ are collinear, which implies that $T$ belongs to the fixed line $TD$.\n\nLet $V$ be the external homothety center of $(\\omega)$ with $(O)$. Since $IH \\parallel MN$, points $V$, $H$, $M$ are collinear. Note that since $U$ is the inner center of $(\\omega)$ and $(O)$, $H$, $U$, $N$ are collinear because $IH \\parallel ON$. Then we have\n\n$$\nA(UH, IN) = A(UM, HN) = M(UH, AN) = M(UV, IO).\n$$\n\nOn the other hand, since $U$, $V$ are the internal and external homothety centers of $(\\omega)$ and $(O)$, $(UV, IO) = -1$. From this it follows that $A(UH, IN) = -1$. Since $AI \\perp AN$, by the property of harmonic bundles, $AI$ is the internal bisector of angle $HAN$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20095,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ be a natural odd number which is not a perfect square. If $m$ and $n$ are strictly positive integers, prove that\n\n$$\n\\begin{align*}\n\\text{a)} \\quad & \\{m(a + \\sqrt{a})\\} \\neq \\{n(a - \\sqrt{a})\\}, \\\\\n\\text{b)} \\quad & [m(a + \\sqrt{a})] \\neq [n(a - \\sqrt{a})].\n\\end{align*}\n$$",
"options": [],
"answer": "See solution",
"solution": "a) As $ma, na$ are natural numbers, the equality implies $\\{m\\sqrt{a}\\} = \\{-n\\sqrt{a}\\}$.\n\nTwo numbers have the same fractional part if and only if their difference is an integer, whence $(m+n)\\sqrt{a} \\in \\mathbb{Z}$, which is absurd.\n\nb) Again, let us suppose that there is a natural number $N$, for which there exist two not equal numbers $m, n$ which are different from zero, such that $N = [m(a + \\sqrt{a})] = [n(a - \\sqrt{a})]$. Then $N \\leq m(a + \\sqrt{a}) < N + 1$ and $N \\leq n(a - \\sqrt{a}) < N + 1$; moreover, the inequalities are strict, because the terms in the middle are irrational numbers.\n\nWe rewrite the inequalities as\n\n$$\n\\frac{N}{a + \\sqrt{a}} < m < \\frac{N + 1}{a + \\sqrt{a}}, \\quad \\frac{N}{a - \\sqrt{a}} < n < \\frac{N + 1}{a - \\sqrt{a}},\n$$\n\nand thus, by addition, we get $N \\frac{2}{a-1} < m + n < (N+1) \\frac{2}{a-1}$.\n\nFrom here, $N < \\frac{a-1}{2}(m+n) < N+1$, which is a contradiction, because the term in the middle is natural ($a$ is odd).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20096,
"subject": "Mathematics (Olympiad)",
"question": "На окружности отмечено $2n+1$ точек ($n \\geq 2$). Два игрока по очереди стирают по одной из оставшихся точек. Проигрывает тот, после чьего хода среди оставшихся точек найдётся остроугольный треугольник. Кто выигрывает при правильной игре?",
"options": [],
"answer": "See solution",
"solution": "*Ответ.* Противник.\n\nПриведём стратегию для второго игрока, позволяющую ему выиграть. Для этого он будет добиваться выполнения следующего условия: перед каждым ходом первого, если осталось $2k+1 \\geq 5$ точек, то на любой полуокружности осталось не менее $k$ отмеченных точек.\n\nПокажем индукцией по числу ходов, что это возможно. В начале игры это условие выполнено. Пусть перед ходом первого оно выполнено; пронумеруем оставшиеся точки по порядку $A_0, \\dots, A_{2k}$. Пусть для определённости первый своим ходом удаляет точку $A_0$; заметим, что треугольник $A_{k+1}A_1A_{k+2}$ остроугольный, так что игра ещё не закончилась. Второму достаточно удалить точку $A_k$. Теперь, если с некоторой полуокружности удалено не более одной точки, то на ней осталось не менее $k-1$ точки; иначе с неё стёрты обе точки $A_0$ и $A_k$, поэтому на ней остались либо точки $A_1, \\dots, A_{k-1}$, либо точки $A_{k+1}, \\dots, A_{2k}$; в любом случае для неё требуемое условие выполнено.\n\nИтак, если первый не проиграет раньше, то после $2n-4$ ходов на доске останется пять точек $A_0, A_1, A_2, A_3, A_4$. Пусть для определённости первый удалит точку $A_0$; тогда ещё останется остроугольный треугольник $A_1A_2A_4$. Второй же последним ходом удалит $A_4$, и оставшийся треугольник $A_1A_2A_3$ будет тупоугольным (иначе нашлась бы полуокружность, содержащая лишь $A_2$). Значит, второй выиграет.\n\n*Замечание 1.* Естественно, вместо точки $A_k$ второй может выбирать также точку $A_{k+1}$. Нетрудно видеть, что при описанной стратегии в конце игры останутся именно три отмеченных точки.\n\n*Замечание 2.* По сути ту же стратегию можно оформить по-другому. Соединим каждую из исходных точек с двумя наиболее удалёнными от неё. Все проведённые отрезки образуют одну $(2n+1)$-звенную ломаную. Тогда второй может ходить так, чтобы после каждого его хода (кроме последнего) все стёртые точки разбивались на пары точек, соединённых отрезком. Можно показать, что соблюдения этого условия также достаточно для выигрыша.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20097,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a quadrilateral with points $P$ on $AB$ and $Q$ on $CD$, and $M$ the intersection of $PQ$ and the diagonals. Consider the conditions under which triangles $APM$ and $BPM$ are similar, and deduce the properties of $ABCD$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Suppose $\\angle APM > 90^\\circ$. Then in $\\triangle BPM$, $\\angle BPM < 90^\\circ$. Since $\\angle PBM + \\angle BMP = \\angle APM$, $\\angle PBM, \\angle BMP < \\angle APM$. So none of the angles of $\\triangle BPM$ can be equal to $\\angle APM$ of $\\triangle APM$. Therefore, $\\triangle APM$ cannot be similar to $\\triangle BPM$. Thus $\\angle APM = 90^\\circ$ and $PQ \\perp AB$. Similarly, $PQ \\perp CD$ and it follows that $AB \\parallel DC$.\n\nIt then follows that $\\triangle APM \\sim \\triangle CQM$ and $\\triangle BMP \\sim \\triangle DQM$.\n\nIf $\\triangle APM \\sim \\triangle BPM$, then $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$. Therefore, $ABCD$ is an isosceles trapezium.\n\nIf $\\triangle APM \\sim \\triangle MPB$, then $\\angle MAP = \\angle BMP$ and $\\angle PMA = \\angle PBM$. It follows that $\\angle AMB = 90^\\circ$. So the diagonals are perpendicular. The quadrilaterals are either isosceles trapezia or trapezia in which the diagonals are perpendicular.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20098,
"subject": "Mathematics (Olympiad)",
"question": "Consider a sequence $\\{a_n\\}$ defined as follows: $a_1 = a$, $a_2 = b$, where $a$ and $b$ are positive integers. For all $n \\geq 2$, $a_{n+1}$ equals the number of indices $i$, $1 \\leq i \\leq n$, such that $a_i = a_n$.\n\nFor example, for $a = 2$, $b = 1$, the sequence starts as $(2, 1, 1, 2, 2, 3, \\ldots)$.\n\nDetermine all pairs $(a, b)$ for which the following condition holds: there exists an index $n_0$ such that the sequence $\\{a_n + a_{n+1}\\}$, for $n \\geq n_0$, is non-decreasing.",
"options": [],
"answer": "See solution",
"solution": "The only solution is $a = b = 1$.\n\nIf $a = b = 1$, then $\\{a_n\\}$ has terms $(1, 1, 2, 1, 3, 1, 4, 1, 5, 1, \\ldots)$, and $\\{a_n + a_{n+1}\\}$ has terms $(2, 3, 3, 4, 4, 5, 5, 6, 6, 7, \\ldots)$. Therefore, $a = b = 1$ satisfies the condition.\n\nSuppose $a_1 = a$ and $a_2 = b$ with $a_1 \\neq 1$ or $a_2 \\neq 1$. Without loss of generality, assume $a_1 \\neq 1$ (otherwise, swap $a_1$ and $a_2$). The sequence $\\{a_n\\}$ is unbounded: if $\\max\\{a_i : i \\in \\mathbb{N}\\} = s$, then among $\\{a_1, a_2, \\ldots, a_{s+1}\\}$ there must be at least $s+1$ equal numbers, so the next term is at least $s+1$, a contradiction.\n\nNow, assume that for $n \\geq n_0$, the sequence $\\{a_n + a_{n+1}\\}$ is non-decreasing. Consider $k \\geq n_0 + 1$ such that $a_k = t \\geq \\max\\{a, b\\} + 1$ and this is the first occurrence of $t$ in the sequence. All previous terms are less than $t$, so $a_{k+1} = 1$. Since $\\{a_n + a_{n+1}\\}$ is non-decreasing, $a_{k-1} = 1$. Thus, there are exactly $t$ indices $1 < i_1 < i_2 < \\ldots < i_t = k-1$ with $a_{i_1} = a_{i_2} = \\ldots = a_{i_t} = 1$. The numbers $\\{a_{i_1-1}, a_{i_2-1}, \\ldots, a_{i_t-1}\\}$ are all $\\leq t-1$, so at least two are equal. Therefore, there are at least two equal pairs $\\{\\ldots, l, 1, \\ldots\\}$ in the sequence, but after the second $l$ the next term must be at least $2$, a contradiction.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20099,
"subject": "Mathematics (Olympiad)",
"question": "Given that\n\n$$\n2 \\sin(x) \\sin((2k + 1)x) = \\cos(2kx) - \\cos((2k + 2)x), \\quad k = 0, 1, 2, \\dots\n$$\n\nderive and prove that\n\n$$\n2 \\sin(x) \\sum_{k=0}^{n} (n + 1 - k) \\sin((2k + 1)x) \\ge 0\n$$\nfor all $x$, with equality if and only if $x = 0$ or $x = \\pi$.",
"options": [],
"answer": "See solution",
"solution": "We start from the identity:\n\n$$\n2 \\sin(x) \\sin((2k + 1)x) = \\cos(2kx) - \\cos((2k + 2)x)\n$$\n\nSumming both sides for $k = 0$ to $n$ with weights $(n + 1 - k)$:\n\n$$\n2 \\sin(x) \\sum_{k=0}^{n} (n + 1 - k) \\sin((2k + 1)x) = \\sum_{k=0}^{n} (n + 1 - k) (\\cos(2kx) - \\cos((2k + 2)x))\n$$\n\nThis telescopes to:\n\n$$\n= \\sum_{k=0}^{n} (n + 1 - k) \\cos(2kx) - \\sum_{k=1}^{n+1} (n + 2 - k) \\cos(2kx)\n$$\n\n$$\n= n + 1 - \\sum_{k=1}^{n} \\cos(2kx) - \\cos((2n + 2)x)\n$$\n\n$$\n= \\sum_{k=1}^{n+1} (1 - \\cos(2kx))\n$$\n\nSince $1 - \\cos(2kx) \\ge 0$ for all $x$, the sum is non-negative, with equality only when $x = 0$ or $x = \\pi$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20100,
"subject": "Mathematics (Olympiad)",
"question": "In triangle $ABC$, let $M_1$, $M_2$, and $M_3$ be the midpoints of sides $BC$, $AC$, and $AB$, respectively. Let $K$ be the point symmetric to $M_2$ with respect to line $BC$. Let $AH$ be the altitude of triangle $ABC$. Prove that the line $KM_3$ bisects the segment $HM_1$.\n\n\n\n**Fig. 42**",
"options": [],
"answer": "See solution",
"solution": "From the right-angled triangles $ABH$ and $ACH$, we have $M_3H = \\frac{1}{2}AB$ and $M_2H = \\frac{1}{2}AC$.\n\nAlso, $M_1M_2 = \\frac{1}{2}AB$ and $M_1M_3 = \\frac{1}{2}AC$ as these are midlines. By symmetry, $M_2M_1 = M_1K$ and $M_2H = HK$. Thus, $M_3H = M_1K$ and $M_3M_1 = HK$, so quadrilateral $HM_3M_1K$ is a parallelogram. Therefore, the line $KM_3$ bisects the segment $HM_1$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20101,
"subject": "Mathematics (Olympiad)",
"question": "Petr and Basil play a game on an initially empty checkered table $100 \\times 100$, making moves in turn. Petr starts. During his turn, the player writes any (capital) letter of the English alphabet into some empty cell (exactly one letter can be written in each cell). When all the cells are filled, Petr is declared the winner if there exist four consecutive cells horizontally in which the word \"PETR\" is written from left to right, or, if there exist four consecutive cells vertically in which the word \"PETR\" is written from top to bottom. Determine if Petr can win (regardless of Basil's actions).",
"options": [],
"answer": "See solution",
"solution": "He won't be able to.\n\nLet's describe Basil's winning strategy. Let Basil always write the letter \"Y\" in a cell according to the following conditions; if the specified cell doesn't exist or is already occupied, or if Petr writes any letter other than \"P\", \"E\", \"T\", or \"R\", then let Basil write \"Y\" in any free cell.\n\nIf Petr writes the letter \"P\" in some cell, then Basil writes \"Y\" in the cell to its right; if Petr writes the letter \"E\" then Basil writes \"Y\" in the cell to its left; if Petr writes the letter \"T\" then Basil writes \"Y\" in the cell below it; if Petr writes the letter \"R\" then Basil writes \"Y\" in the cell above it.\n\nFrom the first two conditions it follows that the sequence \"PE\" read from left to right cannot appear in two horizontally adjacent cells. Indeed, suppose a horizontal \"PE\" appears; then after the first of these two letters appears, Basil, according to the described strategy, will immediately occupy the position of the second letter — a contradiction. Similarly, the sequence \"TR\" read from top to bottom cannot appear in two vertically adjacent cells.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 20102,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $a_1, \\dots, a_n$ are integers whose greatest common divisor is $1$. Let $S$ be a set of integers with the following properties:\n\n(a) For $i = 1, \\dots, n$, $a_i \\in S$.\n\n(b) For $i, j = 1, \\dots, n$ (not necessarily distinct), $a_i - a_j \\in S$.\n\n(c) For any integers $x, y \\in S$, if $x + y \\in S$, then $x - y \\in S$.\n\nProve that $S$ must be equal to the set of all integers.",
"options": [],
"answer": "See solution",
"solution": "We may as well assume that none of the $a_i$ is equal to $0$. We start with the following observations.\n\n(d) $0 = a_1 - a_1 \\in S$ by (b).\n\n(e) $-s = 0 - s \\in S$ whenever $s \\in S$, by (a) and (d).\n\n(f) If $x, y \\in S$ and $x - y \\in S$, then $x + y \\in S$ by (c) and (e).\n\nBy (f) plus strong induction on $m$, we have that $ms \\in S$ for any $m \\ge 0$ whenever $s \\in S$. By (d) and (e), the same holds even if $m \\le 0$, and so we have the following.\n\n(g) For $i = 1, \\dots, n$, $S$ contains all multiples of $a_i$.\n\nWe next verify that\n\n(h) For $i, j \\in \\{1, \\dots, n\\}$ and any integers $c_i, c_j$, $c_i a_i + c_j a_j \\in S$.\n\nWe do this by induction on $|c_i| + |c_j|$. If $|c_i| \\le 1$ and $|c_j| \\le 1$, this follows from (b), (d), (f), so we may assume that $\\max\\{|c_i|, |c_j|\\} \\ge 2$. Suppose without loss of generality (by switching $i$ with $j$ and/or negating both $c_i$ and $c_j$) that $c_i \\ge 2$; then\n\n$$\nc_i a_i + c_j a_j = a_i + ((c_i - 1)a_i + c_j a_j)\n$$\n\nand we have $a_i \\in S$, $(c_i - 1)a_i + c_j a_j \\in S$ by the induction hypothesis, and $(c_i - 2)a_i + c_j a_j \\in S$ again by the induction hypothesis. So $c_i a_i + c_j a_j \\in S$ by (f), and (h) is verified.\n\nLet $e_i$ be the largest integer such that $2^{e_i}$ divides $a_i$; without loss of generality we may assume that $e_1 \\ge e_2 \\ge \\dots \\ge e_n$. Let $d_i$ be the greatest common divisor of $a_1, \\dots, a_i$. We prove by induction on $i$ that $S$ contains all multiples of $d_i$ for $i = 1, \\dots, n$; the case $i = n$ is the desired result. Our base cases are $i = 1$ and $i = 2$, which follow from (g) and (h), respectively.\n\nAssume that $S$ contains all multiples of $d_i$, for some $2 \\le i < n$. Let $T$ be the set of integers $m$ such that $m$ is divisible by $d_i$ and $m + r a_{i+1} \\in S$ for all integers $r$. Then $T$ contains nonzero positive and negative numbers, namely any multiple of $a_i$ by (h). By (c), if $t \\in T$ and $s$ divisible by $d_i$ (so in $S$) satisfy $t - s \\in T$, then $t + s \\in T$. By taking $t = s = d_i$, we deduce ...",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20103,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c > 0$. Find the minimum value of\n$$\n\\frac{a^3 + 8}{a^3(b+c)} + \\frac{b^3 + 8}{b^3(c+a)} + \\frac{c^3 + 8}{c^3(a+b)}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $x = \\frac{1}{a}$, $y = \\frac{1}{b}$, and $z = \\frac{1}{c}$. Then $xyz = 1$.\n\nBy the AM-GM inequality:\n$$\na^3 + 2 = a^3 + 1 + 1 \\ge 3\\sqrt[3]{a^3 \\cdot 1 \\cdot 1} = 3a.\n$$\nTherefore,\n$$\n\\frac{a^3 + 8}{a^3(b+c)} \\ge \\frac{3a + 6}{a^3(b+c)} = \\frac{3a^2bc + 6abc}{a^3(b+c)} = \\frac{3x + 6x^2}{y+z}.\n$$\nSimilarly,\n$$\n\\frac{b^3 + 8}{b^3(c+a)} \\ge \\frac{3y + 6y^2}{z+x}, \\quad \\frac{c^3 + 8}{c^3(a+b)} \\ge \\frac{3z + 6z^2}{x+y}.\n$$\nThus, it remains to minimize\n$$\n3\\left(\\frac{x}{y+z} + \\frac{y}{z+x} + \\frac{z}{x+y}\\right) + 6\\left(\\frac{x^2}{y+z} + \\frac{y^2}{z+x} + \\frac{z^2}{x+y}\\right).\n$$\nBy Nesbitt's inequality,\n$$\n\\frac{x}{y+z} + \\frac{y}{z+x} + \\frac{z}{x+y} \\ge \\frac{3}{2}.\n$$\nBy the Cauchy-Schwarz inequality,\n$$\n((y+z) + (z+x) + (x+y)) \\left( \\frac{x^2}{y+z} + \\frac{y^2}{z+x} + \\frac{z^2}{x+y} \\right) \\ge (x+y+z)^2.\n$$\nThis implies\n$$\n\\frac{x^2}{y+z} + \\frac{y^2}{z+x} + \\frac{z^2}{x+y} \\ge \\frac{x+y+z}{2} \\ge \\frac{3\\sqrt[3]{xyz}}{2} = \\frac{3}{2}\n$$\nby AM-GM. Therefore, the given expression is at least\n$$\n3 \\cdot \\frac{3}{2} + 6 \\cdot \\frac{3}{2} = \\frac{27}{2}.\n$$\nEquality holds when $a = b = c = 1$. Thus, the minimum value is $\\boxed{\\dfrac{27}{2}}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20104,
"subject": "Mathematics (Olympiad)",
"question": "Во квадратна шема со димензии $3 \\times 3$, Димитар може да ги запишува броевите $\\frac{1}{2}$, $\\frac{1}{3}$ и $\\frac{1}{6}$. Дали може Димитар во секое квадратче да запише по еден од овие броеви така што збировите на броевите во трите редици, трите колони и двете дијагонали да се различни меѓу себе?",
"options": [],
"answer": "See solution",
"solution": "Можни збирови на три броја од множеството $\\left\\{ \\frac{1}{2}, \\frac{1}{3}, \\frac{1}{6} \\right\\}$ се:\n\n$$\n\\frac{1}{2} + \\frac{1}{2} + \\frac{1}{2} = \\frac{3}{2}\n$$\n\n$$\n\\frac{1}{2} + \\frac{1}{2} + \\frac{1}{3} = \\frac{4}{3}\n$$\n\n$$\n\\frac{1}{2} + \\frac{1}{2} + \\frac{1}{6} = \\frac{7}{6}, \\quad \\frac{1}{3} + \\frac{1}{3} + \\frac{1}{3} = 1\n$$\n\n$$\n\\frac{1}{3} + \\frac{1}{3} + \\frac{1}{2} = \\frac{7}{6}\n$$\n\n$$\n\\frac{1}{3} + \\frac{1}{3} + \\frac{1}{6} = \\frac{5}{6}\n$$\n\n$$\n\\frac{1}{6} + \\frac{1}{6} + \\frac{1}{6} = \\frac{1}{2}\n$$\n\n$$\n\\frac{1}{6} + \\frac{1}{6} + \\frac{1}{2} = \\frac{5}{6}\n$$\n\n$$\n\\frac{1}{6} + \\frac{1}{6} + \\frac{1}{3} = \\frac{2}{3}\n$$\n\n$$\n\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{6} = 1\n$$\n\nЗначи, можни збирови се $\\frac{1}{2}$, $\\frac{2}{3}$, $\\frac{5}{6}$, $1$, $\\frac{7}{6}$, $\\frac{4}{3}$, $\\frac{3}{2}$, т.е. вкупно можни збирови се 7, а има вкупно 8 хоризонтали, вертикали и дијагонали. Според принципот на Дирихле, при било какво пополнување на квадратната шема ќе има најмалку два еднакви збира. Значи, такво пополнување не е можно.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20105,
"subject": "Mathematics (Olympiad)",
"question": "On a rectangular sheet of paper, several segments are drawn parallel to its sides. These segments divide the sheet into several rectangles (so that there are no parts of drawn segments inside rectangles). Petya wants to draw one of two diagonals in each of these rectangles, dividing it into two triangles, and then color all triangles, each triangle either black or white. Determine if Petya can always do this so that no two triangles of the same color have a common boundary segment.",
"options": [],
"answer": "See solution",
"solution": "Yes, this is always possible.\n\nLet Petya draw a diagonal in each rectangle from the bottom-left corner to the top-right corner. Then, color all triangles adjacent to the top-left corners of the rectangles black, and the rest white.\n\nTo see why this works, consider a common boundary segment of two triangles:\n\n- If the segment is a diagonal, then a black triangle adjoins it from above, and a white one from below.\n- If the segment is horizontal, then a white triangle adjoins it from above, and a black one from below.\n- The case of a vertical segment is similar.\n\nTherefore, such a coloring ensures that no two triangles of the same color share a common boundary segment.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20106,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a parallelogram. Let $M$ be the midpoint of segment $AD$. Given that $\\angle BAD = 84^\\circ$ and $\\angle CDA = 48^\\circ$, find the measure of $\\angle DCM$.",
"options": [],
"answer": "See solution",
"solution": "Looking at the angles of triangle $ABM$, we find that $\\angle MBA = 180^\\circ - 84^\\circ - 48^\\circ = 48^\\circ$, so $ABM$ is an isosceles triangle with the apex at $A$. This implies $|AB| = |AM|$. Since $M$ is the midpoint of segment $AD$, we have $|AM| = |MD|$. Since $ABCD$ is a parallelogram, we have $|AB| = |CD|$. So, $|CD| = |MD|$ and $MCD$ is an isosceles triangle with the apex at $D$. From here, we conclude that $$\\angle DCM = \\frac{180^\\circ - \\angle MDC}{2} = \\frac{\\angle BAD}{2} = 42^\\circ.$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20107,
"subject": "Mathematics (Olympiad)",
"question": "Halla todas las cuaternas $ (a, b, c, d) $ de números enteros positivos que cumplen que\n\n$$\na^2 + b^2 = c^2 + d^2\n$$\n\ny de manera que $ ac + bd $ es divisor de $ a^2 + b^2 $.",
"options": [],
"answer": "See solution",
"solution": "Usando la identidad\n\n$$\n(ac + bd)^2 + (ad - bc)^2 = (a^2 + b^2)(c^2 + d^2) = (a^2 + b^2)^2,\n$$\n\nobservamos que si $k$ es el valor (entero positivo) de $\\frac{a^2+b^2}{ac+bd}$, entonces\n\n$$\n(ad - bc)^2 = (k^2 - 1)(ac + bd)^2,\n$$\n\ny por tanto\n\n$$\n\\frac{ad - bc}{ac + bd} = \\pm\\sqrt{k^2 - 1}.\n$$\n\nEs conocido que si la raíz cuadrada de un entero es racional, entonces es entera. Esto quiere decir que $k^2 - 1$ es un cuadrado perfecto, lo cual solo es posible si $k=1$. En ese caso, $ad - bc = 0$, que reordenando da $\\frac{a}{c} = \\frac{b}{d}$. El cuadrado de esta razón es $\\frac{a^2+b^2}{c^2+d^2} = 1$, por lo que $a=c$ y $b=d$. Concluimos que las soluciones son las cuaternas de la forma $(a,b,a,b)$, que siempre cumplen las condiciones.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20108,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be an integer such that $n \\ge 3$. On a plane, $n$ points are chosen, no three of which are collinear. Consider all triangles with vertices at these points; denote the minimal among the internal angles of these triangles by $\\alpha$. Find the greatest possible value of $\\alpha$ and determine all configurations of $n$ points for which it is achieved.",
"options": [],
"answer": "See solution",
"solution": "Let $m$ be the number of points in the convex hull of the chosen set of points. As the sum of the internal angles of an $m$-gon is $(m-2) \\cdot 180^\\circ$, the minimal internal angle of the convex hull is at most $\\frac{m-2}{m} \\cdot 180^\\circ$. Clearly $m \\le n$, so the minimal internal angle of the convex hull is also bounded by $\\frac{n-2}{n} \\cdot 180^\\circ$.\n\nDenote the chosen points by $A_0, A_1, \\dots, A_{n-1}$ so that the minimal internal angle of the convex hull is at $A_{n-1}$, and a line through this point rotating counterclockwise passes over the remaining points in order $A_0, A_1, \\dots, A_{n-2}$. Then\n\n$$\n\\angle A_0A_{n-1}A_1 + \\angle A_1A_{n-1}A_2 + \\dots + \\angle A_{n-3}A_{n-1}A_{n-2} = \\angle A_0A_{n-1}A_{n-2},\n$$\n\nwhere $\\angle A_0A_{n-1}A_{n-2}$ is the minimal internal angle of the convex hull. Since $\\angle A_0A_{n-1}A_{n-2} \\le \\frac{n-2}{n} \\cdot 180^\\circ$, there must exist an angle among $A_0A_{n-1}A_1, A_1A_{n-1}A_2, \\dots, A_{n-3}A_{n-1}A_{n-2}$ whose size is at most $\\frac{180^\\circ}{n}$. Thus $\\alpha \\le \\frac{180^\\circ}{n}$.\n\nTo achieve $\\alpha = \\frac{180^\\circ}{n}$, all estimations above must be equalities:\n\n* The minimal internal angle of the convex hull must be $\\frac{m-2}{m} \\cdot 180^\\circ$, so all internal angles of the convex hull are equal.\n* $\\frac{m-2}{m} = \\frac{n-2}{n}$, so $m = n$, i.e., the convex hull includes all $n$ points.\n* The angles $A_0A_{n-1}A_1, A_1A_{n-1}A_2, \\dots, A_{n-3}A_{n-1}A_{n-2}$ must all be exactly $\\frac{180^\\circ}{n}$, and by symmetry, this holds for any cyclic reordering of points.\n\n\n\nFig. 28\n\n\n\nFig. 29\n\nFrom this, we obtain $\\angle A_i A_{i+2} A_{i+1} = \\frac{180^\\circ}{n} = \\angle A_{i+2} A_i A_{i+1}$, implying $A_i A_{i+1} = A_{i+1} A_{i+2}$ for every $i = 0, 1, \\dots, n-3$. Consequently, $A_0, A_1, \\dots, A_{n-1}$ are vertices of a regular $n$-gon.\n\nOn the other hand, the internal angles of triangles whose vertices are any three vertices of a regular $n$-gon are inscribed angles subtending an arc of at least $\\frac{360^\\circ}{n}$ of the circumcircle of the $n$-gon (see Fig. 29). Thus all internal angles of these triangles are at least $\\frac{180^\\circ}{n}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20109,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for all real $x, y$,\n\n$$\nx f(x) + y f(xy) = x f(x + y f(y))\n$$",
"options": [],
"answer": "See solution",
"solution": "**Answer:** $f(x) = 0$ and $f(x) = x$.\n\nLet $P(x, y)$ denote the given assertion:\n\n$$\nx f(x) + y f(xy) = x f(x + y f(y))\n$$\n\n**Step 1:** $P(0, 1)$ gives $f(0) = 0$.\n\nAssume there exists $a \\neq 0$ such that $f(a) = 0$. Then $P(x, a)$ gives:\n\n$$\nx f(x) + a f(xa) = x f(x)\n$$\nSo $a f(xa) = 0$ for all $x$, hence $f(xa) = 0$ for all $x$. Thus, $f(x) = 0$ for all $x$ (since $a \\neq 0$), which is one solution.\n\nNow assume $f$ is not identically zero, so $f(x) = 0$ if and only if $x = 0$.\n\n**Step 2:** $P(-1, 1)$ gives:\n\n$$\n- f(-1) + f(-1) = - f(-1 + f(1)) \\implies f(1) = 1\n$$\n\n**Step 3:** $P(x, 1)$ gives:\n\n$$\nx f(x) + f(x) = x f(x + 1)\n$$\nSo $(x + 1) f(x) = x f(x + 1)$ for $x \\neq 0, -1$.\n\nThus,\n$$\n\\frac{f(x)}{x} = \\frac{f(x + 1)}{x + 1}\n$$\nfor $x \\neq 0, -1$.\n\nSince $f(1) = 1$, by induction $f(n) = n$ for all $n \\in \\mathbb{N}$. Similarly, $f(-2) = 2 f(-1)$.\n\n**Step 4:** $P(-1, 2)$ gives:\n\n$$\n- f(-1) + 2 f(-2) = - f(3) = -3 \\implies f(-1) = -1\n$$\n\nSo $f(n) = n$ for all $n \\in \\mathbb{Z}$.\n\n**Step 5:** For all $n \\in \\mathbb{Z}$ and $x \\in \\mathbb{R}$, $(x + n) f(x) = x f(x + n)$. For $x \\notin \\mathbb{Z}$, we can write:\n\n$$\n\\frac{f(x)}{x} = \\frac{f(x + 1)}{x + 1} = \\cdots = \\frac{f(x + n)}{x + n}\n$$\n\n**Step 6:** $P(x, n)$ gives:\n\n$$\nx f(x) + n f(nx) = x f(x + n f(n))\n$$\nBut $f(n) = n$, so $x f(x) + n f(nx) = x f(x + n^2) = (x + n^2) f(x)$.\n\nThus,\n$$\nx f(x) + n f(nx) = x f(x) + n^2 f(x) \\implies f(nx) = n f(x)\n$$\n\n**Step 7:** $P(1, y)$ gives:\n\n$$\n1 + y f(y) = f(1 + y f(y))\n$$\n\nLet $x = y f(y)$ in the earlier ratio, so\n$$\n\\frac{f(y f(y))}{y f(y)} = \\frac{f(y f(y) + 1)}{y f(y) + 1} = 1\n$$\nThus,\n$$\nf(y f(y)) = y f(y)\n$$\n\n**Step 8:** Divide $P(x, y)$ by $x \\neq 0$:\n\n$$\nf(x) + \\frac{y f(xy)}{x} = f(x + y f(y))\n$$\n\nLet $x = x f(x)$ and by symmetry, we get:\n\n$$\nf(x f(x)) + \\frac{y f(x y f(x))}{x f(x)} = f(x f(x) + y f(y)) = f(y f(y)) + \\frac{x f(x y f(y))}{y f(y)}\n$$\n\nSo,\n$$\nf(x f(x)) - \\frac{x f(x y f(y))}{y f(y)} = f(y f(y)) - \\frac{y f(x y f(x))}{x f(x)}\n$$\n\nIf we replace $x$ by $2x$, the left side increases by 4 times (by previous results), but the right side does not change, so both must be zero. Thus,\n$$\nf(x f(x)) = \\frac{x f(x y f(y))}{y f(y)} \\implies f(x y f(y)) = f(x) y f(y)\n$$\n\n**Step 9:** Let $x = x + 1$ in the above:\n\n$$\nf(x + 1) y f(y) = f((x + 1) y f(y)) = f(x y f(y) + y f(y))\n$$\nBut from earlier,\n$$\nf(x y f(y) + y f(y)) = f(x y f(y)) + \\frac{y f(x y^2 f(y))}{x y f(y)} = f(x) y f(y) + \\frac{y^2 f(y) f(x y)}{x y f(y)} = f(x) y f(y) + \\frac{y f(x y)}{x}\n$$\nSo,\n$$\n(f(x + 1) - f(x)) x = \\frac{f(x y)}{f(y)}\n$$\nSubstitute $y = 1$:\n$$\n(f(x + 1) - f(x)) x = f(x)\n$$\nThus, $f(x y) = f(x) f(y)$.\n\n**Step 10:** Using $f(x y) = f(x) f(y)$ in $f(y f(y)) = y f(y)$, we get $f(f(y)) = y$.\n\nNow, $P(f(y), y)$ gives:\n\n$$\nf(y) y + y f(f(y) y) = f(y) f(f(y) + y f(y))\n$$\nBut $f(f(y) y) = f(f(y)) f(y) = y f(y)$, and $f(f(y) + y f(y)) = f(y + y^2) = (y + y^2)$.\n\nSo,\n$$\nf(y) y + y f(y) = f(y) (y + 1) \\implies f(y + 1) = y + 1\n$$\nThus, $f(x) = x$ for all $x$ is the second solution.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20110,
"subject": "Mathematics (Olympiad)",
"question": "There are 16 consecutive positive integers written on the board. Andrew calculates their product and Olesya calculates their sum. Can it happen that in both numbers there coincide:\n\na) the last three digits,\nb) the last four digits?",
"options": [],
"answer": "See solution",
"solution": "**Answer:** a) yes; b) no.\n\n**Solution.** It's obvious that the number Andrew obtains is divisible by $16$ and by $125$, because among 16 consecutive numbers, more than four are divisible by $2$ and at least three are divisible by $5$. This implies that the last three digits of Andrew's number are $0$.\n\na) Let the numbers be $a, a+1, a+2, \\dots, a+15$. Then Olesya's sum is $8(2a+15)$. For $a=55$, the last three digits of this sum are $0$, so the answer to part a) is 'yes'.\n\nb) Since Andrew's number is divisible by $16$, its last four digits are divisible by $16$. If the answer to b) were 'yes', then the last four digits of $8(2a+15)$ would also be divisible by $16$. This would require $8(2a+15)$ to be divisible by $16$, which is not always possible. This contradiction gives the answer 'no'.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20111,
"subject": "Mathematics (Olympiad)",
"question": "Circles $w_1$ and $w_2$ with centers $O_1$ and $O_2$ respectively intersect at points $A$ and $B$. The straight line $O_1O_2$ intersects $w_1$ at a point $Q$ (not inside $w_2$) and $w_2$ at a point $X$ (inside $w_1$). Around triangle $O_1AX$, a circle $w_3$ is circumscribed, which intersects $w_1$ again at a point $T$. The line $QT$ intersects $w_3$ at a point $K$, and the line $QB$ intersects $w_2$ again at a point $H$. Prove that\n\n\n\na) Points $T$, $X$, $B$ are collinear.\n\nb) Points $K$, $X$, $H$ are collinear.",
"options": [],
"answer": "See solution",
"solution": "a) Let $\\angle AO_1X = \\alpha$. Then $\\angle ATX = \\alpha$, because they are subtended by the same arc of the circle $w_3$ (see the figure). Moreover, $\\angle AO_1X = \\angle ATB$, so $\\angle ATX = \\angle ATB$. Therefore, $T$, $X$, $B$ are collinear.\n\nb) Let $K_1 = XH \\cap TQ$. Let $\\angle O_2XH = \\alpha$, $\\angle HXB = \\beta$, $\\angle XHB = \\varphi$. Then:\n\n$$\n\\angle XQB = \\alpha - \\varphi, \\quad \\angle QTB = 90^\\circ - \\alpha + \\varphi,\n$$\n\n$$\n\\angle TQX = 180^\\circ - (\\beta + \\alpha) - (90^\\circ - \\alpha + \\varphi) = 90^\\circ - \\beta - \\varphi = \\alpha,\n$$\n\nIn other words, $\\angle TQX = \\angle QTO_1$. Since $O_1T = QO_1$, then $\\angle K_1XQ = \\alpha$, hence $\\angle K_1XQ = \\angle QTO_1$, that is, $K_1 \\in w_3$. So $K_1 = K$ and points $K$, $X$, $H$ are collinear.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20112,
"subject": "Mathematics (Olympiad)",
"question": "For any positive real numbers $x$, $y$, $z$ with $xyz = 1$, prove the following inequality:\n\n$$\n(-x + y + z)(x - y + z) + (x - y + z)(x + y - z) + (x + y - z)(-x + y + z) \\le 3.\n$$",
"options": [],
"answer": "See solution",
"solution": "Without loss of generality, assume that $z = \\min\\{x, y, z\\}$. The inequality can be rewritten as\n\n$$\n4xy \\le 3 + (x + y - z)^2.\n$$\n\nSince $x + y - z \\ge 2\\sqrt{xy - z} > 0$, we have $(x + y - z)^2 \\ge (2\\sqrt{xy - z})^2$. So, it is sufficient to prove that $3 + (2\\sqrt{xy - z})^2 \\ge 4xy$. From the problem condition $xy = \\frac{1}{z}$, hence, we need to prove the following inequality:\n\n$$\n3 + \\left( \\frac{2}{\\sqrt{z}} - z \\right)^2 \\ge \\frac{4}{z}.\n$$\n\nExpanding the brackets and multiplying by $z$, we obtain the inequality: $z^2 + 3 \\ge 4\\sqrt{z}$, which follows from the AM-GM inequality:\n\n$$\nz^2 + 1 + 1 + 1 \\ge 4\\sqrt{z^2 \\cdot 1 \\cdot 1 \\cdot 1} = 4\\sqrt{z}.\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20113,
"subject": "Mathematics (Olympiad)",
"question": "Consider a sphere and two of its tangent planes in space. Prove that the center of a sphere tangent to all three lies on a fixed ellipse.",
"options": [],
"answer": "See solution",
"solution": "The problem is clear if the two planes are parallel, so assume they intersect on line $l$. Let $r$ denote the radius of the sphere and let $O$ denote its center. Let $a > r$ be the distance from $O$ to $l$.\n\nA sphere with center $P$ and radius $s$ satisfies the condition iff\n\n$$\nr + s = OP \\quad \\text{and} \\quad \\frac{r}{a} = \\frac{s}{b}.\n$$\n\nAssume $r = 1$. Consider a coordinate system where $l$ is the $x$-axis and $O(0, a, 0)$. Then $P(x, y, z)$ satisfies $y = b > 0$ and $z = 0$. Since $a > 1$, the equation\n\n$$\nx^2 + (y - a)^2 = \\left(1 + \\frac{y}{a}\\right)^2\n$$\n\ngives an ellipse.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 20114,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, a_3, b_1, b_2, b_3$ be pairwise distinct positive integers such that\n$$\n(n+1)a_1^2 + n a_2^2 + (n-1)a_3^2 \\mid (n+1)b_1^2 + n b_2^2 + (n-1)b_3^2\n$$\nholds for all positive integers $n$. Prove that there exists a positive integer $k$ such that $b_i = k a_i$ for all $i = 1, 2, 3$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Suppose that $r$ is any positive integer. Since there are infinitely many primes, there is a prime $p$ such that\n$$\np > (a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2).\n$$\nBecause $p$ is prime and the above, we have $(p, a_1^2 + a_2^2 + a_3^2) = 1$. $p$ is coprime to $p-1$; from the Chinese remainder theorem, there is a positive integer $n$ such that\n$$\nn \\equiv r \\pmod{p-1},\n$$\n$$\nn(a_1^r + a_2^r + a_3^r) + a_1^r - a_3^r \\equiv 0 \\pmod{p}.\n$$\nFrom the above and Fermat's theorem,\n$$\n(n+1)a_1^r + n a_2^r + (n-1)a_3^r \\equiv n(a_1^r + a_2^r + a_3^r) + a_1^r - a_3^r \\equiv 0 \\pmod{p}.\n$$\nFrom the assumption of the problem,\n$$\n(n+1)b_1^r + n b_2^r + (n-1)b_3^r \\equiv 0 \\pmod{p}.\n$$\nAgain, from above and Fermat's little theorem,\n$$\nn(b_1^r + b_2^r + b_3^r) + b_1^r - b_3^r \\equiv 0 \\pmod{p}.\n$$\nEliminate $n$ from the two congruences:\n$$\n(a_1^r + a_2^r + a_3^r)(b_1^r - b_3^r) \\equiv (b_1^r + b_2^r + b_3^r)(a_1^r - a_3^r) \\pmod{p}.\n$$\nFrom the choice of $p$, this implies\n$$\n(a_1^r + a_2^r + a_3^r)(b_1^r - b_3^r) = (b_1^r + b_2^r + b_3^r)(a_1^r - a_3^r).\n$$\nThus,\n$$\n(a_2 b_1)^r + 2(a_3 b_1)^r + (a_3 b_2)^r = (a_1 b_2)^r + 2(a_1 b_3)^r + (a_2 b_3)^r.\n$$\nWe then prove the following lemma.\n\n*Lemma*: Assume that $x_1, \\dots, x_s, y_1, \\dots, y_s$ are real numbers,\n$$\n0 < x_1 \\le x_2 \\le \\cdots \\le x_s, \\quad 0 < y_1 \\le y_2 \\le \\cdots \\le y_s,\n$$\nsuch that for any positive integer $r$,\n$$\nx_1^r + x_2^r + \\cdots + x_s^r = y_1^r + y_2^r + \\cdots + y_s^r.\n$$\nThen $x_i = y_i$ for $i = 1, 2, \\dots, s$.\n\n**Proof of the lemma.** We use induction on $s$. If $s=1$, take $r=1$; then $x_1 = y_1$. Assume the lemma holds for $s=t$.\n\nWhen $s=t+1$, if $x_{t+1} \\neq y_{t+1}$, say $x_{t+1} < y_{t+1}$,\n$$\n\\left(\\frac{x_1}{y_{t+1}}\\right)^r + \\cdots + \\left(\\frac{x_{t+1}}{y_{t+1}}\\right)^r = \\left(\\frac{y_1}{y_{t+1}}\\right)^r + \\cdots + \\left(\\frac{y_t}{y_{t+1}}\\right)^r + 1 \\ge 1.\n$$\nBecause $0 < \\frac{x_i}{y_{t+1}} < 1$ ($1 \\le i \\le t+1$), take the limit $r \\to +\\infty$, and we have $0 \\ge 1$, a contradiction.\n\nSo $x_{t+1} = y_{t+1}$, and then $x_1^r + \\cdots + x_t^r = y_1^r + \\cdots + y_t^r$ for all $r$. By induction, the lemma holds for all positive integers $s$.\n\nNow return to the proof of the problem. Since $a_1, a_2, a_3, b_1, b_2, b_3$ are distinct,\n$$\na_2 b_1 \\neq a_3 b_1, \\quad a_3 b_1 \\neq a_3 b_2, \\quad a_1 b_2 \\neq a_1 b_3,\n$$\n$$\na_1 b_3 \\neq a_2 b_3, \\quad a_2 b_1 \\neq a_2 b_3.\n$$\nFrom the previous equation and the lemma, we know that\n$$\na_2 b_1 = a_1 b_2, \\quad a_3 b_1 = a_1 b_3, \\quad a_3 b_2 = a_2 b_3.\n$$\nThen $\\frac{b_1}{a_1} = \\frac{b_2}{a_2} = \\frac{b_3}{a_3}$. Write $\\frac{b_1}{a_1} = \\frac{k}{l}$, with $(k, l) = 1$, $l \\ge 1$; then $b_i = \\frac{k}{l}a_i$, $i = 1, 2, 3$. From $2b_1 + b_2 = \\frac{k}{l}(2a_1 + a_2)$ and the assumption of the problem (with $n=1$), $2a_1 + a_2 \\mid 2b_1 + b_2$ and $\\frac{k}{l}$ is an integer. So $l=1$ and $b_i = k a_i$, $i = 1, 2, 3$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20115,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\alpha$ and $\\beta$ be positive real numbers such that for any positive integers $k_1$ and $k_2$, $\\lfloor k_1\\alpha \\rfloor \\ne \\lfloor k_2\\beta \\rfloor$, where $\\lfloor x \\rfloor$ denotes the greatest integer not exceeding $x$.\n\nProve that there exist positive integers $m_1, m_2$ such that\n$$\n\\frac{m_1}{\\alpha} + \\frac{m_2}{\\beta} = 1.\n$$",
"options": [],
"answer": "See solution",
"solution": "**Solution:**\n\nFirst, $\\frac{\\beta}{\\alpha}$ must be irrational; otherwise, there exist positive integers $k_1, k_2$ with $k_1\\alpha = k_2\\beta$, contradicting the condition.\n\nSuppose $\\alpha$ is rational, say $\\alpha = \\frac{q}{p}$. Then, for some $k_2$, the fractional part of $\\frac{k_2\\beta}{q}$ is less than $\\frac{1}{q}$. Setting $k_1 = p \\lfloor \\frac{k_2\\beta}{q} \\rfloor$ yields\n$$\nk_1\\alpha = q \\cdot \\lfloor \\frac{k_2\\beta}{q} \\rfloor < k_2\\beta < q \\cdot \\lfloor \\frac{k_2\\beta}{q} \\rfloor + 1,\n$$\ncontradicting the condition. Thus, both $\\alpha$ and $\\beta$ are irrational.\n\nDefine a pair $(a, b)$ of positive integers as \"great\" if $0 < b\\beta - a\\alpha < 1$. For such a pair, there is a unique integer $t$ with $a\\alpha < t < b\\beta$. Let $u = t - a\\alpha$, $v = b\\beta - t$.\n\n**Lemma:** For any two great pairs $(a_1, b_1)$ and $(a_2, b_2)$ with differences $(u_1, v_1)$ and $(u_2, v_2)$, $\\frac{u_1}{v_1} = \\frac{u_2}{v_2}$.\n\n*Proof of Lemma:* Suppose not, and $\\frac{u_1}{v_1} > \\frac{u_2}{v_2}$. Let $\\varepsilon = \\frac{u_1v_2 - u_2v_1}{2} > 0$. Since $\\frac{\\beta}{\\alpha}$ is irrational, there exist $a_0, b_0$ with $0 < a_0\\alpha - b_0\\beta < \\varepsilon$. There is $t_0$ with $b_0\\beta < t_0 < a_0\\alpha$, and set $u_0 = a_0\\alpha - t_0$, $v_0 = t_0 - b_0\\beta$.\n\nIf $\\frac{u_1}{u_0} - \\frac{v_1}{v_0} > 1$, set $L = \\lfloor \\frac{v_1}{v_0} \\rfloor + 1$ so $\\frac{u_1}{u_0} > L > \\frac{v_1}{v_0}$. Then\n$$u_1 - Lu_0 = (t_1 + Lt_0) - (a_1 + La_0)\\alpha > 0,$$\n$$v_1 - Lv_0 = (t_1 + Lt_0) - (b_1 + Lb_0)\\beta < 0.$$\n\nChoosing $k_1 = a_1 + La_0$, $k_2 = b_1 + Lb_0$ gives $\\lfloor k_1\\alpha \\rfloor = \\lfloor k_2\\beta \\rfloor = t_1 + Lt_0 - 1$, a contradiction. Thus, $\\frac{u_1}{u_0} - \\frac{v_1}{v_0} \\le 1$; similarly,\n$$\n\\frac{u_2}{u_0} - \\frac{v_2}{v_0} \\geq -1.\n$$\nSo,\n$$\nu_1 - \\frac{u_0}{v_0} v_1 \\leq u_0, \\quad u_2 - \\frac{u_0}{v_0} v_2 \\geq -u_0,$$\nwhich yields $u_1v_2 - u_2v_1 \\leq u_0(v_1 + v_2) < 2\\epsilon$, contradicting the definition of $\\epsilon$. The lemma is proved.\n\nThus, every great pair $(a, b)$ has the same ratio $\\frac{u}{v} = \\frac{t - a\\alpha}{b\\beta - t}$. Let $\\lambda = \\frac{v}{u+v} \\in (0, 1)$, so $\\lambda a\\alpha + (1 - \\lambda) b\\beta = t \\in \\mathbb{Z}$.\n\nA linear combination of great pairs with integer coefficients is called \"nice\". Each nice pair $(c, d)$ satisfies\n$$\n\\lambda c\\alpha + (1 - \\lambda) d\\beta \\in \\mathbb{Z}.\n$$\n\nTake a great pair $(a, b)$ and $\\delta = b\\beta - a\\alpha \\in (0, 1)$. For any $M > 0$, any $(c, d)$ with $0 < d\\beta - c\\alpha < M$ and $c > \\frac{a}{\\delta}M$ is nice. (Choose $L = \\lfloor \\frac{d\\beta - c\\alpha}{\\delta} \\rfloor < \\frac{M}{\\delta} < \\frac{c}{a}$ so $(d-Lb)\\beta - (c-La)\\alpha \\in (0, \\delta) \\subset (0, 1)$, so $(c - La, d - Lb)$ is great and $(c, d)$ is nice.)\n\nSet $M = \\alpha + 2\\beta$, $c_0 > \\frac{a}{\\delta}M + 1$, and $d_0 = \\lfloor \\frac{c_0\\alpha}{\\beta} \\rfloor$ so $0 < d_0\\beta - c_0\\alpha < \\beta$. Then $(c_0, d_0)$, $(c_0, d_0 + 1)$, $(c_0 - 1, d_0)$ are nice, as are their combinations $(0, 1)$ and $(1, 0)$. Thus, both $\\lambda\\alpha$ and $(1 - \\lambda)\\beta$ are positive integers. Set\n$$\nm_2 = \\lambda\\alpha, \\quad m_1 = (1 - \\lambda)\\beta,\n$$\nso\n$$\n\\frac{m_2}{\\alpha} + \\frac{m_1}{\\beta} = \\lambda + (1 - \\lambda) = 1.\n$$\nDone.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20116,
"subject": "Mathematics (Olympiad)",
"question": "Given a triangle $ABC$ with incircle $(I)$ touching the sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$ respectively. Let $I_b$ and $I_c$ be the excenters of triangle $ABC$ with respect to vertices $B$ and $C$. Let $P$ and $Q$ be the midpoints of $I_bE$ and $I_cF$. Suppose that the circumcircle of triangle $PAC$ meets $AB$ at $R$, and the circumcircle of triangle $QAB$ meets $AC$ at $S$ ($R, S \\neq A$).\n\n**a)** Prove that $PR$, $QS$, and $AI$ are concurrent.\n\n**b)** Suppose that $DE$ and $DF$ meet $I_bI_c$ at $K$ and $J$. The line $EJ$ meets $FK$ at $M$, and the lines $PE$ and $QF$ meet the circumcircles of triangles $PAC$ and $QAB$ at $X$ and $Y$ respectively ($X, Y \\neq A$). Prove that $BY$, $CX$, and $AM$ are concurrent.\n\n",
"options": [],
"answer": "See solution",
"solution": "a) Since $EF$ and $I_bI_c$ are both perpendicular to $AI$, $I_bI_cFE$ is a trapezoid. So $PQ$ is the midline of both trapezoid $I_bI_cFE$ and triangle $AEF$. Thus, $P$ and $Q$ belong to the radical axis of the degenerate circle $(A, 0)$ and $(I)$. Similarly, $Q$ belongs to the radical axis of $(B, 0)$ and $(I)$. Hence, $QA^2 = QF^2 \\cdot QY = QB^2$, which implies that $(QAB)$ is tangent to $(I)$ at $Y$. Similarly, $(PAC)$ is also tangent to $(I)$ at $X$.\n\nThus, $(I)$ is the $S$-mixtilinear incircle of triangle $ASB$, so the incenter of triangle $ABS$ is the midpoint $N$ of the segment $EF$, which implies that $SQ$ is the angle bisector of $\\angle ASB$ and thus $SQ$ passes through $N$. Similarly, $RP$ also passes through $N$. Therefore, $PR$, $QS$, and $AI$ are concurrent at $N$.\n\n\n\nb) In the circle $(I)$, the line $I_bI_c$ is the antipole of $N$, so $JE$, $KF$, and $DN$ are concurrent at point $M$ on circle $(I)$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20117,
"subject": "Mathematics (Olympiad)",
"question": "For integer $n \\ge 2$, let $x_1, x_2, \\dots, x_n$ be real numbers satisfying\n$$\nx_1 + x_2 + \\dots + x_n = 0, \\quad \\text{and} \\quad x_1^2 + x_2^2 + \\dots + x_n^2 = 1.\n$$\nFor each subset $A \\subseteq \\{1, 2, \\dots, n\\}$, define\n$$\nS_A = \\sum_{i \\in A} x_i.\n$$\n(If $A$ is the empty set, then $S_A = 0$.)\n\nProve that for any positive number $\\lambda$, the number of sets $A$ satisfying $S_A \\ge \\lambda$ is at most $2^{n-3}/\\lambda^2$. For what choices of $x_1, x_2, \\dots, x_n$, $\\lambda$ does equality hold?",
"options": [],
"answer": "See solution",
"solution": "This problem is a form of Chebyshev's inequality for random variables. For each subset $A \\subseteq \\{1, 2, \\dots, n\\}$, define\n$$\n\\Delta_A = 2S_A = \\sum_{i \\in A} x_i - \\sum_{i \\in \\{1, 2, \\dots, n\\} \\setminus A} x_i = \\sum_{i=1}^{n} \\epsilon_A(i)x_i,\n$$\nwhere $\\epsilon_A(i) = 1$ if $i \\in A$ and $\\epsilon_A(i) = -1$ otherwise. Squaring, we have\n$$\n\\Delta_A^2 = \\sum_{i=1}^{n} x_i^2 + \\sum_{\\substack{i,j \\in \\{1, \\dots, n\\} \\\\ i \\neq j}} \\epsilon_A(i)\\epsilon_A(j)x_i x_j.\n$$\nNow sum the $\\Delta_A^2$'s over all $2^n$ possible choices of $A$. For each pair $i \\neq j$, there are $2^{n-2}$ sets $A$ with $i, j \\in A$, and another $2^{n-2}$ sets with $i, j \\notin A$; these sets each contribute a term of $+x_i x_j$ to the sum above. There are also $2^{n-2}$ sets $A$ with $i \\in A$, $j \\notin A$, and $2^{n-2}$ sets with $i \\notin A$, $j \\in A$. Each of these sets contributes a term of $-x_i x_j$. Hence $x_i x_j$ appears $2^{n-1}$ times with a $+$ sign and $2^{n-1}$ times with a $-$ sign. Therefore all of these terms cancel, and we obtain\n$$\n\\sum_{A \\subseteq \\{1, 2, \\dots, n\\}} \\Delta_A^2 = 2^n (x_1^2 + \\dots + x_n^2) = 2^n.\n$$\nNow let $\\lambda > 0$. There cannot be more than $2^{n-2}/\\lambda^2$ terms $\\Delta_A^2$ whose value is greater than or equal to $4\\lambda^2$. If this were not the case, then the sum of these terms would be greater than $2^n$, so the total sum would exceed $2^n$. Hence, there can be at most $2^{n-2}/\\lambda^2$ sets $A$ such that $|S_A| \\ge \\lambda$. (Recall that $\\Delta_A = 2S_A$.) Moreover, these sets can be arranged into complementary pairs because $S_A = -S_{\\{1, \\dots, n\\} \\setminus A}$. In each of these pairs, exactly one of the two members is positive. Therefore there are at most $2^{n-3}/\\lambda^2$ sets $A$ with $S_A \\ge \\lambda$.\n\nFor equality to hold, it must be the case that all positive values of $\\Delta_A^2$ are equal to $4\\lambda^2$; otherwise the sum of all $\\Delta_A^2$ would exceed $2^n$. In particular, all positive values of $S_A$ must be the same. This will be the case only if at most one of the $x_i$ is positive and at most one of the $x_i$ is negative. Because we must have at least one of each, there must be exactly one positive term and one negative term. Thus, it must be the case that $x_k = \\sqrt{2}/2$ for some $k$, $x_j = -\\sqrt{2}/2$ for some $j \\neq k$, and $x_i = 0$ for $i \\neq j, k$. Then the assumption that every positive $\\Delta_A^2 = 4\\lambda^2$ yields $\\lambda = \\sqrt{2}/2$.\n\nConversely, with the $x_i$ and $\\lambda$ as described, we have exactly $2^{n-2} = 2^{n-3}/\\lambda^2$ sets $A$ such that $S_A \\ge \\lambda$: namely, the sets $A$ that contain the $\\sqrt{2}/2$ term and do not contain the $-\\sqrt{2}/2$ term. Thus, this is indeed the equality case.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20118,
"subject": "Mathematics (Olympiad)",
"question": "A positive integer $N$ is called a *good number* if its decimal representation can be divided into at least five segments, each containing at least one non-zero digit. These segments, when considered as positive integers after ignoring leading zeros, can be divided into two groups, such that the integers in each group form a geometric sequence in appropriate order. (A group of one or two integers is also considered to form a geometric sequence.)\n\nFor example, $20240327$ is a good number because it can be divided into $2|02|403|2|7$, forming two groups of integers $(2, 2, 2)$ and $(7, 403)$ that respectively form geometric sequences.\n\nLet $p = 1 + a + a^2 + \\cdots + a^m$ be a prime number, where $a > 1$ and $m > 2$ are integers. Prove that $\\frac{10^{p-1}-1}{p}$ is a good number.",
"options": [],
"answer": "See solution",
"solution": "Since $p = 1 + a + \\cdots + a^m = \\frac{a^{m+1}-1}{a-1}$ is a prime number, $q = m + 1$ must be a prime number. Given $q = m + 1 > 3$, we have $q \\ge 5$. Note that $p = \\frac{a^q-1}{a-1}$ divides $a^q-1$ and does not divide $a-1$, so the order of $a$ modulo $p$ is exactly $q$. Therefore, $q \\mid p-1$. Let $p-1 = qL$.\n\nConsider the remainder $t$ of $10^L = 10^{(p-1)/q}$ modulo $p$, which satisfies $t^q \\equiv 1 \\pmod{p}$. By Lagrange's theorem, the solutions to the congruence equation $x^q - 1 \\equiv 0 \\pmod{p}$ are exactly $\\{1, a, a^2, \\cdots, a^{q-1}\\}$. Thus, we can assume $t = a^r$, where $r \\in D = \\{0, 1, \\cdots, q-1\\}$. Furthermore, $10^{kL}$ modulo $p$ is $a^{\\langle kr \\rangle}$, where $\\langle \\cdot \\rangle$ denotes the remainder modulo $q$, taking values in $D = \\{0, 1, \\cdots, q-1\\}$.\n\nConsider $N = \\frac{10^{p-1}-1}{p} = \\frac{10^{qL}-1}{p}$ as a $p-1$ digit number, with leading zeros if necessary. Divide $N$ into segments of length $L$ from left to right (the leftmost segment may contain leading zeros), forming $q$ segments corresponding to integers $x_0, x_1, \\cdots, x_{q-1}$:\n\n$$\n\\begin{align*}\nx_0 &= \\left\\lfloor \\frac{10^L}{p} \\right\\rfloor = \\frac{10^L - a^r}{p}, \\\\\nx_1 &= \\left\\lfloor \\frac{10^{2L}}{p} \\right\\rfloor - 10^L \\cdot x_0 = \\frac{10^{2L} - a^{\\langle 2r \\rangle}}{p} - 10^L \\cdot \\frac{10^L - a^r}{p} = \\frac{a^r \\cdot 10^L - a^{\\langle 2r \\rangle}}{p}, \\\\\n\\vdots \\\\\nx_k &= \\frac{a^{\\langle kr \\rangle} \\cdot 10^L - a^{\\langle (k+1)r \\rangle}}{p}, \\quad k = 1, 2, \\ldots, q-1.\n\\end{align*}\n$$\n\nIf $r=0$, then $x_0 = x_1 = \\cdots = x_{q-1} = \\frac{10^L-1}{p}$ can be divided into two groups, each forming a geometric sequence.\n\nIf $r \\in \\{1, 2, \\ldots, q-1\\}$, then $\\{\\langle kr \\rangle \\mid k = 0, 1, \\ldots, q-1\\} = \\{0, 1, 2, \\ldots, q-1\\}$. Let\n\n$$\nA = \\{k \\in D \\mid \\langle kr \\rangle = 0, 1, \\ldots, q-r-1\\}, \\quad B = \\{k \\in D \\mid \\langle kr \\rangle = q-r, \\ldots, q-1\\}.\n$$\n\nWhen $k \\in A$, $\\langle (k+1)r \\rangle = \\langle kr \\rangle + r$, and when $k \\in B$, $\\langle (k+1)r \\rangle = \\langle kr \\rangle + r - q$. Thus,\n\n$$\n\\{x_k \\mid k \\in A\\} = \\frac{10^L - a^r}{p} \\cdot \\{1, a, a^2, \\ldots, a^{q-r-1}\\} \\text{ forms a geometric sequence,}\n$$\n\n$$\n\\{x_k \\mid k \\in B\\} = \\frac{10^L - a^{r-q}}{p} \\cdot \\{a^{q-r}, a^{q-r+1}, \\ldots, a^{q-1}\\} = \\frac{a^{q-r} \\cdot 10^L - 1}{p} \\cdot \\{1, a, \\ldots, a^{r-1}\\}\n$$\n\nalso forms a geometric sequence. This completes the proof. $\\square$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20119,
"subject": "Mathematics (Olympiad)",
"question": "Let $x, y, z$ be positive numbers such that\n\n$$\np(t) = t^3 + a t^2 + b t + c = (t + x)(t + y)(t + z)\n$$\n\nExpress $a$, $b$, and $c$ in terms of $x$, $y$, and $z$. Prove that\n\n$$\n27ac \\leq 9b^2 \\leq a^4.\n$$",
"options": [],
"answer": "See solution",
"solution": "By expanding $(t + x)(t + y)(t + z)$, we have:\n\n$$\na = x + y + z, \\quad b = xy + yz + zx, \\quad c = xyz.\n$$\n\nConsider\n\n$$\nx^2 y^2 + y^2 z^2 + z^2 x^2 - xyz(x + y + z) = x^2 (y - z)^2 + y^2 (z - x)^2 + z^2 (x - y)^2 \\geq 0,\n$$\n\nwith equality iff $x = y = z$. Thus,\n\n$$\n3(x + y + z)xyz \\leq (xy + yz + zx)^2,\n$$\nwhich gives\n$$\n3ac \\leq b^2 \\implies 27ac \\leq 9b^2.\n$$\n\nBy the Cauchy-Schwarz inequality:\n$$\nxy + yz + zx \\leq x^2 + y^2 + z^2,\n$$\nso\n$$\n3b \\leq (x + y + z)^2 = a^2 \\implies 9b^2 \\leq a^4.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20120,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs $(x, y)$ of integers such that\n\n$$\n\\sqrt{x+2019} - \\sqrt{x} = \\sqrt{y}.\n$$",
"options": [],
"answer": "See solution",
"solution": "As $x$ and $y$ occur under square roots, only non-negative solutions can exist. Bringing $\\sqrt{x}$ to the right, squaring both sides, and collecting similar terms gives $2019 = y + 2\\sqrt{xy}$, which is equivalent to the initial equation. Thus, $2\\sqrt{xy}$ is an integer. If $x = 0$, then $y = 2019$; the case $y = 0$ leads to a contradiction. Assume in the rest that both $x$ and $y$ are positive.\n\nLet $a^2$ and $b^2$ be the largest perfect squares dividing $x$ and $y$, respectively; then $x = a^2c$ and $y = b^2c'$, where both $c$ and $c'$ are square-free. As $a^2c \\cdot b^2c'$ is a perfect square, also $c \\cdot c'$ must be a perfect square; this is possible only if $c$ and $c'$ have the same prime factors, i.e., $c = c'$. Hence\n\n$$\n2019 = b^2c + 2abc = bc(2a + b).\n$$\n\nSince $2019 = 3 \\cdot 673$ where both factors are prime, we have the following cases, taking into account that $b < 2a + b$:\n\n- $b=1$, $c=1$, $2a+b=2019$, implying $a=1009$, $x=1009^2$, and $y=1$;\n- $b=1$, $c=3$, $2a+b=673$, implying $a=336$, $x=336^2 \\cdot 3$, and $y=3$;\n- $b=1$, $c=673$, $2a+b=3$, implying $a=1$, $x=673$, and $y=673$;\n- $b=3$, $c=1$, $2a+b=673$, implying $a=335$, $x=335^2$, and $y=9$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20121,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, \\dots, a_{11}$ be 11 distinct positive integers with their sum less than 2007, and write the numbers $1, 2, \\dots, 2007$ in order on the blackboard.\n\nNow we define a group of 22 ordered operations:\n\nThe $i$-th operation is to take any number on the blackboard, and then add $a_i$ to it if $1 \\le i \\le 11$, or subtract $a_{i-11}$ from it if $12 \\le i \\le 22$.\n\nIf the final result after such a group of operations is an even permutation of $1, 2, \\dots, 2007$, then we call it a \"good\" group; if the result is an odd permutation of $1, 2, \\dots, 2007$, then we call it a \"second good\" group.\n\nWhich is greater: the number of \"good\" groups or that of \"second good\" groups? And by how many more?\n\n*Remark*: Suppose $x_1, x_2, \\dots, x_n$ is a permutation of $1, 2, \\dots, n$. We call it an even permutation if $\\prod_{i>j} (x_i - x_j) > 0$, and otherwise an odd permutation.",
"options": [],
"answer": "See solution",
"solution": "The number of \"good\" groups is greater than the number of \"second good\" groups by $\\prod_{i=1}^{11} a_i$.\n\nMore generally, if we write numbers $1, 2, \\dots, n$ in order on the blackboard and define a group of $l$ ordered operations (the $i$-th operation is to take any number and add $b_i$ to it, where $b_i \\in \\mathbb{Z}$), then the difference between the number of \"good\" and \"second good\" groups is $f(b_1, b_2, \\dots, b_l; n)$.\n\nKey properties:\n- Interchanging $b_i$ and $b_j$ does not affect $f$.\n- We only need to count groups with property $P$: after each operation, all numbers on the blackboard remain distinct.\n- The difference between \"good\" and \"second good\" groups without property $P$ is zero (they pair up with opposite parity).\n\nNow, let $a_1, a_2, \\dots, a_m$ be $m$ distinct positive integers with sum less than $n$. By induction,\n\n$$\nf(a_1, a_2, \\dots, a_m, -a_1, -a_2, \\dots, -a_m; n) = \\prod_{i=1}^m a_i.\n$$\n\n**Base case ($m=1$):** The only way to maintain property $P$ is to operate on the last $a_1$ numbers, so the number of \"good\" groups is $a_1$ and \"second good\" groups is $0$.\n\n**Inductive step:** Assume true for $m-1$. For $m$, the first operation must be on the last $a_1$ numbers, and the remaining operations reduce to the $m-1$ case on $n-a_1$ numbers. Thus,\n\n$$\nf(a_1, a_2, \\dots, a_m, -a_1, -a_2, \\dots, -a_m; n) = a_1 f(a_2, \\dots, a_m, -a_2, \\dots, -a_m; n-a_1).\n$$\n\nTherefore, the answer for the original problem is $\\prod_{i=1}^{11} a_i$.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20122,
"subject": "Mathematics (Olympiad)",
"question": "In the interior of the acute-angled triangle $ABC$, the point $Q$ is chosen such that $\\angle QAC = 60^\\circ$, $\\angle QCA = \\angle QBA = 30^\\circ$. Let $M$ and $N$ be the midpoints of sides $AB$ and $BC$, respectively. Find $\\angle QNM$.\n\n",
"options": [],
"answer": "See solution",
"solution": "We have $\\angle AQC = 90^\\circ$. Let $K$ be the midpoint of $QC$. Then $MN \\parallel AB$, $NK \\parallel BQ$. Hence, $\\angle MNK = \\angle ABQ = 30^\\circ$. This implies that $\\angle MNK = \\angle MQK$ since $\\angle MQK = 30^\\circ$. Thus, the points $Q$, $N$, $K$, $M$ lie on the same circle. Therefore, $\\angle QNM = \\angle QKM = 90^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20123,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be positive real numbers such that $ab + bc + ca = 1$.\n\nProve that\n\n$$\n\\sqrt[4]{\\frac{\\sqrt{3}}{a} + 6\\sqrt{3}b} + \\sqrt[4]{\\frac{\\sqrt{3}}{b} + 6\\sqrt{3}c} + \\sqrt[4]{\\frac{\\sqrt{3}}{c} + 6\\sqrt{3}a} \\leq \\frac{1}{abc}.\n$$\n\nWhen does equality hold?",
"options": [],
"answer": "See solution",
"solution": "Using the Power Mean Inequality:\n\n$$\n\\sqrt[4]{\\frac{\\sqrt{3}}{a} + 6\\sqrt{3}b} + \\sqrt[4]{\\frac{\\sqrt{3}}{b} + 6\\sqrt{3}c} + \\sqrt[4]{\\frac{\\sqrt{3}}{c} + 6\\sqrt{3}a} \\leq 3 \\sqrt[4]{\\frac{1}{3} \\left( \\frac{\\sqrt{3}}{a} + 6\\sqrt{3}b + \\frac{\\sqrt{3}}{b} + 6\\sqrt{3}c + \\frac{\\sqrt{3}}{c} + 6\\sqrt{3}a \\right)}\n$$\n\n$$\n= \\frac{3}{\\sqrt[4]{3}} \\sqrt[4]{\\frac{\\sqrt{3}(ab+bc+ca)}{abc} + 6\\sqrt{3}(a+b+c)}\n$$\n\nSince $ab + bc + ca = 1$, we can express $a + b + c$ in terms of $a$, $b$, $c$, but for the bound, we use inequalities:\n\nUsing Cauchy-Schwarz, $3((ab)^2 + (bc)^2 + (ca)^2) \\geq (ab + bc + ca)^2 = 1$.\n\nThus,\n\n$$\n\\sqrt[4]{\\frac{\\sqrt{3}}{a} + 6\\sqrt{3}b} + \\sqrt[4]{\\frac{\\sqrt{3}}{b} + 6\\sqrt{3}c} + \\sqrt[4]{\\frac{\\sqrt{3}}{c} + 6\\sqrt{3}a} \\leq \\frac{3}{\\sqrt[4]{3}} \\sqrt[4]{\\frac{3\\sqrt{3}}{abc}}\n$$\n\nWe need to show:\n\n$$\n\\frac{3}{\\sqrt[4]{3}} \\sqrt[4]{\\frac{3\\sqrt{3}}{abc}} \\leq \\frac{1}{abc}\n$$\n\nThis reduces to $(abc)^3 \\leq \\frac{1}{81\\sqrt{3}}$.\n\nSince $(abc)^2 = (ab)(bc)(ca) \\leq \\left(\\frac{ab+bc+ca}{3}\\right)^3 = \\frac{1}{27}$, so $abc \\leq \\frac{1}{3\\sqrt{3}}$.\n\nTherefore, $(abc)^3 \\leq \\frac{1}{81\\sqrt{3}}$ as required.\n\nEquality holds when $a = b = c$ and $ab + bc + ca = 1$, so $a = b = c = \\frac{1}{\\sqrt{3}}$.\n\n---\n\n*Alternative proof (by Holder's Inequality):*\n\nBy Holder's inequality,\n\n$$\n(\\sqrt{3} + \\sqrt{3} + \\sqrt{3})^{1/4} \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right)^{1/4} (1+1+1)^{1/4} \\left( \\sum_{cyc} (1 + 6ab) \\right)^{1/4} \\geq \\sum_{cyc} \\sqrt[4]{\\frac{\\sqrt{3}}{a} + 6\\sqrt{3}b}\n$$\n\nSince $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = \\frac{1}{abc}$ when $ab + bc + ca = 1$, the left side is $\\sqrt[4]{\\frac{81\\sqrt{3}}{abc}}$.\n\nSo, $\\sqrt[4]{\\frac{81\\sqrt{3}}{abc}} \\leq \\frac{1}{abc}$, which is equivalent to $(abc)^3 \\leq \\frac{1}{81\\sqrt{3}}$, as above.\n\nEquality occurs when $a = b = c = \\frac{1}{\\sqrt{3}}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20124,
"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 line 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 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 two circles $O$ and $O_2$, we have $AC_1 \\perp OO_2$. From this result with the given condition of $OO_2 \\perp AO_1$, it follows that the point $O_1$ is on the line segment $AC_1$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20125,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $x$, and $y$ be real numbers with $a > 4$ and $b > 1$ such that\n\n$$\n\\frac{x^2}{a^2} + \\frac{y^2}{a^2 - 16} = \\frac{(x - 20)^2}{b^2 - 1} + \\frac{(y - 11)^2}{b^2} = 1.\n$$\n\nFind the least possible value for $a + b$.",
"options": [],
"answer": "See solution",
"solution": "The graph of\n\n$$\n\\frac{x^2}{a^2} + \\frac{y^2}{a^2 - 16} = 1\n$$\n\nis an ellipse centered at $(0, 0)$ with major axis parallel to the $x$-axis of length $2a$. The distance from the center to the foci is $\\sqrt{a^2 - (a^2 - 16)} = 4$, so the foci are $F_1 = (-4, 0)$ and $F_2 = (4, 0)$.\n\nSimilarly, the graph of\n\n$$\n\\frac{(x - 20)^2}{b^2 - 1} + \\frac{(y - 11)^2}{b^2} = 1\n$$\n\nis an ellipse centered at $(20, 11)$ with major axis parallel to the $y$-axis of length $2b$. The distance from the center to the foci is $\\sqrt{b^2 - (b^2 - 1)} = 1$, so the foci are $G_1 = (20, 10)$ and $G_2 = (20, 12)$.\n\n\n\nLet $P(x, y)$ be a point that lies on both ellipses, so by a property of ellipses,\n\n$$\n2a = PF_1 + PF_2 \\quad \\text{and} \\quad 2b = PG_1 + PG_2.\n$$\n\nAdding these equations and using the triangle inequality yields\n\n$$\n\\begin{align*}\n2a + 2b &= PF_1 + PF_2 + PG_1 + PG_2 \\\\\n&= (PF_1 + PG_1) + (PF_2 + PG_2) \\\\\n&\\geq F_1G_1 + F_2G_2 \\\\\n&= \\sqrt{24^2 + 10^2} + \\sqrt{16^2 + 12^2} = 46.\n\\end{align*}\n$$\n\nThus $a + b \\geq 23$. Equality is achieved by running this argument in reverse: that is, by taking $P$ to be the intersection of $\\overline{F_1G_1}$ and $\\overline{F_2G_2}$, and then setting $a = \\frac{1}{2}(PF_1 + PF_2)$ and $b = \\frac{1}{2}(PG_1 + PG_2)$. In this case, $a = 16$, $b = 7$, and $P = (14, 7.5)$. Therefore, the least possible value of $a + b$ is $23$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20126,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer and let $a$ and $b$ be positive integers congruent to $1$ modulo $4$. Prove that there exists a positive integer $k$ such that at least one of the numbers $a^k - b$ and $b^k - a$ is divisible by $2^n$.",
"options": [],
"answer": "See solution",
"solution": "First, we use the following lemma:\n\n**Lemma.** For any integer $p \\ge 2$, if $c-1 \\equiv 2^p \\pmod{2^{p+1}}$, then $c^2 - 1 \\equiv 2^{p+1} \\pmod{2^{p+2}}$.\n\n*Proof.* Write $c^2 - 1 = (c-1)(c+1)$. By hypothesis, $c-1$ is divisible by $2^p$. As $c+1$ leaves remainder $2$ upon division by $2^p$, it follows that $2^{p+1}$ is the highest power of $2$ dividing $c^2 - 1$, whence the conclusion of the lemma.\n\nLet $2^\\alpha$ and $2^\\beta$ be the highest powers of $2$ dividing $a-1$ and $b-1$, respectively. By hypothesis, $\\alpha \\ge 2$ and $\\beta \\ge 2$. Note that $a-1 \\equiv 2^\\alpha \\pmod{2^{\\alpha+1}}$ and $b-1 \\equiv 2^\\beta \\pmod{2^{\\beta+1}}$. Assume $\\alpha \\le \\beta$. Apply the lemma $\\beta - \\alpha$ times to get $a^{2^{\\beta-\\alpha}} - 1 \\equiv 2^\\beta \\pmod{2^{\\beta+1}}$, so $a^{2^{\\beta-\\alpha}} - b = (a^{2^{\\beta-\\alpha}} - 1) - (b-1)$ is divisible by $2^{\\beta+1}$.\n\nNow, induct on $n$. By the preceding, if $n \\le \\beta + 1$, then $a^{2^{\\beta-\\alpha}} - b$ is divisible by $2^n$, so $k = 2^{\\beta-\\alpha}$ will do. Let $n > \\beta$.\n\nFor the induction step, let $a^k - b$ be divisible by $2^n$ for some $k \\ge 1$. If it is divisible by $2^{n+1}$ we are done, so let $a^k - b \\equiv 2^n \\pmod{2^{n+1}}$.\n\nLet $m = 2^{n-\\beta} + 1$; note that $m$ is odd, as $n > \\beta$. By the lemma, $b^{m-1} - 1 \\equiv 2^n \\pmod{2^{n+1}}$, so $b^m - b = b(b^{m-1} - 1) \\equiv 2^n \\pmod{2^{n+1}}$, as $b$ is odd.\n\nWrite $a^{km} - b^m = (a^k - b)(a^{k(m-1)} + a^{k(m-2)}b + \\dots + b^{m-1})$. The first factor is $2^n \\pmod{2^{n+1}}$ and the second is odd as a sum of an odd number of odd integers. Consequently,\n\n$$\na^{km} - b^m \\equiv 2^n \\pmod{2^{n+1}}.\n$$\n\nFinally, $a^{km} - b = (a^{km} - b^m) + (b^m - b) \\equiv 2^n + 2^n \\equiv 0 \\pmod{2^{n+1}}$. This completes the inductive step and concludes the proof.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20127,
"subject": "Mathematics (Olympiad)",
"question": "The eight points $A$, $B$, $C$, $D$, $E$, $F$, $G$, and $H$ are placed on five circles as in the figure below. Each of these letters will be replaced with one of the numbers $1, 2, \\ldots, 7, 8$ such that the following two conditions hold:\n\n1. Each of the eight numbers is used exactly once.\n2. The sum of the numbers on each of the five circles is the same.\n\nHow many possibilities are there to replace the letters with numbers in this way?\n\n",
"options": [],
"answer": "See solution",
"solution": "*Answer.* There are eight possibilities.\n\n**Solution.** Let $a, b, \\dots, g, h$ be the numbers that replace the letters $A, B, \\dots, G, H$. Since the sum on each circle has the value $a + b + c + d$, we immediately get $e = c + d$, $f = d + a$, $g = a + b$, and $h = b + c$. This implies $e + f + g + h = 2(a + b + c + d)$. On the other hand,\n\n$$\n36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = (a + b + c + d) + (e + f + g + h) = 3(a + b + c + d),\n$$\n\nand therefore $a + b + c + d = 12$.\n\nConsequently, we get $e + g = (c + d) + (a + b) = 12$ and $f + h = (d + a) + (b + c) = 12$. Therefore, the pairwise distinct numbers $e, f, g, h$ have to come from the set $\\{4, 5, 7, 8\\}$ where $8$ and $4$, as well as $7$ and $5$, lie on opposite points in the figure. There are eight possibilities to replace the letters $E, F, G, H$ in this way with the numbers $4, 5, 7, 8$, because we can freely choose one of the four letters for $8$ and then we have two choices for $7$.\n\nIt remains to show that these choices of $E, F, G, H$ determine the numbers in $A, B, C, D$ uniquely. Without loss of generality, let $g = 8$ and $h = 7$, and therefore $e = 4$ and $f = 5$. From $e = 4 = c + d$ and $f = 5 = d + a$ we get that only $b$ can take the value $6$. The values $a = 2$, $c = 1$, $d = 3$ are now a direct consequence of the circle sums. Therefore, there are $8$ possibilities.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20128,
"subject": "Mathematics (Olympiad)",
"question": "Given a circle $O$ in the plane and two fixed points $B$ and $C$ on the circle such that $BC$ is not a diameter. Let $A$ be a point moving on $O$ such that $AB = AC$ and $A$ does not coincide with $B$ or $C$. Let $D$ and $E$ be the intersections of line $BC$ with the internal and external bisectors of angle $BAC$, respectively. Let $I$ be the midpoint of $DE$. The line passing through the orthocenter of $ABC$ and perpendicular to $AI$ intersects lines $AD$ and $AE$ at $M$ and $N$, respectively.\n\n1. Show that $MN$ always passes through a fixed point.\n\n2. Determine the positions of $A$ such that triangle $ABC$ has the largest area.",
"options": [],
"answer": "See solution",
"solution": "1. First, we show that $MN \\parallel OA$.\n\nAssume without loss of generality that $\\angle ABC > \\angle ACB$. Then:\n\n$$\n\\angle OAD = \\angle OAB - \\angle DAB = \\frac{180^\\circ - \\angle AOB}{2} - \\frac{\\angle BAC}{2} = 90^\\circ - \\left( \\angle C + \\frac{\\angle BAC}{2} \\right) \\quad (1)\n$$\n\nSince $AD$ and $AE$ are the internal and external bisectors of $\\angle BAC$, we have $AD \\perp AE$. With $I$ as the midpoint of $DE$:\n\n$$\n\\angle IAE = \\angle IEA = 90^\\circ - \\angle ADE = 90^\\circ - \\left( \\frac{\\angle BAC}{2} + \\angle C \\right) \\quad (2)\n$$\n\nFrom (1) and (2), $\\angle OAD = \\angle IAE$.\n\nThus, $\\angle OAI = \\angle OAD + \\angle DAI = \\angle IAE + \\angle DAI = 90^\\circ$, so $OA \\perp AI$. Since $MN \\perp AI$ by construction, $MN \\parallel OA$. (3)\n\nLet $H$ be the orthocenter of $\\triangle ABC$ and $K$ the midpoint of $BC$. We have $H \\in MN$ and $AH = 2OK$. Since $O$ and $K$ are fixed, the vector $\\vec{v} = 2\\vec{OK}$ is constant.\n\nBy (3), $MN$ is the image of $OA$ under translation by $\\vec{v}$. Let $O'$ be the image of $O$ under this translation; then $O' \\in MN$. Since $O$ is fixed, $O'$ is also fixed. Thus, $MN$ always passes through a fixed point.\n\n2. From (3), $\\angle AMN = \\angle OAD$. Together with (1) and (2), $\\angle AMN = \\angle IEA$.\n\nBut $\\angle MAH = \\angle IEA$ (angles with parallel sides), so $\\angle AMN = \\angle MAH$. Since $AMN$ is right-angled at $A$, $H$ is the midpoint of $MN$. Thus, $MN = 2AH = 4OK = \\text{const}$. Therefore, $AMN$ has the largest area if and only if it is isosceles at $A$.\n\n$\\triangle AMN$ is isosceles and right-angled at $A$ if and only if $MN \\perp AH \\iff MN \\parallel BC \\iff OA \\parallel BC$.\n\nThus, $\\triangle AMN$ has the largest area if and only if $A$ is at the endpoints of the diameter parallel to $BC$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20129,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{N} \\rightarrow \\mathbb{N}$ such that for any positive integer $n$ and any finite sequence of positive integers $a_0, \\dots, a_n$, whenever the polynomial $a_0 + a_1x + \\dots + a_nx^n$ has at least one integer root, so does $f(a_0) + f(a_1)x + \\dots + f(a_n)x^n$.",
"options": [],
"answer": "See solution",
"solution": "All functions of the form $f(x) = kx$ for some $k \\in \\mathbb{N}$ work. We now prove these are the only possibilities.\n\nSince $x + a$ has an integer root ($-a$) for any $a \\in \\mathbb{N}$, so does $f(1)x + f(a)$, implying $f(1) \\mid f(a)$ for all $a$. Define $g(x) = f(x)/f(1)$, which satisfies the same conditions, so WLOG, assume $f(1) = 1$.\n\nFor any $n \\in \\mathbb{N}$, the polynomial $nx^2 + (n+1)x + 1$ has root $-1$, so $f(n)x^2 + f(n+1)x + 1$ has an integer root, say $-k$:\n\n$$\nf(n+1)k = f(n)k^2 + 1 > f(n)k \\implies f(n+1) > f(n).\n$$\n\nThus, $f(n+1) \\geq f(n) + 1$.\n\nWe prove by induction that $f(n) = n$ for all $n$. The base case is clear. Assume $f(n) = n$; consider $m = n+1 > 1$. The polynomial $x^2 + (n+1)x + n$ has root $-1$, so $x^2 + f(n+1)x + n$ has integer roots. Their sum is $-f(n+1)$, so both roots are integers, negative, and their product is $n$, so they are $-d$ and $-n/d$ for some positive divisor $d$ of $n$. Thus, $f(n+1) = d + \\frac{n}{d}$.\n\nBut\n\n$$\nd + \\frac{n}{d} \\leq n + 1 \\iff (n - d)\\left(1 - \\frac{1}{d}\\right) \\geq 0\n$$\n\nso\n\n$$\nn + 1 \\geq d + \\frac{n}{d} = f(n+1) \\geq f(n) + 1 = n + 1,\n$$\n\nforcing $f(n+1) = n+1$ as desired. $\\square$\n\n**Remark:** There are several ways to finish the induction. For example, considering $(x+1)(x+n)$, $x^2 + f(n+1) + f(n)$ has an integer root, so $f(n+1)^2 - 4f(n)$ is a square. But $f(n+1)^2 > f(n+1)^2 - 4f(n) \\geq (f(n+1) - 2)^2$, so equality must hold, and $f(n) = f(n+1) - 1$.\n\nAlternatively, for $q(x) = (n+1)x^{2k+1} + x^{2k} + \\dots + x^3 + 2x^2 + x + n$ with root $-1$, if $x$ is an integer root of $f(n+1)x^{2k+1} + x^{2k} + \\dots + x^3 + 2x^2 + x + f(n)$,\n\n$$\nf(n+1)|x^{2k+1}| \\leq \\sum_{i=0}^{2k} |x^i| + |x^2| + n - 1.\n$$\n\nIf $|x| \\geq 2$, $\\sum |x^i| < 2^{2k+1}$, so $(f(n+1)-1)|x^{2k}| < f(n)$, which is absurd. Thus, $|x| = 1$ and $-1$ is the root. By induction, $f(n+1) = n+1$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20130,
"subject": "Mathematics (Olympiad)",
"question": "Roos has a chess board, a ruler, and a marker. She chooses two vertices on the edge of the board so that when she draws the straight line between those two points with her marker, the board is divided into two parts. Here, for example, you can see how Roos divides the board into a part with area $49$ and a part with area $15$.\n\n\n\nHow many different values can the area of such a part have?",
"options": [],
"answer": "See solution",
"solution": "$144$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20131,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a convex quadrilateral. Let $X$ be a point on the diagonal $BD$ such that $AX$ and $BC$ are perpendicular, and let $Y$ be a point on the diagonal $AC$ such that $DY$ and $AD$ are perpendicular. Suppose $XY$ and $BC$ are parallel.\n\nProve that $ABCD$ is cyclic.",
"options": [],
"answer": "See solution",
"solution": "Let $\\angle ACB = \\alpha$ and $\\angle ADB = \\beta$.\n\n\n\nSince $AX \\perp BC$, we have $\\angle XAY = \\angle XAC = 90^\\circ - \\alpha$.\n\nAlso, since $DY \\perp AD$, we have $\\angle XDY = 90^\\circ - \\beta$.\n\nNow, since $AX \\perp BC$ and $XY \\parallel BC$, we have $AX \\perp XY$, so $AXYD$ is cyclic.\n\nHence $90^\\circ - \\alpha = 90^\\circ - \\beta$, implying $\\alpha = \\beta$. Therefore $ABCD$ is cyclic.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20132,
"subject": "Mathematics (Olympiad)",
"question": "Let there be 12 scorecards numbered 1 through 12, to be distributed among 3 judges (A, B, C). Each judge holds a subset of the cards. For a given distribution, a score is the sum of three different cards, one from each judge. \n\n**a.** What are the minimum and maximum possible scores, and are all scores in between possible?\n\n**b.** If the only way to express 6 as the sum of three different numbers is $6 = 3 + 2 + 1$, and similarly for 7 and 8, what can be deduced about the distribution of the lowest-numbered cards among the judges?\n\n**c.** List all possible distributions of the scorecards among the judges that allow all scores from 6 to 29 (except possibly 10) to be achieved as the sum of three cards, one from each judge.\n\n**d.** For the distribution(s) found in part (c), which scores are not possible, and why?",
"options": [],
"answer": "See solution",
"solution": "**a.**\n\nThe minimum score is $1 + 5 + 9 = 15$ and the maximum score is $4 + 8 + 12 = 24$. All scores between are also possible. For example:\n\n$$\n\\begin{array}{l l l}\n1 + 5 + 9 = 15 & 1 + 6 + 12 = 19 & 2 + 8 + 12 = 22 \\\\\n1 + 5 + 10 = 16 & 1 + 7 + 12 = 20 & 3 + 8 + 12 = 23 \\\\\n1 + 5 + 11 = 17 & 1 + 8 + 12 = 21 & 4 + 8 + 12 = 24 \\\\\n1 + 5 + 12 = 18 & &\n\\end{array}\n$$\n\n**b.**\n\nThe only way to express 6 as the sum of three different numbers is $6 = 3 + 2 + 1$. So scorecards 1, 2, 3 are with separate judges. (1)\n\nThe only way to express 7 as the sum of three different numbers is $7 = 4 + 2 + 1$. So scorecards 1, 2, 4 are with separate judges. (2)\n\nThe only ways to express 8 as the sum of three different numbers are $8 = 5 + 2 + 1 = 4 + 3 + 1$. From (1) and (2), $4 + 3 + 1$ is impossible.\n\nSo scorecards 1, 2, 5 are with separate judges. (3)\n\nThe only ways to express 9 as the sum of three different numbers are $9 = 6 + 2 + 1 = 5 + 3 + 1 = 4 + 3 + 2$. From (1) and (3), $5 + 3 + 1$ is impossible.\n\nFrom (1) and (2), $4 + 3 + 2$ is impossible.\n\nSo scorecards 1, 2, 6 are with separate judges. (4)\n\nFrom (1), (2), (3), (4), scorecards 3, 4, 5, 6 are with the same judge and scorecards 1 and 2 are held separately by the other two judges.\n\n**c.**\n\nAny one of the following six distributions:\n\n| A | 1 | 9 | 10 | 11 | 1 | 8 | 10 | 11 | 1 | 8 | 9 | 11 |\n|---|---|---|----|----|---|---|----|----|---|---|---|----|\n| B | 2 | 7 | 8 | 12 | 2 | 7 | 9 | 12 | 2 | 7 | 10| 12 |\n| C | 3 | 4 | 5 | 6 | 3 | 4 | 5 | 6 | 3 | 4 | 5 | 6 |\n\n| A | 1 | 9 | 10 | 12 | 1 | 8 | 10 | 12 | 1 | 8 | 9 | 12 |\n|---|---|---|----|----|---|---|----|----|---|---|---|----|\n| B | 2 | 7 | 8 | 11 | 2 | 7 | 9 | 11 | 2 | 7 | 10| 11 |\n| C | 3 | 4 | 5 | 6 | 3 | 4 | 5 | 6 | 3 | 4 | 5 | 6 |\n\n**d.**\n\nAs in Part b, there is only one distribution of scorecards that gives scores 6 to 9:\n\n| A | 1 |\n|---|---|\n| B | 2 |\n| C | 3 4 5 6 |\n\nIf Judge A has scorecard 7, then score 11 is impossible. So Judge B has scorecard 7.\n\nTo achieve score 29, we must place scorecards 11 and 12 with Judges A and B separately. So we have two distributions.\n\n| A | 1 | 11 | 1 | 12 |\n|---|----|---|----|\n| B | 2 7 | 12 | 2 7 | 11 |\n| C | 3 4 5 6 | 3 4 5 6 | 3 4 5 6 | |\n\nIn each case, one of the scorecards 8, 9, 10 must be placed with Judge B. So we have six distributions.\n\n| A | 1 | 9 | 10 | 11 | 1 | 8 | 10 | 11 | 1 | 8 | 9 | 11 |\n|---|---|---|----|----|---|---|----|----|---|---|---|----|\n| B | 2 | 7 | 8 | 12 | 2 | 7 | 9 | 12 | 2 | 7 | 10| 12 |\n| C | 3 | 4 | 5 | 6 | 3 | 4 | 5 | 6 | 3 | 4 | 5 | 6 |\n\n| A | 1 | 9 | 10 | 12 | 1 | 8 | 10 | 12 | 1 | 8 | 9 | 12 |\n|---|---|---|----|----|---|---|----|----|---|---|---|----|\n| B | 2 | 7 | 8 | 11 | 2 | 7 | 9 | 11 | 2 | 7 | 10| 11 |\n| C | 3 | 4 | 5 | 6 | 3 | 4 | 5 | 6 | 3 | 4 | 5 | 6 |\n\nIn each case every score from 6 to 29, except 10, is possible.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20133,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $n$ such that there is a permutation $\\sigma$ of the set $\\{1,2,\\dots,n\\}$ for which $\\sqrt{\\sigma(1)+\\sqrt{\\sigma(2)+\\cdots+\\sqrt{\\sigma(n)}}}$ is a rational number.\n\n*A permutation of the set $\\{1,2,\\dots,n\\}$ is a one-to-one function of this set to itself.*",
"options": [],
"answer": "See solution",
"solution": "Let $n \\in \\mathbb{N}^*$. Suppose there exists a permutation $\\sigma$ such that\n\n$$\n\\sqrt{\\sigma(1)+\\sqrt{\\sigma(2)+\\cdots+\\sqrt{\\sigma(n)}}} = r_1 \\in \\mathbb{Q}.\n$$\n\nSquaring both sides, $\\sqrt{\\sigma(2)+\\cdots+\\sqrt{\\sigma(n)}}$ must also be rational. Repeating this process, for each $k \\in \\{1,2,\\dots,n\\}$, the number\n\n$$\nr_k := \\sqrt{\\sigma(k)+\\sqrt{\\sigma(k+1)+\\cdots+\\sqrt{\\sigma(n)}}}\n$$\nis rational.\n\nLet $a_k := \\sqrt{n+\\sqrt{n+\\cdots+\\sqrt{n}}}$ ($k$ times). By induction, $a_k < \\sqrt{n+1}$ for all $k$, so\n\n$$\nr_1 < a_n < \\sqrt{n+1}.\n$$\n\nLet $\\ell$ be a positive integer with $\\ell^2 \\leq n < (\\ell+1)^2$. Then for some $i$, $\\sigma(i) = \\ell^2$.\n\n**Case 1: $i \\neq n$**\n\nThen\n\n$$\n\\ell < \\sqrt{\\ell^2 + \\sqrt{\\sigma(i+1)+\\cdots+\\sqrt{\\sigma(n)}}} < \\sqrt{n+1} < \\ell+2,\n$$\nso\n\n$$\n\\sqrt{\\ell^2 + \\sqrt{\\sigma(i+1)+\\cdots+\\sqrt{\\sigma(n)}}} = \\ell+1.\n$$\n\nSquaring,\n\n$$\n2\\ell+1 = \\sqrt{\\sigma(i+1)+\\cdots+\\sqrt{\\sigma(n)}} < \\sqrt{n+1} < \\ell+2,\n$$\nwhich implies $\\ell < 1$, a contradiction.\n\n**Case 2: $i = n$**\n\nIf $\\ell > 1$, then $\\ell^2-1$ is among $\\sigma(1),\\dots,\\sigma(n-1)$. For some $j < n$, $\\sigma(j) = \\ell^2-1$.\n\nSimilarly,\n\n$$\n\\ell < \\sqrt{\\ell^2-1 + \\sqrt{\\sigma(j+1)+\\cdots+\\sqrt{\\ell^2}}} < \\sqrt{n+1} < \\ell+2,\n$$\nso\n\n$$\n\\sqrt{\\ell^2-1 + \\sqrt{\\sigma(j+1)+\\cdots+\\sqrt{\\ell^2}}} = \\ell+1.\n$$\n\nSquaring,\n\n$$\n2\\ell+2 = \\sqrt{\\sigma(j+1)+\\cdots+\\sqrt{\\ell^2}} < \\sqrt{n+1} < \\ell+2,\n$$\nwhich is impossible.\n\nIf $\\ell=1$, then $n \\in \\{1,2,3\\}$. Checking these cases:\n- For $n=1$, $\\sqrt{1}=1$ (rational).\n- For $n=3$, $\\sqrt{2+\\sqrt{3}+\\sqrt{1}}=2$ (rational, for a suitable permutation).\n- For $n=2$, no such permutation exists.\n\n**Conclusion:** The only such $n$ are $n=1$ and $n=3$.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20134,
"subject": "Mathematics (Olympiad)",
"question": "Integers $a$, $b$, and $c$ satisfy\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": "Notice that the difference between $100$ and $87$ is $13$, a prime number. This fact will help to simplify the problem. Subtract the second equation from the first to get\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 $c = -8$ satisfies all three equations.\n\nThe requested value is\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": "Intermediate Algebra"
},
{
"id": 20135,
"subject": "Mathematics (Olympiad)",
"question": "Which of the following conditions is sufficient to guarantee that integers $x$, $y$, and $z$ satisfy the equation\n\n$$\nx(x - y) + y(y - z) + z(z - x) = 1?\n$$\n\n(A) $x > y$ and $y = z$\n\n(B) $x = y - 1$ and $y = z - 1$\n\n(C) $x = z + 1$ and $y = x + 1$\n\n(D) $x = z$ and $y - 1 = x$\n\n(E) $x + y + z = 1$",
"options": [],
"answer": "See solution",
"solution": "The given equation is equivalent to $2x(x - y) + 2y(y - z) + 2z(z - x) = 2$, which can be rewritten as $(x - y)^2 + (y - z)^2 + (z - x)^2 = 2$. This equation has an integer solution if and only if two of the squares are $1$ and one is $0$. This in turn means that two of the variables must be equal and the third must differ from this common value by $1$. Choice (D) gives one instance of this, and the other choices do not imply this condition. Specifically, choice (A) fails when $x = 2$, $y = 0$, and $z = 0$ because the left-hand side of the original equation equals $4$; choice (B) fails when $x = 1$, $y = 2$, and $z = 3$ because the left-hand side equals $3$; choice (C) fails when $x = 1$, $y = 2$, and $z = 0$ because the left-hand side equals $3$; and choice (E) fails when $x = 2$, $y = 0$, and $z = -1$ because the left-hand side equals $7$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20136,
"subject": "Mathematics (Olympiad)",
"question": "Find all arrays of prime numbers $(a, b, c)$ satisfying the following conditions:\n\n1. $a < b < c < 100$, where $a$, $b$, and $c$ are all prime numbers.\n2. $a + 1$, $b + 1$, $c + 1$ form a geometric progression.",
"options": [],
"answer": "See solution",
"solution": "From condition (2), we have:\n\n$$\n(a+1)(c+1) = (b+1)^2.\n$$\n\nLet $a+1 = n^2 k$, $c+1 = m^2 k$, where $k$ has no square factor greater than 1. Then:\n\n$$\n(mn)^2 k^2 = (b+1)^2 \\implies b+1 = mnk.\n$$\n\nSo:\n\n$$\n\\begin{cases}\na = k n^2 - 1, \\\\\nb = k m n - 1, \\\\\nc = k m^2 - 1,\n\\end{cases}\n$$\n\nwhere $1 \\leq n < m$, $a < b < c < 100$, and $k > 1$ with no square factor greater than 1.\n\nIf $k = 1$, then $c = m^2 - 1$ is composite for $m \\geq 3$, so $k > 1$.\n\n**Case 1:** $m = 2$, $n = 1$:\n\n$$\n\\begin{cases}\na = k - 1, \\\\\nb = 2k - 1, \\\\\nc = 4k - 1.\n\\end{cases}\n$$\n\nFor $k = 3$, $(a, b, c) = (2, 5, 11)$ (all primes).\n\nFor $k = 6$, $(a, b, c) = (5, 11, 23)$ (all primes).\n\nOther $k$ values do not yield all primes.\n\n**Case 2:** $m = 3$, $n = 2$:\n\n$$\n\\begin{cases}\na = 4k - 1, \\\\\nb = 6k - 1, \\\\\nc = 9k - 1.\n\\end{cases}\n$$\n\nFor $k = 2$, $(a, b, c) = (7, 11, 17)$ (all primes).\n\nOther $k$ values do not yield all primes.\n\n**Conclusion:**\n\nThe arrays of primes $(a, b, c)$ are:\n\n$$\n(2, 5, 11), \\quad (5, 11, 23), \\quad (7, 11, 17).\n$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20137,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c, d$ be positive integers such that $abcd = 2025$ and each of $ab$, $bc$, $cd$, and $da$ is a perfect square. How many ordered quadruples $(a, b, c, d)$ satisfy these conditions?",
"options": [],
"answer": "See solution",
"solution": "Since $2025 = 3^4 \\times 5^2$, each of $a, b, c, d$ has no prime factors other than $3$ and $5$. Thus, we can write\n\n$$\na = 3^{x_1} 5^{y_1}, \\quad b = 3^{x_2} 5^{y_2}, \\quad c = 3^{x_3} 5^{y_3}, \\quad d = 3^{x_4} 5^{y_4}\n$$\n\nfor non-negative integers $x_1, x_2, x_3, x_4, y_1, y_2, y_3, y_4$. Since $abcd = 2025$, we have\n\n$$\nx_1 + x_2 + x_3 + x_4 = 4, \\quad y_1 + y_2 + y_3 + y_4 = 2.\n$$\n\nAlso, since $ab$, $bc$, $cd$, and $da$ are all perfect squares, the following sums must be even:\n\n$$\nx_1 + x_2, \\quad x_2 + x_3, \\quad x_3 + x_4, \\quad x_4 + x_1, \\\\ y_1 + y_2, \\quad y_2 + y_3, \\quad y_3 + y_4, \\quad y_4 + y_1.\n$$\n\nThus, all of $x_1, x_2, x_3, x_4$ must have the same parity, and similarly for $y_1, y_2, y_3, y_4$. The possible $(x_1, x_2, x_3, x_4)$ are the permutations of $(4, 0, 0, 0)$, $(2, 2, 0, 0)$, and $(1, 1, 1, 1)$, and the possible $(y_1, y_2, y_3, y_4)$ are the permutations of $(2, 0, 0, 0)$. Thus, the number of tuples $(a, b, c, d)$ is\n\n$$\n\\left(\\binom{4}{1} + \\binom{4}{2} + 1\\right) \\times \\binom{4}{1} = 44.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20138,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$ be complex numbers such that $|az^2 + bz + c| \\le 1$ for all complex numbers $z$ with $|z| \\le 1$. Find the maximum of $|bc|$.\n",
"options": [],
"answer": "See solution",
"solution": "Let $f(z) = az^2 + bz + c$. We first prove that\n\n$$\n|f(z)| \\le 1 \\text{ for all } z, |z| \\le 1 \\iff |f(z)| \\le 1 \\text{ for all } z, |z| = 1.\n$$\n\nAssume $f(z) = a(z - \\alpha)(z - \\beta)$. For any $z$ with $|z| < 1$, if $\\alpha = \\beta$, one of the two intersection points of the line through $\\alpha$ and the origin with the unit circle is closer to $\\alpha$ than $z$ is; if $\\alpha \\neq \\beta$, the line through $z$ and perpendicular to the line through $\\alpha, \\beta$ intersects the unit circle at two points, one of which is closer to $\\alpha, \\beta$ than $z$ is, respectively. So the equivalence holds.\n\nFor any complex number $z$ with $|z| = 1$, it is clear that\n\n$$\n|f(z)| = |cz^{-2} + bz^{-1} + a|.\n$$\n\nSo $|ab|_{\\max} = |bc|_{\\max}$. Write $a'z^2 + b'z + c' = e^{i\\alpha}f(e^{i\\beta}z)$. One can choose real numbers $\\alpha$, $\\beta$ such that $a'$, $b'$ are positive real numbers, so we may assume $a, b \\ge 0$ without loss of generality.\n\n$$\n1 \\ge |f(e^{i\\theta})| \\ge |\\operatorname{Im} f(e^{i\\theta})| = |a \\sin 2\\theta + b \\sin \\theta + \\operatorname{Im} c|.\n$$\n\nWithout loss of generality, assume $\\operatorname{Im} c \\ge 0$ (otherwise take the map $\\theta \\to -\\theta$). For any $\\theta \\in (0, \\frac{\\pi}{2})$,\n\n$$\n\\begin{align*}\n& 1 \\ge a \\sin 2\\theta + b \\sin \\theta \\ge 2\\sqrt{ab} \\sin 2\\theta \\sin \\theta \\\\\n\\Rightarrow & ab \\le \\frac{1}{4 \\sin 2\\theta \\sin \\theta}, \\quad \\theta \\in (0, \\frac{\\pi}{2}) \\\\\n\\Rightarrow & ab \\le \\min_{\\theta \\in (0, \\frac{\\pi}{2})} \\frac{1}{4 \\sin 2\\theta \\sin \\theta} = \\frac{1}{4 \\max_{\\theta \\in (0, \\frac{\\pi}{2})} (\\sin 2\\theta \\sin \\theta)} = \\frac{3\\sqrt{3}}{16} \\\\\n\\Rightarrow & |bc|_{\\max} = |ab|_{\\max} \\le \\frac{3\\sqrt{3}}{16}.\n\\end{align*}\n$$\n\nAn example of $|bc| = \\frac{3\\sqrt{3}}{16}$ is\n\n$$\nf(z) = \\frac{\\sqrt{2}}{8}z^2 - \\frac{\\sqrt{6}}{4}z - \\frac{3\\sqrt{2}}{8}, \\\\\n|f(e^{i\\theta})|^2 = 1 - \\frac{3}{8} \\left(\\cos \\theta - \\frac{\\sqrt{3}}{3}\\right)^2 \\le 1.\n$$\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20139,
"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 common point of the line $AQ$ and $k_2$, and $D$ is the second common point of $BQ$ with $k_2$. The lines $PB$ and $RC$ intersect at $U$, and $PA$ and $RD$ intersect at $V$. Show that a point $Z$ exists that is common to all possible lines $UV$.",
"options": [],
"answer": "See solution",
"solution": "A homothety with center $Q$ and ratio $-\\frac{r_2}{r_1}$ maps $k_1$ onto $k_2$.\n\n\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": 20140,
"subject": "Mathematics (Olympiad)",
"question": "給定正整數 $k$，求所有整係數多項式 $f(x)$，使得對所有正整數 $n$，$f(n)$ 整除 $(n!)^k$，其中 $n! = 1 \\cdot 2 \\cdot \\cdots n$。\n\nGiven a positive integer $k$, find all polynomials $f(x)$ with integer coefficients such that $f(n)$ divides $(n!)^k$ for all positive integers $n$, where $n! = 1 \\cdot 2 \\cdot \\cdots n$.",
"options": [],
"answer": "See solution",
"solution": "引理：設 $p$ 為質數，若 $p \\mid f(n)$，則 $p \\mid n$。\n\n證明：假設 $p$ 不整除 $n$，則可設 $n = kp + q$，其中 $k$ 為正整數且 $0 < q < p$。\n\n因 $f(x)$ 為整係數多項式，故有\n\n$$\nkp \\mid f(n) - f(n - kp) \\implies p \\mid f(n) - f(n - kp)\n$$\n\n又 $p \\mid f(n)$ 且 $n = kp + q$，故有\n\n$$\np \\mid f(n) - f(n - kp) \\implies p \\mid f(q)\n$$\n\n由題設 $f(q) \\mid (q!)^k$，因此 $p \\mid (q!)^k$。\n\n但 $0 < q < p$ 且 $p$ 為質數，故 $(q!)^k$ 不可能被 $p$ 整除，矛盾。\n\n故引理得證。\n\n由引理知：對於質數 $p$，$f(p)$ 只有質因數 $p$，又 $f(p) \\mid (p!)^k$，故 $f(p)$ 只能為 $1, p, p^2, \\ldots, p^k$。\n\n考慮所有質數在此函數上的取值，由於質數無窮多，依鴿籠原理，必存在非負整數 $r$（$0 \\le r \\le k$）使得有無窮多質數 $p$ 滿足 $f(p) = p^r$。\n\n觀察多項式 $f(x) = x^r$ 滿足條件：\n\n方程 $f(x) - x^r = 0$ 有無窮多根，故 $f(x) = x^r$，其中 $0 \\le r \\le k$，為所有解。",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20141,
"subject": "Mathematics (Olympiad)",
"question": "Let $(x_n)$, $n \\in \\mathbb{N}^*$, be a sequence recursively defined by\n\n$$\nx_{n+1} = 3x_n + x_n,\n$$\n\nwhere $x_1 = \\frac{a}{b}$, and $a, b$ are positive integers such that $3$ does not divide $b$. If for some positive integer $m$ we have that $x_m$ is a perfect square of a rational, prove that $x_1$ is a perfect square of a rational.",
"options": [],
"answer": "See solution",
"solution": "We will prove that if $x_{n+1}$ is a perfect square of a rational, then $x_n$ is also a perfect square of a rational, and the desired result is obtained by a simple induction.\n\nNote first that since $3$ does not divide $b$, it will not divide any of the denominators of the sequence terms.\n\nFrom the recursive relation, we have that\n$$\nx_m = 3x_{m-1}^3 + x_{m-1}.\n$$\nSetting $x_{m-1} = \\frac{p}{q}$ where $q$ is not divisible by $3$ (\\(*\\)) and $(p, q) = 1$, then\n$$\nx_m = 3x_{m-1}^3 + x_{m-1} = \\frac{3p^3 + pq^2}{q^3} = \\frac{p(3p^2 + q^2)}{q^3}.\n$$\nSince $(p, q) = 1$, this is the reduced form of $x_m$. Indeed, the numbers $p(3p^2 + q^2)$ and $q^3$ are coprime, since if $s$ is a prime dividing both, then $s \\mid q^3 \\Rightarrow s \\mid q$ and $s \\mid 3p^2 + q^2$ so $s \\mid 3p^2$. But $s$ does not divide $p$, so $s \\mid 3 \\Rightarrow s = 3$, $3 \\mid q$, which is absurd due to (\\(*)\\)).\n\nMoreover, $x_m$ is a perfect square, so both numerator and denominator in the reduced form should be perfect squares.\n\nSince the denominator is a perfect square, $q$ is a perfect square; let it be $q = a^2$. For the numerator, we have $p(3p^2 + q^2) = \\kappa^2$, so both of them should be perfect squares since they are coprime.\n\nTherefore, $p = b^2$, $3p^2 + q^2 = c^2$, and $x_{m-1} = \\frac{p}{q} = \\frac{b^2}{a^2}$, so $x_{m-1}$ is a perfect square of a rational.\n\nSimilarly, going back, $x_{m-2}$ is also a perfect square of a rational, and so on, till we arrive at $x_1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20142,
"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 $X = W \\cap Y$. Find the sum of the largest and the smallest elements of the \"Xizi\" set $X$.",
"options": [],
"answer": "See solution",
"solution": "If $n \\in W$, then there exist integers $a, b$ in $I$ such that $n = a + b + ab$, or $n + 1 = (a + 1)(b + 1)$. So, $n \\in W$ if and only if $n + 1 \\in I$ is composite. We have the following observations:\n\n1. Every integer in the set $\\{1, 2, 4, 6, 10, 12, \\dots\\} = \\{p - 1 \\mid p \\text{ is prime}\\}$ is not in $W$ (we call such numbers \"pre-primes\"), while all others are in $W$.\n2. For set $Y$: $2 = y(1, 1) \\in Y$; when $(a, b) \\neq (1, 1)$, $y(a, b)$ is composite, and this implies that all odd primes are not in $Y$. For each composite number in $Y$, either it is a product of two consecutive integers $y(1, n) = (n+1) \\cdot n$, or it is a product of three integers, the largest one being the sum of the two smaller ones, $y(a, b) = (a + b) \\cdot a \\cdot b$.\n\nWe claim that the smallest element of $X$ is $20$. On one hand, $20 = w(2, 6) = y(1, 4)$; on the other hand, among all integers less than $20$, excluding the primes and pre-primes, there are four composite numbers $8, 9, 14, 15$, and they all belong to $W$, but not to $Y$.\n\nNow we search for the largest element of $X$. Since $2000 = w(2, 666) = y(10, 10)$, $2000 \\in Y$. For all integers in $I$ larger than $2000$, excluding the three prime numbers $2003, 2011, 2017$ and their corresponding pre-primes, there are $14$ numbers left, and they have the factorizations:\n\n$$\n\\begin{aligned}\n2001 &= 3 \\cdot 23 \\cdot 29, \\\\\n2004 &= 2^2 \\cdot 3 \\cdot 167, \\\\\n2005 &= 5 \\cdot 401, \\\\\n2006 &= 2 \\cdot 17 \\cdot 59, \\\\\n2007 &= 3^2 \\cdot 223, \\\\\n2008 &= 2^3 \\cdot 251, \\\\\n2009 &= 7^2 \\cdot 41, \\\\\n2012 &= 2^2 \\cdot 503, \\\\\n2013 &= 3 \\cdot 11 \\cdot 61, \\\\\n2014 &= 2 \\cdot 19 \\cdot 53, \\\\\n2015 &= 5 \\cdot 13 \\cdot 31, \\\\\n2018 &= 2 \\cdot 1009, \\\\\n2019 &= 3 \\cdot 673, \\\\\n2020 &= 2^2 \\cdot 5 \\cdot 101.\n\\end{aligned}\n$$\n\nNone of them meets the criterion in (2), indicating that they are not in $Y$, and thus the largest element of $X$ is $2000$.\n\nTherefore, the sum of the largest and the smallest elements of $X$ is $2000 + 20 = 2020$.\n\n$$\\boxed{2020}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20143,
"subject": "Mathematics (Olympiad)",
"question": "If real number $x$ satisfies $\\log_2 x = \\log_4(2x) + \\log_8(4x)$, then the value of $x$ is \\_\\_\\_\\_\\_\\_.",
"options": [],
"answer": "See solution",
"solution": "By the given condition, we have\n\n$$\n\\log_2 x = \\log_4 2 + \\log_4 x + \\log_8 4 + \\log_8 x = \\frac{1}{2} + \\frac{1}{2} \\log_2 x + \\frac{2}{3} + \\frac{1}{3} \\log_2 x,\n$$\n\nand its solution is $\\log_2 x = 7$. Therefore, $x = 128$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20144,
"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 $AI = CI$, $\\angle ACI = \\angle CAI$, so $\\angle ACE = \\angle CAD$. Since $ACDE$ is cyclic, $\\angle CAD = \\angle CED$. Hence $\\angle ACE = \\angle CED$, so $AC$ and $DE$ are parallel.\n\nSuppose $CH = AH + DE$. 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, $AH = CH - DE = CH - CA' = HA'$, so $H$ is the midpoint of $AA'$ in isosceles triangle $EAA'$. Therefore $EH$ is orthogonal to $AA'$. It follows that $\\angle BHC = 90^\\circ = \\angle BGC$, so $B$, $C$, $G$, $H$ are cyclic. Let $K$ be the intersection of $BG$ and $CH$. Since $\\triangle BKC \\sim \\triangle HKG$,\n\n$$\n\\frac{BC}{GH} = \\frac{BK}{HK} \\tag{1}\n$$\n\nSince $DE$ is parallel to $KH$, $\\triangle BDE \\sim \\triangle BKH$, so\n\n$$\n\\frac{BK}{HK} = \\frac{BD}{DE} \\tag{2}\n$$\n\nCombining (1) and (2), $\\frac{BC}{GH} = \\frac{BD}{DE}$, hence $GH \\cdot BD = BC \\cdot DE$. This proves sufficiency.\n\nConversely, assume $GH \\cdot BD = BC \\cdot DE$. We show $CH = AH + DE$. Since $AC$ is parallel to $DE$, $\\triangle BKH \\sim \\triangle BDE$, so\n\n$$\n\\frac{BK}{KH} = \\frac{BD}{DE} \\tag{3}\n$$\n\nSince $GH \\cdot BD = BC \\cdot DE$,\n\n$$\n\\frac{BD}{DE} = \\frac{BC}{GH} \\tag{4}\n$$\n\nCombining (3) and (4), $\\frac{BK}{KH} = \\frac{BC}{GH}$.\n\nLet $G'$ be the intersection of $KD$ and the circumcircle of $\\triangle BHC$. Then $\\triangle BCK \\sim \\triangle HG'K$, so\n\n$$\n\\frac{BK}{KH} = \\frac{BC}{G'H}\n$$\n\nBut since $\\angle BKC > 90^\\circ$, only one point $X$ satisfies $\\frac{BK}{KH} = \\frac{BC}{XH}$, so $G' = G$, implying $B$, $C$, $G$, $H$ are cyclic. Thus $\\angle BHC = \\angle BGC = 90^\\circ$. Since $AC$ is parallel to $DE$, $\\angle DEH = 90^\\circ$. Let $D'$ be the foot of the altitude from $D$ to $AC$. Since $H$ is the foot of the altitude from $E$ to $AC$,\n\n$$\nCH = CD' + D'H = CD' + DE = AH + DE.\n$$\n\nThis proves necessity. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20145,
"subject": "Mathematics (Olympiad)",
"question": "It is known that nonzero real numbers $x$, $y$, $z$ satisfy the condition $xy + yz + zx = 0$. What value can the expression\n\n$$\n\\frac{1}{x^2 + 2yz} + \\frac{1}{y^2 + 2zx} + \\frac{1}{z^2 + 2xy}\n$$\nbe equal to?",
"options": [],
"answer": "See solution",
"solution": "Since $xyz \\neq 0$, we can transform the given expression:\n\n$$\n\\begin{aligned}\n\\frac{1}{x^2 + 2yz} + \\frac{1}{y^2 + 2zx} + \\frac{1}{z^2 + 2xy} &= \\frac{1}{x^2 + 2yz - xy - yz - zx} + \\frac{1}{y^2 + 2zx - xy - yz - zx} + \\frac{1}{z^2 + 2xy - xy - yz - zx} \\\\\n&= \\frac{1}{x^2 + yz - xy - zx} + \\frac{1}{y^2 + zx - xy - yz} + \\frac{1}{z^2 + xy - yz - zx} \\\\\n&= \\frac{1}{(x-y)(x-z)} + \\frac{1}{(y-x)(y-z)} + \\frac{1}{(z-x)(z-y)} \\\\\n&= -\\frac{1}{(x-y)(z-x)} - \\frac{1}{(x-y)(y-z)} - \\frac{1}{(z-x)(y-z)} \\\\\n&= -\\frac{(y-z)+(z-x)+(x-y)}{(x-y)(z-x)(y-z)} = 0.\n\\end{aligned}\n$$\n\nThus, none of the three terms on the left-hand side are zero, and the sum is zero.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20146,
"subject": "Mathematics (Olympiad)",
"question": "Let $O$ denote the centre of the circle, and let $\\alpha, \\beta, \\gamma$ be the radian measures of the vertex angles $\\angle A, \\angle B, \\angle C$ of triangle $ABC$. Suppose the circumradius of $ABC$ is $1$.\n\nShow that:\n\n$$\n\\frac{a}{\\sin \\alpha} = \\frac{b}{\\sin \\beta} = \\frac{c}{\\sin \\gamma} = 2,\n$$\n\nand that the area of $ABC$ is given by\n\n$$\n(ABC) = \\frac{1}{2}ab \\sin \\gamma = 2 \\sin \\alpha \\sin \\beta \\sin \\gamma.\n$$",
"options": [],
"answer": "See solution",
"solution": "We now show that the function $f(x) = \\ln(\\sin x)$ is strictly concave on $(0, \\pi)$, which enables us to apply Jensen's inequality. Indeed, if $x, y \\in (0, \\pi)$, then\n\n$$\n\\begin{aligned}\n\\sin x \\cdot \\sin y &= \\frac{1}{2}(\\cos(x-y) - \\cos(x+y)) \\\\\n&= \\sin^2\\left(\\frac{x+y}{2}\\right) - \\sin^2\\left(\\frac{x-y}{2}\\right) \\le \\sin^2\\left(\\frac{x+y}{2}\\right),\n\\end{aligned}\n$$\n\nwith equality iff $x = y$. Taking logarithms, this yields\n\n$$\n\\frac{f(x) + f(y)}{2} \\le f\\left(\\frac{x+y}{2}\\right)\n$$\n\nwith equality iff $x = y$. This shows that $f$ is strictly concave on $(0, \\pi)$, so we can use Jensen's inequality: $f(x) + f(y) + f(z) \\le 3f\\left(\\frac{x+y+z}{3}\\right)$, which translates into\n\n$$\n(ABC) \\le 2 \\sin^3 \\left( \\frac{\\alpha + \\beta + \\gamma}{3} \\right) = 2 \\sin^3 \\left( \\frac{\\pi}{3} \\right) = \\frac{3\\sqrt{3}}{4},\n$$\n\nwith equality iff $\\alpha = \\beta = \\gamma$. That is, $(ABC) \\le \\frac{3\\sqrt{3}}{4}$, with equality iff the triangle is equilateral.\n\nFrom the formula for the circumradius, already employed above, we obtain $a = 2\\sin\\alpha$, $b = 2\\sin\\beta$, and $c = 2\\sin\\gamma$. Hence the perimeter $p$ of $ABC$ is equal to $2(\\sin\\alpha + \\sin\\beta + \\sin\\gamma)$, and Jensen's inequality for the strictly concave function $\\sin(x)$ on $[0, \\pi]$ gives\n\n$$\np = 2(\\sin \\alpha + \\sin \\beta + \\sin \\gamma) \\le 6 \\sin \\left( \\frac{\\alpha + \\beta + \\gamma}{3} \\right) = 6 \\sin \\frac{\\pi}{3} = 3\\sqrt{3},\n$$\n\nwith equality iff $\\alpha = \\beta = \\gamma = \\pi/3$, i.e., equality holds iff $ABC$ is equilateral.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20147,
"subject": "Mathematics (Olympiad)",
"question": "A positive integer whose 1's digit is not 0 is called a *palindromic number* if the number remains the same when its digits are read in reverse order. For example, the number $12321$ is a palindromic number, since the number obtained by reading its digits in reverse order is the same $12321$, while $1234$ is not since the number $4321$ is obtained if its digits are read in reverse order. How many palindromic numbers are there which are less than or equal to $2012$?",
"options": [],
"answer": "See solution",
"solution": "Let us call a palindromic number a p.d. number. We count p.d. numbers less than $2012$ by classifying them according to the number of digits.\n\n* A 4-digit number is a p.d. number if and only if its thousand's digit and one's digit coincide and also its hundred's digit and ten's digit coincide. Therefore, $2002$ is the only p.d. number with its thousand's digit equal to $2$ (and less than $2012$). For a 4-digit number with its thousand's digit $1$ to be a p.d. number, its hundred's digit can be any of the numbers between $0$ and $9$ (inclusive), and therefore there are $10$ such numbers. Thus there are $11$ p.d. numbers with $4$ digits less than $2012$.\n\n* A 3-digit number is a p.d. number if and only if its hundred's digit and one's digit coincide. The hundred's digit of a 3-digit number can be any number between $1$ and $9$ (inclusive), and the ten's digit of a 3-digit p.d. number can be any of the $10$ numbers between $0$ and $9$. Therefore, there are $9 \\times 10 = 90$ p.d. numbers with $3$ digits.\n\n* A 2-digit number is a p.d. number if and only if its ten's digit and one's digit coincide. Since the ten's digit of a 2-digit number can take any number between $1$ and $9$, there are $9$ p.d. numbers with $2$ digits.\n\n* Any 1-digit non-zero number is a p.d. number, so there are $9$ such numbers.\n\nThus the answer we seek is $$11 + 90 + 9 + 9 = 119$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20148,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{R} \\to [0, \\infty)$ such that, for all real $a, b, c, d$ satisfying $ab + bc + cd = 0$, the following equality holds:\n$$\nf(a - b) + f(c - d) = f(a) + f(b + c) + f(d)\n$$\n(Here $\\mathbb{R}$ is the set of real numbers and $[0, \\infty)$ is the set of nonnegative real numbers.)",
"options": [],
"answer": "See solution",
"solution": "We prove the following:\n\n**Lemma**: For all real $p, q, r$ that satisfy $p^2 + q^2 = r^2$, the following equality holds:\n$$\nf(p) + f(q) = f(r)\n$$\n**Proof**: Put $a = \\frac{p-q+r}{2}$, $b = \\frac{p-q-r}{2}$, $c = \\frac{p+q+r}{2}$. Then $ab+bc+cd = \\frac{1}{2}(p^2+q^2-r^2)$. So if $p^2+q^2 = r^2$ then $ab + bc + cd = 0$ and we have\n$$\nf(r) + f\\left(\\frac{p-q-r}{2}\\right) = f\\left(\\frac{p-q-r}{2}\\right) + f(p) + f(q)\n$$\nThus $f(p) + f(q) = f(r)$ and the proof is complete.\n\nFor $(p, q, r) = (0, 0, 0)$, by the Lemma, $f(0) = 0$, and for $(p, q, r) = (p, 0, -p)$ we have $f(-p) = f(p)$, so $f$ is even.\n\nNow for any $t \\ge 0$ define $g : [0, \\infty) \\to [0, \\infty)$ by $g(t) := f(\\sqrt{t})$. Then $g(a+b) = g(a) + g(b)$. For $a \\ge b \\ge 0$ we have $g(a) = g(a-b) + g(b) \\ge g(b)$, so $g$ is monotone increasing. Thus $g(x) = g(1) \\cdot x$ for $x \\ge 0$. Therefore $f(x) = f(1) \\cdot x^2$ for $x \\ge 0$ and $f$ is even, so $f(x) = f(1) \\cdot x^2$ for any real $x$. Since $f(x) \\ge 0$, $f(1) \\ge 0$. Thus the functions satisfying the given equality are of the form $f(x) = \\lambda x^2$ for $\\lambda \\ge 0$, and it is easy to check that these functions satisfy the given equality. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20149,
"subject": "Mathematics (Olympiad)",
"question": "Given a regular 2007-gon. Find the smallest positive integer $k$ such that in every set of $k$ vertices, there exist 4 vertices which form a quadrilateral with 3 edges of the given 2007-gon.",
"options": [],
"answer": "See solution",
"solution": "Denote the vertices of the regular 2007-gon by $A_1, A_2, \\dots, A_{2007}$. Note that a quadrilateral has 3 edges of the polygon if and only if its 4 vertices are consecutive vertices of the polygon.\n\nLet $A$ be the set of all vertices except $A_{4k}$ for $k = 1, 2, \\dots, 501$ and $A_{2007}$; that is, $A = \\{A_1, A_2, A_3, A_5, A_6, A_7, \\dots, A_{2005}, A_{2006}\\}$. Then $|A| = 1505$, and $A$ contains no 4 consecutive vertices of the polygon. Thus, $k \\geq 1506$.\n\nNow, we show that every subset $B$ of 1506 vertices contains 4 consecutive vertices. Let $T$ be any subset of 1506 vertices. By deleting the 501 vertices not in $B$, the set of vertices of the polygon is decomposed into subsets $B_1, B_2, \\dots, B_m$ with $m \\leq 501$. By the Dirichlet principle, some $B_i$ contains at least $\\frac{1506}{501} > 3$ consecutive vertices. Thus, the answer is $k = 1506$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20150,
"subject": "Mathematics (Olympiad)",
"question": "In how many ways can a chessboard $n \\times n$, $n \\ge 3$, from which two diagonally opposite corner cells $1 \\times 1$ were cut out, accommodate $n$ rooks, neither of which attacks another? A rook is a chess piece that attacks all the cells adjacent horizontally or vertically to the cell it is located in.",
"options": [],
"answer": "See solution",
"solution": "**Answer:** $(n-2)!(n^2-3n+3)$.\n\n**Solution.** Without loss of generality, suppose that the left lower cell $A$ and the right upper cell $Z$ were cut out. First, let us see in how many ways rooks can be placed on an $n \\times n$ chessboard. There are $n!$ ways, since for a rook, there are $n$ options in the first column, $n-1$ options in the second column, and so on. Consider the arrangements of rooks in which one of them is located in cell $A$. There are $(n-1)!$ such arrangements, and analogously, $(n-1)!$ ways for when a rook occupies cell $Z$. In the expression $n! - 2(n-1)!$, positions where rooks occupy both cells $A$ and $Z$ are rejected twice. There are $(n-2)!$ such arrangements. Thus, the correct expression would be:\n\n$$\nn! - 2(n-1)! + (n-2)! = (n-2)!(n(n-1) - 2(n-1) + 1) = (n-2)!(n^2 - 3n + 3).\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20151,
"subject": "Mathematics (Olympiad)",
"question": "Given that $a^2 + b^2 + c^2 + ab + bc + ca \\le 2$, prove that\n\n$$\n\\frac{2ab + 2}{(a + b)^2} + \\frac{2bc + 2}{(b + c)^2} + \\frac{2ca + 2}{(c + a)^2} \\ge 6.\n$$",
"options": [],
"answer": "See solution",
"solution": "Adding the last three inequalities gives\n\n$$\n\\frac{ab+1}{(a+b)^2} + \\frac{bc+1}{(b+c)^2} + \\frac{ca+1}{(c+a)^2} \\ge \\frac{3}{2} + \\frac{yz}{2x^2} + \\frac{zx}{2y^2} + \\frac{xy}{2z^2} \\ge 3,\n$$\n\nby the AM-GM inequality. Equality holds if and only if $x = y = z$ or $a = b = c = \\frac{1}{\\sqrt{3}}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20152,
"subject": "Mathematics (Olympiad)",
"question": "There are 11 points equally spaced on a circle. Some of the segments with endpoints among these vertices are drawn and colored in two colors, so that each segment meets at an internal point with at most one other segment of the same color. What is the greatest number of segments that could be drawn?",
"options": [],
"answer": "See solution",
"solution": "Let $n = 11$. Remove the sides of the $n$-gon. Fix a color. The number of diagonals of that color is at most $\\left\\lceil \\frac{3(n-3)}{2} \\right\\rceil$. The same holds for the other color. Adding the sides of the $n$-gon, which can be colored as we want, we get that the greatest number of segments is at most\n\n$$\n2 \\left\\lceil \\frac{3(n-3)}{2} \\right\\rceil + n.\n$$\n\nThe argument uses induction and considers how moving segments does not decrease the maximum possible number, as shown in the images below:\n\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20153,
"subject": "Mathematics (Olympiad)",
"question": "There are 12 chairs aligned and labeled with numbers 1, 2, ..., 12 from left to right. A grasshopper can jump from one chair to another following the rule: from a chair with number $k$, it can jump to the chair with number $n$ if and only if $|k-n| = 5$ or $|k-n| = 8$. It is known that the grasshopper managed to do the jumps so that it visited all chairs exactly once. What chair could be the initial position for the grasshopper?",
"options": [],
"answer": "See solution",
"solution": "Let us write down all possible jumps:\n\n$1 \\leftrightarrow 6$, $1 \\leftrightarrow 9$, $2 \\leftrightarrow 7$, $2 \\leftrightarrow 10$, $3 \\leftrightarrow 8$, $3 \\leftrightarrow 11$, $4 \\leftrightarrow 9$, $4 \\leftrightarrow 12$, $5 \\leftrightarrow 10$, $6 \\leftrightarrow 11$, $7 \\leftrightarrow 12$.\n\nThe chairs with labels $5$ and $8$ occur in this list only once. So if the grasshopper did not start from this chair, it must finish on it. Thus, it has to start from $5$ and finish on $8$, or vice versa. Both options are possible. The following sequence of jumps gives an example for $5$ (and the jumps in the opposite direction give an example for $8$):\n\n$$\n5 \\to 10 \\to 2 \\to 7 \\to 12 \\to 4 \\to 9 \\to 1 \\to 6 \\to 11 \\to 3 \\to 8\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20154,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a quadrangle inscribed in a circle $\\omega$. The lines $AB$ and $CD$ meet at $P$, the lines $AD$ and $BC$ meet at $Q$, and the diagonals $AC$ and $BD$ meet at $R$. Let $M$ be the midpoint of the segment $PQ$, and let $K$ be the common point of the segment $MR$ and the circle $\\omega$. Prove that the circles $KPQ$ and $\\omega$ are tangent to one another.",
"options": [],
"answer": "See solution",
"solution": "Let $O$ be the centre of $\\omega$. Notice that the points $P$, $Q$, and $R$ are the poles (with respect to $\\omega$) of the lines $QR$, $RP$, and $PQ$, respectively. Hence we have $OP \\perp QR$, $OQ \\perp RP$, and $OR \\perp PQ$, thus $R$ is the orthocentre of the triangle $OPQ$.\n\nNow, if $MR \\perp PQ$, then the points $P$ and $Q$ are the reflections of one another in the line $MR = MO$, and the triangle $PQK$ is symmetrical with respect to this line. In this case the statement of the problem is trivial.\n\nOtherwise, let $V$ be the foot of the perpendicular from $O$ to $MR$, and let $U$ be the common point of the lines $OV$ and $PQ$. Since $U$ lies on the polar line of $R$ and $OU \\perp MR$, we obtain that $U$ is the pole of $MR$. Therefore, the line $UK$ is tangent to $\\omega$. Hence it is enough to prove that $UK^2 = UP \\cdot UQ$, since this relation implies that $UK$ is also tangent to the circle $KPQ$.\n\nFrom the rectangular triangle $OKU$, we get $UK^2 = UV \\cdot UO$. Let $\\Omega$ be the circumcircle of triangle $OPQ$, and let $R'$ be the reflection of its orthocentre $R$ in the midpoint $M$ of the side $PQ$. It is well known that $R'$ is the point of $\\Omega$ opposite to $O$, hence $OR'$ is the diameter of $\\Omega$. Finally, since $\\angle OVR' = 90^\\circ$, the point $V$ also lies on $\\Omega$, hence $UP \\cdot UQ = UV \\cdot UO = UK^2$, as required.\n\n\n\n**Remark.** The statement of the problem is still true if $K$ is the other common point of the line $MR$ and $\\omega$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20155,
"subject": "Mathematics (Olympiad)",
"question": "Let $DX = d$. Choose an $xy$-coordinate axis so that $X = (0,0)$, $A = (5,0)$, $B = (0,6)$, $C = (-20,0)$, and $D = (0,-d)$. Let $\\Gamma$ be the circle tangent to each of the circles $O_1, O_2, O_3, O_4$, and let $P = (x, y)$ be the center of $\\Gamma$ and $r$ its radius. Find the value of $d$ such that $\\Gamma$ is tangent to all four circles as described.",
"options": [],
"answer": "See solution",
"solution": "Since the circle $O_1$ touches $\\Gamma$ tangentially from inside, $AP = r - 5$, so $(x - 5)^2 + y^2 = (r - 5)^2$. Simplifying, $$ r^2 - x^2 - y^2 = 10(r - x). $$\n\nSimilarly, for $O_2, O_3, O_4$:\n$$\nr^2 - x^2 - y^2 = 40(r + x), \\quad r^2 - x^2 - y^2 = 12(r - y), \\quad r^2 - x^2 - y^2 = 2d(r + y).\n$$\nLet $t = r^2 - x^2 - y^2$, then:\n$$\nr - x = \\frac{t}{10}, \\quad r + x = \\frac{t}{40}, \\quad r - y = \\frac{t}{12}, \\quad r + y = \\frac{t}{2d}.\n$$\nSince $2r = (r-x) + (r+x) = (r-y) + (r+y)$,\n$$\n2r = t \\left( \\frac{1}{10} + \\frac{1}{40} \\right) = t \\left( \\frac{1}{12} + \\frac{1}{2d} \\right).\n$$\nAs $r \\neq 0$, equate:\n$$\n\\frac{1}{10} + \\frac{1}{40} = \\frac{1}{12} + \\frac{1}{2d}.\n$$\nSolving, $d = 12$ is the desired value.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20156,
"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$$\n\\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_9} = 1.\n$$\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 swapping $x_1$ and $x_2$ pairs them. Thus, for parity, we may assume $x_1 = x_2$. Similarly, among solutions with $x_1 = x_2$ and $x_3 \\neq x_4$, swapping $x_3$ and $x_4$ pairs them, so we may further assume $x_3 = x_4$. Continuing, we may restrict to $x_1 = x_2 = x_3 = x_4$. The same argument 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 $u \\neq v$, swapping the $u$'s and $v$'s pairs the solutions, so we may assume $u = v$, i.e., $x_1 = x_2 = \\dots = x_8 = a$, $x_9 = b$.\n\nThe equation becomes\n\n$$\n\\frac{8}{a} + \\frac{1}{b} = 1.\n$$\n\nThis is equivalent to $8b + a = ab$, or $(a-8)(b-1) = 8$. Since $b \\geq 2$, $b-1$ is a positive divisor of $8$. The possibilities are $b-1 = 1, 2, 4, 8$, giving:\n\n- $a = 16$, $b = 2$\n- $a = 12$, $b = 3$\n- $a = 10$, $b = 5$\n- $a = 9$, $b = 9$\n\nAll four yield valid, distinct solutions. Thus, $N$ is even.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20157,
"subject": "Mathematics (Olympiad)",
"question": "Suppose Bree gives Cala a $h$ metre head start in an 800 m race, and they finish at the same time. Given:\n\n- Asha runs 400 m in the same time as Bree runs 380 m.\n- Asha runs 1500 m in the same time as Cala runs 1254 m.\n\nFind the value of $h$.",
"options": [],
"answer": "See solution",
"solution": "Let $a$, $b$, $c$ denote the running speeds of Asha, Bree, and Cala respectively.\n\nAsha runs 400 m in the same time as Bree runs $400 - 20 = 380$ m.\n\nSo $\\frac{400}{a} = \\frac{380}{b}$, hence $\\frac{a}{b} = \\frac{20}{19}$.\n\nAsha runs 1500 m in the same time as Cala runs $1500 - 246 = 1254$ m.\n\nSo $\\frac{1500}{a} = \\frac{1254}{c}$, hence $\\frac{c}{a} = \\frac{209}{250}$.\n\nSuppose Bree gives Cala a $h$ metre head start in the 800 m race and they finish at the same time. Then $\\frac{800}{b} = \\frac{800 - h}{c}$. Therefore,\n\n$$\n\\frac{800-h}{800} = \\frac{c}{b} = \\left(\\frac{c}{a}\\right)\\left(\\frac{a}{b}\\right) = \\frac{209}{250} \\times \\frac{20}{19} = \\frac{22}{25}\n$$\n\nSo $800 - h = 800 \\times \\frac{22}{25} = 704$, and $h = 800 - 704 = 96$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20158,
"subject": "Mathematics (Olympiad)",
"question": "Let $I \\subset \\mathbb{R}$ be an open interval and consider $f : I \\to \\mathbb{R}$, a function that is twice differentiable on $I$, such that $f(x) \\cdot f''(x) = 0$ for any $x \\in I$. Show that $f''$ is the zero function.",
"options": [],
"answer": "See solution",
"solution": "Consider the set $A = \\{x \\in I \\mid f''(x) \\ne 0\\}$. Suppose, for contradiction, that $A \\ne \\emptyset$. Since $f'' = (f')'$ has the intermediate value property on $I$, $A$ cannot be a singleton.\n\nLet $a, b \\in A$ with $a < b$. Then $f(a) = f(b) = 0$. Define $g : I \\to \\mathbb{R}$ by $g(x) = f(x) f'(x)$. For all $x \\in I$, $g'(x) = (f'(x))^2 + f(x) f''(x) = (f'(x))^2 \\ge 0$, so $g$ is increasing on $I$.\n\nSince $g(a) = g(b) = 0$ and $g$ is increasing, $g(x) = 0$ for all $x \\in [a, b]$, so $g'(x) = 0$ for all $x \\in [a, b]$. Thus, $(f'(x))^2 = 0$ for all $x \\in [a, b]$, so $f'(x) = 0$ and hence $f''(x) = 0$ for all $x \\in [a, b]$, contradicting $f''(a) \\ne 0$. Therefore, $A = \\emptyset$, so $f''$ is the zero function.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20159,
"subject": "Mathematics (Olympiad)",
"question": "In each of six boxes $B_1, B_2, B_3, B_4, B_5, B_6$ there is initially one coin. There are two types of operation allowed:\n\n*Type 1:* Choose a nonempty box $B_j$ with $1 \\leq j \\leq 5$. Remove one coin from $B_j$ and add two coins to $B_{j+1}$.\n\n*Type 2:* Choose a nonempty box $B_k$ with $1 \\leq k \\leq 4$. Remove one coin from $B_k$ and exchange the contents of (possibly empty) boxes $B_{k+1}$ and $B_{k+2}$.\n\nDetermine whether there is a finite sequence of such operations that results in boxes $B_1, B_2, B_3, B_4, B_5$ being empty and box $B_6$ containing exactly $2010^{2010^{2010}}$ coins. (Note that $a^{bc} = a^{(bc)}$.)",
"options": [],
"answer": "See solution",
"solution": "The answer is yes.\n\nLet $(b_1, \\dots, b_n)$ denote the $n$-box configuration where $b_1$ coins are in box $B_1$, $b_2$ coins in $B_2$, etc. Write $(b_1, \\dots, b_n) \\to (b'_1, \\dots, b'_n)$ if we can obtain $(b'_1, \\dots, b'_n)$ from $(b_1, \\dots, b_n)$ by the allowed operations.\n\n**Lemma 1.** Let $a$ be a positive integer. Then $(a, 0, 0) \\to (0, 2^a, 0)$.\n\n*Proof.* We show $(a, 0, 0) \\to (a-k, 2^k, 0)$ for $1 \\leq k \\leq a$ by induction. For $k=1$, a Type 1 operation gives $(a, 0, 0) \\to (a-1, 2, 0) = (a-1, 2^1, 0)$. Assume true for $k < a$. From $(a-k, 2^k, 0)$, apply $2^k$ Type 1 operations at the middle box to get $(a-k, 0, 2^{k+1})$. A Type 2 operation at the first box gives $(a-k-1, 2^{k+1}, 0)$. $\\square$\n\n**Lemma 2.** Define $P_n = \\underbrace{2^{2^{\\cdots}}}_{n}$. Then $(a, 0, 0, 0) \\to (0, P_a, 0, 0)$ for every positive integer $a$.\n\n*Proof.* We show $(a, 0, 0, 0) \\to (a-k, P_k, 0, 0)$ for $1 \\leq k \\leq a$ by induction. For $k=1$, a Type 1 operation at the first box gives $(a-1, P_1, 0, 0)$. Assume $(a, 0, 0, 0) \\to (a-k, P_k, 0, 0)$ for some $k < a$. By Lemma 1 on the last three boxes, $(a-k, P_k, 0, 0) \\to (a-k, 0, P_{k+1}, 0)$. A Type 2 operation at the first box gives $(a-k-1, P_{k+1}, 0, 0)$. $\\square$\n\nNow, for the 6-box setting, let $A = 2010^{2010^{2010}}$.\n\n1. Apply a Type 1 operation to $B_5$:\n $$(1, 1, 1, 1, 1, 1) \\to (1, 1, 1, 1, 0, 3)$$\n2. Apply Type 2 operations to $B_4, B_3, B_2, B_1$ in order:\n $$(1, 1, 1, 1, 0, 3) \\to (1, 1, 1, 0, 3, 0) \\to (1, 1, 0, 3, 0, 0) \\to (1, 0, 3, 0, 0, 0) \\to (0, 3, 0, 0, 0, 0)$$\n3. Apply Lemma 2 twice:\n $$(0, 3, 0, 0, 0, 0) \\to (0, 0, P_3, 0, 0, 0) \\to (0, 0, 0, P_{16}, 0, 0)$$\n Since $P_{16} > A$, $B_4$ now has more than $A$ coins.\n4. Apply Type 2 operations to $B_4$ repeatedly to reduce its coins to $A/4$:\n $$(0, 0, 0, P_{16}, 0, 0) \\to \\dots \\to (0, 0, 0, A/4, 0, 0)$$\n5. Apply Type 1 operations to empty $B_4$ and then $B_5$:\n $$\n (0, 0, 0, A/4, 0, 0) \\to \\dots \\to (0, 0, 0, 0, A/2, 0) \\to \\dots \\to (0, 0, 0, 0, 0, A)\n $$\n\nThus, it is possible to achieve the desired configuration.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20160,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute triangle, where the altitudes $BD$ and $CE$ are drawn. Let $L$ be the point on segment $BD$ such that $AD = DL$, and let $K$ be the point on segment $CE$ such that $AE = EK$. Denote $M$ as the midpoint of segment $KL$. The circumcircle of triangle $ABC$ intersects line $AL$ again at point $T$ and line $AK$ again at point $S$. Prove that the lines $BS$, $CT$, and $AM$ are concurrent.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $P = BS \\cap TC$. Since $M$ is the midpoint of $LK$, it follows that $\\frac{\\sin \\angle LAM}{\\sin \\angle MAK} = \\frac{AK}{AL}$. Since $\\angle BAL = \\angle KAC = \\angle BAC - 45^\\circ$, $\\angle SBC = \\angle TCB$, so $BTSC$ is an isosceles trapezoid. Applying the sine theorem on $\\triangle ATP$:\n\n$$\n\\frac{TP}{\\sin \\angle TAP} = \\frac{AP}{\\sin \\angle ATP}\n$$\n\nApplying the sine theorem on $\\triangle ASP$:\n\n$$\n\\frac{SP}{\\sin \\angle SAP} = \\frac{AP}{\\sin \\angle ASP}\n$$\n\nFrom the above,\n\n$$\n\\frac{\\sin \\angle TAP}{\\sin \\angle SAP} \\cdot \\frac{SP}{TP} = \\frac{\\sin \\angle ATP}{\\sin \\angle ASP} = \\frac{ABC}{ACB} = \\frac{AC}{AB}.\n$$\n\nSince $BTSC$ is an isosceles trapezoid, $SP = TP$, and points $E$ and $D$ are bases, so\n\n$$\n\\frac{AC}{AB} = \\frac{AE}{AD}. \\text{ Therefore } \\frac{\\sin \\angle TAP}{\\sin \\angle SAP} = \\frac{AE}{AD} = \\frac{AK}{AL}.\n$$\n\n$$\n\\angle LAK = \\angle LAM + \\angle MAK = \\angle TAP + \\angle SAP\n$$\n\nand\n\n$$\n\\frac{\\sin \\angle LAM}{\\sin \\angle MAK} = \\frac{\\sin \\angle TAP}{\\sin \\angle SAP} \\text{ so } \\angle LAM = \\angle TAP\n$$\n\nTherefore, the lines $TC$, $BS$, and $AM$ intersect at one point.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20161,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute triangle with $AC \\neq BC$. Let $M$ be the midpoint of segment $AB$. Let $H$ be the orthocenter of triangle $ABC$, $D$ the foot of the altitude from $A$ to $BC$, and $E$ the foot of the altitude from $B$ to $AC$.\n\nProve that lines $AB$, $DE$, and the line through $C$ perpendicular to $MH$ intersect at a single point $S$.",
"options": [],
"answer": "See solution",
"solution": "Let $\\angle ACB = \\gamma$ and let $F$ be the foot of $C$ on $MH$. We first show that $F$ lies on the circumcircle $k$ of triangle $ABC$.\n\nLet $H_1$ be the reflection of $H$ over $M$. The quadrilateral $AH_1BH$ is a parallelogram, and since $\\angle AHB = \\angle AH_1B = 180^\\circ - \\gamma$, the point $H_1$ lies on the circumcircle $k$ of $ABC$. Reflecting $H$ over $AB$ gives point $H_2$, which also lies on $k$. The line $H_1H_2$ is parallel to $AB$, and thus perpendicular to $CH_2$. It follows that $CH_1$ is a diameter of $k$, so $F$ lies on $k$.\n\nIn summary:\n\n- Points $A, B, D, E$ lie on a circle $k_1$.\n- Points $C, E, H, D, F$ lie on a circle $k_2$.\n- Points $A, B, F, C$ lie on the circumcircle $k$.\n\nTherefore, the point $S$ is the radical center of these three circles, completing the proof.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20162,
"subject": "Mathematics (Olympiad)",
"question": "Let $N = 10^{2024}$. Let $S$ be a square in the Cartesian plane with sides parallel to the coordinate axes and side length $N$. Let $P_1, \\dots, P_N$ be $N$ points inside $S$ with distinct $x$-coordinates, such that for any $i \\neq j$, the absolute value of the slope of the line $P_i P_j$ does not exceed $1$.\n\nProve that there exists a line $\\ell$ such that at least $2024$ of the points $P_1, \\dots, P_N$ are within a distance of $1$ from $\\ell$.",
"options": [],
"answer": "See solution",
"solution": "Let the coordinates of the $N$ points be $(x_1, y_1), \\dots, (x_N, y_N)$. By the given conditions, $x_1, \\dots, x_N$ are distinct and for any $i \\neq j$ we have $|y_i - y_j| \\leq |x_i - x_j|$.\n\nFor any $1 \\leq i, j \\leq N$, $i \\neq j$, define\n\n$$\nI_{i,j} = \\left\\{ \\frac{y_j + \\theta - y_i}{x_j - x_i} : -1 \\leq \\theta \\leq 1 \\right\\} \\cap [-1, 1],\n$$\n\nand let $|I_{i,j}|$ be the length of $I_{i,j}$.\n\nFor each $k \\in I_{i,j}$, let $k = \\frac{y_j + \\theta_k - y_i}{x_j - x_i}$, where $\\theta_k \\in [-1, 1]$. The line passing through the point $(x_i, y_i)$ with slope $k$ is denoted as $\\ell_i(k)$. Then $\\ell_i(k)$ also passes through the point $(x_j, y_j + \\theta_k)$. Since $|\\theta_k| \\leq 1$, the distance from the point $(x_j, y_j)$ to $\\ell_i(k)$ does not exceed $1$. If there exists an index $i$ such that\n\n$$\n\\sum_{j \\neq i} |I_{i,j}| > 4044,\n$$\n\nsince $I_{i,j}$ are subsets of $[-1, 1]$, there exists a point $k$ covered by $\\{I_{i,j} : j \\neq i\\}$ at least $2023$ times. Let $k \\in \\bigcap_{r=1}^{2023} I_{i,j_r}$, then let $\\ell = \\ell_i(k)$. There are $2024$ points $(x_i, y_i), (x_{j_1}, y_{j_1}), \\dots, (x_{j_{2023}}, y_{j_{2023}})$ whose distance to $\\ell$ does not exceed $1$, thus proving the statement.\n\nNext, we prove the existence of $i$ satisfying the above inequality. To do so, we need to prove\n\n$$\n\\sum_{1 \\leq i < j \\leq N} |I_{i,j}| > 2022N.\n$$\n\nFirst, estimate the lower bound of $|I_{i,j}|$.\n\n1. When $|x_j - x_i| \\leq \\frac{1}{2}$, we have $|y_j - y_i| \\leq \\frac{1}{2}$. Thus, $y_j + \\theta - y_i$ can take every value in $[-|x_j - x_i|, |x_j - x_i|]$, implying that $I_{i,j} = [-1, 1]$.\n2. When $|x_j - x_i| > \\frac{1}{2}$, assume without loss of generality that $x_j > x_i$. If $y_j \\geq y_i$, let $\\Delta x = x_j - x_i$, $\\Delta y = y_j - y_i$, then\n\n$$\n\\theta \\in [-1, 1] \\text{ and } \\frac{y_j + \\theta - y_i}{x_j - x_i} \\in [-1, 1] \\iff \\max\\{-1, -\\Delta x - \\Delta y\\} \\leq \\theta \\leq \\min\\{1, \\Delta x - \\Delta y\\},\n$$\n\nwhich gives\n\n$$\n|I_{i,j}| = \\frac{\\min\\{1, \\Delta x - \\Delta y\\} + \\min\\{1, \\Delta x + \\Delta y\\}}{\\Delta x} \\geq \\frac{1}{\\Delta x}.\n$$\n\nSimilarly, when $y_j < y_i$ we also have $|I_{i,j}| \\geq \\frac{1}{|x_i - x_j|}$. Therefore, for any $i \\neq j$ we have\n\n$$\n|I_{i,j}| \\geq \\min \\left\\{ 2, \\frac{1}{|x_j - x_i|} \\right\\}.\n$$\n\nNow we estimate the lower bound of $S = \\sum_{1 \\leq i < j \\leq N} |I_{i,j}|$. Without loss of generality, assume $x_1 < x_2 < \\dots < x_N$. For each $1 \\leq d \\leq N-1$, consider $x_{d+1} - x_1, x_{d+2} - x_2, \\dots, x_N - x_{N-d}$. Their total sum does not exceed $dN$. Suppose there are exactly $k$ of these values that do not exceed $\\frac{1}{2}$. By the Cauchy-Schwarz inequality, we get\n\n$$\n\\begin{align*}\n\\sum_{i=1}^{N-d} |I_{i,i+d}| &\\geq 2k + \\sum_{x_{i+d}-x_i > \\frac{1}{2}} \\frac{1}{x_{i+d}-x_i} \\\\\n&\\geq 2k + \\frac{(N-d-k)^2}{dN} \\\\\n&= \\frac{(N-d)^2 + k^2}{dN} + 2k \\frac{dN - N + d}{dN} \\\\\n&\\geq \\frac{(N-d)^2}{dN} = \\frac{N}{d} + \\frac{d}{N} - 2.\n\\end{align*}\n$$\n\nSumming over $d$, we get\n\n$$\n\\begin{align*}\nS &= \\sum_{d=1}^{N-1} \\sum_{i=1}^{N-d} |I_{i,i+d}| \\\\\n &\\geq \\sum_{d=1}^{N-1} \\left( \\frac{N}{d} + \\frac{d}{N} - 2 \\right) \\\\\n &= N \\left( \\frac{1}{1} + \\frac{1}{2} + \\dots + \\frac{1}{N-1} \\right) + \\frac{N-1}{2} - 2(N-1) \\\\\n &> N \\left( \\ln N - \\frac{3}{2} \\right) > 2022N,\n\\end{align*}\n$$\n\nwhich completes the proof.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20163,
"subject": "Mathematics (Olympiad)",
"question": "A sequence of positive real numbers $a_1, a_2, a_3, \\dots$ satisfies $a_n = a_{n-1} + a_{n-2}$ for all $n \\geq 3$.\n\nA sequence $b_1, b_2, b_3, \\dots$ is defined by:\n- $b_1 = a_1$\n- For even $n > 1$: $b_n = a_n + (b_1 + b_2 + \\dots + b_{n-1})$\n- For odd $n > 1$: $b_n = a_n + (b_2 + b_4 + \\dots + b_{n-1})$\n\nProve that if $n \\geq 3$, then\n$$\n\\frac{1}{3} < \\frac{b_n}{n a_n} < 1\n$$",
"options": [],
"answer": "See solution",
"solution": "The definition of the sequence $(b_n)$ implies that $b_n - b_{n-2} = a_n - a_{n-2} + b_{n-1}$ for all $n \\geq 3$. Therefore,\n$$\nb_n - b_{n-2} = a_{n-1} + b_{n-1}\n$$\ni.e.,\n$$\nb_n = a_{n-1} + b_{n-1} + b_{n-2}\n$$\n\nFor the left-hand inequality, $\\frac{1}{3} < \\frac{b_n}{n a_n}$ holds for $n=2$ and $n=3$:\n- $\\frac{b_2}{2a_2} = \\frac{a_2 + b_1}{2a_2} = \\frac{a_1 + a_2}{2a_2} > \\frac{a_2}{2a_2} = \\frac{1}{2} > \\frac{1}{3}$\n- $\\frac{b_3}{3a_3} = \\frac{a_3 + b_2}{3a_3} = \\frac{2(a_1 + a_2)}{3(a_1 + a_2)} = \\frac{2}{3} > \\frac{1}{3}$\n\nAssume $n \\geq 4$ and the statement is true for $n-1$ and $n-2$. Then:\n$$\nb_n = a_{n-1} + b_{n-1} + b_{n-2} > a_{n-1} + \\frac{1}{3}(n-1)a_{n-1} + \\frac{1}{3}(n-2)a_{n-2}\n$$\n$$\n= \\frac{1}{3}n(a_{n-1} + a_{n-2}) + \\frac{2}{3}(a_{n-1} - a_{n-2}) > \\frac{1}{3}n a_n\n$$\nwhere the last inequality holds since $a_{n-1} = a_{n-2} + a_{n-3} > a_{n-2}$. By induction, $b_n > \\frac{1}{3} n a_n$ for all $n \\geq 2$.\n\nFor the right-hand inequality, $\\frac{b_n}{n a_n} < 1$ holds for $n=3$ and $n=4$:\n- $\\frac{b_3}{3a_3} = \\frac{2}{3} < 1$\n- $\\frac{b_4}{4a_4} = \\frac{a_4 + b_3 + b_1}{4a_4} = \\frac{a_4 + a_3 + a_2 + a_1}{4a_4} < 1$\n\nLet $n \\geq 5$ and assume the statement is valid for $n-1$ and $n-2$. Then:\n$$\nb_n = a_{n-1} + b_{n-1} + b_{n-2} < a_{n-1} + (n-1)a_{n-1} + (n-2)a_{n-2} = n(a_{n-1} + a_{n-2}) - 2a_{n-2} < n a_n\n$$\nBy induction, $b_n < n a_n$ holds for all $n \\geq 3$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20164,
"subject": "Mathematics (Olympiad)",
"question": "There is a finite number of lamps in an electrical scheme. Some pairs of lamps are directly connected by a wire. Every lamp is lit either red or blue. With one switch, all lamps that have a direct connection with a lamp of the other colour change their colour (from red to blue or vice versa). Prove that after some number of switches, all lamps have the same colour as two switches before that.",
"options": [],
"answer": "See solution",
"solution": "If some connected lamps are lit in different colours, they both change colour upon switching; hence, they are also lit differently after the switch. The same holds on each following switch. Thus, no pair of connected lamps lit in different colours can disappear, but more such pairs can appear. Since there are only finitely many lamps, the number of connected pairs of differently lit lamps cannot grow infinitely. Therefore, this number stops changing after some number of switches. This means that, at that point, a lamp either changes colour on each switch or never changes colour. Hence, the colours of all lamps are the same after two consecutive switches.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20165,
"subject": "Mathematics (Olympiad)",
"question": "Given positive integers $a$ and $b$, suppose\n\n$$\na^3 + b^3 = 2^c.\n$$\n\nFind all possible values of $a$ and $b$.",
"options": [],
"answer": "See solution",
"solution": "Without loss of generality, let $a \\ge b$. Since $a^3 + b^3 = 2^c$ is a power of two greater than 1, $c \\ge 1$ and $a^3$ and $b^3$ have the same parity, so $a$ and $b$ have the same parity. Let $a + b = 2x$ and $a - b = 2y$ for integers $0 \\le y < x$. Then\n\n$$\na = x + y, \\quad b = x - y.\n$$\n\nSubstituting into the original equation:\n\n$$\n\\begin{aligned}\n(a + y)^3 + (a - y)^3 &= 2^c \\\\\n2x^3 + 6xy^2 &= 2^c \\\\\nx(x^2 + 3y^2) &= 2^{c-1}.\n\\end{aligned}\n$$\n\nLet $x = 2^r$ and $x^2 + 3y^2 = 2^s$ for non-negative integers $r$ and $s$. Then $2r \\le s$ since $x^2 \\le 2^s$. Substituting $x$ into the second equation:\n\n$$\n3y^2 = 2^{2r}(2^{s-2r} - 1).\n$$\n\nSince $\\gcd(3, 2^{2r}) = 1$, $2^{2r} \\mid y^2$, so $2^r \\mid y$. Thus $x \\mid y$, but $0 \\le y < x$, so $y = 0$. Therefore, $a = b$.\n\n$\\boxed{a = b}$ is the only solution.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20166,
"subject": "Mathematics (Olympiad)",
"question": "A circle $\\omega$ with center $O$ is circumscribed about a right triangle $ABC$ ($\\angle A = 90^\\circ$). The tangent to it at $A$ passes through the point $P$ lying on the ray $CB$. Let $M$ denote the midpoint of the minor arc $AB$. The line $PM$ intersects $\\omega$ for the second time at $Q$. Let $X$ be the point on ray $PA$ such that $\\angle XCP = 90^\\circ$.\n\nProve that the line $XQ$ passes through the orthocenter of the triangle $ABO$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let us denote by $H$ the orthocenter of the triangle $ABO$, and by $K$ the point of intersection of the line $CA$ with the tangent to the circle at point $Q$ (see figure). In the 2016 Iranian Geometry Olympiad, it was shown that $\\angle PKC = 90^\\circ$. The proof is based on similarity:\n\n- From the similarity of triangles $\\triangle PMA \\sim \\triangle PAQ$, $\\triangle PMB \\sim \\triangle PCQ$ and $\\triangle PBA \\sim \\triangle PAC$, it follows\n\n$$\n\\frac{AQ}{MA} = \\frac{PQ}{PA}, \\quad \\frac{MB}{QC} = \\frac{PB}{PQ}, \\quad \\text{and} \\quad \\frac{AC}{BA} = \\frac{PA}{PB}\n$$\n\nAccordingly, taking into account the equality $MA = MA$, we obtain $\\frac{AQ}{QC} = \\frac{BA}{AC}$.\n\n- From the similarity of triangles $\\triangle KAQ \\sim \\triangle KQC$ and $\\triangle PBA \\sim \\triangle PAC$, it follows\n\n$$\n\\frac{KA}{KC} = \\left(\\frac{AQ}{QC}\\right)^2, \\quad \\text{and} \\quad \\frac{PB}{PC} = \\left(\\frac{BA}{AC}\\right)^2\n$$\n\nand so $\\frac{KA}{KC} = \\frac{PB}{PC}$, that is, $PK$ is parallel to $BA$.\n\nNext, to prove that the points $X$, $Q$ and $H$ lie on the same line, we can use projective geometry or inversion.\n\n**Method 1 (projective geometry).** On one hand, the pole of the line $PK$ lies on the line $OM$, and on the other, on the polar of the point $P$. Since the polar of the point $P$ is the perpendicular to $PC$ passing through $A$, then the pole of the line $PK$ is the point $H$. Note also that $Q$ is the pole of the line $KQ$, and $X$ is the pole of the line $KC$. Since the lines $PK$, $KQ$ and $KC$ intersect at one point, the points $X$, $Q$ and $H$ lie on the same line.\n\n**Method 2 (inversion).** It is easy to show that under inversion with respect to the circle $\\omega$, the lines $KC$, $KQ$ and $KP$ will turn into three circles passing through the center $O$ and the points $X, Q$ and $H$ respectively. Moreover, the centers of these circles lie on the same line, and the points $X, Q, H$ and the image of the point $K$ lie on a line parallel to it.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20167,
"subject": "Mathematics (Olympiad)",
"question": "Let $s$ be an integer greater than $6$. A solid cube of side $s$ has a square hole of side $x < 6$ drilled directly through from one face to the opposite face (so the drill removes a cuboid). The volume of the remaining solid is numerically equal to the total surface area of the remaining solid. Determine all possible integer values of $x$.",
"options": [],
"answer": "See solution",
"solution": "The volume of the solid is the volume of the cube minus the volume of the hole:\n\n$$\ns^3 - s x^2 = s(s + x)(s - x).\n$$\n\nThe surface area of the solid is the surface area of the cube, minus the surface area of the two missing squares, plus the surface area of the four interior faces created by the hole:\n\n$$\n6s^2 + 4s x - 2x^2 = 2(s + x)(3s - x).\n$$\n\nEquating these two expressions and cancelling $s + x$ (which is nonzero), we find:\n\n$$\n2(3s - x) = s(s - x)\n$$\nwhich rearranges to\n$$\nx = \\frac{s(s - 6)}{s - 2}.\n$$\n\nAs $x$ is an integer, $s - 2$ must divide $s(s - 6)$.\n\nSince $s$ and $s - 2$ differ by $2$, they can only share $\\pm1$ or $\\pm2$ as a factor. Since $s - 6$ and $s - 2$ differ by $4$, they can only share $\\pm1, \\pm2$, or $\\pm4$ as a factor.\n\nTherefore, $s - 2 = \\pm1, \\pm2, \\pm4$, or $\\pm8$. Since $s > 6$, we must have $s = 10$, and\n$$\nx = \\frac{10(10 - 6)}{10 - 2} = 5.\n$$\n\nWith these values of $s$ and $x$, the volume and surface area of the solid both have the numeric value $750$.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 20168,
"subject": "Mathematics (Olympiad)",
"question": "Let $p$ be a prime with $p \\geq 3$, and let $k$ be an odd number not divisible by $p$. Let $K$ be a finite field with $kp + 1$ elements, and let $A = \\{x_1, x_2, \\dots, x_t\\}$ be the set of elements in $K^* = K \\setminus \\{0\\}$ whose order is different from $k$ in the multiplicative group $(K^*, \\cdot)$. Prove that the polynomial\n$$\nP(X) = (X + x_1)(X + x_2)\\dots(X + x_t)\n$$\nhas at least $p$ coefficients equal to $1$.",
"options": [],
"answer": "See solution",
"solution": "Since $k$ and $p$ are odd, $|K|$ is even, so $K$ has characteristic $2$ and is a power of $2$. The elements $x_1, x_2, \\dots, x_t$ are the roots of $P$. Write $P(X) = \\sum_{j=0}^{t} a_j X^j$.\n\nThe multiplicative group $K^*$ is cyclic of order $kp$. Let $a \\in K^*$ be a generator. For any $s \\in \\{1, 2, \\dots, kp\\}$ with $\\gcd(s, kp) = 1$, we have $\\mathrm{ord}(a^s) = kp$.\n\nFor $i \\in \\{1, 2, \\dots, p-1\\}$, $\\gcd(ik + p, kp) = 1$, so $\\mathrm{ord}(a^{ik+p}) = kp$. Thus,\n$$\nP(a^{ik+p}) = 0, \\quad \\text{for } i = 1, \\dots, p-1.\n$$\n\nFor $r = 0, 1, \\dots, p-1$, define\n$$\nA_r = \\sum_{i=0}^{p-1} a^{-irk} P(a^{ik+p}).\n$$\nBy the above, $A_r = P(a^p)$, and since $\\mathrm{ord}(a^p) = k$, $a^p$ is not a root of $P$, so $A_r \\neq 0$.\n\nExpanding,\n$$\nA_r = \\sum_{i=0}^{p-1} a^{-irk} P(a^{ik+p}) = \\sum_{i=0}^{p-1} a^{-irk} \\sum_{j=0}^{t} a_j a^{ikj+pj} = \\sum_{j=0}^{t} a_j a^{pj} \\sum_{i=0}^{p-1} a^{k(j-r)i}.\n$$\n\nSince\n$$\n\\sum_{i=0}^{p-1} a^{kmi} = \\begin{cases} p, & \\text{if } p \\mid m, \\\\ 0, & \\text{if } p \\nmid m, \\end{cases}\n$$\nwe get\n$$\nA_r = p \\cdot \\sum_{j=0}^{\\lfloor \\frac{t-r}{p} \\rfloor} a_{pj+r} a^{p(pj+r)}.\n$$\n\nSince $A_r \\neq 0$, at least one of the coefficients $a_r, a_{p+r}, a_{2p+r}, \\dots$ is nonzero for each $r = 0, \\dots, p-1$.\n\nFor any divisor $d$ not dividing $kp$, let $U_d = \\{x \\in K^* \\mid \\mathrm{ord}(x) = d\\}$ and define\n$$\nQ_d = \\prod_{x \\in U_d} (X + x).\n$$\n\nLet $K_2 = \\{0, 1\\}$ be the prime subfield of $K$, and let $1 = d_1 < d_2 < \\dots < d_n = kp$ be the divisors of $kp$. Since $Q_1 = X + 1 \\in K_2[X]$ and\n$$\nQ_{d_i} = (X^{d_i} + 1) \\cdot \\left( \\prod_{d_j \\mid d_i, d_j \\neq d_i} Q_{d_j} \\right)^{-1} \\in K_2[X],\n$$\nall $Q_{d_i}$ have coefficients $0$ or $1$.\n\nTherefore, $P = (X^{kp} + 1) \\cdot Q_k^{-1} \\in K_2[X]$ also has all coefficients $0$ or $1$, and at least $p$ of them are nonzero, as required.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20169,
"subject": "Mathematics (Olympiad)",
"question": "Label the vertices as shown.\n\n\n\nLet $f(XY)$ denote the number on edge $XY$. According to the condition:\n\n$$\n\\begin{align*}\nf(BA) + f(AD) + f(DB) &= f(AD) + f(DC) + f(AC) \\\\\n&= f(BC) + f(CD) + f(BD) \\\\\n&= f(AB) + f(BC) + f(AC).\n\\end{align*}\n$$\n\nIs it possible to assign the numbers $1$ through $6$ to the edges so that the sum on each face is equal?",
"options": [],
"answer": "See solution",
"solution": "Notice that each edge is counted twice in the sum of the four faces. Therefore, the sum of the four faces equals $2(1+2+3+4+5+6) = 42$, which is not divisible by $4$, a contradiction.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20170,
"subject": "Mathematics (Olympiad)",
"question": "Let $a = 110$. Find the minimum and maximum possible values of $f(23) + f(2011)$, where $f$ is a positive integer-valued function defined on the positive integers, such that for all positive integers $x, y$,\n\n$$\n(x + y)f(x) \\leq x^2 + f(xy) + a.\n$$",
"options": [],
"answer": "See solution",
"solution": "We first show that a necessary and sufficient condition for $f$ to satisfy the given inequality is that\n\n$$\n(\\dagger) \\quad t - a \\leq f(t) \\leq t \\quad \\text{for any positive integer } t.\n$$\n\n**Necessity:**\n- Substitute $(x, y) = (s, 1)$:\n $$(s+1)f(s) \\leq s^2 + f(s) + a \\implies f(s) \\leq s + \\frac{a}{s}.$$\n- Substitute $(x, y) = (t, 2a)$ and use $f(2at) \\leq 2at + \\frac{1}{2t}$:\n $$(t + 2a)f(t) \\leq t^2 + f(2at) + a \\leq t^2 + 2at + \\frac{1}{2t} + a,$$\n $$f(t) \\leq t + \\frac{1}{2t(t + 2a)} + \\frac{a}{t + 2a} < t + 1.$$\n- Since $f$ is integer-valued, $f(t) \\leq t$.\n- Substitute $(x, y) = (1, t)$:\n $$(1 + t)f(1) \\leq 1 + f(t) + a \\implies t - a \\leq f(t).$$\n\n**Sufficiency:**\n- If $f$ satisfies $(\\dagger)$, then for any $x, y$:\n $$(x + y)f(x) \\leq (x + y)x = x^2 + (xy - a) + a \\leq x^2 + f(xy) + a.$$\n\nThus, $f$ satisfies the original inequality if and only if $(\\dagger)$ holds.\n\nNow, for $f(23)$ and $f(2011)$:\n- $1 \\leq f(23) \\leq 23$\n- $1901 = 2011 - 110 \\leq f(2011) \\leq 2011$\n\nTherefore,\n- Minimum: $f(23) + f(2011) = 1 + 1901 = 1902$\n- Maximum: $f(23) + f(2011) = 23 + 2011 = 2034$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20171,
"subject": "Mathematics (Olympiad)",
"question": "Given a fixed natural number $n > 1$, 2019 natural numbers are placed around a circle so that the product of any two neighboring numbers is a perfect $n$-th power. Is it always the case that the product of any (not necessarily neighboring) two numbers is also a perfect $n$-th power?",
"options": [],
"answer": "See solution",
"solution": "Yes, it is.\n\nFor odd $n = 2l + 1$:\n\n$$\nP^2 = (a_1 a_2)(a_2 a_3)(a_3 a_4) \\dots (a_{2018} a_{2019})(a_{2019} a_1) = m^{2l+1},\n$$\nwhere $P = a_1 a_2 \\dots a_{2019}$. Thus, $P = s^{2l+1}$.\n\nTherefore,\n$$\na_1 = \\frac{P}{(a_2 a_3)(a_4 a_5) \\dots (a_{2018} a_{2019})} = \\frac{s^{2l+1}}{u^{2l+1}} = v_1^{2l+1},\n$$\nand similarly for all $a_i = v_i^{2l+1}$. Thus,\n$$\na_i a_j = v_i^{2l+1} v_j^{2l+1} = v^{2l+1}.\n$$\n\nFor even $n = 2l$:\n\n$$\nP^2 = (a_1 a_2)(a_2 a_3)(a_3 a_4) \\dots (a_{2018} a_{2019})(a_{2019} a_1) = m^{2l},\n$$\nwhere $P = a_1 a_2 \\dots a_{2019}$. Thus, $P = m^l$.\n\nTherefore,\n$$\na_1^2 = \\frac{P^2}{(a_2 a_3)(a_4 a_5) \\dots (a_{2018} a_{2019})} = \\frac{m^{2l}}{u^{2l}} = v_1^{2l},\n$$\nand similarly for all $a_i^2 = v_i^{2l}$. Thus,\n$$\nr^{2l} = (a_1 a_2)(a_2 a_3)(a_3 a_4) \\dots (a_{k-1} a_k) = a_1 a_2^2 a_3^2 \\dots a_{k-1} a_k = a_1 a_k w^{2l} \\Rightarrow a_1 a_k = \\frac{r^{2l}}{w^{2l}} = z^{2l}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20172,
"subject": "Mathematics (Olympiad)",
"question": "Prove that the inequality $x^2 + y^2 + 1 \\ge 2(xy - x + y)$ holds for any two real $x$ and $y$. When does the equality hold?",
"options": [],
"answer": "See solution",
"solution": "Rewrite the inequality $x^2 + y^2 + 1 \\ge 2(xy - x + y)$ as $x^2 - 2xy + y^2 + 2x - 2y + 1 \\ge 0$. Rearranging the left-hand side gives $(x - y)^2 + 2(x - y) + 1 \\ge 0$, which is a perfect square. Thus, the inequality becomes $((x - y) + 1)^2 \\ge 0$. This holds for all real $x$ and $y$. Equality occurs if and only if $(x - y) + 1 = 0$, i.e., $x - y = -1$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20173,
"subject": "Mathematics (Olympiad)",
"question": "Suppose that the convex quadrilateral $ABCD$ satisfies $AB = BC$, $AD = DC$. $E$ is a point on $AB$, and $F$ on $AD$, such that $B, E, F, D$ are concyclic. Draw $\\triangle DPE$ directly similar to $\\triangle ADC$, and $\\triangle BQF$ directly similar to $\\triangle ABC$. Prove that $A, P, Q$ are collinear.\n\n\n\n\n\n\n\n",
"options": [],
"answer": "See solution",
"solution": "Denote by $O$ the center of the circle that passes through $B, E, F, D$. Draw lines $OB, OF, BD$.\n\nIn $\\triangle BDF$, $O$ is the circumcenter, so $\\angle BOF = 2\\angle BDA$; and $\\triangle ABD \\sim \\triangle CBD$, so $\\angle CDA = 2\\angle BDA$. Hence, $\\angle BOF = \\angle CDA = \\angle EPD$, which implies that the isosceles triangles\n\n$$\n\\triangle BOF \\sim \\triangle EPD. \\qquad \\textcircled{1}\n$$\n\nOn the other hand, the concyclicity of $B, E, F, D$ implies that\n\n$$\n\\triangle ABF \\sim \\triangle ADE. \\qquad \\textcircled{2}\n$$\n\nCombining ① and ②, we know that the quadrilateral $ABOF \\sim ADPE$, so\n\n$$\n\\angle BAO = \\angle DAP. \\qquad \\textcircled{3}\n$$\n\nThe same argument gives\n\n$$\n\\angle BAO = \\angle DAQ. \\qquad \\textcircled{4}\n$$\n\n③ and ④ imply that $A, P, Q$ are collinear.\n\n**Remark** In fact, when $ABCD$ is not a rhombus, the collinearity of $A, P, Q$ is equivalent to the concyclicity of $B, E, F, D$.\n\nThis can be explained in the following way. Fix the point $E$, and let the point $F$ move along the line $AD$. By similarity, the locus of $Q$ is a line that passes $P$. Now it suffices to show that this locus does not coincide with the line $AP$, i.e., to show that $A$ is not on the locus.\n\nTo this end, we draw $\\triangle BAA' \\sim \\triangle BQF \\sim \\triangle ABC$. Then $\\angle BAA' = \\angle ABC$, which implies that $A'A \\parallel BC$. As $ABCD$ is not a rhombus, $AD$ is not parallel to $BC$, which implies that $A', A, D$ are not collinear, i.e., $A$ is not on the locus.\n\nThe locus. Hence, $Q$ is on the line $AP$ only when $B, E, F, D$ are concyclic. And when $ABCD$ is a rhombus, for any $E$ and $F$ the corresponding points $P$ and $Q$ are always on the diagonal $AC$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20174,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute triangle, $H$ its orthocentre, and $M$ the midpoint of $BC$. Furthermore, let $k_1$ be the circle with diameter $AH$ and $k_2$ be the circle with center $M$ that touches the circumcircle of triangle $ABC$ interiorly. Prove that $k_1$ and $k_2$ are touching circles.",
"options": [],
"answer": "See solution",
"solution": "Let $N$ be the midpoint of $AH$ (and thus the center of $k_1$), and let $X$ be the reflection of $H$ about $M$. Then $X$ lies on the circumcircle of $ABC$, opposite to $A$. Since $OM$ and $AH$ are parallel, by the Intercept Theorem, we have $AH = 2OM$. Hence, $AN = OM$, so $ANMO$ is a parallelogram. Let $r_1$ and $r_2$ be the radii of $k_1$ and $k_2$, respectively, and let $R$ be the radius of the circumcircle of $ABC$. Then $R - r_2 = OM = AN = r_1$, and thus $r_1 + r_2 = R = AO = NM$. This means that the distance between the centers of $k_1$ and $k_2$ is the sum of their radii. Consequently, $k_1$ and $k_2$ touch each other.\n\nRemark: It is easy to show that the touching point of $k_1$ and $k_2$ lies on the bisector of $\\angle BAC$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20175,
"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": "Suppose, for contradiction, that $P(x)$ is not constant. First, consider $P(x)$ linear: $P(x) = ax + b$ with $a \\neq 0$. Let $Q(x) = a x^2 + (b+1)x$. Then\n\n$$\nP(Q(x)) = a(a x^2 + (b+1)x) + b = a^2 x^2 + a(b+1)x + b = (a x + b)(a x + 1)\n$$\n\nHere, both $a x + b$ and $a x + 1$ are non-constant, contradicting the condition.\n\nNow, suppose $\\deg P = n > 1$ and $P(x) = a_n x^n + a_{n-1} x^{n-1} + \\cdots + a_1 x + a_0$ with $a_n \\neq 0$. Let $Q(x) = P(x) + x$. Then\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\nSince $(P(x) + x)^i - x^i$ is divisible by $P(x)$ (by the formula $a^i - b^i = (a-b)(a^{i-1} + a^{i-2}b + \\cdots + b^{i-1})$), $P(Q(x))$ is divisible by $P(x)$. But $\\deg P(Q(x)) > \\deg P(x)$ for $n > 1$, so $P(x)$ is a non-trivial divisor of $P(Q(x))$, contradicting the condition. Thus, $P(x)$ must be constant. $\\square$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20176,
"subject": "Mathematics (Olympiad)",
"question": "As shown in the figure below, in the convex quadrilateral $ABCD$, $I$ and $J$ are the incenters of $\\triangle ABC$ and $\\triangle ADC$, respectively. It is known that $IJ$, $AC$, and $BD$ meet at $P$; the line through $P$ and perpendicular to $BD$ meets the exterior angle bisectors of $\\angle BAD$ and $\\angle BCD$ at $E$ and $F$, respectively. Prove that $PE = PF$.\n\n",
"options": [],
"answer": "See solution",
"solution": "If $AB \\parallel CD$ and $AD \\parallel BC$, then $ABCD$ is a parallelogram, $P$ is the midpoint of $AC$, and $AE \\parallel CF$, so $PE = PF$. In the following, assume $AB$ is not parallel to $CD$.\n\nFirst, we prove $AB + AD = CB + CD$. As illustrated in the figure below, let the extensions of $BA$ and $CD$ meet at $T$; the $B$-escribed circle $\\odot K$ of $\\triangle TBC$ touches the lines $AB$, $BC$, and $CD$ at $X$, $Y$, and $Z$, respectively. Since the internal homothetic center of $\\odot I$ and $\\odot J$ lies on the line $IJ$ and $AC$ is a common internal tangent of the two circles, it follows that the internal homothetic center of $\\odot I$ and $\\odot J$ is $P$.\n\nIt is well known that the internal homothetic center $P$ of $\\odot I$ and $\\odot J$, the external homothetic center $B$ of $\\odot I$ and $\\odot K$, and the internal homothetic center $G$ of $\\odot J$ and $\\odot K$ are collinear. Therefore, $G$ lies on the line $BP$. Furthermore, $CD$ is a common internal tangent of $\\odot J$ and $\\odot K$. We conclude that the internal homothetic center $G$ of $\\odot J$ and $\\odot K$ is $D$, and $AD$ is tangent to $\\odot K$ at $W$.\n\n\n\nBy the theorem of length of tangent,\n\n$$\n\\begin{aligned}\nAB + AD &= (BX - AX) + (AW - DW) \\\\\n&= BX - DW = BY - DZ = (CB + CY) - (CZ - CD) \\\\\n&= CB + CD,\n\\end{aligned}\n$$\n\nthat is,\n\n$$\nAB + AD = CB + CD. \\qquad \\textcircled{1}\n$$\n\nLet $\\odot U$ and $\\odot V$ be the $B$-escribed and $D$-escribed circles of $\\triangle ABD$, respectively. Let $\\odot Q$ and $\\odot R$ be the $B$-escribed and $D$-escribed circles of $\\triangle BCD$, respectively. We prove $UR$ and $VQ$ meet at $P$.\n\nConsider $\\odot K$, $\\odot U$, and $\\odot R$. It is well known that the external homothetic center $A$ of $\\odot K$ and $\\odot U$, the internal homothetic center $C$ of $\\odot K$ and $\\odot R$, and the internal homothetic center of $\\odot U$ and $\\odot R$ are collinear. As $BD$ is a common internal tangent of $\\odot U$ and $\\odot R$, we conclude that the internal homothetic center of $\\odot U$ and $\\odot R$ is $P$, so $UR$ passes through $P$; similarly, $VQ$ passes through $P$ as well.\n\nNow, we show $UQ \\perp BD$ and $VR \\perp BD$. As shown in the figure below, $\\odot U$ and $\\odot Q$ touch the extension of $BD$ at $L$ and $L'$, respectively. Clearly,\n\n$$\nBL = \\frac{1}{2}(AB + AD + BD),\n$$\n\n$$\nBL' = \\frac{1}{2}(CB + CD + BD).\n$$\n\n\n\nFrom (1), it follows that $BL = BL'$, so $L = L'$, and hence $UQ \\perp BD$. Likewise, $VR \\perp BD$. As $EF \\perp BD$, we arrive at $UQ \\parallel EF \\parallel VR$ and\n\n$$\n\\frac{PE}{UQ} = \\frac{VP}{VQ} = \\frac{RF}{RQ} = \\frac{PF}{UQ},\n$$\n\nand thus $PE = PF$.\n\n$\\boxed{}$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20177,
"subject": "Mathematics (Olympiad)",
"question": "Given a real number $\\alpha$, consider the function $\\varphi(x) = x^2 e^{\\alpha x}$ for all real numbers $x$. Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ that satisfy\n\n$$\nf(\\varphi(x) + f(y)) = y + \\varphi(f(x)), \\quad \\forall x, y \\in \\mathbb{R}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $\\varphi(f(0)) = c$. Clearly, $f$ is bijective because\n\n$$\nf(f(y)) = y + c.\n$$\n\nReplacing $y$ by $f(y)$ in the relation, we have\n\n$$\nf(y + c) = f(y) + c.\n$$\n\nSince $f$ is bijective, there exists $d \\in \\mathbb{R}$ such that $f(d) = 0$. Replacing $(x, y)$ by $(d, y + c)$, we get\n\n$$\nf(\\varphi(d) + f(y + c)) = y + c.\n$$\n\nSince $f$ is injective, this implies\n\n$$\n\\varphi(d) + f(y + c) = f(y),\n$$\nwhich means\n$$\n\\varphi(d) + c = 0.\n$$\n\nOn the other hand, since $\\varphi(x) \\ge 0$ and equality only occurs when $x = 0$, we have $f(0) = d = 0$.\n\nHence, $f(f(y)) = y$ and substituting $y = 0$ in the original equation gives\n\n$$\nf(\\varphi(x)) = \\varphi(f(x)).\n$$\n\nBut $\\varphi(x) \\ge 0$ for all $x$, so $f(t) \\ge 0$ for all $t \\ge 0$. Replacing $y$ by $f(y)$ and $\\varphi(x) = t \\ge 0$ for any $t \\ge 0$, we obtain\n\n$$\nf(y + t) = f(y) + f(t), \\quad \\forall t \\ge 0.\n$$\n\nTherefore, for all $x, y \\in \\mathbb{R}$ and $t \\ge \\max(-y, 0)$,\n\n$$\nf(x + y) + f(t) = f(x + y + t) = f(x) + f(y + t) = f(x) + f(y) + f(t),\n$$\n\nso $f$ is additive. Since $f(x) \\ge 0$ for all $x$, it follows that $f(x) = kx$ for some $k \\ge 0$. Substituting $f(x) = kx$ into the original equation, we find $k = 1$. Thus, $f(x) = x$ for all $x \\in \\mathbb{R}$.\n\n$\\boxed{f(x) = x}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20178,
"subject": "Mathematics (Olympiad)",
"question": "Given an integer $n \\ge 3$, determine the maximum value of the product of $n$ non-negative real numbers $x_1, x_2, \\dots, x_n$ subject to\n$$\n\\frac{x_1}{1 + x_1} + \\frac{x_2}{1 + x_2} + \\dots + \\frac{x_n}{1 + x_n} = 1.\n$$",
"options": [],
"answer": "See solution",
"solution": "The required maximum is $1/(n-1)^n$ and is achieved if and only if all $x_i$ are equal to $1/(n-1)$.\n\nThe constraint on the $x_i$ is equivalent to\n$$\n\\sum_{k=1}^{n} (k-1)\\sigma_k = 1,\n$$\nwhere\n$$\n\\sigma_k = \\sum_{1 \\le i_1 < \\dots < i_k \\le n} x_{i_1} \\cdots x_{i_k}, \\quad k = 1, 2, \\dots, n.\n$$\nBy the AM-GM inequality,\n$$\n\\sigma_k \\ge \\binom{n}{k} \\sigma_n^{k/n}, \\quad k = 1, 2, \\dots, n,\n$$\nso, upon substitution $t = \\sigma_n^{1/n}$,\n$$\n\\sum_{k=1}^{n} (k-1) \\binom{n}{k} t^k \\le 1;\n$$\nthat is, $(t+1)^{n-1}((n-1)t-1) \\le 0$. Consequently, $t \\le 1/(n-1)$. Equality clearly forces all the $x_i$ to be equal to $1/(n-1)$. Since these $x_i$ obey the constraint, the conclusion follows.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20179,
"subject": "Mathematics (Olympiad)",
"question": "In a table, there are 2009 coins, each with one side white and the other side black. At the beginning, all coins are aligned in a row with their white sides up, except for one coin, which has its black side up. In each step, we choose a coin with its black side up and flip its two adjacent coins. If we choose an outermost coin, we flip just its one adjacent coin. Find all positions of the black coin at the beginning from which we can reach the state where all coins have black sides up.",
"options": [],
"answer": "See solution",
"solution": "Number the coins from left to right as $1$ to $2009$. Let the initial black coin be at position $s$. \n\nWe show that if the black coin is at $s = 1005$, then it is possible to reach the state where all coins have black sides up. By choosing coins at positions $s$, $s-1$, and $s+1$ in sequence, all five coins from $s-2$ to $s+2$ become black side up. \n\nAssume all coins numbered \n$$\ns-k,\\ s-(k-1),\\ \\dots,\\ s-1,\\ s,\\ s+1,\\ \\dots,\\ s+(k-1),\\ s+k\n$$\nhave black sides up. If $k = 2\\ell$, proceed by choosing coins at positions\n$$\ns-2\\ell,\\ s+2\\ell,\\ s-2(\\ell-1),\\ s+2(\\ell-1),\\ s-2,\\ s+2,\\ s\n$$\nand if $k = 2\\ell-1$, choose\n$$\ns-(2\\ell-1),\\ s+(2\\ell-1),\\ s-(2\\ell-3),\\ s+(2\\ell-3),\\ s-1,\\ s+1\n$$\nso that all coins from $s-k-1$ to $s+k+1$ become black side up.\n\nNow, suppose the coins with black sides up are at positions $T_1, T_2, \\dots, T_k$ in some state. Define\n$$\nM = \\sum_{i=1}^{k} (-1)^{i+1} T_i = T_1 - T_2 + T_3 - \\dots + (-1)^{k-1} T_k.\n$$\nIf we choose the coin at position $T_j$ (with $T_j \\neq 1, 2009$), and $T_{j-1} \\neq T_j - 1$, $T_{j+1} \\neq T_j + 1$, then after the move,\n$$\nM' = \\sum_{i=1}^{j-1} (-1)^{i+1} T_i + (-1)^{j+1} (T_j - 1) + (-1)^{j+2} T_j + (-1)^{j+3} (T_j + 1) + \\sum_{i=j+1}^{k} (-1)^{i+1} T_i,\n$$\nwhich equals $M$. For other cases, $M$ remains invariant in each step. If $j = 1$, $M$ changes by $-M$; if $j = 2009$, $M$ changes by $M + (-1)^{k+1}2010$. Therefore, the remainder $r$ or $2010 - r$ of $M$ divided by $2010$ is invariant under the procedure.\n\nThus, only when the initial black coin is at position $1005$ (the center), can we reach the state where all coins have black sides up.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20180,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer such that $p = 17^{2n} + 4$ is prime. Show that $7^{(p-1)/2} + 1$ is divisible by $p$.",
"options": [],
"answer": "See solution",
"solution": "Write $p = 17^{2n} + 4 \\equiv 4^n + 4 \\pmod{5}$ to deduce that $p \\equiv 0 \\pmod{5}$ if $n$ is even. Since $p$ is prime, $n$ must be $0$, so $p = 5$ and the conclusion follows.\n\nHenceforth, assume $n$ is odd. Rule out the case $n \\equiv 5 \\pmod{6}$ on account of $p = 17^{2n} + 4 \\equiv 3^n + 4 \\pmod{13} \\equiv 0 \\pmod{13}$.\n\nIn the remaining cases, $n \\equiv 1$ or $3 \\pmod{6}$, write $p = 17^{2n} + 4 \\equiv 2^n + 4 \\pmod{7}$ to infer\n$p \\equiv 5$ or $6 \\pmod{7}$, both of which are quadratic nonresidues modulo $7$; that is, $\\left(\\frac{p}{7}\\right) = -1$.\n\nConsequently, $\\left(\\frac{7}{p}\\right) = -1$ by quadratic reciprocity, so $7^{(p-1)/2} \\equiv \\left(\\frac{7}{p}\\right) \\equiv -1 \\pmod{p}$. This ends the proof.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20181,
"subject": "Mathematics (Olympiad)",
"question": "At a round table, several inhabitants of the island \"Loud Mouths\" are sitting at equal distances from one another. 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, each person said that the islander sitting directly opposite them, together with that person's two neighbors, are not all of the same type (i.e., not all three knights and not all three liars).\n\nWhat is the minimum and the maximum number of liars who can sit at the table, if in total there are:\n\na) 2020 islanders;\n\nb) 600 islanders?\n\n\n",
"options": [],
"answer": "See solution",
"solution": "If a knight says \"the islander opposite me and their two neighbors are not all of one type,\" it means that among those three, there are either 1 knight and 2 liars, or 2 knights and 1 liar. If a liar says this, it is false, so among those three, all are either knights or all liars.\n\n**Lemma 1.** There is at least one knight among any two diametrically opposite islanders.\n\n*Proof.* Suppose two opposite islanders, $C$ and $X$, are both liars. Then the neighbors of $X$ (call them $Y$ and $W$) must also be liars, since $C$'s statement is a lie, meaning the three opposite $C$ are all of one type. Similarly, both neighbors of $C$ ($B$ and $D$) must be liars. This creates more opposite pairs of liars, and repeating the argument leads to all islanders being liars, which contradicts the problem statement. Thus, the lemma is proved.\n\n**Lemma 2.** There is at least one liar among any five islanders sitting in a row.\n\n*Proof.* Suppose five knights sit in a row: $L_1, \\dots, L_5$. The people opposite $L_2, L_3, L_4$ must be liars, since otherwise the statement about three of the same type would be true. But then $L_3$ would be lying, which is a contradiction. Thus, the lemma is proved.\n\nFrom Lemma 2, there cannot be a group of more than 4 knights in a row. Similarly, by further analysis:\n\n**Lemma 3.** There cannot be a group of fewer than three knights.\n\n*Proof.* Consider a knight $L$ with $X, Y, Z$ sitting opposite. Among $X, Y, Z$ there is at least one liar. If the liar is opposite $L$, then $A, L, B$ (where $A$ and $B$ are $L$'s neighbors) are all knights, so at least three knights together. If the liar is opposite $A$, then $A$ and its two neighbors (including $L$) are all knights. Thus, every knight is in a group of at least three knights.\n\n**Lemma 4.** There can be no group with more than two liars in a row.\n\n*Proof.* If there are three liars in a row, then five people sitting opposite the first and last of these liars would form a group of five knights, which is impossible by Lemma 2. Thus, the lemma is proved.\n\nNow, consider the possible groupings:\n- A group of 4 knights sits opposite a group of 2 liars.\n- A group of 3 knights sits opposite a single liar, and vice versa.\n\nLet $2N$ be the total number of islanders. Suppose there are $m$ groups of $2+4$ (2 liars, 4 knights) and $k$ groups of $1+3$ (1 liar, 3 knights). Then:\n$$\n6m + 4k = 2N\n$$\nThe number of liars is $2m + k$.\n\n**a) $2N = 2020$**\n\n$6m \\leq 2020 \\implies m \\leq 336$. For $m=336$, $k=1$, so the maximum number of liars is $2 \\times 336 + 1 = 673$.\n\nFor the minimum, $m=0$, $k=505$, so the minimum number of liars is $0 + 505 = 505$.\n\nBoth arrangements are possible by symmetry arguments.\n\n**b) $2N = 600$**\n\n$6m \\leq 600 \\implies m \\leq 100$. For $m=98$, $k=3$, so the maximum number of liars is $2 \\times 98 + 3 = 199$.\n\nFor the minimum, $m=2$, $k=147$, so the minimum number of liars is $2 \\times 2 + 147 = 151$.\n\nThus, the answers are:\n\na) minimum 505, maximum 673;\nb) minimum 151, maximum 199.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 20182,
"subject": "Mathematics (Olympiad)",
"question": "Una configuración es un conjunto finito $S$ de puntos del plano entre los cuales no hay tres colineales y a cada punto se le asigna algún color, de modo que si un triángulo cuyos vértices están en $S$ tiene un ángulo mayor o igual a $120^\\circ$, entonces exactamente dos de sus vértices son de un mismo color. Hallar el número máximo de puntos que puede tener una configuración.",
"options": [],
"answer": "See solution",
"solution": "El número máximo de puntos que puede tener una configuración es $25$.\n\nPrimero observamos que dados $6$ puntos del plano, entre los cuales no hay tres colineales, se determina al menos un triángulo con un ángulo de medida mayor o igual a $120^\\circ$. En efecto, si la envolvente convexa de los puntos es un hexágono, su mayor ángulo tiene medida de al menos $120^\\circ$. En otro caso, hay un punto $A$ interior al triángulo determinado por otros tres puntos $B, C, D$ y uno de los ángulos $\\angle BAC$, $\\angle CAD$, $\\angle DAB$ tiene medida mayor o igual a $120^\\circ$.\n\nPara los puntos de la configuración se usan a lo sumo $5$ colores. Si hay más de $5$ colores, elegimos $6$ puntos con colores diferentes y les aplicamos la observación. Esto nos conduce a un triángulo con un ángulo de medida mayor o igual a $120^\\circ$ y con vértices coloreados con $3$ colores diferentes, lo cual es imposible.\n\nSupongamos ahora que hay una configuración con más de $25$ puntos. Estos están coloreados con a lo sumo $5$ colores, luego algún color se repite al menos $6$ veces. Le aplicamos la observación a los $6$ puntos con dicho color. La conclusión ahora es que hay un triángulo con un ángulo de medida mayor o igual a $120^\\circ$ y vértices monocromáticos, de nuevo una contradicción.\n\nFalta construir un ejemplo de una configuración de $25$ puntos. Tomamos un pentágono regular $P$ de lado muy grande. En cada uno de sus vértices ubicamos un pentágono regular muy pequeño, con centro en los respectivos vértices de $P$. Sea $S$ el conjunto de vértices de los $5$ pentágonos pequeños. Si fuese necesario, pequeñas rotaciones alrededor de sus centros pueden asegurarnos que no haya entre los $25$ puntos de $S$ tres colineales. Coloreamos todos los vértices de cada pentágono pequeño con uno de $5$ colores distintos. Afirmamos que $S$ es una configuración. En efecto, triángulos del mismo color sólo aparecen dentro de los pentágonos regulares pequeños. El ángulo máximo determinado por tres vértices de un pentágono regular tiene medida $108^\\circ$, la que es menor de $120^\\circ$. Tomamos ahora un triángulo $ABC$ con vértices de $3$ colores distintos. Los vértices $A$, $B$, $C$ provienen de diferentes pentágonos pequeños y están en una vecindad muy próxima a tres vértices $A'$, $B'$, $C'$ del pentágono principal $P$. Como $P$ es muy grande, los ángulos de $ABC$ difieren muy poco de los respectivos ángulos de $A'B'C'$, que tienen medidas menores o iguales a $108^\\circ$. Precisamente, eligiendo el tamaño de $P$ tan grande como sea necesario, uno puede estar seguro de que la desviación es tan pequeña como queramos. En particular, ningún ángulo del triángulo $ABC$ con vértices de $3$ colores distintos tendrá medida mayor o igual a $120^\\circ$, lo que completa el objetivo.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20183,
"subject": "Mathematics (Olympiad)",
"question": "There are six boxes $B_1, B_2, B_3, B_4, B_5, B_6$, each of which may contain any number of coins (possibly zero). You may repeatedly perform the following operation: choose an integer $k$ with $1 \\leq k \\leq 4$, remove one coin from each of the boxes $B_k$ and $B_{k+1}$ (if both are nonempty), and add two coins to the box $B_{k+2}$. The boxes $B_{k+1}$ and $B_{k+2}$ may be empty. \n\nDetermine whether there is a finite sequence of such operations that results in the boxes $B_1, B_2, B_3, B_4, B_5$ being empty and the box $B_6$ containing exactly $2010^{2010^{2010}}$ coins. (Note that $a^{b^c} = a^{(b^c)}$.)\n",
"options": [],
"answer": "See solution",
"solution": "The answer is affirmative.\n\nLet $A = 2010^{2010^{2010}}$. Denote by\n\n$$\n(b_i, b_{i+1}, \\dots, b_{i+k}) \\to (b'_i, b'_{i+1}, \\dots, b'_{i+k})\n$$\n\nto mean that there is a finite sequence of operations on the boxes $B_i, B_{i+1}, \\dots, B_{i+k}$ initially containing $b_i, b_{i+1}, \\dots, b_{i+k}$ coins respectively that results in containing $b'_i, b'_{i+1}, \\dots, b'_{i+k}$ coins respectively. Then we shall show that\n\n$$\n(1, 1, 1, 1, 1, 1) \\to (0, 0, 0, 0, 0, A).\n$$\n\n**Lemma 1** For every positive integer $a$, we have $(a, 0, 0) \\to (0, 2^a, 0)$.\n\n*Proof of Lemma 1:* We prove by induction on $k \\leq a$ that $(a, 0, 0) \\to (a-k, 2^k, 0)$.\n\nSince $(a, 0, 0) \\to (a-1, 2, 0) \\to (a-1, 2^1, 0)$, the assertion is true for $k=1$. Suppose that the assertion is true for some $k < a$; then\n\n$$\n\\begin{aligned}\n(a-k, 2^k, 0) &\\to (a-k, 2^k-1, 2) \\to \\dots \\\\\n&\\to (a-k, 0, 2^{k+1}) \\to (a-k-1, 2^{k+1}, 0),\n\\end{aligned}\n$$\n\nand thus\n\n$$\n(a, 0, 0) \\to (a-k, 2^k, 0) \\to (a-k-1, 2^{k+1}, 0).\n$$\n\nThe assertion is also true for $k+1 \\leq a$. By induction, Lemma 1 is proven.\n\n**Lemma 2** For every positive integer $a$, we have $(a, 0, 0, 0) \\to (0, P_a, 0, 0)$, where $P_n = 2^{2^{n^2}}$ (with $n$ 2's) for a positive integer $n$.\n\n*Proof of Lemma 2:* We prove by induction on $k \\leq a$ that $(a, 0, 0, 0) \\to (a-k, P_k, 0, 0)$.\n\nBy the operation of type 1, we have\n\n$$\n(a, 0, 0, 0) \\to (a-2, 2, 0, 0) = (a-1, P_1, 0, 0),\n$$\n\nand the assertion is true for $k=1$. Suppose that the assertion is true for some $k < a$; then\n\n$$\n\\begin{aligned}\n(a-k, P_k, 0, 0) &\\to (a-k, 0, 2^{P_k}, 0) \\\\\n&= (a-k, 0, P_{k+1}, 0) \\to (a-k-1, P_{k+1}, 0, 0),\n\\end{aligned}\n$$\n\nand therefore\n\n$$\n(a, 0, 0, 0) \\to (a-k, P_k, 0, 0) \\to (a-k-1, P_{k+1}, 0, 0),\n$$\n\ni.e., the assertion is also true for $k+1 \\leq a$. By induction, Lemma 2 is proven.\n\nWe have\n\n$$\n\\begin{aligned}\n(1, 1, 1, 1, 1) &\\to (1, 1, 1, 1, 0) \\to (1, 1, 1, 0, 3, 0) \\\\\n&\\to (1, 1, 0, 3, 0, 0) \\to (1, 0, 3, 0, 0, 0) \\to (0, 3, 0, 0, 0, 0) \\\\\n&\\to (0, 0, P_3, 0, 0, 0) = (0, 0, 16, 0, 0, 0) \\to (0, 0, 0, P_{16}, 0, 0),\n\\end{aligned}\n$$\n\nand\n\n$$\n\\begin{aligned}\nA &= 2010^{2010^{2010}} < (2^{11})^{2010^{2010}} \\\\\n&= 2^{11 \\times 2010^{2010}} < 2^{2010^{2011}} < 2^{(2^{11})^{2011}} \\\\\n&= 2^{2^{11 \\times 2011}} < 2^{2^{15}} < P_{16},\n\\end{aligned}\n$$\n\nand thus the number of coins in the box $B_4$ is greater than $A$. Therefore, by performing operations of type 2, we have\n\n$$\n\\begin{aligned}\n(0, 0, 0, P_{16}, 0, 0) &\\to (0, 0, 0, P_{16}-1, 0, 0) \\\\\n&\\to (0, 0, 0, P_{16}-2, 0, 0)\n\\end{aligned}\n$$\n\n$$\n\\rightarrow \\cdots \\rightarrow \\left(0, 0, 0, \\frac{A}{4}, 0, 0\\right).\n$$\n\nThen we get\n\n$$\n\\left(0, 0, 0, \\frac{A}{4}, 0, 0\\right) \\rightarrow \\left(0, 0, 0, 0, \\frac{A}{2}, 0\\right) \\rightarrow \\left(0, 0, 0, 0, 0, A\\right).\n$$\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20184,
"subject": "Mathematics (Olympiad)",
"question": "Find all triples $(a, b, p)$ of positive integers with $p$ prime such that\n\n$$\na^p = b! + p.\n$$",
"options": [],
"answer": "See solution",
"solution": "There are only two such triples: $(2, 2, 2)$ and $(3, 4, 3)$. It is straightforward to check them. We shall prove that no other triples exist. Clearly, $a > 1$. Consider three situations as follows.\n\n1. **Case $a < p$:**\n - If $a \\le b$, then $a \\mid (a^p - b!) = p$, contradicting the assumption $1 < a < p$.\n - If $a > b$, then $b! \\le a! < a^p - p$, the second inequality only requiring $p > a > 1$.\n\n2. **Case $a > p$:**\n - Then $b! = a^p - p > p^p - p \\ge p!$, so $b > p$, and $a^p = b! + p$ is a multiple of $p$.\n - As $b! = a^p - p$, we have $p \\mid b$, and $b < 2p$.\n - If $a < p^2$, then $a/p$ divides $a^p$ and $b!$, hence divides $p$ as well, contradicting $1 < a/p < p$.\n - If $a \\ge p^2$, then $a^p \\ge (p^2)^p > (2p-1)! + p \\ge b! + p$, which is a contradiction.\n\n3. **Case $a = p$:**\n - Then $b! = p^p - p$. Try $p = 2, 3, 5$ to get the two triples $(2, 2, 2)$ and $(3, 4, 3)$.\n - Now assume $p \\ge 7$. From $b! = p^p - p > p!$, it follows $b \\ge p + 1$, and further\n\n $$\n \\begin{align*}\n v_2((p+1)!) &\\le v_2(b!) \\\\\n &= v_2(p^{p-1}-1) = 2v_2(p-1) + v_2(p+1)-1 \\\\\n &= v_2\\left(\\frac{p-1}{2} \\cdot (p-1) \\cdot (p+1)\\right).\n \\end{align*}\n $$\n\n - Since $\\frac{p-1}{2}$, $p-1$, $p+1$ are distinct factors of $(p+1)!$ and $p+1 \\ge 8$, there are four or more even numbers among $1, 2, \\dots, p+1$, which is impossible.\n\n - **Second approach for $a = p \\ge 5$:** By Zsigmondy's theorem, there exists a prime $q$ that divides $p^{p-1}-1$ but not $p^k-1$ for any $k < p-1$.\n - Thus, $p \\neq q$, and $q \\equiv 1 \\pmod{p-1}$. We must have $b \\ge 2p-1$, yet\n\n $$\n b! \\ge (2p-1)! > (2p-1) \\cdot (2p-2) \\cdots (p+1) \\cdot p > p^p > p^p - p\n $$\n\n leads to a contradiction.\n\n - **Third approach for $a = p \\ge 5$:** As $b > p \\ge 5$, the required equation modulo $(p+1)^2$ gives\n\n $$\n \\begin{aligned}\n p^p - p &= (p+1-1)^p - p \\\\\n &\\equiv p \\cdot (p+1)(-1)^{p-1} + (-1)^p - p \\\\\n &= p^2 - 1 \\not\\equiv 0 \\pmod{(p+1)^2}.\n \\end{aligned}\n $$\n\n However, as $p \\ge 5$, $2, \\frac{p+1}{2} < p$ are distinct, and $(p+1) \\mid p!$. It follows that\n\n $$\n (p+1)^2 \\mid (p+1)!\n $$\n\n which is contrary to $(p+1)^2 \\nmid b!$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20185,
"subject": "Mathematics (Olympiad)",
"question": "A bee is moving in three-dimensional space. A fair six-sided die with faces labeled $A^+$, $A^-$, $B^+$, $B^-$, $C^+$, and $C^-$ is rolled. Suppose the bee occupies the point $(a, b, c)$. If the die shows $A^+$, then the bee moves to the point $(a+1, b, c)$, and if the die shows $A^-$, then the bee moves to the point $(a-1, b, c)$. Analogous moves are made with the other four outcomes. Suppose the bee starts at the point $(0, 0, 0)$ and the die is rolled four times. What is the probability that the bee traverses four distinct edges of some unit cube?\n\n(A) $\\frac{1}{54}$ (B) $\\frac{7}{54}$ (C) $\\frac{1}{6}$ (D) $\\frac{5}{18}$ (E) $\\frac{2}{5}$",
"options": [],
"answer": "See solution",
"solution": "**Answer (B):** Without loss of generality, assume that the first roll is $A^+$. In order for the bee to traverse four distinct edges of a cube, the second roll cannot be $A^+$ or $A^-$, so there are $4$ rolls ($B^+$, $B^-$, $C^+$, and $C^-$) that are allowed at this stage. Each new roll must represent a perpendicular direction for the bee, and there are $3$ choices that remain in compliance for the third roll—$2$ of which extend into three dimensions, and $1$ of which creates a “C” shape. In the former case, there are $2$ choices for the fourth roll, while in the latter case, there are $3$ choices (including the one where the bee traverses four edges forming a square). In total, the number of compliant paths is $6 \\cdot 4 \\cdot (2 \\cdot 2 + 1 \\cdot 3) = 2^3 \\cdot 3 \\cdot 7$. The total number of paths is $6^4 = 2^4 \\cdot 3^4$, and the probability that the path represents exactly four edges of a unit cube is\n\n$$\n\\frac{2^3 \\cdot 3 \\cdot 7}{2^4 \\cdot 3^4} = \\frac{7}{2 \\cdot 3^3} = \\frac{7}{54}.\n$$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20186,
"subject": "Mathematics (Olympiad)",
"question": "On the real axis, a bug stands at coordinate $x = 1$. At each step, from position $x = a$, the bug can jump to either $x = a + 2$ or $x = \\frac{a}{2}$. Show that there are precisely $F_{n+4} - (n + 4)$ positions (including the initial one) that the bug can reach in at most $n$ steps.",
"options": [],
"answer": "See solution",
"solution": "Let $M$ be the set of real numbers of the form $\\frac{a}{2^b}$, where $a$ is an odd positive integer and $b$ is a non-negative integer. These are all the positions the bug can reach.\n\nFor each $x \\in M$, let $f(x)$ be the minimum number of steps needed to jump from $1$ to $x$. Clearly, $f(x)$ is well-defined and finite, since for $x_0 = \\frac{a}{2^b} \\in M$, the bug can reach $x_0$ after $\\frac{a-1}{2}$ steps of type \"+2\" and $b$ steps of type \"/2\", so the number of steps does not exceed $\\frac{a-1}{2} + b$.\n\nFor every positive integer $n$, define\n\n$$\nP_n = \\{x \\mid x \\in M \\wedge f(x) = n\\}, \\quad Q_n = \\{x \\mid x \\in P_n \\wedge \\frac{x}{2} \\in P_{n+1}\\}\n$$\n\nand $R_n = P_n \\setminus Q_n$. Note that $Q_n$ is the set of $x$ such that $\\frac{x}{2}$ can only be obtained after $n+1$ steps, and $R_n$ is the set of $x$ such that $\\frac{x}{2} \\in P_n$. By definition, the total number of distinct positions the bug can reach in $n$ steps is\n\n$$\n|P_0| + |P_1| + \\dots + |P_n|.\n$$\n\nWe now find a formula for $|P_n|$. Observe:\n\n- If the bug needs $n$ steps to go from $1$ to $x$, then to reach $x + 2$ from $1$ requires $n + 1$ steps.\n- If the bug needs $n$ steps to go from $1$ to $\\frac{x}{2}$, then to reach $x$ from $1$ also requires $n$ steps.\n\nThus, for $x \\in M$ with $f(x) = n$, we have $f(x+2) = n+1$. Therefore,\n\n$$\n|P_{n+1}| = |P_n| + |Q_n|, \\quad \\forall n \\in \\mathbb{Z}^+.\n$$\n\nNext, we show that $x \\in P_n$ if and only if $x + 4 \\in R_{n+2}$.\n\nIndeed, for $x_0 \\in P_n$, $x_0 + 4 \\in P_{n+2}$, and $\\frac{x_0 + 4}{2} = \\frac{x_0}{2} + 2$, which can be reached in $n + 2$ steps, so $x_0 + 4 \\in R_{n+2}$. Conversely, if $x_0 + 4 \\in R_{n+2}$, then $\\frac{x_0}{2} + 2 \\in P_{n+2} \\implies \\frac{x_0}{2} \\in P_{n+1} \\implies x_0 \\in P_n$. Thus, $|P_n| = |R_{n+2}|$.\n\nSo,\n\n$$\n|P_n| = |Q_n| + |R_n| = |P_{n+1}| - |P_n| + |P_{n-2}|,\n$$\n\nwhich gives $|P_{n+1}| = 2|P_n| - |P_{n-2}|$ for $n \\ge 2$. Note that $|P_0| = 1$, $|P_1| = 2$, $|P_2| = 4$. If we let $u_n = |P_{n+1}| - |P_n|$, then $u_n = u_{n-1} + u_{n-2}$ with $u_0 = 1$, $u_1 = 2$, so $u_n = F_{n+2}$, where $(F_n)$ is the Fibonacci sequence. Thus, $|P_n| = F_{n+2} - 1$ for all $n \\ge 0$.\n\nFinally, we need to show that\n\n$$\n\\sum_{k=0}^{n} (F_{k+2} - 1) = F_{n+4} - (n+4).\n$$\n\nThis can be proved by induction using properties of the Fibonacci sequence. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20187,
"subject": "Mathematics (Olympiad)",
"question": "In a math contest, 50 students participate and 3 problems are given to the contestants. It is known that each student solved at least one problem, and the total number of correct solutions is 100. Prove that at most 25 students solved all three problems.",
"options": [],
"answer": "See solution",
"solution": "Let $a$, $b$, and $c$ be the number of students who solved exactly one, two, and three problems, respectively. Then:\n\n$$a + b + c = 50$$\n$$a + 2b + 3c = 100$$\n\nSubtracting the first equation from the second gives:\n\n$$(a + 2b + 3c) - (a + b + c) = 100 - 50$$\n$$b + 2c = 50$$\n\nThus, $2c \\leq 50$, so $c \\leq 25$.\n\nTherefore, at most 25 students solved all three problems.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20188,
"subject": "Mathematics (Olympiad)",
"question": "Solve the equation $x + y^2 + (\text{gcd}(x, y))^2 = x y \times \text{gcd}(x, y)$ in the set of natural numbers.",
"options": [],
"answer": "See solution",
"solution": "Let $z = \text{gcd}(x, y)$. Then $x = a z$ and $y = b z$ for some natural numbers $a, b$ with $\text{gcd}(a, b) = 1$. Substituting, the equation becomes:\n\n$$\na z + (b z)^2 + z^2 = a b z^3\n$$\n\nwhich simplifies to:\n\n$$\na + b^2 z + z = a b z^2\n$$\n\nSince the right side is divisible by $z$, $a$ must be divisible by $z$, so let $a = c z$ for some $c \\in \\mathbb{N}$. Substituting:\n\n$$\nc z + b^2 z + z = c b z^3\n$$\n\nor\n\n$$\nc + b^2 + 1 = c b z^2\n$$\n\nThus,\n\n$$\nb^2 + 1 = c (b z^2 - 1)\n$$\n\nSince $b z^2 \\neq 1$, we have $c = \\frac{b^2 + 1}{b z^2 - 1}$.\n\nMultiplying by $z^2$:\n\n$$\nc z^2 = \\frac{b^2 z^2 + z^2}{b z^2 - 1} = b + \\frac{b + z^2}{b z^2 - 1}\n$$\n\nFor $c z^2$ to be natural, $\\frac{b + z^2}{b z^2 - 1}$ must be natural, so $b z^2 - 1 \\leq b + z^2$:\n\n$$\n(z^2 - 1)(b - 1) \\leq 2 \\tag{1}\n$$\n\nCase $b = 1$: $c = \\frac{2}{z^2 - 1}$, so $z^2 = 2$ or $3$, impossible.\n\nCase $b = 2$: $c = \\frac{5}{2 z^2 - 1}$. If $2 z^2 - 1 = 1$, $z = 1$, $c = 5$, $a = 5$, so $x = 5$, $y = 2$.\n\nCase $b = 3$: $c = \\frac{10}{3 z^2 - 1}$. If $3 z^2 - 1 = 2$, $z = 1$, $c = 5$, $a = 5$, so $x = 5$, $y = 3$.\n\nFor $b > 3$, (1) implies $z = 1$, but then $c = \\frac{b^2 + 1}{b - 1} = b + 1 + \\frac{2}{b - 1}$, which only works for $b = 2$ or $3$, contradicting $b > 3$.\n\nTherefore, the solutions are $(x, y) = (5, 2)$ and $(5, 3)$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20189,
"subject": "Mathematics (Olympiad)",
"question": "The inscribed circle $\\omega$ of triangle $ABC$ touches its sides $AB$, $BC$, and $CA$ at points $K$, $L$, and $M$, respectively. On the arc $KL$ of the circle $\\omega$ that does not contain the point $M$, a point $S$ is chosen. Let $P$, $Q$, $R$, $T$ be the points of intersection of the lines $AS$ and $KM$, $ML$ and $SC$, $LP$ and $KQ$, $AQ$ and $PC$, respectively. If the points $R$, $S$, and $M$ are collinear, prove that $T$ also belongs to the line $SM$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Consider triangle $SLM$. $LC$ and $MC$ are the tangent lines to the circumscribed circle of this triangle drawn at points $L$ and $M$; therefore, $SC$ is a symmedian, and so $\\frac{MQ}{CL} = \\frac{MS^2}{SL^2}$.\n\nSimilarly, $\\frac{KP}{PM} = \\frac{KS^2}{SM^2}$. Let $E$ be the point of intersection of the lines $KL$ and $MR$. Since the lines $ME$, $KQ$, and $LP$ are concurrent, by Ceva's theorem we have $\\frac{KP}{PM} \\cdot \\frac{MQ}{QL} \\cdot \\frac{LE}{EK} = 1$, hence applying the above equalities we obtain $\\frac{LE}{EK} = \\frac{SL^2}{KS^2}$. This implies that the line $SE$ is a symmedian of triangle $SKL$, therefore it passes through the point of intersection of the tangents to the circumscribed circle of triangle $KSL$ drawn at points $L$ and $K$, that is, through point $B$ (see figure). So, we have proved that points $B$, $S$, $R$, and $M$ are collinear.\n\nNext, we will prove that point $T$ also belongs to this line. By Ceva's theorem, it is sufficient to prove that $\\frac{AP}{\\sin \\angle AMP} = \\frac{AM}{\\sin \\angle APM}$, $\\frac{PS}{\\sin \\angle KMB} = \\frac{MS}{\\sin \\angle SPM}$. Dividing the last two equalities, we obtain $\\frac{AP}{PS} = \\frac{AM \\sin \\angle AMK}{MS \\sin \\angle KMB}$. Similarly, $\\frac{CQ}{QS} = \\frac{MC \\sin \\angle CML}{MS \\sin \\angle LMB}$. Substitute the last two relations into the equality that we need to prove:\n\n$$\n\\frac{AP}{PS} \\cdot \\frac{SQ}{QC} \\cdot \\frac{CM}{MA} = \\frac{AP}{PS} = \\frac{AM \\sin \\angle AMK}{MS \\sin \\angle KMB} \\cdot \\frac{MS \\sin \\angle LMB}{MC \\sin \\angle CML} \\cdot \\frac{CM}{MA} = \\frac{\\sin \\angle AMK \\cdot \\sin \\angle LMB}{\\sin \\angle KMB \\cdot \\sin \\angle CML}\n$$\n\nBy the sine theorem for triangles $BLM$ and $BMK$, we obtain:\n\n$$\n\\frac{BL}{BM} = \\frac{\\sin \\angle LMB}{\\sin \\angle BLM} = \\frac{\\sin \\angle LMB}{\\sin \\angle MLC} = \\frac{\\sin \\angle LMB}{\\sin \\angle CML}, \\text{ since triangle } MLC \\text{ is isosceles. Similarly, } \\frac{BK}{BM} = \\frac{\\sin \\angle KMB}{\\sin \\angle AMK}\n$$\n\nTherefore,\n\n$$\n\\frac{AP}{PS} \\cdot \\frac{SQ}{QC} \\cdot \\frac{CM}{MA} = \\frac{\\sin \\angle AMK \\cdot \\sin \\angle LMB}{\\sin \\angle KMB \\cdot \\sin \\angle CML} = \\frac{BM}{BK} \\cdot \\frac{BL}{BM} = 1,\n$$\n\nwhich completes the solution.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20190,
"subject": "Mathematics (Olympiad)",
"question": "It is given that the bookshelf can fit 9 of the same thick books, but the 10th one will not fit anymore. Similarly, it can hold 15 of the same thin books, but the 16th will not fit anymore. Is it possible for that shelf to hold simultaneously:\n\n- a) 6 thick and 5 thin books?\n- b) 7 thick and 5 thin books?",
"options": [],
"answer": "See solution",
"solution": "Let the length of the shelf be $S$, the width of a thick book be $x$, and the width of a thin book be $y$. Then:\n\n$$\n9x \\leq S < 10x \\quad \\text{and} \\quad 15y \\leq S < 16y\n$$\n\nThis implies:\n$$\n\\frac{1}{10}S < x \\leq \\frac{1}{9}S \\quad \\text{and} \\quad \\frac{1}{16}S < y \\leq \\frac{1}{15}S\n$$\n\na) For 6 thick and 5 thin books:\n$$\n6x + 5y \\leq \\frac{6}{9}S + \\frac{5}{15}S = S\n$$\nSo, it is possible to fit 6 thick and 5 thin books.\n\nb) For 7 thick and 5 thin books:\n$$\n7x + 5y > \\frac{7}{10}S + \\frac{5}{16}S = \\frac{56 + 25}{80}S = \\frac{81}{80}S > S\n$$\nSo, it is not possible to fit 7 thick and 5 thin books.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20191,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be a triangle and let $D$, $E$, $F$ be the midpoints of the sides $BC$, $CA$, and $AB$, respectively. Prove that $\\angle DAC = \\angle ABE$ if and only if $\\angle AFC = \\angle BDA$.",
"options": [],
"answer": "See solution",
"solution": "Let $G$ be the centroid of the triangle. Since $DF \\parallel AC$, we have $\\angle DAC = \\angle GDF$.\n\nIf $\\angle DAC = \\angle ABE$, then $\\angle GDF = \\angle FBG$, so the quadrilateral $BFGD$ is cyclic, which implies $\\angle AFC = \\angle BDA$. The converse follows by the same argument.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20192,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute-angled triangle with $BC > AB$, such that the points $A$, $H$, $I$, and $C$ are concyclic (where $H$ is the orthocenter and $I$ is the incenter of triangle $ABC$). The line $AC$ intersects the circumcircle of triangle $BHC$ at point $T$, and the line $BC$ intersects the circumcircle of triangle $AHC$ at point $P$. If the lines $PT$ and $HI$ are parallel, determine the measures of the angles of triangle $ABC$.",
"options": [],
"answer": "See solution",
"solution": "Let lines $CH$ and $PT$ meet at $U$ and denote by $A'$, $B'$, $C'$ the feet of the altitudes of triangle $ABC$.\n\n$AHIC$ is a cyclic quadrilateral, so $\\angle CHI = \\angle CAI = \\frac{\\angle A}{2}$. Since $PT \\parallel HI$, it follows that $\\angle HUT = \\angle CHI = \\frac{\\angle A}{2}$.\n\nHence, $\\angle CUP = \\angle HUT = \\frac{\\angle A}{2}$. (1)\n\nThe quadrilateral $AHIC$ is also cyclic, so $\\angle A'HI = \\angle ACI = \\frac{\\angle C}{2}$.\n\nFrom triangle $A'HC$ we have $\\angle A'CH = 90^\\circ - \\angle A'HC = 90^\\circ - \\frac{\\angle A + \\angle C}{2} = \\frac{\\angle B}{2}$. From the right-angled triangle $BCC'$, we get $\\angle B + \\frac{\\angle B}{2} = 90^\\circ$, so $\\angle B = 60^\\circ$.\n\nNotice that $\\angle ABB' = \\angle ACC' = 90^\\circ - \\angle A$. The quadrilateral $BCTH$ is cyclic, so $\\angle B'TB = \\angle TCH = 90^\\circ - \\angle A = \\angle ABB'$.\n\nTherefore, $BB'$ is an angle bisector and also an altitude for triangle $ABT$, so $AB = BT$. Since $AHIC$ is cyclic, it follows that $\\angle BPI = \\angle CAI = \\frac{\\angle A}{2}$ and $\\angle API = \\angle ACI = \\frac{\\angle C}{2}$. Hence, $\\angle APB = \\frac{\\angle A + \\angle C}{2} = 60^\\circ = \\angle B$, so triangle $ABP$ is equilateral.\n\nThereby, $BP = AB = BT$, so triangle $BPT$ is isosceles, hence $\\angle BPT = \\frac{180^\\circ - \\angle PBT}{2}$. Triangle $ABT$ is also isosceles, so $\\angle ABT = 180^\\circ - 2\\angle A$. Noticing that $\\angle PBT = \\angle B - \\angle ABT$, it results that $\\angle PBT = 2\\angle A - 120^\\circ$, which leads to $\\angle BPT = 150^\\circ - \\angle A$ (2).\n\nFrom the exterior angle theorem we infer that $\\angle BPT = \\angle CUP + \\angle PCU$. Using (1) and (2), we obtain $150^\\circ - \\angle A = \\frac{\\angle A}{2} + \\frac{\\angle B}{2} = \\frac{\\angle A}{2} + 30^\\circ$, so $\\angle A = 80^\\circ$ and $\\angle C = 40^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20193,
"subject": "Mathematics (Olympiad)",
"question": "Given a positive integer $n$, determine the minimum of\n$$\n\\max \\left\\{ \\frac{x_1}{1 + x_1}, \\frac{x_2}{1 + x_1 + x_2}, \\dots, \\frac{x_n}{1 + x_1 + x_2 + \\dots + x_n} \\right\\},\n$$\nas $x_1, x_2, \\dots, x_n$ run through all non-negative real numbers which add up to $1$.",
"options": [],
"answer": "See solution",
"solution": "Let\n$$\nf(x_1, x_2, \\dots, x_n) = \\max \\left\\{ \\frac{x_k}{1 + x_1 + x_2 + \\dots + x_k} : k = 1, 2, \\dots, n \\right\\},\n$$\nwhere $x_1 \\ge 0, x_2 \\ge 0, \\dots, x_n \\ge 0$ and $x_1 + x_2 + \\dots + x_n = 1$. Notice that\n$$\n\\frac{a_1}{1 + a_1} = \\frac{a_2}{1 + a_1 + a_2} = \\dots = \\frac{a_n}{1 + a_1 + a_2 + \\dots + a_n}\n$$\nfor a unique $n$-tuple $(a_1, a_2, \\ldots, a_n)$ of non-negative real numbers which add up to $1$, namely, $a_k = 2^{k/n} - 2^{(k-1)/n}$ for $k = 1, 2, \\ldots, n$. In this case,\n$$\nf(a_1, a_2, \\ldots, a_n) = 1 - 2^{-1/n}.\n$$\nNow let $(x_1, x_2, \\ldots, x_n) \\neq (a_1, a_2, \\ldots, a_n)$, where the $x_i$ are non-negative real numbers which add up to $1$. Since $x_1 + x_2 + \\cdots + x_n = a_1 + a_2 + \\cdots + a_n$, it follows that $x_k > a_k$ for some index $k$. Let $m = \\min\\{k : x_k > a_k\\}$. Then $x_k \\le a_k$ for $k < m$, so\n$$\nf(x_1, x_2, \\dots, x_n) \\ge \\frac{x_m}{1 + x_1 + x_2 + \\dots + x_m} > \\frac{a_m}{1 + a_1 + a_2 + \\dots + a_m} = f(a_1, a_2, \\dots, a_n).\n$$\nConsequently, the required minimum is $1 - 2^{-1/n}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20194,
"subject": "Mathematics (Olympiad)",
"question": "Let the abscissae of the points $A$, $B$, $C$, $D$, $F$, and $G$ be $a$, $b$, $c$, $d$, $f$, and $g$ respectively. The coordinates of the points $A$, $B$, $C$, and $D$ satisfy the system:\n\n$$\n\\begin{cases}\n(x - \\alpha)^2 + (y - \\beta)^2 = R^2, \\\\\n y = \\frac{1}{x}\n\\end{cases}\n$$\n\nwhere $O_1(\\alpha, \\beta)$ is the center of the circle $S_1$. Eliminating $y$ gives:\n\n$$\nx^4 - 2\\alpha x^3 + (\\alpha^2 + \\beta^2 - R^2)x^2 - 2\\beta x + 1 = 0\n$$\n\nwith solutions $a$, $b$, $c$, and $d$. Similarly, if $O_2(\\lambda, \\mu)$ is the center of the circle $S_2$, then $a + b + f + g = 2\\lambda$ and $abfg = 1$. The radii of $S_1$ and $S_2$ are equal. Show that the quadrilateral $FGCD$ is a parallelogram with center at the origin.",
"options": [],
"answer": "See solution",
"solution": "Since $a + b + c + d = 2\\alpha$ and $abcd = 1$, and $a + b + f + g = 2\\lambda$ and $abfg = 1$, the midpoints of $AB$ and $O_1O_2$ coincide, so $a + b = \\alpha + \\lambda$. Thus,\n\n$$\n2\\alpha + 2\\lambda = (a + b + c + d) + (a + b + f + g) = 2\\alpha + 2\\lambda + c + d + f + g\n$$\n\nwhich implies $f + g = -(c + d)$. Also, $cd = fg = (ab)^{-1}$. Therefore, the pairs $(f, g)$ and $(-c, -d)$ have equal sums and products, so either $f = -c$, $g = -d$ or $f = -d$, $g = -c$. Thus, the points $F, C$ and $G, D$ are symmetric with respect to the origin, and $FGCD$ is a parallelogram centered at the origin.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20195,
"subject": "Mathematics (Olympiad)",
"question": "Suppose the numbers $1, 2, \\dots, n$ are written on the vertices of a regular $n$-gon such that for any three consecutive numbers $(a, b, c)$, we have $a + c \\equiv 2b \\pmod{n}$. How many such arrangements (permutations) are there?",
"options": [],
"answer": "See solution",
"solution": "The answer is $\\phi(n)$.\n\n**Lemma.** If $a, b, c$ are three consecutive numbers, then $a + c \\equiv 2b \\pmod{n}$.\n\n*Proof.* If $a + c \\not\\equiv 2b \\pmod{n}$, then there should be $d \\in \\{a, b, c\\}$ such that $a + c \\equiv b + d \\pmod{n}$, but $ac$ intersects with $bd$, which is a contradiction.\n\n**Lemma.** Suppose that the numbers $1, 2, \\dots, n$ are written on the vertices of a regular $n$-gon such that for any 3 consecutive numbers $(a, b, c)$, we have $a + c \\equiv 2b \\pmod{n}$. Then the numbers satisfy the problem condition.\n\n*Proof.* Let $a_0, a_1, \\dots, a_{n-1}$ be the numbers assigned to the vertices in clockwise order such that $a_0 = 1$. Let $a_1 = a$. Using the previous lemma and induction, one can see that $a_i = i(a-1) + 1$. Now, if $(a_i, a_j)$ intersects $(a_k, a_l)$, we have $k + l \\not\\equiv i + j \\pmod{n}$, hence\n\n$$\na_k + a_l \\equiv (k + l)(a - 1) + 2 \\not\\equiv (i + j)(a - 1) + 2 \\equiv a_i + a_j \\pmod{n}$$\n\nAs we see in the proof of the second lemma, $a$ determines the permutation. So it is necessary and sufficient that $\\{i(a-1) + 1 : 0 \\le i \\le n-1\\}$ forms a complete set of residues modulo $n$. This happens if and only if $\\gcd(a, n) = 1$. Therefore, the number of desired permutations is $\\phi(n)$.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20196,
"subject": "Mathematics (Olympiad)",
"question": "Нехай для деякого парного числа $k$ і натурального числа $l$ виконується рівність $$(k-3)(k-1)(k+1)(k+3) = l^3,$$ тобто $$(k^2-9)(k^2-1) = l^3.$$ Дослідіть, чи існують такі парні $k$ і натуральні $l$, що ця рівність виконується.",
"options": [],
"answer": "See solution",
"solution": "Якщо числа $k^2-9$ і $k^2-1$ мають спільний простий дільник, то він є дільником їхньої різниці, а тому дорівнює $2$. Але числа $k^2-9$ і $k^2-1$ непарні, отже вони взаємно прості і кожне з них є точним кубом. Відтак, $k^2-9 = n^3$ та $k^2-1 = m^3$, де $m, n \\in \\mathbb{N}$. Звідси $m^3-n^3=8$, тобто $$(m-n)(m^2+mn+n^2) = 8.$$ Оскільки $m-n < m^2+mn+n^2$, то можливі лише два випадки:\n\n$$\n\\begin{cases} m-n=1, \\\\ m^2+mn+n^2=8; \\end{cases} \\text{або} \\begin{cases} m-n=2, \\\\ m^2+mn+n^2=4. \\end{cases}\n$$\n\nПерша система не має розв'язків у натуральних числах, бо коли $m-n=1$, то $m$ і $n$ мають різну парність, а тоді $m^2+mn+n^2$ — непарне число.\n\nДруга система також не має розв'язків у натуральних числах, бо коли $m-n=2$, то $m \\geq 3$, а тоді $m^2+mn+n^2 > m^2 \\geq 9$. Отже, вказаний добуток не може бути точним кубом.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20197,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a convex quadrilateral inscribed in a circle $(O, R)$. With centers at the vertices of the quadrilateral and radius $R$, we draw circles $C_A(A, R)$, $C_B(B, R)$, $C_C(C, R)$, $C_D(D, R)$. Circles $C_A$ and $C_B$ meet at $K$, circles $C_B$ and $C_C$ meet at $L$, circles $C_C$ and $C_D$ meet at $M$, and circles $C_D$ and $C_A$ meet at $N$. (Points $K, L, M, N$ are the second common points of the corresponding circles, given that all of them pass through point $O$.) Prove that the quadrilateral $KLMN$ is a parallelogram.",
"options": [],
"answer": "See solution",
"solution": "The line segment $AB$ connects the centers of the circles $C_A$ and $C_B$, and therefore it is the perpendicular bisector of the common chord $OK$. Since the circles $C_A$ and $C_B$ have the same radius, the quadrilateral $AOBK$ is a rhombus. Thus, point $K_1$ is the midpoint of $AB$.\n\n\n\nSimilarly, we can show that $L_1$ is the midpoint of $BC$, $M_1$ is the midpoint of $CD$, and $N_1$ is the midpoint of $AD$.\n\nFrom the triangles $OKL$, $OLM$, $OMN$, and $ONK$, we conclude that: $KL \\parallel K_1L_1$, $LM \\parallel L_1M_1$, $MN \\parallel M_1N_1$, and $NK \\parallel N_1K_1$ (because the line segments $K_1L_1$, $L_1M_1$, $M_1N_1$, and $N_1K_1$ connect the midpoints of the sides of a quadrilateral).\n\nHence, the quadrilaterals $KLMN$ and $K_1L_1M_1N_1$ have their sides parallel. But we know that the midpoints of the sides of a quadrilateral define a parallelogram. So the quadrilateral $KLMN$ is a parallelogram.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20198,
"subject": "Mathematics (Olympiad)",
"question": "Given a positive integer $k$, show that there exists a prime $p$ such that one can choose distinct integers $a_1, a_2, \\dots, a_{k+3} \\in \\{1, 2, \\dots, p-1\\}$ such that $p$ divides $a_i a_{i+1} a_{i+2} a_{i+3} - i$ for all $i = 1, 2, 3, \\dots, k$.",
"options": [],
"answer": "See solution",
"solution": "First, choose distinct positive rational numbers $r_1, \\dots, r_{k+3}$ such that\n$$\nr_i r_{i+1} r_{i+2} r_{i+3} = i \\quad \\text{for } 1 \\le i \\le k\n$$\nLet $r_1 = x$, $r_2 = y$, $r_3 = z$ be some distinct primes greater than $k$; the remaining terms satisfy $r_4 = \\frac{1}{r_1 r_2 r_3}$ and $r_{i+4} = \\frac{i+1}{i} r_i$. It follows that if $r_i$ are represented as irreducible fractions, the numerators are divisible by $x$ for $i \\equiv 1 \\pmod{4}$, by $y$ for $i \\equiv 2 \\pmod{4}$, by $z$ for $i \\equiv 3 \\pmod{4}$, and by none for $i \\equiv 0 \\pmod{4}$. Notice that $r_i < r_{i+4}$; thus the sequences\n$$\nr_1 < r_5 < r_9 < \\dots, \\quad r_2 < r_6 < r_{10} < \\dots, \\quad r_3 < r_7 < r_{11} < \\dots, \\quad r_4 < r_8 < r_{12} < \\dots\n$$\nare increasing and have no common terms, that is, all $r_i$ are distinct.\n\nIf each $r_i$ is represented by an irreducible fraction $\\frac{u_i}{v_i}$, choose a prime $p$ which divides neither $v_i$ ($1 \\le i \\le k+1$), nor $v_i v_j (r_i - r_j) = v_j u_i - v_i u_j$ for $i < j$, and define $a_i$ by the congruence $a_i v_i \\equiv u_i \\pmod{p}$. Since $r_i r_{i+1} r_{i+2} r_{i+3} = i$, we have\n$$\n\\begin{aligned}\niv_i v_{i+1} v_{i+2} v_{i+3} &= r_i v_i \\cdot r_{i+1} v_{i+1} \\cdot r_{i+2} v_{i+2} \\cdot r_{i+3} v_{i+3} \\\\\n&= u_i u_{i+1} u_{i+2} u_{i+3} \\equiv a_i v_i a_{i+1} v_{i+1} a_{i+2} v_{i+2} a_{i+3} v_{i+3} \\pmod{p}\n\\end{aligned}\n$$\nand therefore $a_i a_{i+1} a_{i+2} a_{i+3} \\equiv i \\pmod{p}$ for $1 \\le i \\le k$.\n\nIf $a_i \\equiv a_j \\pmod{p}$, then $u_i v_j \\equiv a_i v_j v_j \\equiv u_j v_i \\pmod{p}$, a contradiction. $\\square$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20199,
"subject": "Mathematics (Olympiad)",
"question": "A rectangle covers 6 whole triangles and 4 half triangles. What is the area of the rectangle, and what is the area of one triangle if the triangle's area is half that of the rectangle?",
"options": [],
"answer": "See solution",
"solution": "The area of the rectangle is $6 + 4 \\times \\frac{1}{2} = 8$ units.\n\nThe area of one triangle is half that of the rectangle, so $\\frac{8}{2} = 4$ units.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20200,
"subject": "Mathematics (Olympiad)",
"question": "In how many ways can 6 juniors and 6 seniors form 3 disjoint teams of 4 people so that each team has 2 juniors and 2 seniors?\n\n(A) 720 (B) 1350 (C) 2700 (D) 3280 (E) 8100",
"options": [],
"answer": "See solution",
"solution": "Select the first junior and call this person A. There are 5 other juniors and $\\binom{6}{2} = 15$ pairs of seniors who could team up with A. Select the next junior not yet on a team, say B. There are 3 other juniors and $\\binom{4}{2} = 6$ pairs of seniors who could team up with B. The third team consists of the people not yet chosen. Thus there are $5 \\cdot 15 \\cdot 3 \\cdot 6 = 1350$ ways to form the teams.\n\n**OR**\n\nThere are $\\binom{6}{2,2,2} = \\frac{6!}{2!2!2!} = 90$ ways to assign the seniors to teams 1, 2, and 3, and, similarly, 90 ways to assign juniors to those teams. But there are $3! = 6$ ways for the three teams to be ordered, so the number of ways to form the teams is $\\frac{90 \\cdot 90}{6} = 1350$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20201,
"subject": "Mathematics (Olympiad)",
"question": "Let $k \\neq 0$ be an integer. Prove that the number of ordered pairs $(x, y)$ of integers satisfying\n$$\nk = \\frac{x^2 - xy + 2y^2}{x + y}\n$$\nis odd if and only if $k$ is divisible by $7$.",
"options": [],
"answer": "See solution",
"solution": "Multiplying both sides by $x + y$ yields\n$$\nx^2 - xy + 2y^2 = k(x + y). \\quad (1)\n$$\nAny solution $(x, y)$ to the original equation is a solution to (1), but (1) can have extra solutions satisfying $x + y = 0$, i.e., $y = -x$.\n\nA pair $(x, -x)$ is a solution to (1) if and only if $x^2 + x^2 + 2x^2 = k \\cdot 0$, that is $x = 0$. Equation (1) therefore has exactly one more solution than the original one, and it suffices to prove that equation (1) has an even number of integer solutions if and only if $7 \\mid k$.\n\nWe rewrite (1) as a quadratic equation\n$$\nx^2 - x(y + k) + 2y^2 - ky = 0 \\quad (2)\n$$\nin $x$. Its discriminant satisfies\n$$\n\\begin{align}\nD(y) &= (y + k)^2 - 4(2y^2 - ky) \\\\\n &= k^2 + 6ky - 7y^2 \\\\\n &= (k - y)(k + 7y) \\\\\n &= -7\\left(y - \\frac{3}{7}k\\right)^2 + \\frac{16}{7}k^2\n\\end{align}\n$$\nwhich is, for every $k$, a quadratic function in $y$ bounded from above. Therefore, for any integer $k$, the discriminant $D(y)$ is non-negative for only finitely many integers $y$ and equation (2) has only finitely many integer solutions $(x, y)$.\n\nIf $D(y) > 0$ for some integer $y$, the equation (2) has precisely two real solutions that can only be integer simultaneously, for their sum $y + k$ is an integer. For any such $y$ we get an even number of solutions to (2).\n\nWe see that $D(y) = 0$ either for $y = k$ or for $y = -\\frac{1}{7}k$. In the first case, equation (2) reduces to $(x - k)^2 = 0$ with double root $x = k$ and the equation (1) has only one solution $(k, k)$ with $y = k$. In the second case, $y$ is integer if and only if $k$ is divisible by $7$ and then equation (2) has a double root $x = \\frac{3}{7}k$, therefore $(\\frac{3}{7}k, -\\frac{1}{7}k)$ is the only solution to equation (1) with $y = -\\frac{1}{7}k$. Moreover, the two solutions $(k, k)$ and $(\\frac{3}{7}k, -\\frac{1}{7}k)$ are different as $k \\neq 0$.\n\nWe see that equation (1) has an even number of integer solutions if and only if $k$ is divisible by $7$. We conclude.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20202,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be the number of lines in a generic configuration in the plane. Let $b_k$ denote the number of bounded $k$-gons (faces with $k$ sides) in the arrangement. Prove that for some $k$ in $\\{3, 4, 5\\}$, at least $\\frac{1}{12}(n-3)(n-2)$ $k$-gons are pairwise non-adjacent (i.e., no two share an edge).",
"options": [],
"answer": "See solution",
"solution": "$$\n2n^2 \\geq \\sum_{k=2}^{m-1} (k-m)f_k + \\frac{1}{2}m(n^2+n+2),\n$$\nwherefrom\n$$\n\\sum_{k=2}^{m-1} (m-k)f_k \\geq \\frac{1}{2}m(n^2+n+2) - 2n^2 = \\frac{1}{2}((m-4)n^2 + mn + 2m).\n$$\nNext, write $f_k = b_k + u_k$, where $b_k$ and $u_k$ are the numbers of bounded and unbounded $k$-faces, respectively. Thus, $b_k$ is the number of $k$-gons. Clearly, $b_2 = 0$ and $\\sum_{k=2}^{n} u_k = 2n$, so\n$$\n\\begin{align*}\n\\sum_{k=2}^{m-1} (m-k)f_k &= \\sum_{k=3}^{m-1} (m-k)b_k + \\sum_{k=2}^{m-1} (m-k)u_k \\\\ &\\leq \\sum_{k=3}^{m-1} (m-k)b_k + (m-2) \\sum_{k=3}^{m-1} u_k \\\\ &\\leq \\sum_{k=3}^{m-1} (m-k)b_k + (m-2) \\sum_{k=2}^{n} u_k = \\sum_{k=3}^{m-1} (m-k)b_k + 2(m-2)n.\n\\end{align*}\n$$\nConsequently,\n$$\n\\begin{align*}\n\\sum_{k=3}^{m-1} (m-k)b_k &\\geq \\sum_{k=2}^{m-1} (m-k)f_k - 2(m-2)n \\\\ &\\geq \\frac{1}{2}((m-4)n^2 + mn + 2m) - 2(m-2)n \\\\ &= \\frac{1}{2}((m-4)n^2 - (3m-8)n + 2m) \\\\ &= \\frac{1}{2}((m-4)n - m)(n-2),\n\\end{align*}\n$$\nso\n$$\n\\max(b_3, \\dots, b_{m-1}) \\geq \\frac{1}{2} \\cdot \\frac{((m-4)n - m)(n-2)}{(m-3) + \\dots + 1} = \\frac{((m-4)n - m)(n-2)}{(m-3)(m-2)}.\n$$\nThe coefficient of $n^2$ in the lower bound is maximized at $m=5$ and $m=6$, so\n$$\n\\max(b_3, b_4) \\geq \\frac{1}{6}(n-5)(n-2) \\quad \\text{and} \\quad \\max(b_3, b_4, b_5) \\geq \\frac{1}{6}(n-3)(n-2).\n$$\nSince each vertex has an even degree (namely, 4), faces are 2-colourable; that is, they can be coloured one of two colours, so that no adjacent faces bear the same colour. Consequently, for some $k$ in $\\{3, 4, 5\\}$, at least $\\frac{1}{2} \\max(b_3, b_4, b_5) \\geq \\frac{1}{12}(n-3)(n-2)$ $k$-gons bear the same colour, and are therefore non-adjacent, as required.\n\n**Appendix.**\n\nFor completeness, here are three different proofs that the faces of the geometric plane graph associated with a generic line configuration in a plane are 2-colourable.\n\n*1st Proof.* Induct on the number of lines. The base case, a single line configuration, is clear. For the induction step, remove one of the lines and consider a valid face 2-colouring of the remaining configuration. Finally, include that line back and swap face colours in one of the half-planes it determines to get a valid face 2-colouring of the initial configuration.\n\n*2nd Proof.* By duality: Cofaces all have an even number of edges, so cocycles all have an even length, and hence the dual graph is bipartite; that is, covertices are 2-colourable.\n\n*3rd Proof (sketch).* Add the degree $2n$ vertex at infinity — alternatively, invert from some point outside the plane. This preserves planarity and the number of edges and faces, respectively. The graph is connected and vertex degrees are all even, so it is Eulerian: it has an Eulerian circuit, i.e., one that visits every edge exactly once (allowing for revisiting vertices).\n\nUse the fact that an Eulerian graph embedded in the plane has an Eulerian circuit that never crosses itself. A *crossing* in an Eulerian circuit $K$ of a plane graph consists of four edges $e, e', f, f'$ at a single vertex $v$ such that $e'$ follows $e$ and $f'$ follows $f$ on $K$, and $\\{e, e'\\}$ alternates with $\\{f, f'\\}$ in the cyclic rotation of edges incident to $v$ in the embedding. A *non-crossing* circuit has no such quadruple of edges. (To show that every Eulerian planar graph has a non-crossing Eulerian circuit, consider one with the fewest crossings.) If each vertex $v$ splits into $\\frac{1}{2} \\deg v$ vertices, each inheriting two edges incident to $v$ that are consecutive in $K$, then the embedding of a non-crossing Eulerian circuit becomes a simple closed curve (a Jordan curve) in the plane.\n\nBack to the goal, let $\\gamma$ be such a simple closed curve. One part of the faces of the initial graph merged to form the interior of $\\gamma$, and the other part merged to form the exterior of $\\gamma$. By the Jordan curve theorem, adjacent initial faces lie on opposite sides of $\\gamma$, so both sides consist of pairwise non-adjacent former faces. Since one of these sides contains at least half of the initial faces, the conclusion follows.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20203,
"subject": "Mathematics (Olympiad)",
"question": "Let $x_1, x_2$ be distinct roots of the equation $a x^2 + b x + c = 0$, i.e., the zeroes of the function $f(x) = a x^2 + b x + c$. Let $g(x) = x^3 + b x^2 + a x + c$. Given that $f(0) = g(0) = c \\ne 0$. Define $F(x) = g(x) - f(x)$. Show that $0, x_1, x_2$ are the distinct zeroes of the polynomial $F(x)$, and find the values of $a$, $b$, and $c$.",
"options": [],
"answer": "See solution",
"solution": "Let $F(x) = g(x) - f(x) = x^3 + b x^2 + a x + c - (a x^2 + b x + c) = x^3 + (b - a) x^2 + (a - b) x$. Since $F(x)$ has roots at $0, x_1, x_2$, we can write $F(x) = x(x - x_1)(x - x_2) = x(x^2 - (x_1 + x_2)x + x_1 x_2) = x^3 - (x_1 + x_2)x^2 + x_1 x_2 x$. Comparing coefficients:\n\n- $x^3$: $1$\n- $x^2$: $b - a = - (x_1 + x_2)$\n- $x$: $a - b = x_1 x_2$\n\nFrom $a x^2 + b x + c = 0$, the sum and product of roots are $x_1 + x_2 = -b/a$, $x_1 x_2 = c/a$.\n\nSo:\n\n$$\nb - a = - ( - \\frac{b}{a} ) = \\frac{b}{a} \\implies b - a = \\frac{b}{a}\n$$\n\n$$\na - b = \\frac{c}{a}\n$$\n\nAlso, $f(0) = c \\ne 0$ and $g(0) = c \\ne 0$.\n\nSolving $b - a = \\frac{b}{a}$:\n\n$$\nb - a = \\frac{b}{a} \\implies (b - a)a = b \\implies a b - a^2 = b \\implies a b - b = a^2 \\implies b(a - 1) = a^2 \\implies b = \\frac{a^2}{a - 1}\n$$\n\nFrom $a - b = \\frac{c}{a}$:\n\n$$\na - b = \\frac{c}{a} \\implies a^2 - b a = c\n$$\n\nSubstitute $b$ from above:\n\n$$\na^2 - \\left(\\frac{a^2}{a - 1}\\right)a = c \\implies a^2 - \\frac{a^3}{a - 1} = c\n$$\n\n$$\na^2 - \\frac{a^3}{a - 1} = \\frac{a^2(a - 1) - a^3}{a - 1} = \\frac{a^3 - a^2 - a^3}{a - 1} = -\\frac{a^2}{a - 1}\n$$\n\nSo $c = -\\frac{a^2}{a - 1}$.\n\nBut $c \\ne 0$, so $a \\ne 0, 1$.\n\nLet $a = 2$:\n\n- $b = \\frac{2^2}{2 - 1} = 4$\n- $c = -\\frac{2^2}{2 - 1} = -4$\n\nThus, $a = 2$, $b = 4$, $c = -4$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20204,
"subject": "Mathematics (Olympiad)",
"question": "We know that for some natural number $n$, the number $n^2 + 2008n$ written in decimal notation ends with 4. Find the digit in the ten's place of this number.",
"options": [],
"answer": "See solution",
"solution": "It's clear that the term $2000n$ does not influence the last two digits, so the relevant digits for $A = n^2 + 2008n$ are the same as for $B = n^2 + 8n$. \n\nConsider $B + 16 = n^2 + 8n + 16 = (n + 4)^2$. Since $(n + 4)^2$ is a perfect square, if $B$ ends with 84, then $(n + 4)^2$ ends with 00. Thus, $B = \\overline{X00} - 16 = \\overline{Y84}$, where $X, Y$ are some natural numbers. Therefore, the last two digits of the number are 84, so the digit in the ten's place is $8$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20205,
"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": "\n\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 $\\frac{AX}{XA_1} = \\frac{AP}{A_1P}$. 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 $\\frac{AY}{YA_1} = \\frac{AP}{A_1P} = \\frac{AX}{XA_1}$. 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": 20206,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute-angled triangle with orthocenter $H$, centroid $G$, and circumcircle $\\omega$. Let $D$ and $M$ be the intersections of lines $AH$ and $AG$ with side $BC$, respectively. Rays $MH$ and $DG$ intersect $\\omega$ again at $P$ and $Q$, respectively. Prove that $PD$ and $QM$ intersect on $\\omega$.",
"options": [],
"answer": "See solution",
"solution": "Note that it is enough to prove that $\\angle DPA + \\angle MQA = 180^\\circ$.\n\nWithout loss of generality, assume that $AB < AC$. Let the reflection of $H$ in point $M$ be $H'$. Since $BHCH'$ is a parallelogram, we get\n\n$$\n\\angle BH'C = \\angle BHC = 180^\\circ - \\angle BAC\n$$\n\nwhich means $H'$ lies on $\\omega$. Also, we get\n\n$$\n\\angle ABH' = \\angle ABC + \\angle CBH' = \\angle ABC + \\angle BCH = 90^\\circ\n$$\n\nsince $CH \\perp AB$. This means $AH'$ is the diameter of $\\omega$. Thus, $\\angle MPA = \\angle H'PA = 90^\\circ$. Since $\\angle MPA = \\angle MDA = 90^\\circ$, we get that $M, D, P, A$ are concyclic.\n\n\n\nThis gives us $\\angle DPA + \\angle AMD = 180^\\circ$. So now it is enough to prove that $\\angle AMB = \\angle MQA$. Taking the homothety with center $G$ and factor $-2$ (the homothety taking the 9-point circle to the circumcircle of $ABC$), we get that $Q$ and $A$ are images of $D$ and $M$, respectively.\n\nThis means $AQ \\parallel DM$, giving $AQ \\parallel BC$. Since $AQ \\parallel BC$ and $A, B, C, Q$ are concyclic, this means $ABCQ$ is an isosceles trapezoid.\n\nThis means $M$ lies on the perpendicular bisector of $AQ$. This gives us $MA = MQ$. Since $AQ \\parallel BC$, this gives us $\\angle AQM = \\angle QAM = \\angle AMB$, which concludes our problem.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20207,
"subject": "Mathematics (Olympiad)",
"question": "Twenty-one girls and twenty-one boys took part in a mathematical competition. It turned out that\n\n(a) each contestant solved at most six problems, and\n(b) for each pair of a girl and a boy, there was at least one problem that was solved by both the girl and the boy.\n\nProve that there is a problem that was solved by at least three girls and at least three boys.",
"options": [],
"answer": "See solution",
"solution": "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 \\textbf{girl-easy} if it was solved by at least three girls; otherwise we call it \\textbf{girl-hard}. Analogously, we define \\textbf{boy-easy} and \\textbf{boy-hard} problems. We want to prove that there is a problem that is both girl-easy and boy-easy.\n\nMost solutions proceed indirectly. Assuming that no problem is both girl-easy and boy-easy, one calculates some particular quantity in two ways to lead to a contradiction. The last two solutions apply the \\textbf{Pigeonhole Principle} to color a $21 \\times 21$ grid to prove the result with very limited arithmetic.\n\n**First Solution.** (By Zhiqiang Zhang, China) We proceed indirectly. Assume that no problem is both girl-easy and boy-easy.\n\n**Lemma 1.** Let $A = \\{p_1, p_2, \\dots, p_k\\}$ denote the set of all girl-hard problems. Let $B = \\{p_{k+1}, p_{k+2}, \\dots, p_{k+m}\\}$ denote the set of all boy-hard problems that are not in $A$. Then $k \\ge 11$, $m \\ge 11$, and consequently,\n\n$$\nk + m \\geq 22. \\tag{1}\n$$\n\n*Proof.* By our assumption, each problem is either girl-hard, boy-hard, or both. Thus, if $P$ is the set of all the problems, then $A \\cup B = P$. A boy contestant $b$ can solve at most 6 problems in set $A$, so there are at most $2 \\cdot 6 = 12$ girls who each solved a problem in $A$ that $b$ solved. By condition (b), there are at least $21 - 12$ girls who each solved a problem in $B$ that $b$ solved. It follows that each boy must solve some problems in set $B$. But each problem in set $B$ can be solved by at most 2 boys, implying that $21 \\le 2m$ and hence $m \\ge 11$. In exactly the same way, we can prove that $k \\ge 11$. ■\n\nFor $1 \\le i \\le k+m$, let $x_i$ be the number of contestants who solved problem $p_i$. In the light of condition (a), we have\n\n$$\nx_1 + x_2 + \\dots + x_{k+m} \\le 6 \\times 42. \\tag{2}\n$$\n\nFor each $p_i \\in P = A \\cup B$, let $q_i$ be the number of girl and boy pairs $(g, b)$ such that girl $g$ and boy $b$ both solved $p_i$. By condition (b), $\\sum_{i=1}^{k+m} q_i \\ge 21^2$. For each $i = 1, 2, \\dots, k+m$, note that $x_i \\ge 2$ because each problem was solved by at least one girl and one boy. Therefore,\n\n$$\nq_i \\le \\max\\{(x_i - 1) \\times 1, (x_i - 2) \\times 2\\} \\le 2x_i - 3.\n$$\n\nHence, by (2) we obtain\n\n$$\n\\begin{align*}\n21^2 &\\le \\sum_{i=1}^{k+m} q_i \\le (2x_1 - 3) + (2x_2 - 3) + \\dots + (2x_{k+m} - 3) \\\\\n&= 2(x_1 + x_2 + \\dots + x_{k+m}) - 3(k+m) \\\\\n&\\le 2 \\times 6 \\times 42 - 3(k+m),\n\\end{align*}\n$$\n\nor\n\n$$k+m \\le 21,$$\n\ncontradicting (1).\n\nThus, our assumption was wrong and there is a problem solved by at least three girls and three boys.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20208,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for any real $x, y$,\n\n$$\nx f(x) + y f(xy) = x f(x + y f(y))\n$$",
"options": [],
"answer": "See solution",
"solution": "*Answer:* $f(x) = 0$ and $f(x) = x$.\n\nLet $P(x, y)$ be the given assertion:\n\n$$\nP(0, 1): f(0) = 0\n$$\n\nAssume there exists $a \\neq 0$ such that $f(a) = 0$. Then $P(x, a)$:\n\n$$\nx f(x) + a f(xa) = x f(x) \\implies \\forall x \\in \\mathbb{R}, f(xa) = 0 \\implies \\forall x \\in \\mathbb{R}, f(x) = 0\n$$\n\nThis gives the first solution.\n\nNow assume $f$ is not identically zero, so $f(x) = 0 \\iff x = 0$.\n\n$$\nP(-1, 1): 0 = -f(-1) + f(-1) = -f(-1 + f(1)) \\implies f(1) = 1\n$$\n\n$$\nP(x, 1): x f(x) + f(x) = x f(x + 1) \\implies (x + 1) f(x) = x f(x + 1)\n$$\n\nFor $x \\neq 0, -1$:\n\n$$\n\\frac{f(x)}{x} = \\frac{f(x + 1)}{x + 1} \\quad (1)\n$$\n\nSince $f(1) = 1$, we get $\\forall n \\in \\mathbb{N}, f(n) = n$. Also, from (1), $f(-2) = 2 f(-1)$.\n\n$$\nP(-1, 2): -f(-1) + 2 f(-2) = -f(-1 + 2 f(2)) = -f(3) = -3 \\implies f(-1) = -1\n$$\n\nCombining with (1), $\\forall n \\in \\mathbb{Z}, f(n) = n$.\n\nNow, for $n \\in \\mathbb{Z}$ and $x \\in \\mathbb{R}$, $(x + n) f(x) = x f(x + n)$. For $x \\in \\mathbb{Z}$ this is clear; otherwise, $\\frac{f(x)}{x} = \\frac{f(x + 1)}{x + 1} = \\cdots = \\frac{f(x + n)}{x + n}$.\n\n$$\n\\begin{align*}\nP(x, n): x f(x) + n f(nx) &= x f(x + n f(n)) \\\\\n&\\implies x f(x) + n f(nx) = x f(x + n^2) = (x + n^2) f(x) \\\\\n&\\implies x f(x) + n f(nx) = x f(x) + n^2 f(x) \\\\\n&\\implies f(nx) = n f(x). \\tag{2}\n\\end{align*}\n$$\n\n$$\nP(1, y): 1 + y f(y) = f(1 + y f(y))\n$$\n\nLet $x = y f(y)$ in (1):\n\n$$\n\\frac{f(y f(y))}{y f(y)} = \\frac{f(y f(y) + 1)}{y f(y) + 1} = 1 \\implies f(y f(y)) = y f(y). \\quad (3)\n$$\n\nDivide $P(x, y)$ by $x \\neq 0$:\n\n$$\nf(x) + \\frac{y f(xy)}{x} = f(x + y f(y)). \\quad (4)\n$$\n\nLet $x = x f(x)$ and by symmetry:\n\n$$\n\\begin{align*}\nf(x f(x)) + \\frac{y f(x y f(x))}{x f(x)} &= f(x f(x) + y f(y)) \\\\\n&= f(y f(y)) + \\frac{x f(x y f(y))}{y f(y)} \\\\\n&\\implies f(x f(x)) - \\frac{x f(x y f(y))}{y f(y)} = f(y f(y)) - \\frac{y f(x y f(x))}{x f(x)}\n\\end{align*}\n$$\n\nIf we substitute $2x$ for $x$, the LHS increases by 4 times (by (2)), but the RHS does not change, so both are zero. Thus,\n\n$$\nf(x y f(y)) = f(x) y f(y) \\quad (5)\n$$\n\nLet $x = x + 1$ in (5):\n\n$$\n\\begin{align*}\nf(x + 1) y f(y) &= f((x + 1) y f(y)) = f(x y f(y) + y f(y)) \\\\\n&= f(x y f(y)) + \\frac{y f(x y^2 f(y))}{x y f(y)} \\\\\n&= f(x) y f(y) + \\frac{y^2 f(y) f(x y)}{x y f(y)} = f(x) y f(y) + \\frac{y f(x y)}{x}\n\\end{align*}\n$$\n\nSo, $(f(x + 1) - f(x)) x = \\frac{f(x y)}{f(y)}$. Substituting $y = 1$:\n\n$$\n(f(x + 1) - f(x)) x = f(x) \\implies f(x y) = f(x) f(y) \\quad (6)\n$$\n\nUsing (6) in (3), $f(f(y)) = y$. So,\n\n$$\n\\begin{align*}\nP(f(y), y): f(y) y + y f(f(y) y) &= f(y) f(f(y) + y f(y)) \\\\\n&\\implies f(y) y + y^2 f(y) = y f(y) f(y + 1) \\\\\n&\\implies f(y + 1) = y + 1 \\\\\n&\\implies \\forall x \\in \\mathbb{R}, f(x) = x \\text{ is the second solution.}\n\\end{align*}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20209,
"subject": "Mathematics (Olympiad)",
"question": "Determine the number of non-negative integers $N < 1000000 = 10^6$ 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 obvious that $N$ and $2012$ must be relatively prime. If $N^k \\equiv 1 \\pmod{n}$ and $N^m \\equiv 1 \\pmod{n}$ both hold, so does $N^d \\equiv 1 \\pmod{n}$ for $d = \\gcd(k, m)$. From $m = \\varphi(n)$, we see that $N^k \\equiv 1 \\pmod{n}$ implies that there exists a divisor $d$ of $\\varphi(n)$ with $N^d \\equiv 1 \\pmod{n}$. Since $2012 = 4 \\cdot 503$ and $\\varphi(503) = 502 = 2 \\cdot 251$ (where 503 and 251 are both prime), the only possible exponents $d$ with $N^d \\equiv 1 \\pmod{503}$ of interest to us are $1$, $2$, $251$, and $502$. We need therefore only consider the exponents $d=1$ and $d=2$. Since $N^1 \\equiv 1$ automatically implies $N^2 \\equiv 1$, we only require the residues $+1$ and $-1$ modulo $503$. Since $N^2 \\equiv 1 \\pmod{4}$ holds for all odd values of $N$, the values of $N$ with the required property are exactly the numbers $N = 1006u + 1$ and $N = 1006v - 1$. Since $1006 \\cdot 995 = 1000970$ and $1006 \\cdot 994 = 999964$, only $0 \\leq u \\leq 994$ and $1 \\leq v \\leq 994$ are possible, and the required number of integers $N$ is equal to $995 + 994 = 1989$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20210,
"subject": "Mathematics (Olympiad)",
"question": "Let $m$ and $n$ be positive integers. Show that if the equation $$(x + m)(x + n - 1) = n$$ has an integer solution $x$, then $\\frac{m}{n} < 2$.",
"options": [],
"answer": "See solution",
"solution": "Since the equation is symmetric in $m$ and $n$, we may assume $m \\geq n$, and it suffices to show $\\frac{m}{n} < 2$.\n\nRewrite the equation as $$(x + m)(x + n - 1) = n.$$ Since $x$ is an integer, this expresses $n$ as a product of two integers, so both factors must lie in one of the intervals $[-n, -1]$ or $[1, n]$. Either way, $$(x + m) - (x + n - 1) \\leq n - 1,$$ which leads to $m \\leq 2n - 2$, so $$\\frac{m}{n} \\leq 2 - \\frac{2}{n} < 2.$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20211,
"subject": "Mathematics (Olympiad)",
"question": "Find all sets of 2014 (not necessarily distinct) rational numbers such that: if an arbitrary number in the set is removed, one can always divide the remaining 2013 numbers into three sets, each with exactly 671 elements, so that the products of all elements in each set are equal.",
"options": [],
"answer": "See solution",
"solution": "Suppose the tuple $$(a_1, a_2, \\dots, a_{2014})$$ satisfies the given condition. We investigate two cases:\n\n*Case 1.* If there exists a zero in this tuple, it is easy to check that there must be at least four zeros. If the tuple has four zeros, then it satisfies the given condition.\n\n*Case 2.* Suppose $a_i \\neq 0$ for all $i$, $1 \\leq i \\leq 2014$. Denote $a_i = x_i/y_i$ where $x_i, y_i \\in \\mathbb{Z}^+$ and $\\gcd(x_i, y_i) = 1$. The tuple $$(a_1, \\dots, a_{2014})$$ satisfies the condition if and only if the sequence $$(ta_1, \\dots, ta_{2014})$$ does, where $t$ is the least common multiple of $y_1, \\dots, y_{2014}$. Thus, we can assume the $a_i$ are nonzero integers.\n\nFor any prime $p$, let $Z_p = (z_1, z_2, \\dots, z_{2014})$ be the set of exponents of $p$ in $$(a_1, a_2, \\dots, a_{2014})$$. Then $Z_p$ is a set of non-negative integers with the following property: if we remove any number, the remaining numbers can be divided into 3 groups of 671 elements each, with equal sums. (\\clubsuit)\n\nWe will show that all elements of $Z_p$ are equal. Otherwise, suppose $z_1, \\dots, z_{2014}$ are not all the same. Choose the set where the sum $z = z_1 + z_2 + \\dots + z_{2014}$ is minimal. If we remove $z_i$, the remaining elements can be divided into three groups of equal sums, so $z \\equiv z_i \\pmod{3}$ for all $i = 1, \\dots, 2014$. Consider three sub-cases:\n\n2.1. If $z_i \\equiv 0 \\pmod{3}$ for all $i$, then the set\n$$ \\left( \\frac{z_1}{3}, \\frac{z_2}{3}, \\dots, \\frac{z_{2014}}{3} \\right) $$\nalso satisfies (\\clubsuit) with a smaller sum, a contradiction.\n\n2.2. If $z_i \\equiv 1 \\pmod{3}$ for all $i$, then consider\n$$ \\left( \\frac{z_1+2}{3}, \\frac{z_2+2}{3}, \\dots, \\frac{z_{2014}+2}{3} \\right), $$\nwhich also leads to a contradiction.\n\n2.3. If $z_i \\equiv 2 \\pmod{3}$ for all $i$, consider\n$$ \\left( \\frac{z_1+1}{3}, \\frac{z_2+1}{3}, \\dots, \\frac{z_{2014}+1}{3} \\right), $$\nwhich again leads to a contradiction.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20212,
"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": 20213,
"subject": "Mathematics (Olympiad)",
"question": "Let $O$ be the circumcenter of an acute triangle $ABC$ in which $\\angle B < \\angle C$. Line $AO$ intersects side $BC$ at $D$. Let $E$ and $F$ be the circumcenters of triangles $ABD$ and $ACD$, respectively. On $[AB]$ produced and $[AC]$ produced, beyond $A$, consider points $G$ and $H$, respectively, such that $AG = AC$ and $AH = AB$. Prove that the quadrilateral $EFGH$ is a rectangle if and only if $\\angle ACB - \\angle ABC = 60^\\circ$.",
"options": [],
"answer": "See solution",
"solution": "Obviously, $EF \\perp AD$. Also, $\\angle BAO = 90^\\circ - \\angle C = 90^\\circ - \\angle AGH$, so $AD \\perp GH$. Thus, $EF \\parallel GH$.\n\nOn the other hand, $\\angle ADC = \\angle B + \\angle BAD = 90^\\circ - \\angle C + \\angle B < 90^\\circ$, which means that triangle $ADC$ is acute and, consequently, $F \\in \\text{int}(\\angle DAC)$. Angle $\\angle ADB$ is obtuse, therefore $B \\in \\text{int}(\\angle EAD)$.\n\nIt is easy to see that $\\angle AFC = 2 \\cdot \\angle ADC = \\angle AEB$. Triangles $AFC$ and $AEB$ are isosceles and have equal angles at their apex, so they are similar. It follows that $\\angle EAF = \\angle A$ and $\\frac{EA}{AB} = \\frac{FA}{AC}$, which shows that triangles $AEF$ and $ABC$ are also similar.\n\nIt follows that $EFGH$ is a parallelogram if and only if $EF = GH$, i.e. if and only if triangles $AEF$ and $ABC$ are equal. This is equivalent to triangles $ABE$ and $ACF$ being equilateral, i.e. to $\\angle ADC = 30^\\circ$, which is equivalent to $\\angle ACB - \\angle ABC = 60^\\circ$.\n\nWe still have to prove that, in the situation described above, $EFGH$ is a rectangle. But $AE = AB = AH$ and $AF = AC = AG$ show that in the parallelogram $EFGH$ the perpendicular bisectors of two opposite sides do meet, which only happens when the parallelogram is a rectangle.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20214,
"subject": "Mathematics (Olympiad)",
"question": "Albert and Betty play the following game. There are two bowls on a table: a red bowl and a blue bowl. At the beginning, there are 100 blue balls in the red bowl and 100 red balls in the blue bowl. On each turn, a player must take one of the following moves:\n\n1. Take two balls of different colors from one bowl and throw the balls away.\n2. Take two red balls from the blue bowl and put them in the red bowl.\n3. Take two blue balls from the red bowl and put them in the blue bowl.\n\nPlayers alternate turns, with Albert starting. The player who first takes the last red ball from the blue bowl or the last blue ball from the red bowl wins. Determine who has a winning strategy.",
"options": [],
"answer": "See solution",
"solution": "Betty has a winning strategy.\n\nBetty's strategy is as follows: If Albert makes move 2, then Betty makes move 3, and vice versa. If Albert makes move 1 from one bowl, Betty makes move 1 from the other bowl. The only exception is if Betty can make a winning move (removing the last blue ball from the red bowl or the last red ball from the blue bowl); in that case, she takes the winning move.\n\nFirst, we prove that it is possible to follow this strategy. Let\n\n$$\nb = (\\# \\textbf{blue balls in the blue bowl}, \\# \\textbf{red balls in the blue bowl})\n$$\n\n$$\nr = (\\# \\textbf{red balls in the red bowl}, \\# \\textbf{blue balls in the red bowl})\n$$\n\nAt the beginning, $b = r = (100, 0)$. If $b = r$ and Albert takes move 2, then Betty can take move 3 and again leave $b = r$ for Albert. The same applies if Albert takes move 3. If $b = r$ and Albert takes move 1 from one bowl, then Betty can take move 1 from the other bowl and again leave $b = r$. Thus, by following this strategy, Betty always leaves $b = r$ to Albert unless she is making a winning move.\n\nNow we prove that using this strategy, Betty wins. Assume that at some point Albert wins by taking a winning move. Since $r = b$ before the move, we must have $b = r = (1, s)$, $s \\geq 1$, or $b = r = (2, t)$ before the move. But that means that either $b$ or $r$ was $(1, s)$, $s \\geq 1$, or $(2, t)$ when Betty made her last move, which is a contradiction because in this situation Betty would have taken a winning move and the game would have stopped. Hence, Betty wins.\n\nNote: There is one situation from which no legal move is possible, namely $b = r = (1, 0)$. When Betty follows the above strategy, this situation will never occur.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20215,
"subject": "Mathematics (Olympiad)",
"question": "Thomas and Nils are playing a game. They have a number of cards, numbered $1, 2, 3, \\ldots$. At the start, all cards are lying face up on the table. They take alternate turns. The person whose turn it is chooses a card that is still lying on the table and decides to either keep the card himself or to give it to the other player. When all cards are gone, each of them calculates the sum of the numbers on his own cards. If the difference between these two outcomes is divisible by $3$, then Thomas wins. If not, then Nils wins.\n\n(a) Suppose they are playing with $2018$ cards (numbered from $1$ to $2018$) and that Thomas starts. Prove that Nils can play in such a way that he will win the game with certainty.\n\n(b) Suppose they are playing with $2020$ cards (numbered from $1$ to $2020$) and that Nils starts. Which of the two players can play in such a way that he wins with certainty?",
"options": [],
"answer": "See solution",
"solution": "(a) Thomas and Nils both make $1009$ moves, and Nils makes the last move. Nils can make sure that the last card on the table contains a number that is *not* divisible by $3$. Indeed, he could start taking cards with numbers that are divisible by $3$, until all these cards are gone. Because there are only $672$ such cards, he has enough turns to achieve that.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20216,
"subject": "Mathematics (Olympiad)",
"question": "As shown in the figure below, $PA$ and $PB$ are tangents to the circle with center $O$ at points $A$ and $B$. Point $C$ (distinct from $A$ and $B$) lies on the minor arc $AB$. The line $l$ through $C$ and perpendicular to $PC$ meets the angle bisectors of $\\angle AOC$ and $\\angle BOC$ at points $D$ and $E$, respectively. Prove that $CD = CE$.\n\n",
"options": [],
"answer": "See solution",
"solution": "As shown in the figure below, line $PC$ meets the circle $\\odot O$ at another point $F$. Join $BC$, $BE$, $BF$, and $OF$. Since $B$ and $C$ are on $\\odot O$, $OE$ bisects $\\angle BOC$, and hence $OE$ perpendicularly bisects $BC$, so $CE = BE$. Since $FO = BO$ and $PC \\perp DE$, we have\n\n$\\angle ECB = 90^\\circ - \\angle FCB = 90^\\circ - \\frac{1}{2} \\angle BOF = \\angle OBF$, so $\\triangle CEB \\sim \\triangle BOF$. Hence,\n\n\n\n$$\n\\frac{CE}{BO} = \\frac{CB}{BF}. \\qquad \\textcircled{1}\n$$\n\nSince $PB$ is tangent to $\\odot O$, $\\angle PBC = \\angle PFB$, and $\\triangle PCB \\sim \\triangle PBF$, so\n\n$$\n\\frac{CB}{BF} = \\frac{PC}{PB}. \\qquad \\textcircled{2}\n$$\n\nBy (1) and (2), $CE = BO \\cdot \\frac{PC}{PB}$. By symmetry, $CD = AO \\cdot \\frac{PC}{PA}$. Since $AO = BO$ and $PA = PB$, it follows that $CD = CE$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20217,
"subject": "Mathematics (Olympiad)",
"question": "老趙當兵太無聊了，因此他在地上畫了 $n+1$ 個格子排成一列（$n$ 為正整數），從左至右依序編號為第 $0$ 格到第 $n$ 格。起始的時候，第 $0$ 格有 $n$ 顆石頭，其他格子則都是空的。\n\n在每一回合，老趙先選擇一格非空的格子，假設其中有 $k$ 顆石頭。接著，他從選定的格子中拿取一顆石頭，並將它向右移動至多 $k$ 格（石頭不得超出最右邊的格子）。\n\n老趙的目標是將所有石頭都放到第 $n$ 格。\n\n試證明，老趙要達成目標，所需的回合數不少於\n\n$$\n\\left\\lceil \\frac{n}{1} \\right\\rceil + \\left\\lceil \\frac{n}{2} \\right\\rceil + \\left\\lceil \\frac{n}{3} \\right\\rceil + \\cdots + \\left\\lceil \\frac{n}{n} \\right\\rceil,\n$$\n\n其中 $\\lceil x \\rceil$ 表示不小於 $x$ 的最小整數。",
"options": [],
"answer": "See solution",
"solution": "讓我們將石頭編號 $1$ 到 $n$，並且不失一般性，假設老趙在移動石頭時，都是移動該格內編號最大的石頭。\n\n現在，注意到當老趙移動編號 $k$ 的石頭時，該格內至多只有 $k$ 顆石頭，因此編號 $k$ 的石頭每次至多只能移動 $k$ 步。\n\n換言之，編號 $k$ 的石頭要移動到第 $n$ 格，至少要 $\\left\\lceil \\frac{n}{k} \\right\\rceil$ 步，從而原命題得證。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20218,
"subject": "Mathematics (Olympiad)",
"question": "Consider the following problem:\n\nFor a prime $p$, let $A_i(n)$ denote the number of subsets of $\\{1, 2, \\dots, n\\}$ whose sum of elements is congruent to $i \\pmod{p}$. Prove that if $c_p = \\frac{2^{p-1} - 1}{p}$ is not divisible by $p$, then for any integer $k$, there exist infinitely many $n$ such that $A_0(n) \\equiv k \\pmod{p}$.\n",
"options": [],
"answer": "See solution",
"solution": "*Proof.* Consider a partitioning of $\\{1, 2, \\dots, (m+1)p\\}$ into $\\{1, 2, \\dots, mp\\}$ and $\\{mp + 1, mp + 2, \\dots, (m+1)p\\}$.\n\nOne can express subsets $S \\subset \\{1, 2, \\dots, (m+1)p\\}$ uniquely as union of two subsets $S_1 \\subset \\{1, 2, \\dots, mp\\}$ and $S_2 \\subset \\{mp+1, \\dots, (m+1)p\\}$. Subtracting $mp$ from all elements of $S_2$ corresponds $S_2$ to a subset of $\\{1, 2, \\dots, p\\}$, and this correspondence also preserves sum of elements mod $p$. Under this correspondence, if $S$ is counted in $A_i((m+1)p)$ and $S_2$ is counted in $A_j(p)$, then $S_1$ should be counted in $A_{i-j}(mp)$.\n\nConsidering those two lemmas and $\\sum_j A_j(mp) = 2^{mp}$ together gives\n\n$$\nA_0((m+1)p) = 2A_0(mp) + 2c_p 2^{mp} \\equiv 2A_0(mp) + 2^{m+1}c_p\n$$\n\nSolving this recurrence in mod $p$ as\n\n$$\n2^{-(m+1)} A_0((m+1)p) \\equiv 2^{-m} A_0(mp) + c_p\n$$\n\ngives\n\n$$\nA_0((m+1)p) \\equiv 2^m A_0(p) + 2^{m+1}mc_p.\n$$\n\nNow we let $m = (p - 1)u$ to obtain\n\n$$\nA_0((1 + (p - 1)u)p) \\equiv 2^{(p-1)u} A_0(p) + 2^{(p-1)u}(p - 1)uc_p \\equiv A_0(p) - 2c_p u.\n$$\n\nFrom the condition that $c_p \\neq 0$, one can see that for any $k$ there exists infinitely many $u$ such that $A_0((1 + (p - 1)u)p) \\equiv k$.\n\n---\n\n**Second Solution.** For an integer $i$, denote $A_i(n)$ as the number of subsets of $\\{1, 2, \\dots, n\\}$ whose sum of elements is $\\equiv i \\pmod{p}$. In a generating function $F_n(x) = (1+x)(1+x^2)\\cdots(1+x^n)$, its $x^m$-coefficient can be considered as number of subsets of $\\{1, 2, \\dots, n\\}$ whose sum of elements is $m$. By substituting any element $z$ to $x$ satisfying $z^p = 1$, we have the following.\n\n$$\n(1 + z)(1 + z^2)\\cdots(1 + z^n) = \\sum_{i=0}^{n-1} A_i(n)z^i\n$$\n\nWe consider $\\zeta = \\exp(2\\pi i/p)$, a primitive $p$-th root of unity.\n\n**Lemma 3.** For integers $k$ not divisible by $p$, we have\n\n$$\n\\prod_{j=1}^{p} (1 + \\zeta^{jk}) = 2.\n$$\n\n*Proof.* The set $\\{jk : j = 1, \\dots, p\\}$ modulo $p$ is always identical to $\\{1, 2, \\dots, p\\}$, so it suffices to prove when $k = 1$. Consider a factorization\n\n$$\nX^p - 1 = \\prod_{j=1}^{p} (X - \\zeta^j)\n$$\n\nand substitute $X = -1$ to obtain\n\n$$\n-2 = (-1)^p - 1 = \\prod_{j=1}^{p} (-1 - \\zeta^j) = (-1)^p \\prod_{j=1}^{p} (1 + \\zeta^j).\n$$\n\nThis gives our desired result for $k = 1$ case.\n\n\n\nNow we consider\n\n$$\n\\sum_{k=0}^{p-1} F_n(\\zeta^k) = \\sum_{i=0}^{n-1} A_i(n) \\left( \\sum_{k=0}^{p-1} \\zeta^{ki} \\right) = pA_0(n)\n$$\n\nand let $n = pm$. We first note that $F_{pm}(1) = 2^{pm}$, and from that $(1 + \\zeta^{kl})$ only depends on $l \\bmod p$ we have\n\n$$\nF_{pm}(\\zeta^k) = F_p(\\zeta^k)^m = 2^m.\n$$\n\nSubstituting these gives\n\n$$\nA_0(pm) = \\frac{2^{pm} + (p-1)2^m}{p} = \\frac{2^m(2^{(p-1)m} + p-1)}{p}.\n$$\n\nNow consider $c_p = \\frac{2^{p-1} - 1}{p}$ to obtain the congruence\n\n$$\n2^{p-1} \\equiv 1 + pc_p \\pmod{p^2}\n$$\n\nand\n\n$$\n2^{(p-1)m} \\equiv (1 + pc_p)^m \\equiv 1 + pmc_p \\pmod{p^2}.\n$$\n\nThis allows us to evaluate $A_0(pm)$ mod $p$ as\n\n$$\nA_0(pm) = \\frac{2^m(2^{(p-1)m} + p-1)}{p} \\equiv \\frac{2^m((1 + pmc_p) + p-1)}{p} \\equiv 2^m(mc_p + 1) \\pmod{p}.\n$$\n\nWe let $m = (p-1)l$ to make $2^m \\equiv 1$, then we have\n\n$$\nA_0(p(p-1)l) \\equiv 1 - lc_p.\n$$\n\nSo under the condition that $c_p$ is not divisible by $p$, for any $k$ one can find infinitely many $l$ such that $A_0(p(p-1)l) \\equiv k$, proving our problem.\n\n\n",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20219,
"subject": "Mathematics (Olympiad)",
"question": "Consider a triangle made up of unit equilateral triangles arranged in a certain pattern (see diagram below).\n\n% \n\nEach small triangle is to be filled with a number so that the sum of the numbers in every unit equilateral triangle is divisible by 3. Let $n$ be the number of zeros among the numbers used to fill the triangle. What are the possible values of $n$?",
"options": [],
"answer": "See solution",
"solution": "Consider the four numbers in any two unit equilateral triangles that share a common edge as shown in the diagram.\n\n$$\n\\begin{matrix} a \\\\ b & c \\\\ d \\end{matrix}\n$$\n\nSince $a + b + c \\equiv 0 \\equiv b + c + d \\pmod{3}$, it follows that $d \\equiv a \\pmod{3}$. Starting with the top two rows as being given by\n\n$$\n\\begin{matrix} a \\\\ b & c \\end{matrix}\n$$\n\nthis implies that if we reduce the entries of the entire triangle modulo 3, it takes the following form:\n\n$$\n\\begin{matrix}\na & & \\\\\n& b & c \\\\\n& & a \\\\\nc & a & b \\\\\n& b & c & a \\\\\nb & c & a & b & c \\\\\n& c & a & b & c & a \\\\\n& & b & c & a & b \\\\\n& & c & a & b & c & a \\\\\n& & & b & c & a & b & c & a\n\\end{matrix}\n$$\n\nNote that $a$, $b$, and $c$ occur 19, 18, and 18 times, respectively, in the triangle. Also, since $a + b + c \\equiv 0 \\pmod{3}$, we see that either $a = b = c$, or $a, b, c$ are equal to 0, 1, 2 in some order. The following table exhibits all the possibilities along with the corresponding values of $n$:\n\n| $a$ | $b$ | $c$ | $n$ |\n|---|---|---|---|\n| 0 | 0 | 0 | 55 |\n| 1 | 1 | 1 | 0 |\n| 2 | 2 | 2 | 0 |\n| 0 | 1 | 2 | 19 |\n| 0 | 2 | 1 | 19 |\n| 1 | 0 | 2 | 18 |\n| 1 | 2 | 0 | 18 |\n| 2 | 0 | 1 | 18 |\n| 2 | 1 | 0 | 18 |\n\nHence, the possible values of $n$ are $0$, $18$, $19$, and $55$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20220,
"subject": "Mathematics (Olympiad)",
"question": "Three consecutive vertices $A$, $B$, and $C$ of a regular octagon (8-gon) are the centres of circles that pass through neighbouring vertices of the octagon. The intersection points $P$, $Q$, and $R$ of the three circles form a triangle (see figure).\n\n\n\nProve that triangle $PQR$ is equilateral.",
"options": [],
"answer": "See solution",
"solution": "An octagon can be subdivided into six triangles (see figure on the left). Together, the angles of those six triangles add up to the same number of degrees as the eight angles of the octagon. Since the angles of any triangle add up to $180$ degrees, this means that the eight angles of the octagon add up to $6 \\cdot 180^\\circ = 1080^\\circ$. Hence, each of the angles of the regular octagon is $\\frac{1}{8} \\cdot 1080^\\circ = 135^\\circ$.\n\nWe now consider the figure from the problem statement (see figure on the right). Line segment *BP* bisects angle *ABC*, so $\\angle ABP = \\angle PBC = 67\\frac{1}{2}^\\circ$. Since triangles *ABP* and *BCP* are isosceles (as $|AB| = |AP|$ and $|BC| = |CP|$), we also have $\\angle APB = \\angle BPC = 67\\frac{1}{2}^\\circ$ and $\\angle BAP = \\angle BCP = 180^\\circ - 135^\\circ = 45^\\circ$.\n\nIn triangles *ABQ* and *BCR* all sides have the same length. These triangles are therefore equilateral and all angles are $60^\\circ$. From this, we deduce that $\\angle PAQ = \\angle BAQ - \\angle BAP = 15^\\circ$. In the same way, we find $\\angle PCR = 15^\\circ$. Furthermore, triangles *PAQ* and *PCR* are isosceles (since $|AP| = |AQ|$ and $|CP| = |CR|$), so $\\angle APQ = \\angle BPC = 67\\frac{1}{2}^\\circ$ and $\\angle CPR = 82\\frac{1}{2}^\\circ$.\n\nBy mirror symmetry, *PQ* and *PR* have the same length, so *PQR* is an isosceles triangle with apex *P*. We have already determined all angles at *P*, except $\\angle QPR$. We deduce that\n\n$$\n\\begin{aligned}\n\\angle QPR &= 360^\\circ - \\angle APQ - \\angle APB - \\angle BPC - \\angle CPR \\\\\n&= 360^\\circ - 2 \\cdot 67\\frac{1}{2}^\\circ - 2 \\cdot 82\\frac{1}{2}^\\circ = 60^\\circ.\n\\end{aligned}\n$$\n\nFrom this and the fact that *PQR* is isosceles, we directly conclude that *PQR* is equilateral. $\\square$\n\n\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20221,
"subject": "Mathematics (Olympiad)",
"question": "Given real numbers $a, b, c, d$ such that $a$ and $b$ are the roots of the equation $x^2 - 10c x - 11d = 0$, and $c$ and $d$ are the roots of the equation $x^2 - 10a x - 11b = 0$, find the value of $a + b + c + d$.",
"options": [],
"answer": "See solution",
"solution": "From Vieta's formulae, we have $a + b = 10c$ and $c + d = 10a$. Adding these gives\n\n$$\na + b + c + d = 10(a + c).\n$$\n\nSince $a$ is a root of $x^2 - 10c x - 11d = 0$ and $d = 10a - c$, we have\n\n$$\n0 = a^2 - 10ac - 11d = a^2 - 10ac - 11(10a - c) = a^2 - 110a + 11c - 10ac.\n$$\n\nSimilarly, for $c$ as a root of $x^2 - 10a x - 11b = 0$ and $b = 10c - a$,\n\n$$\nc^2 - 110c + 11a - 10ac = 0.\n$$\n\nSubtracting these equations gives\n\n$$\n(a - c)(a + c - 121) = 0.\n$$\n\nSince $a \\neq c$, we get $a + c = 121$. Therefore,\n\n$$\na + b + c + d = 10 \\cdot 121 = 1210.\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20222,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers that have exactly 8 positive divisors, among which three are primes of the form $a$, $\\overline{bc}$, and $\\overline{cb}$, given that $a + \\overline{bc} + \\overline{cb}$ is a perfect square and $a$, $b$, $c$ are digits, with $b < c$.",
"options": [],
"answer": "See solution",
"solution": "Suppose $x$ is an integer with the given properties. Then $x$ is a multiple of $y = a \\cdot \\overline{bc} \\cdot \\overline{cb}$. Since $y$ has eight divisors, it follows that $x = y$.\n\nThe numbers $\\overline{bc}$ and $\\overline{cb}$ are distinct primes, so $\\overline{bc} \\in \\{13, 17, 37, 79\\}$. The condition that $a + \\overline{bc} + \\overline{cb}$ is a square and $a$ is a prime forces $a = 5$, $\\overline{bc} = 13$, and $\\overline{cb} = 31$, implying $x = 2015$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20223,
"subject": "Mathematics (Olympiad)",
"question": "A target consists of two circles with the same centre. The larger circle has radius 12 cm. When shots hit the target randomly, 403 out of 900 land in the inner circle. The radius of the inner circle is closest to\n\n\n\n(A) 6 cm \n(B) 7 cm \n(C) 8 cm \n(D) 9 cm \n(E) 10 cm",
"options": [],
"answer": "See solution",
"solution": "If the radius is $r$ cm then\n$$\n\\frac{\\text{area of smaller circle}}{\\text{area of larger circle}} = \\frac{\\pi r^2}{\\pi \\cdot 12^2} = \\frac{r^2}{12^2} = \\frac{403}{900} \\approx \\frac{4}{9}\n$$\nSo $r^2$ is close to $\\frac{2^2 \\cdot 12^2}{3^2}$, which means $r$ is $\\frac{2 \\cdot 12}{3} = 8$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20224,
"subject": "Mathematics (Olympiad)",
"question": "There are $m$ girls and $n$ boys participating in a duet singing contest ($m, n \\ge 2$). At the contest, there will be one show in each section. Each show includes some boy-girl duets where each boy-girl couple will sing no more than one song and each participant will sing at least one song. Two shows are considered different if there exists a boy-girl couple that sings in exactly one of these two shows. The contest will end if and only if every possible show is performed, and each show is performed exactly once.\n\n**a)** A show is called dependent on a participant $X$ if, when we cancel all duets that $X$ performs, there will be at least one participant who is not allowed to sing any song in that show. Prove that among every show that depends on $X$, the number of shows with an odd number of songs equals the number of shows with an even number of songs.\n\n**b)** Prove that the organizers can arrange the shows such that the number of songs in two consecutive shows have different parities.",
"options": [],
"answer": "See solution",
"solution": "a) Label all the girls by $1, 2, \\ldots, m$ and boys by $1, 2, \\ldots, n$. For each show, we represent the performances by an $m \\times n$ table where the entry at the intersection of the $i$th row and $j$th column is:\n\n- $1$ if the $i$th girl performed with the $j$th boy in this show.\n- $0$ otherwise.\n\nCall a table \"nice\" if the sum of all elements in each row and column is positive. All tables corresponding to shows must be nice.\n\nFor a participant $X$, without loss of generality, assume $X$ is a girl. A show depends on $X$ if, in its table, there exists a column with only one cell marked $1$ in $X$'s row. Suppose there are $k$ such columns (with $k < n$). The other $n-k$ cells in $X$'s row can be either $0$ or $1$. Thus, the number of tables depending on $X$ is $2^{n-k}$ times the number of nice tables for the remaining $m-1$ rows and $n-k$ columns (after removing $X$'s row and the $k$ columns). For each such table, flipping the entries in $X$'s row (for columns not depending on $X$) changes the parity of the total number of songs. Therefore, the number of shows depending on $X$ with even and odd numbers of songs is equal.\n\nb) A table is called \"odd\" if it has an odd number of $1$s, and \"even\" otherwise. Let $f(m, n)$ and $g(m, n)$ be the number of nice tables with odd and even numbers of $1$s, respectively. Let $X$ be an arbitrary girl.\n\n- If there exists a column that depends on $X$, then the number of even tables equals the number of odd tables; denote this value by $h(m, n)$.\n- Otherwise, removing $X$'s row gives a $(m-1) \\times n$ nice table.\n\nThe number of ways $X$'s row has an odd or even number of $1$s is:\n\n$$\nL = \\sum_{2 \\mid a-1} \\binom{n}{a}, \\quad C = \\sum_{2 \\mid a,\\ a>0} \\binom{n}{a}.\n$$\n\nIt is well-known that $(x+1)^n = \\sum_{a=0}^n x^a \\binom{n}{a}$, so with $x = -1$, $L = C + 1$. The parity of $X$'s row determines the table's parity, so:\n\n$$\n\\begin{cases}\nf(m,n) = h(m,n) + L \\cdot g(m-1,n) + C \\cdot f(m-1,n), \\\\\ng(m,n) = h(m,n) + L \\cdot f(m-1,n) + C \\cdot g(m-1,n).\n\\end{cases}\n$$\n\nThus,\n\n$$\n\\begin{aligned}\nf(m,n) - g(m,n) &= (L - C)(g(m-1,n) - f(m-1,n)) \\\\\n&= g(m-1,n) - f(m-1,n).\n\\end{aligned}\n$$\n\nBy induction,\n\n$$\nf(m, n) - g(m, n) = (-1)^{m+n-4} (f(2, 2) - g(2, 2)).\n$$\n\nIt is clear that $f(2, 2) = 3$ and $g(2, 2) = 4$, so\n\n$$\nf(m,n) - g(m,n) = (-1)^{m+n-3}.\n$$\n\nSince the difference between odd and even tables is $1$, it is possible to arrange the shows so that the number of songs in two consecutive shows have different parities. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20225,
"subject": "Mathematics (Olympiad)",
"question": "There are three classes, each with $n$ students, and all these $3n$ students have distinct heights. Divide them into $n$ groups of three students, one from each class, and call the tallest person in each group a “leader.” It is known that no matter how the students are divided, there always exist 10 leaders in each class. Prove that $n \\geq 40$.",
"options": [],
"answer": "See solution",
"solution": "First, we show that $n = 40$ is sufficient. Let the three classes be $A$, $B$, and $C$. Label the students from tallest to smallest as $1, 2, \\ldots, 120$. Suppose that $1, 2, \\ldots, 10$ and $71, 72, \\ldots, 100$ are in class $A$; $11, 12, \\ldots, 30$ and $101, 102, \\ldots, 120$ are in class $B$; $31, 32, \\ldots, 70$ are in class $C$. Clearly, the tallest 10 students in class $A$ are all leaders. Among $11, 12, \\ldots, 30$, at most 10 of them are group mates of $1, 2, \\ldots, 10$ and they are not leaders, but the other 10 students must be leaders in their groups. Hence, class $B$ also has 10 leaders. Finally, for class $C$, at most 30 of the students $31, 32, \\ldots, 70$ are group mates of $1, 2, \\ldots, 30$ and they are not leaders, but the other 10 students must be leaders. So, this example meets the conditions and $n = 40$ suffices.\n\nFor necessity, we give two solutions.\n\n**Solution 1**\n\n**Lemma** Suppose the conditions are all satisfied. Then for each class $i$ ($1 \\leq i \\leq 3$), there is a positive integer $k_i$, such that among the tallest $k_i$ students from all classes, the number of those from class $i$ is at least 10 more than those from the other two classes.\n\n**Proof of Lemma** Pick any class, say class $A$, and rank their heights from tallest to smallest as $a_1 < a_2 < \\dots < a_n$. The other classes $B$ and $C$ have $x_1 < x_2 < \\dots < x_{2n}$. For every $1 \\leq i \\leq n-9$, let $a_{i+9}$ (from class $A$) and $x_i$ (from class $B$ or $C$) be in a group. Then add a class $B$ student to every group of class $A$ and class $C$ students; add a class $C$ student to every group of class $A$ and class $B$ students. Now, other than $a_1, a_2, \\dots, a_9$, there must be another leader from class $A$, say $a_m$. We must have $a_m < x_{m-9}$, meaning that among all students taller or equal to $a_m$, at least $m$ of them are from class $A$, and at most $m-10$ from $B$ or $C$. The lemma is verified.\n\nReturn to the original problem. Suppose the classes $A, B, C$ correspond to integers $k_1, k_2, k_3$ as in the lemma, respectively, and $k_1 \\leq k_2 \\leq k_3$. Among $1, 2, \\dots, k_1$, at least 10 students are from class $A$; among $1, 2, \\dots, k_2$, at least $10+10=20$ students are from class $B$; among $1, 2, \\dots, k_3$, at least $10+20+10=40$ students are from class $C$. This implies that each class has at least $n=40$ students.\n\n**Solution 2**\n\nWe show that the conditions are not met if $n < 40$. First, there must be $10 \\cdot 3 = 30$ or more groups so as to have 10 leaders in each class, hence $n \\geq 30$. Rank the students in each class from tallest to smallest as $a_1 < a_2 < \\dots < a_n$, $b_1 < b_2 < \\dots < b_n$ and $c_1 < c_2 < \\dots < c_n$. Consider $a_{n-19}, b_{n-19}, c_{n-19}$, and assume $a_{n-19}$ is the tallest. Since $n \\leq 39$, we infer that $a_1, a_2, \\dots, a_{n-19}$ are all taller than $b_{20}, b_{21}, \\dots, b_n, c_{20}, c_{21}, \\dots, c_n$. For $1 \\leq i \\leq n-19$, make $a_i, b_{i+19}$, and $c_{i+19}$ a group, each with a leader from class $A$; for the others, make groups in an arbitrary way. Then class $B$ and $C$ together have at most 19 leaders, a contradiction. Thus, $n \\geq 40$. $\\square$\n\n**Remark**\n\nIn general, if there always exist $k$ leaders in each class, then $n \\geq 4k$. Furthermore, if there are $m$ classes, then $n \\geq 2^{m-1}k$. In this problem, $(m, k) = (3, 10)$.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 20226,
"subject": "Mathematics (Olympiad)",
"question": "In a plane there is a triangle $ABC$. Line $AC$ is tangent to circle $c_A$ at point $C$ and circle $c_A$ passes through point $B$. Line $BC$ is tangent to circle $c_B$ at point $C$ and circle $c_B$ passes through point $A$. The second intersection point $S$ of circles $c_A$ and $c_B$ coincides with the incenter of triangle $ABC$. Prove that the triangle $ABC$ is equilateral.\n\n",
"options": [],
"answer": "See solution",
"solution": "By the tangent-secant theorem we have $\\angle BCS = \\angle CAS$ and $\\angle ACS = \\angle CBS$ (see fig. 5). The incenter of a triangle is the point of intersection of angle bisectors, therefore $\\angle CAB = 2\\angle CAS = 2\\angle BCS = \\angle BCA$ and $\\angle CBA = 2\\angle CBS = \\angle ACS = \\angle BCA$. Hence $ABC$ is equilateral.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20227,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\alpha$ be a real number larger than $1$. Define a sequence $(s_n)_{n \\geq 1}$ as follows:\n\n- $s_1 = 1$\n- $s_2 = \\alpha$\n- If $s_1, s_2, \\dots, s_{2^n}$ are defined for some $n \\geq 1$, then $s_{2^n+1}, \\dots, s_{2^{n+1}}$ are defined by $s_j = \\alpha s_{j-2^n}$ for $2^n+1 \\leq j \\leq 2^{n+1}$.\n\nThus, the first few terms are $1, \\alpha, \\alpha^2, \\alpha, \\alpha^2, \\alpha^3, \\dots$.\n\nLet $c_n = s_1 + s_2 + \\dots + s_n$.\n\nIf $n = 2^{e_0} + 2^{e_1} + \\dots + 2^{e_k}$, where $e_0 > e_1 > \\dots > e_k \\geq 0$ is the binary representation of a positive integer $n$, prove that\n\n$$\nc_n = (1 + \\alpha)^{e_0} + \\alpha(1 + \\alpha)^{e_1} + \\alpha^2(1 + \\alpha)^{e_2} + \\dots + \\alpha^k(1 + \\alpha)^{e_k}.\n$$",
"options": [],
"answer": "See solution",
"solution": "We prove the formula by induction and properties of the sequence.\n\n**Step A:**\nLet $b(n)$ be the number of $1$'s in the binary representation of $n$. Then for $n \\geq 1$, $s_n = \\alpha^{b(n-1)}$.\n\n*Proof:* Induction shows $s(1) = 1 = \\alpha^{b(0)}$. Assume $s(r) = \\alpha^{b(r-1)}$ for $1 \\leq r \\leq 2^{n-1}$. For $2^{n-1} + 1 \\leq r \\leq 2^n$:\n$$\ns(r) = \\alpha s(r - 2^{n-1}) = \\alpha \\alpha^{b(r-2^{n-1}-1)} = \\alpha^{b(r-2^{n-1}-1)+1} = \\alpha^{b(r-1)}.\n$$\n\n**Step B:**\nFor $m \\geq 1$, $s(2m+1) = s(m+1)$ and $s(2m) = \\alpha s(m)$.\n\n*Proof:* $s(2m+1) = \\alpha^{b(2m)} = \\alpha^{b(m)} = s(m+1)$. For even numbers, $s(2m) = \\alpha s(2m-1)$ by induction, and $s(2m) = \\alpha s(m)$ using $s(2k+1) = s(k+1)$.\n\n**Step C:**\nFor $m \\geq 1$, $c(2m) = (1+\\alpha)c(m)$ and $c(2m+1) = \\alpha c(m) + c(m+1)$.\n\n*Proof:*\n$$\n\\begin{aligned}\nc(2m) &= (s(1) + \\dots + s(m)) + \\alpha(s(1) + \\dots + s(m)) \\\\\n&= (1 + \\alpha)c(m).\n\\end{aligned}\n$$\nSimilarly,\n$$\n\\begin{aligned}\nc(2m+1) &= (s(1) + \\dots + s(m+1)) + \\alpha(s(1) + \\dots + s(m)) \\\\\n&= \\alpha c(m) + c(m+1).\n\\end{aligned}\n$$\n\n**Step D:**\nLet $n = 2^{e_0} + 2^{e_1} + \\dots + 2^{e_k}$, with $e_0 > e_1 > \\dots > e_k \\geq 0$. We prove\n$$\nc(n) = (1 + \\alpha)^{e_0} + \\alpha(1 + \\alpha)^{e_1} + \\alpha^2(1 + \\alpha)^{e_2} + \\dots + \\alpha^k(1 + \\alpha)^{e_k}.\n$$\n\n*Proof by induction:* For $n = 1$, $c(1) = 1 = (1 + \\alpha)^0$. Assume true for $c(1), \\dots, c(n-1)$. If $n = 2m$, then $c(n) = (1 + \\alpha)c(m)$, shifting all exponents up by $1$. If $n = 2m+1$, then $c(n) = \\alpha c(m) + c(m+1)$, which adds the $\\alpha^{k+1}(1+\\alpha)^0$ term. Thus, the formula holds for all $n$ by induction.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20228,
"subject": "Mathematics (Olympiad)",
"question": "The number of distinct positive divisors of an integer $N = \\prod p_i^{e_i}$, where the $p_i$ are distinct primes, is $\\prod (e_i + 1)$. For $N$ with exactly 8 positive divisors, the possible factorizations are:\n\n- $N = p^7$\n- $N = p^3q$\n- $N = pqr$\n\nwhere $p, q, r$ are distinct primes. For each case, consider the possible orderings of the divisors $d_2, d_3, d_4, d_5$ and the corresponding inequalities between the primes. Find all such $N$ for which $2d_2d_5 = d_3d_4 + 3$ holds.",
"options": [],
"answer": "See solution",
"solution": "$$\n\\begin{aligned}\n&\\text{Case A: } N = p^7 \\rightarrow d_k = p^{k-1}.\\ \\text{Equation } 2d_2d_5 = d_3d_4 + 3 \\implies p^5 = 3,\\ \\text{no integer solution.} \\\\\n&\\text{Case B: } N = p^3q.\\ \\text{Sub-case 4: } p^3 < q.\\ \\text{Divisors: } d_2 = p,\\ d_3 = p^2,\\ d_4 = p^3,\\ d_5 = q.\\ \\text{Equation: } 2p q = p^2 p^3 + 3 \\implies 2p q = p^5 + 3.\\ \\text{For } p = 3,\\ q = 41,\\ N = 3^3 \\cdot 41 = 1107. \\\\\n&\\text{Case C: } N = pqr.\\ \\text{Sub-case 2: } pq < r.\\ \\text{Divisors: } d_2 = p,\\ d_3 = q,\\ d_4 = pq,\\ d_5 = r.\\ \\text{Equation: } 2pr = pq^2 + 3 \\implies p(2r - q^2) = 3.\\ \\text{For } p = 3,\\ r = (q^2 + 1)/2.\\ \\text{For } q = 11,\\ r = 61,\\ N = 3 \\cdot 11 \\cdot 61 = 2013. \\\\\n&\\text{Thus, the two smallest solutions are } N = 1107 \\text{ and } N = 2013.\n\\end{aligned}\n$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20229,
"subject": "Mathematics (Olympiad)",
"question": "An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$, which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of the pillars at $A$, $B$, and $C$ are $12$, $9$, and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$?",
"options": [],
"answer": "See solution",
"solution": "Let $M$ be the midpoint of $\\overline{AC}$. The pillars at $A$ and $C$ are $12$ and $10$ meters high, respectively, so the height of the panel at $M$ is $11$ meters. Because $\\triangle BMA$ is a $30$-$60$-$90^\\circ$ triangle, $BM$ is half of the side length of the hexagon. Therefore $BM = \\frac{1}{4}BE$. It follows that the height of the panel at point $E$ is $9 + 4 \\cdot (11 - 9) = 17$ meters.\n\n\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20230,
"subject": "Mathematics (Olympiad)",
"question": "Let $S$ be a set of $n$ points and $P$ a set of $m$ lines such that each line in $P$ is a symmetric axis for some pair of points in $S$, and each point in $S$ belongs to at most one line in $P$.\n\n(b) Prove that $m \\leq n$.\n\n(c) When $m = n$, describe the configuration of $S$ and $P$.\n\n",
"options": [],
"answer": "See solution",
"solution": "As to the array $(x, y, p)$ in $F_1$, since there is only one symmetric axis for the different points $x$ and $y$, then\n\n$$\n|F_1| \\leq \\{(x, y) \\mid x, y \\in S, x \\neq y\\} = 2\\binom{n}{2} = n(n-1).\n$$\n\nAs to the array in $F_2$:\n\nWhen any point in $S$ belongs to no more than one line $p$, then\n\n$$\n|F_2| \\leq \\{x \\mid x \\in S\\} = n.\n$$\n\nFrom the above, we get\n\n$$\nm n \\leq n(n-1) + n,\n$$\nthat is, $m \\leq n$.\n\nWhen there exists one point that belongs to two lines of $P$, then as proved in (a), it must be the center of set $B$. Considering the set $S' = S \\setminus \\{B\\}$, since every line in $P$ is still the symmetric axis of the points in $S'$, then we get\n\n$$\nm \\leq |S'| = n - 1.\n$$\n\nThus, we obtain $m \\leq n$.\n\n(c) When $m = n$, the equalities above hold simultaneously. So, the perpendicular bisector of a line segment joining any two points in $S$ belongs to $P$, and any point in $S$ belongs to one line in $P$, while the 'center of set' $B$ is not in $S$.\n\nNow, we can first prove all the $BA_i$ ($i = 1, 2, \\dots, n$) are equal. Otherwise, if there exist $j, k$ ($1 \\leq j < k \\leq n$) such that $BA_j \\neq BA_k$, then the symmetric axis of $A_jA_k$ does not pass through $B$, which is a contradiction. Thus, $A_1, A_2, \\dots, A_n$ are all on the circle with $B$ as its center, that is, $\\odot B$. We can suppose $A_1, A_2, \\dots, A_n$ are arranged clockwise for convenience.\n\nThen, we can also prove that $A_1, A_2, \\dots, A_n$ are $n$ points dividing $\\odot B$ into equal parts. Otherwise, if there exists $i$ ($1 \\leq i \\leq n$) such that $A_iA_{i+1} \\neq A_{i+1}A_{i+2}$ (let $A_{n+1} = A_1$, $A_{n+2} = A_2$), we can suppose $A_iA_{i+1} < A_{i+1}A_{i+2}$. Then, as shown in the figure, the symmetric axis $l \\in P$, but the symmetric point of $A_{i+1}$ is on the arc $A_{i+1}A_{i+2}$ (excluding the points $A_{i+1}$, $A_{i+2}$). However, this contradicts the condition that $A_{i+1}$ and $A_{i+2}$ are adjacent.\n\nHence, the points in $S$ are the vertices of a regular $n$-gon ($n$-sided polygon), when $m = n$. On the other hand, it is obvious that the regular $n$-gon has exactly $n$ symmetric axes. Therefore, the points in $S$ are the vertices and the lines in $P$ are the symmetric axes of the regular $n$-gon if and only if $m = n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20231,
"subject": "Mathematics (Olympiad)",
"question": "Denote by $P$ and $Q$ the midpoints of $\\overline{BC}$ and $\\overline{AD}$, respectively.\n\n\n\nShow that the quadrilateral $ABNM$ is cyclic. Also, show that the quadrilateral $MNCD$ is cyclic. What can be concluded about the segment $\\overline{MN}$ and the line $O_1O_2$ connecting the centers of the circumscribed circles of triangles $ABN$ and $CDM$?",
"options": [],
"answer": "See solution",
"solution": "We will show that the quadrilateral $ABNM$ is cyclic.\n\nFrom $\\angle MQN = \\angle NPM = 90^\\circ$, we get that the quadrilateral $MNPQ$ is cyclic. This implies $\\angle PQM + \\angle MNP = 180^\\circ$.\n\nThe segment $\\overline{QP}$ is the midsegment of the trapezium $ABCD$, which means that it is parallel to $AB$. From here we conclude that $\\angle MAB = \\angle DQP$. Now we have\n\n$$\n\\angle MAB = \\angle DQP = 180^\\circ - \\angle PQM = \\angle MNP = 180^\\circ - \\angle BNM,\n$$\n\nwhich is enough to conclude that the quadrilateral $ABNM$ is cyclic.\n\nAnalogously, we show that the quadrilateral $MNCD$ is cyclic.\n\nThis shows that the segment $\\overline{MN}$ is simultaneously a chord for both circumscribed circles of triangles $ABN$ and $CDM$. We conclude that the line $O_1O_2$, connecting the centres of these circles, must bisect the segment $\\overline{MN}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20232,
"subject": "Mathematics (Olympiad)",
"question": "Let $f: \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ be a function such that\n\n$$\nf(a) + b \\mid a^2 + f(a)f(b)\n$$\n\nfor all positive integers $a$ and $b$. Find all such functions $f$.",
"options": [],
"answer": "See solution",
"solution": "We prove by induction that $f(n) = n$ for all positive integers $n$.\n\n**Base case:** For $n = 1$, set $a = b = 1$ in (1):\n\n$$\nf(1) + 1 \\mid 1 + f(1)^2.\n$$\n\nBut $f(1) + 1 \\mid f(1)^2 - 1$, so $f(1) + 1 \\mid 2$. Since $f(1)$ is a positive integer, $f(1) = 1$.\n\n**Inductive step:** Assume $f(n) = n$ for some $n$. Set $a = n + 1$, $b = n$ in (1):\n\n$$\nf(n + 1) + n \\mid n^2 + 2n + 1 + n f(n + 1)\n$$\n\nBut $f(n + 1) + n \\mid n^2 + n f(n + 1)$, so $f(n + 1) + n \\mid 2n + 1$.\n\nBut $f(n + 1) + n \\geq n + 1 > \\frac{2n + 1}{2}$, so $f(n + 1) + n = 2n + 1$, hence $f(n + 1) = n + 1$.\n\nThus, by induction, $f(n) = n$ for all $n$. Finally, $f(n) = n$ satisfies (1). $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20233,
"subject": "Mathematics (Olympiad)",
"question": "Let $S$ be the set of vertices of a regular 24-gon. Find the number of ways to draw 12 segments of equal lengths so that each vertex in $S$ is an endpoint of exactly one of the 12 segments.",
"options": [],
"answer": "See solution",
"solution": "Number the vertices of the 24-gon in order from 1 to 24. Given a positive integer $k$, draw all the sides or diagonals of the 24-gon between vertices whose numbers differ by $k$ modulo 24. For example, if $k = 4$, draw the diagonals connecting vertices $1 \\to 5 \\to 9 \\to 13 \\to 17 \\to 21 \\to 1$, $2 \\to 6 \\to 10 \\to 14 \\to 18 \\to 22 \\to 2$, $3 \\to 7 \\to 11 \\to 15 \\to 19 \\to 23 \\to 3$, and $4 \\to 8 \\to 12 \\to 16 \\to 20 \\to 24 \\to 4$. This forms 4 cycles, each of length 6. For each of these cycles, there are two ways to pair vertices so that each vertex on the cycle is paired with a vertex adjacent to it on the cycle. For example, the cycle $1 \\to 5 \\to 9 \\to 13 \\to 17 \\to 21 \\to 1$ can be connected with the three diagonals $1$-$5$, $9$-$13$, $17$-$21$ or with the three diagonals $5$-$9$, $13$-$17$, $21$-$1$.\n\nGiven any positive integer $k$, the vertices of $S$ will break into such cycles of vertices whose numbers differ by $k$ modulo 24. If the cycles have length 2, there is only one way to draw diagonals that connect vertices along the cycle. If the cycles have even length greater than 2, such as in the example above, there are 2 ways to draw diagonals that connect vertices along the cycle. If the cycles have odd length, there is no way to draw diagonals that connect vertices along the cycle.\n\nConsider cases based on the value of $k$ for $k = 1, 2, 3, \\dots, 12$:\n\n- If $k$ is 1, 5, 7, or 11, the vertices belong to 1 cycle of length 24. So, for each of these values of $k$, there are $2^1$ ways to draw the diagonals. In total, this gives $4 \\times 2^1 = 8$ ways.\n- If $k$ is 2 or 10, the vertices belong to 2 cycles of length 12. So, for each of these values of $k$, there are $2^2$ ways to draw the diagonals. In total, this gives $2 \\times 2^2 = 8$ ways.\n- If $k$ is 3 or 9, the vertices belong to 3 cycles of length 8. So, for each of these values of $k$, there are $2^3$ ways to draw the diagonals. In total, this gives $2 \\times 2^3 = 16$ ways.\n- If $k$ is 4, the vertices belong to 4 cycles of length 6. So, for this $k$, there are $2^4 = 16$ ways to draw the diagonals.\n- If $k$ is 6, the vertices belong to 6 cycles of length 4. So, for this $k$, there are $2^6 = 64$ ways to draw the diagonals.\n- If $k$ is 8, the vertices belong to 8 cycles of length 3. So, for this $k$, there are 0 ways to draw the diagonals.\n- If $k$ is 12, the vertices belong to 12 cycles of length 2. So, for this $k$, there is only 1 way to draw the diagonals.\n\nAltogether, the number of ways to draw the diagonals is:\n\n$$\n8 + 8 + 16 + 16 + 64 + 0 + 1 = 113\n$$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20234,
"subject": "Mathematics (Olympiad)",
"question": "Given a positive integer $k$, determine the maximum value of the expression\n$$\na^{3k-1}b + b^{3k-1}c + c^{3k-1}a + k^2 a^k b^k c^k\n$$\nwhere $a, b, c$ are non-negative real numbers such that $a + b + c = 3k$.",
"options": [],
"answer": "See solution",
"solution": "The required maximum is $(3k - 1)^{3k-1}$ and is achieved, for instance, 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 is sufficient to show that $F(a, b, c) \\le (3k-1)^{3k-1}$ for all non-negative real numbers $a, b, c$ with $a+b+c=3k$.\n\nSince the expression is cyclic in $a, b, c$, we may assume that $b$ lies between $a$ and $c$. In this case,\n$$\nb^{3k-1}c + c^{3k-1}a = c(b^{3k-1} + c^{3k-2}a) \\le c(b^{3k-2}a + c^{3k-2}b),\n$$\nso\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$$\nG(a, b, c) = a^{3k-1} + k^2 a^k b^{k-1} c^k + ab^{3k-3}c + c^{3k-1},\n$$\nand notice that\n$$\n\\begin{align*}\nG(a, b, c) &\\le a^{3k-1} + k^2 a^k (a^{k-1} + c^{k-1}) c^k + a(a^{3k-3} + c^{3k-3}) c + c^{3k-1} \\\\\n&\\le a^{3k-1} + \\binom{3k-1}{2} a^k (a^{k-1} + c^{k-1}) c^k + \\binom{3k-1}{1} a(a^{3k-3} + c^{3k-3}) c + c^{3k-1} \\\\\n&\\le a^{3k-1} + \\binom{3k-1}{k} a^k (a^{k-1} + c^{k-1}) c^k + \\binom{3k-1}{1} a(a^{3k-3} + c^{3k-3}) c + c^{3k-1} \\\\\n&\\le (a+c)^{3k-1}, \\quad \\text{provided that } k \\ge 2.\n\\end{align*}\n$$\n\nHence, if $k \\ge 2$, then $F(a, b, c) \\le b(a + c)^{3k-1}$; this is also true if $k = 1$.\n\nConsequently,\n$$\n\\begin{align*}\nF(a, b, c) &\\le b(a + c)^{3k-1} \\\\\n&\\le \\frac{1}{3k-1} \\left( \\frac{(3k-1)b + (3k-1)(a+c)}{3k} \\right)^{3k} \\\\\n&= (3k-1)^{3k-1},\n\\end{align*}\n$$\nfor all non-negative real numbers $a, b, c$ with $a + b + c = 3k$. The conclusion follows.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20235,
"subject": "Mathematics (Olympiad)",
"question": "A rook starts moving on an infinite chessboard, alternating horizontal and vertical moves. The length of the first move is one square, the second is two squares, the third is three squares, and so on.\n\n1. Is it possible for the rook to arrive at its starting point after exactly $2013$ moves?\n\n2. Find all $n$ for which it is possible for the rook to come back to its starting point after exactly $n$ moves.",
"options": [],
"answer": "See solution",
"solution": "Assign integer coordinates to each square, with the starting point at $(0,0)$. After each move, one coordinate changes by an odd number ($\\pm 1, \\pm 3, \\pm 5, \\dots$), and the other by an even number ($\\pm 2, \\pm 4, \\pm 6, \\dots$).\n\n**(a)** If $(0,0)$ could be reached in exactly $2013$ moves, then it would be possible to choose signs so that\n$$\n\\pm 1 \\pm 3 \\pm 5 \\pm \\dots \\pm 2013 = 0\n$$\nand\n$$\n\\pm 2 \\pm 4 \\pm 6 \\pm \\dots \\pm 2012 = 0.\n$$\nBut for any choice of signs, $\\pm 1 \\pm 3 \\pm \\dots \\pm 2013$ has the same parity as $1 + 3 + \\dots + 2013 = 1007^2$, which is odd. Thus, $0$ cannot be obtained, so $(0,0)$ cannot be reached in $2013$ moves.\n\n**(b)** If $(0,0)$ can be reached in $n$ moves, then\n$$\n\\pm 1 \\pm 3 \\pm 5 \\pm \\dots \\pm \\left(2 \\left\\lfloor \\frac{n+1}{2} \\right\\rfloor - 1\\right) = 0 \\quad \\text{and} \\quad \\pm 2 \\pm 4 \\pm 6 \\pm \\dots \\pm 2 \\left\\lfloor \\frac{n}{2} \\right\\rfloor = 0.\n$$\nThe first relation implies $0$ has the same parity as $1 + 3 + \\dots + (2\\lfloor\\frac{n+1}{2}\\rfloor - 1) = \\lfloor\\frac{n+1}{2}\\rfloor^2$, so $\\lfloor\\frac{n+1}{2}\\rfloor$ must be even, i.e., $4 \\mid n$ or $4 \\mid n+1$.\n\nFrom the second, $0$ has the same parity as $1 + 2 + \\dots + \\lfloor\\frac{n}{2}\\rfloor = \\frac{\\lfloor\\frac{n}{2}\\rfloor(\\lfloor\\frac{n}{2}\\rfloor + 1)}{2}$, so $8$ must divide one of $n-1, n, n+1,$ or $n+2$.\n\nCombining, $n$ must be of the form $8k$ or $8k-1$, $k \\in \\mathbb{N}^*$.\n\nFor each $k \\in \\mathbb{N}^*$, $(0,0)$ can be reached in $8k$ moves, because\n$$\n(1-3-5+7) + (9-11-13+15) + \\dots + ((8k-7)-(8k-5)-(8k-3)+(8k-1)) = 0\n$$\nand\n$$\n(2-4-6+8) + (10-12-14+16) + \\dots + ((8k-6)-(8k-4)-(8k-2)+8k) = 0.\n$$\nIt is also possible for $n = 8k-1$ moves, as similar groupings yield $0$.\n\n**Therefore, all $n = 8k$ or $n = 8k-1$, $k \\in \\mathbb{N}^*$, are possible.**",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20236,
"subject": "Mathematics (Olympiad)",
"question": "Suppose you are given $N$ squares, each with side length $a_i$ for $i = 1, 2, \\ldots, N$. Can you always tile a unit square using these $N$ squares, under the condition that $\\sum_{i=1}^N a_i^2 > 1$ and $a_i < 1$ for all $i$?",
"options": [],
"answer": "See solution",
"solution": "If any $a_k > 1$, then the $k$th square alone covers the unit square. Now, assume $a_i < 1$ for all $i$.\n\nEach $a_i$ satisfies $2^{-k_i} \\leq a_i < 2^{-k_i+1}$ for some integer $k_i$. Replace each square by one with side $2^{-k_i}$; its area decreases by at most $4$ times, so the total area of these new squares is still greater than $1$.\n\nWe can tile the unit square with these new squares by recursively subdividing the unit square into $4$ squares of side $1/2$, then $1/4$, and so on. At each stage, place all available squares of the current size into untiled regions. Since the total area exceeds $1$, eventually the unit square will be covered. Replacing each $2^{-k_i}$ square with the original $a_i$ square (which is at least as large), we obtain a tiling of the unit square with the given squares.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20237,
"subject": "Mathematics (Olympiad)",
"question": "Given an arithmetic sequence $\\{a_n\\}$ with common difference $d \\ne 0$ and $a_{2021} = a_{20} + a_{21}$, find the value of $\\frac{a_1}{d}$.",
"options": [],
"answer": "See solution",
"solution": "By the conditions, we have:\n$$a_{2021} = a_{20} + a_{21}$$\nSubstituting the formula for the $n$-th term:\n$$a_1 + 2020d = (a_1 + 19d) + (a_1 + 20d)$$\n$$a_1 + 2020d = 2a_1 + 39d$$\n$$a_1 + 2020d - 2a_1 - 39d = 0$$\n$$-a_1 + 1981d = 0$$\n$$a_1 = 1981d$$\nTherefore,\n$$\\frac{a_1}{d} = 1981$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20238,
"subject": "Mathematics (Olympiad)",
"question": "Find the integer closest to the value of the expression:\n\n$$\n\\left((7 + \\sqrt{48})^{2023} + (7 - \\sqrt{48})^{2023}\\right)^2 - \\left((7 + \\sqrt{48})^{2023} - (7 - \\sqrt{48})^{2023}\\right)^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let's transform the given expression:\n\n$$\n\\begin{align*}\n&\\left((7 + \\sqrt{48})^{2023} + (7 - \\sqrt{48})^{2023}\\right)^2 - \\left((7 + \\sqrt{48})^{2023} - (7 - \\sqrt{48})^{2023}\\right)^2 \\\\\n&= \\left(A + B\\right)^2 - \\left(A - B\\right)^2 \\\\\n&= \\left[(A + B) - (A - B)\\right] \\cdot \\left[(A + B) + (A - B)\\right] \\\\\n&= (2B) \\cdot (2A) = 4AB,\n\\end{align*}\n$$\n\nwhere $A = (7 + \\sqrt{48})^{2023}$ and $B = (7 - \\sqrt{48})^{2023}$.\n\nNow, $A \\cdot B = \\left((7 + \\sqrt{48})(7 - \\sqrt{48})\\right)^{2023} = (49 - 48)^{2023} = 1^{2023} = 1$.\n\nSo, the value is $4 \\times 1 = 4$.\n\nTherefore, the integer closest to the value of the expression is $4$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20239,
"subject": "Mathematics (Olympiad)",
"question": "Let $s = x^2 + y^2$ and $t = x + y$. Show that\n\n$$\ns^3 \\ge 8t(3s - t^2)(t^2 - 2t - s)\n$$\n\nfor $2s \\ge t^2$ and $t \\ge 0$.\n\nNow let $s = rt$. This transforms the inequality to\n\n$$\nr^3 \\ge 8(3r - t)(t - 2 - r)\n$$\n\nfor $2r \\ge t \\ge 0$.",
"options": [],
"answer": "See solution",
"solution": "Since $r^3 - 8(3r - t)(t - 2 - r) = 8(t - (2r + 1))^2 + r^3 - 8r^2 + 16r - 8 \\ge r^3 - 8r^2 + 16r = r(r - 4)^2 \\ge 0$ for $2r \\ge t \\ge 0$, we are done.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20240,
"subject": "Mathematics (Olympiad)",
"question": "a) The digits which rotate to the same digit are: 0, 1, 2, 5, 8.\n\nb) The digits which rotate to a different digit are 6 and 9.\n\nc) Only the digits in parts a) and b) can be used to form numbers whose rotations are numbers. What are the numbers from 10 to 50 whose rotations are also numbers?\n\nd) From parts a) and b), if a number and its rotation are the same, then the first and last digits of the number must both be 1, 2, 5, 8, or the first digit is 6 and the last 9, or the first 9 and the last 6. What are the numbers from 50 to 200 that stay the same when rotated?",
"options": [],
"answer": "See solution",
"solution": "a) $0, 1, 2, 5, 8$\n\nb) $6$ and $9$\n\nc) The numbers from $10$ to $50$ whose rotations are numbers are: $11, 12, 15, 16, 18, 19, 21, 22, 25, 26, 28, 29$.\n\nd) The numbers from $50$ to $200$ that stay the same when rotated are: $55, 69, 88, 96, 101, 111, 121, 151, 181$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20241,
"subject": "Mathematics (Olympiad)",
"question": "The quadrilateral $ABCD$ is inscribed in a circle with diameter $BD$. Points $A'$ and $B'$ are symmetric to $A$ and $B$ with respect to the lines $BD$ and $AC$, respectively. If the lines $A'C$ and $BD$ intersect at $P$, and $AC$ and $B'D$ intersect at $Q$, prove that $PQ$ is perpendicular to $AC$.",
"options": [],
"answer": "See solution",
"solution": "Let $AC$ intersect $BD$ at $R$. Then $\\angle BAR = \\angle BAC = \\angle BA'P = \\angle BAP$. That is, $AB$ bisects $\\angle PAR$. As $\\angle BAD = 90^\\circ$, we also have $AD$ is the external bisector of $\\angle PAR$. By the angle bisector theorem, we have\n$$\n\\frac{BR}{BP} = \\frac{DR}{DP} = \\frac{AR}{AP} \\qquad (1)\n$$\nAs $B$ and $B'$ are symmetric with respect to the line $AC$, we have $\\angle BQR = \\angle B'QR = \\angle DQR$. Thus $QR$ bisects $\\angle BQD$. By the angle bisector theorem and (1), we have\n$$\n\\frac{QD}{QB} = \\frac{RD}{RB} = \\frac{PD}{PB}.\n$$\nThus $QP$ is the external bisector of $\\angle BQD$. Hence $\\angle RQP = 90^\\circ$.\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20242,
"subject": "Mathematics (Olympiad)",
"question": "Let the second and third terms of Žan's sequence be $x$ and $y$, respectively. The sequence $3, x, y$ forms a geometric sequence, and $x^2 = 3y$. Additionally, $x, y, 9$ forms an arithmetic sequence, so $2y = x + 9$. Find Žan's sequence.",
"options": [],
"answer": "See solution",
"solution": "Let $y = \\frac{x+9}{2}$. Substitute into $x^2 = 3y$ to get $x^2 = 3 \\cdot \\frac{x+9}{2}$, so $2x^2 = 3x + 27$, or $2x^2 - 3x - 27 = 0$. Factoring, $(2x - 9)(x + 3) = 0$. Since $x > 0$, $x = \\frac{9}{2}$. Then $y = \\frac{9/2 + 9}{2} = \\frac{27}{4}$. Thus, Žan's sequence is:\n\n$$\n3, \\frac{9}{2}, \\frac{27}{4}, 9\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20243,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $n$ is a positive integer with $3 \\leq n \\leq 2020$. For which values of $n$ does there exist a stable assignment of positive integers $a_1, a_2, \\dots, a_n$ arranged in order on a circle such that for every $i$ (indices modulo $n$), the product of three consecutive numbers satisfies $a_i a_{i+1} a_{i+2} = n$?",
"options": [],
"answer": "See solution",
"solution": "If $n$ is not a multiple of 3, then all $a_i$ must be equal, so $a_1^3 = n$ and $n$ must be a perfect cube. If $n$ is a multiple of 3, we can assign $1, 1, n, 1, 1, n, \\dots$ around the circle, which works. Thus, a stable assignment exists if and only if $n$ is a multiple of 3 or a cube.\n\nCount the values:\n- Multiples of 3 in $[3, 2020]$ are $3, 6, 9, \\dots, 2019$, totaling $673$ numbers.\n- Cubes in $[3, 2020]$ are $2^3, 3^3, \\dots, 12^3$ ($11$ cubes), of which $4$ are divisible by 3, so $7$ cubes are not multiples of 3.\n\nTotal: $673 + 7 = 680$ values of $n$.",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 20244,
"subject": "Mathematics (Olympiad)",
"question": "a) Given the equalities $|PQ| = |PD|$, $|PR| = |PC|$, and $\\angle CPR = \\angle DPQ$, prove that $|CQ| = |DR|$.\n\nb) Let $\\angle APQ = \\angle BPR = \\varphi$. Show that the areas of triangles $APQ$ and $BPR$ are equal if $|AP| \\cdot |PQ| = |BP| \\cdot |PR|$, and relate this to the power of a point theorem for $P$ with respect to the circumcircle of $ABCD$.",
"options": [],
"answer": "See solution",
"solution": "a) Since $|PQ| = |PD|$, $|PR| = |PC|$, and $\\angle CPR = \\angle DPQ$, it follows that $\\angle CPQ = \\angle DPR$. By the S--A--S theorem, triangles $RPD$ and $CPQ$ are congruent. Therefore, $|CQ| = |DR|$.\n\nb) Let $\\angle APQ = \\angle BPR = \\varphi$. Then\n\n$$\n2P(APQ) = |AP| \\cdot |PQ| \\cdot \\sin \\varphi \\quad \\text{and} \\quad 2P(BPR) = |BP| \\cdot |PR| \\cdot \\sin \\varphi.\n$$\n\nSo the areas are equal if $|AP| \\cdot |PQ| = |BP| \\cdot |PR|$.\n\nApplying the power of a point theorem to $P$ and the circumcircle of $ABCD$, we get $|AP| \\cdot |PD| = |BP| \\cdot |PC|$. Since $|PQ| = |PD|$ and $|PR| = |PC|$, the desired statement follows.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20245,
"subject": "Mathematics (Olympiad)",
"question": "(A) Determine the value of the real number $k$ for which the polynomial $P(x) = x^3 - kx + 2$ has $2$ as a root. For that value of $k$, write the polynomial $P(x) = x^3 - kx + 2$ as a product of two polynomials with integer coefficients.\n\n(B) The positive real numbers $a, b$ satisfy the equation $2a + b + \\frac{4}{ab} = 10$. Find the maximal value of $a$.",
"options": [],
"answer": "See solution",
"solution": "(A) Number $2$ is a root of the polynomial $P(x) = x^3 - kx + 2$ if and only if $P(2) = 0$, that is:\n\n$$\n2^3 - 2k + 2 = 0 \\implies 8 - 2k + 2 = 0 \\implies k = 5.\n$$\n\nFor $k = 5$ we get:\n\n$$\n\\begin{aligned}\nP(x) &= x^3 - 5x + 2 \\\\\n&= (x - 2)(x^2 + 2x - 1).\n\\end{aligned}\n$$\n\n(B) From the inequality of arithmetic and geometric means:\n\n$$\nb + \\frac{4}{ab} \\ge 2 \\sqrt{b \\cdot \\frac{4}{ab}} = \\frac{4}{\\sqrt{a}}.\n$$\n\nHence,\n\n$$\n2a + \\frac{4}{\\sqrt{a}} \\le 10.\n$$\n\nLet $\\sqrt{a} = x$, then $a = x^2$ and the inequality becomes:\n\n$$\nx^2 + \\frac{2}{x} \\le 5 \\implies x^3 - 5x + 2 \\le 0.\n$$\n\nFrom part (A), $x^3 - 5x + 2 = (x-2)(x^2 + 2x - 1)$. For $x > 2$, the expression is positive, so $x \\le 2$, thus $a \\le 4$.\n\nIn fact, this is the maximal value of $a$, because for $a = 4$, $b = 1$ satisfies the original equation.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20246,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ be non-negative numbers satisfying the following conditions simultaneously:\n\n$$\n\\sum_{i=1}^{n} (a_i + b_i) = 1\n$$\n\n$$\n\\sum_{i=1}^{n} i(a_i - b_i) = 0\n$$\n\n$$\n\\sum_{i=1}^{n} i^2 (a_i + b_i) = 10\n$$\n\nProve that $\\max\\{a_k, b_k\\} \\le \\frac{10}{10 + k^2}$ for all $1 \\le k \\le n$.",
"options": [],
"answer": "See solution",
"solution": "For any $1 \\le k \\le n$, it follows from the given conditions and Cauchy's Inequality that\n\n$$\n\\begin{align*}\n(ka_k)^2 &\\le \\left(\\sum_{i=1}^n i a_i\\right)^2 = \\left(\\sum_{i=1}^n i b_i\\right)^2 \\\\\n&\\le \\left(\\sum_{i=1}^n i^2 b_i\\right) \\left(\\sum_{i=1}^n b_i\\right) \\\\\n&= (10 - \\sum_{i=1}^n i^2 a_i) (1 - \\sum_{i=1}^n a_i) \\\\\n&\\le (10 - k^2 a_k) (1 - a_k) \\\\\n&= 10 - (10 + k^2)a_k + k^2 a_k^2.\n\\end{align*}\n$$\n\nIt follows that $a_k \\le \\frac{10}{10 + k^2}$. Similarly, $b_k \\le \\frac{10}{10 + k^2}$, and hence the result follows. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20247,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$ lying in the interior. Let $E$ and $F$ be the midpoints of the segments $BC$ and $AD$, respectively. Let $X$ be the point lying on the same side of the line $EF$ as the vertex $C$ such that $\\triangle EXF$ and $\\triangle BOA$ are similar. Prove that $XC = XD$.",
"options": [],
"answer": "See solution",
"solution": "Let $M$ be the midpoint of $CD$. Let $EM$ intersect $AD$ at $Z$ and $FM$ intersect $BC$ at $Y$. As $F$, $M$, and $E$ are midpoints of $AD$, $DC$, and $BC$ respectively, we have that $FM \\parallel AC$ and $EM \\parallel BD$. Thus $\\angle EZF = \\angle BDA = \\angle ACB = \\angle FYE$, hence $FEYZ$ is cyclic.\n\nNote that since $\\triangle EXF \\sim \\triangle BOA$, we have that $XF = XE$ and also that $\\angle EXF = \\angle BOA = 2\\angle ACB = 2\\angle FYE$, hence $X$ is the center of $FEYZ$. Since $MD = MC$, apply the converse of Butterfly Theorem and get that $XM \\perp DC$, thus $XD = XC$.\n\n",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20248,
"subject": "Mathematics (Olympiad)",
"question": "A selection of 3 whatsits, 7 doovers, and 1 thingy costs a total of $329. A selection of 4 whatsits, 10 doovers, and 1 thingy costs a total of $441. What is the total cost, in dollars, of 1 whatsit, 1 doover, and 1 thingy?\n",
"options": [],
"answer": "See solution",
"solution": "Let the required cost be $x$. Let $w$ be the cost of a whatsit, $d$ the cost of a doover, and $t$ the cost of a thingy. Then:\n\n$$\n3w + 7d + t = 329 \\quad (1)\n$$\n\n$$\n4w + 10d + t = 441 \\quad (2)\n$$\n\n$$\nw + d + t = x \\quad (3)\n$$\n\nNow, compute:\n\n$$\n3 \\times (1) - 2 \\times (2):\n$$\n\n$$\n3(3w + 7d + t) - 2(4w + 10d + t) = 3 \\times 329 - 2 \\times 441\n$$\n\n$$\n(9w + 21d + 3t) - (8w + 20d + 2t) = 987 - 882\n$$\n\n$$\n(w + d + t) = 105\n$$\n\nSo, the total cost is $\\boxed{105}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20249,
"subject": "Mathematics (Olympiad)",
"question": "Let $k_1, k_2, \\dots, k_n$ be nonnegative real numbers such that $\\sum k_i = a$ and $\\sum i k_i \\le b$. Show that\n$$\n\\sum k_i^2 \\ge \\frac{a^3}{4b}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Suppose $a$ and $b$ are constants and $k_1, k_2, \\dots, k_n$ are variables satisfying the conditions for which $\\sum k_i^2$ is minimized. The $k_i$'s, $1 \\le i \\le n$, are in decreasing order: if $k_i < k_j$ for some $i < j$, replacing both with $\\frac{1}{2}(k_i + k_j)$ decreases $\\sum k_i^2$.\n\nWe claim that $k_1, k_2, \\dots, k_n$ must be a block of an arithmetic progression unless they are zero. Suppose $k_{j-1}, k_j, k_{j+1}$ are nonzero and $k_j \\ne \\frac{k_{j-1} + k_{j+1}}{2}$. Changing these three to $k_{j-1} + x, k_j - 2x, k_{j+1} + x$ preserves the constraints. The change in $\\sum k_i^2$ is\n$$\n\\Delta = ((k_{j-1} + x)^2 + (k_j - 2x)^2 + (k_{j+1} + x)^2) - (k_{j-1}^2 + k_j^2 + k_{j+1}^2) = 6x^2 + 2x(k_{j-1} + k_j + k_{j+1}).\n$$\nSince the coefficient of $x$ is nonzero, we can choose $x$ so $\\Delta < 0$. Thus, every three consecutive nonzero $k_i$'s form an arithmetic progression. By adjusting $n$ if necessary, we may assume $k_i = r - s i$ for constants $r, s \\ge 0$.\n\nWe have\n$$\n\\begin{aligned}\na &= n r - s \\sum i \\\\\nc &= r \\sum i - s \\sum i^2 \\le b.\n\\end{aligned}\n$$\nThus,\n$$\n\\sum k_i^2 = n r^2 - 2 r s \\sum i + s^2 \\sum i^2 = r (n r - s \\sum i) - s (r \\sum i - s \\sum i^2) = r a - s c.\n$$\nThe values of $r$ and $s$ can be found from the linear system above:\n$$\nr = \\frac{a \\sum i^2 - c \\sum i}{n \\sum i^2 - (\\sum i)^2}, \\quad s = \\frac{a \\sum i - n c}{n \\sum i^2 - (\\sum i)^2}.\n$$\nBy the Cauchy-Schwarz inequality,\n$$\n\\sum k_i^2 \\ge \\frac{(\\sum k_i)^2}{n} = \\frac{a^2}{n}.\n$$\nSince $r - s c$ is negative,\n$$\n\\begin{align*}\n& a \\sum i^2 - c \\sum i - a n \\sum i + n^2 c \\ge 0 \\\\\n\\Rightarrow \\quad & a\\left(\\frac{n^2(n+1)}{2} - \\frac{n(n+1)(2n+1)}{6}\\right) \\le c\\left(n^2 - \\frac{n(n+1)}{2}\\right) \\le \\frac{c n(n+1)}{2} \\\\\n\\Rightarrow \\quad & a\\left(n - \\frac{2n+1}{3}\\right) \\le c \\\\\n\\Rightarrow \\quad & n \\le 3\\frac{c}{a} + 1 \\le 4\\frac{c}{a}.\n\\end{align*}\n$$\nSince $c \\ge a$ (because $c = \\sum i k_i \\ge \\sum k_i = a$),\n$$\n\\sum k_i^2 \\ge \\frac{a^2}{n} = \\frac{a^3}{a n} \\ge \\frac{a^3}{4 c} \\ge \\frac{a^3}{4 b}\n$$\nas desired.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20250,
"subject": "Mathematics (Olympiad)",
"question": "Let $G$ be a group of order $n$ and let $e$ be the identity element. Find all functions $f : G \\to \\mathbb{N}^*$ such that the following conditions both hold:\n\n(a) $f(x) = 1$ if and only if $x = e$,\n\n(b) $f(x^k) = \\dfrac{f(x)}{(f(x), k)}$, for all positive divisors $k$ of $n$. Here $(r, s)$ is the greatest common divisor of $r$ and $s$.",
"options": [],
"answer": "See solution",
"solution": "Let $x$ be an element of $G$ and $\\operatorname{ord} x$ its order. Since $\\operatorname{ord} x$ divides $n$, we get $1 = f(e) = f(x^{\\operatorname{ord} x}) = \\dfrac{f(x)}{(f(x), \\operatorname{ord} x)}$, so $f(x)$ is a divisor of $\\operatorname{ord} x$. Hence $f(x)$ divides $n$, so $f(x^{f(x)}) = \\dfrac{f(x)}{(f(x), f(x))} = 1$ and consequently $x^{f(x)} = e$. Then $\\operatorname{ord} x$ is a divisor of $f(x)$, so $f(x) = \\operatorname{ord} x$.\n\nConversely, the function $f: G \\to \\mathbb{N}^*$, $f(x) = \\operatorname{ord} x$, satisfies the claim. To this end, let $x$ be an element of $G$, $m = \\operatorname{ord} x$, $k \\in \\mathbb{N}^*$, $p = \\operatorname{ord} x^k$ and $d = (m, k)$. Since $(x^k)^{m/d} = (x^m)^{k/d} = e$, it follows that $p$ divides $m/d$. On the other hand, $x^{kp} = (x^k)^p = e$, hence $m$ divides $kp$ and consequently $m/d$ divides $(k/d)p$. Notice that the numbers $m/d$ and $k/d$ are coprime, implying that $m/d$ divides $p$, hence $p = m/d$.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 20251,
"subject": "Mathematics (Olympiad)",
"question": "Is it possible to construct a set of exactly 2024 positive integers, all less than 8000, such that no pair of members of the set is coprime, and yet no prime number divides all integers in the set?",
"options": [],
"answer": "See solution",
"solution": "We can construct such a set as follows:\n\nLet $S = \\{6a, 10b, 15c\\}$, where $a \\leq 1333$, $b < 800$, and $c \\leq 533$, so that all elements are less than 8000. The total number of elements is $1333 + 799 + 533 - 2 \\cdot 266 = 2133$, accounting for the overcount of multiples of 30.\n\nBy removing any 109 elements from $S$, we obtain a set of 2024 elements. No pair in this set is coprime, since every pair shares a factor of 2, 3, or 5. However, no single prime divides all elements, because the greatest common divisor of $6, 10, 15$ is 1, and after removing any 109 elements, at least one pair of consecutive multiples of each type remains. Thus, the answer is yes: such a set exists.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20252,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$, $d$ be positive real numbers satisfying the system of equations\n\n$$\n\\begin{cases}\na^2 + \\frac{1}{b^2} = \\frac{1}{2}, \\\\\nb^2 + \\frac{4}{c^2} = 8, \\\\\nc^2 + \\frac{16}{d^2} = 2, \\\\\nd^2 + \\frac{4}{a^2} = 32.\n\\end{cases}\n$$\n\nDetermine the product $abcd$.",
"options": [],
"answer": "See solution",
"solution": "Multiplying all equations gives\n\n$$\n\\left(a^2 + \\frac{1}{b^2}\\right) \\left(b^2 + \\frac{4}{c^2}\\right) \\left(c^2 + \\frac{16}{d^2}\\right) \\left(d^2 + \\frac{4}{a^2}\\right) = 2^8.\n$$\n\nBy AM-GM, $a^2 + \\frac{1}{b^2} \\ge 2 \\cdot \\frac{a}{b}$, with equality if and only if $a = \\frac{1}{b}$. Similarly, $b^2 + \\frac{4}{c^2} \\ge 4 \\cdot \\frac{b}{c}$ (equality if $b = \\frac{2}{c}$), $c^2 + \\frac{16}{d^2} \\ge 8 \\cdot \\frac{c}{d}$ (equality if $c = \\frac{4}{d}$), and $d^2 + \\frac{4}{a^2} \\ge 4 \\cdot \\frac{d}{a}$ (equality if $d = \\frac{2}{a}$). Multiplying these inequalities gives\n\n$$\n\\left(a^2 + \\frac{1}{b^2}\\right) \\left(b^2 + \\frac{4}{c^2}\\right) \\left(c^2 + \\frac{16}{d^2}\\right) \\left(d^2 + \\frac{4}{a^2}\\right) \\ge 2 \\cdot \\frac{a}{b} \\cdot 4 \\cdot \\frac{b}{c} \\cdot 8 \\cdot \\frac{c}{d} \\cdot 4 \\cdot \\frac{d}{a} = 2^8.\n$$\n\nEquality must hold, so all four inequalities are equalities. Thus,\n\n$$\n\\begin{cases}\na = \\frac{1}{b}, \\\\\nb = \\frac{2}{c}, \\\\\nc = \\frac{4}{d}, \\\\\nd = \\frac{2}{a}.\n\\end{cases}\n$$\n\nMultiplying these equations gives $abcd = \\frac{16}{abcd}$, so $(abcd)^2 = 16$. Since $a, b, c, d > 0$, we have $abcd = 4$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20253,
"subject": "Mathematics (Olympiad)",
"question": "Let $f: \\mathbb{Z}_{>0} \\to \\mathbb{Z}$ be a function with the following properties:\n\n1. $f(1) = 0$\n2. $f(p) = 1$ for all prime numbers $p$\n3. $f(xy) = y f(x) + x f(y)$ for all $x, y \\in \\mathbb{Z}_{>0}$\n\nDetermine the smallest integer $n \\ge 2015$ that satisfies $f(n) = n$.",
"options": [],
"answer": "See solution",
"solution": "We claim that\n\n$$\nf(q_1 \\cdots q_s) = q_1 \\cdots q_s \\left( \\frac{1}{q_1} + \\cdots + \\frac{1}{q_s} \\right)\n$$\n\nholds for (not necessarily distinct) prime numbers $q_1, \\dots, q_s$.\n\nWe prove the claim by induction on $s$. For $s=0$, the claim reduces to $f(1) = 0$, which is true by assumption.\n\nIf the formula holds for some $s$, then\n\n$$\n\\begin{aligned}\nf(q_1 \\cdots q_s q_{s+1}) &= f((q_1 \\cdots q_s)q_{s+1}) \\\\\n&= q_{s+1} f(q_1 \\cdots q_s) + q_1 \\cdots q_s f(q_{s+1}) \\\\\n&= q_1 \\cdots q_{s+1} \\left( \\frac{1}{q_1} + \\cdots + \\frac{1}{q_s} \\right) + q_1 \\cdots q_s \\\\\n&= q_1 \\cdots q_{s+1} \\left( \\frac{1}{q_1} + \\cdots + \\frac{1}{q_s} + \\frac{1}{q_{s+1}} \\right).\n\\end{aligned}\n$$\n\nThus, the formula holds for all $s$ by induction.\n\nIt is easily verified that the function given by this formula fulfills the given functional equation.\n\nLet $p_1, \\dots, p_r$ be distinct primes and $\\alpha_1, \\dots, \\alpha_r$ be positive integers. Then collecting equal primes in the formula leads to\n\n$$\nf(p_1^{\\alpha_1} \\cdots p_r^{\\alpha_r}) = p_1^{\\alpha_1} \\cdots p_r^{\\alpha_r} \\sum_{j=1}^{r} \\frac{\\alpha_j}{p_j}.\n$$\n\nWe now determine all $n \\ge 2015$ with $f(n) = n$. Write $n = p_1^{\\alpha_1} \\cdots p_r^{\\alpha_r}$. Then\n\n$$\n\\frac{\\alpha_1}{p_1} + \\cdots + \\frac{\\alpha_r}{p_r} = 1.\n$$\n\nLet\n\n$$\n\\frac{\\alpha_1}{p_1} + \\cdots + \\frac{\\alpha_{r-1}}{p_{r-1}} = \\frac{a}{p_1 \\cdots p_{r-1}}\n$$\n\nfor some non-negative integer $a$. Then\n\n$$\n\\frac{a}{p_1 \\cdots p_{r-1}} + \\frac{\\alpha_r}{p_r} = 1 \\iff a p_r + \\alpha_r p_1 \\cdots p_{r-1} = p_1 \\cdots p_r.\n$$\n\nAs $p_r$ is coprime to $p_1 \\cdots p_{r-1}$, we conclude that $p_r \\mid \\alpha_r$. Since $\\alpha_r \\le p_r$, we must have $r=1$ and $\\alpha_r = p_r$.\n\nThus $f(n) = n$ holds if and only if $n = p^p$ for some prime $p$.\n\nWe have\n\n$$\n2^2 = 4 < 3^3 = 27 < 2015 < 5^5 = 3125,\n$$\n\nso the smallest such $n$ is $3125$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20254,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be the side length of a hexagonal board. On this board, checkers are placed such that no more than $k$ checkers appear in any row, where rows are defined in three directions: horizontal, and at angles $60^\text{\\degree}$ and $120^\text{\\degree}$. What is the maximum number $q$ of checkers that can be placed under these constraints?\n\nb) Find $T(1, n)$ and $T(2, n)$, where $T(k, n)$ denotes the maximum number of checkers for given $k$ and $n$. Illustrate with examples for $k=1$ and $k=2$.\n\n\n\n\n\n\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $q$ be the total number of checkers placed. Each checker is counted $2n+1$ times across all rows. Let $S$ be the sum over all checkers:\n\n$$\nS = (2n + 1)q.\n$$\n\nLet $x_1, \\dots, x_n$ be the number of checkers in the $n$ horizontal lines, $y_1, \\dots, y_n$ in the $n$ lines at $60^\\text{\\degree}$, and $z_1, \\dots, z_n$ in the $n$ lines at $120^\\text{\\degree}$.\n\nThe $i$-th line has $i$ cells, so:\n\n$$\nS = 1(x_1 + y_1 + z_1) + 2(x_2 + y_2 + z_2) + \\dots + n(x_n + y_n + z_n).\n$$\n\nSince there are $q$ checkers in total and at most $k$ in any row:\n\n$$\nx_1 + \\dots + x_n = y_1 + \\dots + y_n = z_1 + \\dots + z_n = q, \\\\\n0 \\le x_i, y_i, z_i \\le \\min(i, k) \\quad \\text{for } 1 \\le i \\le n.\n$$\n\nLet $q = mk + r$, $0 \\le r < k$. The sum $S$ is maximized when the largest coefficients have the maximum possible sum, i.e., $3k$.\n\nThus,\n\n$$\n\\begin{aligned}\n(2n + 1)q &\\le n \\cdot 3k + (n - 1) \\cdot 3k + \\dots + (n - m + 1) \\cdot 3k + (n - m) \\cdot 3r \\\\\n&\\le \\frac{2n - m + 1}{2} m \\cdot 3k + (n - m) \\cdot 3r \\\\\n&\\le \\frac{6nmk - 3m^2k + 3mk}{2} + 3nr - 3mr \\\\\n&\\le \\frac{6n(mk + r) - 3m^2k + 3(mk + r) - 3r - 6mr}{2} \\\\\n&\\le \\frac{(6n + 3)q - 3m^2k - 3r - 6mr}{2}.\n\\end{aligned}\n$$\n\nMoving $(2n + 1)q$ to the right, multiplying by $2k$, and substituting $mk = q - r$, we get:\n\n$$\n\\begin{aligned}\n0 &\\le (2n + 1)kq - 3(q - r)^2 - 6(q - r)r - 3rk \\\\\n&= (2n + 1)kq - 3q^2 + 6qr - 3r^2 - 6qr + 6r^2 - 3rk \\\\\n&= -3q^2 + (2n + 1)kq + 3r(r - k).\n\\end{aligned}\n$$\n\nThus,\n\n$$\n3q^2 \\le (2n + 1)kq + 3r(r - k) \\le (2n + 1)kq,\n$$\n\nso $q \\le \\frac{2n + 1}{3}k$. Since $q$ is integer, $q \\le \\lfloor \\frac{2n+1}{3}k \\rfloor$.\n\nb) For $k=1$, $T(1, n) = \\lfloor \\frac{2n+1}{3} \\rfloor$; for $k=2$, $T(2, n) = \\lfloor \\frac{4n+2}{3} \\rfloor$. Figures 2, 3, and 4 show placements achieving equality for $k=1$ and $k=2$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20255,
"subject": "Mathematics (Olympiad)",
"question": "Let $m$, $n$, and $r$ be positive integers satisfying $n > m$, where both $m^2 + r$ and $n^2 + r$ are powers of $2$. Prove that:\n\n$$\nn > \\frac{2m^2}{r}.$$",
"options": [],
"answer": "See solution",
"solution": "Let $m^2 + r = 2^k$ and $n^2 + r = 2^l$ with $k, l \\in \\mathbb{Z}_{>0}$.\n\nFirst, we analyze the $2$-adic valuations:\n\n- Since $v_2(m^2) = v_2(2^k - r) = v_2(r)$, we conclude $v_2(r)$ must be even.\n- Let $v_2(r) = 2t$ where $t \\in \\mathbb{Z}_{\\ge 0}$, and write $r = 2^{2t} r_1$, $m = 2^t m_1$.\n- Similarly, $v_2(n^2) = v_2(r)$, so let $n = 2^t n_1$.\n\nDefine $k_1 = k - 2t$ and $l_1 = l - 2t$, which gives the reduced system:\n\n$$\nm_1^2 + r_1 = 2^{k_1},\n$$\n$$\nn_1^2 + r_1 = 2^{l_1}.\n$$\n\n**Claim:** $r_1 \\ge 7$.\n\n- Since $r_1$ is odd, both $m_1$ and $n_1$ must be odd.\n- As $n_1 \\ge 3$ (because $n > m \\ge 1$), we have $l_1 > 3$.\n- Thus $n_1^2 + r_1 \\equiv 1 + r_1 \\equiv 0 \\pmod{8}$, implying $r_1 \\equiv 7 \\pmod{8}$.\n- Therefore $r_1 \\ge 7$.\n\nThis also shows $k_1 \\ge 3$. Now consider the difference:\n\n$$\n2^{l_1} - 2^{k_1} = (n_1 - m_1)(n_1 + m_1).\n$$\n\nSince $2^{k_1}$ divides the product $(n_1 - m_1)(n_1 + m_1)$, and both factors are even but cannot both be divisible by $4$ (which would imply $m_1$ and $n_1$ are both even), we have:\n\n$$\n2^{k_1-1} \\mid (n_1 - m_1) \\quad \\text{or} \\quad 2^{k_1-1} \\mid (n_1 + m_1).\n$$\n\nThis leads to the lower bound:\n\n$$\n\\begin{align*}\nn_1 &\\ge 2^{k_1-1} - m_1 = \\frac{1}{2}(m_1^2 + r_1) - m_1 \\\\\n&= \\frac{3m_1^2}{7} + \\left( \\frac{m_1^2}{14} + \\frac{r_1}{2} - m_1 \\right) \\\\\n&\\ge \\frac{3m_1^2}{r_1} + \\left( \\frac{m_1^2}{14} + \\frac{7}{2} - m_1 \\right) \\quad (\\text{since } r_1 \\ge 7) \\\\\n&= \\frac{3m_1^2}{r_1} + \\frac{1}{14}(m_1 - 7)^2 \\\\\n&\\ge \\frac{3m_1^2}{r_1} = \\frac{3m^2}{r}.\n\\end{align*}\n$$\n\nTherefore, we conclude:\n\n$$\nn = 2^t n_1 \\ge n_1 \\ge \\frac{3m^2}{r} > \\frac{2m^2}{r},$$\n\nwhich completes the proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20256,
"subject": "Mathematics (Olympiad)",
"question": "Let $n, k$ be positive integers satisfying $n \\ge k$.\n\nThere is a group consisting of $n$ people. Each person of this group belongs to one and only one of $k$ clubs, called club $C_1, C_2, \\dots, C_k$. Each club has at least one member. Prove that it is possible to distribute $n^2$ pieces of cake to these $n$ people in such a way that:\n\n* Everyone receives at least 1 piece of cake.\n* For each $i$ ($1 \\le i \\le k$), every member of club $C_i$ receives $a_i$ pieces of cake.\n* If $1 \\le i < j \\le k$, then $a_i > a_j$.",
"options": [],
"answer": "See solution",
"solution": "Let $x_i$ be the number of people in club $C_i$ for $1 \\le i \\le k$.\n\nSet $a_i = x_i + 2(x_{i+1} + x_{i+2} + \\dots + x_k)$ for each $i$. We claim that $a_1, a_2, \\dots, a_k$ satisfy all the conditions.\n\n- Each $a_i > 0$ since $x_i \\ge 1$.\n- For $1 \\le i < k$, $a_i = x_i + x_{i+1} + a_{i+1} > a_{i+1}$, so $a_i > a_j$ for $i < j$.\n- The total number of cakes distributed is:\n\n$$\n\\sum_{i=1}^{k} a_i x_i = \\sum_{i=1}^{k} x_i^2 + 2 \\sum_{i=1}^{k-1} \\sum_{j=i+1}^{k} x_i x_j = (x_1 + x_2 + \\dots + x_k)^2 = n^2.\n$$\n\nThus, such a distribution is possible.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 20257,
"subject": "Mathematics (Olympiad)",
"question": "Sean $a$, $b$, $c$, $d$ cuatro números reales positivos. Si se cumple\n\n$$\na + b + \\frac{1}{ab} = c + d + \\frac{1}{cd} \\quad \\text{y} \\quad \\frac{1}{a} + \\frac{1}{b} + ab = \\frac{1}{c} + \\frac{1}{d} + cd\n$$\n\ndemuestra que al menos dos de los valores $a$, $b$, $c$, $d$ son iguales.",
"options": [],
"answer": "See solution",
"solution": "Sea $u = a + b + \\frac{1}{ab}$ y $v = \\frac{1}{a} + \\frac{1}{b} + ab$. Denotamos $r = \\frac{1}{ab}$. Entonces tenemos que $a + b + r = u$, $ab + br + ra = v$ y $abr = 1$. Por las identidades de Cardano-Vieta, $a$, $b$ y $r$ son las tres raíces del polinomio $p(x) = x^3 - u x^2 + v x - 1$. Por la misma razón, $c$, $d$ y $\\frac{1}{cd}$ son estas mismas tres raíces. Como $p(x)$ solo tiene tres raíces, los valores $a, b, c, d$ no pueden ser todos distintos.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20258,
"subject": "Mathematics (Olympiad)",
"question": "Find the smallest positive integer $n$ with the following property:\n\nFor any sequence $a_1, a_2, \\dots, a_{2021}$ of real numbers satisfying\n\n$0 < a_1, a_2, \\dots, a_{2021} < 2$ and $a_1 + a_2 + \\dots + a_{2021} = 2021,$\n\nthere is a sequence $b_1, b_2, \\dots, b_n$ of real numbers satisfying\n\n$0 < b_1, b_2, \\dots, b_n < 2$ and $b_1 + b_2 + \\dots + b_n = n$\n\nand a permutation $c_1, c_2, \\dots, c_{n+2021}$ of the sequence $a_1, a_2, \\dots, a_{2021}, b_1, b_2, \\dots, b_n$ such that\n\n$$\n\\begin{cases}\n c_1 + c_2 + \\dots + c_l \\le l, & \\text{for all } 1 \\le l \\le n+2021 \\text{ odd} \\\\\n c_1 + c_2 + \\dots + c_l \\ge l, & \\text{for all } 1 \\le l \\le n+2021 \\text{ even}\n\\end{cases}\n$$",
"options": [],
"answer": "See solution",
"solution": "The answer is $n = 2021$.\n\nFirst, we show that $n = 2021$ is the minimum.\n\nLet $N = 2021$ and $m = n + N$. Suppose the sequence $c_1, c_2, \\dots, c_m$ satisfies\n\n$$\n\\begin{cases}\n c_1 + c_2 + \\dots + c_l \\le l, & \\text{for all } 1 \\le l \\le m \\text{ odd} \\\\\n c_1 + c_2 + \\dots + c_l \\ge l, & \\text{for all } 1 \\le l \\le m \\text{ even}\n\\end{cases} \\quad (*)\n$$\n\nThen $c_1 \\le 1$ and $c_2 \\ge 2 - c_1 \\ge 1$. Similarly, $c_l \\le l - (c_1 + \\dots + c_{l-1}) \\le 1$ for any odd $l$, and $c_l \\ge l - (c_1 + \\dots + c_{l-1}) \\ge 1$ for any even $l$.\n\nLet $X = \\{1 \\le l \\le m \\mid c_l \\le 1\\}$ and $Y = \\{1 \\le l \\le m \\mid c_l > 1\\}$. Then $X$ contains all odd $l$, so $|X| \\ge |Y|$. Thus $m = |X| + |Y| \\ge 2|Y|$.\n\nNow let $\\varepsilon = \\frac{1}{N}$ and consider the sequence $a_1 = 1 - (N-1)\\varepsilon$, $a_2 = \\dots = a_N = 1 + \\varepsilon$. Suppose $b_1, \\dots, b_n$ and $c_1, \\dots, c_m$ are chosen to satisfy the problem's conditions. We prove $m \\ge 2N$.\n\n(i) If $b_k > 1$ for some $k$, then $|Y| \\ge N$, so $m \\ge 2N$.\n\n(ii) If $b_k \\le 1$ for all $k$, then $b_1 = \\dots = b_n = 1$ since $b_1 + \\dots + b_n = n$. Thus $c_1, \\dots, c_m$ is a permutation of\n\n$$\n1 - (N - 1)\\varepsilon, \\underbrace{1 + \\varepsilon, \\dots, 1 + \\varepsilon}_{N-1}, \\underbrace{1, \\dots, 1}_{n}\n$$\n\nsatisfying $(*)$. We claim $c_2 = 1$. Suppose $c_2 = 1 + \\varepsilon$. Then $c_1 + c_2 + c_3 \\le 3$ implies either $c_1 = 1 - (N-1)\\varepsilon$ or $c_3 = 1 - (N-1)\\varepsilon$. In either case, $c_1 + c_2 + c_3 + c_4 \\le 4 - (N-4)\\varepsilon < 4$, a contradiction. Hence $c_2 = 1$.\n\nIt follows $X$ contains $2$ and all odd $l$, so $|X| \\ge |Y| + 2$. Since $|Y| = N - 1$, $m = |X| + |Y| \\ge 2N$.\n\nThis proves $n \\ge N$. Now we prove $n = N$ works. Let $0 < a_1, \\dots, a_N < 2$ with $a_1 + \\dots + a_N = N$ and let $b_1 = 2 - a_1, \\dots, b_N = 2 - a_N$. Then $0 < b_1, \\dots, b_N < 2$ and $b_1 + \\dots + b_N = N$.\n\nReindex so\n\n$$\n0 < a_1 \\le a_2 \\le \\dots \\le a_k \\le 1 < a_{k+1} \\le \\dots \\le a_N < 2.\n$$\n\nThen the sequence $a_1, b_1, a_2, b_2, \\dots, a_k, b_k, b_{k+1}, a_{k+1}, \\dots, b_N, a_N$ satisfies the required property.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20259,
"subject": "Mathematics (Olympiad)",
"question": "A chessboard has the form of an $n \\times n$ torus, where $n = 2^{2k} + 1$. Prove that $n$ queens placed in the cells $(i, 2^k \\cdot i)$ for $1 \\leq i \\leq n$ (with multiplication modulo $n$) do not attack each other.\n\n",
"options": [],
"answer": "See solution",
"solution": "Each vertical line contains exactly one queen. Since $2^k$ is invertible modulo $2^{2k} + 1$, each horizontal line also contains exactly one queen. Similarly, each diagonal contains exactly one queen because $2^k \\pm 1$ is invertible modulo $2^{2k} + 1$.",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 20260,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(xf(y) - f(x)) = 2f(x) + xy\n$$\nfor all real numbers $x, y$.",
"options": [],
"answer": "See solution",
"solution": "By taking $x = 1$ in (1), we get\n$$\nf(f(y) - f(1)) = y + 2f(1), \\quad \\forall y \\in \\mathbb{R}.\n$$\nHence $f$ is bijective, so there exists a unique real number $a$ such that $f(a) = 0$. Plugging $x = a$ into (1), we have\n$$\nf(af(y)) = ay, \\quad \\forall y \\in \\mathbb{R}.\n$$\nPlugging $y = 0$ into the above, we have $f(af(0)) = 0 = f(a)$. By injectivity, $af(0) = a$, so $a = 0$ or $f(0) = 1$.\n\nConsider $a = 0$, i.e., $f(0) = 0$. Plugging $y = 0$ into (1), we have $f(-f(x)) = 2f(x)$. Since $f$ is surjective, we conclude $f(x) = -2x$ for all $x$, but this does not satisfy the original equation. Hence, $a \\neq 0$ and $f(0) = 1$.\n\nPlugging $x = 0$ into (1), we have $f(-1) = 2$. Plugging $y = a$ into the earlier result, $a^2 = f(0) = 1$, so $a = 1$ (since $f(-1) = 2$), i.e., $f(1) = 0$. Since $f(1) = 1$, we can write (2) as\n$$\nf(f(y)) = y, \\quad \\forall y \\in \\mathbb{R}.\n$$\nPlugging $f(y)$ for $y$ in (1) and using the above, we get\n$$\nf(xy - f(x)) = 2f(x) + x f(y), \\quad \\forall x, y \\in \\mathbb{R}.\n$$\nFor $x \\neq 0$, put $y = \\frac{f(x)}{x}$:\n$$\n1 = 2f(x) + x f\\left(\\frac{f(x)}{x}\\right),\n$$\nso\n$$\nf\\left(\\frac{f(x)}{x}\\right) = \\frac{1 - 2f(x)}{x}, \\quad x \\neq 0.\n$$\nNow, put $y = \\frac{f(x)}{x}$ into (1) and use the above:\n$$\nf(1 - 3f(x)) = 3f(x), \\quad x \\neq 0.\n$$\nSince $f$ is bijective and $f(0) = 1$, for all $x \\neq 0$, $1 - 3f(x)$ can take all real values except $-2$. Therefore,\n$$\nf(x) = -x + 1, \\quad \\forall x \\neq -2.\n$$\nIn particular, $f(3) = -2$. Plugging $y = 3$ into $f(f(y)) = y$, we have $f(-2) = 3$. Therefore, $f(x) = -x + 1$ for all real numbers $x$. It is easy to check that this function satisfies the original equation. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20261,
"subject": "Mathematics (Olympiad)",
"question": "Prove that for every integer $n$, the number $n^4 - 12n^2 + 144$ is not a perfect cube of an integer.",
"options": [],
"answer": "See solution",
"solution": "Suppose otherwise and let $m \\in \\mathbb{Z}$ be such that $n^4 - 12n^2 + 144 = m^3$. Firstly, $m$ is clearly positive and we can assume that $n$ is a positive integer (since $n = 0$ clearly doesn't work). Note that the polynomial $x^4 - 12x^2 + 144$ may be factored as\n\n$$\nx^4 - 12x^2 + 144 = (x^2 + 12)^2 - (6x)^2 = (x^2 - 6x + 12)(x^2 + 6x + 12).\n$$\n\nBy repeatedly applying Euclid's algorithm, we get\n\n$$\n\\gcd(n^2 - 6n + 12, n^2 + 6n + 12) = \\gcd(n^2 - 6n + 12, 12n), \\quad (1)\n$$\n\nand then\n\n$$\n\\gcd(n^2 - 6n + 12, n) = \\gcd(n^2 - 6n + 12 - (n - 6)n, n) = \\gcd(12, n). \\quad (2)\n$$\n\nWe now distinguish three cases.\n\n**Case I.** $n$ is even. Write $n = 2k$ for some $k \\in \\mathbb{N}$ and we have $16(k^4 - 3k^2 + 9) = m^3$. Hence $m$ is divisible by 4 which means that $m^3$ is divisible by 64. But $k^4 - 3k^2 + 9$ is always odd, so $16(k^4 - 3k^2 + 9)$ cannot be divisible by 64, a contradiction.\n\n**Case II.** $n$ is divisible by 3. Write $n = 3l$ for some $l \\in \\mathbb{N}$ and we have $9(9l^4 - 3l^2 + 16) = m^3$. Hence $m$ is divisible by 3 which means $m^3$ is divisible by 27. But $9l^4 - 3l^2 + 16$ is not divisible by 3, so $9(9l^4 - 3l^2 + 16)$ cannot be divisible by 27, a contradiction.\n\n**Case III.** Suppose that $\\gcd(n, 6) = 1$. Then $\\gcd(n^2 - 6n + 12, n^2 + 6n + 12) = 1$ by (1) and (2) (since $\\gcd(n^2 - 6n + 12, 12) = \\gcd(n(n-6), 12) = 1$, because $\\gcd(n, 6) = 1$), thus $n^2 - 6n + 12 = (n-3)^2 + 3$ and $n^2 + 6n + 12 = (n+3)^2 + 3$ are both perfect cubes of integers. However, we get a contradiction with the following lemma.\n\n**Lemma 1.** For every even integer $x$, the number $x^2+3$ is not a perfect cube of an integer.\n\n*Proof.* Suppose otherwise, namely that there exists a positive integer $y$ such that $x^2+3 = y^3$. The last relation modulo 4 gives $y \\equiv -1 \\pmod 4$. The equation then turns to\n\n$$\nx^2 + 2^2 = y^3 + 1 = (y + 1)(y^2 - y + 1).\n$$\n\nBut we have $y^2 - y + 1 \\equiv (-1)^2 - (-1) + 1 \\equiv -1 \\pmod 4$, hence there exists a prime number $p \\equiv -1 \\pmod 4$ such that $p \\mid y^2 - y + 1 \\mid x^2 + 2^2$. But it is well known that from here we should have $p \\mid x$ and $p \\mid 2$. This means that $p = 2$, which contradicts $p \\equiv -1 \\pmod 4$. $\\Box$\n\nWe conclude that no such integer $m$ can exist, thus $n^4 - 12n^2 + 144$ is never a perfect cube of an integer. $\\Box$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20262,
"subject": "Mathematics (Olympiad)",
"question": "Let $D$ be the point on the side $\\overline{BC}$ such that $\\angle DAC = 90^\\circ$. Let $\\varphi = \\angle CDA$ and $x = |CD|$. Then $\\cos \\angle ACB = \\sin \\varphi$.\n\n\n\nGiven that $|AC| = x \\sin \\varphi$ and $|BD| = |AD| = x \\cos \\varphi$. Also, $\\angle BAD = \\frac{\\varphi}{2}$ and $|AB| = 2x \\cos \\varphi \\cos \\frac{\\varphi}{2}$.\n\nIf $|BC| + |AC| = 2|AB|$, find $\\cos \\angle ACB$.",
"options": [],
"answer": "See solution",
"solution": "Since $|BC| + |AC| = 2|AB|$, we have:\n\n$$\n1 + \\cos \\varphi + \\sin \\varphi = 4 \\cos \\varphi \\cos \\frac{\\varphi}{2}.\n$$\n\nSquaring both sides:\n\n$$\n1 + \\cos^2 \\varphi + \\sin^2 \\varphi + 2 \\cos \\varphi + 2 \\sin \\varphi + 2 \\sin \\varphi \\cos \\varphi = 16 \\cos^2 \\varphi \\cos^2 \\frac{\\varphi}{2}.\n$$\n\nFurther simplification gives:\n\n$$\n\\begin{aligned}\n2(1 + \\cos \\varphi)(1 + \\sin \\varphi) &= 8 \\cos^2 \\varphi(1 + \\cos \\varphi), \\\\\n1 + \\sin \\varphi &= 4(1 - \\sin^2 \\varphi), \\\\\n(4 \\sin \\varphi - 3)(\\sin \\varphi + 1) &= 0, \\\\\n\\sin \\varphi &= \\frac{3}{4}.\n\\end{aligned}\n$$\n\nSince $\\varphi$ is acute, $\\sin \\varphi = \\frac{3}{4}$. Therefore,\n\n$$\n\\cos \\angle ACB = \\sin \\varphi = \\frac{3}{4}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20263,
"subject": "Mathematics (Olympiad)",
"question": "A function $f: (0, +infty) \\to \\mathbb{R}$ satisfies the following conditions:\n\n(a) $f(a) = 1$ for a positive real number $a$,\n\n(b) $f(x)f(y) + f\\left(\\frac{a}{x}\\right)f\\left(\\frac{a}{y}\\right) = 2f(xy)$\n\nfor any positive real numbers $x, y$. Prove that $f(x)$ is constant.",
"options": [],
"answer": "See solution",
"solution": "Setting $x = y = 1$ in (b) gives\n\n$$\nf^2(1) + f^2(a) = 2f(1)\n$$\n\nso\n\n$$\n(f(1) - 1)^2 = 0\n$$\n\nthus $f(1) = 1$.\n\nSetting $y = 1$ in (b) yields\n\n$$\nf(x)f(1) + f\\left(\\frac{a}{x}\\right)f(a) = 2f(x)\n$$\n\nso\n\n$$\nf(x) = f\\left(\\frac{a}{x}\\right), \\quad x > 0\n$$\n\nSetting $y = \\frac{a}{x}$ in (b) yields\n\n$$\nf(x)f\\left(\\frac{a}{x}\\right) + f\\left(\\frac{a}{x}\\right)f(x) = 2f(a)\n$$\n\nso\n\n$$\nf(x)f\\left(\\frac{a}{x}\\right) = 1\n$$\n\nCombining the previous results gives $f^2(x) = 1$ for $x > 0$.\n\nSetting $x = y = \\sqrt{t}$ in (b) gives\n\n$$\nf^2(\\sqrt{t}) + f^2\\left(\\frac{a}{\\sqrt{t}}\\right) = 2f(t)\n$$\n\nSince $f^2(x) = 1$, we have $f(t) > 0$.\n\nSo $f(x) = 1$ for all $x > 0$, as desired.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20264,
"subject": "Mathematics (Olympiad)",
"question": "一個長方形 $R$，其各邊邊長皆為奇整數。我們將之分割為若干個小的長方形，使得每個小長方形的各邊邊長皆為正整數。證明其中至少有一個小長方形，它和 $R$ 各邊的距離的奇偶性相同。\n\n(在此，一個小長方形到 $R$ 某一邊 $L$ 的距離，定義為該小長方形平行 $L$ 的一邊中較靠近 $L$ 者與 $L$ 的垂直距離。)",
"options": [],
"answer": "See solution",
"solution": "令 $R$ 的長寬分別為 $a, b$。將長方形切割為 $ab$ 個單位正方形，並依據西洋棋盤方式將之交替塗成黑白。注意到由於 $a, b$ 皆為奇數，四個角落的正方形必被塗成同色，不失一般性設為黑色。\n\n對於每個長方形，我們稱它為黑的（白的）若且唯若它的四個角落正方形都是黑的（白的），否則稱它為灰的。注意到以下事實：\n\n1. 灰長方形包含相同數量的黑與白正方形；\n2. 黑長方形內的黑正方形比白正方形多一個；\n3. 白長方形內的白正方形比黑正方形多一個。\n\n現在，$R$ 是黑的，所以其內的黑正方形比白正方形多一個。搭配以上事實，我們知道至少要有一個小長方形是黑的。現在令該長方形到四邊的距離，從上方開始依照順時針順序，分別為 $w, x, y, z$。由於小長方形和 $R$ 的右上角都是黑的，易知 $w$ 和 $x$ 必奇偶性相同。同樣的方式我們可知 $x$ 和 $y, y$ 和 $z, z$ 和 $w$ 都同奇偶性，故知此黑長方形到四邊的距離同奇偶。證畢。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20265,
"subject": "Mathematics (Olympiad)",
"question": "A convex quadrilateral $ABCD$ is right-angled at $A$ and satisfies $BC + CD = 1$. Determine its greatest possible area.",
"options": [],
"answer": "See solution",
"solution": "The greatest possible area is $\\frac{1}{8}(1 + \\sqrt{2})$.\n\nReflect the quadrilateral in line $AB$, and reflect the resulting pentagon again in line $AD$; see Figure 1. This produces an octagon of fixed perimeter\n\n$$\n4(BC + CD) = 4\n$$\n\nand area four times that of $ABCD$. The maximal area is obtained for a regular octagon of edge length $\\frac{1}{2}$. Since the area of a regular octagon of edge $a$ is known (or easily verified) to be $2(1 + \\sqrt{2})a^2$, the maximal area of the original quadrilateral is $\\frac{1}{4} \\cdot 2(1 + \\sqrt{2}) \\left(\\frac{1}{2}\\right)^2 = \\frac{1}{8}(1 + \\sqrt{2})$.\n\n\n\nFig. 1. Double reflexion into an octagon.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20266,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$ be the sides of a triangle such that $a = b - 1$ and $c = b + 1$. The inradius is $r = 4$.\n\nFind the values of $a$, $b$, $c$, and the circumradius $R$.",
"options": [],
"answer": "See solution",
"solution": "Recall Heron's formula and two other well-known formulae for the area of a triangle:\n\n$$\n|ABC| = \\sqrt{s(s-a)(s-b)(s-c)} = \\frac{abc}{4R} = rs,\n$$\n\nwhere $r = 4$ is the inradius and $R$ is the circumradius.\n\nWe have:\n\n$$\ns = \\frac{(b - 1) + b + (b + 1)}{2} = \\frac{3b}{2}\n$$\n\nSo,\n\n$$\n|ABC| = rs = 6b\n$$\n\nTherefore,\n\n$$\n|ABC|^2 = (6b)^2 = 36b^2\n$$\n\nBy Heron's formula:\n\n$$\n|ABC|^2 = s(s - a)(s - b)(s - c) = \\frac{3b^2(b^2 - 4)}{16}\n$$\n\nEquate the two expressions:\n\n$$\n36b^2 = \\frac{3b^2(b^2 - 4)}{16}\n$$\n\nMultiply both sides by $16$:\n\n$$\n576b^2 = 3b^2(b^2 - 4)\n$$\n\nDivide both sides by $3b^2$ (assuming $b \\neq 0$):\n\n$$\n192 = b^2 - 4\n$$\n\nSo,\n\n$$\nb^2 = 196 \\implies b = 14\n$$\n\nThen $a = 13$, $c = 15$.\n\nNow, $|ABC| = 6b = 84$.\n\nUsing $|ABC| = \\frac{abc}{4R}$:\n\n$$\n84 = \\frac{13 \\times 14 \\times 15}{4R}\n$$\n\nSo,\n\n$$\n4R = 13 \\times 14 \\times 15 / 84 = 2730 / 84 = 32.5\n$$\n\nThus,\n\n$$\nR = \\frac{65}{8}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20267,
"subject": "Mathematics (Olympiad)",
"question": "At a tennis tournament, 8 girls play in a round-robin (each player plays every other exactly once; no draws). Oksana took second place by points, and no other participant had the same number of points as her. What is the maximum number of games that Olesya, who won the tournament, could have lost?",
"options": [],
"answer": "See solution",
"solution": "Suppose Olesya lost 2 games. Then she scored 5 points. Oksana could not have scored 4 points, because then the total number of wins would be $5 + 4 + 3 \\times 6 = 27$. But there are $\\frac{1}{2}(8 \\times 7) = 28$ games in total, a contradiction. Similarly, Oksana could not have scored 3 or fewer points, nor could Olesya have lost 3 or more games. Thus, Olesya could have lost at most 1 game. This is possible, as shown by the following table:\n\n\n\n| M | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | p |\n|---|---|---|---|---|---|---|---|---|---|\n| 1 | X | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 6 |\n| 2 | 1 | X | 1 | 0 | 1 | 0 | 1 | 0 | 4 |\n| 3 | 0 | 0 | X | 1 | 0 | 1 | 0 | 1 | 3 |\n| 4 | 0 | 1 | 0 | X | 1 | 0 | 1 | 0 | 3 |\n| 5 | 0 | 0 | 1 | 0 | X | 1 | 0 | 1 | 3 |\n| 6 | 0 | 1 | 0 | 1 | 0 | X | 1 | 0 | 3 |\n| 7 | 0 | 0 | 1 | 0 | 1 | 0 | X | 1 | 3 |\n| 8 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | X | 3 |",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20268,
"subject": "Mathematics (Olympiad)",
"question": "For $n = 5$, count the number of exceptional situations, i.e., the permutations for which $S_1'$ and $S_2'$ are completed at the same time. According to the lemma, these must satisfy:\n\n(a) 5 and 4 are not adjacent;\n\n(b) one of the numbers 1, 2, 3 belongs to $S_1'$, while the other two belong to $S_2'$ and are descending.\n\nFix $n = 5$.",
"options": [],
"answer": "See solution",
"solution": "There are 2 ways to fill the number 4 and 3 ways to pick the one number for $S_1'$ in (b). So, there are $2 \\times 3 = 6$ permutations each with 1 additional descending chain; all others have $\\frac{1}{2}$ additional descending chain on average. Hence,\n\n$$\nA(5) = \\frac{2}{4}(A(1) + A(2) + A(3) + A(4)) + \\frac{1}{2} + \\frac{1}{2} \\times \\frac{2 \\times 3}{4!} = \\frac{73}{24},\n$$\n\nand $S(5) = A(1) + \\cdots + A(5) = \\frac{63}{8}$.\n\nPlug $k=5$ into (4) and (5). For $n \\ge 5$,\n\n$$\nS(n) \\ge n \\left( \\frac{n+1}{30} \\times \\frac{83}{8} - \\frac{1}{2} \\right),\n$$\n\n$$\nS(n) \\le n \\left( \\frac{n+1}{30} \\times \\frac{101}{8} - 1 \\right) + \\frac{1}{4}.\n$$\n\nSubstituting them into (1), we find\n\n$$\nA(n) \\ge \\frac{2}{n-1}S(n-1) + \\frac{1}{2} \\ge \\frac{83}{120}n - \\frac{1}{2},\n$$\n\n$$\nA(n) \\le \\frac{101}{120}n - 1 \\le \\frac{101}{120}n - \\frac{1}{2},\n$$\n\nthat is, $\\frac{83}{120}n - \\frac{1}{2} \\le A(n) \\le \\frac{101}{120}n - \\frac{1}{2}$ as required. $\\square$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20269,
"subject": "Mathematics (Olympiad)",
"question": "令 $m$ 與 $n$ 為大於 1 的正整數。在 $m \\times n$ 方格紙上的每一格都有一枚背面向上的硬幣。每一步，我們依次進行以下動作：\n\n1. 選擇一個 $2 \\times 2$ 的區域；\n2. 將該區域左上角與右下角的硬幣翻面；\n3. 將該區域左下角與右上角的硬幣擇一翻面。\n\n試求所有 $(m, n)$，使得我們能透過有限步將硬幣全部翻成正面。",
"options": [],
"answer": "See solution",
"solution": "答案為滿足 $3 \\mid mn$ 的所有 $(m, n)$。\n\n構造：不失一般性假設 $3 \\mid m$。當 $2 \\mid n$ 時，我們可用左圖方式將所有硬幣翻面：\n\n\n\n當 $2 \\nmid n$ 時，先用上述的方法將左邊的 $m \\times (n-1)$ 全部翻成正面，剩下最右邊一排是背面。接著，令 $L(i, j)$ 為將 $(i, j)$、$(i+1, j)$ 和 $(i, j+1)$ 位置翻面的 $L$ 型操作，而 $R(i, j)$ 為將 $(i, j)$、$(i, j-1)$ 和 $(i-1, j)$ 位置翻面的操作。則注意到 $R(i, n)$、$R(i+1, n)$、$L(i, n-1)$ 這連續三個操作會在保持其餘硬幣不動的情況下，將 $(i-1, n)$ 到 $(i+1, n)$ 三個位置的硬幣翻面，故我們可用這個方式將最右邊的一排也翻到正面。\n\n估計：將方陣如右圖賦值，並令 $T(i)$ 為在賦值 $i$ 的格子中的正面硬幣總數。注意到一開始 $T(0) = T(1) = T(2) = 0$，且在每一步中，$T(1) - T(0)$ 與 $T(2) - T(1)$ 的奇偶性都不會變。這表示我們恆有 $T(0) \\equiv T(1) \\equiv T(2) \\pmod{2}$。換言之，要能將硬幣全部翻成正面，賦值 0、1 與 2 的格子數量必須是同奇偶性。然而，直接計算知：\n\n- 當 $mn \\equiv 1 \\pmod{3}$，有 $T(0) - 1 = T(1) = T(2) = \\frac{mn-1}{3}$；\n- 當 $mn \\equiv 2 \\pmod{3}$，有 $T(0) - 1 = T(1) - T(2) = \\frac{mn-2}{3}$。\n\n故 $3 \\mid mn$ 是能全部翻成正面的必要條件。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20270,
"subject": "Mathematics (Olympiad)",
"question": "Prove that there exists a real number $c < \\frac{3}{4}$ such that for each sequence $\\{x_i\\}_{i=1}^\\infty$ satisfying $0 \\le x_i \\le 1$ for all $i$, there are infinitely many pairs $(m, n)$ with $m > n$ such that\n\n$$\n|x_m - x_n| \\le \\frac{c}{m}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Assume, for contradiction, that for every $c < \\frac{3}{4}$ there exists a sequence $\\{x_i\\}_{i=1}^\\infty$ such that there are only finitely many such $(m, n)$ pairs. Thus, there exists a positive integer $N$ so that $|x_m - x_n| > \\frac{c}{m}$ for all $m > n > N$.\n\nConsider $x_{N+1}, x_{N+2}, \\dots, x_{N+K}$, where $K$ is a sufficiently large positive integer congruent to $1$ modulo $8$. Sort these numbers in ascending order:\n\n$$\n0 \\le x_{i_1} \\le x_{i_2} \\le \\dots \\le x_{i_K} \\le 1\n$$\n\nwhere $(x_{i_1}, x_{i_2}, \\dots, x_{i_K})$ is a permutation of $(x_{N+1}, \\dots, x_{N+K})$. Then $x_{i_{j+1}} - x_{i_j} > \\frac{c}{\\max\\{i_j, i_{j+1}\\}}$ for all $j = 1, \\dots, K-1$. Summing these inequalities gives:\n\n$$\n1 \\ge x_{i_K} - x_{i_1} = \\sum_{j=1}^{K-1} (x_{i_{j+1}} - x_{i_j}) > c \\sum_{j=1}^{K-1} \\frac{1}{\\max\\{i_j, i_{j+1}\\}} \\ge c \\sum_{j=1}^{(K-1)/2} \\frac{2}{N+K+1-j}.\n$$\n\nLet $0 < \\epsilon < \\frac{2}{5}$. For $1 \\le j \\le \\frac{K-1}{8}$,\n\n$$\n\\frac{N + K + 1 - j}{N + \\frac{K+1}{2} + j} \\ge 1 + \\epsilon\n$$\n\nbecause this is equivalent to\n\n$$\n(2 + \\epsilon)j \\le \\frac{1 - \\epsilon}{2}(K + 1) - \\epsilon N\n$$\n\nand\n\n$$\n(2 + \\epsilon) \\frac{K - 1}{8} \\le \\frac{1 - \\epsilon}{2}(K + 1) - \\epsilon N\n$$\n\nwhich holds for large $K$ since $\\frac{2+\\epsilon}{8} < \\frac{1-\\epsilon}{2}$.\n\nSince $x + \\frac{1}{x}$ increases for $x > 1$, $\\frac{1}{a} + \\frac{1}{b} \\ge \\frac{3 + \\epsilon + \\frac{1}{1 + \\epsilon}}{a + b}$ for $a, b > 0$ with $\\frac{a}{b} \\ge 1 + \\epsilon$. By AM-GM, $\\frac{1}{a} + \\frac{1}{b} \\ge \\frac{4}{a + b}$ for any $a, b > 0$. Thus,\n\n$$\n\\begin{align*}\n\\sum_{j=1}^{(K-1)/2} \\frac{2}{N+K+1-j} &= 2 \\sum_{j=1}^{(K-1)/4} \\left( \\frac{1}{N+K+1-j} + \\frac{1}{N+\\frac{K+1}{2}+j} \\right) \\\\\n&= 2 \\sum_{j=1}^{(K-1)/8} \\left( \\frac{1}{N+K+1-j} + \\frac{1}{N+\\frac{K+1}{2}+j} \\right) \\\\\n&\\quad + 2 \\sum_{j=(K-1)/8+1}^{(K-1)/4} \\left( \\frac{1}{N+K+1-j} + \\frac{1}{N+\\frac{K+1}{2}+j} \\right) \\\\\n&\\ge 2 \\left( 3 + \\epsilon + \\frac{1}{1+\\epsilon} \\right) \\frac{\\frac{K-1}{8}}{2N+\\frac{3K+3}{2}} + 8 \\frac{\\frac{K-1}{8}}{2N+\\frac{3K+3}{2}} \\\\\n&= \\frac{7+\\epsilon+\\frac{1}{1+\\epsilon}}{4} \\cdot \\frac{K-1}{2N+\\frac{3K+3}{2}}\n\\end{align*}\n$$\n\nSince $7+\\epsilon+\\frac{1}{1+\\epsilon} > 8$ for all $\\epsilon > 0$, if $\\frac{6}{7+\\epsilon+\\frac{1}{1+\\epsilon}} < c < \\frac{3}{4}$, we get a contradiction for large $K$:\n\n$$\n\\frac{7 + \\epsilon + \\frac{1}{1 + \\epsilon}}{4} \\cdot \\frac{K - 1}{2N + \\frac{3K + 3}{2}} > \\frac{1}{c}\n$$\n\n**Remark 1.**\n\n$$\n\\lim_{K \\to \\infty} \\sum_{j=\\frac{K+3}{2}}^{K} \\frac{1}{j} = \\ln 2\n$$\n\nso the result holds for any $c > \\frac{1}{2 \\ln 2} \\approx 0.72135$.\n\n**Remark 2.** Using $\\frac{1}{N+K+1-j} \\ge \\frac{1}{N+K}$ for $1 \\le j \\le (K-1)/2$, we can show there exists $c > 1$ satisfying the condition.\n\n**Remark 3.** Using $\\frac{1}{N+K+1-j} + \\frac{1}{N+\\frac{K+1}{2}+j} \\ge \\frac{4}{2N+\\frac{3(K+1)}{2}}$ for $1 \\le j \\le (K-1)/4$, we can show there exists $c > \\frac{3}{4}$ satisfying the condition.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20271,
"subject": "Mathematics (Olympiad)",
"question": "Find all rational numbers $r$ and all integers $k$ such that the equation\n$$\nr(5k - 7r) = 3\n$$\nis satisfied.",
"options": [],
"answer": "See solution",
"solution": "Obviously, $r \\ne 0$. Let us write $r$ as a reduced fraction $r = \\frac{m}{n}$ and let us assume that $n$ is a positive integer. Then\n$$\n\\frac{m}{n}(5k - 7m) = 3\n$$\nor, equivalently,\n$$\nm(5kn - 7m) = 3n^2.\n$$\nHence, $m$ divides $3n^2$. Since $m$ and $n$ are coprime, we conclude that $m$ divides $3$. Let us consider four cases. If $m = 1$ we have $5kn - 7 = 3n^2$ or $n(5k - 3n) = 7$, which implies that $n$...",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20272,
"subject": "Mathematics (Olympiad)",
"question": "Some of the squares of an $n \\times n$ table are mined. In each square, the number of mined squares among this square and its neighbors (i.e., those which share a common side or vertex with it) is written. Is it always possible to determine which squares are mined if:\n\n- $n = 2000$;\n- $n = 2007$;",
"options": [],
"answer": "See solution",
"solution": "Let the rows be indexed by $i = 1, \\dots, n$ and the columns by $j = 1, \\dots, n$. Let $a(i; j)$ denote the number written in square $(i; j)$.\n\n**a)** No! Consider table $A$ where squares $(i; j)$ are mined if and only if $i \\equiv j \\equiv 1 \\pmod{3}$, and table $B$ where squares $(i; j)$ are mined if and only if $i \\equiv j \\equiv 2 \\pmod{3}$. In both cases, all numbers written in the squares of $A$ and $B$ are equal to $1$, so it is impossible to determine which squares are mined.\n\n**b)** Yes! First, determine the mined squares in the third row. The number $b(j)$ of mined squares among $(3; j-1)$, $(3; j)$, $(3; j+1)$ equals $a(2; j) - a(1; j)$. By comparing $b(1)$ and $b(2)$, we can determine if $(3; 3)$ is mined. Next, compare $b(4)$ and $b(5)$ to decide if $(3; 6)$ is mined. The same argument applies for $(3; 9)$, $(3; 12)$, ..., $(3; 2007)$.\n\nThen, compare $b(2007)$ and $b(2006)$ to decide if $(3; 2005)$ is mined, $b(2004)$ and $b(2003)$ for $(3; 2002)$, and so on. Eventually, we know which squares among $(3; 1999)$, $(3; 1996)$, ..., $(3; 1)$ are mined. Now, $b(1)$ shows if $(3; 2)$ is mined, and comparing $b(3)$ and $b(4)$ determines if $(3; 5)$ is mined, etc., so we find which squares among $(3; 8)$, $(3; 11)$, ..., $(3; 2006)$ are mined. Thus, row 3 is determined.\n\nAnalogously, we can determine the mined squares in rows $6, 9, 12, 15, \\ldots, 2007$. Using a similar argument, we can determine the mined squares in rows $2005, 2002, 1999, \\ldots, 4, 1$ and rows $2, 5, 8, \\ldots, 2006$.\n\n**Remark:** The answer is \"No\" for $n \\equiv 2 \\pmod{3}$ and \"Yes\" otherwise.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20273,
"subject": "Mathematics (Olympiad)",
"question": "$B'$, and $C'$ lie on the circle $c'$ so that segment $AB \\parallel A'B'$, $BC \\parallel B'C'$, and $\\angle ABC = \\angle A'B'C'$. The lines $AA'$, $BB'$, and $CC'$ are all different and intersect in one point $P$, which does not coincide with any of the vertices of the triangles $ABC$ or $A'B'C'$. Prove that $\\angle AOB = \\angle A'O'B'$.\n\n\n\nFig. 1",
"options": [],
"answer": "See solution",
"solution": "The triangles $ABP$ and $A'B'P$ are similar, because their corresponding sides are parallel (Fig. 1). Hence $\\frac{|AB|}{|A'B'|} = \\frac{|BP|}{|B'P|}$. Likewise, the triangles $BCP$ and $B'C'P$ are similar, hence $\\frac{|BC|}{|B'C'|} = \\frac{|BP|}{|B'P|}$. Thus $\\frac{|AB|}{|A'B'|} = \\frac{|BC|}{|B'C'|}$, and since $\\angle ABC = \\angle A'B'C'$, the triangles $ABC$ and $A'B'C'$ are also similar. From the equality of the angles $ACB$ and $A'C'B'$, the equality of the central angles $AOB$ and $A'O'B'$ now follows.\n\n**Remark.** Figure 1 corresponds to the case when the vectors $\\overrightarrow{AB}$ and $\\overrightarrow{A'B'}$ have the same direction. If they have the opposite directions, then the figure is different (the intersection point $P$ lies on the segments $AA'$, $BB'$, and $CC'$), but the argument is still correct.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20274,
"subject": "Mathematics (Olympiad)",
"question": "設數列 $\\{a_n\\}$ 滿足：\n\n$$\na_1 = a_2 = 1, \\quad a_{n+2} = a_{n+1} + a_n \\quad (n \\in \\mathbb{N}).\n$$\n\n當 $n$ 為奇數時，試求出滿足下列方程組的所有實數解 $(x, y)$：\n\n$$\n\\begin{cases}\nx + 2^x a_n + 2^y a_{n+1} = 1 + 2a_{n+2}, \\\\\ny + 2^x a_{n+1} + 2^y a_{n+2} = 1 + 2a_{n+3}.\n\\end{cases}\n$$",
"options": [],
"answer": "See solution",
"solution": "$(x, y) = (1,1)$ 為此方程組的唯一實數解。\n\n顯然，$(x, y) = (1,1)$ 為此方程組的一組實數解。以下證明無其它實數解。\n\n假設 $(x, y) = (x_1, y_1) \\neq (1,1)$ 也是方程組\n\n$$\n\\begin{cases}\nx + 2^x a_n + 2^y a_{n+1} = 1 + 2a_{n+2}, \\\\\ny + 2^x a_{n+1} + 2^y a_{n+2} = 1 + 2a_{n+3}\n\\end{cases}\n$$\n\n的另一組實數解。令 $(x_2, y_2) = (1,1)$，將這兩組解代入方程組並相減，得到：\n\n$$\n(x_1 - x_2) + (2^{x_1} - 2^{x_2})a_n + (2^{y_1} - 2^{y_2})a_{n+1} = 0 \\tag{1}\n$$\n\n$$\n(y_1 - y_2) + (2^{x_1} - 2^{x_2})a_{n+1} + (2^{y_1} - 2^{y_2})a_{n+2} = 0 \\tag{2}\n$$\n\n將 (1) 乘以 $(2^{x_1} - 2^{x_2})$，(2) 乘以 $(2^{y_1} - 2^{y_2})$，然後相加整理得：\n\n$$\n(x_1 - x_2)(2^{x_1} - 2^{x_2}) + (y_1 - y_2)(2^{y_1} - 2^{y_2}) \\\\\n+ (2^{x_1} - 2^{x_2})^2 a_n + 2(2^{x_1} - 2^{x_2})(2^{y_1} - 2^{y_2})a_{n+1} + (2^{y_1} - 2^{y_2})^2 a_{n+2} = 0 \\tag{3}\n$$\n\n現在證明：當 $(x_1, y_1) \\neq (x_2, y_2)$ 時，(3) 不成立，導出矛盾。\n\n注意 $f(x) = 2^x$ 為遞增函數，因此：\n\n$$\n(x_1 - x_2)(2^{x_1} - 2^{x_2}) \\geq 0, \\quad (y_1 - y_2)(2^{y_1} - 2^{y_2}) \\geq 0\n$$\n\n且等號同時成立當且僅當 $(x_1, y_1) = (x_2, y_2)$。\n\n因此，若 $(x_1, y_1) \\neq (x_2, y_2)$，有嚴格不等式：\n\n$$\n(x_1 - x_2)(2^{x_1} - 2^{x_2}) + (y_1 - y_2)(2^{y_1} - 2^{y_2}) > 0 \\tag{4}\n$$\n\n另一方面，利用數學歸納法可證，當 $n$ 為奇數時：\n\n$$\na_{n+1}^2 - a_n a_{n+2} = -1\n$$\n\n（此為費氏數列的基本性質，證明略。）\n\n又 $a_n > 0$，$(x_1, y_1) \\neq (x_2, y_2)$，可得：\n\n$$\n(2^{x_1} - 2^{x_2})^2 a_n + 2(2^{x_1} - 2^{x_2})(2^{y_1} - 2^{y_2})a_{n+1} + (2^{y_1} - 2^{y_2})^2 a_{n+2} \\\\\n= \\frac{1}{a_n} \\left[ (2^{x_1} - 2^{x_2})a_n + (2^{y_1} - 2^{y_2})a_{n+1} \\right]^2 + \\frac{1}{a_n} (2^{y_1} - 2^{y_2})^2 > 0 \\tag{5}\n$$\n\n由 (4) 和 (5) 得：\n\n$$\n(x_1 - x_2)(2^{x_1} - 2^{x_2}) + (y_1 - y_2)(2^{y_1} - 2^{y_2}) \\\\\n+ (2^{x_1} - 2^{x_2})^2 a_n + 2(2^{x_1} - 2^{x_2})(2^{y_1} - 2^{y_2})a_{n+1} + (2^{y_1} - 2^{y_2})^2 a_{n+2} > 0\n$$\n\n與 (3) 矛盾。故此方程組僅有一組實數解 $(x, y) = (1,1)$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20275,
"subject": "Mathematics (Olympiad)",
"question": "Consider a cyclic quadrilateral $ABCD$ and let $M$ and $N$ be the midpoints of the diagonals $AC$ and $BD$, respectively. If $\\angle AMB = \\angle AMD$, prove that $\\angle ANB = \\angle BNC$.",
"options": [],
"answer": "See solution",
"solution": "*Lemma:* Let $ABCD$ be an isosceles trapezoid ($AB \\parallel CD$) inscribed in a circle $C$ of center $O$, $M$ is the intersection point of the diagonals $AC$ and $BD$, and $T$ is the intersection point of lines $AD$ and $BC$. If the line through $M$ parallel to $AB$ meets the circle $C$ at $E$ and $F$, then $TE$ and $TF$ are tangent to the circle.\n\n_Proof of the Lemma:_ Assume $AB < CD$. It is clear that points $T$, $M$ and $O$ are collinear. Also, $\\angle AOT = \\frac{1}{2}\\angle AOB = \\angle ACT$, which means that the quadrilateral $TAOC$ is cyclic. From the power of point $M$ with respect to the circumcircle of triangle $AOC$ we get $MA \\cdot MC = MO \\cdot MT$. But $MA \\cdot MC = ME \\cdot MF$ from the power of $M$ with respect to $C$, so $MO \\cdot MT = ME \\cdot MF$. This means that $TEOF$ is cyclic, so $\\angle TEO + \\angle TFO = 180^\\circ$. But $\\angle TEO = \\angle TFO$, hence $\\angle TEO = 90^\\circ$, i.e., $TE$ is tangent to $C$. $\\square$\n\n\n\nLet us now return to the problem. Denote by $E$ and $F$ the second intersection points with the circle $C$ of lines $BE$, and $DE$, respectively. Because $M$ is the midpoint of the diagonal $AC$, and angles $AMB$ and $AMD$ are equal, it follows that $BDEF$ is an isosceles trapezoid (it is symmetric with respect to the perpendicular bisector of $AC$). Moreover, $BF \\parallel DE \\parallel AC$. If lines $BD$ and $EF$ meet at $T$, from the lemma it follows that $TA$ and $TC$ are tangent to the circumcircle of $ABCD$. If $O$ is the center of this circle, it follows that $OA \\perp TA$ and $OC \\perp TC$. As $N$ is the midpoint of $BD$, we have $ON \\perp BD$. Thus, points $A, N$, and $C$ all lie on the circle of diameter $OT$. Assume $\\angle B > \\angle D$ (if $\\angle B < \\angle D$, simply swap $B$ and $D$; the case $\\angle B = \\angle D$ is easy). In this case, $\\angle ANB = \\angle AOT$, and $\\angle TNC = \\angle TOC$. But $\\angle AOT = \\angle COT$ means $\\angle ANB = \\angle CNT = \\angle CNB$.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20276,
"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$$\na) N = 2016; \\quad b) N = 2017?\n$$",
"options": [],
"answer": "See solution",
"solution": "a) Note that for any integer $x$, the product $x(x+1)$ is even. This suggests the following example:\n\n$$\n1008x(x+1) + 2016 = 1008x^2 + 1008x + 2016.\n$$\n\nb) We have $f(x) = ax^2 + bx + c$. Let us do the following substitutions:\n\n$$x = 0 \\Rightarrow f(0) = c \\equiv 0 \\pmod{2017}$$\n$$x = 1 \\Rightarrow f(1) = a + b + c \\equiv 0 \\pmod{2017}$$\n$$x = -1 \\Rightarrow f(-1) = a - b + c \\equiv 0 \\pmod{2017}$$\n\nThen both $a + b$ and $a - b$ are divisible by 2017, so are $2a$ and $2b$. Since 2017 is odd, all the coefficients are divisible by 2017.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20277,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be an integer greater than or equal to 3. For a permutation $p = (x_1, x_2, \\dots, x_n)$ of $(1, 2, \\dots, n)$, we say that $x_j$ lies in between $x_i$ and $x_k$ if $i < j < k$. (For example, in the permutation $(1, 3, 2, 4)$, 3 lies in between 1 and 4, and 4 does not lie in between 1 and 2.)\n\nSet $S = \\{p_1, p_2, \\dots, p_m\\}$ consists of (distinct) permutations $p_i$ of $(1, 2, \\dots, n)$. Suppose that among every three distinct numbers in $\\{1, 2, \\dots, n\\}$, one of these numbers does not lie in between the other two numbers in every permutation $p_i \\in S$. Determine the maximum value of $m$.",
"options": [],
"answer": "See solution",
"solution": "The answer is $2^{n-1}$.\n\nWe first show that $m \\leq 2^{n-1}$. We induct on $n$. The base case $n=3$ is trivial. (Indeed, say 3 does not lie in between 1 and 2, then we can have $S = \\{(1, 2, 3), (3, 1, 2), (2, 1, 3), (3, 2, 1)\\}$.) Assume that the statement is true for $n=k$ (where $k \\geq 3$). Now consider $n=k+1$ and a set $S_{k+1}$ satisfying the conditions of the problem. Note that if the element $k+1$ is deleted from each permutation $p_i$ in $S_{k+1}$, the resulting permutations $q_i$ form a set $S_k$ that satisfies the conditions of the problem (for $n=k$). It suffices to show the following claim: there are at most two distinct permutations $p$ and $q$ in $S_{k+1}$ that can map to the same permutation $r$ in $S_k$ (by deleting the element $k+1$ in the permutations $p$ and $q$).\n\nIndeed, assume that for\n\n$$\np_1 = (x_1, x_2, \\dots, x_{k+1}), \\quad p_2 = (y_1, y_2, \\dots, y_{k+1}), \\\\\np_3 = (z_1, z_2, \\dots, z_{k+1})\n$$\n\nin $S_{k+1}$, $q_1 = q_2 = q_3 = q$. By symmetry, we may assume that $q = (1, 2, \\dots, k)$. Assume that $x_a = y_b = z_c = k+1$. Again by symmetry, we may assume that $1 \\leq a < b < c \\leq k+1$. (Note that because $q_1 = q_2 = q_3 = q$, $a, b, c$ are distinct.) We consider three numbers $a, b, k+1$. We have $p_1 = (\\dots, k+1, a, \\dots, b, \\dots)$ (in particular, $a$ lies in between $k+1$ and $b$), $p_2 = (\\dots, a, \\dots, k+1, b, \\dots)$ (in particular, $k+1$ lies in between $a$ and $b$), and $p_3 = (\\dots, a, \\dots, b, \\dots, k+1, \\dots)$ (in particular, $b$ lies in between $a$ and $k+1$). Hence each one of the numbers $a, b, k+1$ lies in between the other two numbers in some permutation in $S_k$, violating the conditions of $S_k$. Thus our assumption was wrong and at most two elements in $S_{k+1}$ can be mapped to an element in $S_k$, establishing our claim.\n\nIt remains to be shown that $m = 2^{n-1}$ is achievable. We construct permutation $p$ inductively: (1) place 1; (2) after numbers 1, 2, ..., $l$ are placed, we place $l+1$ either to the left or the right of all the numbers placed so far. Because there are two possible places for each of the numbers 2, 3, ..., $n$, we can construct $2^{n-1}$ such permutations. For any three numbers $1 \\leq a < b < c \\leq n$, $c$ does not lie in between $a$ and $b$. Hence, this set of $2^{n-1}$ permutations satisfies the conditions of the problem, completing our proof. $\\Box$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20278,
"subject": "Mathematics (Olympiad)",
"question": "Find all real numbers $a$ such that there exists a function $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying the following conditions:\n\n1. $f(1) = 2016$;\n2. $f(x + y + f(y)) = f(x) + a y$ for all real numbers $x, y$.",
"options": [],
"answer": "See solution",
"solution": "For $a = 0$, the constant function $f(x) = 2016$ for all $x$ satisfies the conditions.\n\nNow consider $a \\neq 0$. Plugging $x = -f(y)$ into condition 2 gives\n\n$$\nf(y) = f(-f(y)) + a y\n$$\n\nfor all $y$, so $f$ is injective. Setting $y = 0$ in condition 2 yields\n\n$$\nf(x + f(0)) = f(x)\n$$\n\nfor all $x$, so $f(0) = 0$. Setting $y = -\\frac{f(x)}{a}$ in condition 2 and using injectivity, we get\n\n$$\n-\\frac{f(x)}{a} + f\\left(-\\frac{f(x)}{a}\\right) = -x\n$$\n\nfor all $x$. Replacing $y$ by $-\\frac{f(y)}{a}$ in condition 2 and applying the above, we find\n\n$$\nf(x - y) = f(x) - f(y)\n$$\n\nfor all $x, y$, so $f$ is additive. Thus, $f(2016) = 2016 f(1) = 2016^2$.\n\nSince $f$ is additive, condition 2 becomes $f(y) + f(f(y)) = a y$ for all $y$. Setting $y = 1$ gives $a = 2016 \\cdot 2017$. For $a = 2016 \\cdot 2017$, $f(x) = 2016x$ satisfies all conditions.\n\nTherefore, the possible values are $a = 0$ or $a = 2016 \\cdot 2017$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20279,
"subject": "Mathematics (Olympiad)",
"question": "For $n \\in \\mathbb{N}$, $n \\ge 2$, let $A$ be an $n \\times n$ matrix with complex entries such that $A^2 = \\operatorname{tr}(A) \\cdot A$. Prove that the matrices $ABA$ and $ACA$ commute for any $n \\times n$ matrices $B$ and $C$ with complex entries.",
"options": [],
"answer": "See solution",
"solution": "Observe that if $\\operatorname{tr}(A) = 0$, then $A^2 = O_{n \\times n}$ and\n$$\nABA \\cdot ACA = O_{n \\times n} = ACA \\cdot ABA,\n$$\nfor any $B, C \\in \\mathcal{M}_n(\\mathbb{C})$.\n\nIf $\\operatorname{tr}(A) \\neq 0$, we show that $\\operatorname{rank}(A) = 1$. Let $r = \\operatorname{rank}(A)$. Then we can find matrices $X \\in \\mathcal{M}_{n \\times r}(\\mathbb{C})$ and $Y \\in \\mathcal{M}_{r \\times n}(\\mathbb{C})$ with $\\operatorname{rank}(X) = \\operatorname{rank}(Y) = r$ such that $A = XY$. Then $YX \\in \\mathcal{M}_r(\\mathbb{C})$ and, as $\\operatorname{tr}(A) \\neq 0$, we have\n$$\nr \\geq \\operatorname{rank}(YX) \\geq \\operatorname{rank}(X(YX)Y) = \\operatorname{rank}(A^2) = \\operatorname{rank}(A) = r,\n$$\nThat is, $\\operatorname{rank}(YX) = r$ and the matrix $YX$ is non-singular. The equality $A^2 = \\operatorname{tr}(A) \\cdot A$ can be written as $(X)^2 = \\operatorname{tr}(A) \\cdot XY$, obtaining $(YX)^3 = \\operatorname{tr}(A) \\cdot (YX)^2$. As $YX$ is non-singular, we get $YX = \\operatorname{tr}(A) \\cdot I_r$. In conclusion,\n$$\n\\operatorname{tr}(A) = \\operatorname{tr}(XY) = \\operatorname{tr}(YX) = \\operatorname{tr}(A) \\cdot r,\n$$\ngiving $r = 1$. To conclude, let $B, C \\in \\mathcal{M}_n(\\mathbb{C})$. We have $ABA = XYBXY = pXY = pA$, where $p = YBX \\in \\mathcal{M}_1(\\mathbb{C})$. In the same way, $ACA = qA$ with $q \\in \\mathbb{C}$, so $ABA$ and $ACA$ commute.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 20280,
"subject": "Mathematics (Olympiad)",
"question": "Consider a cyclic quadrilateral such that the midpoints of its sides form another cyclic quadrilateral. Prove that the area of the smaller circle is less than or equal to half the area of the bigger circle.",
"options": [],
"answer": "See solution",
"solution": "Let $ABCD$ be a cyclic quadrilateral with $AB = a$, $BC = b$, $CD = c$, $DA = d$, $AC = e$ and $BD = f$. Because the midpoints of the cyclic quadrilateral $ABCD$ form another cyclic quadrilateral, which is a parallelogram, we deduce that this parallelogram is a rectangle and $ABCD$ is orthogonal.\n\nIt follows that\n\n$$\nS = \\sigma(ABCD) = \\frac{ef}{2}\n$$\n\nand\n\n$$\na^2 + c^2 = b^2 + d^2\n$$\n\nLet $R$ and $R_1$ be the circumradii of the cyclic quadrilateral $ABCD$ and the rectangle respectively. We obtain\n\n$$\n4R_1^2 = \\frac{e^2 + f^2}{4}\n$$\n\nand\n\n$$\n16R^2 S^2 = (ac + bd)(ab + cd)(ad + bc) = ef[(ac(b^2 + d^2) + bd(a^2 + c^2))] = (ef)^2(a^2 + c^2),\n$$\n\nso\n\n$$\n4R^2 = a^2 + c^2\n$$\n\nNote that our inequality $R^2 \\geq 2R_1^2$ is equivalent to\n\n$$\n2(a^2 + c^2) \\geq e^2 + f^2\n$$\n\n\n\nFrom Euler's formula for the midpoints of the diagonals we get\n\n$$a^2 + b^2 + c^2 + d^2 - e^2 - f^2 \\geq 0,$$\n\nhence $2(a^2 + c^2) \\geq (e^2 + f^2)$, and we are done. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20281,
"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 $P = 2 \\cdot 3 \\cdot 5 \\cdot \\dots \\cdot 29$ be the product of the first 10 primes $2, 3, 5, 7, 11, 13, 17, 19, 23, 29$. 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 $1 \\le r_1 < P$. 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 $P$. Then, 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 $1 \\le r_2 < P$, 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": 20282,
"subject": "Mathematics (Olympiad)",
"question": "a) What is the smallest and largest 4-digit mid-product number?\n\nb) What is the smallest possible digit sum of a mid-product number?\n\nc) Which mid-product numbers have 5 as their second digit?\n\nd) In which positions can the digit 5 appear in a mid-product number?\n\n*A mid-product number is a 4-digit number $abcd$ such that the 2-digit number $bc$ is the product of the 1-digit numbers $a$ and $d$, and all digits are different and nonzero.*",
"options": [],
"answer": "See solution",
"solution": "a) Since $1$ times a digit 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 2-digit product. Since $2105$ 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\nThe first digit of a mid-product number is at most $8$, since the last digit is larger. If the first digit is $8$, then the last digit must be $9$. Hence the middle 2-digit number is $72$. So the largest mid-product number is $8729$.\n\nb) A mid-product number does not contain $0$ and all its digits must be different. So its digit sum is at least $1+2+3+4 = 10$. The only mid-product number whose digits are $1, 2, 3, 4$ is $3124$, and the digit sum of any other mid-product number will be greater than $10$.\n\nc) The only 2-digit numbers starting with $5$ that have two different 1-digit factors are $54$ and $56$. So the only mid-product numbers with second digit $5$ are $6549$ and $7568$.\n\nd) If the first or last digit of a mid-product number was $5$, then its 2-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 starts or ends in $5$.\n\nIf the third digit is $5$, then the 2-digit middle number is one of $15, 25, 35, 45, 65, 75, 85, 95$. Only the first four of these are products of two single-digit factors, and in each case at least one of those factors must be a $5$. But this means that a $5$ would be repeated. So the third digit cannot be $5$.\n\nSo only the second digit can be $5$, and this results in the two mid-product numbers found in Part c.\n\n*Comment:*\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": 20283,
"subject": "Mathematics (Olympiad)",
"question": "Let $x_1, x_2, \\dots, x_n$ be non-negative real numbers such that $x_1 + x_2 + \\dots + x_n = s$, where $s$ is a fixed non-negative real number. Let $f(x)$ be a strictly increasing function on non-negative real numbers. What is the maximum value of\n\n$$\nF = \\sum_{1 \\le i < j \\le n} \\min\\{f(x_i), f(x_j)\\}?\n$$\n\nShow that the maximum is attained when $x_1 = x_2 = \\cdots = x_n = \\frac{s}{n}$.",
"options": [],
"answer": "See solution",
"solution": "We use induction on $n$ to prove the statement.\n\nSince $F$ is symmetric, assume $x_1 \\le x_2 \\le \\cdots \\le x_n$. Because $f(x)$ is strictly increasing, we have\n\n$$\nF = (n-1)f(x_1) + (n-2)f(x_2) + \\cdots + f(x_{n-1}).\n$$\n\n**Base case ($n=2$):**\n\n$$\nF = f(x_1) \\le f\\left(\\frac{s}{2}\\right),\n$$\n\nequality holds when $x_1 = x_2$.\n\n**Inductive step:**\n\nAssume the statement holds for $n$. For $n+1$, apply the inductive hypothesis to $x_2 + x_3 + \\cdots + x_{n+1} = s - x_1$:\n\n$$\nF \\le n f(x_1) + \\frac{1}{2} n(n-1) f\\left(\\frac{s-x_1}{n}\\right) = g(x_1).\n$$\n\nThe function $g(x_1)$ attains its maximum at $x_1 = \\frac{s}{n+1}$, so the maximum occurs when\n\n$$\nx_1 = x_2 = \\cdots = x_{n+1} = \\frac{s}{n+1}.\n$$\n\nThus, the maximum value of $F$ is achieved when all $x_i$ are equal.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20284,
"subject": "Mathematics (Olympiad)",
"question": "Let $P$ be any inner point of an acute $\\triangle ABC$; $E, F$ be the projection points of $P$ onto lines $AC, AB$, respectively; and the extended lines of $BP, CP$ intersect the circumcircle of $\\triangle ABC$ at points $B_1, C_1$ ($B_1 \\neq B, C_1 \\neq C$), respectively. Let $R$ and $r$ denote the radii of the circumcircle and incircle of $\\triangle ABC$, respectively. Prove that\n$$\n\\frac{EF}{B_1C_1} \\ge \\frac{r}{R},\n$$\nand, when the equality holds, determine completely the positions of $P$.\n\n\n\n",
"options": [],
"answer": "See solution",
"solution": "As seen in the figures, let $PD \\perp BC$ with intersection point $D$, and extend $AP$ to meet the circumcircle of $\\triangle ABC$ at $A_1$. Connect $DE, DF, A_1B_1, A_1C_1$.\n\nSince $P, D, B, F$ are concyclic, $\\angle PDF = \\angle PBF$; and since $P, D, C, E$ are concyclic, $\\angle PDE = \\angle PCE$. Thus,\n$$\n\\begin{aligned}\n\\angle FDE &= \\angle PDF + \\angle PDE = \\angle PBF + \\angle PCE \\\\\n &= \\angle AA_1B_1 + \\angle AA_1C_1 = \\angle C_1A_1B_1.\n\\end{aligned}\n$$\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 $AO', BO', CO'$. We have\n$$\n\\begin{align*}\nS_{\\triangle ABC} &= \\frac{(AB + BC + CA) \\cdot r}{2} \\\\\n&= S_{\\triangle O'AB} + S_{\\triangle O'BC} + S_{\\triangle O'AC} \\\\\n&\\le \\frac{AB \\cdot O'F}{2} + \\frac{BC \\cdot O'D}{2} + \\frac{CA \\cdot O'E}{2} \\\\\n&= \\frac{(AB + BC + CA) \\cdot R'}{2}.\n\\end{align*}\n$$\nTherefore, $R' \\geq r$. Equality holds if and only if\n$$\nO'D \\perp BC, \\quad O'E \\perp CA, \\quad O'F \\perp AB,\n$$\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": 20285,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be real numbers such that $a + b + c = 1$. Prove that\n\n$$\n\\frac{a}{a + b^2} + \\frac{b}{b + c^2} + \\frac{c}{c + a^2} \\leq \\frac{1}{4} \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right).\n$$",
"options": [],
"answer": "See solution",
"solution": "By using the condition $a + b + c = 1$ and the inequality between arithmetic and geometric means, we have:\n\n$$\n\\frac{a}{a + b^2} = \\frac{a}{a(a + b + c) + b^2} = \\frac{a}{a^2 + b^2 + ab + ac} \\leq \\frac{a}{2ab + ab + ac} = \\frac{1}{3b + c}.\n$$\n\nBy applying the inequality between harmonic and arithmetic means, it follows that\n\n$$\n\\frac{4}{\\frac{3}{b} + \\frac{1}{c}} = \\frac{4}{\\frac{1}{b} + \\frac{1}{b} + \\frac{1}{b} + \\frac{1}{c}} \\leq \\frac{3b + c}{4},\n$$\n\ni.e.\n\n$$\n\\frac{a}{a + b^2} \\leq \\frac{1}{3b + c} \\leq \\frac{1}{16} \\left( \\frac{3}{b} + \\frac{1}{c} \\right).\n$$\n\nAnalogously,\n\n$$\n\\frac{b}{b + c^2} \\leq \\frac{1}{16} \\left( \\frac{3}{c} + \\frac{1}{a} \\right) \\quad \\text{and} \\quad \\frac{c}{c + a^2} \\leq \\frac{1}{16} \\left( \\frac{3}{a} + \\frac{1}{b} \\right).\n$$\n\nFinally, by adding the inequalities above, we have\n\n$$\n\\frac{a}{a + b^2} + \\frac{b}{b + c^2} + \\frac{c}{c + a^2} \\leq \\frac{1}{16} \\left( \\frac{4}{a} + \\frac{4}{b} + \\frac{4}{c} \\right) = \\frac{1}{4} \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right).\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20286,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $n$ for which there exists a set of $k \\geq 2$ positive rational numbers $a_1, a_2, \\dots, a_k$ such that both $a_1 + a_2 + \\dots + a_k = n$ and $a_1 \\cdot a_2 \\cdot \\dots \\cdot a_k = n$.\n\n",
"options": [],
"answer": "See solution",
"solution": "If $n = 2k \\geq 4$ is even, set $(a_1, a_2, \\dots, a_k) = (k, 2, 1, \\dots, 1)$:\n$$a_1 + a_2 + \\dots + a_k = k + 2 + 1 \\cdot (k-2) = 2k = n,$$\n$$a_1 \\cdot a_2 \\cdot \\dots \\cdot a_k = 2k = n.$$\n\nIf $n = 2k + 3 \\geq 9$ is odd, set $(a_1, a_2, \\dots, a_k) = \\left(k + \\frac{3}{2}, \\frac{1}{2}, 4, 1, \\dots, 1\\right)$:\n$$a_1 + a_2 + \\dots + a_k = k + \\frac{3}{2} + \\frac{1}{2} + 4 + (k-3) = 2k + 3 = n,$$\n$$a_1 \\cdot a_2 \\cdot \\dots \\cdot a_k = \\left(k + \\frac{3}{2}\\right) \\cdot \\frac{1}{2} \\cdot 4 = 2k + 3 = n.$$\n\nFor $n = 7$, set $(a_1, a_2, a_3) = \\left(\\frac{4}{3}, \\frac{7}{6}, \\frac{9}{2}\\right)$:\n$$a_1 + a_2 + a_3 = a_1 \\cdot a_2 \\cdot a_3 = 7 = n.$$\n\nFor $n \\in \\{1, 2, 3, 5\\}$, suppose there is a set of $k \\geq 2$ positive rational numbers whose sum and product are both $n$. By the Arithmetic-Geometric Mean inequality:\n$$n^{1/k} = \\sqrt[k]{a_1 \\cdot a_2 \\cdot \\dots \\cdot a_k} \\leq \\frac{n}{k},$$\nwhich gives\n$$n \\geq k^{1+\\frac{1}{k-1}}.$$\nFor $k = 3, 4$, or $k \\geq 5$, $n > 5$:\n$$k = 3 \\Rightarrow n \\geq 3\\sqrt{3} = 5.196\\dots > 5;$$\n$$k = 4 \\Rightarrow n \\geq 4\\sqrt[3]{4} = 6.349\\dots > 5;$$\n$$k \\geq 5 \\Rightarrow n \\geq 5^{1+\\frac{1}{k-1}} > 5.$$\nThus, none of $1, 2, 3, 5$ can be represented as the sum and product of three or more positive numbers.\n\nFor $k = 2$, $a_1 + a_2 = a_1 a_2 = n$ implies $n = a_1^2/(a_1 - 1)$, so $a_1$ satisfies\n$$a_1^2 - n a_1 + n = 0.$$\nThe discriminant $n^2 - 4n$ must be a perfect square, but this is not the case for $n \\in \\{1, 2, 3, 5\\}$.\n\n*Note:* Among all positive integers, only $n = 4$ can be represented both as the sum and product of the same two rational numbers. Indeed, $(n - 3)^2 < n^2 - 4n = (n - 2)^2 - 4 < (n - 2)^2$ for $n \\geq 5$; and $n^2 - 4n < 0$ for $n = 1, 2, 3$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20287,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $1 \\leq a_i \\leq 1 + (n-i) \\log_2 k$. Show that for large enough $n$, there must be two different forms of\n$$2^{a_n} k^n + 2^{a_{n-1}} k^{n-1} + \\dots + 2^{a_0}$$\nhaving the same value.",
"options": [],
"answer": "See solution",
"solution": "Given the bounds $1 \\leq a_i \\leq 1 + (n-i) \\log_2 k$, we have $2^{a_i} k^i \\leq 2k^n$. The number of possible forms is at least $\\prod_{i=0}^{n-1} (n-i) \\log_2 k \\leq n! (\\log_2 k)^n$, while the maximum value is $2k^n(n+1)$. For sufficiently large $n$, the number of forms exceeds the number of possible values, so by the pigeonhole principle, two different forms must have the same value.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20288,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. A Steiner tree associated with a finite set $S$ of points in the Euclidean $n$-space is a finite collection $T$ of straight-line segments in that space such that any two points in $S$ are joined by a unique path in $T$; its length is the sum of the segment lengths.\n\nShow that there exists a Steiner tree of length $1 + (2^{n-1} - 1)\\sqrt{3}$ associated with the vertex set of a unit $n$-cube.",
"options": [],
"answer": "See solution",
"solution": "We describe a recursive procedure for constructing the desired Steiner tree.\n\nThe case $n = 1$ is handled by a single line segment.\n\nAssume a Steiner tree of length $1 + (2^{n-1} - 1)\\sqrt{3}$ associated with the vertex set of a unit $n$-cube has been constructed such that each vertex of the $n$-cube is the endpoint of just one segment, and the length of this segment is greater than $\\sqrt{3}/6$.\n\nNow consider a unit $(n+1)$-cube. Select a pair of opposite $n$-faces and consider the $n$-cube whose vertices are the midpoints of the edges joining the corresponding vertices of these faces. Start with the assumed Steiner tree associated with the vertex set of this $n$-cube. For each vertex $v$ of this $n$-cube, delete a segment of length $\\sqrt{3}/6$ from the segment terminating at $v$, then add the segments joining the new endpoint to the two vertices of the $(n + 1)$-cube that are endpoints of the edge containing $v$. By the Pythagorean theorem, these segments each have length $1/\\sqrt{3}$.\n\nThe net effect of these changes is to add $2^n(2/\\sqrt{3} - \\sqrt{3}/6) = 2^{n-1}\\sqrt{3}$ to the length of the tree. Thus the resulting tree has length $1 + (2^n - 1)\\sqrt{3}$. It is clearly a Steiner tree associated with the vertex set of the $(n + 1)$-cube and has the additional properties required for the induction.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20289,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\geq 2$ and consider a table with $n+1$ rows and $n$ columns. On each of the first $n$ rows, Nicușor writes, in some order, the numbers $1, 2, \\dots, n$. Then, he chooses a permutation $a_1, a_2, \\dots, a_n$ of the numbers $1, 2, \\dots, n$ and completes the last row as follows: for each $j \\in \\{1, 2, \\dots, n\\}$, he writes in the cell at column $j$ the number of occurrences of $a_j$ in the cells located at the intersection of column $j$ with the first $n$ rows.\n\nDetermine all $n$ for which Nicușor can complete the table and choose $a_1, a_2, \\dots, a_n$ so that the last row contains, in some order, the numbers $1, 2, \\dots, n$.",
"options": [],
"answer": "See solution",
"solution": "For $n=2$, no matter how Nicușor chooses $a_1$ and $a_2$, on the third row we will have two equal values.\n\nFor $n=3$, suppose without loss of generality that $a_3=3$. Then, in the third column, we have only values equal to 3. Since the set $(1, 2)$ admits only two permutations, among the first three rows there will be two identical ones, so these rows coincide and we cannot have any $a_i = 1$.\n\nNext, we show that for $n \\geq 4$, there exists a completion that works. If $n=4$ and we take $a_1 = 1, a_2 = 2, a_3 = 3, a_4 = 4$, the following construction satisfies the conditions:\n\n\n\n**Construction 1 (induction):**\n\nWe will prove by induction on $n \\geq 4$ that there exists a valid construction in which $a_i = i$ for every $i \\in \\{1, 2, \\dots, n\\}$ and the last row is exactly $1, 2, \\dots, n$. The case $n = 4$ is illustrated above. Suppose there exists a valid table $T_n$ of size $(n+1) \\times n$ for $n \\geq 4$, for which the chosen permutation by Nicușor is $a_1, a_2, \\dots, a_n$ with $a_i = i$ for every $i = 1, 2, \\dots, n$.\n\nWe construct a table $T_{n+1}$ of size $(n+2) \\times (n+1)$, valid for $n+1$, as follows: we add a column at the end of $T_n$ and a new row between the last and the penultimate rows of $T_n$. We fill all the cells of the newly added last column with the value $n+1$, and the first $n$ values of the newly added row with $2, 3, \\dots, n, 1$ (or any permutation that preserves the values for $a_1, a_2, \\dots, a_n$).\n\nIt is easy to verify that the newly constructed table is valid.\n\n**Construction 2 (direct):**\n\nFor $n \\geq 4$, consider $a_i = i$ for every $i = 1, 2, 3, 4, \\dots, n$ and the following construction:\n\n\n\nIt is easy to verify that the newly constructed table is valid.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20290,
"subject": "Mathematics (Olympiad)",
"question": "a) Show that for every positive integer $n$, there exist unique positive integers $x_n$ and $y_n$ such that\n$$\n(1 + \\sqrt{33})^n = x_n + y_n\\sqrt{33}.\n$$\n\nb) Prove that if $x_n, y_n$ are defined as above and $p$ is a positive prime, then at least one of the numbers $y_{p-1}, y_p, y_{p+1}$ is divisible by $p$.",
"options": [],
"answer": "See solution",
"solution": "a) The equality holds for\n$$\nx_n = \\binom{n}{0} + 33\\binom{n}{2} + 33^2\\binom{n}{4} + \\dots\n$$\nand\n$$\ny_n = \\binom{n}{1} + 33\\binom{n}{3} + 33^2\\binom{n}{5} + \\dots.\n$$\n\nIf $a + b\\sqrt{33} = c + d\\sqrt{33}$ with $a, b, c, d \\in \\mathbb{N}$ and $b \\neq d$, then $\\sqrt{33} = \\frac{a-c}{d-b}$ would be rational, which is false. Thus, $b = d$ and $a = c$, proving uniqueness.\n\nb) If $p = 2, 3,$ or $11$, then $y_p$ is divisible by $p$.\n\nFor other primes, note that\n$$\nx_{n+1} = x_n + 33y_n, \\qquad y_{n+1} = x_n + y_n.\n$$\nAlso,\n$$\nx_p \\equiv 1 \\pmod{p}, \\qquad y_p \\equiv 33^{\\frac{p-1}{2}} \\pmod{p},\n$$\nso $y_p^2 \\equiv 1 \\pmod{p}$.\n\nTherefore,\n$$\np \\mid x_p^2 - y_p^2 = (x_p - y_p)(x_p + y_p) = 32y_{p-1}y_{p+1},\n$$\nand since $p$ is prime and $p \\neq 2$, it follows that $p \\mid y_{p-1}$ or $p \\mid y_{p+1}$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20291,
"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 $y = 1$, $x = t \\neq 0$ gives $f(1) = t f(t)$, so $f(t) = 1/t$.\n - If $k \\neq 0$, $x = 1$, $y = t$ in (2): $f(t^k) = t f(t)$. Also, (2) implies $f(xy) = f(x) f(y)$ for $x, y \\neq 0$.\n\n Let $x = t$, $y = t^{k-1}$ ($t \\neq 0$): $f(t^k) = f(t) f(t^{k-1})$, but $f(t^k) = t f(t)$, so $t f(t) = f(t) f(t^{k-1})$, thus $f(t^{k-1}) = t$ for $t \\neq 0$.\n\n Since $k$ is even, every $x \\neq 0$ can be written as $x = t^{k-1}$ for some $t \\neq 0$, so $f(x) = x^{1/(k-1)}$ for $x \\neq 0$.\n\nAll functions found satisfy (2).\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": 20292,
"subject": "Mathematics (Olympiad)",
"question": "Find all real numbers $a$ and all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for all real numbers $x, y$, the following inequality holds:\n\n$$\naf(x) - x \\leq af(f(y)) - y.\n$$\n\n",
"options": [],
"answer": "See solution",
"solution": "We consider three cases for $a$:\n\n1. **Case $a < 0$:**\n - Setting $y = x$ gives $af(x) - x \\leq af(f(x)) - x$, so $af(x) \\leq af(f(x))$, hence $f(x) \\geq f(f(x))$ for all $x$.\n - Setting $x = f(y)$ gives $af(f(y)) - f(y) \\leq af(f(y)) - y$, so $y \\leq f(y)$ for all $y$.\n - For $y = f(x)$, $f(x) \\leq f(f(x))$, so $f(x) = f(f(x))$ for all $x$.\n - The inequality becomes $af(x) - x \\leq af(y) - y$ for all $x, y$, so $af(x) - x$ is constant, i.e., $af(x) = x + c$.\n - Substituting $x$ by $f(x)$ gives $af(f(x)) = f(x) + c$, but also $af(f(x)) = af(x) = x + c$, so $f(x) = x$.\n - Then $ax = x + c$ for all $x$, which is a contradiction for $a < 0$.\n\n2. **Case $a = 0$:**\n - The inequality becomes $-x \\leq -y$ for all $x, y$, which is impossible.\n\n3. **Case $a > 0$:**\n - Let $\\alpha = f(0)$. Setting $x = 0$ gives $af(0) \\leq af(f(y)) - y$, so $y + a\\alpha \\leq af(f(y))$.\n - Setting $y = 0$ gives $af(x) - x \\leq af(\\alpha)$, so $af(x) \\leq x + af(\\alpha)$.\n - Setting $x = f(y)$ in the previous gives $af(f(y)) \\leq f(y) + af(\\alpha)$.\n - Combining, $y + a\\alpha \\leq f(y) + af(\\alpha)$, so $f(y) \\geq y + d$ where $d = a\\alpha - af(\\alpha)$.\n - From above, $x + af(\\alpha) \\geq af(x) \\geq ax + ad$, so $(1-a)x \\geq a(d - f(\\alpha))$ for all $x$.\n - Therefore, $1 - a = 0$, so $a = 1$.\n - The inequality becomes $f(x) - x \\leq f(f(y)) - y$.\n - Setting $x = f(y)$ gives $y \\leq f(y)$; setting $y = x$ gives $f(x) \\leq x$; thus $f(x) = x$.\n\n**Conclusion:** The only solution is $a = 1$ and $f(x) = x$ for all $x$. Verification shows this pair satisfies the original inequality.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20293,
"subject": "Mathematics (Olympiad)",
"question": "Given positive integers $m < n$, let $X_1, \\dots, X_n$ be distinct points in the closed unit disc, at least one of which lies on the boundary. Prove that there are $m$ distinct points $X_{i_1}, \\dots, X_{i_m}$ among the $X_i$ whose centroid is at least\n\n$$\n\\frac{1}{1 + 2m \\left(1 - \\frac{1}{n}\\right)}\n$$\n\nunits away from the centre of the disc.",
"options": [],
"answer": "See solution",
"solution": "The problem is a special case of the following general fact:\n\nLet $n$ be an integer greater than $1$, let $\\mathbf{u}_1, \\dots, \\mathbf{u}_n$ be vectors in some normed vector space over the complex field, and let $\\alpha_1, \\dots, \\alpha_n$ be complex numbers such that $\\alpha = |\\alpha_1 + \\dots + \\alpha_n|$ and $\\delta = \\max_{i,j} |\\alpha_i - \\alpha_j|$ do not vanish simultaneously. Then\n\n$$\n\\max_{\\sigma} \\left| \\sum_{i=1}^{n} \\alpha_i \\mathbf{u}_{\\sigma(i)} \\right| \\geq \\frac{\\alpha \\delta}{2(1 - 1/n)\\alpha + \\delta} \\max_{i} |\\mathbf{u}_i|,\n$$\n\nwhere $\\sigma$ runs through all permutations of $1, 2, \\dots, n$.\n\nTo prove the above inequality, let $M$ denote the maximum over all permutations. Discarding the various trivial cases, we may (and will) assume that\n\n$$\n|\\mathbf{u}_n| = \\max_i |\\mathbf{u}_i| \\neq 0, \\quad |\\mathbf{u}_n - \\mathbf{u}_1| = \\max_i |\\mathbf{u}_n - \\mathbf{u}_i|, \\quad \\alpha \\neq 0 \\quad \\text{and} \\quad |\\alpha_n - \\alpha_1| = \\delta \\neq 0.\n$$\n\nLet\n\n$$\n\\beta = \\frac{|\\mathbf{u}_n - \\mathbf{u}_1|}{|\\mathbf{u}_n|}, \\quad \\mathbf{v} = \\sum_{i=1}^{n} \\alpha_i \\mathbf{u}_i, \\quad \\mathbf{w} = \\alpha_n \\mathbf{u}_1 + \\sum_{i=2}^{n-1} \\alpha_i \\mathbf{u}_i + \\alpha_1 \\mathbf{u}_n,\n$$\n\nto get\n\n$$\n2M \\geq 2 \\max (|\\mathbf{v}|, |\\mathbf{w}|) \\geq |\\mathbf{v} - \\mathbf{w}| = |\\alpha_n - \\alpha_1| \\cdot |\\mathbf{u}_n - \\mathbf{u}_1| = \\beta \\delta |\\mathbf{u}_n|. \\quad (1)\n$$\n\nNext, with the usual notational convention $\\mathbf{u}_{n+i} = \\mathbf{u}_i$, let\n\n$$\n\\mathbf{v}_j = \\sum_{i=1}^{n} \\alpha_i \\mathbf{u}_{i+j}, \\quad j = 1, 2, \\dots, n,\n$$\n\nto obtain\n\n$$\n\\begin{aligned}\nnM &\\geq \\sum_{j=1}^{n} |\\mathbf{v}_j| \\geq \\left| \\sum_{j=1}^{n} \\mathbf{v}_j \\right| = \\alpha \\left| \\sum_{i=1}^{n} \\mathbf{u}_i \\right| = \\alpha \\left| n\\mathbf{u}_n + \\sum_{i=1}^{n-1} (\\mathbf{u}_i - \\mathbf{u}_n) \\right| \\\\ &\\geq \\alpha \\left( n|\\mathbf{u}_n| - \\left| \\sum_{i=1}^{n-1} (\\mathbf{u}_i - \\mathbf{u}_n) \\right| \\right) \\geq \\alpha \\left( n|\\mathbf{u}_n| - \\sum_{i=1}^{n-1} |\\mathbf{u}_i - \\mathbf{u}_n| \\right) \\\\ &\\geq \\alpha(n|\\mathbf{u}_n| - (n-1)|\\mathbf{u}_n - \\mathbf{u}_1|) = \\alpha(n - (n-1)\\beta)|\\mathbf{u}_n|.\n\\end{aligned} \\quad (2)\n$$\n\nHence, by (1) and (2),\n\n$$\nM \\geq \\max\\left(\\frac{\\beta\\delta}{2}, \\alpha(1 - (1 - 1/n)\\beta)\\right).\n$$\n\nTo complete the proof, notice that $\\frac{\\alpha\\delta}{2(1 - 1/n)\\alpha + \\delta}$ is the minimum of the function $t \\mapsto \\max\\left(\\frac{\\delta t}{2}, \\alpha(1 - (1 - 1/n)t)\\right)$, $t \\ge 0$.\n\n**Remark.** In the special case in the problem, the result can be slightly improved. It can be shown by induction on $n$ or a counting argument that the centroid of some $m$-point subsystem of the $X_i$ lies at least $1/(2m - 1) > 1/(1 + 2m(1 - 1/n))$ units away from the centre of the disc.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20294,
"subject": "Mathematics (Olympiad)",
"question": "The game can be reformulated in an equivalent way: Player $A$ chooses an element $x$ from the set $S$ (with $|S| = N$), and player $B$ asks a sequence of questions. The $j$-th question consists of $B$ choosing a set $D_j \\subseteq S$, and player $A$ selecting a set $P_j \\in \\{D_j, D_j^C\\}$. Player $A$ must ensure that for every $j \\ge 1$ the following holds:\n\n$$\nx \\in P_j \\cup P_{j+1} \\cup \\dots \\cup P_{j+k}.\n$$\n\nPlayer $B$ wins if, after a finite number of steps, he can choose a set $X$ with $|X| \\le n$ such that $x \\in X$.\n\n**(a)** Prove that if $N \\ge 2^k + 1$, then player $B$ can determine a set $S' \\subseteq S$ with $|S'| \\le N - 1$ such that $x \\in S'$.\n\n**(b)** Let $p$ and $q$ be real numbers such that $1.99 < p < q < 2$. Show that for sufficiently large $k$, if $|S| \\in (1.99^k, p^k)$, then there is a strategy for player $A$ to select sets $P_1, P_2, \\dots$ (based on sets $D_1, D_2, \\dots$ provided by $B$) such that for each $j$:\n\n$$\nP_j \\cup P_{j+1} \\cup \\dots \\cup P_{j+k} = S.\n$$",
"options": [],
"answer": "See solution",
"solution": "**(a)**\nAssume $N \\ge 2^k + 1$. In the first move, $B$ selects any set $D_1 \\subseteq S$ such that $|D_1| \\ge 2^{k-1}$ and $|D_1^C| \\ge 2^{k-1}$. After receiving $P_1$ from $A$, $B$ makes the second move: $B$ selects $D_2 \\subseteq S$ such that $|D_2 \\cap P_1^C| \\ge 2^{k-2}$ and $|D_2^C \\cap P_1^C| \\ge 2^{k-2}$. $B$ continues: in move $j$, $B$ chooses $D_j$ so that $|D_j \\cap P_j^C| \\ge 2^{k-j}$ and $|D_j^C \\cap P_j^C| \\ge 2^{k-j}$.\n\nAfter $k$ steps, $B$ has obtained sets $P_1, \\dots, P_k$ such that $|(P_1 \\cup \\dots \\cup P_k)^C| \\ge 1$. Then $B$ chooses $D_{k+1}$ to be a singleton containing any element outside $P_1 \\cup \\dots \\cup P_k$.\n\n- *Case 1*: $A$ selects $P_{k+1} = D_{k+1}^C$. Then $B$ can take $S' = S \\setminus D_{k+1}$.\n- *Case 2*: $A$ selects $P_{k+1} = D_{k+1}$. Now $B$ repeats the procedure on $S_1 = S \\setminus D_{k+1}$ to obtain $P_{k+2}, \\dots, P_{2k+1}$. Since $|S_1| \\ge 2^k$, $|(P_{k+1} \\cup \\dots \\cup P_{2k+1})^C| \\ge 1$, and we may take $S' = P_{k+1} \\cup \\dots \\cup P_{2k+1}$.\n\n**(b)**\nLet $p, q$ be real numbers with $1.99 < p < q < 2$. Choose $k_0$ such that\n$$\n\\left(\\frac{p}{q}\\right)^{k_0} \\le 2 \\cdot \\left(1 - \\frac{q}{2}\\right) \\quad \\text{and} \\quad p^k - 1.99^k > 1.\n$$\nFor $k \\ge k_0$ and $|S| \\in (1.99^k, p^k)$, $A$ can select sets $P_1, P_2, \\dots$ so that for each $j$,\n$$\nP_j \\cup P_{j+1} \\cup \\dots \\cup P_{j+k} = S.\n$$\nLet $S = \\{1, 2, \\dots, N\\}$, and define $N$-tuples $\\mathbf{x}^j = (x_1^j, \\dots, x_N^j)$, with $x_i^0 = 1$. After $P_j$ is selected, define\n$$\nx_i^{j+1} = \\begin{cases} 1, & i \\in P_j \\\\ q x_i^j, & i \\notin P_j. \\end{cases}\n$$\nLet $T(\\mathbf{x}) = \\sum_{i=1}^N x_i$. $A$ can keep $B$ from winning if $T(\\mathbf{x}^j) \\le q^k$ for all $j$. Since $T(\\mathbf{x}^0) = N \\le p^k < q^k$, this holds initially.\n\nGiven $\\mathbf{x}^j$ with $T(\\mathbf{x}^j) \\le q^k$ and a set $D_{j+1}$, $A$ can choose $P_{j+1} \\in \\{D_{j+1}, D_{j+1}^C\\}$ so that $T(\\mathbf{x}^{j+1}) \\le q^k$. Let $\\mathbf{y}$ be the sequence if $P_{j+1} = D_{j+1}$, and $\\mathbf{z}$ if $P_{j+1} = D_{j+1}^C$:\n$$\nT(\\mathbf{y}) = \\sum_{i \\in D_{j+1}^C} q x_i^j + |D_{j+1}|,\n$$\n$$\nT(\\mathbf{z}) = \\sum_{i \\in D_{j+1}} q x_i^j + |D_{j+1}^C|.\n$$\nSumming gives\n$$\nT(\\mathbf{y}) + T(\\mathbf{z}) = q T(\\mathbf{x}^j) + N \\le q^{k+1} + p^k.\n$$\nThus,\n$$\n\\min\\{T(\\mathbf{y}), T(\\mathbf{z})\\} \\le \\frac{q}{2} q^k + \\frac{p^k}{2} \\le q^k,\n$$\nbecause of our choice of $k_0$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20295,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $n = 41^2$, where $41$ is a prime number. Consider the general case where $n = p^2$ for an odd prime $p$. A teacher distributes candies to $n$ children in the following way: when the $k$th candy is dropped, the teacher skips $1 + 2 + \\cdots + (k-1) = \\frac{k(k-1)}{2}$ children, so the $k$th candy is given to child $a_i$, where $i \\equiv k + \\frac{k(k-1)}{2} \\pmod{p^2}$. How many children never receive a candy?",
"options": [],
"answer": "See solution",
"solution": "Let $d_i$ be the child who receives the $i$th candy. The sequence $d_1, d_2, \\dots$ is periodic with period $p^2$. We can focus on $d_0, d_1, \\dots, d_{(p^2-1)/2}$. For $0 \\leq i < j \\leq (p^2-1)/2$, $d_i = d_j$ if and only if $i(i+1) \\equiv j(j+1) \\pmod{p^2}$, which leads to $(j-i)(j+i+1) \\equiv 0 \\pmod{p^2}$. This happens for certain $i, j$ values, and it turns out that $(p+1)/2$ candies go to the same child, while the rest go to distinct children. Thus, the number of children who receive at least one candy is $\\frac{p^2+1}{2} - \\frac{p-1}{2} = \\frac{p^2-p+2}{2}$, so the number who never receive a candy is $p^2 - \\frac{p^2-p+2}{2} = \\frac{p^2+p-2}{2}$. For $p = 41$, the answer is $\\frac{41^2+41-2}{2} = 860$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20296,
"subject": "Mathematics (Olympiad)",
"question": "Let $z \\geq 0$ be an integer. Consider the equation:\n$$4^z = 3^x + 7^y$$\nFind all integer solutions $(x, y, z)$ to this equation.",
"options": [],
"answer": "See solution",
"solution": "If $z = 0$, then $4^0 = 1$, so $3^x + 7^y = 1$ has no integer solutions.\n\nIf $z = 1$, then $4^1 = 4$, so $3^x + 7^y = 4$. The only possibility is $y = 0$, $x = 1$ ($3^1 + 7^0 = 3 + 1 = 4$).\n\nIf $z = 2$, then $4^2 = 16$, so $3^x + 7^y = 16$. The only possibility is $y = 1$, $x = 2$ ($3^2 + 7^1 = 9 + 7 = 16$).\n\nLet $z \\geq 3$. Then $3^x + 7^y \\geq 8$. We show that $x$ is even. Indeed, if $x$ is odd, then $3^x \\equiv 3 \\pmod{8}$, and $7^y \\equiv 1$ or $7 \\pmod{8}$, so $3^x + 7^y$ is not divisible by 8. Hence, $x = 2a$ for some nonnegative integer $a$.\n\nThen $7^y = 4^z - 3^x = 2^{2z} - 3^{2a} = (2^z - 3^a)(2^z + 3^a)$. Therefore, $2^z - 3^a$ and $2^z + 3^a$ must be powers of 7. Since $2^z + 3^a > 1$, $2^z + 3^a$ is divisible by 7. If $2^z - 3^a > 1$, then $2^z - 3^a$ is divisible by 7. Therefore, $(2^z + 3^a) + (2^z - 3^a) = 2 \\cdot 2^z$ is divisible by 7, which is impossible. So it suffices to consider the case when $2^z - 3^a = 1$, i.e. $2^z - 1 = 3^a$.\n\nIf $z \\geq 3$, then $2^z - 1 \\neq 3$, so this is only possible if $z$ is even. Let $z = 2c$ for some nonnegative integer $c$. Then $4^c - 3^a = 1$. If $a = 1$, we find $c = 1$ and obtain $x = 1$, $z = 2$, $y = 1$. If $a > 1$, then $4^c = 3^a + 1 \\equiv 1 \\pmod{9}$. The power of 4 has this remainder only if $c$ is divisible by 3. But in this case $4^3 \\equiv 1 \\pmod{7}$ and hence $4^c \\equiv 1 \\pmod{7}$. It follows that $3^a = 4^c - 1 \\neq 7$, which is impossible.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20297,
"subject": "Mathematics (Olympiad)",
"question": "Let $P(x) \\in \\mathbb{R}[x]$ be a monic, non-constant polynomial. Determine all continuous functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(f(P(x)) + y + 2023f(y)) = P(x) + 2024f(y),\n$$\nfor all real numbers $x, y$.",
"options": [],
"answer": "See solution",
"solution": "Because $P(x)$ is a non-constant monic polynomial, there exists a constant $c$ such that $P(x)$ can take all values in $[c, +\\infty)$. From the given equation, for $x \\geq c$ and all $y \\in \\mathbb{R}$:\n\n$$\nf(f(x) + y + 2023f(y)) = x + 2024f(y). \\tag{1}\n$$\n\nSuppose $|f(x)| \\leq M$ for all $x \\geq m \\geq c$. Fix $y_0$. Then $f(x) + y_0 + 2023f(y_0)$ is bounded for $x \\geq m$, so $f$ is bounded on a bounded interval. But from (1), $f(f(x) + y_0 + 2023f(y_0)) = x + 2024f(y_0)$, which is unbounded as $x \\to +\\infty$, a contradiction. Thus, $\\lim_{x \\to +\\infty} |f(x)| = +\\infty$.\n\nThis implies $f$ is monotonic on $[c, +\\infty)$. Consider two cases:\n\n**Case 1:** $\\lim_{x \\to +\\infty} f(x) = -\\infty$.\n\nSuppose $f(x) \\leq L$ for all $x \\leq \\ell$. Fix $y_0$. For large $x_0 \\geq c$, $f(x_0) + y_0 + 2023f(y_0) \\leq \\ell$, so $f(f(x_0) + y_0 + 2023f(y_0)) \\leq L$, but from (1), $f(f(x_0) + y_0 + 2023f(y_0)) = x_0 + 2024f(y_0)$, which can be made arbitrarily large, a contradiction. Thus, $\\lim_{x \\to -\\infty} f(x) = +\\infty$.\n\nBut then $f$ is continuous and surjective from $\\mathbb{R}$ to $\\mathbb{R}$, so for any $x_1 \\geq c$, there exists $y_1$ such that $f(y_1) = -\\frac{f(x_1)}{2023}$. Plugging into (1):\n\n$$\nx_1 - \\frac{2024}{2023} f(x_1) = f(y_1) = -\\frac{f(x_1)}{2023},\n$$\nwhich gives $f(x_1) = x_1$. But this contradicts $\\lim_{x \\to +\\infty} f(x) = -\\infty$.\n\n**Case 2:** $\\lim_{x \\to +\\infty} f(x) = +\\infty$.\n\nSuppose $f(a) = f(b)$ for some $a, b$. For large $d \\geq c$, $f(d) + a + 2023f(a) \\geq c$ and $f(d) + b + 2023f(b) \\geq c$. Then,\n\n$$\nf(f(d) + a + 2023f(a)) = d + 2024f(a) = d + 2024f(b) = f(f(d) + b + 2023f(b)),\n$$\nso $a = b$. Thus, $f$ is injective and, by continuity, strictly increasing on $\\mathbb{R}$.\n\nSuppose $f(x) \\geq V$ for all $x \\leq \\nu$. Substitute $y = -f(x)$ into (1):\n\n$$\nf(2023f(-f(x))) = x + 2024f(-f(x)), \\quad \\forall x \\geq c. \\tag{2}\n$$\n\nFor large $x_2 \\geq c$, $f(x_2) \\geq -\\nu$, so $-f(x_2) \\leq \\nu$ and $f(-f(x_2)) \\geq V$. Then,\n\n$$\nx_2 + 2024f(-f(x_2)) \\geq x_2 + 2024V.\n$$\n\nBut $f(2023f(-f(x_2))) \\leq f(2023f(\\nu))$, so\n\n$$\nf(2023f(\\nu)) \\geq x_2 + 2024V,\n$$\nwhich is impossible for large $x_2$. Thus, $\\lim_{x \\to -\\infty} f(x) = -\\infty$.\n\nNow, for any $x \\geq c$, there exists $y$ such that $f(y) = -\\frac{f(x)}{2023}$. Plug into (1):\n\n$$\nx - \\frac{2024}{2023} f(x) = -\\frac{f(x)}{2023},\n$$\nso $f(x) = x$ for all $x \\geq c$. By continuity and monotonicity, $f(x) = x$ for all $x \\in \\mathbb{R}$.\n\n**Verification:**\n\nPlug $f(x) = x$ into the original equation:\n\n$$\nf(f(P(x)) + y + 2023f(y)) = f(P(x) + y + 2023y) = P(x) + y + 2023y = P(x) + 2024y,\n$$\nwhich matches the right side if and only if $f(y) = y$.\n\n**Conclusion:**\n\nThe only continuous function $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying the equation is $f(x) = x$ for all $x \\in \\mathbb{R}$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20298,
"subject": "Mathematics (Olympiad)",
"question": "For every positive integer $n$, let $\\sigma(n)$ denote the sum of all positive divisors of $n$ (including $1$ and $n$ itself). Show that a positive integer $n$ with at most two distinct prime factors satisfies $\\sigma(n) = 2n - 2$ if and only if $n = 2^k(2^{k+1} + 1)$, where $k$ is a non-negative integer and $2^{k+1} + 1$ is prime.",
"options": [],
"answer": "See solution",
"solution": "Sufficiency is a routine verification. To prove necessity, assume first that $n = 2^k p^l$, where $k$ and $l$ are non-negative integers, and $p$ is an odd prime. Notice that $l$ must be positive, so\n\n$$\n1 + \\frac{1}{p} \\le \\frac{\\sigma(p^l)}{p^l} = \\frac{p - \\frac{1}{p^l}}{p-1} < \\frac{p}{p-1},\n$$\n\nwhence (since $\\sigma$ is multiplicative)\n\n$$\n(2^{k+1} - 1) \\left(1 + \\frac{1}{p}\\right) \\le \\frac{\\sigma(n)}{p^l} = 2^{k+1} - \\frac{2}{p^l} < (2^{k+1} - 1) \\frac{p}{p-1}.\n$$\n\nBy the first inequality, $(2^{k+1}-1)(1+1/p) < 2^{k+1}$, so $p > 2^{k+1}-1$, i.e., $p \\ge 2^{k+1}+1$ since $p$ is odd. On the other hand, $p < 2^{k+1}+2(p-1)/p^l$, by the second inequality, so $2(p-1) > p^l$, and consequently $l=1$ and $p = 2^{k+1}+1$.\n\nTo rule out the case $n = p^k q^l$, where $p$ and $q$ are distinct odd primes, and $k$ and $l$ are positive integers, write\n\n$$\n2 - \\frac{2}{n} = \\frac{\\sigma(n)}{n} = \\frac{\\sigma(p^k)}{p^k} \\cdot \\frac{\\sigma(q^l)}{q^l} < \\frac{p}{p-1} \\cdot \\frac{q}{q-1}.\n$$\n\nAlternatively, but equivalently,\n\n$$\n\\frac{1}{p-1} + \\frac{1}{q-1} + \\frac{1}{(p-1)(q-1)} + \\frac{2}{n} > 1,\n$$\n\nso $\\min(p, q) = 3$, say $p=3$. Then $3/(q-1)+4/n > 1$, and it follows that $q=5$ and $k=l=1$, i.e., $n=15$ which does not satisfy the condition $\\sigma(n) = 2n-2$. This completes the proof.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20299,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a square and $X$ a point such that $A$ and $X$ are on opposite sides of $CD$. The lines $AX$ and $BX$ intersect $CD$ in $Y$ and $Z$ respectively. If the area of $ABCD$ is $1$ and the area of $\\triangle XYZ$ is $\\frac{2}{3}$, determine the length of $YZ$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let the length of $YZ$ be $x$, and let $y$ be the associated height of triangle $XYZ$. Note that triangles $XYZ$ and $XAB$ are similar. Since corresponding sides $YZ$ and $AB$ have lengths $x$ and $1$ respectively, the heights $y$ and $1+y$ have to satisfy\n\n$$\n\\frac{y}{1+y} = \\frac{x}{1} = x.\n$$\n\nHence the area of $XYZ$ is\n\n$$\n\\frac{2}{3} = \\frac{xy}{2} = \\frac{y^2}{2(1+y)}.\n$$\n\nThis yields the quadratic equation\n\n$$\ny^2 - \\frac{4}{3}y - \\frac{4}{3} = \\left(y + \\frac{2}{3}\\right) (y - 2) = 0,\n$$\n\nfrom which it follows that $y = 2$ and thus\n\n$$\nx = \\frac{y}{1+y} = \\frac{2}{3}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20300,
"subject": "Mathematics (Olympiad)",
"question": "The quadrilateral $ABCD$ in the figure has three angles equal to $45\\degree$, at vertices $A$, $B$, and $C$. ($ABCD$ is not convex.) It is allowed to measure the length of exactly one line segment in the figure. Find the area of the quadrilateral.\n\n",
"options": [],
"answer": "See solution",
"solution": "It is enough to measure $BD$ because $(ABCD) = \\frac{1}{2} BD^2$.\n\nIndeed, extend $AD$ to meet $BC$ at $P$. Since $\\angle ABP = \\angle BAP = 45\\degree$, we have $AP = BP$ and $\\angle APB = 90\\degree$. Hence triangle $ABP$ is right and isosceles, so $(ABP) = \\frac{1}{2} BP^2$. Next, triangle $CDP$ has $\\angle PCD = 45\\degree$, $\\angle CPD = 90\\degree$. Therefore it is right and isosceles too, with\n\n$$\n(CDP) = \\frac{1}{2} DP^2. \\text{ It follows that}\n$$\n\n$$\n(ABCD) = (ABP) + (CDP) = \\frac{1}{2}(BP^2 + DP^2) = \\frac{1}{2} BD^2.\n$$\n\nThe last equality follows from Pythagoras' theorem in triangle $BDP$.\n\nComment: An equally good answer is to measure $AC$. This is because $AC = BD$ and so\n\n$$\n(ABCD) = \\frac{1}{2} AC^2.\n$$\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20301,
"subject": "Mathematics (Olympiad)",
"question": "Find the slope $k$ of the angle bisector of the acute angle formed at the origin by the lines $y = x$ and $y = 3x$.",
"options": [],
"answer": "See solution",
"solution": "Applying the Angle Bisector Theorem gives\n\n$$\n\\frac{3-k}{\\sqrt{10}} = \\frac{k-1}{\\sqrt{2}},\n$$\n\nfrom which $k = \\frac{1+\\sqrt{5}}{2}$.\n\n**OR**\n\nLet $\\alpha$ and $\\beta$ be the acute angles from the positive x-axis to the lines $y = x$ and $y = 3x$, respectively. Then the bisector has angle $\\gamma = \\frac{1}{2}(\\alpha + \\beta)$ from the positive x-axis. Because $\\tan \\alpha = 1$ and $\\tan \\beta = 3$, it follows that $\\sin \\alpha = \\cos \\alpha = \\frac{1}{\\sqrt{2}}$, $\\sin \\beta = \\frac{3}{\\sqrt{10}}$, and $\\cos \\beta = \\frac{1}{\\sqrt{10}}$. Using a half-angle identity for tangent and the sum identities for sine and cosine gives\n\n$$\n\\begin{aligned}\nk &= \\tan \\gamma = \\frac{\\sin(\\alpha + \\beta)}{1 + \\cos(\\alpha + \\beta)} \\\\\n&= \\frac{\\sin \\alpha \\cos \\beta + \\sin \\beta \\cos \\alpha}{1 + \\cos \\alpha \\cos \\beta - \\sin \\alpha \\sin \\beta} \\\\\n&= \\frac{\\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{10}} + \\frac{3}{\\sqrt{10}} \\cdot \\frac{1}{\\sqrt{2}}}{1 + \\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{10}} - \\frac{1}{\\sqrt{2}} \\cdot \\frac{3}{\\sqrt{10}}} \\\\\n&= \\frac{2}{\\sqrt{5} - 1} \\\\\n&= \\frac{1 + \\sqrt{5}}{2}.\n\\end{aligned}\n$$\n\n**OR**\n\nThe lines $y = x$ and $y = 3x$ form an acute angle at the origin. The bisector of that angle passes through the origin and the midpoint of a line segment between two points on those lines in the first quadrant that are equally distant from the origin. The point $(1, 1)$ is on the line $y = x$ and is $\\sqrt{2}$ units from the origin. Let $(p, 3p)$ be the point on the line $y = 3x$ in the first quadrant that is $\\sqrt{2}$ units from the origin. Then $p^2 + (3p)^2 = 2$, from which $p = \\frac{1}{\\sqrt{5}} = \\frac{1}{5}\\sqrt{5}$. The midpoint between $(1, 1)$ and $(\\frac{1}{5}\\sqrt{5}, \\frac{3}{5}\\sqrt{5})$ is\n\n$$\n\\left( \\frac{1}{2} \\left( 1 + \\frac{1}{5}\\sqrt{5} \\right), \\frac{1}{2} \\left( 1 + \\frac{3}{5}\\sqrt{5} \\right) \\right) = \\left( \\frac{\\sqrt{5}+5}{10}, \\frac{3\\sqrt{5}+5}{10} \\right).\n$$\n\nThe slope of the line through this midpoint and the origin is\n\n$$\nk = \\frac{3\\sqrt{5} + 5}{\\sqrt{5} + 5} = \\frac{1 + \\sqrt{5}}{2}.\n$$\n\n**Note:** The value $\\frac{1+\\sqrt{5}}{2}$ is the golden ratio.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20302,
"subject": "Mathematics (Olympiad)",
"question": "Prove that the number $A = 7^{2n} - 48n - 1$ is a multiple of $9$, for all positive integers $n$.",
"options": [],
"answer": "See solution",
"solution": "We distinguish three cases modulo $3$:\n\n* If $n = 3k$, then\n $$A = 7^{6k} - 48(3k) - 1 = 49^{3k} - 9 \\cdot 16k - 1.$$ \n Since $49 \\equiv 4 \\pmod{9}$, it follows that $49^3 \\equiv 4^3 \\pmod{9} \\equiv 1 \\pmod{9}$, and hence $9 \\mid 49^{3k} - 1$, so $9 \\mid A$.\n\n* If $n = 3k+1$, then\n $$A = 7^{6k+2} - 48(3k+1) - 1 = 7^2 \\cdot 7^{6k} - 9 \\cdot 16k - 49 = 49(7^{6k} - 1) - 9 \\cdot 16k.$$ \n Since $9 \\mid 7^{6k} - 1$, it follows again that $9 \\mid A$.\n\n* If $n = 3k + 2$, then\n $$A = 7^{6k+4} - 48(3k + 2) - 1 = 7^4 \\cdot 7^{6k} - 9 \\cdot 16k - 97.$$ \n We have $7^{6k} \\equiv 1 \\pmod{9}$, so $7^4 \\cdot 7^{6k} \\equiv 7^4 \\pmod{9} \\equiv 49^2 \\pmod{9} \\equiv 4^2 \\pmod{9} \\equiv 7 \\pmod{9}$, and also $97 \\equiv 7 \\pmod{9}$. Hence $9 \\mid 7^4 \\cdot 7^{6k} - 97$, that is, $9 \\mid A$.\n\n**Note:** Alternatively, we can use induction.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20303,
"subject": "Mathematics (Olympiad)",
"question": "Sea $ABC$ un triángulo isósceles rectángulo con ángulo recto en $A$. Sean $E$ y $F$ puntos en $AB$ y $AC$ respectivamente tales que $ECB = 30^\\circ$ y $FBC = 15^\\circ$. Las rectas $CE$ y $BF$ se cortan en $P$ y la recta $AP$ corta al lado $BC$ en $D$. Calcular la medida del ángulo $FDC$.",
"options": [],
"answer": "See solution",
"solution": "Demostraremos que $FD$ es perpendicular a $BC$.\n\nSea $D'$ en $BC$ tal que $FD'$ es perpendicular a $BC$ y sea $P'$ el punto de intersección de $AD'$ y $BF$.\n\nQueremos demostrar que $\\angle BCP' = 30^\\circ$, lo que implica que $P = P'$ y por lo tanto que $D = D'$.\n\nObservamos que $ABD'F$ es cíclico, ya que $\\angle BAF = \\angle BD'F = 90^\\circ$. Esto implica que $\\angle AD'F = ABF = 30^\\circ$.\n\nPor la suma de los ángulos en el triángulo $BFD'$ resulta que $\\angle BFD' = 180^\\circ - 15^\\circ - 90^\\circ = 75^\\circ$.\n\nAhora en el triángulo $FP'D'$ tenemos que\n\n$$\n\\angle FP'D' = 180^\\circ - \\angle P'D'F - \\angle P'FD' = 180^\\circ - 30^\\circ - 75^\\circ = 75^\\circ\n$$\n\nde donde el triángulo $FP'D'$ es isósceles con $P'D' = FD'$.\n\nDado que $\\angle FD'C = 90^\\circ$ y $\\angle FCD' = 45^\\circ$, tenemos que $\\angle D'FC = 45^\\circ$ y el triángulo $FD'C$ es isósceles con $FD' = CD'$. Esto demuestra que $P'D' = CD'$ y vale que\n\n$$\n\\angle BCP' = \\angle D'CP' = \\frac{1}{2}(180^\\circ - \\angle P'D'C) = \\frac{1}{2}(180^\\circ - 30^\\circ - 90^\\circ) = 30^\\circ\n$$\n\ncomo queríamos demostrar.\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20304,
"subject": "Mathematics (Olympiad)",
"question": "Suppose that the equation $\\lg kx = 2\\lg(x+1)$ has exactly one real root. Then the range of $k$ is ____.",
"options": [],
"answer": "See solution",
"solution": "We have\n\n$$\nkx > 0, \\qquad \\textcircled{1}\n$$\n\n$$\nx + 1 > 0, \\qquad \\textcircled{2}\n$$\n\n$$\nkx = (x + 1)^2. \\qquad \\textcircled{3}\n$$\n\nThe expression $\\textcircled{3}$ can be rewritten as\n\n$$\nx^2 + (2 - k)x + 1 = 0. \\qquad \\textcircled{4}\n$$\n\nThe two roots of $\\textcircled{4}$ are\n\n$$\nx_1, x_2 = \\frac{1}{2}\\left[k - 2 \\pm \\sqrt{k^2 - 4k}\\right], \\qquad \\textcircled{5}\n$$\n\nwhere\n\n$$\n\\Delta = k^2 - 4k \\ge 0 \\iff k \\le 0 \\text{ or } k \\ge 4.\n$$\n\n(i) When $k < 0$, it is easy to see from $\\textcircled{5}$ that $x_1 + 1 > 0$, $x_2 + 1 < 0$, and $kx_1 > 0$. Then the equation has one real root, $x_1 = \\frac{1}{2}\\left[k - 2 + \\sqrt{k^2 - 4k}\\right]$.\n\n(ii) When $k = 4$, the equation has one real root, $x = \\frac{4}{2} - 1 = 1$.\n\n(iii) When $k > 4$, the two roots $x_1, x_2$ are both positive and $x_1 \\ne x_2$. Discarded.\n\nTherefore, the range of $k$ is $k < 0$ or $k = 4$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20305,
"subject": "Mathematics (Olympiad)",
"question": "Five distinct points $A$, $M$, $B$, $C$, and $D$ are on a circle $O$ in this order with $MA = MB$.\n\nLet the lines $AC$ and $MD$ intersect at $P$, and the lines $BD$ and $MC$ at $Q$. Let the line $PQ$ meet the circle $O$ at $X$ and $Y$. Prove that $MX = MY$.",
"options": [],
"answer": "See solution",
"solution": "Since $\\angle ACM = \\angle BDM$ (because $MA = MB$), $\\angle PCQ = \\angle PDQ$, so four points $C$, $D$, $P$, and $Q$ are concyclic. So $\\angle PQD = \\angle PCD = \\angle ACD = \\angle ABD$. Therefore, $AB$ and $PQ$, hence $AB$ and $XY$, are parallel. Since $M$ is the midpoint of the arc $AB$, it is also that of the arc $XY$. Thus $MX = MY$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20306,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $n$ which are not powers of $2$ and which satisfy the equation\n$$\nn = 3D + 5d,\n$$\nwhere $D$ and $d$ denote the greatest and the least numbers among all odd divisors of $n$ which are larger than $1$.",
"options": [],
"answer": "See solution",
"solution": "Let $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}$ be the prime factorization of $n$, with $p_1 < p_2 < \\dots < p_k$ and $\\alpha_i > 0$. The equation implies $p_1 = 2$ (otherwise $D = n$, which contradicts $n = 3D + 5d$) and $k \\ge 2$ (otherwise $n$ is a power of $2$). Thus, $D = p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}$, $d = p_2$, and the equation becomes\n$$\n2^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k} = 3 p_2^{\\alpha_2} \\dots p_k^{\\alpha_k} + 5p_2,\n$$\nor\n$$\n(2^{\\alpha_1} - 3) p_2^{\\alpha_2 - 1} \\dots p_k^{\\alpha_k} = 5.\n$$\nFor $k=2$, this is $(2^{\\alpha_1} - 3)p_2^{\\alpha_2-1} = 5$. Since $5$ has divisors $1$ and $5$, $2^{\\alpha_1} - 3 \\in \\{1, 5\\}$, so $\\alpha_1 = 2$ or $3$.\n\n**Case 1:** $\\alpha_1 = 2$.\nThen $p_2^{\\alpha_2-1} \\dots p_k^{\\alpha_k} = 5$. This gives two solutions: $n = 2^2 5^2 = 100$ and $n = 2^2 3^1 5^1 = 60$.\n\n**Case 2:** $\\alpha_1 = 3$.\nThen $p_2^{\\alpha_2-1} \\dots p_k^{\\alpha_k} = 1$, so $k=2$ and $\\alpha_2 = 1$. Thus, $n = 2^3 p_2^1 = 8p_2$, where $p_2$ is any odd prime.\n\n**Answer:**\nAll such $n$ are $n = 60$, $n = 100$, and $n = 8p$, where $p$ is any odd prime number.\n\n**Remark:**\nAlternatively, write $n = 2^\\alpha p l$, where $2^\\alpha$ is the largest power of $2$ dividing $n$, $p$ is the smallest odd prime divisor, and $l$ is an odd number with no prime divisor smaller than $p$. Then $D = pl$, $d = p$, and the equation becomes $(2^{\\alpha} - 3)l = 5$. This gives $l=1$, $\\alpha=3$ ($n=8p$), or $l=5$, $\\alpha=2$ ($n=20p$ with $p \\in \\{3,5\\}$, so $n=60,100$).\n\n**Another solution:**\nThe equation $n = 3D + 5d$ implies $n > 3D$ and $n \\le 8D$. Since $n/D$ must be a power of $2$, either $n=4D$ or $n=8D$.\n- If $n=4D$, then $D=5d$, $n=20d$, and $d \\in \\{3,5\\}$, so $n=60,100$.\n- If $n=8D$, then $D=d$ is an odd prime, so $n=8D$ for any odd prime $D$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20307,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\Gamma$ be a semicircle with diameter $AB$. The point $C$ lies on the diameter $AB$ and points $E$ and $D$ lie on the arc $BA$, with $E$ between $B$ and $D$. Let the tangents to $\\Gamma$ at $D$ and $E$ meet at $F$. Suppose that $\\angle ACD = \\angle ECB$.\n\n$$\n\\text{Prove that } \\angle EFD = \\angle ACD + \\angle ECB.\n$$\n\n",
"options": [],
"answer": "See solution",
"solution": "Let the circle passing through $C$, $D$, and $E$ meet line $AB$ again at $O$. We will show $O$ is the centre of $\\Gamma$.\n\nAngle $ACD$ is exterior to the cyclic quadrilateral $COED$ so $\\angle ACD = \\angle OED$, while, by equal angles in the same segment, $\\angle ECB = \\angle EDO$. Therefore $\\angle OED = \\angle EDO$ and so $OD = OE$. Thus $O$ is the centre of $\\Gamma$ (it lies on $AB$ and on the perpendicular bisector of $DE$) and so $O$, $C$, $D$, $E$ are concyclic.\n\nLines $DF$ and $EF$ are tangents to $\\Gamma$ so $\\angle OEF = 90^\\circ = \\angle FDO$. Therefore $O$, $D$, $F$, $E$ are concyclic.\n\nWe can finish as in solution 1.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20308,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs of positive integers $a, b$ such that $a - b = 101$ and $ab$ is a perfect square.",
"options": [],
"answer": "See solution",
"solution": "Let $d$ be the greatest common divisor of $a$ and $b$. Then $a = d m$ and $b = d n$, where $m$ and $n$ are coprime positive integers. Since $ab = d^2 m n$ is a perfect square and $m$ and $n$ are coprime, both $m$ and $n$ must be perfect squares. Let $m = x^2$ and $n = y^2$, where $x$ and $y$ are positive integers. Then:\n\n$$\na - b = d(m - n) = d(x^2 - y^2) = d(x + y)(x - y) = 101\n$$\n\nSince $101$ is prime and $x + y \\geq 2$, the only possibility is $d = 1$, $x - y = 1$, and $x + y = 101$. Solving these, $x = 51$ and $y = 50$. Thus,\n\n$$\na = x^2 = 51^2 = 2601, \\quad b = y^2 = 50^2 = 2500\n$$\n\nSo the only solution is $(a, b) = (2601, 2500)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20309,
"subject": "Mathematics (Olympiad)",
"question": "Let $f : \\mathbb{N}_0 \\to \\mathbb{Z}$ be a function such that for all $m, n \\in \\mathbb{N}_0$,\n$$\nf(m + n)^2 = f(m|f(n)|) + f(n^2).\n$$\nHow many possible tuples $(f(0), f(1), \\dots, f(2024))$ are there?",
"options": [],
"answer": "See solution",
"solution": "First, set $m = n = 0$:\n$$\nf(0)^2 = 2f(0)\n$$\nso $f(0) \\in \\{0, 2\\}$.\n\n**Case 1:** $f(0) = 0$\n\nThen $f(m)^2 = f(0) + f(0) = 0$ for all $m$, so $f(m) = 0$ for all $m$. Thus, there is only one possible tuple in this case.\n\n**Case 2:** $f(0) = 2$\n\nSet $m = 0, n = 1$:\n$$\nf(1)^2 = f(1) + 2\n$$\nso $f(1) \\in \\{-1, 2\\}$. If $f(1) = -1$, then $f(2)^2 = -2$, which is impossible. Thus, $f(1) = 2$.\n\nNow, for any $m$, $f(m+1)^2 = f(2m) + 2$ and $f(m)^2 = f(2m) + 2$, so $f(m+1)^2 = f(m)^2$. By induction, $|f(m)| = 2$ for all $m$.\n\nAlso, $f(2m) = f(m)^2 - 2 = 2$ for all $m$.\n\nFor $n$, $f(n^2) = f(n)^2 - f(0) = 2$ for all $n$.\n\nThus, $f(n) = 2$ whenever $n$ is even or a perfect square. For other $n$, $f(n) = -2$ or $2$.\n\nThere are 990 integers between 0 and 2024 that are neither even nor perfect squares, so the number of possible tuples is $2^{990}$ in this case.\n\n**Total:** $2^{990} + 1$ possible tuples.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20310,
"subject": "Mathematics (Olympiad)",
"question": "Los puntos $P$ y $Q$ están en el lado $BC$ del triángulo acutángulo $ABC$ de modo que $\\angle PAB = \\angle BCA$ y $\\angle CAQ = \\angle ABC$. Los puntos $M$ y $N$ están en las rectas $AP$ y $AQ$, respectivamente, de modo que $P$ es el punto medio de $AM$, y $Q$ es el punto medio de $AN$. Demostrar que las rectas $BM$ y $CN$ se cortan en la circunferencia circunscrita del triángulo $ABC$.",
"options": [],
"answer": "See solution",
"solution": "Sean $E, F$ los puntos medios respectivos de $CA, AB$. Claramente, $\\angle MBA = \\angle PFA$ y $\\angle NCA = \\angle QEA$. Ahora bien, como por construcción $ABC, PBA$ y $QAC$ son semejantes, llamando $D$ al punto medio de $BC$, se tiene que $\\angle PFA = \\angle ADC$ y $\\angle QEA = \\angle ADB$, con lo que llamando $R$ al punto de intersección de $BM$ y $CN$, tenemos que\n\n$$\n\\angle RBA + \\angle RCA = \\angle MBA + \\angle NCA = \\angle ADC + \\angle ADB = 180^{\\circ},\n$$\n\ny $ABRC$ es cíclico, como queríamos demostrar.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20311,
"subject": "Mathematics (Olympiad)",
"question": "For which positive integers $k$ can the integers $1, 2, 3, \\dots, (2k)^2$ be arranged as a $2k \\times 2k$ table in such a way that none of the row sums and column sums have the same parity as $k$?",
"options": [],
"answer": "See solution",
"solution": "*Answer*: for all $k \\ge 2$.\n\nSuch an arrangement is impossible for $k = 1$. In order to make all row sums and column sums in a $2 \\times 2$ table even, both odd numbers should occur in the same row and also in the same column, which is impossible.\n\n$$\n\\begin{matrix}\n1 & 1 & 1 & 0 \\\\\n1 & 1 & 0 & 1 \\\\\n1 & 0 & 0 & 0 \\\\\n0 & 1 & 0 & 0\n\\end{matrix}\n$$\n\nFigure 11\n\n\n\nFigure 12\n\nIn the rest, let 0 and 1 denote any even and odd number, respectively. For $k = 2$, one suitable arrangement is shown in Figure 11. A way to obtain a suitable arrangement for $k + 1$ from any suitable arrangement for $k$ is shown in Figure 12. The parity of the sum of each old row and column is inverted; each new column or row contains either $k$ or $k + 2$ odd numbers, so the parity of the row and column sums is the opposite to that of $k + 1$. Hence, the extended table meets the requirements.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20312,
"subject": "Mathematics (Olympiad)",
"question": "Three wheels are pushed together so they don't slip if we turn them. The circumferences of the wheels are 14, 10, and 6 cm, respectively. On each wheel, an arrow is drawn, pointing downwards. Someone turns the big wheel and the other wheels turn with it. This stops at the first moment all arrows point downwards again. Every time one of the arrows is pointing up, a whistle sounds. If two or three arrows point up at the same time, only one whistle sounds. How many whistles sound in total?\n\n",
"options": [],
"answer": "See solution",
"solution": "5",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20313,
"subject": "Mathematics (Olympiad)",
"question": "Find the smallest natural number $n$ for which there exist integers $a_1, \\dots, a_n$ (not necessarily distinct) such that\n$$\na_1^4 + \\dots + a_n^4 = 2013.\n$$",
"options": [],
"answer": "See solution",
"solution": "Note that the fourth powers of even numbers are divisible by $16$, and the fourth powers of odd numbers are congruent to $1$ modulo $16$. Since $2013 \\equiv 13 \\pmod{16}$, the desired representation must contain at least $13$ odd summands.\n\nSuppose that no more summands are needed. Since $7^4 = 2401 > 2013$, each summand must be $1^4 = 1$, $3^4 = 81$, or $5^4 = 625$. There can be at most $3$ summands equal to $625$ since $4 \\cdot 625 > 2013$. Therefore, the number of summands not divisible by $5$ is at least $10$. The fourth power of an integer not divisible by $5$ is congruent to $1$ modulo $5$, whereas $2013 \\equiv 3 \\pmod{5}$. Hence, the number of summands not divisible by $5$ must be at least $13$. This shows that the representation contains only summands $1$ and $81$, but $13$ such numbers sum up to at most $13 \\cdot 81 = 1053$, which is less than $2013$. Thus, representations with $13$ summands are impossible.\n\nOn the other hand, $14$ fourth powers suffice, as\n$$\n6^4 + 5^4 + 3^4 + 11 \\cdot 1^4 = 2013.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20314,
"subject": "Mathematics (Olympiad)",
"question": "Consider a $5 \\times 5$ array filled with the four distinct numbers $a, b, c, d$. Denote by $a_1, b_1, c_1, d_1$ the number of appearances of $a, b, c$, and $d$ in the first row and first column. The total sum of the numbers in the entire array is:\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\nThe possible values for $(a_1, b_1, c_1, d_1)$ are all permutations of $(5,2,2,0)$, $(5,2,1,1)$, $(4,3,2,0)$, and $(4,3,1,1)$. How many different possible total sums $S$ can be obtained by assigning values to $a, b, c, d$ and arranging them in the array as described?\n\n**Alternative version:**\n\nConsider a regular table containing the four distinct numbers $a, b, c, d$. The four $2 \\times 2$ corners each contain all four numbers. If $a_1, b_1, c_1, d_1$ are the numbers of appearances of $a, b, c, d$ 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\nHow many different possible total sums $S$ can be obtained?",
"options": [],
"answer": "See solution",
"solution": "There are $12$ permutations of $(5,2,2,0)$, $12$ of $(5,2,1,1)$, $24$ of $(4,3,2,0)$, and $12$ of $(4,3,1,1)$, totaling $60$ possible combinations for $(a_1, b_1, c_1, d_1)$. By choosing $a = 10^3$, $b = 10^2$, $c = 10$, $d = 1$, all these sums $S$ can be made distinct. Thus, the maximum possible number of different sums is $60$.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20315,
"subject": "Mathematics (Olympiad)",
"question": "Five marks are denoted by $a$, $b$, $c$, $d$, and $e$. The average of all five marks is $80$, and the average of the first four marks is $75$. Find the value of $e$.",
"options": [],
"answer": "See solution",
"solution": "$$\n\\frac{a+b+c+d+e}{5} = 80 \\implies a+b+c+d+e = 400\n$$\n$$\n\\frac{a+b+c+d}{4} = 75 \\implies a+b+c+d = 300\n$$\nTherefore,\n$$\ne = 400 - 300 = 100\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20316,
"subject": "Mathematics (Olympiad)",
"question": "Со $Z^*$ и $N_0$ се означени множеството од сите ненулти цели броеви и множеството од сите ненегативни цели броеви, соодветно. Најди ги сите пресликувања $f: Z^* \\rightarrow N_0$ за кои се исполнети следниве два услови:\n\n1. За секои $a, b \\in Z^*$ за кои $a + b \\in Z^*$ важи $f(a + b) \\geq \\min\\{f(a), f(b)\\}$.\n2. За секои $a, b \\in Z^*$ важи $f(ab) = f(a) + f(b)$.",
"options": [],
"answer": "See solution",
"solution": "Едно тривијално решение е константното пресликување $f = 0$.\n\nНека $f$ е едно нетривијално пресликување за кое важат (1) и (2). Ќе покажеме дека постои природен број $c$ и прост број $p$ за кои е исполнето $f(a) = c v_p(a)$ за секој $a \\in Z^*$, каде $v_p(a)$ е експонентот на $p$ во канонската факторизација на $a$.\n\nДа забележиме најпрво дека $f(1) = f(-1) = 0$ (доказ:\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\nОд ова и од (2) следува дека постои прост број $p$ за кој $f(p) \\neq 0$; за $c := f(p)$ ќе покажеме дека важи $f(a) = c v_p(a)$ за секој $a \\in Z^*$.\n\nИмено, за секој прост број $q \\neq p$ постојат ненулти цели броеви $a, \\beta$ за кои $1 = a p + \\beta q$, па исполнето е неравенството $0 = f(a p + \\beta q) \\geq \\min\\{f(a p), f(\\beta q)\\}$.\n\nОд\n\n$$\nf(a p) = f(a) + f(p) \\geq f(p) = c \\neq 0\n$$\n\nследува $f(\\beta q) = 0$ и $f(q) = 0$.\n\nНека $a = \\pm p^k q^\\beta r^\\gamma \\ldots$ е канонската факторизација на $a$; тогаш\n\n$$\nf(a) = f(\\pm p^k) + f(q^\\beta) + f(r^\\gamma) + \\ldots = f(\\pm 1) + f(p^k) = k f(p) = c v_p(a)\n$$\n\nОстанува да забележиме дека секое вакво пресликување ги исполнува условите (1) и (2), па претставува нетривијално решение на поставениот проблем.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20317,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a parallelogram and let $E$, $F$, $G$, and $H$ be the midpoints of the sides $AB$, $BC$, $CD$, and $DA$, respectively. If $BH \\cap AC = I$, $BD \\cap EC = J$, $AC \\cap DF = K$, and $AG \\cap BD = L$, then prove that the quadrilateral $IJKL$ is a parallelogram.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $AC \\cap BD = O$. Clearly, $AO$ and $BH$ are medians in the triangle $ABD$, hence $I$ is the centroid of $ABD$. Similarly, $K$ is the centroid of $BCD$. If $\\overline{IO} = x$, then $\\overline{AI} = 2x$. Similarly, if $\\overline{KO} = y$, then $\\overline{CK} = 2y$. Therefore, $3x = \\overline{AO} = \\overline{CO} = 3y$, i.e., $x = y$. We analogously prove that $\\overline{JO} = \\overline{LO}$. It follows that $IJKL$ is a parallelogram.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20318,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b$ be real numbers such that the equation $x^3 - a x^2 + b x - a = 0$ has only real roots. Find the minimum of\n$$\n\\frac{2a^3 - 3ab + 3a}{b+1}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $x_1, x_2,$ and $x_3$ be the real roots of the equation $x^3 - a x^2 + b x - a = 0$. By Vieta's formulas, we have:\n\n- $x_1 + x_2 + x_3 = a$\n- $x_1 x_2 + x_2 x_3 + x_1 x_3 = b$\n- $x_1 x_2 x_3 = a$\n\nBy the inequality $(x_1 + x_2 + x_3)^2 \\ge 3(x_1 x_2 + x_2 x_3 + x_1 x_3)$, we get $a^2 \\ge 3b$. Also, since $a = x_1 + x_2 + x_3 \\ge 3 \\sqrt[3]{x_1 x_2 x_3} = 3 \\sqrt[3]{a}$, we have $a \\ge 3 \\sqrt{3}$.\n\nThus,\n$$\n\\begin{aligned}\n\\frac{2a^3 - 3ab + 3a}{b+1} &= \\frac{a(a^2 - 3b) + a^3 + 3a}{b+1} \\\\\n&\\ge \\frac{a^3 + 3a}{b+1} \\ge \\frac{a^3 + 3a}{\\frac{a^2}{3} + 1} \\\\\n&= 3a \\ge 9\\sqrt{3}.\n\\end{aligned}\n$$\nIf $a = 3\\sqrt{3}$ and $b = 9$, then equality holds when each root is equal to $\\sqrt{3}$.\n\nTherefore, the minimum value is $9\\sqrt{3}$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20319,
"subject": "Mathematics (Olympiad)",
"question": "The set of positive integers is divided into subsets in the following way:\n\n$$\n\\{1, 2\\}, \\{3, 4, 5\\}, \\{6, 7, 8, 9\\}, \\{10, 11, 12, 13, 14\\}, \\dots\n$$\n\na) Find the smallest element of the 100th subset.\n\nb) Is 2015 the largest element of such a subset?",
"options": [],
"answer": "See solution",
"solution": "a) The first 99 subsets contain $2 + 3 + \\dots + 100 = 5049$ elements. In the first 99 subsets are written the numbers $1, 2, 3, \\dots, 5049$, so the smallest element of the 100th subset is $5050$.\n\nb) If $2015$ is the largest element of the $n$-th subset, then $2 + 3 + \\dots + (n + 1) = 2015$. Adding $1$ to both sides yields $(n + 1)(n + 2) = 2 \\cdot 2016 = 4032$. Since $63 \\cdot 64 = 4032$, it follows that $2015$ is the largest element of the $62$nd subset.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20320,
"subject": "Mathematics (Olympiad)",
"question": "For what minimum integer $N$ can one change the operation symbols \"/\" to \"+\" and \"-\" in the following expression:\n$$1 \\ast 2 \\ast 3 \\ast \\dots \\ast N$$\nin order to obtain:\n\na) $2010$;\nb) $2011$?",
"options": [],
"answer": "See solution",
"solution": "**a)** $N = 63$; **b)** $N = 65$.\n\n**Explanation:**\n\na) If all operations are $+$, then $1 + 2 + 3 + \\dots + 62 = 1953 < 2010$, and $1 + 2 + 3 + \\dots + 63 = 2016 > 2010$. Thus, $N \\geq 63$. For $N = 63$, we can adjust signs: $1 + 2 - 3 + 4 + \\dots + 62 + 63 = 2016 - 6 = 2010$.\n\nb) For $N = 63$ or $64$, the sum is even, so $2011$ cannot be obtained. For $N = 65$, $1 + 2 + 3 + \\dots + 65 = 2145$. By changing the signs of $2$ and $65$ to minus, $2145 - 2 - 65 - 65 = 2145 - 132 = 2013$, but the solution suggests $-2 \\cdot (2 + 65) + 1 + 2 + 3 + \\dots + 65 = 2145 - 134 = 2011$. Thus, $N = 65$ works.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20321,
"subject": "Mathematics (Olympiad)",
"question": "Given the sum\n\n$$\nS = \\frac{1}{2[\\sqrt{1}] + 1} + \\frac{1}{2[\\sqrt{2}] + 1} + \\dots + \\frac{1}{2[\\sqrt{n}] + 1},\n$$\n\nwhere $[\\sqrt{x}]$ denotes the greatest integer less than or equal to $\\sqrt{x}$, for a positive integer $n$ less than one million, for how many values of $n$ is $S$ an integer?",
"options": [],
"answer": "See solution",
"solution": "For each positive integer $n$, define\n\n$$\nf(n) = \\frac{1}{2\\lfloor\\sqrt{1}\\rfloor + 1} + \\frac{1}{2\\lfloor\\sqrt{2}\\rfloor + 1} + \\dots + \\frac{1}{2\\lfloor\\sqrt{n}\\rfloor + 1}.\n$$\n\nWe claim that $f(k^2 - 1) = k - 1$ for each integer $k \\ge 2$. We prove this by induction.\n\n**Base case ($k = 2$):**\n\n$$\nf(3) = \\frac{1}{3} + \\frac{1}{3} + \\frac{1}{3} = 1.\n$$\n\n**Inductive step:** Assume $f(k^2 - 1) = k - 1$ for some $k \\ge 2$.\n\n$$\n\\begin{aligned}\nf((k+1)^2 - 1) &= f(k^2 - 1) + \\frac{1}{2k+1} + \\frac{1}{2k+1} + \\dots + \\frac{1}{2k+1} \\\\\n&= k - 1 + (2k+1) \\times \\frac{1}{2k+1} \\\\\n&= k.\n\\end{aligned}\n$$\n\nThus, $f(n)$ is an integer if and only if $n = k^2 - 1$ for some integer $k \\ge 2$.\n\nSince $n < 10^6$, the possible values are $n = 2^2 - 1, 3^2 - 1, \\dots, 1000^2 - 1$, giving exactly 999 values of $n$ for which $S$ is an integer.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20322,
"subject": "Mathematics (Olympiad)",
"question": "Let $S(x)$ denote the sum of the digits of $x$ in base $8$. Find all positive integers $x < 100$ such that\n$$\n\\frac{S(x) + S(1)}{S(x+1)} = \\frac{78}{100}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let's observe that $k = 0$ would imply that $S(x) + S(1) = S(x + 1)$ and therefore $k \\neq 0$. $k \\ge 3$ would imply $x \\ge 8^3 - 1 > 100$ and therefore $k = 1$ or $k = 2$.\n\nIf $k = 2$ then $a = 0$ (because $8^2 + 8^2 - 1 > 100$) and therefore $x = 8^2 - 1 = 63$ and\n$$\n\\frac{S(x)+S(1)}{S(x+1)} = \\frac{78}{100}.\n$$\n\n$k = 1$ is also impossible because:\n$$\n\\frac{10S(a) + 8}{10S(a) + 10} < \\frac{8}{10} \\implies 10S(a) + 8 < 8S(a) + 8 \\implies S(a) < 0.\n$$\n\nThus, the only solution is $x = 63$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20323,
"subject": "Mathematics (Olympiad)",
"question": "Let $2n+1$ points be placed on a circle. From each pair of points, draw a directed vector from the point with the smaller label to the point with the larger label. For each triangle formed by three of these points, define it as *regular* if the sum of the vectors along its sides is zero. Let $R$ be the number of regular triangles that can be formed. What are the smallest and largest possible values of $R$?\n\n\n\n*Fig. 9*",
"options": [],
"answer": "See solution",
"solution": "First, we construct an example where $R = 0$, showing that the smallest possible value of $R$ is $0$. In this setup, for each triangle, two vectors originate from the vertex with the smallest label, so the sum of the vectors along the triangle's sides cannot be zero.\n\nTo find the largest possible value of $R$, we use the following lemma:\n\n**Lemma.** Let $m, l$ be positive integers such that $m + l = 2n$. Then $\\binom{m}{2} + \\binom{l}{2}$ is minimized when $m = l = n$.\n\n*Proof.* Since $l = 2n - m$,\n\n$$\n2 \\binom{m}{2} + 2 \\binom{l}{2} = m(m-1) + l(l-1) = m(m-1) + (2n-m)(2n-m-1) = 2(m-n)^2 + 2n^2 - 2n,\n$$\nwhich is minimized when $m = n$.\n\nNow, let $l_i$ be the number of vectors originating at vertex $i$, and $m_i$ the number terminating at $i$. For each vertex $i$, there are $\\binom{l_i}{2} + \\binom{m_i}{2}$ pairs of vectors both originating or both terminating at $i$. The total number of non-regular triangles is\n\n$$\nN = \\frac{1}{2} \\sum_{i=1}^{2n+1} \\left( \\binom{l_i}{2} + \\binom{m_i}{2} \\right),\n$$\n\nsince each non-regular triangle is counted twice. Since $l_i + m_i = 2n$ for each $i$, by the lemma,\n\n$$\nN \\ge \\frac{1}{2}(2n+1) \\left( \\binom{n}{2} + \\binom{n}{2} \\right) = \\frac{1}{2}(2n+1)n(n-1).\n$$\n\nThe total number of triangles is\n\n$$\nM = \\binom{2n+1}{3} = \\frac{1}{3}(2n+1)n(2n-1).\n$$\n\nThus, the maximum number of regular triangles is\n\n$$\nR = M - N \\le \\frac{1}{3}(2n+1)n(2n-1) - \\frac{1}{2}(2n+1)n(n-1) = \\frac{1}{6}(2n+1)n(n+1).\n$$\n\nTo achieve this maximum, construct the following: from each vertex $i$, draw $n$ vectors to the next $n$ vertices clockwise. Then $l_i = m_i = n$ for all $i$, so\n\n$$\nN = \\frac{1}{2}(2n+1)n(n-1),\n$$\n\nand $R = \\frac{1}{6}(2n+1)n(n+1)$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20324,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$, respectively. Rays $BA$ and $ED$ intersect at $F$ while lines $BD$ and $CF$ intersect at $M$. Prove that $MF = MC$ if and only if $MB \\cdot MD = MC^2$.",
"options": [],
"answer": "See solution",
"solution": "Extend segment $DM$ through $M$ to $G$ such that $FG \\parallel CD$.\n\n\n\nThen $MF = MC$ if and only if quadrilateral $CDFG$ is a parallelogram, or, $FD \\parallel CG$. Hence $MC = MF$ if and only if $\\angle GCD = \\angle FDA$, that is, $\\angle FDA + \\angle CGF = 180^\\circ$.\n\nBecause quadrilateral *ABED* is cyclic, $\\angle FDA = \\angle ABE$. It follows that $MC = MF$ if and only if\n\n$$\n180^{\\circ} = \\angle FDA + \\angle CGF = \\angle ABE + \\angle CGF,\n$$\n\nthat is, quadrilateral *CBFG* is cyclic, which is equivalent to\n\n$$\n\\angle CBM = \\angle CBG = \\angle CFG = \\angle DCF = \\angle DCM.\n$$\n\nBecause $\\angle DMC = \\angle CMB$, $\\angle CBM = \\angle DCM$ if and only if triangles *BCM* and *CDM* are similar, that is\n\n$$\n\\frac{CM}{BM} = \\frac{DM}{CM},\n$$\n\nor $MB \\cdot MD = MC^2$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20325,
"subject": "Mathematics (Olympiad)",
"question": "Find the smallest positive integer $n$ such that one can colour each square of an $n \\times n$ grid red, yellow, or blue, satisfying the following three properties:\n\n1. The number of squares coloured in each colour is the same.\n2. If a row contains red square(s), then it also contains blue square(s), but no yellow squares.\n3. If a column contains blue square(s), then it also contains red square(s), but no yellow squares.",
"options": [],
"answer": "See solution",
"solution": "Assume a colouring algorithm exists on an $n \\times n$ grid with all the desired properties. Evidently, $3 \\mid n^2$, so $3 \\mid n$. Let $n = 3k$. There are $3k^2$ squares of each colour on the grid.\n\nNotice that if two rows or two columns are interchanged, it still maintains all required properties. Due to this, we may assume that rows $1, 2, \\dots, u$ and columns $1, 2, \\dots, v$ contain yellow squares.\n\nUse a vertical line and a horizontal line to divide the grid into four rectangular regions $A, B, C$, and $D$, as shown in the figure below: $A, B$ each has $u$ rows; $A, C$ each has $v$ columns. (If $u = 3k$, then $C, D$ do not exist; if $v = 3k$, then $B, D$ do not exist.)\n\nSince rows $1, 2, \\dots, u$ contain yellow squares, by property (ii), $A$ and $B$ do not contain red squares (and $C, D$ must exist). Since columns $1, 2, \\dots, v$ contain yellow squares, by property (iii), $A$ and $C$ do not contain blue squares (and $B, D$ exist). It follows that all squares in $A$ are yellow. On the other hand, all yellow squares are in $A$, so $uv = 3k^2$. Furthermore, $B$ only contains blue squares and $C$ only contains red squares.\n\n\n\nFor $D$, according to properties (ii) and (iii), every row and every column contain red and blue squares. So, the number of blue squares in $B$ is smaller than the total number of blue squares, that is, $u(3k - v) < 3k^2 = uv$; similarly, $(3k - u)v < uv$. After simplification, they become\n\n$$\nu > \\frac{3}{2}k, \\quad v > \\frac{3}{2}k. \\qquad (1)$$\n\nAs $3 \\mid uv$, by symmetry, we may assume $3 \\mid u$ (otherwise, transpose the rows to columns and swap the red and blue squares). Denote $\\frac{u}{3} = rs$ and $v = rt$, in which $r = (\\frac{u}{3}, v)$. Now, $st = \\frac{uv}{3r^2} = \\left(\\frac{k}{r}\\right)^2$ is a perfect square, $(s, t) = 1$, hence $s, t$ are both perfect squares. Let $s = p^2$, $t = q^2$, and we have\n\n$$\nu = 3p^2r, \\quad v = q^2r, \\quad k = pqr.$$ \n\nPlug them into (1) to get $3p^2r > \\frac{3}{2}pqr$, $q^2r > \\frac{3}{2}pqr$, which reduce to $\\frac{3}{2}p < 2p - 1$, or $p \\ge 3$. Since $q > \\frac{3}{2}p$, $q \\ge 5$. It follows that $n = 3k = 3pqr \\ge 45$.\n\nConsider a $45 \\times 45$ grid. From $uv = 675$, $u, v > \\frac{45}{2}$ (by (1)) and $3 \\mid u$, we find $u = 27$, $v = 25$. Colour the upper left $27 \\times 25$ squares in yellow, the upper right $27 \\times 20$ squares in blue, the lower left $18 \\times 25$ squares in red, and for the lower right region, colour 225 squares in red, 135 squares in blue, such that every row contains blue squares and every column contains red squares. Fig. 3.2 shows an algorithm that meets all the requirements.\n\nIn conclusion, the minimum of $n$ is $45$. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20326,
"subject": "Mathematics (Olympiad)",
"question": "*(a)* Given that angles $\\angle AEC$ and $\\angle ABD$ are straight, prove that triangles $\\triangle ABC$ and $\\triangle AED$ are similar.\n\n*(b)* Given that triangles $\\triangle ABC$ and $\\triangle AED$ are similar, and that $|AB| = 5$, $|AE| = 4$, $|AC| = 4 + |EC|$, $|AD| = 5 + 3$, $|BF| = 2$, $|DB| = 3$, $|CE| = 6$, and $|FD| = \\frac{1}{2}|CF|$, find $|CF|$.",
"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": 20327,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be a triangle with $\\angle ABC > \\angle BCA$ and $\\angle BCA \\geq 30^\\circ$. The angle bisectors of $\\angle ABC$ and $\\angle BCA$ meet the opposite sides of the triangle at the points $D$ and $E$, respectively. The line $BD$ intersects the line $CE$ at the point $P$. Assume that $PD = PE$ and that the incircle of the triangle $ABC$ has radius $1$. Determine the largest possible length of $BC$.",
"options": [],
"answer": "See solution",
"solution": "Let $\\hat{A} = \\angle CAB$, $\\hat{B} = \\angle ABC$, $\\hat{C} = \\angle BCA$, and $\\alpha = \\angle BDA$, $\\beta = \\angle AEC$. Since $P$ is the intersection of two angle bisectors, it is the center of the incircle of the triangle $ABC$. Clearly,\n\n$$\n\\alpha = \\hat{C} + \\frac{\\hat{B}}{2}, \\quad \\beta = \\hat{B} + \\frac{\\hat{C}}{2}.\n$$\n\n\n\nFrom $\\hat{B} > \\hat{C}$, we have $\\alpha < \\beta$. Using the sine rule with the triangles $APD$ and $APE$, we get\n\n$$\n\\frac{AP}{\\sin \\alpha} = \\frac{PD}{\\sin(\\hat{A}/2)} = \\frac{PE}{\\sin(\\hat{A}/2)} = \\frac{AP}{\\sin \\beta}\n$$\n\nyielding $\\sin \\alpha = \\sin \\beta$. Since $0 < \\alpha, \\beta < 180^\\circ$ and $\\alpha < \\beta$, we have $\\beta = 180^\\circ - \\alpha$. Substituting into the previous equations, we get $\\hat{B} + \\hat{C} = 120^\\circ$, $\\hat{A} = 60^\\circ$. Applying the sine rule to the triangle $ABC$ gives\n\n$$\n\\frac{a}{\\sin 60^\\circ} = \\frac{b}{\\sin \\hat{B}} = \\frac{c}{\\sin \\hat{C}} = \\frac{b+c-a}{\\sin \\hat{B} + \\sin \\hat{C} - \\sin 60^\\circ}.\n$$\n\nLet $F$ be the tangent point of $CA$ with the incircle of the triangle $ABC$. From the triangle $APF$, we have\n\n$$\n\\tan(\\angle FAP) = \\frac{1}{(b+c-a)/2} = \\frac{1}{\\sqrt{3}},\n$$\n\nso that $b+c-a = 2\\sqrt{3}$. From $\\hat{B} + \\hat{C} = 120^\\circ$, we have\n\n$$\n\\sin \\hat{B} + \\sin \\hat{C} = \\sin(120^\\circ - \\hat{C}) + \\sin \\hat{C} = \\sqrt{3} \\cos(60^\\circ - \\hat{C})\n$$\n\nyielding\n\n$$\na = \\frac{6}{2\\sqrt{3}\\cos(60^\\circ - \\hat{C}) - \\sqrt{3}} \\leq \\frac{6}{2\\sqrt{3}(\\sqrt{3}/2) - \\sqrt{3}} = \\frac{6}{3 - \\sqrt{3}}\n$$\n\nwith equality holding if and only if $\\hat{C} = 30^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20328,
"subject": "Mathematics (Olympiad)",
"question": "When five positive integers $a, b, c, d, e$ satisfy\n$$\na < b < c < d < e < a^2 < b^2 < c^2 < d^2 < e^2 < a^3 < b^3 < c^3 < d^3 < e^3,\n$$\ndetermine the minimum possible value that the sum $a + b + c + d + e$ can take.",
"options": [],
"answer": "See solution",
"solution": "From the given inequalities, it follows that $a + 4 \\leq e$ and $e^2 + 1 \\leq a^3$ must hold. Therefore,\n$$\n(a + 4)^2 \\leq e^2 \\leq a^3 - 1,\n$$\nfrom which it follows that $(a + 4)^2 \\leq a^3 - 1$, i.e.,\n$$\n(a - 4)(a^2 + 3a + 4) \\geq 1 > 0.\n$$\nThis means that we must have $a > 4$. Consequently,\n$$\na + b + c + d + e > 5 + 6 + 7 + 8 + 9 = 35.\n$$\nOn the other hand, we see that the choice $(a, b, c, d, e) = (5, 6, 7, 8, 9)$ satisfies the requirements of the problem. Consequently, we conclude that $35$ is the desired answer.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20329,
"subject": "Mathematics (Olympiad)",
"question": "Find the smallest real number $c > 0$ such that\n$$\n\\{x\\} + \\{y\\} < c + xy\n$$\nfor all real numbers $x, y > 0$ and $xy < 1$.",
"options": [],
"answer": "See solution",
"solution": "First, choose $x = \\frac{n+1}{n+2}$, $y = \\frac{n}{n+1}$ with $n \\in \\mathbb{Z}^+$. Then $xy = \\frac{n}{n+2} < 1$. Substituting into the given inequality, we have\n$$\n\\frac{n+1}{n+2} + \\frac{n}{n+1} < c + \\frac{n}{n+2} \\implies c > \\frac{n}{n+1} + \\frac{1}{n+2}.\n$$\nAs $n \\to +\\infty$, $c \\ge 1$. Next, we prove the inequality with $c = 1$: $\\{x\\} + \\{y\\} < 1 + xy$ for all $x, y > 0$ and $xy < 1$. Indeed,\n$$\n(1 - \\{x\\})(1 - \\{y\\}) > 0 \\implies \\{x\\} + \\{y\\} < 1 + \\{x\\}\\{y\\} \\le 1 + xy.\n$$\nThis inequality is true. Therefore, the minimum value is $c = 1$.\n$\\boxed{1}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20330,
"subject": "Mathematics (Olympiad)",
"question": "On a $5 \\times 5$ square board, a number of tiles consisting of 4 squares, as shown in the figure, are placed along the grid. Note that the placed tiles may be rotated or flipped over. Furthermore, the placed tiles may be overlapped, but must not protrude beyond the board. Assume that every square of the board is covered with at most two tiles. Find the maximum possible number of squares of the board that are covered by at least one tile.\n\n",
"options": [],
"answer": "See solution",
"solution": "Consider the tile placement obtained by superposing the two given arrangements below. In this placement, each square of the board should be covered by at most two tiles. Furthermore, with the exception of the central square, each of the 24 remaining squares are covered by at least one tile.\n\n\n\n\n\nNext, we will prove that the number of squares covered by at least one tile is no more than 24. Let us annotate some squares with the letters A and B as shown in the figure below. Then, no matter how a single tile is placed, exactly one square marked with A and exactly one square marked with B will be covered by the tile. Note that there are 4 squares marked with A. Since each square is covered by at most two tiles, the total number of tiles that can be placed is at most 8. Therefore, it is impossible to cover all of the 9 squares marked with B. Hence, the number of squares covered by at least one tile is $25 - 1 = 24$ or less.\n\n\n\nWith the above argument, we have proved that the maximum value in question is 24.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20331,
"subject": "Mathematics (Olympiad)",
"question": "For a positive integer $n$, two players A and B play the following game: Given a pile of $s$ stones, the players take turns alternately with A going first. On each turn, a player is allowed to take either one stone, a prime number of stones, or a multiple of $n$ stones. The winner is the one who takes the last stone. Assuming both A and B play perfectly, for how many values of $s$ can player A not win?",
"options": [],
"answer": "See solution",
"solution": "Let $k$ be the number of values of $s$ for which player A cannot win, and let $\\{s_1, s_2, \\dots, s_k\\}$ be these values (called *losing numbers*; all other nonnegative integers are *winning numbers*).\n\n1. **Every multiple of $n$ is a winning number.**\n\nSuppose there are two different losing numbers $s_i > s_j$ that are congruent modulo $n$. Then, on A's first turn, A can remove $s_i - s_j$ stones (since $n \\mid s_i - s_j$), leaving $s_j$ stones for B. This contradicts both $s_i$ and $s_j$ being losing numbers.\n\n2. **Thus, there are at most $n-1$ losing numbers, i.e., $k \\leq n-1$.**\n\nSuppose there exists an integer $r \\in \\{1, 2, \\dots, n-1\\}$ such that $mn + r$ is a winning number for every $m \\in \\mathbb{N}_0$. Let $u$ be the greatest losing number (if $k > 0$) or $0$ (if $k = 0$), and let $s = \\mathrm{LCM}(2, 3, \\dots, u + n + 1)$. Note that all numbers $s + 2, s + 3, \\dots, s + u + n + 1$ are composite. Let $m' \\in \\mathbb{N}_0$ be such that $s + u + 2 \\leq m'n + r \\leq s + u + n + 1$. For $m'n + r$ to be a winning number, there must exist $p$ (either $1$, a prime, or a positive multiple of $n$) such that $m'n + r - p$ is a losing number or $0$, and thus $\\leq u$. Since $s + 2 \\leq m'n + r - u \\leq p \\leq m'n + r \\leq s + u + n + 1$, $p$ must be composite, so $p$ is a multiple of $n$ (say $p = qn$). But then $m'n + r - p = (m' - q)n + r$ must be a winning number, by assumption. This is a contradiction.\n\n3. **Therefore, each nonzero residue class modulo $n$ contains a losing number.**\n\n4. **There are exactly $n-1$ losing numbers (one for each residue $r \\in \\{1, 2, \\dots, n-1\\}$).**\n\n**Alternate proof:**\n\n*Lemma:* No pair $(u, n)$ of positive integers satisfies the following property:\n\n(*) In $\\mathbb{N}$, there exists an arithmetic progression $(a_i)_{i=1}^\\infty$ with difference $n$ such that each segment $[a_i - u, a_i + u]$ contains a prime.\n\n*Proof of the lemma:* Suppose such a pair $(u, n)$ and progression $(a_i)_{i=1}^\\infty$ exist. In $\\mathbb{N}$, there exist arbitrarily long patches of consecutive composites. Take such a patch $P$ of length $3un$. Then, at least one segment $[a_i - u, a_i + u]$ is fully contained in $P$, a contradiction.\n\nSuppose such a nonzero residue class modulo $n$ exists (so $n > 1$). Let $u \\in \\mathbb{N}$ be greater than every losing number. Consider the members of the supposed residue class greater than $u$; they form an arithmetic progression with property (*), a contradiction by the lemma.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20332,
"subject": "Mathematics (Olympiad)",
"question": "對於任何正整數 $k$，令 $S(k)$ 表示其在十進制下的各個位數總和（例：$S(209) = 2+0+9=11$）。試求所有整係數多項式 $P(x)$，使得對於所有正整數 $n \\ge 2017$，都有 $P(n) > 0$ 且\n\n$$\nS(P(n)) = P(S(n)).\n$$",
"options": [],
"answer": "See solution",
"solution": "$P(x) = x$ 或 $P(x) = c$，其中 $c \\in \\{1, \\dots, 9\\}$。\n\n假設 $P(x) = a_{d}x^{d} + \\cdots + a_{1}x^{1} + a_{0}$。考慮 $n = 9 \\times 10^{k}$，其中 $k$ 是一個充分大的正整數；注意到 $S(n) = 9 \\Rightarrow P(S(n)) = P(9)$。\n\n1. 我們首先證明所有係數都非負。\n\n假設存在 $0 \\le i < d$ 使得 $a_i < 0$，則易知 $P(n)$ 對應 $10^{ik+m+1}$ 到 $10^{(i+1)k-1}$ 的位數都是 9，故 $S(P(n)) \\ge 9(k-m-1)$。取 $k$ 充分大，則 $9(k-m-1) > P(9) = P(S(n))$，從而 $S(P(n)) \\ne P(S(n))$，矛盾。\n\n2. 接著我們證明 $\\deg(P(x)) \\le 1$。\n\n因所有係數非負，當 $k$ 充分大時，$P(n)$ 將會是由 $a_d \\times 9^d, a_{d-1} \\times 9^{d-1}, \\dots, a_0$ 中間插入一些 0 所組成，也就是說\n\n$$\nS(P(n)) = S(a_d \\times 9^d) + \\cdots + S(a_0).\n$$\n\n又 $S(P(n)) = P(S(n)) = P(9)$，我們有\n\n$$\nS(a_d \\times 9^d) + \\cdots + S(a_0) = P(9) = a_d \\times 9^d + \\cdots + a_0. \\quad (1)\n$$\n\n然而對於所有正整數 $m$，顯然有 $S(m) \\le m$，且等號僅在 $m \\in \\{1, \\dots, 9\\}$ 時成立，故 $\\forall i \\ge 2, a_i = 0$。\n\n3. (1) 同時也告訴我們 $a_1 \\le 1$ 且 $a_0 \\le 9$。\n\n- 若 $a_1 = 1$ 且 $a_0 \\ge 1$，取 $n = 10^k + (10 - a_0)$，得\n\n$$\nS(P(n)) = S(10^k + 10) = 2 \\neq 11 = P(11 - a_0) = P(S(n)).\n$$\n\n故 $P(x) = x$，代入檢驗和。\n\n- 若 $a_1 = 0$，則易知 $a_0 \\in \\{1, \\dots, 9\\}$ 都滿足題意。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20333,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$, $p$, $q$, $r$ be positive integers such that\n$$\na^p + b^q + c^r = a^q + b^r + c^p = a^r + b^p + c^q.\n$$\nProve that $a = b = c$ or $p = q = r$.",
"options": [],
"answer": "See solution",
"solution": "We proceed by considering cases.\n\n**Case 1:** Two of $p, q, r$ are equal, say $q = r$.\n\nSubtract $a^q + b^q + c^q$ from the given equation:\n$$\na^p - a^q = b^p - b^q = c^p - c^q.\n$$\nIf $p > q$, then $x^p - x^q$ is strictly increasing for $x > 0$, so $a = b = c$. If $p < q$, the same argument applies. The only remaining subcase is $p = q = r$.\n\n**Case 2:** Two of $a, b, c$ are equal, say $b = c$.\n\nSubtract $b^p + b^q + b^r$ from the given equation:\n$$\na^p - b^p = a^q - b^q = a^r - b^r.\n$$\nFor $a \\ne b$, the function $a^s - b^s$ is strictly increasing or decreasing depending on $a > b$ or $a < b$, so $p = q = r$. If $a = b$, then $a = b = c$.\n\n**Case 3:** $a, b, c$ are distinct, as are $p, q, r$.\n\nAssume $a$ is the greatest of $a, b, c$ and $p$ is the greatest of $p, q, r$ (by relabeling if necessary), so $a, p \\ge 3$.\n\nWe claim that\n$$\na^p \\ge (a-1)^p + 2a^{p-1}.\n$$\nFor $p = 3$, this becomes $a^3 \\ge (a-1)^3 + 2a^2$, which simplifies to $a^2 - 3a + 1 \\ge 0$, true for $a \\ge 3$.\n\nThus,\n$$\na^p + b^q + c^r > (a-1)^p + 2a^{p-1} \\ge b^p + c^q + a^r,\n$$\ncontradicting the original equality. Therefore, $a = b = c$ or $p = q = r$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20334,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle. An arbitrary circle which passes through the points $B$ and $C$ intersects the sides $AC$ and $AB$ for the second time in $D$ and $E$, respectively. The line $BD$ intersects the circumcircle of triangle $AEC$ at $P$ and $Q$, and the line $CE$ intersects the circumcircle of the triangle $ABD$ at $R$ and $S$, such that $P$ is situated on the segment $BD$, and $R$ lies on the segment $CE$. Prove that:\n\na) the points $P, Q, R$ and $S$ are concyclic;\n\nb) the triangle $APQ$ is isosceles.",
"options": [],
"answer": "See solution",
"solution": "a) Denote by $X$ the intersection of the lines $CE$ and $BD$. Writing the power of $X$ with respect to all three circles, we obtain:\n\n$$\nXR \\cdot XS = XB \\cdot XD = XE \\cdot XC = XP \\cdot XQ.\n$$\n\nSince $\\{X\\} = PQ \\cap RS$, we infer that the points $P, Q, R$ and $S$ are concyclic.\n\nb) We prove that the point $A$ is the center of the circle which passes through $P, Q, R$ and $S$.\n\nThe triangles $ADP$ and $APC$ are similar (A.A.), thus $AP^2 = AD \\cdot AC$. Similarly, we find that $AQ^2 = AE \\cdot AB$.\n\n\n\nFrom the power of the point $A$ with respect to the circumcircle of the quadrilateral $BCDE$, we infer that $AD \\cdot AC = AE \\cdot AB = \\rho(A)$, therefore $AP = AQ = \\sqrt{\\rho(A)}$.\n\nSimilarly, we prove that $AR = AS = \\sqrt{\\rho(A)}$, therefore the triangle $APQ$ is isosceles.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20335,
"subject": "Mathematics (Olympiad)",
"question": "Given any set $A = \\{a_1, a_2, a_3, a_4\\}$ of four distinct positive integers, let $s_A = a_1 + a_2 + a_3 + a_4$. Let $n_A$ denote the number of pairs $(i, j)$ with $1 \\leq i < j \\leq 4$ for which $a_i + a_j$ divides $s_A$. Find all sets $A$ of four distinct positive integers which achieve the largest possible value of $n_A$.\n\n",
"options": [],
"answer": "See solution",
"solution": "For any positive integer $k$, the sets $\\{k, 5k, 7k, 11k\\}$ and $\\{k, 11k, 19k, 29k\\}$ achieve the maximum value of $n_A = 4$.\n\nLet $A = \\{a_1, a_2, a_3, a_4\\}$ with $a_1 < a_2 < a_3 < a_4$. The pairwise sums satisfy:\n\n$$\na_1 + a_2 < a_1 + a_3 < a_1 + a_4 < a_2 + a_3 < a_2 + a_4 < a_3 + a_4.\n$$\n\nIf $a_i + a_j \\mid s_A$, then $a_i + a_j \\mid s_A - (a_i + a_j)$, so $a_i + a_j \\leq s_A - (a_i + a_j)$. Therefore, $a_2 + a_4$ and $a_3 + a_4$ do not divide $s_A$, so $n_A = 4$ is maximal.\n\nTo classify all such $A$, there are integers $2 \\leq m < n$ such that:\n\n$$\n\\begin{cases}\na_1 + a_4 = a_2 + a_3, \\\\\nm(a_1 + a_3) = a_2 + a_4, \\\\\nn(a_1 + a_2) = a_3 + a_4.\n\\end{cases}\n$$\n\nIf $m \\geq 3$, the second equation gives $a_2 + a_4 \\geq 3(a_1 + a_3)$, leading to a contradiction. Thus, $m = 2$. Adding the third equation, twice the second, and three times the first gives $(n+7)a_1 = (5-n)a_2$. Since $n > m = 2$, $n = 3$ or $n = 4$. If $n = 3$, $a_2 = 5a_1$ and $A = \\{k, 5k, 7k, 11k\\}$. If $n = 4$, $a_2 = 11a_1$ and $A = \\{k, 11k, 19k, 29k\\}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20336,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_0, a_1, \\dots, a_N$ be real numbers where $a_0 = a_N = 0$. Prove the inequality\n\n$$\na_1^2 + a_2^2 + \\dots + a_{N-1}^2 \\le C \\left( (a_1 - a_0)^2 + (a_2 - a_1)^2 + \\dots + (a_N - a_{N-1})^2 \\right),\n$$\n\nwhere $C = \\frac{N^2}{4}$.",
"options": [],
"answer": "See solution",
"solution": "Let $b_i = a_i - a_{i-1}$, and note that $a_0 = a_N = 0$ implies $a_i = \\sum_{k=1}^i b_k = -\\sum_{k=i+1}^N b_k$.\n\n**Case 1:** $C = \\frac{N^2}{4}$.\n\nSplit the left hand side of\n\n$$\na_1^2 + a_2^2 + \\dots + a_{N-1}^2 \\le \\frac{N^2}{4} \\left( (a_1 - a_0)^2 + (a_2 - a_1)^2 + \\dots + (a_N - a_{N-1})^2 \\right)\n$$\n\ninto two parts $L_1 = \\sum_{i=1}^{M-1} a_i^2$, and $L_2 = \\sum_{i=M}^{N-1} a_i^2$, where $M = \\lfloor \\frac{N}{2} \\rfloor$ and let $R = \\sum_{i=1}^{N} b_i^2$.\n\nNow, for $i < M$ we have by QM-AM\n\n$$\na_i^2 = \\left( \\sum_{k=1}^{i} b_k \\right)^2 \\le i \\sum_{k=1}^{i} b_k^2 \\le iR.\n$$\n\nTherefore\n\n$$\nL_1 = \\sum_{i=1}^{M-1} a_i^2 \\le \\sum_{i=1}^{M-1} iR = \\frac{M(M-1)}{2} R.\n$$\n\nFor $i \\ge M$ we use the same technique, but from the other side:\n\n$$\na_i^2 = \\left( \\sum_{k=i+1}^{N} b_k \\right)^2 \\le (N-i) \\sum_{k=i+1}^{N} b_k^2 \\le (N-i)R.\n$$\n\nSo\n\n$$\nL_2 = \\sum_{i=M}^{N-1} a_i^2 \\le \\sum_{i=M}^{N-1} (N-i)R = \\frac{(N-M)(N-M+1)}{2} R.\n$$\n\nThe sum of our two inequalities is\n\n$$\nL_1 + L_2 \\le \\frac{(N-M)(N-M+1) + M(M-1)}{2} R.\n$$\n\nRecalling that $M = \\lceil \\frac{N}{2} \\rceil$, and checking the two cases $N$ odd and $N$ even, we see that\n\n$$\n\\frac{(N - M)(N - M + 1) + M(M - 1)}{2} \\le \\frac{N^2}{4}.\n$$\n\n**Case 2:** $C = \\frac{N^2}{8} + \\frac{N}{4}$.\n\nSplit both sides of\n\n$$\na_1^2 + a_2^2 + \\dots + a_{N-1}^2 \\le \\frac{N^2}{4} \\left( (a_1 - a_0)^2 + (a_2 - a_1)^2 + \\dots + (a_N - a_{N-1})^2 \\right)\n$$\n\ninto two parts $L_1 = \\sum_{i=1}^{M-1} a_i^2$, $L_2 = \\sum_{i=M}^{N-1} a_i^2$, $R_1 = \\sum_{i=1}^{M-1} b_i^2$, $R_2 = \\sum_{i=M}^{N} b_i^2$, where $M = \\lceil \\frac{N}{2} \\rceil$.\n\nNow, for $i < M$ we have by QM-AM\n\n$$\na_i^2 = \\left( \\sum_{k=1}^{i} b_k \\right)^2 \\le i \\sum_{k=1}^{i} b_k^2 \\le i R_1.\n$$\n\nTherefore\n\n$$\nL_1 = \\sum_{i=1}^{M-1} a_i^2 \\le \\sum_{i=1}^{M-1} i R_1 = \\frac{M(M-1)}{2} R_1.\n$$\n\nFor $i \\ge M$ we use the same technique, but from the other side:\n\n$$\na_i^2 = \\left( \\sum_{k=i+1}^{N} b_k \\right)^2 \\le (N-i) \\sum_{k=i+1}^{N} b_k^2 \\le (N-i) R_2.\n$$\n\nSo\n\n$$\nL_2 = \\sum_{i=M}^{N-1} a_i^2 \\le \\sum_{i=M}^{N-1} (N-i) R_2 = \\frac{(N-M)(N-M+1)}{2} R_2.\n$$\n\nThe sum of our two inequalities is\n\n$$\nL_1 + L_2 \\le \\frac{M(M-1)}{2} R_1 + \\frac{(N-M)(N-M+1)}{2} R_2 \\\\\n\\le \\max \\left\\{ \\frac{M(M-1)}{2}, \\frac{(N-M)(N-M+1)}{2} \\right\\} (R_1 + R_2)\n$$\n\nRecalling that $M = \\lceil \\frac{N}{2} \\rceil$, and checking the two cases $N$ odd and $N$ even, we see that\n\n$$\n\\max \\left\\{ \\frac{M(M-1)}{2}, \\frac{(N-M)(N-M+1)}{2} \\right\\} \\le \\frac{N^2}{8} + \\frac{N}{4}.\n$$\n\n**Case 3:** $C = (4 \\sin^2(\\pi/2N))^{-1}$ (Sketch of proof).\n\nThe right hand side $\\sum_{i=1}^{N} (a_i - a_{i-1})^2$ expands to\n\n$$\n\\sum_{i=1}^{N-1} 2a_i^2 - a_i a_{i-1} - a_i a_{i+1} = -\\mathbf{a}^\\top B \\mathbf{a},\n$$\n\nwhere $\\mathbf{a} = [a_1, \\dots, a_{N-1}]^\\top$, and $B$ is the $(N-1) \\times (N-1)$ discrete Laplacian matrix. $B$ is symmetric negative definite, and its eigenvalues can be calculated. The smallest (in absolute value) eigenvalue is $-4 \\sin^2(\\pi/2N)$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20337,
"subject": "Mathematics (Olympiad)",
"question": "Find all prime numbers $p$ such that $2p^3 + 4p^2 - 3p + 12$ is the fifth power of an integer.",
"options": [],
"answer": "See solution",
"solution": "Denote $f(n) = 2n^3 + 4n^2 - 3n + 12$. The following table shows the remainders of $n^2$, $n^3$, $n^5$, and $f(n)$ upon division by 11:\n\n| $n$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n|------|---|---|---|---|---|---|---|---|---|---|----|\n| $n^2$ | 0 | 1 | 4 | 9 | 5 | 3 | 3 | 5 | 9 | 4 | 1 |\n| $n^3$ | 0 | 1 | 8 | 5 | 9 | 4 | 7 | 2 | 6 | 3 | 10 |\n| $n^5$ | 0 | 1 | 10| 1 | 1 | 1 | 10| 10| 10| 1 | 10 |\n| $f(n)$| 1 | 4 | 5 | 5 | 5 | 6 | 9 | 4 | 3 | 7 | 6 |\n\nAs one can see from the table, the only remainders upon division by 11 that the fifth power of an integer can give are 0, 1, and 10. On the other hand, integers of the form $f(n)$ give only remainders 1, 3, 4, 5, 6, 7, and 9 upon division by 11, whereby the remainder is 1 only if $n$ is divisible by 11. Consequently, $f(p)$ can be the fifth power of an integer only if $p$ is divisible by 11. As $p$ is prime, the only possibility is $p = 11$. And indeed, $f(11) = 2 \\cdot 11^3 + 4 \\cdot 11^2 - 3 \\cdot 11 + 12 = 3125 = 5^5$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20338,
"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$$\n\\frac{3s^3 + s}{9(1-s)} \\ge \\left( \\frac{s\\sqrt{1+s}}{\\sqrt[4]{3+9s^2}} \\right)^2\n$$\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$$\n\\frac{2}{3} \\ge \\left( \\frac{x\\sqrt{1+x}}{\\sqrt[4]{3+9x^2}} + \\frac{y\\sqrt{1+y}}{\\sqrt[4]{3+9y^2}} + \\frac{z\\sqrt{1+z}}{\\sqrt[4]{3+9z^2}} \\right)^2\n$$\n\nhold.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20339,
"subject": "Mathematics (Olympiad)",
"question": "Prove that there are infinitely many positive integers $m \\geq k$ such that $\\binom{m}{k}$ and $l$ are relatively prime.",
"options": [],
"answer": "See solution",
"solution": "**Solution**\n\nLet $m = k + t \\times l \\times (k!)$, where $t$ is any positive integer. To prove that $\\binom{m}{k}$ and $l$ are relatively prime, we only need to show that for any prime factor $p$ of $l$, $p \\nmid \\binom{m}{k}$.\n\nIf $p \\nmid k!$, we have\n\n$$\nk! \\binom{m}{k} = \\prod_{i=1}^{k} (m - k + i) = \\prod_{i=1}^{k} [i + t l (k!)] = \\prod_{i=1}^{k} i \\equiv k! \\pmod{p}.\n$$\n\nTherefore, $p \\nmid \\binom{m}{k}$.\n\nIf $p \\mid k!$, there exists an integer $\\alpha \\geq 1$ such that $p^\\alpha \\mid k!$ but $p^{\\alpha+1} \\nmid k!$. Then $p^{\\alpha+1} \\mid l (k!)$.\n\nWe have\n\n$$\nk! \\binom{m}{k} = \\prod_{i=1}^{k} (m - k + i) = \\prod_{i=1}^{k} [i + t l (k!)] = \\prod_{i=1}^{k} i \\equiv k! \\pmod{p^{\\alpha+1}}.\n$$\n\nTherefore, $p^\\alpha \\mid k! \\binom{m}{k}$ and $p^{\\alpha+1} \\nmid k! \\binom{m}{k}$. Since $p^\\alpha \\mid k!$, we get $p \\nmid \\binom{m}{k}$. The proof is complete.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20340,
"subject": "Mathematics (Olympiad)",
"question": "Let $x$ be a real number such that $t = x + x^{-1}$ is an integer greater than $2$. Prove that $t_n = x^n + x^{-n}$ is an integer for all positive integers $n$. Determine the values of $n$ for which $t$ divides $t_n$.",
"options": [],
"answer": "See solution",
"solution": "First, $t_1 = t$, which is an integer. Next, $t_2 = x^2 + x^{-2} = (x + x^{-1})^2 - 2 = t^2 - 2$, which is also an integer.\n\nMore generally,\n\n$$\n\\begin{aligned}\nt_k &= x^k + x^{-k} \\\\\n&= (x + x^{-1})(x^{k-1} + x^{-(k-1)}) - (x^{k-2} + x^{-(k-2)}) \\\\\n&= t \\cdot t_{k-1} - t_{k-2}\n\\end{aligned}\n$$\n\nTherefore, if $t_{k-1}$ and $t_{k-2}$ are integers, so is $t_k$. Since $t_1$ and $t_2$ are integers, it follows by induction that $t_n$ is an integer for all $n$.\n\nWe claim that $t$ divides $t_n$ if and only if $n$ is odd.\n\nSince $t > 2$, $t$ does not divide $t_2$. Suppose $k$ is odd, and $t$ divides $t_{k-2}$ but not $t_{k-1}$. Then, since $t_k = t \\cdot t_{k-1} - t_{k-2}$, it follows that $t$ divides $t_k$. Conversely, if $k$ is even and $t$ divides $t_{k-1}$ but not $t_{k-2}$, then $t$ does not divide $t_k$. Thus, by induction, $t$ divides $t_n$ if and only if $n$ is odd.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20341,
"subject": "Mathematics (Olympiad)",
"question": "Determine all pairs of real numbers $a$ and $x$ that satisfy the simultaneous equations\n\n$$\n5x^3 + a x^2 + 8 = 0\n$$\n\nand\n\n$$\n5x^3 + 8x^2 + a = 0.\n$$",
"options": [],
"answer": "See solution",
"solution": "If we subtract the two equations, we obtain\n\n$$\n(a x^2 + 8) - (8 x^2 + a) = (a - 8)(x^2 - 1) = (a - 8)(x + 1)(x - 1) = 0,\n$$\n\nthus either $a = 8$ or $x = -1$ or $x = 1$.\n\nIf $a = 8$, we are left with\n\n$$\n5x^3 + 8x^2 + 8 = (x + 2)(5x^2 - 2x + 4) = 0.\n$$\n\nThe second factor has no real roots, since its discriminant $2^2 - 4 \\cdot 5 \\cdot 4 = -76$ is negative.\nThus $x = -2$ in this case.\n\nIf $x = -1$, we get $a = -5x^3 - 8x^2 = -3$.\n\nIf $x = 1$, we get $a = -5x^3 - 8x^2 = -13$.\n\nIn summary, there are three possible pairs:\n\n- $(a, x) = (8, -2)$\n- $(a, x) = (-3, -1)$\n- $(a, x) = (-13, 1)$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20342,
"subject": "Mathematics (Olympiad)",
"question": "The altitude $AD$ of a triangle $ABC$ ($AB > AC$) intersects its circumcircle $\\omega$ at $P$. Let $H$ be the orthocenter of $ABC$, and $K$ be the point on the segment $BC$ such that $BD = KC$. The circumcircle of the triangle $PKH$ intersects $\\omega$ at $Q$, and the line $BC$ at $N$. Denote by $T$ the point on the line $AD$ such that the lines $TN$ and $PQ$ are perpendicular. Prove that the line $TK$ passes through the center of the circle $\\omega$.\n\n",
"options": [],
"answer": "See solution",
"solution": "We extend the line $PQ$ to the intersection with $BC$ at a point $S$, and the line $PN$ to the intersection with $\\omega$ at a point $L$ (see the figure above). Let $F$ be the point on $\\omega$ diametrically opposite to $P$. Let us prove that the points $S$, $L$, $T$, $F$ lie on the same line.\n\nFirst, it is obvious that $AF \\parallel BC$, which means that $BAFC$ is an isosceles trapezoid. Therefore, $\\angle FKN = 90^\\circ$. At the same time, $\\angle NLF = 90^\\circ$, so $LNKF$ is inscribed. Since $PQ$, $KN$, $LF$ are pairwise radical axes for circles $(HPK)$, $(ABC)$, $(LFK)$, then these three lines must intersect at one point, implying that the points $S$, $L$, $F$ are collinear.\n\nSince the pairs of lines $PN$, $SL$ and $SN$, $TD$ are perpendicular, the point $N$ is the orthocenter of the triangle formed by the lines $SL$, $SP$, $PT$, and since the lines $TN$ and $PS$ are perpendicular, the points $S$, $L$, $T$ are collinear. This means that $S$, $L$, $T$, $F$ lie on the same line.\n\nIt is known that $PD = DH$, and so the circle $(PKH)$ is symmetrical with respect to the line $BC$, whence $\\angle NPK = 90^\\circ$, from which follows that $LF \\parallel PK$. Since the lines $FK$ and $TP$ are also parallel, $PKFT$ is a parallelogram. Since $PF$ is the diameter of the circle $\\omega$, the midpoint of the diagonal $PF$ of the parallelogram $PKFT$ is its center and $TK$ passes through it, as required.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20343,
"subject": "Mathematics (Olympiad)",
"question": "設三角形 $ABC$ 的內心為 $I$，角 $A$ 內的旁心為 $J$。令 $\\overline{AA'}$ 為 $\\triangle ABC$ 外接圓的直徑，點 $H_1, H_2$ 分別為 $\\triangle BIA', \\triangle CJA'$ 的垂心。證明：$H_1H_2$ 平行於 $BC$。\n\n",
"options": [],
"answer": "See solution",
"solution": "(∠ 代表有向角，$\\pm$ 代表正向相似。)\n\n平移 $\\triangle CJA' \\cup H_2$ 至 $\\triangle C_1BA'_1 \\cup H'_2$（即使得 $J$ 平移至與 $B$ 重合）。由 $\\overline{IJ}$ 中點位於 $\\overline{BC}$ 的中垂線上，可得 $C_1I \\perp BC$。令 $I'$ 為 $I$ 關於 $\\odot(BIA')$ 的對徑點、$C'_1$ 為 $C_1$ 關於 $\\odot(C_1BA'_1)$ 的對徑點。則由 $\\overline{BI'} = \\overline{H_1A'}$，$\\overline{BC'_1} = \\overline{H_2A'}$，知 $BH_1A'I'$、$BH_2A'C'_1$ 為平行四邊形。因此原命題等價於證明 $I'C'_1$ 平行於 $BC$。\n\n令 $M, T, T'$ 分別為 $AI, A'I, A'J$ 與 $\\odot(ABC)$ 的另一個交點，點 $D, D'$ 分別為內切圓、$A$-旁切圓與 $BC$ 的切點，則熟知 $M, D, T$ 及 $M, D', T'$ 分別共線。注意到 $D, D'$ 關於 $\\overline{BC}$ 中點對稱，所以\n\n$$\n\\begin{align*}\n\\angle BI'I &= \\angle BA'I = \\angle BA'T = \\angle BMT = \\angle T'MC \\\\\n&= \\angle T'A'C = \\angle JA'C = \\angle BA'_1C_1 = \\angle BC'_1C_1\n\\end{align*}\n$$\n\n因此，由 $\\angle IBI' = 90^\\circ = \\angle C_1BC'_1$，可得 $\\triangle BII' \\sim \\triangle BC_1C'_1$，或 $\\triangle BIC_1 \\sim \\triangle BI'C'_1$。又 $BI \\perp BI'$，故 $I'C'_1 \\perp IC_1 \\perp BC$，即 $I'C'_1$ 平行於 $BC$，從而原命題得證。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20344,
"subject": "Mathematics (Olympiad)",
"question": "Let $p > 10$ be a prime. Show that there exist positive integers $m, n$ with $m + n < p$ such that $p \\mid 5^m 7^n - 1$.",
"options": [],
"answer": "See solution",
"solution": "If $5$ is a primitive root modulo $p$, then for some $1 \\le m \\le p-1$, $5^m \\equiv 7^{-1} \\pmod{p}$. Since $m \\ne p-1$ (otherwise $1 \\equiv 7^{-1} \\pmod{p}$, contradicting $p > 10$), we can take $n = 1$, so $m + n \\le p-1 < p$ and $p \\mid 5^m 7 - 1$.\n\nSimilarly, if $7$ is a primitive root modulo $p$, the argument is analogous.\n\nIf the orders $d_1$ and $d_2$ of $5$ and $7$ modulo $p$ are both less than $p-1$, then $d_1, d_2 \\le \\frac{p-1}{2}$. Take $m = d_1$, $n = d_2$, so $m + n \\le p-1 < p$ and\n\n$$\n5^{d_1} 7^{d_2} - 1 \\equiv (1)(1) - 1 \\equiv 0 \\pmod{p}.\n$$\n\nThus, such $m, n$ exist in all cases.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20345,
"subject": "Mathematics (Olympiad)",
"question": "a) Prove that if $n = 2015$, then, starting from any distribution of signs, by an appropriate succession of moves, one can get \"+\" in all vertices.\n\nb) Prove that, if $n = 2016$, then there exists a choice of signs such that no succession of moves can turn all the signs into \"+\".",
"options": [],
"answer": "See solution",
"solution": "a) The key is to obtain the \"atomic movement\", that is, a succession of moves whose final result is to change the sign of one single (arbitrary) vertex. Combining atomic movements, one can obtain any desired final configuration starting from any initial one. In particular, one can turn all the signs into \"+\".\n\nFirst, let us change the signs in the 671 groups of three consecutive vertices $(1, 2, 3), (4, 5, 6), \\ldots, (2011, 2012, 2013)$. This provides a succession of moves that changes all the signs but two adjacent ones. With one additional move, we achieve the change of all the signs with one exception. Now we change all the signs, performing all the possible 2015 moves exactly once, so that each vertex changes sign three times. Combining all the moves made until now provides the \"atomic-movement\".\n\nAnother approach: label the vertices from $1$ to $n$, then, for $k = \\lceil n/2 \\rceil$, successively change the signs in the group $(k, k+1, k+2)$ if the sign of $k$ is \"-\". Thus we turn all of the first $n-2$ signs into \"+\". If the remaining two signs are \"+\", we are done. If they are \"-\", a move turns them into \"+\", another vertex turning into \"-\". Finally, we reduce all the situations to one single \"-\". From here we can continue as above.\n\nb) We color the vertices with 3 colors, periodically: red, green, blue, red, green, blue, etc. As $2016$ is a multiple of $3$, this coloring will 'close' well. At any move, the number of red \"+\"-es changes its parity, and so does the number of green \"+\"-es. It follows that the difference between the number of red \"+\"-es and the number of green ones does not change its parity. In the final configuration this difference should be $0$. But if in the initial configuration this difference is odd, one can not get from that position to the desired one.\n\n**Remark:** In fact, one can prove that, in the case when $n$ is not a multiple of $3$, any configuration can be obtained from any initial one.\n\nIn the case when $n$ is a multiple of $3$, a given final position can be obtained from an initial one if and only if the parity of the number of red, green and blue \"+\"-es are respectively the same for the initial and the final configuration.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20346,
"subject": "Mathematics (Olympiad)",
"question": "Call a pair of positive integers $(a, b)$ *carroty* if $S(a^{b+1}) = a^b$, where $S(m)$ is the digit sum of $m$. Find all *carroty* pairs $(a, b)$.",
"options": [],
"answer": "See solution",
"solution": "The *carroty* pairs are $$(a, b) \\in \\{(1, b) \\mid b \\in \\mathbb{Z}^+\\} \\cup \\{(3, 2), (9, 1)\\}.$$ \n\nLet $k$ be the number of digits of $a$. Then $10^{k-1} \\le a < 10^k$, so $10^{(k-1)b} \\le a^b$ and $a^{b+1} < 10^{k(b+1)}$. Since digits are at most 9, $S(a^{b+1}) \\le 9k(b+1)$. Thus,\n\n$$\n10^{(k-1)b} \\le a^b = S(a^{b+1}) \\le 9k(b+1) \\implies 10^{(k-1)b} \\le 9k(b+1).\n$$\n\nIf $k \\ge 2$, then $k \\le 2(k-1)$ and $b+1 \\le 2b$ for $b \\ge 1$. Let $x = (k-1)b$, then $k(b+1) \\le 4x$, so $10^x \\le 36x$. For $x \\ge 2$, $10^x$ grows faster, so only $x = 0$ or $x = 1$ are possible. Thus, either $k = 1$ or $k = 2$ and $b = 1$.\n\n**Case 1:** $b = 1$, $k = 2$. We need $S(a^2) = a$ for $10 \\le a < 100$. Since $a^2 < 10^4$, $a = S(a^2) \\le 36$. Also, $a^2 \\equiv a \\pmod{9}$, so $a(a-1) \\equiv 0 \\pmod{9}$, i.e., $a \\equiv 0$ or $1 \\pmod{9}$. Checking $a \\in \\{10, 18, 19, 27, 28, 36\\}$, none work.\n\n**Case 2:** $k = 1$. $a^{b+1} \\equiv a^b \\pmod{9}$ implies $a^b(a-1) \\equiv 0 \\pmod{9}$, so $a = 1$ or $a$ divisible by 3. $a = 1$ works for all $b$. For $a \\in \\{3, 6, 9\\}$, $3^b \\le S(a^{b+1}) \\le 9(b+1)$, so $b \\le 3$. Checking these, only $(3, 2)$ and $(9, 1)$ work.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20347,
"subject": "Mathematics (Olympiad)",
"question": "Натурал тоон $\\{a_n\\}_{n \\ge 1}$ дарааллын хувьд $\\forall n \\ge 1: 0 < a_{n+1} - a_n \\le m$ (энд $m \\in \\mathbb{N}$ тогтмол тоо) бол $a_r \\mid a_s$ ба $1 \\le r < s \\in \\mathbb{N}$ байх гишүүд олдоно гэдгийг батал.",
"options": [],
"answer": "See solution",
"solution": "Ямар ч дараалсан $m$ ширхэг тооны хувьд, $a_1$-ээс их буюу тэнцүү тооны дотор ядаж нэг нь $\\{a_n\\}$ дарааллын гишүүн байх нь илэрхий.\n\n$$(m+1) \\times m$$ хүснэгт байгуулъя:\n\n$x_{0,1} = a_1$, $x_{0,j} = x_{0,j-1} + 1$ $(j = 2, \\ldots, m)$\n\n$$\n\\begin{array}{l}\nx_{0,1},\\ x_{0,2},\\ \\dots,\\ x_{0,m} \\\\\nx_{1,1},\\ x_{1,2},\\ \\dots,\\ x_{1,m} \\\\\n\\vdots \\\\\nx_{m,1},\\ x_{m,2},\\ \\dots,\\ x_{m,m}\n\\end{array}\n$$\n\n$1 \\le i, j \\le m$ үед $x_{i,j} = x_{i-1,1} \\cdots x_{i-1,m} + x_{i-1,j}$ гэж байгуулъя.\n\nӨөрөөр хэлбэл, энэ хүснэгтийн мөр бүр нь дараалсан $m$ ширхэг гишүүн бөгөөд багана бүрийн их дугаартай нь бага дугаартайдаа хуваагддаг хүснэгт юм.\n\nТэгэхээр мөр бүрт дарааллын ядаж нэг гишүүн байгаа. $m+1$ мөртэй учраас нэг баганад орсон ялгаатай хоёр гишүүн дарааллаас олдоно. Тэр хоёр гишүүний их нь багадаа хуваагдана.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20348,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle with circumcentre $O$ and centroid $G$. Let $M$ be the midpoint of $BC$ and $N$ be the reflection of $M$ across $O$. Prove that $NO = NA$ if and only if $\\angle AOG = 90^\\circ$.",
"options": [],
"answer": "See solution",
"solution": "Let $H$ be the orthocenter of $\\triangle ABC$ and let $X$ be the midpoint of $AH$. Then we know that $AXON$ is a parallelogram.\n\nNow, observe that $\\angle AOH = \\angle AOG$. Now,\n\n$$\nNO = NA \\iff XA = XO \\iff \\angle AOH = 90^\\circ\n$$\n\nThus, we are done.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20349,
"subject": "Mathematics (Olympiad)",
"question": "Mike and Nick play a game with a heap of 330 stones. On each turn, a player can remove either $n$ or $m$ stones from the heap, where $n$ and $m$ are two distinct numbers chosen from the set $\\{2, 3, 4, 5, 6, 7, 8, 9\\}$. Mike moves first. For each possible pair $(n, m)$, determine whether Mike can guarantee a win, and explain the strategy.",
"options": [],
"answer": "See solution",
"solution": "We separate the numbers from 2 to 9 into the pairs $(2, 7)$, $(3, 8)$, $(5, 6)$, $(4, 9)$. To win, Mike can use the following rule: if Nick fixes one number from any pair, Mike fixes the other number from the same pair.\n\nNow, we analyze the game for each pair $(n, m)$ by marking the numbers from 1 to 330 with \"+\" (winning) or \"-\" (losing). If $k$ stones remain and the player can win, we write $+k$; otherwise, $-k$.\n\n**Case 1:** $n, m = 2, 7$\n\nThe table shows a period of 3:\n\n| $+1$ | $+2$ | $-3$ | $+4$ | $+5$ | $-6$ | $+7$ | $+8$ | $-9$ | $+10$ | $+11$ | $-12$ | $+13$ | $+14$ | $-15$ | $+16$ | ... |\n|------|------|------|------|------|------|------|------|------|-------|-------|-------|-------|-------|-------|-------|-----|\n\nSince $330 \\div 3 = 110$, 330 is marked with \"-\". Thus, 330 is a losing position for the first player.\n\n**Case 2:** $n, m = 3, 8$\n\nThe table shows a period of 11:\n\n| $+1$ | $-2$ | $+3$ | $-4$ | $+5$ | $-6$ | $+7$ | $+8$ | $+9$ | $+10$ | $-11$ | ... |\n|------|------|------|------|------|------|------|------|------|-------|-------|-----|\n\nSince $330 \\div 11 = 30$, 330 is marked with \"-\". Thus, 330 is a losing position for the first player.\n\n**Case 3:** $n, m = 5, 6$\n\nThe table shows a period of 11:\n\n| $+1$ | $-2$ | $+3$ | $-4$ | $+5$ | $+6$ | $+7$ | $+8$ | $+9$ | $+10$ | $-11$ | ... |\n|------|------|------|------|------|------|------|------|------|-------|-------|-----|\n\nSince $330 \\div 11 = 30$, 330 is marked with \"-\". Thus, 330 is a losing position for the first player.\n\n**Case 4:** $n, m = 4, 9$\n\nThe table shows a period of 10:\n\n| $+1$ | $-2$ | $+3$ | $+4$ | $-5$ | $+6$ | $-7$ | $+8$ | $+9$ | $-10$ | ... |\n|------|------|------|------|------|------|------|------|------|-------|-----|\n\nSince $330 \\div 10 = 33$, 330 is marked with \"-\". Thus, 330 is a losing position for the first player.\n\n**Conclusion:**\n\nFor all possible pairs $(n, m)$, 330 is a losing position for the first player. Therefore, Mike cannot guarantee a win if both play optimally.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20350,
"subject": "Mathematics (Olympiad)",
"question": "Given a positive integer $n$ and $n^3$ integers $a_{ijk} \\in \\{1, -1\\}$ for $1 \\leq i, j, k \\leq n$, prove that there exist\n\n$x_1, \\dots, x_n, y_1, \\dots, y_n, z_1, \\dots, z_n \\in \\{1, -1\\}$\n\nsuch that\n\n$$\n\\left| \\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sum_{k=1}^{n} a_{ijk} x_i y_j z_k \\right| > \\frac{n^2}{3}.\n$$",
"options": [],
"answer": "See solution",
"solution": "*Proof 1.* For any $(x_i)$ and $(y_j)$ satisfying the given conditions, define\n\n$$\nX_k = \\sum_{i=1}^{n} \\sum_{j=1}^{n} a_{ijk} x_i y_j.\n$$\n\nSince we can always choose $z_k$ with the same sign as $X_k$, we have\n\n$$\n\\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sum_{k=1}^{n} a_{ijk} x_i y_j z_k = \\sum_{k=1}^{n} |X_k|.\n$$\n\nNote that, summing over all possible $(x_i)$ and $(y_j)$,\n\n$$\n\\begin{aligned}\n\\sum_{(x_i),(y_j)} |X_k|^2 &= \\sum_{(x_i),(y_j)} \\sum_{i_1,i_2=1}^{n} \\sum_{j_1,j_2=1}^{n} a_{i_1j_1k} a_{i_2j_2k} x_{i_1} x_{i_2} y_{j_1} y_{j_2} \\\\\n&= \\sum_{i_1,i_2=1}^{n} \\sum_{j_1,j_2=1}^{n} a_{i_1j_1k} a_{i_2j_2k} \\sum_{(x_i),(y_j)} x_{i_1} x_{i_2} y_{j_1} y_{j_2},\n\\end{aligned}\n$$\n\nand the last line is non-zero only when $i_1 = i_2$ and $j_1 = j_2$, so\n\n$$\n\\sum_{(x_i),(y_j)} |X_k|^2 = (2^n)^2 \\sum_{i=1}^{n} \\sum_{j=1}^{n} a_{ijk}^2 = 2^{2n} n^2.\n$$\n\nTo obtain a lower bound for $\\sum_{(x_i),(y_j)} |X_k|$, we estimate $\\sum_{(x_i),(y_j)} |X_k|^n$ for some large $n > 2$. Similarly,\n\n$$\n\\sum_{(x_i),(y_j)} |X_k|^4 = \\sum_{i_1,i_2,i_3,i_4=1}^{n} \\sum_{j_1,j_2,j_3,j_4=1}^{n} a_{i_1j_1k} a_{i_2j_2k} a_{i_3j_3k} a_{i_4j_4k} \\sum_{(x_i),(y_j)} x_{i_1} x_{i_2} x_{i_3} x_{i_4} y_{j_1} y_{j_2} y_{j_3} y_{j_4}.\n$$\n\nIn the last summation, it is non-zero only when $(i_1, i_2, i_3, i_4)$ and $(j_1, j_2, j_3, j_4)$ are paired up, and each term is repeated when all 4 items are the same, so\n\n$$\n\\sum_{i_1,i_2,i_3,i_4=1}^{n} \\sum_{j_1,j_2,j_3,j_4=1}^{n} a_{i_1j_1k} a_{i_2j_2k} a_{i_3j_3k} a_{i_4j_4k} \\sum_{(x_i),(y_j)} x_{i_1} x_{i_2} x_{i_3} x_{i_4} y_{j_1} y_{j_2} y_{j_3} y_{j_4} < 9n^4(2^n)^2.\n$$\n\nNow, by H\"older's inequality,\n\n$$\n\\left( \\sum_{(x_i),(y_j)} \\left( |X_k|^{2/3} \\right)^{3/2} \\right)^{2/3} \\left( \\sum_{(x_i),(y_j)} \\left( |X_k|^{4/3} \\right)^3 \\right)^{1/3} \\geq \\sum_{(x_i),(y_j)} |X_k|^2 = 2^{2n} n^2,\n$$\n\nso\n\n$$\n\\left( \\sum_{(x_i),(y_j)} |X_k| \\right)^{2/3} \\cdot (2^{2n} \\cdot 9n^4)^{1/3} > 2^{2n} n^2.\n$$\n\nHence,\n\n$$\n\\sum_{(x_i),(y_j)} |X_k| > \\left( (2^{2n})^{2/3} \\frac{n^{2/3}}{3^{2/3}} \\right)^{3/2} = 2^{2n} \\cdot \\frac{n}{3},\n$$\n\nsumming over $k$ yields\n\n$$\n(*) \\qquad \\sum_{(x_i),(y_j)} \\sum_{k=1}^{n} |X_k| > 2^{2n} \\cdot \\frac{n^2}{3},\n$$\n\nwhich implies the existence of $(x_i, y_j)$ such that $\\sum_{k=1}^{n} |X_k| > \\frac{n^2}{3}$. The proposition holds.\n\n*Note:* The last part using H\"older's inequality can also be proved using the following lemma. When $X \\geq 0$, $(X - 3)^2(X + 6)X = X^4 - 27X^2 + 54X \\geq 0$. Hence, $X^4 - 27n^2X^2 + 54n^3X \\geq 0$. Thus,\n\n$$\n\\sum_{(x_i),(y_j)} |X_k|^4 - 27n^2 \\sum_{(x_i),(y_j)} |X_k|^2 + 54n^3 \\sum_{(x_i),(y_j)} |X_k| \\geq 0.\n$$\n\nTherefore,\n\n$$\n\\sum_{(x_i),(y_j)} |X_k| \\geq \\frac{1}{54} \\cdot 2^{2n} \\cdot \\frac{(27n^2 \\cdot n^2 - 9n^4)}{n^3} = \\frac{1}{3} \\cdot 2^{2n} n.\n$$\n\nSumming over $k$ yields $(*)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20351,
"subject": "Mathematics (Olympiad)",
"question": "Given an acute non-isosceles triangle $ABC$, let $AK$ and $CN$ be its angle bisectors, and let $I$ be their intersection point. Let $X$ be the second point of intersection of the circumcircles of $\\triangle ABC$ and $\\triangle KBN$. Let $M$ be the midpoint of $AC$. Show that the Euler line of $\\triangle ABC$ is perpendicular to $BI$ if and only if points $X$, $I$, and $M$ are collinear.",
"options": [],
"answer": "See solution",
"solution": "*The Euler line of a non-isosceles triangle is a line that passes through the orthocenter, centroid, and circumcenter.*\n\nFirst, we show that the Euler line of $\\triangle ABC$ is perpendicular to $BI$ if and only if $\\angle ABC = 60^\\circ$. Let $H$ and $O$ be the orthocenter and circumcenter of $\\triangle ABC$, respectively. Then $\\angle HBI = \\angle IBO$, and $BI \\perp OH$ if and only if $BO = BH$, which is equivalent to $\\sin \\angle BAH = \\frac{1}{2}$. Since $\\triangle ABC$ is acute-angled, this is equivalent to $\\angle ABC = 60^\\circ$.\n\n**Lemma.** If $O$ and $H$ are the circumcenter and orthocenter of $\\triangle ABC$, then\n$$\nBO = BH \\iff \\angle ABC = 60^\\circ \\text{ or } \\sin \\angle BAH = \\frac{1}{2}.\n$$\n\n\n\n**Proof.** Consider a triangle whose sides are parallel to those of $\\triangle ABC$ and pass through its vertices (see figure above). The new triangle is similar to $\\triangle ABC$ with ratio $2$. The point $H$ becomes the circumcenter of the larger triangle, and $B$ corresponds to a point $K$, the midpoint of $AC$. Then $BH = 2OK = 2OC \\cos \\angle COK = 2R \\cos \\angle ABC$. Thus, $BH = BO = R$ if and only if $\\cos \\angle ABC = \\frac{1}{2}$, i.e., $\\angle ABC = 60^\\circ$.\n\n*End of Lemma proof.*\n\nNow, it suffices to show that $\\angle ABC = 60^\\circ$ if and only if points $X$, $I$, and $M$ are collinear. Let the line $XI$ intersect $AC$ at $M_0$. By the Law of Sines for triangles $AIM_0$ and $CIM_0$:\n\n$$\n\\frac{\\sin \\angle AIM_0}{\\sin \\angle M_0 IC} = \\frac{AM_0}{M_0 C} \\cdot \\frac{\\sin \\angle IAM_0}{\\sin \\angle ICM_0}.\n$$\n\nThus,\n$$\n\\frac{\\sin \\angle AIM_0}{\\sin \\angle CIM_0} = \\frac{\\sin \\angle KIX}{\\sin \\angle NIX}.\n$$\n\nBy the Law of Sines for triangles $XIK$ and $XIN$:\n$$\n\\frac{\\sin \\angle KIX}{\\sin \\angle NIX} = \\frac{XK}{XN} \\cdot \\frac{NI}{KI} \\cdot \\frac{\\sin \\angle KXI}{\\sin \\angle NXI}.\n$$\n\nNote that $\\triangle XNA \\sim \\triangle XKC$, so $\\frac{KX}{NX} = \\frac{KC}{NA}$. Therefore,\n$$\n\\frac{AM_0}{CM_0} = \\frac{KC}{NA} \\cdot \\frac{NI}{KI} \\cdot \\frac{\\sin \\angle KXI}{\\sin \\angle NXI} \\cdot \\frac{\\sin \\angle ICA}{\\sin \\angle IAC}.\n$$\n\nFrom the Law of Sines for triangles $ANI$ and $KIC$:\n$$\n\\frac{\\sin \\angle ICB}{KI} \\cdot \\frac{KC}{\\sin \\angle KIC} = 1, \\quad \\frac{\\sin \\angle KIC}{NA} \\cdot \\frac{NI}{\\sin \\angle IAB} = 1.\n$$\nThus,\n$$\n\\frac{AM_0}{CM_0} = \\frac{\\sin \\angle KXI}{\\sin \\angle NXI}.\n$$\n\nTherefore, $X$, $I$, and $M$ are collinear if and only if $\\angle KXI = \\angle NXI$.\n\nLet $Y$ and $Z$ be the intersections of $XI$ and $BI$ with the circumcircle of $\\triangle BNK$. Since $BI$ is a bisector, $Z$ is the midpoint of arc $NK$. For $\\angle KXI = \\angle NXI$, we must have $Y \\equiv Z$, so either $X \\equiv B$ (contradicting non-isoscelesness) or $I$ lies on the circumcircle of $\\triangle KBN$.\n\n\n\nThus, $X$, $I$, and $M$ are collinear if and only if $B$, $N$, $K$, and $I$ are concyclic, which is equivalent to $\\angle ABC = 60^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20352,
"subject": "Mathematics (Olympiad)",
"question": "We say that a natural number $n \\geq 3$ is *almost square-free* if there exists a prime number $p$, with $p \\equiv 1 \\pmod{3}$, such that $n$ is divisible by $p^2$, and the number $\\frac{n}{p}$ is square-free (i.e., not divisible by the square of any prime number).\n\nShow that, for any natural number $n$ that is almost square-free, the ratio between twice the sum of the divisors of $n$ and the number of divisors of $n$ is a natural number.",
"options": [],
"answer": "See solution",
"solution": "Consider $n = p_1 \\cdots p_{k-1} \\cdot p_k^2 \\cdot p_{k+1} \\cdots p_s$, with $p_1 < p_2 < \\dots < p_s$ prime numbers, $1 \\leq k \\leq s$, and $p = p_k \\equiv 1 \\pmod{3}$. Moreover, $p$ is odd and satisfies $p^2 + p + 1 \\equiv 0 \\pmod{3}$.\n\nThe number of divisors of $n$, $\\tau(n)$, is\n\n$$\n\\tau(n) = \\underbrace{(1+1) \\cdot (1+1) \\cdots (1+1)}_{\\text{applied } s-1 \\text{ times}} \\cdot (2+1) = 3 \\cdot 2^{s-1},\n$$\n\nand the sum of the divisors, $\\sigma(n)$, is:\n\n$$\n\\begin{aligned}\n\\sigma(n) &= \\frac{p_1^2-1}{p_1-1} \\cdots \\frac{p_{k-1}^2-1}{p_{k-1}-1} \\cdot \\frac{p_k^3-1}{p_k-1} \\cdots \\frac{p_{k+1}^2-1}{p_{k+1}-1} \\cdots \\frac{p_s^2-1}{p_s-1} \\\\\n&= (p_1+1) \\cdots (p_{k-1}+1) \\cdot (p_k^2+p_k+1) \\cdot (p_{k+1}+1) \\cdots (p_s+1).\n\\end{aligned}\n$$\n\nConsider two cases:\n\n*Case I.* If $n$ is even, then $p_1 = 2$, and $p_2 < p_3 < \\dots < p_s$ are odd primes. Hence,\n$$\n2\\sigma(n) = 2 \\cdot (2+1) \\cdot (p_2+1) \\cdots (p_{k-1}+1) \\cdot (p_k^2+p_k+1) \\cdot (p_{k+1}+1) \\cdots (p_s+1)\n$$\nso $2\\sigma(n)$ is divisible by $2 \\cdot 3^2 \\cdot 2^{s-2} = 3 \\cdot \\tau(n)$.\n\n*Case II.* If $n$ is odd, then $p_1 < p_2 < \\dots < p_s$ are odd primes, so\n$$\n\\sigma(n) = (p_1 + 1) \\cdots (p_{k-1} + 1) \\cdot (p_k^2 + p_k + 1) \\cdot (p_{k+1} + 1) \\cdots (p_s + 1)\n$$\nis divisible by $3 \\cdot 2^{s-1} = \\tau(n)$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20353,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $A, B \\in \\mathcal{M}_n(\\mathbb{C})$ are such that $A + B = AB + BA$.\n\n(a) If $n$ is odd, then $\\det(AB - BA) = 0$.\n\n(b) If $\\det(A) \\ne \\det(B)$, then $\\det(AB - BA) = 0$.",
"options": [],
"answer": "See solution",
"solution": "**(a)** Define $C = 2A - I_n$ and $D = 2B - I_n$. We have $CD - I_n = -(DC - I_n) = 2(AB - BA)$. As $n$ is odd, $\\det(CD - I_n) = -\\det(DC - I_n)$.\n\nFor any two square matrices $X, Y$, $\\det(XY - I_n) = \\det(YX - I_n)$, so $\\det(CD - I_n) = -\\det(CD - I_n)$, which implies $\\det(CD - I_n) = 0$. Consequently, $\\det(AB - BA) = 0$.\n\n**(b)** Define $E = AB - BA$. Suppose $\\det(E) \\ne 0$. We have $AE + EA = A^2B - BA^2 = A(A + B - BA) - (A + B - AB)A = AB - BA = E$. As $E$ is invertible, $E^{-1}AE + A = I_n$. Using $\\operatorname{tr}(E^{-1}AE) = \\operatorname{tr}(A)$, we get $\\operatorname{tr}(A) = \\frac{n}{2}$. By symmetry, $\\operatorname{tr}(B) = \\frac{n}{2}$, implying $\\operatorname{tr}(A) = \\operatorname{tr}(B)$, which contradicts condition (b). Thus, $\\det(AB - BA) = 0$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20354,
"subject": "Mathematics (Olympiad)",
"question": "Let $f(x)$ be an odd function on $\\mathbb{R}$, and $f(x) = x^2$ for $x \\ge 0$. Suppose for any $x \\in [a, a+2]$, $f(x+a) \\ge 2f(x)$. Find the range of real numbers $a$.",
"options": [],
"answer": "See solution",
"solution": "Given $f(x)$ is odd and $f(x) = x^2$ for $x \\ge 0$, so:\n\n$$\nf(x) = \\begin{cases} x^2 & \\text{if } x \\ge 0, \\\\ -x^2 & \\text{if } x < 0. \\end{cases}\n$$\n\nThe condition $f(x+a) \\ge 2f(x)$ for $x \\in [a, a+2]$ can be rewritten as $f(x+a) \\ge f(\\sqrt{2}x)$, since $2f(x) = f(\\sqrt{2}x)$ for $x \\ge 0$.\n\nSince $f(x)$ is increasing for $x \\ge 0$, $x + a \\ge \\sqrt{2}x$, so:\n\n$$a \\ge (\\sqrt{2} - 1)x.$$\n\nFor $x \\in [a, a+2]$, the maximum of $(\\sqrt{2}-1)x$ is at $x = a+2$:\n\n$$a \\ge (\\sqrt{2}-1)(a+2).$$\n\nSolving for $a$:\n\n$$a \\ge \\sqrt{2}$$\n\nSo the range is $[\\sqrt{2}, +\\infty)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20355,
"subject": "Mathematics (Olympiad)",
"question": "Let $G$ be the multiplicative group of invertible $2 \\times 2$ matrices with complex entries.\n\na) For every integer $n \\geq 3$, exhibit a finite non-Abelian subgroup of $G$ of order $2n$.\n\nb) Show that every finite subgroup of $G$ of even order greater than $2$ has a proper normal subgroup.",
"options": [],
"answer": "See solution",
"solution": "a) The group of rotational and reflectional symmetries of a regular $n$-gon, called the dihedral group $D_n$, is an example. It is generated by the matrices\n\n$$\nR = \\begin{pmatrix} \\cos \\frac{2\\pi}{n} & \\sin \\frac{2\\pi}{n} \\\\ -\\sin \\frac{2\\pi}{n} & \\cos \\frac{2\\pi}{n} \\end{pmatrix}, \\quad S = \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix}.\n$$\n\n$R$ and $S$ have orders $n$ and $2$, respectively, and $RSR = S$. The $2n$ matrices\n\n$$\nI, R, R^2, \\dots, R^{n-1}, S, SR, SR^2, \\dots, SR^{n-1}\n$$\n\nare pairwise distinct and exhaust $D_n$. For $n \\geq 3$, $D_n$ is non-Abelian.\n\nb) Let $H$ be a finite subgroup of $G$ of even order greater than $2$. $H$ has a proper subgroup of order $2$.\n\nConsider the kernel $K$ of the homomorphism $\\Delta: H \\to \\mathbb{C}^*$, $\\Delta(A) = \\det A$. $K$ is normal in $H$.\n\nIf $K$ is trivial, $\\Delta$ is injective, so $H$ is commutative, and every subgroup is normal, including the proper subgroup of order $2$.\n\nIf $K = H$, then $-I$ is in $H$ (the only $2 \\times 2$ matrix of order $2$ with determinant $1$), and the group it generates is a proper normal subgroup of $H$.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 20356,
"subject": "Mathematics (Olympiad)",
"question": "A real sequence $\\{a_n\\}_{n=0}^\\infty$ is defined recursively by $a_0 = 2$ and the recursion formula\n$$\na_n = \\begin{cases} a_{n-1}^2 & \\text{if } a_{n-1} < \\sqrt{3} \\\\ \\frac{a_{n-1}^2}{3} & \\text{if } a_{n-1} \\ge \\sqrt{3}. \\end{cases}\n$$\nAnother real sequence $\\{b_n\\}_{n=1}^\\infty$ is defined in terms of the first by the formula\n$$\nb_n = \\begin{cases} 0 & \\text{if } a_{n-1} < \\sqrt{3} \\\\ \\frac{1}{2^n} & \\text{if } a_{n-1} \\ge \\sqrt{3}, \\end{cases}\n$$\nvalid for each $n \\ge 1$. Prove that\n$$\nb_1 + b_2 + \\dots + b_{2020} < \\frac{2}{3}.\n$$",
"options": [],
"answer": "See solution",
"solution": "The first step is to prove, using induction, the formula\n$$\na_n = \\frac{2^{2^n}}{3^{2^n}(b_1 + b_2 + \\cdots + b_n)}.\n$$\nThe base case $n = 0$ is trivial. Assume the formula is valid for $a_{n-1}$, that is,\n$$\na_{n-1} = \\frac{2^{2^{n-1}}}{3^{2^{n-1}}(b_1 + b_2 + \\cdots + b_{n-1})}.\n$$\nIf $a_{n-1} < \\sqrt{3}$, then $b_n = 0$, and so\n$$\na_n = a_{n-1}^2 = \\frac{2^{2^n}}{3^{2^n}(b_1 + b_2 + \\cdots + b_{n-1})} = \\frac{2^{2^n}}{3^{2^n}(b_1 + b_2 + \\cdots + b_{n-1} + b_n)},\n$$\nwhereas if $a_{n-1} \\ge \\sqrt{3}$, then $b_n = \\frac{1}{2^n}$, and so\n$$\na_n = \\frac{a_{n-1}^2}{3} = \\frac{2^{2^n}}{3^{2^n}(b_1 + b_2 + \\cdots + b_{n-1}) + 1} = \\frac{2^{2^n}}{3^{2^n}(b_1 + b_2 + \\cdots + b_{n-1} + b_n)}.\n$$\nThis completes the induction.\n\nNext, we inductively establish the inequality $a_n \\ge 1$. The base case $n = 0$ is again trivial. Suppose $a_{n-1} \\ge 1$. If $a_{n-1} < \\sqrt{3}$, then\n$$\na_n = a_{n-1}^2 \\ge 1^2 = 1,\n$$\nwhereas if $a_{n-1} \\ge \\sqrt{3}$, then\n$$\na_n = \\frac{a_{n-1}^2}{3} \\ge \\frac{(\\sqrt{3})^2}{3} = 1,\n$$\nand the induction is complete.\n\nFrom\n$$\n1 \\le a_n = \\frac{2^{2^n}}{3^{2^n}(b_1 + b_2 + \\cdots + b_n)} = \\left( \\frac{2}{3^{b_1 + b_2 + \\cdots + b_n}} \\right)^{2^n},\n$$\nwe may then draw the conclusion\n$$\n3^{b_1 + b_2 + \\cdots + b_n} \\le 2.\n$$\nSince $3^{2/3} > 2$ (and the function $x \\mapsto 3^x$ is strictly increasing), we must have\n$$\nb_1 + b_2 + \\dots + b_n < \\frac{2}{3}\n$$\nfor all $n$, and we are finished.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20357,
"subject": "Mathematics (Olympiad)",
"question": "Let $a = x^2$, $b = y^2$, and $c = z^2$ for $x \\ge y \\ge z \\ge 0$. Prove that\n\n$$\n2(xA_z + yA_y + zA_x) \\ge (y+z)A_x + (x+z)A_y + (x+y)A_z,\n$$\n\nwhere\n\n$$\nA_x = \\sqrt{y^2 + z^2 - yz}, \\quad A_y = \\sqrt{z^2 + x^2 - zx}, \\quad A_z = \\sqrt{x^2 + y^2 - xy}.\n$$",
"options": [],
"answer": "See solution",
"solution": "We have $x^2 \\le y^2 + z^2 \\le (y+z)^2 \\Rightarrow x \\le y+z$, so $x, y, z$ are sides of a triangle.\n\nConsider the orderings:\n\n$$\nA_x \\le A_y \\Leftrightarrow (x-y)(x+y-z) \\ge 0\n$$\n$$\nA_y \\le A_z \\Leftrightarrow (y-z)(y+z-x) \\ge 0\n$$\n\nThus, $A_x \\le A_y \\le A_z$.\n\nNow,\n$$\nxA_z + yA_y \\ge yA_z + xA_y \\Leftrightarrow (A_z - A_y)(x - y) \\ge 0\n$$\n$$\nyA_y + zA_x \\ge zA_y + yA_x \\Leftrightarrow (A_y - A_x)(y - z) \\ge 0\n$$\n\nFor the main inequality:\n\n$$\n2(xA_z + yA_y + zA_x) = (xA_z + yA_y) + (yA_y + zA_x) + xA_z + zA_x \\ge \\\\\n(yA_z + xA_y) + (zA_y + yA_x) + xA_z + zA_x = (y+z)A_x + (x+z)A_y + (x+y)A_z.\n$$\n\nBy Cauchy-Schwarz:\n\n$$\n(x+y)A_z = (x+y)\\sqrt{x^2+y^2-xy} = \\sqrt{(x+y)(x^3+y^3)} \\ge x^2+y^2,\n$$\n\nand similar inequalities hold for $(x+z)A_y$ and $(y+z)A_x$. Summing these three inequalities proves the assertion. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20358,
"subject": "Mathematics (Olympiad)",
"question": "Suppose that $f(x)$ is defined on $\\mathbb{R}$, satisfying $f(0) = 2008$, and for any $x \\in \\mathbb{R}$\n\n$$\n\\begin{aligned}\nf(x+2) - f(x) &\\leq 3 \\times 2^x, \\\\\nf(x+6) - f(x) &\\geq 63 \\times 2^x.\n\\end{aligned}\n$$\n\nThen $f(2008) = \\underline{\\hspace{2cm}}$.",
"options": [],
"answer": "See solution",
"solution": "We have\n\n$$\n\\begin{aligned}\nf(x+2) - f(x) &= -(f(x+4) - f(x+2)) - (f(x+6) - f(x+4)) + (f(x+6) - f(x)) \\\\\n&\\geq -3 \\times 2^{x+2} - 3 \\times 2^{x+4} + 63 \\times 2^x = 3 \\times 2^x.\n\\end{aligned}\n$$\n\nThis means that $f(x+2) - f(x) = 3 \\times 2^x$. So we have\n\n$$\n\\begin{aligned}\nf(2008) &= f(2008) - f(2006) + f(2006) - f(2004) + \\dots \\\\\n&\\quad + f(2) - f(0) + f(0) \\\\\n&= 3 \\times (2^{2006} + 2^{2004} + \\dots + 2^2 + 1) + f(0) \\\\\n&= 3 \\times \\frac{4^{1003} + 1}{4 - 1} + 2008 \\\\\n&= 2^{2008} + 2007.\n\\end{aligned}\n$$\n\nWe define $g(x) = f(x) - 2^x$. Then we have\n\n$$\n\\begin{aligned}\ng(x+2) - g(x) &= f(x+2) - f(x) - 2^{x+2} + 2^x \\\\\n&\\leq 3 \\times 2^x - 3 \\times 2^x = 0, \\\\\ng(x+6) - g(x) &= f(x+6) - f(x) - 2^{x+6} + 2^x \\\\\n&\\geq 63 \\times 2^x - 63 \\times 2^x = 0.\n\\end{aligned}\n$$\n\nThis means that $g(x) \\leq g(x+6) \\leq g(x+4) \\leq g(x+2) \\leq g(x)$, and it implies that $g(x)$ is a periodic function with 2 as a period. So\n\n$$\n\\begin{aligned}\nf(2008) &= g(2008) + 2^{2008} = g(0) + 2^{2008} \\\\\n&= 2007 + 2^{2008}.\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20359,
"subject": "Mathematics (Olympiad)",
"question": "Let the function $f: \\mathbb{N} \\to \\mathbb{R}$ be such that for every natural number $n > 1$, there is a prime divisor $p$ of $n$ such that:\n\n$$\nf(n) = f\\left(\\frac{n}{p}\\right) - f(p)\n$$\n\nIf $f(2^{2007}) + f(3^{2008}) + f(5^{2009}) = 2006$, compute $f(2007^2) + f(2008^3) + f(2009^5)$.",
"options": [],
"answer": "See solution",
"solution": "If $n = p$ is a prime number, then\n$$\nf(p) = f\\left(\\frac{p}{p}\\right) - f(p) = f(1) - f(p)\n$$\nso\n$$\nf(p) = \\frac{f(1)}{2} \\qquad (1)\n$$\nIf $n = pq$ (where $p$ and $q$ are primes), then\n$$\nf(n) = f\\left(\\frac{n}{p}\\right) - f(p) = f(q) - f(p) \\text{ or } f(n) = f\\left(\\frac{n}{q}\\right) - f(q) = f(p) - f(q)\n$$\nso\n$$\nf(n) = 0 \\text{ (by (1))}.\n$$\nIf $n$ is a product of three primes,\n$$\nf(n) = f\\left(\\frac{n}{p}\\right) - f(p) = 0 - f(p) = -f(p) = -\\frac{f(1)}{2}.\n$$\nBy induction on the number of prime divisors, if $n$ is a product of $k$ primes,\n$$\nf(n) = (2 - k) \\frac{f(1)}{2} \\qquad (2)\n$$\nGiven $f(2^{2007}) + f(3^{2008}) + f(5^{2009}) = 2006$ and (2):\n$$\n\\begin{aligned}\n2006 &= f(2^{2007}) + f(3^{2008}) + f(5^{2009}) \\\\\n&= \\frac{2 - 2007}{2} f(1) + \\frac{2 - 2008}{2} f(1) + \\frac{2 - 2009}{2} f(1) \\\\\n&= -\\frac{3 \\cdot 2006}{2} f(1)\n\\end{aligned}\n$$\nSo\n$$\nf(1) = -\\frac{2}{3} \\qquad (3)\n$$\nNow, $2007 = 3^2 \\cdot 223$, $2008 = 2^3 \\cdot 251$, $2009 = 7^2 \\cdot 41$. Using (3):\n$$\n\\begin{aligned}\nf(2007^2) + f(2008^3) + f(2009^5) &= \\frac{2 - 6}{2} f(1) + \\frac{2 - 12}{2} f(1) + \\frac{2 - 15}{2} f(1) \\\\\n&= -\\frac{27}{2} f(1) = -\\frac{27}{2} \\cdot \\left(-\\frac{2}{3}\\right) = 9\n\\end{aligned}\n$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20360,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be the number of co-funders and $d_i$ be the value of the profit of the $i$-th director, $i = 1, \\ldots, n$. By condition,\n\n$$\nd_i = 3 \\cdot \\frac{d_{i+1} + d_{i+2} + \\dots + d_n}{n-i}.\n$$\n\nIf $\\frac{d_1}{d_n} = 120$, find $n$.",
"options": [],
"answer": "See solution",
"solution": "We have\n\n$$\n\\begin{aligned}\nd_{i-1} &= 3 \\cdot \\frac{d_i + d_{i+1} + \\dots + d_n}{n-i+1} \\\\\n&= 3 \\cdot \\frac{3 \\cdot \\frac{d_{i+1} + d_{i+2} + \\dots + d_n}{n-i} + d_{i+1} + \\dots + d_n}{n-i+1} \\\\\n&= 3 \\cdot \\frac{(n-i+3)(d_{i+1} + d_{i+2} + \\dots + d_n)}{(n-i)(n-i+1)}.\n\\end{aligned}\n$$\n\nTherefore,\n$$\n\\frac{d_{i-1}}{d_i} = \\frac{n-i+3}{n-i+1}, \\quad i = 2, \\ldots, n.\n$$\n\nMultiplying these equalities, we obtain\n$$\n\\frac{d_1}{d_n} = \\frac{d_1}{d_2} \\cdot \\frac{d_2}{d_3} \\cdots \\frac{d_{n-1}}{d_n} = \\frac{n+1}{n-1} \\cdot \\frac{n}{n-2} \\cdots \\frac{4}{2} \\cdot \\frac{3}{1} = \\frac{(n+1)n}{2}.\n$$\n\nBy condition, $\\frac{d_1}{d_n} = 120$, so $(n+1)n = 240$, which gives $n = 20$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20361,
"subject": "Mathematics (Olympiad)",
"question": "Let $f(x) = \\frac{1 + (4 - 4x + x^2)}{2-x}$ for $x < 2$. What is the minimum value of $f(x)$?\n\n(A) 0\n\n(B) 1\n\n(C) 2\n\n(D) 3",
"options": [],
"answer": "See solution",
"solution": "Let $x < 2$, so $2 - x > 0$. Then\n\n$$\n\\begin{aligned}\nf(x) &= \\frac{1 + (4 - 4x + x^2)}{2-x} \\\\\n&= \\frac{1}{2-x} + (2-x) \\\\\n&\\geq 2 \\times \\sqrt{\\frac{1}{2-x} \\cdot (2-x)} = 2.\n\\end{aligned}\n$$\n\nEquality holds if and only if $\\frac{1}{2-x} = 2-x$, which occurs when $x = 1$ (since $x < 2$). Thus, the minimum value of $f(x)$ is $2$ at $x = 1$.\n\nAnswer: C",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20362,
"subject": "Mathematics (Olympiad)",
"question": "Let $b$ be the number of black socks and $w$ the number of white ones. If $b$ or $w$ is $0$, then the probability of withdrawing a pair of socks of the same colour would be $1$. So $b$ and $w$ are positive. From symmetry, we may assume that $b \\ge w$.\n\nThe number of pairs of black socks is $\\frac{b(b-1)}{2}$. The number of pairs of white socks is $\\frac{w(w-1)}{2}$. The number of pairs of socks with one black and the other white is $bw$.\n\nThe probability of selecting a pair of socks of the same colour is the same as the probability of selecting a pair of socks of different colour. Hence,\n\n$$\nb(b - 1) + w(w - 1) = 2bw\n$$\n\nWhat is the largest possible value of $b$?",
"options": [],
"answer": "See solution",
"solution": "Let $d = b - w$. Then $w = b - d$ and\n\n$$\n\\begin{aligned}\nb(b-1) + (b-d)(b-d-1) &= 2b(b-d) \\\\\nb^2 - b + b^2 - bd - b - bd + d^2 + d &= 2b^2 - 2bd \\\\\n-2b + d^2 + d &= 0 \\\\\nd(d+1) &= 2b\n\\end{aligned}\n$$\n\nThe following table shows all possible values of $d$. Note that $b + w = 2b - d = d^2$.\n\n| $d$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $\\ge 8$ |\n|-----|-----|-----|-----|-----|-----|-----|-----|-----|----------|\n| $b$ | $0$ | $1$ | $3$ | $6$ | $10$ | $15$ | $21$ | $28$ | |\n| $b + w$ | $0$ | $1$ | $4$ | $9$ | $16$ | $25$ | $36$ | $49$ | $\\ge 64$ |\n\nThus the largest value of $b$ is **28**.\n\n% IMAGE: \n\nSo the probability that Justin draws a pair of socks of different colours is $\\frac{2bw}{(b+w)(b+w-1)}$.\n\nHence $4bw = b^2 + 2bw + w^2 - b - w$ and $b^2 - 2bw + w^2 - b - w = 0$.\n\nWe have $b^2 - (2w + 1)b + (w^2 - w) = 0$.\n\nThe quadratic formula gives\n\n$$\nb = \\frac{2w + 1 \\pm \\sqrt{(2w + 1)^2 - 4(w^2 - w)}}{2} = \\frac{2w + 1 \\pm \\sqrt{8w + 1}}{2}\n$$\n\nIf $b = \\frac{2w + 1 - \\sqrt{8w + 1}}{2} = w + \\frac{1}{2} - \\frac{1}{2}\\sqrt{8w + 1}$, then $b \\le w + \\frac{1}{2} - \\frac{1}{2}\\sqrt{9} = w - 1 < w$.\n\nSo $b = \\frac{2w + 1 + \\sqrt{8w + 1}}{2}$.\n\nNow $w < 25$ otherwise $b + w \\ge 2w \\ge 50$.\n\nSince $b$ increases with $w$, we want the largest value of $w$ for which $8w + 1$ is a square. Thus $w = 21$ and the largest value of $b$ is\n\n$$\n\\frac{43 + 13}{2} = \\mathbf{28}.\n$$\n\nWe have $b + w = (b - w)^2$. Thus $b + w$ is a square number less than $50$ and greater than $1$.\n\nThe following table gives all values of $b + w$ and the corresponding values of $b - w$ and $b$.\n\n| $b + w$ | $4$ | $9$ | $16$ | $25$ | $36$ | $49$ |\n|---------|-----|-----|------|------|------|------|\n| $b - w$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ |\n| $2b$ | $6$ | $12$| $20$ | $30$ | $42$ | $56$ |\n| $b$ | $3$ | $6$ | $10$ | $15$ | $21$ | $28$ |\n\nThus the largest value of $b$ is **28**.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20363,
"subject": "Mathematics (Olympiad)",
"question": "Max has 2015 jars labelled with the numbers 1 to 2015 and an unlimited supply of coins.\n\nConsider the following starting configurations:\n\n(a) All jars are empty.\n\n(b) Jar 1 contains 1 coin, jar 2 contains 2 coins, and so on, up to jar 2015 which contains 2015 coins.\n\n(c) Jar 1 contains 2015 coins, jar 2 contains 2014 coins, and so on, up to jar 2015 which contains 1 coin.\n\nNow Max selects in each step a number $n$ from 1 to 2015 and adds $n$ coins to each jar except to the jar $n$.\n\nDetermine for each starting configuration in (a), (b), (c), if Max can use a finite, strictly positive number of steps to obtain an equal number of coins in each jar.",
"options": [],
"answer": "See solution",
"solution": "Max can achieve his goal in all three cases by the procedures described below.\n\nLet $N = 2015$ be the number of jars.\n\n(a) Let Max select jar $j$ exactly $\\frac{N!}{j}$ times. Then jar $j$ will contain\n\n$$\n\\sum_{k \\neq j} k \\cdot \\frac{N!}{k} = (N-1) \\cdot N!\n$$\n\ncoins, which does not depend on $j$ as desired and has clearly needed at least one step.\n\n(b) Let Max select each jar $j$ exactly once. Then jar $j$ will contain $j + \\sum_{k \\neq j} k = \\sum_{k} k$ coins, which does not depend on $j$ as desired.\n\n(c) Let Max select jar $j$ exactly $\\left( \\frac{N!}{j} - 1 \\right)$ times. Then jar $j$ will contain\n\n$$\nN + 1 - j + \\sum_{k \\neq j} k \\cdot \\left( \\frac{N!}{k} - 1 \\right) = N + 1 - j + \\sum_{k \\neq j} (N! - k) = (N - 1)N! + (N + 1) - \\sum_{k} k\n$$\n\ncoins, which does not depend on $j$ as desired.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20364,
"subject": "Mathematics (Olympiad)",
"question": "Let $M \\subseteq \\{1, 2, \\dots, 2011\\}$ be a subset satisfying the following condition: For any three elements in $M$, there exist two of them $a$ and $b$, such that $a \\mid b$ or $b \\mid a$. Determine, with proof, the maximum value of $|M|$, where $|M|$ denotes the number of elements of $M$.\n",
"options": [],
"answer": "See solution",
"solution": "One can check that $M = \\{1, 2, 2^2, 2^3, \\dots, 2^{10}, 3, 3 \\times 2, 3 \\times 2^2, \\dots, 3 \\times 2^9\\}$ satisfies the condition, and $|M| = 21$.\n\nSuppose that $|M| \\geq 22$, and let $a_1 < a_2 < \\dots < a_k$ be the elements of $M$, where $|M| = k \\geq 22$. We first prove that $a_{n+2} \\geq 2a_n$ for all $n$; otherwise, we have $a_n < a_{n+1} < a_{n+2} < 2a_n$ for some $n < k + 2$, then any two of these three integers $a_n, a_{n+1}, a_{n+2}$ do not have any multiple relationship, which contradicts the assumption.\n\nIt follows from the inequality above that $a_4 \\geq 2a_2 \\geq 4$, $a_6 \\geq 2a_4 \\geq 8$, $\\dots$, $a_{22} \\geq 2a_{20} \\geq 2^{11} > 2011$, which is a contradiction!\n\nHence, the maximum value of $|M|$ is $21$. $\\square$\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20365,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute triangle with perimeter $2s$. We are given three pairwise disjoint circles with pairwise disjoint interiors, centered at $A$, $B$, and $C$, respectively. Prove that there exists a circle of radius $s$ which contains all three circles.",
"options": [],
"answer": "See solution",
"solution": "To simplify the formulation, we say that a point lies inside a circle if it lies on or within the circle. Assume we are given a circle $\\omega$ with center $O$ and radius $r$. A circle $\\omega'$ with center $O'$ contains $\\omega$ if and only if its radius is at least $O'O + r$.\n\n\n\nLet $r_a$, $r_b$, $r_c$ be the radii of the circles centered at $A$, $B$, and $C$, respectively. Using the above observation, the center $X$ of the desired circle must satisfy $AX \\leq s - r_a$, $BX \\leq s - r_b$, and $CX \\leq s - r_c$.\n\nNotice that $s - r_a$, $s - r_b$, and $s - r_c$ are positive. For example, since the circles are disjoint with disjoint interiors, $r_a < b$ and $r_a < c$, so $r_a < \\frac{b + c}{2} < \\frac{a + b + c}{2} = s$, thus $s - r_a > 0$.\n\nNow consider three circles centered at $A$, $B$, and $C$ with radii $s - r_a$, $s - r_b$, and $s - r_c$, respectively. If there is a point $X$ lying inside all three, we are done.\n\nEach pair of these circles intersects at two points, because $(s - r_a) + (s - r_b) > 2s - c = a + b > c$ and $c > |(s - r_a) - (s - r_b)|$. Suppose, for contradiction, that there is no point lying inside all three. Then, as in the picture, there exists a point $X$ inside triangle $ABC$ but outside all three circles:\n\n\n\nFor such $X$, $AX + BX + CX > s - r_a + s - r_b + s - r_c > 2s$. This is impossible. Let $Y$ be the intersection of $BX$ and $AC$. Using triangle inequalities for $CXY$ and $ABY$:\n\n$$\nBX + CX < BX + XY + CY = BY + CY < AB + AY + CY = AB + AC.\n$$\n\nSimilarly, $AX + BX < AC + BC$ and $CX + AX < BC + AB$. Summing, $AX + BX + CX < AB + BC + AC = 2s$, a contradiction.\n\n**Remark.** If three circles $\\omega_a$, $\\omega_b$, $\\omega_c$ centered at $A$, $B$, $C$ each pairwise intersect but have no common interior point, then there exists a point inside triangle $ABC$ lying outside all three circles.\n\n\n\nTo see this, consider the intersection point $P$ of $\\omega_b$ and $\\omega_c$ in the half-plane determined by $BC$ and $A$. The intersection $Q$ of ray $BA$ with $\\omega_b$ lies inside $\\omega_a$, since it is the closest point of $\\omega_b$ to $A$ (even if $A$ is inside $\\omega_b$, since $\\omega_a \\cap \\omega_b \\neq \\emptyset$). Therefore, $A$ cannot lie in angle $CBP$ (otherwise $Q$ would lie inside all three circles). Thus, $P$ lies in the interior of angle $CBA$ and also in angle $BCA$, so $P$ is inside triangle $ABC$. Since $P$ does not lie inside $\\omega_a$, there is a neighborhood of $P$ outside all three circles.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20366,
"subject": "Mathematics (Olympiad)",
"question": "We consider the following operation applied to a positive integer: The integer is represented in an arbitrary base $b \\ge 2$, in which it has exactly two digits and in which both digits are different from $0$. Then the two digits are swapped and the result in base $b$ is the new number.\n\nIs it possible to transform every number $> 10$ to a number $\\le 10$ with a series of such operations?",
"options": [],
"answer": "See solution",
"solution": "We show that each number $> 10$ can be transformed to a smaller number. In that way, we will eventually reach a number $\\le 10$.\n\nIf the number $n = 2k + 1$ is odd, we choose base $b = k$ with $n = (21)_k$. Swapping the two digits, we obtain the new number $(12)_k = k + 2$. Since $k \\ge 5$, the choice of $b = k$ as base is admissible (the digits are smaller than the base) and we have $k + 2 \\le 2k - 5 + 2 < 2k + 1$ as desired.\n\nIf the number $n = 2k$ is even, we choose the base $b = 2k - 2$ with $n = (12)_{2k-2}$ and obtain the new number $(21)_{2k-2} = 4k - 3$. Now we choose the base $k - 1$ with $4k - 3 = (41)_{k-1}$ and obtain the new number $(14)_{k-1} = k + 3$. Since $k > 5$, both bases are admissible, and we have $k + 3 < 2k$ as desired.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20367,
"subject": "Mathematics (Olympiad)",
"question": "In a far, far galaxy there are 225 inhabited planets. Between some pairs of inhabited planets there is a two-way space connection, and from each planet you can get to any other planet (possibly with several transfers). The influence of a planet is defined as the number of other planets with which this planet has a direct connection. It is known that if two planets are not connected by a direct space flight, then they have different influence. What is the smallest number of connections possible under these conditions?",
"options": [],
"answer": "See solution",
"solution": "The smallest number of connections is $1593$.\n\nLet's reformulate the problem in terms of graphs: planets are vertices, direct flights are edges. The condition is that any two vertices not connected by an edge have different degrees. We seek the minimum number of edges in such a graph with 225 vertices.\n\n**Lemma 1.** In a graph, there are no more than $k + 1$ vertices of degree $k$ for any natural number $k$.\n\n*Proof.* Suppose there are at least $k + 2$ vertices of degree $k$. If any two are not connected, the condition is violated. Thus, all are mutually connected, so each has degree at least $k + 1$, a contradiction.\n\n**Lemma 2.** For any $k \\geq 3$, in a simple graph with the minimum number of edges, there are at most $k$ vertices of degree $k$.\n\n*Proof.* By Lemma 1, there can be at most $k + 1$ vertices of degree $k$. If there are fewer, the statement holds. If there are $k + 1$, they form a complete graph $K_{k+1}$.\n\nConsider $K_k$ plus one vertex connected to one of the $k$ vertices. Then, degrees are: one vertex of degree 1, one of degree $k$, and $k-1$ of degree $k-1$. The number of edges in $K_{k+1}$ is $\\frac{k(k+1)}{2}$, and in the second case $\\frac{(k-1)k}{2} + 1$. Since $k > 1$, the second case has fewer edges.\n\n**Lemma 3.** Decreasing the degree of any vertex decreases the total number of edges.\n\nLet $S$ be the sum of degrees and $R$ the number of edges: $S = 2R$. If $S$ decreases, so does $R$.\n\nNow, for 225 vertices, find $k$ such that:\n\n$$\n1 + 2 + \\cdots + k = \\frac{k(k+1)}{2} \\leq 225 < \\frac{(k+1)(k+2)}{2}\n$$\n\n$k = 20$, since $1 + 2 + \\cdots + 20 = 210$ and $1 + 2 + \\cdots + 21 = 231$.\n\nLet $l = 225 - 210 = 15$.\n\nFor the minimum, have 1 vertex of degree 1, 2 of degree 2, ..., 20 of degree 20, and the remaining 15 vertices of degree $21$ (ideally). The total degree sum:\n\n$$\nL = 1 \\cdot 1 + 2 \\cdot 2 + \\cdots + 20 \\cdot 20 + 15 \\cdot 21 = 3185\n$$\n\nEach edge is counted twice, so the number of edges is $\\frac{3185}{2}$, but since 3185 is odd, the minimum is $\\frac{3186}{2} = 1593$.\n\nThus, the minimum number of connections is $1593$.\n\nTo construct such a graph: have 1 vertex of degree 1, 2 of degree 2, ..., 20 of degree 20, 14 of degree 21, and 1 of degree 22. Build complete graphs for each degree group, and connect the groups to ensure the graph is connected and the degree conditions are met.\n\n\n\nThe remaining edges can be distributed to maintain the required degree sequence and connectivity.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20368,
"subject": "Mathematics (Olympiad)",
"question": "令 $Q_{>0}$ 表示所有正有理数所成之集合。求所有函数 $f: Q_{>0} \\to Q_{>0}$ 满足\n\n$$\nf(x^2 f(y)^2) = f(x)^2 f(y), \\text{ 对所有 } x, y \\in Q_{>0} \\text{ 均成立。}\n$$",
"options": [],
"answer": "See solution",
"solution": "$f(x) = 1$ 对所有 $x \\in Q_{>0}$。\n\n取任意 $a, b \\in Q_{>0}$。令 $x = f(a)$, $y = b$ 以及 $x = f(b)$, $y = a$ 代入题设条件，得到\n\n$$\n(f(f(a)))^2 f(b) = f(f(a)^2 f(b)^2) = f(f(b))^2 f(a),\n$$\n\n从而\n\n$$\n\\frac{f(f(a))^2}{f(a)} = \\frac{f(f(b))^2}{f(b)} \\text{ 对所有 } a, b \\in Q_{>0}。\n$$\n\n即存在常数 $C \\in Q_{>0}$ 使得\n$$f(f(a))^2 = C f(a),$$\n\n$$\n\\left(\\frac{f(f(a))}{C}\\right)^2 = \\frac{f(a)}{C} \\quad \\text{对所有 } a \\in Q_{>0}。 \\qquad (1)\n$$\n\n记 $f^n(x)$ 为 $f$ 的 $n$ 次迭代。由上式可得\n\n$$\n\\frac{f(a)}{C} = \\left(\\frac{f^2(a)}{C}\\right)^2 = \\left(\\frac{f^3(a)}{C}\\right)^4 = \\dots = \\left(\\frac{f^{n+1}(a)}{C}\\right)^{2^n}\n$$\n\n对所有正整数 $n$ 成立。因此，$f(a)/C$ 必须是任意 $2^n$ 次方的有理数，这只有 $f(a)/C = 1$ 时才可能，否则素因子的指数无法被任意大的 $2^n$ 整除。故 $f(a) = C$ 对所有 $a \\in Q_{>0}$。\n\n将 $f \\equiv C$ 代入题设条件，得 $C = C^3$，所以 $C = 1$。因此 $f(x) \\equiv 1$ 是唯一满足题设条件的函数。\n\n*注1.* 也可以先求 $f(1) = 1$。设 $d = f(1)$，分别令 $x = y = 1$ 和 $x = d^2, y = 1$ 代入题设，得 $f(d^2) = d^3$ 和 $f(d^6) = f(d^2)^2 \\cdot d = d^7$。再令 $x = 1, y = d^2$，得 $f(d^6) = d^2 \\cdot d^3 = d^5$，所以 $d^7 = f(d^6) = d^5$，即 $d = 1$。\n\n此后，剩下的证明简化，因为已知 $C = \\frac{f(f(1))^2}{f(1)} = 1$，于是上式变为 $f(f(a))^2 = f(a)$，同理可得 $f(a) = 1$。\n\n*注2.* 存在非常数函数 $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$ 满足题设条件，例如 $f(x) = \\sqrt{x}$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20369,
"subject": "Mathematics (Olympiad)",
"question": "Let $P, N$ be the projections of $O, K$ on $AE$ and $M, T$ be the midpoints of $BC$ and the minor arc $BC$ of $(O)$. Let $AH = h$ and $R, R'$ be the radii of $(O), (K)$ respectively.\n\n\n\nProve that $AHKT$ is a parallelogram and therefore $AD \\parallel HK$.",
"options": [],
"answer": "See solution",
"solution": "By Thales' theorem, we have\n\n$$\n\\frac{HE}{HA} = \\frac{MO}{MT} = \\frac{MO}{OT - OM} = \\sqrt{2} + 1 \\implies HE = (\\sqrt{2} + 1)h.\n$$\n\nAccording to the Pythagorean theorem, $BK^2 - BO^2 = KM^2 - OM^2$ so\n\n$$\nR'^2 - R^2 = KM^2 - OM^2 \\implies KM^2 = R'^2 - \\frac{R^2}{2}.\n$$\n\nSimilarly,\n\n$$\n\\begin{align*}\nAO^2 - (AH - OM)^2 &= OP^2 = KN^2 = KE^2 - (EH - MK)^2 \\\\\n\\Leftrightarrow R^2 - \\left(h - \\frac{R}{\\sqrt{2}}\\right)^2 &= R'^2 - \\left((\\sqrt{2} + 1)h - MK\\right)^2 \\\\\n\\Leftrightarrow MK^2 + h R\\sqrt{2} &= R'^2 - \\frac{R^2}{2} - (\\sqrt{2} + 1)^2 h^2 + (2 + 2\\sqrt{2})h MK \\\\\n\\Leftrightarrow R\\sqrt{2} &= (2 + 2\\sqrt{2})h + (2 + 2\\sqrt{2})MK \\\\\n\\Leftrightarrow MK &= h + \\frac{R\\sqrt{2}}{2 + 2\\sqrt{2}} = h + MT.\n\\end{align*}\n$$\n\nThus $KT = MK - MT = h = AH$, proving that $AHKT$ is a parallelogram. Therefore, $AD \\parallel HK$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20370,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ 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 $Q$, $R$, $O$, and $T$ 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\n\n**Lemma 2.** $OK \\perp BC$.\n\n*Proof.* Let $D$ and $D'$ denote the foot of perpendiculars from $H$ and $A'$ onto $BC$, respectively. 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": 20371,
"subject": "Mathematics (Olympiad)",
"question": "設三角形 $ABC$ 的內心、重心、外心分別為點 $I, G, O$。令點 $X, Y, Z$ 分別落在射線 $BC, CA, AB$ 上，並且滿足 $BX = CY = AZ$。設點 $F$ 為三角形 $XYZ$ 的重心。\n\n試證：直線 $FG$ 與 $IO$ 垂直。",
"options": [],
"answer": "See solution",
"solution": "先來證明一個引理。\n\n**Lemma 1.** 設 $I, O$ 分別為 $\\triangle ABC$ 的內心與外心。設點 $E, F$ 分別在射線 $CA, BA$ 上，並滿足 $BF = BC = CE$。則 $EF \\perp IO$。\n\n**引理證明.** 設 $M$ 為 $\\triangle ABC$ 的外接圓上包含 $A$ 的 $BC$ 弧中點，且令 $X, Y, Z$ 分別為三角形 $AEF, CIA, AIB$ 的外心。因為 $MB = MC$, $BF = CE$ 且 $\\angle MCE = \\angle MBF$，所以 $\\triangle MBF$ 與 $\\triangle MCE$ 正向全等。由此知\n\n$$\n\\angle MEA = \\angle MFA,\n$$\n\n即 $M$ 位於 $\\triangle AEF$ 的外接圓上，故 $OX \\perp YZ$。另一方面，由 $\\angle CFA = \\angle CIA$，得知 $F$ 位於 $\\triangle CIA$ 的外接圓上。同理有 $E$ 位於 $\\triangle AIB$ 的外接圓上。故\n\n$$\nXY \\perp AB, \\quad XZ \\perp CA \\implies OY \\parallel XZ, \\quad OZ \\parallel XY,\n$$\n\n故 $OYXZ$ 為菱形。因為 $YZ$ 平行於 $\\angle EAF$ 的角平分線，且 $AX, IO$ 關於 $YZ$ 對稱，故 $EF \\perp IO$，引理得證。\n\n\n\n回到原題，設 $M, N, P$ 分別為 $AB, CZ, XY$ 的中點。令 $U$ 為 $Z$ 關於 $M$ 的對稱點，$V$ 為 $C$ 關於 $P$ 的對稱點，並令 $AB$ 與 $VX$ 交於 $W$。因為 $G$ 在 $CM$ 上且 $\\frac{GM}{CG} = \\frac{1}{2}$，所以 $G$ 為 $\\triangle CUZ$ 的重心，知 $G, N, U$ 共線，且 $\\frac{GN}{UG} = \\frac{1}{2}$。因為 $F$ 在 $ZP$ 上且 $\\frac{FP}{ZF} = \\frac{1}{2}$，所以 $F$ 為 $\\triangle CVZ$ 的重心，知 $F, N, V$ 共線，且 $\\frac{FN}{VF} = \\frac{1}{2}$。另一方面，注意到 $\\triangle ABC$ 與 $\\triangle WBX$ 位似，且 $BU = BX = XV$，故由引理得 $UV \\perp IO$。證明完畢。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20372,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer and $x_1, x_2, \\dots, x_n > 0$ be real numbers such that\n$$\nx_1 + x_2 + \\dots + x_n = \\frac{1}{x_1^2} + \\frac{1}{x_2^2} + \\dots + \\frac{1}{x_n^2}.\n$$\nShow that for each positive integer $k \\le n$, there are $k$ numbers among $x_1, x_2, \\dots, x_n$ whose sum is at least $k$.",
"options": [],
"answer": "See solution",
"solution": "Arguing by contradiction, suppose that every sum of $k$ numbers from $x_1, x_2, \\dots, x_n$ is strictly less than $k$. Then the numbers\n$$\na_j = x_j + x_{j+1} + \\dots + x_{j+k-1}, \\quad j = 1, 2, \\dots, n\n$$\n(where the indices in the sums $a_j$ are taken modulo $n$) are also less than $k$.\n\nAdding up yields $a_1 + a_2 + \\dots + a_n < nk$ and, since\n$$\na_1 + a_2 + \\dots + a_n = k(x_1 + x_2 + \\dots + x_n),\n$$\nwe get\n$$\nx_1 + x_2 + \\dots + x_n < n.\n$$\nBut $n^2 \\le \\left(\\sum_{i=1}^n x_i\\right) \\left(\\sum_{i=1}^n \\frac{1}{x_i}\\right) < n \\sum_{i=1}^n \\frac{1}{x_i}$, so $\\sum_{i=1}^n \\frac{1}{x_i} > n$. It follows that\n$$\n\\sum_{i=1}^{n} x_i = \\sum_{i=1}^{n} \\frac{1}{x_i^2} \\ge \\frac{1}{n} \\left( \\sum_{i=1}^{n} \\frac{1}{x_i} \\right)^2 > \\frac{1}{n} \\cdot n^2 = n,\n$$\ncontradiction.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20373,
"subject": "Mathematics (Olympiad)",
"question": "In an acute triangle $ABC$, points $D$ and $E$ lie on sides $AB$ and $AC$ respectively such that $BD = CE$. Point $P$ is on line segment $DE$ and point $Q$ lies on the arc $BC$ (not containing $A$) of the circumcircle of triangle $ABC$. These points satisfy $\\displaystyle \\frac{BP}{PC} = \\frac{EQ}{QD}$, and points $A, B, C, D, E, P, Q$ are all distinct.\n\nShow that $\\angle BPC = \\angle BAC + \\angle EQD$.\n\nIn the above, denote by $XY$ the length of line segment $XY$.",
"options": [],
"answer": "See solution",
"solution": "Let $F$ and $G$ be the intersection points of lines $QD$ and $QE$ with triangle $ABC$ (other than $Q$), respectively. Let lines $BG$ and $CF$ meet at $R$. By applying Pascal's theorem to the six points $A, B, G, Q, F, C$ (which are concyclic), we see that $D, E, R$ are collinear. These six points are on the same circle in the order $A, F, B, Q, C, G$. In particular, $R$ lies on segment $DE$. We will show that $P = R$.\n\nFirst, we show that $\\displaystyle \\frac{BP}{PC} = \\frac{BR}{RC}$. Applying the sine rule to triangle $BRC$ gives:\n\n$$\n\\frac{BR}{RC} = \\frac{\\sin \\angle RCB}{\\sin \\angle CBR} = \\frac{\\sin \\angle DQB}{\\sin \\angle CQE}.\n$$\n\nOn the other hand, applying the sine rule to triangles $DQB$ and $CQE$ gives:\n\n$$\n\\frac{BP}{PC} = \\frac{QE}{QD} = \\frac{CE \\cdot \\sin \\angle ECQ}{BD \\cdot \\sin \\angle QBD} \\cdot \\frac{\\sin \\angle QBD}{\\sin \\angle DQB} \\cdot \\frac{\\sin \\angle ECQ}{\\sin \\angle CQE}.\n$$\n\nFrom $\\angle ECQ + \\angle QBD = 180^\\circ$ and $BD = CE$, we obtain:\n\n$$\n\\frac{BP}{PC} = \\frac{\\sin \\angle DQB}{\\sin \\angle CQE} = \\frac{BR}{RC}.\n$$\n\nAssume that $P$ and $R$ are distinct. If $D, P, R, E$ lie in that order, then:\n\n$$\n\\angle CBR < \\angle CBP < \\angle CBA < 90^\\circ, \\\\\n\\angle PCB < \\angle RCB < \\angle ACB < 90^\\circ.\n$$\n\nThis shows that $\\sin \\angle CBR < \\sin \\angle CBP$ and $\\sin \\angle PCB < \\sin \\angle RCB$. By the sine rule, we obtain:\n\n$$\n\\frac{PC}{BP} = \\frac{\\sin \\angle CBP}{\\sin \\angle PCB} > \\frac{\\sin \\angle CBR}{\\sin \\angle RCB} = \\frac{RC}{BR},\n$$\n\nwhich contradicts $\\displaystyle \\frac{BP}{PC} = \\frac{BR}{RC}$. If $D, R, P, E$ lie in that order, we have a similar contradiction. Thus, $P = R$ by contradiction.\n\nFrom above, we have:\n\n$$\n\\angle BPC = \\angle BRC = 180^\\circ - \\angle RCB - \\angle CBR = 180^\\circ - \\angle DQB - \\angle CQE.\n$$\n\nIn addition,\n\n$$\n\\angle DQB + \\angle CQE = \\angle CQB - \\angle EQD = 180^\\circ - \\angle BAC - \\angle EQD.\n$$\n\nPutting these together, we obtain $\\angle BPC = \\angle BAC + \\angle EQD$, as desired.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20374,
"subject": "Mathematics (Olympiad)",
"question": "We call a set of three numbers \"arithmetic\" if one of its elements is the arithmetic mean of the other two. Similarly, we call a set of three numbers \"harmonic\" if one of its elements is the harmonic mean of the other two. How many three-element subsets of the set\n$$\n\\{z \\mid -2011 < z < 2011\\}\n$$\nof integers are both arithmetic and harmonic?",
"options": [],
"answer": "See solution",
"solution": "Let $\\{u, v, w\\}$ be an arithmetic set. Assume $u < v < w$, so $u = a - d$, $v = a$, $w = a + d$ with $d > 0$. We want this set to also be harmonic. If $q$ is the harmonic mean of $p$ and $r$, then $\\frac{1}{p} + \\frac{1}{r} = \\frac{2}{q}$, which is equivalent to $qr - 2rp + pq = 0$.\n\nIf $v$ is the harmonic mean of $u$ and $w$, this gives:\n$$\na(a + d) - 2(a + d)(a - d) + (a - d)a = 2d^2 = 0\n$$\nwhich implies $d = 0$, a contradiction since $d > 0$.\n\nIf $w$ is the harmonic mean of $u$ and $v$:\n$$\n(a + d)a - 2a(a - d) + (a - d)(a + d) = 3ad - d^2 = d(3a - d) = 0\n$$\nSo $d = 3a$ (since $d = 0$ is not possible). Thus, any set of the form $\\{-2a, a, 4a\\}$ works.\n\nIf $u$ is the harmonic mean of $v$ and $w$:\n$$\n(a - d)(a + d) - 2(a + d)a + a(a - d) = -3ad - d^2 = -d(3a + d) = 0\n$$\nSo $d = -3a$, which gives the same sets as before.\n\n$a$ can be any integer $\\neq 0$ such that $-2011 < 4a < 2011$, so $-502 \\leq a \\leq 502$, $a \\neq 0$. Thus, there are $1004$ such subsets.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20375,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\alpha \\neq 0$ be a real number. Consider the sequence of real numbers $\\{x_n\\}$, $n = 1, 2, 3, \\dots$, defined by:\n\n$$\nx_1 = 0, \\quad x_{n+1}(x_n + \\alpha) = \\alpha + 1 \\quad \\text{for all } n = 1, 2, 3, \\dots\n$$\n\n1. Find the general term of the sequence $\\{x_n\\}$.\n2. Prove that the sequence $\\{x_n\\}$ has a finite limit as $n \\to +\\infty$. Find this limit.",
"options": [],
"answer": "See solution",
"solution": "1. *For* $\\alpha = -1$, *it is easily seen that* $x_n = 0$ *for all* $n = 1, 2, 3, \\dots$.\n\n*For* $\\alpha \\neq -1$, $x_n \\neq -\\alpha$ *for all* $n$, *so we can write the recurrence as:*\n\n$$\nx_{n+1} = \\frac{\\alpha + 1}{x_n + \\alpha} \\quad \\forall n = 1, 2, 3, \\dots\n$$\n\n*With* $x_1 = 0$, $x_2 = \\frac{\\alpha + 1}{\\alpha}$.\n\n*Let* $p_1 = 0$, $p_2 = \\alpha + 1$, $q_1 = 1$, $q_2 = \\alpha$, *and suppose* $x_k = \\frac{p_k}{q_k}$. *Then*\n\n$$\np_{k+1} = (\\alpha + 1)q_k, \\quad q_{k+1} = \\alpha q_k + p_k\n$$\n\n*By induction,* $x_n = \\frac{p_n}{q_n}$ *for all* $n$, *where* $\\{p_n\\}$ *and* $\\{q_n\\}$ *are defined by:*\n\n$$\np_1 = 0, \\quad q_1 = 1, \\quad q_2 = \\alpha, \\quad p_{n+1} = (\\alpha + 1)q_n, \\quad q_{n+1} = \\alpha q_n + p_n\n$$\n\n*The sequence* $\\{q_n\\}$ *satisfies:*\n\n$$\nq_1 = 1, \\quad q_2 = \\alpha, \\quad q_{n+1} = \\alpha q_n + (\\alpha + 1)q_{n-1}\n$$\n\n*The characteristic equation is*\n\n$$\nx^2 - \\alpha x - (\\alpha + 1) = 0\n$$\n\n*with roots* $-1$ *and* $\\alpha + 1$.\n\n- *If* $\\alpha = -2$, $q_n = (-1)^{n-1} + (-1)^{n-1}(n-1)$\n- *If* $\\alpha \\neq -2$, $q_n = \\frac{(-1)^{n-1} + (\\alpha+1)^n}{\\alpha+2}$\n\n*Similarly:*\n\n- *If* $\\alpha = -2$, $p_1 = 0$, $p_n = -[((-1)^{n-2} + (-1)^{n-2})(n-2)]$ for $n \\ge 2$\n- *If* $\\alpha \\neq -2$, $p_1 = 0$, $p_n = \\frac{(-1)^{n-2} + (\\alpha + 1)^{n-1}}{\\alpha + 2} (\\alpha + 1)$ for $n \\geq 2$\n\n*Thus:*\n\n- *If* $\\alpha = -2$, $x_n = \\frac{n-1}{n}$\n- *If* $\\alpha \\neq -2$, $x_n = \\frac{((-1)^{n-2} + (\\alpha + 1)^{n-1})(\\alpha + 1)}{(-1)^{n-1} + (\\alpha + 1)^n}$\n\n2. *It is easy to see that:*\n\n- *If* $\\alpha = -1$, $\\lim x_n = 0$\n- *If* $\\alpha = -2$, $\\lim x_n = 1$\n\n*For* $\\alpha \\neq -2$:\n\n$$\nx_{2k-1} = \\frac{(\\alpha + 1)^{2k-1} - (\\alpha + 1)}{1 + (\\alpha + 1)^{2k-1}}\n$$\n\n$$\nx_{2k} = \\frac{(\\alpha + 1) + (\\alpha + 1)^{2k}}{(\\alpha + 1)^{2k} - 1}\n$$\n\n*Thus:*\n\n- *If* $|\\alpha + 1| > 1$, $\\lim x_n = 1$\n- *If* $|\\alpha + 1| < 1$, $\\lim x_n = -(\\alpha + 1)$\n\n*Therefore, the sequence* $\\{x_n\\}$ *is convergent and its limit is as above.*",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20376,
"subject": "Mathematics (Olympiad)",
"question": "Let $S$ be a set consisting of $n$ points in space, with no four among them lying on the same plane. What is the number of subsets of $S$ that can be obtained as the intersection of $S$ with some half-space?\n\nWhat is the value for $n = 10$?",
"options": [],
"answer": "See solution",
"solution": "Let $P(n) = \\frac{n^3 - 3n^2 + 8n}{3}$. For $n = 10$:\n\n$$\nP(10) = \\frac{10^3 - 3 \\times 10^2 + 8 \\times 10}{3} = \\frac{1000 - 300 + 80}{3} = \\frac{780}{3} = 260.\n$$\n\nThus, the answer is $260$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20377,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. Prove that the numbers\n\n$$\n1^1, 3^3, 5^5, \\dots, (2^n - 1)^{2^n - 1}\n$$\n\nare in different residue classes modulo $2^n$.",
"options": [],
"answer": "See solution",
"solution": "We proceed by induction on $n$.\n\nFor $n=1$, the only number in the sequence is $1^1$, so the statement is trivially true.\n\nFor the induction step, assume that for $n \\geq 1$, the numbers $1^1, 3^3, 5^5, \\dots, (2^n - 1)^{2^n - 1}$ are in different residue classes modulo $2^n$.\n\nWe want to show that $1^1, 3^3, 5^5, \\dots, (2^{n+1} - 1)^{2^{n+1} - 1}$ are in different residue classes modulo $2^{n+1}$.\n\nSplit the numbers into two groups:\n- *Lesser group*: $1^1, 3^3, 5^5, \\dots, (2^n - 1)^{2^n - 1}$\n- *Greater group*: $(2^n + 1)^{2^n + 1}, (2^n + 3)^{2^n + 3}, \\dots, (2^{n+1} - 1)^{2^{n+1} - 1}$\n\nThe numbers in the lesser group are also all in different residue classes modulo $2^{n+1}$.\n\nNote that $\\varphi(2^{n+1}) = 2^n$, so for odd $a$, $a^k \\equiv a^\\ell \\pmod{2^{n+1}}$ if $k \\equiv \\ell \\pmod{2^n}$.\n\nWrite the numbers in the greater group as $(2^n + m)^{2^n + m}$ for odd $m$ with $1 \\leq m \\leq 2^n - 1$. Expanding $(2^n + m)^{2^n + m}$ using the binomial theorem, any term with at least two factors of $2^n$ is congruent to $0$ modulo $2^{n+1}$. Thus, modulo $2^{n+1}$:\n\n$$\n\\begin{align}\n(2^n + m)^{2^n + m} &\\equiv m^{2^n + m} + 2^n \\cdot m^{2^n + m - 1} \\pmod{2^{n+1}} \\\\\n&\\equiv m^m + 2^n \\pmod{2^{n+1}}\n\\end{align}\n$$\n\nSince the numbers $m^m$ from the lesser group are different modulo $2^{n+1}$, the numbers from the greater group are also different modulo $2^{n+1}$. Moreover, the numbers from the lesser group are different from those from the greater group modulo $2^{n+1}$.\n\nSuppose for contradiction that $(2^n + m)^{2^n + m} \\equiv k^k \\pmod{2^{n+1}}$ for $1 \\leq k, m \\leq 2^n - 1$. Then $m^m + 2^n \\equiv k^k \\pmod{2^{n+1}}$, so $m^m \\equiv k^k \\pmod{2^n}$, which by the induction hypothesis implies $m = k$. But then $(2^n + m)^{2^n + m} \\equiv m^m + 2^n \\not\\equiv m^m \\pmod{2^{n+1}}$, a contradiction.\n\nTherefore, all numbers are in different residue classes modulo $2^n$ for all positive integers $n$.\n\n$\\square$\n\n",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 20378,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle. A circle through $B$ and $C$ crosses the sides $AB$ and $AC$ at $P$ and $Q$, respectively. Points $X$ and $Y$ on segments $BQ$ and $CP$, respectively, satisfy $\\angle ABY = \\angle AXP$ and $\\angle ACX = \\angle AYQ$. Prove that $XY$ and $BC$ are parallel.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $BY$ and $CX$ cross at $S$ and let circles $APX$ and $AQY$ cross again at $T$. We first prove that $A$, $S$, $T$ are collinear. Invert from $A$ with power $AP \\cdot AB = AQ \\cdot AC$. As $\\angle ABY = \\angle AXP$, the circle $APX$ is mapped to line $BY$. Similarly, the circle $AQY$ is mapped to line $CX$, so the inversion switches $S$ and $T$. Hence $A$, $S$, $T$ are collinear, as stated.\n\nNext, we show that the lines $AT$, $PX$ and $QY$ are projectively concurrent.\n\nLet $PX$ and $QY$ cross projectively at $R$. Apply Pappus' theorem to the hexagram $BPX C QY$ to deduce that $A = BP \\cap CQ$, $R = PX \\cap QY$ and $S = XC \\cap YB$ are collinear. The desired concurrence now follows by the preceding paragraph.\n\nWe now prove that $PQYX$ is cyclic. If $AT$, $PX$ and $QY$ are parallel, then $PQYX$ is an isosceles trapezoid, so it is cyclic. Otherwise, read the power of $R$ from circles $APX$ and $AQY$ to write $RX \\cdot RP = RT \\cdot RA = RY \\cdot RQ$, so $PQYX$ is cyclic.\n\nFinally, read angles from circle $PQYX$ and the given circle through $B$ and $C$ to write $\\angle PYX = \\angle PQX \\equiv \\angle PQB = \\angle PCB$. Consequently, $XY$ and $BC$ are parallel, as required.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20379,
"subject": "Mathematics (Olympiad)",
"question": "For a given positive integer $k$, we call an integer $n$ a $k$-number if both of the following conditions are satisfied:\n\n1. The integer $n$ is the product of two positive integers which differ by $k$.\n2. The integer $n$ is $k$ less than a square number.\n\nFind all $k$ such that there are infinitely many $k$-numbers.",
"options": [],
"answer": "See solution",
"solution": "Note that $n$ is a $k$-number if and only if the equation\n\n$$\nn = m^2 - k = r(r + k)\n$$\n\nhas solutions in integers $m, r$ with $k \\geq 0$.\n\nThe right-hand equality can be rewritten as\n\n$$\nk^2 - 4k = (2r + k)^2 - (2m)^2,\n$$\n\nso $k$-numbers correspond to ways of writing $k^2 - 4k$ as a difference of two squares, $N^2 - M^2$ with $N > r$ and $M$ even (which forces $N$ to have the same parity as $k$).\n\nAny non-zero integer can only be written as a difference of two squares in finitely many ways (because each gives a factorization, and a number has only finitely many factors).\n\nIf $k \\neq 4$ then $k^2 - 4k \\neq 0$, and as a result, if $k \\neq 4$ then there are only finitely many $k$-numbers.\n\nConversely, if $k = 4$ then setting $m = r + 2$ for $r \\geq 0$ shows that there are infinitely many 4-numbers.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20380,
"subject": "Mathematics (Olympiad)",
"question": "For every non-negative integer $a$, consider the set\n$$\nA_a = \\{ n \\in \\mathbb{N} \\mid \\sqrt{n^2 + an} \\in \\mathbb{N} \\}.\n$$\n\na) Prove that the set $A_a$ is finite if and only if $a \\neq 0$.\n\nb) Find the largest element of the set $A_{40}$.",
"options": [],
"answer": "See solution",
"solution": "a) If $a = 0$, then $A = \\mathbb{N}$, which is infinite.\n\nIf $a \\neq 0$, then there exists $p \\in \\mathbb{N}$ such that $n^2 + an = p^2$. This leads to $4n^2 + 4an = 4p^2$, or $4n^2 + 4an + a^2 = 4p^2 + a^2$, so $(2n + a - 2p)(2n + a + 2p) = a^2$.\n\nThus, $2n + a + 2p$ is a divisor of $a^2 \\neq 0$, so $2n < 2n + a + 2p \\leq a^2$, which shows that $n$ can assume only finitely many values; that is, $A$ is finite.\n\nb) We must find the largest integer $n$ such that $n^2 + 40n$ is a perfect square. Let $p \\in \\mathbb{N}$ be such that $p^2 = n^2 + 40n$. Then $p^2 + 400 = (n + 20)^2$, so $(n + 20 - p)(n + 20 + p) = 400$.\n\nThe numbers $n + 20 - p$ and $n + 20 + p$ have the same parity, so they are both even. The even divisors of $400$ are $2 \\cdot 200 = 400$, $4 \\cdot 100 = 400$, $8 \\cdot 50 = 400$, etc. This gives $n \\in \\{81, 32, 9\\}$. Thus, the answer is $81$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20381,
"subject": "Mathematics (Olympiad)",
"question": "Let two polynomials be given:\n\n$$\nP(x) = 4x^3 - 2x^2 - 15x + 9\n$$\n\nand\n\n$$\nQ(x) = 12x^3 + 6x^2 - 7x + 1.\n$$\n\n1. Prove that each of these polynomials has three distinct real roots.\n\n2. Let $\\alpha$ and $\\beta$ be respectively the greatest roots of $P(x)$ and $Q(x)$. Prove that $\\alpha^2 + 3\\beta^2 = 4$.",
"options": [],
"answer": "See solution",
"solution": "1. We have:\n\n$$\nP(-2) = -1; \\quad P(-1) = 18; \\quad P\\left(\\frac{3}{2}\\right) = -\\frac{9}{2}; \\quad P\\left(\\frac{\\sqrt{33}}{3}\\right) = \\frac{15 - \\sqrt{33}}{9}.\n$$\n\n$$\nQ(-2) = -57; \\quad Q(-1) = 2; \\quad Q\\left(\\frac{1}{3}\\right) = -\\frac{2}{9}; \\quad Q(1) = 12.\n$$\n\nFrom these values, it follows that $P(x)$ and $Q(x)$ each have three distinct real roots.\n\n2. As $\\alpha$ is a root of $P(x)$,\n\n$$\n4\\alpha^3 - 2\\alpha^2 - 15\\alpha + 9 = 0.\n$$\n\nTherefore,\n$$\n4\\alpha^3 - 15\\alpha = 2\\alpha^2 - 9\n$$\nso\n$$\n16\\alpha^6 - 120\\alpha^4 + 225\\alpha^2 = 4\\alpha^4 - 36\\alpha^2 + 81\n$$\nwhich gives\n$$\n16\\alpha^6 - 124\\alpha^4 + 261\\alpha^2 - 81 = 0.\n$$\n\nSince $\\alpha$ is the greatest root of $P(x)$, from above we get:\n$$\n\\frac{\\sqrt{33}}{3} > \\alpha > \\frac{3}{2}\n$$\nand so $4 - \\alpha^2 > 0$. We shall prove that $\\frac{\\sqrt{3(4 - \\alpha^2)}}{3}$ is a root of $Q(x)$.\n\nIndeed,\n$$\nQ\\left(\\frac{\\sqrt{3(4 - \\alpha^2)}}{3}\\right) = 0 \\iff \\frac{4}{3}(4 - \\alpha^2)\\sqrt{3(4 - \\alpha^2)} + 2(4 - \\alpha^2) - \\frac{7}{3}\\sqrt{3(4 - \\alpha^2)} + 1 = 0\n$$\nwhich simplifies to\n$$\n\\left(3 - \\frac{4\\alpha^2}{3}\\right)\\sqrt{3(4 - \\alpha^2)} + 9 - 2\\alpha^2 = 0\n$$\nso\n$$\n(9 - 2\\alpha^2)^2 = \\left(3 - \\frac{4\\alpha^2}{3}\\right)^2 \\cdot 3(4 - \\alpha^2)\n$$\nwhich leads to\n$$\n3(81 - 36\\alpha^2 + 4\\alpha^4) = (81 - 72\\alpha^2 + 16\\alpha^4)(4 - \\alpha^2)\n$$\nso\n$$\n16\\alpha^6 - 124\\alpha^4 + 261\\alpha^2 - 81 = 0.\n$$\n\nThus, $x_0 = \\frac{\\sqrt{3(4 - \\alpha^2)}}{3}$ is a root of $Q(x)$.\n\nMoreover, from above it is easy to see that $x_0 \\in \\left(\\frac{1}{3}, \\frac{\\sqrt{21}}{6}\\right) \\subset \\left(\\frac{1}{3}, 1\\right)$.\n\nOn the other hand, since $\\beta$ is the greatest root of $Q(x)$, the previous values show that $\\beta$ is the unique root of $Q(x)$ in the interval $\\left(\\frac{1}{3}, 1\\right)$.\n\nFrom these results, it follows that $\\frac{\\sqrt{3(4 - \\alpha^2)}}{3} = \\beta$, thus $\\alpha^2 + 3\\beta^2 = 4$.\n\n**Remark:** The values $P\\left(\\frac{3}{2}\\right)$ and $P\\left(\\frac{\\sqrt{33}}{3}\\right)$ in the first part are used in the proof of the second part. If we are interested only in proving the first part, we can instead study $P(1)$ and $P(2)$; in fact, $P(1) = -4$ and $P(2) = 3$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20382,
"subject": "Mathematics (Olympiad)",
"question": "Given a natural number $N$, we append two different nonzero digits to its right. It turns out that the new number is divisible by $N$. What is the maximal value that $N$ can be?",
"options": [],
"answer": "See solution",
"solution": "Assume the two digits form the number $\\overline{ab}$. The new number is $100N + \\overline{ab}$, which must be divisible by $N$. This means $\\overline{ab}$ is also divisible by $N$. Since $\\overline{ab}$ consists of two distinct digits, its maximum value is $98$, so $N \\leq 98$. The number $N = 98$ satisfies all the requirements.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20383,
"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 $n - km \\in B$ for $0 \\leq k \\leq q$ (because no more moves are possible, these numbers must be on the blackboard). Thus $r = n - qm \\in B$, which contradicts the minimality of $m$.\n\nWe conclude that $m \\mid 1 = \\gcd(2016, 47) \\in B$. By induction on $c$, we have $n - c \\in B$ for $0 \\leq c \\leq 2015$. 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": 20384,
"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 are among the five consecutive odd numbers $p, p + 2, p + 4, p + 6,$ and $p + 8$.\n\nWe must choose 4 out of these 5 numbers, omitting exactly one. If we omit $p$, $p + 2$, $p + 6$, or $p + 8$, three consecutive odd numbers remain, one of which must be divisible by 3, which is not allowed since all are primes. 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 $p, q, r, s$ can be divisible by 5, $p + 4$ must be divisible by 5.\n\nTherefore,\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 and by 3 (as shown above), $4(p + 4)$ 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": 20385,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be a right-angled triangle with $\\angle B = 90^\\circ$. Let $D$ be a point on $AC$ such that the inradii of triangles $ABD$ and $CBD$ are equal. If this common value is $r'$ and if $r$ is the inradius of triangle $ABC$, prove that\n\n$$\n\\frac{1}{r'} = \\frac{1}{r} + \\frac{1}{BD}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $E$ and $F$ be the incenters of triangles $ABD$ and $CBD$ respectively. Let the incircles of triangles $ABD$ and $CBD$ touch $AC$ at $P$ and $Q$ respectively. If $\\angle BDA = \\theta$, we see that\n\n\n\n$$\nr' = PD \\tan\\left(\\frac{\\theta}{2}\\right) = QD \\cot\\left(\\frac{\\theta}{2}\\right).\n$$\n\nHence,\n\n$$\nPQ = PD + QD = r' \\left( \\cot \\frac{\\theta}{2} + \\tan \\frac{\\theta}{2} \\right) = \\frac{2r'}{\\sin \\theta}.\n$$\n\nBut we observe that\n\n$$\nDP = \\frac{BD + DA - AB}{2}, \\quad DQ = \\frac{BD + DC - BC}{2}.\n$$\n\nThus $PQ = \\frac{b - c - a + 2BD}{2}$. We also have\n\n$$\n\\begin{aligned}\n\\frac{ac}{2} &= [ABC] = [ABD] + [CBD] = r' \\frac{AB + BD + DA}{2} + r' \\frac{CB + BD + DC}{2} \\\\\n&= r' \\frac{c + a + b + 2BD}{2} = r'(s + BD).\n\\end{aligned}\n$$\n\nBut\n\n$$\nr' = \\frac{PQ \\sin \\theta}{2} = \\frac{PQ \\cdot h}{2BD},\n$$\n\nwhere $h$ is the altitude from $B$ onto $AC$. But we know that $h = \\frac{ac}{b}$. Thus we get\n\n$$\nac = 2 r'(s + BD) = 2 \\cdot \\frac{PQ \\cdot h}{2 BD} (s + BD) = \\frac{(b - c - a + 2BD)ca(s + BD)}{2 BD b}.\n$$\n\nThus we get\n\n$$\n2 BD b = 2 (BD - (s-b))(s+BD).\n$$\n\nThis gives $BD^2 = s(s-b)$. Since $ABC$ is a right-angled triangle, $r = s-b$. Thus we get $BD^2 = rs$. On the other hand, we also have $[ABC] = r'(s+BD)$. Thus we get\n\n$$\nrs = [ABC] = r'(s + BD).\n$$\n\nHence,\n\n$$\n\\frac{1}{r'} = \\frac{1}{r} + \\frac{BD}{rs} = \\frac{1}{r} + \\frac{1}{BD}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20386,
"subject": "Mathematics (Olympiad)",
"question": "Consider two games, each with towns labeled by natural numbers: the first game has towns $1$ through $N$, and the second has towns $N+1$ through $N+M$. The initial and terminal towns are $1$ and $N$ in the first game, and $N+1$ and $N+M$ in the second. For every direction $d$, let $A_d$ be the $(N+M) \\times (N+M)$ matrix with entries\n\n$$\na_d(i, j) = \\begin{cases} 2^p, & \\text{if starting from town } i \\text{ and following direction } d \\\\ 0, & \\text{otherwise.} \\end{cases}\n$$\n\nFor any sequence of directions $\\mathbf{d} = (d_1, d_2, \\dots, d_n)$, the $(i, j)$ entry of $A_{d_1}A_{d_2}\\dots A_{d_n}$ is $2^p$ if the hero, following $\\mathbf{d}$ from town $i$, arrives at town $j$ and wins $p$ points. Let $e_i$ be the standard unit vector in $\\mathbb{R}^{N+M}$ with $1$ in the $i$-th entry and $0$ elsewhere. Define $s = e_1 - e_{N+1}$ and $f = e_N + e_{N+M}$.\n\nSince there are no connections between towns in the two games, prove that $s^T A f = 0$ for all $A = A_{d_1} \\cdots A_{d_n}$ if and only if $s^T A f = 0$ for all $A = A_{d_1} \\cdots A_{d_n}$ with $n \\le N + M$.",
"options": [],
"answer": "See solution",
"solution": "This is a straightforward implication of the minimisation algorithm for (sub)sequential transducers. It consists of two steps, each of which contains an interesting idea:\n\n1. Pushing forward the costs – in this case, awarding the hero the maximum amount of points that he has already guaranteed. This is also related to potentials in weighted graphs, often used to accelerate the Minimum Cost Paths Problem. However, in this case the issue is not only of algorithmic flavour, but is conceptual.\n2. The bisimulation and Nerode-Myhill relation that stabilises in at most $O(|Q|)$ steps.\n\nAlternatively, consider the sequence of vector spaces $V_0 = \\{s\\}$ and $V_{n+1} = V_n \\cup \\{v^T A_d : v \\in V_n,\\ d \\le k\\}$. Each $V_n$ spans a space of dimension at most $N+M$, so for some $n < N+M$, $\\operatorname{span} V_n = \\operatorname{span} V_{n+1}$. By induction, all vectors in $V_m$ for $m \\ge n$ are linear combinations of those in $V_n$. Thus, if all vectors in $V_n$ are orthogonal to $f$, so are those in $V_m$ for any $m \\in \\mathbb{N}$.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 20387,
"subject": "Mathematics (Olympiad)",
"question": "Find the integer that is closest to the value of the expression:\n\n$$\n\\left( (3 + \\sqrt{1})^{2023} - \\left( \\frac{1}{3 - \\sqrt{1}} \\right)^{2023} \\right) \\cdot \\left( (3 + \\sqrt{2})^{2023} - \\left( \\frac{1}{3 - \\sqrt{2}} \\right)^{2023} \\right) \\cdot \\left( (3 + \\sqrt{3})^{2023} - \\left( \\frac{1}{3 - \\sqrt{3}} \\right)^{2023} \\right) \\dots \\left( (3 + \\sqrt{8})^{2023} - \\left( \\frac{1}{3 - \\sqrt{8}} \\right)^{2023} \\right).\n$$",
"options": [],
"answer": "See solution",
"solution": "**Answer:** 0.\n\n**Solution.** Let's consider the last factor:\n\n$$\n\\frac{1}{3 - \\sqrt{8}} = \\frac{3 + \\sqrt{8}}{9 - 8} = 3 + \\sqrt{8} \\Rightarrow (3 + \\sqrt{8})^{2023} = \\left(\\frac{1}{3 - \\sqrt{8}}\\right)^{2023},\n$$\n\nwhich means that the last factor equals 0, and therefore the whole product equals 0.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20388,
"subject": "Mathematics (Olympiad)",
"question": "Suppose there are $b$ blue amoebas and $r$ red amoebas. Consider the following operations:\n\n- When a blue amoeba appears or disappears, two red amoebas disappear or appear, respectively.\n- Blue amoebas can be fused with red amoebas to produce three red amoebas.\n- Red amoebas can be changed back into blue amoebas: if $2k$ red amoebas are changed, $0 < k \\leq b + \\lfloor r/2 \\rfloor$, we obtain $k$ blue amoebas and $2b + r - 2k$ red amoebas, totaling $2b + r - k$ amoebas.\n\nIf there is only one amoeba, it remains unchanged. Otherwise, describe the process and invariants when transforming blue and red amoebas using these operations.",
"options": [],
"answer": "See solution",
"solution": "$2b + r$ is invariant: whenever one blue amoeba appears or disappears, two red amoebas disappear or appear, respectively. If there is only one amoeba, it will never change. Otherwise, change all blue amoebas into red amoebas by repeatedly fusing a blue amoeba with a red amoeba to get three red amoebas. At this point, there are $2b + r$ amoebas. Then, change red amoebas back into blue amoebas: if $2k$ red amoebas are changed, $0 < k \\leq b + \\lfloor r/2 \\rfloor$, we get $k$ blue amoebas and $2b + r - 2k$ red amoebas, totaling $2b + r - k$ amoebas.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20389,
"subject": "Mathematics (Olympiad)",
"question": "Find all $x, y, z$ which satisfy:\n\n$$\n\\begin{cases}\nx^2 + xy + xz = y, \\\\\ny^2 + yz + yx = z, \\\\\nz^2 + zx + zy = x.\n\\end{cases}\n$$",
"options": [],
"answer": "See solution",
"solution": "*Answer:* $x = y = z = \\frac{1}{3}$, $x = y = z = 0$.\n\n*Solution.* Summing up the equalities, we get\n\n$$\nx^2 + y^2 + z^2 + 2xy + 2xz + 2yz = x + y + z \\text{ or } (x + y + z)^2 = x + y + z.\n$$\n\nNow we face two cases.\n\n*Case 1.* $x + y + z = 1$. Then from the first equality,\n\n$$\nx^2 + xy + xz = x(x + y + z) = x = y,\n$$\n\nAnalogously, $x = y = z$, so the only solution is $x = y = z = \\frac{1}{3}$.\n\n*Case 2.* $x + y + z = 0$. Again from the first equality,\n\n$$\nx^2 + xy + xz = x(x + y + z) = 0 = y,\n$$\n\nhence $x = y = z = 0$ – the second solution.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20390,
"subject": "Mathematics (Olympiad)",
"question": "A sequence $a_n$ satisfies $a_1 = 2$, $a_2 = 3$, $a_3 = 5$, and $a_n = a_{n-1}^2$ for any $n \\geq 4$.\n\nA sequence $b_n$ satisfies $b_1 = 2$, $b_2 = 3$, $b_3 = 5$, and $b_n = b_{n-1} \\cdot b_{n-2} \\cdot b_{n-3}^2$ for any $n \\geq 4$.\n\nA sequence $c_n$ satisfies $c_1 = 2$, $c_2 = 3$, $c_3 = 5$, and $c_n = c_1 \\cdot c_2 \\cdot \\dots \\cdot c_{n-1}$ for any $n \\geq 4$.\n\nOrder the numbers $a_{1000}$, $b_{1000}$, and $c_{1000}$ by size.",
"options": [],
"answer": "See solution",
"solution": "First, for any $n \\geq 5$:\n\n$$\nc_n = (c_1 \\cdots c_{n-2}) \\cdot c_{n-1} = c_{n-1}^2.\n$$\n\nSince $a_4 = 5^2 = 25$ and $c_4 = 2 \\cdot 3 \\cdot 5 = 30$, we have $a_n < c_n$ for any $n \\geq 4$.\n\nNext, $c_1 = b_1 = 2$, $c_2 = b_2 = 3$, $c_3 = b_3 = 5$, and\n\n$$\nc_4 = 30 < 60 = b_4, \\quad c_5 = 900 < 2700 = b_5, \\quad c_6 = 810000 < 4050000 = b_6.\n$$\n\nNow, prove by induction that $c_n < b_n$ for any $n \\geq 7$. Assume the statement holds for every $m$ with $4 \\leq m < n$. As $c_{n-3} = c_{n-4} \\cdot c_{n-5} \\cdot \\dots \\cdot c_1$, we get $c_n = c_{n-1}c_{n-2}c_{n-3}c_{n-4}\\dots c_1 = c_{n-1}c_{n-2}c_{n-3}^2 < b_{n-1}b_{n-2}b_{n-3}^2 = b_n$, as desired.\n\nHence $a_n < c_n < b_n$ for any $n \\geq 4$.\n\nAlternatively, by induction, $a_n = 5^{2^{n-3}}$ for any $n \\geq 3$.\n\nAlso, $c_n = 2^{2^{n-4}} 3^{2^{n-4}} 5^{2^{n-4}}$ for any $n \\geq 4$.\n\nFor $b_n$, the powers of $2$, $3$, and $5$ in $b_n$ grow even faster (see table below):\n\n\n\nThus, $a_{1000} < c_{1000} < b_{1000}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20391,
"subject": "Mathematics (Olympiad)",
"question": "A positive integer $n$ is called _divisor primary_ if for every positive divisor $d$ of $n$, at least one of the numbers $d-1$ and $d+1$ is prime. For example, $8$ is divisor primary, because its positive divisors $1$, $2$, $4$, and $8$ each differ by $1$ from a prime number ($2$, $3$, $5$, and $7$, respectively), while $9$ is not divisor primary, because the divisor $9$ does not differ by $1$ from a prime number (both $8$ and $10$ are composite).\n\nDetermine the largest divisor primary number.",
"options": [],
"answer": "See solution",
"solution": "Suppose $n$ is divisor primary. Then $n$ cannot have an odd divisor $d \\ge 5$. Indeed, for such a divisor, both $d-1$ and $d+1$ are even. Because $d-1 > 2$, these are both composite numbers and that would contradict the fact that $n$ is divisor primary. The odd divisors $1$ and $3$ can occur, because the integer $3$ itself is divisor primary.\n\nBecause of the unique factorisation in primes, the integer $n$ can now only have some factors $2$ and at most one factor $3$. The number $2^6 = 64$ and all its multiples are not divisor primary, because both $63 = 7 \\cdot 9$ and $65 = 5 \\cdot 13$ are not prime. Hence, a divisor primary number has at most five factors $2$. Therefore, the largest possible number that could still be divisor primary is $3 \\cdot 2^5 = 96$.\n\nWe now check that $96$ is indeed divisor primary: its divisors are $1$, $2$, $3$, $4$, $6$, $8$, $12$, $16$, $24$, $32$, $48$, and $96$, and these numbers are next to $2$, $3$, $2$, $3$, $5$, $7$, $11$, $17$, $23$, $31$, $47$, and $97$, which are all prime. Therefore, the largest divisor primary number is $96$. $\\square$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20392,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be positive real numbers. Prove that\n\n$$\n\\frac{a}{b+c} + \\frac{b}{c+a} + \\frac{c}{a+b} \\ge \\frac{3}{2}.\n$$",
"options": [],
"answer": "See solution",
"solution": "We have\n\n$$\n\\begin{align*}\n\\frac{a}{b+c} + \\frac{b}{c+a} + \\frac{c}{a+b} &\\ge \\frac{3}{2} \\\\\n\\Leftrightarrow \\frac{a+b+c}{b+c} + \\frac{a+b+c}{c+a} + \\frac{a+b+c}{a+b} &\\ge \\frac{9}{2} \\\\\n\\Leftrightarrow (a+b+c) \\left( \\frac{1}{b+c} + \\frac{1}{c+a} + \\frac{1}{a+b} \\right) &\\ge \\frac{9}{2} \\\\\n\\Leftrightarrow \\left[ (b+c) + (c+a) + (a+b) \\right] \\left( \\frac{1}{b+c} + \\frac{1}{c+a} + \\frac{1}{a+b} \\right) &\\ge 9.\n\\end{align*}\n$$\n\nThis is true by the Cauchy-Schwarz inequality. Equality holds when $a = b = c$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20393,
"subject": "Mathematics (Olympiad)",
"question": "Given a triangle $ABC$, where $O$ is its circumcenter, $M$ is the midpoint of $BC$, and $W$ is the point of the second intersection of the angle bisector of $C$ with the circumcircle. The line parallel to $BC$ that passes through $W$ intersects $AB$ at point $K$, so that $BK = BO$. Find the angle $WMB$.",
"options": [],
"answer": "See solution",
"solution": "**Answer:** $45^\\circ$.\n\n**Solution.** Let the line that passes through $W$ parallel to $AB$ intersect\n\n\n\nFig. 6\n\nthe line $BC$ at point $T$ (Fig. 6). Then, $KWTB$ is a parallelogram and:\n$$WT = BK = BO = WO.$$\n\nNotice that $WO \\perp AB$, since $\\triangle ABO$ is isosceles, and $CW$ is a bisector of $\\angle BCA$, thus $\\angle OWT = 90^\\circ$. It is also clear that $\\angle OMT = 90^\\circ$, thus, $OWTM$ is inscribed with diameter $OT$. Therefore, since $WT = WO$, $MW$ is a bisector of $\\angle OMT$, hence $\\angle WMT = 45^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20394,
"subject": "Mathematics (Olympiad)",
"question": "Each student in a class has a finite number of cards. Each card has a number on it from the interval $[0,1]$. Find the smallest possible constant $c > 0$ such that, regardless of how the cards are distributed among the students:\n\nEach student whose total sum of card numbers is less than $1000$ can divide their cards into $100$ boxes so that the sum of the card numbers in each box is at most $c$.",
"options": [],
"answer": "See solution",
"solution": "Consider all possible ways to arrange the cards into boxes, and select an arrangement where the largest sum in any box is minimized. If there are multiple such arrangements, choose one where the number of boxes attaining this maximum is minimized.\n\nLet the sums in the boxes be $10 + x_1 \\geq 10 + x_2 \\geq \\dots \\geq 10 + x_{100}$. Since the total sum is at most $1000$,\n\n$$\n10 + x_1 + 10 + x_2 + \\dots + 10 + x_{100} \\leq 1000 \\implies x_1 + x_2 + \\dots + x_{100} \\leq 0.\n$$\n\nSince $x_{100}$ is the smallest, $x_1 + 99x_{100} \\leq x_1 + x_2 + \\dots + x_{100} \\leq 0$. Suppose, for contradiction, that $x_1 > \\frac{90}{91}$.\n\nSince the first box has sum greater than $10$, it contains at least $11$ cards (since each card is at most $1$). Thus, there is a card in the first box with value at most $\\frac{10 + x_1}{11}$.\n\nRemove this card from the first box and place it in the $100$th box. The $100$th box must then have sum at least $10 + x_1$, otherwise we would have a configuration with a smaller maximal sum. However, after the move, the $100$th box has sum at most\n\n$$\n10 + x_{100} + \\frac{10 + x_1}{11} \\leq 10 - \\frac{x_1}{99} + \\frac{10 + x_1}{11} = 10 + x_1 + \\frac{90 - 91x_1}{99} < 10 + x_1,\n$$\n\nwhich is a contradiction.\n\nNow, we show that $c = 11 - 11a$ with $1 > a > \\frac{1}{1001}$ does not work. Take $r \\in [\\frac{1}{1001}, a)$ and let $n = \\lfloor \\frac{1000}{1 - r} \\rfloor$. Since $r \\geq \\frac{1}{1001}$, $n \\geq 1001$. Take $n$ cards each of value $1 - r$. Their sum is $n(1 - r) \\leq 1000$. No matter how we distribute them into $100$ boxes, since $n \\geq 1001$, at least one box contains $11$ cards. But $11(1 - r) > 11 - 11a$, so $c = 11 - 11a$ does not work.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20395,
"subject": "Mathematics (Olympiad)",
"question": "Given a right square pyramid $P$-$ABCD$ with $\\angle APC = 60^\\circ$, as shown in the figure, prove that the cosine of the plane angle of the dihedral angle $A$-$PB$-$C$ is ( ):\n\n(A) $\\frac{1}{7}$\n\n(B) $-\\frac{1}{7}$\n\n(C) $\\frac{1}{2}$\n\n(D) $-\\frac{1}{2}$",
"options": [],
"answer": "See solution",
"solution": "On $PAB$, draw $AM \\perp PB$ with $M$ as the foot of the perpendicular, connecting $CM$ and $AC$, as seen in the figure. Then $\\angle AMC$ is the plane angle of the dihedral angle $A$-$PB$-$C$. Assume $AB = 2$. Then $PA = AC = 2\\sqrt{2}$, and the vertical height of $\\triangle PAB$ with $AB$ as base is $\\sqrt{7}$. So $2 \\times \\sqrt{7} = AM \\cdot 2\\sqrt{2}$, which means $AM = \\sqrt{\\frac{7}{2}} = CM$. By the Cosine Rule:\n\n$$\n\\cos \\angle AMC = \\frac{AM^2 + CM^2 - AC^2}{2 \\cdot AM \\cdot CM} = -\\frac{1}{7}.\n$$\n\n**Answer:** B.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20396,
"subject": "Mathematics (Olympiad)",
"question": "Let $d_1 < d_2 < d_3$ be proper divisors of a positive integer $n$ such that $d_1 + d_2 + d_3 = 1001$. Find the smallest possible value of $n$.",
"options": [],
"answer": "See solution",
"solution": "Let $d_1 < d_2 < d_3$ be the divisors given in the condition, so $1 < d_1 < d_2 < d_3 < n$. Since $d_1, d_2,$ and $d_3$ are divisors of $n$, there exist positive integers $a, b,$ and $c$ such that $d_1 a = d_2 b = d_3 c = n$. Clearly, $1 < c < b < a < n$, so $c \\geq 2$, $b \\geq 3$, $a \\geq 4$. The equation $d_1 + d_2 + d_3 = 1001$ can be rewritten as:\n\n$$\n1001 = \\frac{n}{a} + \\frac{n}{b} + \\frac{n}{c} = n \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) \\leq n \\left( \\frac{1}{4} + \\frac{1}{3} + \\frac{1}{2} \\right) = \\frac{13n}{12}\n$$\n\nSo $12 \\cdot 1001 \\leq 13n$, i.e., $n \\geq 924$. On the other hand, $n = 924$ satisfies the condition, since 924 has proper divisors 462, 308, and 231, and $462 + 308 + 231 = 1001$. Thus, the smallest possible value of $n$ is $\\boxed{924}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20397,
"subject": "Mathematics (Olympiad)",
"question": "All cards are lying on the table with the yellow side facing up. Sil tries to discover what is on the blue side, without turning over the cards. For example, cards with a 6 on the yellow side have a 5 on the blue side, because the yellow expression $2 \\times 3 = 6$ must have $2+3=5$ on the blue back. Cards with a 20 on the yellow side have a 9 on the blue back side, because the yellow expression $2 \\times 2 \\times 5 = 20$ becomes $2+2+5=9$ in blue. The back of a yellow card containing a fraction, for example $\\frac{5}{3}$, can be determined using $\\frac{5}{3} \\times 3 = 5$, which becomes $2+3=5$ when flipped; hence on the blue side is a 2.\n\nFor which of the following numbers on the yellow side will there be a negative number on the blue side?\n\nA) $\\frac{9}{8}$\nB) $\\frac{25}{27}$\nC) $\\frac{32}{27}$\nD) $\\frac{64}{81}$\nE) $\\frac{128}{125}$",
"options": [],
"answer": "See solution",
"solution": "E) $\\frac{128}{125}$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20398,
"subject": "Mathematics (Olympiad)",
"question": "Let $O$ be an interior point of the triangle $ABC$. Prove that there exist positive integers $p, q$, and $r$ such that\n$$\n|p \\cdot \\vec{OA} + q \\cdot \\vec{OB} + r \\cdot \\vec{OC}| < \\frac{1}{2007}.\n$$",
"options": [],
"answer": "See solution",
"solution": "It is well-known that there are positive real numbers $\\beta, \\gamma$ such that\n$$\n\\vec{OA} + \\beta \\vec{OB} + \\gamma \\vec{OC} = \\vec{0}.\n$$\nSo for any positive integer $k$, we have\n$$\nk \\vec{OA} + k\\beta \\vec{OB} + k\\gamma \\vec{OC} = \\vec{0}.\n$$\nLet $m(k) = [k\\beta]$, $n(k) = [k\\gamma]$, where $[x]$ is the greatest integer less than or equal to $x$, and $\\{x\\} = x - [x]$.\nAssume $T$ is an integer larger than $\\max\\left\\{\\frac{1}{\\beta}, \\frac{1}{\\gamma}\\right\\}$. Then the sequences $\\{m(kT) \\mid k = 1, 2, \\dots\\}$ and $\\{n(kT) \\mid k = 1, 2, \\dots\\}$ are increasing, and\n$$\n\\begin{aligned}\n& \\left|kT \\vec{OA} + m(kT) \\vec{OB} + n(kT) \\vec{OC}\\right| \\\\\n&= \\left|-\\{kT\\beta\\} \\vec{OB} - \\{kT\\gamma\\} \\vec{OC}\\right| \\\\\n&\\leq |\\vec{OB}| \\cdot \\{kT\\beta\\} + |\\vec{OC}| \\cdot \\{kT\\gamma\\} \\\\\n&\\leq |\\vec{OB}| + |\\vec{OC}|.\n\\end{aligned}\n$$\nThis shows there exist infinitely many vectors of the form\n$$\nkT \\vec{OA} + m(kT) \\vec{OB} + n(kT) \\vec{OC},\n$$\nwhose endpoints lie in a circle centered at $O$ of radius $|\\vec{OB}| + |\\vec{OC}|$. By the pigeonhole principle, there are two such vectors whose endpoints are less than $\\frac{1}{2007}$ apart. Thus, there exist integers $k_1 < k_2$ such that\n$$\n\\begin{aligned}\n& \\left| (k_2 T \\vec{OA} + m(k_2 T) \\vec{OB} + n(k_2 T) \\vec{OC}) \\\\\n&\\quad - (k_1 T \\vec{OA} + m(k_1 T) \\vec{OB} + n(k_1 T) \\vec{OC}) \\right| \\\\\n&< \\frac{1}{2007}.\n\\end{aligned}\n$$\nLet $p = (k_2 - k_1)T$, $q = m(k_2T) - m(k_1T)$, $r = n(k_2T) - n(k_1T)$. Then $p, q, r$ are positive integers, and\n$$\n|p \\vec{OA} + q \\vec{OB} + r \\vec{OC}| < \\frac{1}{2007}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20399,
"subject": "Mathematics (Olympiad)",
"question": "A convex polygon of $n$ sides is divided into triangles by diagonals which do not intersect inside the polygon. The triangles are painted black and white so that any two triangles with a common side are painted in different colors. For each $n$, determine the maximal difference between the number of black and the number of white triangles.",
"options": [],
"answer": "See solution",
"solution": "Construct a graph by representing each triangle by a vertex of a graph. Two vertices are joined by an edge if their corresponding triangles share a common side.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20400,
"subject": "Mathematics (Olympiad)",
"question": "考慮一排寶寶，每個寶寶有若干片餅乾。設 $M$ 為某個寶寶手上的餅乾數，且至少有 3 個寶寶沒有餅乾。設在 $M$ 左右兩側最靠近的兩位寶寶，其間的餅乾分佈為：\n\n$$\n0, \\overbrace{1, 1, \\dots, 1}^{x}, M, \\overbrace{1, 1, \\dots, 1}^{y}, 0.\n$$\n\n證明：對於任何 $(x, y) \\in \\mathbb{Z}_{\\ge 0} \\times \\mathbb{Z}_{\\ge 0}$，都可以讓 $M$ 傳遞一片餅乾到至少一個 $0$ 的位置，從而使分佈降成 $(M-1)$ 或 $(M-2)$-uniform。\n\n進一步，對於滿足條件 (1) 的起始配置，證明可以讓所有寶寶達到 1-uniform（每人 1 片餅乾）的狀態。\n\n\n\n",
"options": [],
"answer": "See solution",
"solution": "我們用數學歸納法證明上述命題。\n\n不失一般性，假設 $x \\ge y$。\n\n- **(0,0)：** 只需對 $M$ 操作，讓 $0, M, 0$ 變成 $1, M-2, 1$，即得證。\n- **$(x,0)$：** 先對 $M$ 操作，然後從 $M$ 左邊的寶寶開始依次向左操作一次，直到左邊的 $0$ 變成 $1$，此時分佈為：\n $$\n 1, 0, \\overbrace{1, 1, \\dots, 1}^{x-1}, M-1, 1\n $$\n 故為 $(M-1)$-uniform，得證。\n- **$(x, y)$：** 假設命題對所有 $y < N$ 成立。當 $y = N$ 時，先對 $M$ 操作，然後從 $M$ 左邊的寶寶開始依次向左操作一次，直到左邊的 $0$ 變成 $1$，再從 $M$ 右邊的寶寶開始依次向右操作一次，直到右邊的 $0$ 變成 $1$。此時分佈為：\n $$\n 1, 0, \\overbrace{1, 1, \\dots, 1}^{x-1}, M, \\overbrace{1, 1, \\dots, 1}^{y-1}, 0, 1.\n $$\n 由歸納假設，對於中間從 $0$ 到 $0$ 的子列，必可讓 $M$ 傳遞一片餅乾到至少一個 $0$ 的位置，故得證。\n\n**引理三：** 對於滿足 (1) 的起始配置，可以讓寶寶們 1-uniform，從而滿足題意。\n\n*證明*：由引理二知可以讓寶寶們 1-uniform 或 2-uniform。但對於 2-uniform，必然存在唯一的一個 $c_j = 0$，從而\n\n$$\n\\sum_{i=1}^{n} ic_{i} \\equiv \\sum_{i=1}^{n-1} ic_{i} + n \\equiv \\sum_{i=1}^{n} i - j \\equiv \\frac{n(n+1)}{2} - j \\mod n\n$$\n\n這與 (1) 不合。故必會將寶寶操作成 1-uniform。證明完畢。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20401,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer such that $n = \\tau(n) + 2023$, where $\\tau(n)$ is the number of positive divisors of $n$. Find all such $n$.\n\nb) Let $k$ be a prime number greater than $6996$. Show that for any positive integer $n$, the equation $\\tau(kn) + 2023 = n$ has no solution.",
"options": [],
"answer": "See solution",
"solution": "We have $n = \\tau(n) + 2023 \\ge 2025$, so $2025 \\le n \\le 2114$. 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 becomes:\n\n$$\n(e_1 + 1)(e_2 + 1)\\cdots (e_k + 1) + 2023 = p_1^{e_1} p_2^{e_2} \\cdots p_k^{e_k}\n$$\n\nIf all $e_i$ are even, $n$ is a perfect square, but $n = 2025$ does not satisfy the equation. Thus, at least one $e_i$ is odd, so all $p_i$ are odd primes. If $k \\ge 5$, then $n \\ge 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 > 2114$, a contradiction. So $k \\le 4$.\n\n**Case 1:** $k = 1$. $n = p_1^{e_1} \\le 2114$, so $e_1 \\le 6$. No $p_1$ and $e_1$ satisfy the equation.\n\n**Case 2:** $k = 2$. $n = p_1^{e_1} p_2^{e_2} \\le 2114$, so $e_1 + e_2 \\le 6$. $\\tau(n) \\le 9$. No $n$ satisfies the equation.\n\n**Case 3:** $k = 3$. $n = p_1^{e_1} p_2^{e_2} p_3^{e_3} \\le 2114$, so $e_1 + e_2 + e_3 \\le 5$. Possible $(e_1, e_2, e_3)$ are permutations of $(1,1,1)$, $(1,1,2)$, $(1,1,3)$, $(1,2,2)$. $\\tau(n) \\in \\{8,12,16,18\\}$. No $n$ satisfies the equation.\n\n**Case 4:** $k = 4$. $n = p_1^{e_1} p_2^{e_2} p_3^{e_3} p_4^{e_4} \\le 2114$. If any $e_i > 1$, $n > 2114$. So $e_1 = e_2 = e_3 = e_4 = 1$, $\\tau(n) = 16$. No $n$ satisfies the equation.\n\nThus, the equation $\\tau(n) + 2023 = n$ has no positive integer solution.\n\nb) For any prime $k > 6996$, and any $n$, $\\tau(kn) \\le 2\\tau(n)$. If $n$ is not divisible by $k$, $\\tau(kn) = 2\\tau(n)$. For any solution $n$ to $\\tau(kn) + 2023 = n$, we have:\n\n$$\nn = \\tau(kn) + 2023 \\le 2\\tau(n) + 2023 \\le 4\\sqrt{n} + 2023.\n$$\n\nThis inequality cannot hold for large $n$, so there are no solutions for $k > 6996$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20402,
"subject": "Mathematics (Olympiad)",
"question": "Let $n > 10$ be an integer, and let $A_1, A_2, \\dots, A_n$ be distinct points in the plane such that the distances between the points are pairwise different. Define $f_{10}(j, k)$ to be the 10th smallest of the distances from $A_j$ to $A_1, A_2, \\dots, A_k$, excluding $A_j$ if $k \\ge j$. Suppose that for all $j$ and $k$ satisfying $11 \\leq j \\leq k \\leq n$, we have $f_{10}(j, j-1) \\geq f_{10}(k, j-1)$. Prove that $f_{10}(j, n) \\geq \\frac{1}{2} f_{10}(n, n)$ for all $j$ in the range $1 \\leq j \\leq n-1$.",
"options": [],
"answer": "See solution",
"solution": "For every $i$, denote $a_i = f_{10}(i, i-1)$ and $b_i = f_{10}(i, n)$. So, we need to show that $b_n \\leq 2b_i$ for all $i$. Notice that $a_i \\geq b_i$ for all $i$.\n\nTo prove this, choose an arbitrary $i < n$, and let $A_iA_{j_1}, A_iA_{j_2}, \\dots, A_iA_{j_{10}}$ be the ten smallest numbers among the $A_iA_j$ with $j \\neq i$, ordered so that $j_1 < j_2 < \\dots < j_{10}$.\n\nIf $j_{10} < i$, then $i > 10$, and the problem condition yields\n\n$$\nb_i = \\max_{1 \\leq k \\leq 10} A_iA_{j_k} = a_i \\geq a_n \\geq b_n,\n$$\n\nwhich is even stronger than we need.\n\nOtherwise, set $j = j_{10} > i$ (in this case we also have $j_{10} > 10$), and denote $m = b_i = \\max_{1 \\leq k \\leq 10} A_iA_{j_k}$. By the problem condition, we have $a_j \\geq a_n = b_n$. On the other hand, we have\n\n$$a_j \\leq \\max \\left( A_jA_i, \\max_{1 \\leq k \\leq 9} A_jA_{j_k} \\right) \\leq \\max \\left( A_jA_i, \\max_{1 \\leq k \\leq 9} (A_jA_i + A_iA_{j_k}) \\right) \\leq 2m,$$\nas $A_jA_i, A_iA_{j_k} \\leq m$. So $b_n \\leq a_j \\leq 2m = 2b_i$, as desired.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20403,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integer pairs $\\{a, b\\}$ such that\n$$\n14\\varphi^2(a) - \\varphi(ab) + 22\\varphi^2(b) = a^2 + b^2,\n$$\nwhere $\\varphi(n)$ is the number of positive integers less than $n$ and relatively prime to $n$.",
"options": [],
"answer": "See solution",
"solution": "We denote by $p^{\\alpha}\\|b$ that the highest power of $p$ dividing $b$ is $p^{\\alpha}$, i.e., $p^{\\alpha}|b$ but $p^{\\alpha+1} \\nmid b$.\n\nIf $\\{a, b\\}$ is a solution, and $p$ is a prime with $p^2|a, p^2|b$, then $\\{\\frac{a}{p}, \\frac{b}{p}\\}$ is also a solution. Thus, for each prime $p$, at least one of $a, b$ is either coprime to $p$ or divisible exactly by $p$.\n\nProperties of Euler's function:\n- (a) For $x > 2$, $\\varphi(x)$ is even.\n- (b) If $2^{\\alpha}\\|\\varphi(x)$, then $x$ has at most $\\alpha$ distinct odd prime factors.\n\nIf $a, b \\leq 3$, the left side of the equation is larger than the right, so no solutions exist. Thus, at least one of $a, b$ exceeds 3, and $2|\\varphi(ab)$, so the left side is even and $a, b$ have the same parity.\n\n**Case 1: Both $a, b$ are odd.**\nThe right side is congruent to 2 mod 4. If $ab$ has two or more distinct prime factors, then $4|\\varphi(ab)$, and one of $\\varphi(a), \\varphi(b)$ must be 1, so one of $a, b$ is 1. Suppose $b = 1$, then the equation becomes\n$$\n14\\varphi^2(a) - \\varphi(a) + 22 = a^2 + 1.\n$$\nIf $8|\\varphi(a)$, then $22 \\equiv 2 \\pmod{8}$, contradiction. Thus, $a$ has at most two distinct prime factors, say $p, q$, but then\n$$\n13\\varphi^2(a) = 13\\left(\\frac{p-1}{p} \\cdot \\frac{q-1}{q}\\right)^2 a^2 \\geq 3a^2 > a^2,\n$$\ncontradicting the equation. So no solutions in this case.\n\n**Case 2: Both $a, b$ are even.**\nLet $a = 2c$, $b = 2d$. If $c$ or $d = 1$, the equation has no integer solutions. Suppose $c, d > 1$.\n\nIf $c, d$ have different parities, $4\\|(a^2 + b^2)$, but $8|14\\varphi^2(a)+22\\varphi^2(b)$, and $4\\|\\varphi(ab)$. Since $c$ or $d$ is even, $8|ab$. Let $ab = 2^\\delta m$, $\\delta \\geq 3$, $m$ odd. Then $\\varphi(ab) = 2^{\\delta-1}\\varphi(m)$ is divisible exactly by 4, so $\\varphi(m)$ is odd, $m=1, \\delta=3$, $ab=8$, but $ab = 4cd \\geq 16$, contradiction.\n\nIf $c, d$ have the same parity (both odd),\n$$\n7\\varphi^2(c) - \\varphi(cd) + 11\\varphi^2(d) = 2(c^2 + d^2),\n$$\nwith $c, d > 2$ odd. If $2|\\varphi(c)$, $2|\\varphi(d)$,\n$$\n7\\varphi^2(c) \\geq 7 \\left(\\frac{2}{3}\\right)^2 c^2 > 2.5c^2,\n$$\n$$\n11\\varphi^2(d) \\geq 11 \\left(\\frac{2}{3}\\right)^2 d^2 > 2.5d^2.\n$$\nSince $\\varphi(cd) < cd \\leq 0.5c^2 + 0.5d^2$, the left side is always larger, so $4|\\varphi(c)$ or $4|\\varphi(d)$.\n\nAssume $4|\\varphi(c)$. If $16|\\varphi(cd)$, modulo 16 gives $2(c^2 + d^2) \\equiv 11\\varphi^2(d) \\pmod{16}$, which implies $2|\\varphi(d)$, but $11\\varphi^2(d) \\equiv 12 \\pmod{16}$, contradiction. So $16 \\nmid \\varphi(cd)$. Similar for $4|\\varphi(d)$.\n\nThus, either $4\\|\\varphi(cd)$ or $8\\|\\varphi(cd)$. In any case, $cd$ has at most three distinct prime factors, and if three, all are $4k+3$ primes.\n\n- If $cd$ is a prime power $p$, $4|\\varphi(cd)$ implies $p=4k+1$, $p|c, p|d$, but then $7\\varphi^2(c) > 2.5c^2$, $11\\varphi^2(d) > 2.5d^2$, contradiction.\n- If $cd$ has two distinct prime factors $p, q$, $p$ or $q$ must be 3, else contradiction. If $c, d$ not coprime, and $pq|c, pq|d$, then $p=3$, $q=7$, but this leads to contradiction as above.\n\nIf $c, d$ share a common factor $p$, and $pq|d$, then $c$ has only one prime factor, contradiction. Thus $pq|c$, $p|d$, $(q, c)=1$, $p=3$, $q=5$ or 7.\n\n- If $q=5$, $3\\nmid(q-1)$. If $3^2|a$ or $3^2|b$, contradiction. So $3\\|a, 3\\|b$, $d=3$. If $5^2|a$, modulo 5 gives $11\\varphi^2(b) = 2b^2 \\pmod{5}$, but 2 is not a quadratic residue mod 5, contradiction. So $5\\|a$, $a=15$. Checking, $(a, b) = (15, 3)$ solves the equation, so $(30, 6)$ is a solution to the original equation.\n\n**Final answer:**\nThe only positive integer solution is $(a, b) = (30, 6)$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20404,
"subject": "Mathematics (Olympiad)",
"question": "Natural numbers $a, b, c, d$ satisfy\n$$\n0 < |ad - bc| < \\min\\{c, d\\}.\n$$\nProve that for any coprime natural numbers $x, y > 1$, the number $x^a + y^b$ is not divisible by $x^c + y^d$.",
"options": [],
"answer": "See solution",
"solution": "We will prove this by contradiction.\nLet $s = x^c + y^d$. Then $x^c \\equiv -y^d \\pmod{s}$, and if $x^a + y^b$ is divisible by $s$, then $x^a \\equiv -y^b \\pmod{s}$.\n\nThis implies:\n$$\nx^{ad} \\equiv (-1)^d y^{bd} \\pmod{s}\n$$\nand\n$$\nx^{bc} \\equiv (-1)^b y^{bd} \\pmod{s}.\n$$\nThus,\n$$\n(-1)^d x^{ad} \\equiv y^{bd} \\equiv (-1)^b x^{bc} \\pmod{s} \\implies x^{ad} \\equiv (-1)^{b-d} x^{bc} \\pmod{s}.\n$$\nSince $x$ and $s$ are coprime, we can divide by $x^{\\min\\{ad, bc\\}}$ to get:\n$$\nx^{\\max\\{ad, bc\\} - \\min\\{ad, bc\\}} \\equiv (-1)^{b-d} \\pmod{s}.\n$$\nSimilarly, for $y$:\n$$\ny^{\\max\\{ad, bc\\} - \\min\\{ad, bc\\}} \\equiv (-1)^{a-c} \\pmod{s}.\n$$\nTherefore, $y^{|ad-bc|} \\pm x^{|ad-bc|}$ is divisible by $s$.\n\nBut since $0 < |ad-bc| < \\min\\{c, d\\}$, we have:\n$$\n|y^{|ad-bc|} - x^{|ad-bc|}| < y^{|ad-bc|} + x^{|ad-bc|} < y^d + x^c = s.\n$$\nSo $|y^{|ad-bc|} \\pm x^{|ad-bc|}|$ can only be divisible by $s$ if it is zero, which is impossible since $x$ and $y$ are coprime. Thus, $x^a + y^b$ is not divisible by $x^c + y^d$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20405,
"subject": "Mathematics (Olympiad)",
"question": "證明存在一個正實數 $c > 0$，使得有無窮多組正整數 $(n, N)$ 滿足下列條件：\n\n(a) $N > c \\cdot 2^k n \\log(n)$，其中 $k$ 為 $N$ 的質因數個數；\n\n(b) 對 $N$ 的任一質因數 $p$，設 $e$ 為 $N$ 的標準質因數分解中 $p$ 的次方數，皆有 $n$ 被 $\\varphi(p^e)$ 整除，其中 $\\varphi$ 為歐拉函數，即 $\\varphi(s)$ 為不超過 $s$ 且與 $s$ 互質的正整數個數。",
"options": [],
"answer": "See solution",
"solution": "我們構造無窮多組滿足條件的 $(n, N)$。固定整數 $x \\ge 3$，令 $N = \\prod_{p \\le x} p$，即所有不超過 $x$ 的質數之積。設 $n = 2^{2-\\pi(x)} \\prod_{p \\le x} (p-1)$，其中 $\\pi(x)$ 表示不超過 $x$ 的質數個數。這樣的 $n$ 顯然是偶數，且 $(n, N)$ 滿足條件 (b)。\n\n估算 $N$ 的大小：\n\n$$\n\\frac{4N}{2^{\\pi(x)}n} = \\prod_{p \\le x} \\frac{p}{p-1} = \\prod_{p \\le x} \\left(1 + p^{-1} + p^{-2} + \\dots\\right) > \\sum_{r=1}^{x!} \\frac{1}{r} = \\log(x!) + O(1)\n$$\n\n利用調和數的近似 $\\sum_{r=1}^{m} \\frac{1}{r} = \\log(m) + O(1)$。\n由定義 $x! \\ge N \\ge n$，整理得：\n\n$$\nN > \\frac{1}{4} \\cdot 2^{\\pi(x)} n \\log(n) + O(2^{\\pi(x)} n)\n$$\n\n因此，若取 $c < 1/4$，條件 (a) 亦成立，證畢。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20406,
"subject": "Mathematics (Olympiad)",
"question": "Все веса в решении будут измеряться в граммах. Назовём кусок яблока (или само яблоко) *большим*, если его вес не меньше $25$.\n\nДокажите индукцией по $n$, что $n$ больших яблок суммарного веса $100n$ можно разрезать на большие куски и раздать $n$ детям поровну.",
"options": [],
"answer": "See solution",
"solution": "База при $n = 1$ очевидна. Пусть $n > 1$. Рассмотрим два самых тяжёлых яблока; пусть их веса $a \\ge b$. Заметим, что $a + b \\ge 200$ (иначе средний вес одного яблока будет меньше, чем $200/2 = 100$). Выкинем эти два яблока из набора и добавим в него яблоко веса $c = a + b - 100 \\ge 100$. По предположению индукции, полученный набор можно разрезать на большие куски и раздать $n-1$ детям поровну. Если при этом какой-то кусок нового яблока оказался больше $50$, разрежем его на два больших куска. Через несколько таких разрезаний мы придём к ситуации, когда новое яблоко разделено на куски весов $c_1, c_2, \\dots, c_k$, не превосходящих $50$. Обозначим $s_d = c_1 + \\dots + c_d$ при $d = 1, 2, \\dots, k$ и положим $s_0 = 0$.\n\nПокажем теперь, как разрезать исходный набор. Все яблоки, кроме $a$ и $b$, разрежем так же, как и в новом наборе. Заметим, что $a \\ge 200/2 = 100$. Обозначим через $t$ минимальный индекс такой, что $a - s_t \\le 75$ и отрежем от $a$ куски $c_1, \\dots, c_t$, а от $b$ — куски $c_{t+1}, \\dots, c_k$. Заметим, что $a - s_{t-1} > 75$, поэтому от $a$ остался кусок $a' = a - s_t = (a - s_{t-1}) - c_t$ такой, что $75 \\ge a' > 75 - c_t \\ge 25$. От $b$ же остался кусок $b'$ такой, что $a' + b' = a + b - c = 100$, поэтому $25 \\le b' \\le 75$. Итак, можно $a'$ и $b'$ отдать одному ребёнку, а остальные куски распределить между остальными детьми так же, как это делалось в новом наборе. Утверждение доказано.\n\n**Замечание.** В доказанном общем утверждении число $25$ нельзя заменить на большее, не зависящее от $n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20407,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x^2 + x f(y)) = x f(x + y)\n$$\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$$\nf(x^2 + x f(\\alpha)) = x f(x + \\alpha),\n$$\nfor all $x \\in \\mathbb{R}$. Therefore,\n$$\nx f(x + \\alpha) = f(x^2) = x f(x),\n$$\nfor all $x \\in \\mathbb{R}$. For $x \\neq 0$, we get $f(x) = f(x + \\alpha)$. Note that this is also valid for $x = 0$.\n\nSuppose $f(1) = 0$. Then\n$$\nf(1 + f(x)) = f(x + 1),\n$$\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$$\nf(s^2 + s f(\\lambda - s)) = s f(s + \\lambda - s) = s f(\\lambda) = t,\n$$\nwhich shows that $f$ is onto. Hence there exists $x_0$ such that $f(x_0) = 1$. This gives\n$$\n1 = f(x_0) = f(f(x_0)) = f(1) = 0,\n$$\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$$\n\\begin{aligned}\nf(\\alpha^2 + 1) &= f((1 + \\alpha)^2 + (1 + \\alpha) f(\\alpha)) = (\\alpha + 1) f(1 + 2\\alpha) = (\\alpha + 1) f(1), \\\\\nf(\\alpha^2 + 1) &= f((1 - \\alpha)^2 + (1 - \\alpha) f(\\alpha)) = (1 - \\alpha) f(1).\n\\end{aligned}\n$$\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": 20408,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_n$ be the maximum possible number of sections of a set of $n$ points in the plane.\n\nWhat is the maximum possible number of sections of a set of $n$ points in the plane? Express your answer in terms of $n$.",
"options": [],
"answer": "See solution",
"solution": "Let $S$ be a set of $n+1$ points in the plane and let us consider one of these points, call it $T$, on the convex hull of that set. Note that each section of $S$ restricts to a section of the set $S \\setminus \\{T\\}$.\n\nLet $\\{A, B\\}$ and $\\{C, D\\}$ be two different sections of the set $S$ that restrict to the same section of the set $S \\setminus \\{T\\}$. This means that $\\{A \\setminus \\{T\\}, B \\setminus \\{T\\}\\} = \\{C \\setminus \\{T\\}, D \\setminus \\{T\\}\\}$. Furthermore, it means, without loss of generality, that the sets $A$ and $C$ are the same and the sets $B$ and $D$ are the same, up to the point $T$. Hence, for the section $\\{A \\setminus \\{T\\}, B \\setminus \\{T\\}\\}$ of the set $S \\setminus \\{T\\}$ we can take a line passing through the point $T$ such that all the points of the set $A \\setminus \\{T\\}$ are on one side of the line, while all the points of the set $B \\setminus \\{T\\}$ are on the other.\n\nLet us consider all the lines passing through the point $T$ and none of the points of the set $S \\setminus \\{T\\}$. Two such lines could give two different sections of the set $S \\setminus \\{T\\}$ only if there is at least one point of the set $S \\setminus \\{T\\}$ between them. Hence, the number of different sections of the set $S \\setminus \\{T\\}$ corresponding to a line passing through $T$ is at most $n$, i.e. as many as there are points in the set $S \\setminus \\{T\\}$.\n\nFinally, we note that\n\n$$\na_{n+1} \\le n + a_n \\le n + (n-1) + a_{n-1} \\le \\dots \\le n + (n-1) + \\dots + 2 + 1 + a_1 = \\binom{n+1}{2} + 1.$$ \n\nSo, the set of $n$ points in the plane can have at most $\\binom{n}{2} + 1$ sections. That number is achievable, e.g. in the case of a regular $n$-gon.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20409,
"subject": "Mathematics (Olympiad)",
"question": "Determine all finite sets $S$ of points in the plane that have the following closure property: If $x, x', y, y' \\in S$ and the closed line segments $xy$ and $x'y'$ meet at a single point $z$, then $z \\in S$.",
"options": [],
"answer": "See solution",
"solution": "It is easiest to describe an $n$-element planar set $S$ with the closure property in terms of the number $m$ of vertices of its convex hull $[S]$.\n\nWe will prove that $m \\leq 4$ and:\n\n1. If $m = 1$, then $n = 1$.\n2. If $m = 2$, then the points of $S$ are all collinear.\n3. If $m = 3$, then either $n = 6$ and the configuration is as in Figure 1, or there is a line $\\ell$ through a vertex of $[S]$ such that all non-vertices of $S$ lie on $\\ell$.\n4. If $m = 4$, then $n \\geq 5$; a fifth point is the intersection of the diagonals of $[S]$, and all other points of $S$ lie on the same diagonal.\n\n\n\nAny set of points satisfying one of (1) through (4) is a geometrically closed finite planar set.\n\nTo show that $m \\leq 4$, note that among any five points in $S$ in convex position, no four may be in strictly convex position. Otherwise, by the closure property, the convex hull of a 5-point configuration in $S$ in convex position, containing four points in strictly convex position, would strictly contain the convex hull of another such 5-point configuration, and $S$ would be infinite (a minimality argument would work as well)—see Figure 2.\n\n\n\nThus, $m \\leq 4$. Cases (1) and (2) are obvious.\n\nIf $m = 3$, then any two points of $S$ that are not vertices of $[S]$ lie on a line through a vertex. Otherwise, $S$ would contain a forbidden 5-point configuration—Figure 3 shows one such if $x$ and $y$ were two points of $S$ such that the line $xy$ passes through no vertex of $[S]$. This implies (3).\n\n\n\nIf $m = 4$, then any two points of $S$ that are not vertices of $[S]$ lie on a common diagonal. Otherwise, $S$ would again contain a forbidden 5-point configuration—Figure 4 shows one such if: (a) a point $x$ of $S$ were on neither diagonal; and (b) two points $x$ and $y$ of $S$ were not on the same diagonal (in this case, two non-vertices of $S$, each on one diagonal, would yield a third on neither diagonal). This establishes (4) and completes the proof.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20410,
"subject": "Mathematics (Olympiad)",
"question": "For an integer $m \\ge 3$, set\n$$\nS(m) = 1 + \\frac{1}{3} + \\dots + \\frac{1}{m}\n$$\n(the fraction $1/m$ does not participate in the sum). Let $n \\ge 3$ and $k \\ge 3$. Compare the numbers $S(nk)$ and $S(n) + S(k)$.",
"options": [],
"answer": "See solution",
"solution": "We show that $S(nk) < S(n) + S(k)$. Cancel the summand of $S(k)$ on both sides of this inequality, then add $\\frac{1}{2}$ to both sides and rearrange. This yields the equivalent inequality\n\n$$\n\\frac{1}{k+1} + \\frac{1}{k+2} + \\dots + \\frac{1}{nk} + \\frac{1}{2} < 1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n}. \\quad (*)\n$$\n\nDivide the first $(n-1)k$ numbers on the left-hand side into $n-1$ sums of $k$ fractions with consecutive denominators: $A_1 = \\frac{1}{k+1} + \\dots + \\frac{1}{2k}$, $A_2 = \\frac{1}{2k+1} + \\dots + \\frac{1}{3k}$, ..., $A_{n-1} = \\frac{1}{(n-1)k+1} + \\dots + \\frac{1}{nk}$.\n\nThen compare $A_j$ with $\\frac{1}{j}$ for $j=1, \\dots, n-1$. Denoting $d_j = \\frac{1}{j} - A_j = \\frac{1}{j} - \\frac{1}{jk+1} - \\frac{1}{jk+2} - \\dots - \\frac{1}{jk+k}$,\n\nwe have\n$$\nd_j = \\left( \\frac{1}{jk} - \\frac{1}{jk+1} \\right) + \\left( \\frac{1}{jk} - \\frac{1}{jk+2} \\right) + \\dots + \\left( \\frac{1}{jk} - \\frac{1}{jk+k} \\right) =\n$$\n$$\n= \\frac{1}{jk(jk+1)} + \\frac{2}{jk(jk+2)} + \\dots + \\frac{k}{jk(jk+k)} >\n$$\n$$\n> \\frac{1+2+\\dots+k}{jk(jk+k)} = \\frac{k+1}{2k} \\cdot \\frac{1}{j(j+1)} > \\frac{1}{2j} - \\frac{1}{2(j+1)}\n$$\n\nIt follows that\n\n$$\n\\left(1+\\frac{1}{2}+\\frac{1}{3}+\\dots+\\frac{1}{n}\\right)-\\left(\\frac{1}{k+1}+\\frac{1}{k+2}+\\dots+\\frac{1}{nk}\\right)=d_1+\\dots+d_{n-1}+\\frac{1}{n} > \\left(\\frac{1}{2}-\\frac{1}{4}\\right)+\\left(\\frac{1}{4}-\\frac{1}{6}\\right)+\\dots+\\left(\\frac{1}{2n-2}-\\frac{1}{2n}\\right)+\\frac{1}{n} = \\frac{1}{2}+\\frac{1}{2n} > \\frac{1}{2}\n$$\n\nThis proves (*), hence $S(nk) < S(n) + S(k)$ holds true.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20411,
"subject": "Mathematics (Olympiad)",
"question": "Given any set $A = \\{a_1, a_2, a_3, a_4\\}$ of four distinct positive integers, let $s_A = a_1 + a_2 + a_3 + a_4$. Let $n_A$ denote the number of pairs $(i, j)$ with $1 \\leq i < j \\leq 4$ for which $a_i + a_j$ divides $s_A$. Find all sets $A$ of four distinct positive integers which achieve the largest possible value of $n_A$.",
"options": [],
"answer": "See solution",
"solution": "Let $A = \\{a_1, a_2, a_3, a_4\\}$ with $a_1 < a_2 < a_3 < a_4$. Since\n\n$$\n\\frac{1}{2}s_A = \\frac{1}{2}(a_1 + a_2 + a_3 + a_4) < a_2 + a_4 < a_3 + a_4 < s_A,\n$$\n\nthe sums $a_2 + a_4$ and $a_3 + a_4$ do not divide $s_A$. Therefore,\n\n$$\nn_A \\leq \\binom{4}{2} - 2 = 4.\n$$\n\nFor example, if $A = \\{1, 5, 7, 11\\}$, then $n_A = 4$. Thus, the largest possible value of $n_A$ is $4$.\n\nNext, we find all such sets $A$ with $n_A = 4$.\n\nWe observe that $a_2 + a_4$ and $a_3 + a_4$ do not divide $s_A$, and\n\n$$\n\\frac{1}{2}s_A \\leq \\max\\{a_1 + a_4, a_2 + a_3\\} < s_A.\n$$\n\nThus, $\\frac{1}{2}s_A = \\max\\{a_1 + a_4, a_2 + a_3\\}$, so $a_1 + a_4 = a_2 + a_3$.\n\nSuppose $a_1 + a_3 \\mid s_A$, so $s_A = k(a_1 + a_3)$ for some integer $k > 2$ (since $a_1 + a_3 < a_2 + a_3$). Also, $2(a_2 + a_3) = s_A = k(a_1 + a_3)$, so $a_2 = \\frac{1}{2}(k a_1 + (k-2)a_3)$. Since $a_2 < a_3$, $k < 4$, so $k = 3$.\n\nTherefore,\n\n$$\n2(a_2 + a_3) = 2(a_1 + a_4) = 3(a_1 + a_3) = s_A,\n$$\n\nwhich gives $a_2 = \\frac{1}{2}(3a_1 + a_3)$ and $a_4 = \\frac{1}{2}(a_1 + 3a_3)$.\n\nNow, let $s_A = l(a_1 + a_2)$ for some integer $l$. Then,\n\n$$\n3(a_1 + a_3) = l\\left(a_1 + \\frac{1}{2}(3a_1 + a_3)\\right),\n$$\n\nwhich simplifies to $(6 - l)a_3 = (5l - 6)a_1$.\n\nSince $a_1 < a_3$, $l = 4$ or $5$.\n\n- If $l = 4$, $a_3 = 7a_1$, so $a_2 = 5a_1$, $a_4 = 11a_1$.\n- If $l = 5$, $a_3 = 19a_1$, so $a_2 = 11a_1$, $a_4 = 29a_1$.\n\nIt is easy to verify that for $l = 4, 5$, 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": 20412,
"subject": "Mathematics (Olympiad)",
"question": "The sequence $\\{a_n\\}$ of positive integers is defined by $a_1 = 1$ and $a_n = (a_{n-1} + 1)^2$ for $n \\geq 2$. Let $p$ be an odd prime. Prove that $a_{2p} - a_p$ has at least $p$ pairwise distinct prime divisors.",
"options": [],
"answer": "See solution",
"solution": "Let $b_0 = 1$ and define $b_n = a_n + 1$ for $n \\geq 1$. Clearly, $b_{n+1} = b_n^2 + 1$ for $n \\geq 0$. If $n > m \\geq 1$, then\n\n$$\na_n - a_m = b_n - b_m = (b_{n-m} - b_0) \\prod_{k=0}^{m-1} (b_{n-m+k} + b_k). \\quad (*)\n$$\n\nIn particular,\n\n$$\na_{2p} - a_p = b_{2p} - b_p = (b_p - b_0) \\prod_{k=0}^{p-1} (b_{p+k} + b_k).\n$$\n\nAs the parities of the $b_n$ alternate and $p$ is odd, each of the above factors is odd. We will prove that the $p$ factors $b_{p+k} + b_k$ are pairwise coprime, which implies the conclusion.\n\nLet $0 \\leq k < \\ell \\leq p-1$ and let $d = \\gcd(b_{p+k} + b_k, b_{p+\\ell} + b_\\ell)$. By the above, $d$ is odd. Suppose, for contradiction, that $d > 1$ and let $\\equiv$ denote congruence modulo $d$. Then $b_{p+k} \\equiv -b_k$, so $b_{p+k+1} = b_{p+k}^2 + 1 \\equiv b_k^2 + 1 = b_{k+1}$. Continuing, $b_{p+k+2} = b_{p+k+1}^2 + 1 \\equiv b_{k+1}^2 + 1 = b_{k+2}$, and so on, up to $b_{p+\\ell} \\equiv b_\\ell$. On the other hand, $b_{p+\\ell} \\equiv -b_\\ell$, so $2b_\\ell \\equiv 0$. Since $d$ is odd, $b_\\ell \\equiv 0$, so $b_{\\ell+1} = b_\\ell^2 + 1 \\equiv b_0$.\n\nConsider any index $j$ in the range $0$ through $p-1$ and use $(*)$ to get\n\n$$\nb_{j+\\ell+1} - b_j = (b_{\\ell+1} - b_0) \\prod_{i=0}^{j-1} (b_{\\ell+1+i} + b_i) \\equiv 0,\n$$\n\nas $b_{\\ell+1} \\equiv b_0$. Hence $b_{j+\\ell+1} \\equiv b_j$. Similarly, $b_{j+2(\\ell+1)} \\equiv b_{j+\\ell+1}$, and so on, so the sequence $\\{b_n\\}$ is periodic modulo $d$.\n\nLet $t$ be the smallest period of $b_n$ modulo $d$. Recall that $b_{p+k+1} \\equiv b_{k+1}$, so $t$ divides $p$. As $p$ is prime, either $t=1$ or $t=p$. The former is impossible, since $b_0 = 1 \\neq 2 = b_1$, so $t = p$.\n\nHence $b_k \\equiv b_{p+k} \\equiv -b_k$, so $b_k \\equiv 0$. Recalling that $b_\\ell \\equiv 0$, it follows that $\\ell - k$ is divisible by $p$, which is a contradiction since $0 < \\ell - k < p$.\n\nConsequently, the $p$ numbers $b_{p+k} + b_k$ for $k = 0, 1, \\dots, p-1$ are pairwise coprime, as claimed. This completes the proof.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20413,
"subject": "Mathematics (Olympiad)",
"question": "The equilateral triangle $ABC$ has sides of integer length $N$. The triangle is completely divided (by drawing lines parallel to the sides of the triangle) into equilateral triangular cells of side length $1$.\n\nA continuous route is chosen, starting inside the cell with vertex $A$ and always crossing from one cell to another through an edge shared by the two cells. No cell is visited more than once. Find, with proof, the greatest number of cells which can be visited.",
"options": [],
"answer": "See solution",
"solution": "Colour the cells black and white alternately, as in the diagram below.\n\n\n\nSince no cells are adjacent to a cell of the same colour, the route must alternate from white to black.\n\nIn the $r$th row, there are $r$ black cells and $r-1$ white cells. Therefore, there are $$\\sum_{r=1}^{N} r = \\frac{(N+1)N}{2}$$ black cells, and $$\\sum_{r=1}^{N} (r-1) = \\frac{N(N-1)}{2}$$ white cells.\n\nAs the route starts on a black cell, you can visit at most twice the number of white cells plus one (by starting on black, alternating through all the white cells, and ending on black). This bound is\n\n$$2 \\cdot \\frac{N(N-1)}{2} + 1 = N^2 - N + 1.$$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20414,
"subject": "Mathematics (Olympiad)",
"question": "Let $x = 2^m$ and $y = 2^n$ with $m > n$ and $xy = 2^{m+n} = 2048$. Given $\\frac{x}{y} - \\frac{y}{x} = 7.875$, find the value of $x$.",
"options": [],
"answer": "See solution",
"solution": "Let $x = 2^m$ and $y = 2^n$. Since $m > n$ and $xy = 2^{m+n}$, we have $m + n = 11$.\n\nGiven $\\frac{x}{y} - \\frac{y}{x} = 7.875 = 7\\frac{7}{8}$, so:\n\n$$\n\\frac{x}{y} - \\frac{y}{x} = 2^{m-n} - 2^{n-m} = \\frac{63}{8}\n$$\n\nLet $m - n = t$. Then:\n\n$$\n2^t - 2^{-t} = \\frac{63}{8}\n$$\n\nMultiply both sides by $8(2^t)$:\n\n$$\n8(2^t)^2 - 63(2^t) - 8 = 0\n$$\n\nFactor:\n\n$$\n(2^t - 8)(8(2^t) + 1) = 0\n$$\n\nSo $2^t = 8 = 2^3$, thus $m - n = 3$.\n\nSince $m + n = 11$ and $m - n = 3$, adding gives $2m = 14$, so $m = 7$.\n\nTherefore, $x = 2^7 = \\boxed{128}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20415,
"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 $c$.",
"options": [],
"answer": "See solution",
"solution": "**First Solution:**\n\nWe rewrite $$(\\ddagger)$$ as\n\n$$\n\\frac{9(x + y)}{M} \\leq \\left( \\frac{x^2 + y^2 + (x + y)^2 + 1}{\\sqrt{xy}} \\right)^2 = \\left( \\frac{2(x + y)^2 + 1}{\\sqrt{xy}} - 2\\sqrt{xy} \\right)^2.\n$$\n\nSetting\n\n$$\nA = \\frac{2(x + y)^2 + 1}{\\sqrt{xy}}, \\quad B = 2\\sqrt{xy},\n$$\n\nthe 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 $x + y$, increasing $\\sqrt{xy}$ decreases $A$ and increases $B$, so $A - B$ decreases. Thus, the right-hand side decreases, and the inequality is strongest when $x = y$.\n\nSo, set $x = y$ in $(\\ddagger)$:\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\nLet\n\n$$\nf(x) = 36x + \\frac{12}{x} + \\frac{1}{x^3}, \\quad x > 0.\n$$\n\nCompute the derivative:\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\nThe only critical value in the domain is $x = \\frac{1}{\\sqrt{2}}$. It is easy to check that $f(x)$ attains its global minimum $32\\sqrt{2}$ at $x = \\frac{1}{\\sqrt{2}}$.\n\nTherefore, 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": 20416,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs of positive integers $ (x, y) $ for which\n\n$$\nx(x+1) = y(y+1)(y^2+1).\n$$",
"options": [],
"answer": "See solution",
"solution": "The only solution is $ (x, y) = (5, 2) $.\n\n**Solution 1:**\n\nConsider the equation as a quadratic in $x$. The discriminant is\n$$\nD = 1 + 4y(y+1)(y^2+1) = 4y^4 + 4y^3 + 4y^2 + 4y + 1.\n$$\nFor integer solutions, $D$ must be a perfect square. Note that\n$$\n(2y^2 + y)^2 = 4y^4 + 4y^3 + y^2 < D\n$$\nand\n$$\n(2y^2 + y + 1)^2 = D + y^2 - 2y.\n$$\nIf $y^2 - 2y > 0$, then $D$ is between two consecutive squares and cannot be a perfect square. The case $y^2 - 2y = 0$ gives $y = 2$, and then\n$$\nx = \\frac{-1 + \\sqrt{D}}{2} = \\frac{-1 + (2y^2 + y + 1)}{2} = 5.\n$$\nIf $y^2 - 2y < 0$, then $y = 1$, but $x$ is not an integer.\n\n**Solution 2:**\n\nRewrite the equation as\n$$\nx(x+1) = (y^2 + y)(y^2 + 1).\n$$\nIf $x \\leq y^2$, then $x(x+1) < (y^2 + y)(y^2 + 1)$. If $x \\geq y^2 + y$, then $x(x+1) > (y^2 + y)(y^2 + 1)$. Thus, $x = y^2 + a$ with $0 < a < y$. Substitute into the equation:\n$$\n(y^2 + a)(y^2 + a + 1) = (y^2 + y)(y^2 + 1).\n$$\nExpanding and simplifying gives\n$$\n2a y^2 + a^2 + a = y^3 + y,\n$$\nwhich is equivalent to\n$$\n2a(y^2 + 1) + a^2 - a = y(y^2 + 1).\n$$\nThus, $y^2 + 1$ divides $a^2 - a$. But $a^2 - a < y^2 + 1$, so $a^2 - a = 0$, i.e., $a = 1$. Substituting $a = 1$ gives $2(y^2 + 1) = y(y^2 + 1)$, so $y = 2$. Then $x = 5$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20417,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle with incenter $I$ and circumcenter $O$, and let $M$ be the midpoint of $BC$. The bisector of angle $A$ intersects lines $BC$ and $OM$ at $L$ and $Q$, respectively. Prove that\n\n$$\nAI \\cdot LQ = IL \\cdot IQ.\n$$",
"options": [],
"answer": "See solution",
"solution": "\n\nThe bisector of $\\widehat{BAC}$ and the perpendicular bisector of side $BC$ intersect at $Q$, the midpoint of arc $\\widehat{BC}$ not containing $A$.\n\nWe have $BQ = IQ$, since $\\widehat{QBI} = \\widehat{QIB} = 90^\\circ - \\frac{1}{2}\\widehat{C}$. On the other hand, the triangles $BLQ$ and $ALC$ are similar, meaning that\n\n$$\n\\frac{BQ}{LQ} = \\frac{AC}{LC}.\n$$\n\nUsing the angle bisector theorem in the triangles $ABC$ and $ABL$, we get\n\n$$\n\\frac{IQ}{LQ} = \\frac{BQ}{LQ} = \\frac{AC}{LC} = \\frac{AB}{BL} = \\frac{AI}{IL}.\n$$\n\nSo $AI \\cdot LQ = IQ \\cdot IL$, as desired.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20418,
"subject": "Mathematics (Olympiad)",
"question": "Prove that for all real $x$,\n$$\n\\sqrt[3]{\\frac{x^6 + 1}{2}} \\le \\frac{3x^2 - 4x + 3}{2},\n$$\nwith equality if and only if $x = 1$.",
"options": [],
"answer": "See solution",
"solution": "Note that for all real $x$,\n$$\nx^6 + 1 = (x^2 + 1)(x^4 - x^2 + 1) = (x^2 + 1)(x^2 + \\sqrt{3}x + 1)(x^2 - \\sqrt{3}x + 1),\n$$\nwhich is a product of three positive numbers, since $x^2 \\pm \\sqrt{3}x + 1 = (x \\pm \\frac{\\sqrt{3}}{2})^2 + \\frac{1}{4}$.\n\nWe use the AGM inequality for three positive variables: $\\sqrt[3]{ABC} \\le \\frac{A+B+C}{3}$, with $A = \\frac{x^2+1}{2}$, $B = a(x^2 + \\sqrt{3}x + 1)$, and $C = b(x^2 - \\sqrt{3}x + 1)$, where $a, b$ are positive numbers such that $ab = 1$. Then,\n$$\n\\begin{aligned}\n\\sqrt[3]{\\frac{x^6 + 1}{2}} &\\le \\frac{\\frac{x^2+1}{2} + a(x^2 + \\sqrt{3}x + 1) + b(x^2 - \\sqrt{3}x + 1)}{3} \\\\\n&= \\frac{(1 + 2(a + b))x^2 + 2\\sqrt{3}(a - b)x + 1 + 2(a + b)}{6} \\\\\n&= \\frac{3x^2 - 4x + 3}{2},\n\\end{aligned}\n$$\nif, in addition, $a, b$ satisfy the equations $1 + 2(a + b) = 9$, $\\sqrt{3}(a - b) = -6$. The numbers $a = 2 - \\sqrt{3}$, $b = 2 + \\sqrt{3}$ satisfy these and $ab = 1$. Hence,\n$$\n\\sqrt[3]{\\frac{x^6 + 1}{2}} \\le \\frac{3x^2 - 4x + 3}{2},\n$$\nfor all real $x$, with equality iff\n$$\n\\frac{x^2 + 1}{2} = a(x^2 + \\sqrt{3}x + 1) = b(x^2 - \\sqrt{3}x + 1),\n$$\ni.e., $x = 1$.\n\n_Remark_: This can also be solved with the substitution used in Problem 22.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20419,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABQP$ be a convex quadrilateral. Let the angle bisectors of $\\angle PAQ$ and $\\angle PBQ$ intersect at $C$. The circumcircle of $APQ$ intersects $AB$ and $AC$ at $M$ and $N \\ne A$, respectively. Lines $PQ$ and $BC$ intersect at $S$. Prove that if $AC$ is perpendicular to $BC$, then $M$, $S$, $N$ are collinear.\n\n",
"options": [],
"answer": "See solution",
"solution": "We divide our solution into two main steps:\n\n*Step 1:* Let $O$ be the midpoint of $AB$ and $R$ the midpoint of $PQ$, then $C$, $R$, $O$ are collinear.\n\n**Proof.** We will denote the complex number of each point by its lowercase letter.\n\nLet $AP \\cap BQ = V$, $AQ \\cap BP = U$. Let $\\angle PAC = \\alpha$, $\\angle PBC = \\beta$, then $\\angle PVQ = 90^\\circ - \\alpha - \\beta$ and $\\angle PUQ = 90^\\circ + \\alpha + \\beta$, therefore quadrilateral $PUQV$ is cyclic.\n\nLet $AC \\cap BP = W$, $AQ \\cap BC = X$, $AC \\cap BQ = Y$, and $AP \\cap BC = Z$. Since $C$ is on the angle bisector of both $\\angle PAQ$ and $\\angle PBQ$, and since $\\angle ACB = 90^\\circ$, $WC = CY$ and $XC = CZ$. Thus $WXYZ$ is a rhombus, hence $c = \\frac{w+x+y+z}{4}$.\n\n\n\nSince $PUQV$ is cyclic, $\\frac{AU}{AP} = \\frac{AV}{AQ}$. By the angle bisector theorem, $\\frac{UW}{PW} = \\frac{AU}{AP}$ and $\\frac{AV}{AQ} = \\frac{VY}{YQ}$. Therefore $\\frac{UW}{UP} = \\frac{VY}{VQ} = d$ for some $d \\in \\mathbb{R}$. It follows that $w = dp + (1-d)u$ and $y = dq + (1-d)v$. Similarly, there exists $e \\in \\mathbb{R}$ such that $x = eq + (1-e)u$ and $z = ep + (1-e)v$. Therefore\n\n$$\nc = \\frac{(d+e)(p+q) + (2-d-e)(u+v)}{4}.\n$$\n\nLet $T$ be the midpoint of $UV$. Clearly $t = \\frac{u+v}{2}$, and $r = \\frac{p+q}{2}$. Therefore\n\n$$\nc = \\left(\\frac{d+e}{2}\\right)t + \\left(1 - \\frac{d+e}{2}\\right)r\n$$\n\nwith $\\frac{d+e}{2} + \\left(1 - \\frac{d+e}{2}\\right) = 1$, so $C$, $R$, $T$ are collinear.\n\nUsing the well-known fact that in a complete quadrilateral, the midpoints of three diagonals are collinear, we have $R$, $T$, $O$ are collinear. Hence $C$, $R$, $O$ are collinear. $\\square$\n\n*Step 2:* Showing that $M$, $S$, $N$ are collinear.\n\n**Proof.** First, since $\\angle PAN = \\angle NAQ$, $N$ is the midpoint of arc $PQ$, so $\\angle NRQ = 90^\\circ$, hence quadrilateral $NCRS$ is cyclic.\n\nLet $\\angle OBC = \\theta$. We have $\\angle MNQ = \\angle MAQ = 90^\\circ - \\theta - \\alpha$. Since $\\angle SNQ = 90^\\circ - \\angle RNS - \\angle PQN = 90^\\circ - \\angle OCB - \\angle PAN = 90^\\circ - \\theta - \\alpha$, so $\\angle MNQ = \\angle SNQ$, therefore $M$, $S$, $N$ are collinear. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20420,
"subject": "Mathematics (Olympiad)",
"question": "Divide $\\overrightarrow{A_1BA_6}$ into five equal parts, with the corresponding points $A_2, A_3, A_4, A_5$ (see the figure). The length of each segment $\\overrightarrow{A_iA_{i+1}}$ is exactly $\\frac{1}{7}$ of the perimeter for $i = 1, 2, 3, 4, 5$, and the distance between any two points is not more than $a_7 < 9$.\n\nSuppose there are $m_i$ given points on the arc $\\overrightarrow{A_iA_{i+1}}$. Then, the number of lines from the given points whose lengths are more than $9$ is at most\n\n$$\nl = \\sum_{1 \\le i < j \\le 5} m_i m_j$$\n\nwhere\n\n$$\nm_1 + m_2 + m_3 + m_4 + m_5 = 5m + r.$$\n\nSince there are finitely many non-negative integer groups $(m_1, m_2, m_3, m_4, m_5)$, the maximum value of $l$ exists. Now, we prove that when the maximum is attained, the inequality\n\n$$\n|m_i - m_j| \\le 1 \\quad (1 \\le i < j \\le 5)\n$$\n\nmust hold.\n\nIf there exist $i, j$ ($1 \\le i < j \\le 5$) such that $|m_i - m_j| \\ge 2$ when the maximum is attained, suppose $m_1 - m_2 \\ge 2$. Then let\n\n$$\nm_1' = m_1 - 1, \\quad m_2' = m_2 + 1, \\quad m' = m,\n$$\n\nand let the corresponding integer be $l'$. We have\n\n$$\n\\begin{align*}\nm_1' + m_2' &= m_1 + m_2, \\\\\nm_1' + m_2' + m_3' + m_4' + m_5' &= m_1 + m_2 + m_3 + m_4 + m_5, \\\\\nl' - l &= (m_1'm_2' - m_1m_2) + [(m_1' + m_2') \\\\\n&\\quad -(m_1 + m_2)](m_3 + m_4 + m_5) \\\\\n&= m_1 - m_2 - 1 \\ge 1.\n\\end{align*}\n$$\n\nThis is a contradiction.\n\nTherefore, when $l$ reaches the maximum value, the number of $m+1$ is $r$ and the number of $m$ is $5-r$. Thus, the number of lines from the given points whose lengths are more than $9$ is at most\n\n$$\n\\begin{aligned}\n& \\binom{r}{2}(m+1)^2 + \\binom{r}{1}(5-r)(m+1)m + \\binom{5-r}{2}m^2 \\\\\n&= 10m^2 + 4rm + \\frac{1}{2}r(r-1).\n\\end{aligned}\n$$\n\n*Lemma 3*: Take arbitrary $n$ points on the circle $C$ with radius $10$ to form set $M$, where $n = 6m + r$ ($m, r$ are non-negative integers, $0 \\le r < 6$). Assume that there are $S_n$ triangles whose vertices are from $M$ and each side is longer than $9$. Prove\n\n$$\nS_n \\le 20m^3 + 10rm^2 + 2r(r-1)m + \\frac{1}{6}r(r-1)(r-2).\n$$",
"options": [],
"answer": "See solution",
"solution": "We prove Lemma 3 by mathematical induction.\n\n**Base case:** When $n = 1, 2$, $S_n = 0$, which satisfies the inequality.\n\n**Inductive step:** Suppose for $n = k = 6m + r$ ($0 \\le r < 6$),\n$$\nS_k \\le 20m^3 + 10rm^2 + 2r(r-1)m + \\frac{1}{6}r(r-1)(r-2).\n$$\n\nFor $n = k + 1$, the $k + 1$ points must include a point $P$ such that at least $\\lfloor \\frac{k+1+5}{6} \\rfloor = m + 1$ points are in the $\\frac{2}{7}$ arc $A_1PA_6$. The distances from these points to $P$ are $\\le PA_1 = PA_6 = a_7 < 9$. Thus, at most $(k+1)-(m+1) = 5m + r$ points have distances to $P$ greater than $9$, and these are in the other $\\frac{5}{7}$ arc.\n\nFrom Lemma 2, the number of lines from such points whose lengths are more than $9$ is at most\n$$\n10m^2 + 4rm + \\frac{1}{2}r(r-1).\n$$\n\nTherefore, the number of triangles with vertex $P$ and each side longer than $9$ is at most\n$$\nS_p = 10m^2 + 4rm + \\frac{1}{2}r(r-1).\n$$\n\nWithout $P$, there are $k = 6m + r$ points, and by the induction hypothesis,\n$$\nS_k \\le 20m^3 + 10rm^2 + 2r(r-1)m + \\frac{1}{6}r(r-1)(r-2).\n$$\n\nThus,\n$$\n\\begin{align*}\nS_{k+1} &= S_k + S_p \\\\\n&\\le 20m^3 + 10rm^2 + 2r(r-1)m + \\frac{1}{6}r(r-1)(r-2) \\\\\n&\\quad + 10m^2 + 4rm + \\frac{1}{2}r(r-1) \\\\\n&= 20m^3 + 10(r+1)m^2 + 2r(r+1)m + \\frac{1}{6}r(r-1)(r+1).\n\\end{align*}\n$$\n\nWhen $r = 5$, $n = k + 1 = 6(m+1)$ and $S_{k+1} = 20(m+1)^3$, which also satisfies the inequality.\n\nTherefore, Lemma 3 is proved.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20421,
"subject": "Mathematics (Olympiad)",
"question": "Let $l$, $b$, and $h$ denote the numbers of cubes in the length, breadth, and height, respectively, of a cuboid made up of unit cubes. Given that $lbh = 120$, and that removing a layer at both ends of the length, breadth, and height leaves $(l-2)(b-2)(h-2) = 24$ unpainted cubes, find the surface area of the cuboid.",
"options": [],
"answer": "See solution",
"solution": "By trying various factorizations of $120$, we find that $l$, $b$, and $h$ are $4$, $5$, and $6$ in some order. The surface area is $$2(lb + lh + bh) = 2(20 + 30 + 24) = 148.$$ (Solving the equations algebraically is harder than using trial and error.)",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 20422,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(x + f(y)) - f(x) = (x + f(y))^4 - x^4\n$$\nfor all $x, y \\in \\mathbb{R}$.",
"options": [],
"answer": "See solution",
"solution": "We rewrite the given equation as\n\n$$\nf(x + f(y)) = (x + f(y))^4 - x^4 + f(x).\n$$\n\nSetting $x = -f(z)$ and $y = z$, we obtain\n\n$$\nf(0) = -(f(z))^4 + f(-f(z)) \\quad \\text{for all } z \\in \\mathbb{R}.\n$$\n\nNow, setting $x = -f(z)$ in the previous equation and using the above, we get\n\n$$\nf(f(y) - f(z)) = (f(y) - f(z))^4 - (f(z))^4 + f(-f(z)) = (f(y) - f(z))^4 + f(0)\n$$\nfor all $y, z \\in \\mathbb{R}$. This means that if a number $t$ can be expressed as the difference of two values of $f$, that is $t = f(y) - f(z)$, then $f(t) = t^4 + f(0)$.\n\nIf $f$ takes any nonzero value, then every real number is a difference of two values of $f$. Let $f(a) = b \\ne 0$. Putting $y = a$ in the original equation gives\n\n$$\nf(x + b) - f(x) = (x + b)^4 - x^4.\n$$\n\nSince $b \\ne 0$, the right-hand side is a polynomial of degree 3 and takes every real value as $x$ varies. Thus, the left-hand side, which is the difference of two values of $f$, can take any real value. Therefore, $f(t) = t^4 + f(0)$ for all $t \\in \\mathbb{R}$.\n\nAll functions of the form $f(x) = x^4 + k$ satisfy the given functional 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": 20423,
"subject": "Mathematics (Olympiad)",
"question": "坐標平面上有 2020 個點 $\\{A_i = (x_i, y_i) : i = 1, \\dots, 2020\\}$，滿足\n\n$$\n0 = x_1 < x_2 < \\cdots < x_{2020},\n$$\n\n$$\n0 = y_{2020} < y_{2019} < \\cdots < y_1.\n$$\n\n又令 $O = (0,0)$ 為座標原點。我們依序連邊 $OA_1, A_1A_2, \\dots, A_{2019}A_{2020}, A_{2020}O$，構成一個 2021 邊形 $C$。\n\n濤哥想要將 $C$ 整個塗黑。他每一次可以指定一個點 $(x, y)$，支付 $xy$ 元，並將 $\\{(x', y') : 0 \\le x' \\le x, 0 \\le y' \\le y\\}$ 的區域塗黑。證明：濤哥總是可以用至多 $4|C|$ 元將整個 $C$ 塗黑，其中 $|C|$ 為 $C$ 的面積。",
"options": [],
"answer": "See solution",
"solution": "題目可以擴張為：在第一象限中，有一個連續有界遞減函數 $f$，並令 $C$ 為此曲線與 $x$、$y$ 兩軸所夾區域，則我們可以用題設之方式，在 $4|C|$ 元內將此區域塗黑。我們可以直接構造此塗法：\n\n\n\n以上構造中，$x$ 為讓 $f(x) \\ge f(0)/2$ 之正實數（由連續性此 $x$ 必存在）。則我們有：\n\n- 1 號彩色區域面積 $\\times 4 = 4x f(x) \\ge x f(0) + 2x f(x) = $ 選 $(x, f(0))$ 加上選 $(2x, f(x))$ 時需付金額。\n- 2 號彩色區域面積 $\\times 4 = 4x f(2x) = $ 選 $(4x, f(2x))$ 時需付金額。\n- 3 號彩色區域面積 $\\times 4 = 8x f(4x) = $ 選 $(8x, f(4x))$ 時需付金額。\n- 以下依此類推。\n\n因此若我們依序選擇 $(x, f(0)), (2x, f(x)), (4x, f(2x)), (8x, f(4x)), \\dots$，則我們必然將 $f$ 以下區域（也就是 $C$）全數塗黑，且有\n\n$$\n\\text{所需支付金額} \\le 4 \\times \\text{彩色區域面積} \\le 4 \\times f \\text{以下面積} = 4|C|\n$$\n\n從而完成構造。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20424,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\mathbb{N}$ be the set of positive integers. Determine all functions $g: \\mathbb{N} \\to \\mathbb{N}$ such that $$(g(m) + n)(m + g(n))$$ is a perfect square for all $m, n \\in \\mathbb{N}$.\n\n",
"options": [],
"answer": "See solution",
"solution": "The answer is $g(n) = n + C$, where $C$ is a nonnegative integer.\n\nClearly, the function $g(n) = n + C$ satisfies the required property since\n\n$$\n(g(m) + n)(m + g(n)) = (n + m + C)^2\n$$\n\nis a perfect square. We first prove a lemma.\n\n**Lemma**: If a prime number $p$ divides $g(k) - g(l)$ for some positive integers $k$ and $l$, then $p \\mid k-l$.\n\n_Proof of lemma_: If $p^2 \\mid g(k) - g(l)$, let $g(l) = g(k) + p^2 a$, where $a$ is an integer. Choose an integer\n\n$$\nD > \\max\\{g(k), g(l)\\},\n$$\n\nand $D$ is not divisible by $p$. Set $n = pD - g(k)$; then $n + g(k) = pD$, and thus\n\n$$\nn + g(l) = pD + (g(l) - g(k)) = p(D + pa)\n$$\n\nis divisible by $p$, but not divisible by $p^2$.\n\nBy assumption, $(g(k) + n)(g(n) + k)$ and $(g(l) + n)(g(n) + l)$ are both perfect squares, and therefore they are divisible by $p^2$ since they are divisible by $p$. Hence,\n\n$$\np \\mid ((g(n) + k) - (g(n) + l)),\n$$\n\ni.e. $p \\mid k-l$.\n\nIf $p \\mid g(k) - g(l)$ but $p^2$ does not divide $g(k) - g(l)$, choose an integer $D$ as above and set $n = p^3 D - g(k)$. Then $g(k) + n = p^3 D$ is divisible by $p^3$, but not by $p^4$, and\n\n$$\ng(l) + n = p^3 D + (g(l) - g(k))\n$$\n\nis divisible by $p$, but not by $p^2$. As with the above argument, we have $p \\mid g(n) + k$ and $p \\mid g(n) + l$, and therefore\n\n$$\np \\mid ((g(n) + k) - (g(n) + l)),\n$$\n\ni.e. $p \\mid k-l$. This completes the proof of the lemma.\n\nBack to the original problem: if there exist positive integers $k$ and $l$ such that $g(k) = g(l)$, then the lemma implies that $k-l$ is divisible by any prime number. Hence, $k-l=0$, i.e. $k=l$, and thus $g$ is injective.\n\nNow consider $g(k)$ and $g(k+1)$. Since $(k+1)-k=1$, once again the lemma implies that $g(k+1) - g(k)$ is not divisible by any prime number, and therefore\n\n$$\n|g(k+1) - g(k)| = 1.\n$$\n\nLet $g(2) - g(1) = q$, where $|q| = 1$. It follows easily by induction that\n\n$$\ng(n) = g(1) + (n - 1)q.\n$$\n\nIf $q = -1$, then $g(n) \\le 0$ for $n \\ge g(1) + 1$, a contradiction. Therefore, we must have $q = 1$ and\n\n$$\ng(n) = n + (g(1) - 1)\n$$\n\nfor any $n \\in \\mathbb{N}$, where $g(1) - 1 \\ge 0$. Set $g(1) - 1 = C$ (a constant). Then $g(n) = n + C$, where $C$ is a nonnegative integer.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20425,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs $ (x, y) $ of integers such that $ y^3 - 1 = x^4 + x^2 $.",
"options": [],
"answer": "See solution",
"solution": "If $x = 0$, we get the solutions $(x, y) = (0, \\pm 1)$. These solutions will turn out to be the only ones. From now on, assume $x \\neq 0$. We add 1 to both sides and factor:\n\n$$\ny^3 = x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)\n$$\n\nWe show that the factors $x^2 + x + 1$ and $x^2 - x + 1$ are co-prime. Assume that a prime $p$ divides both of them. Then $p \\mid x^2 + x + 1 - (x^2 - x + 1) = 2x$. Since $x^2 + x + 1$ is always odd, $p \\mid x$. But then $p$ does not divide $x^2 + x + 1$, a contradiction. Since $x^2 + x + 1$ and $x^2 - x + 1$ have no prime factors in common and their product is a cube, both of them are cubes by a consequence of the fundamental theorem of arithmetic. Therefore, $x^2 + x + 1 = a^3$ and $x^2 - x + 1 = b^3$ for some non-negative integers $a$ and $b$.\n\nIf $x < 0$, we may write $x = -x'$ and obtain $x'^2 - x' + 1 = a^3$, $x'^2 + x' + 1 = b^3$, which is the same pair of equations with $a$ and $b$ interchanged. Therefore we only need to consider the case $x > 0$. The first equation implies that $a > x^{2/3}$. But since clearly $b < a$ we get\n\n$$\nx^2 - x + 1 = b^3 \\leq (a-1)^3 = a^3 - 3a^2 + 3a - 1 \\leq a^3 - 3a^2 + 3a \\leq a^3 - 2a^2 = x^2 + x + 1 - 2a^2 < x^2 + x + 1 - 2x^{4/3}\n$$\n\nwhen $2a^2 \\leq 3a^2 - 3a$, i.e. $a^2 \\geq 3a$ which holds for $a \\geq 3$. Clearly $a = 2$ is impossible and $a = 1$ means $x = 0$. We got $x^2 - x + 1 < x^2 + x + 1 - 2x^{4/3}$ which means $0 \\leq 2x - 2x^{4/3}$. Hence $x = 1$, but then $3$ would be a cube, a contradiction.\n\n**Remark.** There are many ways to guess the factorization $x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)$. Writing $x^4 + x^2 + 1 = (x^2 + 1)^2 - x^2$ we immediately obtain it. Another way to see it is to write $p(x) = x^4 + x^2 + 1$ and notice that $p(0) = 1 \\cdot 1$, $p(1) = 3 \\cdot 1$, $p(2) = 7 \\cdot 3$, so there probably is a quadratic factor $q(x)$ for which $q(0) = 1$, $q(1) = 3$, $q(2) = 7$ (clearly, there cannot be linear factors). It is easy to see then that $q(x) = x^2 + x + 1$ and to complete the factorization by long division. One more way is to write $x^4+x^2+1 = (x^2+ax+b)(x^2+cx+d)$ and compare the coefficients.\n\nIt is well-known that if $s$ and $t$ are co-prime integers whose product is a perfect $k$\\text{th}$ power, then $s$ and $t$ both are perfect $k$\\text{th}$ powers. The proof goes like this. Let $a = p_1^{\\alpha_1} \\dots p_h^{\\alpha_h}$, $b = q_1^{\\beta_1} \\dots q_l^{\\beta_l}$ and $x = r_1^{\\gamma_1} \\dots r_m^{\\gamma_m}$ be the prime factorizations of $a$, $b$ and $x$. In the prime factorization of $x^k$, every exponent is divisible by $k$, so the same must hold for the factorization of $st$. But $s$ and $t$ are co-prime, so the exponents of primes in $s$ and $t$ must be divisible by $k$. Therefore $s$ and $t$ are perfect $k$\\text{th}$ powers.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20426,
"subject": "Mathematics (Olympiad)",
"question": "Find the smallest real number $\\alpha$ such that for any convex polygon $P$ of area $1$, there exists a point $M$ in the plane such that the area of the convex hull of $P \\cup Q$ is at most $\\alpha$, where $Q$ is the centrally symmetric figure of $P$ about $M$.",
"options": [],
"answer": "See solution",
"solution": "$\\alpha = 2$.\n\nFirst, we prove $\\alpha \\ge 2$. For a convex polygon $R$, let $S(R)$ be the area of $R$. For a plane set $X$, let $\\bar{X}$ be the convex hull of $X$. Take $P$ as $\\triangle ABC$ of area $1$, $M$ as any point in the plane, and $Q$ as the centrally symmetric figure of $P$ about $M$. There are two situations.\n\n(i) If $M$ is outside $P$ or on the boundary. Draw a line $l$ through $M$ such that $P$, $Q$ lie on different sides of $l$ (they have no common interior points). Then $S(P \\cup Q) \\ge S(P) + S(Q) \\ge 2$.\n\n(ii) If $M$ is inside $P$. Let $A', B', C'$ be the respective symmetric points of $A, B, C$ about $M$. Then $R = P \\cup Q = \\{A, B, C, A', B', C'\\}$, $R$ has a centre of symmetry, $R$ is a parallelogram or a hexagon. If $R$ is a parallelogram, say $R = ABA'B'$, then $C$ is inside or on the boundary, and\n\n$$\nS(R) = S(ABA'B') \\ge 2S(ABC) = 2.\n$$\n\nIf $R$ is a hexagon, $R = AC'BA'CB'$, then\n\n$$\n\\begin{aligned}\nS(R) &= S(AC'BM) + S(BA'CM) + S(CB'AM) \\\\\n&= (S(AMC) + S(BMC)) + (S(BMA) + S(CMA)) \\\\\n&\\quad + (S(CMB) + S(AMB)) = 2.\n\\end{aligned}\n$$\n\nNext, we prove $\\alpha = 2$ satisfies the problem statement in two ways. $\\square$\n\n**Method 1** We begin with a lemma.\n\n**Lemma** As shown in Fig. 5.1, let $l_1 \\parallel l_2$ and $P_1, P_2$ be two convex polygons both contained in the shaded region, both including the points $A$ and $B$. Then $S(\\overline{P_1 \\cup P_2}) \\le S(P_1) + S(P_2)$.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p213_data_ad582868fb.png \"\")\n\nFig. 5.1\n\n**Proof of lemma** Assume $\\overline{P_1 \\cup P_2} = C_0C_1\\cdots C_n$, where $C_0 = A, C_n = B$. If there exists $0 < i < n-1$ such that $C_i, C_{i+1} \\in P_1$, let\n\n$$\nP_1 = \\dots UC_iC_{i+1}V\\dots\n$$\n\nExtend $UC_i$ and $VC_{i+1}$ to meet at point $W$, as shown in Fig. 5.2. Change $P_1$ to $P'_1 = \\dots UWV\\dots$.\n\nObviously, the area of $P_1$ has increased by $S(C_iC_{i+1}W)$, and the area of $P_1 \\cup P_2$ has increased by $S(C_1C_{i+1}W)$. It suffices to justify the lemma for $P'_1$ and $P_2$. Notice that $P'_1$ has one fewer vertex than $P_1$, and thus the above operation cannot continue infinitely. We may assume $C_1, C_3, \\dots \\in P_1$, $C_2, C_4, \\dots \\in P_2$.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p213_data_ffed15c937.png \"\")\n\nFig. 5.2\n\nFor $1 \\le i \\le n-1$, draw a parallel line of $l_1$ through $C_i$ which meets $C_{i-1}C_{i+1}$ at $D_i$; for $1 \\le i \\le n-2$, let $C_{i-1}C_{i+1}$ and $C_iC_{i+2}$ intersect at $E_i$, as shown in Fig. 5.3. Observe that for $1 \\le i \\le n-2$,\n\n$$\n\\begin{align*}\n\\overline{P_1 \\cup P_2} \\setminus (P_1 \\cup P_2) &\\subseteq \\bigcup_{i=1}^{n-2} (\\triangle C_i C_{i+1} E_i), \\\\\n\\triangle D_i D_{i+1} E_i &\\subseteq (P_1 \\cap P_2), \\\\\nS(C_i C_{i+1} E_i) &= S(D_i D_{i+1} E_i).\n\\end{align*}\n$$\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p214_data_2c4389d460.png \"\")\n\nFig. 5.3\n\nAlso, observe that for $1 \\le i < j \\le n-2$, $\\triangle D_i D_{i+1} E_i \\cap \\triangle D_j D_{j+1} E_i = \\emptyset$. Therefore,\n\n$$\nS(P_1 \\cap P_2) \\ge \\sum_{i=1}^{n-2} S(D_i D_{i+1} E_i) = \\sum_{i=1}^{n-2} S(C_i C_{i+1} E_i) \\\\\n\\ge S(\\overline{P_1 \\cup P_2} \\setminus (P_1 \\cup P_2)),\n$$\n\nand the lemma conclusion $S(\\overline{P_1 \\cup P_2}) \\le S(P_1) + S(P_2)$ follows immediately.\n\nReturn to the original problem. Along an arbitrary direction, draw parallel lines $l_1, l_2$ that bound $P$ such that points $A, B$ of $P$ lie on $l_1, l_2$, respectively, and take the midpoint of $AB$ as the centre of symmetry $M$.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p214_data_ad267a5c1d.png \"\")\n\nFig. 5.4\n\nSuppose that $P$ is divided by $AB$ into two convex polygons $P_1, P_2$, as shown in Fig. 5.4. Let $P'_1, P'_2$ be their centrally symmetric figures about $M$. Then $P_1, P'_2$ are on the same side of $AB$, and likewise for $P_2, P'_1$. We infer that\n\n$$\n\\overline{P \\cup Q} = \\overline{P_1 \\cup P'_2} \\cup \\overline{P'_1 \\cup P_2}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20427,
"subject": "Mathematics (Olympiad)",
"question": "Во овоштарникот на дедо Ване растат овошни дрвца: 240 дрвца кајсии, 260 дрвца праски и 130 дрвца јаболка. Учениците од одделението на малиот Ване, на денот на дрвото, засадиле уште 75 дрвца кајсии, 126 дрвца праски и 142 дрвца јаболка.\n\nа) Колку ученици има во одделението на малиот Ване, ако секој ученик засадил по 7 овошни дрвца?\n\nб) По колку овошни дрвца од секој вид има во овоштарникот на дедо Ване?\n\nв) Колку вкупно овошни дрвца има во овоштарникот на дедо Ване?",
"options": [],
"answer": "See solution",
"solution": "а) На денот на дрвото се засадени вкупно $75 + 126 + 142 = 343$ овошни дрвца. Бидејќи секој ученик засадил по 7 дрвца, во одделението има $343 \\div 7 = 49$ ученици.\n\nб) Има: $240 + 75 = 315$ дрвца кајсии, $260 + 126 = 386$ дрвца праски и $130 + 142 = 272$ дрвца јаболка.\n\nв) Во овоштарникот има вкупно $315 + 386 + 272 = 973$ овошни дрвца.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20428,
"subject": "Mathematics (Olympiad)",
"question": "Let $x, y, z > 0$. Prove that\n$$\n\\frac{x^3}{z^3 + x^2 y} + \\frac{y^3}{x^3 + y^2 z} + \\frac{z^3}{y^3 + z^2 x} \\geq \\frac{3}{2}.\n$$",
"options": [],
"answer": "See solution",
"solution": "From the AM-GM inequality, we get $x^2 y \\leq \\frac{x^3 + x^3 + y^3}{3}$, which leads to\n\n$$\n\\frac{x^3}{z^3 + x^2 y} \\geq \\frac{x^3}{z^3 + \\frac{x^3 + x^3 + y^3}{3}} = \\frac{3x^3}{2x^3 + y^3 + 3z^3}.\n$$\n\nSumming this with the other two similar inequalities and letting $x^3 = a$, $y^3 = b$, $z^3 = c$, it is sufficient to prove that\n\n$$\n\\frac{a}{2a + b + 3c} + \\frac{b}{3a + 2b + c} + \\frac{c}{a + 3b + 2c} \\geq \\frac{1}{2},\n$$\n\nwhich follows from\n\n$$\n\\sum \\frac{a}{2a + b + 3c} = \\sum \\frac{a^2}{2a^2 + ab + 3ac} \\geq \\frac{(a + b + c)^2}{2(a^2 + b^2 + c^2 + 2ab + 2ac + 2bc)} = \\frac{1}{2}.\n$$\n\nEquality holds for $x = y = z$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20429,
"subject": "Mathematics (Olympiad)",
"question": "Prove that if, for each integer $n$, the polynomial $P(x) + nQ(x)$ has an integer root, then $Q(x)$ divides $P(x)$, and $P(x)/Q(x)$ is a linear polynomial with integer coefficients.\n\n*Remark.* You can prove the last statement either by addressing the growth rate of polynomials of degree at least two or using divisibility properties of polynomials with integer coefficients.",
"options": [],
"answer": "See solution",
"solution": "Suppose $P(x) + nQ(x)$ has an integer root $r_n$ for each $n$. If $m \\ne n$, then $P + mQ$ and $P + nQ$ have no common integer root, otherwise $P$ and $Q$ would have a common root. For $r_n$, $P(r_n) + nQ(r_n) = 0$, so $P(r_n)/Q(r_n) = -n$.\n\nDivide $P$ by $Q$: $P(x) = T(x)Q(x) + R(x)$, with $\\deg R < \\deg Q$ and $T, R$ integer-coefficient polynomials. Plug $x = r_n$:\n\n$$T(r_n) + \\frac{R(r_n)}{Q(r_n)} = -n$$\n\nSince $T(r_n)$ and $-n$ are integers, $R(r_n)/Q(r_n)$ is integer. For large $r_n$, $|R(r_n)/Q(r_n)| < 1$ by degree, so $R(r_n)/Q(r_n) = 0$ for infinitely many $r_n$, implying $R(x) = 0$ and $P(x)$ is divisible by $Q(x)$.\n\nLet $P(x) = Q(x)S(x)$, $S(x)$ integer-coefficient. Then $P(x) + nQ(x) = Q(x)(S(x) + n)$, so $r_n$ is an integer root of $S(x) + n$. Thus, for each $n$, $S(x) = -n$ has an integer solution, which is only possible if $S(x)$ is linear: $S(x) = Ax + B$ for integers $A, B$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20430,
"subject": "Mathematics (Olympiad)",
"question": "Let $M$ be a positive integer, and $A = \\{1, 2, \\dots, M+1\\}$. Show that if $f$ is a bijection from $A$ to $A$, then\n$$\n\\sum_{n=1}^{M} \\frac{1}{f(n) + f(n+1)} > \\frac{M}{M+3}.\n$$",
"options": [],
"answer": "See solution",
"solution": "From the AM-HM inequality,\n$$\n\\sum_{n=1}^{M} \\frac{1}{f(n) + f(n+1)} > \\frac{M^2}{\\sum_{n=1}^{M} (f(n) + f(n+1))} = \\frac{M^2}{2 \\sum_{n=1}^{M+1} f(n) - f(1) - f(M+1)}.\n$$\nSince $f$ is a bijection, we have\n$$\n\\sum_{n=1}^{M+1} f(n) = \\sum_{n=1}^{M+1} n = \\frac{(M+1)(M+2)}{2}.\n$$\nNote that\n$$\nf(1) + f(M+1) > 2\n$$\nhence,\n$$\n\\sum_{n=1}^{M} \\frac{1}{f(n) + f(n+1)} > \\frac{M^2}{(M+1)(M+2) - 2} = \\frac{M}{M+3}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20431,
"subject": "Mathematics (Olympiad)",
"question": "令 $a, b, c$ 為非負實數，且 $(a+b)(b+c)(c+a) \\neq 0$。求下式的最小值：\n\n$$\n(a+b+c)^{2016} \\left( \\frac{1}{a^{2016} + b^{2016}} + \\frac{1}{b^{2016} + c^{2016}} + \\frac{1}{c^{2016} + a^{2016}} \\right).\n$$\n\nLet $a, b, c$ be non-negative real numbers such that $(a+b)(b+c)(c+a) \\neq 0$. Find the minimum of\n\n$$\n(a+b+c)^{2016} \\left( \\frac{1}{a^{2016} + b^{2016}} + \\frac{1}{b^{2016} + c^{2016}} + \\frac{1}{c^{2016} + a^{2016}} \\right).\n$$",
"options": [],
"answer": "See solution",
"solution": "解：設 $a, b, c$ 為非負實數且 $(a+b)(b+c)(c+a) \\neq 0$。\n\n不妨設 $a^{2016}b^{2016} + b^{2016}c^{2016} + c^{2016}a^{2016} = 1$。則\n\n$$\n\\frac{1}{a^{2016} + b^{2016}} + \\frac{1}{b^{2016} + c^{2016}} + \\frac{1}{c^{2016} + a^{2016}} \\geq \\frac{5}{2}.\n$$\n\n又，當 $a = b \\to 1, c \\to 0$ 時，$(a+b+c) \\to 2$，$a^{2016} + b^{2016} \\to 2$，$b^{2016} + c^{2016} \\to 1$，$c^{2016} + a^{2016} \\to 1$，所以\n\n$$\n(a+b+c)^{2016} \\left( \\frac{1}{a^{2016} + b^{2016}} + \\frac{1}{b^{2016} + c^{2016}} + \\frac{1}{c^{2016} + a^{2016}} \\right) \\to 2^{2016} \\left( \\frac{1}{2} + 1 + 1 \\right) = 2^{2016} \\times \\frac{5}{2} = 5 \\times 2^{2015}.\n$$\n\n因此，最小值為 $5 \\times 2^{2015}$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20432,
"subject": "Mathematics (Olympiad)",
"question": "Let the five integers be $a, b, c, d, e$, with $a < b < c < d < e$.\n\nWhat is the value of $a + b + c + d + e$ if the ten pairwise sums $a+b, a+c, a+d, a+e, b+c, b+d, b+e, c+d, c+e, d+e$ are exactly the numbers $0, 1, 2, 4, 7, 8, 9, 10, 11, 12$?",
"options": [],
"answer": "See solution",
"solution": "Let the five integers be $a, b, c, d, e$, with $a < b < c < d < e$.\n\nThen $a + b < a + c$ and all the other pair sums are larger. So $a + b = 0$, $a + c = 1$, $c = b + 1$.\n\nSince $c + e < d + e$ and all the other pair sums are smaller, $d + e = 12$, $c + e = 11$, $d = c + 1$.\n\nThus $b, c, d$ are consecutive integers. Therefore $a + d = 2$ and $b + e = 10$.\n\nSince $b, c, d$ are consecutive integers, so are $b + c$, $b + d$ and $c + d$. The only consecutive integers left are $7, 8, 9$, so $b + c = 7$, $b + d = 8$, $c + d = 9$. The only pair sum left is $a + e = 4$.\n\nAdding $a + b = 0$ and $a + c = 1$ gives $2a + b + c = 1$. But $b + c = 7$, so $2a = -6$ and $a = -3$. Then $b = 3$, $c = 4$, $d = 5$, $e = 7$. Therefore $a + b + c + d + e = -3 + 3 + 4 + 5 + 7 = \\mathbf{16}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20433,
"subject": "Mathematics (Olympiad)",
"question": "Let $S$ be a subset of $\\{1, 2, 3, \\dots, 2024\\}$ such that the following two conditions hold:\n\n- If $x$ and $y$ are distinct elements of $S$, then $|x - y| > 2$.\n- If $x$ and $y$ are distinct odd elements of $S$, then $|x - y| > 6$.\n\nWhat is the maximum possible number of elements in $S$?\n\n(A) 436 (B) 506 (C) 608 (D) 654 (E) 675",
"options": [],
"answer": "See solution",
"solution": "If $S$ consists of the positive integers less than or equal to 2024 that are congruent to $1$, $4$, or $8$ modulo $10$, then every pair of elements in $S$ differ by at least $|4 - 1| = |11 - 8| = 3$, and every pair of odd elements of $S$ differ by at least $|11 - 1| = 10$. This set,\n\n$$\n\\{1, 4, 8, 11, 14, 18, \\dots, 2011, 2014, 2018, 2021, 2024\\},\n$$\n\nsatisfies the given conditions and has $3 \\cdot \\left(\\frac{2020}{10}\\right) + 2 = 608$ elements. To see that no larger set satisfies the given conditions, note that if a set satisfies the first condition and some block of $10$ consecutive integers contains $4$ elements of the set, then those $4$ elements would need to be the $1$st, $4$th, $7$th, and $10$th elements in that block, and the two odd numbers among them would differ by $6$, in violation of the second condition. Therefore, there are at most $3 \\cdot 202 = 606$ elements of $S$ among the first $2020$ positive integers, and at most $2$ elements of $S$ can be among $\\{2021, 2022, 2023, 2024\\}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20434,
"subject": "Mathematics (Olympiad)",
"question": "Consider a graph with $n$ vertices, $A_1, A_2, \\dots, A_n$. If row $i$ and column $j$ intersect at a white cell filled by a number $a > 0$, connect $A_i$ and $A_j$ with an edge of weight $a$. The graph is well-defined due to symmetry, and from the given condition, pairs of non-connected vertices correspond to edges with different weights. Show that for all $n$-vertex graphs, there is a way to assign at most $\\frac{n^2}{4}$ distinct weights to the edges.",
"options": [],
"answer": "See solution",
"solution": "We prove the statement by induction. For $n = 1, 2, 3$, the claim holds. Assume it is true for a graph with $n-3$ vertices. Consider a graph with $n \\ge 4$ vertices:\n\n1. If the number of edges is at most $\\frac{n^2}{4}$, assign a unique weight to each edge.\n\n2. If the number of edges exceeds $\\frac{n^2}{4}$, Mantel-Turán's theorem guarantees a triangle, i.e., three vertices $A_i, A_j, A_k$ are pairwise connected. Assign the same weight to the edges $A_iA_j$, $A_jA_k$, and $A_kA_i$. Assign at most $n-3$ distinct weights to edges connecting one of $A_i, A_j, A_k$ to the remaining $n-3$ vertices. By induction, assign at most $\\frac{(n-3)^2}{4}$ distinct weights among the $n-3$ other vertices. Thus,\n\n$$\n1 + (n-3) + \\frac{(n-3)^2}{4} \\le \\frac{n^2}{4}.\n$$\n\nTherefore, the statement holds for all $n \\ge 1$. For $n = 2017$, $\\left\\lfloor \\frac{2017^2}{4} \\right\\rfloor = \\frac{2017^2 - 1}{4}$ is the minimum value of $k$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20435,
"subject": "Mathematics (Olympiad)",
"question": "Find all polynomials $P(x)$ with real coefficients which satisfy the equality\n\n$$\nP(a-b) + P(b-c) + P(c-a) = 2P(a+b+c)\n$$\n\nfor all triples $(a, b, c)$ of real numbers such that $ab+bc+ca = 0$.\n\n*Note:* We call a triple $(a, b, c)$ of real numbers *good* if $ab + bc + ca = 0$. A key observation is that if $(a, b, c)$ is good, so is $(at, bt, ct)$ for any real $t$.",
"options": [],
"answer": "See solution",
"solution": "A polynomial $P(x)$ satisfies the condition if and only if $P(x) = c_1x^2 + c_2x^4$, where $c_1$ and $c_2$ are arbitrary real numbers.\n\nAssume $P(x)$ is a polynomial satisfying the condition. Write\n\n$$\nP(x) = \\sum_{i=0}^{n} p_i x^i\n$$\n\nfor real $p_0, \\ldots, p_n$ with $p_n \\neq 0$. For a good triple $(at, bt, ct)$,\n\n$$\n\\sum_{i=0}^{n} p_i t^i \\big[(a-b)^i + (b-c)^i + (c-a)^i - 2(a+b+c)^i\\big] = 0.\n$$\n\nThis must hold for all $t$, so for each $i$,\n\n$$\np_i\\big[(a-b)^i + (b-c)^i + (c-a)^i - 2(a+b+c)^i\\big] = 0.\n$$\n\nBy considering specific good triples, we find that $p_i = 0$ unless $i = 2$ or $i = 4$. Thus, $P(x) = c_1x^2 + c_2x^4$.\n\nTo check sufficiency, let $P_1(x) = x^2$ and $P_2(x) = x^4$.\n\nFor $P_1(x) = x^2$ and a good triple $(a, b, c)$:\n\n$$\n\\begin{align*}\n& (a-b)^2 + (b-c)^2 + (c-a)^2 \\\\\n= 2(a^2 + b^2 + c^2) - 2(ab + bc + ca) \\\\\n= 2(a^2 + b^2 + c^2) \\\\\n= 2(a+b+c)^2 = 2P_1(a+b+c).\n\\end{align*}\n$$\n\nFor $P_2(x) = x^4$:\n\n$$\n\\begin{align*}\n& (a-b)^4 + (b-c)^4 + (c-a)^4 \\\\\n= 2(a^4 + b^4 + c^4) - 4(a^3b + b^3c + c^3a) \\\\\n& \\quad +6(a^2b^2 + b^2c^2 + c^2a^2) - 4(ab^3 + bc^3 + ca^3).\n\\end{align*}\n$$\n\nAlso,\n\n$$\n2(a+b+c)^4 = 2(a^2+b^2+c^2)^2 = 2(a^4+b^4+c^4) + 4(a^2b^2+b^2c^2+c^2a^2).\n$$\n\nSubtracting, the difference is zero for good triples. Thus, all such $P(x)$ work.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20436,
"subject": "Mathematics (Olympiad)",
"question": "Suppose that $P$ is a point inside a convex hexagon $ABCDEF$ such that $PABC$, $PCDE$, and $PEFA$ are parallelograms of equal area.\n\nProve that $PBCD$, $PDEF$, and $PFAB$ are also parallelograms of equal area.",
"options": [],
"answer": "See solution",
"solution": "A diagonal of a parallelogram divides it into two triangles of equal area. Hence, the lines connecting $P$ to the vertices of the hexagon partition it into six triangles of equal area.\n\nFrom this, it follows that $PBCD$, $PDEF$, and $PFAB$ are quadrilaterals of equal area.\n\nWe also have\n\n$$\n|APC| = \\frac{1}{2}|PABC| = \\frac{1}{2}|PEFA| = |APE|.\n$$\n\nLet point $Q$ be the intersection of lines $AP$ and $CE$.\n\n\n\nWe have\n\n$$\n\\frac{|CPQ|}{|EPQ|} = \\frac{CQ}{QE} = \\frac{|CAQ|}{|EAQ|}.\n$$\n\nUsing addendo, this implies\n\n$$\n\\frac{CQ}{QE} = \\frac{|CAQ| - |CPQ|}{|EAQ| - |EPQ|} = \\frac{|APC|}{|APE|} = 1.\n$$\n\nThus, $Q$ is the midpoint of $CE$.\n\nSince $PCDE$ is a parallelogram, its diagonals bisect each other. In particular, $PQD$ is a straight line. Hence, points $A$, $P$, $Q$, and $D$ are collinear. Thus, $PD \\parallel FE$. A similar argument shows that $PF \\parallel DE$. Hence, $PDEF$ is a parallelogram. Similarly, $PFAB$ and $PBCD$ are also parallelograms, as required.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20437,
"subject": "Mathematics (Olympiad)",
"question": "Teacher wrote on the board 5 distinct numbers. After that, Petrik counted the sums of each two of these numbers and wrote them on the left half of the board. Vasyl did the same for the sums of each three of these numbers and wrote them on the right half of the board. Could the teacher write such numbers so that the sets of numbers written on the left and right halves of the board are the same (counting multiplicity)?",
"options": [],
"answer": "See solution",
"solution": "It's enough to choose the following numbers: $-2$, $-1$, $0$, $1$, $2$. For any two numbers, say, $a$ and $b$, selected by Petrik, there exists a pair $(-a, -b)$ whose sum is the opposite to the initial, and there also exists a triple of numbers except $a$ and $b$, whose sum is $-a - b$, since the sum of all five numbers is zero. Thus, there is a correspondence between the numbers from the left and right parts of the board.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20438,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$, and $d$ be real numbers with $a^2 + b^2 + c^2 + d^2 = 4$. Prove the inequality\n\n$$(a+2)(b+2) \\geq cd$$\n\nand give four numbers $a$, $b$, $c$, and $d$ such that equality holds.",
"options": [],
"answer": "See solution",
"solution": "The claimed inequality is equivalent to $2ab + 4a + 4b + 8 \\geq 2cd$, which can be written as\n\n$$\n2ab + 4a + 4b + a^2 + b^2 + c^2 + d^2 + 4 \\geq 2cd\n$$\n\nusing the condition $a^2 + b^2 + c^2 + d^2 = 4$. By the identity\n\n$$\na^2 + b^2 + 2ab + 4a + 4b + 4 = (a + b + 2)^2\n$$\n\nwe arrive at the equivalent and obvious inequality\n\n$$(a+b+2)^2 + (c-d)^2 \\geq 0$$\n\nEquality occurs when\n\n$$a + b = -2 \\quad \\text{and} \\quad c = d$$\n\ntogether with $a^2 + b^2 + c^2 + d^2 = 4$.\n\nFor instance, $a = b = c = d = -1$ gives equality.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20439,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $k$ such that there exist positive integers $a, b$ satisfying\n\n$$\na^2 + 4 = (k^2 - 4)b^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "Suppose $k, a, b$ satisfy the equation. Rewrite the equation as\n\n$$\na = \\sqrt{(k^2 - 4)b^2 - 4} = \\sqrt{(kb)^2 - 4(b^2 + 1)} \\quad (1)\n$$\n\nConsider the quadratic equation\n\n$$\nx^2 - kbx + (b^2 + 1) = 0 \\quad (2)\n$$\n\nIts solutions are\n\n$$\nc = \\frac{kb \\pm \\sqrt{(kb)^2 - 4(b^2 + 1)}}{2} = \\frac{kb \\pm a}{2} \\quad (3)\n$$\n\nwhere $a > 0$, $k \\ge 3$. Since $a$ and $kb$ have the same parity, $c \\in \\mathbb{Z}^+$. If one of the roots is $b$, then (2) becomes $(2-k)b^2 + 1 = 0$ which gives $k = 3$, $b = 1$ and hence $a = 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20440,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle. The moving points $M$ on the half-line $BC$, $N$ on the half-line $CA$, and $P$ on the half-line $AB$ start simultaneously from vertices $B$, $C$, and $A$, respectively, and move with constant speeds $v_1, v_2, v_3 > 0$, expressed using the same unit.\n\n**a)** Knowing that there are three distinct moments in which the triangle $MNP$ is equilateral, prove that the triangle $ABC$ is also equilateral and $v_1 = v_2 = v_3$.\n\n**b)** Prove that if $v_1 = v_2 = v_3$ and there exists a moment in which the triangle $MNP$ is equilateral, then the triangle $ABC$ is also equilateral.",
"options": [],
"answer": "See solution",
"solution": "Let $a, b, c \\in \\mathbb{C}$ be the affixes of the vertices of triangle $ABC$. For $t \\ge 0$, the affixes of the points $M(t), N(t)$, and $P(t)$ are:\n\n$$\n\\begin{cases}\nm(t) = b (1 - v_1 t) + c v_1 t \\\\\nn(t) = c (1 - v_2 t) + a v_2 t \\\\\np(t) = a (1 - v_3 t) + b v_3 t\n\\end{cases}\n$$\n\nThe condition that triangle $MNP$ is equilateral at some time $t \\ge 0$ can be written as:\n\n$$\nm(t) + \\varepsilon n(t) + \\bar{\\varepsilon} p(t) = 0,\n$$\n\nwhere $\\varepsilon \\in \\left\\{ \\frac{-1 + i \\sqrt{3}}{2}, \\frac{-1 - i \\sqrt{3}}{2} \\right\\}$.\n\nThis is equivalent to:\n\n$$\nt(-b v_1 + c v_1 - \\varepsilon c v_2 + \\varepsilon a v_2 - \\bar{\\varepsilon} a v_3 + \\bar{\\varepsilon} b v_3) + b + \\varepsilon c + \\bar{\\varepsilon} a = 0 \\tag{*}\n$$\n\nfor any $t \\ge 0$ for which $MNP$ is equilateral.\n\n**a)** Since (*) holds for three distinct values $t_1, t_2, t_3 \\ge 0$, by the pigeonhole principle, there exists $\\varepsilon$ for which (*) is satisfied at two distinct moments $t_i, t_j \\ge 0$, $1 \\le i < j \\le 3$, so:\n\n$$\n\\begin{cases}\n(b - c) v_1 + \\varepsilon (c - a) v_2 + \\bar{\\varepsilon} (a - b) v_3 = 0 \\\\\nb + \\varepsilon c + \\bar{\\varepsilon} a = 0\n\\end{cases}\n$$\n\nThe second equation is equivalent to $ABC$ being equilateral. For the first, since $ABC$ is equilateral, we can write $b = \\varepsilon a$, $c = \\bar{\\varepsilon} a$ (or vice versa). Substituting, we get:\n\n$$\n(\\varepsilon a - \\bar{\\varepsilon} a) v_1 + \\varepsilon (\\bar{\\varepsilon} a - a) v_2 + \\bar{\\varepsilon} (a - \\varepsilon a) v_3 = 0\n$$\n\nwhich simplifies to:\n\n$$\n(\\varepsilon - \\varepsilon^2) v_1 + (1 - \\varepsilon) v_2 + (\\varepsilon^2 - 1) v_3 = 0\n$$\n\nDividing by $1 - \\varepsilon \\neq 0$ and noting $v_1, v_2, v_3 \\in \\mathbb{R}$, we obtain $v_1 = v_2 = v_3$.\n\n*Remark:* Alternatively, using $\\bar{\\varepsilon} = -1 - \\varepsilon$, we can deduce $b - a = \\varepsilon(a - c)$. Substituting into the first equation and dividing by $c - a \\neq 0$, we get $v_1 - v_3 = \\varepsilon (v_2 - v_1)$, and since $v_i \\in \\mathbb{R}$, $v_1 = v_2 = v_3$.\n\n**b)** If $v_1 = v_2 = v_3 = v$, relation (*) becomes:\n\n$$\nt v \big[(c - b) + \\varepsilon (a - c) + \\bar{\\varepsilon} (b - a)\\big] + b + \\varepsilon c + \\bar{\\varepsilon} a = 0\n$$\n\nThis is invariant under translations, rotations, and homotheties, so we can fix $a = 1$, $c = \\bar{\\varepsilon}$, and obtain:\n\n$$\n(b - \\varepsilon) (t v (\\bar{\\varepsilon} - 1) + 1) = 0\n$$\n\nSince the second factor cannot be zero, we have $b = \\varepsilon$, so triangle $ABC$ is equilateral.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20441,
"subject": "Mathematics (Olympiad)",
"question": "For positive $a, b, c$ that satisfy $ab + bc + ca = 3$, prove the inequality:\n\n$$\n\\frac{1}{2a^3+1} + \\frac{1}{2b^3+1} + \\frac{1}{2c^3+1} \\ge 1.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let us make the following transformation:\n\n$$\n1 = \\frac{ab+bc+ca}{3} \\ge \\sqrt[3]{(abc)^2} \\implies abc \\le 1.\n$$\n\nHence, $a \\le \\frac{1}{bc}$, $b \\le \\frac{1}{ac}$, $c \\le \\frac{1}{ab}$, so $a+b+c \\le \\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca}$.\n\nNow, use the well-known inequality:\n\n$$\n\\frac{a_1^2}{b_1} + \\frac{a_2^2}{b_2} + \\dots + \\frac{a_n^2}{b_n} \\ge \\frac{(a_1+a_2+\\dots+a_n)^2}{b_1+b_2+\\dots+b_n}.\n$$\n\nWe make the following transformation:\n\n$$\n\\begin{align*}\n\\frac{1}{2a^3+1} + \\frac{1}{2b^3+1} + \\frac{1}{2c^3+1} &= \\frac{\\frac{1}{a^2}}{2a+\\frac{1}{a^2}} + \\frac{\\frac{1}{b^2}}{2b+\\frac{1}{b^2}} + \\frac{\\frac{1}{c^2}}{2c+\\frac{1}{c^2}} \\\\\n&\\ge \\frac{\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)^2}{2a+2b+2c+\\frac{1}{a^2}+\\frac{1}{b^2}+\\frac{1}{c^2}} \\\\\n&\\ge \\frac{\\frac{1}{2}\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)^2}{\\frac{1}{bc}+\\frac{1}{ca}+\\frac{1}{ab}+\\frac{1}{a^2}+\\frac{1}{b^2}+\\frac{1}{c^2}} \\\\\n&= \\frac{\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)^2}{\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)^2} = 1.\n\\end{align*}\n$$\n\nThat is what we had to prove.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20442,
"subject": "Mathematics (Olympiad)",
"question": "Positive real numbers $a, b, c, d$ satisfy the relations\n$$\nabcd = 4, \\quad a^2 + b^2 + c^2 + d^2 = 10.\n$$\nDetermine the largest possible value of the expression $ab + bc + cd + da$.",
"options": [],
"answer": "See solution",
"solution": "Let $V = ab + bc + cd + da$. We will find the maximum value of\n$$\nV^2 = (a + c)^2 (b + d)^2 = (a^2 + c^2 + 2ac)(b^2 + d^2 + 2bd). \\quad (1)\n$$\nAll the given expressions do not change under the simultaneous replacement of $a$ by $b$, $b$ by $c$, $c$ by $d$, and $d$ by $a$. Since $ac \\cdot bd = 4$, at least one of the numbers $ac$ and $bd$ is at least $2$. We may assume $bd \\ge 2$.\n\nSimple manipulations yield $ac = 4 / bd$ and $a^2 + c^2 = 10 - b^2 - d^2$. We plug these expressions into (1) and get\n$$\n\\begin{aligned}\nV^2 &= \\left(10 - b^2 - d^2 + \\frac{8}{bd}\\right) (b^2 + d^2 + 2bd) \\\\\n&= 10(b^2 + d^2) + 20bd + \\frac{8(b^2 + d^2)}{bd} + 16 - (b^2 + d^2)^2 - 2bd(b^2 + d^2).\n\\end{aligned} \\quad (2)\n$$\nLet $P = b^2 + d^2$ and $Q = bd$; then $P \\ge 2Q$ and $Q \\ge 2$. Thus (2) becomes\n$$\n\\begin{aligned}\nV^2 &= 10P + 20Q + \\frac{8P}{Q} + 16 - P^2 - 2PQ \\\\\n&= -(P^2 - 10P + 25) + \\left(41 - 2PQ + 20Q + \\frac{8P}{Q}\\right) \\\\\n&= -(P - 5)^2 + \\left[P\\left(\\frac{8}{Q} - 2Q\\right) + 41 + 20Q\\right].\n\\end{aligned}\n$$\nObviously, $-(P-5)^2 \\le 0$. The condition $Q \\ge 2$ implies $\\frac{8}{Q} - 2Q \\le 0$, which means the expression in the brackets is linear in $P$ with non-positive slope and it achieves its maximum for the smallest possible $P$. By $P \\ge 2Q$ we obtain\n$$\nV^2 \\le 2Q \\left( \\frac{8}{Q} - 2Q \\right) + 41 + 20Q = -4Q^2 + 20Q + 57 = -(2Q - 5)^2 + 82 \\le 82.\n$$\nFinally, we shall show that there are positive real numbers $a, b, c, d$ such that $V = \\sqrt{82}$. The equality occurs if $P = 2Q = 5$, which is true for $b = d = \\frac{1}{2}\\sqrt{10}$. The numbers $a$ and $c$ satisfy $a^2 + c^2 = 5$, $ac = 8/5$. Thus\n$$\n\\{a, c\\} = \\left\\{ \\frac{\\sqrt{41}-3}{2\\sqrt{5}}, \\frac{\\sqrt{41}+3}{2\\sqrt{5}} \\right\\}.\n$$\n**Answer:** The maximum possible value of $ab + bc + cd + da$ is $\\sqrt{82}$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20443,
"subject": "Mathematics (Olympiad)",
"question": "Each unit square in a $2016 \\times 2016$ grid contains a positive integer. You play a game on the grid in which the following two types of moves are allowed:\n\n- Choose a row and multiply every number in the row by $2$.\n- Choose a positive integer, choose a column, and subtract the positive integer from every number in the column.\n\nYou win if all of the numbers in the grid are $0$. Is it always possible to win after a finite number of moves?",
"options": [],
"answer": "See solution",
"solution": "We will prove that it is possible to make every number in an $m \\times n$ grid equal to $0$ after a finite number of moves. The proof will be by induction on $n$, the number of columns in the grid.\n\nConsider the base case, in which the number of columns is $1$. Let the difference between the maximum and minimum numbers in the column be $D$. We will show that if $D > 0$, then it is possible to reduce the value of $D$ after a finite number of moves. First, we subtract a positive integer from every number in the column to make the minimum number in the column $1$. Now multiply all of the entries equal to $1$ by $2$, which reduces $D$ by $1$. Therefore, after a finite number of moves, it is possible to make all numbers in the column equal to each other. By subtracting this number from each entry of the column, we have made every number in the column equal to $0$.\n\nNow consider an $m \\times n$ grid with $n \\ge 2$. Suppose that we can make every number in a grid with $n-1$ columns equal to $0$ after a finite number of moves. We simply apply the construction of the previous paragraph to the leftmost column in the grid. Note that the entries in the remaining columns may change, but remain positive. Therefore, after a finite number of moves, we obtain a grid whose leftmost column contains only $0$. By the inductive hypothesis, we can make every number in the remaining $m \\times (n-1)$ grid equal to $0$ after a finite number of moves. Furthermore, observe that these moves do not change the $0$ entries in the leftmost column. Therefore, it is always possible to make every number in an $m \\times n$ grid equal to $0$ after a finite number of moves.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20444,
"subject": "Mathematics (Olympiad)",
"question": "Find distinct numbers $a, b, c, d$ such that:\n\n$$\n\\frac{1}{2011} = \\frac{a}{a+1} + \\frac{b}{b+1} - \\frac{c}{c+1} - \\frac{d}{d+1}\n$$",
"options": [],
"answer": "See solution",
"solution": "We use the following formulas:\n\n$$\n\\frac{1}{n} = \\frac{n-1}{n(n-1)} = \\frac{1}{n-1} - \\frac{1}{n(n-1)}, \\quad \\frac{1}{n} = \\frac{n+1}{n(n+1)} = \\frac{1}{n+1} + \\frac{1}{n(n+1)}\n$$\n\nUsing these, we arrive at:\n\n$$\n\\frac{1}{2011} = \\frac{1}{2010} - \\frac{1}{2010 \\cdot 2011} = \\frac{1}{2011} + \\frac{1}{2011 \\cdot 2012} - \\frac{1}{2010 \\cdot 2011 + 1}\n$$\n\nSubtracting $\\frac{1}{2010 \\cdot 2011(2010 \\cdot 2011+1)}$, we get:\n\n$$\n\\frac{1}{a+1} + \\frac{1}{b+1} - \\frac{1}{c+1} - \\frac{1}{d+1}\n$$\n\nAdding 1 to the first two fractions and subtracting one from the last two, we get:\n\n$$\n\\frac{1}{2011} = \\frac{2010 \\cdot 2011 \\cdot 2012 - 1}{2011 \\cdot 2011 \\cdot 2012} + \\frac{2010 \\cdot 2011}{2010 \\cdot 2011 + 1} + \\frac{2010 \\cdot 2011(2010 \\cdot 2011 + 1) - 1}{2010 \\cdot 2011(2010 \\cdot 2011 + 1)}\n$$\n\nThus, we have expressed $\\frac{1}{2011}$ in the required form.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20445,
"subject": "Mathematics (Olympiad)",
"question": "Let $S$ be the set of positive integers. Determine all functions $f: S \\to S$ such that $x^2 + f(y)$ divides $f(x)^2 + y$ for every pair of positive integers $x$ and $y$.",
"options": [],
"answer": "See solution",
"solution": "Consider $x = y = 1$. Then we get that $1 + f(1)$ divides $f(1)^2 + 1$. So $1 + f(1)$ divides $f(1)^2 + 1$. This implies $1 + f(1) \\leq f(1)^2 + 1$, and checking possible values, we find $f(1) = 1$.\n\nFor $y = 1$, we obtain $x^2 + 1$ divides $f(x)^2 + 1$. Therefore, $x^2 \\leq f(x)^2$, which means $x \\leq f(x)$ for every positive integer $x$.\n\nFor $x = 1$, we get $1 + f(y)$ divides $1 + y$, and hence $f(y) \\leq y$ for every positive integer $y$.\n\nFrom the above inequalities, it follows that the unique function is $f(x) = x$ for all $x \\in S$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20446,
"subject": "Mathematics (Olympiad)",
"question": "Square $ABCD$ of size $2019 \\times 2019$ is divided by two lines into four rectangles with integer side lengths, and some of the rectangles might be squares. Turns out, the area of the rectangle containing vertex $A$ equals the perimeter of the rectangle containing vertex $B$. What is the area of the smallest of the four rectangles (or squares)?\n\n",
"options": [],
"answer": "See solution",
"solution": "Let us denote the sides of the rectangle containing vertex $A$ as $a$ and $b$ (see figure). Then the sides of the rectangle containing vertex $B$ are $a$ and $2019-b$. The equality of the area of the first rectangle and perimeter of the second yields:\n\n$$\nab = 2(a + 2019 - b) \\Rightarrow ab - 2a + 2b = 4038 \\Rightarrow (a+2)(b-2) = 4034 = 2017 \\cdot 2.$$ \n\nSince one of the factors is greater than $2$ and $2017$ is prime, the only solution is $a+2=2017$ and $b-2=2$. Hence, $a=2015$ and $b=4$. Clearly, the smallest rectangle of the four would be the square with area $4 \\cdot 4 = 16$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20447,
"subject": "Mathematics (Olympiad)",
"question": "Prove that, among any $n$ vertices of a regular $2n - 1$ polygon ($n \\ge 3$), there are three of them which are the vertices of an isosceles triangle.",
"options": [],
"answer": "See solution",
"solution": "Since it is easy to verify directly the assertion for the cases $n = 3$ (a pentagon) and $n = 4$ (a heptagon), we may assume that $n > 4$ in the following discussion.\n\nBy reduction to absurdity, assume we can select $n$ vertices in a regular $2n-1$ polygon $A_1A_2A_3\\cdots A_{2n-1}$ such that no three of them constitute an isosceles triangle. We mark these points with color red and the remaining $n-1$ ones with blue, respectively.\n\nWe may let $A_1$ be red, and divide the other $2n-2$ points into $n-1$ pairs (see the figure): $(A_2, A_{2n-1})$, $(A_3, A_{2n-2})$, ..., $(A_n, A_{n+1})$.\n\nThe two points in each pair cannot both be red as they, together with $A_1$, constitute an isosceles triangle, and must be one red and one blue since there are exactly $n$ red points.\n\n\n\nAssuming $A_2$ is red, then $A_{2n-1}$ is blue, and $A_3$ must be blue as $\\triangle A_1A_2A_3$ is an isosceles triangle. Therefore $A_{2n-2}$ is red, from which we infer that $A_5, A_{2n-4}$ are both blue, as $\\triangle A_{2n-2}A_2A_5$ and $\\triangle A_{2n-4}A_{2n-2}A_1$ are two isosceles triangles. But $A_5, A_{2n-4}$ are the two points in a pair, so they must be one red and one blue. This is a contradiction! The assertion for $n > 4$ is then also true. The proof is complete. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20448,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c, d$ be the roots of the equation $x^4 + x + 1 = 0$. Compute\n$$\np + 2q + 4r + 8s,\n$$\nwhere $p, q, r, s$ are the coefficients of the monic quartic polynomial whose roots are $a^5 + 2a + 1$, $b^5 + 2b + 1$, $c^5 + 2c + 1$, and $d^5 + 2d + 1$.",
"options": [],
"answer": "See solution",
"solution": "Firstly, since $a^4 + a + 1 = 0$, we have\n$$\na^5 + 2a + 1 = a(a^4 + a + 1) - a^2 + a + 1 = -a^2 + a + 1.\n$$\nLet $y = -x^2 + x + 1$. Then $x = \\frac{1 \\pm \\sqrt{-4y+5}}{2}$. Now,\n$$\n\\begin{aligned}\n0 &= x^4 + x + 1 \\\\\n &= \\left( \\frac{3 - 2y \\pm \\sqrt{-4y+5}}{2} \\right)^2 + \\frac{1 \\pm \\sqrt{-4y+5}}{2} + 1 \\\\\n &= \\frac{2y^2 - 8y + 7 \\pm (3 - 2y)\\sqrt{-4y+5}}{2} + \\frac{1 \\pm \\sqrt{-4y+5}}{2} + 1 \\\\\n &= y^2 - 4y + 5 \\pm (2 - y)\\sqrt{-4y+5}.\n\\end{aligned}\n$$\nThis implies $(y^2 - 4y + 5)^2 = (y - 2)^2(-4y + 5)$, which is a degree 4 polynomial equation. Therefore, $a^5 + 2a + 1, b^5 + 2b + 1, c^5 + 2c + 1$ and $d^5 + 2d + 1$ are roots of the equation\n$$\nf(y) = (y^2 - 4y + 5)^2 - (y - 2)^2(-4y + 5) = 0.\n$$\nNote that $f$ is monic. So it is exactly $x^4 + px^3 + qx^2 + rx + s$. Thus,\n$$\np + 2q + 4r + 8s = 8f\\left(\\frac{1}{2}\\right) - \\frac{1}{2} = 30.\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20449,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\in \\mathbb{N}$ and $\\mathcal{A}$ be a nonempty family of nonempty subsets of $\\{1, 2, \\dots, n\\}$ with the following property: if $A \\in \\mathcal{A}$ and $A \\subset B \\subseteq \\{1, 2, \\dots, n\\}$, then $B \\in \\mathcal{A}$. Prove that the function\n\n$$\nf(x) := \\sum_{A \\in \\mathcal{A}} x^{|A|} (1-x)^{n-|A|}\n$$\n\nis strictly increasing in the interval $(0, 1)$.",
"options": [],
"answer": "See solution",
"solution": "Let $0 < p < q < 1$. Notice that $p^* = \\frac{q-p}{1-p} \\in (0, 1)$. We construct the sets $X$ and $Y$ as follows: For each element $i \\in \\{1, 2, \\dots, n\\}$, we put $i$ in $X$ with probability $p$ (independently of each other), and for each element $j \\in \\{1, 2, \\dots, n\\} \\setminus X$, we put $j$ in $Y$ with probability $p^*$. Then $P(x \\in X \\cup Y) = p + (1-p)p^* = q$. It is easy to see that $f(p) = P(X \\in \\mathcal{A})$, and $f(q) = P(X \\cup Y \\in \\mathcal{A})$. Since $\\mathcal{A}$ has the property from the condition, we have $X \\subseteq X \\cup Y \\implies f(p) < f(q)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20450,
"subject": "Mathematics (Olympiad)",
"question": "Find the least number of buttons that can be placed on the squares of a $4 \\times 4$ grid so that no two buttons are on the same square or on squares with a common side (buttons may be on squares with a common vertex), and no buttons can be added to the grid under the same conditions.",
"options": [],
"answer": "See solution",
"solution": "If there are 3 or fewer buttons placed, then it is always possible to add more, as each button blocks up to 5 squares: the square it is in and its 4 neighbors. Thus, 3 buttons cover up to $3 \\cdot 5 = 15$ of the $4 \\times 4 = 16$ squares of the grid, meaning that a button can be placed in at least one of the squares. It remains to observe that by placing 4 buttons as shown below, no more buttons can be added.\n\nWe say that a button covers a square if the button lies on either the square itself or one of its neighbors. Observe that one button can cover at most one of the 4 corner squares of the grid, meaning that at least 4 buttons are needed to cover the grid. To show that 4 buttons are sufficient, we can use the same placement as in Solution 1.\n\n\n\nFig. 2",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20451,
"subject": "Mathematics (Olympiad)",
"question": "由於疫情持續擴大，政府決定對交通實施管制，於是在平面上建築了 $n$ 條無限長直線作為牆壁，其中任兩線不平行，任三線不共點；這些牆壁的集合被記為 $W_n$。\n\n政府將 $W_n$ 中每一面牆的其中一面塗上藍色，另一面則塗上綠色。這讓任兩面牆的交叉點都分為四個角落：兩面牆都是藍色的同色角落，兩面牆都是綠色的同色角落，以及兩面牆一藍一綠的異色角落。\n\n為了保持最起碼的交通，政府在任兩牆的交叉點開了一扇門，使得人們可以從其異色角落通到另一個異色角落。除此之外，人們沒有其他方式可以跨越牆。\n\n給定 $W_n$，令 $k(W_n)$ 為最大的正整數 $k$，使得不論政府如何塗色，我們都可以在平面上放置 $k$ 個人，讓其中任兩個人永遠無法在平面上碰頭。對於所有 $n$，試求 $k(W_n)$ 的所有可能值。",
"options": [],
"answer": "See solution",
"solution": "對於所有 $n$，$k(W_n)$ 的唯一可能值是 $k = n + 1$。\n\n由數學歸納法易知 $W_n$ 將平面分為 $\\binom{n+1}{2} + 1$ 區。我們將此題轉化為一個圖 $G$，其中每一點對應一區，而可由門相通的兩區以邊連線。\n\n首先我們證明 $k(W_n) \\geq n + 1$。注意到牆的交點必為 $\\binom{n}{2}$ 個，因此 $G$ 的邊數必為 $\\binom{n}{2}$ 個。讓我們將 $G$ 中所有的邊先移除，然後一條一條加回去，則我們每加回一條邊時，$G$ 中的連通區域數至多減一，從而最終 $G$ 的連通區域數至少為 $\\binom{n+1}{2} + 1 - \\binom{n}{2} = n + 1$。若我們在每個連通區域各擺一個人，則顯然這些人永無法碰頭，從而 $k(W_n) \\geq n + 1$。\n\n接下來我們證明，不論 $W_n$ 為何，以下的塗色方法將迫使我們最多只能擺 $n + 1$ 個人：\n\n- 首先，選擇一個直角座標系統，使得沒有一座牆是南北向或東西向；\n- 將每座牆的西面塗成綠色，東面塗成藍色。\n\n此外，我們計算每座牆的東進值 $E$，即“該區域在多少牆的東邊”。顯然 $E$ 介於 $0$ 到 $n$ 之間，且這之間每個值都恰有一個“北邊沒有牆”的區域有對應的 $E$ 值。因此，若我們能夠證明：\n\n**Claim. 所有有相同 $E$ 值的區域互相連通。**\n\n則我們便證明至多只能擺 $n + 1$ 個人，從而原命題得證。\n\n**Proof of Claim.** 固定 $i \\in \\{0, \\dots, n\\}$，並將一個人放入一個 $E = i$ 的區域。讓這個人往其北方前進。注意到每個有限區域皆為凸多邊形，且其最北邊必為一交叉點，必有一門，故此人最終必達到一北方無牆的區域。此外，注意到當人經過一門，抵達一新的區域時，恰有一牆從其東邊換到西邊，且恰有一牆從其西邊換到東邊，因此此人路徑上的所有區域的 $E$ 值皆相等。因此，所有 $E = i$ 的區域，皆與北邊無牆且 $E = i$ 的區域相連通，故證畢。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20452,
"subject": "Mathematics (Olympiad)",
"question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function satisfying the functional equation\n$$\nf(\\lfloor x \\rfloor y) = f(x) \\lfloor f(y) \\rfloor \\quad \\text{for all } x, y \\in \\mathbb{R}.\n$$\nFind all such functions $f$.",
"options": [],
"answer": "See solution",
"solution": "Set $x = 0$ in the original functional equation to get $f(0) = f(0) \\lfloor f(y) \\rfloor$ for all $y \\in \\mathbb{R}$.\n\n**Case 1.** $f(0) \\neq 0$. Then $\\lfloor f(y) \\rfloor = 1$ for all $y$, so $f(\\lfloor x \\rfloor y) = f(x)$ for all $x, y$. Setting $x = 1$ gives $f(y) = f(1)$ for all $y$, so $f$ is constant. The only possible constant values are $f(x) = c$ with $1 \\leq c < 2$ (since $\\lfloor c \\rfloor = 1$ and $c \\neq 0$).\n\n**Case 2.** $f(0) = 0$. Set $x = 1$ to get $f(y) = f(1) \\lfloor f(y) \\rfloor$.\n\n- **Case 2a.** $f(1) = 0$. Then $f(y) = 0$ for all $y$, which is a valid solution.\n- **Case 2b.** $f(1) \\neq 0$. Let $g(x) = \\frac{f(x)}{f(1)}$. Then $g(0) = 0$, $g(1) = 1$, and $g(y) = \\lfloor f(y) \\rfloor$ for all $y$. Substituting into the original equation gives\n $$\ng(\\lfloor x \\rfloor y) = g(x)g(y) \\quad (*)\n $$\n for all $x, y$. Setting $x = y = -1$ gives $g(-1)^2 = 1$, so $g(-1) \\neq 0$. Setting $y = 1$ gives $g(\\lfloor x \\rfloor) = g(x)$. Setting $x = 1, y = \\frac{1}{2}$ gives $g(-\\frac{1}{2}) = g(\\frac{1}{2})g(-1)$, but $g(\\frac{1}{2}) = g(0) = 0$ and $g(-\\frac{1}{2}) = g(-1) \\neq 0$, a contradiction. So this case does not occur.\n\n**Conclusion:** The solutions are $f(x) = 0$ for all $x$, or $f(x) = c$ for all $x$ with $1 \\leq c < 2$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20453,
"subject": "Mathematics (Olympiad)",
"question": "Consider $a \\in (0,1)$ and let $C$ be the set of all increasing functions $f: [0,1] \\to [0, \\infty)$ such that $$\\int_0^1 f(x)\\,dx = 1.$$ \nDetermine:\n\n(a) $\\max_{f \\in C} \\int_0^a f(x)\\,dx$\n\n(b) $\\max_{f \\in C} \\int_0^a (f(x))^2\\,dx$",
"options": [],
"answer": "See solution",
"solution": "The required maximum is $a$ if $a \\leq \\frac{1}{2}$, and $\\frac{1}{4(1-a)}$ if $a > \\frac{1}{2}$. In both cases, it is achieved at an essentially unique function.\n\nFor $a \\leq \\frac{1}{2}$, the maximum is achieved at $f(x) \\equiv 1$ on $(0,1)$; at $x=0$ the function may take any non-negative value less than or equal to $1$, and at $x=1$ it may take any value greater than or equal to $1$.\n\nFor $a > \\frac{1}{2}$, the maximum is achieved at\n$$\nf(x) = \\begin{cases} 0 & \\text{if } 0 \\leq x < 2a - 1, \\\\ \\frac{1}{2(1-a)} & \\text{if } 2a - 1 < x < 1. \\end{cases}\n$$\nAt $x = 2a - 1$ the function may take any non-negative value less than or equal to $\\frac{1}{2(1-a)}$, and at $x=1$ it may take any value greater than or equal to $\\frac{1}{2(1-a)}$.\n\nLet $f \\in C$. Notice that\n$$\n\\begin{aligned}\n\\int_{0}^{a} (f(x))^2\\,dx &\\leq f(a) \\int_{0}^{a} f(x)\\,dx \\\\\n&\\leq \\left( \\frac{1}{1-a} \\int_{a}^{1} f(x)\\,dx \\right) \\left( \\int_{0}^{a} f(x)\\,dx \\right) \\\\\n&= \\frac{1}{1-a} \\left( 1 - \\int_{0}^{a} f(x)\\,dx \\right) \\left( \\int_{0}^{a} f(x)\\,dx \\right).\n\\end{aligned}\n$$\n\nNext, we show that\n$$\n\\int_{0}^{a} f(x)\\,dx \\leq a.\n$$\nTo see this,\n$$\n\\begin{aligned}\na - \\int_{0}^{a} f(x)\\,dx &= a \\int_{0}^{1} f(x)\\,dx - \\int_{0}^{a} f(x)\\,dx \\\\\n&= a \\int_{a}^{1} f(x)\\,dx - (1-a) \\int_{0}^{a} f(x)\\,dx \\\\\n&\\geq a \\int_{a}^{1} f(a)\\,dx - (1-a) \\int_{0}^{a} f(a)\\,dx = 0.\n\\end{aligned}\n$$\nEquality holds if and only if $f(x) = 1$ for $0 < x < 1$; at $x=0$ the function may take any non-negative value less than or equal to $1$, and at $x=1$ it may take any value greater than or equal to $1$.\n\nFurther, note that\n$$\n\\max \\{ t(1-t) : t \\leq a \\} = \\begin{cases} a(1-a) & \\text{if } a \\leq \\frac{1}{2}, \\\\ \\frac{1}{4} & \\text{if } a > \\frac{1}{2}. \\end{cases}\n$$\nIn both cases, the maximum is achieved at a single point: $t=a$ for $a \\leq \\frac{1}{2}$, and $t=\\frac{1}{2}$ for $a > \\frac{1}{2}$.\n\nConsequently,\n$$\n\\int_{0}^{a} (f(x))^2\\,dx \\leq \\begin{cases} a & \\text{if } a \\leq \\frac{1}{2}, \\\\ \\frac{1}{4(1-a)} & \\text{if } a > \\frac{1}{2}. \\end{cases}\n$$\n\nIf $a \\leq \\frac{1}{2}$, then\n$$\n\\int_{0}^{a} (f(x))^2\\,dx = a\n$$\nforces\n$$\n\\int_{0}^{a} f(x)\\,dx = a,\n$$\nso $f(x) = 1$ for $0 < x < 1$; at $x=0$ the function may take any non-negative value less than or equal to $1$, and at $x=1$ it may take any value greater than or equal to $1$.\n\nIf $a > \\frac{1}{2}$, then\n$$\n\\int_{0}^{a} (f(x))^2\\,dx = \\frac{1}{4(1-a)}\n$$\nforces\n$$\n\\int_{0}^{a} f(x)\\,dx = \\frac{1}{2} \\qquad (1)\n$$\nand\n$$\n\\int_{0}^{a} (f(x))^2\\,dx = f(a) \\int_{0}^{a} f(x)\\,dx \\qquad (2)\n$$\nso\n$$\n\\int_{a}^{1} f(x)\\,dx = \\frac{1}{2}, \\quad f(a) = \\frac{1}{2(1-a)}.\n$$\nThis forces $f(x) = \\frac{1}{2(1-a)}$ for $a \\leq x < 1$; at $x=1$ the function may take any value greater than or equal to $\\frac{1}{2(1-a)}$.\n\nWe now show that (1) and (2) force\n$$\nf(x) = \\begin{cases} 0 & \\text{if } 0 < x < 2a - 1, \\\\ \\frac{1}{2(1-a)} & \\text{if } 2a - 1 < x < a. \\end{cases}\n$$\nOf course, $f(0) = 0$, and at $x = 2a - 1$ the function may take any non-negative value less than or equal to $\\frac{1}{2(1-a)}$. To prove this, consider the sets:\n$$\nA = \\{ x : 0 < x < a, f(x) = 0 \\},\n$$\n$$\nB = \\{ x : 0 < x < a, 0 < f(x) < f(a) \\},\n$$\n$$\nC = \\{ x : 0 < x < a, f(x) = f(a) \\}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20454,
"subject": "Mathematics (Olympiad)",
"question": "Determine all real-valued functions $f$ that satisfy\n\n$$\n2f(xy + xz) + 2f(xy - xz) \\geq 4f(x)f(y^2 - z^2) + 1\n$$\n\nfor all real numbers $x, y, z$.",
"options": [],
"answer": "See solution",
"solution": "Suppose that $f$ satisfies the given inequality. For $x = y = z = 0$, we obtain\n\n$$\n2f(0) + 2f(0) \\geq 4f(0)^2 + 1,\n$$\n\ni.e. $0 \\geq (2f(0) - 1)^2$, whence $f(0) = \\frac{1}{2}$.\n\nFor $x = y = 1$, $z = 0$, we obtain\n\n$$\n2f(1) + 2f(1) \\geq 4f(1)^2 + 1,\n$$\n\ni.e. $0 \\geq (2f(1) - 1)^2$, whence $f(1) = \\frac{1}{2}$.\n\nFor $y = z = 0$, we obtain\n\n$$\n2f(0) + 2f(0) \\geq 4f(x)f(0) + 1,\n$$\n\nwhence $2 \\geq 2f(x) + 1$, i.e. $\\frac{1}{2} \\geq f(x)$ for all $x$.\n\nFor $y = 1$, $z = 0$, we obtain\n\n$$\n2f(x) + 2f(x) \\geq 4f(x)f(1) + 1,\n$$\n\nwhence $4f(x) \\geq 2f(x) + 1$, i.e. $f(x) \\geq \\frac{1}{2}$ for all $x$.\n\nIt follows that $f(x) = \\frac{1}{2}$ for all $x$.\n\nConversely, the function $f$ defined by $f(x) = \\frac{1}{2}$ for all $x$ satisfies the given inequality as can be seen by inspection.\n\n**Answer:** $f(x) = \\frac{1}{2}$ for all $x$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20455,
"subject": "Mathematics (Olympiad)",
"question": "Suppose a bounded number sequence $\\{a_n\\}$ satisfies\n$$\na_n < \\sum_{k=n}^{2n+2006} \\frac{a_k}{k+1} + \\frac{1}{2n+2007}, \\quad n = 1, 2, 3, \\dots\n$$\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$$\nb_n < \\sum_{k=n}^{2n+2006} \\frac{b_k}{k+1}, \\quad n = 1, 2, 3, \\dots. \\qquad \\textcircled{1}\n$$\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$$\n\\begin{align*}\nb_n &< \\sum_{k=n}^{2n+2006} \\frac{b_k}{k+1} \\\\\n&< M \\sum_{k=n}^{2n+2006} \\frac{1}{k+1} \\\\\n&= M \\sum_{k=n}^{\\lfloor \\frac{3n}{2} \\rfloor} \\frac{1}{k+1} + M \\sum_{k=\\lfloor \\frac{3n}{2} \\rfloor+1}^{2n+2006} \\frac{1}{k+1} \\\\\n&< M \\cdot \\frac{1}{2} + M \\cdot \\frac{\\frac{n}{2} + 2006}{\\frac{3n}{2} + 1} \\\\\n&< \\frac{6}{7} M,\n\\end{align*}\n$$\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$$\nb_n < \\left(\\frac{6}{7}\\right)^m M,\n$$\nwhich implies that $b_n \\le 0$ for $n \\ge 100\\,000$. Substitute this into (1) to 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 $b_N < 0$. That means\n\n$b_n < 0$ for $n = 1, 2, 3, \\dots$.\n\nThis implies that $a_n < \\frac{1}{n}$ for $n = 1, 2, 3, \\dots$.\n\nThe proof is complete.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20456,
"subject": "Mathematics (Olympiad)",
"question": "Determine whether there exist two different sets $A, B$, each consisting of at most $2011^2$ positive integers, such that for every $x$ with $0 < x < 1$,\n\n$$\n\\left| \\sum_{a \\in A} x^a - \\sum_{b \\in B} x^b \\right| < (1-x)^{2011}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Yes, such sets exist. We construct them as follows:\n\nRewrite the inequality using $y = 1 - x$:\n\n$$\n\\left| \\sum_{a \\in A} (1-y)^a - \\sum_{b \\in B} (1-y)^b \\right| < y^{2011}\n$$\nfor all $0 < y < 1$.\n\n**Step 1:** There exist two different sets $A', B'$ of $2011^2$ positive integers each such that for $k = 0, 1, \\dots, 2011$,\n$$\n\\sum_{a \\in A'} \\binom{a}{k} = \\sum_{b \\in B'} \\binom{b}{k}.\n$$\nThis follows by the pigeonhole principle: For large $N$, the number of possible $2011^2$-element subsets of $\\{1,2,\\dots,N\\}$ exceeds the number of possible $2011$-tuples of binomial sums, so two distinct sets $A', B'$ must have the same tuple.\n\n**Step 2:** Expanding $(1-y)^a$ and $(1-y)^b$, all terms of degree $\\leq 2011$ cancel, so only terms of degree $\\geq 2012$ remain. Let $M$ be the sum of the absolute values of the coefficients of these terms. Then\n$$\n\\left| \\sum_{a \\in A'} (1-y)^a - \\sum_{b \\in B'} (1-y)^b \\right| \\leq M y^{2012}\n$$\nfor all $0 < y < 1$.\n\n**Step 3:** For all $y$, $(1-y)^M < \\frac{1}{My}$ (by AM-GM inequality). Thus, multiplying the previous bound by $(1-y)^M$ gives the desired inequality.\n\n**Step 4:** Let $A = \\{a + M \\mid a \\in A'\\}$ and $B = \\{b + M \\mid b \\in B'\\}$. Then $A$ and $B$ satisfy the original inequality.\n\n**Remark:** If only finiteness is required, an alternate solution uses the expansion of $(1-x)(1-x^2)\\cdots(1-x^{2011})$; assign $A'$ to exponents with $+1$ coefficients and $B'$ to those with $-1$, then shift by $M$ as above.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20457,
"subject": "Mathematics (Olympiad)",
"question": "Define the sequence $a_1, a_2, a_3, \\dots$ by $a_1 = 1$ and\n\n$$\na_n = a_{\\lfloor n/2 \\rfloor} + a_{\\lfloor n/3 \\rfloor} + \\dots + a_{\\lfloor n/n \\rfloor} + 1\n$$\n\nfor $n > 1$. Prove that there are infinitely many $n$ such that\n\n$$\na_n \\equiv n \\pmod{2^{2010}}.\n$$",
"options": [],
"answer": "See solution",
"solution": "*Second proof of Lemma 1.* Let us interpret the sequence $a_n$ combinatorially as follows. Call a finite increasing sequence of integers $k_1 < k_2 < \\dots < k_r$ good if $k_1 = 1$ and $k_j \\mid k_{j+1}$ for each $1 \\le j < r$. Let $c_n$ be the number of good sequences whose last term is at most $n$. Then, we claim that $a_n = c_n$.\n\nIndeed, for any $1 < k \\le n$, good sequences with second term $k$ and last term at most $n$ are in bijection with good sequences with last term at most $\\lfloor n/k \\rfloor$; here, the bijection is provided by dividing each term after the first by $k$. Counting also the sequence consisting of the single term 1, we obtain the recurrence\n\n$$\nc_n = c_{\\lfloor n/2 \\rfloor} + c_{\\lfloor n/3 \\rfloor} + \\dots + c_{\\lfloor n/n \\rfloor} + 1.\n$$\n\nObserve further that $c_1 = a_1 = 1$, meaning that the sequence $\\{c_n\\}$ satisfies the same recurrence and initial conditions as $\\{a_n\\}$. Therefore, we obtain $a_n = c_n$.\n\nNow, note that $a_n - a_{n-1}$ is the number of good sequences whose last term is exactly $n$. It suffices therefore to show that if $p^s$ is the highest power of $p$ dividing $n$, then the number of good sequences ending in $n$ is divisible by $2^{s-1}$. Let the $p$-skeleton of a good sequence $k_1, \\dots, k_r$ be the sub-sequence consisting of all the terms $k_j$ such that $k_j/k_{k-1}$ is not a power of $p$ (including the initial 1). It suffices for us to show that the number of good sequences ending in $n$ with a given $p$-skeleton is divisible by $2^{s-1}$.\n\nTake any $p$-skeleton, which we may write in the form\n\n$$\np^{t_1} k'_1, p^{t_2} k'_2, \\dots, p^{t_m} k'_m\n$$\n\nfor some $t_1, \\dots, t_m$ and $k'_1, \\dots, k'_m$ satisfying $t_1 = 0$, $k'_1 = 1$, $p \\nmid k'_j$, $k'_1 \\mid k'_2 \\mid \\dots \\mid k'_m$, and $t_1 \\le t_2 \\le \\dots \\le t_m$. Now, to form a good sequence ending in $n$ that has this $p$-skeleton, between any two consecutive terms $p^{t_j} k'_j, p^{t_{j+1}} k'_{j+1}$ of the $p$-skeleton we can insert any subset of the set\n\n$$\n\\{p^{t_{j+1}} k'_j, p^{t_{j+2}} k'_j, \\dots, p^{t_{j+1}} k'_j\\}.\n$$\n\nAlso, if the last term $p^{t_{r'}} k'_{r'}$ is not equal to $n$, we can insert any subset containing $n$ of\n\n$$\n\\{p^{t_{r'}+1} k'_{r'}, p^{t_{r'}+2} k'_{r'}, \\dots, p^{s-1} k'_{r'}\\}\n$$\nafter it. Hence, for each $t = 1, \\dots, s-1$, there is exactly one number of the form $p^t k$ with $p \\nmid k$ that we can choose to include or not in our good sequence. For $t = s$, we can choose to include the number if $n$ is not in the skeleton; otherwise including it is obligatory. So the number of good sequences ending in $n$ that have the given $p$-skeleton is either $2^{s-1}$ or $2^s$, depending whether or not $n$ is part of the skeleton; in any case, it is divisible by $2^{s-1}$. It follows that the total number of good sequences ending in $n$ is divisible by $2^{s-1}$, establishing the lemma. $\\square$\n\nWe now consider the problem proper. For any positive integer $m$, choose distinct primes $p_1, p_2, \\dots, p_m$. By the Chinese Remainder Theorem, there exists some $k$ such that $k + i$ is divisible by $p_i^s$ for $1 \\le i \\le m$. By Lemma 1, this implies that $a_{k+i} - a_{k+i-1}$ is divisible by $2^{s-1}$ for $1 \\le i \\le m$, meaning that $a_k, a_{k+1}, \\dots, a_{k+m}$ are all congruent modulo $2^{s-1}$. For any $N$, if we take $m > N + 2^{s-1} - 1$, there exist positive integers $k$ and $n \\in \\{k+N, \\dots, k+N+2^{s-1}-1\\}$ such that $a_n \\equiv n \\pmod{2^{s-1}}$. In particular, this $n$ satisfies $n > N$. Therefore, for any positive integer $s$, the set of $n$ such that $a_n \\equiv n \\pmod{2^{s-1}}$ is unbounded, hence infinite. Taking $s = 2011$ gives the desired result.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20458,
"subject": "Mathematics (Olympiad)",
"question": "A table is filled in steps, where each step $k$ fills $4k + 1$ cells. After 31 steps, $2016$ numbers are filled. In which step is the number $2022$ written, and what are its coordinates (column and row) in the table?",
"options": [],
"answer": "See solution",
"solution": "After 31 steps, $2016$ numbers are filled. Step 32 fills $4 \\cdot 32 + 1 = 129$ more numbers, so $2022$ is written in step 32. The last number of step 31, $2016$, is in column $50 - 31 = 19$ and row $1$ (since step 31 is odd, filled anti-clockwise). Step 32 is filled clockwise, so $2022$ is in column $18$ and row $6$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20459,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\alpha$ and $\\beta$ be real numbers with $\\beta \\neq 0$. Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(\\alpha f(x) + f(y)) = \\beta x + f(y)\n$$\n\nholds for all real $x$ and $y$.",
"options": [],
"answer": "See solution",
"solution": "The functional equation only has solutions for $\\alpha = \\beta$, namely:\n\n- For $\\alpha = \\beta = -1$, the functions $f(x) = x + C$ with $C \\in \\mathbb{R}$.\n- For $\\alpha = \\beta \\neq 0, -1$, the function $f(x) = x$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20460,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer for which $5n + 1$ is a perfect square. Show that $n + 1$ is a sum of 5 perfect squares.",
"options": [],
"answer": "See solution",
"solution": "Let $5n + 1 = m^2 \\equiv 1 \\pmod{5}$. Thus $m = 5k \\pm 1$ for some integer $k$. We have\n\n$$\nn + 1 = \\frac{(5k \\pm 1)^2 + 4}{5} = 5k^2 \\pm 2k + 1 = 4k^2 + (k \\pm 1)^2\n$$\n\nwhich can be written as a sum of 5 perfect squares as desired. $\\blacksquare$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20461,
"subject": "Mathematics (Olympiad)",
"question": "Find all non-negative real numbers $c \\ge 0$ such that there exists a function $f : (0, +\\infty) \\to (0, +\\infty)$ with the property\n\n$$\nf(y^2 f(x) + y + c) = x f(x + y^2)\n$$\n\nfor all $x, y > 0$.",
"options": [],
"answer": "See solution",
"solution": "\n\n$$\n\\therefore \\text{يوجد و يحقق التساوي } f(x+\\frac{1}{x}) \\ge f(x) \\text{ ($x \\ge 0$)} \\\\\n\\text{لما } x > 0 \\text{ نريد } \\therefore \\text{ أن } f(x) \\ge 0\n$$\n\n$$\n\\therefore \\text{如果 } \\left( \\frac{1-f(x)}{x} \\right)^{\\frac{1}{1-f(x)}} \\le \\frac{1}{4} \\text{ と } \\left( \\frac{1-f(x)}{x} \\right)^{\\frac{1}{1-f(x)}} > \\frac{1}{4} \\\\\n\\text{那么 } f(x) \\ge \\frac{1+\\sqrt{1-4x(1-f(x))}}{\\left(\\frac{1-f(x)}{x}\\right)^{\\frac{1}{1-f(x)}}} \\text{ と } \\therefore \\text{ يحقق التساوي } f(x) \\ge 0\n$$\n\n$$\n\\text{لما } \\left( \\frac{1-f(x)}{x} \\right)^{\\frac{1}{1-f(x)}} \\le \\frac{1}{4} \\text{ と } \\left( \\frac{1-f(x)}{x} \\right)^{\\frac{1}{1-f(x)}} = \\frac{1}{4} \\text{ と } \\text{يحقق التساوي } f(x) = 0 \\text{ (لما نحقق قسمة f(x) في x)}\n$$\n\n$$\n\\text{لما } \\left( \\frac{1-f(x)}{x} \\right)^{\\frac{1}{1-f(x)}} > \\frac{1}{4} \\text{ と } \\left( \\frac{1-f(x)}{x} \\right)^{\\frac{1}{1-f(x)}} \\le \\frac{1}{4} \\text{ と } \\text{يحقق قسمة } f(x) \\text{ في x} \\text{، و نستنتج أن } C > 0 \\text{ يحقق}\n$$\n\n\n\nالآن ليكن $C > 0$، لربما\n\n$$\nf(x) = f(x+\\frac{1}{x}) \\text{ يحقق قسمة f(x) في x} \\text{، و يحقق قسمة } f(x) = C \\text{ في x}\n$$\n\nنستعمل جدراً\n\nليكن\n\n$$\n\\text{إنما لو } f(x) = 1 \\text{ فن ن查閱 } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} \\le f(x) \\text{ يحقق قسمة } f(x) = 1 \\text{ بالعواملة}\n$$\n\nوما يrest في إيجاد العواملة و كن لك في باقي المات\n\n$$\n\\text{لما لو } f(x) = 1 \\text{ نريد } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} \\le \\frac{1}{4} \\text{ と } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} = \\frac{1}{4} \\text{ يحقق قسمة } f(x) = 0 \\text{ (لما نحقق قسمة } f(x) = 1 \\text{ في x)}\n$$\n\n$$\n\\text{لما مرتب على قسمات } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} \\le \\frac{1}{4} \\text{ と } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} = \\frac{1}{4} \\text{ يحقق قسمة } f(x) = 0 \\text{ (لما نحقق قسمة } f(x) = 1 \\text{ في x)}\n$$\n\n$$\n\\text{لما لو } f(x) = 1 \\text{ نريد } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} \\le \\frac{1}{4} \\text{ と } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} = \\frac{1}{4} \\text{ يحقق قسمة } f(x) = 0 \\text{ (لما نحقق قسمة } f(x) = 1 \\text{ في x)}\n$$\n\n$$\n\\text{و ما مرتب على قسمات } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} \\ge \\frac{1}{4} \\text{ と } \\left( \\frac{1}{x} \\right)^{\\frac{1}{1-x}} = \\frac{1}{4} \\text{ يحقق قسمة } f(x) = 0 \\text{ (لما نحقق قسمة } f(x) = 1 \\text{ في x)}\n$$\n\n$$\n\\text{لما نستنتج أن } C > 0 \\text{ يحقق قسمة } f(x) = C \\text{ (لما نحقق قسمة } f(x) = 0 \\text{ في x)} \\\\\n\\text{لما نستنتج أن } C < 0 \\text{ يحقق قسمة } f(x) = C \\text{ (لما نحقق قسمة } f(x) = 0 \\text{ في x)}\n$$\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20462,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for all real $x, y$,\n\n$$\nf(f(x) - y^2) = f(x^2) + y^2 f(y) - 2f(xy).\n$$",
"options": [],
"answer": "See solution",
"solution": "**Answer:** $f(x) = x^2$ and $f(x) = 0$.\n\n**Solution.**\n\nSubstitute $y=1$ and $y=-1$:\n\n$$\nf(f(x)-1) = f(x^2) + f(1) - 2f(x) \\quad \\text{and} \\quad f(f(x)-1) = f(x^2) + f(-1) - 2f(-x). \\tag{*}\n$$\n\nCombining both equalities, we get $f(1) - 2f(x) = f(-1) - 2f(-x)$. Hence, if $x=1$: $f(1) = f(-1)$, so $f(x) = f(-x)$; the function $f$ is even.\n\nIf $x = y = 1$, $f(f(1) - 1) = 0$, so there exists a number $b$ such that $f(b) = 0$. If $x = b$ in $(*)$ then\n\n$$\nf(f(b) - 1) = f(b^2) + f(1), \\quad f(-1) = f(b^2) + f(1), \\quad \\text{so } f(b^2) = 0.\n$$\n\nSubstitute $x = b$ and $y = 0$ in the original equation: $f(f(b)) = f(b^2) - 2f(0)$, so $3f(0) = f(b^2) = 0$. If $x = 0$ then\n\n$$\nf(y^2) = y^2 f(y). \\tag{**}\n$$\n\nThere are two cases:\n\n*Case 1:* There exists $b \\in \\mathbb{R}$, $b \\neq 0$ such that $f(b) = 0$. As shown above, $f(b^2) = 0$. Substitute $x = b$ in the original equation:\n\n$$\nf(f(b) - y^2) = f(b^2) + y^2 f(y) - 2f(by), \\quad f(y^2) = y^2 f(y) - 2f(by),\n$$\n\nand using $(**)$, this means $f(x) = 0$ for all $x \\in \\mathbb{R}$.\n\n*Case 2:* $f(b) = 0$ if and only if $b = 0$. If $x = y$ then\n\n$$\nf(f(x) - x^2) = f(x^2) + x^2 f(x) - 2f(x^2) = x^2 f(x) - f(x^2) = 0,\n$$\n\nso $f(x) - x^2 = 0$, thus another answer is $f(x) = x^2$ for all $x \\in \\mathbb{R}$.\n\nTesting shows that both variants fulfill the conditions.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20463,
"subject": "Mathematics (Olympiad)",
"question": "Given the functional equation:\n\n$$\nf(2a) + 2f(b) = f(f(a + b)).\n$$\n\nFind all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ satisfying this equation.",
"options": [],
"answer": "See solution",
"solution": "Let $g(x) = f(x) - f(0)$. Then $g: \\mathbb{Z} \\to \\mathbb{Z}$ and $f(x) = g(x) + f(0)$. Substituting into the derived equation $f(a) + f(b) = f(a + b) + f(0)$ gives:\n\n$$\ng(a + b) = g(a) + g(b)\n$$\nfor all integers $a, b$. This is Cauchy's functional equation on $\\mathbb{Z}$, whose solutions are $g(x) = mx$ for some integer $m$. Thus, $f(x) = mx + c$ for constants $m, c \\in \\mathbb{Z}$. The values of $m$ and $c$ can be determined by substituting back into the original equation.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20464,
"subject": "Mathematics (Olympiad)",
"question": "Consider a convex quadrilateral $ABCD$ with $\\angle BCD = 120^\\circ$, $\\angle CBA = 45^\\circ$, $\\angle CBD = 15^\\circ$, and $\\angle CAB = 90^\\circ$. Show that $AB = AD$.",
"options": [],
"answer": "See solution",
"solution": "Consider the mirror image $E$ of point $C$ with respect to $A$. Since $\\angle CDB = \\angle CEB = 45^\\circ$, the quadrilateral $BCDE$ is cyclic, and $CE$ is a diameter of its circumcircle. Hence, $A$ is its circumcenter, implying $AB = AD$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20465,
"subject": "Mathematics (Olympiad)",
"question": "Determine the number which can be represented in the form $n^2 + 4n$ for some positive integer $n$ and for which the absolute value of its difference from the number $10000$ is the smallest.",
"options": [],
"answer": "See solution",
"solution": "The answer is $9996$.\n\nNote that $n^2 + 4n = (n + 2)^2 - 4$. If $m < n$, then $m^2 + 4m < n^2 + 4n$.\n\nSince $100^2 = 10000$, the closest values are either $100^2 - 4 = 9996$ or $101^2 - 4 = 10197$. Comparing their differences from $10000$:\n\n- $|9996 - 10000| = 4$\n- $|10197 - 10000| = 197$\n\nThus, $9996$ is the closest.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20466,
"subject": "Mathematics (Olympiad)",
"question": "A grid consists of all points of the form $(m, n)$ where $m$ and $n$ are integers with $|m| \\leq 2019$, $|n| \\leq 2019$, and $|m| + |n| < 4038$. We call the points $(m, n)$ of the grid with either $|m| = 2019$ or $|n| = 2019$ the *boundary points*. The four lines $x = \\pm 2019$ and $y = \\pm 2019$ are called *boundary lines*. Two points in the grid are called *neighbours* if the distance between them is equal to $1$.\n\nAnna and Bob play a game on this grid.\n\nAnna starts with a token at the point $(0, 0)$. They take turns, with Bob playing first.\n\n1. On each of his turns, Bob deletes at most two boundary points on each boundary line.\n2. On each of her turns, Anna makes exactly three *steps*, where a *step* consists of moving her token from its current point to any neighbouring point which has not been deleted.\n\nAs soon as Anna places her token on some boundary point which has not been deleted, the game is over and Anna wins.\n\nDoes Anna have a winning strategy?\n\n",
"options": [],
"answer": "See solution",
"solution": "Anna does not have a winning strategy. We will provide a winning strategy for Bob. It is enough to describe his strategy for the deletions on the line $y = 2019$.\n\nBob starts by deleting $(0, 2019)$ and $(-1, 2019)$. Once Anna completes her turn, he deletes the next two available points on the left if Anna decreased her $x$-coordinate, the next two available points on the right if Anna increased her $x$-coordinate, and the next available point to the left and the next available point to the right if Anna did not change her $x$-coordinate. The only exception to the above rule is on the very first time Anna decreases $x$ by exactly $1$. In that turn, Bob deletes the next available point to the left and the next available point to the right.\n\nBob's strategy guarantees the following: If Anna makes a sequence of steps reaching $(-x, y)$ with $x > 0$ and the exact opposite sequence of steps in the horizontal direction reaching $(x, y)$, then Bob deletes at least as many points to the left of $(0, 2019)$ in the first sequence than points to the right of $(0, 2019)$ in the second sequence.\n\nSo we may assume for contradiction that Anna wins by placing her token at $(k, 2019)$ for some $k > 0$.\n\nDefine $\\Delta = 3m - (2x + y)$, where $m$ is the total number of points deleted by Bob to the right of $(0, 2019)$, and $(x, y)$ is the position of Anna's token.\n\nFor each sequence of steps performed first by Anna and then by Bob, $\\Delta$ does not decrease. This can be seen by looking at the following table exhibiting the changes in $3m$ and $2x + y$. We have excluded the cases where $2x + y < 0$.\n\n| Turn | (0,3) | (1,2) | (-1,2) | (2,1) | (0,1) | (3,0) | (1,0) | (2,-1) | (1,-2) |\n|-----------|-------|-------|--------|-------|-------|-------|-------|--------|--------|\n| $m$ | 1 | 2 | 0 (or 1) | 2 | 1 | 2 | 2 | 2 | 2 |\n| $3m$ | 3 | 6 | 0 (or 3) | 6 | 3 | 6 | 6 | 6 | 6 |\n| $2x + y$ | 3 | 4 | 0 | 5 | 1 | 6 | 2 | 3 | 0 |\n\nThe table also shows that, if in this sequence of turns Anna changes $y$ by $+1$ or $-2$, then $\\Delta$ is increased by $1$. Also, if Anna changes $y$ by $+2$ or $-1$, then the first time this happens $\\Delta$ is increased by $2$. (This also holds if her turn is $(0, -1)$ or $(-2, -1)$, which are not shown in the table.)\n\nSince Anna wins by placing her token at $(k, 2019)$ we must have $m \\leq k - 1$ and $k \\leq 2018$. So at that exact moment we have:\n\n$$\n\\Delta = 3m - (2k + 2019) = k - 2022 \\leq -4.\n$$\n\nSo in her last turn she must have decreased $\\Delta$ by at least $4$. So her last turn must have been $(1, 2)$ or $(2, 1)$, which give a decrease of $4$ and $5$ respectively. (It could not be $(3, 0)$ because then she must have already won. Also she could not have done just one or two steps in her last turn since this is not enough for the required decrease in $\\Delta$.)\n\nIf her last turn was $(1, 2)$, then just before doing it we had $y = 2017$ and $\\Delta = 0$. This means that in one of her turns the total change in $y$ was not $0 \\mod 3$. However, in that case we have seen that $\\Delta > 0$, a contradiction.\n\nIf her last turn was $(2, 1)$, then just before doing it we had $y = 2018$ and $\\Delta = 0$ or $\\Delta = 1$. So she must have made at least two turns with the change of $y$ being $+1$ or $-2$ or at least one step with the change of $y$ being $+2$ or $-1$. In both cases, consulting the table, we get an increase of at least $2$ in $\\Delta$, a contradiction.\n\n**Note 1:** If Anna is allowed to make *at most* three steps at each turn, then she actually has a winning strategy.\n\n**Note 2:** If $2019$ is replaced by $N > 1$, then Bob has a winning strategy if and only if $3 \\mid N$. $\\Box$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 20467,
"subject": "Mathematics (Olympiad)",
"question": "Let $AB$, $BC$, $CD$, $DE$, $EF$, and $FA$ be the sides of a hexagon, each tangent to a circle at points $U$, $V$, $W$, $X$, $Y$, and $Z$ respectively.\n\n\n\nFind the perimeter of hexagon $ABCDEF$ if $AB = 6$, $CD = 7$, and $EF = 8$.",
"options": [],
"answer": "See solution",
"solution": "Since the two tangents from a point to a circle have equal length, we have:\n\n$UB = BV$, $VC = CW$, $WD = DX$, $XE = EY$, $YF = FZ$, $ZA = AU$.\n\nThe perimeter of hexagon $ABCDEF$ is:\n\n$$\n\\begin{align*}\n& AU + UB + BV + VC + CW + WD + DX + XE + EY + YF + FZ + ZA \\\\\n&= AU + UB + UB + CW + CW + WD + WD + EY + EY + YF + YF + AU \\\\\n&= 2(AU + UB + CW + WD + EY + YF) \\\\\n&= 2(AB + CD + EF) = 2(6 + 7 + 8) = 2(21) = \\mathbf{42}.\n\\end{align*}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20468,
"subject": "Mathematics (Olympiad)",
"question": "Solve the equation\n$$\n\\sqrt{1 + 3\\sin^3 x} = 3 - \\sqrt{1 - \\cos^4 x}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Since\n$$\n\\sqrt{1 + 3\\sin^3 x} + \\sqrt{1 - \\cos^4 x} \\leq \\sqrt{1 + 3} + 1 = 3,\n$$\nthe equality can be attained only if\n$$\n\\begin{cases}\n\\sin x = 1, \\\\\n\\cos x = 0\n\\end{cases}\n$$\nwhich means that\n$$\nx = \\frac{\\pi}{2} + 2k\\pi, \\quad k \\in \\mathbb{N}.\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20469,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be an integer greater than $6$. Prove that if $n - 1$ and $n + 1$ are both prime, then $n^2(n^2 + 16)$ is divisible by $720$. Is the converse true?",
"options": [],
"answer": "See solution",
"solution": "As $n-1$ and $n+1$ are both prime, $n$ must be divisible by $2$ (since $n > 6$). Thus $n^2(n^2 + 16)$ is divisible by $2^4$, as $n^4$ and $16n^2$ both are.\n\nOne of $n$, $n-1$, and $n+1$ is divisible by $3$. However, $n-1$ and $n+1$ are prime, so $n$ must be divisible by $3$ (again, since $n > 6$). Therefore $n^2(n^2 + 16)$ is divisible by $9$, as $n^2$ is.\n\nOne of $n-2$, $n-1$, $n$, $n+1$, and $n+2$ is divisible by $5$. However, $n-1$ and $n+1$ are prime, and $n > 6$, so it is neither of these. Thus $(n-2)n(n+2) = n^3 - 4n$ is a multiple of $5$. Hence $n(n^3 - 4n)$ is also a multiple of $5$.\n\nSince $20n^2$ is clearly also divisible by $5$, so is $n^4 - 4n^2 + 20n^2 = n^2(n^2 + 16)$. Thus $n^2(n^2 + 16)$ is a multiple of $2^4 \\times 3^2 \\times 5 = 720$.\n\nThe converse, \"If $n^2(n^2 + 16)$ is divisible by $720$, then $n-1$ and $n+1$ are prime\" is not true.\n\nFor example, if $n = 78$, then $78^2(78^2 + 16)$ is divisible by $720$. However, $77 = 7 \\times 11$ is not prime.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20470,
"subject": "Mathematics (Olympiad)",
"question": "Point $P$ lies inside triangle $ABC$, $M$ is the midpoint of side $BC$, and $P'$ is symmetric to $P$ with respect to $M$. $K$ and $H$ are projections of $P$ onto $AB$ and $AC$ respectively, and $KM = HM$. Prove that $\\angle PAB = \\angle CAP'$.\n\n*Remark:* Solution 2 shows that the statement of the problem is not ideal. The point $M$ does not need to be a midpoint, and the solution using that fact is more complicated than the presented solution. The following reformulation of the problem is suggested.",
"options": [],
"answer": "See solution",
"solution": "Denote by $A'$, $K'$, $H'$ the points symmetric to $A$, $K$, $H$ with respect to $M$, as in Figure 13.\n\nIt is clear that $KH'H'K$ is a rectangle and the quadrilateral $AKPH$ is cyclic; denote its circumcircle by $\\omega$. Then\n\n$$\n\\angle CHK' = 90^\\circ - \\angle AHK = \\angle PHK = \\angle PAK.\n$$\n\nLet $L$ be the second intersection point of line $HK'$ and $\\omega$. The above equality of angles means that $\\angle LHA = \\angle PAK$, therefore $AL = KP$, and then $AKPL$ is a rectangle. Hence $AL \\parallel KP \\parallel K'P'$ and $AP'K'L$ is a parallelogram. Thus, $HK' \\parallel AP'$ and then $\\angle CHK' = \\angle CAP'$.\n\n\n\nFigure 13",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20471,
"subject": "Mathematics (Olympiad)",
"question": "Niek has 16 square cards that are white on one side and black on the other. He puts down the cards to form a $4 \\times 4$ square. Some of the cards show their white side and some show their black side. For a colour pattern he calculates the *monochromaticity* as follows. For every pair of adjacent cards that share a side he counts $+1$ or $-1$ according to the following rule: $+1$ if the adjacent cards show the same colour, and $-1$ if the adjacent cards show different colours. Adding this all together gives the monochromaticity (which might be negative).\n\nFor example, if he lays down the cards as below, there are 15 pairs of adjacent cards showing the same colour, and 9 such pairs showing different colours.\n\n\n\nThe monochromaticity of this pattern is thus $15 \\cdot (+1) + 9 \\cdot (-1) = 6$.\n\nNiek investigates all possible colour patterns and makes a list of all possible numbers that appear at least once as a value of the monochromaticity. That is, Niek makes a list with all numbers such that there exists a colour pattern that has this number as its monochromaticity.\n\n(a) What are the three largest numbers on his list?\n\n(Explain your answer. If your answer is, for example, 12, 9 and 6, then you have to show that these numbers do in fact appear on the list by giving a colouring for each of these numbers, and furthermore prove that the numbers 7, 8, 10, 11 and all numbers bigger than 12 do not appear.)\n\n(b) What are the three smallest (most negative) numbers on his list?\n\n(c) What is the smallest positive number (so, greater than 0) on his list?",
"options": [],
"answer": "See solution",
"solution": "(a) First, note that there are $3 \\cdot 4 = 12$ horizontal borders between two cards, and also 12 vertical borders. Suppose that $k$ of these borders count as $-1$, then there are $24 - k$ borders counting as $+1$. This gives a monochromaticity of $(24 - k) \\cdot (+1) + k \\cdot (-1) = 24 - 2k$. Hence, the monochromaticity is always an even number.\n\nIf all cards have the same colour, then all borders count as $+1$, and we get the maximal monochromaticity of $24$. Can $22$ also occur as the monochromaticity? No, and we will prove that by contradiction. Suppose there is an assignment of cards having monochromaticity $22$. Then there has to be one border with $-1$ and the rest must count as $+1$. In other words, all adjacent cards have the same colour, except for one border. Consider the two cards at this border, and choose two adjacent cards so that you obtain a $2 \\times 2$ square. For each pair of cards, you can find such a $2 \\times 2$ square. If you start on the left top and go around the four cards in a circle (left top – right top – right bottom – left bottom – left top), then you cross four borders. Since you are starting and ending in the same colour, you must have crossed an even number of borders where the colour is changing. This, however, is in contradiction with the assumption that there is only one border at which the two cards have different colours. We conclude that the monochromaticity can never be $22$.\n\nThe next possibilities for large monochromaticities are $20$ and $18$. Then there have to be $2$ or $3$ borders between cards of different colours. This can be achieved by the following colourings:\n\n\n\nThe three largest numbers on Niek's list are $24$, $20$, and $18$. $\\Box$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 20472,
"subject": "Mathematics (Olympiad)",
"question": "Given a triangle $ABC$, let $D$ be the midpoint of the side $AC$ and let $M$ be the point that divides the segment $BD$ in the ratio $1/2$; that is, $MB/MD = 1/2$. The rays $AM$ and $CM$ meet the sides $BC$ and $AB$ at points $E$ and $F$, respectively. Assume the two rays are perpendicular: $AM \\perp CM$. Show that the quadrangle $AFED$ is cyclic if and only if the line of support of the median from $A$ in triangle $ABC$ meets the line $EF$ at a point situated on the circle $ABC$.",
"options": [],
"answer": "See solution",
"solution": "Denote by $a$, $b$, $c$ the sidelengths, and by $m_a$, $m_b$, $m_c$ the lengths of the medians of the triangle $ABC$. Since $MD$ is median in the right-angled triangle $AMC$, it follows that $2m_b/3 = MD = AD = CD = b/2$, so $m_b = 3b/4$, whence $(3b/4)^2 = m_b^2 = (a^2 + c^2)/2 - b^2/4$; that is, $13b^2 = 8(a^2 + c^2)$.\n\nNext, apply the Menelaus theorem to get $EC/EB = 4 = FA/FB$ and deduce thereby that the lines $AC$ and $EF$ are parallel. The quadrangle $AFED$ is therefore a trapezium; it is cyclic if and only if $AF = DE$.\n\nExpress the two in terms of $a$, $b$ and $c$. Recall that $FA/FB = 4$ to obtain $AF = 4c/5$. Next, apply Stewart's theorem in triangle $BCD$ to get $DE^2 = b^2/2 - 4a^2/25$. By the preceding, the quadrangle $AFED$ is cyclic if and only if $25b^2 - 8a^2 = 32c^2$. Recall that $13b^2 = 8(a^2 + c^2)$ to express $b$ and $c$ in terms of $a$: $b = 2a\\sqrt{2}/3$ and $c = 2a/3$.\n\n\n\nFinally, let $N$ be the midpoint of the side $BC$ and let the lines $AN$ and $EF$ meet at $P$. Notice that $EN = a/2 - a/5 = 3a/10$, and the triangles $ANC$ and $PNE$ are similar, to obtain $NP = 3m_a/5$, so\n\n$$\nNA \\cdot NP = 3m_a^2/5 = 3(2(b^2 + c^2) - a^2)/20 = a^2/4 = NB \\cdot NC.\n$$\n\nThe conclusion follows.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20473,
"subject": "Mathematics (Olympiad)",
"question": "Let $f$ and $g$ be two polynomials with integer coefficients such that $\\deg f > \\deg g$ and $\\deg f \\geq 2$. If the polynomial $pf + g$ has a rational root for infinitely many primes $p$, prove that $f$ has a rational root.",
"options": [],
"answer": "See solution",
"solution": "Since $\\deg f > \\deg g$, $|g(z)/f(z)| < 1$ for all complex numbers $z$ of large enough absolute value. Consequently, as $p$ runs through the infinite set of primes under consideration, the roots of $pf + g$ all lie in some disc $|z| < R$, where $R$ does not depend on $p$; for if $|z|$ is large enough, then\n\n$$\n|pf(z) + g(z)| \\geq |f(z)| (p - |g(z)/f(z)|) > 0.\n$$\n\nFor each prime $p$ such that $pf + g$ has a rational root, by Gauss' lemma, $pf + g$ is the product of two integral polynomials, one of degree $1$ and the other of degree $\\deg f - 1$. Now the condition that $p$ be prime comes in to imply that the leading coefficient of one of these factors is a divisor of the leading coefficient of $f$.\n\nFor each such prime $p$, it is therefore possible to choose an integral factor $h_p$ of $pf + g$, $\\deg h_p = 1$ or $\\deg h_p = \\deg f - 1$, whose leading coefficient divides the leading coefficient of $f$. Hence the leading coefficients of the $h_p$ form a bounded set. Since the roots of the $h_p$ all lie in the same disc, $|z| < R$, Vieta's relations imply that the coefficients of the $h_p$ all form a bounded set; and since they are all integral, this set is finite, so $h_p = h$ for infinitely many of these primes $p$.\n\nFinally, if $p$ and $q$ are two such, then $h$ is a divisor of $(p-q)f$. Since $\\deg h_p = 1$ or $\\deg h_p = \\deg f - 1$, the conclusion follows.\n\n**Remarks.** Clearly, $h$ is a common factor of $f$ and $g$ over the rationals. If $\\deg h = 1$, then $f$ and $g$ share a common rational root. Otherwise, $\\deg h = \\deg f - 1$ forces $\\deg g = \\deg f - 1$ and $g$ divides $f$ over the rationals.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20474,
"subject": "Mathematics (Olympiad)",
"question": "For any nonzero natural number $n$, consider the set\n$$\nA = \\{n^2,\\ n^2 + 1,\\ n^2 + 2,\\ \\dots,\\ (n+1)^2\\}.\n$$\nFind the numbers $a, b, c \\in A$, with $a < b < c$, such that $b$ is the geometric mean of $a$ and $c$.",
"options": [],
"answer": "See solution",
"solution": "Since $b$ is the geometric mean of $a$ and $c$, we have $b^2 = ac$.\n\nLet $\\frac{b}{a} = \\frac{c}{b} = \\frac{x}{y}$, where $x, y \\in \\mathbb{N}^*$, $(x, y) = 1$, and $x > y$. Then $b = \\frac{a x}{y}$ and $c = a \\left(\\frac{x}{y}\\right)^2$.\n\nFor $c$ to be a natural number, $\\frac{a}{y^2} \\in \\mathbb{N}^*$, so $a = p y^2$ for some $p \\in \\mathbb{N}^*$. Thus, $a = p y^2$, $b = p x y$, $c = p x^2$.\n\nSince $a, b, c \\in A$, we require $n^2 \\leq a < b < c \\leq (n+1)^2$. This gives $n \\leq y < x \\leq n+1$.\n\nThe only possibility is $y = n$, $x = n+1$, and $p = 1$. Therefore, the solution is:\n$$\na = n^2, \\quad b = n(n+1), \\quad c = (n+1)^2.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20475,
"subject": "Mathematics (Olympiad)",
"question": "試證：在等差數列 $11, 21, 31, 41, 51, 61, \\ldots$ 中，存在無窮多個質數。",
"options": [],
"answer": "See solution",
"solution": "考慮整數 $N = (11 \\cdot 31 \\cdot 41 \\cdot 61 \\cdots p_n)^5 - 1$。則 $N$ 與 $11, 41, \\cdots, p_n$ 互質。令 $a = 11 \\cdot 31 \\cdot 41 \\cdot 61 \\cdots p_n$，則\n\n$$\nN = a^5 - 1 = (a-1)(a^4 + a^3 + a^2 + a + 1).\n$$\n\n顯然，$2 \\nmid (a^4 + a^3 + a^2 + a + 1)$，且 $5 \\nmid (a^4 + a^3 + a^2 + a + 1)$。令 $p \\neq 5$ 是 $a^4 + a^3 + a^2 + a + 1$ 的一個質因數。因此，$p \\nmid (a-1)$。理由如下：\n\n若 $p \\mid (a-1)$，則 $a = kp + 1$，對某個整數 $k$，即得\n\n$$\na^2 = (kp + 1)^2, \\quad a^3 = (kp + 1)^3, \\quad a^4 = (kp + 1)^4,\n$$\n\n且\n\n$$\na^4 + a^3 + a^2 + a + 1 = (kp+1)^4 + (kp+1)^3 + (kp+1)^2 + (kp+1) + 1 \\equiv 5 \\pmod{p}.\n$$\n\n事實上，$(p-1) \\equiv 4 \\pmod{5}$，即 $p-1 = 5k + 4$。\n\n由費馬小定理，$p \\mid (a^{p-1} - 1)$。但此時\n\n$$\na^{p-1} - 1 = a^{5k+4} - 1 = a^4(a^{5k} - 1) + (a^4 - 1),\n$$\n\n且因 $(a^5 - 1) \\mid (a^{5k} - 1) = (a^5)^k - 1^k$，其意為 $p \\mid (a^{5k} - 1)$。即 $p \\mid (a^4 - 1)$。然而\n\n$$\na^5 - 1 = a(a^4 - 1) + (a - 1)\n$$\n\n因此，若 $p \\mid (a^5 - 1)$ 且 $p \\mid (a^4 - 1)$，則 $p \\mid (a - 1)$。上述不可能發生。類似地作法，得 $(p-1)$ 除以 5 所得之餘數不會是 1, 2, 3。\n\n故 $5 \\mid (p-1)$ 且 $p-1$ 是偶數。得 $10 \\mid (p-1)$，即 $p = 10k + 1$，得 $p$ 為給定的等差數列之其中一項。故 $a^4 + a^3 + a^2 + a + 1$ 的質因數為 $5$ 與形如 $10k + 1$ 之質數。\n\n然而 $a^4 + a^3 + a^2 + a + 1 > 5$，且 $5^2 \\nmid (a^4 + a^3 + a^2 + a + 1)$。事實上，整數 $a$ 的末位數字為 $1$，即 $a = 5k + 1$。由二項式定理，得\n\n$$\n\\begin{aligned}\na^4 + a^3 + a^2 + a + 1 &= (5k + 1)^4 + (5k + 1)^3 + (5k + 1)^2 + (5k + 1) + 1 \\\\\n&= 5 \\cdot [5(25k^4 + 25k^3 + 10k^2 + 2k) + 1].\n\\end{aligned}\n$$\n\n故 $N = a^5 - 1$ 至少有一個形如 $10k + 1$ 的質數。但由上述，$N$ 與所有形如 $10k + 1$ 之數互質。此為矛盾。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20476,
"subject": "Mathematics (Olympiad)",
"question": "Prove that if $$(x + \\sqrt{x^2 + 1}) \\cdot (y + \\sqrt{y^2 + 1}) = 1$$ then $x + y = 0$.",
"options": [],
"answer": "See solution",
"solution": "Multiplying $$(x + \\sqrt{x^2 + 1}) \\cdot (y + \\sqrt{y^2 + 1}) = 1$$ by $x - \\sqrt{x^2 + 1}$, we have:\n\n$$\n(x - \\sqrt{x^2 + 1})(x + \\sqrt{x^2 + 1}) \\cdot (y + \\sqrt{y^2 + 1}) = (x - \\sqrt{x^2 + 1})\n$$\n\nThat is,\n\n$$\n- y - \\sqrt{y^2 + 1} = x - \\sqrt{x^2 + 1} \\quad (1)\n$$\n\nSimilarly, multiplying by $y - \\sqrt{y^2 + 1}$ gives:\n\n$$\n- x - \\sqrt{x^2 + 1} = y - \\sqrt{y^2 + 1} \\quad (2)\n$$\n\nAdding (1) and (2):\n\n$$\n- y - x - \\sqrt{y^2 + 1} - \\sqrt{x^2 + 1} = x - \\sqrt{x^2 + 1} + y - \\sqrt{y^2 + 1}\n$$\n\nThis simplifies to $2(x + y) = 0$, i.e., $x + y = 0$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20477,
"subject": "Mathematics (Olympiad)",
"question": "Which of the following conditions is sufficient to guarantee that integers $x$, $y$, and $z$ satisfy the equation\n\n$$\nx(x - y) + y(y - z) + z(z - x) = 1?\n$$\n\n(A) $x > y$ and $y = z$\n(B) $x = y - 1$ and $y = z - 1$\n(C) $x = z + 1$ and $y = x + 1$\n(D) $x = z$ and $y - 1 = x$\n(E) $x + y + z = 1$",
"options": [],
"answer": "See solution",
"solution": "The given equation can be rewritten as:\n\n$$\nx(x - y) + y(y - z) + z(z - x) = 1\n$$\n\nThis is equivalent to:\n\n$$\n(x - y)^2 + (y - z)^2 + (z - x)^2 = 2\n$$\n\nThis equation has integer solutions only if two of the squares are $1$ and one is $0$, which means two variables are equal and the third differs by $1$. Choice (D) ($x = z$ and $y - 1 = x$) satisfies this condition. The other choices do not guarantee this:\n\n- (A) fails when $x = 2$, $y = 0$, $z = 0$ (left side equals $4$)\n- (B) fails when $x = 1$, $y = 2$, $z = 3$ (left side equals $3$)\n- (C) fails when $x = 1$, $y = 2$, $z = 0$ (left side equals $3$)\n- (E) fails when $x = 2$, $y = 0$, $z = -1$ (left side equals $7$)",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20478,
"subject": "Mathematics (Olympiad)",
"question": "Find the minimum and maximum possible values of $f(23) + f(2011)$, where $f$ is a positive integer-valued function defined on the positive integers, subject to the condition that for all positive integers $x$ and $y$,\n\n$$(x + y)f(x) \\leq x^2 + f(xy) + 110.$$",
"options": [],
"answer": "See solution",
"solution": "Let $a = 110$. We will show that a necessary and sufficient condition for a positive integer-valued function $f$ defined on the positive integers to satisfy the given inequality is that $f$ satisfies:\n\n$$\n(\\dagger) \\quad t - a \\leq f(t) \\leq t \\quad \\text{for any positive integer } t.\n$$\n\nFirst, for any positive integer $s$, substitute $(x, y) = (s, 1)$ into the given inequality:\n\n$$(s+1)f(s) \\leq s^2 + f(s) + a,$$\n\nwhich gives $f(s) \\leq s + \\frac{a}{s}$. Next, for any positive integer $t$, substitute $(x, y) = (t, 2a)$ and use $f(2at) \\leq 2at + \\frac{1}{2t}$ (from above):\n\n$$(t + 2a)f(t) \\leq t^2 + f(2at) + a \\leq t^2 + 2at + \\frac{1}{2t} + a,$$\n\nwhich yields\n\n$$\nf(t) \\leq t + \\frac{1}{2t(t + 2a)} + \\frac{a}{t + 2a} < t + 1.\n$$\n\nSince $f$ is integer-valued, $1 \\leq f(t) \\leq t$. In particular, $f(1) = 1$. Substituting $(x, y) = (1, t)$ gives $(1+t) \\cdot 1 \\leq 1 + f(t) + a$, so $t - a \\leq f(t)$. Thus, $f$ satisfies $(\\dagger)$ for all $t$.\n\nConversely, if $f$ satisfies $(\\dagger)$, then for any $x, y$:\n\n$$(x + y)f(x) \\leq (x + y)x = x^2 + (xy - a) + a \\leq x^2 + f(xy) + a,$$\n\nso $f$ satisfies the original inequality.\n\nTherefore,\n\n$$\n1 \\leq f(23) \\leq 23, \\quad 1901 = 2011 - 110 \\leq f(2011) \\leq 2011.\n$$\n\nThus, the minimum possible value for $f(23) + f(2011)$ is $1 + 1901 = 1902$ and the maximum is $23 + 2011 = 2034$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20479,
"subject": "Mathematics (Olympiad)",
"question": "During three hours, a driver has driven $180\\,\\text{km}$. During the first hour, he drove $0.375$ of the total distance, and during the second hour, he drove $0.9$ of the distance that he had driven during the first hour. What distance did the driver drive during the third hour?",
"options": [],
"answer": "See solution",
"solution": "The distance driven during the third hour is:\n\n$$\n180 - 0.375 \\times 180 - 0.9 \\times 0.375 \\times 180 = 51.75\\,\\text{km} = 51\\,\\text{km}\\ 75\\text{m}\n$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 20480,
"subject": "Mathematics (Olympiad)",
"question": "We consider 111 mutually distinct points on the interior or on the circle of a unit disk. Prove that we can find at least 1998 segments with ends from these points and length less than $\\sqrt{3}$.",
"options": [],
"answer": "See solution",
"solution": "We divide the circle into three equal sectors of $120^\\circ$ such that none of the points belong to their border, except possibly the center of the circle. If the center is one of the points, we assign it to only one sector. This is possible because the number of points is finite and there are infinitely many ways to choose the sector boundaries.\n\n\n\nLet $A, B$ belong to the same sector defined by the radii $OK$ and $O\\Lambda$. If $OA, OB$ intersect the circle at $A', B'$ and $\\angle A'OB = \\omega < 60^\\circ$, then $AB \\leq A'B' = 2R\\sin(\\omega/2) < 2R\\sin(60^\\circ/2) = 2 \\times \\frac{1}{2} = 1$. However, the original argument uses $2R\\sin(\\omega/2) < \\sqrt{3}$, so for $\\omega < 60^\\circ$, $AB < \\sqrt{3}$. Equality is not possible because the points do not lie on the sector border. Thus, any two points in the same sector have distance less than $\\sqrt{3}$.\n\nLet the three sectors contain $x, y, z$ points, respectively, with $x + y + z = 111$. The number of segments of length less than $\\sqrt{3}$ is at least:\n\n$$\n\\binom{x}{2} + \\binom{y}{2} + \\binom{z}{2} = \\frac{x(x-1) + y(y-1) + z(z-1)}{2} = \\frac{x^2 + y^2 + z^2 - 111}{2}.\n$$\n\nBy the Cauchy-Schwarz inequality: $x^2 + y^2 + z^2 \\geq \\frac{(x + y + z)^2}{3} = \\frac{111^2}{3}$, so\n\n$$\n\\binom{x}{2} + \\binom{y}{2} + \\binom{z}{2} \\geq \\frac{111^2}{3} - 111 = 1998.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20481,
"subject": "Mathematics (Olympiad)",
"question": "A communications network consisting of some terminals is called a 3-connector if among any three terminals, some two of them can directly communicate with each other.\n\nA communications network contains a windmill with $n$ blades if there exist $n$ pairs of terminals $\\{x_1, y_1\\}, \\dots, \\{x_n, y_n\\}$ such that each $x_i$ can directly communicate with the corresponding $y_i$ and there is a hub terminal that can directly communicate with each of the $2n$ terminals $x_1, y_1, \\dots, x_n, y_n$.\n\nDetermine the minimum value of $f(n)$, in terms of $n$, such that a 3-connector with $f(n)$ terminals always contains a windmill with $n$ blades.",
"options": [],
"answer": "See solution",
"solution": "The answer is\n\n$$\nf(n) = \\begin{cases} 6 & \\text{if } n = 1; \\\\ 4n + 1 & \\text{if } n \\ge 2. \\end{cases}\n$$\n\nWe will use *connected* as a synonym for directly communicating, call a set of $k$ terminals for which each of the $\\binom{k}{2}$ pairs of terminals is connected *complete* and call a set of $2k$ terminals forming $k$ disjoint connected pairs a *k-matching*.\n\nWe first show that $f(n) = 4n + 1$ for $n > 1$. The $4n$-terminal network consisting of two disconnected complete sets of $2n$ terminals clearly does not contain an $n$-bladed windmill (henceforth called an $n$-mill), since such a windmill requires a set of $2n+1$ connected terminals. So we need only demonstrate that $f(n) = 4n + 1$ is sufficient.\n\nNote that we can inductively create a $k$-matching in any subnetwork of $2k + 1$ elements, as there is a connected pair in any set of three or more terminals. Also, the set of terminals that are not connected to a given terminal $x$ must be complete, as otherwise there would be a set of three mutually disconnected terminals. We now proceed by contradiction and assume that there is a $(4n + 1)$-terminal network without an $n$-mill. Any terminal $x$ must then be connected to at least $2n$ terminals, for otherwise there would be a complete set of size at least $2n + 1$, which includes an $n$-mill. In addition, $x$ cannot be directly connected to more than $2n$ terminals, for otherwise we could construct an $n$-matching among these, and therefore an $n$-mill. Therefore every terminal is connected to precisely $2n$ others.\n\nIf we take two terminals $u$ and $v$ that are not connected we can then note that at least one must be connected to the $4n - 1$ remaining terminals, and therefore there must be exactly one, $w$, to which both are connected. The rest of the network now consists of two complete sets of terminals $A$ and $B$ of size $2n - 1$, where every terminal in $A$ is connected to $u$ and not connected to $v$, and every terminal in $B$ is not connected to $u$ and connected to $v$. If $w$ were connected to any terminal in $A$ or $B$, it would form a blade with this element and hub $u$ or $v$ respectively, and we could fill out the rest of an $n$-mill with terminals in $A$ or $B$ respectively. Hence $w$ is only connected to two terminals, and therefore $n = 1$.\n\n\n\nExamining the preceding proof, we can find the only 5-terminal network with no 1-mill: With terminals labeled A, B, C, D, and E, the connected pairs are (A, B), (B, C), (C, D), (D, E), and (E, A). (As indicated in the figure above, a pair of terminals are connected if and only if the edge connecting them are darkened.) To show that any 6-terminal network has a 1-mill, we note that any complete set of three terminals is a 1-mill. We again work by contradiction. Any terminal *a* would have to be connected to at least three others, *b*, *c*, and *d*, or the terminals not connected to *a* would form a 1-mill. But then one of the pairs (*b*, *c*), (*c*, *d*), and (*b*, *d*) must be connected, and this creates a 1-mill with that pair and *a*.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20482,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $x$, $y$, $z$ such that\n\n$$\n7^x + 13^y = 8^z.\n$$",
"options": [],
"answer": "See solution",
"solution": "Reducing modulo $3$ and modulo $4$, we must have that $x$ and $z$ are odd. Let $z = 2m + 1$ and $x = 2n + 1$, where $m$ and $n$ are natural numbers; we have then $7^{2m+1} + 13^y = 8^{2m+1}$.\n\n**a)** If $n = 3s$, we have $7^{6s+1} + 13^y \\equiv 7 \\cdot 343^{2s} \\equiv 7 \\cdot 25^s \\equiv 7 \\cdot (-1)^s \\pmod{13}$ and $8^{2m+1} = 8 \\cdot 64^m \\equiv 8 \\cdot (-1)^m \\pmod{13}$, so this case yields no solutions.\n\n**b)** If $n = 3s + 2$, then $7^{6s+5} + 13^y \\equiv 49 \\cdot 343^{2s+1} \\equiv 50 \\cdot 25^s \\equiv 11 \\cdot (-1)^s \\pmod{13}$ and $8^{2m+1} = 8 \\cdot 64^m \\equiv 8 \\cdot (-1)^m \\pmod{13}$; again, no solutions.\n\n**c)** If $n = 3s + 1$, we have $7^{6s+3} + 13^y = 8^{2m+1}$, which leads to\n\n$$\n(2^{2m+1} - 7^{2s+1})(4^{2m+1} + 2^{2m+1} \\cdot 7^{2s+1} + 49^{2s+1}) = 13^y.\n$$\n\nTaking $a = 2^{2m+1}$, $b = 7^{2s+1}$, it is easy to show that $(a-b, a^2+ab+b^2) = 1$, so $a-b=1$ (and $a^2+ab+b^2 = 13^y$). It follows that $2^{2m+1} = 7^{2s+1} + 1$, so $2^{2m+1} = 8 \\cdot (7^{2s} - 7^{2s-1} + \\dots - 7 + 1)$.\n\nIf $m \\ge 2$, we get $2^{2m-2} = 7^{2s} - 7^{2s-1} + \\dots - 7 + 1$, a contradiction, because the right hand sum is an odd number. For $m=1$ we get $s=0$ and the solution $x=3$, $y=2$, $z=3$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20483,
"subject": "Mathematics (Olympiad)",
"question": "Let $x = k^6$, $y = 2k^{10}$, and $z = 3k^{15}$ for some natural number $k$. Show that $x^5 + y^3 = z^2$ and that there are infinitely many such triples $(x, y, z)$.",
"options": [],
"answer": "See solution",
"solution": "$$\nx^5 + y^3 = (k^6)^5 + (2k^{10})^3 = k^{30} + 8k^{30} = 9k^{30} = (3k^{15})^2 = z^2\n$$\n\nThus, $x^5 + y^3 = z^2$ for all natural $k$. Since there are infinitely many possible values for $k$, there are infinitely many such triples $(x, y, z)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20484,
"subject": "Mathematics (Olympiad)",
"question": "設三角形 $ABC$ 的外接圓為 $\\Omega$，外心為 $O$，垂心為 $H$。令 $S$ 為 $\\Omega$ 上一點，點 $P$ 在 $BC$ 上使得 $\\angle ASP = 90^\\circ$，直線 $SH$ 與 $\\triangle APS$ 的外接圓交於 $X \\ne S$。設 $OP$ 分別與 $CA, AB$ 交於 $Q, R$，$QY, RZ$ 為 $\\triangle AQR$ 的高。\n\n證明：$X, Y, Z$ 共線。\n\n",
"options": [],
"answer": "See solution",
"solution": "令 $AD, BE, CF$ 為 $\\triangle ABC$ 的高，$\\Gamma_A, \\Gamma_B, \\Gamma_C$ 分別為以 $\\overline{AP}, \\overline{BQ}, \\overline{CR}$ 為直徑的圓，則 $H$ 關於 $\\Gamma_A, \\Gamma_B, \\Gamma_C$ 的幂分別為 $HA \\cdot HD, HB \\cdot HE, HC \\cdot HF$，因此 $H$ 關於三個圓的幂相等。\n\n令 $A', B'$ 分別為 $A, B$ 關於 $O$ 的對稱點，$S'$ 為 $A'P$ 與 $B'Q$ 的交點，則由 $O, P, Q$ 共線與帕斯卡逆定理知 $A, A', S', B', B, C$ 共圓錐曲線（考慮折線 $AA'S'B'BC$），故 $S'$ 在 $\\Omega$ 上，即 $S = S'$。\n\n由 $\\angle BSQ = \\angle BSB' = 90^\\circ$ 知 $S$ 在 $\\Gamma_B$ 上，同理有 $S$ 在 $\\Gamma_C$ 上。所以 $\\Gamma_A, \\Gamma_B, \\Gamma_C$ 共軸且 $SH$ 為根軸，因此 $\\Gamma_A, \\Gamma_B, \\Gamma_C$ 的另一個交點為 $X$，所以由 $Y, Z$ 分別位於 $\\Gamma_B, \\Gamma_C$ 上可得\n\n$$\n\\angle YXZ = \\angle YXS + \\angle SXZ = \\angle ABS + \\angle SCA = 180^{\\circ}\n$$\n\n即 $X, Y, Z$ 共線，證畢。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20485,
"subject": "Mathematics (Olympiad)",
"question": "A set of points in the plane is called *obtuse* when it contains no three collinear points, and every triangle with its vertices in this set has one angle greater than $91\\degree$. Is it true that every finite obtuse set can be extended to an infinite obtuse set?",
"options": [],
"answer": "See solution",
"solution": "It is true that every finite obtuse set can be thus extended.\n\nIt suffices to show that any finite obtuse set $\\{P_0, P_1, \\dots, P_n\\}$ can be enlarged to an obtuse set $\\{P_0, P_1, \\dots, P_n, Q\\}$. This is trivial for the empty set and a set with one element, so we may suppose $n \\ge 1$.\n\nNow we construct the new point $Q$ as follows: let $r$ be a ray from $P_0$ at an angle $\\epsilon$ from the ray $P_0P_1$, and let $Q$ be the point a distance $d$ from $P_0$ along this ray $r$. We claim that for $\\epsilon, d$ small enough, this gives an obtuse set $\\{P_0, P_1, \\dots, P_n, Q\\}$.\n\n\n\nFigure 1: Placing $Q$ very close to $P_0$ ensures that all the triangles $QP_iP_j$ have one angle $> 91\\degree$, for $i, j \\neq 0$.\n\nIn order to see this, we need only consider the triangles containing $Q$ as a vertex. For the triangles $QP_iP_j$ where $i, j \\neq 0$, its angles are close to those of $P_0P_iP_j$, so that for $d$ sufficiently small (independent of $\\epsilon$), they must also have one angle $> 91\\degree$. For the remaining triangles $QP_0P_i$ we split into two cases.\n\nIf $|\\angle P_1P_0P_i| > 91\\degree$ then, for $\\epsilon$ sufficiently small (independent of $d$), $|\\angle QP_0P_i| > 91\\degree$ as well.\n\nIn the other case, either $i = 1$ or $i > 1$. For $i = 1$: in triangle $QP_0P_1$ the angle at $P_0$ is $\\epsilon$ which was taken close to $0\\degree$, and for $d$ small enough the angle at $P_1$ is close to $0\\degree$, thus the angle at $Q$ is greater than $91\\degree$. For $i > 1$, notice that the large angle of triangle $P_1P_0P_i$ is located at a vertex other than $P_0$, so that $|\\angle P_1P_0P_i| < 89\\degree$. Also, in triangle $QP_0P_i$ for $i \\neq 1$ we have $|\\angle QP_0P_i| = |\\angle P_1P_0P_i| \\pm \\epsilon < 89\\degree$ for $\\epsilon$ sufficiently small (independent of $d$) and $|\\angle QP_iP_0|$ can be made arbitrarily small by choosing $d$ small enough (maybe depending on $\\epsilon$). Hence, for $\\epsilon$ and $d$ sufficiently small, we know that $|\\angle P_0QP_i|$ can be made arbitrarily close to $180\\degree - |\\angle QP_0P_i| > 91\\degree$ as desired.\n\n\n\nFigure 2: Placing $Q$ on a ray very close to $P_0P_1$ ensures that all the triangles $QP_0P_i$ have one angle $> 91\\degree$ at either $Q$ or $P_0$. On the shown diagram, $QP_0P_1$ will have its obtuse angle at $Q$, while both $QP_0P_2$ and $QP_0P_3$ will have theirs at $P_0$.\n\nHence for sufficiently small $\\epsilon$ and $d$, our set $\\{P_0, P_1, \\dots, P_n, Q\\}$ is obtuse, concluding the proof. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20486,
"subject": "Mathematics (Olympiad)",
"question": "Call a convex polygon on a plane *correct* if for its every side there exists a unique vertex of the polygon that lies farther from that side than any other vertex of the polygon. Call the perpendicular drawn from the vertex farthest from side $XY$ to side $XY$ an *altitude* of the correct polygon. Find all natural numbers $n$ for which there exists a correct $n$-gon whose all $n$ altitudes meet in one point.",
"options": [],
"answer": "See solution",
"solution": "Let one of the vertices be $O(0,0)$ and let the other vertices $A_1, \\dots, A_{n-1}$ lie on a circle with radius $1$ and centre $O$ in such a way that $A_1(1,0)$, $A_{n-1}(0,1)$ and $A_2, \\dots, A_{n-2}$ are all on the shorter arc $A_1A_{n-1}$. The vertex farthest from line $OA_1$ is $A_{n-1}$, the vertex farthest from line $OA_{n-1}$ is $A_1$. The vertex farthest from any other line determined by a side of the polygon is $O$ because the line passing through $O$ parallel to such a side lies in the second and fourth quadrants while the other vertices of the polygon lie above it in the first quadrant. Thus the polygon is correct. The altitudes drawn to sides $OA_1$ and $OA_{n-1}$ are $OA_{n-1}$ and $OA_1$, respectively; they meet at point $O$. As $O$ is the vertex farthest from any other side, all other altitudes meet in $O$, too.\n\n\n\nFig. 16",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20487,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a fixed natural number and\n\n$$\nS_n = \\{\\overline{c_n c_{n-1} \\dots c_1}_{(10)} \\mid c_1, \\dots, c_{n-1}, c_n \\in \\{1, 2, 3, 4\\}\\}.\n$$\n\nAre there distinct numbers $x$ and $y$, $x, y \\in S_n$, such that $4^n \\mid x - y$?",
"options": [],
"answer": "See solution",
"solution": "For $n = 1$, the answer is negative. For $n > 1$ we will show that the answer is positive. Suppose, for contradiction, that no such $x$ and $y$ exist. Since $|S_n| = 4^n$, the set $S_n$ forms a complete system of remainders modulo $4^n$. Thus,\n\n$$\n\\sum_{x \\in S_n} x^3 \\equiv \\sum_{i=1}^{4^n} i^3 \\pmod{4^n} \\equiv \\left(\\frac{4^n(4^n+1)}{2}\\right)^2 \\pmod{4^n} \\equiv 0 \\pmod{4^n}.\n$$\n\nLet us calculate $\\sum_{x \\in S_n} x^3$. Denote $A_n = \\sum_{x \\in S_n} x$, $B_n = \\sum_{x \\in S_n} x^2$, and $C_n = \\sum_{x \\in S_n} x^3$, for all $n \\in \\mathbb{N}$.\n\nThen:\n\n$$\nA_n = \\sum_{k=1}^{n} 10^{k-1} 4^{n-1} (1+2+3+4) \\equiv 2 \\cdot 4^{n-1} \\pmod{4^n},\n$$\n\n$$\n\\begin{aligned}\nB_{n+1} &= \\sum_{(c,x_n) \\in \\{1,2,3,4\\} \\times S_n} (10^n c + x_n)^2 \\\\\n&= \\sum_{(c,x_n) \\in \\{1,2,3,4\\} \\times S_n} (10^{2n} c^2 + 2 \\cdot 10^n c x_n + x_n^2) \\\\\n&= 4^n \\cdot 10^{2n} \\cdot \\sum_{c=1}^{4} c^2 + 2 \\cdot 10^n \\cdot \\sum_{c=1}^{4} c \\cdot A_n + 4 \\cdot B_n \\equiv 4B_n \\pmod{4^{n+1}}.\n\\end{aligned}\n$$\n\nSince $B_1 = 1^2 + 2^2 + 3^2 + 4^2 = 30 \\equiv 2 \\pmod{4}$, by induction, it follows that\n\n$$\nB_n \\equiv 2 \\cdot 4^{n-1} \\pmod{4^n}\n$$\n\nholds for all $n \\in \\mathbb{N}$.\n\nSo, using the above,\n\n$$\n\\begin{aligned}\nC_{n+1} &= \\sum_{(c,x_n) \\in \\{1,2,3,4\\} \\times S_n} (10^n c + x_n)^3 \\\\\n&= \\sum_{(c,x_n) \\in \\{1,2,3,4\\} \\times S_n} (10^{3n} c^3 + 3 \\cdot 10^{2n} c^2 x_n + 3 \\cdot 10^n c x_n^2 + x_n^3) \\\\\n&= 4^n \\cdot 10^{3n} \\cdot \\sum_{c=1}^{4} c^3 + 3 \\cdot 10^{2n} \\cdot \\sum_{c=1}^{4} c^2 \\cdot A_n + 3 \\cdot 10^n \\cdot \\sum_{c=1}^{4} c \\cdot B_n + 4 \\cdot C_n \\\\\n&\\equiv 0 + 0 + 3 \\cdot 10^{n+1} \\cdot B_n + 4C_n \\pmod{4^{n+1}} \\\\\n&\\equiv 3 \\cdot 10^{n+1} \\cdot (k \\cdot 4^n + 2 \\cdot 4^{n-1}) + 4C_n \\pmod{4^{n+1}} \\\\\n&\\equiv 3 \\cdot 5^{n+1} \\cdot 2^{n+2} \\cdot 4^{n-1} + 4C_n \\pmod{4^{n+1}}.\n\\end{aligned}\n$$\n\nSince $C_1 = 100$, from above, $C_2 \\equiv 1000 \\pmod{16} \\equiv 8 \\pmod{16}$ and for $n \\ge 2$, since $4^2 \\mid 2^{n+2}$, we get\n\n$$\nC_{n+1} \\equiv 4C_n \\equiv 4^{n+1}.\n$$\n\nFinally, by induction, for all $n \\in \\mathbb{N} \\setminus \\{1\\}$,\n\n$$\nC_n \\equiv 2 \\cdot 4^{n-1} \\pmod{4^n}\n$$\n\nholds, and this leads to a contradiction with the earlier result. Therefore, for $n > 1$, there exist distinct $x, y \\in S_n$ such that $4^n \\mid x - y$.\n\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20488,
"subject": "Mathematics (Olympiad)",
"question": "1. 令 $\\triangle ABC$ 為銳角三角形，且 $AB < AC$。令 $\\Omega$ 為 $\\triangle ABC$ 的外接圓。令 $B_0$ 為 $AC$ 的中點，$C_0$ 為 $AB$ 的中點，$\\triangle AB_0C_0$ 的外接圓為 $\\Omega_1$。令 $\\omega$ 為一過 $B_0$ 和 $C_0$，且與 $\\Omega$ 切於異於 $A$ 的點 $X$ 的圓。令 $a$ 為 $\\Omega$ 和 $\\Omega_1$ 的公切線，$x$ 則為 $\\Omega$ 和 $\\omega$ 的公切線。試證：$a, x$ 和 $B_0C_0$ 三線共點。\n\n2. 承上，令 $D$ 為 $A$ 對 $BC$ 的垂足，$G$ 為 $\\triangle ABC$ 的重心。試證：$D, G, X$ 三點共線。",
"options": [],
"answer": "See solution",
"solution": "1. 注意到 $a$ 是 $\\Omega$ 和 $\\Omega_1$ 的根軸（radical axis），$x$ 是 $\\Omega$ 和 $\\omega$ 的根軸，$B_0C_0$ 則為 $\\Omega_1$ 和 $\\omega$ 的根軸。根據三圓的根軸定理，三根軸必共點，因此 $a, x$ 和 $B_0C_0$ 三線共點。\n\n2. 令 $O$ 為 $\\triangle ABC$ 的外心，$A_0$ 為 $BC$ 的中點，$Q$ 為 $A_0$ 對 $B_0C_0$ 的垂足。注意到 $\\angle WAO = \\angle WQO = \\angle WXO = 90^\\circ$，故 $A, W, X, O, Q$ 五點共圓。此外，關於 $B_0C_0$ 的鏡射會將 $A$ 映到 $D$，關於 $OW$ 的鏡射則會將 $A$ 映到 $X$。因此：\n\n$$\n\\angle WQD = \\angle WQA = \\angle WXA = \\angle WAX = \\angle WQX.\n$$\n\n因此 $Q, D, X$ 三點共線。\n\n最後，注意到以重心 $G$ 為中心，位似比 $1:2$，旋轉 $180^\\circ$ 的位似旋轉變換可將 $\\triangle ABC$ 映到 $\\triangle A_0B_0C_0$，同時將 $AD$ 映到 $A_0Q$。\n\n故 $D, G, Q$ 三點共線，從而 $D, G, X$ 三點共線。得證。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20489,
"subject": "Mathematics (Olympiad)",
"question": "Define an _arithmetic permutation_ as a bijective function $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ which satisfies $m \\mid n \\Leftrightarrow f(m) \\mid f(n)$ for all $m, n \\in \\mathbb{Z}^+$. Show that each of the following conditions on bijective functions gives an equivalent definition of an arithmetic permutation:\n\n1. $f(m) \\cdot f(n) = f(mn)$ for all $m, n \\in \\mathbb{Z}^+$.\n2. $\\operatorname{lcm}(f(m), f(n)) = f(\\operatorname{lcm}(m, n))$ for all $m, n \\in \\mathbb{Z}^+$.\n\n% ",
"options": [],
"answer": "See solution",
"solution": "We first establish a lemma characterizing arithmetic permutations.\n\n**Lemma.** Let $\\mathbb{P}$ denote the set of prime numbers. An arithmetic permutation $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ maps $\\mathbb{P}$ onto itself bijectively. Furthermore, given a bijection $\\tilde{f} : \\mathbb{P} \\to \\mathbb{P}$, there is a unique arithmetic permutation $f$ extending $\\tilde{f}$, given by:\n$$\nf(p_1^{\\alpha_1} \\cdots p_k^{\\alpha_k}) = \\tilde{f}(p_1)^{\\alpha_1} \\cdots \\tilde{f}(p_k)^{\\alpha_k}.\n$$\n\n*Proof.* Let $D(n)$ denote the set of positive divisors of $n$. An arithmetic permutation $f$ maps $D(n)$ onto $D(f(n))$ bijectively, so $|D(n)| = |D(f(n))|$. Since primes are characterized by $|D(p)| = 2$, $p$ is prime if and only if $f(p)$ is prime.\n\nWe prove $f(p^n) = f(p)^n$ by induction on $n$. Assume $f(p^{n-1}) = f(p)^{n-1}$. Since $f(p^{n-1}) \\mid f(p^n)$, let $f(p^n) = f(p)^{n-1+a} \\cdot A$ with $a \\ge 0$ and $f(p) \\nmid A$. Then $|D(f(p^n))| = (n+a) \\cdot |D(A)|$. But $|D(f(p^n))| = |D(p^n)| = n+1$, so $(n+a) \\cdot |D(A)| = n+1$. This forces $A=1$, $a=1$, so $f(p^n) = f(p)^n$.\n\nFor $n = p_1^{a_1} \\cdots p_k^{a_k}$, similar reasoning shows $f(p_1^{a_1} \\cdots p_k^{a_k}) = f(p_1)^{a_1} \\cdots f(p_k)^{a_k}$.\n\nGiven any bijection $\\tilde{f}: \\mathbb{P} \\to \\mathbb{P}$, the function $f$ defined by $f(p_1^{a_1} \\cdots p_k^{a_k}) = \\tilde{f}(p_1)^{a_1} \\cdots \\tilde{f}(p_k)^{a_k}$ is an arithmetic permutation.\n\nNow, consider the condition $f(m) \\cdot f(n) = f(mn)$. If $a \\mid b$, then $b = a \\cdot s$, so $f(b) = f(a) \\cdot f(s)$, hence $f(a) \\mid f(b)$. Conversely, if $f(a) \\mid f(b)$, then $f(b) = f(a) \\cdot f(t) = f(at)$, so $b = at$, thus $a \\mid b$. Therefore, a bijection $f$ with $f(m) \\cdot f(n) = f(mn)$ is an arithmetic permutation, and vice versa.\n\nSimilarly, for $\\operatorname{lcm}(f(m), f(n)) = f(\\operatorname{lcm}(m, n))$, if $a \\mid b$, then $b = \\operatorname{lcm}(a, b)$, so $f(b) = f(\\operatorname{lcm}(a, b)) = \\operatorname{lcm}(f(a), f(b))$, hence $f(a) \\mid f(b)$. Conversely, if $f(a) \\mid f(b)$, then $f(b) = \\operatorname{lcm}(f(a), f(b)) = f(\\operatorname{lcm}(a, b))$, so $b = \\operatorname{lcm}(a, b)$, thus $a \\mid b$. Therefore, a bijection $f$ with $\\operatorname{lcm}(f(m), f(n)) = f(\\operatorname{lcm}(m, n))$ is an arithmetic permutation, and vice versa.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20490,
"subject": "Mathematics (Olympiad)",
"question": "In triangle $ABC$, point $P$ divides side $AB$ in the ratio $\\frac{AP}{PB} = \\frac{1}{4}$. The perpendicular bisector of segment $PB$ intersects side $BC$ at point $Q$. If $\\text{area}(PQC) = \\frac{4}{25}\\,\\text{area}(ABC)$ and $AC = 7$, find $BC$.",
"options": [],
"answer": "See solution",
"solution": "If $\\text{area}(ABC) = S$, then $\\text{area}(APC) = \\frac{AP}{AB} S = \\frac{1}{5} S$. As $\\text{area}(PQC) = \\frac{4}{25} S$, we have $\\text{area}(PQB) = S - \\frac{1}{5} S - \\frac{4}{25} S = \\frac{16}{25} S$.\n\nOn the other hand,\n\n$$\n\\text{area}(PBQ) = \\frac{BQ}{BC} \\cdot \\text{area}(PBC) = \\frac{BQ}{BC} \\cdot \\frac{BP}{BA} \\cdot S = \\frac{BQ}{BC} \\cdot \\frac{4}{5} S.\n$$\n\nHence $\\frac{16}{25} S = \\frac{BQ}{BC} \\cdot \\frac{4}{5} S$, which implies $\\frac{BQ}{BC} = \\frac{4}{5}$. Because $\\frac{BP}{BA} = \\frac{4}{5}$, it follows that $PQ \\parallel AC$, so triangles $ABC$ and $PBQ$ are similar. But $PQ = BQ$ as $Q$ lies on the perpendicular bisector of $PB$. Therefore, $BC = AC = 7$.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20491,
"subject": "Mathematics (Olympiad)",
"question": "Prove the following inequality for positive numbers $x, y, z$:\n\n$$\n\\frac{x}{x+y+z} \\cdot \\frac{(x+y)(x+z)}{(y+z)^2} + \\frac{y}{x+y+z} \\cdot \\frac{(y+z)(y+x)}{(z+x)^2} + \\frac{z}{x+y+z} \\cdot \\frac{(z+x)(z+y)}{(x+y)^2} \\ge 1.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let us rewrite the inequality:\n\n$$\n\\frac{(x+y)(y+z)(z+x)}{x+y+z} \\left( \\frac{x}{(y+z)^3} + \\frac{y}{(z+x)^3} + \\frac{z}{(x+y)^3} \\right) \\ge 1\n$$\n\nThis is equivalent to:\n\n$$\n\\frac{(x+y)(y+z)(z+x)}{2(x+y+z)(xy+yz+zx)} \\left( \\frac{x}{(y+z)^3} + \\frac{y}{(z+x)^3} + \\frac{z}{(x+y)^3} \\right) (x(y+z)+y(x+z)+z(x+y)) \\ge 1.\n$$\n\nBy the Cauchy-Bunyakovsky inequality:\n\n$$\n\\left( \\frac{x}{(y+z)^3} + \\frac{y}{(z+x)^3} + \\frac{z}{(x+y)^3} \\right) (x(y+z)+y(x+z)+z(x+y)) \\ge \\left( \\frac{x}{y+z} + \\frac{y}{z+x} + \\frac{z}{x+y} \\right)^2.\n$$\n\nLet us prove the auxiliary inequality for arbitrary positive numbers $a, b, c$:\n\n$$\n\\frac{a}{b+c} + \\frac{b}{c+a} + \\frac{c}{a+b} \\ge \\frac{3}{2}.\n$$\n\nSince the left side is homogeneous, we may assume $a+b+c=1$. The inequality $\\frac{t}{1-t} \\ge \\frac{9t-1}{4}$ holds for $0 < t < 1$. Indeed,\n\n$$\n4t \\ge (1-t)(9t-1) \\Leftrightarrow 9t^2 - 6t + 1 \\ge 0 \\Leftrightarrow (3t-1)^2 \\ge 0.\n$$\n\nThus,\n\n$$\n\\frac{a}{b+c} + \\frac{b}{c+a} + \\frac{c}{a+b} = \\frac{a}{1-a} + \\frac{b}{1-b} + \\frac{c}{1-c} \\ge \\frac{9a-1}{4} + \\frac{9b-1}{4} + \\frac{9c-1}{4} = \\frac{9(a+b+c)-3}{4} = \\frac{3}{2}.\n$$\n\nTherefore, $\\frac{x}{y+z} + \\frac{y}{z+x} + \\frac{z}{x+y} \\ge \\frac{3}{2}$, so\n\n$$\n\\left( \\frac{x}{(y+z)^3} + \\frac{y}{(z+x)^3} + \\frac{z}{(x+y)^3} \\right) (x(y+z)+y(x+z)+z(x+y)) \\ge \\frac{9}{4}.\n$$\n\nIt suffices to show that\n\n$$\n\\frac{(x+y)(y+z)(z+x)}{2(x+y+z)(xy+yz+zx)} \\cdot \\frac{9}{4} \\ge 1.\n$$\n\nMultiplying both sides by the denominator and expanding, we get:\n\n$$\nx^2 y + xy^2 + y^2 z + yz^2 + z^2 x + zx^2 \\ge 6xyz\n$$\n\nwhich follows from the AM-GM inequality for six numbers:\n\n$$\nx^2 y + xy^2 + y^2 z + yz^2 + z^2 x + zx^2 \\ge 6 \\sqrt{x^2 y \\cdot xy^2 \\cdot y^2 z \\cdot yz^2 \\cdot z^2 x \\cdot zx^2} = 6xyz.\n$$\n\nThus, the inequality is proved.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20492,
"subject": "Mathematics (Olympiad)",
"question": "Let $O$ be the circumcentre of an acute scalene triangle $ABC$. Line $OA$ intersects the altitudes of $ABC$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcentre of triangle $PQH$ lies on the median drawn from vertex $A$ of triangle $ABC$.",
"options": [],
"answer": "See solution",
"solution": "Let $AD$ be the altitude of triangle $ABC$ and the midpoints of sides $BC$, $CA$, and $AB$ be $K$, $L$, and $M$, respectively. Let $O_1$ be the circumcentre of triangle $HPQ$.\n\nAs $O$ is the circumcentre of $ABC$, we have $\\angle AOL = \\frac{1}{2}\\angle AOC = \\angle ABC$ and $\\angle AOM = \\frac{1}{2}\\angle AOB = \\angle ACB$. Therefore $\\angle QPH = \\angle AOL = \\angle ABC$ and $\\angle PQH = \\angle AOM = \\angle ACB$. Thus the triangles $ABC$ and $HPQ$ are similar. From this, $\\angle O_1HQ = \\angle OAC = 90^\\circ - \\angle AOL = 90^\\circ - \\angle ABC = \\angle HCB$. Therefore $HO_1 \\parallel BC$.\n\nNow let $R$ and $S$ be the points of intersection of $AO$ with $HO_1$ and $BC$, respectively. Because of the similarity of triangles $ABC$ and $HPQ$ we must have $\\frac{O_1R}{HR} = \\frac{OS}{AS}$. Lines $OK$ and $AD$ both being orthogonal to $BC$ implies $OK \\parallel AD$, from which $\\frac{OS}{AS} = \\frac{KS}{DS}$. In summary, $\\frac{O_1R}{HR} = \\frac{KS}{DS}$. As $HR$ and $DS$ are parallel, we conclude from this equation that $A$, $O_1$, and $K$ are collinear.\n\n\n\n\n\n\n\nFrom the vertex of an isosceles triangle bisects the base, from which we get that *MK* is the perpendicular bisector of *DX* and *LK* is the perpendicular bisector of *DY*. The point of intersection of those bisectors, *K*, is the circumcentre of triangle *DXY*.\n\nFrom parallel lines $\\frac{AH}{AD} = \\frac{AP}{AX} = \\frac{AQ}{AY}$ (see figure). Therefore, homothety centred at $A$ with ratio $\\frac{AH}{AD}$ converts triangle $DXY$ into triangle $HPQ$ and point $K$ into the circumcentre of triangle $HPQ$. Therefore, the circumcentre of triangle $HPQ$ is located on the segment $AK$.\n",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20493,
"subject": "Mathematics (Olympiad)",
"question": "Find all real-valued functions $f$ defined on the real numbers such that\n$$\nf(f(x) + f(y)) = f(x) + y\n$$\nfor all real $x, y$.",
"options": [],
"answer": "See solution",
"solution": "Let $z_1, z_2$ be real numbers such that $f(z_1) = f(z_2)$. Substituting $y = z_1$ and $y = z_2$ into the given equation, we get:\n$$\nf(f(x) + f(z_1)) = f(x) + z_1,\n$$\n$$\nf(f(x) + f(z_2)) = f(x) + z_2.\n$$\nSince the left-hand sides are equal, $f(x) + z_1 = f(x) + z_2$, so $z_1 = z_2$. Thus, $f$ is one-to-one.\n\nSubstituting $y = 0$ into the original equation gives:\n$$\nf(f(x) + f(0)) = f(x)\n$$\nfor any real $x$. Since $f$ is one-to-one, $f(x) + f(0) = x$. Setting $x = 0$ gives $2f(0) = 0$, so $f(0) = 0$. Therefore, $f(x) = x$.\n\nFinally, $f(x) = x$ satisfies the original equation.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20494,
"subject": "Mathematics (Olympiad)",
"question": "Each point of the plane is coloured in one of two colours. Given an odd integer $n \\geq 3$, prove that there exist (at least) two similar triangles whose similitude ratio is $n$, each of which has a monochromatic vertex set.",
"options": [],
"answer": "See solution",
"solution": "We first show that there exists a rectangle with monochromatic vertices, then subdivide it into $n^2$ rectangles by subdividing each side into $n$ congruent segments, and show that (at least) one of these smaller rectangles has (at least) three monochromatic vertices—here is where the assumption that $n$ be odd comes in.\n\nTo find a rectangle with monochromatic vertices, consider a line $a$ in the plane and a five-element monochromatic subset $A$ of $a$. Project $A$ orthogonally onto another line $b$ which is parallel to $a$, and consider a three-element monochromatic subset $B$ of the image of $A$ under projection. If the colours of $A$ and $B$ agree, we are done. Otherwise, project $B$ onto a third line $c$ which is parallel to both $a$ and $b$, and consider a two-element monochromatic subset $C$ of the image of $B$ under projection. Clearly, the two points of $C$ together with their orthogonal projections on either $A$ or $B$ are monochromatic.\n\nWithout loss of generality, we may assume that the vertices of the square $[0, n] \\times [0, n]$ are monochromatic. Since $n$ is odd, and the points $(0, 0)$ and $(n, 0)$ (respectively, $(0, n)$) are monochromatic, there exists $p$ (respectively, $q$) in $\\{0, 1, \\dots, n-1\\}$ such that the points $(p, 0)$ and $(p+1, 0)$ (respectively, $(0, q)$ and $(0, q+1)$) are monochromatic. We now show that (at least) one of the unit squares\n\n$$\n[i, i+1] \\times [q, q+1], \\quad i = 0, 1, \\dots, p,\n$$\n\n$$\n[p, p+1] \\times [j, j+1], \\quad j = 0, 1, \\dots, q,\n$$\n\nhas (at least) three monochromatic vertices. Suppose that each of the first $p$ (respectively, $q$) horizontal (respectively, vertical) unit squares above has two vertices of each colour. Then the colours of the lattice points alternate in the same way along both horizontal segments $[0, p] \\times \\{q\\}$ and $[0, p] \\times \\{q+1\\}$, according to the parity of the abscissae, and in the same way along both vertical segments $\\{p\\} \\times [0, q]$ and $\\{p+1\\} \\times [0, q]$, according to the parity of the ordinates. Consequently, the vertical pair $((p, q), (p, q+1))$ and the horizontal pair $((p, q), (p+1, q))$ are both monochromatic; that is, the unit square $[p, p+1] \\times [q, q+1]$ has (at least) three monochromatic vertices.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20495,
"subject": "Mathematics (Olympiad)",
"question": "Determine all positive integers $a$ for which there exist exactly $2014$ positive integers $b$ such that $2 \\le \\frac{a}{b} \\le 5$.",
"options": [],
"answer": "See solution",
"solution": "Rewrite $2 \\le \\frac{a}{b} \\le 5$ as $2 \\le \\frac{a}{b} \\le 5$, which is equivalent to $a/5 \\le b \\le a/2$. The number of positive integers $b$ satisfying this is $\\left\\lfloor \\frac{a}{2} \\right\\rfloor - \\left\\lceil \\frac{a}{5} \\right\\rceil + 1$. We want this to equal $2014$.\n\nSet up the equation:\n$$\n\\left\\lfloor \\frac{a}{2} \\right\\rfloor - \\left\\lceil \\frac{a}{5} \\right\\rceil + 1 = 2014\n$$\n\nSolving for $a$, we find that the required values are $a = 6710, 6712, 6713$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20496,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c \\ge -1$ be real numbers with $a^3 + b^3 + c^3 = 1$. Prove that\n$$\na + b + c + a^2 + b^2 + c^2 \\le 4.\n$$\n\nWhen does equality hold?",
"options": [],
"answer": "See solution",
"solution": "We note that\n$$\n1 - x - x^2 + x^3 = (1-x)^2(1+x) \\ge 0\n$$\nholds for all real numbers $x \\ge -1$. Here, equality holds if and only if $x = \\pm 1$.\n\nThus, we have $x + x^2 \\le 1 + x^3$ for $x = a, b, c$, and therefore\n$$\na + a^2 + b + b^2 + c + c^2 \\le 1 + a^3 + 1 + b^3 + 1 + c^3 = 3 + (a^3 + b^3 + c^3) = 3 + 1 = 4.\n$$\n\nAs equality in the previous inequality holds for $x = \\pm 1$ and the sum of the cubes equals $1$, equality in the original inequality holds if and only if $(a, b, c)$ is a permutation of $(1, 1, -1)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20497,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\ell$ be a line, and let $\\gamma$ and $\\gamma'$ be two circles. The line $\\ell$ meets $\\gamma$ at points $A$ and $B$, and $\\gamma'$ at points $A'$ and $B'$. The tangents to $\\gamma$ at $A$ and $B$ meet at point $C$, and the tangents to $\\gamma'$ at $A'$ and $B'$ meet at point $C'$. The lines $\\ell$ and $CC'$ meet at point $P$. Let $\\lambda$ be a variable line through $P$ and let $X$ be one of the points where $\\lambda$ meets $\\gamma$, and $X'$ be one of the points where $\\lambda$ meets $\\gamma'$. Prove that the point of intersection of the lines $CX$ and $C'X'$ lies on a fixed circle.",
"options": [],
"answer": "See solution",
"solution": "Let the lines $CX$ and $C'X'$ meet at point $Q$. The line $CX$ meets $\\ell$ at $D$, and $\\gamma$ a second time at $Y$; similarly, the line $C'X'$ meets $\\ell$ at $D'$, and $\\gamma'$ a second time at $Y'$. Notice that the cross-ratios $(CDXY)$ and $(C'D'X'Y')$ are both harmonic, for $C$ and $C'$ are the poles of $\\ell$ relative to $\\gamma$ and $\\gamma'$, respectively. Since the lines $CC'$, $DD'$ and $XX'$ all pass through $P$, so does the line $YY'$. Consequently, $(QCXD) = (QC'X'D')$ and $(QCYD) = (QC'Y'D')$. Let $\\varrho$ and $\\varrho'$ be the powers of $Q$ relative to $\\gamma$ and $\\gamma'$, respectively, let $\\omega$ be the power of $C$ relative to $\\gamma$, and let $\\omega'$ be the power of $C'$ relative to $\\gamma'$. Multiply the last two equalities involving cross-ratios and apply Menelaus' theorem to triangle $QCC'$ and transversal $PDD'$ to get\n\n$$\n\\frac{\\varrho}{\\varrho'} = \\frac{\\omega}{\\omega'} \\left( \\frac{C'D'}{QD'} \\cdot \\frac{QD}{CD} \\right)^2 = \\frac{\\omega}{\\omega'} \\left( \\frac{PC'}{PC} \\right)^2 = \\text{constant}\n$$\n\nand infer that $Q$ lies on a fixed circle of the pencil of circles generated by $\\gamma$ and $\\gamma'$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20498,
"subject": "Mathematics (Olympiad)",
"question": "Let $x_n = a \\cdot b^n + c \\cdot d^n$, with $\\gcd(a, c) = e$ and $\\gcd(b, d) = f$. Then $a = e a_1$, $c = e c_1$, $b = f b_1$, $d = f d_1$, where $\\gcd(a_1, c_1) = 1$ and $\\gcd(b_1, d_1) = 1$.\n\nLet $P$ be the set of all primes such that each element of $P$ divides at least one element of the sequence $(y_n)_{n \\ge 1}$, where $y_n = a_1 b_1^n + c_1 d_1^n$. Show that $P$ is a finite set and deduce that $b_1 = d_1 = 1$.",
"options": [],
"answer": "See solution",
"solution": "Define the following subsets of $P$:\n\n$$\n\\begin{align*}\nP_{ad} &= \\{p \\in P : p \\mid a_1 d_1\\}, \\\\\nP_{bc} &= \\{p \\in P : p \\mid b_1 c_1\\}, \\\\\nP_{a+c} &= \\{p \\in P : p \\mid a_1 + c_1\\}, \\\\\nP_0 &= \\{p \\in P : p \\nmid a_1 b_1 c_1 d_1 \\text{ and } p \\nmid a_1 + c_1\\}\n\\end{align*}\n$$\n\nIt can be seen that $P = P_{ad} \\cup P_{bc} \\cup P_{a+c} \\cup P_0$.\n\n- For $p \\in P_{ad}$: If $p \\mid a_1$, then $p \\mid c_1 d_1^n$ for all $n$, so $p \\mid d_1$. Similarly, if $p \\mid d_1$, then $p \\mid a_1$. For sufficiently large $n$, $v_p(y_n) = v_p(a_1)$.\n- For $p \\in P_{bc}$: Similarly, for large $n$, $v_p(y_n) = v_p(c_1)$.\n- For $p \\in P_{a+c}$: Here $p \\nmid b_1 d_1$. Let $v_p(a_1 + c_1) = \\alpha$. By Euler's theorem, for each $n$ divisible by $\\phi(p^{\\alpha+1}) = p^{\\alpha}(p-1)$, we have\n $$\n y_n \\equiv a_1 + c_1 \\pmod{p^{\\alpha+1}}\n $$\n so $v_p(y_n) = \\alpha = v_p(a_1 + c_1)$.\n- For $p \\in P_0$: By Fermat's theorem, for $n$ divisible by $p-1$, $y_n \\equiv a_1 + c_1 \\pmod{p}$, so $p \\nmid y_n$.\n\nTherefore, for sufficiently large $n$ of the form\n$$\nn = k \\cdot \\prod_{p \\in P_{a+c}} (p-1) p^{v_p(a_1+c_1)} \\cdot \\prod_{p \\in P_0} (p-1)\n$$\nfor integer $k$, the $p$-adic valuations $v_q(y_n)$ for all prime divisors $q$ of $y_n$ are bounded above. Since all four sets are finite, $y_n$ is bounded above for large $n$. The only possibility is $b_1 = d_1 = 1$. Thus, $b = d$ and the result follows.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20499,
"subject": "Mathematics (Olympiad)",
"question": "A convex polygon $\\mathcal{P}$ in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon $\\mathcal{P}$ are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers.",
"options": [],
"answer": "See solution",
"solution": "Let $\\mathcal{P} = A_1A_2\\dots A_n$, where $n \\ge 3$. The problem is trivial for $n=3$ because there are no diagonals and thus no dissections. Assume $n \\ge 4$. Our proof is based on the following lemma.\n\n**Lemma:** Let $ABCD$ be a convex quadrilateral such that all its sides and diagonals have rational lengths. If segments $AC$ and $BD$ meet at $P$, then segments $AP$, $BP$, $CP$, $DP$ all have rational lengths.\n\n\n\nBy the lemma, the desired result holds when $\\mathcal{P}$ is a convex quadrilateral. Let $A_iA_j$ ($1 \\le i < j \\le n$) be a diagonal of $\\mathcal{P}$. Assume $C_1, C_2, \\dots, C_m$ are the consecutive division points on diagonal $A_iA_j$ (where $C_1$ is closest to $A_i$ and $C_m$ is closest to $A_j$). Then the segments $C_\\ell C_{\\ell+1}$, $1 \\le \\ell \\le m-1$, are the sides of all polygons in the dissection. Let $C_\\ell$ be the point where diagonal $A_iA_j$ meets diagonal $A_sA_t$. Then quadrilateral $A_iA_sA_jA_t$ satisfies the lemma. Consequently, segments $A_iC_\\ell$ and $C_\\ell A_j$ have rational lengths. Therefore, segments $A_iC_1, A_iC_2, \\dots, A_jC_m$ all have rational lengths. Thus, $C_\\ell C_{\\ell+1} = AC_{\\ell+1} - AC_\\ell$ is rational. Because $i, j, \\ell$ are arbitrarily chosen, we have proved that all sides of all polygons in the dissection are also rational numbers.\n\nNow we present two proofs of the lemma:\n\n* **First approach:** We show only that segment $AP$ is rational; the proof for the others is similar. Introduce Cartesian coordinates with $A = (0, 0)$ and $C = (c, 0)$. Put $B = (a, b)$ and $D = (d, e)$. Then by hypothesis, the numbers\n\n$$\n\\begin{aligned}\nAB &= \\sqrt{a^2 + b^2}, & AC &= c, & AD &= \\sqrt{d^2 + e^2}, \\\\\nBC &= \\sqrt{(a-c)^2 + b^2}, & BD &= \\sqrt{(a-d)^2 + (b-e)^2}, \\\\\nCD &= \\sqrt{(d-c)^2 + e^2}\n\\end{aligned}\n$$\n\nare rational. In particular,\n\n$$\nBC^2 - AB^2 - AC^2 = (a-c)^2 + b^2 - (a^2 + b^2) - c^2 = -2ac\n$$\n\nis rational. Because $c \\neq 0$, $a$ is rational. Likewise, $d$ is rational.\n\n\n\nNow $b^2 = AB^2 - a^2$, $e^2 = AD^2 - d^2$, and $(b-e)^2 = BD^2 - (a-d)^2$ are rational, so $2be = b^2 + e^2 - (b-e)^2$ is rational. Because quadrilateral $ABCD$ is convex, $b$ and $e$ are nonzero and have opposite sign. Hence $b/e = 2be/2b^2$ is rational.\n\nWe now calculate\n\n$$\nP = \\left( \\frac{bd - ae}{b - e}, 0 \\right)\n$$\nso\n$$\nAP = \\frac{\\frac{b}{e} \\cdot d - a}{\\frac{b}{e} - 1}\n$$\nis rational.\n\n* **Second approach:** To prove the lemma, set $\\angle DAP = A_1$ and $\\angle BAP = A_2$. Applying the Law of Cosines to triangles $ADC$, $ABC$, $ABD$ shows that angles $A_1$, $A_2$, $A_1+A_2$ all have rational cosine values. By the addition formula,\n\n$$\n\\sin A_1 \\sin A_2 = \\cos A_1 \\cos A_2 - \\cos(A_1 + A_2)\n$$\n\nimplying that $\\sin A_1 \\sin A_2$ is rational.\n\nThus\n$$\n\\frac{\\sin A_2}{\\sin A_1} = \\frac{\\sin A_2 \\sin A_1}{\\sin^2 A_1} = \\frac{\\sin A_2 \\sin A_1}{1 - \\cos^2 A_1}\n$$\nis rational.\n\n\n\nNote that the ratio between the areas of triangles *ADP* and *ABP* is equal to $\\frac{PD}{BP}$. Therefore\n\n$$\n\\frac{BP}{PD} = \\frac{[ABP]}{[ADP]} = \\frac{\\frac{1}{2}AB \\cdot AP \\cdot \\sin A_2}{\\frac{1}{2}AD \\cdot AP \\cdot \\sin A_1} = \\frac{AB}{AD} \\cdot \\frac{\\sin A_2}{\\sin A_1}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20500,
"subject": "Mathematics (Olympiad)",
"question": "Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\\angle XOZ$, where $O$ is the midpoint of segment $AB$.",
"options": [],
"answer": "See solution",
"solution": "Let $T$ be the foot of the perpendicular from $Y$ to line $AB$. We note that $P, Q, T$ are the feet of the perpendiculars from $Y$ to the sides of triangle $ABX$. Because $Y$ lies on the circumcircle of triangle $ABX$, points $P, Q$, and $T$ are collinear by Simson's theorem. Likewise, points $S, R$, and $T$ are collinear.\n\n\n\nWe need to show that $\\angle XOZ = 2\\angle PTS$. Notice that\n\n$$\n\\frac{\\angle XOZ}{2} = \\frac{\\widehat{XZ}}{2} = \\frac{\\widehat{XY}}{2} + \\frac{\\widehat{YZ}}{2} = \\angle XAY + \\angle ZBY = \\angle PAY + \\angle SBY\n$$\n\nand that $\\angle PTS = \\angle PTY + \\angle STY$. Therefore, it suffices to prove that\n\n$$\n\\angle PTY = \\angle PAY \\text{ and } \\angle STY = \\angle SBY.\n$$\n\nFor this, it is enough to show that quadrilaterals $APYT$ and $BSYT$ are cyclic. This follows because $\\angle APY = \\angle ATY = 90^\\circ$ and $\\angle BTY = \\angle BSY = 90^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20501,
"subject": "Mathematics (Olympiad)",
"question": "Nonzero real numbers $a, b, c$ satisfy the equation $ab + bc + ac = 0$. Prove that the numbers $a + b + c$ and $\\frac{1}{a+b} + \\frac{1}{b+c} + \\frac{1}{c+a}$ have the same sign.",
"options": [],
"answer": "See solution",
"solution": "**Solution.** If we assume that $b + c = 0$, then we have\n\n$$\nab + bc + ac = a(b + c) + ac = ac = 0,\n$$\n\nwhich contradicts the condition. Thus $(a+b)(b+c)(c+a) \\neq 0$.\n\nSince\n\n$$\n(a + b + c)(a + b) = a^2 + ab + b^2 + ab + bc + ac = a^2 + ab + b^2,\n$$\n\nwe have $\\frac{1}{a+b} = (a+b+c) \\cdot \\frac{1}{a^2+ab+b^2}$. Then, adding three similar equations, we get\n\n$$\n\\frac{1}{a+b} + \\frac{1}{b+c} + \\frac{1}{c+a} = (a+b+c) \\left( \\frac{1}{a^2+ab+b^2} + \\frac{1}{b^2+bc+c^2} + \\frac{1}{a^2+ac+c^2} \\right).\n$$\n\nIt is easy to see that each of the expressions of the form $a^2 + ab + b^2 = \\frac{a^2+(a+b)^2+b^2}{2} > 0$, which implies the desired statement.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20502,
"subject": "Mathematics (Olympiad)",
"question": "Given a triangle $ABC$, let the bisectors of the angles $BAC$, $CBA$, $ACB$ intersect the circumcircle of the triangle $ABC$ at the points $M$, $N$, $K$ respectively. Let $P$ be the intersection of the segments $AB$ and $MK$, and $Q$ be the intersection of the segments $AC$ and $MN$. Prove that the lines $PQ$ and $BC$ are parallel.",
"options": [],
"answer": "See solution",
"solution": "From $\\vec{AN} = \\vec{NC}$, we see that $MQ$ is a bisector of $\\triangle AMC$.\n\n\n\nBy the angle bisector theorem, we have\n\n$$\n\\frac{AQ}{QC} = \\frac{AM}{MC}. \\qquad (1)\n$$\n\nSimilarly, we have\n\n$$\n\\frac{AP}{PB} = \\frac{AM}{MB}. \\qquad (2)\n$$\n\nSince $\\vec{BM} = \\vec{MC}$, we have $MB = MC$. Thus from (1) and (2), we get\n\n$$\n\\frac{AQ}{QC} = \\frac{AP}{PB}.\n$$\n\nBy Thales' theorem, we have $PQ \\parallel BC$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20503,
"subject": "Mathematics (Olympiad)",
"question": "Let $P(x)$ be a real polynomial of degree at most $n$, and let $a_1, a_2, \\dots, a_m$ be $m > n$ distinct real numbers. Suppose that for all $1 \\leq i < j \\leq m$, we have\n$$\n|P(a_i) - P(a_j)| = |a_i - a_j|.\n$$\nFind all such polynomials $P(x)$.\n\n(b) Prove that if $Q(x) = x^n$ and $|a_i| < \\frac{1}{n}$ for all $1 \\leq i \\leq m$, then for all $i \\neq j$,\n$$\n|Q(a_i) - Q(a_j)| < |a_i - a_j|.\n$$",
"options": [],
"answer": "See solution",
"solution": "**Case 1:** The numbers $d_1, d_2, \\dots, d_{m-1}$ are positive. Then from the given condition,\n$$\nP(a_1) - P(a_2) = a_1 - a_2 \\implies P(a_1) - a_1 = P(a_2) - a_2.\n$$\nSimilarly, $P(a_2) - a_2 = P(a_3) - a_3$, and so on, so\n$$\nP(a_1) - a_1 = P(a_2) - a_2 = \\dots = P(a_m) - a_m = k.\n$$\nThus, $P(x) = x + k$ for some $k \\in \\mathbb{R}$.\n\n**Case 2:** The numbers $d_1, d_2, \\dots, d_{m-1}$ are negative. Then,\n$$\n-P(a_1) + P(a_2) = a_1 - a_2 \\implies P(a_1) + a_1 = P(a_2) + a_2.\n$$\nSo,\n$$\nP(a_1) + a_1 = P(a_2) + a_2 = \\dots = P(a_m) + a_m = \\lambda.\n$$\nThus, $P(x) = -x + \\lambda$ for some $\\lambda \\in \\mathbb{R}$.\n\nTherefore, the only polynomials satisfying the condition are $P(x) = x + k$ and $P(x) = -x + \\lambda$.\n\n**Alternate approach:**\nSuppose $m \\geq 3$ and there exist $p, q, r \\in \\{a_1, \\dots, a_m\\}$ such that\n$$\nP(p) - P(q) = p - q, \\quad P(p) - P(r) = r - p.\n$$\nSubtracting gives $P(r) - P(q) = 2p - q - r$. But from the condition, $P(r) - P(q) = r - q$ or $q - r$. Both cases lead to a contradiction unless $p = r$ or $p = q$, which is not possible for distinct $p, q, r$. Thus, only the linear polynomials above are possible.\n\n**Part (b):**\nLet $Q(x) = x^n$ and $|a_i| < \\frac{1}{n}$ for all $i$. Then\n$$\n|Q(a_i) - Q(a_j)| = |a_i^n - a_j^n| = |a_i - a_j| \\cdot |a_i^{n-1} + a_i^{n-2}a_j + \\dots + a_j^{n-1}|.\n$$\nBut\n$$\n|a_i^{n-1} + a_i^{n-2}a_j + \\dots + a_j^{n-1}| < 1\n$$\nsince each term is less than $\\frac{1}{n^{n-1}}$ and there are $n$ terms. Therefore,\n$$\n|Q(a_i) - Q(a_j)| < |a_i - a_j|.\n$$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20504,
"subject": "Mathematics (Olympiad)",
"question": "Let $H$ be the orthocentre of $\\triangle ABC$. Reflect $A$ with respect to the altitudes $CH$ and $BH$, and let $D'$ and $E'$ be the respective points thus obtained. Then $D'$ lies on $AB$ and $E'$ lies on $AC$, and $H$ is the circumcentre of the triangle $AD'E'$. Show that $AF$ is the radical axis of the circumcircles of $ABC$ and $AD'E'$, where $F$ is the intersection of $D'E'$ and $BC$.",
"options": [],
"answer": "See solution",
"solution": "It is sufficient to show that $AF$ is the radical axis of the circumcircles of $ABC$ and $AD'E'$. The former circle has centre $O$, the circumcentre of $ABC$, and the latter has centre $H$. The radical axis of two circles is perpendicular to the line joining their centres. It is clear that $A$ lies on the radical axis since it is a common point of the circles.\n\nFor $F$, we first show that $B, C, E', D'$ are concyclic. This follows from\n\n$$\n\\angle BD'C = \\angle BAC = \\angle BE'C.\n$$\n\nNext, we observe that $BE$ and $CD$ are tangent to the circle passing through $B, C, E', D'$, since\n\n$$\n\\angle BCD = \\angle BAC = \\angle BD'C\n$$\n\nand similarly $\\angle CBE = \\angle CE'B$. By applying Pascal's theorem to $BBCCE'D'$, we get that $D, E$ and the intersection point of $BC$ and $D'E'$ are collinear. In other words, $D', E', F$ are collinear. Hence the power of $F$ with respect to the circumcircle of $ABC$ is $FB \\times FC$ and the power of $F$ with respect to the circumcircle of $AD'E'$ is $FD' \\times FE'$; they are equal because $B, C, D', E'$ are concyclic. Hence $F$ lies on the radical axis of these circles. The proof is complete.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20505,
"subject": "Mathematics (Olympiad)",
"question": "Suppose that the digits of $2^a$ can be rearranged to form $2^b$, with $a > b$. Does such a pair $(a, b)$ exist?",
"options": [],
"answer": "See solution",
"solution": "Suppose that the digits of $2^a$ can be rearranged to form $2^b$, with $a > b$. Then, since the two numbers have the same digit set, it follows that they're congruent modulo $9$. Hence $9 \\mid 2^a - 2^b = 2^b(2^{a-b} - 1)$ and so $2^{a-b} \\equiv 1 \\pmod{9}$. However, the smallest positive power of $2$ with this property is $2^6$, so $a-b \\ge 6$. But in that case, $2^a \\ge 2^{b+6} = 64 \\cdot 2^b > 10 \\cdot 2^b$, which means that $2^a$ and $2^b$ can't have the same number of digits, a contradiction. So no such pair exists.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20506,
"subject": "Mathematics (Olympiad)",
"question": "Solve the following system of equations:\n\n$$\n\\begin{cases}\nx - y + z - w = 2, \\\\\nx^2 - y^2 + z^2 - w^2 = 6, \\\\\nx^3 - y^3 + z^3 - w^3 = 20, \\\\\nx^4 - y^4 + z^4 - w^4 = 66.\n\\end{cases}\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $p = x + z$, $q = xz$. The second to fourth equations become:\n\n$$\n\\begin{aligned}\np^2 &= x^2 + z^2 + 2q, \\\\\np^3 &= x^3 + z^3 + 3pq, \\\\\np^4 &= x^4 + z^4 + 4p^2q - 2q^2.\n\\end{aligned}\n$$\n\nSimilarly, let $s = y + w$, $t = yw$. Then:\n\n$$s^2 = y^2 + w^2 + 2t$$\n$$s^3 = y^3 + w^3 + 3st$$\n$$s^4 = y^4 + w^4 + 4s^2t - 2t^2$$\n\nThe first equation gives:\n\n$$p = s + 2 \\qquad \\textcircled{1}$$\n\nTherefore:\n\n$$p^2 = s^2 + 4s + 4$$\n$$p^3 = s^3 + 6s^2 + 12s + 8$$\n$$p^4 = s^4 + 8s^3 + 24s + 32s + 16$$\n\nSubstituting these into the original system, we get:\n\n$$x^2 + z^2 + 2q = y^2 + w^2 + 2t + 4s + 4$$\n$$x^3 + z^3 + 3pq = y^3 + w^3 + 3st + 6s^2 + 12s + 8$$\n$$x^4 + z^4 + 4p^2q - 2q^2 = y^4 + w^4 + 4s^2t - 2t^2 + 8s^3 + 24s + 32s + 16$$\n\nUsing the system, we simplify to:\n\n$$q = t + 2s - 1 \\qquad \\textcircled{2}$$\n$$pq = st + 2s^2 + 4s - 4 \\qquad \\textcircled{3}$$\n$$2p^2q - q^2 = 2s^2t - t^2 + 4s^3 + 12s^2 + 16s - 25 \\qquad \\textcircled{4}$$\n\nSubstituting ① and ② into ③:\n\n$$t = \\frac{s}{2} - 1 \\qquad \\textcircled{5}$$\n\nSubstituting ⑤ into ②:\n\n$$q = \\frac{5}{2}s - 2 \\qquad \\textcircled{6}$$\n\nSubstituting ①, ⑤, ⑥ into ④, we get $s = 2$. Therefore $t = 0$, $p = 4$, $q = 3$.\n\nThus, $x, z$ and $y, w$ are roots of $X^2 - 4X + 3 = 0$ and $Y^2 - 2Y = 0$ respectively:\n\n$$\n\\begin{cases}\nx = 3, \\\\\nz = 1\n\\end{cases}\n\\text{ or }\n\\begin{cases}\nx = 1, \\\\\nz = 3\n\\end{cases}\n$$\n\nand\n\n$$\n\\begin{cases}\ny = 2, \\\\\nw = 0\n\\end{cases}\n\\text{ or }\n\\begin{cases}\ny = 0, \\\\\nw = 2\n\\end{cases}\n$$\n\nSo, the system has 4 solutions:\n\n$$x = 3,\\ y = 2,\\ z = 1,\\ w = 0$$\n$$x = 3,\\ y = 0,\\ z = 1,\\ w = 2$$\n$$x = 1,\\ y = 2,\\ z = 3,\\ w = 0$$\n$$x = 1,\\ y = 0,\\ z = 3,\\ w = 2$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20507,
"subject": "Mathematics (Olympiad)",
"question": "For every $n \\ge 3$, determine all configurations of $n$ distinct points $X_1, X_2, \\dots, X_n$ in the plane such that for any pair of distinct points $X_i, X_j$, there exists a permutation $\\sigma$ of $\\{1, \\dots, n\\}$ with $d(X_i, X_k) = d(X_j, X_{\\sigma(k)})$ for all $1 \\le k \\le n$. Here, $d(X, Y)$ denotes the distance between points $X$ and $Y$.",
"options": [],
"answer": "See solution",
"solution": "Let us first prove that the points must be concyclic. Assign to each point $X_k$ the vector $x_k$ in a system of orthogonal coordinates whose origin is the centroid of the configuration, so $\\frac{1}{n}\\sum_{k=1}^{n} x_k = 0$.\n\nThen $d^2(X_i, X_k) = \\|x_i - x_k\\|^2 = \\langle x_i - x_k, x_i - x_k \\rangle = \\|x_i\\|^2 - 2 \\langle x_i, x_k \\rangle + \\|x_k\\|^2$. Thus,\n$$\n\\sum_{k=1}^{n} d^2(X_i, X_k) = n\\|x_i\\|^2 - 2\\left\\langle x_i, \\sum_{k=1}^{n} x_k \\right\\rangle + \\sum_{k=1}^{n} \\|x_k\\|^2 = n\\|x_i\\|^2 + \\sum_{k=1}^{n} \\|x_k\\|^2\n$$\nSimilarly,\n$$\n\\sum_{k=1}^{n} d^2(X_j, X_{\\sigma(k)}) = n\\|x_j\\|^2 + \\sum_{k=1}^{n} \\|x_{\\sigma(k)}\\|^2 = n\\|x_j\\|^2 + \\sum_{k=1}^{n} \\|x_k\\|^2\n$$\nTherefore, $\\|x_i\\| = \\|x_j\\|$ for all pairs $(i, j)$, so the points are concyclic (lie on a circle centered at $O(0, 0)$).\n\nLet $m$ be the least angular distance between any two points. Two points at angular distance $m$ must be adjacent on the circle. Connect each such pair with an edge. The resulting graph $G$ is regular, with degree $\\deg(G) = 1$ or $2$. If $n$ is odd, since $\\sum_{k=1}^{n} \\deg(X_k) = n \\deg(G) = 2|E|$, we must have $\\deg(G) = 2$, so the configuration is a regular $n$-gon.\n\nIf $n$ is even, we may have a regular $n$-gon, or $\\deg(G) = 1$. In that case, let $M$ be the next least angular distance; such points are also adjacent. Connect each such pair to get graph $G'$. Similarly, $\\deg(G') = 1$, so the configuration is an equiangular $n$-gon (with alternating equal side-lengths).",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20508,
"subject": "Mathematics (Olympiad)",
"question": "Is it possible to place 2014 points on the plane so that, using these points as vertices, one can construct $1006^2$ parallelograms of area $1$?",
"options": [],
"answer": "See solution",
"solution": "We shall prove that it is possible.\n\nConsider two parallel lines $\\varepsilon_1$ and $\\varepsilon_2$ at a distance $1$ apart. Place $1007$ points on each line so that every two successive points are $1$ unit apart. On each line, there are $1006$ segments of length $1$.\n\nAny pair of unit segments, one from $\\varepsilon_1$ and one from $\\varepsilon_2$, forms a parallelogram of area $1$. Therefore, there are $1006^2$ parallelograms of area $1$ in total.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20509,
"subject": "Mathematics (Olympiad)",
"question": "Let $(x_n)$ be a sequence defined by\n$$\nx_0 = 2, \\quad x_1 = 1, \\quad x_{n+2} = x_{n+1} + x_n, \\quad \\forall n \\ge 0.\n$$\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 $\\lambda^2 - \\lambda - 1 = 0$.\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 $x_n$ is prime for $n \\ge 1$, then $n$ is a prime or $n$ has 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 $n$ 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 all $k \\in \\mathbb{N}$.\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\nSince $x_n$ is divisible by $x_m$, $x_n \\ge x_m \\ge 3$, so $n \\ge m > 1$. Let $n = qm + r$ with $q \\in \\mathbb{N}^*$, $r \\in \\mathbb{N}$, $0 \\le r \\le m-1$. Consider:\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 $x_m \\mid x_r$ only if $r = 0$. Thus, $n$ is a multiple of $m$ with even quotient.\n\n- If $q$ is odd, similar reasoning applies, but the divisibility depends on the recurrence and may require further analysis.\n\nTherefore, all pairs $(m, n)$ with $m = 0$ and $n$ divisible by 3, $m = 1$ and any $n$, and for $m > 1$, $n$ a multiple of $m$ with even quotient, satisfy $x_m \\mid x_n$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20510,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\omega_1$ and $\\omega_3$ be two circles, touching externally at a common point $P$. Let $\\omega_2$ and $\\omega_4$ be two circles also touching externally at $P$. Suppose that for $i \\in \\{1, 2, 3, 4\\}$, $\\omega_i$ intersects $\\omega_{(i \\bmod 4)+1}$ again at $A_i$. Let $\\ell_1$ be the common tangent of $\\omega_1$ and $\\omega_3$, and $\\ell_2$ be the common tangent of $\\omega_2$ and $\\omega_4$. Show that $A_1, A_2, A_3,$ and $A_4$ are concyclic if and only if $\\ell_1$ and $\\ell_2$ are orthogonal.",
"options": [],
"answer": "See solution",
"solution": "Let the points $B_1$ and $B_3$ be the intersections of the line $\\ell_1$ with $A_1A_2$ and $A_3A_4$, respectively, as depicted in the figure. Let $B_2$ and $B_4$ be the intersections of $\\ell_2$ with $A_2A_3$ and $A_4A_1$, respectively. Since $\\ell_1 = B_1B_3$ is tangent to $\\omega_1$, we have $\\angle A_4A_1P = \\angle A_4PB_3$. Similarly,\n\n$$\n\\begin{align*}\n\\angle PA_1A_2 &= \\angle B_2PA_2 \\\\\n\\angle A_2A_3P &= \\angle A_2PB_1, \\quad \\text{and} \\\\\n\\angle PA_3A_4 &= \\angle B_4PA_4.\n\\end{align*}\n$$\n\nLet $\\theta = \\angle B_2PB_1$ be the angle between the lines $\\ell_1$ and $\\ell_2$.\n\nWe have",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 20511,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $x$, $y$, $z$, and $t$ such that\n$$\n2^x \\cdot 3^y + 5^z = 7^t.\n$$",
"options": [],
"answer": "See solution",
"solution": "Reducing modulo $3$, we get $5^z \\equiv 1$, so $z$ is even: $z = 2c$, $c \\in \\mathbb{N}$.\n\nNext, we prove that $t$ is even. Suppose $t$ is odd, $t = 2d+1$, $d \\in \\mathbb{N}$. The equation becomes $2^x \\cdot 3^y + 25^c = 7 \\cdot 49^d$. If $x \\ge 2$, reducing modulo $4$ gives $1 \\equiv 3$, a contradiction. If $x=1$, then $2 \\cdot 3^y + 25^c = 7 \\cdot 49^d$. Reducing modulo $24$:\n$$\n2 \\cdot 3^y + 1 \\equiv 7 \\pmod{24} \\implies 24 \\mid 2(3^y - 3), \\text{ i.e. } 4 \\mid 3^{y-1} - 1\n$$\nwhich means $y-1$ is even, so $y = 2b+1$, $b \\in \\mathbb{N}$. We get $6 \\cdot 9^b + 25^c = 7 \\cdot 49^d$, and reducing modulo $5$ gives $(-1)^b = 2(-1)^d$, which is false for all $b, d \\in \\mathbb{N}$. Hence $t$ is even: $t=2d$, $d \\in \\mathbb{N}$.\n\nNow the equation is\n$$\n2^x \\cdot 3^y + 25^d = 49^d \\iff 2^x \\cdot 3^y = (7^d - 5^c)(7^d + 5^c).\n$$\nSince $\\gcd(7^d - 5^c, 7^d + 5^c) = 2$ and $7^d + 5^c > 2$, there are three cases:\n\n1. $\\begin{cases} 7^d - 5^d = 2^{x-1} \\\\ 7^d + 5^d = 2 \\cdot 3^y \\end{cases}$\n2. $\\begin{cases} 7^d - 5^d = 2 \\cdot 3^y \\\\ 7^d + 5^d = 2^{x-1} \\end{cases}$\n3. $\\begin{cases} 7^d - 5^d = 2 \\\\ 7^d + 5^d = 2^{x-1} \\cdot 3^y \\end{cases}$\n\n**Case 1:**\n$7^d = 2^{x-2} + 3^y$. Reducing modulo $3$ gives $2^{x-2} \\equiv 1 \\pmod{3}$, so $x-2$ is even: $x=2a+2$, $a>0$. Also, $7^d - 5^d \\equiv 2 \\cdot 4^a \\pmod{4}$, so $7^d \\equiv 1 \\pmod{4}$, so $d=2e$, $e \\in \\mathbb{N}$. Similarly, $5^c \\equiv 1 \\pmod{8}$, so $c=2f$, $f \\in \\mathbb{N}$. But $49^c - 25^f = 2 \\cdot 4^a \\implies 0 \\equiv 2 \\pmod{3}$, which is false. No solutions in this case.\n\n**Case 2:**\n$2^{x-1} = 7^d + 5^c \\ge 12 \\implies x \\ge 5$. $7^d + 5^c \\equiv 0 \\pmod{4}$, so $3^d + 1 \\equiv 0 \\pmod{4}$, so $d$ is odd. $7^d = 5^c + 2 \\cdot 3^y \\ge 11 \\implies d \\ge 2$, so $d=2e+1$. As before, $x=2a+2$. Then $7^d = 4^a + 3^y$, i.e., $7 \\cdot 49^c = 4^a + 3^y$. Reducing modulo $8$ gives $7 \\equiv 3^y$, which is false. No solutions in this case.\n\n**Case 3:**\n$7^d = 5^c + 2$; the last digit of $7^d$ is $7$, so $d=4k+1$. If $c \\ge 2$, reducing modulo $25$ gives $7 \\equiv 2 \\pmod{2}$, which is false. For $c=1$, $d=1$, and the solution is $x=3$, $y=1$, $z=t=2$.\n\n**Final answer:**\nThe only solution in positive integers is $x=3$, $y=1$, $z=2$, $t=2$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20512,
"subject": "Mathematics (Olympiad)",
"question": "Let $I$ be the incenter of triangle $ABC$. The circle passing through vertex $B$ and touching the line $AI$ at $I$ intersects sides $AB$ and $BC$ at points $P$ and $Q$, respectively. Let $R$ be the intersection point of the line $QI$ and side $AC$. Prove that\n\n$$\n|AR| \\cdot |BQ| = |PI|^2.\n$$\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $\\alpha, \\beta, \\gamma$ denote the measures of the interior angles at vertices $A$, $B$, $C$ of triangle $ABC$, and let $J$ be the intersection point of line $AI$ and side $BC$.\n\nThe inscribed angle $PBI$ corresponds to chord $PI$, while $QBI$ corresponds to chord $IQ$, and since both angles measure $\\frac{1}{2}\\beta$, the chords $PI$ and $IQ$ have equal length.\n\nFig. 2\n\nSince the inscribed angle $JIQ$ also corresponds to chord $IQ$, its measure is $\\frac{1}{2}\\beta$. By congruence of vertical angles, $|\\angle RIA| = \\frac{1}{2}\\beta$. This measure is also shared by inscribed angle $PIA$ as it corresponds to chord $PI$ of equal length as $IQ$. Further, $|\\angle RAI| = |\\angle PAI| = \\frac{1}{2}\\alpha$. The triangles $RIA$ and $PIA$ are thus congruent by ASA, hence $|RI| = |PI|$.\n\nTherefore, the measure of angle $QIB$ is\n\n$$\n\\begin{aligned}\n|\\angle QIB| &= 180^\\circ - |\\angle AIB| - |\\angle JIQ| \\\\\n&= 180^\\circ - (90^\\circ + \\frac{\\gamma}{2}) - \\frac{\\beta}{2} \\\\\n&= 90^\\circ - (\\frac{\\beta}{2} + \\frac{\\gamma}{2}) = \\frac{\\alpha}{2} = |\\angle RAI|.\n\\end{aligned}\n$$\n\nAlternatively, since inscribed angles $AIP$ and $IPQ$ correspond to chords of equal length, $|\\angle AIP| = \\frac{1}{2}\\beta = |\\angle IPQ|$. The congruence of alternate angles implies $AI \\parallel PQ$. Hence $|\\angle QPB| = |\\angle IAB| = \\frac{1}{2}\\alpha$, and so $|\\angle QIB| = \\frac{1}{2}\\alpha$ as well, since both are inscribed angles corresponding to chord $QB$.\n\nSince $|\\angle QIB| = |\\angle RAI|$ and $|\\angle QBI| = |\\angle RIA|$, it follows that $AIR \\sim IBQ$. Therefore, $|AR|/|RI| = |IQ|/|QB|$, so\n\n$$\n|AR| \\cdot |QB| = |RI| \\cdot |IQ| = |PI|^2.\n$$\n\nTo prove similarity of triangles $AIR$ and $IBQ$, we could also use the fact that triangle $CRQ$ is isosceles, so its median coincides with its angle bisector.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20513,
"subject": "Mathematics (Olympiad)",
"question": "When you line up 4020 stones, 2010 of them white and 2010 black, along a horizontal straight line, how many ways are there to arrange them so that the number of pairs of stones with a white stone lying to the right of a black stone is odd?\n\n",
"options": [],
"answer": "See solution",
"solution": "$$\n\\frac{\\binom{4020}{2010} - \\binom{2010}{1005}}{2}\n$$\n\nConsider all possible arrangements of 4020 stones (2010 white, 2010 black). There are $\\binom{4020}{2010}$ such arrangements. Grouping stones into pairs at positions $(2i-1)$ and $2i$, if both stones in each pair are the same color, there are $\\binom{2010}{1005}$ such arrangements. These always yield an even number of white-right-of-black pairs, so do not satisfy the problem's condition. For the remaining arrangements, exactly half have an odd number of such pairs. Thus, the answer is $\\frac{\\binom{4020}{2010} - \\binom{2010}{1005}}{2}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20514,
"subject": "Mathematics (Olympiad)",
"question": "Suppose there are 6 problems and $n$ contestants. Each contestant solved either 4 or 5 problems, and for every pair of problems, more than $\\frac{2}{5}$ of the contestants solved both problems. Prove that at least two contestants solved 5 problems.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let\n$$\nd_i = \\sum_{j \\neq i} p_{ij}, \\quad i = 1, 2, \\dots, 6.\n$$\nWe have $d_s = d_t = 5m + 1$ and $d_i = 5m$ otherwise, where $m = \\frac{2n+1}{5}$. Building up the 6-tuple $(d_1, d_2, \\dots, d_6)$ one contestant at a time, starting with the winner, we get $(4, 4, 4, 4, 4, 0)$, and each subsequent contestant adds a permutation of $(3, 3, 3, 3, 0, 0)$. Thus,\n$$\n(d_1, d_2, \\dots, d_6) \\equiv (1, 1, 1, 1, 1, 0) \\pmod{3},\n$$\ncontradicting the earlier conclusion. Hence, at least two contestants solved five problems.\n\nAlternatively, suppose a minimal counterexample exists. If every 4-tuple of problems was solved by a non-winner, we could remove $\\binom{6}{4} = 15$ contestants, one for each 4-tuple, and still have more than $\\frac{2}{5}$ of contestants solving each pair, contradicting minimality. Considering a multigraph where vertices are problems and edges represent students solving both, we analyze the edge counts for sets $S$ and $T$ (as defined above). Calculations show that $n \\le 2$ or $n \\le 7$, both contradictions. Thus, at least two contestants must have solved five problems.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20515,
"subject": "Mathematics (Olympiad)",
"question": "Does there exist a function $f : \\mathbb{N} \\to \\mathbb{Q}$ such that for any rational number $r$, there exists exactly one ordered pair $(m, n)$ of positive integers satisfying the equation\n$$\nr = f(m) + \\frac{1}{n}?\n$$",
"options": [],
"answer": "See solution",
"solution": "Because every infinite subset of $\\mathbb{Q}$ is countable, the problem is equivalent to asking whether there exists a subset $A \\subseteq \\mathbb{Q}$ with the property that\n$$\n\\mathbb{Q} = \\bigsqcup_{a \\in A} \\left\\{ a + \\frac{1}{n} : n \\in \\mathbb{N} \\right\\}.\n$$\nHere, $\\bigsqcup$ denotes a disjoint union. We will prove a stronger statement: there exists a subset $A' \\subseteq \\mathbb{Q}$ such that\n$$\n\\mathbb{Q} = \\bigsqcup_{a \\in A'} \\left\\{ a + m + \\frac{1}{n} : m \\in \\mathbb{Z}, n \\in \\mathbb{N} \\right\\}.\n$$\nThe latter statement implies the former, since we may take $A := \\{a + m : a \\in A', m \\in \\mathbb{Z}\\}$.\n\nFor each rational number $r$, let $S(r)$ denote the set $\\{r + m + \\frac{1}{n} : m \\in \\mathbb{Z}, n \\in \\mathbb{N}\\}$. We claim as follows.\n\n*Claim.* Let $r_1, r_2, \\dots, r_n$ be rational numbers for which $S(r_i) \\cap S(r_j) = \\emptyset$ for all $i \\neq j$. Let $r' \\in \\mathbb{Q} \\setminus \\bigcup_{i=1}^n S(r_i)$. Then, there exists a rational number $r$ such that $r' \\in S(r)$ and $S(r) \\cap S(r_i) = \\emptyset$ for all $i$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20516,
"subject": "Mathematics (Olympiad)",
"question": "Let $(a_0, a_1, a_2, \\ldots)$ and $(b_0, b_1, b_2, \\ldots)$ be two infinite sequences of integers such that\n\n$$\n(a_n - a_{n-1})(a_n - a_{n-2}) + (b_n - b_{n-1})(b_n - b_{n-2}) = 0,\n$$\n\nfor all integers $n \\ge 2$. Prove that there exists a positive integer $K$ such that\n\n$$\na_{K+2011} = a_{K+(2011)^{2011}}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Consider points $P_j = (a_j, b_j)$ in the plane. The slopes of the lines $P_nP_{n-1}$ and $P_nP_{n-2}$ are\n\n$$\nr = \\frac{b_n - b_{n-1}}{a_n - a_{n-1}}, \\quad s = \\frac{b_n - b_{n-2}}{a_n - a_{n-2}},\n$$\n\nrespectively. The given condition implies that $rs = -1$. Hence, the lines $P_nP_{n-1}$ and $P_nP_{n-2}$ are perpendicular to each other. This implies that $P_n$ lies on the circle $C_n$ with diameter $P_{n-1}P_{n-2}$. Let $f(n) = |P_{n-1} - P_{n-2}|^2$. Then $f(n)$ is an integer. The observation that $P_n$ lies on the circle $C_n$ shows that $f(n) \\leq f(n-1)$. Thus, we get a non-increasing sequence of positive integers. This must be constant after a certain stage. Thus, $f(n)$ is constant for $n \\geq N$, for some positive integer $N$. This implies that the diameter of $C_n$ is constant for all $n \\geq N$. Hence, $C_n$ are all equal circles for $n \\geq N$. This implies that $P_n = P_{n+2}$ for all $n \\geq N$. But then $a_n = a_{n+2}$ for all $n \\geq N$. Hence\n\n$$\na_{N+2011} = a_{N+(2011)^{2011}}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20517,
"subject": "Mathematics (Olympiad)",
"question": "For how many positive three-digit numbers is the hundreds digit smaller than the units digit?",
"options": [],
"answer": "See solution",
"solution": "Systematic counting reveals a pattern:\n\n- In the 900s, there are no such numbers.\n- In the 800s, there are 10: $809, 819, 829, \\ldots, 899$.\n- In the 700s, there are 20: $708, 718, 728, \\ldots, 798$ and $709, 719, 729, \\ldots, 799$.\n- In the 600s, there are 30: $607, 617, 627, \\ldots, 697$; $608, 618, 628, \\ldots, 698$; $609, 619, 629, \\ldots, 699$.\n\nThis pattern continues until the 100s, where there are 80 such numbers.\n\nThe total is:\n$$\n0 + 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 = 360\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20518,
"subject": "Mathematics (Olympiad)",
"question": "Find all integer pairs $(a, n)$ with $a \\geq 2$ and $n \\geq 3$ such that\n$$\na^{\\phi(n)} \\leq 4n.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $f(n) = \\frac{2^{\\phi(n)}}{n}$, which is multiplicative. For any prime $p$ and $k \\geq 2$,\n$$\nf(p^k) = \\frac{2^{p^{k-1}(p-1)}}{p^k} = \\frac{2^{p^{k-2}(p-1)^2}}{p} f(p^{k-1}) \\geq \\frac{2^{(p-1)^2}}{p} f(p^{k-1}) \\geq f(p^{k-1}),\n$$\nsince $2^{(p-1)^2} \\geq p$ for $p \\geq 2$. For any prime $p \\geq 7$, $f(p) > 4$. Also, $f(n) > 4$ for $n = 2^4, 3^2, 5^2$. Thus, if $f(n) \\leq 4$, then $n$ must divide $2^3 \\cdot 3 \\cdot 5$. Noting that $f(2) = 1$, $f(4) = 1$, $f(8) = 2$, $f(3) = \\frac{4}{3}$, and $f(5) = \\frac{16}{5}$, and using multiplicativity, the only $n$ with $f(n) \\leq 4$ are $n = 2, 3, 4, 5, 6, 8, 12, 24$.\n\nTherefore, for $a = 2$ and $n \\geq 3$, the solutions are $n \\in \\{3, 4, 5, 6, 8, 12, 24\\}$. For $a \\geq 3$ and $\\phi(n) \\geq 4$, we have $\\frac{a^{\\phi(n)}}{n} = f(n) \\cdot (a/2)^{\\phi(n)} > 4$, so only $n \\in \\{3, 4\\}$ are possible. Checking these, the only other solutions are $(a, n) = (4, 4), (3, 3), (3, 4)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20519,
"subject": "Mathematics (Olympiad)",
"question": "Find the greatest positive integer $k$ such that the following inequality holds for all positive real numbers $a, b, c$ satisfying $abc = 1$:\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{k}{a+b+c+1} \\geq 3 + \\frac{k}{4}\n$$",
"options": [],
"answer": "See solution",
"solution": "Replace $a = b = x$ and $c = \\frac{1}{x^2}$ where $0 < x \\neq 1$ into the original inequality:\n\n$$\nx^2 + \\frac{2}{x} + \\frac{k}{2x + \\frac{1}{x^2} + 1} \\geq 3 + \\frac{k}{4}\n$$\n\nor\n\n$$\n\\frac{k}{4} \\leq x^2 + 2x + \\frac{2}{x} - \\frac{3}{2x+1}\n$$\n\nNow, for $x = \\frac{2}{3}$, we get $k \\leq \\frac{880}{63} < 14$. Since $k$ is a positive integer, $k \\leq 13$. We prove that $k = 13$ satisfies the requirement.\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{13}{a+b+c+1} \\geq \\frac{25}{4}\n$$\n\nDenote\n\n$$\nf(a, b, c) = \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{13}{a+b+c+1}\n$$\n\nWithout loss of generality, assume $a = \\max\\{a, b, c\\}$. Then:\n\n$$\n\\begin{aligned}\nf(a, b, c) - f(\\sqrt{a}, \\sqrt{b}, \\sqrt{c}) &= \\left(\\frac{1}{\\sqrt{b}} + \\frac{1}{\\sqrt{c}} - \\frac{2}{\\sqrt{bc}}\\right) + 13 \\left(\\frac{1}{a+b+c+1} - \\frac{1}{a+2\\sqrt{bc}+1}\\right) \\\\\n&= (\\sqrt{b} - \\sqrt{c})^2 \\left[ \\frac{1}{bc} - \\frac{13}{(a+b+c+1)(a+2\\sqrt{bc}+1)} \\right]\n\\end{aligned}\n$$\n\nSince $a = \\max\\{a, b, c\\}$ and $abc = 1$, it follows that $bc \\le 1$, so $\\frac{1}{bc} \\ge 1$. Using the AM-GM inequality:\n\n$$\n\\frac{13}{(a+b+c+1)(a+2\\sqrt{bc}+1)} \\le \\frac{13}{(3\\sqrt[3]{abc}+1)^2} = \\frac{13}{16}\n$$\n\nThus, $f(a, b, c) \\ge f(\\sqrt{a}, \\sqrt{b}, \\sqrt{c})$. It suffices to prove that\n\n$$\nf\\left(\\frac{1}{x^2}, x, x\\right) \\ge \\frac{25}{4} \\quad \\text{where} \\quad x = \\sqrt{bc},\\ 0 < x \\le 1.\n$$\n\nIf $x = 1$, the inequality becomes equality. If $0 < x < 1$, this is equivalent to\n\n$$\n\\frac{(x+2)(2x^3 + x^2 + 1)}{x(2x+1)} \\ge \\frac{13}{4}\n$$\n\nor\n\n$$\n8x^4 + 20x^3 - 18x^2 - 9x + 8 \\ge 0.\n$$\n\nThe last inequality is true since the left side can be written as\n\n$$\n\\begin{aligned}\n& (8x^4 - 8x^2 + 2) + (20x^3 - 20x^2 + 5x) + (10x^2 - 14x + 6) \\\\\n&= 2(2x^2 - 1)^2 + 5x(2x - 1)^2 + 2(5x^2 - 7x + 3) > 0.\n\\end{aligned}\n$$\n\nTherefore, $k = 13$ is the desired value. $\\square$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20520,
"subject": "Mathematics (Olympiad)",
"question": "Determine all positive integers $k$ for which there exist a positive integer $m$ and a set $S$ of positive integers such that any integer $n > m$ can be written as a sum of distinct elements of $S$ in exactly $k$ ways.",
"options": [],
"answer": "See solution",
"solution": "We claim that $k = 2^a$ for all $a \\ge 0$.\n\nLet $A = \\{1, 2, 4, 8, \\dots\\}$ and $B = \\mathbb{N} \\setminus A$. For any set $T$, let $s(T)$ denote the sum of the elements of $T$. (If $T$ is empty, we let $s(T) = 0$.)\n\nWe first show that any positive integer $k = 2^a$ satisfies the desired property. Let $B'$ be a subset of $B$ with $a$ elements, and let $S = A \\cup B'$. Recall that any nonnegative integer has a unique binary representation. Hence, for any integer $t > s(B')$ and any subset $B'' \\subseteq B'$, the number $t - s(B'')$ can be written as a sum of distinct elements of $A$ in a unique way. This means that $t$ can be written as a sum of distinct elements of $S$ in exactly $2^a$ ways.\n\nNext, assume that some positive integer $k$ satisfies the desired property for a positive integer $m \\ge 2$ and a set $S$. Clearly, $S$ is infinite.\n\n**Lemma:** For all sufficiently large $x \\in S$, the smallest element of $S$ larger than $x$ is $2x$.\n\n**Proof of Lemma:** Let $x \\in S$ with $x > 3m$, and let $x < y < 2x$. We will show that $y \\notin S$. Suppose first that $y > x + m$. Then $y - x$ can be written as a sum of distinct elements of $S$ not including $x$ in $k$ ways. If $y \\in S$, then $y$ can be written as a sum of distinct elements of $S$ in at least $k+1$ ways, a contradiction. Suppose now that $y \\le x + m$. We consider $z \\in (2x - m, 2x)$. Similarly as before, $z - x$ can be written as a sum of distinct elements of $S$ not including $x$ or $y$ in $k$ ways. If $y \\in S$, then since $m < z - y < x$, $z - y$ can be written as a sum of distinct elements of $S$ not including $x$ or $y$. This means that $z$ can be written as a sum of distinct elements of $S$ in at least $k+1$ ways, a contradiction.\n\nWe now show that $2x \\in S$; assume for contradiction that this is not the case. Observe that $2x$ can be written as a sum of distinct elements of $S$ including $x$ in exactly $k-1$ ways. This means that $2x$ can also be written as a sum of distinct elements of $S$ not including $x$. If this sum includes any number less than $x - m$, then removing this number, we can write some number $y \\in (x+m, 2x)$ as a sum of distinct elements of $S$ not including $x$. Now if $y = y' + x$ where $y' \\in (m, x)$ then $y'$ can be written as a sum of distinct elements of $S$ including $x$ in exactly $k$ ways. Therefore $y$ can be written as a sum of distinct elements of $S$ in at least $k+1$ ways, a contradiction. Hence the sum only includes numbers in the range $[x-m, x)$. Clearly two numbers do not suffice. On the other hand, three such numbers sum to at least $3(x-m) > 2x$, a contradiction.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20521,
"subject": "Mathematics (Olympiad)",
"question": "For each positive integer $n$ let\n\n$$\nf(n) = \\frac{1}{2\\lfloor\\sqrt{1}\\rfloor + 1} + \\frac{1}{2\\lfloor\\sqrt{2}\\rfloor + 1} + \\dots + \\frac{1}{2\\lfloor\\sqrt{n}\\rfloor + 1}.\n$$\n\nHow many positive integers $n$ less than one million make $f(n)$ an integer?",
"options": [],
"answer": "See solution",
"solution": "We claim that $f(k^2 - 1) = k - 1$ for each integer $k \\ge 2$. We prove this by induction.\n\n*Base case*: For $k = 2$,\n\n$$\nf(3) = \\frac{1}{3} + \\frac{1}{3} + \\frac{1}{3} = 1.\n$$\n\n*Inductive step*: Assume $f(k^2 - 1) = k - 1$ for some $k \\ge 2$.\n\nThen,\n\n$$\n\\begin{align*}\nf((k+1)^2 - 1) &= f(k^2 - 1) + \\frac{1}{2k+1} + \\frac{1}{2k+1} + \\dots + \\frac{1}{2k+1} \\\\&= k - 1 + \\underbrace{\\frac{1}{2k+1} + \\dots + \\frac{1}{2k+1}}_{2k+1\\ \\text{terms}} \\\\&= k.\n\\end{align*}\n$$\n\nThis is because $k^2, k^2+1, \\dots, k^2+2k$ all have $\\lfloor\\sqrt{m}\\rfloor = k$ for $m$ in this range, so each term is $\\frac{1}{2k+1}$ and there are $2k+1$ such terms.\n\nThus, $f(n)$ is an integer if and only if $n = k^2 - 1$ for some $k \\ge 2$.\n\nFor $n < 10^6$, $k^2 - 1 < 10^6$ gives $k \\le 1000$. So $k = 2, 3, \\dots, 1000$, which is $999$ values.\n\nTherefore, there are exactly $999$ such $n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20522,
"subject": "Mathematics (Olympiad)",
"question": "Let the lines $KL$ and $NP$, $LJ$ and $PM$, $JK$ and $MN$ meet at points $Q$, $R$, $S$, respectively. By Desargues' theorem on perspective triangles, the lines $JM$, $KN$, and $LP$ are concurrent if and only if the points $Q$, $R$, and $S$ are collinear.\n\n\n\nLet $ABC$ be a triangle and let $\\gamma$ be a conic tangent at points $D$, $E$, $F$ to the lines $BC$, $CA$, $AB$, respectively. Let further $J$, $K$, $L$ be points on the lines $EF$, $FD$, $DE$, respectively, and let $M$, $N$, $P$ be points on the lines $BC$, $CA$, $AB$, respectively. If triangles $JKL$ and $DEF$ are perspective from some point on $\\gamma$, and triangles $MNP$ and $ABC$ are perspective, then triangles $JKL$ and $MNP$ are perspective.",
"options": [],
"answer": "See solution",
"solution": "Without loss of generality, we may assume that $X$ lies on the arc $FD$ of the incircle that does not contain $E$. Consequently, $K$ lies on the side $FD$, while $L$ and $J$ lie on the respective extensions of the sides $DE$ and $EF$.\n\nConsider the cyclic quadrangle $DEFX$: The diagonals meet at $K$, and the extensions of the opposite sides meet at $L$ and $J$, respectively, so the line $LJ$ is the polar of $K$ with respect to the incircle – in what follows, all polar lines are considered with respect to the incircle. Since the line $FD$ is the polar of $B$, and $K$ lies on the line $FD$, it follows that $B$ lies on the line $LJ$. Similarly, $C$ lies on the line $JK$, and $A$ lies on the line $KL$. Consequently, the lines $AQ$ and $CS$ meet at $K$.\n\nProjectively, the lines $FD$ and $LJ$ meet at some point $T$. Notice that $T$ lies on the polar lines of $B$ and $K$ to deduce that the line $BK$ is the polar of $T$, so the cross-ratio $(TFKD)$ is harmonic. Let further the lines $BK$ and $PM$ meet at $U$. Read from $B$, the cross-ratio $(RPUM)$ equals $(TFKD)$, so it is harmonic; that is, $U$ is the harmonic conjugate of $R$ relative to $M$ and $N$. Consequently, the points $Q$, $R$, and $S$ are collinear if and only if the lines *MQ*, *NU*, and *PS* are concurrent.\n\nTo prove the lines *MQ*, *NU*, and *PS* are concurrent, simply check that\n\n$$\n\\frac{QN}{QP} \\cdot \\frac{UP}{UM} \\cdot \\frac{SM}{SN} = 1.\n$$\n\nTo this end, write\n\n$$\n\\begin{aligned}\nQN &= NA \\cdot \\sin(\\angle NAQ), & QP &= PA \\cdot \\sin(\\angle PAQ), \\\\\nUP &= PB \\cdot \\sin(\\angle PBU), & UM &= MB \\cdot \\sin(\\angle MBU), \\\\\nSM &= MC \\cdot \\sin(\\angle MCS), & SN &= NC \\cdot \\sin(\\angle NCS)\n\\end{aligned}\n$$\n\nto get upon rearrangement of factors\n\n$$\n\\frac{QN}{QP} \\cdot \\frac{UP}{UM} \\cdot \\frac{SM}{SN} = \\left( \\frac{NA}{NC} \\cdot \\frac{PB}{PA} \\cdot \\frac{MC}{MB} \\right) \\cdot \\left( \\frac{\\sin(\\angle NAQ)}{\\sin(\\angle PAQ)} \\cdot \\frac{\\sin(\\angle PBU)}{\\sin(\\angle MBU)} \\cdot \\frac{\\sin(\\angle MCS)}{\\sin(\\angle NCS)} \\right).\n$$\n\nFinally, notice that the lines in both triples\n\n$$\n(AM, BN, CP) \\text{ and } (AQ, BU, CS)\n$$\n\nare concurrent (the former by hypothesis, and the latter concur at *K*) to infer that the products in the parentheses above both equal $1$ and thereby conclude the proof.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20523,
"subject": "Mathematics (Olympiad)",
"question": "Let $t_k = a_1^k + a_2^k + \\dots + a_n^k$, where $a_1, a_2, \\dots, a_n$ are positive real numbers and $k \\in \\mathbb{N}$. Prove that\n$$\n\\frac{t_5^2 t_1^6}{15} - \\frac{t_4^4 t_2^2 t_1^2}{6} + \\frac{t_2^3 t_4^5}{10} \\ge 0.\n$$",
"options": [],
"answer": "See solution",
"solution": "The inequality we need to prove is equivalent to\n$$\n2t_5^2 t_1^6 - 5t_4^4 t_2^2 t_1^2 + 3t_2^3 t_4^5 \\ge 0,\n$$\nor\n$$\n2t_5^2 t_1^6 + 3t_2^3 t_4^5 \\ge 5t_4^4 t_2^2 t_1^2.\n$$\nUsing the inequality between arithmetic and geometric means, we have\n$$\n2t_5^2 t_1^6 + 3t_2^3 t_4^5 \\ge 5\\left(t_5^4 t_1^{12} t_2^9 t_4^{15}\\right)^{\\frac{1}{5}}.\n$$\nBy the last inequality, it is enough to prove that\n$$\nt_5^4 t_1^{12} t_2^9 t_4^{15} \\ge t_4^{20} t_2^{10} t_1^{10},\n$$\ni.e.\n$$\nt_5^4 t_1^2 \\ge t_4^5 t_2.\n$$\nNote that $t_5 t_3 \\ge t_4^2$, since $\\sum_{i=1}^n a_i^8$ appears on both sides of the inequality, and moreover,\n$$\na_i^5 a_j^3 + a_i^3 a_j^5 = a_i^3 a_j^3 (a_i^2 + a_j^2) \\ge a_i^3 a_j^3 (2a_i a_j) = 2 a_i^4 a_j^4, \\text{ for all } i < j.\n$$\nFurthermore, $t_5 t_1 \\ge t_3^2$. Indeed, $\\sum_{i=1}^n a_i^6$ appears on both sides, and moreover,\n$$\na_i^5 a_j + a_i a_j^5 = a_i a_j (a_i^4 + a_j^4) \\ge 2 a_i a_j (a_i^2 a_j^2) = 2 a_i^3 a_j^3, \\text{ for all } i < j.\n$$\nIt also holds that $t_5 t_1 \\ge t_4 t_2$, since\n$$\n(a_i^5 a_j + a_i a_j^5) - (a_i^4 a_j^2 + a_i^2 a_j^4) = a_i a_j (a_i - a_j)^2 (a_i^2 + a_i a_j + a_j^2) \\ge 0.\n$$\nFinally,\n$$\nt_5^4 t_3^2 t_1^2 \\ge t_5^2 t_4^4 t_1^2 \\ge t_5 t_4^4 t_3^2 t_1 \\ge t_4^5 t_3^2 t_2,\n$$\nso we obtain\n$$\nt_5^4 t_1^2 \\ge t_4^5 t_2.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20524,
"subject": "Mathematics (Olympiad)",
"question": "Given three cubes with integer edge lengths, if the sum of their surface areas is $564\\ \\text{cm}^2$, then the sum of their volumes is ( ):\n\n(A) $764\\ \\text{cm}^3$ or $586\\ \\text{cm}^3$\n\n(B) $764\\ \\text{cm}^3$\n\n(C) $586\\ \\text{cm}^3$ or $564\\ \\text{cm}^3$\n\n(D) $586\\ \\text{cm}^3$",
"options": [],
"answer": "See solution",
"solution": "Denote the edge lengths of the three cubes as $a$, $b$, and $c$, respectively. Then:\n\n$$\n6(a^2 + b^2 + c^2) = 564\n$$\n\ni.e., $a^2 + b^2 + c^2 = 94$. Assume $1 \\leq a \\leq b \\leq c < 10$.\n\nThen:\n$$\n3c^2 \\geq a^2 + b^2 + c^2 = 94\n$$\n\nIt follows that $c^2 > 31$, so $6 \\leq c < 10$, i.e., $c$ can be $6$, $7$, $8$, or $9$.\n\n- If $c = 9$:\n $$a^2 + b^2 = 94 - 9^2 = 13$$\n $a = 2$, $b = 3$ is a solution. So $(a, b, c) = (2, 3, 9)$.\n\n- If $c = 8$:\n $$a^2 + b^2 = 94 - 8^2 = 30$$\n $b = 4$ or $5$, but $a^2 = 5$ or $14$ has no integer solution.\n\n- If $c = 7$:\n $$a^2 + b^2 = 94 - 7^2 = 45$$\n $a = 3$, $b = 6$ is the only solution. So $(a, b, c) = (3, 6, 7)$.\n\n- If $c = 6$:\n $$a^2 + b^2 = 94 - 6^2 = 58$$\n $b = 6$, $a^2 = 22$ has no integer solution.\n\nIn summary, two solutions: $(2, 3, 9)$ and $(3, 6, 7)$. The possible volumes are:\n\n$$\nV_1 = 2^3 + 3^3 + 9^3 = 764\\ \\text{cm}^3\n$$\n$$\nV_2 = 3^3 + 6^3 + 7^3 = 586\\ \\text{cm}^3\n$$\n\n*Answer: (A)*",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20525,
"subject": "Mathematics (Olympiad)",
"question": "A circle, inscribed into a triangle $ABC$, touches its sides $AB$, $BC$, and $CA$ at the points $N$, $P$, and $K$, respectively. A segment $BK$ intersects the inscribed circle a second time at a point $L$.\n\nLet us define points $T = AL \\cap NK$, $Q = CL \\cap KP$. Prove that straight lines $BK$, $NQ$, and $PT$ intersect in one single point.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let us denote the angles as shown in Fig. 26. For triangles $AKT$ and $ATN$, apply the Law of Sines:\n\n$$\n\\frac{NT}{TK} = \\frac{AN \\cdot \\sin \\alpha_1}{AK \\cdot \\sin \\alpha_2} = \\frac{\\sin \\alpha_1}{\\sin \\alpha_2}\n$$\n\nSimilarly,\n\n$$\n\\frac{NF}{FP} = \\frac{NB \\cdot \\sin \\beta_1}{PB \\cdot \\sin \\beta_2} = \\frac{\\sin \\beta_1}{\\sin \\beta_2}, \\quad \\frac{PQ}{QK} = \\frac{PC \\cdot \\sin \\gamma_1}{KC \\cdot \\sin \\gamma_2} = \\frac{\\sin \\gamma_1}{\\sin \\gamma_2}\n$$\n\nMultiplying all these expressions gives:\n\n$$\n\\frac{NT \\cdot KQ \\cdot PF}{TK \\cdot QP \\cdot FN} = \\frac{\\sin \\alpha_1 \\cdot \\sin \\beta_2 \\cdot \\sin \\gamma_2}{\\sin \\alpha_2 \\cdot \\sin \\beta_1 \\cdot \\sin \\gamma_1} = 1\n$$\n\nSince $AL$, $BL$, and $CL$ intersect in one point, Cheva's theorem in its trigonometric form is fulfilled. Therefore, by Cheva's theorem in its standard form, $BK$, $NQ$, and $PT$ intersect in one single point, as required.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20526,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $101$ people are seated in a circle, each holding a number of cards so that the total number of cards is $101 \\times 51 = 5151$. At each step, a person may pass a card to their neighbor. What is the minimum number of such transitions required so that, regardless of the initial distribution, each person ends up with exactly $51$ cards?",
"options": [],
"answer": "See solution",
"solution": "**Solution.**\n\nLet the initial total potential be\n\n$$\nS = 101 \\times 50 + (100 + 99) \\times 49 + \\cdots + (2 + 1) \\times 0.\n$$\n\nAt the end, the total potential is\n\n$$\nT = 51 \\times 50 + (51 + 51) \\times 49 + \\cdots + (51 + 51) \\times 0.\n$$\n\nThe difference is $S - T = 42,925$.\n\nSince each transition changes the total potential by at most $1$, at least $42,925$ transitions are needed.\n\nTo show that $42,925$ transitions suffice, we use two lemmas.\n\n**Lemma 1.** Let $c, a_0, a_1, \\dots, a_{n-1}$ be integers with sum zero, $c \\ge 0$, $a_0 \\le a_1 \\le \\dots \\le a_{n-1}$. If $n+1$ people $[c], [a_0], \\dots, [a_{n-1}]$ have $N+c, N+a_0, \\dots, N+a_{n-1}$ cards, with $N+a_0 > 0$, and are placed at $0, 1, \\dots, n$ on the number line (with $[c]$ at $n$), then there is a way to reach $N$ cards each in at most $cn + \\sum_{i=0}^{n-1} ia_i$ transitions.\n\n*Proof of Lemma 1.*\n\nSuppose $a_{n-1} \\ge \\cdots \\ge a_s > 0 \\ge a_{s-1} \\ge \\cdots \\ge a_0$, and let $M = a_{n-1} + \\cdots + a_s$. If $M=0$, $[c]$ passes $c$ cards to $[a_i]$ so that $[a_i]$ gets $-a_i$ cards ($0 \\le i \\le s-1$). If $[a_i]$ is at $x_i$, then $x_0, \\dots, x_{n-1}$ is a permutation of $0, \\dots, n-1$. Each card from $[c]$ to $[a_i]$ needs $n-x_i$ transitions. So the total is\n\n$$\n\\sum_{i=0}^{s-1} (n - x_i)(-a_i) = cn + \\sum_{i=0}^{s-1} x_i a_i \\le cn + \\sum_{i=0}^{s-1} ia_i = cn + \\sum_{i=0}^{n-1} ia_i.\n$$\n\nFor $M > 0$, use induction. Suppose\n\n$$\na_{n-1} = \\cdots = a_{n-s} > a_{n-s-1}, \\quad a_l > a_{l-1} = \\cdots = a_0.\n$$\n\nThere is a person in $[a_{n-1}], \\dots, [a_{n-s}]$ and one in $[a_0], \\dots, [a_{l-1}]$ with distance at most $n-s-l+1$. Let $[a_{n-s}]$ pass a card $u$ to $[a_{l-1}]$ in at most $n-s-l+1$ transitions. By induction, the remaining transitions are at most\n\n$$\n\\begin{align*}\nL &= cn + (n-1)a_{n-1} + \\dots + (n-s+1)a_{n-s+1} + \\\\\n & \\quad (n-s)(a_{n-s}-1) + (n-s-1)a_{n-s-1} + \\dots + \\\\\n & \\quad la_l + (l-1)(a_{l-1}+1) + (l-2)a_{l-2} + \\dots + 0 \\cdot a_0\n\\end{align*}\n$$\n\nAdding the transition for $u$, the total is at most\n\n$$\nL + (n - s - l + 1) = cn + \\sum_{i=0}^{n-1} ia_i.\n$$\n\n**Lemma 2.** For any permutation of $[-50], [-49], \\dots, [49], [50]$ on a circle, there exists a person $[c]$ such that the sum of each side of the line through $[c]$ and the origin (including $c$) has the same sign as $c$.\n\n*Proof of Lemma 2.*\n\nLet the permutation be $[a_1], \\dots, [a_{101}]$ clockwise. There is a diameter $l$ such that $[a_1], \\dots, [a_{50}]$ are on one side, $[a_{51}], \\dots, [a_{101}]$ on the other. If $\\sum_{i=1}^{50} a_i = 0$, take $c = a_{51}$. If not, as $l$ rotates $180^\\circ$, the sum changes sign, so some $[c]$ meets the requirement.\n\nTake $[c]$ from Lemma 2. Suppose $c \\ge 0$ (otherwise, reverse all signs and directions). Let $c = c_1 + c_2$, $c_1, c_2 \\ge 0$, so that $c_1$ plus the numbers on one side (excluding $c$) sum to zero. Denote these $50$ numbers as $a_0 \\le \\cdots \\le a_{49}$, and the other $50$ as $b_0 \\le \\cdots \\le b_{49}$. The sum of $c_2$ and $b_0, \\dots, b_{49}$ is also zero. Applying Lemma 1 to $c_1, a_0, \\dots, a_{49}$ and $c_2, b_0, \\dots, b_{49}$, the total transitions are at most\n\n$$\nL = 50c_1 + \\sum_{j=0}^{49} ja_j + 50c_2 + \\sum_{j=0}^{49} jb_j.\n$$\n\nSince $c, a_0, \\dots, a_{49}, b_0, \\dots, b_{49}$ is a permutation of $-50, \\dots, 50$, by the order inequality,\n\n$$\n\\begin{align*}\nL &= 50c + \\sum_{j=0}^{49} j(a_j + b_j) \\\\\n&\\le 50^2 + \\sum_{j=0}^{49} j(2j - 50 + 2j - 49) \\\\\n&= 42,925.\n\\end{align*}\n$$\n\nTherefore, the minimum number of transitions required is $k = 42,925$. $\\Box$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20527,
"subject": "Mathematics (Olympiad)",
"question": "Let $D$ be a point inside triangle $ABC$. The circle $S_1$ inscribed in triangle $ABD$ touches the circle $S_2$ inscribed in triangle $CBD$. Prove that the intersection point of the outer common tangent lines of circles $S_1$ and $S_2$ lies on the line $AC$.",
"options": [],
"answer": "See solution",
"solution": "Let the rays $CD$ and $AD$ intersect sides $AB$ and $BC$ at points $X$ and $Y$, respectively. Denote by $K, L, M, P, Q$ the points where circles $S_1$ and $S_2$ are tangent to segments $BD, AD, CD, AB, BC$ (see the picture).\n\n$$\nAD + BC = AL + LD + BQ + CQ = AP + DM + BP + CM = AB + CD.\n$$\n\nSo, the sums of opposite sides of quadrilateral $ABCD$ are equal. Though the quadrilateral is not convex, this means it is circumscribed; in other words, $DXBY$ is a circumscribed quadrilateral. Let $S_3$ be its incircle. Monge's theorem claims that the outer center of similarity of $S_1$ and $S_2$ lies on the line passing through the outer centers of similarity of $S_1$, $S_3$ and $S_2$, $S_3$, i.e., it lies on the line $AC$. This observation is equivalent to the problem statement.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20528,
"subject": "Mathematics (Olympiad)",
"question": "Let $I_b$ and $I_d$ be the incenters of triangles $ABC$ and $ACD$, respectively, so that the lines $AI_b$ and $AI_d$ bisect the angles $\\angle BAC$ and $\\angle CAD$. Let $O_a$ be the center of the circle passing through $A$, $I_b$, and $I_d$. Show that $O_a$ lies on the angle bisector of $\\angle BAD$, that is, on the line $AI$.\n\n\n\nLet $\\omega_a$ and $\\omega_c$ be the circles centered at $O_a$ and $O_c$ (the center of the circle through $C$, $I_b$, $I_d$), respectively. Let $X$ be the external similitude center and $U$ the internal similitude center of $\\omega_a$ and $\\omega_c$. Prove that the lines $IX$ and $IY$ (defined analogously for $B, D$) are the internal and external angle bisectors of $\\angle BID$, and thus are perpendicular.",
"options": [],
"answer": "See solution",
"solution": "We have:\n\n$$\n\\begin{aligned}\n\\angle O_a AD &= \\angle O_a AI_d + \\angle I_d AD = \\angle I_b AT + \\angle I_d AD \\\\\n&= \\frac{1}{2} \\angle BAC + \\frac{1}{2} \\angle CAD = \\frac{1}{2} \\angle BAD,\n\\end{aligned}\n$$\n\nso $O_a$ lies on the angle bisector of $\\angle BAD$, that is, on the line $AI$.\n\nThe points $O_a$ and $O_c$ lie on the perpendicular bisector of the common chord $I_bI_d$ of $\\omega_a$ and $\\omega_c$, and the two similitude centers $X$ and $U$ lie on the same line, which is parallel to $AC$.\n\nFrom the similarity of the circles $\\omega_a$ and $\\omega_c$, and the relations $O_aI_b = O_aI_d = O_aA = r_a$ and $O_cI_b = O_cI_d = O_cC = r_c$, and from $AC \\parallel O_aO_c$, we see that\n\n$$\n\\begin{aligned}\n\\frac{O_a X}{O_c X} &= \\frac{O_a U}{O_c U} = \\frac{r_a}{r_c} = \\frac{O_a I_b}{O_c I_b} \\\\\n&= \\frac{O_a I_d}{O_c I_d} = \\frac{O_a A}{O_c C} = \\frac{O_a I}{O_c I}.\n\\end{aligned}\n$$\n\nThus, the points $X, U, I_b, I_d$ lie on the Apollonius circle of $O_a, O_c$ with ratio $r_a : r_c$. In this circle, $XU$ is the diameter, and the lines $IU$ and $IX$ are the internal and external bisectors of $\\angle O_a IO_c = \\angle AIC$, by the angle bisector theorem. Moreover, the diameter $UX$ is the perpendicular bisector of $I_b I_d$, so $IX$ and $IU$ are the internal and external bisectors of $\\angle I_b II_d = \\angle BID$, respectively.\n\nRepeating the argument for $B, D$ instead of $A, C$, we get that $IY$ is the internal bisector of $\\angle AIC$ and the external bisector of $\\angle BID$. Therefore, $IX$ and $IY$ are the internal and external bisectors of $\\angle BID$, so they are perpendicular.\n\n*Comment.* In fact, the points $O_a, O_b, O_c, O_d$ lie on the line segments $AI, BI, CI, DI$, respectively. For $O_a$, for example:\n\n$$\n\\begin{aligned}\n\\angle I_d O_a A + \\angle A O_a I_b &= (180^\\circ - 2\\angle O_a A I_d) + (180^\\circ - 2\\angle I_b A O_a) \\\\\n&= 360^\\circ - \\angle BAD \\\\\n&= \\angle ADI + \\angle DIA + \\angle AIB + \\angle IBA \\\\\n&> \\angle I_d I A + \\angle A I I_b.\n\\end{aligned}\n$$\n\nThe solution also shows that the line $IY$ passes through $U$, and analogously, $IX$ passes through the internal similitude center of $\\omega_b$ and $\\omega_d$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20529,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, \\dots, a_n$, $b_1, \\dots, b_n$ be real numbers and $c_1, \\dots, c_n$ be positive real numbers. Prove that\n\n$$\n\\left( \\sum_{i,j=1}^{n} \\frac{a_i a_j}{c_i + c_j} \\right) \\left( \\sum_{i,j=1}^{n} \\frac{b_i b_j}{c_i + c_j} \\right) \\ge \\left( \\sum_{i,j=1}^{n} \\frac{a_i b_j}{c_i + c_j} \\right)^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $d_1, \\dots, d_n$ be real numbers and define\n\n$$\nf(x) = \\sum_{i,j=1}^{n} \\frac{d_i d_j}{c_i + c_j} x^{c_i + c_j}, \\quad x > 0.\n$$\n\nSince $x f'(x) = \\left( \\sum_{i=1}^n d_i x^{c_i} \\right)^2 \\ge 0$, it follows that $f(x) \\ge f(0^+) = 0$. In particular,\n\n$$\nf(1) = \\sum_{i,j=1}^{n} \\frac{d_i d_j}{c_i + c_j} \\ge 0.\n$$\n\nThen for any real number $t$,\n\n$$\n0 \\le \\sum_{i,j=1}^{n} \\frac{(a_i t + b_i)(a_j t + b_j)}{c_i + c_j} = A t^2 + 2C t + B,\n$$\nwhere\n\n$$\nA = \\sum_{i,j=1}^{n} \\frac{a_i a_j}{c_i + c_j}, \\quad B = \\sum_{i,j=1}^{n} \\frac{b_i b_j}{c_i + c_j}, \\quad C = \\sum_{i,j=1}^{n} \\frac{a_i b_j}{c_i + c_j}.\n$$\n\nThis implies that $AB \\ge C^2$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20530,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be an acute-angled triangle. Consider the points $M, N \\in BC$, $Q \\in AB$, and $P \\in AC$ such that $MNPQ$ is a rectangle. Prove that if the centre of the rectangle $MNPQ$ is also the centroid of the triangle $ABC$, then $AB = AC = 3AP$.",
"options": [],
"answer": "See solution",
"solution": "Let $D$ be the midpoint of side $BC$ and let $G$ be the centre of the rectangle $MNPQ$. Notice that points $A$, $G$, and $D$ are collinear and $AG = 2 \\cdot GD$.\n\nLet the parallel line from $E$ to $BC$ meet the line segments $AB$, $QM$, $PN$, and $AC$ at the points $E$, $R$, $S$, and $F$, respectively. Then $GE = GF$ and $GR = GS$, hence $ER = SF$. Notice that the triangles $QRE$ and $PSF$ are congruent, so $\\angle QER = \\angle PFS$, $\\angle ABC = \\angle ACB$, and $AB = AC$.\n\nOn the other hand, since $G$ is the midpoint of the segment $PM$, the segment $GF$ is a midline in triangle $PMC$ and therefore $PF = FC$. Apply Thales' theorem to get $\\dfrac{CF}{FA} = \\dfrac{DG}{GA} = \\dfrac{1}{2}$, implying $AP = PF = FC$, hence $AB = AC = 3AP$, as claimed.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20531,
"subject": "Mathematics (Olympiad)",
"question": "Prove that the value of the expression\n\n$$\n\\frac{\\sqrt{n + \\sqrt{0}} + \\sqrt{n + \\sqrt{1}} + \\sqrt{n + \\sqrt{2}} + \\dots + \\sqrt{n + \\sqrt{n^2 - 1}} + \\sqrt{n + \\sqrt{n^2}}}{\\sqrt{n - \\sqrt{0}} + \\sqrt{n - \\sqrt{1}} + \\sqrt{n - \\sqrt{2}} + \\dots + \\sqrt{n - \\sqrt{n^2 - 1}} + \\sqrt{n - \\sqrt{n^2}}}\n$$\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, for a positive integer $n$, sum over all integers $m$ from $0$ to $n^2$:\n\n$$\n\\sum_{m=0}^{n^2} \\sqrt{n + \\sqrt{m}} = \\sum_{m=0}^{n^2} \\sqrt{\\frac{n + \\sqrt{n^2 - m}}{2}} + \\sum_{m=0}^{n^2} \\sqrt{\\frac{n - \\sqrt{n^2 - m}}{2}}.\n$$\n\nBy changing variables and symmetry, this leads to\n\n$$\n\\sum_{m=0}^{n^2} \\sqrt{n + \\sqrt{m}} = (1 + \\sqrt{2}) \\sum_{m=0}^{n^2} \\sqrt{n - \\sqrt{m}}.\n$$\n\nTherefore,\n\n$$\n\\frac{\\sum_{m=0}^{n^2} \\sqrt{n + \\sqrt{m}}}{\\sum_{m=0}^{n^2} \\sqrt{n - \\sqrt{m}}} = 1 + \\sqrt{2}.\n$$\n\nThus, the value is constant for all positive integers $n$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20532,
"subject": "Mathematics (Olympiad)",
"question": "Let $BE$ be an altitude of an acute triangle $ABC$ and let $P$ be the point on side $AB$ such that $AP = AE$. Let $N$ be the point for which $BCEN$ is a parallelogram. The areas of the triangles $AEP$ and $BNP$ are equal. Lines $NE$ and $AB$ intersect at $Q$.\n\na) Prove that the median of triangle $ABC$ drawn from the vertex $C$ intersects the line segment $PQ$.\n\nb) Prove that in the triangle $ABC$, the bisector of the angle $A$, the altitude drawn from the vertex $B$ and the median drawn from the vertex $C$ meet in one point.\n\n\n\n",
"options": [],
"answer": "See solution",
"solution": "Denote $AC = b$, $AB = c$, $\\angle BAC = \\alpha$, and $AE = AP = u$. As $BCEN$ is a parallelogram, $BN = CE = b-u$ and $BN \\parallel CE$. The latter implies $\\angle NBP = \\alpha$. As $\\angle QAE = \\angle QBN$ and $\\angle AQE = \\angle BQN$, triangles $AQE$ and $BQN$ are similar.\n\na) The triangles $AEP$ and $BNP$ have areas $\\frac{1}{2}u^2 \\sin \\alpha$ and $\\frac{1}{2}(b-u)(c-u) \\sin \\alpha$, respectively. Thus $u^2 = (b-u)(c-u)$, whence $\\frac{c-u}{u} = \\frac{u}{b-u}$. Similarity of triangles $AQE$ and $BQN$ implies $\\frac{AQ}{BQ} = \\frac{AE}{BN}$ which, after defining $BQ = x$, rewrites to $\\frac{c-x}{x} = \\frac{u}{b-u}$. This equality reduces to a linear equation of $x$, meaning that it has only one root. By equality $\\frac{c-u}{u} = \\frac{u}{b-u}$, $x = u$ must be the only root. Thus $BQ = u = AP$, whence the midpoint of the side $AB$ coincides with the midpoint of the line segment $PQ$.\n\nb) Let $F$ be the midpoint of the side $AB$ and let $D$ be the point of intersection of the bisector of angle $A$ with side $BC$. By angle bisector theorem, $\\frac{BD}{CD} = \\frac{AB}{AC}$. Since $NE$ and $BC$ are parallel, triangles $ABC$ and $AQE$ are similar, whence also triangles $ABC$ and $BQN$ are similar. This and part a) of the problem together imply $\\frac{CE}{EA} = \\frac{BN}{BQ} = \\frac{AC}{AB}$. Consequently, $\\frac{AF}{FB} \\cdot \\frac{BD}{DC} \\cdot \\frac{CE}{EA} = 1 \\cdot \\frac{AB}{AC} \\cdot \\frac{AC}{AB} = 1$, giving the desired result by Ceva's theorem.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20533,
"subject": "Mathematics (Olympiad)",
"question": "Find the least number of buttons that can be placed on the squares of a $5 \\times 5$ grid so that no two buttons are on the same square or on squares with a common side (buttons may be on squares with a common vertex), and no buttons can be added to the grid under the same conditions.",
"options": [],
"answer": "See solution",
"solution": "We say that a button *covers* a square if the button lies on either the square itself or one of its neighbors. Consider the $2 \\times 2$ corner areas and the central cross consisting of 5 squares (colored with green and red respectively in the figure below).\n\nEvery button covering a corner has to be located in the corresponding $2 \\times 2$ corner area, thus there is at least one button in each of them. Such a button covers exactly 3 squares in its area, meaning there exists a square in each area covered by other buttons.\n\nIf there is a button in the central square, then it covers no squares in the corner areas. One button can cover the missing squares of only 2 corner areas at once. Thus we would need at least 3 other buttons in addition to the 4 buttons located in the corner areas, meaning a total of at least 7 buttons.\n\nIf there is no button in the central square, then each button can cover at most 2 squares in the central cross, meaning at least 3 buttons are needed to cover it. However, no such button can cover a corner square, meaning 4 other buttons to cover them for a total of at least 7 buttons.\n\nThus we need at least 7 buttons in each case. One possibility for this is shown below.\n\n\n\n",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20534,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a convex quadrilateral. Prove that $ABCD$ is circumscribed (i.e., has an inscribed circle) if and only if the angle bisectors of $A$, $B$, $C$, and $D$ are concurrent.",
"options": [],
"answer": "See solution",
"solution": "First, consider the case where $ABCD$ is not circumscribed. Then the incenter $I$ does not lie on the internal bisector of $A$, so the bisectors do not all meet at a single point. Assume the bisector of $A$ and the line $DI$ meet at $I'$. Suppose $M \\equiv B$ and $N \\equiv I'$, satisfying $M \\in [BI)$, $N \\in [DI)$, and $\\angle MAN = \\angle BAI' = \\frac{1}{2}\\angle A$. However, $\\angle MCN \\neq \\frac{1}{2} \\angle C$. If we assume the contrary, $\\angle MCN = \\frac{1}{2} \\angle C = \\angle BCI$, so $CI'$ is the bisector of angle $C$, which leads to a contradiction. Thus, $ABCD$ must be circumscribed.\n\nNow, consider $ABCD$ as a circumscribed and convex quadrilateral. The bisectors of angles $A$, $B$, $C$, and $D$ meet at $I$. In the figure, angles $\\alpha, \\beta, \\gamma, \\varphi, x, y, z$ are marked. By the Law of Sines:\n\n$$\n\\begin{align*}\n\\frac{BM}{MI} \\cdot \\frac{DN}{NI} &= \\frac{BM}{MA} \\cdot \\frac{MA}{MI} \\cdot \\frac{DN}{NA} \\cdot \\frac{NA}{NI} \\\\\n&= \\frac{\\sin x}{\\sin \\frac{\\beta}{2}} \\cdot \\frac{\\sin\\left(\\frac{\\alpha}{2} + \\frac{\\beta}{2}\\right)}{\\sin\\left(\\frac{\\alpha}{2} - x\\right)} \\cdot \\frac{\\sin\\left(\\frac{\\alpha}{2} - x\\right)}{\\sin \\frac{\\varphi}{2}} \\cdot \\frac{\\sin\\left(\\frac{\\alpha+\\varphi}{2}\\right)}{\\sin x} \\\\\n&= \\frac{\\sin \\frac{\\alpha+\\beta}{2} \\cdot \\sin \\frac{\\alpha+\\beta}{2}}{\\sin \\frac{\\beta}{2} \\cdot \\sin \\frac{\\varphi}{2}}. \\tag{1}\n\\end{align*}\n$$\n\nAnalogously,\n\n$$\n\\frac{BM}{MI} \\cdot \\frac{DN}{NI} = \\frac{BM}{MC} \\cdot \\frac{MC}{MI} \\cdot \\frac{DN}{NC} \\cdot \\frac{NC}{NI} =\n$$\n$$\n= \\frac{\\sin(\\frac{\\gamma}{2} - y)}{\\sin \\frac{\\beta}{2}} \\cdot \\frac{\\sin(\\frac{\\beta+\\gamma}{2})}{\\sin y} \\cdot \\frac{\\sin(\\frac{\\gamma}{2} - z)}{\\sin \\varphi} \\cdot \\frac{\\sin(\\frac{\\alpha+\\beta}{2})}{\\sin z}. \\quad (2)\n$$\n\nIf we observe that $\\frac{\\alpha + \\beta}{2} = \\frac{\\pi}{2} - \\frac{\\gamma + \\varphi}{2}$ and $\\frac{\\alpha + \\varphi}{2} = \\frac{\\pi}{2} - \\frac{\\beta + \\gamma}{2}$, then from (1) and (2) we get:\n\n$$\n\\frac{\\sin(\\frac{\\gamma}{2} - y)}{\\sin y} = \\frac{\\sin z}{\\sin(\\frac{\\gamma}{2} - z)} = \\frac{\\sin(\\frac{\\gamma}{2} - (\\frac{\\gamma}{2} - z))}{\\sin(\\frac{\\gamma}{2} - z)}\n$$\n\nIf and only if $\\operatorname{ctg} y = \\operatorname{ctg}(\\frac{\\gamma}{2} - z)$, and since $f(x) = \\operatorname{ctg} x$ is $\\pi k$ periodic, $y = \\frac{\\gamma}{2} - z$. In other words, $\\angle MCN = y + z = \\frac{\\gamma}{2} = \\frac{1}{2} \\cdot \\angle C$.\n\n\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20535,
"subject": "Mathematics (Olympiad)",
"question": "A square with area 4 is divided into two grey and two transparent squares, each having an area of 1; see the left figure. Another such square is put on top of this square. The side of the second square is lying exactly on the middle of the diagonal of the first square; see the right figure.\n\n\n\nWhat is the area of the grey part in the right figure? Give your answer as a reduced fraction.",
"options": [],
"answer": "See solution",
"solution": "$\\frac{7}{2}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20536,
"subject": "Mathematics (Olympiad)",
"question": "Circumcircle $\\omega$ of triangle $ABC$ with $AB < BC < AC$ has centre $O$. Let $I$ be the incenter of $\\triangle ABC$ and $M$ be the midpoint of $BC$. Let $Q$ be the point symmetric to $I$ with respect to $M$. Suppose that $OM$ intersects $\\omega$ at point $D$ and $QD$ intersects $\\omega$ at $T$. Prove that $\\angle ACT = \\angle DOI$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $DL$ be the diameter of $\\omega$ and let $K$ be symmetric to $D$ with respect to $BC$. Note that $QDIK$ is a parallelogram (see figure). In the right-angled triangle $DBL$, $DB^2 = DM \\cdot DL$. Observe that $DB = DI$. Thus, $DI^2 = DM \\cdot DL = 2DM \\cdot DO = DK \\cdot DO \\Rightarrow \\frac{DI}{DK} = \\frac{DO}{DI}$. Therefore, triangles $DOI$ and $DIK$ are similar and $\\angle IOD = \\angle KID$. Thus, we get:\n$$\\angle DOI = \\angle KID = \\angle KIQ + \\angle QID = \\angle DQI + \\angle QID = 180^\\circ - \\angle IDQ = \\angle ADT = \\angle ACT.$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20537,
"subject": "Mathematics (Olympiad)",
"question": "Find all $a \\in \\mathbb{N}$ such that $n(a+n)$ is not a perfect square for any $n \\in \\mathbb{N}$.",
"options": [],
"answer": "See solution",
"solution": "The answer is $a = 1, 2, 4$. For all $n \\in \\mathbb{N}$, we have:\n\n$$\nn^2 < n(n+1) < n(n+2) < (n+1)^2 \\text{ and } n^2 < n(n+4) < (n+2)^2, \\\\ n(n+4) \\neq (n+1)^2.\n$$\n\nHence, $n(a+n)$ is never a perfect square for $a \\in \\{1, 2, 4\\}$.\n\nOn the contrary, for each $a \\ne 1, 2, 4$, there is an $n \\in \\mathbb{N}$ such that $n(a+n)$ is a perfect square. Suppose first that such an $a$ is a power of $2$. Since $a \\ne 1, 2, 4$, $a$ is divisible by $8$; let $a=8k$. To obtain $n(a+n)$ as a perfect square, take $n=k$, because then $n(a+n) = k(8k+k) = (3k)^2$.\n\nIf $a$ is not a power of $2$, it has an odd prime divisor greater than $1$; let $a = (2k+1)l$ with $k \\ge 1$, $l \\ge 1$. Take $n = k^2l$ to obtain $n(a+n) = k^2l((2k+1)l+k^2l) = k^2l^2(k^2+2k+1) = (kl(k+1))^2$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20538,
"subject": "Mathematics (Olympiad)",
"question": "Determine all prime numbers $p, q < 2023$ such that $q \\mid p^2 + 8$ and $p \\mid q^2 + 8$.",
"options": [],
"answer": "See solution",
"solution": "If one of the numbers is $2$, then $p = q = 2$. We can thus assume $p, q \\ge 3$.\n\nSince $(p, p^2 + 8) = (q, q^2 + 8) = 1$, we have:\n\n$q \\mid p^2 + 8$ and $p \\mid q^2 + 8 \\implies pq \\mid (p^2 + 8)(q^2 + 8) \\implies pq \\mid 8(p^2 + q^2 + 8) \\implies pq \\mid (p^2 + q^2 + 8)$.\n\nFor a fixed $k \\in \\mathbb{N}^*$, we determine the solutions in $\\mathbb{N}^* \\times \\mathbb{N}^*$ of the equation $p^2 + q^2 + 8 = k p q$ with $p, q < 2023$.\n\nAssume $(p_0, q_0)$ is a solution for which $p_0 + q_0$ is minimal and $p_0 \\ge q_0$. If $p_0 = q_0$, since $p_0^2 \\mid 8$, we have $p_0 = q_0 = 1$ or $p_0 = q_0 = 2$ (cases to analyze later). Now, assume $p_0 \\ge 3$ and $p_0 > q_0$.\n\nIf $p_0 \\ge 5$ and $p'$ is the second solution of $x^2 - (q_0 k)x + q_0^2 + 8 = 0$, then $p' = \\frac{q_0^2 + 8}{p_0} \\le \\frac{p_0^2 - 2p_0 + 9}{p_0} < p_0$. Since $p' + q_0 < p_0 + q_0$, it follows $p_0 \\in \\{3, 4\\}$.\n\nIf $p_0 = 3$, then $q_0 \\mid 17$, so $q_0 = 1$ and $k = 6$. Using Vieta jumping, we obtain the sequence $p_0 = 3, q_0 = 1, q_{n+1} = p_n, p_{n+1} = 6p_n - q_n$. Thus, the solutions are $(3, 1)$, $(17, 3)$, $(99, 17)$, $(577, 99)$, with the remaining solutions $p \\ge 2023$. The only solution with both $p, q$ prime and $<2023$ is $(17, 3)$.\n\nIf $p_0 = 4$, then $q_0 \\mid 24$ and $q_0 \\le 3$, so $q_0$ is $2$ or $1$, but both cases are impossible.\n\nNow, consider $p_0 = 2$ and $p_0 = 1$.\n\nIf $p_0 = 2$, then $q_0 \\mid 12$ and $q_0 \\le 2$, so $q_0 = 2$ (the case $q_0 = 1$ is not possible) and $k = 4$. Using Vieta jumping, all solutions have both components even. The acceptable solution is $(2, 2)$.\n\n**Final answer:** The pairs of primes $p, q < 2023$ are $(2, 2)$ and $(17, 3)$ (and by symmetry, $(3, 17)$).",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20539,
"subject": "Mathematics (Olympiad)",
"question": "For a given positive integer $k$, find, in terms of $k$, the minimum value of $N$ for which there is a set of $2k+1$ distinct positive integers that has sum greater than $N$ but every subset of size $k$ has sum at most $N/2$.",
"options": [],
"answer": "See solution",
"solution": "The minimum is $N = 2k^3 + 3k^2 + 3k$.\n\nThe set\n$$\n\\{k^2+1,\\ k^2+2,\\ \\dots,\\ k^2+2k+1\\}\n$$\nhas sum $2k^3 + 3k^2 + 3k + 1 = N + 1$, which exceeds $N$, but the sum of the $k$ largest elements is only $(2k^3 + 3k^2 + 3k)/2 = N/2$. Thus, this $N$ is such a value.\n\nSuppose $N < 2k^3 + 3k^2 + 3k$ and there are positive integers $a_1 < a_2 < \\dots < a_{2k+1}$ with $a_1 + a_2 + \\dots + a_{2k+1} > N$ and $a_{k+2} + \\dots + a_{2k+1} \\le N/2$. Then\n$$\n(a_{k+1} + 1) + (a_{k+1} + 2) + \\dots + (a_{k+1} + k) \\le a_{k+2} + \\dots + a_{2k+1} \\le \\frac{N}{2} < \\frac{2k^3 + 3k^2 + 3k}{2}.\n$$\nThis rearranges to give $2k a_{k+1} \\le N - k^2 - k$ and $a_{k+1} < k^2 + k + 1$. Hence $a_{k+1} \\le k^2 + k$.\n\nCombining these we get\n$$\n2(k+1)a_{k+1} \\le N + k^2 + k.\n$$\nWe also have\n$$\n(a_{k+1} - k) + \\dots + (a_{k+1} - 1) + a_{k+1} \\ge a_1 + \\dots + a_{k+1} > \\frac{N}{2}\n$$\nor $2(k+1)a_{k+1} > N + k^2 + k$. This contradicts the previous inequality, hence no such set exists for $N < 2k^3 + 3k^2 + 3k$ and the stated value is the minimum.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20540,
"subject": "Mathematics (Olympiad)",
"question": "Find all real values of $\\lambda$ such that for all $a, b \\ge 0$, the inequality\n$$\n\\frac{a+b}{2} \\ge \\lambda\\sqrt{ab} + (1-\\lambda)\\sqrt{\\frac{a^2+b^2}{2}}\n$$\nholds.",
"options": [],
"answer": "See solution",
"solution": "The answer is $[\\frac{1}{2}, +\\infty)$. \n\nIf $a = b \\ge 0$, the inequality holds for all $\\lambda \\in \\mathbb{R}$. For $a \\ne b$, we analyze the inequality:\n$$\n\\frac{a+b}{2} \\ge \\lambda\\sqrt{ab} + (1-\\lambda)\\sqrt{\\frac{a^2+b^2}{2}}\n$$\nRewriting and simplifying, we obtain:\n$$\n2\\lambda \\ge 1 - \\frac{(a-b)^2}{(\\sqrt{2a^2+2b^2}+(a+b))(\\sqrt{2a^2+2b^2}+(a+b)+2\\sqrt{ab})}\n$$\nFor $a = 1$, $b = 1+\\epsilon$ ($\\epsilon > 0$), the denominator is at least $16$, so $2\\lambda \\ge 1 - \\frac{\\epsilon^2}{16}$. As $\\epsilon \\to 0$, $2\\lambda \\ge 1$, so $\\lambda \\ge \\frac{1}{2}$.\n\nFor $\\lambda = \\frac{1}{2}$, the inequality holds (see solution of Problem D5). If $\\lambda$ increases, the right-hand side decreases, so the inequality holds for all $\\lambda \\ge \\frac{1}{2}$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20541,
"subject": "Mathematics (Olympiad)",
"question": "Every day at a railway station, there is just one train arriving between 8:00 am and 9:00 am and between 9:00 am and 10:00 am, respectively. The arrival times and their probabilities for the two trains are shown in the following table:\n\n\n\nSuppose that these random events are independent of each other. Now, a traveler comes into the station at 8:20. Then the mathematical expectation of his waiting time is ______ (round to minute).",
"options": [],
"answer": "See solution",
"solution": "The distribution table for the waiting times of the traveler is shown below.\n\n\n\nTherefore, the mathematical expectation of his waiting time is\n\n$$\n10 \\times \\frac{1}{2} + 30 \\times \\frac{1}{3} + 50 \\times \\frac{1}{36} + 70 \\times \\frac{1}{12} + 90 \\times \\frac{1}{18} \\approx 27\\;\\text{min}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20542,
"subject": "Mathematics (Olympiad)",
"question": "Denote the number of all positive divisors of a positive integer $n$ by $\\delta(n)$ and the sum of all positive divisors of a positive integer $n$ by $\\sigma(n)$. Prove that\n$$\n\\sigma(n) > \\frac{\\delta(n)^2}{2}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $a_1, a_2, \\dots, a_{\\delta(n)}$ be the positive divisors of $n$ in increasing order. We have:\n\n$$\n\\sigma(n) = a_1 + a_2 + \\dots + a_{\\delta(n)} \\geq 1 + 2 + \\dots + \\delta(n) = \\frac{\\delta(n) (\\delta(n) + 1)}{2} > \\frac{\\delta(n)^2}{2}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20543,
"subject": "Mathematics (Olympiad)",
"question": "An 8-element set is colored using two colors (say, red and black). A subset is called *monochromatic* if all its elements are the same color, and *bichromatic* otherwise. What is the greatest possible number of bichromatic subsets?",
"options": [],
"answer": "See solution",
"solution": "Let $N$ be the number of bichromatic subsets, and $M$ the number of monochromatic subsets. Since every nonempty subset is either monochromatic or bichromatic, $N = 2^8 - 1 - M$.\n\nIf at least 5 elements are colored the same, then $M \\geq 2^5 - 1$. But if exactly 4 elements are colored one color and the other 4 the other color, then $M = (2^4 - 1) + (2^4 - 1) = 2^5 - 2$. This is less than $2^5 - 1$, so $N$ is maximized when the set is split evenly.\n\nThus, $N = 2^8 - 1 - (2^5 - 2) = 255 - 30 = 225$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20544,
"subject": "Mathematics (Olympiad)",
"question": "There are $n$ ($n \\geq 3$) players in a table tennis tournament, in which any two players have a match. Player $A$ is called *not out-performed* by player $B$ if at least one of player $A$'s losers is not a $B$'s loser.\n\nDetermine, with proof, all possible values of $n$ such that the following case could happen: after finishing all the matches, every player is not out-performed by any other player.",
"options": [],
"answer": "See solution",
"solution": "The answer is $n = 3$ or $n \\geq 5$.\n\n1. **For $n = 3$:**\n Suppose $A$, $B$, and $C$ are three players, and the results of the matches are as follows: $A$ wins $B$, $B$ wins $C$, and $C$ wins $A$. These results obviously satisfy the condition.\n\n2. **For $n = 4$:**\n Suppose the condition holds, i.e., after all matches, every player is not out-performed by any other player. It is obvious that none of these four players wins all three games, otherwise the other three players would be out-performed by this player. Similarly, none of these players loses all three games. It follows that each player wins one or two matches.\n\n For player $A$, assume $A$ wins $B$ and $D$, but loses to $C$. Then both $B$ and $D$ win $C$; otherwise, they would be out-performed by $A$. For the loser in the match $B$ vs. $D$, he only wins $C$, and so the loser cannot be not out-performed by the winner. Consequently, for $n = 4$, the given condition cannot happen.\n\n3. **For $n = 6$:**\n One can construct the tournament results by means of the following directed graph, in which each black dot represents a player, and $\\bullet \\to \\circ$ represents a match with the result that player $\\bullet$ wins player $\\circ$.\n\n\n\n4. **Inductive Step:**\n If there exist tournament results such that each of $n$ players $A_i$ ($1 \\leq i \\leq n$) is not out-performed by any other player, we will prove that the same holds for $n + 2$ players as follows. Suppose $M$ and $N$ are the additional players. Construct the game results of $M$ and $N$ as follows:\n\n $$\n A_i \\to M, \\quad M \\to N, \\quad N \\to A_i\n $$\n for all $i = 1, 2, \\dots, n$, and the game results among $A_i$'s are still the original ones. Now we want to check that these $n+2$ players satisfy the given condition. For any player $G \\in \\{A_1, A_2, \\dots, A_n\\}$, it suffices to consider the players $G$, $M$, $N$, and it reduces to the case $n = 3$.\n\n\n\nOne can check that each of these three players $G$, $M$, $N$ is not out-performed by any one of the other two players. Hence, the tournament result of these $n+2$ players satisfies the required condition.\n\nIn particular, it follows from (1) and the induction step of (4) that the required condition holds for any odd $n$ with $n \\geq 3$. Moreover, it follows from (3) and the induction step of (4) that the required condition holds for any even $n$ with $n \\geq 6$, and this completes the proof. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20545,
"subject": "Mathematics (Olympiad)",
"question": "The six-digit number $20210A$ 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 (4) 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 (2) 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": 20546,
"subject": "Mathematics (Olympiad)",
"question": "Con baldosas cuadradas de lado un número exacto de unidades se ha podido embaldosar una habitación de superficie $18144$ unidades cuadradas de la siguiente manera: el primer día se puso una baldosa, el segundo dos baldosas, el tercero tres, etc. ¿Cuántas baldosas fueron necesarias?",
"options": [],
"answer": "See solution",
"solution": "Supongamos que fueron necesarias $n$ baldosas y que su tamaño es $k \\times k$. Entonces $n k^2 = 18144 = 2^5 \\times 3^4 \\times 7$. Hay nueve casos posibles para $n$, a saber, $2 \\times 7$, $2^3 \\times 7$, $2^5 \\times 7$, $2 \\times 3^2 \\times 7$, $2^3 \\times 3^2 \\times 7$, $2^5 \\times 3^2 \\times 7$, $2 \\times 3^4 \\times 7$, $2^3 \\times 3^4 \\times 7$, $2^5 \\times 3^4 \\times 7$. Además, este número tiene que poderse expresar en la forma $1 + 2 + 3 + \\cdots + N = \\frac{N(N+1)}{2}$ y esto sólo es posible en el caso sexto: $2^5 \\times 3^2 \\times 7 = 63 \\times 64 / 2 = 2016$. Para descartar los otros casos rápidamente observamos que $N$ y $N+1$ son números primos entre sí. Si por ejemplo $\\frac{N(N+1)}{2} = 2^3 \\times 7$, tendría que ser $N+1 = 2^4$ y $N = 7$, que es imposible, etc. Por tanto, se necesitaron $2016$ baldosas.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20547,
"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, and let $v$ be the length of the altitude to side $\\overline{AB}$. Let $\\alpha = \\angle BAC$ and $\\beta = \\angle CBA$.\n\n% \n\nFind the ratio $P(A'B'C') : P(ABC)$, where $A'$, $B'$, and $C'$ are the midpoints of the sides of $ABC$.",
"options": [],
"answer": "See solution",
"solution": "Since $T$ is the centroid of triangle $ABC$, we have:\n\n$$\n|TA'| = |B'C| = \\frac{1}{3}b, \\quad |TB'| = |A'C| = \\frac{1}{3}a, \\quad |TC'| = \\frac{1}{3}v.\n$$\n\nWe compute the area:\n\n$$\n\\begin{align*}\nP(A'B'C')\n&= P(A'B'T) + P(B'C'T) + P(C'A'T) \\\\\n&= \\frac{1}{2} (|TA'| \\cdot |TB'| + |TB'| \\cdot |TC'| \\sin(\\pi - \\alpha) + |TC'| \\cdot |TA'| \\sin(\\pi - \\beta)) \\\\\n&= \\frac{1}{18} (ab + av \\sin \\alpha + bv \\sin \\beta).\n\\end{align*}\n$$\n\nUsing $v = a \\sin \\beta$, $v = b \\sin \\alpha$, $a = c \\sin \\alpha$, $b = c \\sin \\beta$, 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\nHence, $P(A'B'C') : P(ABC) = 2 : 9$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20548,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle. Let $M$ be a variable point interior to the segment $AB$, and let $\\gamma_B$ be the circle through $M$ and tangent at $B$ to $BC$. Let $P$ and $Q$ be the touch points of $\\gamma_B$ and its tangents from $A$, and let $X$ be the midpoint of the segment $PQ$.\n\nSimilarly, let $N$ be a variable point interior to the segment $AC$, and let $\\gamma_C$ be the circle through $N$ and tangent at $C$ to $BC$. Let $R$ and $S$ be the touch points of $\\gamma_C$ and its tangents from $A$, and let $Y$ be the midpoint of the segment $RS$.\n\nProve that the line through the centres of the circles $AMN$ and $AXY$ passes through a fixed point.",
"options": [],
"answer": "See solution",
"solution": "We show that the line through the centres of the circles $AMN$ and $AXY$ passes through the centre of the circle $ABC$. Alternatively, but equivalently, we prove that the three circles share a point different from $A$.\n\nInvert the whole configuration from $A$ and let $Z'$ denote the image of $Z$ under inversion: The circles $ABC$, $AMN$ and $AXY$ are transformed into the lines $B'C'$, $M'N'$ and $X'Y'$, respectively; the line $BC$ is transformed into the circle $\\gamma$ through $A$, $B'$ and $C'$; the circle $\\gamma_B$ is transformed into a circle $\\gamma'_B$ through $B'$ and $M'$, centred at $X'$ and externally tangent to $\\gamma$ at $B'$; and the circle $\\gamma_C$ is transformed into a circle $\\gamma'_C$ through $C'$ and $N'$, centred at $Y'$ and externally tangent to $\\gamma$ at $C'$. In this setting, we are to prove that the lines $B'C'$, $M'N'$ and $X'Y'$ are (projectively) concurrent.\n\nTo this end, let the pair of external common tangents of $\\gamma'_B$ and $\\gamma'_C$ meet at $O$, and let $\\theta$ be the homothety centred at $O$ mapping $\\gamma'_B$ onto $\\gamma'_C$; clearly, $O$ lies on the line $X'Y'$ through the centres of the two circles. Let further $\\theta_B$ and $\\theta_C$ be the homotheties centred at $B'$ and $C'$, respectively, mapping $\\gamma$ onto $\\gamma'_B$ and $\\gamma'_C$, respectively.\n\nThe centres of the three homotheties are collinear, so $B'C'$ passes through $O$.\n\nFinally, $\\theta = \\theta_C\\theta_B^{-1}$, so $\\theta M' = \\theta_C\\theta_B^{-1}M' = \\theta_C A = N'$, showing that the line $M'N'$ passes through $O$ as well. This ends the proof.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20549,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs of prime numbers $p$ and $q$ such that\n$$\nq^3 = p^2 - p + 1.\n$$",
"options": [],
"answer": "See solution",
"solution": "We have\n$$\nq^3 = p^2 - p + 1 \\Leftrightarrow (q-1)(q^2+q+1) = p(p-1). \\quad (1)\n$$\n\nIf $(q-1) \\nmid p$, then $q \\ge p+1$, so $q^3 > p^2 - p + 1$.\n\nHence, $(q^2+q+1) \\nmid p$, i.e.\n$$\nq^2 + q + 1 = k p \\quad (2)\n$$\nfor some $k \\in \\mathbb{N}$. It follows that $k(q-1) = p-1$, or $p = kq - k + 1$. Substituting this expression in (2) we obtain $q^2 + (1-k^2)q + (k^2-k+1) = 0$. The discriminant of this quadratic in $q$ is\n$$\nD = (k^2 - 1)^2 - (4k^2 - k + 1) = k^4 - 6k^2 + 4k - 3\n$$\nwhich must be a perfect square. But it is easy to check that\n$$\n(k^2 - 3)^2 < k^4 - 6k^2 + 4k - 3 < (k^2 - 1)^2 \\quad \\text{for any } k > 3.\n$$\nMoreover, the equation $k^4 - 6k^2 + 4k - 3 = (k^2 - 2)^2$ has no solutions.\n\nHence, $k \\le 3$. For $k = 1$ we have $D = -4$, and for $k = 2$ we have $D = -3$, which is impossible. For $k = 3$ we obtain $D = 6^2$, so $q = (8 \\pm 6)/2$. Thus $q = 7$ and $p = k(q-1) + 1 = 19$. It remains to note the primes $p = 19, q = 7$ do satisfy the equation.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20550,
"subject": "Mathematics (Olympiad)",
"question": "The angles of a triangle are $22.5^\\circ$, $45^\\circ$, and $112.5^\\circ$. Prove that inside this triangle there exists a point that is located on the median through one vertex, the angle bisector through another vertex, and the altitude through the third vertex.",
"options": [],
"answer": "See solution",
"solution": "Consider triangle $ABC$, where $\\angle CAB = 22.5^\\circ$, $\\angle ABC = 45^\\circ$, and $\\angle BCA = 112.5^\\circ$.\n\nLet $D$ be the point where the median from vertex $A$ meets $BC$, $E$ the intersection of the angle bisector from $B$ with $AC$, and $F$ the intersection of the altitude from $C$ with $AB$.\n\n\n\nLet $X$ and $Y$ be the points where the line through $D$ parallel to $CF$ meets $AC$ and $AF$, respectively. We have $\\angle YAX = 22.5^\\circ$, $\\angle XYA = 90^\\circ$, and $\\angle AXY = 67.5^\\circ$. Therefore,\n\n$$\n\\angle XCD = 180^\\circ - \\angle BCA = 67.5^\\circ = \\angle AXY = \\angle CXD,\n$$\n\nso $|XD| = |CD| = |DB|$.\n\nSince $DY$ and $CF$ are parallel, $DY$ is the midsegment of triangle $BCF$. Thus,\n\n$$\n\\frac{|XD|}{|DY|} = \\frac{|DB|}{|DY|} = \\frac{|CB|}{|CF|} = \\frac{|CB|}{|FB|}.\n$$\n\nLet $K$ be the intersection of $BE$ and $CF$. By the angle bisector property,\n\n$$\n\\frac{|CK|}{|KF|} = \\frac{|CB|}{|FB|},\n$$\n\nso $\\frac{|XD|}{|DY|} = \\frac{|CK|}{|KF|}$, which implies $\\angle FAK = \\angle YAD$. Therefore, $AD$ passes through $K$, as required.\n\n",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20551,
"subject": "Mathematics (Olympiad)",
"question": "Let $m_1 < m_2 < \\dots < m_s$ be a sequence of $s \\geq 2$ positive integers, none of which can be written as the sum of two or more distinct other numbers in the sequence. For every integer $r$ with $1 \\leq r < s$, prove that\n$$\nr m_r + m_s \\geq (r+1)(s-1).\n$$",
"options": [],
"answer": "See solution",
"solution": "For $k, \\ell$ with $0 \\leq k \\leq r$ and $k+1 \\leq \\ell \\leq s$, introduce the auxiliary value\n$$\nT(k, \\ell) := m_\\ell + \\sum_{i=1}^{k} m_i.\n$$\nWe claim these $\\frac{1}{2}(r+1)(2s - r)$ values are all distinct: Suppose $T(k, \\ell) = T(u, v)$. Without loss of generality, $k \\leq u$, so\n$$\nm_\\ell = m_v + \\sum_{i=k+1}^{u} m_i.\n$$\nBut then $m_\\ell$ is a sum of distinct other sequence elements unless $\\ell = v$ and $k = u$. Thus, all $T(k, \\ell)$ are distinct. The largest value $T(r, s)$ must be at least $\\frac{1}{2}(r+1)(2s - r)$, so\n$$\n\\frac{1}{2}(r+1)(2s - r) \\leq T(r, s) = m_s + \\sum_{i=1}^{r} m_i \\leq m_s + \\sum_{i=1}^{r} (m_r - r + i).\n$$\nHere, $m_i \\leq m_r - r + i$ since the sequence is increasing. This gives\n$$\nr s + s - r \\leq r m_r + m_s,\n$$\nwhich is the desired inequality.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20552,
"subject": "Mathematics (Olympiad)",
"question": "Each unit square in a $2017 \\times 2019$ grid is coloured black or white such that, in each row and in each column, the number of black squares minus the number of white squares is either $1$ or $-1$.\n\nWhat is the maximum possible difference between the number of black squares and the number of white squares in the entire grid?",
"options": [],
"answer": "See solution",
"solution": "For every row and column, label it with the letter $B$ if it contains more black squares and with the letter $W$ if it contains more white squares. Observe that the difference between the number of black squares and white squares in the entire grid is equal to the difference between the number of $B$-rows and the number of $W$-rows. It is clear that this cannot be greater than $2017$.\n\nWe will now show that it is possible for the difference between the number of black squares and white squares in the entire grid to be $2017$. Start with a $2017 \\times 2017$ grid of unit squares. We colour the top $1009$ squares of the leftmost column black and the bottom $1008$ squares white. For the next column along, we cyclically shift the colouring by one square down. We keep moving to the next column along and cyclically shifting the colouring by one square down until we have coloured the entire grid. In this colouring of the $2017 \\times 2017$ grid, every row and column has one more black square than white square. We then add two more columns to the right of the grid. We colour the top $1009$ squares of one column black and the bottom $1008$ squares white, while we colour the top $1009$ squares of the other column white and the bottom $1008$ squares black. In this particular colouring of the $2017 \\times 2019$ grid, the difference between the number of black squares and white squares in each row and column is $1$. Furthermore, the difference between the number of black squares and white squares in the entire grid is $2017$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20553,
"subject": "Mathematics (Olympiad)",
"question": "Each of the lattice points $(x, y)$, where $x$ and $y$ are integers, in the plane can be coloured black or white. A single strike by an L-shaped punch changes the colour of the four lattice points $(a, b)$, $(a+1, b)$, $(a, b+1)$, and $(a, b+2)$. All lattice points are initially coloured white. Prove that after any number of strikes, the number of black lattice points will be either zero or greater than or equal to four.",
"options": [],
"answer": "See solution",
"solution": "When the lattice points $(a, b)$, $(a+1, b)$, $(a, b+1)$, and $(a, b+2)$ change colour after a single strike by the punch, let us call the point $(a, b)$ the corner point of the strike.\n\nWe first show (by induction on the number of strikes) that, after any finite number of strikes, there will always be an even number of black lattice points. After the first strike, four points (an even number) will be black. Suppose that after $n$ strikes, for some $n \\geq 1$, an even number, say $2m$, of black lattice points $X = \\{(x_1, y_1), (x_2, y_2), \\dots, (x_{2m}, y_{2m})\\}$ remain. For the $(n+1)$-th strike, with corner point $(a, b)$, put $Y = \\{(a, b), (a+1, b), (a, b+1), (a, b+2)\\}$. Let $|X \\cap Y| = k \\in \\{0, 1, 2, 3, 4\\}$, where $|S|$ denotes the size of a set $S$. Then, after strike number $n+1$, the number of black points will be $|(X \\cup Y) \\setminus (X \\cap Y)|$, which is $|X| + |Y| - 2k = 2m + 4 - 2k$, an even number. This completes the induction.\n\nWe now show that it never happens that exactly two black points remain after any finite number of strikes, which will solve the problem.\n\nFirst, we observe that if the same corner point $(a, b)$ appears twice in a series of strikes, we can ignore these two strikes, since the colours of the points in the set $\\{(a, b), (a+1, b), (a, b+1), (a, b+2)\\}$ will remain the same after two strikes with corner point $(a, b)$, irrespective of where they appear in the series. Hence, we may assume that all the corner points in the finite series of strikes are different (each appearing exactly once).\n\nLet $X = \\{(x_1, y_1), (x_2, y_2), \\dots, (x_{2m}, y_{2m})\\}$ be the total number of black lattice points that remain after these strikes.\n\nAmongst the $y$-coordinates $y_1, y_2, \\dots, y_{2m}$, let $y'$ be the smallest. Say $y_{i_1}, y_{i_2}, \\dots, y_{i_k}$ are all the $y$-coordinates equal to $y'$. We assume them to be ordered such that $x_{i_1} < x_{i_2} < \\dots < x_{i_k}$. It follows that, amongst these points, the point $(x_{i_1}, y')$ is the corner point of a strike, while $(x_{i_k}, y')$ is not (with $(x_{i_k} - 1, y')$ the corner point of the strike producing $(x_{i_k}, y')$). So the two points $(x_{i_1}, y')$ and $(x_{i_k}, y')$ cannot be the same, since $(x_{i_1}, y')$ is a corner point of a strike, while $(x_{i_k}, y')$ is not. Hence, there are at least two black points with $y$-coordinates equal to $y'$.\n\nAnalogously, amongst the $x$-coordinates $x_1, x_2, \\dots, x_{2m}$, let $x'$ be the smallest. Say $x_{j_1}, x_{j_2}, \\dots, x_{j_m}$ are all the $x$-coordinates equal to $x'$. We assume them to be ordered such that $y_{j_1} < y_{j_2} < \\cdots < y_{j_m}$. It follows that, amongst these points, the point $(x', y_{j_1})$ is the corner point of a strike, while $(x', y_{j_m})$ is not (with $(x', y_{j_m} - 2)$ the corner point of the strike producing $(x', y_{j_m})$). So the two points $(x', y_{j_1})$ and $(x', y_{j_m})$ cannot be the same, since $(x', y_{j_1})$ is a corner point of a strike, while $(x', y_{j_m})$ is not. Hence, there are at least two black points with $x$-coordinates equal to $x'$.\n\nWe see that there are at least three black points in the subset $\\{(x_{i_1}, y'), (x_{i_k}, y'), (x', y_{j_1}), (x', y_{j_m})\\}$ of $X$. (There is a possibility that $(x_{i_1}, y') = (x', y_{j_1})$.) Therefore, since $X$ contains an even number of points, we conclude that $|X| \\geq 4$.\n\nRecall that we can strike the punch twice at corner point $(0, 0)$, say, to obtain a scenario where there are no black points after a series of strikes.\n\nOur conclusion is that the number of black lattice points after a finite number of strikes is an even number that is either $0$, or greater than or equal to $4$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20554,
"subject": "Mathematics (Olympiad)",
"question": "Тооны дараалал $\\{1, 2, \\ldots, 10\\}$-ийг $B$ ба $C$ гэсэн хоёр салангид ($B \\cap C = \\varnothing$) олонлогт хуваахад, $B$-ийн элементүүдийн нийлбэр нь $C$-ийн элементүүдийн үржвэртэй тэнцүү байхаар хуваах боломжтой бүх хуваалтыг ол.",
"options": [],
"answer": "See solution",
"solution": "Нийлбэр: $1 + 2 + \\cdots + 10 = 55$.\n\n$C$ олонлогийн элементүүдийн тоог $|C|$ гэж тэмдэглэе. $C$-ийн үржвэр $\\leq 120$ ($1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 = 120$), тиймээс $|C| \\leq 4$.\n\n1. $|C| = 2$:\n - $C = \\{6, 7\\}$, $B = \\{1, 2, 3, 4, 5, 8, 9, 10\\}$.\n2. $|C| = 3$:\n - $C = \\{1, 4, 10\\}$, $B = \\{2, 3, 5, 6, 7, 8, 9\\}$.\n3. $|C| = 4$:\n - $C = \\{1, 2, 3, 7\\}$, $B = \\{4, 5, 6, 8, 9, 10\\}$.\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": 20555,
"subject": "Mathematics (Olympiad)",
"question": "As shown in Fig. 6.1, $\\triangle ABC$ is a right-angled triangle with $\\angle C = 90^\\circ$. Draw a circle centered at $B$ with radius $BC$. Let $D$ be a point on the side $AC$, and $DE$ be tangent to the circle at $E$.\n\n\n\nThe line through $C$ perpendicular to $AB$ meets line $BE$ at point $F$. Line $AF$ meets $DE$ at point $G$. The line through $A$ parallel to $BG$ meets $DE$ at $H$. Prove that $GE = GH$.",
"options": [],
"answer": "See solution",
"solution": "Let $K$ be the intersection of $AB$ and $DE$, and $M$ be the intersection of $AB$ and $CF$. Join $FK$, $AE$, and $ME$.\n\nIn $\\triangle ABC$, since $CM \\perp AB$, we have $BM \\cdot BA = BC^2 = BE^2$, so $\\triangle BEM \\sim \\triangle BAE$, and $\\angle BEM = \\angle BAE$.\n\n\n\nAs $\\angle FMK = \\angle FEK = 90^\\circ$, $MFEK$ is cyclic, and $\\angle BEM = \\angle FKM$. It follows that $\\angle BAE = \\angle BEM = \\angle FKM$, so $FK \\parallel AE$, and hence\n\n$$\n\\frac{KA}{KB} = \\frac{EF}{BF}, \\quad \\text{that is, } \\frac{KA}{KB} \\cdot \\frac{BF}{FE} = 1. \\qquad \\textcircled{1}\n$$\n\nAs the line $EGA$ intersects $\\triangle EBK$, we have\n\n$$\n\\frac{EG}{GK} \\cdot \\frac{KA}{AB} \\cdot \\frac{BF}{FE} = 1. \\qquad \\textcircled{2}\n$$\n\nAs $BG \\parallel AH$, we have $\\frac{HK}{KG} = \\frac{AK}{KB}$, so\n\n$$\n\\frac{HG}{GK} = \\frac{AB}{BK}. \\qquad \\textcircled{3}\n$$\n\nFrom (1)–(3), we have $\\frac{EG}{HG} = 1$, and so $EG = HG$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20556,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute, non-isosceles triangle with circumcircle $(O)$. $BE$ and $CF$ are the altitudes of triangle $ABC$, and they intersect at $H$. Let $M$ be the midpoint of $AH$, and $K$ be the point on $EF$ such that $HK \\perp EF$. A line not passing through $A$ and parallel to $BC$ intersects the minor arcs $AB$ and $AC$ of $(O)$ at $P$ and $Q$, respectively. Show that the tangent line to the circumcircle of $CQE$ at $E$, the tangent line to the circumcircle of $BPF$ at $F$, and $MK$ concur.",
"options": [],
"answer": "See solution",
"solution": "We state the following familiar lemma:\n\n**Lemma.** Given triangle $ABC$ and two points $X, Y$. Suppose $BX$ cuts $CY$ at $Z$ and $BY$ cuts $CX$ at $T$. Then if $AX$ and $AY$ are isogonal in $\\angle BAC$, so are $AZ$ and $AT$.\n\nWe first reduce the problem to a simpler form through pairs of similar triangles. Let $D$ be the projection of $H$ onto $BC$. It is easy to see that\n\n$$\n\\triangle MFE \\sim \\triangle OBC\n$$\n\nso we think of constructing triangle $TBC$ similar to triangle $RFE$ where $RE$ and $RF$ are tangent to $(PBF)$ and $(QCE)$.\n\nLet $X, Y$ be the intersections of $AP$ and $AQ$ with $EF$. Firstly,\n\n$$\n\\angle XPF = 90^\\circ - \\angle OAX = \\angle PBF\n$$\n\nso $X$ is on $(BPF)$, or\n\n$$\n\\angle RFE = \\angle BXA \\quad \\text{and} \\quad \\angle REF = \\angle YCA.\n$$\n\n\n\nTherefore, if $\\triangle TBC \\sim \\triangle RFE$ then $BT, BX$ are isogonal in $\\triangle ABC$ and $CY, CT$ are isogonal in $\\triangle ACB$. In short, if $S$ is the intersection of $BX$ and $CY$, then $S$ and $T$ are isogonal conjugates in triangle $ABC$.\n\nLet $L$ be the intersection of $BY$ with $CX$ and $J$ the isogonal conjugate of $L$ in triangle $ABC$. We will prove that $O$ and $D$ lie on $JT$. Indeed,\n\n$$\n\\begin{aligned}\nB(JT, OC) &= B(LS, HA) = (YX, EF) = C(YX, EF) \\\\\n&= C(XY, FE) = C(LS, HA) = C(JT, OB)\n\\end{aligned}\n$$\n\nso $J, O, T$ are collinear.\n\n\n\nNext, since $AX$ and $AY$ are isogonal in $\\angle BAC$, according to the lemma, $AS$ and $AL$ are isogonal in $\\angle BAC$, which leads to $A, L, T$ and $A, J, S$ being collinear. Furthermore, in the complete quadrilateral $BLSC.XY$, if $XY$ intersects $BC$ at $D'$, then\n\n$$\nL(CB, SD') = -1 = L(CB, DD'),\n$$\n\nin other words, $SL$ passes through $D$. Finally, to prove that $JT$, $SL$, and $BC$ are concurrent, we use Desargues' theorem for the pair of triangles $BLT$ and $CSJ$. Since $BT, BX$ and $CJ, CX$ are isogonal pairs in triangle $ABC$, the intersection $Z$ of $BT$ and $CJ$ is the isogonal conjugate of $X$ in triangle $ABC$, thus $Z$ is on $AY$. Therefore, the intersection points $(BT, CJ)$, $(BL, CS)$, and $(TL, JS)$ are collinear, or $JT$ passes through the point $D$.\n\nTherefore, $T$ belongs to $OD$ so $R$ belongs to $MK$. This finishes the proof. $\\Box$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20557,
"subject": "Mathematics (Olympiad)",
"question": "Let $f(m) = \\lfloor \\frac{m}{1} \\rfloor + \\lfloor \\frac{m}{2} \\rfloor + \\dots + \\lfloor \\frac{m}{m} \\rfloor$.\n\nShow that there exists an integer $a$ such that the equation $f(m) = n^2 + a$ has at least one million different solutions in positive integers $(m, n)$.",
"options": [],
"answer": "See solution",
"solution": "**Answer:** Yes.\n\nLet $f(m) = \\lfloor \\frac{m}{1} \\rfloor + \\lfloor \\frac{m}{2} \\rfloor + \\dots + \\lfloor \\frac{m}{m} \\rfloor$.\n\n$$\n\\begin{align*}\nf(m) &= \\lfloor \\frac{m}{1} \\rfloor + \\lfloor \\frac{m}{2} \\rfloor + \\dots + \\lfloor \\frac{m}{m} \\rfloor \\\\\n&\\le \\lfloor \\frac{m}{1} \\rfloor + \\lfloor \\frac{m}{2} \\rfloor + \\lfloor \\frac{m}{2} \\rfloor + \\lfloor \\frac{m}{4} \\rfloor + \\lfloor \\frac{m}{4} \\rfloor + \\lfloor \\frac{m}{4} \\rfloor + \\lfloor \\frac{m}{4} \\rfloor + \\dots \\\\\n&= \\sum_{i=0}^{\\infty} 2^i \\lfloor \\frac{m}{2^i} \\rfloor.\n\\end{align*}\n$$\n\nIf $i \\ge \\lfloor \\log_2(m) \\rfloor + 1$, then $0 < \\frac{m}{2^i} < 1$ and so $\\lfloor \\frac{m}{2^i} \\rfloor = 0$. Thus,\n\n$$\nf(m) \\le \\sum_{i=0}^{\\lfloor \\log_2(m) \\rfloor} 2^i \\lfloor \\frac{m}{2^i} \\rfloor \\le \\sum_{i=0}^{\\lfloor \\log_2(m) \\rfloor} 2^i \\frac{m}{2^i} \\le m (\\log_2(m) + 1).\n$$\n\nMoreover, $f$ is strictly increasing:\n\n$$\n\\begin{align*}\nf(m+1) &= \\lfloor \\frac{m+1}{1} \\rfloor + \\lfloor \\frac{m+1}{2} \\rfloor + \\dots + \\lfloor \\frac{m+1}{m} \\rfloor + \\lfloor \\frac{m+1}{m+1} \\rfloor \\\\\n&\\ge \\lfloor \\frac{m}{1} \\rfloor + \\lfloor \\frac{m}{2} \\rfloor + \\dots + \\lfloor \\frac{m}{m} \\rfloor + 1 \\\\\n&= f(m) + 1.\n\\end{align*}\n$$\n\nConsider pairs $(m, n)$ where $n$ is the largest integer with $n^2 \\le f(m)$. For each such pair,\n\n$$\n\\begin{align*}\nf(m) < n^2 + 2n + 1 = (n + 1)^2 &\\implies 0 \\le f(m) - n^2 < 2n + 1 \\\\\n&\\implies 0 \\le f(m) - n^2 \\le 2n.\n\\end{align*}\n$$\n\nLet $N$ be a large integer. Then $0 \\le f(m) - n^2 \\le 2n \\le 2\\sqrt{f(m)} \\le 2\\sqrt{N}$ when $f(m) \\le N$. So $f(m) - n^2$ can take at most $2\\sqrt{N}$ values for $n^2 \\le f(m) \\le N$.\n\nNow, the number of $m$ with $f(m) \\le N$ is at least $\\lfloor N^{2/3} \\rfloor \\ge N^{2/3} - 1 \\ge \\frac{N^{3/2}}{2}$ for large $N$ (since $f(m) \\le m^{3/2}$ for large $m$).\n\nThus, there are at least $\\frac{N^{3/2}}{2}$ pairs $(m, n)$ with $f(m) \\le N$, and $f(m) - n^2$ can take at most $2\\sqrt{N}$ values. By the pigeonhole principle, some value $a$ is shared by at least $\\left\\lfloor \\frac{N^{2/3}}{2N^{1/2}} \\right\\rfloor = \\left\\lfloor \\frac{N^{1/6}}{4} \\right\\rfloor$ pairs. For $N > 10^{72}$, this is more than $10^6$.\n\nTherefore, there exists an integer $a$ such that the equation $f(m) = n^2 + a$ has at least one million solutions in positive integers $(m, n)$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20558,
"subject": "Mathematics (Olympiad)",
"question": "三角形 $ABC$ 不是正三角形，$I$ 為其內心，$I_A$ 為角 $A$ 內的旁心，$I'_A$ 為 $I_A$ 對直線 $BC$ 的對稱點，且 $\\ell_A$ 為直線 $AI'_A$ 對 $AI$ 的對稱線。依此類推，可定義 $I_B, I'_B$ 與 $\\ell_B$。令 $P$ 為 $\\ell_A$ 與 $\\ell_B$ 的交點。\n\n(a) 證明：若 $O$ 為三角形 $ABC$ 的外心，則 $P$ 會落在直線 $OI$ 上。\n\n(b) 設由 $P$ 對三角形 $ABC$ 的內切圓所引的某一條切線，交 $ABC$ 的外接圓於 $X, Y$ 兩點。證明 $\\angle XIY = 120^\\circ$。",
"options": [],
"answer": "See solution",
"solution": "解：\n\n(a) 令 $A'$ 為 $A$ 對 $BC$ 的對稱點，$M$ 為直線 $AI$ 與 $ABC$ 的外接圓 $\\Gamma$ 的另一個交點。由於 $\\triangle ABA'$ 與 $\\triangle AOC$ 都是等腰三角形，且 $\\angle ABA' = 2\\angle ABC = \\angle AOC$，它們互為相似三角形。同理 $\\triangle ABI_A$ 與 $\\triangle AIC$ 也相似。所以有\n\n$$\n\\frac{AA'}{AI_A} = \\frac{AA'}{AB} \\cdot \\frac{AB}{AI_A} = \\frac{AC}{AO} \\cdot \\frac{AI}{AC} = \\frac{AI}{AO}.\n$$\n\n與 $\\angle A'AI_A = \\angle IAO$ 一起考慮，知 $\\triangle AA'I_A$ 與 $\\triangle AIO$ 相似。\n\n令 $P'$ 為直線 $AP$ 與 $OI$ 的交點。使用有向角（記為 $\\angle^*$），計算得：\n\n$$\n\\begin{aligned}\n\\angle^* MAP' &= \\angle^* I'_A AI_A = \\angle^* I'_A AA' - \\angle^* I_A AA' \\\\\n&= \\angle^* AA' I_A - \\angle^* (AM, OM) \\\\\n&= \\angle^* AIO - \\angle^* AMO = \\angle^* MOP'.\n\\end{aligned}\n$$\n\n所以 $M, O, A, P'$ 四點共圓。\n\n令 $R, r$ 分別代表 $\\triangle ABC$ 的外接圓半徑與內切圓半徑。則有\n\n$$\nIP' = \\frac{IA \\cdot IM}{IO} = \\frac{IO^2 - R^2}{IO},\n$$\n\n此值明顯與 $A$ 無關。所以，$BP$ 與 $OI$ 的交點也是 $P'$，故有 $P = P'$。\n得證點 $P$ 在直線 $OI$ 上。\n\n\n\n(b) 由 Poncelet's Porism， 由 $X, Y$ 分別對 $ABC$ 的內切圓再引的切線，會交於 $\\Gamma$ 上的一點 $Z$。令 $XY$ 與內切圓的切點為 $T$，$XY$ 線段的中點為 $D$。計算得\n\n$$\n\\begin{aligned}\nOD &= IT \\cdot \\frac{OP}{IP} = r\\left(1 + \\frac{OI}{IP}\\right) \\\\\n&= r\\left(1 + \\frac{OI^2}{OI \\cdot IP}\\right) \\\\\n&= r\\left(1 + \\frac{R^2 - 2Rr}{R^2 - IO^2}\\right) \\\\\n&= r\\left(1 + \\frac{R^2 - 2Rr}{2Rr}\\right) = \\frac{R}{2} = \\frac{OX}{2}\n\\end{aligned}\n$$\n\n由此知 $\\angle XZY = 60^\\circ$，故 $\\angle XIY = 120^\\circ$。\n\n證明完畢。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20559,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. Two people, $P$ and $Q$, play a game in which they alternately call an integer $m$ ($1 \\leq m \\leq n$). $P$ calls the first number. No number can be called more than once by either player. The game ends when neither can call a number. If the sum of the numbers that $P$ has called is divisible by $3$, $P$ wins; otherwise, $Q$ wins. \nFind all $n$ such that $P$ can win the game no matter what $Q$ does.",
"options": [],
"answer": "See solution",
"solution": "Let the number called by a player in the $m$th turn be $N_m$. Then the sequence $(N_1, \\dots, N_l)$ is called the \"history up to the $l$th turn\". A number $j$ is \"free in the $(l+1)$th turn\" if $j \\neq N_1, \\dots, N_l$.\n\nWe will prove that $P$ can always win if and only if $n \\equiv 0, 4, 5 \\pmod{6}$.\n\n**Case 1: $0 \\leq n \\leq 5$**\n\n- For $n = 0, 1, 2$, the result is clear.\n- For $n = 3$: If $Q$ calls $1$ or $2$ in the second turn, the sum of the numbers that $P$ has called is $5$ or $4$, so $Q$ can always win.\n- For $n = 4$: If $P$ calls $2$ first, and then $1$ or $4$ in the third turn, the sum of the numbers that $Q$ has called is $3$ or $6$, so $P$ can always win.\n- For $n = 5$: Pair $(1,4)$ and $(2,5)$. If $P$ calls $3$ first, and then always calls the other number of the pair that $Q$ just called, $P$ can always win.\n\n**Case 2: Induction Step**\n\nSuppose the proposition holds for $n = k$. Consider $n = k + 6$. Let $M = \\{k+1, k+2, k+3, k+4, k+5, k+6\\}$, and pair $(k+1, k+4)$, $(k+2, k+5)$, $(k+3, k+6)$.\n\n- If there are free numbers outside $M$, $A$ (the player with the winning strategy for $n = k$) acts as follows:\n - If $A$ is $P$ and $l=1$, call the number as in the winning strategy for $n=k$.\n - If $B$ called $i \\in M$ last turn, $A$ calls the other number in the pair.\n - If $B$ called $i \\notin M$, reduce the history to $n=k$ and play accordingly.\n- If all free numbers are in $M$, $A$ always calls the pair of the number $B$ just called, or any free number if $B$ called outside $M$.\n\nBy induction, when the game ends, the sum of numbers called by $A$ is congruent modulo $3$ to a winning sum for $n=k$. Thus, $A$ wins.\n\n**Conclusion:**\n\n$P$ has a winning strategy if and only if $n \\equiv 0, 4, 5 \\pmod{6}$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20560,
"subject": "Mathematics (Olympiad)",
"question": "Let $T$ be any triangulation graph of a convex $2n$-sided polygon $P$. Let $f(T)$ denote the number of perfect matchings of $T$ (i.e., sets of $n$ pairwise non-intersecting edges covering all vertices).\n\nLet $F_1 = 1$, $F_2 = 2$, and $F_{k+2} = F_{k+1} + F_k$ for $k \\ge 2$ (the Fibonacci sequence).\n\n**Prove that for any triangulation graph $T$ of a convex $2n$-gon, $f(T) \\leq F_n$, and determine the maximum value of $f(T)$ for a convex 20-sided polygon $P$.**",
"options": [],
"answer": "See solution",
"solution": "Suppose there was an odd chord $e_1$ in a perfect matching. Since a perfect matching of $T$ gives a pairwise division of the set of vertices of $P$, and there are an odd number of vertices on each side of $e_1$, in this perfect matching there must be another edge $e_2$ of $T$, with endpoints on each side of $e_1$, respectively. And since $P$ is a convex polygon, $e_1$ and $e_2$ intersect in the interior of $P$. This contradicts the fact that $T$ is a triangulation graph.\n\nLet us prove by induction on $n$ that $f(T) \\leq F_n$ for any triangulation $T$ of a convex $2n$-gon.\n\n**Base cases:**\n- For $n=2$, $T$ is a triangulation of a quadrilateral. The only perfect matchings use the polygon's edges, so $f(T) = 2 = F_2$.\n- For $n=3$, $T$ is a triangulation of a hexagon. If $T$ has no even chord, $f(T) = 2$. If $T$ contains one even chord (say $A_1A_4$), the perfect matching using $A_1A_4$ is unique, and the other two edges must be $A_2A_3, A_5A_6$, so $f(T) = 3 = F_3$.\n\n**Inductive step:**\nSuppose the result holds for all $k < n$, $n \\geq 4$. Consider a triangulation $T$ of a convex $2n$-gon $P = A_1A_2 \\cdots A_{2n}$.\n\nIf $T$ has no even chords, then $f(T) = 2$ as above.\n\nIf $T$ contains even chords, pick an even chord $e$ minimizing the smaller number of vertices on either side, say $w(e) = 2k$. Set $e = A_{2n}A_{2k+1}$. Then, by minimality, no $A_i$ with $i = 1, \\dots, 2k$ can be incident to an even chord, and by the properties of triangulations and perfect matchings, $A_1$ can only be paired with $A_2$ or $A_{2n}$.\n\n- **Case 1:** Choose $A_1A_2$, then $A_3A_4, \\dots, A_{2k-1}A_{2k}$ must also be chosen. The remaining vertices form a convex $2n-2k$-gon, whose triangulation graph $T_1$ is induced by $T$. By induction, $f(T_1) \\leq F_{n-k}$.\n- **Case 2:** Choose $A_1A_{2n}$, then $A_2A_3, \\dots, A_{2k}A_{2k+1}$ must be chosen. The remaining vertices form a convex $2n-2k-2$-gon, with triangulation $T_2$ induced by $T$. By induction, $f(T_2) \\leq F_{n-k-1}$.\n\nThus,\n$$\nf(T) \\leq f(T_1) + f(T_2) \\leq F_{n-k} + F_{n-k-1} = F_{n-k+1} \\leq F_n.\n$$\n\n**Sharpness:**\nConsider the triangulation $\\Delta_n$ of a convex $2n$-gon constructed by adding diagonals as described. By induction, $f(\\Delta_n) = F_n$.\n\n**Conclusion:**\nFor a convex 20-sided polygon ($n=10$), the maximum number of perfect matchings is $F_{10} = 89$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20561,
"subject": "Mathematics (Olympiad)",
"question": "Nick and Mary play a game with a heap of 360 stones. On each turn, a player can remove between 1 and $m$ stones from the heap, where $m$ is a fixed integer. Nick moves first. For which values of $m$ can Nick guarantee a win, and what strategy should he use?",
"options": [],
"answer": "See solution",
"solution": "Nick can win if he sets $n = 2$. Consider all possibilities for $m$:\n\nLet $m = 3$. If Mary removes $k$ stones, then Nick removes $4 - k$. In this case, exactly 4 stones are removed from the heap after each pair of moves (Mary and Nick). Since $360 \\div 4 = 90$, Nick wins.\n\nNow, let's solve the problem by moving backward. Write all numbers from 1 to 360 and mark them with \"+\" or \"-\". If $k$ stones remain in the heap before a player's move and they can win, mark $+k$; otherwise, mark $-k$.\n\nFor $m = 4, 5, 7, 8$, the table is:\n\n| +1 | +2 | -3 | +4 | +5 | -6 | +7 | +8 | -9 | +10 | +11 | -12 | +13 | +14 | -15 | +16 | ... |
\n\nThe signs repeat with period 3 (provable by induction). Since $360 \\div 3 = 120$, 360 is marked with \"-\", so 360 is a losing position for the first player.\n\nFor $m = 6$:\n\n| +1 | +2 | -3 | +4 | +5 | +6 | -7 | +8 | +9 | -10 | +11 | +12 | +13 | -14 | ... |
| 1 | 2 | - | 1 | 2 | 6 | - | 1 | 2 | - | 1 | 2 | 6 | - | ... |
\n\nThe signs repeat with period 7. Since $360 \\equiv 3 \\pmod{7}$, 360 and 3 are marked with \"-\", so 360 is a losing position for the first player. To win, Nick should remove the number of stones indicated in the second row of the table.\n\nFor $m = 9$:\n\n| +1 | +2 | -3 | +4 | +5 | -6 | +7 | +8 | +9 | -10 |
| 1 | 2 | - | 1 | 2 | - | 1 | 2 | 9 | - |
\n\n| +11 | +12 | -13 | +14 | +15 | -16 | +17 | +18 | +19 | -20 | ... |
| 1 | 2 | - | 1 | 2 | - | 1 | 2 | 9 | - | ... |
\n\nThe signs repeat with period 10. Since $360 \\div 10 = 36$, 360 and 10 are marked with \"-\", so 360 is a losing position for the first player. To win, Nick should remove the number of stones indicated in the second row of the table.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20562,
"subject": "Mathematics (Olympiad)",
"question": "Suppose every equilateral triangle in the plane has at least two vertices of different colours. Consider a regular hexagon $ABCDEF$ with centre $S$ (with $A$ at the bottom left, vertices labelled anticlockwise).\n\n\n\nProve that there must exist an equilateral triangle whose vertices are all the same colour.",
"options": [],
"answer": "See solution",
"solution": "Assume, for contradiction, that every equilateral triangle has at least two vertices of different colours. Colour $S$ black. In triangle $BDF$, one vertex must be black; let it be $B$. Considering triangles $SBA$ and $SBC$, $A$ and $C$ must be white. In triangle $ACE$, $E$ must be black, and in $SEF$, $F$ must be white. Let $G$ be the intersection of $BA$ and $EF$. If $G$ is white, triangle $GAF$ is equilateral with all vertices white; if $G$ is black, triangle $GBE$ is equilateral with all vertices black. This contradicts our assumption. Thus, there exists an equilateral triangle with all vertices the same colour.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20563,
"subject": "Mathematics (Olympiad)",
"question": "Let $A'$ be the reflection of vertex $A$ of triangle $ABC$ across line $BC$, and let $B'$ be the reflection of vertex $B$ across line $AC$. Given that $\\angle BA'C = \\angle BB'C$, can the largest angle of triangle $ABC$ be located at:\n\na) Vertex $A$;\nb) Vertex $B$;\nc) Vertex $C$?",
"options": [],
"answer": "See solution",
"solution": "a) No; b) Yes; c) Yes.\n\nLet the foot of the altitude from vertex $B$ of triangle $ABC$ be $K$. By symmetry, $\\angle BA'C = \\angle BAC$ and $\\angle BB'C = \\angle KB'C = \\angle KBC$.\n\n**a)** By assumption, $\\angle BAC = \\angle KBC$. Since the angle at vertex $K$ of triangle $KBC$ is right, $\\angle KBC < 90^\\circ$. If the largest angle of triangle $ABC$ were at vertex $A$, the triangle would have to be acute. This would mean that $K$ would lie on segment $AC$, so $\\angle ABC > \\angle KBC = \\angle BAC$. Thus, the largest angle cannot be at vertex $A$.\n\n**b)** If the angle at vertex $B$ of triangle $ABC$ is right, then it is the largest angle of the triangle. Triangles $BAC$ and $KBC$ are similar, so $\\angle BAC = \\angle KBC$ and $\\angle BA'C = \\angle BB'C$.\n\n**c)** Let $ABP$ be an equilateral triangle with midpoint $C$. Clearly, $P$ is the reflection of $A$ across $BC$ and the reflection of $B$ across $AC$. Thus, defining $A' = B' = P$ fulfills the conditions. The angle at vertex $C$ of triangle $ABC$ is $120^\\circ$, so it is the largest.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20564,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. Show that 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 $4a^2 + 9b^2 - 1 = (2k + 1)(2k - 1)$, so $n$ 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 $4a^2 + 9b^2 - 1 = (3k + 1)(3k - 1)$, so $n$ 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, we get\n\n$$\n2^{2r}k^2 + 3^{2s}m^2l^2 + 2 \\cdot 2^{2r}3^s m k l = 1,\n$$\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 $4a^2 + 9b^2 - 1$, which finishes the proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20565,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be non-negative real numbers, no two of which are equal. Prove that\n$$\n\\frac{a^2}{(b-c)^2} + \\frac{b^2}{(c-a)^2} + \\frac{c^2}{(a-b)^2} > 2.\n$$",
"options": [],
"answer": "See solution",
"solution": "The left-hand side is symmetric with respect to $a$, $b$, and $c$. Hence, we may assume that $a > b > c \\ge 0$. Note that replacing $(a, b, c)$ with $(a-c, b-c, 0)$ lowers the value of the left-hand side, since the numerators of each of the fractions would decrease and the denominators remain the same. Therefore, to obtain the minimum possible value of the left-hand side, we may assume that $c=0$.\n\nThen the left-hand side becomes\n$$\n\\frac{a^2}{b^2} + \\frac{b^2}{a^2},\n$$\nwhich yields, by the Arithmetic Mean–Geometric Mean Inequality,\n$$\n\\frac{a^2}{b^2} + \\frac{b^2}{a^2} \\ge 2\\sqrt{\\frac{a^2}{b^2} \\cdot \\frac{b^2}{a^2}} = 2,\n$$\nwith equality if and only if $a^2/b^2 = b^2/a^2$, or equivalently, $a^4 = b^4$. Since $a, b \\ge 0$, $a = b$. But since no two of $a, b, c$ are equal, $a \\ne b$. Hence, equality cannot hold. This yields\n$$\n\\frac{a^2}{b^2} + \\frac{b^2}{a^2} > 2.\n$$\nUltimately, this implies the desired inequality. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20566,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $x$, $y$ be non-negative numbers, and let $k > 0$. Suppose the following inequalities hold:\n\n$$\na + kx \\le 1; \\quad a + ky \\le 1; \\quad b + \\frac{x}{k} \\le 1; \\quad b + \\frac{y}{k} \\le 1.\n$$\n\nProve that:\n\n$$\n(a + x)^2 + (b + y)^2 \\le 2.\n$$\n\nUnder what conditions does equality hold?",
"options": [],
"answer": "See solution",
"solution": "From $a + kx \\le 1$, we have $x \\le \\frac{1-a}{k}$. From $b + \\frac{x}{k} \\le 1$, we get $x \\le k(1-b)$.\n\nThus, $x^2 \\le (1-a)(1-b)$ and $x \\le \\frac{1}{2}(2-a-b)$ (by the arithmetic-geometric mean inequality for $1-a$ and $1-b$).\n\nSimilarly, $y^2 \\le (1-a)(1-b)$ and $y \\le \\frac{1}{2}(2-a-b)$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n(a+x)^2 + (b+y)^2 &= a^2 + 2ax + x^2 + b^2 + 2by + y^2 \\\\\n&\\le a^2 + 2a \\cdot \\frac{1}{2}(2-a-b) + (1-a)(1-b) + b^2 + 2b \\cdot \\frac{1}{2}(2-a-b) + (1-a)(1-b) \\\\\n&= a^2 + b^2 + (a+b)(2-a-b) + 2(1-a)(1-b) = 2.\n\\end{aligned}\n$$\n\nEquality holds when $a = b$, $x = y = 1 - a$, and $k = 1$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20567,
"subject": "Mathematics (Olympiad)",
"question": "En el pizarrón están escritos los 18 números enteros desde 1 hasta 18. Determina la menor cantidad de números que hay que borrar para que entre los números restantes no haya dos tales que su suma sea un cuadrado perfecto.",
"options": [],
"answer": "See solution",
"solution": "Borramos los siguientes 9 números: 3, 5, 7, 8, 10, 12, 14, 15, 16. Los números restantes son: 1, 2, 4, 6, 9, 11, 13, 17, 18. Se verifica que la suma de cualesquiera dos de ellos no es un cuadrado perfecto. Por lo tanto, es posible lograr el objetivo borrando 9 números.\n\nVeamos que es necesario borrar al menos 9. Consideremos los siguientes pares de enteros: $(1,15)$; $(2,14)$; $(3,13)$; $(4,12)$; $(5,11)$; $(6,10)$; $(7,18)$; $(8,17)$; $(9,16)$. En todos los pares la suma es un cuadrado perfecto y todos los números están exactamente una vez. Por lo tanto, es necesario borrar al menos 9 números, uno en cada pareja.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20568,
"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.",
"options": [],
"answer": "See solution",
"solution": "Suppose that there exists 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 even, and specifically, since all odd primes are congruent to $1$ or $3$ modulo $4$, the sequence of primes modulo $4$ must alternate between $1$ and $3$. However, since $n$ is odd, this is impossible.\n\nIf one of $p_1, p_2, \\dots, p_n$ is $2$, without loss of generality, let $p_1 = 2$. Then both $p_1 + p_2$ and $p_n + p_1$ are perfect squares and both are odd, so $p_2$ and $p_n$ are odd. By similar reasoning, the sequence $p_2, p_3, \\dots, p_n$ modulo $4$ must alternate between $1$ and $3$, so $n-1$ is odd, which is again impossible.\n\nTherefore, there are no odd integers $n \\ge 3$ and $n$ distinct primes satisfying the given conditions.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20569,
"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 $E$, $K$, $H$, and $G$ lie on the circle with diameter $GE$, so the angles $EHK$ and $EGK$ are equal.\n\n\n\nAlso, from $\\angle DCA = \\angle DBA$ and $\\dfrac{CE}{CD} = \\dfrac{BE}{BA}$, it follows that\n\n$$\n\\frac{CE}{CG} = \\frac{2CE}{CD} = \\frac{2BE}{BA} = \\frac{BE}{BF},\n$$\n\nso the triangles $CGE$ and $BFE$ are similar. In particular, the angles $EGC$ and $BFE$ are equal, and therefore so are the angles $EHK$ and $BFE$.\n\nBut the lines $EH$ and $BF$ are perpendicular, and so, since $EF$ and $HK$ are obtained by rotations of these lines by the same (directed) angle, the lines $EF$ and $HK$ are also perpendicular.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20570,
"subject": "Mathematics (Olympiad)",
"question": "a) Calculate $(-3ab + a^2 - 1) - (-4a^2 + 5ab - 7) + 2(a - 2b)(a + b)$\n\nb) If $A = 7x^2$ and $B = -2y$, calculate $(A+B)^2$ and $(A+B)(A-B)$.",
"options": [],
"answer": "See solution",
"solution": "a)\n\n$$\n\\begin{align*}\n&(-3ab + a^2 - 1) - (-4a^2 + 5ab - 7) + 2(a - 2b)(a + b) \\\\\n&= -3ab + a^2 - 1 + 4a^2 - 5ab + 7 + 2(a^2 + ab - 2ab - 2b^2) \\\\\n&= -3ab + a^2 - 1 + 4a^2 - 5ab + 7 + 2a^2 + 2ab - 4ab - 4b^2 \\\\\n&= (a^2 + 4a^2 + 2a^2) + (-3ab - 5ab + 2ab - 4ab) + (-1 + 7) - 4b^2 \\\\\n&= 7a^2 - 10ab + 6 - 4b^2\n\\end{align*}\n$$\n\nb)\n\n$$\n\\begin{align*}\n(A+B)^2 &= A^2 + 2AB + B^2 \\\\\n&= (7x^2)^2 + 2 \\cdot 7x^2 \\cdot (-2y) + (-2y)^2 \\\\\n&= 49x^4 - 28x^2y + 4y^2\n\\end{align*}\n$$\n\n$$\n\\begin{align*}\n(A+B)(A-B) &= (A)^2 - (B)^2 \\\\\n&= (7x^2)^2 - (-2y)^2 \\\\\n&= 49x^4 - 4y^2\n\\end{align*}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20571,
"subject": "Mathematics (Olympiad)",
"question": "Players $A$ and $B$ play a game: given a positive integer $n$, one has to choose a divisor $m$ of $n$ which is greater than $1$ and smaller than $n$, and replace $n$ with $n - m$. Player $A$ makes the first move, and players move alternately. The player who can't make a move loses the game. For which starting positive numbers $n$ does player $B$ have a winning strategy?",
"options": [],
"answer": "See solution",
"solution": "First, note that for a given $n$, exactly one player has a winning strategy. We'll show by induction that $B$ has a winning strategy if $n$ is odd.\n\n**Case 1: $n$ is odd**\n\nBase case is clear. Assume $n$ is odd and $B$ has a winning strategy for all odd integers smaller than $n$. If player $A$ can't make a move, $B$ wins. Otherwise, $A$ chooses a divisor $m$. Note that $m \\mid n - m$ and $m < n - m$, because $m \\leq \\frac{n}{3}$ as $n$ is odd. Therefore, $B$ may choose $m$ (in particular, can make a move) and pass the number $n - 2m$ to player $A$. The number $n - 2m$ is odd and smaller than $n$, so the claim holds by induction.\n\n**Case 2: $n$ is even, but not a power of $2$**\n\nIn this case, $n$ has an odd divisor greater than $1$. Player $A$ may choose an odd divisor and pass an odd integer to player $B$. Then $B$ starts with an odd integer, so $A$ has a winning strategy.\n\n**Case 3: $n = 2^k$ for some positive integer $k$**\n\nWe'll prove by induction: for odd $k$, player $B$ has a winning strategy; for even $k$, player $A$ has a winning strategy.\n\n- *Base cases*: For $k=1$, $n=2$, player $B$ has a winning strategy as $A$ can't make the first move. For $k=2$, $n=4$, player $A$ may win by passing $2$ to player $B$.\n- *Inductive step*:\n - Assume $A$ has a winning strategy for $2^k$, then $B$ has one for $2^{k+1}$. Let $n = 2^{k+1}$. Player $A$ must choose a divisor $2^l$ for $1 \\leq l \\leq k$. If $A$ chooses $2^k$, he passes $n - 2^k = 2^k$ to $B$. By induction, $B$ has a winning strategy. If $A$ chooses a smaller divisor, he passes an even integer which is not a power of $2$ (since $2^k < n - 2^l < n = 2^{k+1}$). We have already proved that the starting player (in this case $B$) has a winning strategy for such a number.\n - Assume $B$ has a winning strategy for $2^k$, then $A$ has one for $2^{k+1}$. Let $n = 2^{k+1}$. It is sufficient for $A$ to choose a divisor $2^k$, then he passes $2^k$ to $B$. By induction, the second player (in this case $A$) has a winning strategy.\n\n**Conclusion:** Player $B$ has a winning strategy for odd $n$ and for $n = 2 \\cdot 4^k$ for non-negative integers $k$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20572,
"subject": "Mathematics (Olympiad)",
"question": "The nonnegative integers 2000, 17, and $n$ are written on a blackboard. Alice and Bob play the following game: Alice begins, then they play in turns. A move consists of replacing one of the three numbers by the absolute difference of the other two. No moves are allowed where all three numbers remain unchanged. A player in turn who cannot make a legal move loses the game.\n\n*Prove that the game will end for every number $n$.*\n\n*Who wins the game in the case $n = 2017$?*",
"options": [],
"answer": "See solution",
"solution": "If three numbers are written on the blackboard and one of them is replaced by the (positive) difference of the other two, then after this move one number on the blackboard will be the sum of the other two. Let $a$, $b$, and $a+b$ be the numbers on the blackboard; without loss of generality, assume $b > a$. Because $a+b-b = a$ and $a+b-a = b$, there is only one possible move. After it, the numbers $a$, $b$, and $b-a$ are written on the blackboard. Again, one number (namely $b$) is the sum of the other two, and there exists only one possible move.\n\nThis means that, at the latest from the second turn on, there is no choice of moves and all moves are inevitable. Furthermore, from the second move on, the largest of the three numbers is decreased, and since no number can become negative, after a finite number of moves one of the numbers will be $0$. Since $0$ is the difference of the other two numbers, we must have $0$, $a$, $a$ on the blackboard. Now $a-0 = a$ and $a-a = 0$, therefore no further move is possible. Thus, the player writing $0$, $a$, $a$ onto the blackboard is the winner.\n\nIf the game starts with the numbers $2000$, $17$, and $2017$ on the blackboard, the course of the game is as follows:\n\n1st move (A): $2000$, $17$, $1983$\n\n2nd move (B): $1966$, $17$, $1983$\n\n3rd move (A): $1966$, $17$, $1949$\n\n... (since $2000 \\div 17 = 117.6\\ldots$)\n\n117th move (A): $28$, $17$, $11$\n\n118th move (B): $6$, $17$, $11$\n\n119th move (A): $6$, $5$, $11$\n\n120th move (B): $6$, $5$, $1$\n\n121st move (A): $4$, $5$, $1$\n\n122nd move (B): $4$, $3$, $1$\n\n123rd move (A): $2$, $3$, $1$\n\n124th move (B): $2$, $1$, $1$\n\n125th move (A): $0$, $1$, $1$\n\nand Alice wins the game.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20573,
"subject": "Mathematics (Olympiad)",
"question": "Suppose the line passing through $C$ and parallel to $AB$ meets $PN$ and $PM$ at $X$ and $Y$ respectively. Prove that $PC \\perp XY$ if and only if $CX = CY$.\n\n",
"options": [],
"answer": "See solution",
"solution": "By Ceva's theorem, we have\n\n$$\n\\frac{AP}{PB} \\times \\frac{BM}{MC} \\times \\frac{CN}{NA} = 1.\n$$\n\nNote that $\\frac{BM}{MC} = \\frac{PB}{YC}$ and $\\frac{CN}{NA} = \\frac{XC}{PA}$ by similar triangles. It follows that $CX = CY$ as desired.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20574,
"subject": "Mathematics (Olympiad)",
"question": "Let $x$, $y$, $z$ be positive real numbers such that $x + y + z = 9xyz$. Prove that:\n\n$$\n\\frac{x}{\\sqrt{x^2 + 2yz + 2}} + \\frac{y}{\\sqrt{y^2 + 2zx + 2}} + \\frac{z}{\\sqrt{z^2 + 2xy + 2}} \\geq 1.\n$$\n\nWhen does equality hold?",
"options": [],
"answer": "See solution",
"solution": "From the inequality $2yz \\leq y^2 + z^2$, we have $x^2 + 2yz + 2 \\leq x^2 + y^2 + z^2 + 2$, so\n\n$$\n\\frac{x}{\\sqrt{x^2 + 2yz + 2}} \\geq \\frac{x}{\\sqrt{x^2 + y^2 + z^2 + 2}}.\n$$\n\nSimilarly for the other terms, so adding up:\n\n$$\n\\frac{x}{\\sqrt{x^2 + 2yz + 2}} + \\frac{y}{\\sqrt{y^2 + 2zx + 2}} + \\frac{z}{\\sqrt{z^2 + 2xy + 2}} \\geq \\frac{x + y + z}{\\sqrt{x^2 + y^2 + z^2 + 2}}.\n$$\n\nTherefore, it suffices to prove that\n\n$$\n\\frac{x + y + z}{\\sqrt{x^2 + y^2 + z^2 + 2}} \\geq 1 \\iff (x + y + z)^2 \\geq x^2 + y^2 + z^2 + 2 \\iff xy + yz + zx \\geq 1.\n$$\n\nFrom the given condition, $x + y + z = 9xyz$, so $\\frac{1}{xy} + \\frac{1}{yz} + \\frac{1}{zx} = 9$. By the Cauchy-Schwarz inequality:\n\n$$\n(xy + yz + zx) \\left( \\frac{1}{xy} + \\frac{1}{yz} + \\frac{1}{zx} \\right) \\geq 9,\n$$\nso $xy + yz + zx \\geq 1$, which is the desired result.\n\n**Alternative approach:** Using Hölder's inequality:\n\n$$\n\\left( \\sum_{cyc} \\frac{x}{\\sqrt{x^2 + 2yz + 2}} \\right)^2 \\left( \\sum_{cyc} x(x^2 + 2yz + 2) \\right) \\geq (x + y + z)^3.\n$$\n\nTherefore, it suffices to prove that\n\n$$\n\\frac{(x + y + z)^3}{\\sum_{cyc} x(x^2 + 2yz + 2)} \\geq 1 \\iff (x + y + z)^3 \\geq x^3 + y^3 + z^3 + 6xyz + 2(x + y + z).\n$$\n\nThis reduces to\n\n$$\n(x + y)(y + z)(z + x) \\geq 8xyz,\n$$\nwhich holds since\n\n$$\nx + y \\geq 2\\sqrt{xy}, \\quad y + z \\geq 2\\sqrt{yz}, \\quad z + x \\geq 2\\sqrt{zx}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20575,
"subject": "Mathematics (Olympiad)",
"question": "Let $P(x) = x^2 + a x + b$ be a real polynomial with $a < 2$. Suppose $P(P(x)) = 0$ has four distinct real roots and the sum of some two of them is $\\leq -1$. Prove that $P(x+y) \\geq P(x) + P(y)$ for all non-negative real numbers $x, y$.",
"options": [],
"answer": "See solution",
"solution": "Let $\\alpha, \\beta$ be the roots of $P(x) = 0$. Let $x_1, x_2$ be the roots of $P(x) = \\alpha$ and $x_3, x_4$ be those of $P(x) = \\beta$. The roots of $P(P(x)) = 0$ are precisely those of the equations $P(x) = \\alpha$ and $P(x) = \\beta$.\n\nWe consider two possibilities: $x_1 + x_2 \\leq -1$ or $x_1 + x_3 \\leq -1$. Suppose $x_1 + x_2 \\leq -1$. Then $-a = x_1 + x_2$ shows that $a > 0$. Since $P(x) = \\alpha$ and $P(x) = \\beta$ have distinct real roots,\n\n$$\n4(b - \\alpha) < a^2, \\quad 4(b - \\beta) < a^2.\n$$\n\nThus $4b < a^2 + 2(\\alpha + \\beta) = a^2 - 2a = a(a - 2) < 0$, so $b < 0$ in this case.\n\nSuppose $x_1 + x_3 \\leq -1$. We have $P(x_1) = \\alpha$, $P(x_3) = \\beta$, so\n\n$$\nx_1^2 + x_3^2 + a(x_1 + x_3) + 2b = \\alpha + \\beta = -a.\n$$\n\nIf $a > 0$,\n\n$$\n\\left(x_1 + \\frac{a}{2}\\right)^2 + \\left(x_3 + \\frac{a}{2}\\right)^2 + 2b = -a + \\frac{a^2}{2} = \\frac{a(a-2)}{2},\n$$\n\nso $2b \\leq \\frac{a(a - 2)}{2} < 0$. If $a \\leq 0$,\n\n$$\nx_1^2 + x_3^2 + 2b = -a(x_1 + x_3 + 1) \\leq 0,\n$$\n\nso $b \\leq 0$. Thus $b \\leq 0$ in all cases.\n\nNow,\n\n$$\nP(x+y) = P(x) + P(y) + 2xy - b \\geq P(x) + P(y),\n$$\n\nfor all non-negative $x, y$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20576,
"subject": "Mathematics (Olympiad)",
"question": "The Perseverance, NASA's Mars rover, starts each day at its home base on Mars. Each day, it moves north, south, east, or west, making a 90° turn every 1 kilometer. The rover does not visit the same place twice in a day, except for the home base, which is both the starting and ending point of its trip. What are the possible lengths of the rover's path?",
"options": [],
"answer": "See solution",
"solution": "Let's define a coordinate system with the origin at the home base and vertical-horizontal axes. Without loss of generality, assume the first move is east and the path has length $n$. Each odd move changes the $x$ coordinate by 1, and each even move changes the $y$ coordinate by 1.\n\nAt the end of the day, both coordinates must return to zero, so there must be an even number of both odd and even moves. This implies that only $n$ divisible by 4 can fulfill the conditions.\n\nFor $n = 4$, we have a square path. For $n = 8$, there are 4 changes in $x$ and 4 in $y$, so the path is inside a $2 \\times 2$ square. However, this is not possible without revisiting a point.\n\nNow, we prove that all $n > 8$ divisible by 4 are possible. For $n = 12$, there is a path in the shape of a \"+\" with the first 4 moves as $(\\rightarrow, \\uparrow, \\rightarrow, \\uparrow)$. We can modify the middle $(\\uparrow, \\rightarrow)$ sequence to $(\\downarrow, \\rightarrow, \\uparrow, \\rightarrow, \\uparrow, \\leftarrow)$, allowing the robot to explore new territory southeast of the previous path. This adds 4 to the path length. Repeating this process, we can achieve any length of $4k + 8$ for all $k \\in \\mathbb{Z}^+$.\n\n*Remark:* It should be clarified that the problem is not on a sphere. (But does the story make sense if the rover drives on an infinite plane?)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20577,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a cyclic quadrilateral. Let $O$ be the circumcenter of the quadrilateral $ABCD$. The diagonals $AC$ and $BD$ intersect at $G$. Let $P$, $Q$, $R$, and $S$ be the circumcenters of triangles $AGB$, $BGC$, $CGD$, and $DGA$ respectively. The lines $PR$ and $QS$ intersect at $M$. Show that $M$ is the midpoint of $G$ and $O$.",
"options": [],
"answer": "See solution",
"solution": "First we show that $PORG$ is a parallelogram.\n\n\n\nSince $R$ and $O$ lie on the perpendicular bisector of chord $CD$, we have $OR$ is perpendicular to $CD$.\n\nLet $L$ be the intersection of $PG$ and $CD$, and $X$ be the midpoint of $BG$.\n\nSince $\\angle(LP, PX) = \\angle(GP, PX) = \\angle(GA, AB) = \\angle(CA, AB) = \\angle(CD, DB) = \\angle(LD, DX)$, the points $L$, $P$, $D$, $X$ are concyclic. From $PX \\perp XG$, we deduce $LP \\perp CD$ and thus $GP \\parallel OR$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20578,
"subject": "Mathematics (Olympiad)",
"question": "Fedir and Mykhailo have three piles of stones: the first contains 100 stones, the second 101, and the third 102. They play a game with these rules:\n\nOn each turn, a player chooses any two piles containing $a$ and $b$ stones, and removes from each pile the number of stones equal to the greatest common divisor of $a$ and $b$. The winner is the player whose move first causes any pile to become empty. If Fedir moves first and both play optimally, who wins?",
"options": [],
"answer": "See solution",
"solution": "Mykhailo wins.\n\nHere is Mykhailo's winning strategy:\n\nSuppose before Fedir's move, the piles contain $(2n, 2n+1, 2n+2)$ stones for some integer $n > 1$.\n\n- If Fedir chooses the first two piles, $(2n, 2n+1)$, their gcd is $1$. After his move: $(2n-1, 2n, 2n+2)$. Mykhailo then chooses the last two piles, $(2n, 2n+2)$, whose gcd is $2$, resulting in $(2n-2, 2n-1, 2n)$—the initial configuration with $n$ decreased by $1$.\n- If Fedir chooses the first and third piles, $(2n, 2n+2)$, their gcd is $2$. After his move: $(2n-2, 2n, 2n+1)$. Mykhailo chooses the last two piles, $(2n, 2n+1)$, whose gcd is $1$, resulting in $(2n-2, 2n-1, 2n)$ again.\n- If Fedir chooses the second and third piles, $(2n+1, 2n+2)$, their gcd is $1$. After his move: $(2n, 2n, 2n+1)$. Mykhailo then chooses the first two equal piles, $(2n, 2n)$, whose gcd is $2n$, leaving $(0, 0, 2n+1)$, and wins immediately.\n\nThe initial configuration is $n = 50$. After each pair of moves, $n$ decreases by $1$, or Mykhailo wins earlier. Eventually, the piles reach $(4, 5, 6)$. Fedir's move leads to one of $(3, 4, 6)$, $(2, 5, 4)$, or $(4, 4, 5)$, and in all cases, Mykhailo wins on his next turn.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20579,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, \\dots, a_n$ be distinct positive integers and let $M$ be a set of $n-1$ positive integers not containing $s = a_1 + a_2 + \\dots + a_n$. A grasshopper is to jump along the real axis, starting at the point $0$ and making $n$ jumps to the right with lengths $a_1, a_2, \\dots, a_n$ in some order. Prove that the order can be chosen in such a way that the grasshopper never lands on any point in $M$.",
"options": [],
"answer": "See solution",
"solution": "We proceed by induction on $n$.\n\n**Base case ($n=1$):** $M$ is empty, so the result is obvious.\n\n**Case $n=2$:** $M$ does not contain at least one of $a_1, a_2$, so the grasshopper can first jump the number not in $M$, then the other.\n\n**Inductive step:** Let $m \\geq 3$ and assume the result is true for all $n < m$. Suppose, for contradiction, that there exist $m$ distinct positive integers $a_1, a_2, \\dots, a_m$ and a set $M$ of $m-1$ numbers such that the grasshopper cannot finish the jumping as required.\n\nSuppose the grasshopper can jump at most $k$ steps to the right with lengths of distinct numbers in $a_1, a_2, \\dots, a_m$, such that the landing points are never in $M$. Since $M$ contains only $m-1$ numbers, the grasshopper can always take its first step. If the grasshopper can jump $m-1$ steps, it can also jump the last step since $M$ does not contain $a_1 + a_2 + \\dots + a_n$, so $1 \\leq k \\leq m-2$.\n\nChoose such $k$ steps with a minimum total length (if there is a tie, choose any one that attains the minimum), denote the lengths of these $k$ steps by $b_1, b_2, \\dots, b_k$, and let $b_{k+1}, b_{k+2}, \\dots, b_m$ be the remaining numbers in $a_1, a_2, \\dots, a_m$ in increasing order. Clearly,\n\n$$\n\\{a_1, a_2, \\dots, a_m\\} = \\{b_1, b_2, \\dots, b_m\\}.\n$$\n\nBy assumption, for any $k+1 \\leq j \\leq m$, $b_1 + b_2 + \\dots + b_k + b_j \\in M$. Thus, there are at least $m-k$ elements of $M$ that are greater than or equal to $b_1 + b_2 + \\dots + b_{k+1}$, and hence at most $k-1$ elements of $M$ that are less than $b_1 + b_2 + \\dots + b_{k+1}$.\n\nLet\n\n$$\nA = \\{b_i \\mid 1 \\leq i \\leq k+1,\\ b_1 + b_2 + \\dots + b_{k+1} - b_i \\notin M\\}.\n$$\n\nThen $b_{k+1} \\in A$. For any $b_i \\in A$, if we remove $b_i$ from $b_1, b_2, \\dots, b_{k+1}$, the sum of the remaining $k$ numbers is not in $M$, and since\n\n$$\n|M \\cap \\{1, 2, \\dots, b_1 + b_2 + \\dots + b_{k+1} - b_i\\}| \\leq k-1,\n$$\n\nand $k < m$, by the inductive hypothesis there exists an ordering $c_1, c_2, \\dots, c_k$ of these remaining numbers, such that\n\n$$\nc_1,\\ c_1 + c_2,\\ \\dots,\\ c_1 + c_2 + \\dots + c_k\n$$\n\nare not in $M$. Thus, $c_1, c_2, \\dots, c_k$ is also a possible length of $k$ jumps, and by the choice of $b_1, b_2, \\dots, b_k$ we have\n\n$$\nb_1 + b_2 + \\dots + b_k \\leq c_1 + c_2 + \\dots + c_k,\n$$\n\ni.e., $b_i \\leq b_{k+1}$ for any $b_i \\in A$.\n\nLet $A = \\{x_1, x_2, \\dots, x_t\\}$, where $x_1 < x_2 < \\dots < x_t = b_{k+1}$. By definition of $A$, there are at least $k+1-t$ numbers of $M$ that are less than $b_1 + b_2 + \\dots + b_{k+1}$. Since for any $k+1 \\leq j \\leq m$, $b_1 + b_2 + \\dots + b_k + b_j \\in M$, there are at least $m-k$ numbers of $M$ in the interval\n\n$$\n[b_1 + b_2 + \\dots + b_{k+1},\\ b_1 + b_2 + \\dots + b_k + b_m].\n$$\n\nOn the other hand, for any $x_i \\in A$, the number\n\n$$\nb_1 + b_2 + \\dots + b_{k+1} - x_i + b_m\n$$\n\nbelongs to $M$ (otherwise, since $k \\leq m-2$, $b_m$ is different from $b_1, b_2, \\dots, b_{k+1}$, and the grasshopper can take $k+1$ jumps). For $1 \\leq i \\leq t-1$, the numbers\n\n$$\nb_1 + b_2 + \\dots + b_{k+1} - x_i + b_m\n$$\n\nare pairwise distinct and greater than $b_1 + b_2 + \\dots + b_k + b_m$, so there are at least $t-1$ numbers of $M$ greater than $b_1 + b_2 + \\dots + b_k + b_m$. Summing up, $M$ contains at least\n\n$$\n(k+1-t) + (m-k) + (t-1) = m\n$$\n\nnumbers, which contradicts the fact that $M$ contains only $m-1$ numbers.\n\nThus, our assumption is false, and the result also holds for $n = m$. By induction, the result holds for every positive integer $n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20580,
"subject": "Mathematics (Olympiad)",
"question": "Two players, A and B, remove stones alternately from a heap initially containing $n \\geq 2$ stones. The first player, A, takes at least one and at most $n-1$ stones. Each subsequent player must take at least one stone and at most as many stones as their opponent took in the previous move. The winner is the player who takes the last stone. Which player has a winning strategy?",
"options": [],
"answer": "See solution",
"solution": "Let $(a, b)$ denote a position where the player to move faces $a$ stones in the heap and can take at most $b$ stones.\n\nWe claim that the losing positions are those of the forms $(0, r)$ and $(2^k(2m+1), r)$, where $k, m \\in \\mathbb{N}$ and $r \\in \\{1, 2, \\dots, 2^k - 1\\}$.\n\nThe initial position is $(n, n-1)$, so the losing initial positions are those with $m=0$, i.e., when $n$ is a power of $2$. Thus, if $n$ is a power of $2$, the second player has a winning strategy; otherwise, the first player does.\n\nKey observations:\n\n- The game ends only in positions of the form $(0, r)$.\n- From a position $(2^k(2m+1), r)$ with $0 < r < 2^k$, there is no move that leaves the opponent in a similar form.\n- From any position not of the above form, there exists a move that leaves the opponent in a position $(2^k(2m+1), r)$ with $0 < r < 2^k$.\n\nSuppose from $(2^k(2m+1), r)$ with $0 < r < 2^k$, one could move to $(2^s(2p+1), t)$ with $0 < t < 2^s$. This move would remove $t$ stones, so $2^k(2m+1) - 2^s(2p+1) = t$. Since $t \\leq r < 2^k$, $s < k$, so $2^s$ divides $2^k(2m+1) - 2^s(2p+1)$, and thus $2^s$ divides $t$, which contradicts $0 < t < 2^s$. Therefore, such a move does not exist.\n\nThere is also no move to a position $(0, x)$.\n\nFrom any remaining position $(2^k(2m+1), r)$ with $r \\geq 2^k$, there is a move to $(2^k \\cdot 2m, 2^k)$, which is a losing position, by taking $2^k$ stones.\n\nTherefore, the winning strategy is: always take $2^k$ stones, where $2^k$ is the largest power of $2$ dividing the number of remaining stones.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20581,
"subject": "Mathematics (Olympiad)",
"question": "Prove that there is an angle $x$ such that\n$$\nsin x = \\frac{\\sin \\beta \\cdot \\sin \\gamma}{1 - \\cos \\alpha \\cdot \\cos \\beta \\cdot \\cos \\gamma}\n$$\nfor every angle $\\alpha$, and $\\beta$ and $\\gamma$ acute angles.",
"options": [],
"answer": "See solution",
"solution": "We will prove that $\\frac{\\sin \\beta \\cdot \\sin \\gamma}{1 - \\cos \\alpha \\cdot \\cos \\beta \\cdot \\cos \\gamma}$ belongs in $[-1, 1]$.\n\nSince $\\beta$ and $\\gamma$ are acute angles, we have $\\cos \\beta > 0$, $\\cos \\gamma > 0$, so $\\cos \\beta \\cdot \\cos \\gamma > 0$.\n\nBecause $\\cos \\alpha \\le 1$, we have $\\cos \\alpha \\cdot \\cos \\beta \\cdot \\cos \\gamma \\le \\cos \\beta \\cdot \\cos \\gamma$, i.e., $-\\cos \\alpha \\cdot \\cos \\beta \\cdot \\cos \\gamma \\ge -\\cos \\beta \\cdot \\cos \\gamma$.\n\nFrom the sum identities, we have\n$$\n\\sin \\beta \\cdot \\sin \\gamma + \\cos \\beta \\cdot \\cos \\gamma = \\cos(\\beta - \\gamma) \\le 1.\n$$\nThen\n$$\n0 < \\sin \\beta \\cdot \\sin \\gamma \\le 1 - \\cos \\beta \\cdot \\cos \\gamma \\le 1 - \\cos \\alpha \\cdot \\cos \\beta \\cdot \\cos \\gamma.\n$$\nSo\n$$\n0 < \\frac{\\sin \\beta \\cdot \\sin \\gamma}{1 - \\cos \\alpha \\cdot \\cos \\beta \\cdot \\cos \\gamma} \\le 1.\n$$\nThere is an angle $x$ such that\n$$\n\\sin x = \\frac{\\sin \\beta \\cdot \\sin \\gamma}{1 - \\cos \\alpha \\cdot \\cos \\beta \\cdot \\cos \\gamma}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20582,
"subject": "Mathematics (Olympiad)",
"question": "The lengths of the sides of a quadrilateral are $a$, $b$, $c$, $d$ and its area is $S$. Prove that\n$$\na^2 + b^2 + c^2 + d^2 \\ge 4S.\n$$\nFor which quadrilaterals does the equality hold?",
"options": [],
"answer": "See solution",
"solution": "Without loss of generality, assume $a$, $b$, $c$, and $d$ are the lengths of consecutive sides of the quadrilateral. A diagonal divides the quadrilateral into two triangles. From one partition, we get the inequality\n$$\n\\frac{ab}{2} + \\frac{cd}{2} \\ge S,\n$$\nwhich gives $ab + cd \\ge 2S$. From the other partition,\n$$\n\\frac{bc}{2} + \\frac{da}{2} \\ge S,\n$$\nso $bc + da \\ge 2S$. Therefore,\n$$\nab + bc + cd + da \\ge 4S.\n$$\n\nOn the other hand, by adding the inequalities $a^2 + b^2 \\ge 2ab$, $b^2 + c^2 \\ge 2bc$, $c^2 + d^2 \\ge 2cd$, and $d^2 + a^2 \\ge 2da$, and dividing by 2, we get\n$$\na^2 + b^2 + c^2 + d^2 \\ge ab + bc + cd + da,\n$$\nwhich implies the required inequality.\n\nEquality holds if all the inequalities used are equalities. In the first step, equality holds if all angles are right angles. In the second step, equality holds if all sides are of equal length. Thus, equality holds if and only if the quadrilateral is a square.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20583,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be non-negative real numbers such that\n$$\n2(a^2 + b^2 + c^2) + 3(ab + bc + ca) = 5(a + b + c).\n$$\nProve that\n$$\n4(a^2 + b^2 + c^2) + 2(ab + bc + ca) + 7abc \\le 25.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $p = a + b + c$, $q = ab + bc + ca$, and $r = abc$. We have\n$$\n2(p^2 - 2q) + 3q = 5p \\quad \\text{or} \\quad 2p^2 = 5p + q. \\tag{1}\n$$\nWe need to prove that $4(p^2 - 2q) + 2q + 7r \\le 25$, or equivalently, $4p^2 + 7r \\le 25 + 6q$.\nSince $q = p^2 - 5p$, the inequality becomes:\n$$\n7r + 30p \\le 8p^2 + 25.\n$$\nNotice that $(ab + bc + ca)^2 \\ge 3abc(a + b + c)$ implies $q^2 \\ge 3pr$. We consider two cases for $p$:\n\n* If $p = 0$, then $a = b = c = 0$, so the inequality holds.\n* If $p > 0$, then $r \\le \\dfrac{q^2}{3p}$, so we need to prove\n$$\n7\\frac{q^2}{3p} + 30p \\le 8p^2 + 25.\n$$\nSubstituting $q = p^2 - 5p$ gives\n$$\n7(2p^2 - 5p)^2 + 90p^2 \\le 24p^3 + 75p,\n$$\nor\n$$\np(p-3)(2p-5)(14p-5) \\le 0. \\tag{2}\n$$\nOn the other hand, $2p^2 - 5p = q \\le \\dfrac{p^2}{3}$ implies $\\dfrac{5}{2} \\le p \\le 3$, so inequality (2) holds. Therefore, the original inequality is true.\n\nEquality holds when $(a, b, c)$ is a permutation of $(0, 0, 0)$, $(1, 1, 1)$, or $(0, 0, 5/2)$.\n\n$\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20584,
"subject": "Mathematics (Olympiad)",
"question": "Can real numbers $x$, $y$, $z$ satisfy\n\n$$\n\\frac{1}{(x-y)(x+y)} + \\frac{1}{(y-z)(y+z)} + \\frac{1}{(z-x)(z+x)} = 0?\n$$",
"options": [],
"answer": "See solution",
"solution": "**Answer:** no.\n\n**Solution.** Denote $a = x^2 - y^2$, $b = y^2 - z^2$, then $-a - b = z^2 - x^2$ and the equation can be rewritten as:\n\n$$\n\\frac{1}{a} + \\frac{1}{b} = \\frac{1}{a+b} \\Leftrightarrow (a+b)^2 = ab \\Leftrightarrow a^2 - ab + b^2 = 0 \\Leftrightarrow \\left(a - \\frac{b}{2}\\right)^2 + \\frac{3b^2}{4} = 0.\n$$\n\nThe last equality holds only with $a = b = 0$, which is impossible. Therefore, the equation has no solutions.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20585,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge 5$ be an integer. Consider $n$ squares with side lengths $1, 2, \\dots, n$, respectively. The squares are arranged in the plane with their sides parallel to the $x$ and $y$ axes. Suppose that no two squares touch, except possibly at their vertices.\n\nShow that it is possible to arrange these squares in a way such that every square touches exactly two other squares.",
"options": [],
"answer": "See solution",
"solution": "Set aside the squares with side lengths $n-3, n-2, n-1$, and $n$. Suppose we can split the remaining squares into two sets $A$ and $B$ such that the sum of the side lengths of the squares in $A$ is 1 or 2 units larger than the sum of the side lengths of the squares in $B$.\n\nString the squares of each set $A$ and $B$ along two parallel diagonals, one for each diagonal. Now use the four largest squares along two perpendicular diagonals to finish the construction: one will have side lengths $n$ and $n-3$, and the other, side lengths $n-1$ and $n-2$. If the sum of the side lengths of the squares in $A$ is 1 unit larger than the sum of the side lengths of the squares in $B$, attach the squares with side lengths $n-3$ and $n-1$ to the $A$-diagonal, and the other two squares to the $B$-diagonal. The resulting configuration, in which the $A$ and $B$-diagonals are represented by unit squares, and the side lengths $a_i$ of squares from $A$ and $b_j$ of squares from $B$ are indicated within each square, follows:\n\n\n\nSince $$(a_1 + a_2 + \\dots + a_k)\\sqrt{2} + \\frac{((n-3)+(n-2))\\sqrt{2}}{2} = (b_1 + b_2 + \\dots + b_\\ell + 2)\\sqrt{2} + \\frac{(n+(n-1))\\sqrt{2}}{2},$$ this case is done.\n\nIf the sum of the side lengths of the squares in $A$ is 2 units larger than the sum of the side lengths of the squares in $B$, attach the squares with side lengths $n-3$ and $n-2$ to the $A$-diagonal, and the other two squares to the $B$-diagonal. The resulting configuration follows:\n\n\n\nSince $$(a_1 + a_2 + \\dots + a_k)\\sqrt{2} + \\frac{((n-3)+(n-1))\\sqrt{2}}{2} = (b_1 + b_2 + \\dots + b_\\ell + 1)\\sqrt{2} + \\frac{(n+(n-2))\\sqrt{2}}{2},$$ this case is also done.\n\nIn both cases, the distance between the $A$-diagonal and the $B$-diagonal is $$\\frac{((n-3)+n)\\sqrt{2}}{2} = \\frac{(2n-3)\\sqrt{2}}{2}.$$\n\nSince $a_i, b_j \\le n-4$, $$\\frac{(a_i+b_j)\\sqrt{2}}{2} < \\frac{(2n-4)\\sqrt{2}}{2} < \\frac{(2n-3)\\sqrt{2}}{2},$$ and therefore the $A$- and $B$-diagonals do not overlap.\n\nFinally, we prove that it is possible to split the squares of side lengths $1$ to $n-4$ into two sets $A$ and $B$ such that the sum of the side lengths of the squares in $A$ is 1 or 2 units larger than the sum of the side lengths in $B$. One can do that in several ways; we present two possibilities:\n\n* **Direct construction:** Split the numbers from $1$ to $n-4$ into several sets of four consecutive numbers $\\{t, t+1, t+2, t+3\\}$, beginning with the largest numbers; put squares of side lengths $t$ and $t+3$ in $A$ and squares of side lengths $t+1$ and $t+2$ in $B$. Notice that $t + (t+3) = (t+1) + (t+2)$. In the end, at most four numbers remain.\n - If only $1$ remains, put the corresponding square in $A$, so the sum of the side lengths of the squares in $A$ is one unit larger than those in $B$;\n - If $1$ and $2$ remain, put the square of side length $2$ in $A$ and the square of side length $1$ in $B$ (the difference is $1$);\n - If $1$, $2$, and $3$ remain, put the squares of side lengths $1$ and $3$ in $A$, and the square of side length $2$ in $B$ (the difference is $2$);\n - If $1$, $2$, $3$, and $4$ remain, put the squares of side lengths $2$ and $4$ in $A$, and the squares of side lengths $1$ and $3$ in $B$ (the difference is $2$).\n* **Indirect construction:** Starting with $A$ and $B$ as empty sets, add the squares of side lengths $n-4, n-5, \\dots, 2$ to either $A$ or $B$ in that order such that at each stage the difference between the sum of the side lengths in $A$ and the sum of the side lengths of $B$ is minimized. By induction it is clear that after adding an integer $j$ to one of the sets, this difference is at most $j$. In particular, the difference is $0$, $1$ or $2$ at the end. Finally, adding the final $1$ to one of the sets can ensure that the final difference is $1$ or $2$. If necessary, flip $A$ and $B$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20586,
"subject": "Mathematics (Olympiad)",
"question": "Find all integers $n \\ge 3$ such that among any $n$ positive real numbers $a_1, a_2, \\dots, a_n$ with\n\n$$\n\\max(a_1, a_2, \\dots, a_n) \\le n \\cdot \\min(a_1, a_2, \\dots, a_n),\n$$\n\nthere exist three that are the side lengths of an acute triangle.",
"options": [],
"answer": "See solution",
"solution": "The answer is $n \\ge 13$.\n\nFirst, we show that any $n \\ge 13$ satisfies the desired condition. Suppose for the sake of contradiction that $a_1 \\le a_2 \\le \\dots \\le a_n$ are integers such that $\\max(a_1, a_2, \\dots, a_n) \\le n \\cdot \\min(a_1, a_2, \\dots, a_n)$ and no three are the side lengths of an acute triangle. We conclude that\n\n$$\na_{i+2}^2 \\ge a_i^2 + a_{i+1}^2 \\qquad (2)\n$$\n\nfor all $i \\le n-2$. Letting $\\{F_n\\}$ be the Fibonacci numbers, defined by $F_1 = F_2 = 1$ and $F_{n+1} = F_n + F_{n-1}$ for $n \\ge 2$, repeated application of (2) and the ordering of the $\\{a_i\\}$ implies that\n\n$$\na_i^2 \\ge F_i \\cdot a_1^2 \\qquad (3)\n$$\n\nfor all $i \\le n$. Noting that $F_{12} = 12^2$, an easy induction shows that $F_n > n^2$ for $n > 12$. Hence, if $n \\ge 13$, (3) implies $a_n^2 > n^2 \\cdot a_1^2$, a contradiction. This shows that any $n \\ge 13$ satisfies the condition of the problem.\n\nOn the other hand, for any $n < 13$, we may take $a_i = \\sqrt{F_i}$ for $1 \\le i \\le n$, so that\n\n$$\n\\max(a_1, a_2, \\dots, a_n) \\le n \\cdot \\min(a_1, a_2, \\dots, a_n)\n$$\n\nholds because $F_n \\le n^2$ for $n \\le 12$. Further, for $i < j$, we have $F_i + F_j \\le F_{j+1}$, which shows that for $i < j < k$, we have $a_k^2 \\ge a_i^2 + a_j^2$. 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 $n \\ge 13$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20587,
"subject": "Mathematics (Olympiad)",
"question": "Given triangle $ABC$ with $AB > AC$. Circles $o_B$ and $o_C$ are inscribed in angle $BAC$ with $o_B$ tangent to $AB$ at $B$ and $o_C$ tangent to $AC$ at $C$. The tangent to $o_B$ from $C$ (different from $AC$) intersects $AB$ at $K$, and the tangent to $o_C$ from $B$ (different from $AB$) intersects $AC$ at $L$. Line $KL$ and the angle bisector of $BAC$ intersect $BC$ at points $P$ and $M$, respectively. Prove that $BP = CM$.",
"options": [],
"answer": "See solution",
"solution": "Note that the length of the segment of the common tangent to $o_B$ and $o_C$ joining the tangency points is equal to\n$$\nAB - AC = LB - LC = KC - KB,\n$$\nwhich means that points $A$, $L$ lie on one, and point $K$ on the other leg of some hyperbola $\\eta$ with foci $B$, $C$.\n\nDenote by $S$ the midpoint of the segment $BC$ (the center of symmetry of $\\eta$) and denote by $A'$, $K'$, $M'$ the central reflections in $S$ of $A$, $K$, $M$, respectively. Then $A'$, $K' \\in \\eta$ and $A'$, $K'$, $C$ are collinear.\n\n\n\nFrom the optical property of a hyperbola, it follows that $A'M'$ is tangent to $\\eta$ (as it is the bisector of $BA'C$). Therefore, Pascal's theorem applied to the degenerate hexagon $AA'A'K'KL$ inscribed in $\\eta$ gives the collinearity of points\n$$\nAA' \\cap KK' = S, \\quad A'A' \\cap KL, \\quad A'K' \\cap LA = C.\n$$\nTherefore, the three lines $A'A'$ (tangent to $\\eta$ in $A'$), $KL$, $BC$ are concurrent, which means that $M' = A'A' \\cap BC = KL \\cap BC = P$ and in consequence $CM = BP$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20588,
"subject": "Mathematics (Olympiad)",
"question": "We want to cover a table of size $4 \\times 4$ with dominoes of the following shape\n\n\n\n(the dominoes can also be reflected or rotated).\n\nThe dominoes may overlap or extend over the edges of the table.\n\nAt least how many dominoes do we need?",
"options": [],
"answer": "See solution",
"solution": "We need at least 5 dominoes. If it were possible to cover the table with four dominoes, then they could not overlap or extend over the edges, since four dominoes would cover exactly 16 squares. In this case, there are only two possible ways of covering the top left corner, as shown in the figures below. But in both cases, the field marked with * cannot be covered.\n\n\n\n\n\n\n\nIt is easy to see that 5 dominoes suffice, as demonstrated by the figure above.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20589,
"subject": "Mathematics (Olympiad)",
"question": "Given a $5 \\times 5$ square filled with the numbers $1$ to $5$ in each row and column, the sum of all the numbers in any row or column is\n\n$$\n1 + 2 + 3 + 4 + 5 = 15.\n$$\n\nIf the sum of the numbers in the dark squares of column 4 is $9$, what is the sum of the numbers in the light squares of column 4?",
"options": [],
"answer": "See solution",
"solution": "The sum of the numbers in the light squares of column 4 is $15 - 9 = 6$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20590,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs of positive integers $(n, p)$ such that\n$$\nn(n-1)(n+1)(n^2+1) = p^4(p^4-1).\n$$",
"options": [],
"answer": "See solution",
"solution": "It is clear that $p \\ne n$. If $p = 2$, then $n \\ge 3$ and we have $2^8 - 2^4 = 240 = 3^5 - 3$, i.e., $p = 2$, $n = 3$ is a solution. On the other hand, if $n > 3$, then $n^5 - n = n(n^4 - 1) > 3(3^4 - 1) = 240$, so for $p = 2$ there are no $n$ different from $3$ satisfying the initial equality.\n\nNow let $p > 2$. Then $p$ is an odd prime number and $n \\ge 3$. We rewrite the initial equality as\n$$\nn(n-1)(n+1)(n^2+1) = p^4(p^4-1).\n$$\nNote that exactly one of the four co-factors on the left-hand side can be divisible by $p$. Indeed, $n$ is coprime with any of $n-1$, $n+1$, $n^2+1$. The greatest common divisor of any two of $n-1$, $n+1$, $n^2+1$ is $1$ or $2$, so any two of them do not share $p$ as a common divisor.\n\nThus, exactly one of the four co-factors on the left is divisible by $p$, and so it must be divisible by $p^4$. Therefore, this co-factor is at least $p^4$.\n\nIn any case, $n^2 + 1 \\ge p^4$ or $n^2 \\ge p^4 - 1$. So $p^4(p^4 - 1) = n(n^2 - 1)(n^2 + 1) \\ge n(p^4 - 2)p^4$, whence $p^4 - 1 \\ge n(p^4 - 2) > 2(p^4 - 1)$, which is impossible. Therefore, the pair $(n, p) = (3, 2)$ is the unique solution of the given equation.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20591,
"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 in the interval $[1011, 2021]$. Given that $P(x)$ 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 $\\sum_{k=0}^{10} (a_{2k+1} - a_{2k})^2 \\leq 440c^2$.",
"options": [],
"answer": "See solution",
"solution": "**a)** Let $\\alpha$ be an integer root of $P$. We 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 integer root of $P$ is $\\alpha = -1$.\n\n**b)** 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": 20592,
"subject": "Mathematics (Olympiad)",
"question": "Find all possible triads of non-negative integers $x, y, z$ with $x \\leq y$, which satisfy the equation:\n\n$$x^2 + y^2 = 3 \\cdot 2016^z + 77$$",
"options": [],
"answer": "See solution",
"solution": "We distinguish the cases:\n\nIf $z = 0$, then the equation becomes $x^2 + y^2 = 80$.\n\nThen $x, y$ must be multiples of 4, that is $x = 4a$, $y = 4b$, $0 \\leq a \\leq b$, and the equation becomes $a^2 + b^2 = 5$, hence $(a, b) = (1, 2)$, that is $(x, y) = (4, 8)$.\n\nIf $z > 0$, then $7 \\mid 2016^z$ (since $7 \\mid 2016$) and $7 \\mid 77$, and therefore $7$ must divide the left part, that is $7 \\mid x^2 + y^2$. The possible remainders of the division of a square by $7$ are $0, 1, 2, 4$. Therefore, in order to have $7 \\mid x^2 + y^2$ we need $7 \\mid x$, $7 \\mid y$. We write $x = 7x_1$ and $y = 7y_1$, with $0 \\leq x_1 \\leq y_1$.\n\nBy substitution, we get $49(x_1^2 + y_1^2) = 3 \\cdot 2016^z + 77$.\n\nIf $z \\geq 2$, then $49 \\mid 2016^z$, and then $49 \\mid 77$, which is absurd.\n\nIf $z = 1$, then $49(x_1^2 + y_1^2) = 3 \\cdot 7 \\cdot 288 + 77 = 7(3 \\cdot 288 + 11) = 7 \\cdot 7 \\cdot 125$.\n\nHence $x_1^2 + y_1^2 = 125$. Since $x_1 \\leq y_1$, we have $2y_1^2 \\geq 125 \\Rightarrow y_1 \\geq 8$.\n\nIf $y_1 = 8, 9, 10, 11$, we find the solutions $(x_1, y_1) \\in \\{(5, 10), (2, 11)\\}$. Therefore, we have the solutions: $(x, y, z) \\in \\{(4, 8, 0), (35, 70, 1), (14, 77, 1)\\}$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20593,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. Real numbers $a_1, a_2, \\dots, a_{2n}$ satisfy the following conditions:\n\n1. For every $i = 1, 2, \\dots, 2n - 1$, one has $0 < a_{i+1} - a_i \\le 1$.\n2. Rounding the numbers $a_1, a_2, \\dots, a_{2n}$ to the closest integer (numbers equidistant from two closest integers are rounded up) gives pairwise distinct positive integers.\n\nNumbers $a_1, a_2, \\dots, a_{2n}$ are placed as the numerators and denominators of $n$ fractions. Prove that the sum of the obtained fractions is greater than $\\frac{n}{4}$.",
"options": [],
"answer": "See solution",
"solution": "Suppose that a numerator is greater than a denominator. Then interchanging these two numbers makes both fractions smaller. Thus, we can assume without loss of generality that all numerators are less than all denominators.\n\nFor every $i = 1, 2, \\dots, 2n$, define $x_i = a_i - i + \\frac{1}{2}$. As $a_1 < a_2 < \\dots < a_{2n}$ and rounding the numbers $a_i$ produces pairwise distinct positive integers, we must have $a_i \\ge i - \\frac{1}{2}$, which implies $x_i \\ge 0$. From $a_{i+1} - a_i \\le 1$, we have $x_{i+1} - x_i = a_{i+1} - a_i - 1 \\le 0$, which implies $x_1 \\ge x_2 \\ge \\dots \\ge x_{2n}$.\n\nHence, $i < j$ always implies\n$$\n\\frac{a_i - x_j}{a_j - x_j} \\ge \\frac{a_i - x_i}{a_j - x_j} = \\frac{2i-1}{2j-1}\n$$\n(the first inequality holds because $a_i < a_j$ and $x_j \\ge 0$, while the second holds because $x_i \\ge x_j$). Thus, it suffices to prove the desired inequality for the case where the numerators are integers $1, 3, \\dots, 2n - 1$ and the denominators are integers $2n + 1, 2n + 3, \\dots, 4n - 1$ in some order.\n\nDenote the sum of all fractions by $s$. Applying AM-GM to the fractions gives\n$$\n\\frac{s}{n} \\ge \\sqrt[n]{\\frac{1 \\cdot 3 \\cdots (2n-1)}{(2n+1)(2n+3)\\cdots(4n-1)}}.\n$$\nThus, it suffices to prove for every $n$ the inequality\n$$\n\\frac{1 \\cdot 3 \\cdots (2n-1)}{(2n+1)(2n+3)\\cdots(4n-1)} > \\frac{1}{4^n}.\n$$\nWe can do this by induction on $n$. The claim holds for $n = 1$ since $\\frac{1}{3} > \\frac{1}{4}$. For the induction step, it suffices to show that\n$$\n\\frac{(2n+1)^2}{(4n+1)(4n+3)} > \\frac{1}{4}\n$$\nwhich is equivalent to $(4n+2)^2 > (4n+1)(4n+3)$. The latter follows from AM-GM for $4n+1$ and $4n+3$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20594,
"subject": "Mathematics (Olympiad)",
"question": "Four teams, A, B, C, and D, each play a single game against each of the other teams. There are no draws, and teams A, B, and C have the same number of wins. If team A beats team D, how many wins does team D have?",
"options": [],
"answer": "See solution",
"solution": "The total number of games is $\\frac{4 \\times 3}{2} = 6$. Since team A scored at least 1 win, each of teams A, B, and C has either 1 or 2 wins. Since team A beats team D, team D does not have 3 wins, so the scores of A, B, and C must total more than 3. Hence, each of team A, B, and C has 2 wins, and therefore team D has 0 wins.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20595,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle. Line $\\ell$ is parallel to $BC$ and it intersects side $AB$ at point $D$, side $AC$ at point $E$, and the circumcircle of triangle $ABC$ at points $F$ and $G$, where points $F, D, E, G$ lie in this order on $\\ell$. The circumcircles of triangles $FEB$ and $DGC$ intersect at points $P$ and $Q$. Prove that points $A, P, Q$ are collinear.",
"options": [],
"answer": "See solution",
"solution": "Let $\\omega_B$ and $\\omega_C$ be the circumcircles of triangles $FEB$ and $DGC$, respectively. Since $PQ$ is the radical axis of $\\omega_B$ and $\\omega_C$, it is sufficient to prove that the powers of $A$ with respect to $\\omega_B$ and $\\omega_C$ are equal. If we denote by $X$ the second intersection of $\\omega_B$ and $AC$, and by $Y$ the second intersection of $\\omega_C$ and $AB$, then this is equivalent to\n\n$$\nAX \\cdot AE = AY \\cdot AD.\n$$\n\nLines $DE$ and $BC$ are parallel, yielding $\\frac{AB}{AD} = \\frac{AC}{AE}$, so the above is equivalent to\n\n$$\nAX \\cdot AC = AY \\cdot AB.\n$$\n\nThis, in turn, is equivalent to $B, Y, X, C$ being concyclic. From angles in $\\omega_B$ and $\\omega_C$ we have\n\n$$\n\\angle BYC = \\angle DYC = \\angle DGC \\quad \\text{and} \\quad \\angle BXC = \\angle BXE = \\angle BFE.\n$$\n\nOn the other hand, trapezoid $BFGC$ is inscribed in the circumcircle of $ABC$, so it is isosceles, hence $\\angle BFE = \\angle DGC$. We conclude that $\\angle BYC = \\angle BXC$, so $B, Y, X, C$ are indeed concyclic and we are done.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20596,
"subject": "Mathematics (Olympiad)",
"question": "Let $S$ be the set of all $2007$-digit decimal integers of the form $\\overline{2a_1a_2a_3\\cdots a_{2006}}$ such that the sequence $a_1, a_2, a_3, \\ldots, a_{2006}$ contains an odd number of digits equal to $9$. What is the cardinality of $S$?",
"options": [],
"answer": "See solution",
"solution": "Let $A$ be the number of elements in $S$.\n\nThe number of ways to choose an odd number of positions for the digit $9$ among $2006$ places is:\n\n$$\nA = \\binom{2006}{1} 9^{2005} + \\binom{2006}{3} 9^{2003} + \\dots + \\binom{2006}{2005} 9.\n$$\n\nRecall the binomial expansions:\n\n$$\n(9+1)^{2006} = \\sum_{k=0}^{2006} \\binom{2006}{k} 9^{2006-k}\n$$\n\nand\n\n$$\n(9-1)^{2006} = \\sum_{k=0}^{2006} \\binom{2006}{k} (-1)^k 9^{2006-k}.\n$$\n\nSubtracting the second from the first and dividing by $2$ gives the sum over odd $k$:\n\n$$\n\\begin{aligned}\nA &= \\frac{1}{2} \\left( (10^{2006}) - (8^{2006}) \\right).\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20597,
"subject": "Mathematics (Olympiad)",
"question": "A positive integer $N$ is called a *lighthouse number* if it has the following property: any positive integer not exceeding $N$ which is relatively prime to it has at most two prime factors (possibly equal). Determine whether the lighthouse numbers are infinitely or finitely many and, in the latter case, find the greatest among them.",
"options": [],
"answer": "See solution",
"solution": "*Answer:* The greatest lighthouse number is $1260$.\n\nDenote by $p_n$ the $n$-th prime number. We first set out to prove that, for $n \\ge 5$, the inequality\n\n$$\np_{n+1}^3 \\le p_1 \\cdots p_n, \\quad n \\ge 5.\n$$\n\nAs is readily verified by computation, the inequality holds for $n = 5, 6$. Consider now the case $n \\ge 7$. By Bertrand's Postulate, $p_k < 2p_{k-1}$ for each $k$. Consequently,\n\n$$\n\\begin{aligned}\np_{n+1}^3 &< 2^3 p_n^3 = 2^3 p_n \\cdot p_n^2 \\\\\n&< 2^3 p_n \\cdot 2^2 p_{n-1}^2 = 2^3 p_n \\cdot 2^2 p_{n-1} \\cdot p_{n-1} \\\\\n&< 2^3 p_n \\cdot 2^2 p_{n-1} \\cdot 2 p_{n-2} = 64 p_n p_{n-1} p_{n-2} \\\\\n&\\le p_n p_{n-1} \\cdots p_1.\n\\end{aligned}\n$$\n\nThis holds as long as $64 \\le p_1 p_2 \\cdots p_{n-3}$, which is true for $n \\ge 7$.\n\nNow, consider a number $N$ divisible by the primes $p_1, \\dots, p_n$, but not by $p_{n+1}$. Then $N$ is a lighthouse number precisely when $N < p_{n+1}^3$, since $p_{n+1}^3$ is the least number relatively prime to $N$ with more than two prime factors. But $p_1 \\cdots p_n \\le N < p_{n+1}^3$, which, together with the previous inequality, enforces $1 \\le n \\le 4$, so the only possible prime divisors of $N$ are $2, 3, 5, 7$. We examine these cases in detail:\n\nSuppose $n \\le 3$. Then\n\n$$\nN < p_4^3 = 7^3 = 343.\n$$\n\nSuppose $n = 4$. Writing $N = p_1 p_2 p_3 p_4 M$, we find\n\n$$\n210M = p_1 p_2 p_3 p_4 M = N < p_5^3 = 11^3 = 1331,\n$$\nimplying $M \\le 6$, and\n\n$$\nN \\le 210 \\cdot 6 = 1260.\n$$\n\nConsequently, the lighthouse numbers are bounded by $1260$, hence finite in number. Further, the number $1260$ has the lighthouse property, since\n\n$$\np_5^3 = 11^3 = 1331 > 1260.\n$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20598,
"subject": "Mathematics (Olympiad)",
"question": "Each of 4 players rolls a standard 6-sided die. The winner is the player who rolls the highest number. If there is a tie for the highest roll, those involved in the tie roll again, repeating until one player wins. Hugo is one of the players. What is the probability that Hugo's first roll was a 5, given that he won the game?\n\n\n\n(A) $\\frac{61}{216}$ (B) $\\frac{367}{1296}$ (C) $\\frac{41}{144}$ (D) $\\frac{185}{648}$ (E) $\\frac{11}{36}$",
"options": [],
"answer": "See solution",
"solution": "Let $N$ be the value of Hugo's first roll.\n\nIf $q$ players tie on the initial roll, the probability that any one of these $q$ players will ultimately win is $\\frac{1}{q}$.\n\nConsider four cases for Hugo's first roll:\n\n- Hugo rolls higher than all others:\n $$\n \\frac{(N - 1)^3}{6^3} = \\frac{N^3 - 3N^2 + 3N - 1}{216}\n $$\n- Hugo ties one other player, beats two, and wins:\n $$\n 3 \\cdot \\frac{1}{6} \\cdot \\frac{(N - 1)^2}{6^2} \\cdot \\frac{1}{2} = \\frac{N^2 - 2N + 1}{144}\n $$\n- Hugo ties two, beats one, and wins:\n $$\n 3 \\cdot \\frac{1}{6^2} \\cdot \\frac{N-1}{6} \\cdot \\frac{1}{3} = \\frac{N-1}{216}\n $$\n- Hugo ties all three and wins:\n $$\n \\frac{1}{6^3} \\cdot \\frac{1}{4} = \\frac{1}{864}\n $$\n\nSum:\n$$\n\\frac{4N^3 - 6N^2 + 4N - 1}{864}\n$$\n\nFor $N = 1$ to $6$, this yields $\\frac{1}{864}$, $\\frac{15}{864}$, $\\frac{65}{864}$, $\\frac{175}{864}$, $\\frac{369}{864}$, $\\frac{671}{864}$.\n\nThus, the probability Hugo rolled a 5 given he won is:\n$$\n\\frac{369}{1 + 15 + 65 + 175 + 369 + 671} = \\frac{369}{1296} = \\frac{41}{144}\n$$\n\nAlternatively, by Bayes' Theorem:\n$$\nP(5 | W) = \\frac{P(W | 5) \\cdot P(5)}{P(W)} = \\frac{\\frac{41}{96} \\cdot \\frac{1}{6}}{\\frac{1}{4}} = \\frac{41}{144}\n$$\n\nGeneral formula for $k$ players and $n$-sided die:\n$$\nP(\\text{Hugo's first roll was } m \\mid \\text{Hugo won}) = \\frac{m^k - (m-1)^k}{n^k}\n$$\nFor this problem: $\\frac{5^4-4^4}{6^4} = \\frac{41}{144}$.\n\n**Proof 1: Algebra**\nFor $0 \\le j \\le k-1$, let $A_j$ be the event that $j$ of the other $k-1$ players rolled $m$, others less than $m$:\n$$\nP(A_j) = \\binom{k-1}{j} \\left(\\frac{1}{n}\\right)^j \\left(\\frac{m-1}{n}\\right)^{k-1-j}\n$$\nWinning probability:\n$$\n\\sum_{j=0}^{k-1} \\frac{1}{j+1} \\binom{k-1}{j} \\left(\\frac{1}{n}\\right)^j \\left(\\frac{m-1}{n}\\right)^{k-1-j}\n$$\nBy binomial theorem:\n$$\n\\frac{m^k - (m-1)^k}{k n^{k-1}}\n$$\nBayes' theorem gives:\n$$\n\\frac{m^k - (m-1)^k}{n^k}\n$$\n\n**Proof 2: Combinatorics**\nGiven Hugo won, his roll was at most $m$ iff all rolls were at most $m$:\n$$\nP(\\text{all first rolls } \\le m) = \\left(\\frac{m}{n}\\right)^k\n$$\nSo,\n$$\nP(\\text{Hugo's first roll was } m | \\text{Hugo won}) = \\frac{m^k - (m-1)^k}{n^k}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20599,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge 2$ be an integer. Find all sets of real numbers $(a_1, a_2, \\dots, a_n)$ such that $a_1 - 2a_2, a_2 - 2a_3, \\dots, a_{n-1} - 2a_n, a_n - 2a_1$ is a permutation of $a_1, a_2, \\dots, a_n$.\n\nNote that $a_1, a_2, \\dots, a_n$ itself is also a permutation of $a_1, a_2, \\dots, a_n$.",
"options": [],
"answer": "See solution",
"solution": "Let $a_{n+1} = a_1$, $a_0 = a_n$, and let $M$ and $m$ denote the maximum and minimum values of $a_1, a_2, \\dots, a_n$, respectively. Remark that the maximum and minimum values of $a_1 - 2a_2, a_2 - 2a_3, \\dots, a_n - 2a_1$ are also $M$ and $m$, respectively. Take $s$ satisfying $a_s = m$. Then, from $M \\ge a_{s-1} - 2a_s \\ge m - 2m$, we have $M + m \\ge 0$. Therefore, if we choose $t$ satisfying $a_t = M$, then from $a_{t-1} - 2a_t \\ge m$, it follows that $a_{t-1} \\ge 2a_t + m = 2M + m \\ge M$, implying $a_{t-1} = M$. Hence, by induction, we conclude that $a_i = M$ for any $i$. Consequently, from the assumption, we have $M = M - 2M$, which implies $M = 0$. Thus, it is necessary for $a_i = 0$ for any $i$. Conversely, the problem assumption is satisfied for this case. Therefore, the solution is $(a_1, a_2, \\dots, a_n) = (0, 0, \\dots, 0)$.\n\n**Another Solution.** Let $a_{n+1} = a_1$, then from the assumption, we have:\n\n$$\n\\begin{align*}\n0 &= \\sum_{i=1}^{n} (a_i - 2a_{i+1})^2 - \\sum_{i=1}^{n} a_i^2 \\\\\n&= \\sum_{i=1}^{n} a_i^2 - 4 \\sum_{i=1}^{n} a_i a_{i+1} + 4 \\sum_{i=1}^{n} a_{i+1}^2 - \\sum_{i=1}^{n} a_i^2 \\\\\n&= 2 \\left( \\sum_{i=1}^{n} a_i^2 - 2 \\sum_{i=1}^{n} a_i a_{i+1} + \\sum_{i=1}^{n} a_{i+1}^2 \\right) \\\\\n&= 2 \\sum_{i=1}^{n} (a_i - a_{i+1})^2\n\\end{align*}\n$$\n\nHence, $a_1 = a_2 = \\dots = a_n$. The subsequent steps are the same as the main solution.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20600,
"subject": "Mathematics (Olympiad)",
"question": "In a triangle $ABC$, denote by $D$, $E$, and $F$ the points where the angle bisectors of $\\angle CAB$, $\\angle ABC$, and $\\angle BCA$ meet its circumcircle, respectively.\n\n1. Prove that the orthocenter of triangle $DEF$ coincides with the incenter of triangle $ABC$.\n\n2. Prove that if $AD + BE + CF = 0$, then the triangle $ABC$ is equilateral.",
"options": [],
"answer": "See solution",
"solution": "1. Let $I$ be the incenter of triangle $ABC$. The points $D$, $E$, and $F$ are the midpoints of the arcs $\\widehat{BC}$, $\\widehat{CA}$, and $\\widehat{AB}$, respectively.\n\nThe angle between lines $AD$ and $EF$ is $\\frac{1}{2}(\\widehat{AE} + \\widehat{DF}) = \\frac{1}{4}(\\widehat{AB} + \\widehat{BC} + \\widehat{CA}) = 90^\\circ$, so $AD \\perp EF$. Similarly, $BE \\perp DF$. Therefore, the orthocenter of $DEF$ is $I$.\n\n2. Let $O$ be the circumcenter of triangle $ABC$. The given relation implies $OA + OB + OC = OD + OE + OF$.\n\nBy Sylvester's theorem, triangles $ABC$ and $DEF$ share the same orthocenter. Thus, in triangle $ABC$, point $I$ is both the incenter and the orthocenter, so $ABC$ is equilateral.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20601,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge 2$ be a positive integer. Alice distributes $2n$ candies into $4n$ boxes $B_1, B_2, \\dots, B_{4n}$. After checking the number of candies Alice puts in each box, Bob chooses $2n$ boxes $B_{k_1}, B_{k_2}, \\dots, B_{k_{2n}}$ out of the $4n$ boxes satisfying the following:\n\n$k_i - k_{i-1} \\in \\{1, 3\\}$ for each $i = 1, 2, \\dots, 2n$, and $k_{2n} = 4n$ (with $k_0 = 0$).\n\nAlice gets all candies in the $2n$ boxes Bob did not choose. If Alice and Bob use their best strategies to take as many candies as possible, how many candies can Alice take?",
"options": [],
"answer": "See solution",
"solution": "The answer is $n$.\n\nIf Alice puts one candy in each of boxes $B_1, B_2, \\dots, B_{4n-1}$, then Bob can choose at most $n$ out of the $2n$ boxes, so Alice can get exactly $n$ candies.\n\nNow we prove that Bob can take at least $n$ candies. Let $b_i$ be the number of candies Alice puts in $B_i$ for $i = 1, 2, \\dots, 4n$.\n\nSuppose there exists $m = 1, 2, \\dots, n$ such that $b_{4m-2} < 2$. We consider the following two sequences:\n\n$$\np_i = \\begin{cases} 1 & i=1 \\\\ 1 & i=2, 4, \\dots, 2m-2 \\\\ 3 & i=3, 5, \\dots, 2m-1 \\\\ 3 & i=2m, 2m+2, \\dots, 2n \\\\ 1 & i=2m+1, 2m+3, \\dots, 2n-1 \\end{cases}, \\quad q_i = \\begin{cases} 3 & i=1, 3, \\dots, 2m-1 \\\\ 1 & i=2, 4, \\dots, 2m-2 \\\\ 3 & i=2m, 2m+2, \\dots, 2n-2 \\\\ 1 & i=2m+1, 2m+3, \\dots, 2n-1 \\\\ 1 & i=2n \\end{cases}\n$$\n\nLet $P_i = p_1 + p_2 + \\dots + p_i$ and $Q_i = q_1 + q_2 + \\dots + q_i$ for $i = 1, 2, \\dots, 2n$. Then, $\\{P_i\\}_{i=1}^{2n}$ and $\\{Q_i\\}_{i=1}^{2n}$ satisfy the condition in the problem, and furthermore $\\{P_1, P_2, \\dots, P_{2n}\\} \\cup \\{Q_1, Q_2, \\dots, Q_{2n}\\} = \\{1, 2, \\dots, 4n\\} \\setminus \\{4m-2\\}$.\n\nSo,\n\n$$\n\\sum_{i=1}^{2n} (b_{P_i} + b_{Q_i}) \\ge 2n - b_{4m-2} \\ge 2n - 1\n$$\n\nwhich implies that either $\\sum_{i=1}^{2n} b_{P_i}$ or $\\sum_{i=1}^{2n} b_{Q_i}$ is at least $n$, so Bob can take at least $n$.\n\nNow we assume that $b_{4m-2} = 2$ for every $m = 1, 2, \\dots, n$. Then, Bob chooses $B_{4k-3}, B_{4k-2}$ for $k = 1, 2, \\dots, n-1$ and $B_{4n-3}, B_{4n}$, where Bob takes $2n - 2 \\ge n$ candies.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20602,
"subject": "Mathematics (Olympiad)",
"question": "Petro has to plant 8 trees in a row: apple trees or oak trees. There is one restriction: there must be no apple trees between any two oak trees. For example, plantings like $\\text{AAOOAOAA}$ or $\\text{OAOAAAAA}$ are not allowed, but $\\text{AAOOAAAA}$ is allowed. How many different plantings are possible?\n\n",
"options": [],
"answer": "See solution",
"solution": "All oak trees must be planted together as a single group, regardless of where this group is placed. Let's count the possible numbers of oaks:\n\n- If there are no oak trees, there is only one way to plant.\n- If there are $k$ oaks, where $1 \\leq k \\leq 8$, there are $9 - k$ possible positions for the group of oaks.\n\nFor example: $\\text{OOOAAAAA}$, $\\text{AOOOAAAA}$, $\\ldots$, $\\text{AAAAAOOO}$.\n\nThe total number of plantings is:\n$$1 + 1 + 2 + 3 + \\ldots + 8 = 37.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20603,
"subject": "Mathematics (Olympiad)",
"question": "Does there exist an integer $a$ such that\n\n$$\n\\frac{1}{\\sqrt{2024}} < \\frac{1}{\\sqrt{a+1}} + \\frac{1}{\\sqrt{a+2}} + \\dots + \\frac{1}{\\sqrt{a+2023}} < \\frac{1}{\\sqrt{2023}}?\n$$",
"options": [],
"answer": "See solution",
"solution": "For each $i = 1, 2, \\dots, 2023$, we have\n\n$$\n\\frac{1}{\\sqrt{2023^3 + i}} < \\frac{1}{\\sqrt{2023^3}} = \\frac{1}{2023\\sqrt{2023}},\n$$\n\nthus\n\n$$\n\\frac{1}{\\sqrt{2023^3+1}} + \\frac{1}{\\sqrt{2023^3+2}} + \\dots + \\frac{1}{\\sqrt{2023^3+2023}} < 2023 \\cdot \\frac{1}{2023\\sqrt{2023}} = \\frac{1}{\\sqrt{2023}}.\n$$\n\nSimilarly, by estimating all fractions from below, we see that\n\n$$\n\\begin{aligned}\n& \\frac{1}{\\sqrt{2023^3+1}} + \\frac{1}{\\sqrt{2023^3+2}} + \\dots + \\frac{1}{\\sqrt{2023^3+2023}} \\\\\n& > 2023 \\cdot \\frac{1}{\\sqrt{2023^3+2023^2}} = \\frac{2023}{2023\\sqrt{2024}} = \\frac{1}{\\sqrt{2024}}\n\\end{aligned}\n$$\n\nTherefore, $a = 2023^3$ is suitable.\n\n**Note:** The value $a = 2023^3$ used in the solution is not the only possibility. The same solution works whenever $2023^3 \\le a \\le 2023^3 + 2023^2 - 2023$. There are even more integers that satisfy these inequalities.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20604,
"subject": "Mathematics (Olympiad)",
"question": "Two externally tangent unit circles are given in the plane. Consider any rectangle (or square) containing both circles such that each side of the rectangle is tangent to at least one circle. Find the largest and the smallest possible area of such a rectangle.",
"options": [],
"answer": "See solution",
"solution": "Denote the circles by $k_1, k_2$, their radius by $r = 1$, and their centers by $O_1, O_2$, respectively. Let $ABCD$ be one such rectangle (or square), and without loss of generality, assume that the sides $AB, BC$ are tangent to $k_1$ while the sides $CD, DA$ are tangent to $k_2$. Let $P$ be the intersection of a line through $O_1$ parallel to $AB$ and a line through $O_2$ parallel to $BC$. Finally, let $\\phi = \\angle PO_1O_2$ with $\\phi \\in [0, \\frac{1}{4}\\pi]$.\n\n$$\n[ABCD] = AB \\cdot BC = (2r + 2r \\cos \\phi)(2r + 2r \\sin \\phi) = 4(1 + \\sin \\phi)(1 + \\cos \\phi).\n$$\n\nIt remains to analyze the expression $V(\\phi) = (1 + \\sin \\phi)(1 + \\cos \\phi)$ for $\\phi \\in [0, \\frac{1}{4}\\pi]$. Multiplying out, this rewrites as\n\n$$\nV(\\phi) = 1 + \\sin \\phi + \\cos \\phi + \\sin \\phi \\cos \\phi = \\frac{1}{2} + (\\sin \\phi + \\cos \\phi) + \\frac{1}{2}(\\sin \\phi + \\cos \\phi)^2.\n$$\n\n\n\nNow, let $u = \\sin \\phi + \\cos \\phi$. Squaring and using the formula $\\sin 2\\phi = 2\\sin \\phi \\cos \\phi$, we obtain $1 \\leq u \\leq \\sqrt{2}$. Since the function $V(\\phi) = \\frac{1}{2} + u + \\frac{1}{2}u^2$ is increasing on the interval $[1, \\sqrt{2}]$, we get $2 \\leq V(\\phi) \\leq \\frac{3}{2} + \\sqrt{2}$ and finally\n\n$$\n8 \\leq [ABCD] \\leq 6 + 4\\sqrt{2}.\n$$\n\nThe first inequality is sharp for $\\phi = 0$, that is, if both *AB* and *CD* are tangent to both circles. The second inequality is sharp for $\\phi = \\frac{1}{4}\\pi$, that is, if $ABCD$ is a square.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20605,
"subject": "Mathematics (Olympiad)",
"question": "Find the inverse of the matrix equation\n\n$$\n\\frac{A}{4} = I - X,\n$$\n\nwhere all entries in $X$ are either $0$ or $\\frac{1}{4}$.",
"options": [],
"answer": "See solution",
"solution": "We use the series\n\n$$\n(I - X)^{-1} = I + X + X^2 + X^3 + \\dots\n$$\n\nFirst, we prove that this series converges. It suffices to show that the maximum $M$ such that $\\|Xw\\| \\leq M\\|w\\|$ for all column vectors $w$ of $X$ is less than $1$. This ensures the sum of the entries always decreases by a factor smaller than $1$ when multiplying a vector by $X^2$; then we sum the series as $(I+X)(I+X^2+X^4+\\dots)$. Every row of $X$ has at most four nonzero entries, all equal to $\\frac{1}{4}$. If $w = (a_1, a_2, \\dots, a_n)$, every entry of $Xw$ is of the form $\\frac{a_{r_1}+a_{r_2}+\\dots+a_{r_s}}{4}$, $s \\leq 4$. By the Cauchy-Schwarz inequality, its square is at most $\\frac{s}{16}(a_{r_1}^2+a_{r_2}^2+\\dots+a_{r_s}^2) \\leq \\frac{a_{r_1}^2+a_{r_2}^2+\\dots+a_{r_s}^2}{4}$, with equality if and only if $a_{r_1} = a_{r_2} = \\dots = a_{r_s}$ and $s=4$. Summing over all rows, the sum of squares of the coordinates of $Xw$ is at most $a_1^2 + a_2^2 + \\dots + a_n^2 = \\|w\\|^2$, because all columns of $X$ have at most four nonzero entries. But equality would only happen if all entries $a_i$ are equal and $s=4$ always, which does not happen for, say, the first row. So $M < 1$ and the series converges.\n\nConsider the graph whose vertices are $v_1, v_2, \\dots, v_n$ and connect $v_i$ and $v_j$ if and only if the entry $x_{ij}$ in $X$ is $\\frac{1}{4}$. By the definition of $A$, this graph has a lattice-like configuration: it can be split into several paths $v_1v_2 \\dots v_{n_1-1}$, $v_{k_1}v_{k_1+1} \\dots v_{(k+1)n_1-1}$, $1 \\leq k \\leq \\frac{n}{2}-1$, $v_r v_{r+n_1} v_{r+2n_1} \\dots v_{r+n-n_1}$, $1 \\leq r \\leq n_1$. This graph is connected. Since the entry $m_{ij}$ in $X^k$ is nonzero if and only if there exists a circuit from $i$ to $j$ with $k$ edges, for all $i, j$ there is $k$ such that the corresponding entry $m_{ij}$ in $X^k$ is nonzero. This proves that all entries in the inverse of $A$ are positive.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 20606,
"subject": "Mathematics (Olympiad)",
"question": "Do there exist positive integers $n, k$ such that\n$$\n\\frac{n}{11^k - n}\n$$\nis a square of an integer?",
"options": [],
"answer": "See solution",
"solution": "Such numbers don't exist. For the sake of contradiction, assume that there exist positive integers $n, k, a$ such that\n$$\n\\frac{n}{11^k - n} = a^2\n$$\nwhich rewrites as\n$$\nn(a^2 + 1) = a^2 \\cdot 11^k.\n$$\nFrom $\\text{GCD}(a^2, a^2+1) = 1$ we deduce $a^2+1 \\mid 11^k$ and hence $a^2+1 = 11^t$ for $1 \\le t \\le k$. In particular, $a^2 \\equiv 10 \\pmod{11}$. However, this is impossible as the squares of integers give remainders $0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1, \\dots$ upon division by $11$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20607,
"subject": "Mathematics (Olympiad)",
"question": "Se considera un polígono regular de 90 vértices, numerados del 1 al 90 de manera aleatoria. Probar que siempre podemos encontrar dos vértices consecutivos cuyo producto es mayor o igual que $2014$.",
"options": [],
"answer": "See solution",
"solution": "Consideremos el primer par de números consecutivos cuyo producto es mayor o igual que $2014$, que son el $45$ y el $46$. Por lo tanto, para que no se cumpliera el enunciado, los números que deben ir a izquierda y derecha de los vértices numerados del $46$ al $90$ tendrían que ser menores o iguales que $44$. Sin embargo, entre los vértices numerados del $46$ al $90$ hay, al menos, $45$ vértices.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20608,
"subject": "Mathematics (Olympiad)",
"question": "In a mathematical olympiad, students received marks in each of four areas: algebra, geometry, number theory, and combinatorics. Any two students have distinct marks in all four areas. A group of students is called *nice* if all students in the group can be ordered in increasing order simultaneously in at least two of the four areas. Find the least positive integer $N$ such that, among any $N$ students, there exists a nice group of ten students.",
"options": [],
"answer": "See solution",
"solution": "The answer is $730$.\n\n**Lemma.** A sequence $A = a_1, a_2, \\dots, a_k$ of distinct numbers does not possess a 10-term increasing subsequence if and only if its terms can be colored in 9 colors such that the members of the same color form a decreasing sequence.\n\n*Proof.* First, assume such a coloring exists. Any 10-term subsequence of $A$ contains two terms of the same color, so the sequence is not increasing.\n\nSuppose now $A$ does not contain a 10-term increasing subsequence. Color in color $i$ any term $a$ of $A$ such that the longest increasing subsequence of $A$ ending at $a$ has length $i$. This coloring has the desired properties. This completes the proof of the lemma.\n\nWe prove that among any $730$ students, there exist $10$ that form a nice group.\n\nLet $M_1, M_2, \\dots, M_{730}$ be the students ordered by increasing algebra marks. Let $a_i$ be the geometry mark of $M_i$. If the sequence $a_1, a_2, \\dots, a_{730}$ has a 10-term increasing subsequence, then we have a nice group.\n\nOtherwise, by the lemma, we can color all students in 9 colors so that the geometry marks for any color form a decreasing sequence. There are at least $82$ students of the same color. Let $N_1, N_2, \\dots, N_{82}$ be these students, ordered by increasing algebra and decreasing geometry marks.\n\nLet $b_i$ be the number theory mark for $N_i$. If $b_1, b_2, \\dots, b_{82}$ has a 10-term decreasing subsequence, we have the desired nice group. Otherwise, we can color $N_1, \\dots, N_{82}$ in 9 colors so that the number theory marks for any color form an increasing sequence. There are at least $10$ students of the same color, and they form a nice group with respect to algebra and number theory.\n\nIt remains to show an example of $729$ students without a 10-term nice group.\n\nLet $k$ be an integer between $0$ and $728$. For $0 \\leq i < j \\leq 2$, let $f_{ij}(k)$ be the integer from the base-9 representation of $k$ when the digits in the $i$-th and $j$-th places are replaced by their complements to $8$. (If $k \\leq 80$, pad with zeros on the left.)\n\nConsider $729$ students with algebra marks $0, 1, \\ldots, 728$, and the student with mark $k$ has geometry mark $f_{01}(k)$, number theory mark $-f_{02}(k)$, and combinatorics mark $-f_{12}(k)$.\n\nFor any two areas, there exist $0 \\leq i < j \\leq 2$ such that, for any student, one of the marks for these two areas is obtained from the other via $f_{ij}$.\n\nWe show that a 10-term nice group for algebra and geometry does not exist. For the remaining pairs, the proof is similar.\n\nLet $M_1, M_2, \\dots, M_{729}$ be the students ordered by algebra marks. Let $a_i$ be the geometry mark of $M_i$. It suffices to show that $a_1, a_2, \\dots, a_{729}$ has no 10-term increasing subsequence.\n\nFor any $i$, color $a_i$ by the second digit of its base-9 representation. Any monochromatic subsequence of $a_1, \\dots, a_{729}$ is decreasing. Hence, by the lemma, the proof is complete.",
"topic": "Number Theory",
"subtopic": "Other"
},
{
"id": 20609,
"subject": "Mathematics (Olympiad)",
"question": "Гурвалжин $ABC$ нь хурц өнцөгт бөгөөд $w$ тойрог нь $BC$ талыг $K$ цэгт шүргэнэ. $AD$ нь өндөр бөгөөд $AD$-ийн дундаж цэгийг $M$ гэе. $N$ нь $w$ болон $KM$ шулууны огтлолцлын хоёр дахь цэг бөгөөд $BCN$ гурвалжныг багтаасан тойрог ба $w$ тойргууд $N$ цэгт огтлолцоно гэдгийг батал.",
"options": [],
"answer": "See solution",
"solution": "Хэрэв $AB = AC$ бол батлах зүйл илт үнэн. $AB < AC$ гэж үзье. $BC$-ийн дундаж цэгийг $A'$ гэе. $BC$ дээр босгосон перпендикуляр шулуун $NK$-тай $P$ цэгт огтлолцог. $BCN$ гурвалжныг багтаасан тойргийн төвийг $S$ гэе. Хэрэв $N$, $I$, $S$ цэгүүд нэг шулуун дээр оршиж байгааг баталбал бодлого шийдэгдэнэ (энд $I$ нь $\triangle ABC$-д багтсан тойргийн төв). Энэ нь мөн $P$ цэг $\triangle ABC$-ийг багтаасан тойрог дээр оршиж байгааг батлахад хангалттай. Учир нь $SP = SN$, $IK \bot BC$, $SP \bot BC$ тул $\triangle PSN = \triangle NPS = \triangle NKI = \triangle PNI$ учраас $N$, $I$, $S$ шулуун дээр оршино. Иймд $NK \\cdot KP = BK \\cdot KC$ гэдгийг батлахад хангалттай.\n\n$$\nNK = p - b, \\quad KC = p - c \\text{ байх нь илэрхий.}\n$$\n\n$$\n\\angle D = c \\cdot \\cos \\beta = \\frac{c^2 + a^2 - b^2}{2a}, \\quad KA' = BA' - BK = \\frac{1}{2}(b-c),\n$$\n\n$$\nDK = BK - BD = \\frac{(b-c)(p-a)}{a}, \\quad \\angle MKD = \\varphi \\text{ гэе.}\n$$\n\n$$\n\\tan \\varphi = \\frac{MD}{DK} = \\frac{\\frac{1}{2}AD \\cdot a}{(b-c)(p-a)} = \\frac{S_{ABC}}{(b-c)(p-a)}\n$$\n\n$\\angle NIK = \\varphi$ тул $NK = 2\\tau \\sin \\varphi$, $\\triangle A'KP$-аас $KP = KA' \\cdot \\sec \\varphi$.\n\nИймд\n\n$$\nNK \\cdot KP = 2\\tau (KA') \\tan \\varphi = \\frac{r \\cdot S_{ABC}}{p-a}\n$$\n\n$$\n= \\frac{(S_{ABC})^2}{p(p-a)} = (p-b)(p-c) = BK \\cdot KC.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20610,
"subject": "Mathematics (Olympiad)",
"question": "The positive real numbers $a$, $b$, $c$ satisfy $a + b + c = 1$. For every natural number $n$, find the minimal possible value of the expression\n\n$$\nE = \\frac{a^{-n} + b}{1 - a} + \\frac{b^{-n} + c}{1 - b} + \\frac{c^{-n} + a}{1 - c}\n$$",
"options": [],
"answer": "See solution",
"solution": "We transform the first term of $E$ as follows:\n\n$$\n\\frac{a^{-n} + b}{1 - a} = \\frac{1 + a^n b}{a^n (b + c)} = \\frac{a^{n+1} + a^n b + 1 - a^{n+1}}{a^n (b + c)} = \\frac{a^n (a + b) + (1 - a)(1 + a + a^2 + \\dots + a^n)}{a^n (b + c)}\n$$\n\n$$\n= \\frac{a^n (a + b)}{a^n (b + c)} + \\frac{(b + c)(1 + a + a^2 + \\dots + a^n)}{a^n (b + c)} = \\frac{a + b}{b + c} + 1 + \\frac{1}{a} + \\frac{1}{a^2} + \\dots + \\frac{1}{a^n}\n$$\n\nAnalogously,\n\n$$\n\\frac{b^{-n} + c}{1 - b} = \\frac{b + c}{c + a} + 1 + \\frac{1}{b} + \\frac{1}{b^2} + \\dots + \\frac{1}{b^n}\n$$\n\n$$\n\\frac{c^{-n} + a}{1 - c} = \\frac{c + a}{a + b} + 1 + \\frac{1}{c} + \\frac{1}{c^2} + \\dots + \\frac{1}{c^n}\n$$\n\nSo,\n\n$$\nE = \\frac{a+b}{b+c} + \\frac{b+c}{c+a} + \\frac{c+a}{a+b} + 3 + \\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right) + \\left(\\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{c^2}\\right) + \\dots + \\left(\\frac{1}{a^n} + \\frac{1}{b^n} + \\frac{1}{c^n}\\right)\n$$\n\nBy the inequality\n\n$$\n\\frac{a+b}{b+c} + \\frac{b+c}{c+a} + \\frac{c+a}{a+b} \\ge 3\\sqrt[3]{\\frac{(a+b)(b+c)(c+a)}{(b+c)(c+a)(a+b)}} = 3\n$$\n\nand using the power mean inequality for $k = -1, -2, \\dots, -n$,\n\n$$\n\\frac{1}{a^m} + \\frac{1}{b^m} + \\frac{1}{c^m} \\ge 3^{m+1} \\quad \\text{for } m = 1, 2, \\dots, n\n$$\n\nThus,\n\n$$\nE \\ge 3 + 3 + 3^2 + \\dots + 3^{n+1} = 2 + \\frac{3^{n+2} - 1}{2} = \\frac{3^{n+2} + 3}{2}\n$$\n\nFor $a = b = c = \\frac{1}{3}$, we obtain $E = \\frac{3^{n+2} + 3}{2}$. Therefore, the minimal value is\n\n$$\n\\min E = \\frac{3^{n+2} + 3}{2}.\n$$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20611,
"subject": "Mathematics (Olympiad)",
"question": "Find the maximum value of $k$ such that for all $a, b, c$ which are the side lengths of a triangle,\n\n$$\nk \\cdot \\frac{3(a^2 + b^2 + c^2)}{(a + b + c)^2} + \\sqrt{\\frac{ab + bc + ca}{a^2 + b^2 + c^2}} \\leq k + 1.\n$$",
"options": [],
"answer": "See solution",
"solution": "The given inequality can be rewritten as\n\n$$\n2k \\cdot \\frac{(a^2 + b^2 + c^2) - (ab + bc + ca)}{(a + b + c)^2} \\leq \\frac{\\sqrt{a^2 + b^2 + c^2} - \\sqrt{ab + bc + ca}}{\\sqrt{a^2 + b^2 + c^2}}\n$$\n\nor\n\n$$\n2k \\cdot \\frac{(a^2 + b^2 + c^2) - (ab + bc + ca)}{(a + b + c)^2} \\leq \\frac{(a^2 + b^2 + c^2) - (ab + bc + ca)}{\\sqrt{a^2 + b^2 + c^2}((a^2 + b^2 + c^2) + (ab + bc + ca))}.\n$$\n\nNote that\n\n$$\n(a^2 + b^2 + c^2) - (ab + bc + ca) = \\frac{1}{2} \\left[ (a - b)^2 + (b - c)^2 + (c - a)^2 \\right] \\geq 0\n$$\n\nso\n\n$$\n2k\\sqrt{a^2 + b^2 + c^2} \\left( \\sqrt{a^2 + b^2 + c^2} + \\sqrt{ab + bc + ca} \\right) \\leq (a + b + c)^2.\n$$\n\nConsider $a = b = 1$ and $c \\to 0^+$ (isosceles triangle with the base arbitrarily small), then $\\text{LHS} \\to 2k \\cdot \\sqrt{2}(\\sqrt{2} + 1)$ and $\\text{RHS} \\to 4$. Thus,\n\n$$\nk \\leq \\frac{4}{2\\sqrt{2}(\\sqrt{2} + 1)} = 2 - \\sqrt{2}.\n$$\n\nFor $k = 2 - \\sqrt{2}$, we need to prove that\n\n$$\n(4 - 2\\sqrt{2})\\sqrt{a^2 + b^2 + c^2} \\left( \\sqrt{a^2 + b^2 + c^2} + \\sqrt{ab + bc + ca} \\right) \\leq (a + b + c)^2.\n$$\n\nLet $x = \\sqrt{a^2 + b^2 + c^2}$ and $y = \\sqrt{ab + bc + ca}$, then the above can be written as\n\n$$\n(4 - 2\\sqrt{2})x(x + y) \\leq x^2 + 2y^2\n$$\n\n$$\nx^2 + 2(2 + \\sqrt{2})xy - (2 + \\sqrt{2})^2 y^2 \\leq 0.\n$$\n\nThis is equivalent to $x \\leq y\\sqrt{2}$ or $a^2 + b^2 + c^2 \\leq 2(ab + bc + ca)$. The last inequality is true for $a, b, c$ as triangle side lengths since it can be written as\n\n$$\na(b + c - a) + b(c + a - b) + c(a + b - c) \\geq 0.\n$$\n\nHence, $k_{\\max} = 2 - \\sqrt{2}$.\n\n_Remark_: We can use Ravi's substitution for $a, b, c$ and then apply derivatives to find the upper bound for $k$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20612,
"subject": "Mathematics (Olympiad)",
"question": "令 $\\mathcal{X}$ 為正整數集 $\\mathbb{N}$ 的所有非空子集（不一定有限）所組成的集合。試求所有函數 $f: \\mathcal{X} \\to \\mathbb{R}^+$ 滿足以下性質：\n\n1. 若 $S \\subseteq T$ 皆為 $\\mathbb{N}$ 的非空子集，則 $f(T) \\le f(S)$；\n2. 對於所有 $S, T \\in \\mathcal{X}$，\n\n$$\nf(S) + f(T) \\le f(S + T), \\quad f(S)f(T) = f(S \\cdot T),\n$$\n\n其中 $S + T = \\{s + t \\mid s \\in S, t \\in T\\}$，$S \\cdot T = \\{s \\cdot t \\mid s \\in S, t \\in T\\}$。",
"options": [],
"answer": "See solution",
"solution": "設 $f(S) = (\\min S)^\\alpha$，對所有 $S \\in \\mathcal{X}$，其中 $\\alpha \\ge 1$。易知所有這類函數都是解。以下證明僅有這樣的函數滿足所有條件。\n\n因為 $\\{1\\} \\cdot \\{1\\} = \\{1\\}$，$\\mathbb{N} \\cdot \\mathbb{N} = \\mathbb{N}$，所以\n\n$$\nf(\\{1\\})^2 = f(\\{1\\}), \\quad f(\\mathbb{N})^2 = f(\\mathbb{N}) \\implies f(\\{1\\}) = f(\\mathbb{N}) = 1.\n$$\n\n對於所有包含 1 的集合 $S$，由條件 (1)，$1 = f(\\mathbb{N}) \\le f(S) \\le f(\\{1\\}) = 1$，即 $f(S) = 1$。\n\n對於正整數 $n$，定義 $g(n) = f(\\{n\\})$，那麼條件 (2) 告訴我們\n\n$$\ng(m) + g(n) \\le g(m + n), \\quad g(mn) = g(m)g(n).\n$$\n\n前式會得到 $g$ 是一個 $\\mathbb{N}$ 上的嚴格遞增函數。取 $\\alpha$ 使得 $g(2) = 2^\\alpha$，則 $2^\\alpha = g(2) \\ge g(1) + g(1) = 2$，會告訴我們 $\\alpha \\ge 1$。\n\n以下證明 $g(n) = n^\\alpha$：當 $n = 1, 2$ 時是顯然的。對於任意 $n \\ge 3$ 及正整數 $k$，取正整數 $\\ell$ 使得 $2^\\ell \\le n^k < 2^{\\ell+1}$。因為 $g$ 是完全積性函數，所以\n\n$$\n2^{\\ell\\alpha} \\le g(n)^k < 2^{(\\ell+1)\\alpha} \\implies 2^{\\ell/k} \\le g(n)^{1/\\alpha} < 2^{(\\ell+1)/k}.\n$$\n\n由於 $2^{\\ell/k} \\le n < 2^{(\\ell+1)/k}$，我們知道\n\n$$\n2^{-1/k} < \\frac{2^{\\ell/k}}{n} \\le \\frac{g(n)^{1/\\alpha}}{n} < \\frac{2^{(\\ell+1)/k}}{n} \\le 2^{1/k}.\n$$\n\n因為這對於所有的正整數 $k$ 都成立，$g(n)^{1/\\alpha}$ 一定要是 $n$，即 $f(\\{n\\}) = g(n) = n^\\alpha$。\n\n對於任意不包含 1 的集合 $S$，令 $m = \\min S$。因為 $\\{m\\} \\subseteq S$，所以由條件 (1) 及上式，\n\n$$\nf(S) \\le f(\\{m\\}) = m^{\\alpha}. \\qquad (1)\n$$\n\n由於 $1 = m - (m-1) \\in S - \\{m-1\\} := \\{s - (m-1) \\mid s \\in S\\}$，所以\n\n$$\nf(S) \\ge f(\\{m-1\\}) + f(S - \\{m-1\\}) = (m-1)^{\\alpha} + 1.\n$$\n\n在上式中將 $S$ 換成 $S^k := \\{s_1 \\cdots s_k \\mid s_1, \\dots, s_k \\in S\\}$，我們得到\n\n$$\nf(S)^k = f(S^k) \\ge (\\min(S^k) - 1)^\\alpha + 1 = (m^k - 1)^\\alpha + 1.\n$$\n\n因此當 $k > \\alpha$ 時，\n\n$$\nf(S) > (m^k - 1)^{\\alpha/k} = m^{\\alpha} \\left(1 - \\frac{1}{m^k}\\right)^{\\alpha/k} \\ge m^{\\alpha} \\left(1 - \\frac{1}{m^k}\\right).\n$$\n\n因為這對於所有的正整數 $k > \\alpha$ 都成立，所以 $f(S) \\ge m^\\alpha$。結合 (1) 便有\n\n$$\nf(S) = m^{\\alpha} = (\\min S)^{\\alpha}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20613,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b$ be real numbers such that $a + b > 2$. Prove that the system of inequalities\n\n$$\n(a-1)x + b < x^2 < a x + (b-1)\n$$\n\nhas infinitely many real solutions $x$.",
"options": [],
"answer": "See solution",
"solution": "We rewrite the system as\n\n$$\nF(x) > 0 \\quad \\text{and} \\quad G(x) < 0,\n$$\n\nwhere $F(x) = x^2 - (a-1)x - b$ and $G(x) = x^2 - a x - b + 1$. Observe that $F(x) - G(x) = x - 1$.\n\nThe condition $a + b > 2$ implies that\n\n$$\nF(1) = G(1) = 2 - a - b < 0,\n$$\n\nso $x = 1$ is not a solution. However, $G(1) < 0$ implies that the quadratic equation $G(x) = 0$ has a root $x_0 > 1$. Then\n\n$$\nF(x_0) = F(x_0) - G(x_0) = x_0 - 1 > 0.\n$$\n\nFrom $F(1) < 0$ and $F(x_0) > 0$, there exists a root $x_1$ of $F(x) = 0$ in the open interval $(1, x_0)$. Since\n\n$$\nF(1) < 0, \\quad F(x_1) = 0, \\quad G(1) < 0, \\quad G(x_0) = 0,\n$$\n\nany $x \\in (x_1, x_0)$ is a solution to the original system.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20614,
"subject": "Mathematics (Olympiad)",
"question": "Prove that, for any natural number $n$, either $3^{2n} - 3^{n+1} + 3^n - 3$ or $3^{2n} - 3^{n+1} + 3^n + 1$ is divisible by $32$.",
"options": [],
"answer": "See solution",
"solution": "Note that $3^{2n} - 3^{n+1} + 3^n - 3 = (3^n - 3)(3^n + 1)$. If $n$ is odd, then $3^n - 3 = 3(3^{n-1} - 1) = 3\\left(3^{\\frac{n-1}{2}} - 1\\right)\\left(3^{\\frac{n-1}{2}} + 1\\right)$. As all powers of $3$ are odd, $3^{\\frac{n-1}{2}} - 1$ and $3^{\\frac{n-1}{2}} + 1$ are consecutive even numbers. One of these numbers must be divisible by $4$, whence their product is divisible by $8$. Thus $8 \\mid 3^n - 3$, implying that $4 \\mid 3^n + 1$. Consequently, $32 \\mid 3^{2n} - 3^{n+1} + 3^n - 3$.\n\nAssume now $n$ is even. Note that\n\n$$\n3^{2n} - 3^{n+1} + 3^n + 1 = 3^{2n} - 3 \\cdot 3^n + 3^n + 1 = (3^n)^2 - 2 \\cdot 3^n + 1 = (3^n - 1)^2.\n$$\n\nSimilarly to the previous case, $3^n - 1 = (3^{\\frac{n}{2}} - 1)(3^{\\frac{n}{2}} + 1)$, where the factors on the right-hand side are consecutive even numbers. Hence $8 \\mid 3^n - 1$, implying that $64 \\mid 3^{2n} - 3^{n+1} + 3^n + 1$. Consequently, $32 \\mid 3^{2n} - 3^{n+1} + 3^n + 1$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20615,
"subject": "Mathematics (Olympiad)",
"question": "Consider 729 students with algebra marks $0, 1, \\dots, 728$, and the student with mark $k$ has geometry mark $f_{01}(k)$, number theory mark $-f_{02}(k)$, and combinatorics mark $-f_{12}(k)$.\n\nFor any two areas, there exist two numbers $0 \\leq i < j \\leq 2$ such that for any student, one of the marks for these two areas is obtained from the other through the function $f_{ij}$.\n\nProve that a 10-term nice sequence for algebra and geometry does not exist. (For the remaining pairs, the proof is the same.)",
"options": [],
"answer": "See solution",
"solution": "Let $M_1, M_2, \\dots, M_{729}$ be the sequence in increasing order according to algebra marks. Let $a_i$ be the geometry mark of $M_i$. It suffices to show that a 10-term increasing subsequence of $a_1, a_2, \\dots, a_{729}$ does not exist.\n\nFor any $i$, color $a_i$ with color $s$, where $s$ is the second digit of the base 9 representation of $a_i$. It is easy to see that any monochromatic subsequence of $a_1, a_2, \\dots, a_{729}$ is decreasing. Hence, by the Lemma, the proof is complete.",
"topic": "Number Theory",
"subtopic": "Number-Theoretic Functions"
},
{
"id": 20616,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle and a circle $\\omega$ through $C$ and its incenter $I$ meets $CA$, $CB$ at $P$, $Q$. The circumcircles $(CPQ)$ and $(ABC)$ meet at $L$. The angle bisector of $\\angle ALB$ meets $AB$ at $K$. Show that, as $\\omega$ varies, $\\angle PKQ$ is constant.",
"options": [],
"answer": "See solution",
"solution": "Clearly $IP = IQ$ from the bisector $CI$ in $\\omega$. Our aim is to show that $IK = IP$, as it would follow that $I$ is the circumcenter of triangle $PKQ$, whence $\\angle PKQ = \\frac{1}{2}\\angle PIQ = 90^\\circ - \\frac{1}{2}\\angle ACB$.\n\nLet the angle bisectors $CI$ and $LK$ intersect at the midpoint $T$ of the arc $\\widehat{AB}$ from $k$, not containing $C$. We have $\\angle TAK = \\angle TAB = \\angle BCT = \\angle ACT = \\angle ALT$, thus $\\triangle AKT \\sim \\triangle LAT$ and $TA^2 = TK \\cdot TL$. Since $TA = TI$ from the trillium lemma, we deduce $TI^2 = TK \\cdot TL$ and $\\triangle IKT \\sim \\triangle LIT$. Hence\n\n$$\nIK = IT \\cdot \\frac{LI}{LT} = AT \\cdot \\frac{LI}{LT}.\n$$\n\nOn the other hand, the circumcircles $k$ and $\\omega$ yield $\\angle LPI = 180^\\circ - \\angle LCI = \\angle LAT$ and $\\angle PLI = \\angle PCI = \\angle ALT$. Hence $\\triangle LAT \\sim \\triangle LPI$ and so\n\n$$\n\\frac{LI}{LT} = \\frac{PI}{AT}\n$$\n\nwhich completes the proof. $\\Box$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20617,
"subject": "Mathematics (Olympiad)",
"question": "Пусть $f = \\frac{n}{d}$ и $f' = \\frac{n}{d'}$ — целые числа, причем $f > f'$. Докажите, что $d' - d > \\frac{d^2}{n}$.",
"options": [],
"answer": "See solution",
"solution": "Поскольку $f = \\frac{n}{d}$ и $f' = \\frac{n}{d'}$ — целые, а $f > f'$, имеем $f - f' \\ge 1$, или\n\n$$\n1 \\le \\frac{n}{d} - \\frac{n}{d'} = \\frac{(d' - d)n}{dd'} < \\frac{(d' - d)n}{d^2}.\n$$\n\nДомножая на $\\frac{d^2}{n}$, получаем $d' - d > \\frac{d^2}{n}$, что и требовалось доказать.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20618,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. Show that 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 $4a^2 + 9b^2 - 1 = (2k + 1)(2k - 1)$, so $n$ 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 $4a^2 + 9b^2 - 1 = (3k + 1)(3k - 1)$, so $n$ 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\n$$\n2^{2r}k^2 + 3^{2s}m^2l^2 + 2 \\cdot 2^{r}3^{s}m k l = 1,\n$$\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 $4a^2 + 9b^2 - 1$, which finishes the proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20619,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be a triangle. Find all points $P$ on segment $BC$ satisfying the following property: If $X$ and $Y$ are the intersections of line $PA$ with the common external tangent lines of the circumcircles of triangles $PAB$ and $PAC$, then\n\n$$\n\\left(\\frac{PA}{XY}\\right)^2 + \\frac{PB \\cdot PC}{AB \\cdot AC} = 1.\n$$",
"options": [],
"answer": "See solution",
"solution": "We consider the configuration shown below. Let $O_B$ and $\\omega_B$ ($O_C$ and $\\omega_C$) denote the circumcenter and circumcircle of triangle $ABP$ ($ACP$) respectively. Line $ST$, with $S$ on $\\omega_B$ and $T$ on $\\omega_C$, is one of the common tangent lines of the two circumcircles. Point $X$ lies on segment $ST$. Point $Y$ lies on the other common tangent line.\n\n\n\nLet $M$ be the intersection of segments $XY$ and $O_B O_C$. By symmetry, $M$ is the midpoint of both segments $AP$ and $XY$, and line $O_B O_C$ is the perpendicular bisector of segments $XY$ and $AP$. By the power of a point theorem,\n\n$$\nXS^2 = XA \\cdot XP = XT^2 \\quad \\text{and} \\quad X \\text{ is the midpoint of segment } ST.\n$$\n\nWe claim that triangles $ABC$ and $AO_B O_C$ are similar to each other, which is the *Salmon theorem*. Indeed, $\\angle ABC = \\angle MO_B A = \\angle O_C O_B A$, because each angle is equal to half of the angular size of arc $\\widehat{AP}$ of $\\omega_B$. Likewise, $\\angle O_B O_C A = \\angle C$. In particular, we have\n\n$$\n\\frac{AB}{AO_B} = \\frac{BC}{O_B O_C} = \\frac{CA}{O_C A}.\n$$\n\nSet $AB = c$, $BC = a$, and $CA = b$. We claim that it suffices to show the following key fact:\n\n$$\n1 - \\left(\\frac{PA}{XY}\\right)^2 = \\frac{BC^2}{(AB + AC)^2} = \\frac{a^2}{(b+c)^2}.\n$$\n\nAssuming this, the given condition in the problem becomes\n\n$$\n\\frac{PB \\cdot PC}{AB \\cdot AC} = \\frac{a^2}{(b+c)^2} \\quad \\text{or} \\quad PB \\cdot PC = \\frac{a^2bc}{(b+c)^2}.\n$$\n\nThere are precisely two points $P_1$ and $P_2$ on segment $BC$ satisfying this. Write\n\n$$\nPB \\cdot PC = PB \\cdot (a - PB) = \\frac{a^2bc}{(b+c)^2},\n$$\n\nwhich is a quadratic equation in $PB$, so there are at most two solutions. Construct $P_1$ so that $AP_1$ is the bisector of $\\angle BAC$, and let $P_2$ be the reflection of $P_1$ across the midpoint of segment $BC$. By the angle-bisector theorem, $P_2C = P_1B = \\frac{ac}{b+c}$ and $P_2B = P_1C = \\frac{ab}{b+c}$, from which the condition follows for $P_1$ and $P_2$.\n\nIt remains to establish the key fact. One approach (by Titu Andreescu and Cosmin Pohoata): Rays $O_B X$ and $O_C T$ meet in $W$. Because $O_B S \\parallel O_C T$, triangles $O_B S X$ and $W T X$ are congruent. Hence $O_B X = X W$ and triangles $O_B X O_C$ and $W X O_C$ have the same area. Note that $X M$ and $X T$ are altitudes in triangles $O_B X O_C$ and $W X O_C$, respectively. Thus,\n\n$$\n\\frac{XY \\cdot O_B O_C}{4} = \\frac{XM \\cdot O_B O_C}{2} = \\frac{XT \\cdot O_C W}{2} = \\frac{ST \\cdot (O_C T + TW)}{4} = \\frac{ST \\cdot (O_C T + O_B S)}{4}.\n$$\n\nBy similarity,\n\n$$\n\\frac{XY}{ST} = \\frac{O_C T + O_B S}{O_B O_C} = \\frac{O_C A + O_B A}{O_B O_C} = \\frac{AB + AC}{BC} \\quad \\text{or} \\quad \\frac{XY^2}{ST^2} = \\frac{(b+c)^2}{a^2}.\n$$\n\nNote that $O_B S T O_C$ is a right trapezoid. Let $U$ be the foot of the perpendicular from $O_C$ on $O_B S$. We have\n\n$$\nST^2 = UO_C^2 = O_B O_C^2 - O_S U^2 = O_B O_C^2 - (O_B S - O_C T)^2 = O_B O_C^2 - (O_B A - O_C A)^2.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20620,
"subject": "Mathematics (Olympiad)",
"question": "In the first barrel there is $5\\ \\text{hl}$ and $25\\ \\text{l}$ of wine. In the second barrel there is 3 times more wine than in the first one, and in the third there is $1\\ \\text{hl}$ and $75\\ \\text{l}$ less wine than in the first one. How much wine is there in each of the barrels?",
"options": [],
"answer": "See solution",
"solution": "In the first barrel there is $5\\ \\text{hl} + 25\\ \\text{l} = 525\\ \\text{l}$ of wine. \n\nSo in the second barrel there is $3 \\times 525\\ \\text{l} = 1575\\ \\text{l}$ of wine.\n\nIn the third barrel: $5\\ \\text{hl} + 25\\ \\text{l} - (1\\ \\text{hl} + 75\\ \\text{l}) = 525\\ \\text{l} - 175\\ \\text{l} = 350\\ \\text{l}$ of wine.\n\nIn total, the three barrels contain $525\\ \\text{l} + 1575\\ \\text{l} + 350\\ \\text{l} = 2450\\ \\text{l} = 24\\ \\text{hl} + 50\\ \\text{l}$ of wine.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20621,
"subject": "Mathematics (Olympiad)",
"question": "Let $p \\neq 2, 5$. Let $m = 5^{4n} - 5^{3n} + 5^{2n} - 5^n + 1$.\n\nShow that if $p$ divides $m$, then $p \\equiv 1 \\pmod{4}$.",
"options": [],
"answer": "See solution",
"solution": "We have\n\n$$\n(2 \\cdot 5^{2n} - 5^n + 2)^2 - 5 \\cdot 5^{2n} = 4m \\equiv 0 \\pmod{p}.\n$$\n\nThis gives $5 \\equiv (5^{-n}(2 \\cdot 5^{2n} - 5^n + 2))^2 \\pmod{p}$. Using the Legendre symbol, $\\left(\\frac{5}{p}\\right) = 1$.\n\nOn the other hand,\n\n$$\n(5^{2n} - 5^n + 1)^2 + 5^n(5^n - 1)^2 = m \\equiv 0 \\pmod{p}.\n$$\n\nAs above, this implies $\\left(\\frac{-5^n}{p}\\right) = 1$. It follows that\n\n$$\n\\left(\\frac{-1}{p}\\right) = \\left(\\frac{-5^n}{p}\\right) \\left(\\frac{5^n}{p}\\right) = 1 \\cdot 1^n = 1.\n$$\n\nTherefore, $p \\equiv 1 \\pmod{4}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20622,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a cyclic quadrilateral such that $AB \\cdot CD = BC \\cdot DA$, so $ABCD$ is harmonic. Let $P$ be a point such that the lines $PA, PB, PC, PD$ form a harmonic bundle. The intersections of these lines with any other line yield four harmonic points. In particular, intersecting $PA, PB, PC, PD$ with the line $BY$ gives that $B, X, D, Y$ are harmonic.\n\n\n\nLet $M$ be the midpoint of $XY$. Prove that if $MP = MX = MY$, then $MP$ is tangent to the circle $K$ at $P$.",
"options": [],
"answer": "See solution",
"solution": "Since $M$ is the midpoint of $XY$ and $B, X, D, Y$ are harmonic, a well-known calculation gives\n\n$$\nMX \\cdot MY = MD \\cdot MB. \\tag{1}\n$$\n\nTo derive (1), let $MD = a$, $DX = b$, and $XB = c$. Since $MX = MY$, we have $MY = a + b$. Since $B, X, D, Y$ are harmonic in that order,\n\n$$\n\\begin{aligned}\nDX \\cdot YB &= YD \\cdot XB \\\\\n\\Leftrightarrow b(2a + 2b + c) &= c(2a + b) \\\\\n\\Leftrightarrow 2ab + 2b^2 + bc &= 2ac + bc \\\\\n\\Leftrightarrow ab + b^2 &= ac \\\\\n\\Leftrightarrow a^2 + 2ab + b^2 &= a^2 + ab + ac \\\\\n\\Leftrightarrow (a+b)^2 &= a(a+b+c) \\\\\n\\Leftrightarrow MX \\cdot MY &= MD \\cdot MB\n\\end{aligned}\n$$\n\nSince $MP = MX = MY$, it follows that\n\n$$\nMP^2 = MD \\cdot MB\n$$\n\nTherefore, $MP$ is tangent to $K$ at $P$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20623,
"subject": "Mathematics (Olympiad)",
"question": "Real numbers $a_1, a_2, \\dots, a_n$ satisfy the following conditions: $a_1 + a_2 + \\dots + a_n = n$ and $a_1 \\ge a_2 \\ge \\dots \\ge a_n \\ge 0$. Prove the inequality:\n\n$$\nna_1 \\ge a_1^2 + a_2^2 + \\dots + a_n^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "The proof follows from the following transformations:\n\n$$\na_1^2 + a_2^2 + \\dots + a_n^2 \\le a_1(a_1 + a_2 + \\dots + a_n) = n a_1.\n$$\n\nThus, the inequality holds.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20624,
"subject": "Mathematics (Olympiad)",
"question": "Pablo will decorate each of 6 identical white balls with either a striped or a dotted pattern, using either red or blue paint. He will decide on the color and pattern for each ball by flipping a fair coin for each of the 12 decisions he must make. After the paint dries, he will place the 6 balls in an urn. Frida will randomly select one ball from the urn and note its color and pattern. The events “the ball Frida selects is red” and “the ball Frida selects is striped” may or may not be independent, depending on the outcome of Pablo’s coin flips. The probability that these two events are independent can be written as $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m$?\n\n(Recall that two events $A$ and $B$ are independent if $P(A \\text{ and } B) = P(A) \\cdot P(B)$.)\n\n(A) 243 (B) 245 (C) 247 (D) 249 (E) 251",
"options": [],
"answer": "See solution",
"solution": "**Answer (A):** There are $4^6$ ways to paint the balls, each equally likely. It remains to count the number of paintings for which the two given events are independent.\n\n- If all the balls are red, then $P(\\text{red}) = 1$ and the events are independent regardless of $P(\\text{striped})$. The same reasoning applies if all the balls are blue. This accounts for $2 \\cdot 2^6 = 128$ paintings.\n- If 5 of the balls are red and $s$ balls are striped, then\n $$\n P(\\text{red}) \\cdot P(\\text{striped}) = \\frac{5}{6} \\cdot \\frac{s}{6}.\n $$\n On the other hand, $P(\\text{red and striped})$ is one of the fractions $\\frac{0}{6}$, $\\frac{1}{6}$, $\\frac{2}{6}$, $\\frac{3}{6}$, $\\frac{4}{6}$, or $\\frac{5}{6}$, depending on how many red balls are striped. These are equal if and only if $s = 0$ or $s = 6$. There are $2 \\cdot \\binom{6}{5} = 12$ ways for this to happen. A similar argument handles the case in which 5 balls are blue, giving 24 paintings in all.\n- Suppose 4 balls are red. In order for the two given events to be independent, the fraction of red balls that are striped must equal the fraction of blue balls that are striped. This happens when all the balls are striped, or none of them are striped, or 2 of the red balls and 1 of the blue balls are striped. There are $\\binom{4}{2} \\cdot \\binom{2}{1} = 12$ ways to choose the patterns in the last case, so this accounts for $\\binom{6}{4} \\cdot (1 + 1 + 12) = 210$ situations. There are another 210 for which 2 balls are red.\n- Suppose 3 balls are red. Again, in order for the two given events to be independent, the fraction of red balls that are striped must equal the fraction of blue balls that are striped. This happens when $s$ red balls are striped and $s$ blue balls are striped, where $s \\in \\{0, 1, 2, 3\\}$. The calculation in this case is\n $$\n \\binom{6}{3} \\left( \\binom{3}{0}^2 + \\binom{3}{1}^2 + \\binom{3}{2}^2 + \\binom{3}{3}^2 \\right) = 20 \\cdot 20 = 400.\n $$\n\nIn all there are $128 + 24 + 210 + 210 + 400 = 972$ cases in which the color and the pattern of the drawn ball are independent, so the required probability is\n$$\n\\frac{972}{2^{12}} = \\frac{243}{1024},\n$$\nand the requested numerator is 243.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20625,
"subject": "Mathematics (Olympiad)",
"question": "The polynomial $x^3 + px + q$, where $p$ and $q$ are real numbers and at least one of them is nonzero, has a real root $a$ that satisfies\n$$a^2 \\le -\\frac{4}{3}p.$$ \nProve that this polynomial has a real root different from $a$.",
"options": [],
"answer": "See solution",
"solution": "The assumption $a^3 + pa + q = 0$ implies $q = -a(a^2 + p)$, so\n$$x^3 + px + q = (x - a)(x^2 + a x + a^2 + p).$$\nThe discriminant of $x^2 + a x + a^2 + p$ is\n$$D = a^2 - 4(a^2 + p) = -(3a^2 + 4p).$$\nThe assumption $a^2 \\le -\\frac{4}{3}p$ implies $D \\ge 0$. Hence, there are real numbers $b$ and $c$ such that $x^2 + a x + a^2 + p = (x - b)(x - c)$, so the polynomial $x^3 + p x + q$ has roots $b$ and $c$.\n\nIf $a = b = c$, then $x^3 + p x + q = (x - a)^3 = x^3 - 3a x^2 + 3a^2 x + a^3$. Thus, $-3a = 0$, $3a^2 = p$, and $a^3 = q$, implying $p = q = 0$. This contradicts the assumption that at least one of $p$ or $q$ is nonzero. Consequently, $x^3 + p x + q$ has a real root different from $a$.\n\n*Remark:* The problem can also be solved by standard means of calculus.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20626,
"subject": "Mathematics (Olympiad)",
"question": "At a party, each guest has exactly four friends among the other guests. What are the possible numbers $n$ of guests at the party?",
"options": [],
"answer": "See solution",
"solution": "We first consider the friends of one guest, say Marieke. Marieke has exactly four friends at the party: Aad, Bob, Carla, and Demi. Any other guests are not friends with Marieke, so they can only be friends with Aad, Bob, Carla, and Demi. Since everyone has exactly four friends, each of these other guests must be friends with Aad, Bob, Carla, and Demi (and with no one else).\n\nSince Aad also has exactly four friends (including Marieke), the group of guests not friends with Marieke can consist of no more than three people. If the group consists of zero, one, or three people, we have the following solutions (two guests are connected by a line if they are friends):\n\n\n\n\n\n*Solutions with five, six, and eight guests in total.*\n\nNow we show that it is not possible for this group to consist of two people. In that case, Aad would have exactly one friend among Bob, Carla, and Demi. Assume, without loss of generality, that Aad and Bob are friends. Similarly, Carla must be friends with one of Aad, Bob, and Demi. Since Aad and Bob already have four friends, Carla and Demi must be friends. However, since they are both not friends with Aad, this contradicts the requirement in the problem statement.\n\nWe conclude that there can be five, six, or eight guests at the party. Hence, the possible values for $n$ are $5$, $6$, and $8$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20627,
"subject": "Mathematics (Olympiad)",
"question": "Prove that there exists a polygon on a square grid that can be tiled with dominoes ($1 \\times 2$ or $2 \\times 1$ figures) in exactly 2020 ways.",
"options": [],
"answer": "See solution",
"solution": "Let $A_n$ be a figure that consists of a $2 \\times 2$ square combined with $n$ L-shapes ($A_3$ is shown in the figure below).\n\nWe will prove by induction that $A_n$ can be tiled with dominoes in exactly $2n + 2$ ways.\n\nThe base case for $A_0$ is evident. Now assume that we have proved it for $A_k$ and consider the figure $A_{k+1}$. Its bottom rightmost square can be tiled in two ways. If it is tiled with a horizontal domino (see figure (b)), then this leads to a unique tiling for all $A_{k+1}$.\n\nIf it is tiled with a vertical domino, then consider the bottom rightmost square of the remaining figure. If it is tiled with a vertical domino (see figure (c)), then this again leads to a unique tiling for all $A_{k+1}$. But if it is tiled with a horizontal domino (see figure (d)), then the remaining figure is $A_k$ that can be tiled in $2k + 2$ ways.\n\nTherefore, $A_{k+1}$ can be covered with dominoes in exactly $1 + 1 + (2k + 2) = 2k + 4$ ways, which finishes the induction step.\n\nNow we see that $A_{1009}$ can be covered in exactly $2020$ ways.\n\n_Note_: If we start with a $1 \\times 2$ horizontal rectangle instead of a $2 \\times 2$ square, then we get a sequence of figures for which the number of tilings are all odd numbers.\n\n\n\nFigure 1:",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20628,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle with $\\angle ABC = 120^\\circ$ and angle bisectors $AA_1$, $BB_1$, $CC_1$. Let $B_1F \\perp A_1C_1$, where $F \\in A_1C_1$. Let $R$, $I$, and $S$ be the centers of the circles inscribed in triangles $C_1B_1F$, $C_1B_1A_1$, and $A_1B_1F$, respectively. Let $B_1S \\cap A_1C_1 = \\{Q\\}$. Show that $R$, $I$, $S$, and $Q$ are concyclic.",
"options": [],
"answer": "See solution",
"solution": "First, we will show that $\\angle C_1B_1A_1 = 90^\\circ$. Let $K \\in BC$ so that $B \\in (KA_1)$, then $\\angle ABK = 60^\\circ$. Point $C_1$ is on the bisector of $\\angle ACB$ and this implies that $d(C_1, BC) = d(C_1, AC)$ or $C_1F_1 = C_1F_3$, where $F_1$ is the projection of $C_1$ on $BC$ and $F_3$ is the projection of $C_1$ on $AC$. Segment $BA$ is the bisector of $\\angle KBB_1$ implies that $d(C_1, KB) = d(C_1, BB_1)$ or $C_1F_1 = C_1F_2$, where $F_2$ is the projection of $C_1$ on $BB_1$. So, $C_1F_2 = C_1F_3$ and $C_1B_1$ is the bisector of $\\angle BB_1A$. Let us denote $\\angle BB_1C_1 = \\alpha$. Likewise, we prove that $B_1A_1$ is the bisector of $\\angle BB_1C$. Let $\\angle BB_1A_1 = \\angle CB_1A_1 = \\beta$. Then, from $\\angle AB_1C = 180^\\circ$ we have $2\\alpha + 2\\beta = 180^\\circ$ and $\\alpha + \\beta = 90^\\circ$.\n\n\n\nLet $r_1$ be the radius of the inscribed circle to $\\triangle A_1B_1C_1$, $r_2$ the radius of the inscribed circle to $\\triangle C_1B_1F$, and $r_3$ the radius of the inscribed circle to $\\triangle A_1B_1F$. Considering the properties of right triangles, we have\n\n$$\n\\triangle C_1FB_1 \\sim \\triangle C_1B_1A_1 \\implies \\frac{r_2}{r_1} = \\frac{B_1C_1}{C_1A_1} = \\cos C_1\n$$\nfrom which follows $r_2 = r_1 \\cos C_1 = r_1 \\sin A_1$. Likewise,\n\n$$\n\\frac{r_3}{r_1} = \\frac{A_1B_1}{C_1A_1} = \\cos A_1 \\implies r_3 = r_1 \\cos A_1 = r_1 \\sin C_1\n$$\n\nNow, we will see that $I_1R \\parallel A_1B_1$ and $I_1S \\parallel C_1B_1$, where $I_1$ is the projection of $I$ on $C_1A_1$. Let $I_1R_2 \\perp C_1B_1$, $R_2 \\in C_1B_1$ and $I_1R_2 \\cap C_1I = \\{R^*\\}$. Since $\\triangle C_1II_1 \\sim \\triangle C_1R^*R_2$, then\n\n$$\n\\frac{R^*R_2}{II_1} = \\frac{C_1R_2}{C_1I_1} = \\cos C_1 \\implies R^*R_2 = r_1 \\cos C_1 = r_2\n$$\n\nfrom which follows $R^* = R$, $I_1R \\perp C_1B_1 \\implies I_1R \\parallel A_1B_1$. Likewise, we get $I_1S \\parallel C_1B_1$.\n\n\n\nIn triangle $I_1RR_1$ we have $r_2 = I_1R \\sin A_1 = r_1 \\cos C_1 = r_1 \\sin A_1$ and $I_1R = r_1$. In triangle $I_1SS_1$ we have $r_3 = I_1S \\sin C_1 = r_1 \\cos A_1 = r_1 \\sin C_1$ from which follows $I_1S = r_1$. Finally, we get $I_1R = II_1 = I_1S = r_1$. Since $\\triangle QSI_1 \\sim \\triangle QB_1C_1$, then\n\n$$\n\\frac{SI_1}{B_1C_1} = \\frac{I_1Q}{C_1Q} \\implies SI_1 = I_1Q\n$$\n\nsince $B_1C_1 = C_1Q$. Now, we can conclude that points $R$, $I$, $S$, and $Q$ lie on the same circle. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20629,
"subject": "Mathematics (Olympiad)",
"question": "Let $f(x) = 3\\sin x + 2\\cos x + 1$. If real numbers $a, b, c$ are such that $af(x) + bf(x-c) = 1$ holds for any $x \\in \\mathbb{R}$, then $\\frac{b\\cos c}{a}$ equals ( )\n\n(A) $-\\frac{1}{2}$\n\n(B) $\\frac{1}{2}$\n\n(C) $-1$\n\n(D) $1$",
"options": [],
"answer": "See solution",
"solution": "**Solution**\n\nLet $c = \\pi$. Then $f(x) + f(x-c) = 2$ for any $x \\in \\mathbb{R}$.\n\nNow let $a = b = \\frac{1}{2}$, and $c = \\pi$. We have\n\n$$\naf(x) + bf(x-c) = 1\n$$\n\nfor any $x \\in \\mathbb{R}$. Consequently, $\\frac{b\\cos c}{a} = -1$. So the answer is (C).\n\nMore generally, we have\n\n$$\nf(x) = \\sqrt{13}\\sin(x + \\varphi) + 1,\n$$\n\n$$\nf(x-c) = \\sqrt{13}\\sin(x + \\varphi - c) + 1,\n$$\n\nwhere $0 < \\varphi < \\frac{\\pi}{2}$ and $\\tan \\varphi = \\frac{2}{3}$. Then $af(x) + bf(x-c) = 1$ becomes\n\n$$\n\\sqrt{13}a\\sin(x + \\varphi) + \\sqrt{13}b\\sin(x + \\varphi - c) + a + b = 1.\n$$\n\nThat is,\n\n$$\n\\sqrt{13}a\\sin(x + \\varphi) + \\sqrt{13}b\\sin(x + \\varphi)\\cos c - \\sqrt{13}b\\sin c\\cos(x + \\varphi) + (a + b - 1) = 0.\n$$\n\nTherefore,\n\n$$\n\\sqrt{13}(a + b\\cos c)\\sin(x + \\varphi) - \\sqrt{13}b\\sin c\\cos(x + \\varphi) + (a + b - 1) = 0.\n$$\n\nSince the equality above holds for any $x \\in \\mathbb{R}$, we must have\n\n$$\n\\begin{cases} a + b\\cos c = 0, \\\\ b\\sin c = 0, \\end{cases}\n$$\n\n$$\n\\begin{cases} b\\sin c = 0, \\\\ a + b - 1 = 0. \\end{cases}\n$$\n\n$$\n\\begin{cases} a + b - 1 = 0. \\end{cases}\n$$\n\nIf $b=0$, then $a=0$ from the first system, and this contradicts the last equation. So $b \\neq 0$, and $\\sin c = 0$ from the second system. Therefore $c = 2k\\pi + \\pi$ or $c = 2k\\pi$ ($k \\in \\mathbb{Z}$).\n\nIf $c = 2k\\pi$, then $\\cos c = 1$, and it leads to a contradiction between the first and last equations. So $c = 2k\\pi + \\pi$ ($k \\in \\mathbb{Z}$) and $\\cos c = -1$. From the first and last equations, we get $a = b = \\frac{1}{2}$. Consequently, $\\frac{b\\cos c}{a} = -1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20630,
"subject": "Mathematics (Olympiad)",
"question": "Find the sum of all products $a_1 a_2 \\cdots a_{50}$ where $a_1, a_2, \\ldots, a_{50}$ are distinct positive integers not exceeding 101 and such that no two of them have sum 101.",
"options": [],
"answer": "See solution",
"solution": "We distinguish between two cases for an admissible 50-tuple $a_1, a_2, \\ldots, a_{50}$.\n\n**Case 1:** If no $a_i$ is equal to 101, then $a_1, a_2, \\ldots, a_{50}$ contains exactly one number from every pair $(i, 101-i)$, $1 \\leq i \\leq 50$. Hence, there are $2^{50}$ choices for $a_1, a_2, \\ldots, a_{50}$, and each respective product $a_1 a_2 \\cdots a_{50}$ appears exactly once in the expansion of the product\n\n$$\nP = (1 + 100)(2 + 99) \\cdots (50 + 51) = 101^{50}.\n$$\n\n**Case 2:** If one of the $a_i$ is 101, then the remaining ones come from 49 different pairs $(i, 101-i)$, $1 \\leq i \\leq 50$. Suppose that pair $(1, 100)$ is not present. There are $2^{49}$ such products $a_1 a_2 \\cdots a_{50}$, the summands in the expansion of\n\n$$\nP_1 = 101 (2 + 99) \\cdots (50 + 51) = 101^{50}.\n$$\n\nAnalogously, if the non-represented pair is $(2, 99)$, $(3, 98)$, \\dots, $(50, 51)$, the respective products appear once in the expansions of\n\n$$\n\\begin{align*}\nP_2 &= (1 + 100) 101 (3 + 98) \\cdots (50 + 51) = 101^{50}, \\\\\nP_3 &= (1 + 100)(2 + 99) 101 \\cdots (50 + 51) = 101^{50}, \\\\\n&\\quad\\vdots \\\\\nP_{50} &= (1 + 100)(2 + 99) \\cdots (49 + 52) 101 = 101^{50}.\n\\end{align*}\n$$\n\nBy both cases, the sum in question is $101^{50} + 50 \\cdot 101^{50} = 51 \\cdot 101^{50}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20631,
"subject": "Mathematics (Olympiad)",
"question": "Prueba que las sumas de las primeras, segundas y terceras potencias de las raíces del polinomio $p(x) = x^3 + 2x^2 + 3x + 4$ valen lo mismo.\n\nSean $r, s$ y $t$ las raíces, reales o complejas, del polinomio $p(x)$ y sea $S_n$ la suma de sus $n$-ésimas potencias, esto es, $S_n = r^n + s^n + t^n$.",
"options": [],
"answer": "See solution",
"solution": "Por las fórmulas de Vièta:\n\n- $S_1 = r + s + t = -2$\n- $rs + st + tr = 3$\n- $rst = -4$\n\nPara $S_2$:\n$$\nS_2 = r^2 + s^2 + t^2 = (r + s + t)^2 - 2(rs + st + tr) = (-2)^2 - 2 \\times 3 = 4 - 6 = -2\n$$\n\nPara $S_3$, usando que $p(r) = 0$ (y análogamente para $s$ y $t$):\n$$\nr^3 = -2r^2 - 3r - 4\n$$\nSumando para $r, s, t$:\n$$\nS_3 = r^3 + s^3 + t^3 = -2S_2 - 3S_1 - 12\n$$\nSustituyendo $S_1$ y $S_2$:\n$$\nS_3 = -2(-2) - 3(-2) - 12 = 4 + 6 - 12 = -2\n$$\n\nPor lo tanto, $S_1 = S_2 = S_3 = -2$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20632,
"subject": "Mathematics (Olympiad)",
"question": "На републичкиот натпревар по математика, познато е дека секој натпреварувач има не повеќе од тројца познаници (познанството е симетрична релација). Покажи дека е можно натпреварувачите да се распоредат во две простории така што секој од нив на натпреварот има не повеќе од еден познаник во просторијата во која е сместен.",
"options": [],
"answer": "See solution",
"solution": "Да ги распоредиме натпреварувачите во две простории сосема произволно. Со $S_1$ нека го означиме вкупниот број познанства во првата просторија, а со $S_2$ вкупниот број познанства во втората просторија. Нека $S = S_1 + S_2$ (доколку два натпреварувачи распоредени во иста соба се познаници, тоа познанство го броиме еднаш). Тогаш $S$ е ненегативен цел број.\n\nПретпоставуваме дека ваквиот распоред не е задоволителен, т.е. постои натпреварувач $A$ така што во неговата просторија има барем двајца негови познаници $B$ и $C$. Го префрламе $A$ во другата просторија; со ова вредноста на $S$ се намалува барем за $1$. Во другата просторија $A$ има не повеќе од еден познаник. Значи, едниот од броевите $S_1$ и $S_2$ се намалил за $2$, а другиот се зголемил за $1$, па според тоа $S$ се намалил за $1$.\n\nПостапката ја продолжуваме се додека моменталниот распоред не е поволен. Бидејќи $S$ е ненегативен цел број, после конечен број вакви префрлања ќе стигнеме до поволен распоред, т.е. распоред во кој натпреварувачите се распоредени во две простории така што секој натпреварувач има не повеќе од еден познаник во просторијата во која е сместен.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20633,
"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. Demuestra 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 $$n - k = 1001B = 7 \\cdot 11 \\cdot 13 B$$\nPor lo tanto, $n$ y $k$ son congruentes módulo $7$, $11$ y $13$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20634,
"subject": "Mathematics (Olympiad)",
"question": "Is it possible to cover a rectangle with an odd number of L-tetrominoes and several straight tetrominoes? Tetrominoes have to fit in the rectangle completely and each unit square of the rectangle has to be covered by exactly one tetromino.\n\n",
"options": [],
"answer": "See solution",
"solution": "The area of the rectangle must be divisible by $4$, so the number of rows or columns is even. Without loss of generality, let the number of columns be even. Color every other row in black. There is an even number of black squares.\n\nEvery straight tetromino covers an even number of black squares, while every L-tetromino covers an odd number of black squares. Therefore, the number of L-tetrominoes must be even.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20635,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ be a real number in the open interval $(0, 1)$, let $n$ be a positive integer, and let $f_n: \\mathbb{R} \\to \\mathbb{R}$ defined by $f_n(x) = x + \\frac{x^2}{n}$. Show that\n\n$$\n\\frac{a(1-a)n^2 + 2a^2n + a^3}{(1-a)^2n^2 + a(2-a)n + a^2} < \\underbrace{(f_n \\circ \\dots \\circ f_n)}_{n} (a) < \\frac{an + a^2}{(1-a)n + a}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $a_k = \\underbrace{(f_n \\circ \\cdots \\circ f_n)}_{k}(a)$ for $k \\in \\mathbb{N}$. Notice that\n\n$$\n\\frac{1}{a_{k+1}} = \\frac{1}{a_k} - \\frac{1}{a_k + n}, \\quad k \\in \\mathbb{N},\n$$\n\nto deduce that\n$$\n\\frac{1}{a_n} = \\frac{1}{a} - \\sum_{k=0}^{n-1} \\frac{1}{a_k + n}.\n$$\nSo,\n$$\n\\frac{1}{a} - \\frac{n}{a + n} < \\frac{1}{a_n} < \\frac{1}{a} - \\frac{n}{a_n + n}, \\quad (*).\n$$\n\nSince the $a_k$ form an increasing sequence of positive real numbers, the first inequality above yields the required upper bound:\n$$\na_n < \\frac{an + a^2}{(1-a)n + a}.\n$$\n\nPlugging this upper bound into the rightmost expression in $(*)$ yields the required lower bound:\n$$\na_n > \\frac{a(1-a)n^2 + 2a^2n + a^3}{(1-a)^2n^2 + a(2-a)n + a^2}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20636,
"subject": "Mathematics (Olympiad)",
"question": "Peter has several equal white squares with dimensions $4 \\times 4$ sectors. He paints each of the $1 \\times 1$ sectors red or blue so that for any two squares, the first, second, third, and fourth columns are respectively different, and also the first, second, third, and fourth rows are respectively different. Rotating the squares is forbidden. How many squares can Peter paint in that way?",
"options": [],
"answer": "See solution",
"solution": "Since every two squares differ in each row and column, they already differ in the first column. There are altogether $2^4 = 16$ differently painted columns, therefore Peter cannot paint more than $2^4 = 16$ squares that satisfy the condition.\n\nLet us prove that this number is the answer. Peter takes 16 white squares and paints their first columns in all possible patterns. The received squares already differ in the first columns. To paint the rest according to the problem statement, he adheres to a cyclic diagonal pattern:\n\nPeter takes the first column of each square, repositions its uppermost sector to the lowest, and paints the second column accordingly. Similarly, he obtains the third column from the second one, and so on. For example:\n\nABCD\n\nBCDA\n\nCDAB\n\nDABC\n\nLet's consider two different squares $X \\neq Y$. They differ in the first column because they were constructed that way. Let them differ in sector number $(4-j)$ counting from the bottom. If they differ in two sectors, you can take any of them. Then their second columns differ in the sector $(5-j)$, etc. Thus, all the columns are different, but the sector they differ in appears in a certain row, in which another two squares differ from each other. Thus, the sector in which the squares differ changes its position in columns as well as in rows. Therefore, we obtain that squares differ in rows too.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20637,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $A = \\{x \\mid 5x - a \\le 0\\}$, $B = \\{x \\mid 6x - b > 0\\}$, $a, b \\in \\mathbb{N}$, and $A \\cap B \\cap \\mathbb{N} = \\{2, 3, 4\\}$. The number of such pairs $(a, b)$ is $\\boxed{\\phantom{0}}$.\n\n(A) 20 \n(B) 25 \n(C) 30 \n(D) 42",
"options": [],
"answer": "See solution",
"solution": "Since $5x - a \\le 0 \\implies x \\le \\frac{a}{5}$, and $6x - b > 0 \\implies x > \\frac{b}{6}$. To have $A \\cap B \\cap \\mathbb{N} = \\{2, 3, 4\\}$, we require:\n\n$$\n\\begin{cases}\n2 > \\frac{b}{6} \\geq 1 \\\\\n4 \\leq \\frac{a}{5} < 5\n\\end{cases}\n$$\n\nThis gives $6 < b < 12$ (i.e., $b = 7, 8, 9, 10, 11$) and $20 \\leq a < 25$ (i.e., $a = 20, 21, 22, 23, 24$). Thus, there are $5 \\times 6 = 30$ such pairs $(a, b)$. The answer is (C).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20638,
"subject": "Mathematics (Olympiad)",
"question": "There is an isosceles obtuse triangle $ABC$ with vertex at point $B$. The perpendicular bisector to side $BC$ intersects lines $AC$ and $AB$ at points $K$ and $M$, respectively. Prove that the point symmetric to $A$ with respect to line $BK$ lies on line $CM$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $A_1$ be the point symmetric to $A$ with respect to $BK$ (see the figure). First, $\\angle BA_1K = \\angle BAK = \\angle BCK$, so quadrilateral $BA_1CK$ is cyclic. Then,\n\n$$\n\\angle A_1CB = \\angle A_1KB = \\angle AKB = 2\\angle BCA = \\angle MBC = \\angle MCB,\n$$\n\nso we get that points $C$, $A_1$, and $M$ are collinear.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20639,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\mathbb{R}^+$ be the set of positive real numbers. Determine all functions $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$ such that\n\n$$\nf(x^2 + x f(y)) = f(f(x))(x + y)\n$$\n\nfor all positive real numbers $x$ and $y$.",
"options": [],
"answer": "See solution",
"solution": "The only solution is $f(x) = x$.\n\nSince $f(f(x)) > 0$, by varying $y$, we see that $f$ is injective. Consider $x \\in (0, 1)$ and substitute $y = 1 - x$. Applying injectivity, we get $x^2 + x f(1 - x) = f(x)$. Further, substituting $1 - x$ into $x$ gives the simultaneous equations:\n\n$$\n\\begin{aligned}\nx^2 + x f(1 - x) &= f(x) \\\\\n(1 - x)^2 + (1 - x) f(x) &= f(1 - x)\n\\end{aligned}\n$$\n\nSolving these, we obtain:\n\n$$\nf(x) = \\frac{x (1 - x)^2 + x^2}{1 - x (1 - x)} = x\n$$\n\nfor all $x \\in (0, 1)$.\n\nNow, we prove by induction on $n$ that $f(x) = x$ for $x \\in (0, n)$. Suppose this is true for $n = k$, then substitute $x, y \\in (0, k)$. This gives:\n\n$$\nf(x(x + y)) = f(x^2 + x f(y)) = f(f(x))(x + y) = x(x + y).\n$$\n\nHowever, $x(x + y)$ covers all values in $(0, k + 1) \\subseteq (0, 2k^2)$ as $x, y \\in (0, k)$ vary. This completes the induction. Thus, $f(x) = x$ for all $x > 0$.\n\nFinally, the solution $f(x) = x$ can be easily verified to satisfy the problem condition.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20640,
"subject": "Mathematics (Olympiad)",
"question": "Find the smallest positive integer $N$ with the following property:\n\nFor any degree five polynomial $P(x)$ with integer coefficients, there is $0 \\le x \\le 1$ such that $|P(x)| > \\frac{1}{N}$.",
"options": [],
"answer": "See solution",
"solution": "The answer is $N = 56$.\n\nFirst, we prove that $N = 56$ has the property. Suppose, on the contrary, that for a degree five polynomial $P(x)$ with integer coefficients, we have $|P(x)| \\le \\frac{1}{56}$ for all $0 \\le x \\le 1$.\n\nThen the nonnegative integers $|P(0)|$, $|P(1)|$, and $32|P(1/2)|$ are strictly smaller than $1$, thus are equal to $0$. Let $Q(x) = x(1-x)(2x-1)$. By the Gauss lemma, there is a quadratic polynomial $D(x)$ with integer coefficients such that $P(x) = Q(x)D(x)$.\n\nNow consider\n\n$$\n\\alpha = \\frac{1}{2} - \\frac{1}{2\\sqrt{5}} \\quad \\text{and} \\quad \\beta = \\frac{1}{2} + \\frac{1}{2\\sqrt{5}}.\n$$\n\nClearly $0 < \\alpha < \\beta < 1$, $\\alpha + \\beta = 1$, $\\alpha\\beta = \\frac{1}{5}$, and $Q(\\alpha) = -\\frac{1}{5\\sqrt{5}}$, $Q(\\beta) = \\frac{1}{5\\sqrt{5}}$.\n\nMoreover, $25D(\\alpha)D(\\beta)$ is a polynomial expression of $5\\alpha\\beta$ and $\\alpha + \\beta$ with integer coefficients, thus is an integer. It follows that\n\n$$\n3125P(\\alpha)P(\\beta) = -25D(\\alpha)D(\\beta)\n$$\n\nis an integer. On the other hand, $|3125P(\\alpha)P(\\beta)| \\le \\frac{3125}{3136} < 1$, hence $D(\\alpha)D(\\beta) = 0$. It follows that $(5x^2 - 5x + 1)$ divides $D(x)$, and we may assume that $D(x) = 5x^2 - 5x + 1$. This leads to the contradiction:\n\n$$\nP(0.9) = 0.9 \\cdot 0.1 \\cdot 0.8 \\cdot 0.55 > 0.036 > 0.02 > \\frac{1}{56}.\n$$\n\nThis proves that for any degree five polynomial $P(x)$ with integer coefficients, there is $0 \\le x \\le 1$ such that $|P(x)| > \\frac{1}{56}$.\n\nNow let $S(x) = x^2(1-x)^2(2x-1)$. Then it is easy to check that for any $0 \\le x \\le 1$, we have\n\n$$\n|S(x)| \\le S(\\beta) = \\frac{Q(\\beta)^2}{2\\beta - 1} = \\frac{1}{25\\sqrt{5}} < \\frac{1}{55}.\n$$\n\nThis shows $N = 55$ does not have the property.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20641,
"subject": "Mathematics (Olympiad)",
"question": "Solve in real numbers the equation\n\n$$\n2(5^x + 6^x - 3^x) = 7^x + 9^x.\n$$",
"options": [],
"answer": "See solution",
"solution": "We claim that the only solutions are $x = 0$ and $x = 1$.\n\nThe given equation can be rewritten as follows:\n\n$$\n5^x \\left( \\left(\\frac{7}{5}\\right)^x + \\left(\\frac{3}{5}\\right)^x - 2 \\right) + 6^x \\left( \\left(\\frac{9}{6}\\right)^x + \\left(\\frac{3}{6}\\right)^x - 2 \\right) = 0. \\tag{\\star}\n$$\n\nConsider the functions $f_a : \\mathbb{R} \\to \\mathbb{R}$, $f_a(x) = a^x + (2-a)^x - 2$, with $a \\in (0, 2) \\setminus \\{1\\}$. These functions are strictly convex and satisfy $f_a(0) = f_a(1) = 0$. Moreover,\n\n$$\nf_a(x) < x f_a(1) + (1-x) f_a(0) = 0,\n$$\nfor all $x \\in (0, 1)$, and\n\n$$\nf_a(1) < \\frac{x-1}{x} f_a(0) + \\frac{1}{x} f_a(x),\n$$\nfor all $x \\in (1, \\infty)$,\n\n$$\nf_a(0) < \\frac{1}{1-x} f_a(x) + \\frac{-x}{1-x} f_a(1),\n$$\nfor all $x \\in (-\\infty, 0)$.\n\nThus, $f_a(x) > 0$ for $x \\in (-\\infty, 0) \\cup (1, \\infty)$. The equivalent equation from $(\\star)$ is:\n\n$$\n5^x f_{\\frac{7}{5}}(x) + 6^x f_{\\frac{9}{6}}(x) = 0,\n$$\n\nso the left side is strictly less than zero for $x \\in (0, 1)$ and strictly greater than zero for $x \\in (-\\infty, 0) \\cup (1, \\infty)$, thus proving our claim.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20642,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute, non-isosceles triangle and $D$, $E$, and $F$ be the midpoints of $AB$, $BC$, and $AC$. Denote by $(O)$ and $(O')$ the circumcircle and Euler circle of triangle $ABC$. Consider an arbitrary point $P$ lying inside triangle $DEF$; $DP$, $EP$, and $FP$ intersect $(O')$ at $D'$, $E'$, and $F'$ respectively. Let $A'$ be the reflection of $A$ through $D$. Points $B'$, $C'$ are defined similarly.\n\na) Prove that if $PO = PO'$, then the circumcircle of triangle $A'B'C'$ passes through $O$.\n\nb) Let $X$ be the reflection of $A'$ through $OD$; $Y$, $Z$ are defined similarly. $H$ is the orthocenter of $ABC$; $XH$, $YH$, and $ZH$ intersect $BC$, $AC$, and $AB$ at $M$, $N$, and $K$ respectively. Prove that the three points $M$, $N$, and $K$ are collinear.",
"options": [],
"answer": "See solution",
"solution": "a) Let $I$ be the reflection of $O$ over $P$. We have $O'$ is the midpoint of $OH$ so $O'P \\parallel IH$. Note that $PO = PO'$, then $IO = IH$. Let $S$ and $G$ be the midpoints of $AI$ and $AH$. We have\n$$\nSP = \\frac{1}{2}AO = \\frac{1}{2}R = O'D\n$$\nand $SP \\parallel AO \\parallel O'D$, so $O'SPD$ is a parallelogram. It follows that $DP \\parallel O'S$. Note that we just have proved $SP = \\frac{1}{2}R = O'D'$, so $SD'PO'$ is an isosceles trapezoid. It implies that $O'P = SD'$ and\n$$\nIH = 2O'P = 2SD' = IA'\n$$\n\nThus, $IA' = IH = IO$, so $A'$ lies on the circle $(I, IO)$. Similarly, we also have $B'$, $C'$ lie on the circle $(I, IO)$ and a) is proved.\n\nb) Let $R$ be the radius of the circle $(O)$. Clearly, $GD = R$. Consider the homothety with center $A$ and ratio $\\frac{1}{2}$; it turns $B$, $C$, $A'$, $X$, $H$, $M$ and $BC$ into $F$, $E$, $D'$, $U$, $G$, $M'$ and $EF$ respectively. Therefore, $\\frac{MB}{MC} = \\frac{M'F}{M'E}$ and $U$ is the reflection of $D'$ through $EF$. Thus,\n$$\n\\frac{MB}{MC} = \\frac{M'F}{M'E} = \\frac{GF}{GE} \\cdot \\frac{UF}{UE} = \\frac{\\sqrt{R^2 - DF^2}}{\\sqrt{R^2 - DE^2}} \\cdot \\frac{D'E}{D'F}\n$$\nSimilarly, we can also calculate $\\frac{NC}{NA}$ and $\\frac{KA}{KB}$.\n\nSince $DD'$, $EE'$, $FF'$ are concurrent, we have\n$$\n\\frac{D'F}{D'E} \\cdot \\frac{F'E}{F'D} \\cdot \\frac{E'D}{E'F} = 1.\n$$\nFrom there,\n$$\n\\frac{MB}{MC} \\cdot \\frac{NC}{NA} \\cdot \\frac{KA}{KB} = 1,\n$$\nand then applying Menelaus' theorem, one could get $M$, $N$, and $K$ are collinear. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20643,
"subject": "Mathematics (Olympiad)",
"question": "Let $(a_i)_{i \\in \\mathbb{N}_0}$ be a periodic sequence of positive integers with period $d$. Let $b_0 = a_0$ and recursively define $b_{i+1} = a_{i+1}^{b_i}$ for $i = 0, 1, 2, \\dots$. Show that $(b_i)_{i \\in \\mathbb{N}_0}$ is eventually periodic modulo $n$ with period $d$ for all $n \\in \\mathbb{N}$.",
"options": [],
"answer": "See solution",
"solution": "First, $(b_i)_{i \\in \\mathbb{N}_0}$ is bounded if and only if $a_i = 1$ for some $i = 0, \\dots, d-1$. In this case, $b_{i+j d} = 1$ for all $j \\in \\mathbb{N}$, so by the recursion, $(b_i)_{i \\in \\mathbb{N}_0}$ is actually periodic with period $d$, which is stronger than required.\n\nAssume now that $(b_i)_{i \\in \\mathbb{N}_0}$ is unbounded. Then $b_{i+1} = a_{i+1}^{b_i} > b_i$ for all $i$. We proceed by induction on $n$.\n\nFor $n=1$, the sequence is constant. Let $n > 1$ and assume $(b_i)_{i \\in \\mathbb{N}_0}$ is eventually periodic modulo $n'$ with period $d$ for all $n' < n$. Fix $i \\in \\{1, \\dots, d\\}$; we show $(b_{i+j d})_{j \\in \\mathbb{N}}$ is eventually constant modulo $n$, which implies eventual $d$-periodicity.\n\nLet $p_1, \\dots, p_k$ be the primes dividing both $a_i$ and $n$, and write $n = p_1^{e_1} \\dots p_k^{e_k} \\cdot m$ with $p_l \\nmid m$ for all $l$. Since $(b_i)$ is strictly increasing, for large $j$, $b_{i-1+j d} \\ge \\max(e_1, \\dots, e_k)$, so $p_1^{e_1} \\dots p_k^{e_k}$ divides $b_{i+j d} = a_i^{b_i}$.\n\nWe show that eventually\n\n$$\nb_{i+j d} \\equiv b_{i+(j+1) d} \\pmod{m}\n$$\n\nwhich implies $(b_i)$ is eventually periodic modulo $n$.\n\nBy induction, from some $k_0$, $(b_{i-1+j d})_{j \\in \\mathbb{N}_0}$ is constant modulo $\\varphi(m)$ (Euler's totient), since $\\varphi(m) < n$. By Euler-Fermat, for $k > k_0$,\n\n$$\nb_{i-1+j d} \\equiv b_{i-1+(j+1) d} \\pmod{\\varphi(m)} \\implies a_i^{b_{i-1+j d}} \\equiv a_i^{b_{i-1+(j+1) d}} \\pmod{m}\n$$\n\nBy the $d$-periodicity of $(a_i)$, $a_i = a_{i-1+j d} = a_{i-1+(j+1) d}$, so\n\n$$\na_{i+j d}^{b_{i-1+j d}} \\equiv a_{i+(j+1) d}^{b_{i-1+(j+1) d}} \\pmod{m} \\iff b_{i+j d} \\equiv b_{i+(j+1) d} \\pmod{m}\n$$\n\nThis proves the desired $d$-periodicity, completing the induction.",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 20644,
"subject": "Mathematics (Olympiad)",
"question": "The positive real numbers $a$, $b$, $c$ satisfy $a^2 + b^2 + c^2 = 3$. Prove that\n\n$$\n\\frac{a^2 + b^2}{2ab} + \\frac{b^2 + c^2}{2bc} + \\frac{c^2 + a^2}{2ca} + \\frac{2(ab + bc + ca)}{3} \\geq 5.\n$$\n\nWhen does equality hold?",
"options": [],
"answer": "See solution",
"solution": "Since $a^2 + b^2 + c^2 = 3$, the inequality is equivalent to\n\n$$\n\\left( \\frac{a^2 + b^2}{2ab} + 1 \\right) + \\left( \\frac{b^2 + c^2}{2bc} + 1 \\right) + \\left( \\frac{c^2 + a^2}{2ca} + 1 \\right) + \\left( \\frac{2(ab + bc + ca)}{a^2 + b^2 + c^2} + 1 \\right) \\geq 9.\n$$\n\nwhich can be written as\n\n$$\n\\frac{(a+b)^2}{2ab} + \\frac{(b+c)^2}{2bc} + \\frac{(c+a)^2}{2ca} + \\frac{(a+b+c)^2}{a^2 + b^2 + c^2} \\geq 9.\n$$\n\nLet $L$ be the left-hand side of the last inequality. By the Cauchy-Schwarz inequality (Engel or Andreescu form),\n\n$$\nL \\geq \\frac{((a+b) + (b+c) + (c+a) + (a+b+c))^2}{2ab + 2bc + 2ca + a^2 + b^2 + c^2} = \\frac{9(a+b+c)^2}{(a+b+c)^2} = 9.\n$$\n\nEquality holds if and only if\n\n$$\n\\frac{a+b}{2ab} = \\frac{b+c}{2bc} = \\frac{c+a}{2ca} = \\frac{a+b+c}{3}.\n$$\n\nSince $a, b, c > 0$, from the first two equations we get $a = b = c$, and using the last equation we find $a = b = c = 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20645,
"subject": "Mathematics (Olympiad)",
"question": "Given the equations:\n\n- $ab + c = 13$\n- $a + bc = 23$\n\nwhere $a, b, c$ are positive integers, find all possible ordered triples $(a, b, c)$ that satisfy both equations.",
"options": [],
"answer": "See solution",
"solution": "If $b = 2$, solving $a + c = 12$ and $c - a = 10$ gives $(a, b, c) = (1, 2, 11)$.\n\nIf $b = 3$, solving $a + c = 9$ and $c - a = 5$ gives $(a, b, c) = (2, 3, 7)$.\n\nIf $b = 11$, solving $a + c = 3$ and $c - a = 1$ gives $(a, b, c) = (1, 11, 2)$.\n\nAll these triples satisfy the original equations, so the solutions are $(1, 2, 11)$, $(1, 11, 2)$, and $(2, 3, 7)$.\n\n**Alternate Solution:**\n\nSince $ab + c = 13$, $abc = (13 - c)c = -(c - \\frac{13}{2})^2 + \\frac{169}{4} \\leq \\frac{169}{4}$. As $a, b, c$ are positive integers, $abc \\leq 42$. Also, $a + bc = 23$, so possible $a$ values are $1, 2, 21, 22$.\n\nIf $a = 21$ or $a = 22$, $ab + c > 21$, so $ab + c = 13$ is not possible.\n\nIf $a = 1$, solving $b + c = 13$, $bc = 22$ gives $(b, c) = (2, 11)$ or $(11, 2)$.\n\nIf $a = 2$, solving $2b + c = 13$, $bc = 21$ gives $(b, c) = (3, 7)$.\n\nThus, the solutions are $(1, 2, 11)$, $(1, 11, 2)$, and $(2, 3, 7)$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20646,
"subject": "Mathematics (Olympiad)",
"question": "For integral $m$, let $p(m)$ be the greatest prime divisor of $m$. By convention, we set $p(\\pm 1) = 1$ and $p(0) = \\infty$. Find all polynomials $f$ with integer coefficients such that the sequence $\\{p(f(n^2)) - 2n\\}_{n \\ge 0}$ is bounded above. (In particular, this requires $f(n^2) \\ne 0$ for $n \\ge 0$.)",
"options": [],
"answer": "See solution",
"solution": "The polynomial $f$ has the required properties if and only if\n$$\nf(x) = c(4x - a_1^2)(4x - a_2^2)\\cdots(4x - a_k^2),\n$$\nwhere $a_1, a_2, \\dots, a_k$ are odd positive integers and $c$ is a nonzero integer. It is straightforward to verify that polynomials given by this form have the required property. If $p$ is a prime divisor of $f(n^2)$ but not of $c$, then $p \\mid (2n - a_j)$ or $p \\mid (2n + a_j)$ for some $j \\leq k$. Hence $p - 2n \\leq \\max\\{a_1, a_2, \\dots, a_k\\}$. The prime divisors of $c$ form a finite set and do not affect whether or not the given sequence is bounded above. The rest of the proof is devoted to showing that any $f$ for which $\\{p(f(n^2)) - 2n\\}_{n \\ge 0}$ is bounded above is given by this form.\n\nLet $\\mathbb{Z}[x]$ denote the set of all polynomials with integer coefficients. Given $f \\in \\mathbb{Z}[x]$, let $\\mathcal{P}(f)$ denote the set of those primes that divide at least one of the numbers in the sequence $\\{f(n)\\}_{n \\ge 0}$. The solution is based on the following lemma.\n\n**Lemma**: If $f \\in \\mathbb{Z}[x]$ is a nonconstant polynomial then $\\mathcal{P}(f)$ is infinite.\n\n*Proof*: Repeated use will be made of the following basic fact: if $a$ and $b$ are distinct integers and $f \\in \\mathbb{Z}[x]$, then $a-b$ divides $f(a)-f(b)$. If $f(0)=0$, then $p$ divides $f(p)$ for every prime $p$, so $\\mathcal{P}(f)$ is infinite. If $f(0)=1$, then every prime divisor $p$ of $f(n!)$ satisfies $p>n$. Otherwise $p$ divides $n!$, which in turn divides $f(n!)-f(0)=f(n!)-1$. This yields $p \\mid 1$, which is false. Hence $f(0)=1$ implies that $\\mathcal{P}(f)$ is infinite. To complete the proof, set $g(x) = f(f(0)x)/f(0)$ and observe that $g \\in \\mathbb{Z}[x]$ and $g(0)=1$. The preceding argument shows that $\\mathcal{P}(g)$ is infinite, and it follows that $\\mathcal{P}(f)$ is infinite. $\\blacksquare$\n\nSuppose $f \\in \\mathbb{Z}[x]$ is nonconstant and there exists a number $M$ such that $p(f(n^2)) - 2n \\leq M$ for all $n \\geq 0$. Application of the lemma to $f(x^2)$ shows that there is an infinite sequence of distinct primes $\\{p_j\\}$ and a corresponding infinite sequence of nonnegative integers $\\{k_j\\}$ such that $p_j \\mid f(k_j^2)$ for all $j \\geq 1$. Consider the sequence $\\{r_j\\}$ where $r_j = \\min\\{k_j \\bmod p_j, p_j - k_j \\bmod p_j\\}$. Then $0 \\leq r_j \\leq (p_j-1)/2$ and $p_j \\mid f(r_j^2)$. Hence $2r_j+1 \\leq p_j \\leq p(f(r_j^2)) \\leq M+2r_j$, so $1 \\leq p_j - 2r_j \\leq M$ for all $j \\geq 1$. It follows that there is an integer $a_1$ such that $1 \\leq a_1 \\leq M$ and $a_1 = p_j - 2r_j$ for infinitely many $j$. Let $m = \\deg f$. Then $p_j \\mid 4^m f\\left(\\left(\\frac{p_j-a_1}{2}\\right)^2\\right)$ and $4^m f\\left(\\left(\\frac{x-a_1}{2}\\right)^2\\right) \\in \\mathbb{Z}[x]$. Consequently, $p_j \\mid f\\left(\\left(\\frac{a_1}{2}\\right)^2\\right)$ for infinitely many $j$, which shows that $\\left(\\frac{a_1}{2}\\right)^2$ is a zero of $f$. Since $f(n^2) \\ne 0$ for $n \\geq 0$, $a_1$ must be odd. Then $f(x) = (4x-a_1^2)g(x)$ where $g \\in \\mathbb{Z}[x]$. (See the note below.) Observe that $\\{p(g(n^2)) - 2n\\}_{n \\geq 0}$ must be bounded above. If $g$ is constant, we are done. If $g$ is nonconstant, the argument can be repeated to show that $f$ is given by the required form.\n\n**Note:** The step that gives $f(x) = (4x - a_1^2)g(x)$ where $g \\in \\mathbb{Z}[x]$ follows immediately using a lemma of Gauss. The use of such an advanced result can be avoided by first writing $f(x) = r(4x - a_1^2)g(x)$ where $r$ is rational and $g \\in \\mathbb{Z}[x]$. Then continuation gives $f(x) = c(4x - a_1^2) \\cdots (4x - a_k^2)$ where $c$ is rational and the $a_i$ are odd. Consideration of the leading coefficient shows that the denominator of $c$ is $2^s$ for some $s \\geq 0$ and consideration of the constant term shows that the denominator is odd. Hence $c$ is an integer.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20647,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be positive real numbers such that $a > b + c$. Prove that\n\n$$\n(a^2 - b^2 - c^2)(a^5 - b^5 - c^5) \\leq (a^3 - b^3 - c^3)(a^4 - b^4 - c^4).\n$$",
"options": [],
"answer": "See solution",
"solution": "We shall apply the following inequality due to Aczél:\n\n*Lemma.* Let $n \\in \\mathbb{Z}_+$, $t, u \\in \\mathbb{R}$, and $x, y \\in \\mathbb{R}^n$ so that $t^2 > |x|^2$ and $u^2 > |y|^2$. Then\n\n$$\n(tu - x \\cdot y)^2 \\geq (t^2 - |x|^2)(u^2 - |y|^2).\n$$\n\nAssume this lemma is true. Since $a > b + c$, we have $a^k > (b + c)^k > b^k + c^k$ for each $k \\in \\{2, 3, 4, 5\\}$, so we may apply the lemma twice to estimate:\n\n$$\n\\begin{aligned}\n(a^2 - b^2 - c^2)(a^5 - b^5 - c^5) &\\leq \\frac{(a^2 - b^2 - c^2)(a^4 - b^4 - c^4)^2}{a^3 - b^3 - c^3} \\\\\n&\\leq \\frac{(a^2 - b^2 - c^2)(a^4 - b^4 - c^4)}{a^3 - b^3 - c^3} \\cdot \\frac{(a^3 - b^3 - c^3)^2}{a^2 - b^2 - c^2} \\\\\n&= (a^3 - b^3 - c^3)(a^4 - b^4 - c^4).\n\\end{aligned}\n$$\n\nThus, it remains to prove the lemma.\n\n*Proof of the Lemma.* By homogeneity, it suffices to prove the lemma for $t = u = 1$, $|x| < 1$, and $|y| < 1$. Let $\\varphi \\in \\mathbb{R}$ such that $x \\cdot y = |x| \\cdot |y| \\cos \\varphi$. The lemma becomes:\n\n$$\n|x|^2 + |y|^2 + |x|^2 |y|^2 \\cos^2 \\varphi \\geq 2|x| |y| \\cos \\varphi + |x|^2 |y|^2.\n$$\n\nSince $|x|^2 \\geq |x|^2 |y|^2 |\\cos \\varphi|$ and $1 \\geq |\\cos \\varphi|$, by the rearrangement inequality:\n\n$$\n|x|^2 + |x|^2 |y|^2 \\cos^2 \\varphi \\geq |x|^2 |\\cos \\varphi| + |x|^2 |y|^2 |\\cos \\varphi|.\n$$\n\nSimilarly, since $|y|^2 \\geq |x|^2 |y|^2$ and $1 \\geq |\\cos \\varphi|$:\n\n$$\n|y|^2 + |x|^2 |y|^2 |\\cos \\varphi| \\geq |y|^2 \\cos \\varphi + |x|^2 |y|^2.\n$$\n\nCombining these estimates gives:\n\n$$\n\\begin{aligned}\n|x|^2 + |y|^2 + |x|^2 |y|^2 \\cos^2 \\varphi &\\geq |x|^2 |\\cos \\varphi| + |y|^2 + |x|^2 |y|^2 |\\cos \\varphi| \\\\\n&\\geq |x|^2 |\\cos \\varphi| + |y|^2 |\\cos \\varphi| + |x|^2 |y|^2 \\\\\n&\\geq 2|x| |y| \\cos \\varphi + |x|^2 |y|^2,\n\\end{aligned}\n$$\n\nas required.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20648,
"subject": "Mathematics (Olympiad)",
"question": "Given a regular polygon $A_1A_2\\ldots A_{2010}$ centered at $O$. On each segment $OA_k$, for $k = 1, 2, \\ldots, 2010$, there is a point $B_k$ such that\n$$\n\\frac{OB_k}{OA_k} = \\frac{1}{k}.\n$$\nDetermine the ratio between the area of the polygon $B_1B_2\\ldots B_{2010}$ and that of $A_1A_2\\ldots A_{2010}$.",
"options": [],
"answer": "See solution",
"solution": "Let $S$ be the area of the polygon $A_1A_2\\ldots A_{2010}$.\n\nSince the polygon is regular, $[OA_1A_2] = [OA_2A_3] = \\ldots = [OA_{2010}A_1] = \\frac{1}{2010}S$.\n\nFor each $k = 1, 2, \\ldots, 2009$,\n$$\n\\frac{[OB_k B_{k+1}]}{[OA_k A_{k+1}]} = \\frac{OB_k \\cdot OB_{k+1}}{OA_k \\cdot OA_{k+1}} = \\frac{1}{k(k+1)},\n$$\nand\n$$\n\\frac{[OB_{2010}B_1]}{[OA_{2010}A_1]} = \\frac{OB_{2010} \\cdot OB_1}{OA_{2010} \\cdot OA_1} = \\frac{1}{2010}.\n$$\n\nLet $T$ be the area of the polygon $B_1B_2\\ldots B_{2010}$. Then\n$$\n\\begin{align*}\nT &= \\sum_{k=1}^{2009} [OB_k B_{k+1}] + [OB_{2010} B_1] \\\\\n&= \\frac{S}{2010} \\left( \\sum_{k=1}^{2009} \\frac{1}{k(k+1)} + \\frac{1}{2010} \\right) \\\\\n&= \\frac{S}{2010} \\left( \\sum_{k=1}^{2009} \\left( \\frac{1}{k} - \\frac{1}{k+1} \\right) + \\frac{1}{2010} \\right) \\\\\n&= \\frac{S}{2010} \\left( 1 - \\frac{1}{2010} + \\frac{1}{2010} \\right) = \\frac{S}{2010}.\n\\end{align*}\n$$\n\nTherefore, $\\frac{T}{S} = \\frac{1}{2010}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20649,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ and $b$ be two distinct positive integers having the same parity.\n\nProve that $\\frac{a! + b!}{2^a}$ is not an integer.\n\n(Note: $n! = 1 \\cdot 2 \\cdot \\dots \\cdot n$, for any positive integer $n$.)",
"options": [],
"answer": "See solution",
"solution": "First, we prove that for any positive integer $n$, $2^n$ does not divide $n!$.\n\nAssume the contrary. In the prime factorization of $n!$, the exponent of $2$ is $\\lfloor \\frac{n}{2} \\rfloor + \\lfloor \\frac{n}{2^2} \\rfloor + \\dots + \\lfloor \\frac{n}{2^k} \\rfloor$, where $k$ is the largest integer such that $2^k \\leq n$.\n\nBut $\\frac{n}{2} + \\frac{n}{2^2} + \\dots + \\frac{n}{2^k} \\geq \\lfloor \\frac{n}{2} \\rfloor + \\lfloor \\frac{n}{2^2} \\rfloor + \\dots + \\lfloor \\frac{n}{2^k} \\rfloor \\geq n$.\n\nSo $\\frac{n}{2} + \\frac{n}{2^2} + \\dots + \\frac{n}{2^k} \\geq n$, which implies $1 - \\frac{1}{2^k} \\geq 1$, a contradiction.\n\nNow, suppose there exists $n \\in \\mathbb{N}^*$ such that $a! + b! = n \\cdot 2^a$.\n\nIf $a \\geq b + 2$, then $a! + b! = b! \\cdot \\left(1 + (b+1) \\cdot (b+2) \\cdots a\\right) = n \\cdot 2^a$. Since $1 + (b+1) \\cdot (b+2) \\cdots a$ is odd, $2^a$ must divide $b!$. But $b!$ divides $a!$, so $2^a$ divides $a!$, which is false.\n\nIf $b \\geq a + 2$, a similar argument shows $2^a$ divides $a!$, also false.\n\nTherefore, $\\frac{a! + b!}{2^a}$ is not an integer.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20650,
"subject": "Mathematics (Olympiad)",
"question": "Find all real $x > 0$ and integer $n > 0$ such that $$\\lfloor x \\rfloor + \\left\\{ \\frac{1}{x} \\right\\} = 1.005 n.$$",
"options": [],
"answer": "See solution",
"solution": "Rewrite the equation as $$\\lfloor x \\rfloor + \\left\\{ \\frac{1}{x} \\right\\} = n + \\frac{n}{200}.$$ Let $n = 200q + r$ with $0 \\leq r < 200$. Then $$\\lfloor x \\rfloor + \\left\\{ \\frac{1}{x} \\right\\} = 200q + r + q + \\frac{r}{200}.$$ The integer part of the left side is $\\lfloor x \\rfloor$, and of the right side is $200q + r + q = 201q + r$, so $\\lfloor x \\rfloor = 201q + r$ and $\\left\\{ \\frac{1}{x} \\right\\} = \\frac{r}{200}$.\n\nIf $x < 1$, then $\\lfloor x \\rfloor + \\left\\{ \\frac{1}{x} \\right\\} = \\left\\{ \\frac{1}{x} \\right\\} < 1 < 1.005 n$ for any $n > 0$. If $x = 1$, then $1 = 1.005 n$, which is impossible. Thus, $x > 1$, so $0 < \\frac{1}{x} < 1$, and $\\left\\{ \\frac{1}{x} \\right\\} = \\frac{1}{x}$. Therefore, $\\frac{1}{x} = \\frac{r}{200}$, so $r \\neq 0$ and $x = \\frac{200}{r}$.\n\nThe inequality $\\lfloor x \\rfloor \\leq x < \\lfloor x \\rfloor + 1$ gives $201q + r \\leq \\frac{200}{r} < 201q + r + 1$. Thus, $201qr + r^2 \\leq 200 < 201qr + r(r+1)$. Since $q, r \\in \\mathbb{N}$, these cannot hold for $qr \\geq 1$, so $qr = 0$. Since $r \\neq 0$, $q = 0$, so $r^2 \\leq 200 < r(r+1)$, which leads to $r = 14$, $n = 14$, and $x = \\frac{200}{14} = \\frac{100}{7}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20651,
"subject": "Mathematics (Olympiad)",
"question": "Let $c_n$ be the number of coins on the $n$th minute. Consider the ordered pair $(x_n, y_n) = (c_n, n - c_n)$, noting that each minute either $x_n$ or $y_n$ increases by $1$. The number written on the $n$th minute is\n\n$$\n2c_n^2 - n^2 = 2x_n^2 - (x_n + y_n)^2 = x_n^2 - 2x_n y_n - y_n^2.\n$$\n\nIt thus suffices to prove the slightly more general problem:\n\nAlice starts at an arbitrary integer lattice point $(x, y) = (x_0, y_0)$. Every minute, she walks one unit right or up (corresponding to increasing $x$ or $y$ by $1$), then writes down the number $f(x, y) = x^2 - 2 x y - y^2$. Show that for any positive integer $m$, there will eventually be two numbers on the board that sum to a multiple of $m$.\n\nThe function $f$ has the special property of $90^\\circ$ rotational anti-symmetry, i.e., taken mod $m$:\n\n$$\nf(x, y) + f(y, -x) \\equiv 0 \\pmod{m}\n$$",
"options": [],
"answer": "See solution",
"solution": "**Claim:** Among the numbers written over $4m$ minutes, there is either (a) two numbers that sum to a multiple of $m$, or (b) a number which is a multiple or half-multiple of $m$.\n\n**Proof of Claim:** Suppose Bob starts at integer lattice point $(u, v) = (u_0, v_0) \\equiv (y_0, -x_0) \\pmod{m}$. Each minute, if Alice moves right, Bob moves down; if Alice moves up, Bob moves right.\n\nIn this manner, Alice's and Bob's coordinates are always related by $(u, v) \\equiv (y, -x) \\pmod{m}$. In particular, whenever Alice writes down the number $f(x, y)$, Bob writes down the number $f(u, v) \\equiv -f(x, y) \\pmod{m}$.\n\nIt suffices to show that Alice's and Bob's paths coincide on some lattice point $(x^*, y^*)$. If this occurs, then either\n\n1. (a) Alice and Bob reach $(x^*, y^*)$ at different times, thus Alice has written down $f(x^*, y^*)$ and $-f(x^*, y^*)$ (mod $m$) at two different times; or\n2. (b) Alice and Bob reach $(x^*, y^*)$ at the same time; thus $f(x^*, y^*) \\equiv -f(x^*, y^*)$ (mod $m$), so Alice has written down a multiple or half-multiple of $m$.\n\nWLOG among Alice's first $2m-1$ moves, she moves right at least $m$ times (the alternative is that she moves up at least $m$ times, in which case a similar argument holds by swapping the axes).\n\nThen, within Alice's first $2m-1$ moves, she passes through some lattice point $(u_0, w_0)$, where $x_0 < u_0 \\le x_0 + m$ and $u_0 \\equiv y_0 \\pmod{m}$. We then select Bob's starting point as $(u_0, v_0)$ where $w_0 < v_0 \\le w_0 + m$ and $v_0 \\equiv -x_0 \\pmod{m}$.\n\nConsequently, the path $\\mathcal{L}_{\\text{Bob}}$ traversed by Bob within his first $2m-1$ moves starts vertically above $(u_0, w_0)$, and at distance $\\le m$ from $(u_0, w_0)$. Since Bob moves down at least $m$ times and moves right at most $m-1$ times within his first $2m-1$ moves, his path $\\mathcal{L}_{\\text{Bob}}$ traverses within the upper-right quadrant $\\{(x, y) : x \\ge u_0, y \\ge w_0\\}$ with respect to $(u_0, w_0)$, and leaves the quadrant from the bottom edge at a distance $\\le m$ from $(u_0, w_0)$ (see Figure 1).\n\nThus, a further $2m$ moves by Alice starting from $(u_0, w_0)$ will take her from below $\\mathcal{L}_{\\text{Bob}}$ to above it, guaranteeing a lattice point intersection with $\\mathcal{L}_{\\text{Bob}}$.\n\nSince two multiples of $m$ or two half-multiples of $m$ sum to a multiple of $m$, thus by the claim, Alice is guaranteed to write two numbers that sum to a multiple of $m$ within $3 \\cdot 4m = 12m$ moves.\n\n\n\nFigure 1: Sample game for $m=6$. Alice's path is lightly shaded and starts at $(x_0, y_0) = (2, 1)$ which is at the bottom. Bob's path is heavily shaded and starts at $(u_0, v_0) = (7, 10)$ at the top. Alice's path after $(u_0, w_0)$ which is at the bottom left corner of the box, is guaranteed to intersect Bob's path; the intersection is the square with 2. When Bob is at the intersection, Alice has also made 5 moves and is at the square with 4. The numbers written by Alice in these 2 squares sum to a multiple of $m=6$.",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 20652,
"subject": "Mathematics (Olympiad)",
"question": "Some unit squares of a $2007 \\times 2007$ square board are colored. Let $(i, j)$ be a unit square belonging to the $i$-th row and $j$-th column, and let $S_{i,j}$ be the set of all colored unit squares $(x, y)$ satisfying $x \\le i$ and $y \\le j$. \n\nAt the first step, in each colored unit square $(i, j)$, we write the number of colored unit squares in $S_{i,j}$. In each subsequent step, in each colored unit square $(i, j)$, we write the sum of all numbers written in $S_{i,j}$ in the previous step. \n\nProve that after a finite number of steps, all numbers in the colored unit squares will be odd.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $f_{(i,j)}$ be the number written on $(i, j)$ modulo $2$. We can suppose that at the $0$th step, $f_{(i,j)} = 1$ for all colored unit squares $(i, j)$. We prove the statement by induction with respect to $n$, the total number of colored unit squares.\n\nIf $n=1$, then after the first step $f_{(i,j)} = 1$ for the only colored unit square $(i, j)$.\n\nSuppose that the statement is proved for $n=k$ and consider the case $n=k+1$. Define a partial order between colored unit squares: we say that $(i, j) \\le (k, l)$ if $i \\le k$ and $j \\le l$. Let $(p, q)$ be any maximal element with respect to this order (there is at least one maximal element). The unit square $(p, q)$ has no influence on other colored unit squares at any step.\n\nIf we remove $(p, q)$, then by the inductive hypothesis, after $N$ steps, the number $1$ will be written in each of the remaining unit squares. Therefore, if we do not remove $(p, q)$, after $N$ steps $f_{(i,j)} = 1$ for all unit squares except possibly for $f_{(p,q)}$. If $f_{(p,q)} = 1$, we are done. Suppose that $f_{(p,q)} = 0$. This means that the first $N$ steps have added $1$ to $f_{(p,q)}$ and it has changed from $1$ to $0$. Therefore, after $2N$ steps, $f_{(i,j)} = 1$ for all colored unit squares.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20653,
"subject": "Mathematics (Olympiad)",
"question": "For which positive integers $n$ can one exactly cover an equilateral triangle of side length $n$ by trapeziums of the shape shown in the figure, consisting of three equilateral triangles of side length 1? Trapeziums are allowed to be rotated but not to cover each other.\n\n",
"options": [],
"answer": "See solution",
"solution": "An equilateral triangle of side length 3 can be covered by three trapeziums (see the figure below). All equilateral triangles with side length divisible by 3 can be partitioned into equilateral triangles of side length 3. Hence, all equilateral triangles with side length divisible by 3 can be covered by trapeziums of the given shape.\n\n\n\nOn the other hand, whenever one partitions an equilateral triangle of side length $n$ into equilateral triangles of side length 1, the number of the small triangles is $n^2$, since multiplying the side length by $n$ causes the area to increase by $n^2$ times. Consequently, the desired covering is possible only if the number $n^2$ is divisible by 3. This condition implies that $n$ must be divisible by 3, since 3 is prime.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20654,
"subject": "Mathematics (Olympiad)",
"question": "For a positive integer $n$, consider all nonincreasing functions $f: \\{1, \\dots, n\\} \\to \\{1, \\dots, n\\}$. Some of them have a fixed point ($c = f(c)$), some don't. Which are more numerous? Evaluate the difference between the sizes of the two sets of functions.",
"options": [],
"answer": "See solution",
"solution": "Lemma: Let $A$ be a set of $a$ consecutive integers and $B$ a set of $b$ consecutive integers. There are exactly $\\binom{a+b-1}{a}$ nonincreasing functions $f: A \\to B$.\n\n*Proof.* Let $A = \\{1, \\dots, a\\}$, $B = \\{1, \\dots, b\\}$. A function $f: A \\to B$ is nonincreasing if and only if the function $g: A \\to C = \\{1, \\dots, a+b-1\\}$ given by $g(x) = f(x) + (a-x)$ is strictly decreasing. Any such $g$ is uniquely identified with its set of values $g(A)$, which has size $a$. So there are as many $f$ as $a$-element subsets of $C$; thus, the lemma follows.\n\nNow, count the functions of the two types:\n\nA function $f$ with a fixed point $c \\in \\{1, \\dots, n\\}$ maps $\\{1, \\dots, c-1\\}$ (nonincreasingly) into $\\{c, \\dots, n\\}$, and $\\{c+1, \\dots, n\\}$ (nonincreasingly) into $\\{1, \\dots, c\\}$. By the lemma, for the left side ($a = c-1$, $b = n-c+1$) there are $\\binom{n-1}{c-1}$ choices; for the right side ($a = n-c$, $b = c$) also $\\binom{n-1}{c-1}$ choices. Thus, the number of nonincreasing functions with a fixed point is\n\n$$\nF_n = \\sum_{c=1}^{n} \\binom{n-1}{c-1}^2.\n$$\n\nIf $f$ has no fixed point, then there exists $c \\in \\{1, \\dots, n-1\\}$ such that $f(x) > x$ for $x \\leq c$ and $f(x) < x$ for $x > c$. Similarly, $f$ maps $\\{1, \\dots, c\\}$ into $\\{c+1, \\dots, n\\}$ and $\\{c+1, \\dots, n\\}$ into $\\{1, \\dots, c\\}$, both nonincreasingly. For the left ($a = c$, $b = n-c$) there are $\\binom{n-1}{c}$ choices; for the right ($a = n-c$, $b = c$) $\\binom{n-1}{c-1}$ choices. Thus, the number of nonincreasing functions without a fixed point is\n\n$$\nG_n = \\sum_{c=1}^{n-1} \\binom{n-1}{c} \\binom{n-1}{c-1}.\n$$\n\nConsider $P_n(x) = (1+x)^{n-1}$. We recognize $F_n$ and $G_n$ as the coefficients of $x^{n-1}$ and $x^n$ in $P_n(x)P_n(x) = (1+x)^{2n-2}$. Thus,\n\n$$\nF_n = \\binom{2n-2}{n-1}, \\quad G_n = \\binom{2n-2}{n},\n$$\n\nand the difference is\n\n$$\nF_n - G_n = \\binom{2n-2}{n-1} - \\binom{2n-2}{n} = \\binom{2n-2}{n-1} - \\binom{2n-2}{n-1} = 0.\n$$\n\nThus, the two sets are equally numerous.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20655,
"subject": "Mathematics (Olympiad)",
"question": "Consider a semicircle with center $O$ and diameter $AB$. Let $C$ be an arbitrary point on the line segment $OB$. The perpendicular to the line $AB$ at $C$ meets the semicircle at $D$. A circle centered at $P$ is tangent to the arc $BD$ at point $F$ and to the segments $AB$ and $CD$ at points $G$ and $E$, respectively. Prove that the triangle $ADG$ is isosceles.",
"options": [],
"answer": "See solution",
"solution": "Notice that the lines $PE$ and $AB$ are parallel and $\\frac{FP}{FO} = \\frac{PE}{OA}$, so the points $A$, $E$, and $F$ are collinear. By the power of point $A$ with respect to the circle $C(P, PG)$, we have $AE \\cdot AF = AG^2$.\n\nOn the other hand, $\\angle DFA = \\angle DBA = 90^\\circ - \\angle DAC = \\angle ADC$. Thus, triangles $DFA$ and $EDA$ are similar, so $AE \\cdot AF = AD^2$. Combining $AE \\cdot AF = AG^2$ and $AE \\cdot AF = AD^2$ gives $AD = AG$, and the conclusion follows.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20656,
"subject": "Mathematics (Olympiad)",
"question": "A positive integer $n$ is *downhill* if its decimal representation $\\overline{a_k a_{k-1} \\dots a_0}$ satisfies $a_k \\ge a_{k-1} \\ge \\dots \\ge a_0$. A real-coefficient polynomial $P$ is *integer-valued* if $P(n)$ is an integer for all integer $n$, and *downhill-integer-valued* if $P(n)$ is an integer for all downhill positive integers $n$. Is it true that every downhill-integer-valued polynomial is also integer-valued?",
"options": [],
"answer": "See solution",
"solution": "No, it is not.\n\nA downhill number can always be written as $a - b_1 - b_2 - \\dots - b_9$, where $a$ is of the form $\\overline{99\\dots99}$ and each $b_i$ either equals $0$ or is of the form $\\overline{11\\dots11}$.\n\nLet $n$ be a positive integer. The numbers of the form $\\overline{99\\dots99}$ yield at most $n$ different remainders upon division by $2^n$, as do the numbers of the form $\\overline{11\\dots11}$. Therefore, downhill numbers yield at most $n(n+1)^9$ different remainders upon division by $2^n$.\n\nLet $n$ be so large that $n(n+1)^9 < 2^n$. ($n = 63$ works: $63 \\times 64^9 < 64^{10} = 2^{60} < 2^{63}$.) Let $0 \\le r < 2^n$ be such that no downhill number is congruent to $r$ modulo $2^n$.\n\nConsider the polynomial\n\n$$\nP(x) = \\frac{1}{2 \\times (2^n - 1)!} \\prod_{1 \\le i < 2^n} (x - r + i).\n$$\n\nWe have that $P(r) = \\frac{1}{2}$ is not an integer.\n\nLet, then, $x$ be a downhill number. The number $(x - r + 1) \\dots (x - r + 2^n - 1)$ is a multiple of $(2^n - 1)!$ (as a product of $2^n - 1$ consecutive integers); therefore, $2P(x)$ is an integer. On the other hand, the number $(x - r)(x - r + 1) \\dots (x - r + 2^n - 1)$ is a multiple of $2^n!$ (as a product of $2^n$ consecutive integers); therefore, $2(x - r)P(x)$ is an integer multiple of $2^n$. Since $x$ is downhill, $x - r$ is not divisible by $2^n$. Therefore, $2P(x)$ is even and $P(x)$ is an integer.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20657,
"subject": "Mathematics (Olympiad)",
"question": "Let $N_i$ be the number written on the card placed on the $i$$^{\\text{th}}$ position from the left, for $1 \\leq i \\leq 5$. If $N_i \\geq i$ is satisfied for all $i$, what is the probability that the cards are arranged in this way, given that the cards are numbered $1$ through $6$ and no two cards have the same number?",
"options": [],
"answer": "See solution",
"solution": "If $N_5 \\geq 5$, then $N_5$ can be $5$ or $6$ (2 possibilities). For each $N_5$, $N_4 \\geq 4$ and $N_4 \\neq N_5$, so $N_4$ can be $4$, $5$, or $6$ except $N_5$ (2 possibilities). For each $N_4, N_5$, $N_3 \\geq 3$ and $N_3 \\neq N_4, N_5$, so $N_3$ can be $3$, $4$, $5$, or $6$ except $N_4, N_5$ (2 possibilities). $N_2 \\geq 2$ and $N_1 \\geq 1$ are always satisfied. Thus, the probability is\n\n$$\n\\frac{2 \\times 2 \\times 2 \\times 2 \\times 1}{5 \\times 4 \\times 3 \\times 2 \\times 1} = \\frac{2}{15}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20658,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c$ be real numbers such that $a + b + c = 0$ and $a^2 + b^2 + c^2 = 1$. Show that $|abc| \\leq \\frac{1}{\\sqrt{54}}$.\n",
"options": [],
"answer": "See solution",
"solution": "First, $0 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$, so\n$$\nab + bc + ca = -\\frac{1}{2}.\n$$\nThus, $a, b, c$ are roots of $x^3 - \\frac{1}{2}x - abc = 0$, which has real roots. The product of the local extrema of this cubic is non-positive. These extrema occur at $x = \\pm\\sqrt{6}$, so the requirement is\n$$\n\\left(\\frac{1}{6\\sqrt{6}} - \\frac{1}{2\\sqrt{6}} - abc\\right)\\left(-\\frac{1}{6\\sqrt{6}} + \\frac{1}{2\\sqrt{6}} - abc\\right) \\le 0,\n$$\nwhich simplifies to\n$$\n\\left(-\\frac{1}{3\\sqrt{6}} - abc\\right)\\left(\\frac{1}{3\\sqrt{6}} - abc\\right) \\le 0, \\quad \\text{i.e.}\\quad (abc)^2 \\le \\frac{1}{54},\n$$\nthe desired result.\n\nIf equality occurs, then $0$ is a local max or min, so the cubic has a double root. Say $a = b$, $c = -2a$, whence $6a^2 = 1$, so equality happens iff two of $a, b, c$ are $\\pm\\frac{1}{\\sqrt{6}}$ and the third is $\\mp\\frac{2}{\\sqrt{6}}$.\n\nAlternatively, if $a, b, c$ are roots of $x^3 - px - q$, then\n$$\n(a-b)^2(b-c)^2(c-a)^2 = 4p^3 - 27q^2.\n$$\nThis implies $4p^3 - 27q^2 \\ge 0$. With $2p = a^2 + b^2 + c^2 = 1$, the result follows.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20659,
"subject": "Mathematics (Olympiad)",
"question": "Given a $p \\times p$ matrix $A$ with distinct entries, you may 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 take a finite series of such operations resulting in 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 final result of a series of operations can be realized as subtracting integer $x_i$ from each number of the $i$-th row and subtracting integer $y_j$ from each number of the $j$-th column. Thus, the matrix $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 consider only the case that $x_1 < x_2 < \\cdots < x_p$ since swapping the value of $x_i$ and $x_j$ results in swapping the $i$-th row and $j$-th row, which is again a good matrix. Similarly, we may consider only the case that $y_1 < y_2 < \\cdots < y_p$, thus the matrix is increasing from left to right, also from top to bottom.\n\nFrom the assumptions above, we have $a_{11} = 1$, and $a_{12}$ or $a_{21}$ equals $2$. We may consider only the case that $a_{12} = 2$ since the transpose of the matrix is again good. Now we argue by contradiction that the first row is $1, 2, \\dots, p$. Assume on the contrary that $1, 2, \\dots, k$ is on the first row, but $k+1$ is not, $2 \\leq k < p$, therefore $a_{21} = k+1$. We call $k$ consecutive integers a *block*, and we shall prove that the first row consists of several blocks, that is, the first $k$ numbers is a block, the next $k$ numbers is again a block, and so on.\n\nIf it is not so, assume the first $n$ groups of $k$ numbers are *blocks*, but the next $k$ numbers is not a *block* (or there are no $k$ numbers remaining). It follows that for $j = 1, 2, \\dots, n$,\n\n$y_{(j-1)k+1}, y_{(j-1)k+2}, \\dots, y_{jk}$ is a *block*, the first $nk$ columns of the matrix can be divided into $pn \\times k$ submatrices $a_{i, (j-1)k+1}, a_{i, (j-1)k+2}, \\dots, a_{i, jk}$, $i = 1, 2, \\dots, p$, $j = 1, 2, \\dots, n$, each submatrix is a *block*. Now assume $a_{1, nk+1} = a$, let $b$ be the smallest positive integer such that $a+b$ is not on the first row, then $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$, therefore $a+b$ lies in the first $nk$ columns. Therefore, $a+b$ is contained in one of the $1 \\times k$ submatrices mentioned above, which is a *block*, however $a, a+k$ are not in this *block*, which is a contradiction.\n\nWe showed that the first row is formed by blocks, in particular $k \\mid p$, however, $1 < k < p$, and $p$ is a prime, which is impossible. So we conclude that the first row is $1, 2, \\dots, p$, the $k$-th row must be $(k-1)p+1, (k-1)p+2, \\dots, kp$. Thus up to interchanging rows, columns and transpose, the good matrix is unique, the answer is therefore $2(p!)^2$. $\\square$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 20660,
"subject": "Mathematics (Olympiad)",
"question": "Let $P(x)$ and $Q(x)$ be polynomials with non-negative real coefficients, and let $P'(x)$ denote the derivative of $P(x)$. Suppose that $P(0) = Q(0) = 0$ and $Q(1) \\le 1 \\le P'(0)$.\n\n1. Prove that $0 \\le Q(x) \\le x \\le P(x)$ for all $0 \\le x \\le 1$.\n2. Prove that $P(Q(x)) \\le Q(P(x))$ for all $0 \\le x \\le 1$.\n\nIt is *not* necessary to study the conditions for equality.",
"options": [],
"answer": "See solution",
"solution": "Since $P(0) = Q(0) = 0$ and the coefficients of $P(x)$ and $Q(x)$ are non-negative, the functions $P(x)/x$ and $Q(x)/x$ are increasing for $x > 0$.\n\nLet $0 \\le x \\le 1$.\n\n1. For $x = 0$, $Q(0) = 0 = P(0)$. For $0 < x \\le 1$,\n$$\n0 \\le \\frac{Q(x)}{x} \\le \\frac{Q(1)}{1} \\le 1 \\le P'(0) \\le \\frac{P(x)}{x},\n$$\nso $0 \\le Q(x) \\le x \\le P(x)$.\n\n2. If $Q(x) = 0$, then $P(Q(x)) = P(0) = 0 \\le Q(P(x))$. For $Q(x) > 0$, $\\frac{P(Q(x))}{Q(x)} \\le \\frac{P(x)}{x}$ and $\\frac{Q(x)}{x} \\le \\frac{Q(P(x))}{P(x)}$, so\n$$\nP(Q(x)) \\le \\frac{P(x) Q(x)}{x} \\le Q(P(x)).\n$$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20661,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f$ defined on all real numbers and taking real values such that\n\n$$\nf(f(y)) + f(x - y) = f(xf(y) - x)\n$$\n\nfor all real numbers $x, y$.",
"options": [],
"answer": "See solution",
"solution": "Substituting $x = y = 0$ into the original equation gives $f(f(0)) + f(0) = f(0)$, implying\n\n$$\nf(f(0)) = 0.\n$$\n\nTaking $x = \\frac{f(0)}{2}$ and $y = f(0)$ in the original equation gives\n\n$$\nf(f(f(0))) + f\\left(-\\frac{f(0)}{2}\\right) = f\\left(\\frac{f(0)}{2} \\cdot f(f(0)) - \\frac{f(0)}{2}\\right).\n$$\n\nApplying the previous result, $f(f(0)) = 0$, leads to $f(0) + f\\left(-\\frac{f(0)}{2}\\right) = f\\left(-\\frac{f(0)}{2}\\right)$, which implies\n\n$$\nf(0) = 0.\n$$\n\nNext, substitute $y = 0$ into the original equation. Using $f(0) = 0$, we obtain $f(f(0)) + f(x) = f(xf(0) - x)$, which reduces to\n\n$$\nf(x) = f(-x)\n$$\nfor all $x$.\n\nNow, substitute $x = 0$ into the original equation. With $f(0) = 0$ and $f(x) = f(-x)$, we get $f(f(y)) + f(y) = 0$, i.e.,\n\n$$\nf(f(y)) = -f(y)\n$$\nfor all real numbers $y$.\n\nNow, for all $y$,\n\n$$\n\\begin{align*}\nf(y) &= -f(f(y)) && \\text{(from above)} \\\\\n&= f(f(f(y))) && \\text{(applying the same result)} \\\\\n&= f(-f(y)) && \\text{(since $f(x) = f(-x)$)} \\\\\n&= f(f(y)) && \\text{(since $f(x) = f(-x)$)} \\\\\n&= -f(y), && \\text{(from above)}\n\\end{align*}\n$$\n\nimplying $f(y) = 0$ for all real $y$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20662,
"subject": "Mathematics (Olympiad)",
"question": "Determine all pairs of real numbers $a$ and $x$ that satisfy the simultaneous equations\n\n$$\n5x^3 + a x^2 + 8 = 0\n$$\n\nand\n\n$$\n5x^3 + 8x^2 + a = 0.\n$$",
"options": [],
"answer": "See solution",
"solution": "If we subtract the two equations, we obtain\n\n$$\n(a x^2 + 8) - (8 x^2 + a) = (a - 8)(x^2 - 1) = (a - 8)(x + 1)(x - 1) = 0.\n$$\n\nThus, either $a = 8$, $x = -1$, or $x = 1$.\n\nIf $a = 8$, we are left with\n\n$$\n5x^3 + 8x^2 + 8 = (x + 2)(5x^2 - 2x + 4) = 0.\n$$\n\nThe second factor has no real roots, since its discriminant $2^2 - 4 \\cdot 5 \\cdot 4 = -76$ is negative. Thus $x = -2$ in this case.\n\nIf $x = -1$, we get $a = -5(-1)^3 - 8(-1)^2 = -5(-1) - 8(1) = 5 - 8 = -3$.\n\nIf $x = 1$, we get $a = -5(1)^3 - 8(1)^2 = -5(1) - 8(1) = -5 - 8 = -13$.\n\nIn summary, there are three possible pairs:\n\n- $(a, x) = (8, -2)$\n- $(a, x) = (-3, -1)$\n- $(a, x) = (-13, 1)$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20663,
"subject": "Mathematics (Olympiad)",
"question": "Four teams, A, B, C, and D, play a round-robin tournament (each team plays every other team exactly once). Each game results in a win for one team and a loss for the other (no ties). If team A beats team D and each team wins at least one game, how many games does team D win?",
"options": [],
"answer": "See solution",
"solution": "The total number of games is $\\frac{4 \\times 3}{2} = 6$. Since team A scored at least 1 win, each of teams A, B, and C has either 1 or 2 wins. Since team A beats team D, team D does not have 3 wins, so the scores of A, B, and C must total more than 3. Hence, each of teams A, B, and C has 2 wins and therefore team D has 0 wins.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20664,
"subject": "Mathematics (Olympiad)",
"question": "How many ways are there to assign $1$'s and $-1$'s to the unit squares of a $2007 \\times 2007$ chessboard so that the sum of the numbers in every $k \\times k$ square (for all $1 \\leq k \\leq 2007$) is either $0$ or has absolute value $1$?",
"options": [],
"answer": "See solution",
"solution": "Assigning $1$'s and $-1$'s alternately to the unit squares of the chessboard gives two ways to satisfy the condition.\n\nNow, consider an assignment different from these two. Then there must be two adjacent unit squares with the same number assigned. Assume these squares are in the same row. Let $x$ denote the number assigned to these squares, and $y$ the other number. Considering a $2 \\times 2$ square containing these two squares, $y$ is assigned to the squares directly above and below. Thus, $1$'s and $-1$'s alternate in horizontal pairs along the columns containing these squares.\n\nIn this assignment, we cannot also have two adjacent squares in the same column with the same number assigned. Otherwise, a similar argument would force an alternating pattern along the two rows, leading to a contradiction at their intersection. Hence, $1$'s and $-1$'s alternate along all columns, and the first row determines the entire pattern.\n\nNow, we determine which $2007$-term sequences of $1$'s and $-1$'s in the first row yield an assignment satisfying the condition. For even $k$, the sum in any $k \\times k$ square is $0$; for odd $k$, the absolute value of the sum equals the absolute value of the sum in any row of the square. Therefore, the necessary and sufficient condition is:\n\n**K:** The sum of any odd number of consecutive terms has absolute value $1$.\n\nAssociate a $2006$-term sequence of $=$'s and $\\neq$'s to any $2007$-term sequence by writing $=$ between two consecutive terms if they are the same, and $\\neq$ if they are different.\n\nFor this sequence, the equivalent condition is:\n\n**L:** All $=$'s occur at indices of the same parity.\n\nThe number of such sequences with at least one $=$ is $2 \\cdot (2^{1003} - 1)$. Each such sequence comes from two different assignments (starting with $1$ or $-1$), so the number of assignments of this type is $2^{1005} - 4$. Switching rows and columns gives $2^{1005} - 4$ more. Including the two alternating assignments, the total number is:\n\n$$\n2^{1006} - 6\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20665,
"subject": "Mathematics (Olympiad)",
"question": "Find all primes $p$, $q$, $r$ such that\n$$\n p(p-7) + q(q-7) = r(r-7).\n$$",
"options": [],
"answer": "See solution",
"solution": "The given equality rewrites as\n$$\np^2 + q^2 - r^2 = 7(p + q - r).\n$$\nSince $(p+q-r)(p+q-r) = p^2 + q^2 - r^2 + 2pq$, it follows that $p+q-r$ divides $2pq$.\n\nIf $p, q, r > 2$, then $p+q-r$ is odd, so $p+q-r = p$, $q$, or $pq$. The first case gives $r = q$, then $p = 7$, so $(p, q, r) = (7, q, q)$, $q$ being an arbitrary prime. Likewise, the second case gives $(p, q, r) = (p, 7, p)$, $p$ being an arbitrary prime. If $p+q-r = pq$, then $1-r = (p-1)(q-1)$, impossible.\n\nIf $p=2$ (or $q=2$), the equality becomes $(q-r)(q+r-7) = 10$, implying $q=7$ and $r=5$, with two new solutions obtained: $(p, q, r) = (2, 7, 5)$ and $(p, q, r) = (7, 2, 5)$. If $r=2$, no further solutions are obtained.\n\nThe solution triplets therefore are $(2, 7, 5)$, $(7, 2, 5)$, $(7, t, t)$, $(t, 7, t)$, where $t$ is an arbitrary prime. (Notice the fact $r$ was known to be a prime turned out to be inconsequential.)",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20666,
"subject": "Mathematics (Olympiad)",
"question": "Prove that for every nonnegative integer $n$, the number $7^{7^n} + 1$ is the product of at least $2n+3$ (not necessarily distinct) primes.",
"options": [],
"answer": "See solution",
"solution": "The proof is by induction.\n\n**Base case:** For $n=0$, $7^{7^0} + 1 = 7^1 + 1 = 8 = 2^3$, which is the product of $3 = 2 \\cdot 0 + 3$ primes.\n\n**Inductive step:** Assume the statement holds for $n = k$. For $n = k+1$, let $x = 7^{2^{m-1}}$ for some positive integer $m$. We show that $\\dfrac{x^7+1}{x+1}$ is composite. Thus, $x^7+1$ has at least two more prime factors than $x+1$.\n\nTo see that $\\dfrac{x^7+1}{x+1}$ is composite, observe:\n\n$$\n\\begin{aligned}\n\\frac{x^7+1}{x+1} &= \\frac{(x+1)^7 - ((x+1)^7 - (x^7+1))}{x+1} \\\\\n&= (x+1)^6 - \\frac{7x(x^5 + 3x^4 + 5x^3 + 5x^2 + 3x + 1)}{x+1} \\\\\n&= (x+1)^6 - 7x(x^4 + 2x^3 + 3x^2 + 2x + 1) \\\\\n&= (x+1)^6 - 7^{2m}(x^2 + x + 1)^2 \\\\\n&= \\left((x+1)^3 - 7^m(x^2 + x + 1)\\right)\\left((x+1)^3 + 7^m(x^2 + x + 1)\\right)\n\\end{aligned}\n$$\n\nEach factor exceeds $1$. For the smaller factor, since $\\sqrt{7x} \\le x$:\n\n$$\n\\begin{aligned}\n(x+1)^3 - 7^m(x^2+x+1) &= (x+1)^3 - \\sqrt{7x}(x^2+x+1) \\\\\n&\\geq x^3 + 3x^2 + 3x + 1 - x(x^2+x+1) \\\\\n&= 2x^2 + 2x + 1 \\geq 113 > 1.\n\\end{aligned}\n$$\n\nTherefore, $\\dfrac{x^7 + 1}{x + 1}$ is composite, and the induction is complete.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20667,
"subject": "Mathematics (Olympiad)",
"question": "Points $P$ and $Q$ are chosen uniformly and independently at random on sides $\\overline{AB}$ and $\\overline{AC}$, respectively, of equilateral triangle $\\triangle ABC$. Which of the following intervals contains the probability that the area of $\\triangle APQ$ is less than half the area of $\\triangle ABC$?\n\n(A) $[\\frac{3}{8}, \\frac{1}{2}]$ \n(B) $(\\frac{1}{2}, \\frac{2}{3}]$ \n(C) $(\\frac{2}{3}, \\frac{3}{4}]$ \n(D) $(\\frac{3}{4}, \\frac{7}{8}]$ \n(E) $(\\frac{7}{8}, 1]$",
"options": [],
"answer": "See solution",
"solution": "**Answer (D):** Without loss of generality let $AB = AC = BC = 1$; then the area of $\\triangle ABC$ is $\\frac{1}{4}\\sqrt{3}$. Let $x = AP$ and $y = AQ$. Then the area of $\\triangle APQ$ is\n\n$$\n\\frac{1}{2}xy \\cdot \\sin 60^\\circ = \\frac{1}{4}\\sqrt{3} \\cdot xy.\n$$\n\nThe probability that the area of $\\triangle APQ$ is less than half the area of $\\triangle ABC$ is therefore the probability that $xy < \\frac{1}{2}$. Graph the curve $xy = \\frac{1}{2}$ in the unit square whose lower left corner is at the origin, as shown. Note that the curve passes through the points $(\\frac{1}{2}, 1)$ and $(1, \\frac{1}{2})$ and is concave up on the interval $\\frac{1}{2} < x < 1$.\n\n\n\nThe probability that $xy > \\frac{1}{2}$ is the area of the upper right \"fat triangular\" region with curved longest side, which is less than $\\frac{1}{4}$ but greater than $\\frac{1}{8}$. Therefore the probability that $xy < \\frac{1}{2}$ lies between $1 - \\frac{1}{4} = \\frac{3}{4}$ and $1 - \\frac{1}{8} = \\frac{7}{8}$.\n\n**Note:** The required area can be calculated to be $\\frac{1}{2} + \\frac{1}{2}\\ln 2 \\approx 0.85$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20668,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ that satisfy\n\n$$\nf(f(x) + x f(y)) = x f(y + 1), \\quad \\forall x, y \\in \\mathbb{R}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $f(0) = a$, where $a \\in \\mathbb{R}$. Set $x = 0$ to obtain $f(a) = 0$. Next, set $y = a$:\n\n$$\nf(f(x)) = x f(a + 1) \\quad (1).\n$$\n\nAssume first that $f(a + 1) \\neq 0$. Then $f$ is injective. Indeed, if $f(x_1) = f(x_2)$ for $x_1 \\neq x_2$, plugging into (1) gives a contradiction. Set $x = 1$ in the original condition. Since $f$ is injective,\n\n$$\nf(1) + f(y) = y + 1 \\quad (2)\n$$\n\nSetting $y = 1$ yields $2f(1) = 2$, so $f(1) = 1$. Thus, $f(y) = y + 1$. Now, (2) gives $f(y) = y$ for all $y \\in \\mathbb{R}$, which is a solution.\n\nNow consider $f(a + 1) = 0$. Then\n\n$$\nf(f(x)) = 0, \\quad \\forall x \\in \\mathbb{R}.\n$$\n\nThis means $f(y) = 0$ for all $y$ in $\\mathrm{Im}(f)$. Assume there exists $y_0$ with $f(y_0 + 1) \\neq 0$. Setting $y = y_0$ in the original condition gives\n\n$$\nf(f(x) + x f(y_0)) = x f(y_0 + 1) \\quad (3).\n$$\n\nFor any $x_0 \\in \\mathbb{R}$, set $x = x_0 / f(y_0 + 1)$ in (3):\n\n$$\nf(f(x) + x f(y_0)) = x_0.\n$$\n\nSince $x_0$ is arbitrary, $f$ is surjective, so $\\mathrm{Im}(f) = \\mathbb{R}$. Therefore, $f(y) = 0$ for all $y \\in \\mathbb{R}$, contradicting $f(y_0 + 1) \\neq 0$. Thus, $f(x) = 0$ for all $x \\in \\mathbb{R}$ is also a solution. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20669,
"subject": "Mathematics (Olympiad)",
"question": "It is known that $\\triangle ABC$ satisfies $AB = 1$, $AC = 2$, and $\\cos B + \\sin C = 1$. Find the length of side $BC$.",
"options": [],
"answer": "See solution",
"solution": "Let $a = BC$, $b = AC = 2$, and $c = AB = 1$.\n\nBy the law of sines:\n$$\n\\frac{\\sin B}{\\sin C} = \\frac{b}{c} = 2 \\implies \\sin B = 2 \\sin C.\n$$\nGiven $\\cos B + \\sin C = 1$, so $\\cos B = 1 - \\sin C$.\n\nNow,\n$$\n\\sin^2 B + \\cos^2 B = 1 \\\\\n(2 \\sin C)^2 + (1 - \\sin C)^2 = 1 \\\\\n4\\sin^2 C + 1 - 2\\sin C + \\sin^2 C = 1 \\\\\n5\\sin^2 C - 2\\sin C = 0\n$$\nSince $\\sin C \\neq 0$, $\\sin C = \\frac{2}{5}$.\n\nThen $\\cos B = 1 - \\sin C = \\frac{3}{5}$.\n\nBy the law of cosines:\n$$\n\\cos B = \\frac{a^2 + c^2 - b^2}{2ac} \\\\\n\\frac{3}{5} = \\frac{a^2 + 1 - 4}{2a} = \\frac{a^2 - 3}{2a}\n$$\nSo,\n$$\n3 \\cdot 2a = 5(a^2 - 3) \\\\\n6a = 5a^2 - 15 \\\\\n5a^2 - 6a - 15 = 0 \\\\\na^2 - \\frac{6}{5}a - 3 = 0\n$$\nSolving for $a > 0$:\n$$\na = \\frac{3 + 2\\sqrt{21}}{5}\n$$\nThus, the length of side $BC$ is $\\boxed{\\dfrac{3 + 2\\sqrt{21}}{5}}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20670,
"subject": "Mathematics (Olympiad)",
"question": "設 $P$ 為銳角三角形 $ABC$ 內部一點，且 $P$ 到三頂點的距離分別為 $d_A, d_B, d_C$，到三邊的垂直距離分別為 $d_1, d_2, d_3$。試證：\n\n$$\nd_A + d_B + d_C \\geq 2(d_1 + d_2 + d_3).\n$$",
"options": [],
"answer": "See solution",
"solution": "如圖所示，$PD \\perp BC$，$PE \\perp CA$，$PF \\perp AB$。\n\n\n\n因為 $\\angle AEP + \\angle AFP = 90^\\circ + 90^\\circ = 180^\\circ$，所以 $A, E, P, F$ 共圓，且 $AP = d_A$ 為該圓直徑，於是可得：$\\angle EPF = 180^\\circ - \\angle A = \\angle B + \\angle C$。利用正弦定理 $EF = d_A \\sin A$，及餘弦定理可推得\n\n$$\n\\begin{aligned}\n(d_A \\sin A)^2 &= EF^2 = d_2^2 + d_3^2 - 2d_2 d_3 \\cos \\angle EPF \\\\\n&= d_2^2 + d_3^2 - 2d_2 d_3 \\cos(\\angle B + \\angle C) \\\\\n&= d_2^2 (\\cos^2 C + \\sin^2 C) + d_3^2 (\\cos^2 B + \\sin^2 B) - 2d_2 d_3 (\\cos B \\cos C - \\sin B \\sin C) \\\\\n&= (d_2 \\cos C - d_3 \\cos B)^2 + (d_2 \\sin C + d_3 \\sin B)^2 \\\\\n&\\geq (d_2 \\sin C + d_3 \\sin B)^2.\n\\end{aligned}\n$$\n\n因此有 $d_A \\sin A > d_2 \\sin C + d_3 \\sin B$，即 $d_A \\geq \\dfrac{d_2 \\sin C + d_3 \\sin B}{\\sin A}$。同理可證\n\n$$\nd_B \\geq \\dfrac{d_3 \\sin A + d_1 \\sin C}{\\sin B}, \\quad d_C \\geq \\dfrac{d_1 \\sin B + d_2 \\sin A}{\\sin C}.\n$$\n\n於是\n\n$$\n\\begin{aligned}\nd_A + d_B + d_C &\\geq \\dfrac{d_2 \\sin C + d_3 \\sin B}{\\sin A} + \\dfrac{d_3 \\sin A + d_1 \\sin C}{\\sin B} + \\dfrac{d_1 \\sin B + d_2 \\sin A}{\\sin C} \\\\\n&= d_1 \\left( \\dfrac{\\sin C}{\\sin B} + \\dfrac{\\sin B}{\\sin C} \\right) + d_2 \\left( \\dfrac{\\sin A}{\\sin C} + \\dfrac{\\sin C}{\\sin A} \\right) + d_3 \\left( \\dfrac{\\sin B}{\\sin A} + \\dfrac{\\sin A}{\\sin B} \\right) \\\\\n&\\geq 2(d_1 + d_2 + d_3).\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20671,
"subject": "Mathematics (Olympiad)",
"question": "Let $p_1, p_2, \\dots$ be the sequence of positive integers defined by $p_1 = 2$, and for all positive integers $n$, $p_{n+1}$ is defined to be the least prime number dividing $n p_1^{1!} p_2^{2!} \\cdots p_n^{n!} + 1$.\n\nProve that all prime numbers appear in this sequence.",
"options": [],
"answer": "See solution",
"solution": "Let $q_1, q_2, \\dots$ be the prime numbers in ascending order.\n\nAssume to the contrary that there exists a prime number not appearing in the sequence. Let $q_s$ be the least such prime.\n\nNote that by Fermat's little theorem we have $p_k^{k!} \\equiv 1 \\pmod{q_s}$ for all $k \\geq q_s - 1$.\n\nLet $r$ be the remainder of $n p_1^{1!} p_2^{2!} \\cdots p_{q_s-2}^{(q_s-2)!}$ when divided by $q_s$. Since $q_s$ does not appear in the sequence, we have $r \\neq 0$. Let $r'$ be an integer where $r r' \\equiv -1 \\pmod{q_s}$.\n\nPick a sufficiently large integer $N$ such that $\\{q_1, q_2, \\dots, q_{s-1}\\} \\subseteq \\{p_1, p_2, \\dots, p_N\\}$, and $N \\equiv r' \\pmod{q_s}$. By our choice of $N$ we have\n\n$$\nN p_1^{1!} p_2^{2!} \\cdots p_N^{N!} + 1 \\equiv r' \\cdot r + 1 \\equiv 0 \\pmod{q_s}.\n$$\n\nThus, $q_s$ divides $N p_1^{1!} p_2^{2!} \\cdots p_N^{N!} + 1$. Furthermore, it is the least such prime, since $q_i \\mid N p_1^{1!} p_2^{2!} \\cdots p_N^{N!}$ for all $i < s$, and thus $q_i \\nmid N p_1^{1!} p_2^{2!} \\cdots p_N^{N!} + 1$. This contradicts our assumption that $q_s$ does not appear in $p_1, p_2, \\dots$. Therefore, all primes appear in the sequence.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20672,
"subject": "Mathematics (Olympiad)",
"question": "Andriy has cards with numbers $1, 3, 5, \\ldots, 2009$ (odd numbers from 1 to 2009). Lesya has cards with numbers $2, 4, 6, \\ldots, 2010$ (even numbers from 2 to 2010). Lesya places her cards face down in a sequence: she starts at some even number, arranges the cards in increasing order up to 2010, then continues from 2 in increasing order up to the number before her starting point. Andriy then places his cards on top of Lesya's, so each Lesya's card is covered by exactly one Andriy's card, forming 1005 pairs. In each pair, the person with the higher number gets 1 point. What is the maximum number of points Andriy can guarantee to get, regardless of Lesya's arrangement?",
"options": [],
"answer": "See solution",
"solution": "**Answer:** 502 points.\n\nAndriy can guarantee at least 502 points by placing his cards in decreasing order: $2009, 2007, 2005, \\ldots, 3, 1$. For any arrangement Lesya chooses, this ensures that in each possible pairing, Andriy wins exactly 502 times. This can be shown by induction: for any cyclic shift of Lesya's sequence, the number of points Andriy gets remains 502. Lesya can always choose the arrangement that minimizes Andriy's score, so he cannot guarantee more than 502 points. The total number of points over all possible arrangements is $1004 + 1003 + \\ldots + 1 + 0 = \\frac{1005 \\cdot 1004}{2} = 502 \\cdot 1005$, so the average is 502 points per arrangement, which is the maximum Andriy can guarantee.\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20673,
"subject": "Mathematics (Olympiad)",
"question": "Suppose that the geometric sequence $\\{a_n\\}$ satisfies $a_1 - a_2 = 3$, $a_1 - a_3 = 2$. What is the common ratio of $\\{a_n\\}$?",
"options": [],
"answer": "See solution",
"solution": "Let the common ratio of $\\{a_n\\}$ be $q$. Then\n$$\na_1(1 - q) = a_1 - a_2 = 3, \\\\\na_1(1 - q^2) = a_1 - a_3 = 2.\n$$\nThus,\n$$\n\\frac{a_1(1 - q^2)}{a_1(1 - q)} = \\frac{2}{3}.\n$$\nBut $1 - q^2 = (1 - q)(1 + q)$, so\n$$\n1 + q = \\frac{2}{3}.\n$$\nTherefore, $q = -\\frac{1}{3}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20674,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle with a right angle at $A$, and let $D$ be the foot of the altitude from $A$. A variable point $M$ traces the interior of the minor arc $AB$ of the circumcircle of $ABC$. The internal angle bisector of $\\angle DAM$ meets $CM$ at $N$. The line through $N$ perpendicular to $CM$ meets $AD$ at $P$. Determine the locus of the point where the line $BN$ meets the line $CP$.",
"options": [],
"answer": "See solution",
"solution": "The required locus is the interior of the minor arc $AC$ of the circle $\\gamma$ passing through $A$, $B$, and $C$. More precisely, as $M$ traces the interior of the minor arc $AB$ of $\\gamma$ from $A$ to $B$, the point $Q$ where $BN$ meets $CP$ traces the interior of the minor arc $AC$ of $\\gamma$ from $A$ to $C$.\n\nWe first show that the lines $BN$ and $CP$ are perpendicular. It then follows that $Q$ is an interior point of the minor arc $AC$ of $\\gamma$. To this end, reflect $A$ across $D$ to obtain $A'$. Clearly, $A'$ lies on $\\gamma$, and $N$ is the incenter of triangle $AA'M$.\n\nThe circle $\\omega$ through $A$, $A'$, and $N$ is centered at $C$—this is the so-called trillium lemma for the incenter; it is readily proved by an angle chase.\n\nSince $BA$ and $BA'$ are tangents from $B$ to $\\omega$, the line $AA'$ is the polar of $B$ with respect to $\\omega$, so it passes through the pole of $BN$ relative to $\\omega$.\n\nThis pole also lies on the tangent to $\\omega$ at $N$, which is the line $NP$, so $P$ is the pole of $BN$ relative to $\\omega$.\n\nRecall that $\\omega$ is centered at $C$, so the lines $BN$ and $CP$ are indeed perpendicular.\n\nConversely, let $Q$ be an interior point of the minor arc $AC$ of $\\gamma$; let $CQ$ meet $AD$ at $P$; let the circle with diameter $CP$ meet $BQ$ at $N$; and finally, let $CN$ meet $\\gamma$ again at $M$.\n\n\n\n\n\nWe show that the ray $AN$, emanating from $A$, bisects $\\angle DAM$ internally. Alternatively, $N$ is the incenter of triangle $AA'M$. (Recall that $A'$ is the reflection of $A$ across $D$, so $MC$ bisects $\\angle AMA'$ internally.)\n\nWith reference again to the trillium lemma, the $C$-centered circle through $A$ and $A'$ meets $CM$ at the incenter of triangle $AA'M$.\n\nIt is therefore sufficient to show that $CA = CN$. To this end, notice that $CA^2 = CD \\cdot CB = \\overrightarrow{CP} \\cdot \\overrightarrow{CB}$, and $CN^2 = CQ \\cdot CP = \\overrightarrow{CB} \\cdot \\overrightarrow{CP}$. The conclusion follows.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20675,
"subject": "Mathematics (Olympiad)",
"question": "A number of $N$ children are at a party, and they sit in a circle to play a game of Pass the Parcel. The parcel has infinitely many layers. On turn $i$ (starting with $i = 1$), the following two things happen in order:\n\n1. The parcel is passed $i^2$ positions clockwise.\n2. The child currently holding the parcel unwraps a layer and claims the prize inside.\n\nFor what values of $N$ will every child receive a prize?",
"options": [],
"answer": "See solution",
"solution": "Every child receives a prize if and only if $N = 2^a 3^b$ for some non-negative integers $a$ and $b$. For convenience, say $N$ is *good* if every child receives a prize.\n\nNumber the children $0, \\ldots, N-1$ clockwise around the circle, with child number $0$ starting with the parcel. After $n$ turns, the parcel will have been passed $1^2 + 2^2 + \\cdots + n^2 = \\frac{n(n+1)(2n+1)}{6}$ places around the circle. For convenience, write $s_n = \\frac{n(n+1)(2n+1)}{6}$. Thus, child $m$ receives the parcel (and a prize) if and only if $m \\equiv s_n \\pmod{N}$ for some $n$, so $N$ is good if and only if $s_n$ assumes every possible value modulo $N$.\n\nTo rule out the case where $N$ is divisible by a prime $p > 3$, it is sufficient to show that $s_n$ misses some value modulo $p$. This follows from the fact that $6$ has a multiplicative inverse modulo $p$, and $s_n \\equiv 0 \\pmod{p}$ if $n \\equiv 0 \\pmod{p}$ or $-1 \\pmod{p}$, so $s_n$ assumes at most $p-1$ values modulo $p$. Consequently, such an $N$ is not good.\n\nWe now show that each $N$ of the form $2^a 3^b$ is good. We do this by showing that, if $N$ is good, then so are both $2N$ and $3N$; since $1$ is clearly good, this is sufficient to prove goodness of $2^a 3^b$ inductively on $a + b$.\n\nTo show that, if $N$ is good, then so is $2N$, refer to goodness of the former to infer that, for each $m$ modulo $2N$, there exists an $n$ such that $s_n \\equiv m \\pmod{2N}$. Only the case $s_n \\equiv m + 2N \\pmod{2N}$ remains. In this case,\n\n$$\ns_{n+6N} = s_n + (6n^2 + 6n + 1)N + 18(2n + 1)N^2 + 72N^3 \\equiv s_n + N \\equiv m \\pmod{2N}.\n$$\n\nConsequently, $2N$ is indeed good.\n\nTo show that, if $N$ is good, then so is $3N$, refer again to goodness of the former to infer that, for each $m$ modulo $3N$, there exists an $n$ such that $s_n \\equiv m \\pmod{3N}$ or $m + 2N \\pmod{3N}$. Only the last two cases remain. In the former case,\n\n$$\ns_{n+12N} = s_n + 2(6n^2 + 6n + 1)N + 72(2n + 1)N^2 + 2^6 3^2 N^3 \\equiv s_n + 2N \\equiv m \\pmod{3N},\n$$\n\nand in the latter,\n\n$$\ns_{n+6N} = s_n + (6n^2 + 6n + 1)N + 18(2n + 1)N^2 + 72N^3 \\equiv s_n + N \\equiv m \\pmod{3N}.\n$$\n\nConsequently, $3N$ is indeed good.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20676,
"subject": "Mathematics (Olympiad)",
"question": "The incircle of a triangle $A_0B_0C_0$ touches the sides $B_0C_0$, $C_0A_0$, $A_0B_0$ at the points $A$, $B$, $C$, respectively, and the incircle of the triangle $ABC$ with incenter $I$ touches the sides $BC$, $CA$, $AB$ at the points $A_1$, $B_1$, $C_1$, respectively. Let $\\sigma(ABC)$ and $\\sigma(A_1B_1C_1)$ be the areas of the triangles $ABC$ and $A_1B_1C_1$ respectively. Show that if $\\sigma(ABC) = 2\\sigma(A_1B_1C_1)$, then the lines $AA_0$, $BB_0$, $IC_1$ pass through a common point.",
"options": [],
"answer": "See solution",
"solution": "Let $BC = a$, $CA = b$, $AB = c$ and $2u = a + b + c$. Then $CA_1 = CB_1 = u - c$, $AC_1 = u - a$, $BC_1 = u - b$.\n\nWe have $\\sigma(ABC) = \\frac{1}{2}ab \\sin \\angle C$ and\n\n$$\n\\sigma(A_1B_1C) = \\frac{1}{2}(u - c)(u - c) \\sin \\angle C = \\frac{1}{8}(a + b - c)^2 \\sin \\angle C.\n$$\n\nTherefore $\\sigma(ABC) = 2\\sigma(A_1B_1C)$ implies $(a + b - c)^2 = 2ab$.\n\nLet $AA_0 \\cap BC = A_2$ and $BB_0 \\cap AC = B_2$. The Law of Sines in the triangles $ABA_0$ and $ACA_0$ gives\n\n$$\n\\frac{AA_0}{BA_0} = \\frac{\\sin(\\angle A + \\angle B)}{\\sin(\\angle BAA_0)}, \\quad \\frac{AA_0}{CA_0} = \\frac{\\sin(\\angle A + \\angle C)}{\\sin(\\angle CAA_0)}.\n$$\n\nHence\n\n$$\n\\frac{\\sin(\\angle BAA_0)}{\\sin(\\angle CAA_0)} = \\frac{c}{b}\n$$\n\nand\n\n$$\n\\frac{BA_2}{CA_2} = \\frac{AB \\sin(\\angle BAA_0)}{AC \\sin(\\angle CAA_0)} = \\frac{c^2}{b^2}.\n$$\n\nIn particular,\n\n$$\nBA_2 = \\frac{ac^2}{b^2 + c^2}, \\quad \\frac{CB_2}{AB_2} = \\frac{a^2}{c^2}.\n$$\n\nLet $C_1I \\cap BC = D$. Then\n\n$$\nBD = \\frac{u - b}{\\cos \\angle B}\n$$\n\n$$\nA_2D = BD - BA_2 = \\frac{u - b}{\\cos \\angle B} - \\frac{ac^2}{b^2 + c^2}\n$$\n\nLet $AA_0 \\cap BB_0 = E$ and $AA_0 \\cap C_1I = F$. We want to show that $E = F$. It suffices to show that\n\n$$\n\\frac{EA_2}{AE} = \\frac{FA_2}{AF}\n$$\n\nBy Menelaus' theorem we have\n\n$$\n\\frac{FA_2}{AF} = \\frac{DA_2}{BD} \\cdot \\frac{BC_1}{AC_1}\n$$\n\nand\n\n$$\n\\frac{A_2E}{AE} = \\frac{BA_2}{BC} \\cdot \\frac{CB_2}{B_2A}\n$$\n\nTherefore\n\n$$\n\\frac{EA_2}{AE} = \\frac{FA_2}{AF}\n$$\n\nif and only if\n\n$$\n\\frac{BA_2}{BC} \\cdot \\frac{CB_2}{AB_2} = \\frac{DA_2}{BD} \\cdot \\frac{BC_1}{AC_1}\n$$\n\nNow substituting these lengths and using the Law of Cosines $2ac \\cos \\angle B = a^2 + c^2 - b^2$, we find that\n\n$$\n\\frac{EA_2}{AE} = \\frac{FA_2}{AF}\n$$\n\nif and only if\n\n$$\n\\frac{a^2}{b^2 + c^2} = \\frac{a + c - b}{b + c - a} - \\frac{(a^2 + c^2 - b^2)c}{(b^2 + c^2)(b + c - a)}\n$$\n\nThis equality is equivalent to $(a - b)((a + b - c)^2 - 2ab) = 0$, and we are done. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20677,
"subject": "Mathematics (Olympiad)",
"question": "Find all natural numbers $n > 2010$ for which the following statement is true:\n\nFor all $k \\in \\{1, 2, \\dots, n - 2010\\}$, the numbers $n + k$ and $2010 + k$ are coprime.",
"options": [],
"answer": "See solution",
"solution": "Let $m = n - 2010$. For all $k \\in \\{1, 2, \\dots, m\\}$, we have:\n\n$$(2010 + k, n + k) = (2010 + k, n + k - 2010 - k) = (2010 + k, m)$$\n\nSo, $2010 + k$ and $m$ must be coprime for all $k$ in the range. However, among $m$ consecutive numbers ($2011, 2012, \\dots, 2010 + m$), there is always one divisible by $m$, unless $m = 1$. Thus, the only solution is $n = 2011$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20678,
"subject": "Mathematics (Olympiad)",
"question": "Let $f(m, n)$ be a function defined recursively as follows:\n\n- $f(1, n) = n + 1$ for all $n \\geq 0$,\n- $f(m, 0) = f(m-1, 1)$ for all $m > 1$,\n- $f(m, n) = f(m-1, f(m, n-1))$ for all $m > 1$ and $n > 0$.\n\n(a) Find $f(3, 2005)$.\n\n(b) Let $g(1) = 2$ and $g(n+1) = 2^{g(n)}$ for $n \\in \\mathbb{Z}^+$. Find $f(4, 2005)$ in terms of $g$.",
"options": [],
"answer": "See solution",
"solution": "By induction, we obtain\n\n$$\nf(2, y) = f(2, 0) + 2y = f(1, 1) + 2y = 2y + 3.\n$$\n\nNext, we have\n\n$$\nf(3, y) = f(2, f(3, y - 1)) = 2f(3, y - 1) + 3.\n$$\n\nAdding\n\n$$\n\\begin{aligned}\nf(3, y) &= 2f(3, y - 1) + 3, \\\\\n2f(3, y - 1) &= 2^2 f(3, y - 2) + 2 \\cdot 3, \\\\\n2^2 f(3, y - 2) &= 2^3 f(3, y - 3) + 2^2 \\cdot 3, \\\\\n& \\vdots, \\\\\n2^{y-1} f(3, 1) &= 2^y f(3, 0) + 2^{y-1} \\cdot 3,\n\\end{aligned}\n$$\n\nwe obtain\n\n$$\n\\begin{aligned}\nf(3, y) &= 2^y f(3, 0) + 3(1 + 2 + 2^2 + \\cdots + 2^{y-1}) \\\\\n&= 2^y f(2, 1) + 3(2^y - 1) = 2^{y+3} - 3.\n\\end{aligned}\n$$\n\nIn particular, we have $f(3, 2005) = 2^{2008} - 3$.\n\n(b) The answer is $f(4, 2005) = g(2008) - 3$, where $g(n)$ is defined by $g(1) = 2$ and $g(n + 1) = 2^{g(n)}$ for any $n \\in \\mathbb{Z}^+$.\n\nWe prove by induction that $f(4, y) = g(y + 3) - 3$. Firstly, by (2), we have\n\n$$\nf(4, 0) = f(3, 1) = 2^4 - 3 = 2^{2^2} - 3 = g(3) - 3.\n$$\n\nThis proves the base case.\n\nAssume $f(4, y) = g(y + 3) - 3$ for some $y \\in \\mathbb{Z}^+$. Using (3), we find that\n\n$$\nf(4, y + 1) = f(3, f(4, y)) = 2^{f(4,y)+3} - 3 = 2^{g(y+3)} - 3 = g(y + 4) - 3.\n$$\n\nThis proves the inductive step.\n\nTherefore, we have $f(4, 2005) = g(2008) - 3$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20679,
"subject": "Mathematics (Olympiad)",
"question": "Let $A$, $B$, $C$, $D$, $E$, $F$, and $G$ be different points in the plane such that the line segments $AB$, $BC$, $CD$, $DE$, $EF$, $FA$, $AG$, $CG$, and $EG$ have equal lengths.\n\nProve that the lines $AD$, $BE$, and $CF$ all pass through the same point.",
"options": [],
"answer": "See solution",
"solution": "Since $AB = BC = CG = GA$, quadrilateral $ABCG$ is a rhombus.\n\nIn particular, $AB \\parallel CG$ and $BC \\parallel GA$. By the same token, $CDEG$ and $EFAG$ are rhombuses, and\n\n$$\nCD \\parallel EG, \\qquad (1)\n$$\n\n$DE \\parallel GC$, $EF \\parallel AG$, and\n\n$$\nFA \\parallel GE. \\qquad (2)\n$$\n\nNow (1) and (2) imply that $CD \\parallel AF$. Since $CD$ and $AF$ have equal length, quadrilateral $AFDC$ is a parallelogram.\n\nDiagonals of a parallelogram bisect one another. Hence the midpoint, $M$ say, of $AD$ is also the midpoint of $CF$.\n\n\n\nA similar argument shows that quadrilateral $ABDE$ is a parallelogram and that $AD$ and $BE$ bisect each other. But $M$ is the midpoint of $AD$, hence $M$ is also the midpoint of $BE$. Therefore $AD$, $BE$, and $CF$ meet in $M$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20680,
"subject": "Mathematics (Olympiad)",
"question": "On Thursday, 1st January 2015, Anna buys one book and one shelf. For the next two years, she buys one book every day and one shelf on alternate Thursdays, so she next buys a shelf on 15th January 2015. On how many days in the period Thursday, 1st January 2015 until (and including) Saturday, 31st December 2016 is it possible for Anna to put all her books on all her shelves, so that there is an equal number of books on each shelf?",
"options": [],
"answer": "See solution",
"solution": "2016 is a leap year, so the total number of days in the two years is $365 + 366 = 731$. We split these days up by how many shelves Anna had on each day: for the first 14 days she had 1 shelf, for days 15 to 28 she had 2, and so on up to days 729 to 731, when she had 53 shelves. For $s \\leq 52$, she had $s$ shelves from days $14s - 13$ to $14s$ inclusive.\n\nIf at some time she has $b$ books and $s$ shelves, she can split her books equally among her shelves if and only if $b$ is a multiple of $s$. Thus, we count the number of multiples of $s$ between $14s - 13$ and $14s$, for $s \\leq 52$. (We can ignore the last three days when $s = 53$ since none of 729, 730, or 731 is a multiple of 53.) Note that $s$ dividing $14s - k$ is equivalent to $s$ dividing $k$, where $k$ ranges from 0 to 13. Counting the number of such $k$ for a given $s$:\n\nNumber of $0 \\leq k \\leq 13$ which are multiples of $s$:\n\n\n\nNote that there is one value of $k$ (namely $k = 0$) which all $s$ divide, which is why the 1 continues all the way down to $s = 52$. Summing the bottom row, we find that the answer is\n\n$$\n14 + 7 + 5 + 4 + 3 + 3 + (2 \\times 7) + (1 \\times 39) = 89.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20681,
"subject": "Mathematics (Olympiad)",
"question": "Let $t > 0$. Find all $t$ such that for any infinite set $X$ of positive real numbers, for any $x, y, z \\in X$ (not necessarily distinct), for all real numbers $a$ and all positive real numbers $d$, we have\n$$\n\\max\\{|x - (a - d)|, |y - a|, |z - (a + d)|\\} > td.\n$$",
"options": [],
"answer": "See solution",
"solution": "The answer is $0 < t < \\frac{1}{2}$.\n\n**Proof:**\n\n*Case 1: $0 < t < \\frac{1}{2}$.*\n\nLet $\\lambda \\in \\left(0, \\frac{1-2t}{2(1+t)}\\right)$, and define $x_i = \\lambda^i$, $X = \\{x_1, x_2, \\dots\\}$. We claim that for all (not necessarily distinct) $x, y, z \\in X$, all real $a$, and all $d > 0$,\n$$\n\\max\\{|x - (a - d)|, |y - a|, |z - (a + d)|\\} > td.\n$$\nSuppose not: there exist $a \\in \\mathbb{R}$, $d > 0$, and $x_i, x_j, x_k$ such that\n$$\n\\max\\{|x_i - (a - d)|, |x_j - a|, |x_k - (a + d)|\\} \\le td.\n$$\nThis gives\n$$\n\\begin{cases}\n-td \\le x_i - (a - d) \\le td, \\\\\n-td \\le x_j - a \\le td, \\\\\n-td \\le x_k - (a + d) \\le td,\n\\end{cases}\n$$\ni.e.,\n$$\n\\begin{cases}\nx_i + (1-t)d \\le a \\le x_i + (1+t)d, \\\\\nx_j - td \\le a \\le x_j + td, \\\\\nx_k - (1+t)d \\le a \\le x_k - (1-t)d,\n\\end{cases} \\quad (*)\n$$\nwhich implies\n$$\n\\begin{cases}\nx_k - (1+t)d \\le a \\le x_i + (1+t)d, \\\\\nx_i + (1-t)d \\le a \\le x_j + td, \\\\\nx_j - td \\le a \\le x_k - (1-t)d,\n\\end{cases}\n$$\nSince $0 < t < \\frac{1}{2}$, it follows that\n$$\n\\begin{cases}\nd \\ge \\frac{x_k - x_i}{2(1+t)}, \\tag{1} \\\\\nd \\le \\frac{x_j - x_i}{1-2t}, \\tag{2} \\\\\nd \\le \\frac{x_k - x_j}{1-2t}. \\tag{3}\n\\end{cases}\n$$\nFrom (2), (3), and $d > 0$, we get $x_i < x_j < x_k$. Thus,\n$$\n\\frac{x_j - x_i}{x_k - x_i} = \\frac{\\lambda^j - \\lambda^i}{\\lambda^k - \\lambda^i} \\le \\lambda. \\tag{4}\n$$\nBut from (1) and (2),\n$$\n\\frac{x_j - x_i}{1 - 2t} \\ge \\frac{x_k - x_i}{2(1 + t)} \\implies \\frac{x_j - x_i}{x_k - x_i} \\ge \\frac{1 - 2t}{2(1 + t)} > \\lambda,\n$$\ncontradicting (4). Thus, the claim holds for $0 < t < \\frac{1}{2}$.\n\n*Case 2: $t \\ge \\frac{1}{2}$.*\n\nFor any infinite $X$ and any $x < y < z$ in $X$, set $d = \\frac{z-x}{2}$ and $a = \\max\\{x + (1-t)d, y - td\\}$. Since $t \\ge \\frac{1}{2}$, we have\n$$\n\\max\\{|x - (a - d)|, |y - a|, |z - (a + d)|\\} \\le td.\n$$\nSo $t \\ge \\frac{1}{2}$ does not satisfy the requirement.\n\n**Conclusion:** The set of all required $t$ is $(0, \\frac{1}{2})$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20682,
"subject": "Mathematics (Olympiad)",
"question": "We call a natural number a _twin_ if it has two natural divisors whose difference is equal to 2. Determine whether there are more twin numbers or numbers that are not twin among the first $20112012$ natural numbers.",
"options": [],
"answer": "See solution",
"solution": "Let $M = 20112012$. Denote by $M_K$ the set of all natural numbers not exceeding $M$ and divisible by $K$, and by $N_K$ the number of elements in $M_K$, i.e., $N_K = |M_K|$. It is well-known that $N_K = \\left[ \\frac{M}{K} \\right]$.\n\nLet $N$ be the number of twin numbers not exceeding $M$. Then $N \\geq N_3 + N_4 - N_{12}$, since all numbers in $M_3$ and $M_4$ are twin (they have divisors $1,3$ and $2,4$ respectively), and $M_{12}$ consists of numbers belonging to both $M_3$ and $M_4$. There are also twin numbers not in $M_3$ or $M_4$ (e.g., $35 = 5 \\cdot 7$), so $N > N_3 + N_4 - N_{12}$.\n\nSince $M$ is divisible by $12$:\n\n$$\nN > \\frac{M}{3} + \\frac{M}{4} - \\frac{M}{12} = \\frac{M}{2}\n$$\n\nThus, there are more twin numbers than numbers that are not twin among the first $M = 20112012$ natural numbers.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20683,
"subject": "Mathematics (Olympiad)",
"question": "設 $\\triangle ABC$ 為銳角三角形，$AB > BC$。$\\Omega$ 與 $O$ 分別是 $\\triangle ABC$ 的外接圓與外心。$\\angle ABC$ 的角平分線再交 $\\Omega$ 於 $M$ 點。令 $\\Gamma$ 是以 $BM$ 為直徑的圓。$\\angle AOB$ 與 $\\angle BOC$ 的角平分線分別交 $\\Gamma$ 於 $P, Q$ 兩點。設 $R$ 點落在直線 $PQ$ 上，並滿足 $BR = MR$。證明：$BR \\parallel AC$。\n\n(註：所有的角平分線都當成是射線。)\n\nLet $\\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $ABC$ with $AB > BC$. The angle bisector of $\\angle ABC$ intersects $\\Omega$ at $M \\neq B$. Let $\\Gamma$ be the circle with diameter $BM$. The angle bisectors of $\\angle AOB$ and $\\angle BOC$ intersect $\\Gamma$ at points $P$ and $Q$, respectively. The point $R$ is chosen on the line $PQ$ so that $BR = MR$. Prove that $BR \\parallel AC$.\n\n(Here we always assume that an angle bisector is a ray.)",
"options": [],
"answer": "See solution",
"solution": "令 $K$ 為 $BM$ 的中點，亦即 $\\Gamma$ 的圓心。注意到 $AB \\neq BC$ 可推得 $K \\neq O$。\n\n易知直線 $OM$ 與 $OK$ 分別是 $AC$ 及 $BM$ 的中垂線。因此，$R$ 是 $PQ$ 和 $OK$ 的交點。\n\n\n\n設 $N$ 為 $OM$ 與 $\\Gamma$ 的第二個交點。因為 $BM$ 是 $\\Gamma$ 的直徑，$BN$ 與 $AC$ 都與 $OM$ 垂直。所以 $BN \\parallel AC$，而我們只需再證明 $BN$ 通過 $R$ 點即可。\n\n須證此事，我們將說明，直線 $BN$, $OK$, $PQ$ 是某三個圓的根軸。\n\n令 $\\omega$ 是以 $BO$ 為直徑的圓。因為 $\\angle BNO = \\angle BKO = 90^\\circ$，點 $N$, $K$ 都在圓 $\\omega$ 上。\n\n接下來我們證明 $O$, $K$, $P$, $Q$ 四點共圓。設 $D$ 與 $E$ 分別是 $BC$, $AB$ 的中點。易知 $D$, $E$ 分別落在射線 $OQ$, $OP$ 上。由題目對於 $\\triangle ABC$ 的設定，點 $B$, $E$, $O$, $K$, $D$ 依序落在 $\\omega$ 上。於是 $\\angle EOR = \\angle EBK = \\angle KBD = \\angle KOD$，故 $KO$ 是 $\\angle POQ$ 的外角平分線。因為 $K$ 是圓 $\\Gamma$ 的圓心，$K$ 也會落在 $PQ$ 的中垂線上。所以，$K$ 就是三角形 $POQ$ 的外接圓 $\\gamma$ 上、弧 $POQ$ 的中點。\n\n由上述可知，直線 $OK$, $BN$, $PQ$ 兩兩是圓 $\\omega$, $\\gamma$, $\\Gamma$ 的根軸，因此這三線共點於 $R$。證畢。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20684,
"subject": "Mathematics (Olympiad)",
"question": "Let $p$ and $q$ be real numbers such that the quadratic equation\n\n$$\nx^2 + px + q = 0\n$$\n\nhas two real solutions $x_1$ and $x_2$.\n\nThe following two conditions hold:\n\n1. The numbers $x_1$ and $x_2$ differ by $1$.\n2. The numbers $p$ and $q$ differ by $1$.\n\nShow that $p$, $q$, $x_1$, and $x_2$ are integers.",
"options": [],
"answer": "See solution",
"solution": "Without loss of generality, assume $x_1 = x_2 + 1$. By Vieta's formulas, $p = -(x_1 + x_2) = -2x_2 - 1$ and $q = x_1 x_2 = x_2^2 + x_2$.\n\nTherefore, it is enough to check that $x_2$ must be an integer.\n\n**Case 1:** $q = p - 1$\n\nThen $x_2^2 + x_2 = -2x_2 - 1 - 1$, so $x_2^2 + 3x_2 + 2 = 0$. Thus, $x_2 = -1$ or $x_2 = -2$, both integers.\n\n**Case 2:** $q = p + 1$\n\nThen $x_2^2 + x_2 = -2x_2 - 1 + 1$, so $x_2^2 + 3x_2 = 0$. Thus, $x_2 = 0$ or $x_2 = -3$, both integers.\n\nTherefore, $x_2$, $x_1 = x_2 + 1$, $p$, and $q$ are all integers.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20685,
"subject": "Mathematics (Olympiad)",
"question": "Determine all polynomials $P(x)$ with real coefficients such that\n\n$$\n(x + 1)P(x - 1) - (x - 1)P(x)\n$$\n\nis a constant polynomial.",
"options": [],
"answer": "See solution",
"solution": "The answer is that $P(x)$ can be any constant polynomial, or $P(x) = kx^2 + kx + c$ for any constant $k$ and $c$.\n\nLet $\\Lambda = (x+1)P(x-1) - (x-1)P(x)$.\n\nSubstitute $x = -1$ into $\\Lambda$ to get $2P(-1)$, and $x = 1$ to get $2P(1)$. Since $\\Lambda$ is constant, $2P(-1) = 2P(1)$, so $P(-1) = P(1)$.\n\nLet $c = P(-1) = P(0)$ and define $Q(x) = P(x) - c$. Then $Q(-1) = Q(0) = 0$, so $Q(x)$ has roots at $x = -1$ and $x = 0$. Thus, $Q(x) = x(x+1)R(x)$ for some polynomial $R(x)$, so $P(x) = x(x+1)R(x) + c$.\n\nSubstitute into $\\Lambda$:\n\n$$\n(x+1)[(x-1)xR(x-1) + c] - (x-1)[x(x+1)R(x) + c]\n$$\n\nThis simplifies to\n\n$$\nx(x-1)(x+1)(R(x-1) - R(x)) + 2c.\n$$\n\nFor this to be constant, $R(x-1) - R(x) = 0$, so $R(x)$ is constant, say $k$. Thus, $P(x) = kx(x+1) + c = kx^2 + kx + c$.\n\nVerifying, substitute $P(x) = kx^2 + kx + c$ into $\\Lambda$:\n\n$$\n(x+1)(k(x-1)^2 + k(x-1) + c) - (x-1)(kx^2 + kx + c)\n$$\n\nThis simplifies to $2c$, a constant. Thus, all such $P(x)$ work, including constant polynomials ($k=0$).",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20686,
"subject": "Mathematics (Olympiad)",
"question": "Denote the intersection of the lines $CT$ and $AB$ by $U$. Prove that\n$$\n\\frac{|AU|}{|UB|} = \\frac{|AC|}{|CB|},\n$$\nsince it will follow that $CT$ is the angle bisector of $\\angle ACB$.",
"options": [],
"answer": "See solution",
"solution": "\n\nSince the points $C, E, D$, and $L$ are concyclic, and the lines $AB$ and $CL$ are parallel, we have $\\angle AED = \\pi - \\angle DEC = \\angle CLD = \\pi - \\angle DMA$. Thus the points $A, M, D$, and $E$ are concyclic.\n\nMoreover, we have $\\angle EDN = \\angle EDA - \\angle NDA = \\angle EMA - \\angle MEB = \\pi - \\angle BME = \\prec MEB = \\prec EBN$. Hence the points $N, B, D$, and $E$ are also concyclic.\n\nThis gives $\\prec NED = \\pi - \\prec DBN = \\pi - \\prec DCL = \\pi - \\prec DEL$. Hence the points $N, E$, and $L$ are collinear.\n\nSince the lines $AB$ and $CL$ are parallel the triangles $MBD$ and $LCD$ have the same angles. Therefore they are similar and we have\n$$\n\\frac{|MD|}{|BD|} = \\frac{|DL|}{|DC|}.\n$$\nFor the same reason the triangles $NAE$ and $LCE$ also have the same angles. Therefore they are similar and we have\n$$\n\\frac{|NE|}{|AE|} = \\frac{|EL|}{|EC|}.\n$$\nSince the lines $AD, BE$, and $CU$ intersect in the same point the Ceva's theorem gives\n$$\n\\frac{|AU|}{|UB|} \\frac{|BD|}{|DC|} \\frac{|CE|}{|EA|} = 1.\n$$\nInserting the previous ratios into this equation we get\n$$\n\\frac{|AU|}{|UB|} \\frac{|MD|}{|DL|} \\frac{|LE|}{|EN|} = 1.\n$$\nUsing the condition $|DM| = |EN|$ we get\n$$\n\\frac{|AU|}{|UB|} \\frac{|LE|}{|DL|} = 1.\n$$\nWe also have $\\prec ACB = \\prec ECD = \\prec ELD$ and $\\prec DEL = \\prec DCL = \\prec CBA$. Therefore the triangles $ABC$ and $DEL$ are similar and we get\n$$\n\\frac{|LE|}{|DL|} = \\frac{|CB|}{|AC|}.\n$$\nInserting this into the previous equation we get exactly the desired result. Therefore the line $CT$ is indeed the angle bisector of $\\prec ACB$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20687,
"subject": "Mathematics (Olympiad)",
"question": "Let $S(k)$ denote the sum of all positive integers less than $k$ and relatively prime to $k$. For $k = 1$, $S(1) = 1$. For $k > 1$, show that $S(k) = \\frac{k\\varphi(k)}{2}$, where $\\varphi(k)$ is Euler's totient function.\n\nLet $m$ be a positive integer with largest prime factor $q$, and let $2 = p_1 < p_2 < \\dots < p_s = q$ be the consecutive prime numbers up to $q$. Suppose $m = p_1^{a_1} \\dots p_s^{a_s}$ (with some $a_i$ possibly zero). Construct a number $x = p_1^{b_1} \\dots p_s^{b_s}$ such that $2S(x)$ is a perfect $n$th power, and $b_i \\ge a_i$ for all $i$.",
"options": [],
"answer": "See solution",
"solution": "For $k > 1$, each $a < k$ with $\\gcd(a, k) = 1$ can be paired with $k - a$, which is also coprime to $k$. Each pair sums to $k$, and there are $\\varphi(k)$ such numbers less than $k$, so:\n\n$$\nS(k) = \\frac{k\\varphi(k)}{2}.\n$$\n\nLet $x = p_1^{b_1} \\dots p_s^{b_s}$. Then:\n\n$$\n2S(x) = x\\varphi(x) = p_1^{2b_1-1}(p_1-1) \\dots p_s^{2b_s-1}(p_s-1).\n$$\n\nExpress $(p_1-1)\\dots(p_s-1) = p_1^{c_1} \\dots p_s^{c_s}$ for suitable $c_i$. Thus:\n\n$$\n2S(x) = p_1^{2b_1+c_1-1} \\dots p_s^{2b_s+c_s-1}.\n$$\n\nTo make $2S(x)$ a perfect $n$th power, require $2b_i + c_i - 1$ divisible by $n$ and $b_i \\ge a_i$. Since $n$ is odd, for each $i$ choose $k_i$ so that $k_i n \\equiv c_i - 1 \\pmod{2}$ and\n\n$$\nb_i = \\frac{k_i n - c_i + 1}{2} \\ge a_i.\n$$\n\nThen $2S(x) = (p_1^{k_1} \\dots p_s^{k_s})^n$, so $x$ meets the required condition.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20688,
"subject": "Mathematics (Olympiad)",
"question": "Computing aggregate works according to the principle: every new minute, given integers $x_1$, $x_2$, and $x_3$ in order, it computes a value $x_4$ that satisfies the equality:\n\n$$\nx_1(x_4 + x_2) = x_3(x_3 + x_2).\n$$\n\nIf $x_4$ is not an integer, the aggregate stops its work.\n\nIf $x_4$ is an integer, then a new set of three numbers is given: $x'_1 = x_2$, $x'_2 = x_3$, and $x'_3 = x_4$, and in a minute the aggregate computes $x'_4$.\n\nIf at first $x_1 = 1$, $x_2 = 2$, $x_3 = 3$, will the aggregate work without stops for at least 2018 minutes?",
"options": [],
"answer": "See solution",
"solution": "Let $x_4, x_5, x_6, \\dots$ denote the sequence of numbers computed by the aggregate. For every natural $n$, the following equality holds:\n\n$$\nx_{n-2}(x_{n+1} + x_{n-1}) = x_n(x_n + x_{n-1}).\n$$\n\nAdd $x_n x_{n-2}$ to both sides:\n\n$$\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\\Rightarrow \\frac{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} x_{n-2}} = \\frac{x_n + x_{n-1} + x_{n-2}}{x_{n-1} x_{n-2}}.\n\\end{aligned}\n$$\n\nThis ratio holds for every natural $n$. Extending to the initial values:\n\n$$\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 \\\\\n\\Rightarrow x_{n+1} + x_n + x_{n-1} &= 3x_n x_{n-1} \\\\\n\\Rightarrow x_{n+1} = 3x_n x_{n-1} - x_n - x_{n-1}.\n\\end{aligned}\n$$\n\nSo, $x_{n+1} \\in \\mathbb{Z}$ for any natural $n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20689,
"subject": "Mathematics (Olympiad)",
"question": "Find all natural numbers $n$ such that $n^2 - 10n + 23$, $n^2 - 9n + 31$, and $n^2 - 12n + 46$ are primes.",
"options": [],
"answer": "See solution",
"solution": "Let $n$ be a natural number that satisfies the condition. First, calculate the sum of the three expressions:\n\n$$\n(n^2 - 10n + 23) + (n^2 - 9n + 31) + (n^2 - 12n + 46) = 3n^2 - 31n + 100\n$$\n\nThis sum is even, so at least one of the numbers must be even, i.e., equal to $2$ (the only even prime). Solve:\n\n$$\n\\begin{aligned}\nn^2 - 10n + 23 &= 2 \\\\\nn^2 - 9n + 31 &= 2 \\\\\nn^2 - 12n + 46 &= 2\n\\end{aligned}\n$$\n\nSolving these, we get $n = 3$ or $n = 7$. Checking:\n\nFor $n = 3$:\n- $n^2 - 10n + 23 = 2$\n- $n^2 - 9n + 31 = 13$\n- $n^2 - 12n + 46 = 19$\n\nAll are primes.\n\nFor $n = 7$:\n- $n^2 - 10n + 23 = 2$\n- $n^2 - 9n + 31 = 17$\n- $n^2 - 12n + 46 = 11$\n\nAll are primes.\n\nThus, the solutions are $n = 3$ and $n = 7$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20690,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $f(x) = |\\lg(x+1)|$ and real numbers $a, b$ with $a < b$ satisfy\n\n- $f(a) = f\\left(-\\frac{b+1}{b+2}\\right)$,\n- $f(10a + 6b + 21) = 4\\lg 2$.\n\nFind the values of $a$ and $b$.",
"options": [],
"answer": "See solution",
"solution": "As $f(a) = f\\left(-\\frac{b+1}{b+2}\\right)$, we have\n$$\n|\\lg(a+1)| = \\left|\\lg\\left(-\\frac{b+1}{b+2} + 1\\right)\\right| = \\left|\\lg\\left(\\frac{1}{b+2}\\right)\\right| = |\\lg(b+2)|.\n$$\n\nThen either $a+1 = b+2$ or $(a+1)(b+2) = 1$. Since $a < b$, $a+1 \\neq b+2$, so $(a+1)(b+2) = 1$.\n\nFrom $f(a) = |\\lg(a+1)|$ we know $0 < a+1 < 1$. Then\n$$\n0 < a + 1 < b + 1 < b + 2,\n$$\nwhich implies\n$$\n0 < a + 1 < 1 < b + 2.\n$$\n\nNow,\n$$\n(10a + 6b + 21) + 1 = 10(a + 1) + 6(b + 2)\n$$\nBut since $(a+1)(b+2) = 1$, $a+1 = \\frac{1}{b+2}$.\nSo,\n$$\n10(a+1) + 6(b+2) = 10\\left(\\frac{1}{b+2}\\right) + 6(b+2) = 6(b+2) + \\frac{10}{b+2}\n$$\n\nTherefore,\n$$\nf(10a + 6b + 21) = \\left| \\lg \\left[ 6(b + 2) + \\frac{10}{b+2} \\right] \\right| = \\lg \\left[ 6(b + 2) + \\frac{10}{b+2} \\right]\n$$\n\nGiven $f(10a + 6b + 21) = 4\\lg 2$, so\n$$\n\\lg\\left[6(b+2) + \\frac{10}{b+2}\\right] = 4\\lg 2\n$$\nwhich means\n$$\n6(b+2) + \\frac{10}{b+2} = 2^4 = 16\n$$\nLet $x = b+2$, then\n$$\n6x + \\frac{10}{x} = 16\n$$\nMultiply both sides by $x$:\n$$\n6x^2 - 16x + 10 = 0\n$$\nSolve the quadratic:\n$$\nx = \\frac{16 \\pm \\sqrt{256 - 240}}{12} = \\frac{16 \\pm 4}{12}\n$$\nSo $x = \\frac{20}{12} = \\frac{5}{3}$ or $x = 1$.\n\nIf $x = 1$, $b = -1$, but then $a+1 = 1/(b+2) = 1/1 = 1$, which is not in $(0,1)$.\nIf $x = \\frac{5}{3}$, $b = \\frac{5}{3} - 2 = -\\frac{1}{3}$.\nThen $a+1 = 1/(b+2) = 1/(\\frac{5}{3}) = \\frac{3}{5}$, so $a = -\\frac{2}{5}$.\n\nTherefore,\n$$\na = -\\frac{2}{5}, \\quad b = -\\frac{1}{3}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20691,
"subject": "Mathematics (Olympiad)",
"question": "Given a natural number $n \\ge 3$, find the smallest real number $k > 0$ such that for every 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.\n\n",
"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$ there are two vertices connected by an edge in $G$. Removing these two vertices leaves the tree (and thus $G$) connected.\n- *Case 2*: Among $v_1, \\dots, v_k$ there are two vertices that 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 each of $v_2, \\dots, v_k$, connect it 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$$\n\nDeleting $y$ edges suffices, so $k \\le \\frac{1}{2}$.\n\nTo show $k \\ge \\frac{1}{2}$, consider the complete graph $K_n$. To make it bipartite, we must delete all edges within each part. For $n = 2n_1$:\n\n$$\n\\text{Edges to delete} = 2 \\binom{n_1}{2} = n_1^2 - n_1\n$$\n\nBut\n\n$$\nn_1^2 - n_1 \\le k \\left( \\frac{2n_1(2n_1-1)}{2} - n_1 \\right ) \\implies k \\ge \\frac{1}{2}\n$$\n\nFor $n = 2n_1 + 1$:\n\n$$\n\\text{Edges to delete} = \\binom{n_1+1}{2} + \\binom{n_1}{2} = n_1^2\n$$\n\nand\n\n$$\nn_1^2 \\le k \\left( \\frac{2n_1(2n_1+1)}{2} - n_1 \\right ) \\implies k \\ge \\frac{1}{2}\n$$\n\nThus, the minimal $k$ is $\\boxed{\\frac{1}{2}}$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20692,
"subject": "Mathematics (Olympiad)",
"question": "Given the functional equation:\n\n$$\nf(2a) + 2f(b) = f(f(a + b)).\n$$\n\nFind all functions $f$ that satisfy this equation for all integers $a$ and $b$.",
"options": [],
"answer": "See solution",
"solution": "Swapping $a$ and $b$ in the equation gives:\n\n$$\nf(2b) + 2f(a) = f(f(a + b)).\n$$\n\nSetting $b = 0$ yields:\n\n$$\nf(2a) = 2f(a) - f(0).\n$$\n\nSo the original equation can be rewritten as:\n\n$$\n2f(a) + 2f(b) = f(f(a + b)) + f(0).\n$$\n\nLetting $(a, b) = (c-1, c+1)$ and $(a, b) = (c, c)$, we find:\n\n$$\n2f(c-1) + 2f(c+1) = f(f(2c)) + f(0) = 2f(c) + 2f(c).\n$$\n\nThis implies:\n\n$$\nf(c+1) - f(c) = f(c) - f(c-1),\n$$\n\nso $f$ is an arithmetic progression: $f(x) = mx + c$ for some constants $m$ and $c$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20693,
"subject": "Mathematics (Olympiad)",
"question": "There are exactly three positive real numbers $k$ such that the function\n\n$$\nf(x) = \\frac{(x - 18)(x - 72)(x - 98)(x - k)}{x}\n$$\n\ndefined over the positive real numbers $x$ achieves its minimum value at exactly two positive real numbers $x$. Find the sum of these three values of $k$.",
"options": [],
"answer": "See solution",
"solution": "Let $a, b, c$, and $d$ be positive real numbers such that the function\n\n$$\nf(x) = \\frac{(x - a)(x - b)(x - c)(x - d)}{x}\n$$\n\nachieves its minimum value at exactly two positive real numbers $x$.\n\nThe function $f$ is continuous and tends to infinity as $x \\to 0$ and $x \\to \\infty$. Since the numerator is negative for some $x > 0$, $f$ must achieve a negative minimum value on $(0, \\infty)$. Denote this minimum by $-\\lambda$ for some $\\lambda > 0$. Then $f(x) + \\lambda \\ge 0$ for all $x > 0$, so the positive roots of\n\n$$\nP(x) = (x - a)(x - b)(x - c)(x - d) + \\lambda x\n$$\n\nare exactly the points where $f$ achieves its minimum $-\\lambda$, and $P$ has exactly two positive roots. Since $P(x) \\ge 0$ for $x > 0$, any positive real roots must have even multiplicity. Thus, $P(x) = (x - r)^2 (x - s)^2$ for some $r, s > 0$.\n\nLet $Q(x) = (x - r)(x - s)$. Then\n\n$$\n(x - a)(x - b)(x - c)(x - d) + \\lambda x = (Q(x))^2.\n$$\n\nReplacing $x$ with $y^2$ and factoring gives\n\n$$\n\\begin{aligned}\n& (y - \\sqrt{a})(y - \\sqrt{b})(y - \\sqrt{c})(y - \\sqrt{d})(y + \\sqrt{a})(y + \\sqrt{b})(y + \\sqrt{c})(y + \\sqrt{d}) \\\\\n& \\qquad = (Q(y^2) - \\sqrt{\\lambda}y)(Q(y^2) + \\sqrt{\\lambda}y).\n\\end{aligned}\n$$\n\nBy symmetry, there must be an identity of the form\n\n$$\n(y \\pm \\sqrt{a})(y \\pm \\sqrt{b})(y \\pm \\sqrt{c})(y \\pm \\sqrt{d}) = Q(y^2) + \\sqrt{\\lambda}y,\n$$\n\nwhere the $\\pm$ signs do not necessarily correspond. Comparing the $y^3$ coefficients gives\n\n$$\n\\pm\\sqrt{a} \\pm\\sqrt{b} \\pm\\sqrt{c} \\pm\\sqrt{d} = 0.\n$$\n\nSince $a, b, c, d > 0$, the all-plus and all-minus cases are impossible. Thus, at least one of the following must hold:\n\n$$\n\\begin{aligned}\n\\sqrt{a} + \\sqrt{b} &= \\sqrt{c} + \\sqrt{d} \\\\\n\\sqrt{a} + \\sqrt{c} &= \\sqrt{b} + \\sqrt{d} \\\\\n\\sqrt{a} + \\sqrt{d} &= \\sqrt{b} + \\sqrt{c}\n\\end{aligned}\n$$\n\nReturning to the original problem, the possible values of $k$ must satisfy\n\n$$\n\\begin{aligned}\n\\sqrt{k} &= \\sqrt{18} + \\sqrt{72} - \\sqrt{98} = 2\\sqrt{2} \\\\\n\\sqrt{k} &= \\sqrt{18} + \\sqrt{98} - \\sqrt{72} = 4\\sqrt{2} \\\\\n\\sqrt{k} &= \\sqrt{72} + \\sqrt{98} - \\sqrt{18} = 10\\sqrt{2}\n\\end{aligned}\n$$\n\nThus, $k = (2\\sqrt{2})^2 = 8$, $k = (4\\sqrt{2})^2 = 32$, and $k = (10\\sqrt{2})^2 = 200$. The requested sum is $8 + 32 + 200 = 240$.\n\nAlternatively, by equating coefficients in\n\n$$\n(x - 18)(x - 72)(x - 98)(x - k) + \\lambda x = (x - r)^2(x - s)^2,\n$$\n\nand letting $y = \\sqrt{\\frac{k}{2}}$, we obtain\n\n$$\ny^4 - 188y^2 + 1008y - 1280 = 0.\n$$\n\nFactoring gives $(y - 2)(y - 4)(y - 10)(y + 16) = 0$, so the positive solutions are $y = 2, 4, 10$, corresponding to $k = 8, 32, 200$. Their sum is $240$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20694,
"subject": "Mathematics (Olympiad)",
"question": "Start with a $3 \\times n$ rectangle. What is the largest integer $N$ for which it is not possible to cover exactly $N$ unit squares by overlaying any number of $3 \\times n$ rectangles (possibly overlapping, in any orientation and position)?\n\nDescribe all such $N$ in terms of $n$ and its residue modulo $3$.",
"options": [],
"answer": "See solution",
"solution": "We produce three infinite families of coverings:\n\n1. Overlay the original $3 \\times n$ rectangle with one shifted both across 1 unit and down 1 unit. This covers $4n + 2$ unit squares. Adding rectangles shifted to the right, we can cover $4n + 2 + 3k$ unit squares for any $k \\ge 0$. These numbers are congruent to $n + 2$ modulo $3$.\n\n2. Overlay the original rectangle with one shifted down 1 unit and one shifted across 1 unit. This covers $4n + 3$ unit squares. Adding rectangles shifted to the right, we can cover $4n + 3 + 3k$ unit squares for any $k \\ge 0$. These numbers are congruent to $n$ modulo $3$.\n\n3. Overlay the original rectangle with one shifted down 1 unit, one shifted across 1 unit, and one shifted both across 1 and down 1 unit. This covers $4(n + 1) = 4n + 4$ squares. Adding rectangles shifted to the right, we can cover $4n + 4 + 3k$ unit squares for any $k \\ge 0$. These numbers are congruent to $n + 1$ modulo $3$.\n\nSince $n, n+1, n+2$ are congruent modulo $3$ to $0, 1, 2$ in some order, each number greater than $4n+1$ is in one of the families according to its remainder modulo $3$. Thus, any number of squares greater than $4n+1$ can be covered.\n\nAdditionally, $4n$ squares are covered by two overlapping rectangles with one shifted down one unit. If $n \\equiv 2 \\pmod{3}$, then $4n+1$ is divisible by $3$, so $4n+1$ squares can be covered with some rectangles overlapping horizontally.\n\n**Impossibility for $4n+1$ and $4n-1$:**\n\n- If $n \\equiv 0$ or $1 \\pmod{3}$, it is not possible to cover exactly $4n+1$ unit squares. Suppose otherwise. If two rectangles are in different orientations, the overlap is at most $9$ squares, so the total covered is at least $6n-9 > 4n+1$ for $n > 5$, a contradiction. If all rectangles are in the same orientation, the overlap is at most $2(n-1)$, so the total is at least $4n+2 > 4n+1$, again a contradiction. Thus, the number of squares covered must be a multiple of $3$ or $n$, but $4n+1$ is not.\n\n- If $n \\equiv 2 \\pmod{3}$, it is not possible to cover exactly $4n-1$ unit squares. The same argument applies: the number of squares covered must be a multiple of $3$ or $n$, but $4n-1$ is not.\n\n**Conclusion:**\n\n- If $n \\equiv 0$ or $1 \\pmod{3}$, the largest impossible $N$ is $4n+1$.\n- If $n \\equiv 2 \\pmod{3}$, the largest impossible $N$ is $4n-1$.",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 20695,
"subject": "Mathematics (Olympiad)",
"question": "Let $A_i$ be the segment with exactly $3n$ points from $S$, the leftmost being the $i$-th point from left to right of $S$, for $i = 1, 2, \\dots, 3n + 1$. Define $f(i)$ as the number of blue points in $A_i$.\n\nProve that $f(i) = 2n$ for some $i$.",
"options": [],
"answer": "See solution",
"solution": "Notice that $|f(i + 1) - f(i)| \\le 1$ since $A_i$ and $A_{i+1}$ have $3n - 1$ common points. Also, $f(1) + f(3n+1) = 4n$ because the disjoint segments $A_1$ and $A_{3n+1}$ cover $S$.\n\nIf $f(1) = 2n$, we are done. Suppose, without loss of generality, that $f(1) < 2n$. Then $f(3n + 1) > 2n$, and since $f(i)$ changes by at most $1$ at each step, there must exist some $k$ such that $f(k) = 2n$ by the intermediate value property.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20696,
"subject": "Mathematics (Olympiad)",
"question": "Do there exist distinct positive integers $x$ and $y$ such that the number $x + y$ is divisible by $2016$, the number $x - y$ is divisible by $2017$, and the number $xy$ is divisible by $2018$?",
"options": [],
"answer": "See solution",
"solution": "Yes. For example, the numbers $x = 2016 \\cdot 2015 - 2018$ and $y = 2018$ meet the conditions. Since $2016 \\cdot 2015 > 4 \\cdot 1009 = 2 \\cdot 2018$, we have $x > y$, so they are distinct.\n\nThe sum $x + y = 2016 \\cdot 2015$ is divisible by $2016$, and the product $xy$ is obviously divisible by $2018$. Furthermore,\n\n$$\nx - y = 2016 \\cdot 2015 - 2 \\cdot 2018 = (2017 - 1)(2017 - 2) - 2 \\cdot (2017 + 1) = 2017^2 - 3 \\cdot 2017 + 2 - 2 \\cdot 2017 - 2 = 2017 \\cdot 2012,\n$$\nso the difference $x - y$ is divisible by $2017$.\n\n*Remark.* This choice is not unique. All suitable numbers are of the form $x = 2016k - 2018m$ and $y = 2018m$, where $k$ and $m$ are integers such that $k + 2m$ is divisible by $2017$. Indeed, since $2018 = 2 \\cdot 1009$ and $1009$ is prime, one of $x$ and $y$ must be divisible by $1009$. Also, one of $x$ and $y$ must be even, but since $x + y$ is divisible by the even number $2016$, both must be even. Consequently, one of these numbers must be divisible by $2018$. Without loss of generality, let $y = 2018m$. Since $x + y = 2016k$, we must have $x = 2016k - 2018m$. Then\n$$\nx - y = 2016k - 2 \\cdot 2018m = 2017(k - 2m) - (k + 2m),\n$$\nso $x - y$ is divisible by $2017$ if and only if $k + 2m$ is divisible by $2017$. The pair in the solution is obtained by taking $k = 2015$ and $m = 1$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20697,
"subject": "Mathematics (Olympiad)",
"question": "Solve the inequality:\n\n$$\n\\frac{x^2 - |x-1| - 4}{x-4} \\ge 2x-1.\n$$",
"options": [],
"answer": "See solution",
"solution": "Note that $x \\ne 4$ and $|x-1| = x-1$ for $x > 1$, otherwise $|x-1| = 1-x$.\n\nIf $x > 1$ we get:\n\n$$\n\\frac{x^2 - 8x + 7}{x - 4} \\le 0 \\Leftrightarrow \\frac{(x-7)(x-1)}{x-4} \\le 0 \\Leftrightarrow x \\in (4, 7]\n$$\n\nIf $x \\le 1$ we get:\n\n$$\n\\frac{x^2 - 10x + 9}{x - 4} \\le 0 \\Leftrightarrow \\frac{(x-9)(x-1)}{x-4} \\le 0 \\Leftrightarrow x \\in (-\\infty, 1]\n$$\n\nFinally, $x \\in (-\\infty, 1] \\cup (4, 7]$. $\\Box$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20698,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b \\in \\mathbb{R}$ with $a < b$ be two arbitrary real numbers. We say that a function $f : [a, b] \\to \\mathbb{R}$ has property $(\\mathcal{P})$ if it is an integrable function on $[a, b]$ such that\n\n$$\nf(x) - f\\left(\\frac{x+a}{2}\\right) = f\\left(\\frac{x+b}{2}\\right) - f(x) \\quad \\text{for any } x \\in [a, b].\n$$\n\nShow that for any real number $t$ there is a unique function $f : [a, b] \\to \\mathbb{R}$ with property $(\\mathcal{P})$ such that\n$$\n\\int_a^b f(x)\\,dx = t.\n$$",
"options": [],
"answer": "See solution",
"solution": "We will show that the functions with property $(\\mathcal{P})$ are precisely the constant functions on the interval $[a, b]$.\n\nThe equality in the statement can be rewritten as\n\n$$\nf(x) = \\frac{1}{2} \\left( f\\left(\\frac{x+a}{2}\\right) + f\\left(\\frac{x+b}{2}\\right) \\right). \\tag{1}\n$$\n\nWe now show that, for any $n \\in \\mathbb{N}^*$ and any $x \\in [a, b]$, the following holds:\n\n$$\nf(x) = \\frac{1}{2^n} \\sum_{k=0}^{2^n-1} f\\left(\\frac{x + (2^n - 1 - k)a + k b}{2^n}\\right). \\tag{2}\n$$\n\nFor $n = 1$, this is just relation (1). Assume the equality holds for some $n \\in \\mathbb{N}^*$ and any $x \\in [a, b]$. Then:\n\n$$\n\\begin{aligned}\nf(x) &= \\frac{1}{2^n} \\sum_{k=0}^{2^n-1} f\\left(\\frac{x + (2^n - 1 - k)a + k b}{2^n}\\right) \\\\\n&= \\frac{1}{2^n} \\sum_{k=0}^{2^n-1} \\Bigg( f\\left(\\frac{1}{2} \\cdot \\frac{x + (2^n - 1 - k)a + k b}{2^n} + \\frac{1}{2} a\\right) \\\\\n&\\qquad + f\\left(\\frac{1}{2} \\cdot \\frac{x + (2^n - 1 - k)a + k b}{2^n} + \\frac{1}{2} b\\right) \\Bigg) \\\\\n&= \\frac{1}{2^{n+1}} \\sum_{k=0}^{2^{n+1}-1} f\\left(\\frac{x + (2^{n+1} - 1 - k)a + k b}{2^{n+1}}\\right).\n\\end{aligned}\n$$\n\nConsider, for some $n \\in \\mathbb{N}^*$ and arbitrary $x \\in [a, b]$, the division $\\Delta_n = (x_0 = a < x_1 = \\frac{(2^n-1)a + b}{2^n} < \\dots < x_k = \\frac{(2^n - k)a + k b}{2^n} < \\dots < x_{2^n} = b)$ with norm $|\\Delta_n| = \\frac{b-a}{2^n}$ and the system of intermediate points\n\n$$\n\\xi_{(n)}(x) = \\left( \\xi_k(x) = \\frac{x + (2^n - k)a + (k-1) b}{2^n} \\mid k = 1, \\dots, 2^n \\right).\n$$\n\nRelation (2) can then be written as\n\n$$\nf(x) = \\frac{1}{b-a} \\cdot \\sigma(f; \\Delta_n, \\xi_{(n)}(x)),\n$$\n\nwhere $\\sigma(f; \\Delta_n, \\xi_{(n)}(x))$ denotes the Riemann sum associated with $f$, the division $\\Delta_n$, and the system of intermediate points $\\xi_{(n)}(x)$. Since $f$ is Riemann integrable on $[a, b]$, we have $\\lim_{n \\to \\infty} \\sigma(f; \\Delta_n, \\xi_{(n)}(x)) = \\int_a^b f(s)\\,ds$, so $f(x) = \\frac{1}{b-a} \\int_a^b f(s)\\,ds$ for any $x \\in [a, b]$. Thus, any function with property $(\\mathcal{P})$ is constant.\n\nConversely, any constant function on $[a, b]$ is Riemann integrable and satisfies the given equality.\n\nFor any $t \\in \\mathbb{R}$, there is then a unique function with property $(\\mathcal{P})$, namely $f(x) = \\frac{t}{b-a}$ for all $x \\in [a, b]$, such that $\\int_a^b f(x)\\,dx = t$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20699,
"subject": "Mathematics (Olympiad)",
"question": "There are ten red cards, numbered $1, 2, \\ldots, 10$, and ten blue cards, also numbered $1, 2, \\ldots, 10$. How many ways are there to choose three from these twenty cards so that the sum of the numbers on the cards chosen is $16$ or less?",
"options": [],
"answer": "See solution",
"solution": "$570$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20700,
"subject": "Mathematics (Olympiad)",
"question": "We have 15 cards numbered $1, 2, \\ldots, 15$. How many ways are there to choose some (at least 1) cards so that all numbers on these cards are greater than or equal to the number of cards chosen?",
"options": [],
"answer": "See solution",
"solution": "Consider a general problem with cards $1, 2, \\dots, n$. Let $F_n$ be the number of choices when there are $n$ cards. $F_1 = 1$ and $F_2 = 2$ are trivial.\n\nLet $k \\geq 3$. We will consider $F_k$.\n\n- If card $k$ is not chosen, the number of ways is $F_{k-1}$.\n- If card $k$ is chosen:\n - If any card other than $k$ is chosen, we cannot choose $1$. If we remove card $k$ and decrease the numbers on other cards by $1$, we get a proper choice from cards $1, \\dots, k-2$.\n - If no card other than $k$ is chosen, there is only $1$ way.\n\nTherefore, the number of proper choices with card $k$ is $F_{k-2} + 1$.\n\nSo we get the recurrence:\n$$F_k = F_{k-1} + F_{k-2} + 1$$\n\nCalculating the sequence by this relation, we get $F_{15} = 1596$.\n\n*If we allow a choice with no cards, we get the Fibonacci sequence.*",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20701,
"subject": "Mathematics (Olympiad)",
"question": "Let $X$, $Y$ be points on $AB$, $AC$ of triangle $ABC$, respectively, such that $B$, $C$, $X$, $Y$ lie on one circle. The median of triangle $ABC$ from $A$ intersects the perpendicular bisector of $XY$ at $P$. Find $\\angle BAC$, if $\\triangle PXY$ is equilateral.",
"options": [],
"answer": "See solution",
"solution": "Since $AP$ is the symmedian of $AXY$ and $P$ lies on the perpendicular bisector of $XY$, then $PX$ and $PY$ are tangent to the circumcircle of triangle $AXY$. Therefore, we can easily find that $\\angle YAX = 60^\\circ$ or $\\angle YAX = 120^\\circ$.\n\nSince $B$, $C$, $X$, $Y$ lie on a circle, $XY$ is antiparallel to $BC$ in angle $BAC$. So, the $A$-median of $ABC$ is the $A$-symmedian of $AXY$. Hence, $PX$, $PY$ are tangent to the circumcircle of $AXY$. Let $O$ be the circumcenter of $AXY$.\n\nIf $A$ and $P$ lie on different sides of the line $XY$, then we obtain\n\n$$\n\\angle XAY = \\frac{1}{2} \\angle XOY = \\frac{1}{2} (180^\\circ - \\angle YPX) = \\frac{1}{2} (180^\\circ - 60^\\circ) = 60^\\circ.\n$$\n\nIf $A$ and $P$ lie on the same side of the line $XY$, then $\\angle XAY = 180^\\circ - \\frac{1}{2} \\angle XOY = 180^\\circ - \\frac{1}{2} \\cdot 120^\\circ = 120^\\circ$.\n\nHence, there are only two possible values of angle $XAY$: $60^\\circ$ and $120^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20702,
"subject": "Mathematics (Olympiad)",
"question": "Given a regular pentagon $ABCDE$, determine the least value of the expression\n\n$$\n\\frac{PA + PB}{PC + PD + PE}\n$$\n\nwhere $P$ is an arbitrary point lying in the plane of the pentagon $ABCDE$.",
"options": [],
"answer": "See solution",
"solution": "Without loss of generality, assume that the pentagon $ABCDE$ has side length $1$. Then the length of its diagonal is\n\n$$\n\\lambda = \\frac{1 + \\sqrt{5}}{2}.\n$$\n\nSet $a = PA$, $b = PB$, $c = PC$, $d = PD$, $e = PE$.\n\n\n\nApplying the Ptolemy inequality for the (not necessarily convex) quadrilaterals $APDE$, $BPDC$, and $PCDE$, we obtain (respectively):\n\n$$\na + d \\geq e\\lambda, \\quad b + d \\geq c\\lambda, \\quad e + c \\geq d\\lambda.\n$$\n\nMultiply the third inequality by $\\frac{\\lambda+2}{\\lambda+1}$ and add it to the first and second inequalities. We get:\n\n$$\na + b + 2d + e \\cdot \\frac{\\lambda+2}{\\lambda+1} + c \\cdot \\frac{\\lambda+2}{\\lambda+1} \\geq e\\lambda + c\\lambda + d \\cdot \\frac{\\lambda(\\lambda+2)}{\\lambda+1}.\n$$\n\nGrouping terms, this reduces to\n\n$$\na + b \\geq \\frac{\\lambda^2 - 2}{\\lambda + 1}(c + d + e).\n$$\n\nTherefore,\n\n$$\n\\frac{a + b}{c + d + e} \\geq \\frac{\\lambda^2 - 2}{\\lambda + 1} = \\sqrt{5} - 2.\n$$\n\nEquality holds if and only if the quadrilaterals $APDE$, $BPDC$, and $PCDE$ are cyclic, which occurs if and only if $P$ lies on the minor arc $AB$ of the circumcircle of pentagon $ABCDE$.\n\nTherefore, the smallest possible value of the given expression is $\\sqrt{5} - 2$.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20703,
"subject": "Mathematics (Olympiad)",
"question": "A farmer has two rectangular lands of size $120\\ \\text{m} \\times 100\\ \\text{m}$.\n\n(a) On the first land, there are 9 circular gardens of diameter $5\\ \\text{m}$. Prove that, regardless of the positions of the gardens, he can always build a rectangular garden of size $25\\ \\text{m} \\times 35\\ \\text{m}$.\n\n(b) On the second land, he builds a convex polygonal lake such that the shortest distance from any point on the boundary of the land to the lake is at most $5\\ \\text{m}$. Prove that the perimeter of the lake is at least $440 - 20\\sqrt{2}$ m.",
"options": [],
"answer": "See solution",
"solution": "(a) Consider the rectangle $ABCD$ with $AB = CD = 120$ and $AD = BC = 100$. Divide it into 10 subrectangles of size $30 \\times 40$ as follows.\n\n\n\nConsider the 9 centers of the given gardens. By the pigeonhole principle, there is some subrectangle that does not contain any of these points. Suppose it is the rectangle $XYZT$ with $XY = ZT = 40$, $XT = YZ = 30$. Consider another rectangle $X'Y'Z'T'$ inside $XYZT$ such that the sides of the two rectangles are pairwise parallel and the gap is $2.5$; then $X'Y'Z'T'$ has size $25 \\times 35$. It is clear that $X'Y'Z'T'$ does not intersect any garden, so the desired result follows.\n\n(b) Consider the rectangle $ABCD$ with $AB = CD = 120$ and $AD = BC = 100$. Let $L$ be the boundary of the lake. From the hypothesis, there are four points $A', B', C', D'$ on $L$ such that\n\n$$\nAA',\\ BB',\\ CC',\\ DD' \\leq 5.\n$$\n\nSince the lake is a convex polygon, the segments $A'B',\\ B'C',\\ C'D',\\ D'A'$ do not overlap. Hence, the length of $L$ is at least\n\n$$\nA'B' + B'C' + C'D' + D'A'.\n$$\n\nLet $A_1$ be the projection of $A'$ onto $AD$, $A_2$ the projection onto $AB$, and similarly define $B_1, B_2, C_1, C_2, D_1, D_2$. We have\n\n$$\nA_1A' + A'B' + B'B_1 \\geq A_1B_1 \\geq AB = 120.\n$$\n\nSimilarly,\n\n$$\n\\begin{aligned}\nB_2B' + B'C' + C'C_2 &\\geq 100, \\\\\nC_1C' + C'D' + D'D_1 &\\geq 120, \\\\\nD_2D' + D'A' + A'A_2 &\\geq 100.\n\\end{aligned}\n$$\n\nThese imply that\n\n$$\nA'B' + B'C' + C'D' + D'A' + (A'A_1 + A'A_2 + B'B_1 + B'B_2 + C'C_1 + C'C_2 + D'D_1 + D'D_2) \\geq 440.\n$$\n\nBy the Cauchy-Schwarz inequality,\n\n$$\nA'A_1 + A'A_2 \\leq \\sqrt{2(A'A_1^2 + A'A_2^2)} \\leq 5\\sqrt{2}.\n$$\n\nSimilarly, $B'B_1 + B'B_2 \\leq 5\\sqrt{2}$, $C'C_1 + C'C_2 \\leq 5\\sqrt{2}$, $D'D_1 + D'D_2 \\leq 5\\sqrt{2}$.\n\nFrom these inequalities, it is clear that the length of $L$ is at least $440 - 20\\sqrt{2}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20704,
"subject": "Mathematics (Olympiad)",
"question": "On a coordinate plane there are two regions, $M$ and $N$:\n\n$M$ is confined by\n$$\n\\begin{cases}\ny \\ge 0, \\\\\ny \\le x, \\\\\ny \\le 2 - x\n\\end{cases}\n$$\n\nand $N$ is determined by the inequalities $t \\le x \\le t+1$, $0 \\le t \\le 1$.\n\nThen the size of the common area of $M$ and $N$ is given by $f(t) = \\underline{\\hspace{2cm}}$.",
"options": [],
"answer": "See solution",
"solution": "As shown in the figure,\n\n$$\n\\begin{aligned}\nf(t) &= S_{\\text{shaded area}} \\\\\n&= S_{\\triangle AOB} - S_{\\triangle OCD} - S_{\\triangle BEF} \\\\\n&= 1 - \\frac{1}{2}t^2 - \\frac{1}{2}(1-t)^2 \\\\\n&= -t^2 + t + \\frac{1}{2}, \\quad 0 \\le t \\le 1.\n\\end{aligned}\n$$\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20705,
"subject": "Mathematics (Olympiad)",
"question": "Randomly select three vertices from the six vertices of a regular hexagon with side length 1. What is the probability that two of the three vertices are at a distance of $\\sqrt{3}$?",
"options": [],
"answer": "See solution",
"solution": "Let the vertices of the regular hexagon be $A_1, A_2, A_3, A_4, A_5, A_6$.\n\nConsider any selection of three vertices:\n\n- If two of the selected vertices are adjacent (e.g., $A_1$ and $A_2$), the third vertex must be one of the remaining vertices. Note that $A_1A_3 = A_1A_5 = A_2A_4 = A_2A_6 = \\sqrt{3}$, so the third vertex will always be at a distance of $\\sqrt{3}$ from one of the chosen vertices.\n- If no two selected vertices are adjacent, the only possibilities are $A_1, A_3, A_5$ or $A_2, A_4, A_6$. In these cases, there are always two vertices at a distance of $\\sqrt{3}$.\n\nTherefore, in every possible selection of three vertices, there are always two vertices at a distance of $\\sqrt{3}$.\n\nThe desired probability is $1$.\n\n$\\boxed{1}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20706,
"subject": "Mathematics (Olympiad)",
"question": "Determine all polynomials $P$ with integer coefficients, satisfying $0 \\leq P(n) \\leq n!$ for all non-negative integers $n$.",
"options": [],
"answer": "See solution",
"solution": "The required polynomials are $P = 0$, $P = 1$, $P = (X - 1)^2$, $P = X(X - 1)\\cdots(X - k)$, and $P = X(X - 1)\\cdots(X - k)(X - k - 2)^2$ for some non-negative integer $k$.\n\nLet $P$ be a polynomial satisfying the condition in the statement. Clearly, $P(0) = 0$ or $P(0) = 1$.\n\nWe first deal with the case $P(0) = 1$. The polynomials $P_1 = 1$ and $P_2 = (X-1)^2$ both satisfy the condition in the statement and $P_1(0) = P_2(0) = 1$.\n\nWe will prove that either $P = P_1$ or $P = P_2$. Consider an index $i$ such that $P(1) = P_i(1)$ and let $\\tilde{P} = P - P_i$.\n\nInduct on $n$ to show that $\\tilde{P}(n) = 0$ for all non-negative integers $n$. The base cases $n = 0$ and $n = 1$ are clear. For the inductive step, assume $\\tilde{P}(m) = 0$ for all non-negative integers $m < n$. Then $X(X-1)\\cdots(X-(n-1))$ divides $\\tilde{P}$, so $n!$ divides $\\tilde{P}(n)$. As $0 < P_i(n) < n!$, it follows that $|\\tilde{P}(n)| = |P(n) - P_i(n)| < n!$, so $\\tilde{P}(n) = 0$.\n\nConsequently, $\\tilde{P}$ has infinitely many roots, so it vanishes identically; that is, $P = P_i$, as desired.\n\nFinally, we deal with the case $P(0) = 0$. Assume $P$ is non-zero. Let $P(X) = X Q(X - 1)$, where $Q$ has integer coefficients. Then $0 \\leq Q(n) \\leq n!$ for all non-negative integers $n$. If $Q(0) = 0$, repeat the argument for $Q$ and so on, all the way down to some polynomial with a non-zero constant term—this is clearly the case, as $P$ is non-zero and degrees strictly decrease in the process. By the preceding, such a polynomial is either $1$ or $(X - 1)^2$. An obvious induction then shows that $P$ has one of the last two forms mentioned in the beginning.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20707,
"subject": "Mathematics (Olympiad)",
"question": "Obtén los dos valores enteros de $x$ más próximos a $2013\\degree$, tanto por defecto como por exceso, que cumplen esta ecuación trigonométrica:\n\n$$\n2^{\\sin^2 x} + 2^{\\cos^2 x} = 2\\sqrt{2}\n$$",
"options": [],
"answer": "See solution",
"solution": "Aplicando la desigualdad entre las medias aritmética y geométrica, resulta:\n\n$$\n2^{\\sin^2 x} + 2^{\\cos^2 x} \\geq 2\\sqrt{2^{\\sin^2 x} \\cdot 2^{\\cos^2 x}} = 2\\sqrt{2^{\\sin^2 x + \\cos^2 x}} = 2\\sqrt{2}\n$$\n\nLa igualdad se alcanza cuando $2^{\\sin^2 x} = 2^{\\cos^2 x}$, es decir, cuando $\\sin^2 x = \\cos^2 x$ o $\\sin x = \\pm \\cos x$. Los valores de $x$ que satisfacen la igualdad anterior son $x = 45\\degree + 90\\degree k$ con $k \\in \\mathbb{Z}$. Los valores pedidos se obtienen para:\n\n$$\nk_1 = \\left[ \\frac{2013\\degree - 45\\degree}{90} \\right] = 21 \\quad \\text{y} \\quad k_2 = \\left[ \\frac{2013\\degree + 45\\degree}{90} \\right] = 22\n$$\n\ny son $x_1 = 1935\\degree$ y $x_2 = 2025\\degree$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20708,
"subject": "Mathematics (Olympiad)",
"question": "Let $P$ be a point lying inside triangle $ABC$. Let $Q$ be a point on the segment $AB$, and let $R$ be a point on the segment $AC$ such that both circles $(BPQ)$ and $(CPR)$ are tangent to line $AP$. Through $B$ and $C$ we draw the lines passing through the center of the circle $(BPC)$, and through $Q$ and $R$ we draw the lines passing through the center of the circle $(PQR)$. Prove that there exists a circle tangent to the four drawn lines.",
"options": [],
"answer": "See solution",
"solution": "Since $AB \\cdot AQ = AP^2 = AC \\cdot AR$, quadrilateral $BCRQ$ is cyclic. Let $O$ be the center of circle $(BCRQ)$. Denote by $O_1$ and $O_2$ the centers of circles $(BPC)$ and $(QPR)$. We will show that lines $BO_1$, $CO_1$, $QO_2$, $RO_2$ are equidistant from $O$. Since $OB = OC = OQ = OR$, it suffices to establish the equality of (directed) angles $\\angle OCO_1 = \\angle O_1BO = \\angle OQO_2 = \\angle O_2RO$. Here the first and last equalities are obvious from symmetry about the perpendicular bisectors of $BC$ and $QR$.\n\n\n\nIt remains to prove the equality $\\angle O_1BO = \\angle OQO_2$ (*). By angle chasing we obtain $\\angle OQO_2 = \\angle OQR - \\angle O_2QR = (90^\\circ - \\angle RCQ) - (90^\\circ - \\angle RPQ) = \\angle RPQ - \\angle RCQ$. Similarly $\\angle O_1BO = \\angle BPC - \\angle BQC$. Thus, (*) is equivalent to the equality $\\angle RPQ - \\angle RCQ = \\angle BPC - \\angle BQC$ or $\\angle BQC - \\angle RCQ = \\angle BPC - \\angle RPQ$ (**). From the tangency of circles $(BPQ)$ and $(CPR)$ it follows that $\\angle RPQ = \\angle RCP + \\angle PBQ$, which equals (from the sum of angles in quadrilateral $BPCA$) $\\angle BPC - \\angle BAC$. Therefore, (**), transforms into $\\angle BQC - \\angle RCQ = \\angle BAC$, which holds true. The problem is solved.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20709,
"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$$\n\nAll digits of $A + B$ are odd. To find the digits of $A + B$, we must know if $a + d$ and $b + c$ are less than $10$.\n\n**Case 1:** $a + d \\geq 10$ and $b + c \\geq 10$.\n\nSince $a > b > c > d > 0$:\n$$\na + d = 10 + k,\\quad k = 0,1,2,\\ldots,5 \\\\\nb + c = 10 + \\ell,\\quad \\ell = 0,1,2,\\ldots,5\n$$\nThen:\n$$\nA + B = (10 + k) \\cdot 10^3 + (10 + \\ell) \\cdot 10^2 + (10 + \\ell) \\cdot 10 + (10 + k)\n$$\nThis results in digits $1, k+1, \\ell+1, \\ell+1, k$, which cannot all be odd. Contradiction.\n\n**Case 2:** $a + d \\geq 10$ and $b + c < 10$.\n\n$a + d = 10 + k,\\quad k = 0,1,2,\\ldots,5$\n\n$$\nA + B = (10 + k) \\cdot 10^3 + (b + c) \\cdot 10^2 + (b + c) \\cdot 10 + (10 + k)\n$$\nThis results in digits $b + c$ and $b + c + 1$, which cannot both be odd. Contradiction.\n\n**Case 3:** $a + d < 10$ and $b + c \\geq 10$.\n\n$b + c = 10 + \\ell,\\quad \\ell = 0,1,2,\\ldots,5$\n\n$$\nA + B = (a + d) \\cdot 10^3 + (10 + \\ell) \\cdot 10^2 + (10 + \\ell) \\cdot 10 + (a + d)\n$$\nThis results in digits $\\ell$ and $\\ell + 1$, which cannot both be odd. Contradiction.\n\n**Case 4:** $a + d < 10$ and $b + c < 10$.\n\nBoth $a + d$ and $b + c$ must be odd. Since $a > b > c > d > 0$ and $a \\geq 7$, it follows that $a + d = 9$. For $b + c < 10$ and $b > c > 0$, possible values for $b + c$ are $5, 7, 9$.\n\nPossible combinations:\n- $a + d = 9$ with $a = 8, d = 1$ and $b + c = 9$ ($b = 7, c = 2$; $b = 6, c = 3$; $b = 5, c = 4$): $A = 8721$, $A = 8631$, $A = 8541$\n- $a + d = 9$ with $a = 7, d = 2$ and $b + c = 9$ ($b = 6, c = 3$; $b = 5, c = 4$): $A = 7632$, $A = 7542$\n- $a + d = 9$ with $a = 8, d = 1$ and $b + c = 7$ ($b = 5, c = 2$; $b = 4, c = 3$): $A = 8521$, $A = 8431$\n- $a + d = 9$ with $a = 7, d = 2$ and $b + c = 7$ ($b = 4, c = 3$): $A = 7432$\n- $a + d = 9$ with $a = 8, d = 1$ and $b + c = 5$ ($b = 3, c = 2$): $A = 8321$\n\n**Final possible values:**\n$$\nA = 8721,\\ 8631,\\ 8541,\\ 7632,\\ 7542,\\ 8521,\\ 8431,\\ 7432,\\ 8321\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20710,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $n$ such that $2^n + 7^n$ is a perfect square.",
"options": [],
"answer": "See solution",
"solution": "Clearly, $n = 1$ is a solution. We show that there is no solution for $n > 1$. Suppose that\n\n$$\n2^n + 7^n = m^2\n$$\nfor some positive integers $n$ and $m$.\n\n**Case 1:** $n$ is odd and $n > 1$\n\nConsider the equation modulo $4$:\n\n$$\n2^n + 7^n \\equiv 0 + (-1)^n \\pmod{4} \\equiv 3 \\pmod{4}.\n$$\n\nHowever, $m^2 \\equiv 0$ or $1 \\pmod{4}$ for any integer $m$. Thus, there are no solutions in this case.\n\n**Case 2:** $n$ is even\n\nConsider the equation modulo $3$:\n\n$$\n2^n + 7^n \\equiv (-1)^n + 1^n \\pmod{3} \\equiv 2 \\pmod{3}.\n$$\n\nHowever, $m^2 \\equiv 0$ or $1 \\pmod{3}$ for any integer $m$. Thus, there are no solutions in this case either.\n\nHaving covered all possible cases, the proof is complete.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20711,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ and $b$ be positive real numbers with $a \\le 2b \\le 4a$.\n\nProve that\n\n$$\n4ab \\le 2(a^2 + b^2) \\le 5ab.\n$$",
"options": [],
"answer": "See solution",
"solution": "The terms in the left inequality can be rewritten as a square:\n\n$$\n2(a^2 + b^2) \\ge 4ab \\Leftrightarrow (a - b)^2 \\ge 0.\n$$\n\nThe square in the last inequality is clearly weakly positive.\n\nIn the right inequality, we multiply by $8$ and complete the square to obtain\n\n$$\n16a^2 - 40ab + 16b^2 \\le 0 \\Leftrightarrow (4a - 5b)^2 - 9b^2 \\le 0.\n$$\n\nFactorisation of the difference of the two squares gives\n\n$$\n(4a - 5b - 3b)(4a - 5b + 3b) \\le 0 \\Leftrightarrow (4a - 8b)(4a - 2b) \\le 0 \\Leftrightarrow (a - 2b)(4a - 2b) \\le 0.\n$$\n\nThe hypotheses ensure that the first factor is weakly negative, the second weakly positive, and therefore, the product is weakly negative. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20712,
"subject": "Mathematics (Olympiad)",
"question": "Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(2x + f(y)) = x + y + f(x)\n$$\n\nfor all $x, y \\in \\mathbb{R}$.",
"options": [],
"answer": "See solution",
"solution": "Let us choose $x$ so that the arguments on both sides become equal, i.e., solve $2x + f(y) = x$. For this value, $x = -f(y)$, we get:\n\n$$\nf(-f(y) + f(y)) = -f(y) + y + f(-f(y))\n$$\nwhich simplifies to\n$$\nf(0) = -f(y) + y + f(-f(y))\n$$\nBut $f(0)$ is constant, so rearranging gives $f(-f(y)) = f(y) - y + f(0)$. Now, try $f(y) = y$:\n\n$$\nf(2x + f(y)) = f(2x + y) = x + y + f(x) = x + y + x = 2x + y\n$$\nBut $f(2x + y) = 2x + y$ if $f(y) = y$, so $f(y) = y$ satisfies the equation. Thus, the only solution is $f(y) = y$ for all $y \\in \\mathbb{R}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20713,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute-angled triangle with $AB < AC < BC$, inscribed in the circle $c(O, R)$. The circle $c_1$ with center $A$ and passing through $AC$ intersects the circle $c(O, R)$ at point $D$ and the extension of the side $CB$ at $E$. The line $AE$ intersects the circle $c(O, R)$ at point $F$, and $G$ is the symmetric point of $E$ with respect to $B$. Prove that the quadrilateral $FEDG$ is cyclic.",
"options": [],
"answer": "See solution",
"solution": "Since the quadrilateral $AFBC$ is inscribed in the circle $c$, we have: $\\hat{F}_1 = \\hat{A}\\hat{C}B = \\hat{C}$. Since triangle $AEC$ is isosceles, we have $\\hat{E}_1 = \\hat{A}\\hat{C}B = \\hat{C}$. Therefore, $\\hat{F}_1 = \\hat{E}_1$, and hence the triangle $BEF$ is isosceles and thus\n\n$$\nBE = BF \\qquad (1).\n$$\n\n\n\nWe put $\\hat{C}_1 = x$. Then from the circle $c_1$ we get $E\\hat{A}D = 2x$, and hence\n\n$$\nE\\hat{A}B + B\\hat{A}D = 2x \\qquad (2)\n$$\n\nMoreover, from the circle $c$ we have:\n\n$$\nB\\hat{A}D = \\hat{C}_1 = x \\qquad (3)\n$$\n\nFrom (2) and (3) we find $E\\hat{A}B = B\\hat{A}D = x$, which means that $AB$ is the bisector of the isosceles triangle $EAD$. Hence, it is the perpendicular bisector of $ED$, and\n\n$$\nBE = BD. \\qquad (4)\n$$\n\nFrom (1) and (4), and from the equality $BE = BG$, we conclude that $BE = BF = BG = BD$, and hence the quadrilateral $FEDG$ is inscribed in a circle with center $B$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20714,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ be a given integer. Prove that there exist infinitely many prime numbers $p$ such that\n\n$$\np \\mid n^2 + 3, \\quad p \\mid m^3 - a\n$$\n\nfor some integers $n$ and $m$.",
"options": [],
"answer": "See solution",
"solution": "Let $k$ be an arbitrary integer. Note that\n\n$$\n(9a^2k^3)^2 + 3 = 3(27a^4k^6 + 1)\n$$\n\nand\n\n$$\n(9a^3k^4)^3 - a = a(3^6a^8k^{12} - 1) = a(27a^4k^6 - 1)(27a^4k^6 + 1)\n$$\n\nIt follows that for every $k \\in \\mathbb{Z}$, the number $27a^4k^6 + 1$ is a common divisor of the numbers $n^2 + 3$ and $m^3 - a$ with $n = 9a^2k^3$ and $m = 9a^3k^4$. So it is enough to prove that there are infinitely many primes $p$ such that $p \\mid 27a^4k^6 + 1$ for some integer $k$.\n\nSuppose that there are only finitely many such primes and these are $p_1, p_2, \\dots, p_r$. If we take $k = p_1p_2\\dots p_r + 1$, then it is clear that the number $27a^4k^6 + 1$ is not divisible by any $p_i$ for $1 \\le i \\le r$ and that it is also greater than 1. It follows that it has a prime divisor $p$, which is different from every $p_i$ for $1 \\le i \\le r$. We have obtained a contradiction, which finishes the proof. $\\Box$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20715,
"subject": "Mathematics (Olympiad)",
"question": "Find all three-digit multiples of 7 whose final digital sum (FDS) is 7.",
"options": [],
"answer": "See solution",
"solution": "Let $n$ be a positive integer. Any $n$ can be written as:\n\n$$\nn = a + 10b + 100c + \\dots\n$$\n\nwhere $a, b, \\dots$ are the digits of $n$. Thus,\n\n$$\n\\begin{aligned}\nn &= a + (1+9)b + (1+99)c + \\dots \\\\\n&= (a + b + c + \\dots) + 9(b + 11c + \\dots)\n\\end{aligned}\n$$\n\nSo $n$ and its digital sum have the same remainder when divided by 9. Therefore, all digital sums in the digital sum sequence for $n$, including its FDS, have the same remainder when divided by 9. Hence, if $n$ is a multiple of 9, then the FDS of $n$ is 9, and if $n$ is not a multiple of 9, then the FDS of $n$ is the remainder when $n$ is divided by 9.\n\nSo, the FDS of $n$ is 7 if and only if $n$ has the form $n = 7 + 9r$. Such an $n$ is a multiple of 7 if and only if 7 divides $r$. So $n$ is a multiple of 7 and has FDS 7 if and only if $n$ has the form $n = 7 + 63t$. Hence, the only three-digit multiples of 7 that have FDS 7 are the integers $n = 7 + 63t$ where $t = 2, 3, \\dots, 15$. These are:\n\n$$\n133,\\ 196,\\ 259,\\ 322,\\ 385,\\ 448,\\ 511,\\ 574,\\ 637,\\ 700,\\ 763,\\ 826,\\ 889,\\ 952.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20716,
"subject": "Mathematics (Olympiad)",
"question": "設 $x, y, z$ 為正整數，且 $z(xz+1)^2 = (5z+2y)(2z+y)$。試證：$z$ 必為奇數且 $z$ 為完全平方數。",
"options": [],
"answer": "See solution",
"solution": "1. 利用反證法，假設 $z$ 為偶數，即 $z = 2z_1$。原式改寫為：\n$$z_1(2xz_1+1)^2 = (5z_1+y)(4z_1+y)$$\n令 $w = \\gcd(y, z_1)$，所以令 $y = w y_0, z_1 = w z_0$，其中 $(y_0, z_0) = 1$。\n$$z_0(2xwz_0+1)^2 = w(5z_0+y_0)(4z_0+y_0)$$\n因為 $(z_0, 5z_0+y_0) = (z_0, 4z_0+y_0) = 1$ 且 $(w, (2xwz_0+1)^2) = 1$，所以 $w \\mid z_0, z_0 \\mid w$，故 $w = z_0$。\n因此 $(2xwz_0+1)^2 = (5z_0+y_0)(4z_0+y_0)$。\n又 $(5z_0+y_0, 4z_0+y_0) = (z_0, y_0) = 1$，可設 $5z_0+y_0 = m^2, 4z_0+y_0 = n^2$，其中 $m, n$ 為正整數，且 $m > n$，即 $m-n \\ge 1$。\n$$w = z_0 = m^2-n^2$$\n$$2xw^2+1 = 2xwz_0+1 = mn$$\n因此 $mn = 1+2xw^2 = 1+2x(m^2-n^2)^2 = 1+2x(m-n)^2(m+n)^2 \\ge 1+2x(m+n)^2 \\ge 1+8xmn \\ge 1+8mn$，即 $7mn \\le -1$（矛盾），所以 $z$ 不為偶數。\n\n2. 令 $w = \\gcd(y, z)$，且 $y = w y_0, z = w z_0$，其中 $(y_0, z_0) = 1$。\n$$z_0(xwz_0+1)^2 = w(5z_0+2y_0)(2z_0+y_0)$$\n因為 $(y_0, z_0) = 1$ 且 $z_0$ 為奇數，\n所以 $(z_0, 2z_0+y_0) = 1 = (z_0, 5z_0+2y_0)$，故 $z_0 \\mid w$，又 $(w, (xwz_0+1)^2) = 1$，所以 $w \\mid z_0$。\n因此 $z_0 = w, z = w z_0 = w^2$，得證。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20717,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be a triangle with circumcentre $O$ and centroid $G$. Let $M$ be the midpoint of $BC$ and $N$ be the reflection of $M$ across $O$. Prove that $NO = NA$ if and only if $\\angle AOG = 90^\\circ$.",
"options": [],
"answer": "See solution",
"solution": "Let $H$ be the orthocenter of $\\triangle ABC$ and let $X$ be the midpoint of $AH$. Then we know that $AXON$ is a parallelogram.\n\nNow, observe that $\\angle AOH = \\angle AOG$. Now,\n\n$$\nNO = NA \\iff XA = XO \\iff \\angle AOH = 90^\\circ\n$$\n\nThus, we are done. $\\Box$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20718,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, \\dots, a_n$ and $b_1, \\dots, b_n$ be $2n$ real numbers. Prove that there exists an integer $k$ with $1 \\leq k \\leq n$ such that\n$$\n\\sum_{i=1}^{n} |a_i - a_k| \\leq \\sum_{i=1}^{n} |b_i - a_k|.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $A_{\\min}$ and $A_{\\max}$ denote respectively the minimum and maximum value among $a_1, a_2, \\dots, a_n$. Then the triangle inequality implies for every $i$ with $1 \\leq i \\leq n$ that\n$$\n\\begin{aligned}\n|A_{\\max} - a_i| + |a_i - A_{\\min}| &= (A_{\\max} - a_i) + (a_i - A_{\\min}) = A_{\\max} - A_{\\min} \\\\\n&= |A_{\\max} - A_{\\min}| \\leq |A_{\\max} - b_i| + |b_i - A_{\\min}|.\n\\end{aligned}\n$$\nSumming these inequalities up over all $i = 1, 2, \\dots, n$ yields\n$$\n\\sum_{i=1}^{n} |A_{\\max} - a_i| + \\sum_{i=1}^{n} |a_i - A_{\\min}| \\leq \\sum_{i=1}^{n} |A_{\\max} - b_i| + \\sum_{i=1}^{n} |b_i - A_{\\min}|.\n$$\nThis implies that the integer $k$ with $a_k = A_{\\max}$ or the integer $k$ with $a_k = A_{\\min}$ must satisfy the desired inequality. $\\square$\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20719,
"subject": "Mathematics (Olympiad)",
"question": "Michael prints the net in the figure twice on cardboard and makes it into two identical dice, such that the pips are visible on the outside of the dice. He puts one die on top of the other to make a small tower. The front face of the lower die shows 3 pips. The total number of pips on the two faces touching in the middle is equal to 9. The total number of pips on the back of the small tower is three times the total number of pips on the right side of the small tower.\n\n\n\nHow many pips are on the face that touches the ground?\n\nA) 1 \nB) 2 \nC) 4 \nD) 5 \nE) 6",
"options": [],
"answer": "See solution",
"solution": "A) 1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20720,
"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) \\ge f(xy)$.\n2. For all $x, y \\in \\mathbb{Q}_{>0}$, we have $f(x + y) \\ge 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) \\ge f(nx) \\ge n f(x)$ for positive integer $n$. Setting $x = a$, we see that $f(n) \\ge 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) \\le 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) \\ge x$ for all $x \\ge 1$. First, note that $f(x) \\ge f(\\lfloor x \\rfloor) \\ge \\lfloor x \\rfloor > x - 1$. From (1) we know that $f(x)^n \\ge f(x^n)$, so $f(x)^n \\ge 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 \\ge (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) \\ge x$ for all $x > 1$, and we already know that $f(1) \\ge 1$, yielding the claim.\n\nWe now show $f(x) = x$ for $x \\ge 1$. Note that $a^k = f(a)^k \\ge f(a^k)$ for positive integers $k$ by (1). We also have $f(a^k) \\ge a^k$, so $f(a^k) = a^k$ for positive integers $k$. For $x \\ge 1$ and $k$ with $a^k > 2x$, we have $a^k = f(a^k) \\ge f(x) + f(a^k - x) \\ge x + (a^k - x) = a^k$. Equality thus holds, so $f(x) = x$ for $x \\ge 1$.\n\nFinally, for any integer $n$, we have $f(n) = n$ and $f(n)f(x) \\ge f(nx) \\ge 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": 20721,
"subject": "Mathematics (Olympiad)",
"question": "設 $P$ 為三角形 $ABC$ 內一點，直線 $AP, BP, CP$ 分別與三角形 $ABC$ 的外接圓交於 $T, S, R$ 點（$T \\neq A, S \\neq B, R \\neq C$）。設 $U$ 為線段 $PT$ 內一點。過 $U$ 並與 $AB$ 平行的直線與 $CR$ 交於 $W$ 點，過 $U$ 並與 $AC$ 平行的直線與 $BS$ 交於 $V$ 點。最後，過 $B$ 並與 $CP$ 平行的直線與過 $C$ 並與 $BP$ 平行的直線交於 $Q$ 點。已知 $RS$ 與 $VW$ 平行，證明 $\\angle CAP = \\angle BAQ$。\n\nLet $P$ be a point inside triangle $ABC$. Suppose that the lines $AP, BP, CP$ intersect the circumcircle of triangle $ABC$ at points $T, S, R$ respectively ($T \\neq A, S \\neq B, R \\neq C$). Let $U$ be a point interior to segment $PT$. The line through $U$ parallel to $AB$ meets $CR$ at $W$. The line through $U$ parallel to $AC$ meets $BS$ at $V$. Finally, the line through $B$ parallel to $CP$ meets the line through $C$ parallel to $BP$ at $Q$. Given that $RS$ is parallel to $VW$, prove that $\\angle CAP = \\angle BAQ$.",
"options": [],
"answer": "See solution",
"solution": "1. 設過 $U$ 並與 $AC$ 平行的直線交 $CR$ 於 $X$，過 $U$ 並與 $AB$ 平行的直線交 $BS$ 於 $Y$。因 $UY \\parallel AB$，$\\triangle PUT \\sim \\triangle PAB$，由此得 $\\frac{PU}{AP} = \\frac{PY}{BP}$。同理 $\\frac{PU}{AP} = \\frac{PX}{CP}$，故 $\\frac{PY}{BP} = \\frac{PX}{CP}$，因此 $XY \\parallel BC$。\n\n2. 因為 $\\angle VWP = \\angle XRS = \\angle PBC = \\angle BYX$，故有 $R, S, X, Y$ 共圓，以及 $V, W, X, Y$ 共圓。\n\n3. 因 $V, W, X, Y$ 共圓，$\\angle BYU = \\angle CYU$，並由此得 $\\angle ABP = \\angle ACP$（因為 $UY \\parallel AB$, $UX \\parallel AC$）。\n\n4. 設 $AP$ 交 $CQ$ 於 $D$，$BP$ 交 $AC$ 於 $E$，$CP$ 交 $AB$ 於 $F$。要證 $\\angle CAP = \\angle BAQ$，只須證 $\\triangle BAQ \\sim \\triangle CAD \\sim \\triangle EAP$。\n\n而因 $BPCQ$ 為平行四邊形，且 $\\angle ABP = \\angle ACP$，$\\angle ABQ = \\angle ACD = \\angle AEP$，又 $PC = BQ$。故若 $\\frac{BQ}{AB} = \\frac{EP}{AE}$ 或 $\\frac{PC}{AB} = \\frac{EP}{AE}$，可得 $\\triangle BAQ \\sim \\triangle EAP$。\n\n5. 由正弦定律得 $\\frac{AB}{AE} = \\frac{\\sin \\angle AEP}{\\sin \\angle ABP}$，$\\frac{PC}{PE} = \\frac{\\sin \\angle PEC}{\\sin \\angle ACP}$。因 $\\angle ABP = \\angle ACP$ 且 $\\angle PEC, \\angle AEP$ 互補，得 $\\frac{AB}{AE} = \\frac{PC}{PE}$。證畢！",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20722,
"subject": "Mathematics (Olympiad)",
"question": "Consider $n$ similar isosceles triangles and their reflection in $AO$. Draw the dotted radii as shown. Note that $2\\alpha + \\beta = 180^{\\circ}$.\n\n\n\nFind the value of $\\alpha$ in terms of $n$.",
"options": [],
"answer": "See solution",
"solution": "In isosceles triangle $AOB_n$, $\\angle AB_nO = \\alpha$, so $\\angle AOB_n = \\beta$.\n\nIn cyclic quadrilateral $AB_nB_{n-1}B'_n$, $\\angle AB_nB_{n-1} + \\angle AB'_nB_{n-1} = 180^{\\circ}$.\n\nThus, $\\angle OB_nB_{n-1} = (180^{\\circ} - \\beta) - \\alpha = \\alpha$.\n\nIn isosceles triangle $B_nOB_{n-1}$, $\\angle B_nB_{n-1}O = \\alpha$, so $\\angle B_nOB_{n-1} = \\beta$.\n\nIn cyclic quadrilateral $B_nB_{n-1}B_{n-2}B'_{n-1}$, $\\angle B_nB'_{n-1}B_{n-2} + \\angle B_nB_{n-1}B_{n-2} = 180^{\\circ}$.\n\nThus, $\\angle OB_{n-1}B_{n-2} = (180^{\\circ} - \\beta) - \\alpha = \\alpha$.\n\nIn isosceles triangle $B_{n-1}OB_{n-2}$, $\\angle B_{n-1}OB_{n-2}O = \\alpha$, so $\\angle B_{n-1}OB_{n-2} = \\beta$.\n\nSimilarly, $\\angle B_{k+1}OB_k = \\beta$ for $k = n-3, n-4, \\dots, 1$.\n\nSo $\\alpha = \\angle AOB_1 = n\\beta$.\n\nGiven $2\\alpha + \\beta = 180^{\\circ}$, substitute $\\alpha = n\\beta$:\n\n$$2n\\beta + \\beta = 180^{\\circ}$$\n$$ (2n+1)\\beta = 180^{\\circ} $$\n$$ \\beta = \\frac{180^{\\circ}}{2n+1} $$\n$$ \\alpha = n\\beta = \\frac{180n}{2n+1} $$\n\nThus, $\\boxed{\\alpha = \\dfrac{180n}{2n+1}}$ degrees.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20723,
"subject": "Mathematics (Olympiad)",
"question": "設 $k$ 為一給定實數。試找出所有從實數映至實數的函數 $f(x)$ 滿足對任意實數 $x, y$，均有\n\n$$\nf(x) + (f(y))^2 = k f(x + y^2).\n$$",
"options": [],
"answer": "See solution",
"solution": "若 $k \\neq 1$，則將 $y = 0$ 代入原式得到\n\n$$\n(f(0))^2 = (k - 1)f(x).\n$$\n\n因此 $f(x)$ 為常數函數，代入原式後可解出 $f(x) = 0$ 或 $f(x) = \\frac{1}{k-1}$。\n\n若 $k = 1$，則將 $x = 0$ 代入原式得到\n\n$$\nf(y)^2 = f(y^2).\n$$\n\n藉由上式我們可以知道對任意 $y > 0$ 均有 $f(y) > 0$ 並且 $f(1) = 0$ 或 $1 > f(x)$。將上式代回原題中：\n\n$$\n\\begin{align*}\n& f(x) + (f(y))^2 = f(x + y^2), \\quad \\forall x, y \\in \\mathbb{R} \\\\\n\\Rightarrow \\quad & f(x) + f(y^2) = f(x + y^2), \\quad \\forall x, y \\in \\mathbb{R} \\\\\n\\Rightarrow \\quad & f(x) + f(y) = f(x + y), \\quad \\forall x, y \\in \\mathbb{R},\\ y \\ge 0\n\\end{align*}\n$$\n\n此即為一標準柯西方程，因此 $f(x) = cx$。由 $f(1) = 0$ 或 $1 > f(1) = 0$ 或 $f(x) = x$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20724,
"subject": "Mathematics (Olympiad)",
"question": "Alice and Bob play a game with the following rules:\n\n- There are two playing numbers: $19$ and $20$.\n- There are two possible starting numbers: $9$ and $10$.\n- Alice chooses one playing number for herself and assigns the other to Bob.\n- Bob independently chooses the starting number.\n- The game proceeds as follows: Alice adds her playing number to the starting number, then Bob adds his playing number to the sum, then Alice again adds her playing number, and so on, alternating turns.\n- The game continues until the number $2019$ is reached or exceeded.\n- A player who obtains exactly $2019$ wins. If $2019$ is exceeded, the game ends in a draw.\n\n**Questions:**\n\n1. Show that Bob cannot win.\n2. Which starting number does Bob have to choose in order to prevent Alice from winning?",
"options": [],
"answer": "See solution",
"solution": "Let Alice's playing number be $a$ and Bob's be $b$, with $a, b \\in \\{19, 20\\}$ and $a \\neq b$. Let the starting number be $s \\in \\{9, 10\\}$.\n\nThe sequence of numbers is:\n\n- After Alice's first move: $s + a$\n- After Bob's move: $s + a + b$\n- After Alice's second move: $s + a + b + a = s + 2a + b$\n- After Bob's second move: $s + 2a + 2b$\n- ...\n\nAfter $n$ full rounds (each round is Alice then Bob), the sum is:\n$$\ns_n = s + n(a + b)\n$$\nIf Alice starts a round, her move is:\n$$\ns_{A, n} = s + n(a + b) + a\n$$\nIf Bob starts a round, his move is:\n$$\ns_{B, n} = s + n(a + b) + b\n$$\n\nTo win, a player must reach exactly $2019$ on their turn.\n\nSuppose Bob wins. Then for some $n$:\n$$\ns_{B, n} = 2019 \\implies s + n(a + b) + b = 2019\n$$\n$$\nn(a + b) = 2019 - s - b\n$$\nBut $a + b = 39$ (since $a, b$ are $19$ and $20$), so:\n$$\nn = \\frac{2019 - s - b}{39}\n$$\nBut $2019 - s - b$ is not divisible by $39$ for any $s \\in \\{9, 10\\}$ and $b \\in \\{19, 20\\}$ (check all cases), so Bob cannot win.\n\nTo prevent Alice from winning, Bob must choose $s$ so that $2019 - s - a$ is not divisible by $39$ (so Alice cannot reach $2019$ on her turn). For each possible $a$, Bob can always choose $s$ to block Alice:\n\n- If Alice picks $a = 19$, Bob picks $s = 10$.\n- If Alice picks $a = 20$, Bob picks $s = 9$.\n\nThus, Bob should choose the starting number so that $2019 - s - a$ is not divisible by $39$.\n\n**Summary:**\n- Bob cannot win.\n- Bob can always choose the starting number to prevent Alice from winning, resulting in a draw.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20725,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$ be positive real numbers such that\n$$\na^2 + b^2 + c^2 + (a+b+c)^2 \\le 4.\n$$\nProve that\n$$\n\\frac{ab+1}{(a+b)^2} + \\frac{bc+1}{(b+c)^2} + \\frac{ca+1}{(c+a)^2} \\ge 3.\n$$",
"options": [],
"answer": "See solution",
"solution": "The given condition is equivalent to $a^2 + b^2 + c^2 + ab + bc + ca \\le 2$. We will prove that\n$$\n\\frac{2ab+2}{(a+b)^2} + \\frac{2bc+2}{(b+c)^2} + \\frac{2ca+2}{(c+a)^2} \\ge 6.\n$$\nIndeed, we have\n$$\n\\frac{2ab+2}{(a+b)^2} \\ge \\frac{2ab+a^2+b^2+c^2+ab+bc+ca}{(a+b)^2} = 1 + \\frac{(c+a)(c+b)}{(a+b)^2}.\n$$\nAdding the last inequality with its analogous cyclic forms yields\n$$\n\\frac{2ab+2}{(a+b)^2} + \\frac{2bc+2}{(b+c)^2} + \\frac{2ca+2}{(c+a)^2} \\ge 3 + \\frac{(c+a)(c+b)}{(a+b)^2} + \\frac{(a+b)(a+c)}{(b+c)^2} + \\frac{(b+c)(b+a)}{(c+a)^2}.\n$$\nHence it remains to prove that\n$$\n\\frac{(c+a)(c+b)}{(a+b)^2} + \\frac{(a+b)(a+c)}{(b+c)^2} + \\frac{(b+c)(b+a)}{(c+a)^2} \\ge 3.\n$$\nThis follows directly from the AM-GM inequality.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20726,
"subject": "Mathematics (Olympiad)",
"question": "В гостинице 100 комнат вместимостью от 101 до 200 человек каждая (все вместимости целые и различны). Какое наибольшее число постояльцев может быть в гостинице, чтобы директор мог освободить любую комнату, переселив её жильцов по другим комнатам, не превышая их вместимости?",
"options": [],
"answer": "See solution",
"solution": "Предположим, что при 8824 постояльцах директор не может осуществить переселение. Разобьём комнаты на пары по вместимости: 101–200, 102–199, ..., 150–151. Для каждой пары сумма жильцов больше вместимости большей комнаты, иначе всех можно было бы переселить в неё. Значит, общее число жильцов не меньше $201 + 200 + 199 + \\dots + 152 = 353 \\cdot 25 = 8825$. Поэтому при 8824 постояльцах директор может освободить любую комнату.\n\nТеперь пример, показывающий, что при 8825 и более постояльцах можно расселить так, что ни одну комнату освободить нельзя:\n\nПусть в первых 50 комнатах живёт по 76 человек, а в комнате вместимости $k$ при $151 \\leq k \\leq 200$ — $k-75$ человек. Тогда всего:\n\n$$\n76 \\cdot 50 + (76 + 77 + \\dots + 125) = 3800 + 201 \\cdot 25 = 3800 + 5025 = 8825.\n$$\n\nВ комнате вместимости $b$ живёт не меньше $b-75$ человек, а в любой меньшей — не меньше 76. Переселить жильцов из одной в другую нельзя, значит, пример подходит. Если $n > 8825$, оставшихся можно расселить по свободным местам.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20727,
"subject": "Mathematics (Olympiad)",
"question": "The plane is divided into unit squares by two sets of parallel lines. The unit squares are coloured in 1201 colours so that no rectangle of perimeter 100 contains two squares of the same colour. Show that no rectangle of size $1 \\times 1201$ contains two squares of the same colour.",
"options": [],
"answer": "See solution",
"solution": "Let the centers of the unit squares be the integer points in the plane, and denote each unit square by the coordinates of its center.\n\nConsider the set $D$ of all unit squares $(x, y)$ such that $|x| + |y| \\leq 24$. Any translate of $D$ is called a *diamond*.\n\nSince any two unit squares that belong to the same diamond also belong to some rectangle of perimeter 100, a diamond cannot contain two unit squares of the same colour. Since a diamond contains exactly $24^2 + 25^2 = 1201$ unit squares, a diamond must contain every colour exactly once.\n\nChoose one colour, say, green, and let $a_1, a_2, \\dots$ be all green unit squares. Let $P_i$ be the diamond of center $a_i$. We will show that no unit square is covered by two $P_i$'s and that every unit square is covered by some $P_i$.\n\nIndeed, suppose first that $P_i$ and $P_j$ contain the same unit square $b$. Then their centers lie within the same rectangle of perimeter 100, a contradiction.\n\nLet, on the other hand, $b$ be an arbitrary unit square. The diamond of center $b$ must contain some green unit square $a_i$. The diamond $P_i$ of center $a_i$ will then contain $b$.\n\nTherefore, $P_1, P_2, \\dots$ form a covering of the plane in exactly one layer. It is easy to see, though, that, up to translation and reflection, there exists a unique such covering. (Indeed, consider two neighbouring diamonds. Unless they fit neatly, uncoverable spaces of two unit squares are created near the corners: see Fig. 1.)\n\n\n\nFigure 1:\n\nWithout loss of generality, then, this covering is given by the diamonds of centers $(x, y)$ such that $24x + 25y$ is divisible by 1201. (See Fig. 2 for an analogous covering with smaller diamonds.) It follows from this that no rectangle of size $1 \\times 1201$ can contain two green unit squares, and analogous reasoning works for the remaining colours. $\\Box$\n\n\n\nFigure 2:\n\nRemark: The number of the unit squares in a diamond can be evaluated alternatively with the formula\n\n$$\n2 \\times (1 + 3 + 5 + \\dots + 47) + 49 = 2 \\times 24^2 + 49 = 1201\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20728,
"subject": "Mathematics (Olympiad)",
"question": "Let $f(x)$ and $g(x)$ be strictly increasing linear functions from $\\mathbb{R}$ to $\\mathbb{R}$ such that $f(x)$ is an integer if and only if $g(x)$ is an integer. Prove that for any real number $x$, $f(x) - g(x)$ is an integer.",
"options": [],
"answer": "See solution",
"solution": "We can write $f(x) = a x + b$ and $g(x) = c x + d$ for some real numbers $a, b, c, d$ with $a, c > 0$.\n\nBy symmetry, we may assume that $a \\geq c$.\n\nWe claim that $a = c$. Assume on the contrary that $a > c$. Because $a > c > 0$, the ranges of $f$ and $g$ are both $\\mathbb{R}$. There is an $x_0$ such that $f(x_0) = a x_0 + b$ is an integer. Hence $g(x_0) = c x_0 + d$ is also an integer. But then,\n\n$$\nf\\left(x_0 + \\frac{1}{a}\\right) = a x_0 + b + 1\n$$\n\nand\n\n$$\ng\\left(x_0 + \\frac{1}{a}\\right) = c x_0 + d + \\frac{c}{a}.\n$$\n\nBut this is impossible because we cannot have two integers $g(x_0)$ and $g(x_0 + \\frac{1}{a})$ that have positive difference $\\frac{c}{a}$ which is less than $1$.\n\nTherefore, we can write $f(x) = a x + b$ and $g(x) = a x + d$ for some real numbers $a, b, d$ with $a > 0$.\n\nThen $b - d = f(x_0) - g(x_0)$ must be an integer, that is,\n\n$$\nf(x) - g(x) = b - d\n$$\n\nis an integer for all $x$.\n\n$\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20729,
"subject": "Mathematics (Olympiad)",
"question": "A base 7 three-digit number has its digits reversed when written in base 9. Find the decimal representation of the number.",
"options": [],
"answer": "See solution",
"solution": "Let the number be $\\left(abc\\right)_7$, where $a, b, c$ are digits less than 7. Then $\\left(abc\\right)_7 = \\left(cba\\right)_9$.\n\nSo:\n$$\n49a + 7b + c = 81c + 9b + a\n$$\n\nRearrange:\n$$\n49a - a + 7b - 9b + c - 81c = 0 \\\\\n48a - 2b - 80c = 0 \\\\\n48a = 80c + 2b \\\\\n24a = 40c + b\n$$\n\nSince $b$ is a digit less than 7, and $b = 24a - 40c$, $b$ must be a multiple of 8 and less than 7, so $b = 0$.\n\nNow:\n$$\n24a = 40c \\\\\n3a = 5c\n$$\n\nWith $a, c < 7$, the only solution is $a = 5$, $c = 3$.\n\nSo the number is $\\left(503\\right)_7$.\n\nConvert to decimal:\n$$\n5 \\times 49 + 0 \\times 7 + 3 = 245 + 0 + 3 = 248\n$$\n\n**Answer:** $248$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20730,
"subject": "Mathematics (Olympiad)",
"question": "Determine the natural numbers $v$ for which $2007 + 4v$ is a perfect square.",
"options": [],
"answer": "See solution",
"solution": "Let $2007 + 4v = k^2$, where $k \\in \\mathbb{N}$. Then\n$$\nk^2 - 4v = 2007.\n$$\nWe can rewrite this as\n$$\nk^2 - 2007 = 4v \\implies v = \\frac{k^2 - 2007}{4}.\n$$\nFor $v$ to be a natural number, $k^2 - 2007$ must be divisible by $4$ and $v > 0$.\n\nAlternatively, set $k^2 - 4v = 2007$ and factor $2007 = 1 \\cdot 2007 = 3 \\cdot 669 = 9 \\cdot 223$.\n\nSet $k - 2v_1 = d$, $k + 2v_1 = \\frac{2007}{d}$ for divisors $d < \\frac{2007}{d}$:\n\n- $d = 1$: $k - 2v_1 = 1$, $k + 2v_1 = 2007$ \\implies $2k = 2008$ \\implies $k = 1004$, $2v_1 = 2006$ \\implies $v_1 = 1003$ (not a power of $2$).\n- $d = 3$: $k - 2v_1 = 3$, $k + 2v_1 = 669$ \\implies $2k = 672$ \\implies $k = 336$, $2v_1 = 666$ \\implies $v_1 = 333$ (not a power of $2$).\n- $d = 9$: $k - 2v_1 = 9$, $k + 2v_1 = 223$ \\implies $2k = 232$ \\implies $k = 116$, $2v_1 = 214$ \\implies $v_1 = 107$ (not a power of $2$).\n\nIn all cases, $v$ is not a natural number. Therefore, there are no natural numbers $v$ such that $2007 + 4v$ is a perfect square.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20731,
"subject": "Mathematics (Olympiad)",
"question": "求所有正整數組 $(x, y, z)$ 滿足：\n\n$$\nx^2 + 4^y = 5^z.\n$$",
"options": [],
"answer": "See solution",
"solution": "所有的解為 $(1, 1, 1)$、$(11, 1, 3)$、$(3, 2, 2)$。\n\n**證明：**\n\n考慮 $y \\geq 2$ 的情形：\n\n$$\nx^2 \\equiv 5^z \\pmod{8}\n$$\n\n因為 $x^2 \\equiv 0, 1, 4$，$5^z \\equiv 1, 5$，所以 $z$ 為偶數。設 $z = 2z'$，則：\n\n$$\nx^2 + (2^y)^2 = (5^{z'})^2\n$$\n\n$5^{z'}$ 與 $2^y$ 互質且 $2 \\mid 2^y$，由畢氏三元數公式：\n\n$$\nx = m^2 - n^2, \\quad 2^y = 2mn, \\quad 5^{z'} = m^2 + n^2\n$$\n\n其中 $m, n$ 互質且 $m > n$。由 $2^y = 2mn$ 得 $m = 2^{y-1}, n = 1$，所以 $5^{z'} = 2^{2(y-1)} + 1$。\n\n若 $y \\geq 3$，則：\n\n$$\n5^{z'} \\equiv 1 \\pmod{8} \\implies 2 \\mid z'\n$$\n\n設 $z' = 2z''$，則：\n\n$$\n(5^{z''} + 1)(5^{z''} - 1) = 2^{2(y-1)} \\implies 5^{z''} + 1 = 2^a,\\ 5^{z''} - 1 = 2^b\n$$\n\n但 $4$ 不整除 $5^{z''} + 1$ 或 $5^{z''} - 1$，所以 $a = 1$ 或 $b = 1$，無解。\n\n若 $y = 2$，得 $(x, y, z) = (3, 2, 2)$。\n\n考慮 $y = 1$：\n\n$$\nx^2 + 4 \\equiv 5^z \\pmod{8}\n$$\n\n$x^2 + 4 \\equiv 0, 4, 5$，$5^z \\equiv 1, 5$，所以 $z$ 為奇數。設 $z = 2z' + 1$，考慮佩爾方程：\n\n$$\ns^2 - 5t^2 = -4\n$$\n\n設 $(s, t) = (x, 5^{z'})$。若 $(s, t)$ 是正整數解，則 $(\\frac{3s-5t}{2}, \\frac{3t-s}{2})$ 也是整數解（$s, t$ 同奇偶），可遞降至 $3s \\leq 5t$ 或 $3t \\leq s$。但 $3t > s$ 永遠成立。對 $3s \\leq 5t$：\n\n$$\n-20t^2 = 25t^2 - 45t^2 \\geq 9(s^2 - 5t^2) = -36 \\implies t = 1 \\implies s = 1\n$$\n\n所以所有正整數解 $(s_n, t_n)$ 滿足：\n\n$$\n(s_0, t_0) = (1, 1), \\quad s_{n+1} = \\frac{3s_n + 5t_n}{2}, \\quad t_{n+1} = \\frac{3t_n + s_n}{2}\n$$\n\n觀察前幾項 $(s_1, t_1) = (4, 2)$，$(s_2, t_2) = (11, 5)$，並考慮費氏數列：\n\n$$\nF_0 = 0, F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, F_5 = 5, \\dots\n$$\n\n由遞迴式可得：\n\n$$\n(s_n, t_n) = (F_{2n+2} + F_{2n}, F_{2n+1})\n$$\n\n若 $t_n = 5^{z'}$，則 $z' = 0$ 或 $5 \\mid F_{2n+1}$。前者得 $(x, y, z) = (1, 1, 1)$，後者由費氏數列模 $5$ 的規律知 $5 \\mid n' := 2n + 1$。若 $p \\neq 5$ 為 $n'$ 的質因數，則 $F_p \\mid F_{n'} = 5^{z'}$，但 $F_p \\neq 1$ 且 $5 \\nmid F_p$，矛盾。因此 $n'$ 為 $5$ 的冪次。若 $z' > 1$，則 $n' > 5$，故 $25 \\mid n'$，但 $F_{25} = 75025$ 非 $5$ 的冪次，矛盾。因此 $z' = 1$，得 $(x, y, z) = (11, 1, 3)$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20732,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\lfloor x \\rfloor$ denote the largest integer less than or equal to $x$, and let $\\lceil x \\rceil$ denote the smallest integer greater than or equal to $x$.\n\nFor every given pair $(a, b)$ of positive natural numbers, find all natural numbers $n$ such that\n\n$$\nb + \\lfloor \\frac{n}{a} \\rfloor = \\lfloor \\frac{n+b}{a} \\rfloor.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $k := \\lfloor \\frac{n}{a} \\rfloor$ and $l := \\lfloor \\frac{n+b}{a} \\rfloor$. We seek all nonnegative integers $n$ such that there exist integers $k$ and $l$ with\n\n$$\nb + k = l \\quad \\text{and} \\quad k \\le \\frac{n}{a} < k+1 \\quad \\text{and} \\quad l-1 < \\frac{n+b}{a} \\le l.\n$$\n\nSubstituting $l = b + k$ gives\n\n$$\nka \\le n < (k+1)a \\quad \\text{and} \\quad (b+k-1)a < n + b \\le a(b+k).\n$$\n\nSince all variables are integers, this is equivalent to\n\n$$\nka \\le n \\le (k+1)a - 1 \\quad \\text{and} \\quad (b+k-1)a + 1 - b \\le n \\le a(b+k) - b.\n$$\n\nNote that\n\n$$\n(b+k-1)a + 1 - b = ka + (a-1)(b-1) \\ge ka\n$$\nand\n$$\na(b+k) - b = (k+1)a - 1 + (a-1)(b-1) \\ge (k+1)a - 1.\n$$\n\nTherefore, the desired values of $n$ are those satisfying\n\n$$\n(b+k-1)a + 1 - b \\le n \\le (k+1)a - 1 \\quad (1)\n$$\nfor some $k$.\n\nFrom (1), for some $k$ we must have $(b+k-1)a + 1 - b \\le (k+1)a - 1$, i.e., $0 \\ge ab - 2a + 2 - b = (a-1)(b-2)$. Thus, consider the following cases:\n\n- If $a = 1$, (1) gives $k \\le n \\le k$, so all natural numbers $n \\ge 0$ are solutions.\n- If $b = 1$, (1) gives $ka \\le n \\le (k+1)a - 1$, which is always satisfied for $k = \\lfloor \\frac{n}{a} \\rfloor$. Thus, all $n \\ge 0$ are solutions.\n- If $b = 2$, (1) gives $(k+1)a - 1 \\le n \\le (k+1)a - 1$, so $n = (k+1)a - 1$. Such $k$ exists if and only if $n \\equiv -1 \\pmod a$, so all such $n$ are solutions. (For $a = 1$, all $n \\ge 0$ are solutions as above.)\n- For $a \\ge 2$ and $b \\ge 3$, there is no solution.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20733,
"subject": "Mathematics (Olympiad)",
"question": "在銳角三角形 $ABC$ 中，令點 $F$ 是通過 $A$ 的高的垂足，而點 $P$ 位於線段 $AF$ 上。過點 $P$ 分別作平行 $AC$ 和 $AB$ 的直線，設它們分別交 $BC$ 於點 $D$ 和 $E$。在圓 $ABD$ 及圓 $ACE$ 上分別取點 $X \\ne A$、$Y \\ne A$，滿足 $DA = DX$、$EA = EY$。證明 $B, C, X, Y$ 共圓。",
"options": [],
"answer": "See solution",
"solution": "解法一：令 $A'$ 為直線 $BX$ 與 $CY$ 的交點。由圓幂性質，我們只需證明 $A'B \\cdot A'X = A'C \\cdot A'Y$，或等價於證出 $A'$ 位於圓 $ABDX$ 及 $ACEY$ 的根軸上。\n\n\n\n由 $DA = DX$，知在圓 $ABDX$ 上，點 $D$ 平分以 $A, X$ 為端點的兩弧之一。所以，根據點的順序，直線 $BC$ 是 $\\angle ABX$ 的內角平分線或外角平分線。不論在哪種情形，直線 $BX$ 皆是直線 $BA$ 關於直線 $BC$ 的對稱線。同理可知，直線 $CY$ 是直線 $CA$ 關於直線 $BC$ 的對稱線。於是 $A'$ 為 $A$ 關於直線 $BC$ 的對稱點，由此得 $A, F, A'$ 三點共線。\n\n由 $PD \\parallel AC$ 及 $PE \\parallel AB$，知 $\\frac{FD}{FC} = \\frac{FP}{FA} = \\frac{FE}{FB}$，可得 $FD \\cdot FB = FE \\cdot FC$。\n因此點 $F$ 到圓 $ABDX$ 及圓 $ACEY$ 等幂。\n\n點 $A$ 在這兩圓上，所以到它們也等幂。所以圓 $ABDX$ 與圓 $ACEY$ 的根軸是高 $AF$，其經過 $A'$。得證。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20734,
"subject": "Mathematics (Olympiad)",
"question": "Let the area of quadrilateral $ABCD$ be maximized given $AB = x$, $BD = y$, $CD = z$, with $x + y + z = L$. What is the maximal area $S(ABCD)$ in terms of $L$?",
"options": [],
"answer": "See solution",
"solution": "Let $S(ABCD) = S(ABD) + S(DBC) = \\frac{1}{2} AB \\cdot BD \\sin \\angle ABD + \\frac{1}{2} BD \\cdot DC \\sin \\angle BDC$. The area is maximized when $\\angle ABD = \\angle CDB = 90^\\circ$.\n\nThus,\n$$\nS(ABCD) = \\frac{1}{2} x y + \\frac{1}{2} y z = \\frac{1}{2} y (x + z)\n$$\nSince $x + y + z = L$, $x + z = L - y$, so\n$$\nS(ABCD) = \\frac{1}{2} y (L - y)\n$$\nThe product $y(L - y)$ is maximized when $y = \\frac{L}{2}$, giving\n$$\nS_{\\text{max}} = \\frac{1}{2} \\cdot \\frac{L}{2} \\cdot \\left(L - \\frac{L}{2}\\right) = \\frac{1}{2} \\cdot \\frac{L}{2} \\cdot \\frac{L}{2} = \\frac{L^2}{8}\n$$\nTherefore, the maximal area is $\\boxed{\\dfrac{L^2}{8}}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20735,
"subject": "Mathematics (Olympiad)",
"question": "In cyclic quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at $P$. Let $E$ and $F$ be the respective feet of the perpendiculars from $P$ to lines $AB$ and $CD$. Segments $BF$ and $CE$ meet at $Q$. Prove that lines $PQ$ and $EF$ are perpendicular to each other.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $G$, $X$, and $Y$ be the respective feet of the perpendiculars from $P$ to $EF$, $EC$, and $FB$. Note that $EPYB$ and $FPXC$ are cyclic quadrilaterals, so\n\n$$\n\\begin{aligned}\n\\angle EYF &= 90^\\circ + \\angle EYP = 90^\\circ + \\angle EBP = 90^\\circ + \\angle ABP \\\\\n&= 90^\\circ + \\angle DCP = 90^\\circ + \\angle FCP = 90^\\circ + \\angle FXP = \\angle FXE.\n\\end{aligned}\n$$\n\nThus, $EXYF$ is a cyclic quadrilateral. Note that $EGPX$ and $FGPY$ are also cyclic. The radical axes of the circumcircles of $EXYF$, $EGPX$, and $FGPY$ with each other are $GP$, $EX$, and $FY$. Thus, by the radical axis theorem, $GP$, $EX$, and $FY$ concur. Since $Q$ is the intersection of $EX$ and $FY$, by the construction of $G$ we have $GP \\perp EF$, so it follows that $QP \\perp EF$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20736,
"subject": "Mathematics (Olympiad)",
"question": "Let the tick be placed in any one of the 16 blocks of a $4 \\times 4$ grid. Then the cross can go in any block not in the same row or column as the tick. How many ways can this be done?",
"options": [],
"answer": "See solution",
"solution": "The tick can be placed in any of the $16$ blocks. For each tick position, the cross can be placed in any of the $3$ other rows and $3$ other columns, giving $3 \\times 3 = 9$ possible positions for the cross. Thus, the total number of ways is $16 \\times 9 = 144$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20737,
"subject": "Mathematics (Olympiad)",
"question": "Let $k \\geq 1$ be a positive integer, and let $p_1, p_2, \\dots, p_k$ be distinct primes. Denote $n = p_1 p_2 \\dots p_k$.\n\nFor a function $f : \\{1, 2, \\dots, n\\} \\to \\{1, 2, \\dots, n\\}$, define $p(f) = f(1) f(2) \\dots f(n)$.\n\n**a)** Determine the number of functions $f$ such that $p(f)$ divides $n$.\n\n**b)** For $n = 6$, determine the number of functions $f$ such that $p(f)$ divides $36$.",
"options": [],
"answer": "See solution",
"solution": "a) If $p(f)$ divides $n$, then $p(f) = p_1^{a_1} p_2^{a_2} \\dots p_k^{a_k}$, where $a_i \\in \\{0, 1\\}$. This means that for each prime $p_i$, it either does not divide $p(f)$, or it appears in the prime decomposition of exactly one of the numbers $f(1), \\dots, f(n)$. Thus, for each $p_i$ there are $n + 1$ choices, so the total number of such functions is $(n + 1)^k$.\n\nb) For $n = 6$, $p(f)$ divides $36$ if $p(f) = 2^a 3^b$, where $a, b \\in \\{0, 1, 2\\}$. There are $1 + \\binom{6}{2} + 2 \\binom{6}{1}$ such functions for $b = 0$, $\\binom{6}{1}(1 + \\binom{6}{1} + \\binom{5}{1} + \\binom{6}{2})$ functions for $b = 1$, and $\\binom{6}{2}(1 + \\binom{6}{1} + \\binom{4}{1} + \\binom{6}{2})$ functions for $b = 2$. Summing these, we obtain $580$ functions with the given property.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20738,
"subject": "Mathematics (Olympiad)",
"question": "For any positive integer $k$, define\n\n$$\nH_k = 1 + \\frac{1}{2} + \\dots + \\frac{1}{k}.\n$$\n\nProve the relation\n\n$$\n1 + \\frac{1}{n+1} \\sum_{k=1}^{n} H_k = H_{n+1}.\n$$",
"options": [],
"answer": "See solution",
"solution": "We have\n\n$$\n\\begin{align*}\n\\sum_{k=1}^{n} H_k &= 1 + \\left(1 + \\frac{1}{2}\\right) + \\left(1 + \\frac{1}{2} + \\frac{1}{3}\\right) + \\dots + \\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{n}\\right) \\\\\n&= n + \\frac{n-1}{2} + \\frac{n-2}{3} + \\dots + \\frac{1}{n} \\\\\n&= (n+1) - 1 + \\frac{n+1-2}{2} + \\frac{n+1-3}{3} + \\dots + \\frac{n+1-n}{n} \\\\\n&= (n+1) \\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{n}\\right) - n = (n+1)H_n - n.\n\\end{align*}\n$$\n\nUsing the above relation we get\n\n$$\n\\begin{align*}\n1 + \\frac{1}{n+1} \\sum_{k=1}^{n} H_k &= 1 + \\frac{1}{n+1} [(n+1)H_n - n] \\\\\n&= H_n + 1 - \\frac{n}{n+1} = H_n + \\frac{1}{n+1} = H_{n+1}.\n\\end{align*}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20739,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle and $D$, $E$, $F$ be points on the sides $BC$, $CA$, $AB$ respectively such that $AD = BE = CF$. Suppose the line segments $AD$, $BE$, $CF$ are not concurrent and enclose an equilateral triangle. Is $ABC$ necessarily equilateral?",
"options": [],
"answer": "See solution",
"solution": "Yes. Draw $AK$ parallel to $BC$ and $AK = BD$. Join $K$ to $E$ and $F$. Choose $L$ on $BC$ with $L$ between $B$ and $C$, such that $BD = LC$. Note that $KB$ is parallel to $AD$ and $KB = AD$. Hence $\\angle KBE = \\angle APE = 60^\\circ$. On the other hand, $KB = AD = BE$. Thus $KBE$ is an equilateral triangle. This gives $\\angle BEK = 60^\\circ = \\angle BQR$. Thus $CF$ is parallel to $EK$. But $CF = BE = EK$. We conclude that $FCEK$ is a parallelogram. Since $CL$ is parallel to $AK$ and $CF$ is parallel to $KE$ we have $\\angle LCF = \\angle AKE$.\n\nConsider the triangles $LCF$ and $AKE$. We have $LC = AK$, $CF = KE$ and the included angles are also the same. Hence $LCF$ is congruent to $AKE$. Moreover $LF$ is parallel to $CA$. Thus\n\n$$\n\\frac{BF}{FA} = \\frac{BL}{LC} = \\frac{CD}{BD} = \\frac{1}{\\lambda}, \\text{ say.}\n$$\n\nConsider the transversal $FQC$ of the triangle $ABD$. By Menelaus' theorem\n\n$$\n\\frac{BF}{FA} \\cdot \\frac{AP}{PD} \\cdot \\frac{CD}{CB} = 1.\n$$\n\nThus\n\n$$\n\\frac{AP}{PD} = \\frac{FA}{BF} \\cdot \\frac{CB}{CD} = \\lambda \\left( \\frac{CD + DB}{CD} \\right) = \\lambda(1+\\lambda).\n$$\n\nSimilarly, we get\n\n$$\n\\frac{BQ}{QE} = \\frac{CR}{RF} = \\lambda(1+\\lambda).\n$$\n\nWe thus have\n\n$$\n\\frac{AP}{PD} = \\frac{BQ}{QE} = \\frac{CR}{RF}.\n$$\n\nThis equality implies\n\n$$\n\\frac{AD}{PD} = \\frac{BE}{QE} = \\frac{CF}{RF}.\n$$\n\nUsing $AD = BE = CF$, we get $PD = QE = RF$. But then $AP = BQ = CR$ and hence $AR = BP = CQ$. Consider the triangles $ABP$ and $BCQ$. Observe that $\\angle BPA = 120^\\circ = \\angle CQB$, $BP = CQ$ and $AP = BQ$. Thus $ABP$ and $BCQ$ are congruent triangles. This gives $AB = BC$. Similarly, we obtain $BC = CA$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20740,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $k$ and $n$ are positive integers such that $k \\leq n \\leq 2k - 1$. Julian has a large pile of rectangular $k \\times 1$ tiles. Merlijn picks a positive integer $m$, and receives from Julian $m$ tiles to place on an $n \\times n$ board. On each tile, Julian writes whether this tile should be placed horizontally or vertically. Tiles may not overlap on the board, and they must fit entirely inside the board. What is the largest number $m$ that Merlijn can pick while still guaranteeing he can put all tiles on the board according to Julian's instructions?",
"options": [],
"answer": "See solution",
"solution": "We show that the largest $m$ Merlijn can pick is $\\min(n,\\ 3(n-k)+1)$. \n\nFirst, we show that $m \\leq \\min(n,\\ 3(n-k)+1)$. If Merlijn asks for $n+1$ tiles, Julian can instruct Merlijn to place them all horizontally. As $n \\leq 2k-1$, it is impossible to place more than one tile horizontally on a single row, so Merlijn would need at least $n+1$ rows, which is a contradiction. Therefore, $m \\leq n$.\n\nNow suppose that Merlijn asks for $3(n-k)+2$ tiles. Julian can instruct Merlijn to place $n-k+1$ tiles vertically and $2n-2k+1$ tiles horizontally.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20741,
"subject": "Mathematics (Olympiad)",
"question": "A function $f$ defined on the set of real numbers $\\mathbb{R}$ and taking nonnegative real values satisfies the condition\n\n$$\nf(x + y) \\le 2 \\max\\{f(x), f(y)\\}\n$$\n\nfor all $x, y \\in \\mathbb{R}$. Is it true that for each positive integer $k$ the inequality\n\n$$\nf(x_1 + \\cdots + x_k) \\le 2(f(x_1) + \\cdots + f(x_k))\n$$\n\nholds for all $x_1, \\dots, x_k \\in \\mathbb{R}$?",
"options": [],
"answer": "See solution",
"solution": "**Answer:** Yes.\n\n**Solution.** The required inequality obviously holds for $k = 1$. For $k \\ge 2$ we will prove the next (stronger) inequality\n\n$$\n\\frac{f(x_1 + \\cdots + x_k)}{2} \\le f(x_1) + \\cdots + f(x_k) - \\min\\{f(x_1), \\ldots, f(x_k)\\} \\quad (1)\n$$\n\nfor arbitrary $x_1, \\dots, x_k \\in \\mathbb{R}$.\n\nThe proof is by induction on $k$. For $k = 2$ we have\n\n$$\n\\frac{f(x_1 + x_2)}{2} \\le \\max\\{f(x_1), f(x_2)\\} = f(x_1) + f(x_2) - \\min\\{f(x_1), f(x_2)\\},\n$$\n\nwhich is (1). Suppose that (1) holds for some $k \\ge 2$. Take any $x_1, \\dots, x_{k+1} \\in \\mathbb{R}$. Without restriction of generality we may assume that\n\n$$\nf(x_1) \\le f(x_2) \\le f(x_3) \\le \\dots \\le f(x_{k+1}). \\quad (2)\n$$\n\nAs the minimum among $f(x_j)$, $1 \\le j \\le k+1$, is $f(x_1)$, in order to prove (1) for $k+1$ (and so complete the induction step) it remains to show that\n\n$$\n\\frac{f(x_1 + x_2 + x_3 + \\cdots + x_{k+1})}{2} \\le f(x_2) + f(x_3) + \\cdots + f(x_{k+1}). \\quad (3)\n$$\n\nApplying (1) to the list of $k$ numbers $x_1 + x_2, x_3, \\dots, x_{k+1}$ we obtain\n\n$$\n\\frac{f(x_1 + x_2 + x_3 + \\cdots + x_{k+1})}{2} \\le f(x_1 + x_2) + f(x_3) + \\cdots + f(x_{k+1}) - M, \\quad (4)\n$$\n\nwhere, by (1) and (2),\n\n$$\nM = \\min\\{f(x_1 + x_2), f(x_3), \\dots, f(x_{k+1})\\} = \\min\\{f(x_1 + x_2), f(x_3)\\}.\n$$\n\nIn case $f(x_1 + x_2) \\le f(x_3)$, we have $M = f(x_1 + x_2)$, so the right hand side of (4) equals $f(x_3) + \\dots + f(x_{k+1})$. This implies (3). In the alternative case, $f(x_1 + x_2) > f(x_3)$, we have $M = f(x_3)$. Then the right hand side of (4) equals $f(x_1 + x_2) + f(x_4) + \\dots + f(x_{k+1})$ (the terms $f(x_4), \\dots, f(x_{k+1})$ appear only for $k \\ge 3$). By the condition on $f$ and (2), we obtain\n\n$$\nf(x_1 + x_2) \\le 2 \\max\\{f(x_1), f(x_2)\\} = 2f(x_2) \\le f(x_2) + f(x_3),\n$$\n\nwhich yields (3) too.\n\n",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20742,
"subject": "Mathematics (Olympiad)",
"question": "Let $f(x) = x^{2020} + \\sum_{i=0}^{2019} c_i x^i$, where $c_i \\in \\{-1, 0, 1\\}$, be a polynomial, and let $N$ be the number of positive integer roots of $f(x)$ (counted with multiplicities). Given that $f(x)$ has no negative integer roots, find the maximum value of $N$.",
"options": [],
"answer": "See solution",
"solution": "The maximum value of $N$ is $10$.\n\nNotice that $c_i \\in \\{-1, 0, 1\\}$, so any real root of $f(x)$ must have absolute value less than $2$. Thus, the only possible integer roots are $-1$, $0$, and $1$. Since $f(x)$ has no negative integer roots, only $0$ and $1$ can be roots.\n\nConsider the monic polynomial:\n$$\n f(x) = (x-1)(x^3-1)(x^5-1)(x^{11}-1)(x^{21}-1)(x^{43}-1)(x^{85}-1)(x^{171}-1)(x^{341}-1)(x^{683}-1)x^{656}\n$$\nThis polynomial has degree $2020$ and integer roots $x = 0, 1$ only. The multiplicity of $x = 1$ is $N = 10$.\n\nLet $f_i = x^{q_i} - 1$, where $q_1 = 1$, $q_2 = 3$, $q_3 = 5$, $q_4 = 11$, $q_5 = 21$, $q_6 = 43$, $q_7 = 85$, $q_8 = 171$, $q_9 = 341$, $q_{10} = 683$. Since each $q_i$ is larger than the sum of all previous $q_j$, by induction, in the expanded form of $\\prod_{i=1}^{J} f_i$, each coefficient $c_k$ satisfies $c_k \\in \\{-1, 0, 1\\}$ for $J = 1, 2, \\dots, 10$.\n\nTo prove that $N = 10$ is maximal, suppose $N \\geq 11$. Then $(x-1)^{11}$ divides $x^{2020} + \\sum_{i=0}^{2019} c_i x^i$. Setting $x = -1$, we get $2^{11}$ divides $f(-1)$. However, $|f(-1)| \\neq 0$, so $|f(-1)| \\geq 2^{11}$. On the other hand, $|f(-1)| \\leq 1 + \\sum_{i=0}^{2019} |c_i| \\leq 2021 < 2^{11}$, a contradiction. Therefore, the maximum value of $N$ is $10$.\n\n$\\square$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20743,
"subject": "Mathematics (Olympiad)",
"question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ and suppose\n\n$$\nf(f(x) + f(y)) + f(f(x) - f(y)) = x^k f(x) + y^k f(y)\n$$\n\nholds for all $x, y \\in \\mathbb{R}$, where $k$ is a given natural number. What values can $f(1)$ have? (Easier case: $k=2$).",
"options": [],
"answer": "See solution",
"solution": "For $x = y = 0$, let $a = f(0)$:\n\n$$\nf(2a) = -a\n$$\n\nFor $x = 0$ and $y = 2a$:\n\n$$\na - a = -a (2a)^k \\implies a = 0\n$$\n\nFor $x = y = 1$, let $b = f(1)$:\n\n$$\nf(2b) = 2b\n$$\n\nFor $x = 2b$ and $y = 0$:\n\n$$\n4b = (2b)^{k+1} \\implies b = 0 \\text{ or } b = \\frac{k\\sqrt{2}}{2}\n$$\n\nFor $b = 0$, $f(x) = 0$ for all $x$ satisfies the equation.\n\nFor $k = 2$ and $b = \\frac{\\sqrt{2}}{2}$, $f(x) = \\frac{x^2}{\\sqrt{2}}$ satisfies the equation.\n\nFor $k \\neq 2$, for $x = 1$ and $y = 0$:\n\n$$\n2f(b) = b \\implies f(b) = \\frac{b}{2}\n$$\n\nFor $x = y = b$:\n\n$$\nf(b) = 2b^k \\frac{b}{2} \\implies b^k = \\frac{1}{2}\n$$\n\nBut this contradicts $b^k = \\frac{1}{2^{k-1}}$ for $k \\neq 2$.\n\n**Conclusion:**\n- For $k = 2$, $f(1)$ can be $0$ or $\\frac{\\sqrt{2}}{2}$.\n- For $k \\neq 2$, $f(1) = 0$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20744,
"subject": "Mathematics (Olympiad)",
"question": "Find the least real $m$ such that there exist real numbers $a$ and $b$ for which the inequality\n\n$$\n|x^2 + a x + b| \\leq m\n$$\n\nholds for all $x \\in (0, 2)$.",
"options": [],
"answer": "See solution",
"solution": "No negative value of $m$ can satisfy the problem, since the absolute value is always non-negative.\n\nGeometrically, the graph of $y = x^2 + a x + b$ must lie within the horizontal strip between $y = m$ and $y = -m$ for $x \\in (0, 2)$. We seek the smallest such $m$.\n\n\n\nConsider the function\n\n$$\nf(x) = (x - 1)^2 - \\frac{1}{2} = x^2 - 2x + \\frac{1}{2}.\n$$\n\nThis function has $a = -2$, $b = \\frac{1}{2}$, and satisfies $-\\frac{1}{2} \\leq f(x) \\leq \\frac{1}{2}$ for $x \\in (0, 2)$. These inequalities are equivalent to $0 \\leq (x-1)^2 \\leq 1$, which holds for $x \\in (0, 2)$. Thus, $f(x)$ meets the problem's conditions for $m = \\frac{1}{2}$.\n\nNow, we show that no quadratic function can satisfy the condition for $m < \\frac{1}{2}$.\n\nFor any $f(x) = x^2 + a x + b$, at least one of the differences $f(0) - f(1)$ or $f(2) - f(1)$ is greater than or equal to $1$. This implies the strip's width must be at least $1$, so $m \\geq \\frac{1}{2}$. Using the triangle inequality:\n\n$$\n1 \\leq |f(0) - f(1)| \\leq |f(0)| + |f(1)| \\leq 2m.\n$$\n\nSimilarly for $f(2) - f(1)$. Now, compute:\n\n$$\nf(0) = b, \\quad f(1) = 1 + a + b, \\quad f(2) = 4 + 2a + b,\n$$\n\nSo,\n\n$$\nf(0) - f(1) = -1 - a \\geq 1 \\iff a \\leq -2,\n$$\n\n$$\nf(2) - f(1) = 3 + a \\geq 1 \\iff a \\geq -2.\n$$\n\nThus, for any $a$, at least one of these inequalities holds.\n\n*Conclusion.* The minimal value of $m$ is $\\frac{1}{2}$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20745,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a convex quadrilateral such that the circles with diameters $AB$ and $CD$ are tangent externally at a point $M$, different from the intersection point of the diagonals of the quadrilateral.\n\nLet $K$ be the second point of intersection of the circumcircle of triangle $AMC$ with the line through $M$ and the midpoint of $AB$, and let $L$ be the second point of intersection of the circumcircle of triangle $BMD$ with the line through $M$ and the midpoint of $CD$.\n\nProve that $|MK - ML| = |AB - CD|$.",
"options": [],
"answer": "See solution",
"solution": "Since $\\angle DMC = 90^\\circ$ (because $DC$ is a diameter) and $O_2O_3 \\perp MD$, $O_2O_4 \\perp MC$, it follows that $\\angle O_4O_2O_3 = 90^\\circ$. Similarly, one shows that $\\angle O_4O_1O_3 = 90^\\circ$ and thus the quadrilateral $O_4O_1O_2O_3$ is cyclic.\n\nLet $F$ and $E$ be the midpoints of chords $ML$ and $MK$, respectively. Then $O_3F \\perp ML$ and $O_4E \\perp MK$. Since $O_4O_1O_2O_3$ is cyclic with $\\angle O_4O_2O_3 = \\angle O_4O_1O_3 = 90^\\circ$, the midpoint $X$ of segment $O_3O_4$ is the center of the circumscribed circle of $O_4O_1O_2O_3$. Moreover, if $T$ denotes the foot of the perpendicular from $X$ onto $O_1O_2$, then $TO_1 = TO_2$.\n\nIn the right trapezoid $O_4EFO_3$, since $X$ is the midpoint of $O_3O_4$ and $XT \\perp EF$, it follows that $XT \\parallel EO_4 \\parallel FO_3$, so $XT$ is a midline and therefore $ET = TF$.\n\nConsequently, using the last two equalities we get\n\n$$\nTE - TO_1 = TF - TO_2 \\implies EO_1 = FO_2.\n$$\n\nThus, $|MK - ML| = |2ME - 2MF| = 2|ME - MF| = 2|EO_1 + O_1M - MO_2 - O_2F| = |2O_1M - 2O_2M|$.\n\nBut $AB = 2O_1M$, $CD = 2O_2M$, so substituting into the last equality we obtain $|MK - ML| = |AB - CD|$, which was to be proven.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20746,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. Find all possible values of $n$ for which the equation\n$$\nx^3 + y^3 + z^3 = n x^2 y^2 z^2\n$$\nhas positive integer solutions $(x, y, z)$.",
"options": [],
"answer": "See solution",
"solution": "We analyze possible values of $n$:\n\nWLOG, assume $x \\ge y \\ge z$. Then\n$$\n3x^3 \\ge x^3 + y^3 + z^3 = n x^2 y^2 z^2,\n$$\nwhich gives $x \\ge \\frac{n}{3} y^2 z^2$.\n\nRewrite as $y^3 + z^3 = x^2 (n y^2 z^2 - x)$. Since the left side is positive, $n y^2 z^2 - x \\ge 1$. Thus,\n$$\n2y^3 \\ge y^3 + z^3 \\ge \\left(\\frac{n}{3} y^2 z^2\\right)^2 (1) = \\frac{n^2}{9} y^4 z^4.\n$$\nSo $18 \\ge n^2 y z^4$. There is no positive integer solution when $n \\ge 5$.\n\nFor $n = 4$, $y = z = 1$. The equation becomes $x^3 + 2 = 4x^2$, so $x^2 \\mid 2$, which is impossible for positive integer $x$.\n\nFor $n = 3$, $x = y = z = 1$ is a solution.\n\nFor $n = 2$, $18 \\ge 4 y z^4$ implies $z = 1$ and $y \\le 4$. For $x^2 \\mid y^3 + 1$ and $x \\ge y$, possible $(x, y)$ are $(1, 1)$ and $(3, 2)$, but neither solves $x^3 + y^3 + 1 = 2x^2 y^2$.\n\nFor $n = 1$, $(x, y, z) = (3, 2, 1)$ is a solution.\n\nThus, $n$ can only be $1$ or $3$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20747,
"subject": "Mathematics (Olympiad)",
"question": "Does there exist a sequence $a_1, a_2, \\ldots, a_n, \\ldots$ of positive real numbers satisfying both of the following conditions:\n\n1. $\\sum_{i=1}^{n} a_i \\le n^2$ for every positive integer $n$;\n2. $\\sum_{i=1}^{n} \\frac{1}{a_i} \\le 2008$ for every positive integer $n$?",
"options": [],
"answer": "See solution",
"solution": "The answer is no. It is enough to show that if $\\sum_{i=1}^{n} a_i \\le n^2$ for any $n$, then $\\sum_{i=2}^{2n} \\frac{1}{a_i} > \\frac{n}{4}$ (or any other precise estimate).\n\nFor this, we use that $\\sum_{i=2^k+1}^{2^{k+1}} a_i \\le \\sum_{i=2^k+1}^{2^{k+1}} \\frac{1}{a_i} \\ge 2^{2k}$ for any $k \\ge 0$ by the arithmetic-harmonic mean inequality.\n\nSince $\\sum_{i=2^k+1}^{2^{k+1}} a_i < \\sum_{i=1}^{2^{k+1}} a_i \\le 2^{2k+2}$, it follows that $\\sum_{i=2^k+1}^{2^{k+1}} \\frac{1}{a_i} > \\frac{1}{4}$ and hence\n\n$$\n\\sum_{i=2}^{2n} \\frac{1}{a_i} > \\sum_{k=0}^{n-1} \\sum_{i=2^k+1}^{2^{k+1}} \\frac{1}{a_i} > \\frac{n}{4}.\n$$\n\n*Remark:* No points for using some inequality that doesn't lead to a solution.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20748,
"subject": "Mathematics (Olympiad)",
"question": "令 $\\Omega$ 為三角形 $ABC$ 的 $A$-旁切圓，並設其分別切直線 $BC, CA, AB$ 於點 $D, E, F$。設 $M$ 為線段 $EF$ 的中點。在 $\\Omega$ 上再取兩點 $P, Q$ 使得 $EP$ 與 $FQ$ 皆平行於 $DM$。令 $BP$ 與 $CQ$ 交於點 $X$。證明：直線 $AM$ 是 $\\angle XAD$ 的內角平分線。\n\n註：$A$-旁切圓指的是在 $\\angle A$ 內部的旁切圓。",
"options": [],
"answer": "See solution",
"solution": "延長 $DM$ 交 $\\Omega$ 另一點於 $R$。設 $AR$ 分別交 $EF$ 與 $\\Omega$ 於 $U, V \\neq R$，那麼由 $FREV$ 為調和四邊形知\n\n$$\n\\frac{\\sin \\angle BAR}{\\sin \\angle RAC} = \\frac{\\sin \\angle FAU}{\\sin \\angle UAE} = \\frac{\\overline{FU}}{\\overline{UE}} = \\frac{\\overline{FR} \\cdot \\overline{FV}}{\\overline{RE} \\cdot \\overline{VE}} = \\left(\\frac{\\overline{FR}}{\\overline{RE}}\\right)^2 = \\left(\\frac{\\overline{FM}}{\\overline{ME}} \\cdot \\frac{\\overline{DE}}{\\overline{FD}}\\right)^2\n$$\n\n延長 $PE, QF$ 分別交 $FD, DE$ 於 $Y, Z$，那麼同理有\n\n$$\n\\begin{align*}\n\\frac{\\sin \\angle CBP}{\\sin \\angle PBA} &= -\\frac{\\sin \\angle DBP}{\\sin \\angle PBF} = -\\left(\\frac{\\overline{DP}}{\\overline{PF}}\\right)^2 = -\\left(\\frac{\\overline{DY}}{\\overline{YF}} \\cdot \\frac{\\overline{EF}}{\\overline{DE}}\\right)^2 \\\\\n\\frac{\\sin \\angle ACQ}{\\sin \\angle QCB} &= -\\frac{\\sin \\angle ECQ}{\\sin \\angle QCD} = -\\left(\\frac{\\overline{EQ}}{\\overline{QD}}\\right)^2 = -\\left(\\frac{\\overline{EZ}}{\\overline{ZD}} \\cdot \\frac{\\overline{FD}}{\\overline{EF}}\\right)^2\n\\end{align*}\n$$\n\n將三式相乘，並注意到 $EFZY$ 為平行四邊形，可得\n\n$$\n\\begin{align*}\n\\frac{\\sin \\angle BAR}{\\sin \\angle RAC} \\cdot \\frac{\\sin \\angle CBP}{\\sin \\angle PBA} \\cdot \\frac{\\sin \\angle ACQ}{\\sin \\angle QCB} &= \\left( \\frac{\\overline{FM}}{\\overline{ME}} \\cdot \\frac{\\overline{DE}}{\\overline{FD}} \\cdot \\frac{\\overline{DY}}{\\overline{YF}} \\cdot \\frac{\\overline{EF}}{\\overline{DE}} \\cdot \\frac{\\overline{EZ}}{\\overline{ZD}} \\cdot \\frac{\\overline{FD}}{\\overline{EF}} \\right)^2 \\\\\n&= \\left( \\frac{\\overline{FM}}{\\overline{ME}} \\cdot \\frac{\\overline{DY}}{\\overline{YF}} \\cdot \\frac{\\overline{EZ}}{\\overline{ZD}} \\right)^2 \\\\\n&= \\left( 1 \\cdot \\frac{1}{2} \\cdot 2 \\right)^2 = 1\n\\end{align*}\n$$\n\n因此由角元西瓦定理可得 $AR, BP, CR$ 共點，即 $A, X, R$ 共線。\n\n令 $J$ 為 $\\triangle ABC$ 的 $A$-旁心，因為 $A, M, J$ 共線，所以只需證明 $AJ$ 為 $\\angle RAD$ 的內角平分線。由\n\n$$\n\\overline{MA} \\cdot \\overline{MJ} = \\overline{ME} \\cdot \\overline{MF} = \\overline{MD} \\cdot \\overline{MR}\n$$\n\n知 $A, J, D, R$ 共圓。又 $\\overline{JD} = \\overline{JR}$，所以 $AJ$ 為 $\\angle RAD$ 的內角平分線，從而原命題成立。 $\\Box$\n\n\n\n註：難度約為 G2。本題的關鍵是作出 $R$ 並證明 $A, X, R$ 共線，證明共線的方法很多。事實上，有個更一般的結論：\n\n設 $\\triangle ABC$ 的三邊分別與圓錐曲線 $C$ 切於 $D, E, F$，$P$ 為任意一點，$DP, EP, FP$ 分別交 $C$ 另一點於 $X, Y, Z$，則 $AX, BY, CZ$ 共點。\n\n而本題為取 $C$ 為旁切圓，$P$ 為 $DM$ 上的無窮遠點之特例。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20749,
"subject": "Mathematics (Olympiad)",
"question": "For each real-coefficient polynomial $f(x) = a_0 + a_1x + \\dots + a_nx^n$, let\n$$\n\\Gamma(f(x)) = a_0^2 + a_1^2 + \\dots + a_n^2.\n$$\nGiven a polynomial $P(x) = (x+1)(x+2)\\dots(x+2020)$, prove that there exist at least $2^{2019}$ pairwise distinct polynomials $Q_k(x)$ with $1 \\le k \\le 2^{2019}$, each satisfying the following two conditions:\n\ni) $\\deg Q_k(x) = 2020$,\nii) $\\Gamma(Q_k(x)^n) = \\Gamma(P(x)^n)$ for all positive integers $n$.",
"options": [],
"answer": "See solution",
"solution": "First, consider the following lemma.\n\n*Lemma.* $\\Gamma(f(x))$ is equal to the constant term in the expansion of $f(x)f\\left(\\frac{1}{x}\\right)$.\n\n*Proof.* Indeed, $f(x) = a_nx^n + a_{n-1}x^{n-1} + \\dots + a_1x + a_0$, so\n$$\n\\text{constant term of } f(x)f\\left(\\frac{1}{x}\\right) = (a_nx^n + \\dots + a_0) \\left(\\frac{a_n}{x^n} + \\dots + a_0\\right)\n$$\nThe constant term is $a_n^2 + a_{n-1}^2 + \\cdots + a_1^2 + a_0^2$.\n\n\n\nFor every polynomial $f(x)$ and every positive integer $n$,\n$$\n\\Gamma((ax+b)^n f(x)) = \\text{constant term of } (ax+b)^n \\left(\\frac{a}{x}+b\\right)^n f(x) f\\left(\\frac{1}{x}\\right)\n$$\n$$\n= \\text{constant term of } (a+bx)^n \\left(a+\\frac{b}{x}\\right)^n f(x) f\\left(\\frac{1}{x}\\right)\n$$\n$$\n= \\Gamma((bx+a)^n f(x)).\n$$\nThus, in each binomial of $P(x) = (x+1)(x+2)\\cdots(x+2020)$, the exchange $x+k \\rightarrow kx+1$ for $2 \\le k \\le 2020$ does not change the value of $\\Gamma(P(x)^n)$. Since each binomial can be chosen to keep or change, there are $2^{2019}$ ways to modify the polynomial, and all these $2^{2019}$ polynomials satisfy the problem's conditions. $\\square$\n\n",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20750,
"subject": "Mathematics (Olympiad)",
"question": "In triangle $ABC$, $AB = AC$. Point $D$ is the midpoint of side $BC$. Point $E$ lies outside the triangle $ABC$ such that $CE \\perp AB$ and $BE = BD$. Let $M$ be the midpoint of segment $BE$. Point $F$ lies on the minor arc $\\widehat{AD}$ of the circumcircle of triangle $ABD$ such that $MF \\perp BE$. Prove that $ED \\perp FD$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Construct point $F_1$ such that $EF_1 = BF_1$ and ray $DF_1$ is perpendicular to line $ED$. It suffices to show that $F = F_1$ or $ABDF_1$ is cyclic; that is,\n\n$$\n\\angle BAD = \\angle BF_1 D. \\qquad \\textcircled{1}\n$$\n\nSet $\\angle BAD = \\angle CAD = x$. Because $EC \\perp AB$ and $AD \\perp BC$,\n$$ \\angle ECB = 90^\\circ - \\angle ABD = \\angle BAD = x. $$\n\nNote that $MD$ is a midline of triangle $BCE$. In particular, $MD \\parallel EC$ and\n\n$$\n\\angle MDB = \\angle ECD = x. \\qquad \\textcircled{2}\n$$\n\nIn isosceles triangle $EF_1M$, we may set $\\angle EF_1M = \\angle BF_1M = y$. Because $EM \\perp MF_1$ and $MD \\perp DF_1$,\n\n$$\n\\angle EMF_1 = \\angle EDF_1 = 90^\\circ,\n$$\n\nimplying that $EMDF_1$ is cyclic. Consequently, we have\n\n$$\n\\angle EDM = \\angle EF_1 M = y. \\qquad \\textcircled{3}\n$$\n\nCombining ② and ③, we obtain\n\n\n\n$$\n\\angle BDE = \\angle EDM + \\angle MDB = x + y.\n$$\n\nBecause $BE = BD$, we conclude that triangle $BED$ is isosceles with $\\angle MED = \\angle BED = \\angle BDE = x + y$. Because $EMDF_1$ is cyclic, we have $\\angle MF_1 D = \\angle MED = x + y$. It is then clear that\n\n$$\n\\angle BF_1 D = \\angle MF_1 D - \\angle MF_1 B = x = \\angle BAD,\n$$\n\nwhich is ①.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20751,
"subject": "Mathematics (Olympiad)",
"question": "In a group of $m$ girls and $n$ boys, any two of them either know each other or do not know each other. For any two boys and two girls, at least one boy and one girl do not know each other. Prove that the number of boy-girl pairs that know each other is at most $m + \\frac{n(n-1)}{2}$.",
"options": [],
"answer": "See solution",
"solution": "From the hypothesis, for any two boys, there is at most one girl that knows both of them. Let $x_i$ be the number of girls that know exactly $i$ boys, $1 \\leq i \\leq n$. So $\\sum_{i=1}^{n} x_i = m$.\n\nBy counting the number of the above two boys–one girl combinations, we have\n\n$$\n\\sum_{i \\geq 2} \\frac{i(i-1)}{2} x_i \\leq \\frac{n(n-1)}{2}.\n$$\n\nThe number of boy-girl pairs that know each other is then\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} i x_i &= m + \\sum_{i=2}^{n} (i-1)x_i \\\\\n&\\leq m + \\sum_{i=2}^{n} \\frac{i(i-1)}{2} x_i \\\\\n&\\leq m + \\frac{n(n-1)}{2}.\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 20752,
"subject": "Mathematics (Olympiad)",
"question": "a. Let $b_n = \\frac{2^{(k+1)n}k - 1}{2^{(k+1)n} - 1} = 2^{(k+1)n(k-1)} + \\dots + 2^{(k+1)n} + 1$ for $n \\ge 0$ and\n$$a_n = \\frac{b_n}{b_{n-1}} = \\frac{(2^{(k+1)n}k - 1)(2^{(k+1)n-1} - 1)}{(2^{(k+1)n} - 1)(2^{(k+1)n-1}k - 1)} \\quad \\text{for } n \\ge 1.$$ \n\nb. Show that $t(n(2^r - 1)) \\ge r$ for all positive integers $n$ and $r$.",
"options": [],
"answer": "See solution",
"solution": "* For $n = 1$, $t(n(2^r - 1)) = t(2^r - 1) = r$.\n\n* Let $n > 1$. If $n$ is even, then $t(n(2^r - 1)) = t\\left(\\frac{n}{2}(2^r - 1)\\right) \\ge r$ by the induction hypothesis. Assume that $n = 2j + 1$ where $j$ is a positive integer. Then\n\n$$\n\\begin{align*}\nt(n(2^r - 1)) &= t((2j + 1)(2^r - 1)) \\\\\n&= t((2j + 2)(2^r - 1) - 2^r + 1) \\\\\n&= t((2j + 2)(2^r - 1) - 2^r) + 1 \\\\\n&\\ge t((2j + 2)(2^r - 1)) - 1 + 1 \\\\\n&= t((j + 1)(2^r - 1)) \\\\\n&\\ge r\n\\end{align*}\n$$\n\nwhere we used the induction hypothesis and the fact that $t(i - 2^r) \\ge t(i) - 1$ for $i > 2^r$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20753,
"subject": "Mathematics (Olympiad)",
"question": "At a certain orphanage, every pair of orphans are either friends or enemies of each other. For every three of an orphan's friends, an even number of pairs of them are enemies. Prove that it's possible to assign each orphan two parents such that every pair of friends shares exactly one parent, but no pair of enemies does, and no three parents are in a love triangle (where each pair of them has a child).",
"options": [],
"answer": "See solution",
"solution": "Form a graph $G$ with the orphans as vertices and with an edge between two orphans if they are friends. Let $A_1, A_2, \\dots$ be the maximal cliques of $G$; that is, every pair of orphans in $A_i$ is friends, and every orphan outside $A_i$ is an enemy of some orphan in $A_i$. We now have a sequence of lemmas.\n\n**Lemma 1.** Any two maximal cliques intersect in at most one orphan.\n\n*Proof.* Suppose for the sake of contradiction that $A_i$ and $A_j$ had intersection of size at least 2. We may choose orphans $u \\in A_i$ and $v \\in A_j$ such that $u$ is not friends with $v$; in particular, this means that $u \\notin A_j$ and $v \\notin A_i$. Then, taking two orphans $w, w' \\in A_i \\cap A_j$, notice that $w$ is friends with $u$, $v$, and $w'$, and that $\\{u, w'\\}$ and $\\{v, w'\\}$ are pairs of friends, but $\\{u, v\\}$ is not, contradicting the given. This proves the lemma. $\\Box$\n\n**Lemma 2.** Every orphan $v$ is in either 1 or 2 of the $A_i$.\n\n*Proof.* No orphan is in none of the $A_i$, as any orphan forms a clique of size 1, which may be grown into a maximal clique. Suppose for the sake of contradiction that some orphan $v$ were in at least 3 maximal cliques. Without loss of generality, let these cliques include $A_1, A_2$, and $A_3$, and choose friends $v_1 \\in A_1, v_2 \\in A_2$, and $v_3 \\in A_3$ of $v$.\n\nFor $i \\neq j$, notice that $v_i \\notin A_j$, as $A_j$ intersects $A_i$ in at most one orphan, which is $v$. Thus, $v_1, v_2$, and $v_3$ are distinct. If, say, $v_1$ and $v_2$ are friends, then $\\{v, v_1, v_2\\}$ forms a clique, hence is contained in some maximal clique $A$; then $A$ intersects both $A_1$ and $A_2$ in at least two orphans, implying by Lemma 1 that $A_1 = A = A_2$, a contradiction. Thus, there are no friendships between $v_1, v_2$, and $v_3$, which contradicts the given because $v$ is friends with each of them. $\\Box$\n\nForm a graph $H'$ with vertices $P_A$ the maximal cliques of $G$ and with an edge between two vertices if the corresponding cliques intersect. From $H'$, form the graph $H$ by attaching a leaf $P_{A,v}$ to the maximal clique $A$ for each orphan $v$ for which $A$ is the unique maximal clique containing $v$. Assign a parent to each vertex of $H$ and assign orphans to parents in the following way.\n\n- If an orphan $v$ lies in only one maximal clique $A$, assign it to $P_A$ and $P_{A,v}$.\n- If an orphan $v$ lies in two maximal cliques $A$ and $A'$, assign it to $P_A$ and $P_{A'}$.\n\nThis construction covers all cases by Lemma 2 and assigns each pair of parents at most one orphan by Lemma 1. By construction, if an orphan is assigned to two parents, there is an edge between those parents in $H$. Further, two orphans share a parent if and only if they share a maximal clique, meaning that they are friends. Finally, if $H$ contained a love triangle, then we would have three maximal cliques $A, A', A''$ with non-empty pairwise intersections at $v, v', v''$. Then $\\{v, v', v''\\}$ is a clique, hence lies in a maximal clique $A^*$ which intersects each of $A, A', A''$ with cardinality at least 2. By Lemma 1, this implies that $A^* = A = A' = A''$, a contradiction. Thus, $H$ has no love triangles, completing the proof.\n\n**Remark.** We may recast the problem in more advanced language as follows. Recall that the *girth* of a graph $G$ is the length of the shortest cycle in $G$, and define the *line graph* $L(G)$ of a graph $G$ to be the graph whose vertices are the edges $e_i$ of $G$ and which has an edge between $e_i$ and $e_j$ if they are incident to a common vertex in $G$. In this language, the statement becomes:\n\nIf $G$ has no induced subgraph of the form $K_{1,3}$ or $K_4 - e$, then $G$ is the line graph of a graph of girth at least 4.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20754,
"subject": "Mathematics (Olympiad)",
"question": "Let $s_k(n) = 1^k + 2^k + \\dots + n^k$. A positive integer $k$ has the property $T(m)$ if $s_k(n)$ covers the complete residue system modulo $m$ as $n$ ranges over the positive integers.\n\n1. For $m = 20$, determine all positive integers $k$ with property $T(20)$.\n2. For $m = 20^{15}$, determine the minimum positive integer $k$ with property $T(20^{15})$.",
"options": [],
"answer": "See solution",
"solution": "a) If $k > 1$ has property $T(20)$, then so does $k + 4$, since $n^{k+4} - n^k$ is divisible by $20$ for all $n$, $k > 1$. Thus, it suffices to check $k = 1, 2, 3, 4, 5$. By direct computation, only $k = 4$ has property $T(20)$; $k = 1, 2, 3, 5$ do not. Therefore, $k$ has property $T(20)$ if and only if $k$ is divisible by $4$.\n\nb) From part (a), $k = 1, 2, 3$ do not have property $T(20^{15})$. We show $k = 4$ does. It suffices to show $s_4(n)$ covers all residues modulo $20^{15}$ as $n$ varies. Let $S(n) = 30 s_4(n)$. We need to show that for any integer $a$, there exists $n$ such that $s_4(n) \\equiv a \\pmod{20^{15}}$, or equivalently, $S(n) \\equiv 30a \\pmod{3 \\times 2^{31} \\times 5^{16}}$.\n\nBy the Chinese Remainder Theorem, it suffices to solve:\n\n$$\n\\begin{align*}\nS(n) &\\equiv 30a \\pmod{3} \\\\\nS(n) &\\equiv 30a \\pmod{2^{31}} \\\\\nS(n) &\\equiv 30a \\pmod{5^{16}}\n\\end{align*}\n$$\n\nFor the first, $S(0) \\equiv 30a \\pmod{3}$.\n\nFor the second, we use induction on $r$ to show $S(n) \\equiv 30a \\pmod{2^r}$ is solvable for all $r$. For $r=1$, $n=0$ works. Suppose true for $r$, so $S(n_0) \\equiv 30a \\pmod{2^r}$. For $n = n_0 + 2^r q$,\n\n$$\nS(n) \\equiv S(n_0) - 2^r q \\pmod{2^{r+1}}.\n$$\n\nChoosing $q$ appropriately, $S(n) \\equiv 30a \\pmod{2^{r+1}}$. Thus, by induction, the result holds for all $r$.\n\nThe third congruence is similar. Therefore, the minimum $k$ with property $T(20^{15})$ is $k = 4$.",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 20755,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, \\dots, a_n$ be real numbers, and let $d = \\max\\{a_j : 1 \\leq j \\leq g\\} - \\min\\{a_j : g \\leq j \\leq n\\}$ for some $1 \\leq g \\leq n$.\n\n1. Prove that for any real numbers $x_1 \\leq x_2 \\leq \\dots \\leq x_n$,\n\n$$\n\\max\\{|x_i - a_i| : 1 \\leq i \\leq n\\} \\geq \\frac{d}{2}.\n$$\n\n2. Show that there exist real numbers $x_1 \\leq x_2 \\leq \\dots \\leq x_n$ such that equality holds in the above inequality.",
"options": [],
"answer": "See solution",
"solution": "(1) Define\n\n$$\nd = a_p - a_r,\n$$\nwhere $a_p = \\max\\{a_j : 1 \\leq j \\leq g\\}$ and $a_r = \\min\\{a_j : g \\leq j \\leq n\\}$ for some $1 \\leq p \\leq g \\leq r \\leq n$.\n\nFor any real numbers $x_1 \\leq x_2 \\leq \\dots \\leq x_n$,\n\n$$\n(a_p - x_p) + (x_r - a_r) = (a_p - a_r) + (x_r - x_p) \\geq a_p - a_r = d.\n$$\n\nThus,\n\n$$\na_p - x_p \\geq \\frac{d}{2} \\quad \\text{or} \\quad x_r - a_r \\geq \\frac{d}{2}.\n$$\n\nTherefore,\n\n$$\n\\max\\{|x_i - a_i| : 1 \\leq i \\leq n\\} \\geq \\frac{d}{2}.\n$$\n\n(2) Define the sequence $\\{x_k\\}$ as follows:\n\n- $x_1 = a_1 - \\frac{d}{2}$\n- $x_k = \\max\\{x_{k-1}, a_k - \\frac{d}{2}\\}$ for $2 \\leq k \\leq n$\n\nThis sequence is non-decreasing, and $x_k - a_k \\geq -\\frac{d}{2}$ for all $k$.\n\nFor all $k$, let $l \\leq k$ be the smallest integer such that $x_k = x_l$. Then $x_k = a_l - \\frac{d}{2}$.\n\nSince $a_l - a_k \\leq d$, we have\n\n$$\nx_k - a_k = a_l - a_k - \\frac{d}{2} \\leq d - \\frac{d}{2} = \\frac{d}{2}.\n$$\n\nThus,\n\n$$\n-\\frac{d}{2} \\leq x_k - a_k \\leq \\frac{d}{2}\n$$\nfor all $1 \\leq k \\leq n$, so\n\n$$\n\\max\\{|x_i - a_i| : 1 \\leq i \\leq n\\} \\leq \\frac{d}{2}.\n$$\n\nBy part (1), equality holds for this sequence.\n\nAlternatively, for each $i$, define $M_i = \\max\\{a_j : 1 \\leq j \\leq i\\}$ and $m_i = \\min\\{a_j : i \\leq j \\leq n\\}$. Set $x_i = \\frac{M_i + m_i}{2}$ and $d_i = M_i - m_i$. Then\n\n$$\n-\\frac{d}{2} \\leq x_i - a_i \\leq \\frac{d_i}{2} \\leq \\frac{d}{2}.\n$$\n\nThus,\n\n$$\n\\max\\{|x_k - a_k| : 1 \\leq k \\leq n\\} \\leq \\frac{d}{2},\n$$\nso equality is achieved.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20756,
"subject": "Mathematics (Olympiad)",
"question": "Prove that there exists a positive integer $n$ such that the decimal representation of\n$$\n\\sum_{k=1}^{\\lfloor \\frac{n}{3} \\rfloor} \\binom{n}{3k} 8^k\n$$\nends in 2023 digits 8.",
"options": [],
"answer": "See solution",
"solution": "Let $f(n) = \\sum_{k=1}^{\\lfloor \\frac{n}{3} \\rfloor} \\binom{n}{3k} 8^k$ and let $\\omega \\neq 1$ be a third root of unity. Using the fact that for every integer $k \\geq 0$:\n$$\n1 + \\omega^k + \\omega^{2k} = \\begin{cases} 3, & \\text{if } 3 \\mid k \\\\ 0, & \\text{otherwise} \\end{cases}\n$$\nwe get:\n$$\n\\begin{aligned}\nf(n) + 1 &= \\sum_{k=0}^{\\lfloor \\frac{n}{3} \\rfloor} \\binom{n}{3k} 8^k \\\\\n&= \\frac{1}{3} \\sum_{k=0}^{n} (1 + \\omega^k + \\omega^{2k}) \\binom{n}{k} 8^k \\\\\n&= \\frac{1}{3} 9^n + \\frac{1}{3} (1 + 8\\omega)^n + \\frac{1}{3} (1 + 8\\omega^2)^n\n\\end{aligned}\n$$\nNow, $9$, $1+8\\omega$, and $1+8\\omega^2$ are roots of the polynomial:\n$$\nP(x) = (x-9)(x-1-8\\omega)(x-1-8\\omega^2) = (x-1)^3 - 512 = x^3 - 3x^2 + 3x - 513\n$$\nwhich is the characteristic polynomial of the recursive sequence $(a_i)_{i \\geq 0}$:\n$$\na_{i+3} = 3a_{i+2} - 3a_{i+1} + 513a_i \\text{ for } i \\geq 0.\n$$\nSet $a_i = f(i) + 1$ for $0 \\leq i \\leq 2$. Then $f(n) + 1 = a_n$ for every $n \\geq 0$. Let $b_i = a_i \\pmod{10^{2023}}$. Since $\\gcd(3, 10^{2023}) = 1$, any three consecutive terms of $(b_i)$ uniquely determine the next and previous terms. Since there are finitely many residues modulo $10^{2023}$, $(b_i)$ is periodic with some period $d > 3$. Therefore:\n$$\n9(f(d-1)+1) = 9a_{d-1} = a_{d+2} - 3a_{d+1} + 3a_d \\equiv a_2 - 3a_1 + 3a_0 \\pmod{10^{2023}} = 1 \\pmod{10^{2023}}\n$$\nSince $9 \\mid 8 \\cdot 10^{2023} + 1$, we have $9^{\\frac{8 \\cdot 10^{2023} + 1}{9}} \\equiv 1 \\pmod{10^{2023}}$, so:\n$$\nf(d-1) + 1 = a_{d-1} \\equiv \\frac{8 \\cdot 10^{2023} + 1}{9} = \\underbrace{88\\ldots89}_{2022\\text{ digits}} \\pmod{10^{2023}}\n$$\nThus $f(d-1) \\equiv \\underbrace{88\\ldots8}_{2022\\text{ digits}} \\pmod{10^{2023}}$. Therefore, $n = d-1$ has the desired property. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20757,
"subject": "Mathematics (Olympiad)",
"question": "Докажите, что не существует такого $a > 0$, для которого для любого $x > 0$ выполняется неравенство\n$$\n|\\cos x| + |\\cos a x| > \\sin x + \\sin a x.\n$$",
"options": [],
"answer": "See solution",
"solution": "**Первое решение.** Предположим, что $0 < a \\le 1$. Тогда при $x = \\pi/2$ левая часть примет значение $|\\cos(a \\pi/2)|$, то есть будет не больше 1, в то время как правая часть будет равна $1 + \\sin(a \\pi/2)$, то есть она больше 1. Итак, неравенство не выполнено.\n\nЕсли же $a > 1$, то, обозначив $a x = t$ и $b = 1/a$, мы приведём неравенство из условия к виду\n$$\n|\\cos b t| + |\\cos t| > \\sin b t + \\sin t,\n$$\nсводя задачу к предыдущему случаю.\n\n**Второе решение.** Выберем такое $x$, что $x(a+1) = \\pi/2$. Тогда $x$ и $a x$ лежат в интервале $(0, \\pi/2)$, поэтому $|\\cos x| = \\cos x = \\sin a x$ и $|\\cos a x| = \\cos a x = \\sin x$. Значит, для выбранного $x$ имеем\n$$\n|\\cos x| + |\\cos a x| = \\sin a x + \\sin x.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20758,
"subject": "Mathematics (Olympiad)",
"question": "Given four numbers $x_1, x_2, x_3, x_4$ in $\\left[0, \\frac{\\pi}{2}\\right]$, prove that there exist two, say $x$ and $y$, such that\n$$\n8 \\cos x \\cos y (\\cos x \\cos y + \\sin x \\sin y) + 1 > 4 \\cos^2 x + 4 \\cos^2 y.\n$$",
"options": [],
"answer": "See solution",
"solution": "The given inequality is successively equivalent to\n$$\n8 \\cos x \\cos y (\\cos x \\cos y + \\sin x \\sin y) + 1 > 4 \\cos^2 x + 4 \\cos^2 y,\n$$\n$$\n8 \\cos^2 x \\cos^2 y - 4 \\cos^2 x - 4 \\cos^2 y + 2 \\sin 2x \\sin 2y + 1 > 0,\n$$\n$$\n2 (2 \\cos^2 x - 1) (2 \\cos^2 y - 1) + 2 \\sin 2x \\sin 2y - 1 > 0,\n$$\n$$\n\\cos 2x \\cos 2y + \\sin 2x \\sin 2y > \\frac{1}{2},\n$$\n$$\n\\cos (2x - 2y) > \\frac{1}{2}.\n$$\n\nBy the pigeonhole principle, in one of the sets\n\n$$\n\\langle 0, \\frac{\\pi}{6} \\rangle, \\quad \\left[ \\frac{\\pi}{6}, \\frac{\\pi}{3} \\right), \\quad \\left[ \\frac{\\pi}{3}, \\frac{\\pi}{2} \\right]\n$$\n\nthere are two out of four given numbers. Let these be $x$ and $y$.\n\nNow we have $|2x - 2y| < \\frac{\\pi}{3}$ and $\\cos(2x - 2y) > \\frac{1}{2}$, which proves the claim.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20759,
"subject": "Mathematics (Olympiad)",
"question": "We are to paint $n$ seats in a row, each either red or green. A painting is called *odd* if every monochromatic sequence has odd length. A *monochromatic sequence* is a sequence of seats of the same color, bounded by seats of the other color or by a wall. How many odd paintings are there?",
"options": [],
"answer": "See solution",
"solution": "Let $g_k$ and $r_k$ be the numbers of possible odd paintings of $k$ seats where the first seat is painted green or red, respectively. Clearly, $g_k = r_k$ for any $k$. Note that\n$$\ng_k = r_{k-1} + g_{k-2} = g_{k-1} + g_{k-2}\n$$\nsince $r_{k-1}$ counts odd paintings with the first seat green and the second red, and $g_{k-2}$ counts those with the first two seats green. Also, $g_1 = g_2 = 1$, so $g_k$ is the $k$-th Fibonacci number. Therefore, the total number of odd paintings is $g_n + r_n = 2f_n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20760,
"subject": "Mathematics (Olympiad)",
"question": "A graph $G$ is called a *divisibility graph* if the vertices can be assigned distinct positive integers such that between two vertices assigned $u, v$ there is an edge if and only if $\\frac{u}{v}$ or $\\frac{v}{u}$ is a positive integer.\n\nShow that for any positive integer $n$ and $0 \\leq e \\leq \\frac{n(n-1)}{2}$, there is a divisibility graph with $n$ vertices and $e$ edges.",
"options": [],
"answer": "See solution",
"solution": "We reason inductively on $n$, not assigning the number $1$ to any vertex.\n\nFor $n=1$, the claim is clear. For $n=2$, an example with $e=1$ is $(2, 4)$, and with $e=0$ is $(2, 3)$. For $n=3$, an example with $e=0$ is $(3, 5, 7)$; with $e=1$ is $(2, 4, 7)$; with $e=2$ is $(2, 4, 10)$; and with $e=3$ is $(2, 4, 8)$.\n\nFor $n \\geq 4$, we have $n-1 \\leq \\frac{(n-1)(n-2)}{2}$, so at least one of $e \\geq n-1$ and $e \\leq \\frac{(n-1)(n-2)}{2}$ is satisfied.\n\nLet $e \\geq n-1$ first. Given an example with $n-1$ vertices and $e - (n-1)$ edges, we add a vertex (of degree $n-1$) by assigning it a prime number $p$ greater than the numbers in the other vertices, then multiply the numbers in the remaining vertices by $p$—this does not create new edges between the remaining vertices.\n\nNow let $e \\leq \\frac{(n-1)(n-2)}{2}$. Given an example with $n-1$ vertices and $e$ edges, we add a vertex (of degree $0$) by assigning it a prime number $p$ greater than the numbers in the other vertices, and do not change the numbers in the other vertices—this does not create new edges. $\\square$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20761,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ be a positive integer such that $np + 1 = a^2$, where $p$ is a prime and $n \\geq p - 1$. Note that $a > 1$.\n\nShow that $n + 1$ can be written as $x^2 + (p-1)y^2$ for some positive integers $x$ and $y$.",
"options": [],
"answer": "See solution",
"solution": "We have $a^2 - 1 = np$, i.e., $(a-1)(a+1) = np$.\n\nSince $p$ is prime, $p$ divides either $a-1$ or $a+1$. We consider both cases:\n\n**Case 1:** $p \\mid a-1$. Let $a = kp + 1$ for some integer $k > 0$ (since $a > 1$).\n\nThen:\n$$\n1 + np = (kp + 1)^2 = k^2p^2 + 2kp + 1\n$$\nSo,\n$$\nn = k^2p + 2k\n$$\nThus,\n$$\nn + 1 = k^2p + 2k + 1 = k^2 + 2k + 1 + (p-1)k^2 = (k+1)^2 + (p-1)k^2\n$$\nwith $k+1$ and $k$ both positive integers.\n\n**Case 2:** $p \\mid a+1$. Let $a = kp - 1$ for some integer $k \\geq 2$ (since $a > 1$).\n\nAnalogously,\n$$\nn + 1 = (k-1)^2 + (p-1)k^2\n$$\nwith $k-1$ and $k$ both positive integers.\n\nIf $k = 1$, then $n + 1 = p - 1$, so $n = p - 2$, which contradicts $n \\geq p - 1$.\n\nTherefore, in both cases, $n + 1$ can be written as $x^2 + (p-1)y^2$ for positive integers $x$ and $y$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20762,
"subject": "Mathematics (Olympiad)",
"question": "Consider a tetrahedron $ABCD$ with $\\widehat{BAC} + \\widehat{CAD} + \\widehat{DAB} = 180^\\circ$ and $\\widehat{ABC} \\equiv \\widehat{DAB}$. If the projection of the vertex $D$ on the plane $(ABC)$ is the orthocenter of the triangle $ABC$, prove that $AB = AC$ and $DB = DC$.",
"options": [],
"answer": "See solution",
"solution": "Let $BE$ and $CF$ be the altitudes from $B$ and $C$, respectively, of triangle $ABC$, and denote by $H$ their intersection point. The three perpendiculars theorem implies that $DE \\perp AC$ and $DF \\perp AB$.\n\nUnfold the tetrahedron onto the plane $(ABC)$ and denote by $D_1$ the image of vertex $D$ in triangle $DAB$ and by $D_2$ the image of vertex $D$ in triangle $DAC$ after unfolding.\n\n\n\nThe perpendicularities $DE \\perp AC$ and $DF \\perp AB$ are preserved in the unfolding. Therefore, $D_1$, $F$, and $C$ are collinear. Similarly, $D_2$, $E$, and $B$ are collinear.\n\nSince $\\widehat{BAC} + \\widehat{CAD} + \\widehat{DAB} = 180^\\circ$, the points $D_1$, $A$, and $D_2$ are collinear; clearly, $A$ is the midpoint of segment $D_1D_2$.\n\n$\\widehat{ABC} \\equiv \\widehat{DAB}$, so the lines $D_1D_2$ are parallel to $BC$, making $D_1D_2CB$ a trapezium whose diagonals meet at $H$. The midpoint of the larger base lies on line $AH$, so $AH$ also contains the midpoint of the smaller base $BC$. Since $AH$ is both an altitude and a median of triangle $ABC$, it follows that triangle $ABC$ is isosceles, with $AB = AC$.\n\nThe line $AH$ is the common perpendicular bisector of the bases of trapezium $D_1D_2CB$, so $D_1D_2CB$ is an isosceles trapezium, hence $D_1B = D_2C$ and thus $DB = DC$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20763,
"subject": "Mathematics (Olympiad)",
"question": "a) Let $a$ and $b$ be arbitrary positive integers of equal parity. Can we always find noninteger numbers $x$ and $y$ such that $x + y$ and $ax + by$ are integers?\n\nb) The same question when $a$ and $b$ have different parities.",
"options": [],
"answer": "See solution",
"solution": "a) *Yes.*\n\nFor example, take $x = y = \\frac{1}{2}$. Then $x + y = 1$ is an integer, and $ax + by = \\frac{1}{2}(a + b)$ is also an integer since $a + b$ is even.\n\nAlternatively, if $a = b$, any noninteger $x$ and $y$ with integer sum work, since $ax + ay = a(x + y)$ is an integer. If $a \\neq b$, take $x = \\frac{1}{a-b}$ and $y = \\frac{a-b-1}{a-b}$, so $x + y = 1$ and\n$$\nax + by = \\frac{a}{a-b} + \\frac{b(a-b-1)}{a-b} = b+1.\n$$\n\nb) *No.*\n\nSuppose $x + y$ and $ax + by$ are integers. Then $ax + by = a(x + y) + (b - a)y$. Since $a(x + y)$ is integer, $(b - a)y$ must also be integer. If $b - a = 1$, this forces $y$ to be integer, contradicting the requirement for noninteger $y$. Thus, it is not always possible when $a$ and $b$ have different parities.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20764,
"subject": "Mathematics (Olympiad)",
"question": "Let $A_1, A_2, \\dots, A_{160}$ be sets such that $|A_i| = i$ for $i = 1, 2, \\dots, 160$. Using the elements of these sets, we construct new sets $M_1, M_2, \\dots, M_n$ with the following procedure:\n\nAt each step, we choose some of the sets $A_1, A_2, \\dots, A_{160}$ and subtract from each of them the same number of elements. All these elements form the set $M_1$ at the first step, $M_2$ at the second step, and so on. We continue this process until all sets $A_1, A_2, \\dots, A_{160}$ become empty. Find the minimal value of $n$.",
"options": [],
"answer": "See solution",
"solution": "Suppose that at each step we select $k_j$ elements from some sets, for $j = 1, 2, \\dots, n$. After all sets $A_1, A_2, \\dots, A_{160}$ are emptied, each $i = |A_i|$ must be the sum of some of the numbers $k_1, k_2, \\dots, k_n$.\n\nThe number of possible sums formed by subsets of $\\{k_1, k_2, \\dots, k_n\\}$ is $2^n$. Since we need to represent all integers from $1$ to $160$, we require $2^n \\geq 160$, so $n \\geq 8$. Thus, the minimal possible value of $n$ is $8$.\n\nTo show that $n = 8$ is achievable, consider the following procedure:\n\n- At the first step, select sets $A_{81}, \\dots, A_{160}$ and subtract $80$ elements from each. The set $M_1$ consists of $80 \\times 80 = 6400$ elements.\n- After this, the remaining sets $A_{81}^1, \\dots, A_{160}^1$ each have $i$ elements for $i = 1, 2, \\dots, 80$.\n- At the second step, select sets $A_{41}, \\dots, A_{80}$ and $A_{121}^1, \\dots, A_{160}^1$, and subtract $40$ elements from each. The set $M_2$ consists of $80 \\times 40 = 3200$ elements.\n- Continue similarly, halving the number of elements subtracted at each step.\n\nThus, the process can be completed in $8$ steps.\n\n**Comment:** The choice of sets at each step is not unique; other valid selections are possible.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20765,
"subject": "Mathematics (Olympiad)",
"question": "In an acute triangle $ABC$, let $H$ be the orthocenter and $A_0$, $B_0$, $C_0$ be the midpoints of the sides $BC$, $CA$, and $AB$, respectively. Consider three circles passing through $H$: $\\omega_a$ centered at $A_0$, $\\omega_b$ centered at $B_0$, and $\\omega_c$ centered at $C_0$. The circle $\\omega_a$ intersects the line $BC$ at $A_1$ and $A_2$; $\\omega_b$ intersects $CA$ at $B_1$ and $B_2$; $\\omega_c$ intersects $AB$ at $C_1$ and $C_2$. Show that the points $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ lie on a circle.",
"options": [],
"answer": "See solution",
"solution": "Remark: The problem can be solved in many ways without much difficulty using calculation, trigonometry, coordinate geometry, or vectors. As an example, we present a short proof using complex numbers in our third solution.\n\nThe perpendicular bisectors of the segments $A_1A_2$, $B_1B_2$, $C_1C_2$ are also the perpendicular bisectors of segments $BC$, $CA$, $AB$. So they meet at $O$, the circumcenter of $ABC$. Thus $O$ is the only point that can possibly be the center of the desired circle.\n\n**Solution 1.** We are going to show that the circumcenter $O$ is equidistant from the six points in question.\n\nLet $A'$ be the second intersection point of $\\omega_b$ and $\\omega_c$. The line $B_0C_0$, which is the line of centers of circles $\\omega_b$ and $\\omega_c$, is a midline in triangle $ABC$, parallel to $BC$ and perpendicular to the altitude $AH$. The points $A'$ and $H$ are symmetric with respect to the line of centers. Therefore $A'$ lies on the line $AH$.\n\nFrom the two circles $\\omega_b$ and $\\omega_c$ we obtain $AC_1 \\cdot AC_2 = AA' \\cdot AH = AB_1 \\cdot AB_2$. So the quadrilateral $B_1B_2C_1C_2$ is cyclic. The perpendicular bisectors of the sides $B_1B_2$ and $C_1C_2$ meet at $O$. Hence $O$ is the circumcenter of $B_1B_2C_1C_2$ and so $OB_1 = OB_2 = OC_1 = OC_2$.\n\nAnalogous arguments yield $OA_1 = OA_2 = OB_1 = OB_2$ and $OA_1 = OA_2 = OC_1 = OC_2$. Thus $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ lie on a circle centered at $O$.\n\n\n\n**Solution 2.** We show again $B_1$, $B_2$, $C_1$, $C_2$ are cyclic. Note that\n\n$$\nAC_1 \\cdot AC_2 = (AC_0 + C_0H)(AC_0 - C_0H) = AC_0^2 - C_0H^2\n$$\n\nand, likewise,\n\n$$\nAB_1 \\cdot AB_2 = AB_0^2 - B_0H^2.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20766,
"subject": "Mathematics (Olympiad)",
"question": "Let $k$ be a positive integer. Find all positive integers $k$ such that there exist positive integers $a$ and $b$ with exactly $k$ positive divisors each, and $2a + 3b$ also has exactly $k$ positive divisors.",
"options": [],
"answer": "See solution",
"solution": "For $i \\ge 0$, let $a = 2 \\cdot 5^i$ and $b = 3 \\cdot 5^i$. Then both $a$ and $b$ have $2(i+1)$ divisors. Moreover, $2a + 3b = 4 \\cdot 5^i + 9 \\cdot 5^i = 13 \\cdot 5^i$, which also has $2(i+1)$ divisors. Therefore, all even values of $k$ satisfy the condition.\n\nNow suppose $k$ is odd. Then $a$ has an odd number of divisors and therefore is a square, say $a = x^2$. Similarly, $b = y^2$, and $2a + 3b = z^2$. Thus,\n\n$$\n2x^2 + 3y^2 = z^2.\n$$\n\nWe show this equation has no positive integer solutions. Suppose, for contradiction, that it does. Let $(x, y, z)$ be a solution with minimal $x + y + z$. Then $2x^2 + 3y^2 = z^2$. Modulo $3$, $2x^2 \\equiv z^2$. If $x$ is not divisible by $3$, $x^2 \\equiv 1 \\pmod{3}$, so $z^2 \\equiv 2 \\pmod{3}$, which is impossible. Thus, $x$ is divisible by $3$, so $z$ is as well. Then $2x^2$ and $z^2$ are divisible by $9$, so $3y^2$ is divisible by $9$, so $y$ is divisible by $3$. But then $(x/3, y/3, z/3)$ is a smaller solution, contradicting minimality. Therefore, no positive integer solutions exist.\n\nThus, no odd $k$ satisfies the condition. The positive integers $k$ that satisfy the condition are the even integers.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20767,
"subject": "Mathematics (Olympiad)",
"question": "Let $P$ and $Q$ be two points inside parallelogram $ABCD$, symmetric with respect to the intersection point of the diagonals. Prove that the circumcircles of triangles $ABP$, $CDP$, $BCQ$, and $ADQ$ have a common point.",
"options": [],
"answer": "See solution",
"solution": "Denote by $X$ the second point of intersection of the circumcircles of $\\triangle ADQ$ and $\\triangle BCQ$. Let $QE$ be a ray in the same direction as ray $CB$.\n\nThen $\\angle EQB = \\angle QBC$, $\\angle EQA = \\angle QAD$, so $\\angle AQB = \\angle QAD + \\angle QBC = 180^\\circ - \\angle QXD + 180^\\circ - \\angle QXC = \\angle CXD$.\n\nMoreover, $\\angle AQB = \\angle CPD$, because triangles $AQB$ and $CPD$ are symmetric with respect to the center of the parallelogram.\n\nHence, points $D$, $P$, $X$, $C$ are cyclic. By analogy, $A$, $P$, $X$, $B$ also lie on the same circle, so all four circles pass through the common point $X$.\n\nIt is worth noting that this proof is valid for the specific configuration of the given points only. For a complete solution, one has to consider at least three more configurations. However, it is possible to handle all these cases simultaneously using oriented angles.\n\nIndeed:\n\n$$\n\\angle(QB; QA) = \\angle(QB; BC) + \\angle(BC; QA) = \\angle(QB; BC) + \\angle(AD; QA) = \\angle(QX; XC) + \\angle(DX; XQ) = \\angle(DX; XC)\n$$\n\nThe rest coincides with what we have done before, because $\\angle(QB; QA) = \\angle(PD; PC)$ (since $AQB$ and $CPD$ are symmetric with respect to the center of the parallelogram). Hence, $D$, $P$, $X$, $C$ are cyclic as well as $A$, $P$, $X$, $B$. Therefore, all circles pass through $X$, and we are done.\n\n\n\nFig.12",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20768,
"subject": "Mathematics (Olympiad)",
"question": "Define\n$$\n\\begin{align*}\nf_n(x_1, \\dots, x_n) &= (n-2) \\sum_{i=1}^{n} x_i + n \\left( \\prod_{i=1}^{n} x_i \\right)^{\\frac{1}{n}} - 2 \\sum_{i a_1$, $x_2 = \\cdots = x_n$, $x_1 = 0$, we have\n\n$$\ng_n(x_1, \\dots, x_n) = \\frac{1}{n a_n} h_n(x_1, \\dots, x_n).\n$$\n\nProof of (i). By induction.\n\nWhen $n = 2$, $f_n(x_1, \\dots, x_n) = 0$. Suppose for $n > 2$, $f_n(x_1, \\dots, x_n) \\ge 0$. Now for $n + 1$, without loss of generality, assume $x_1 \\le x_2 \\le \\dots \\le x_n \\le x_{n+1}$.\n\nLet $x_1' = x_1$, $x_i' = (x_2 \\cdots x_{n+1})^{\\frac{1}{n}} = a$, for $i = 2, \\dots, n+1$. We have\n\n$$\nf_{n+1}(x') = (n-1)x_1 + (n+1)x_1^{\\frac{1}{n+1}} a^{\\frac{n}{n+1}} - 2n(x_1 a)^{\\frac{1}{2}}.\n$$\n\nWhen $x_1 = 0$, $f_n(x_1, \\dots, x_n) = 0$.\n\nWhen $x_1 > 0$,\n$$\n\\frac{f_{n+1}(x')}{2n x_1^{\\frac{1}{1+n}}} = \\frac{n-1}{2n} x_1 + \\frac{n+1}{2n} a^{\\frac{n}{n+1}} - x_1^{\\frac{n-1}{2(n+1)}} a^{\\frac{1}{2}} \\ge 0,\n$$\nand $f_n = 0 \\Leftrightarrow x_1 = 0$. We find that\n\n$$\nf_{n+1}(x) - f_{n+1}(x') = (n-1) \\sum_{i=2}^{n+1} x_i - 2x_1^{\\frac{1}{2}} \\left( \\sum_{i=2}^{n+1} \\sqrt{x_i} - n\\sqrt{a} \\right) - 2 \\sum_{2 \\le i < j \\le n} \\sqrt{x_i x_j} = f_n(x_2, \\dots, x_{n+1}) + \\sum_{i=2}^{n+1} x_i - 2x_1^{\\frac{1}{2}} \\left( \\sum_{i=2}^{n+1} \\sqrt{x_i} - n\\sqrt{a} \\right) - n \\prod_{i=2}^{n+1} x_i^{\\frac{1}{n}}.\n$$\n\nSince $x_1 \\le a$, $a = (x_2 \\cdots x_{n+1})^{\\frac{1}{n}}$, and $f_n(x_2, \\dots, x_{n+1}) \\ge 0$, it follows that\n\n$$\nf_{n+1}(x) - f_{n+1}(x') \\ge \\sum_{i=2}^{n+1} x_i - 2\\sqrt{a} \\left( \\sum_{i=2}^{n+1} \\sqrt{x_i} \\right) + n a = \\sum_{i=2}^{n+1} (\\sqrt{x_i} - \\sqrt{a})^2 \\ge 0\n$$\n\nand thereby $f_{n+1}(x) \\ge f_{n+1}(x') \\ge 0$, yielding the desired inequality. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20769,
"subject": "Mathematics (Olympiad)",
"question": "In a trapezoid $ABCD$ with bases $AD$ and $BC$, a point $F$ is chosen on the side $CD$. Let $E$ be the point of intersection of the lines $AF$ and $BD$. A point $G$ is chosen on the side $AB$ so that $EG \\perp AD$. Let $H$ be the point of intersection of the lines $CG$ and $BD$, and let $I$ be the point of intersection of the lines $FH$ and $AB$. Prove that the lines $CI$, $FG$, and $AD$ are concurrent.\n\n",
"options": [],
"answer": "See solution",
"solution": "Without loss of generality, assume that $BC < AD$. Let $S$ be the point of intersection of the lines $AB$ and $CD$, and $T$ the point of intersection of the lines $AF$ and $DG$ (see the figure). First, we prove that the points $S$, $H$, and $T$ are collinear. To this end, we use Menelaus' theorem for triangle $ABE$ and the three points $S$, $H$, $T$ that lie on the lines containing its sides: the points $S$, $H$, $T$ will be collinear if and only if\n\n$$\n\\frac{AT}{TE} \\cdot \\frac{EH}{HB} \\cdot \\frac{BS}{SA} = 1.\n$$\n\nBecause $EG \\parallel AD$, $GE \\parallel BC$, and $AD \\parallel BC$, we have that $\\triangle ATD \\sim \\triangle ETG$, $\\triangle GHE \\sim \\triangle CHB$, and $\\triangle ASD \\sim \\triangle BSC$. This implies that\n\n$$\n\\frac{AT}{TE} = \\frac{AD}{GE}, \\quad \\frac{EH}{HB} = \\frac{GE}{BC}, \\quad \\frac{BS}{SA} = \\frac{BC}{AD}.\n$$\n\nSo,\n\n$$\n\\frac{AT}{TE} \\cdot \\frac{EH}{HB} \\cdot \\frac{BS}{SA} = \\frac{AD}{GE} \\cdot \\frac{GE}{BC} \\cdot \\frac{BC}{AD} = 1,\n$$\n\nwhich proves that the points $S$, $H$, $T$ are collinear.\n\nNext, consider the triangles $AFI$ and $DGC$. Because $IF \\cap CG = H$, $FA \\cap GD = T$, $AI \\cap CD = S$, and these points are collinear, Desargues' theorem implies that the lines $CI$, $FG$, and $AD$ are concurrent.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20770,
"subject": "Mathematics (Olympiad)",
"question": "Find all triples $ (a, b, c) $ of positive real numbers that satisfy the system:\n\n$$\n\\begin{aligned}\n11bc - 36b - 15c &= abc \\\\\n12ca - 10c - 28a &= abc \\\\\n13ab - 21a - 6b &= abc.\n\\end{aligned}\n$$",
"options": [],
"answer": "See solution",
"solution": "Consider each of the equalities:\n\n$$\n\\begin{aligned}\nabc &= 11bc - 36b - 15c \\\\\nabc &= 12ac - 10c - 28a \\\\\nabc &= 13ab - 21a - 6b\n\\end{aligned}\n$$\n\nDividing the first by $bc > 0$, the second by $ac$, and the third by $ab$, we obtain:\n\n$$\n\\begin{aligned}\na &= 11 - \\frac{36}{c} - \\frac{15}{b} \\\\\nb &= 12 - \\frac{10}{a} - \\frac{28}{c} \\\\\nc &= 13 - \\frac{21}{b} - \\frac{6}{a}.\n\\end{aligned}\n$$\n\nSumming all three and rearranging, we get:\n\n$$\na + \\frac{16}{a} + b + \\frac{36}{b} + c + \\frac{64}{c} = 36.\n$$\n\nSince $a, b, c > 0$, by AM-GM: $a + \\frac{16}{a} \\geq 8$, $b + \\frac{36}{b} \\geq 12$, $c + \\frac{64}{c} \\geq 16$. Since $8 + 12 + 16 = 36$, equality holds only if $a = 4$, $b = 6$, $c = 8$.\n\nFor $(a, b, c) = (4, 6, 8)$:\n\n$$\na = 4 = 11 - \\frac{36}{8} - \\frac{15}{6} = 11 - \\frac{9}{2} - \\frac{5}{2}.\n$$\n\nSimilarly,\n\n$$\nb = 6 = 12 - \\frac{10}{4} - \\frac{28}{8} = 12 - \\frac{5}{2} - \\frac{7}{2}.\n$$\n\nSince two equalities are satisfied and the sum of the left and right sides match, the third also holds. Thus, $(a, b, c) = (4, 6, 8)$ is the unique positive solution.\n\n_Comment._ Dividing by $abc$ and summing is crucial for untangling the problem. Without the positivity assumption, there are five solutions:\n\n$$\n(a, b, c) = \\left(-11, \\frac{21}{11}, \\frac{28}{11}\\right),\\ (0, 0, 0),\\ \\left(\\frac{15}{28}, -\\frac{140}{9}, \\frac{63}{20}\\right),\\ \\left(\\frac{10}{13}, \\frac{15}{13}, -13\\right),\\ (4, 6, 8)\n$$\n\nA brute-force approach leads to a non-trivial quartic polynomial, e.g., for $c$:\n\n$$\n220c^5 - 153c^4 - 27381c^3 + 139132c^2 - 183456c = 0\n$$\n\nand the ones for $a$, $b$ are even more intractable.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20771,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a quadrilateral with equal angles $B$ and $D$. Circles $\\omega_1$ and $\\omega_2$ are symmetric with respect to $AC$; $\\omega_1$ passes through $B$ and intersects $AB$ and $BC$ for the second time at $K$ and $L$, respectively, and $\\omega_2$ passes through $D$ and intersects $CD$ and $AD$ for the second time at $M$ and $N$, respectively.\n\nProve that $KM$ and $LN$ intersect on $AC$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $D'$ be symmetric to $D$ with respect to $AC$. If $D'$ coincides with $B$, then $N$ and $M$ are symmetric to $K$ and $L$, respectively, with respect to $AC$, and the problem clearly holds. Let $D' \\neq B$. Since $\\omega_2$ is symmetric to $\\omega_1$ with respect to $AC$, if we denote by $M'$ and $N'$ the points symmetric to $M$ and $N$, respectively, with respect to $AC$, we will have that $B, D', M', L, N', K$ all lie on $\\omega_1$.\n\nSince $\\angle AD'C = \\angle ADC = \\angle ABC$, points $A, B, D', C$ are concyclic. We get $\\angle (CA, AD') = \\angle (CB, BD') = \\angle (LN', N'D')$, so $LN' \\parallel AC$, and similarly $M'K \\parallel AC$. Then $LM'KN'$ is a cyclic trapezoid, so it's isosceles, and $KL = M'N' = MN$. We also get $\\angle (MN, AC) = \\angle (AC, M'N') = \\angle (N'L, M'N') = \\angle (KL, N'L) = \\angle (KL, AC)$, so $MN \\parallel KL$. Since segments $KL$ and $MN$ are equal, $KLMN$ is a parallelogram, so $KM$ intersects $NL$ at its midpoint, and since $N'L \\parallel AC$ and $N$ is symmetric to $N'$ with respect to $AC$, the midpoint of $NL$ lies on $AC$, as desired.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20772,
"subject": "Mathematics (Olympiad)",
"question": "任選橢圓 $C: x^2 + 2y^2 = 2098$ 上的一個有理點 $P_0 = (x_p, y_p)$。我們將依以下方式遞迴決定 $P_1, P_2, \\dots$：對於所有 $i = 0, 1, \\dots$：\n\n1. 選取一個不在 $C$ 上的整點 $Q_i = (x_i, y_i)$，使得 $|x_i| < 50$ 且 $|y_i| < 50$。\n2. 連接 $\\overline{P_iQ_i}$，並令其與 $C$ 的另一交點為 $P_{i+1}$。\n\n試證：對於任何 $P_0$，我們都可以適當選取 $Q_0, Q_1, \\dots$，使得存在某個非負整數 $k$，讓 $\\overline{OP_k} = 2017$。\n\n（我們稱 $(x, y)$ 為整點，若且唯若 $x$ 和 $y$ 都是整數。我們稱 $(x, y)$ 為有理點，若且唯若 $x$ 和 $y$ 都是有理數。）",
"options": [],
"answer": "See solution",
"solution": "易知 $C$ 上的所有整數點為 $(\\pm44, \\pm9)$，且 $44^2 + 9^2 = 2017$，故我們只要證明經過適當的操作後，某個 $P_k$ 是整點即可。\n\n若 $P_0$ 是整點，由上述知取 $k=0$ 即可，故假設 $P_0 = (a/m, b/m)$ 不為整點，其中 $a, b, m \\in \\mathbb{Z}$ 且 $m > 0$。\n\n顯然存在整數 $s$ 和 $t$ 滿足 $|s - \\frac{a}{m}| \\le 1/2$ 及 $|t - \\frac{b}{m}| \\le 1/2$。又，$|s| < \\sqrt{2098} + 1 < 50$，同理 $|t| < 50$，故可取 $Q_0 = (s, t)$。\n\n注意到 $(s - \\frac{a}{m})^2 + 2(t - \\frac{b}{m})^2 = 2098 + s^2 + 2t^2 - 2(sa + 2tb)/m = m'/m$，其中 $m' \\in \\mathbb{N}$。但由定義，我們有\n\n$$\n\\left| \\left(s - \\frac{a}{m}\\right)^2 + 2\\left(t - \\frac{b}{m}\\right)^2 \\right| \\le 1/4 + 2/4 < 1,\n$$\n\n故 $m' = m \\left( \\left(s - \\frac{a}{m}\\right)^2 + 2\\left(t - \\frac{b}{m}\\right)^2 \\right) < m$。\n\n現在，假設 $P_1 = (s+z(a-sm), t+z(b-tm))$，則有 $(s+z(a-sm))^2 + 2(t+z(b-tm))^2 = 2098$，展開得\n\n$$\n\\frac{m'}{m}z^2 + 2(s(a - sm) + 2t(b - tm))z + (s^2 + 2t^2 - 2098) = 0.\n$$\n\n上式的一解為 $z = 1/m$（對應 $P_0$ 的解），故由根與係數，另一解為 $(s^2 + 2t^2 - 2098)/m'$。因此 $P_1 = (a'/m', b'/m')$，其中 $a', b', m' \\in \\mathbb{Z}$ 且 $0 < m' < m$。\n\n由以上討論得知，若每次討論都用以上方式選取 $Q_i$，則得到的 $P_{i+1}$ 其座標分母將比 $P_i$ 的座標分母小。這表示存在充分大的 $k$，使得 $P_k$ 是一個整點，$\\Rightarrow \\overline{OP_k} = 2017$。證畢。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20773,
"subject": "Mathematics (Olympiad)",
"question": "Let $D$ be the set of all natural divisors of a number except $1$ and the number itself.\n\n**a)** Are there more twin numbers or more numbers that are not twin?\n\n**b)** Justify your answer.",
"options": [],
"answer": "See solution",
"solution": "Let $M = 20112012$. Denote by $M_K$ the set of all natural numbers not exceeding $M$ and divisible by $K$, and by $N_K = |M_K|$ their count. 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$$\nn < N_4 + N_{3.5} + N_{5.7} + N_{7.9} + \\dots + N_{(M-3)(M-1)},\n$$\n\nsince all twin numbers are counted, possibly more than once (e.g., $945 = 3 \\cdot 5 \\cdot 7 \\cdot 9$ is counted three times). Thus,\n\n$$\n\\begin{align*}\nn < & \\left[ \\frac{M}{4} \\right] + \\left[ \\frac{M}{3.5} \\right] + \\left[ \\frac{M}{5.7} \\right] + \\left[ \\frac{M}{7.9} \\right] + \\dots + \\left[ \\frac{M}{(M-3)(M-1)} \\right] \\\n& < \\frac{M}{4} + \\frac{M}{3.5} + \\frac{M}{5.7} + \\frac{M}{7.9} + \\dots + \\frac{M}{(M-3)(M-1)} \\\n& = \\frac{M}{4} + M \\left( \\frac{1}{3.5} + \\frac{1}{5.7} + \\frac{1}{7.9} + \\dots + \\frac{1}{(M-3)(M-1)} \\right) \\\n& = \\frac{M}{4} + \\frac{M}{2} \\left( \\frac{5-3}{3.5} + \\frac{7-5}{5.7} + \\frac{9-7}{7.9} + \\dots + \\frac{(M-1)-(M-3)}{(M-3)(M-1)} \\right) \\\n& = \\frac{M}{4} + \\frac{M}{2} \\left( \\frac{1}{3} + \\frac{1}{5} + \\frac{1}{5} + \\frac{1}{7} + \\frac{1}{7} + \\frac{1}{9} + \\dots + \\frac{1}{M-3} + \\frac{1}{M-1} \\right) \\\n& = \\frac{M}{4} + \\frac{M}{2} \\left( \\frac{1}{3} - \\frac{1}{M-1} \\right) < \\frac{M}{4} + \\frac{M}{6} = \\frac{5M}{12} < \\frac{M}{2},\n\\end{align*}\n$$\n\nwhich proves that there are more numbers that are not twin.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20774,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a cyclic quadrilateral with no parallel sides and circumcircle $(O)$. Let $E$ be the intersection of the two diagonals, and let the bisector of $\\angle AEB$ cut lines $AB$, $BC$, $CD$, and $DA$ at $M$, $N$, $P$, and $Q$ respectively.\n\n**a)** Prove that the four circumcircles of $AQM$, $BMN$, $CNP$, and $DPQ$ have a common point $K$.\n\n**b)** Let $\\min\\{AC, BD\\} = m$. Prove that $OK \\le \\dfrac{2R^2}{\\sqrt{4R^2 - m^2}}$.",
"options": [],
"answer": "See solution",
"solution": "Let $R$ be the intersection of $AD$ and $BC$, and $S$ the intersection of $AB$ and $CD$ (since the opposite sides of the quadrilateral $ABCD$ are not parallel, these points are completely determined). Suppose $B$ lies between $A$, $S$ and between $C$, $R$ as shown in the figure.\n\n\n\nLet $K$ be the intersection of the circumcircles of triangles $RAB$ and $SBC$. It is clear that\n$$\n\\angle BKR + \\angle BKS = \\angle BAD + \\angle BCD = 180^{\\circ},\n$$\nso $R$, $K$, and $S$ are collinear. It follows that\n$$\nRK \\cdot RS = RB \\cdot RC = RA \\cdot RD \\text{ and } SK \\cdot SR = SB \\cdot SA = SC \\cdot SD\n$$\nso the quadrilaterals $ADSK$ and $CDRK$ are cyclic, and thus $K$ also lies on the circles $(RCD)$ and $(SDA)$. Hence, we have $\\angle AKD = \\angle ASD = \\angle BSC = \\angle BKC$ and $\\angle ADK = \\angle ASK = \\angle BSK = \\angle BCK$, so the triangles $KAD$ and $KBC$ are similar. Thus,\n$$\n\\frac{KA}{KB} = \\frac{AD}{BC} = \\frac{AE}{BE} = \\frac{AM}{BM}\n$$\nwhich means $KM$ is the bisector of angle $AKB$. Otherwise, we have\n$$\n\\angle RNQ = \\angle BNE = \\angle CBD - \\angle BEN = \\angle CAD - \\angle AEQ = \\angle RQN\n$$\nHence, $\\angle ARB = 2 \\cdot \\angle BNM$. From here we have $\\angle BKM = \\frac{1}{2} \\cdot \\angle AKB = \\frac{1}{2} \\cdot \\angle ARB = \\angle BNM$, so $BMNK$ is cyclic, i.e., $K$ lies on $(BMN)$. Similarly, $K$ also belongs to $(AQM)$, $(CNP)$, $(DPQ)$.\n\nNext, we prove that $K$ is the only common point of these circles. Indeed, the circles $(AMP)$ and $(BMQ)$ have two common points, $K$ and $M$, and the circles $(DNP)$ and $(CNP)$ have two common points, $K$ and $N$. Therefore, if these four circles had two common points, then $M$ and $N$ would coincide, which is a contradiction.\n\n**b)** By power of a point, we get\n$$\nRK \\cdot RS = RB \\cdot RC = RO^2 - R^2, \\quad SK \\cdot SR = SB \\cdot SA = SO^2 - R^2\n$$\nTherefore,\n$$\nRO^2 - SO^2 = RK \\cdot RS - SK \\cdot SR = RK^2 - SK^2\n$$\nwhich implies $OK \\perp RS$. Furthermore, applying Brocard's theorem in the cyclic quadrilateral $ABCD$, we obtain that $E$ is the orthocenter of the triangle $ORS$, so $OE \\perp RS$. Therefore, $O$, $E$, and $K$ are collinear.\n\nWe also have\n$$\n\\angle RKA + \\angle SKC = \\angle RBA + \\angle SBC = 2\\angle ADC = \\angle AOC\n$$\nso $\\angle AKC + \\angle AOC = 180^\\circ$, i.e., $AOCK$ is cyclic. Note that\n$$\nEO \\cdot EK = EA \\cdot EC = R^2 - OE^2\n$$\nTherefore, $EO \\cdot (EO + EK) = R^2$, or $OK = \\dfrac{R^2}{EO}$. We obtain that\n$$\n\\begin{aligned}\nEO &\\ge \\max \\{d(O, AC), d(O, BD)\\} \\\\\n&= \\max \\left\\{ \\frac{1}{2}\\sqrt{4R^2 - AC^2}, \\frac{1}{2}\\sqrt{4R^2 - BD^2} \\right\\} \\\\\n&= \\frac{1}{2}\\sqrt{4R^2 - m^2}.\n\\end{aligned}\n$$\nSo we get $OK \\le \\dfrac{2R^2}{\\sqrt{4R^2 - m^2}}$ as desired. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20775,
"subject": "Mathematics (Olympiad)",
"question": "Let $p$ be a prime. Consider the sequence $b_n$ defined by\n$$\n\\begin{cases}\na_n = (a^2 + 2b^2)a_{n-1} + 4ab b_{n-1} \\\\\nb_n = 2ab a_{n-1} + (a^2 + 2b^2) b_{n-1}.\n\\end{cases}\n$$\nwhere $a$ and $b$ are relatively prime integers. For which primes $p$ is there a positive integer $n$ such that $b_n$ is divisible by $p$?",
"options": [],
"answer": "See solution",
"solution": "Suppose $p$ is an odd prime dividing $a^2 - 2b^2$. Since $a$ and $b$ are relatively prime, $b_1 = 2ab$ is not divisible by $p$. Assume there is a positive integer $n$ such that $b_n$ is divisible by $p$, and let $r$ be the smallest such integer. Note that $(a - b\\sqrt{2})^{2n} = a_n - b_n\\sqrt{2}$, and\n$$\n\\begin{cases}\na_n = (a^2 + 2b^2)a_{n-1} + 4ab b_{n-1} \\\\\nb_n = 2ab a_{n-1} + (a^2 + 2b^2) b_{n-1}.\n\\end{cases}\n$$\nThen,\n$$\n\\begin{aligned}\n0 \\equiv b_r &= 2ab((a^2 + 2b^2)a_{r-2} + 4ab b_{r-2} + (a^2 + 2b^2) b_{r-1}) \\\\\n&= 2(a^2 + 2b^2) b_{r-1} - (a^2 - 2b^2)^2 b_{r-2} \\equiv 2(a^2 + 2b^2) b_{r-1} \\pmod{p},\n\\end{aligned}\n$$\nwhich is a contradiction. Therefore, $b_n$ is not divisible by $p$ for any $n$.\n\nNow let $p$ be a prime not dividing $a^2 - 2b^2$. Since $b_1 = 2ab$, we may assume $p$ is odd and $ab$ is not divisible by $p$. Note that\n$$\nb_n = \\frac{(a + b\\sqrt{2})^{2n} - (a - b\\sqrt{2})^{2n}}{\\sqrt{2}} = \\sum_{k=0,\\ k \\equiv 1 \\pmod{2}}^{2n} 2 \\binom{2n}{k} a^{2n-k} b^k 2^{\\frac{k-1}{2}}\n$$\nIf there is $m$ such that $m^2 \\equiv 2 \\pmod{p}$, then\n$$\n\\begin{aligned}\nb_n &= \\sum_{k=0,\\ k \\equiv 1 \\pmod{2}}^{2n} 2 \\binom{2n}{k} a^{2n-k} b^k 2^{\\frac{k-1}{2}} \\\\\n&= \\sum_{k=0,\\ k \\equiv 1 \\pmod{2}}^{2n} m^2 \\binom{2n}{k} a^{2n-k} b^k m^{k-1} \\\\\n&= \\frac{(a + bm)^{2n} - (a - bm)^{2n}}{m} \\pmod{p}.\n\\end{aligned}\n$$\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\nIf $x^2 \\equiv 2 \\pmod{p}$ has no solution (i.e., 2 is a quadratic non-residue modulo $p$), then\n$$\n\\binom{p+1}{k} \\equiv 0 \\pmod{p} \\text{ for } 2 \\leq k \\leq p-1.\n$$\nBy Euler's criterion,\n$$\n\\begin{aligned}\nb_{\\frac{p+1}{2}} &= \\sum_{k=0,\\ k \\equiv 1 \\pmod{2}}^{p+1} 2 \\binom{p+1}{k} a^{p+1-k} b^k 2^{\\frac{k-1}{2}} \\\\\n&= 2(p+1)a^p b + (p+1)ab^p 2^{\\frac{p+1}{2}} \\\\\n&= 2ab(1 + 2^{\\frac{p-1}{2}}) \\equiv 0 \\pmod{p}.\n\\end{aligned}\n$$\n\nTherefore, such a prime is exactly 2 or relatively prime to $a^2 - 2b^2$. $\\square$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20776,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be distinct positive integers such that the sum of any two of them is a perfect square. Assume $a < b < c$. Write $a + b = x^2$, $b + c = y^2$, and $c + a = z^2$. Find the values of $a$, $b$, and $c$ that minimize $x^2 + y^2 + z^2$ under the conditions $x < y < z$, $z^2 < x^2 + y^2$, and $x^2 + y^2 + z^2$ is even.",
"options": [],
"answer": "See solution",
"solution": "We analyze possible values for $z$:\n\n- If $z > 5$, since otherwise $z^2 \\geq x^2 + y^2$.\n- If $z = 6$, $x$ and $y$ must both be even or both odd, but neither works since $6^2 > 5^2 + 3^2 > 4^2 + 2^2$.\n- If $z = 7$, only $(x, y, z) = (5, 6, 7)$ satisfies all conditions.\n- If $z \\geq 8$, $x^2 + y^2 + z^2 > 2z^2 \\geq 2 \\times 8^2 > 7^2 + 6^2 + 5^2$.\n\nTherefore, $(x, y, z) = (5, 6, 7)$ minimizes $x^2 + y^2 + z^2$.\n\nSolving for $a$, $b$, $c$:\n\n\\begin{align*}\na + b &= 25 \\\\\nb + c &= 36 \\\\\nc + a &= 49\n\\end{align*}\n\nAdding all three: $2(a + b + c) = 25 + 36 + 49 = 110 \\implies a + b + c = 55$.\n\nSubtracting:\n- $c = 55 - (a + b) = 55 - 25 = 30$\n- $a = 55 - (b + c) = 55 - 36 = 19$\n- $b = 55 - (c + a) = 55 - 49 = 6$\n\nThus, $(a, b, c) = (6, 19, 30)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20777,
"subject": "Mathematics (Olympiad)",
"question": "Prove or disprove: For every integer $n > 0$, there is a polynomial $a_n x^n + a_{n-1} x^{n-1} + \\dots + a_1 x + a_0$ satisfying the following conditions:\n\n1. All coefficients $a_n, a_{n-1}, \\dots, a_1, a_0$ are positive real numbers.\n2. At least one of the coefficients is $\\frac{1}{n}$.\n3. The value of the polynomial is integer for every integer $x$.",
"options": [],
"answer": "See solution",
"solution": "The claim is true.\n\nThe product of any $n$ consecutive integers is divisible by $n$. Thus, the conditions are satisfied by the polynomial obtained by expanding and collecting terms in the expression $$\\frac{1}{n}(x+1)\\dots(x+n).$$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20778,
"subject": "Mathematics (Olympiad)",
"question": "Let $XY$ and $AP$ intersect at $K$. Let $R$ be the circumradius of $\\triangle ABC$.\n\nShow that $A$, $P$, $X$, $Y$ are concyclic.",
"options": [],
"answer": "See solution",
"solution": "*Claim 1.* $XI^2 = XB \\cdot XC$\n\n*Proof.* $\\triangle XIB \\sim \\triangle XCI$ because $\\angle XIB = 90^\\circ - \\angle BIP = \\frac{1}{2} \\angle C = \\angle XCI$.\n\n*Claim 2.* $KI^2 = KA \\cdot KP$\n\n*Proof.* By Claim 1 and power of a point, $XI^2 = XB \\cdot XC = XO^2 - R^2$. Also, $XO^2 - XI^2 = YO^2 - YI^2 = KO^2 - KI^2$. Hence, $KI^2 = KO^2 - R^2$. By power of point $K$, $KO^2 - R^2 = KA \\cdot KP$.\n\nBy Euclidean Theorem, $KI^2 = KX \\cdot KY$ which by Claim 2 is also equal to $KA \\cdot KP$. Therefore, $A$, $P$, $X$, $Y$ are concyclic, as desired.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20779,
"subject": "Mathematics (Olympiad)",
"question": "Given a triangle $ABC$, with $AB \\perp BC$. The altitude $BH$ intersects the bisectors $AD$ and $CE$ at $Q$ and $P$, respectively. Prove that the straight line through the midpoints of the segments $QD$ and $PE$ is parallel to $AC$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Denote $a$, $b$, $c$ as the lengths of the sides of the triangle. Then\n\n$$\n\\frac{HA}{HC} = \\frac{c^2}{a^2}, \\quad \\frac{DB}{DC} = \\frac{c}{b}, \\quad \\frac{QA}{QD} = \\frac{c^2 b + c}{a^2 c} = \\frac{c}{b-c},\n$$\n\nwhence\n\n$$\n\\overrightarrow{BQ} = \\frac{b-c}{b} \\overrightarrow{BA} + \\frac{c}{b} \\overrightarrow{BD}, \\quad \\overrightarrow{BM} = \\frac{b-c}{2b} \\overrightarrow{BA} + \\frac{c+b}{2b} \\overrightarrow{BD} = \\frac{b-c}{2b} \\overrightarrow{BA} + \\frac{c}{2b} \\overrightarrow{BC}\n$$\n\nand, in the same way,\n\n$$\n\\overrightarrow{BN} = \\frac{b-a}{2b}\\overrightarrow{BC} + \\frac{a}{2b}\\overrightarrow{BA}.\n$$\n\nIt follows\n\n$$\n\\overrightarrow{MN} = \\frac{1}{2b}((b-a-c)\\overrightarrow{BC} + (a-c-b)\\overrightarrow{BA}) = \\frac{a+c-b}{2b}\\overrightarrow{CA},\n$$\n\nwhich proves the conclusion.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20780,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ be an odd positive integer that is not a perfect square, and let $b$ and $c$ be odd primes. Suppose\n\n$$\na^2 + a + 1 = 3(b^2 + b + 1)(c^2 + c + 1).\n$$\n\nProve that at least one of $b^2 + b + 1$ and $c^2 + c + 1$ is composite.",
"options": [],
"answer": "See solution",
"solution": "Assume that both $b^2 + b + 1$ and $c^2 + c + 1$ are primes. Then\n\n$$\na^2 + a + 1 \\equiv 0 \\pmod{b^2 + b + 1}.\n$$\n\nThis quadratic equation modulo a prime has at most two roots. Thus, $a \\equiv b \\pmod{b^2 + b + 1}$ or $a \\equiv b^2 \\pmod{b^2 + b + 1}$. In the latter case,\n\n$$\na^2 + a + 1 \\equiv b^4 + b^2 + 1 = (b^2 + b + 1)(b^2 - b + 1) \\equiv 0.\n$$\n\nWe can rule out the smallest such values for $a$: since $a > b$ and $a$ is not a perfect square, $a \\neq b^2$. The next possible value is $a = b^2 + b + 1 + b = (b + 1)^2$, which is even, as is $a = b^2 + b + 1 + b^2$. Thus, $a \\ge 2(b^2 + b + 1) + b > 2(b^2 + b + 1)$. Similarly, $a > 2(c^2 + c + 1)$. But then\n\n$$\n3(b^2 + b + 1)(c^2 + c + 1) = a^2 + a + 1 > a^2 > 4(b^2 + b + 1)(c^2 + c + 1),\n$$\n\nwhich is a contradiction.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20781,
"subject": "Mathematics (Olympiad)",
"question": "The diagonals of a tangential quadrilateral $ABCD$ intersect at point $P$. The side $AB$ is longer than any other side of $ABCD$. Prove that the angle $APB$ is obtuse.",
"options": [],
"answer": "See solution",
"solution": "Let $a, b, c, d$ be the lengths of the tangent line segments at vertices $A, B, C, D$, respectively, and $\\alpha = \\angle APB$.\n\nWe have $a + b > b + c$ and $a + b > a + d$ as $AB$ is the longest side, implying $a > c$ and $b > d$.\n\nBy the law of cosines:\n\n$$\n(a+b)^2 = PA^2 + PB^2 - 2 \\cdot PA \\cdot PB \\cdot \\cos \\alpha,\n$$\n$$\n(b+c)^2 = PB^2 + PC^2 + 2 \\cdot PB \\cdot PC \\cdot \\cos \\alpha,\n$$\n$$\n(c+d)^2 = PC^2 + PD^2 - 2 \\cdot PC \\cdot PD \\cdot \\cos \\alpha,\n$$\n$$\n(d+a)^2 = PD^2 + PA^2 + 2 \\cdot PD \\cdot PA \\cdot \\cos \\alpha.\n$$\n\nAdding the first and the third equation and subtracting the second and the fourth equation gives\n\n\n\n$2(ab + cd - bc - da) = -2(PA \\cdot PB + PB \\cdot PC + PC \\cdot PD + PD \\cdot PA) \\cos \\alpha$.\n\nThe left-hand side can be expressed as $2(a - c)(b - d)$, which is positive since $a > c$ and $b > d$. The parenthesized expression on the right-hand side is also positive. Hence $\\cos \\alpha$ must be negative. This means that the angle $APB$ is obtuse.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20782,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute triangle with $AB \\neq AC$ and circumcircle $\\Gamma$. The angle bisector of $\\angle BAC$ intersects $BC$ and $\\Gamma$ at $D$ and $E$ respectively. The circle with diameter $DE$ intersects $\\Gamma$ again at $F \\neq E$. Point $P$ is on $AF$ such that $PB = PC$, and $X$ and $Y$ are the feet of the perpendiculars from $P$ to $AB$ and $AC$ respectively. Let $H$ and $H'$ be the orthocenters of $ABC$ and $AXY$ respectively. $AH$ meets $\\Gamma$ again at $Q$. If $AH'$ and $HH'$ intersect the circle with diameter $AH$ again at points $S$ and $T$, respectively, prove that the lines $AT$, $HS$, and $FQ$ are concurrent.",
"options": [],
"answer": "See solution",
"solution": "WLOG, assume $AB < AC$. Let $M$ be the midpoint of side $BC$ and let the circumcircle of $DFE$ intersect $AF$ again at $K$. Since\n\n$$\n90^\\circ + \\angle MED = 180^\\circ - \\angle MDE = \\angle ABC + \\frac{\\angle BAC}{2} = \\angle AFE = \\angle DFE + \\angle AFD = 90^\\circ + \\angle AFD\n$$\n\nit follows that\n\n$$\n\\angle AFD = \\frac{\\angle ABC - \\angle ACB}{2} = \\angle MED\n$$\n\nBecause\n\n$$\n\\angle DKE = \\angle DME = 90^\\circ\n$$\n\nand\n\n$$\n\\angle KED = \\angle KFE = \\angle MED\n$$\n\nwe get $\\triangle KDE \\cong \\triangle MDE$ from which it follows that $DE$ is the perpendicular bisector of $MK$ and here we get\n\n$$\n\\angle FAD = \\angle KAD = \\angle MAD\n$$\n\nIt is obvious that $P$ is the intersection of $ME$ and $AF$. Let $ME$ intersect $\\Gamma$ again at $L$. From the angle bisector theorem in triangle $AMP$ we get\n\n$$\n\\frac{PE}{ME} = \\frac{AP}{AM} = \\frac{LP}{LM} \\quad (1)\n$$\n\n($LA$ is the external angle bisector of $\\angle MAP$ since $LE$ is the diameter of $\\Gamma$). Now we prove that $CE$ and $CL$ are angle bisectors of $\\angle MCP$. Let $M'$ be the point on $LE$ such that\n\n$\\angle M'CE = \\angle ECP$. From the angle bisector theorem we get\n\n$$\n\\frac{M'E}{PE} = \\frac{CM'}{CP} = \\frac{M'L}{PL} \\quad (2)\n$$\n\nMultiplying (1) and (2) we get $\\frac{ME}{LM} = \\frac{M'E}{M'L}$ adding 1 on both sides we get $LM = LM'$ from which it follows that $M = M'$ and thus $CE$ and $CL$ are the bisectors of $\\angle MCP$. Now we have\n\n$$\n\\angle MPC = 90^\\circ - \\angle MCP = 90^\\circ - 2\\angle MCE = 90^\\circ - 2\\angle EAC = 90^\\circ - \\angle BAC\n$$\n\n\n\nSince $X$ and $Y$ are perpendicular to $AB$ and $AC$ we have $BXPM$ and $CYPM$ are concyclic. Here we get\n\n$$\\angle MYC = \\angle MPC = 90^\\circ - \\angle BAC$$\n\nand it follows that $YM \\perp AX$. Similarly we get $XM \\perp AY$ and so $M$ is the orthocenter of $\\triangle AXY$ giving us $M = H'$.\n\nSince $ATHS$ and $ATQF$ are both concyclic it is enough to prove that $HSFQ$ is concyclic. Since\n\n$$\n\\angle BQC = 180^\\circ - \\angle BAC = \\angle BHC\n$$\n\nand $HQ \\perp BC$ it follows that $BC$ is the perpendicular bisector of $HQ$. It is enough to prove that $BC$ is the perpendicular bisector of $SF$. Let $AM$ and $TH$ meet $\\Gamma$ again at points $A'$ and $N$ respectively.\n\nSince $HN$ passes through the midpoint of side $BC$ and\n\n$$\n\\angle BHC = 180^\\circ - \\angle BAC = \\angle BNC\n$$\n\nit follows that $BNCH$ is a parallelogram. From here we get that\n\n$$\n\\angle NCB = \\angle HBC = 90^\\circ - \\angle ACB\n$$\n\ngiving us $\\angle NCA = 90^\\circ$ and similarly $\\angle NBA = 90^\\circ$. This means $AN$ is the diameter of $\\Gamma$, so\n\n$$\n\\angle NA'A' = \\angle NA'A = 90^\\circ = \\angle HSA = \\angle HSA'\n$$\n\nand from here we have $HS \\parallel A'N$. Now since $HS \\parallel A'N$ and $M$ is the midpoint of $HN$ (because $BHCN$ is a parallelogram) we get\n\n\n\nthat $HSNA'$ is a parallelogram. Since\n\n$$\n\\angle FAE = \\angle EAM = \\angle EAA'\n$$\n\n\n\nHS, AT are concurrent.\n\nwe get that $FA'BC$ is an isosceles trapezoid which means that $ME$ is the perpendicular bisector of $FA'$ (since it is the perpendicular bisector of $BC$).\n\nThis gives us $BF = CA' = BS$ and $CF = BA' = CS$ giving us that $SBFC$ is a deltoid, meaning that $BC$ is the perpendicular bisector of $FS$. This means that $HSFQ$ is an isosceles trapezoid. Now from the radical axis theorem of the circumcircles of $HSFQ$, $HSAT$ and $ATQF$ we get that $QF$,\n\n\n\nHS, AT are concurrent.\n\nwe get that $FA'BC$ is an isosceles trapezoid which means that $ME$ is the perpendicular bisector of $FA'$ (since it is the perpendicular bisector of $BC$).\n\nThis gives us $BF = CA' = BS$ and $CF = BA' = CS$ giving us that $SBFC$ is a deltoid, meaning that $BC$ is the perpendicular bisector of $FS$. This means that $HSFQ$ is an isosceles trapezoid. Now from the radical axis theorem of the circumcircles of $HSFQ$, $HSAT$ and $ATQF$ we get that $QF$,\n\n\n\nare concurrent.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20783,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle. Let $A'$ be the centre of the circle through the midpoint of the side $BC$ and the orthogonal projections of $B$ and $C$ on the lines of support of the internal bisectors of the angles $ACB$ and $ABC$, respectively; the points $B'$ and $C'$ are defined similarly. Prove that the nine-point circle of the triangle $ABC$ and the circumcircle of $A'B'C'$ are concentric.\n\n\n\n\n\n",
"options": [],
"answer": "See solution",
"solution": "All the angles in the solution are directed modulo $\\pi$. The following notation is used throughout the proof:\n\n- $2\\alpha$, $2\\beta$, $2\\gamma$: measures of the angles $BAC$, $CBA$, $ACB$, respectively.\n- $2a$, $2b$, $2c$: lengths of the sides $BC$, $CA$, $AB$, respectively.\n- $I$: incenter of the triangle $ABC$.\n- $M_A$, $M_B$, $M_C$: midpoints of the sides $BC$, $CA$, $AB$, respectively.\n- $X_Y$: orthogonal projection of $X$ on the line $YI$, for all $X, Y \\in \\{A, B, C\\}$.\n- $\\omega_A$, $\\omega_B$, $\\omega_C$: circumcircles of the triangles $M_A B_C C_B$, $M_B C_A A_C$, $M_C A_B B_A$, respectively, centered at $A'$, $B'$, $C'$, respectively.\n\nSince the angle $AA_B B$ is right, the segments $M_C A = M_C B = M_C A_B = c$ (see Fig. 3), so $\\angle M_C A_B B = \\angle A_B B M_C C = \\angle C B A_B = \\beta$; this means that the lines $M_C A_B$ and $BC$ are parallel, so $A_B$ lies on the line $M_B M_C$. So, $(M_A, M_B, C_A, C_B)$, $(M_B, M_C, A_B, A_C)$, $(M_C, M_A, B_C, B_A)$ are quartets of collinear points.\n\nLet $A''$ be the incenter of the triangle $AM_C M_B$ (in other words, $A''$ is the midpoint of $AI$). We show that $A''$ lies on both $\\omega_B$ and $\\omega_C$. For that, notice first that the points $A, B, A_B$, and $B_A$ lie on the circle on diameter $AB$; hence $\\angle A_B B_A A = \\angle A_B B A = \\beta$. Next, the points $A_B, A_C, M_B$, and $M_C$ lie on a line parallel to $BC$, so $\\angle A_B M_C A'' = \\angle M_B M_C A'' = \\beta = \\angle A_B B_A A$. This means that $A''$ lies on $\\omega_C$. Similarly, $A''$ lies on $\\omega_B$.\n\nLet $X$ be the second point of intersection of $\\omega_B$ and $\\omega_C$. By the preceding, $\\angle M_B X A'' = \\angle M_B C_A A'' = \\alpha$ and similarly $\\angle A'' X M_C = \\alpha$. This yields $\\angle M_B X M_C = \\angle M_B X A'' + \\angle A'' X M_C = 2\\alpha = \\angle M_B M_A M_C$, which shows that $X$ lies also on the circumcircle $\\omega$ of the triangle $M_A M_B M_C$, which is the nine-point circle of the triangle $ABC$. Denote the center of $\\omega$ by $O'$.\n\nNow the lines $A''X, M_BX$, and $M_CX$ are the radical axes of the circles $\\omega_B, \\omega_C$, and $\\omega$. Since $XA''$ forms equal angles with $M_BX$ and $M_CX$, the triangle $O'B'C'$ formed by the centers of these circles has equal angles at $B'$ and $C'$; therefore, $O'B' = O'C'$. A similar argument shows that $O'B' = O'A'$, and $O'$ is consequently the circumcenter of the triangle $A'B'C'$.\n\n**Remark.** The point $X$ in the solution is the Feuerbach point of the triangle $ABC$ — the point at which the incircle is internally tangent to the nine-point circle.\n\n**Alternative Solution.** With reference to the notation in Solution 1, let $O$ and $O'$ be the circumcenter and the center of the nine-point circle of the triangle $ABC$, respectively. As in the previous solution, usage is made of the fact that $(M_A, M_B, C_A, C_B)$, $(M_B, M_C, A_B, A_C)$, and $(M_C, M_A, B_C, B_A)$ are quartets of collinear points, and $M_A B_C = M_A C_B = a$, $M_B A_C = M_B C_A = b$, and $M_C A_B = M_C B_B = c$.\n\nWe claim that $A'O' = IO/2$; similarly, $B'O' = IO/2 = C'O'$, whence the required result. To prove the claim, notice that each vector $v$ is uniquely determined by its projections on the lines $AB$ and $AC$. The signed lengths of these projections will be denoted $\\mathrm{pr}_c v$ and $\\mathrm{pr}_b v$, respectively, the rays $AB$ and $AC$ emanating from $A$ being considered positive.\n\nThe points $O'$ and $A'$ are the circumcenters of the triangles $M_A M_B M_C$ and $M_A B_C C_B$, respectively. Project onto $M_A M_C$, to get $\\mathrm{pr}_b A'O' = \\mathrm{pr}_b (M_A O' - \\ldots$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20784,
"subject": "Mathematics (Olympiad)",
"question": "Assume $S = \\{1, 2, \\dots, N\\}$. Player A maintains a sequence of $N$-tuples: $$(\\mathbf{x})_{j=0}^{\\infty} = (x_1^j, x_2^j, \\dots, x_N^j).$$ Initially, $x_1^0 = x_2^0 = \\dots = x_N^0 = 1$. After a set $P_j$ is selected, define $\\mathbf{x}^{j+1}$ from $\\mathbf{x}^j$ as:\n\n$$\nx_i^{j+1} = \\begin{cases} 1, & \\text{if } i \\in P_j \\\\ q \\cdot x_i^j, & \\text{if } i \\notin P_j. \\end{cases}\n$$\n\nPlayer A can keep B from winning if $x_i^j \\leq q^k$ for all $(i, j)$. For a sequence $\\mathbf{x}$, define $T(\\mathbf{x}) = \\sum_{i=1}^N x_i$. It suffices for A to ensure $T(\\mathbf{x}^j) \\leq q^k$ for all $j$.\n\nNotice $T(\\mathbf{x}^0) = N \\leq p^k < q^k$.\n\nProve: Given $\\mathbf{x}^j$ with $T(\\mathbf{x}^j) \\leq q^k$, and a set $D_{j+1}$, player A can choose $P_{j+1} \\in \\{D_{j+1}, D_{j+1}^C\\}$ so that $T(\\mathbf{x}^{j+1}) \\leq q^k$.",
"options": [],
"answer": "See solution",
"solution": "Let $\\mathbf{y}$ be the sequence if $P_{j+1} = D_{j+1}$, and $\\mathbf{z}$ if $P_{j+1} = D_{j+1}^C$. Then:\n\n$$\nT(\\mathbf{y}) = \\sum_{i \\in D_{j+1}^C} q x_i^j + |D_{j+1}|\n$$\n\n$$\nT(\\mathbf{z}) = \\sum_{i \\in D_{j+1}} q x_i^j + |D_{j+1}^C|\n$$\n\nSumming gives:\n\n$$\nT(\\mathbf{y}) + T(\\mathbf{z}) = q \\cdot T(\\mathbf{x}^j) + N \\leq q^{k+1} + p^k\n$$\n\nThus,\n\n$$\n\\min \\{T(\\mathbf{y}), T(\\mathbf{z})\\} \\leq \\frac{q}{2} \\cdot q^k + \\frac{p^k}{2} \\leq q^k,\n$$\n\nbecause of our choice of $k_0$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20785,
"subject": "Mathematics (Olympiad)",
"question": "Let $g$ be a convex function. Consider the inequality:\n\n$$\ng(ty + (1-t)z) \\leq t g(y) + (1-t) g(z)\n$$\n\nfor $t \\in [0,1]$ and $y, z$ real numbers.\n\n(a) Show that for $c \\in [0,1]$ and integer $n$,\n\n$$\nc [g(n) - g(n-1)] \\leq g(n+c) - g(n) \\leq c [g(n+1) - g(n)].\n$$",
"options": [],
"answer": "See solution",
"solution": "(b) Suppose $g(x) = \\log f(x)$. Then the inequality in part (a) becomes:\n\n$$\nt (\\log f(n) - \\log f(n-1)) \\leq \\log f(n+t) - \\log f(n) \\leq t (\\log f(n+1) - \\log f(n)).\n$$\n\nExponentiating both sides gives:\n\n$$\n\\left( \\frac{f(n)}{f(n-1)} \\right)^t \\leq \\frac{f(n+t)}{f(n)} \\leq \\left( \\frac{f(n+1)}{f(n)} \\right)^t.\n$$\n\nGiven $f(n) = (n-1)f(n-1)$ and $f(n+1) = n f(n)$, this reduces to:\n\n$$\n(n-1)^t \\leq \\frac{f(n+t)}{f(n)} \\leq n^t.\n$$\n\nNow, with $f(2) = 1 \\cdot f(1) = 1$, set $n = 2$ and $t = \\frac{1}{2}$:\n\n$$\n1 \\leq f\\left(\\frac{5}{2}\\right) \\leq \\sqrt{2}.\n$$\n\nSince $f\\left(\\frac{5}{2}\\right) = \\frac{3}{2} f\\left(\\frac{3}{2}\\right) = \\frac{3}{4} f\\left(\\frac{1}{2}\\right)$, we get:\n\n$$\n\\frac{4}{3} \\leq f\\left(\\frac{1}{2}\\right) \\leq \\frac{4}{3} \\sqrt{2}.\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20786,
"subject": "Mathematics (Olympiad)",
"question": "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\nProve that for any $i \\in \\{1,2,\\dots,n\\}$ there exists at least one good path from point $A_i$.",
"options": [],
"answer": "See solution",
"solution": "*Proof.* We will prove the claim for point $A_1$, and it holds analogously for all points $A_i$. Define the distance between a point $A$ and a wall $w$ as\n\n$$\n\\min_{T \\in w} |AT|.\n$$\n\nSince all walls are mutually disjoint, none of the walls $\\overline{A_2B_2}, \\dots, \\overline{A_nB_n}$ contain point $A_1$. Let $\\overline{A_kB_k}$ be the wall closest to $A_1$. We claim that the path $A_1A_k$ is good. If we assume the contrary, that means that there exists a wall which intersects $A_1A_k$, but then this wall is closer to $A_1$ than wall $\\overline{A_kB_k}$, which is a contradiction. $\\square$\n\n*Claim 2.* There exists at least one irrelevant wall.\n\n*Proof.* Without loss of generality, assume that points $A_1, A_2, \\dots, A_k$ are the vertices of the convex hull of $\\{A_1, A_2, \\dots, A_n\\}$, labelled clockwise as the vertices of polygon $P = A_1A_2\\dots A_k$. If the wall $w_1 = \\overline{A_1B_1}$ is irrelevant, we are done. Otherwise, assume $w_1$ intersects some path. That means this wall goes through the inner points of polygon $P$, and since the walls are mutually disjoint, $w_1$ must also intersect a path which is a side of $P$, because it cannot pass through any of the vertices of $P$ except $A_1$. Let $C$ be the intersection of wall $w_1$ and a side of $P$. Consider the arc $\\overarc{A_1C}$ (clockwise from $A_1$ to $C$), which contains points $A_2, \\dots, A_l$. Repeat this inference for wall $A_2B_2$. Since all walls are mutually disjoint, its corresponding arc is strictly smaller than the arc of $w_1$. Therefore, if all of the walls $\\overline{A_2B_2}, \\dots, \\overline{A_{l-1}B_{l-1}}$ are relevant, then wall $\\overline{A_lB_l}$ must be irrelevant. $\\square$\n\nFinally, we prove the problem statement by mathematical induction on $n$. The claim obviously holds for $n=1$. Assume the claim holds for some positive integer $n$. For $n+1$, it follows from Claim 2 that there exists a point $A_i$ such that the wall $\\overline{A_iB_i}$ is irrelevant. By the inductive assumption, the claim holds for all points except possibly $A_i$. However, by Claim 1, point $A_i$ is connected by a good path with at least one of the remaining points, which proves the claim for $n+1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20787,
"subject": "Mathematics (Olympiad)",
"question": "Suppose we have a large circle and wish to fit seven smaller circles of radius $r$ inside it, arranged in the standard way (one in the center, six around it). What is the minimal possible radius $R$ of the large circle so that all seven small circles fit inside without overlapping?",
"options": [],
"answer": "See solution",
"solution": "Assume, for contradiction, that the radius $R$ of the large circle is less than $3r$.\n\nLet $O$ be the center of the large circle, and $A_1, A_2, \\dots, A_6$ the centers of the six outer small circles. None of the $A_i$ can coincide with $O$, since then no other circle would fit. Each $A_i$ must be at least a distance $r$ from the boundary, so $OA_i < 2r$ for all $i$.\n\nAmong the six angles $A_1OA_2, A_2OA_3, \\dots, A_6OA_1$, at least one is at most $60^\\circ$. Suppose $\\angle A_1OA_2 \\leq 60^\\circ$. In triangle $A_1OA_2$, $A_1A_2$ cannot be the longest side, so $A_1A_2 \\leq \\max(OA_1, OA_2) < 2r$. But the centers of two tangent small circles must be $2r$ apart, so the circles at $A_1$ and $A_2$ would overlap—a contradiction. Thus, $R \\geq 3r$ is necessary.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20788,
"subject": "Mathematics (Olympiad)",
"question": "Given the functional equation $f(f(f(x))) = x$ for all $x$, and $f(x)$ is a linear function with integer coefficients, find the value of $1 + 2 + \\cdots + 2014$.",
"options": [],
"answer": "See solution",
"solution": "So we must have $a = -1$. Then $f(x) = -x + b$, and hence $f(f(x)) = -(-x + b) + b = x$. It follows that $f(f(f(x))) = f(f(x)) = x$ for all $x$, and thus the answer is\n\n$$\n1 + 2 + \\cdots + 2014 = \\frac{2014 \\times 2015}{2} = 2029105.\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20789,
"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**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.",
"options": [],
"answer": "See solution",
"solution": "and $\\angle QDB = \\angle QCA$. Hence triangles $ACQ$ and $BDQ$ are similar to each other. Because $AC = BD$, we conclude that triangle $ACQ$ and $BDQ$ are congruent to each other, implying that $QA = QB$ and $QC = QD$. Because $\\angle BQD = \\angle AQC$, we have $\\angle AQB = \\angle CQD$ and isosceles triangles $ABQ$ and $CDQ$ are similar to each other. In particular, $\\angle QBA = \\angle QAB = \\angle QCD = \\angle QDC$.\n\n\n\n**Solution 1.** Let $Q$ be the second intersection of $\\omega_1$ and $\\omega_2$. Because $PQ \\perp O_1O_2$, it suffices to show that $PQ \\perp MN$.\n\nBecause $ABPQ$ and $DCPQ$ are cyclic, it follows that $\\angle QPA = \\angle QBA = \\angle QCD = \\angle QPD$; that is, $PQ$ bisects $\\angle CPB$. Because $M$ and $N$ are midpoints of arcs $PS$ and $PT$, $CN$ and $BM$ are bisectors of $\\angle PCB$ and $\\angle PBC$, respectively. Therefore, lines $PQ, BM, CN$ are interior angle bisectors of triangle $PBC$, and then are concurrent at the incenter $I$ of triangle $PBC$.\n\nBy Power of a Point, we have $IN \\cdot IC = IP \\cdot IQ = IM \\cdot IB$, and consequently, $CNMB$ is cyclic by the converse of the theorem. Because $BCNM$ is cyclic, $\\angle MNI = \\angle MBC = \\angle IBC$. Because $\\angle INP$ is an exterior angle of triangle $CNP$, we have $\\angle INP = \\angle ICP + \\angle CPN$. Therefore, we have\n\n$$\n\\begin{align*}\n\\angle MNP + \\angle NPI &= \\angle MNI + \\angle INP + \\angle NPI = \\angle IBC + \\angle ICP + \\angle CPN + \\angle NPI \\\\\n&= \\angle IBC + \\angle ICP + \\angle CPI = \\frac{\\angle PCB + \\angle CPB + \\angle PBC}{2} = 90^{\\circ},\n\\end{align*}\n$$\n\nand so $PQ \\perp MN$, as desired.\n\n\n\n**Solution 2.** For point $X$ and line $\\ell$, let $d(X, \\ell)$ denote the distance from $X$ to $\\ell$. Because $O_1MNO_2$ is a convex quadrilateral by the note prior to Solution 1, it suffices to show that\n\n$$\nd(M, O_1O_2) = d(N, O_1O_2).\n$$\n\nWorking on arcs along $\\omega_1$, we have\n\n$$\n\\begin{align*}\n\\angle MO_1O_2 &= \\angle MO_1P + \\angle PO_1O_2 = \\widehat{MP} + \\frac{\\angle PO_1Q}{2} = \\frac{\\widehat{SP}}{2} + \\angle PBQ \\\\\n&= \\angle PBS + \\angle PBQ = \\angle SBQ = \\angle CBQ.\n\\end{align*}\n$$\n\nThus, we have\n\n$$\n\\frac{d(M, O_1O_2)}{O_1M} = \\sin \\angle MO_1O_2 = \\sin \\angle QBC = \\frac{d(Q, BC)}{BQ} \\quad \\text{or} \\quad \\frac{d(M, O_1O_2)}{d(Q, BC)} = \\frac{BQ}{O_1M}.\n$$\n\nIn exactly the same way, we can show that $\\angle NO_2O_1 = \\angle BCQ$ and\n\n$$\n\\frac{d(N, O_1O_2)}{d(Q, BC)} = \\frac{CQ}{O_2N}.\n$$\n\nIt suffices to show that\n\n$$\n\\frac{BQ}{O_1M} = \\frac{CQ}{O_2N},\n$$\n\nwhich holds because $BQ$ and $CQ$ are two corresponding sides of two similar (isosceles) triangles (namely, $BAQ$ and $CDQ$) inscribed in circles $\\omega_1$ and $\\omega_2$, with radii $O_1M$ and $O_2N$, respectively.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20790,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, \\dots, a_n$ ($n \\ge 3$) be real numbers. Prove that\n$$\n\\sum_{i=1}^{n} a_i^2 - \\sum_{i=1}^{n} a_i a_{i+1} \\le \\left[ \\frac{n}{2} \\right] (M-m)^2,\n$$\nwhere $a_{n+1} = a_1$, $M = \\max_{1 \\le i \\le n} a_i$, $m = \\min_{1 \\le i \\le n} a_i$. $[x]$ is the largest integer not exceeding $x$.",
"options": [],
"answer": "See solution",
"solution": "If $n = 2k$ (where $k$ is a positive integer), then\n$$\n2 \\left( \\sum_{i=1}^{n} a_i^2 - \\sum_{i=1}^{n} a_i a_{i+1} \\right) = \\sum_{i=1}^{n} (a_i - a_{i+1})^2 \\le n(M-m)^2,\n$$\ntherefore,\n$$\n\\sum_{i=1}^{n} a_i^2 - \\sum_{i=1}^{n} a_i a_{i+1} \\le \\frac{n}{2} (M-m)^2 = \\left[ \\frac{n}{2} \\right] (M-m)^2.\n$$\nIf $n = 2k + 1$ (where $k$ is a positive integer), then for $2k + 1$ numbers arranged cyclically, one can always find three consecutive increasing or decreasing terms, so it is not possible that for every $i$, $a_i - a_{i-1}$ and $a_{i+1} - a_i$ have opposite signs. Without loss of generality, assume $a_1, a_2, a_3$ are monotonic, then\n$$\n(a_1 - a_2)^2 + (a_2 - a_3)^2 \\le (a_1 - a_3)^2.\n$$\nHence,\n$$\n2 \\left( \\sum_{i=1}^{n} a_i^2 - \\sum_{i=1}^{n} a_i a_{i+1} \\right) = \\sum_{i=1}^{n} (a_i - a_{i+1})^2 \\\\ \\le (a_1 - a_3)^2 + \\sum_{i=3}^{n} (a_i - a_{i+1})^2,\n$$\nwhich reduces the question to the case of $2k$ numbers. We have\n$$\n2 \\left( \\sum_{i=1}^{n} a_i^2 - \\sum_{i=1}^{n} a_i a_{i+1} \\right) \\le (a_1 - a_3)^2 + \\sum_{i=3}^{n} (a_i - a_{i+1})^2 \\le 2k (M - m)^2,\n$$\ni.e.,\n$$\n\\sum_{i=1}^{n} a_i^2 - \\sum_{i=1}^{n} a_i a_{i+1} \\le k (M-m)^2 = \\left[ \\frac{n}{2} \\right] (M-m)^2.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20791,
"subject": "Mathematics (Olympiad)",
"question": "Around a round table, $n \\geq 3$ players are sitting. The game leader divides $n$ coins among the players, in such a way that not everyone gets exactly one coin. Any player can see the number of coins of each other player. Every 10 seconds, the game leader rings a bell. At that moment, each player looks at how many coins their two neighbours have. Then they all do the following at the same time:\n\n- If a player has more coins than at least one of their neighbours, the player gives away exactly one coin. They give this coin to the neighbour with the smallest number of coins. If both of their neighbours have the same number of coins, they give the coin to the neighbour on the left.\n- If a player does not have more coins than at least one of their neighbours, the player does nothing and waits for the next round.\n\nThe game ends if everyone has exactly one coin.\n\n(a) For each $n \\geq 3$, find a distribution of the coins at the start such that the game will never stop (and prove that the game does not stop for your starting distribution).\n\n(b) For each $n \\geq 4$, find a distribution of the coins at the start of the game such that the game will stop (and prove that the game stops for your starting distribution).",
"options": [],
"answer": "See solution",
"solution": "(a) Consider the situation where the first player has 2 coins, the second player has 0 coins, and all other players have 1 coin. This situation looks as follows:\n\n$$\n\\underbrace{2011\\cdots11}_{n-2 \\text{ ones}}\n$$\n\nFor example, for $n=3$ the starting distribution is 201. We see that the first and the third player both give a coin to the second player. This gives the distribution 120. This is exactly the distribution 201 if you shift all players by one place. We see that the game never stops. In this case, the first player has to give a coin to the second player, the third player has to give a coin to the left, and all other players keep their coin. We end with the following situation:\n\n$$\n\\underbrace{12011\\cdots11}_{n-3 \\text{ ones}}\n$$\n\nThis is exactly the same distribution as the starting distribution, except now it is player 2 that has 2 coins and player 3 that has 0 coins. If we continue playing, there will always be a player with 2 coins and thus the game never stops. $\\Box$\n\n(b) For $n \\geq 4$ we can consider the following starting distribution:\n\n$$\n2002\\underbrace{11\\cdots11}_{n-4 \\text{ ones}}\n$$\n\nFor example, for $n = 4$ the starting distribution is 2002. In this case, there is no player with exactly one coin. The first and the last player give a coin to the second and third player, respectively. Then the game stops.\n\nThe first player has to give a coin to the right and the fourth player has to give a coin to the left. All other players keep their coin. This gives a situation where all players have 1 coin, thus the game stops. $\\Box$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 20792,
"subject": "Mathematics (Olympiad)",
"question": "Suppose there are red, green, and blue candies, with one of each color weighing a total of 32 grams. There are equal numbers of each color in the pack, so the total weight of the candies is a multiple of 32.\n\nMary ate $k$ candies. What is the least possible value of $k$ such that the remaining candies' total weight is not a multiple of 32?",
"options": [],
"answer": "See solution",
"solution": "Suppose $k = 4$ is possible. For example, if there were 27 candies of each color, the total weight is $27 \\times 3 \\times 32 = 864$ grams. Mary could have eaten three blue candies and one red candy (total weight $3b + r = 77$ grams), leaving $864 - 77 = 787$ grams, which is not a multiple of 32.\n\nLet us show that $k$ cannot be less than 4:\n\n- If $k = 1$, the remaining weights would be 789, 792, or 812 grams, none divisible by 32.\n- If $k = 2$, possible remaining weights are 791, 797, 837, 794, 817, or 814 grams, none divisible by 32.\n- If $k = 3$, possible remaining weights are 793, 802, 862, 799, 839, 796, 842, 816, 822, or 819 grams, none divisible by 32.\n\nTherefore, the least number of candies Mary could have eaten is $k = 4$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20793,
"subject": "Mathematics (Olympiad)",
"question": "Given acute triangles $ACD$ and $BCD$, placed in different planes. Denote $G$ and $H$ as the barycenter and orthocenter of triangle $BCD$, respectively, and $G'$, $H'$ as the barycenter and orthocenter of triangle $ACD$, respectively. It is known that the straight line $HH'$ is perpendicular to the plane $(ACD)$.\n\nProve that the straight line $GG'$ is perpendicular to the plane $(BCD)$.",
"options": [],
"answer": "See solution",
"solution": "The relation $HH' \\perp (ACD)$ implies $HH' \\perp CD$, and since $CD \\perp BH$, we get $CD \\perp (BHH')$. Now, from $AH' \\perp CD$ it follows that $AH' \\subset (BHH')$, so the points $A, H, H', B$ are coplanar and $CD \\perp AB$. (1)\n\nOn the other hand, $AC \\perp DH'$ and $AC \\perp HH'$ yield $AC \\perp (DHH')$, so $DH \\perp AC$. This, combined with $DH \\perp BC$, leads to $DH \\perp (ABC)$; therefore, $AB \\perp DH$. (2)\n\n\n\nDenote $M$ as the midpoint of $[CD]$. Then $\\frac{MG'}{MA} = \\frac{MG}{MB} = \\frac{1}{3}$, hence $GG' \\parallel AB$. Finally, from (1) and (2) it follows that $AB \\perp (BCD)$, and $GG' \\parallel AB$ leads now to $GG' \\perp (BCD)$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20794,
"subject": "Mathematics (Olympiad)",
"question": "a) Prove that there are infinitely many positive integers $t$ such that both $2012t + 1$ and $2013t + 1$ are perfect squares.\n\nb) Suppose $m, n$ are positive integers such that both $mn + 1$ and $mn + n + 1$ are perfect squares. Prove that $8(2m + 1)$ divides $n$.",
"options": [],
"answer": "See solution",
"solution": "a) Let $d = \\gcd(2012t + 1, 2013t + 1)$. We can easily prove that $d = 1$. Therefore, $2012t + 1$ and $2013t + 1$ are both perfect squares if and only if $(2012t + 1)(2013t + 1) = y^2$ for some integer $y$. We can rewrite this equation as\n\n$$\n(2 \\cdot 2012 \\cdot 2013 t + 4025)^2 - 1 = 4 \\cdot 2012 \\cdot 2013 \\cdot y^2.\n$$\n\nLet $x = 2 \\cdot 2012 \\cdot 2013 t + 4025$, so we have\n\n$$\nx^2 - 4 \\cdot 2012 \\cdot 2013 \\cdot y^2 = 1.\n$$\n\nSince $4 \\cdot 2012 \\cdot 2013$ is not a perfect square, this Pell equation has infinitely many solutions. The smallest solution is $(x, y) = (4025, 1)$, and the solutions are given by:\n\n$$\n\\begin{cases}\nx_0 = 1,\\ x_1 = 4025,\\ x_{n+2} = 8050 x_{n+1} - x_n & (n \\ge 0) \\\\\ny_0 = 1,\\ y_1 = 1,\\ y_{n+2} = 8050 y_{n+1} - y_n & \\end{cases}\n$$\n\nBy induction, $x_{2i+1}$ has remainder $4025$ when divided by $2 \\cdot 2012 \\cdot 2013$ for every $i$, so $t = \\dfrac{x_{2i+1} - 4025}{2 \\cdot 2012 \\cdot 2013}$ will satisfy the problem. Thus, there are infinitely many positive integers $t$ such that $2012t + 1$ and $2013t + 1$ are both perfect squares.\n\nb) Let $d = \\gcd(mn + 1, mn + n + 1)$. We get $d \\mid (mn + n + 1 - mn - 1) = n$, which means $d \\mid n$. Also, $d \\mid (mn + 1 - mn) = 1$, so $d = 1$. Thus, $mn + 1$ and $(m + 1)n + 1$ are coprime. These are perfect squares if and only if $(mn + 1)[(m + 1)n + 1] = y^2$ for some integer $y$. Rewriting:\n\n$$\n[2m(m + 1)n + (2m + 1)]^2 - 1 = 4m(m + 1)y^2.\n$$\n\nLet $x = 2m(m + 1)n + (2m + 1)$, so\n\n$$\nx^2 - 4m(m + 1)y^2 = 1. \\quad (1)\n$$\n\nSince $4m(m + 1)$ is not a perfect square for all positive integers $m$, Pell's equation (1) has infinitely many solutions. The smallest solution is $(x, y) = (2m + 1, 1)$, and the formula for $(x_i, y_i)$ is\n\n$$\n\\begin{cases}\nx_0 = 1,\\ x_1 = 2m + 1,\\ x_{i+2} = 2(2m + 1)x_{i+1} - x_i & (i \\ge 0) \\\\\ny_0 = 0,\\ y_1 = 1,\\ y_{i+2} = 2(2m + 1)y_{i+1} - y_i & \\end{cases}\n$$\n\nBy induction, $x_{2i}$ has remainder $1$ when divided by $2m(m + 1)$ and $x_{2i+1}$ has remainder $2m + 1$ when divided by $2m(m + 1)$ for all natural numbers $i$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20795,
"subject": "Mathematics (Olympiad)",
"question": "Determine all twice differentiable functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that $$(f'(x))^2 + f''(x) \\le 0$$ for all $x \\in \\mathbb{R}$.",
"options": [],
"answer": "See solution",
"solution": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a twice differentiable function satisfying the given condition. Define $g(x) = e^{f(x)}$ for $x \\in \\mathbb{R}$. Then,\n$$g''(x) = e^{f(x)} \\left( (f'(x))^2 + f''(x) \\right) \\le 0$$\nfor all $x \\in \\mathbb{R}$. Thus, $g'$ is a nonincreasing function. Therefore, the limits $\\ell_1 = \\lim_{x \\to -\\infty} g'(x)$ and $\\ell_2 = \\lim_{x \\to \\infty} g'(x)$ exist and are finite.\n\nSince $g(x) > 0$ for all $x$, we have $\\ell_1 \\le 0$ and $\\ell_2 \\ge 0$. Because $g'$ is nonincreasing, it follows that $g'(x) = 0$ for all $x \\in \\mathbb{R}$. Therefore, $g$ is a constant positive function, so $f(x) = \\ln(g)$ is constant. Conversely, any constant function $f$ satisfies the original inequality.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20796,
"subject": "Mathematics (Olympiad)",
"question": "ABCD is a cyclic quadrilateral. The lines $AD$ and $BC$ meet at $X$, and the lines $AB$ and $CD$ meet at $Y$. The line joining the midpoints $M$ and $N$ of the diagonals $AC$ and $BD$, respectively, meets the internal bisector of angle $AXB$ at $P$ and the external bisector of angle $BYC$ at $Q$. Prove that $PXQY$ is a rectangle.",
"options": [],
"answer": "See solution",
"solution": "The interior angle bisectors of $AXB$ and $DYA$ are perpendicular. This can be shown by a simple angle chase. Let $\\alpha = \\angle DAB$, $\\beta = \\angle ABC$, $\\delta = \\angle CDA$ so that $\\beta + \\delta = \\pi$. Let $P'$ be the intersection of the interior angle bisectors of $AXB$ and $DYA$.\n\nThen we have that $\\angle AYD = \\pi - \\alpha - \\delta$, so $\\angle AYP' = \\frac{1}{2}(\\pi - \\alpha - \\delta)$. Similarly, $\\angle P'XA = \\frac{1}{2}(\\pi - \\alpha - \\beta)$. Thus, by summing interior angles around the concave quadrilateral $AYP'X$, we see\n\n$$\n\\alpha + \\frac{1}{2}(\\pi - \\alpha - \\delta) + (2\\pi - \\angle XP'Y) + \\frac{1}{2}(\\pi - \\alpha - \\beta) = 2\\pi\n$$\n\nThus $\\angle XP'Y = \\frac{1}{2}(\\beta + \\delta) = \\frac{\\pi}{2}$.\n\nLetting the interior and external bisectors of $AXB$ and $DYA$ meet at $P'$ and $Q'$, respectively, we will show that $P = P'$ and $Q = Q'$.\n\nLet $T_X$ represent the transformation given by reflection in $XP'$ followed by a suitable dilation about $X$ such that $T_X(A) = B$. Note that by power of a point about $X$, $T_X(C) = D$. Let $T_Y$ represent the transformation given by reflection in $YP'$ followed by a suitable dilation about $Y$ such that $T_Y(B) = C$, $T_Y(D) = A$.\n\nNote that the composition $T_Y \\circ T_X$ interchanges $A$ and $C$. As it is an orientation-preserving similarity transformation, it is rotation about $M$ by $\\pi$. Similarly, $T_X \\circ T_Y$ is rotation about $N$ by $\\pi$. Let $T_X, T_Y$ have scale factors $\\lambda, \\mu$ respectively. These are both positive, and the composition preserves lengths, so $\\lambda\\mu = 1$.\n\nNow choose $P'$ as an origin of Cartesian coordinates so that $X, Y$ have coordinates $(X, 0), (0, Y)$ respectively. The transforms are then represented by:\n\n$$\nT_X : (x, y) \\mapsto (X + \\lambda(x - X), -\\lambda y)\n$$\n\n$$\nT_Y : (x, y) \\mapsto (-\\mu x, Y + \\mu(y - Y))\n$$\n\nSo by finding $M, N$ as the fixed points of the compositions $T_Y \\circ T_X, T_X \\circ T_Y$ we see that they have coordinates $\\frac{1-\\mu}{2}(X, Y)$, $\\frac{1-\\lambda}{2}(X, Y)$. Now $P'$ has coordinates $(0, 0)$ and the exterior angle bisectors (perpendicular to the interior bisectors) have equations $x = X, y = Y$ so intersect at $Q' = (X, Y)$. Thus clearly $M, N, P', Q'$ are collinear.\n\nThus as $P'$ lies on both $MN$ and the interior angle bisector at $X$, $P' = P$. Similarly $Q' = Q$. We showed in the course of this proof that $P'XQ'Y$ is a rectangle, so we are done.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20797,
"subject": "Mathematics (Olympiad)",
"question": "Consider complex numbers $a$, $b$, and $c$ such that $a + b + c = 0$ and $|a| = |b| = |c| = 1$. Prove that for any complex number $z$ with $|z| \\le 1$, we have\n$$\n3 \\leq |z - a| + |z - b| + |z - c| \\leq 4.\n$$",
"options": [],
"answer": "See solution",
"solution": "Consider points $A$, $B$, $C$, and $M$ with complex coordinates $a$, $b$, $c$, and $z$, respectively. The triangle $ABC$ is equilateral and inscribed in the unit circle centered at the origin $O$ in the complex plane.\n\nTo prove the left inequality, we have successively:\n$$\n\\sum |z - a| = \\sum |\\bar{a}| |z - a| = \\sum |\\bar{a}z - \\bar{a}a| \\geq \\sum |\\bar{a}(az - 1)| = |z(\\sum \\bar{a}) - 3| = 3.\n$$\n\nFor the right inequality, consider a chord containing $M$ and denote by $P$, $Q$ its points of intersection with the unit circle. Let $p$ and $q$ be the complex coordinates of $P$ and $Q$. Let $\\alpha \\in [0, 1]$ be such that $m = \\alpha p + (1-\\alpha)q$. We get\n$$\n\\sum |z-a| = \\sum |\\alpha p + (1-\\alpha)q - a| \\leq \\alpha \\sum |p-a| + (1-\\alpha) \\sum |q-a|,\n$$\nso $\\sum |z-a| \\leq \\max \\{\\sum |p-a|, \\sum |q-a|\\}$.\n\nWe can suppose without loss of generality that $\\max \\{\\sum |p-a|, \\sum |q-a|\\} = \\sum |p-a|$ and that $P$ is on the arc from $A$ to $C$. By Ptolemy's inequality, $PA + PC = PB$, that is, $|p-a| + |p-c| = |p-b|$. Then $\\sum |z-a| \\leq \\sum |p-a| = 2|p-b| \\leq 4$, which concludes the proof.\n\n**Remark.** Equality holds in the left inequality when $z=0$, and in the right one when $z \\in \\{-a, -b, -c\\}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20798,
"subject": "Mathematics (Olympiad)",
"question": "Let $f, g: \\mathbb{R} \\to \\mathbb{R}$ be continuous, non-constant functions satisfying\n\n$$\nf(x-y) = f(x)f(y) + g(x)g(y)\n$$\n\nfor all $x, y \\in \\mathbb{R}$.\n\n(i) Show that for any $x, y \\in \\mathbb{R}$, we have $g(x+y) = f(x)g(y) + g(x)f(y)$.\n\n(ii) Find all pairs $f, g$ satisfying the conditions.",
"options": [],
"answer": "See solution",
"solution": "Let $f$ and $g$ be functions satisfying the functional equation\n\n$$\nf(x-y) = f(x)f(y) + g(x)g(y)\n$$\n\nTaking $x \\to y$ in (1), we get\n\n$$\nf(0) = f(x)^2 + g(x)^2.\n$$\n\nTaking $y \\to 0$ in (1), we get\n\n$$\nf(x)(1-f(0)) = g(x)g(0).\n$$\n\nCombining the above, since $g$ is non-constant, we have\n\n$$\nf(0) = 1 \\quad \\text{and} \\quad g(0) = 0.\n$$\n\nHence\n\n$$\nf(x)^2 + g(x)^2 = 1\n$$\n\nfor all $x \\in \\mathbb{R}$. Now we show that $f$ is even and $g$ is odd. Taking $(x, y) \\to (0, x)$ in the original equation, we get $f(-x) = f(x)$. Similarly, $g(-x) = -g(x)$.\n\nNow we prove (i).\n\nFrom the above, we have\n\n$$\nf(x+y) = f(x)f(y) - g(x)g(y).\n$$\n\nUsing this, we derive\n\n$$\ng(x+y) = f(x)g(y) + g(x)f(y).\n$$\n\n(ii) Let $h: \\mathbb{R} \\to \\{z \\in \\mathbb{C} : |z| = 1\\}$ be defined by $h(x) = f(x) + i g(x)$. Then\n\n$$\nh(x+y) = h(x)h(y).\n$$\n\nSince $h$ is continuous, there exists $c \\in \\mathbb{R}$ such that $h(x) = \\exp(icx)$. Thus,\n\n$$\n\\begin{cases}\nf(x) = \\cos(cx) \\\\\ng(x) = \\sin(cx)\n\\end{cases}\n$$\n\nfor some $c \\in \\mathbb{R}$. Conversely, any such pair solves the problem.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20799,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a quadrilateral and $P$ a point on segment $AB$. Let $X$, $Y$, $Z$ be points on $BC$, $CD$, $AD$ respectively such that $PX \\parallel AC$, $XY \\parallel BD$, $YZ \\parallel AC$. Show that $PZ \\parallel BD$.\n\n",
"options": [],
"answer": "See solution",
"solution": "*Proof.* From $PX \\parallel AC$, $XY \\parallel BD$, $YZ \\parallel AC$, we conclude that\n\n$$\n\\begin{aligned}\n\\frac{BP}{AB} &= \\frac{BX}{BC} \\\\\n\\frac{BX}{BC} &= \\frac{DY}{DC} \\\\\n\\frac{DY}{DC} &= \\frac{DZ}{AD}\n\\end{aligned}\n$$\n\nTherefore,\n$$\n\\frac{BP}{AB} = \\frac{DZ}{AD} \\implies PZ \\parallel BD\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20800,
"subject": "Mathematics (Olympiad)",
"question": "Determine all prime numbers $p$ such that the number\n\n$$\n\\binom{p}{1}^2 + \\binom{p}{2}^2 + \\dots + \\binom{p}{p-1}^2\n$$\n\nis divisible by $p^3$.",
"options": [],
"answer": "See solution",
"solution": "We start from the observation that for $k = 1, 2, \\dots, p$ we have\n\n$$\n\\binom{p-1}{k-1} \\equiv \\pm 1 \\pmod{p}.\n$$\n\nTo see this, note that\n\n$$\np-1 \\equiv -1 \\pmod{p},\n$$\n$$\np-2 \\equiv -2 \\pmod{p},\n$$\n$$\n\\vdots\n$$\n$$\nk \\equiv -(p-k) \\pmod{p},\n$$\n\nhence, multiplying,\n\n$$\n\\frac{(p-1)!}{(k-1)!} \\equiv \\pm (p-k)! \\pmod{p}\n$$\n\nand the congruence follows. This can be written in an equivalent form:\n\n$$\n\\frac{k}{p} \\binom{p}{k} \\equiv \\pm 1 \\pmod{p},\n$$\n\nwhich implies that\n\n$$\n\\binom{p}{k} = p \\cdot \\frac{a_k p \\pm 1}{k},\n$$\n\nfor some integer $a_k$. Therefore, if $p$ satisfies the given conditions, we have\n\n$$\np \\mid \\sum_{k=1}^{p-1} \\frac{(a_k p \\pm 1)^2}{k^2},\n$$\n\nor $p \\mid \\sum_{k=1}^{p-1} 1/k^2$, where by $1/m$ we understand the unique integer $1 \\le l \\le p-1$ satisfying $ml \\equiv 1 \\pmod{p}$. Now observe that $1/k^2 \\equiv (1/k)^2 \\pmod{p}$ and\n\n$$\n\\sum_{k=1}^{p-1} \\frac{1}{k^2} \\equiv \\sum_{k=1}^{p-1} k^2 = \\frac{p(p-1)(2p-1)}{6} \\pmod{p},\n$$\n\nwhich is divisible by $p$ if and only if $p \\ge 5$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20801,
"subject": "Mathematics (Olympiad)",
"question": "A circle, inscribed into a triangle $ABC$, touches its sides $AB$, $BC$, and $CA$ at points $N$, $P$, and $K$, respectively. A segment $BK$ intersects the inscribed circle a second time at point $L$.\n\nDefine points $T = AL \\cap NK$ and $Q = CL \\cap KP$. Prove that the straight lines $BK$, $NQ$, and $PT$ intersect at a single point.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let us denote the angles as shown in Fig. 26. For triangles $AKT$ and $ATN$, apply the Law of Sines:\n\n$$\n\\frac{NT}{TK} = \\frac{AN \\cdot \\sin \\alpha_1}{AK \\cdot \\sin \\alpha_2} = \\frac{\\sin \\alpha_1}{\\sin \\alpha_2}\n$$\n\nSimilarly,\n\n$$\n\\frac{NF}{FP} = \\frac{NB \\cdot \\sin \\beta_1}{PB \\cdot \\sin \\beta_2} = \\frac{\\sin \\beta_1}{\\sin \\beta_2}, \\quad \\frac{PQ}{QK} = \\frac{PC \\cdot \\sin \\gamma_1}{KC \\cdot \\sin \\gamma_2} = \\frac{\\sin \\gamma_1}{\\sin \\gamma_2}\n$$\n\nMultiplying these expressions gives:\n\n$$\n\\frac{NT \\cdot KQ \\cdot PF}{TK \\cdot QP \\cdot FN} = \\frac{\\sin \\alpha_1 \\cdot \\sin \\beta_2 \\cdot \\sin \\gamma_2}{\\sin \\alpha_2 \\cdot \\sin \\beta_1 \\cdot \\sin \\gamma_1} = 1\n$$\n\nSince $AL$, $BL$, and $CL$ intersect at a single point, Cheva's theorem in its trigonometric form is satisfied. Therefore, by Cheva's theorem in its standard form, $BK$, $NQ$, and $PT$ intersect at a single point, as required.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20802,
"subject": "Mathematics (Olympiad)",
"question": "Пусть $n = 10^{1000}$. Обозначим исходные числа (в порядке обхода) через $a_1, \\dots, a_n$; будем считать, что $a_{n+1} = a_1$. Положим $b_i = \\text{НОК}(a_i, a_{i+1})$. Могут ли числа $b_1, \\dots, b_n$ быть $n$ подряд идущими натуральными числами?",
"options": [],
"answer": "See solution",
"solution": "*Ответ.* Не могут.\n\nРассмотрим наибольшую степень двойки $2^m$, на которую делится хотя бы одно из чисел $a_i$. Заметим, что ни одно из чисел $b_1, \\dots, b_n$ не делится на $2^{m+1}$. Пусть для определённости $a_1 \\ge 2^m$; тогда $b_1 \\ge 2^m$ и $b_n \\ge 2^m$. Значит, $b_1 = 2^m x$ и $b_n = 2^m y$ при некоторых нечётных $x$ и $y$. Без ограничения общности можно считать, что $x < y$. Тогда, поскольку $b_1, \\dots, b_n$ образуют $n$ последовательных чисел, среди них должно быть и число $2^m(x+1)$ (так как $2^m x < 2^m(x+1) < 2^m y$). Но это число делится на $2^{m+1}$ (так как $x+1$ чётно), что невозможно. Противоречие.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20803,
"subject": "Mathematics (Olympiad)",
"question": "Let $P(x) \\in \\mathbb{Q}[x]$ be a polynomial with rational coefficients and degree $d \\ge 2$. Prove that there is no infinite sequence $a_0, a_1, \\dots$ of rational numbers such that $P(a_i) = a_{i-1} + i$ for all $i \\ge 1$.",
"options": [],
"answer": "See solution",
"solution": "Assume, for the sake of contradiction, that such a sequence exists.\n\nWe first show that all $a_i$ must be of the form $\\frac{k_i}{n}$ for some $k_i \\in \\mathbb{Z}$ and fixed $n$ depending only on $P$ and $a_1$.\n\nWrite $P(x) = \\frac{Q(x)}{N}$, where $Q(x) = \\sum_{j=0}^d b_j x^j$ is an integer polynomial, and $N$ is the least common multiple of the denominators of the coefficients.\n\n**Claim 1.** For any prime $p$ and $i \\in \\mathbb{N}$,\n$$\n\\nu_p(a_i) \\ge \\min(-\\nu_p(b_d), \\nu_p(a_1))\n$$\n*Proof.* Suppose $\\nu_p(a_i) < -\\nu_p(b_d)$. Then,\n$$\n\\nu_p(b_d a_i^d) = \\nu_p(b_d) + d\\nu_p(a_i) < j\\nu_p(a_i) \\text{ for } j < d\n$$\nso $\\nu_p(b_d a_i^d)$ dominates $Q(a_i)$, and $\\nu_p(P(a_i)) < \\nu_p(a_i) < 0$. Since $\\nu_p(i) \\ge 0$,\n$$\n\\nu_p(a_{i-1}) = \\nu_p(P(a_i) - i) = \\nu_p(P(a_i)) < \\nu_p(a_i)\n$$\nIterating, $\\nu_p(a_1) < \\nu_p(a_2) < \\dots < \\nu_p(a_i)$, so $\\nu_p(a_i) > \\nu_p(a_1)$, a contradiction. Thus, the claim holds.\n\n**Lemma 1.** There exists $N$ such that $N a_i$ is integral for all $i$.\n\n*Proof.* Follows from Claim 1.\n\n**Claim 2.** The sequence $a_i$ is unbounded.\n\n*Proof.* If $a_i$ were bounded, then $P(a_i)$ would be bounded, so $P(a_i) - a_{i-1} = i$ would be bounded, which is impossible. Thus, $a_i$ is unbounded.\n\nNow, we proceed by cases.\n\n**Case 1: $d > 2$**\n\nBy bounding arguments and the pigeonhole principle, we show that the number of distinct $a_i$ among the first $n$ terms is $O(n^{1/d})$, but the recurrence structure forces a contradiction for $d > 2$ (see detailed argument above).\n\n**Case 2: $d = 2$**\n\nWrite $P(x) = c(x - a)^2 + b$. By similar bounding and recurrence arguments, we show that the sequence cannot remain unbounded and satisfy the recurrence, leading to a contradiction.\n\n**Case 3: $d$ odd**\n\nFor large $|a_{i+1}|$, the difference $|a_{i+1} - a_i|$ becomes arbitrarily small, but since all $a_i$ are rational with bounded denominators, this forces $a_{i+1} = a_i$, which is impossible given the recurrence.\n\nTherefore, in all cases, no such infinite sequence exists.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20804,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $n$ and $m$ such that $n! + 2^{n-1} = 2^m$.",
"options": [],
"answer": "See solution",
"solution": "We first check that $n = 1, 2, 4$ are solutions among $n \\le 4$, with $m = 1, 2, 5$ respectively.\n\nNow assume $n \\ge 5$. By Legendre's formula, $v_2(n!) = n - s_2(n)$, where $s_2(n)$ is the number of nonzero digits in the binary representation of $n$. Thus $v_2(n!) \\le n - 1$.\n\nIf $v_2(n!) = a < n - 1$, letting $\\mathrm{odd}(x)$ denote the odd part of $x$, we have\n\n$$\n\\mathrm{odd}(n!) + 2^{n-1-a} = 2^{m-a}\n$$\n\nwhich cannot happen as $\\mathrm{odd}(n!)$ is odd and the other two terms are even. Hence $v_2(n!) = n - 1$ and we must have $s_2(n) = 1$, so $n$ is a power of $2$.\n\nLet $n = 2^a$ with $a \\ge 3$. We claim that $\\mathrm{odd}(n!) \\equiv 3 \\pmod 8$. To see this, pair $i$ with $n-i = 2^a - i$. If $v_2(i) < a-2$, then $\\mathrm{odd}(2^a-i) \\equiv -\\mathrm{odd}(i) \\pmod 8$, so\n\n$$\n\\mathrm{odd}(i) \\cdot \\mathrm{odd}(2^a - i) \\equiv -\\mathrm{odd}(i)^2 \\equiv -1 \\pmod 8\n$$\n\nand there are an even number of such pairs, giving a product of $1$. The remaining $i$ are $2^{a-2}, 2^{a-1}$, and $3 \\cdot 2^{a-2}$, whose odd parts multiply to $3 \\pmod 8$. Hence the total product is $3 \\pmod 8$ as desired.\n\nThen\n\n$$\nn! + 2^{n-1} = 2^m \\implies \\mathrm{odd}(n!) + 1 = 2^{m-n-1} \\implies 2^{m-n-1} \\equiv 4 \\pmod 8 \\implies 2^{m-n-1} \\le 4\n$$\n\nThus $\\mathrm{odd}(n!) \\le 4$, which leads to $n \\le 4$. So there are no other solutions.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20805,
"subject": "Mathematics (Olympiad)",
"question": "Determine all 2017-tuples of integers $n_1, n_2, \\dots, n_{2017}$ that are *29-good*, where:\n\n- A 5-tuple of integers is *k*-arrangeable if its elements can be labelled $a, b, c, d, e$ in some order such that $a - b + c - d + e = k$.\n- A 2017-tuple $n_1, n_2, \\dots, n_{2017}$ is *k*-good if every 5-tuple $n_i, n_{i+1}, n_{i+2}, n_{i+3}, n_{i+4}$ ($i = 1, 2, \\dots, 2017$, with subscripts modulo 2017) is *k*-arrangeable.",
"options": [],
"answer": "See solution",
"solution": "Let $n_1 = n_2 = \\dots = n_{2017} = 29$. This is a valid 2017-tuple.\n\nWe claim there are no others. All subscripts are taken modulo 2017.\n\nSuppose $n_1, n_2, \\dots, n_{2017}$ is 29-good. Define $m_i = n_i - 29$ for $i = 1, 2, \\dots, 2017$. Then $n_1, \\dots, n_{2017}$ is 29-good if and only if $m_1, \\dots, m_{2017}$ is 0-good.\n\nSuppose there exists a 0-good sequence $m_1, \\dots, m_{2017}$ not all zero, minimizing $|m_1| + |m_2| + \\dots + |m_{2017}|$.\n\nFor any 5-tuple, $a - b + c - d + e = 0$ implies $a + b + c + d + e = 2(b + d) \\equiv 0 \\pmod{2}$. Thus, for each $i$:\n$$\nm_i + m_{i+1} + m_{i+2} + m_{i+3} + m_{i+4} \\equiv 0 \\pmod{2}.\n$$\nReplacing $i$ with $i+1$ and subtracting, we get:\n$$\nm_i \\equiv m_{i+5} \\pmod{2}.\n$$\nSince $\\gcd(5, 2017) = 1$, all $m_i$ are congruent modulo 2. From above, all $m_i$ are even.\n\nConsider $\\frac{m_1}{2}, \\dots, \\frac{m_{2017}}{2}$, which is also 0-good, and\n$$\n0 < \\left|\\frac{m_1}{2}\\right| + \\dots + \\left|\\frac{m_{2017}}{2}\\right| < |m_1| + \\dots + |m_{2017}|.\n$$\nThis contradicts minimality unless all $m_i = 0$.\n\nThus, the only 29-good 2017-tuple is $n_1 = n_2 = \\dots = n_{2017} = 29$. $\\square$",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 20806,
"subject": "Mathematics (Olympiad)",
"question": "Let $n = p_2 + p_3 - 1$. Ana and Banana play the following game: Ana selects $n$ integers $x_1, \\dots, x_n$, and Banana writes down $y_1, \\dots, y_n$ in order, where each $y_i$ is determined by a function $f$ with period $p$ (a prime). Ana's goal is to reconstruct $p$ from the sequence $y_1, \\dots, y_n$. Show that Ana has a winning strategy with the selection $x_i = p_1(i-1)$, where $p_1, p_2, p_3$ are distinct primes.",
"options": [],
"answer": "See solution",
"solution": "We claim Ana can always reconstruct $p$ with $x_i = p_1(i-1)$.\n\n**Case 1:** If $y_1 = \\dots = y_n$, then $f$ is constant, so $p = p_1$.\n\n**Case 2:** If $y_1, \\dots, y_n$ are not all equal, $p \\ne p_1$. Suppose for contradiction there exist two primes $q, r < p_1$ and nonconstant functions $f_q, f_r$ with periods $q$ and $r$ such that $f_q(x_i) = f_r(x_i) = y_i$ for all $i$.\n\nConsider the graph $G$ on vertices $\\{0, \\dots, q + r - 2\\}$ with edges $i \\sim j$ if $|i - j| \\in \\{q, r\\}$. $G$ has $q + r - 2$ edges. If $G$ had a cycle $c_1, \\dots, c_k$ ($k \\ge 3$), analyzing consecutive differences shows that either all are in $\\{q, -r\\}$ or $\\{-q, r\\}$. Let $a$ be the number of $q$'s and $b$ the number of $-r$'s; then $a + b = k$ and $qa - rb = 0$. This forces $a \\ge r$ and $b \\ge q$, so $k \\ge q + r$, contradicting the fact that $G$ has only $q + r - 1$ vertices. Thus, $G$ is connected, and the functions must be constant, which is a contradiction.\n\nTherefore, Ana can always reconstruct $p$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20807,
"subject": "Mathematics (Olympiad)",
"question": "As seen in Fig. 1.1, $AB$ is a chord of circle $\\omega$, $P$ is a point on arc $AB$, and $E, F$ are two points on $AB$ satisfying $AE = EF = FB$. Connect $PE$ and $PF$ and extend them to intersect with $\\omega$ at $C$ and $D$, respectively. Prove\n\n$$\nEF \\cdot CD = AC \\cdot BD.\n$$\n\n\n\n",
"options": [],
"answer": "See solution",
"solution": "**Solution.** As shown in Fig. 1.2, we connect $AD$, $BC$, $CF$, and $DE$. Since $AE = EF = FB$, we have\n\n$$\n\\frac{BC \\cdot \\sin \\angle BCE}{AC \\cdot \\sin \\angle ACE} = \\frac{\\text{distance between } B \\text{ and } CP}{\\text{distance between } A \\text{ and } CP} = \\frac{BE}{AE} = 2. \\qquad \\textcircled{1}\n$$\n\nIn the same way,\n\n$$\n\\frac{AD \\cdot \\sin \\angle ADF}{BD \\cdot \\sin \\angle BDF} = \\frac{\\text{distance between } A \\text{ and } PD}{\\text{distance between } B \\text{ and } PD} = \\frac{AF}{BF} = 2. \\qquad \\textcircled{2}\n$$\n\nOn the other hand, since\n\n$$\n\\angle BCE = \\angle BCP = \\angle BDP = \\angle BDF,\n$$\n\n$$\n\\angle ACE = \\angle ACP = \\angle ADP = \\angle ADF,\n$$\n\nmultiplying $\\textcircled{1}$ by $\\textcircled{2}$, we have $\\frac{BC \\cdot AD}{AC \\cdot BD} = 4$, or\n\n$$\nBC \\cdot AD = 4AC \\cdot BD. \\qquad \\textcircled{3}\n$$\n\nBy Ptolemy's Theorem, we have\n\n$$\nAD \\cdot BC = AC \\cdot BD + AB \\cdot CD. \\qquad \\textcircled{4}\n$$\n\nCombining $\\textcircled{3}$ and $\\textcircled{4}$, we get $AB \\cdot CD = 3AC \\cdot BD$, and that is\n\n$$\nEF \\cdot CD = AC \\cdot BD.\n$$\n\nThe proof is complete.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20808,
"subject": "Mathematics (Olympiad)",
"question": "令 $R^+$ 表示所有正實數所成的集合。給定正整數 $n \\ge 3$。試找出所有函數 $f: R^+ \\to R^+$ 使得對任意 $n$ 個正實數 $a_1, \\cdots, a_n$，都滿足\n\n$$\n\\sum_{i=1}^{n} (a_i - a_{i+1}) f(a_i + a_{i+1}) = 0,\n$$\n\n其中 $a_{n+1} = a_1$。",
"options": [],
"answer": "See solution",
"solution": "首先，在原式中令 $a_4 = a_5 = \\cdots = a_n = a_1$，即得\n\n$$\n\\sum_{i=1}^{3} (a_i - a_{i+1}) f(a_i + a_{i+1}) = 0, \\quad (1)\n$$\n\n其中 $a_4 = a_1$。\n\n接著證明：若 $x, y$ 為相異正實數，且\n\n$$\nm = \\frac{f(x) - f(y)}{x - y}, \\quad l = \\frac{x f(y) - y f(x)}{x - y},\n$$\n\n那麼對於任一個滿足 $|x - y| < z < x + y$ 的 $z$ 都有\n\n$$\nf(z) = m z + l \\quad (2)\n$$\n\n事實上由 $|x - y| < z < x + y$ 知：存在正實數 $a_1, a_2, a_3$ 滿足 $a_1 + a_2 = x,\\ a_2 + a_3 = y,\\ a_3 + a_1 = z$。代入 (1) 式得到\n\n$$\n\\begin{aligned}\n0 &= \\sum_{i=1}^{3} (a_i - a_{i+1}) f(a_i + a_{i+1}) \\\\\n &= (a_1 - a_2)(m x + l) + (a_2 - a_3)(m y + l) + (a_3 - a_1) f(z) \\\\\n &= (a_1 - a_2)(m(a_1 + a_2) + l) + (a_2 - a_3)(m(a_2 + a_3) + l) + (a_3 - a_1) f(z) \\\\\n &= m(a_1^2 - a_3^2) + l(a_1 - a_3) + (a_3 - a_1) f(z) \\\\\n &= (a_1 - a_3)(m(a_1 + a_3) + l - f(z))\n\\end{aligned}\n$$\n\n又 $x \\neq y$，所以 $a_1 - a_3 \\neq 0$，因此 $f(z) = m(a_1 + a_3) + l = m z + l$，得證。\n\n由 (2) 式立即得到若 $x, y$ 為相異正實數，則 $t > |x - y|$ 時，\n\n$(t, f(t))$ 在通過 $(x, f(x)), (y, f(y))$ 兩點的直線上。 (3)\n\n若 $t \\le |x - y|$，不妨設 $x > y$。則由 (3) 式知：$(x, f(x))$ 在通過 $(y, f(y)), (t, f(t))$ 兩點的直線上，所以 $(t, f(t))$ 仍在通過 $(x, f(x)), (y, f(y))$ 兩點的直線上。\n\n由此得到 $f$ 為線型函數，也就是存在 $a, b$ 使得 $f(t) = a t + b$，對所有 $t \\in R^+$。但 $f$ 是正實數到正實數的函數，所以 $a, b > 0$。\n\n代回原式易檢驗 $f(t) = a t + b$，其中 $a, b > 0$，是原方程的解，因此 $f(t) = a t + b$，其中 $a, b > 0$，是方程的所有解。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20809,
"subject": "Mathematics (Olympiad)",
"question": "On an infinite sheet of grid paper, 999 grid squares are colored black. Call *special* a rectangle with sides on grid lines if it has two black opposite corner cells (rectangles with side 1 are included). Let $N$ be the maximum number of black cells in a special rectangle for a given configuration. Find the minimum of $N$ over all configurations.",
"options": [],
"answer": "See solution",
"solution": "The minimum of $N$ is 201.\n\nIn an arbitrary configuration, take the minimal rectangle $R$ that contains all black cells. Choose a black cell on every side (there is at least one by minimality) and label these $A, B, C, D$ as in the first figure. Consider the special rectangles $[AB], [BC], [CD], [DA]$, and $[AC]$. Here, $[XY]$ denotes the grid rectangle with opposite corner cells $X$ and $Y$ (they determine the rectangle uniquely).\n\nNote that if the part of $R$ to the left of rectangle $[AC]$ is ignored, the remainder is covered by $[AB], [BC]$, and $[AC]$. Likewise, if the part of $R$ to the right of $[AC]$ is ignored, the remainder is covered by $[CD], [DA]$, and $[AC]$. Hence, the entire $R$ is covered by the five rectangles, with certain overlaps. Each of cells $A$ and $C$ belongs to 3 of the 5 rectangles; each of $B$ and $D$ belongs to 2 of them. (There may be other black cells contained in more than one rectangle.)\n\nHence, if $x_1, \\dots, x_5$ are the numbers of black cells in the 5 rectangles, we obtain\n$$\nx_1 + \\dots + x_5 \\ge 995 + 3 \\cdot 2 + 2 \\cdot 2 = 1005.\n$$\nThis is because the sum counts each black cell except $A, B, C, D$ at least once, each of $B$ and $D$ at least twice, and each of $A$ and $C$ at least thrice. It follows that one of $x_1, \\dots, x_5$ is at least $\\frac{1005}{5} = 201$. So there is always a special rectangle with at least 201 black cells. Therefore $N \\ge 201$ for every configuration.\n\nWe tacitly assumed that $A, B, C, D$ are distinct. This can be ensured unless there are no black cells on some two adjacent sides of $R$ except their common corner cell. But then it is clear that $R$ can be covered by at most 3 special rectangles like the ones above, so $N > 201$. (We ignore the trivial case where $R$ has a side 1.)\n\nNow we show an example where $N = 201$. In the second figure, the 999 black squares are inside a big $3k \\times 3k$ grid square, with $k > 400$. Four groups 1, 2, 3, 4 of 200 black cells each are placed in the corner $k \\times k$ squares as shown, in a diagonal-like manner.\n\nNote that adjacent corner groups are separated by a horizontal line and also by a vertical line. The remaining 199 black cells are in the central $k \\times k$ square. Their exact location is irrelevant.\n\n\n\n\nLet $X$ and $Y$ be opposite black corner cells of a special rectangle $T$ which contains $m$ black cells. If $X$ and $Y$ are in the same group, it is clear that $m \\le 200$. If $X$ and $Y$ are in adjacent groups among 1, 2, 3, 4, then $T$ intersects only these two groups. In addition, observe that one of the two groups has exactly one cell in $T$ (this is $X$ or $Y$). Thus $m \\le 200 + 1 = 201$.\n\nLet $X$ and $Y$ be in opposite corner groups, say 1 and 3. Then $T$ does not intersect groups 2 and 4. Moreover, $X$ and $Y$ are the only black cells of $T$ from groups 1 and 3. The remaining ones are all in the central part. Hence $m \\le 199 + 2 = 201$. Finally, let one of $X$ and $Y$ be in the central group, say $X$. If $Y$ is also there, then $m \\le 199$. And if $Y$ is in a corner group, then $Y$ is the only black cell of $T$ out of the central part. So $m \\le 199 + 1 = 200$, which completes the solution.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20810,
"subject": "Mathematics (Olympiad)",
"question": "Given $n$ as a positive integer, find all points $(x_1, x_2, \\dots, x_n) \\in \\mathbb{R}^n$ that minimize the function $f: \\mathbb{R}^n \\to \\mathbb{R}$,\n\n$$\nf(x_1, x_2, \\dots, x_n) = \\sum_{k_1=0}^{2} \\sum_{k_2=0}^{2} \\dots \\sum_{k_n=0}^{2} \\left| k_1 x_1 + k_2 x_2 + \\dots + k_n x_n - 1 \\right|.\n$$",
"options": [],
"answer": "See solution",
"solution": "$f$ is minimized when $x_1 = \\dots = x_n = \\frac{1}{n+1}$.\n\nPartition $A = \\{0, 1, 2\\}^n$ into $2n+1$ subsets, $A = A_0 \\cup A_1 \\cup \\dots \\cup A_{2n}$ where\n\n$$A_k = \\{\\beta = (i_1, i_2, \\dots, i_n) \\in A : i_1 + i_2 + \\dots + i_n = k\\}, \\quad k = 0, 1, \\dots, 2n.$$ \n\nLet $A_k$ have $|A_k| = a_k$ elements. Clearly,\n\n$$(1 + t + t^2)^n = a_0 + a_1 t + \\dots + a_n t^n + \\dots + a_{2n} t^{2n}, \\quad a_{2n-k} = a_k.$$ \n\nDenote $X = (x_1, \\dots, x_n)$, $y = (x_1 + \\dots + x_n)/n$. Then\n\n$$f(X) = \\sum_{k=0}^{2n} \\sum_{\\beta \\in A_k} |\\beta \\cdot X - 1|.$$ \n\nFirst, remove the absolute value signs and sum by partition, where\n\n$$B_k = \\sum_{\\beta \\in A_k} (\\beta \\cdot X - 1) = \\frac{k|A_k|}{n} (x_1 + \\dots + x_n) - |A_k| = k a_k y - a_k.$$ \n\nWe estimate $\\sum_{k=0}^{2n} |B_k|$ by canceling positive and negative terms. To identify large and small $k a_k$s, observe that\n\n$$\\sum_{k=0}^{2n} k a_k t^{k-1} = \\frac{d}{dt}[(1+t+t^2)^n] = n(1+2t)(1+t+t^2)^{n-1}$$\n\nand in\n\n$$(1 + t + t^2)^{n-1} = \\sum_{k=0}^{2n-2} c_k t^k$$\n\nthe coefficients $(c_0, c_1, \\dots, c_{n-1}, \\dots, c_{2n-2})$ are unimodal and symmetric.\n\nFrom\n\n$$k a_k = \\frac{c_{k-1} + 2c_{k-2}}{n}$$\n\nwe have\n\n$$\n\\begin{aligned}\n(n+1)a_{n+1} &\\ge n a_n \\ge (n+2)a_{n+2} \\ge (n-1)a_{n-1} \\\\\n&\\ge (n+3)a_{n+3} \\ge \\dots \\ge 2a_2 \\ge 2n a_{2n} \\ge 1a_1,\n\\end{aligned}\n$$\n\nwhich means $U_1 := \\sum_{k=0}^{n} k a_k < \\sum_{k=n+1}^{2n} k a_k$, $U_2 := \\sum_{k=n+2}^{2n} k a_k < \\sum_{k=0}^{n+1} k a_k$.\n\nDefine $\\lambda = \\frac{U_1 - U_2}{(n+1)a_{n+1}} \\in (-1, 1)$. Then\n\n$$f(X) \\ge \\sum_{k=0}^{2n} |B_k| \\ge \\left[ \\sum_{k=0}^{n} (-B_k) + \\lambda B_{n+1} + \\sum_{k=n+2}^{2n} B_k \\right] + (1 - |\\lambda|) |B_{n+1}|.$$ \n\nIn the square brackets, the coefficient of $y$ equals $-U_1 + \\lambda(n+1)a_{n+1} + U_2 = 0$; hence it is a constant\n\n$$C = \\sum_{k=0}^{n} a_k - \\lambda a_{n+1} + \\sum_{k=n+2}^{2n} a_k.$$ \n\nWe infer that $f(X) \\ge C$ always holds, with equality if and only if $x_1 = \\dots = x_n = \\frac{1}{n+1}$. In the meantime, the $a_{n+1}$ terms in the sum of $B_{n+1}$ must all vanish, that is, the inner product of $X = (x_1, \\dots, x_n)$ and any vector in $A_{n+1}$ equals $1$, so $x_1 = \\dots = x_n = \\frac{1}{n+1}$. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20811,
"subject": "Mathematics (Olympiad)",
"question": "In triangle $ABC$, $AB = AC$, and $D$ is on $BC$. A point $E$ is chosen on $AC$, and a point $F$ is chosen on $AB$, such that $DE = DC$ and $DF = DB$. It is given that $\\frac{DC}{BD} = 2$ and $\\frac{AF}{AE} = 5$. Determine the value of $\\frac{AB}{BC}$.",
"options": [],
"answer": "See solution",
"solution": "Note that triangles $ABC$ and $DFB$ are both isosceles, and they share the angle at $B$. Therefore, they must be similar. Likewise, triangle $DCE$ is similar to these two. Let $BF = x$ and $BD = FD = y$. Then we have $DC = 2BD = 2y$, and since $DFB$ and $DCE$ are similar, it also follows that $CE = 2x$.\n\nNext, we let $AB = AC = a$, so $AF = AB - BF = a - x$ and $AE = AC - CE = a - 2x$. We are given that $AF = 5AE$, so\n\n$$\na - x = 5(a - 2x)\n$$\nwhich is equivalent to $4a = 9x$, i.e. $a = \\frac{9x}{4}$.\n\nFinally, since $ABC$ and $DFB$ are similar, we have\n\n$$\n\\frac{\\frac{9x}{4}}{3y} = \\frac{AB}{BC} = \\frac{BD}{BF} = \\frac{y}{x}.\n$$\n\nThus\n\n$$\n\\left(\\frac{y}{x}\\right)^2 = \\frac{3}{4},\n$$\n\nand finally\n\n$$\n\\frac{AB}{BC} = \\frac{y}{x} = \\frac{\\sqrt{3}}{2}.\n$$\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20812,
"subject": "Mathematics (Olympiad)",
"question": "Find all integers $n$ satisfying $n \\geq 2$ and $\\frac{\\sigma(n)}{p(n)-1} = n$, where $\\sigma(n)$ denotes the sum of all positive divisors of $n$, and $p(n)$ denotes the largest prime divisor of $n$.",
"options": [],
"answer": "See solution",
"solution": "Let $n = p_1^{\\alpha_1} \\cdots p_k^{\\alpha_k}$ be the prime factorization of $n$ with $p_1 < \\dots < p_k$, so that $p(n) = p_k$ and $\\sigma(n) = (1 + p_1 + \\cdots + p_1^{\\alpha_1}) \\cdots (1 + p_k + \\cdots + p_k^{\\alpha_k})$. Hence\n\n$$\np_k - 1 = \\frac{\\sigma(n)}{n} = \\prod_{i=1}^{k} \\left(1 + \\frac{1}{p_i} + \\cdots + \\frac{1}{p_i^{\\alpha_i}}\\right) < \\prod_{i=1}^{k} \\frac{1}{1 - \\frac{1}{p_i}} = \\prod_{i=1}^{k} \\left(1 + \\frac{1}{p_i - 1}\\right) \\leq \\prod_{i=1}^{k} \\left(1 + \\frac{1}{i}\\right) = k + 1,\n$$\n\nthat is, $p_k - 1 < k + 1$, which is impossible for $k \\geq 3$, because in this case $p_k - 1 \\geq 2k - 2 \\geq k + 1$. Then $k \\leq 2$ and $p_k < k + 2 \\leq 4$, which implies $p_k \\leq 3$.\n\nIf $k = 1$ then $n = p^\\alpha$ and $\\sigma(n) = 1 + p + \\cdots + p^\\alpha$, and in this case $n \\nmid \\sigma(n)$, which is not possible. Thus $k = 2$, and $n = 2^\\alpha 3^\\beta$ with $\\alpha, \\beta > 0$. If $\\alpha > 1$ or $\\beta > 1$,\n\n$$\n\\frac{\\sigma(n)}{n} > \\left(1 + \\frac{1}{2}\\right) \\left(1 + \\frac{1}{3}\\right) = 2.\n$$\n\nTherefore $\\alpha = \\beta = 1$ and the only answer is $n = 6$.\n\n*Comment:* There are other ways to deal with the case $n = 2^\\alpha 3^\\beta$. For instance, we have $2^{\\alpha+2}3^\\beta = (2^{\\alpha+1}-1)(3^{\\beta+1}-1)$. Since $2^{\\alpha+1}-1$ is not divisible by 2, and $3^{\\beta+1}-1$ is not divisible by 3, we have\n\n$$\n\\begin{cases} 2^{\\alpha+1} - 1 = 3^{\\beta} \\\\ 3^{\\beta+1} - 1 = 2^{\\alpha+2} \\end{cases} \\iff \\begin{cases} 2^{\\alpha+1} - 1 = 3^{\\beta} \\\\ 3 \\cdot (2^{\\alpha+1} - 1) - 1 = 2 \\cdot 2^{\\alpha+1} \\end{cases} \\iff \\begin{cases} 2^{\\alpha+1} = 4 \\\\ 3^{\\beta} = 3 \\end{cases},\n$$\n\nand $n = 2^\\alpha 3^\\beta = 6$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20813,
"subject": "Mathematics (Olympiad)",
"question": "Let $M(a, 1/a)$ be a point on the upper-right branch of the hyperbola $H_1: y = 1/x$. Let $M_1(b_1, -1/b_1)$ and $M_2(b_2, -1/b_2)$ be the points of tangency with the hyperbola $H_2: y = -1/x$ as described (see the figure below).\n\n\n\nShow that the line $M_1 M_2$ is tangent to $H_1$ at the point $N(-a, -1/a)$.",
"options": [],
"answer": "See solution",
"solution": "The derivative of $y(x) = -1/x$ is $y'(x) = \\frac{1}{x^2}$, so the tangent to $H_2$ at $(b, -1/b)$ is\n\n$$\ny = \\frac{1}{b^2}(x - b) - \\frac{1}{b}.\n$$\n\nSince $M(a, 1/a)$ lies on this tangent,\n\n$$\n\\frac{1}{a} = \\frac{1}{b^2}(a - b) - \\frac{1}{b}.\n$$\n\nThis leads to the quadratic equation for $b$:\n\n$$\nb^2 + 2ab - a^2 = 0.\n$$\n\nLet $b_1$ and $b_2$ be the roots. The line through $M_1$ and $M_2$ has equation\n\n$$\ny = \\frac{1}{b_1 b_2}(x - b_1) - \\frac{1}{b_1} = \\frac{x - (b_1 + b_2)}{b_1 b_2}.\n$$\n\nBy Vieta's formulas, $b_1 b_2 = -a^2$ and $b_1 + b_2 = -2a$, so\n\n$$\ny = \\frac{x + 2a}{-a^2}.\n$$\n\nFor $x = -a$, $y = -1/a$, so $N(-a, -1/a)$ lies on this line. The slope of the tangent to $H_1$ at $N$ is $y'(-a) = -\\frac{1}{(-a)^2} = -\\frac{1}{a^2}$, which matches the slope of $M_1 M_2$. Thus, $M_1 M_2$ is tangent to $H_1$ at $N$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20814,
"subject": "Mathematics (Olympiad)",
"question": "Given any finite string made up of $A$s and $B$s, an occurrence of $AB$ may be replaced with $BBAA$. Starting from any finite string of $A$s and $B$s, is it always possible to perform a number of such replacements so that all $B$s appear to the left of all $A$s?",
"options": [],
"answer": "See solution",
"solution": "No, it's not always possible to move all $B$s to the left of all $A$s.\n\nWe will prove that if a string contains at least one of $AAB$ or $ABB$, then it will always contain at least one of $AAB$ or $ABB$.\n\nIf a string has $AAB$, then either a move doesn't affect this, or it changes $AAB$ to $ABBAAB$, which contains $ABB$. Similarly, if a string has $ABB$, then either a move doesn't affect this, or it changes $ABB$ to $BBAAB$, which contains $AAB$. This completes the proof.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 20815,
"subject": "Mathematics (Olympiad)",
"question": "設一個三角形的三邊長分別為 $a, b, c$，而 $a, b, c$ 三邊所對應的高分別為 $h_a, h_b, h_c$。\n\n證明\n$$\n\\left(\\frac{a}{h_a}\\right)^2 + \\left(\\frac{b}{h_b}\\right)^2 + \\left(\\frac{c}{h_c}\\right)^2 \\ge 4.\n$$\n\nIn a triangle, let $a, b, c$ be the lengths of sides, and $h_a, h_b, h_c$ be the lengths of corresponding heights. Prove that\n$$\n\\left(\\frac{a}{h_a}\\right)^2 + \\left(\\frac{b}{h_b}\\right)^2 + \\left(\\frac{c}{h_c}\\right)^2 \\ge 4.\n$$",
"options": [],
"answer": "See solution",
"solution": "將三角形的面積記為 $A$。易知 $a h_a = b h_b = c h_c = 2A$。故原命題等價於證明\n\n$$\na^4 + b^4 + c^4 \\ge 16A^2. \\quad (1)\n$$\n\n將 (1) 式的左邊減去右邊，並由 Heron 公式得\n\n$$\n\\begin{aligned}\n& a^4 + b^4 + c^4 - 16A^2 \\\\\n=\\ & a^4 + b^4 + c^4 - (a+b+c)(a+b-c)(a-b+c)(-a+b+c) \\\\\n=\\ & 2a^4 + 2b^4 + 2c^4 - 2a^2b^2 - 2a^2c^2 - 2b^2c^2 \\\\\n=\\ & (a^2 - b^2)^2 + (b^2 - c^2)^2 + (c^2 - a^2)^2 \\ge 0.\n\\end{aligned}\n$$\n\n故 (1) 式及原命題得證。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20816,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge 3$ 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, such that $x - y + z - c$ is divisible by $p$.",
"options": [],
"answer": "See solution",
"solution": "Let $p$ be an arbitrary prime. 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*. Our solution is based on the following lemmas.\n\n**Lemma 1.** The set $S$ of $n \\ge 3$ 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 in $S$. Choose $c = i + j - k > 0$. For any triple $(x, y, z) \\in S^3$ with distinct elements,\n\n$$\n0 \\ge x - y + z - c > -\\frac{p}{3} - c > -\\frac{p}{3} - \\frac{2p}{3} = -p.\n$$\n\nHence,\n\n$$\np \\mid x - y + z - c \\iff x - y + z = c \\iff \\{x, z\\} = \\{i, j\\} \\text{ and } y = k.\n$$\n\n**Lemma 2.** If $p > 5 \\cdot 6^{n-2}$, for any set $S$ of $n \\ge 3$ 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 any integer $b_0$ so 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 $x_i \\ne 0$. 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 $[a x] \\in \\left[0, \\frac{p}{6}\\right) \\cup \\left(\\frac{5p}{6}, p\\right)$ for all $x \\in S$.\n\nIt's easy to verify that if $b = \\left\\lfloor \\frac{p}{6} \\right\\rfloor$ then $aS + b \\subset \\left[0, \\frac{p}{3}\\right]$.\n\nSince $p > 6^{n-1} - 2^n + 1$, then $p \\ge 6^{n-1} - 2^n + 3 > 5 \\cdot 6^{n-2}$, which holds for all integers $n \\ge 3$. Hence, our proof is complete. $\\square$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20817,
"subject": "Mathematics (Olympiad)",
"question": "For nonnegative real numbers $a$, $b$, $c$ with $a + b + c = 1$, prove that\n\n$$\n\\sqrt{a + \\frac{(b-c)^2}{4}} + \\sqrt{b} + \\sqrt{c} \\le \\sqrt{3}.\n$$",
"options": [],
"answer": "See solution",
"solution": "**Proof I**\n\nWithout loss of generality, assume $b \\ge c$.\n\nSet $\\sqrt{b} = x + y$ and $\\sqrt{c} = x - y$ for some nonnegative real numbers $x$ and $y$. Then $b - c = 4xy$ and $a = 1 - 2x^2 - 2y^2$. Thus,\n\n$$\n\\sqrt{a + \\frac{(b-c)^2}{4}} + \\sqrt{b} + \\sqrt{c} = \\sqrt{1 - 2x^2 - 2y^2 + 4x^2y^2} + 2x.\n$$\n\nNote that $2x = \\sqrt{b} + \\sqrt{c}$, so\n\n$$\n4x^2 = (\\sqrt{b} + \\sqrt{c})^2 \\le 2b + 2c \\le 2,\n$$\n\nby the AM-GM inequality. Thus, $4x^2y^2 \\le 2y^2$ and\n\n$$\n1 - 2x^2 - 2y^2 + 4x^2y^2 \\le 1 - 2x^2.\n$$\n\nSubstituting this into the previous expression gives\n\n$$\n\\begin{aligned}\n\\sqrt{a + \\frac{(b-c)^2}{4}} + \\sqrt{b} + \\sqrt{c} &\\le \\sqrt{1-2x^2} + 2x \\\\\n&= \\sqrt{1-2x^2} + x + x \\\\\n&\\le \\sqrt{3},\n\\end{aligned}\n$$\n\nby the Cauchy-Schwarz inequality.\n\n**Proof II**\n\nLet $a = u^2$, $b = v^2$, $c = w^2$. Then $u^2 + v^2 + w^2 = 1$ and the inequality becomes\n\n$$\n\\sqrt{u^2 + \\frac{(v^2 - w^2)^2}{4}} + v + w \\le \\sqrt{3}.\n$$\n\nNote that\n\n$$\n\\begin{aligned}\nu^2 + \\frac{(v^2 - w^2)^2}{4} &= 1 - (v^2 + w^2) + \\frac{(v^2 - w^2)^2}{4} \\\\\n&= \\frac{4 - 4(v^2 + w^2) + (v^2 - w^2)^2}{4} \\\\\n&= \\frac{4 - 4(v^2 + w^2) + (v^2 + w^2)^2 - 4v^2w^2}{4} \\\\\n&= \\frac{(2 - v^2 - w^2)^2 - 4v^2w^2}{4} \\\\\n&= \\frac{(2 - v^2 - w^2 - 2vw)(2 - v^2 - w^2 + 2vw)}{4} \\\\\n&= \\frac{[2 - (v+w)^2][2 - (v-w)^2]}{4} \\\\\n&\\le 1 - \\frac{(v+w)^2}{2}.\n\\end{aligned}\n$$\n\n(Note: $(v+w)^2 \\le 2(v^2+w^2) \\le 2$.) Substituting gives\n\n$$\n\\sqrt{1 - \\frac{(v+w)^2}{2}} + v + w \\le \\sqrt{3}.\n$$\n\nSet $\\frac{v+w}{2} = x$. Then the inequality becomes\n\n$$\n\\sqrt{1 - 2x^2} + 2x \\le \\sqrt{3},\n$$\n\nwhich can be completed as in the first proof.\n\n**Note:** The second proof reveals the motivation for the substitution used in the first proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20818,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$ be real numbers such that\n\n$$\n|a - b| \\ge |c|, \\quad |b - c| \\ge |a|, \\quad |c - a| \\ge |b|.\n$$\n\nProve that one of the numbers $a$, $b$, $c$ equals the sum of the other two.",
"options": [],
"answer": "See solution",
"solution": "Squaring the first inequality gives $(a-b)^2 \\ge c^2$, hence $(a-b+c)(b+c-a) \\ge 0$. Multiplying this with the other two similar inequalities implies $$(a+b-c)^2 (b+c-a)^2 (c+a-b)^2 \\le 0$$, hence one of $a$, $b$, $c$ is the sum of the other two.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20819,
"subject": "Mathematics (Olympiad)",
"question": "Determine all positive integers $n$ with the following property: For any integer $k$, the polynomial $x^n + k$ is either irreducible or has an integer root.",
"options": [],
"answer": "See solution",
"solution": "The answer is $n = 1$ or $n$ prime.\n\nIt is clear that $n = 1$ has the desired property. Assume that $n > 1$. To see that $n$ must be a prime, assume that $n$ is composite, and let $d$ be a non-trivial divisor of $n$.\n\nWe consider $k = -2^d$. Since $2$ is a prime number and $d < n$, $2^d$ is not a perfect $n$th power, so $x^n - 2^d = 0$ has no integer solutions. On the other hand,\n$$\nx^n - 2^d = (x^{n/d})^d - 2^d\n$$\nis divisible by $x^{n/d} - 2$, which has degree less than $n$ since $d > 1$. Therefore, $x^n + k$ has no integer root and is not irreducible, so $n$ does not have the desired property.\n\nAssume that $n$ is a prime. If $n = 2$, $n$ has the desired property, so assume that $n$ is odd. Let $k$ be given, and define $a = -k^{1/n}$, i.e., $a$ is the real root of $x^n + k$. If $k$ is a perfect $n$th power, $a$ is an integer, and $x^n + k$ has an integer root. Assume that $k$ is not a perfect $n$th power. Assume that $f(x)$ is a non-constant polynomial with integer coefficients dividing $x^n + k$. Any root $r \\in \\mathbb{C}$ of $f$ satisfies $r^n = -k$, and hence $|r|^n = |a|^n$ since also $a^n = -k$. Therefore, $|r| = |a|$.\n\nAssume that $\\deg f = m$, and $f$ has roots $r_1, \\dots, r_m$. The constant term of $f$ is an integer since $f(x)$ divides $x^n + k$, which has integer coefficients. But the constant term is also given by $(-1)^m r_1 \\cdots r_m$, so $|(-1)^m r_1 \\cdots r_m| = |a^m|$ is an integer, and hence $a^m$ is an integer. But $a$ was given by $a = -k^{1/n}$, so since $n$ is a prime and $k$ is not a perfect $n$th power, $a^m$ is an integer if and only if $n \\mid m$. Therefore, $n \\mid m$, and $\\deg f = m \\geq n$, so we must have $f(x) = x^n + k$, which shows that $x^n + k$ is irreducible, and $n$ has the desired property.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20820,
"subject": "Mathematics (Olympiad)",
"question": "Alice and Bob play the following game: They start with two non-empty piles of coins. Taking turns, with Alice playing first, each player chooses a pile with an even number of coins and moves half of this pile to the other pile. The game ends if a player cannot move, in which case the other player wins.\n\nDetermine all pairs $\\left(a, b\\right)$ of positive integers such that if initially the two piles have $a$ and $b$ coins respectively, then Bob has a winning strategy.",
"options": [],
"answer": "See solution",
"solution": "Let $\\nu_2(n)$ denote the largest nonnegative integer $r$ such that $2^r \\mid n$.\n\nA position $(a, b)$ (i.e., two piles of sizes $a$ and $b$) is called $k$-happy if $\\nu_2(a) = \\nu_2(b) = k$ for some integer $k \\geq 0$, and $k$-unhappy if $\\min\\{\\nu_2(a), \\nu_2(b)\\} = k < \\max\\{\\nu_2(a), \\nu_2(b)\\}$.\n\nWe claim that Bob has a winning strategy if and only if the initial position is $k$-happy for some even $k$.\n\n- In a $0$-happy position, the player whose turn it is and is unable to play loses.\n- In a $k$-happy position $(a, b)$ with $k \\geq 1$, the player in turn can transform it into one of the positions $(a + \\frac{1}{2}b, \\frac{1}{2}b)$ or $(b + \\frac{1}{2}a, \\frac{1}{2}a)$, both of which are $(k-1)$-happy because\n $$\n \\nu_2(a + \\frac{1}{2}b) = \\nu_2(\\frac{1}{2}b) = \\nu_2(b + \\frac{1}{2}a) = \\nu_2(\\frac{1}{2}a) = k-1.\n $$\n Therefore, if the starting position is $k$-happy, after $k$ moves they will reach a $0$-happy position, so Bob will win if and only if $k$ is even.\n\n- In a $k$-unhappy position $(a, b)$ with $k$ odd and $\\nu_2(a) = k < \\nu_2(b) = l$, Alice cannot play to position $(\\frac{1}{2}a, b + \\frac{1}{2}a)$, because the new position is $(k-1)$-happy and will lead to Bob's victory. Thus, she must play to position $(a + \\frac{1}{2}b, \\frac{1}{2}b)$. We claim that this position is also $k$-unhappy. Indeed, if $l > k + 1$, then $\\nu_2(a + \\frac{1}{2}b) = k < \\nu_2(\\frac{1}{2}b) = l - 1$, whereas if $l = k + 1$, then $\\nu_2(a + \\frac{1}{2}b) > \\nu_2(\\frac{1}{2}b) = k$.\n\nTherefore, a $k$-unhappy position is winning for Alice if $k$ is odd, and drawing if $k$ is even.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20821,
"subject": "Mathematics (Olympiad)",
"question": "A positive integer is *bold* if it has 8 positive divisors that sum up to 3240. For example, 2006 is bold because its 8 positive divisors, $1, 2, 17, 34, 59, 118, 1003,$ and $2006$, sum up to 3240. Find the smallest positive bold number.",
"options": [],
"answer": "See solution",
"solution": "Let $n = p_1^{\\alpha_1} \\cdots p_k^{\\alpha_k}$. Then $8 = (\\alpha_1 + 1) \\cdots (\\alpha_k + 1)$, and $3240 = \\frac{p_1^{\\alpha_1+1}-1}{p_1-1} \\cdots \\frac{p_k^{\\alpha_k+1}-1}{p_k-1}$. Hence, there are three cases:\n\n(a) $8 = \\alpha_1 + 1 \\implies n = p^7$.\n\n(b) $8 = (\\alpha_1 + 1)(\\alpha_2 + 1) \\implies n = p_1 p_2^3$.\n\n(c) $8 = (\\alpha_1 + 1)(\\alpha_2 + 1)(\\alpha_3 + 1) \\implies n = p_1 p_2 p_3$.\n\nNow, check $\\sigma(n)$:\n\n(a) $1 + p^2 + \\cdots + p^7 = 3240$. We have $2 < p < 5$, so $p = 3$. But substituting yields no solution.\n\n(b) $(p_1 + 1)(p_2^3 + p_2^2 + p_2 + 1) = 3240 \\iff (p_1 + 1)(p_2 + 1)(p_2^2 + 1) = 3240$. Since $3240 = 2^3 \\cdot 3^4 \\cdot 5$, the only primes that can divide $p_2^2 + 1$ are 2 and 5. This leaves $p_2 = 2$ and $p_2 = 3$, none of which yield a solution.\n\n(c) $(p_1+1)(p_2+1)(p_3+1) = 3240$. Consider some cases:\n\n(c.1) One of the primes $p_i$ is 2. Suppose $p_1 = 2$. Then $(p_2+1)(p_3+1) = 1080 \\implies \\left(\\frac{p_2+1}{2}\\right)\\left(\\frac{p_3+1}{2}\\right) = 270$.\n\nLet $x = \\frac{p_2+1}{2}$ and $y = \\frac{p_3+1}{2}$. Then $xy = 270$ is fixed and we want to minimize $2p_2p_3 = 2(2x-1)(2y-1) = 8 \\cdot 270 - 4(x+y) + 2$, i.e., maximize $x+y$. This happens when $|x-y|$ is maximum. Since $p_2$ and $p_3$ are primes, the optimal values are $x=2$ and $y=135$, i.e., $p_2 = 3$ and $p_3 = 269$, leading to the minimal solution $n = 1614$.\n\n(c.2) All primes $p_i$ are odd. Then $\\left(\\frac{p_1+1}{2}\\right)\\left(\\frac{p_2+1}{2}\\right)\\left(\\frac{p_3+1}{2}\\right) = 405 = 3^3 \\cdot 5$. One prime, say $p_1$, is $2 \\cdot 3^k \\cdot 5 - 1 = 10 \\cdot 3^k - 1$; since $\\frac{p_i+1}{2} \\geq 3$, $k=1$ yields $p_1 = 29$ and $\\left(\\frac{p_2+1}{2}\\right)\\left(\\frac{p_3+1}{2}\\right) = 27 \\implies p_2 = 5$ and $p_3 = 17$, leading to $n = 2465$.\n\nHence, the smallest value for $n$ is $1614$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20822,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge 1$ be an integer. Consider a permutation $(a_1, a_2, \\ldots, a_{2n})$ of the first $2n$ positive integers such that the numbers $|a_{i+1} - a_i|$ for $i = 1, 2, \\ldots, 2n-1$ are all distinct.\n\nProve that $a_1 - a_{2n} = n$ if and only if $1 \\leq a_{2k} \\leq n$ for every $k = 1, 2, \\ldots, n$.",
"options": [],
"answer": "See solution",
"solution": "a) *Sufficient condition.*\n\nAssume $1 \\leq a_{2k} \\leq n$ for all $k = 1, 2, \\ldots, n$. Then:\n\n$$\nT = \\sum_{i=1}^{2n-1} |a_{i+1} - a_i| = 2(a_1 + a_3 + \\cdots + a_{2n-1}) - 2(a_2 + a_4 + \\cdots + a_{2n}) + a_{2n} - a_1 = 2n^2 + a_{2n} - a_1.\n$$\n\nOn the other hand, since $1 \\leq |a_{i+1} - a_i| \\leq 2n - 1$ for all $i = 1, 2, \\ldots, 2n-1$, and the numbers $|a_{i+1} - a_i|$ are distinct, it follows that $T = 2n^2 - n$. Therefore, $a_{2n} - a_1 = n$.\n\nb) *Necessary condition.*\n\nWe have:\n\n$$\nT_0 = \\sum_{i=1}^{2n-1} |a_{i+1} - a_i| + a_1 - a_{2n} = 2n^2. \\quad (1)\n$$\n\nBut $T_0$ can be written as $T_0 = \\sum_{i=1}^{2n} \\delta_i a_i$, where $\\delta_i \\in \\{-2, 0, 2\\}$ for $i = 1, 2, \\ldots, 2n$. It is easy to see that:\n\n1. $\\delta_1 + \\delta_2 + \\cdots + \\delta_{2n} = 0$,\n2. If we erase all zeros in the sequence $\\delta_1, \\delta_2, \\ldots, \\delta_{2n}$ and retain all nonzero numbers, the numbers $-2$ and $2$ alternate.\n\nFrom (1), we can show that\n\n$$\nT_0 = \\sum_{i=1}^{2n} \\delta_i a_i = \\sum_{i=1}^{2n} \\delta_i (a_i - n) \\leq 2n^2,\n$$\n\nand equality occurs if and only if\n\n$$\n\\delta_i = 2 \\text{ for all } i \\text{ such that } a_i > n, \\quad \\delta_i = -2 \\text{ for all } i \\text{ such that } a_i \\leq n. \\quad (2)\n$$\n\nSince $a_1 - a_{2n} = n$, we have $a_{2n} \\leq n$.\n\nTherefore, from (1), (2), and the alternation property, we have $1 \\leq a_{2k} \\leq n$ for all $k = 1, 2, \\ldots, n$.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20823,
"subject": "Mathematics (Olympiad)",
"question": "A $12 \\times 12$ grid is to be colored with black and white squares so that every $1 \\times 3$ rectangle contains exactly one black square, and every $3 \\times 4$ rectangle contains exactly four black squares. How many ways are there to color the grid to satisfy these conditions?\n\n\n\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $0 \\leq a \\leq 9$ be the number of black squares among those that are marked with an asterisk. Then, there are $18r + 9 \\cdot (4 - r) + a$ black squares in this $12 \\times 12$ square. On the other hand, this $12 \\times 12$ grid can be covered by twelve $3 \\times 4$ rectangles and so it has to contain $12 \\cdot 4 = 48$ black squares. Therefore, $18r + 9 \\cdot (4 - r) + a = 48$, i.e. $9r + a = 12$. There is only one solution to this equation satisfying the constraints on $a$ and $r$, namely $r = 1$ and $a = 3$.\n\nThis shows that each $1 \\times 3$ rectangle must contain exactly one black square.\n\n(B) After choosing one square to be colored black, there are exactly two possibilities to complete the coloring.\n\nConsider the $3 \\times 4$ rectangle with upper left corner the chosen black square. It follows from (A) that the squares A and B have to be black and that the remaining two black squares can only be among C, D, E, F.\n\n\n\n\n\n\nBecause of (A), either C and F or D and E are the black squares. Both patterns can be completed in a unique way, using (A) again, and it is easy to see that both satisfy the requirements of the problem:\n\n\n\n\n(C) If we fix a $1 \\times 3$ rectangle, there are three choices of the black square in it. For each of these choices we have seen in (B) that there are two ways to complete the coloring. Thus there are $6$ ways of coloring the grid.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20824,
"subject": "Mathematics (Olympiad)",
"question": "Pete and Basil play the following game on a checkered $n \\times n$ board. Initially, the whole board is white except for one corner square, which is black; a rook is placed on this square. The players move alternately, with Pete moving first. On each turn, a player moves the rook to another square horizontally or vertically. Immediately after that, all the squares passed by the rook (including the destination square) become black. It is prohibited to move the rook to or across any black square. The player who cannot move loses. Which player has a winning strategy?",
"options": [],
"answer": "See solution",
"solution": "Pete.\n\nA winning strategy for Pete is to make the largest possible vertical move on each turn. After each of his moves, the reachable part of the board consists of at most two rectangles, each with the horizontal side smaller than the vertical one, and the rook stands next to a corner square of each of these rectangles.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20825,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be positive integers such that their least common multiple is $720$. The prime factorization of $720$ is $2^4 \\times 3^2 \\times 5^1$. Thus, none of $a$, $b$, or $c$ has prime factors other than $2$, $3$, or $5$, so we can write\n\n$$\na = 2^{p_1} 3^{q_1} 5^{r_1}, \\quad b = 2^{p_2} 3^{q_2} 5^{r_2}, \\quad c = 2^{p_3} 3^{q_3} 5^{r_3},\n$$\n\nwhere, for each $i$, $p_i$, $q_i$, $r_i$ are nonnegative integers. The fact that the least common multiple of $a$, $b$, and $c$ is $720 = 2^4 \\times 3^2 \\times 5^1$ is equivalent to the validity of all of the following three conditions:\n\n1. The largest number among $p_1, p_2, p_3$ is $4$.\n2. The largest number among $q_1, q_2, q_3$ is $2$.\n3. The largest number among $r_1, r_2, r_3$ is $1$.\n\nHow many ordered triplets $(a, b, c)$ satisfy these conditions?",
"options": [],
"answer": "See solution",
"solution": "Let us find the number of triplets $(p_1, p_2, p_3)$ which satisfy requirement (1). (We distinguish triplets obtained from the same three numbers by changing their order in the arrangement.)\n\nThere are $5^3 = 125$ triplets $(p_1, p_2, p_3)$ for which every $p_i$ ($i = 1, 2, 3$) satisfies $0 \\leq p_i \\leq 4$, and $4^3 = 64$ triplets for which $0 \\leq p_i \\leq 3$ for each $i$. The number of those triplets $(p_1, p_2, p_3)$ satisfying condition (1) is given by the difference of these two numbers, so it is $125 - 64 = 61$.\n\nSimilarly, the number of triplets $(q_1, q_2, q_3)$ satisfying condition (2) is $3^3 - 2^3 = 19$, and the number of triplets $(r_1, r_2, r_3)$ satisfying (3) is $2^3 - 1^3 = 7$.\n\nTherefore, the number of ordered triplets $(a, b, c)$ satisfying the conditions is:\n\n$$\n61 \\times 19 \\times 7 = 8113.\n$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20826,
"subject": "Mathematics (Olympiad)",
"question": "Suppose positive real numbers $a, b, c, d$ satisfy $abcd = 1$. Prove\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} + \\frac{9}{a+b+c+d} \\ge \\frac{25}{4}.\n$$",
"options": [],
"answer": "See solution",
"solution": "First, we will prove that, whenever there are two numbers among $a, b, c, d$ that are equal, the inequality holds. We may assume that $a = b$ and let $s = a + b + c + d$. Then we have\n\n$$\n\\begin{aligned}\n& \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} + \\frac{9}{a+b+c+d} \\\\\n&= \\frac{2}{a} + \\frac{c+d}{cd} + \\frac{9}{s} = \\frac{2}{a} + a^2(s - 2a) + \\frac{9}{s} \\\\\n&= \\frac{2}{a} - 2a^3 + \\left(a^2s + \\frac{9}{s}\\right).\n\\end{aligned}\n$$\n\nWe define the expression above as $f(s)$. Then we see that $f(s)$ reaches the minimum for $s = \\frac{3}{a}$.\n\nWhen $a \\ge \\frac{\\sqrt{2}}{2}$, however, we have\n\n$$\ns = a + b + c + d \\ge 2a + \\frac{2}{a} \\ge \\frac{3}{a}.\n$$\n\nAt this time, $f(s)$ reaches the minimum for $s = 2a + \\frac{2}{a}$.\n\nWe then have\n\n$$\n\\begin{aligned}\n& \\frac{2}{a} - 2a^3 + \\left(a^2s + \\frac{9}{s}\\right) = \\frac{2}{a} - 2a^3 + a^2\\left(2a + \\frac{2}{a}\\right) + \\frac{9}{s} \\\\\n&= \\frac{2}{a} + 2a + \\frac{9}{s} = s + \\frac{9}{s} \\\\\n&= \\frac{7}{16}s + \\frac{9}{16}s + \\frac{9}{s} \\ge \\frac{7}{16} \\times 4 + 2\\sqrt{\\frac{9}{16}s \\cdot \\frac{9}{s}} \\\\\n&= \\frac{7}{4} + \\frac{9}{2} = \\frac{25}{4} \\quad (\\therefore s = 2a + \\frac{2}{a} \\ge 4).\n\\end{aligned}\n$$\n\nWhen $0 < a < \\frac{\\sqrt{2}}{2}$, we have\n\n$$\n\\begin{aligned}\n\\frac{2}{a} - 2a^3 + \\left(a^2 s + \\frac{9}{s}\\right) &\\ge \\frac{2}{a} - 2a^3 + 6a = \\frac{2}{a} + 5a + (a - 2a^3) \\\\\n&> \\frac{2}{a} + 5a \\ge 2\\sqrt{\\frac{2}{a} \\cdot 5a} = 2\\sqrt{10} > \\frac{25}{4}.\n\\end{aligned}\n$$\n\nSecond, we consider the case that $a, b, c, d$ are different from each other. We may assume that $a > b > c > d$. If $\\frac{ad}{c} \\cdot b \\cdot c \\cdot c = abcd = 1$, by using the result above, we have\n\n$$\n\\frac{1}{ad} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{c} + \\frac{9}{ad} \\ge \\frac{25}{4}.\n$$\n\nTherefore, we only need to prove that\n\n$$\n\\begin{aligned}\n& \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} + \\frac{9}{a + b + c + d} \\\\\n\\ge & \\frac{1}{ad} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{c} + \\frac{9}{ad} + b + c + d.\n\\end{aligned}\n\\qquad \\textcircled{1}\n$$\n\nWe have\n\n$$\n\\begin{aligned}\n\\textcircled{1} &\\Leftrightarrow \\frac{1}{a} + \\frac{1}{d} + \\frac{9}{a+b+c+d} \\ge \\frac{c}{ad} + \\frac{1}{c} + \\frac{9}{ad} + \\frac{b+2c}{c} \\\\\n&\\Leftrightarrow \\frac{ac+cd-c^2-ad}{acd} \\ge \\frac{9}{(a+b+c+d)\\left(\\frac{ad}{c} + b + 2c\\right)} \\\\\n&\\qquad \\left(a+d-\\frac{ad}{c}-c\\right) \\\\\n&\\Leftrightarrow \\frac{(a-c)(c-d)}{acd} \\ge \\frac{9}{(a+b+c+d)\\left(\\frac{ad}{c} + b + 2c\\right)} \\\\\n&\\qquad \\frac{(a-c)(c-d)}{c} \\\\\n&\\Leftrightarrow \\frac{1}{ad} \\ge \\frac{9}{(a+b+c+d)\\left(\\frac{ad}{c} + b + 2c\\right)}\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20827,
"subject": "Mathematics (Olympiad)",
"question": "Show that the sum of the decimal digits of $2^{2^{2023}}$ is greater than $2023$.",
"options": [],
"answer": "See solution",
"solution": "We will prove the more general statement that, for every positive integer $n$, the sum of decimal digits of $2^{2^{2n}}$ is greater than $n$.\n\nLet $m = 2^{2n} = 4^n$, so that we need to consider the digits of $2^m$. It will suffice to prove that at least $n$ of these digits are different from 0, since the last digit is at least 2.\n\nLet $0 = e_0 < e_1 < \\dots < e_k$ be the positions of non-zero digits, so that $2^m = \\sum_{i=0}^k d_i \\cdot 10^{e_i}$ with $1 \\leq d_i \\leq 9$. Considering this number modulo $10^{e_j}$, for some $0 < j \\leq k$, the residue $\\sum_{i=0}^{j-1} d_i \\cdot 10^{e_i}$ is a multiple of $2^{e_j}$, hence at least $2^{e_j}$, but on the other hand it is bounded by $10^{e_{j-1}+1}$.\n\nIt follows that $2^{e_j} < 10^{e_{j-1}+1} < 16^{e_{j-1}+1}$, and hence $e_j < 4(e_{j-1} + 1)$. With $e_0 = 4^0 - 1$ and $e_j \\leq 4(e_{j-1} + 1) - 1$, it follows that $e_j \\leq 4^j - 1$, for all $0 \\leq j \\leq k$. In particular, $e_k \\leq 4^k - 1$ and hence\n\n$$\n2^m = \\sum_{i=0}^{k} d_i \\cdot 10^{e_i} < 10^{4^k} < 16^{4^k} = 2^{4 \\cdot 4^k} = 2^{4^{k+1}},\n$$\n\nwhich yields $4^n = m < 4^{k+1}$, i.e., $n - 1 < k$. In other words, $2^m$ has $k \\geq n$ non-zero decimal digits, as claimed.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20828,
"subject": "Mathematics (Olympiad)",
"question": "There are $n \\ge 3$ positive real numbers $a_1, a_2, \\dots, a_n$. For each $1 \\le i \\le n$ we let $b_i = \\frac{a_{i-1} + a_{i+1}}{a_i}$ (here we define $a_0$ to be $a_n$ and $a_{n+1}$ to be $a_1$). Assume that for all $i$ and $j$ in the range $1$ to $n$, we have $a_i \\le a_j$ if and only if $b_i \\le b_j$.\n\nProve that $a_1 = a_2 = \\dots = a_n$.",
"options": [],
"answer": "See solution",
"solution": "*Alternative 1*\n\nLet $r$ and $s$ be indices such that $a_r$ and $a_s$ are maximal and minimal, respectively, amongst $a_1, a_2, \\dots, a_n$. Then\n\n$$\nb_r = \\frac{a_{r-1} + a_{r+1}}{a_r} \\le \\frac{2a_r}{a_r} = 2,\n$$\n\nand\n\n$$\nb_s = \\frac{a_{s-1} + a_{s+1}}{a_s} \\ge \\frac{2a_s}{a_s} = 2.\n$$\n\nSo $b_r \\le 2 \\le b_s$, and hence $a_r \\le a_s$. Since $a_r$ is maximal and $a_s$ is minimal, all $a_i$ must be equal.\n\n*Alternative 2*\n\nAs in Solution 1, $b_r \\le 2$ where $r$ is an index such that $a_r$ is maximal amongst the $a_i$. Hence for all $i$ we have $b_i \\le b_r \\le 2$ (as $b_r$ is maximal amongst the $b_i$). Summing these gives\n\n$$\n2n \\ge \\sum_{i=1}^{n} b_i = \\sum_{i=1}^{n} \\frac{a_{i-1}}{a_i} + \\sum_{i=1}^{n} \\frac{a_{i+1}}{a_i}.\n$$\n\nWe can apply the AM-GM inequality to both groups of $n$ terms on the right:\n\n$$\n\\sum_{i=1}^{n} \\frac{a_{i-1}}{a_i} \\ge n \\sqrt[n]{\\prod_{i=1}^{n} \\frac{a_{i-1}}{a_i}} = n,\n$$\n\nand\n\n$$\n\\sum_{i=1}^{n} \\frac{a_{i+1}}{a_i} \\ge n \\sqrt[n]{\\prod_{i=1}^{n} \\frac{a_{i+1}}{a_i}} = n,\n$$\n\nthus\n\n$$\n2n \\ge \\sum_{i=1}^{n} b_i = \\sum_{i=1}^{n} \\frac{a_{i-1}}{a_i} + \\sum_{i=1}^{n} \\frac{a_{i+1}}{a_i} \\ge n + n.\n$$\n\nIt follows that $\\sum_{i=1}^{n} b_i = 2n$, and since each $b_i \\le 2$ we have all $b_i$ are equal (to 2), so all $a_i$ are equal.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20829,
"subject": "Mathematics (Olympiad)",
"question": "A positive integer $n$ is called *good* if there is a set of divisors of $n$ whose members sum to $n$ and include $1$. Prove that every positive integer has a multiple which is good.",
"options": [],
"answer": "See solution",
"solution": "Firstly, we show that if $m > 1$ is good, then so is $2m$. This is true since some proper divisors of $m$, including $1$ (and hence not including $m$ itself), sum to $m$; if we consider all these together with $m$, they will all be factors of $2m$ which sum to $2m$.\n\nThis means that it suffices to prove the claim for odd numbers. The claim holds for $1$ since good numbers exist (such as $6$, which is $1 + 2 + 3$, for example).\n\nIf $a > 1$ is odd and $n = 2^k a$ for some $k$, then $a + 2a + 4a + \\dots + 2^{k-1}a = (2^k - 1)a = n - a$, which is close to $n$. This value of $n$ will be good if we can find some other factors of $n$, including $1$, which sum to $a$.\n\nTo do this, we write $a$ as a sum of powers of $2$, including $1$, by writing $a$ in binary, and then choose $k$ to be large enough for all those powers of $2$ to be factors of $n$. (We may take $k$ to be $\\lceil \\log_2(a) \\rceil$, the smallest integer greater than or equal to the base-$2$ logarithm of $a$.) None of these powers of $2$ are multiples of $a$, so there is no risk that we are using the same factor of $n$ twice.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20830,
"subject": "Mathematics (Olympiad)",
"question": "In a plane rectangular coordinate system $xOy$, the focus of the parabola $\\Gamma: y^2 = 2px$ ($p > 0$) is $F$. A tangent line to $\\Gamma$ passes through point $P$ (different from $O$) on $\\Gamma$ and intersects the $y$-axis at point $Q$. If $|FP| = 2$ and $|FQ| = 1$, then the dot product of vectors $\\overrightarrow{OP}$ and $\\overrightarrow{OQ}$ is \\_\\_\\_\\_\\_.",
"options": [],
"answer": "See solution",
"solution": "Let $P\\left(\\dfrac{t^2}{2p},\\ t\\right)$ ($t \\neq 0$). The equation of the tangent line to $\\Gamma$ at $P$ is $yt = p\\left(x + \\dfrac{t^2}{2p}\\right)$.\n\nLet $x = 0$, then $yt = \\dfrac{t}{2}$, so $y = \\dfrac{1}{2}$.\n\nThe coordinates of $F$ are $\\left(\\dfrac{p}{2},\\ 0\\right)$.\n\n$$\n|FP| = \\sqrt{\\left(\\dfrac{p}{2} - \\dfrac{t^2}{2p}\\right)^2 + t^2} = \\dfrac{p}{2} + \\dfrac{t^2}{2p}, \\\\\n|FQ| = \\dfrac{\\sqrt{p^2 + t^2}}{2}.\n$$\n\nGiven $|FP| = 2$ and $|FQ| = 1$, we have $p^2 + t^2 = 4p$ and $p^2 + t^2 = 4$, so $p = 1$, $t^2 = 3$.\n\nTherefore,\n$$\n\\overrightarrow{OP} \\cdot \\overrightarrow{OQ} = \\frac{t^2}{2} = \\frac{3}{2}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20831,
"subject": "Mathematics (Olympiad)",
"question": "Prove that for each integer $k$ satisfying $2 \\leq k \\leq 100$, there exist positive integers $b_2, b_3, \\dots, b_{101}$ such that\n\n$$\nb_2^2 + b_3^3 + \\dots + b_k^k = b_{k+1}^{k+1} + b_{k+2}^{k+2} + \\dots + b_{101}^{101}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Consider the equation\n\n$$\na_2^2 + \\dots + a_k^k - a_{k+1}^{k+1} - \\dots - a_{100}^{100} = L.\n$$\n\nFirst, choose $a_2, \\dots, a_{100}$ arbitrarily so that $L$ is positive (for example, by making $a_2$ very large). Since $101$ is coprime to $100!$, there exist positive integers $c$ and $d$ such that $100!c + 1 = 101d$ (for example, by setting $c$ to be the inverse of $-100!$ modulo $101$). In fact, by Wilson's theorem, $100! + 1$ is divisible by $101$, so $c = 1$ works.\n\nMultiplying both sides by $L^{100!c}$, we have\n\n$$\n(a_2 L^{\\frac{100!c}{2}})^2 + \\dots + (a_k L^{\\frac{100!c}{k}})^k - (a_{k+1} L^{\\frac{100!c}{k+1}})^{k+1} - \\dots - (a_{100} L^{\\frac{100!c}{100}})^{100} = (L^d)^{101}.\n$$\n\nTherefore, setting $b_i = a_i L^{\\frac{100!c}{i}}$ for $2 \\leq i \\leq 100$ and $b_{101} = L^d$ satisfies the required condition.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20832,
"subject": "Mathematics (Olympiad)",
"question": "The sequence $a_1, a_2, a_3, \\dots$ is defined by $a_1 = a_2 = 1$, $a_{2n+1} = 2a_{2n} - a_n$, and $a_{2n+2} = 2a_{2n+1}$ for $n \\in \\mathbb{N}$. Prove that if $n > 3$ and $n-3$ is divisible by $8$, then $a_n$ is divisible by $5$.",
"options": [],
"answer": "See solution",
"solution": "First, for $k \\in \\mathbb{N}$, $k \\ge 2$, we have\n\n$$\na_{2k+1} + a_{2k-1} = 2a_{2k} - a_k + a_{2k-1} = 5a_{2k-1} - a_k \\equiv -a_k \\pmod{5}. \\quad (1)\n$$\n\nWe prove the assertion by induction on $k$, where $n = 8k+3$. For the base case $k=1$, we compute $a_3 = 1$, $a_4 = 2$, $a_5 = 3$, $a_6 = 6$, $a_7 = 11$, $a_8 = 22$, $a_9 = 42$, $a_{10} = 84$, and $a_{11} = 165$.\n\nBy repeatedly applying (1), we have\n\n$$\n\\begin{align*}\na_{8(k+1)+3} - a_{8k+3} &= (a_{8k+11} + a_{8k+9}) - (a_{8k+9} + a_{8k+7}) \\\\\n&\\quad +(a_{8k+7} + a_{8k+5}) - (a_{8k+5} + a_{8k+3}) \\\\\n&\\equiv -a_{4k+5} + a_{4k+4} - a_{4k+3} + a_{4k+2} \\\\\n&\\quad = -a_{4k+5} + a_{4k+3} + 2a_{4k+1} \\\\\n&\\quad = -(a_{4k+5} + a_{4k+3}) + 2(a_{4k+3} + a_{4k+1}) \\\\\n&\\equiv a_{2k+2} - 2a_{2k+1} = 0 \\pmod{5}.\n\\end{align*}\n$$\n\nTherefore, if $a_{8k+3}$ is divisible by $5$, so is $a_{8(k+1)+3}$. This completes the proof.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20833,
"subject": "Mathematics (Olympiad)",
"question": "Consider all numbers of the form $a + b k$ where $a, b$ are integers with $0 \\leq a, b < \\sqrt{p}$. Prove that there exist integers $m, n$, not both zero, such that $m^2 + 5 n^2 = r p$ for some integer $r$ with $1 \\leq r \\leq 5$.",
"options": [],
"answer": "See solution",
"solution": "Since there are more than $(\\sqrt{p})^2 = p$ numbers of the form $a + b k$, by the pigeonhole principle, two of them must be congruent modulo $p$. Their difference, which is of the form $c + d k$ with $|c|, |d| < \\sqrt{p}$ and $(c, d) \\neq (0, 0)$, is divisible by $p$. Now, note that\n\n$$\n(c + d k)(c - d k) = c^2 - d^2 k^2 \\equiv c^2 + 5 d^2 \\pmod{p}.\n$$\n\nTherefore, $p \\mid c^2 + 5 d^2$. Since\n\n$$\n0 < c^2 + 5 d^2 < p + 5p = 6p,\n$$\n\nwe must have $c^2 + 5 d^2 = r p$ with $r = 1, 2, 3, 4, 5$.\n\n- If $c^2 + 5 d^2 = p$, then\n\n$$\n(c^2 - 5 d^2)^2 + 5 (2 c d)^2 = (c^2 + 5 d^2)^2 = p^2.\n$$\n\nClearly, $c, d \\neq 0$. So we can take $(m, n) = (|c^2 - 5 d^2|, 2|c d|)$.\n\n- If $c^2 + 5 d^2 = 2p$, then $c, d$ are odd as $2p \\equiv 2 \\pmod{4}$. Therefore,\n\n$$\n\\left(\\frac{c^2 - 5 d^2}{2}\\right)^2 + 5 (c d)^2 = \\left(\\frac{c^2 + 5 d^2}{2}\\right)^2 = p^2.\n$$\n\nSo we can take $(m, n) = \\left(\\left|\\frac{c^2 - 5 d^2}{2}\\right|, |c d|\\right)$.\n\n- If $c^2 + 5 d^2 = 3p$, then $3 \\mid c \\pm d$ as $9 \\mid 3p$. Therefore, $3 \\mid c \\pm d$ for some choice of the sign. Then\n\n$$\n\\left( \\frac{c \\mp 5 d}{3} \\right)^2 + 5 \\left( \\frac{c \\pm d}{3} \\right)^2 = \\frac{2(c^2 + 5 d^2)}{3} = 2p\n$$\n\nwith $\\frac{c \\mp 5 d}{3}$ and $\\frac{c \\pm d}{3}$ nonzero. Thus, this is reduced to the second case.\n\n- If $c^2 + 5 d^2 = 4p$, then $c, d$ are even. Therefore,\n\n$$\n\\left(\\frac{c}{2}\\right)^2 + 5 \\left(\\frac{d}{2}\\right)^2 = p,\n$$\n\nwhich is reduced to the first case.\n\n- If $c^2 + 5 d^2 = 5p$, then $5 \\mid c$. Therefore,\n\n$$\nd^2 + 5 \\left( \\frac{c}{5} \\right)^2 = p,\n$$\n\nwhich is reduced to the first case.\n\nSince all cases are exhausted, we are done.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20834,
"subject": "Mathematics (Olympiad)",
"question": "Let $A$ be the set of all permutations $X = (x_1, x_2, \\dots, x_9)$ of $1, 2, \\dots, 9$. For each $X \\in A$, define $f(X) = x_1 + 2x_2 + 3x_3 + \\dots + 9x_9$. Let $M = \\{f(X) \\mid X \\in A\\}$. Find the value of $|M|$.",
"options": [],
"answer": "See solution",
"solution": "We generalize to $n \\geq 4$. Let $A$ be the set of all permutations $X_n = (x_1, x_2, \\dots, x_n)$ of $1, 2, \\dots, n$, and define $f(X_n) = x_1 + 2x_2 + 3x_3 + \\dots + nx_n$. Let $M_n = \\{f(X) \\mid X \\in A\\}$. We claim that:\n\n$$\n|M_n| = \\frac{n^3 - n + 6}{6}\n$$\n\nWe use induction on $n$.\n\nFor $n=4$, the smallest value is $f(4,3,2,1) = 20$ and the largest is $f(1,2,3,4) = 30$. Listing all possible values, we find $|M_4| = 11 = \\frac{4^3 - 4 + 6}{6}$.\n\nAssume the result holds for $n-1$ ($n \\geq 5$). For $n$, consider permutations where $x_n = n$; then:\n\n$$\n\\sum_{k=1}^n kx_k = n^2 + \\sum_{k=1}^{n-1} kx_k\n$$\n\nBy the induction hypothesis, $\\sum_{k=1}^{n-1} kx_k$ ranges over an interval, so $\\sum_{k=1}^n kx_k$ ranges over:\n\n$$\n\\left[ n^2 + \\frac{(n-1)n(n+1)}{6},\\ n^2 + \\frac{(n-1)n(2n-1)}{6} \\right] = \\left[ \\frac{n(n^2+5)}{6},\\ \\frac{n(n+1)(2n+1)}{6} \\right]\n$$\n\nSimilarly, for $x_n = 1$, we get:\n\n$$\n\\sum_{k=1}^n kx_k = \\frac{n(n+1)}{2} + \\sum_{k=1}^{n-1} k(x_k - 1)\n$$\n\nSo $\\sum_{k=1}^n kx_k$ ranges over:\n\n$$\n\\left[ \\frac{n(n+1)(n+2)}{6},\\ \\frac{2n(n^2+2)}{6} \\right]\n$$\n\nSince $\\frac{2n(n^2+2)}{6} \\geq \\frac{n(n^2+5)}{6}$, the possible values of $f(X)$ fill the interval:\n\n$$\n\\left[ \\frac{n(n+1)(n+2)}{6},\\ \\frac{n(n+1)(2n+1)}{6} \\right]\n$$\n\nThe length of this interval is:\n\n$$\n\\frac{n(n+1)(2n+1)}{6} - \\frac{n(n+1)(n+2)}{6} = \\frac{n^3 - n + 6}{6}\n$$\n\nSo $|M_n| = \\frac{n^3 - n + 6}{6} + 1$.\n\nFor $n=9$:\n\n$$\n|M_9| = \\frac{9^3 - 9 + 6}{6} + 1 = \\frac{729 - 9 + 6}{6} + 1 = \\frac{726}{6} + 1 = 121\n$$\n\n**Answer:** $|M| = 121$.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20835,
"subject": "Mathematics (Olympiad)",
"question": "設 $I$ 為三角形 $ABC$ 的內心，$D$ 為 $I$ 關於邊 $BC$ 的垂足。設 $D'$ 為 $D$ 關於 $I$ 的對稱點並且滿足 $\\overline{AD'} = \\overline{ID'}$。以 $D'$ 為圓心，作圓 $\\Gamma$ 過 $A, I$ 並交 $AB, AC$ 於 $X, Y$。設 $Z$ 為 $\\Gamma$ 上一點滿足 $AZ \\perp BC$。\n\n證明：$AD, D'Z, XY$ 共點。\n\n",
"options": [],
"answer": "See solution",
"solution": "注意到 $AI$ 平分 $\\angle XAY$，所以 $D'I \\perp XY$，因此 $\\triangle AXY$ 與 $\\triangle ABC$ 位似。令 $E$ 為 $I$ 關於 $CA$ 的垂足，$I'$ 為 $\\triangle AXY$ 的內心，注意到 $AD$ 與 $XY$ 交於 $I'$ 關於 $XY$ 的垂足 $T$，所以只需證明 $D', T, Z$ 共線。由 $\\overline{IY} = \\overline{II'}$，$\\overline{IE} = \\overline{ID'}$ 及\n\n$$\n\\angle YIE = \\frac{1}{2} |\\angle B - \\angle C| = \\angle D'II'\n$$\n\n可得 $\\triangle YIE \\sim \\triangle I'ID'$，所以 $I'D' \\parallel XY$。令 $M, M'$ 分別為 $\\overline{BC}, \\overline{XY}$ 中點，那麼 $I'TM'D'$ 為長方形，又熟知 $AD' \\parallel IM \\parallel I'M'$，可得\n\n$$\n\\angle ITD' = \\angle IM'D' = \\angle D'AZ = \\angle AZD'\n$$\n\n結合 $IT \\parallel AZ$ 即可得 $D', T, Z$ 共線。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20836,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, \\dots, a_{100}$ be a sequence of integers such that $a_1 = a_{100} = 0$ and for all $2 \\leq i \\leq 99$,\n\n$$\n\\frac{a_{i-1} + a_{i+1}}{2} < a_i.\n$$\n\nWhat is the minimum possible value of $a_{19}$? Prove your answer.",
"options": [],
"answer": "See solution",
"solution": "Define the sequence $\\Delta_1, \\Delta_2, \\dots, \\Delta_{99}$ by $\\Delta_i = 50 - i$. Let $b_1 = 0$ and $b_{i+1} = b_i + \\Delta_i$ for $i = 1, 2, \\dots, 99$. Note that\n\n$$\nb_{100} = b_1 + \\Delta_1 + \\Delta_2 + \\dots + \\Delta_{99} = 0 + 49 + 48 + \\dots + (-47) + (-48) + (-49) = 0.\n$$\n\nIf $a_i = b_i$ for $i = 1, 2, \\dots, 100$, then\n\n$$\n\\frac{a_{i-1} + a_{i+1}}{2} < a_i \\Leftrightarrow a_{i+1} - a_i < a_i - a_{i-1} \\Leftrightarrow \\Delta_i < \\Delta_{i-1},\n$$\n\nwhich is true. Moreover, $a_1 = a_{100} = 0$ and\n\n$$\na_{19} = b_{19} = 0 + 49 + 48 + \\dots + 32 = \\frac{18(32 + 49)}{2} = 729.\n$$\n\nLet $c_i = a_i - b_i$ for $i = 1, 2, \\dots, 100$. Note $c_1 = c_{100} = 0$. Also,\n\n$$\n\\begin{align*}\n& \\frac{1}{2}(a_{i-1} + a_{i+1}) < a_i \\\\\n\\Leftrightarrow & a_{i+1} - a_i < a_i - a_{i-1} \\\\\n\\Leftrightarrow & b_{i+1} + c_{i+1} - b_i - c_i < b_i + c_i - b_{i-1} - c_{i-1} \\\\\n\\Leftrightarrow & c_{i+1} - c_i + \\Delta_i < c_i - c_{i-1} + \\Delta_{i-1} \\\\\n\\Leftrightarrow & c_i - c_{i-1} + 1 < c_{i+1} - c_i \\\\\n\\Leftrightarrow & c_{i+1} - c_i \\leq c_i - c_{i-1} \\tag{1}\n\\end{align*}\n$$\n\nbecause the $c_i$ are integers.\n\nFor each $i$, inequality (1) implies inductively that\n\n$$\nc_k - c_{k-1} \\leq c_i - c_{i-1} \\quad \\text{for all } k \\geq i. \\qquad (2)\n$$\n\nSuppose, for contradiction, that some $c_i < 0$. Let $i$ be the smallest such index. Then $i \\geq 2$ and $c_{i-1} \\geq 0$, so $c_i - c_{i-1} < 0$. By (2), $c_k - c_{k-1} < 0$ for all $k \\geq i$, so $c_{100} < c_{99} < \\dots < c_i < 0$, contradicting $c_{100} = 0$.\n\nThus $c_i \\geq 0$ for all $i$, so $a_i = b_i + c_i \\geq b_i$. In particular, $a_{19} \\geq b_{19} = 729$.\n\nIt is straightforward to compute\n\n$$\nb_n = \\sum_{i=1}^{n-1} \\Delta_i = \\sum_{i=1}^{n-1} (50 - i) = \\frac{(n-1)(100-n)}{2}\n$$\n\nfor $n \\leq 100$.\n\nTherefore, the minimum possible value of $a_{19}$ is $729$, and this is achieved uniquely when $a_n = b_n$ for all $n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20837,
"subject": "Mathematics (Olympiad)",
"question": "在三角形 $ABC$ 中，$BC > AB$。設 $L$ 為 $\\angle ABC$ 的內角平分線。由 $A, C$ 分別對 $L$ 引垂線，設垂足分別為 $P, Q$。令 $M, N$ 分別是 $AC$ 邊與 $BC$ 邊的中點。設三角形 $PQM$ 的外接圓圓心為 $O$，且該圓與 $AC$ 的另一個交點為 $H$。證明：$O, M, N, H$ 共圓。",
"options": [],
"answer": "See solution",
"solution": "延伸 $AP$，交 $BC$ 於 $D$ 點，則 $P$ 為 $AD$ 中點。因為 $M$ 是 $AC$ 中點，故 $PM$ 與 $CD$（即 $BC$）平行，且\n\n$$\n\\angle QPM = \\angle QBC = \\frac{1}{2} \\angle ABC.\n$$\n\n同理可知 $\\angle MQP = \\frac{1}{2}\\angle ABC$，故 $PM = QM$。\n\n\n\n由於 $P, H, Q, M$ 共圓，知\n\n$$\n\\angle QHC = \\angle QHM = \\angle QPM,\n$$\n即 $\\angle QHC = \\angle QBC$。所以 $Q, H, B, C$ 共圓，且\n\n$$\n\\angle BHC = \\angle BQC = 90^\\circ,\n$$\n\n故得 $HN = \\frac{1}{2}BC = NQ$。\n\n由於 $OH = OQ$，可知 $ON$ 為線段 $HQ$ 的中垂線。同時，$\\angle MQP = \\frac{1}{2}\\angle ABC$ 且 $N$ 為 $BC$ 中點，可得 $Q, M, N$ 三點共線。於是\n\n$$\n\\angle NHO = \\angle NQO = \\angle MQO = \\angle OMQ,\n$$\n\n知 $O, M, N, H$ 共圓。證畢。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20838,
"subject": "Mathematics (Olympiad)",
"question": "Let $g : \\mathbb{R} \\to \\mathbb{R}$ be a continuous, decreasing function such that $g(\\mathbb{R}) = (-\\infty, 0)$. Prove that there are no continuous functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that the equality $f \\circ f \\circ \\dots \\circ f = g$ holds for some integer $k \\ge 2$.",
"options": [],
"answer": "See solution",
"solution": "Suppose such a function $f$ exists. The injectivity of $g$ implies that $f$ is also injective, and since $f$ is continuous, it must be strictly monotone. As $g$ is decreasing, $f$ must be increasing, so $k$ is odd. Moreover, $f$ is not surjective.\n\nLet $f^{[k]} = f \\circ f \\circ \\dots \\circ f$ ($k$ times). Since $f(\\mathbb{R})$ is an interval and $g(\\mathbb{R}) = (-\\infty, 0) = f(f^{[k-1]}(\\mathbb{R})) \\subset f(\\mathbb{R})$, we deduce that $f$ is bounded above.\n\nLet $m \\in \\mathbb{R}$ be such that $f(x) < m$ for all $x \\in \\mathbb{R}$. Then $f^{[k-1]}(x) < m$ for all $x$, so $g(x) = f(f^{[k-1]}(x)) > f(m)$ for all $x$, which contradicts $g(\\mathbb{R}) = (-\\infty, 0)$. Thus, such a function $f$ cannot exist.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20839,
"subject": "Mathematics (Olympiad)",
"question": "Let $x$ and $y$ be positive real numbers with $x + y = 1$. Prove that\n$$\n\\frac{x+1}{y} + \\frac{y+1}{x} \\geq 6.\n$$\n\n*When does equality hold?*",
"options": [],
"answer": "See solution",
"solution": "Equality holds exactly for $x = y = \\frac{1}{2}$.\n\nWe have\n$$\n\\frac{x+1}{y} + \\frac{y+1}{x} = \\frac{x+x+y}{y} + \\frac{y+x+y}{x} = 2\\left(\\frac{x}{y} + \\frac{y}{x}\\right) + 2.\n$$\nFor $x, y > 0$, the AM-GM inequality gives\n$$\n\\frac{\\frac{x}{y} + \\frac{y}{x}}{2} \\geq \\sqrt{\\frac{x}{y} \\cdot \\frac{y}{x}} = 1,\n$$\nwhich immediately implies the desired inequality.\n\nEquality holds for $\\frac{x}{y} = \\frac{y}{x}$, i.e. $x = y = \\frac{1}{2}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20840,
"subject": "Mathematics (Olympiad)",
"question": "A number $a$ from a subset of a partition of the set $M = \\{n, n+1, \\dots, 24\\}$ is called **major** if $a$ is equal to the sum of all other numbers in its subset. All numbers in each subset must be distinct, and since one of them is major, each subset must contain at least three numbers. For which values of $n$ does there exist a partition of $M$ into subsets, each containing a major number as defined?",
"options": [],
"answer": "See solution",
"solution": "Let $k$ be the number of subsets in the desired partition. Since $M$ has $25-n$ numbers, $k \\le \\frac{25-n}{3}$. If $a$ is the major number of a subset, the sum of that subset is $2a$. Thus, the total sum $S(n)$ of $M$ must satisfy:\n\n$$\nS(n) \\le \\frac{(122+n)(25-n)}{9}.\n$$\n\nCalculating $S(n) = \\frac{(n+24)(25-n)}{2}$, we get the necessary condition:\n\n$$\n\\frac{(n+24)(25-n)}{2} \\le \\frac{(122+n)(25-n)}{9} \\implies n \\le 4.\n$$\n\nFor $n=2$ and $n=3$, $S(2) = 13 \\cdot 23$ and $S(3) = 27 \\cdot 11$ are odd, so no partition exists. For $n=1$ and $n=4$, examples of such partitions exist:\n\n\n\nThus, the desired partitions exist only for $n=1$ and $n=4$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20841,
"subject": "Mathematics (Olympiad)",
"question": "Find all $n \\in \\mathbb{N}$ divisible by $11$, such that all numbers that can be obtained from $n$ by an arbitrary rearrangement of its digits are again divisible by $11$.",
"options": [],
"answer": "See solution",
"solution": "Let $n$ be a natural number divisible by $11$ with at least two digits, written as $n = \\overline{a_k a_{k-1} \\dots a_0}$, where $a_i$ are digits and $a_k \\ne 0$.\n\nSuppose any rearrangement of the digits of $n$ yields a number divisible by $11$. Consider swapping two adjacent digits $a_{i-1}$ and $a_i$ to form $n'$. Then $n'$ must also be divisible by $11$, so $11 \\mid n - n'$. The difference $n - n'$ is $10^{i-1}(a_{i-1} - a_i) + 10^{i-2}(a_i - a_{i-1}) = (10^{i-1} - 10^{i-2})(a_{i-1} - a_i) = 9 \\cdot 10^{i-2}(a_{i-1} - a_i)$. Thus, $11$ divides $9 \\cdot 10^{i-2}(a_{i-1} - a_i)$, so $a_{i-1} = a_i$.\n\nSince this holds for any adjacent pair, all digits of $n$ are equal. Thus, $n = a \\cdot \\overline{11\\ldots11}_{k+1}$, where $a$ is a digit and the number has $k+1$ digits. For $n$ to be divisible by $11$, the number of digits must be odd. Therefore, all such $n$ are numbers with all digits equal and an odd number of digits, and divisible by $11$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20842,
"subject": "Mathematics (Olympiad)",
"question": "Each vertex of the complete bipartite graph $K_{128,128}$ is occupied by a person. Initially, all persons are unfamiliar with each other. If persons are on adjacent vertices, they immediately become acquainted. It is allowed to select several edges without common vertices and swap the persons at the endpoints of each edge in one operation. What is the minimum number of such operations required so that all persons become acquainted?",
"options": [],
"answer": "See solution",
"solution": "The minimum number of operations required is $6$.\n\nLet $r = 6$. Assign each person a string $x = (x_0, x_1, \\dots, x_r) \\in \\{0, 1\\}^{r+1}$, where all persons who started on the same side have the same first bit $x_0$. We describe an $r$-round strategy:\n\nIn the $i$-th round, move all persons with $x_i = 0$ to side A and all with $x_i = 1$ to side B. If two persons have strings $x$ and $x'$ with $x_i \\neq x'_i$, then in the $i$-th round they will be on different sides and thus become acquainted.\n\nTo show $r$ rounds are necessary, suppose there is a $t$-round strategy. Assign each person a string $x = (x_0, x_1, \\dots, x_t) \\in \\{0, 1\\}^{t+1}$, where $x_i = 0$ if and only if in the $i$-th round the person was in A. Two agents meet during the $t$ rounds if and only if their strings differ. Thus, $2^{t+1} \\ge 2n$, so $t \\ge r$ as required.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20843,
"subject": "Mathematics (Olympiad)",
"question": "A square with side length $3$ is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length $2$ has two vertices on the other square and the other two on sides of the triangle, as shown. What is the area of the triangle?\n\n(A) $19\\frac{1}{4}$\n\n(B) $20\\frac{1}{4}$\n\n(C) $21\\frac{3}{4}$\n\n(D) $22\\frac{1}{2}$\n\n(E) $23\\frac{3}{4}$",
"options": [],
"answer": "See solution",
"solution": "Label the vertices as shown in the diagram.\n\n\n\nThen $\\triangle ABC$, $\\triangle CDE$, and $\\triangle EFG$ are similar. Because $CD = 2$ and $DE = \\frac{3-2}{2} = \\frac{1}{2}$, the lengths of the legs of each of these triangles are in the ratio of $4$ to $1$. It follows that $FG = \\frac{3}{4}$, so the base of the isosceles triangle has length $\\frac{3}{4} + 3 + \\frac{3}{4} = \\frac{9}{2}$. Similarly, because $BC = 1$, similar triangles give $AB = 4$. It follows that the altitude of the isosceles triangle is $4 + 2 + 3 = 9$. The area of the triangle is then given by $\\frac{1}{2} \\cdot \\frac{9}{2} \\cdot 9 = 20\\frac{1}{4}$.\n\n**OR**\n\nPlace the triangle in a coordinate plane with the base on the $x$-axis and the apex on the positive $y$-axis. The upper right vertex of the large square is $(\\frac{3}{2}, 3)$, and the upper right vertex of the small square is $(1, 5)$. The side of the triangle in the first quadrant has slope\n\n$$\n\\frac{3-5}{\\frac{3}{2}-1} = -4,\n$$\n\nand its equation is $y = 9 - 4x$. Thus the side intersects the $x$-axis at $(\\frac{9}{4}, 0)$ and the $y$-axis at $(0, 9)$. Therefore the triangle has base $2 \\cdot \\frac{9}{4} = \\frac{9}{2}$ and altitude $9$, so its area is $\\frac{1}{2} \\cdot \\frac{9}{2} \\cdot 9 = 20\\frac{1}{4}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20844,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs of nonnegative integers $(m, n)$ such that\n\n$$\n(m + n - 5)^2 = 9mn.\n$$",
"options": [],
"answer": "See solution",
"solution": "The equation is symmetric in $m$ and $n$. The solutions are the unordered pairs\n\n$$\n\\{5F_{2k}^2, 5F_{2k+2}^2\\}, \\quad \\{L_{2k-1}^2, L_{2k+1}^2\\},\n$$\n\nwhere $k$ is a nonnegative integer and $\\{F_j\\}$, $\\{L_j\\}$ are the **Fibonacci** and **Lucas sequences**, respectively — that is, the sequences defined by $F_1 = F_2 = 1$, $L_1 = 1$, $L_2 = 3$, and the recursive relations $F_{j+2} = F_{j+1} + F_j$ and $L_{j+2} = L_{j+1} + L_j$ for $j \\ge 1$. Note that we amended the Lucas sequence by considering $L_1 = -1$ and $L_0 = 2$.\n\nLet $g = \\gcd(m, n)$ and write $m = gm_1$ and $n = gn_1$. Because $9mn$ is a perfect square, $m_1$ and $n_1$ are perfect squares. Let $m_1 = x^2$ and $n_1 = y^2$. The given condition becomes\n\n$$\n(gx^2 + gy^2 - 5)^2 = 9g^2x^2y^2.\n$$\n\nTaking the square root on both sides yields\n\n$$\ng(x^2 + y^2) - 5 = \\pm 3gxy,\n$$\n\nor\n\n$$\ng(x^2 + y^2 \\pm 3xy) = 5.\n$$\n\nIf $g(x^2 + y^2 + 3xy) = 5$, then $x^2 + y^2 + 3xy \\le 5$, implying that $x = y = g = 1$ and $(m, n) = (1, 1)$.\n\nOtherwise, $g(x^2 + y^2 - 3xy) = 5$ and $g = 1$ or $5$. Fix $g$ equal to one of these values, so that\n\n$$\nx^2 - 3xy + y^2 = \\frac{5}{g}. \\qquad (1)\n$$\n\nLet $\\{a, b\\}$ denote an unordered pair of numbers. We call $\\{a, b\\}$ a *g-pair* if $\\{x, y\\} = \\{a, b\\}$ satisfies (1) and *a* and *b* are positive integers. Also, we call $\\{p, q\\}$ *smaller* (respectively, *larger*) than $\\{r, s\\}$ if $p + q$ is smaller (respectively, larger) than $r + s$.\n\nSuppose that $\\{a, b\\}$ is a *g-pair*. View (1) as a monic quadratic in $x$ with $y = b$ constant. The coefficient of $x$ in a monic quadratic function $(x - r_1)(x - r_2)$ equals $-(r_1 + r_2)$, implying that $\\{3b - a, b\\}$ should also satisfy (1). Indeed,\n\n$$\nb^2 - 3b(3b - a) + (3b - a)^2 = a^2 - 3ab + b^2 = \\frac{5}{g}.\n$$\n\nAlso, if $b > 2$, note that\n\n$$\na^2 - 3ab + b^2 = \\frac{5}{g} < b^2.\n$$\n\nIt follows that $a^2 - 3ab < 0$ and so $3b - a > 0$. Thus, if $\\{a, b\\}$ is a $g$-pair with $b > 2$, then $\\{b, 3b - a\\}$ is a $g$-pair as well. Also note that for $a' = b$ and $b' = 3b - a$, $\\{a', 3b' - a'\\} = \\{a, b\\}$.\n\nFurthermore, if $a \\ge b$, note that $a \\ne b$ because otherwise $-a^2 = 5/g > 0$, which is impossible. Thus, $a > b$ and\n\n$$\na^2 - 3ab + b^2 = \\frac{5}{g} > b^2 - a^2,\n$$\n\nwhich implies that $a(2a - 3b) > 0$ and hence $a + b > b + (3b - a)$ and also $3b - a > b$. Thus, $\\{a', b'\\}$ is a smaller $g$-pair than $(a, b)$ with $a' \\ge b'$.\n\nGiven any $g$-pair $\\{a, b\\}$ with $b \\le a$, if $b \\le 2$ then $a$ must equal $r(g)$, where $r(5) = 3$ if $r(1) = 4$. Otherwise, according to the above observation we can repeatedly reduce it to a smaller $g$-pair until $\\min(a, b) \\le 2$ — that is, to the $g$-pair $(r(g), 1)$.\n\nBeginning with $\\{r(g), 1\\}$, we reverse the reducing process so that $\\{x, y\\}$ is replaced by the larger $g$-pair $\\{3x - y, x\\}$. Moreover, this must generate all $g$-pairs since all $g$-pairs can be reduced to $\\{r(g), 1\\}$. We may express these possible pairs in terms of the Fibonacci and Lucas numbers; for $g = 1$, observe that $L_2 = 1$, $L_3 = 4 = r(1)$, and that\n\n$$\n\\begin{aligned}\nL_{2k+3} &= L_{2k+2} + L_{2k+1} = (L_{2k+1} + L_{2k}) + L_{2k+1} \\\\\n&= (L_{2k+1} + (L_{2k+1} - L_{2k-1})) + L_{2k+1} \\\\\n&= 3L_{2k+1} - L_{2k-1}\n\\end{aligned}\n$$\n\nfor $k \\ge 0$. For $g = 5$, the Fibonacci numbers satisfy an analogous recursive relation, and $F_2 = 1$, $F_4 = 3 = r(5)$. Therefore, $\\{m, n\\} = \\{L_{2k-1}^2, L_{2k+1}^2\\}$ and $\\{m, n\\} = \\{5F_{2k}^2, 5F_{2k+2}^2\\}$ for $k \\ge 0$.\n\n**Note.** The next solution is based on some knowledge of solving *Pell's Equation*. For interested readers, please check [26], [29], and [33] for related materials.\n\nAs we have seen in the first solutions, it suffices to solve the equations\n\n$$\nx^2 + y^2 - 3xy = 1 \\qquad (2)\n$$\n\nand\n\n$$\nx^2 + y^2 - 3xy = 5. \\qquad (3)\n$$\n\nThe first equation is equivalent to $(2x - 3y)^2 - 5y^2 = 4$, which is a *Pell's* equation of the form\n\n$$\nu^2 - dv^2 = 4. \\qquad (4)\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20845,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. Determine, in terms of $n$, the greatest integer which divides every number of the form $p+1$, where $p \\equiv 2 \\pmod{3}$ is a prime number which does not divide $n$.",
"options": [],
"answer": "See solution",
"solution": "Let $k$ be the greatest such integer. We will show that $k = 3$ when $n$ is odd and $k = 6$ when $n$ is even.\n\nWe will say that a number $p$ is *nice* if $p$ is a prime number of the form $2 \\pmod{3}$ which does not divide $n$.\n\nNote first that $3 \\mid p+1$ for every nice number $p$, so $k$ is a multiple of $3$.\n\nIf $n$ is odd, then $p = 2$ is nice, so $p+1 = 3$, and thus $k = 3$.\n\nIf $n$ is even, then $p = 2$ is not nice, so every nice $p$ is of the form $5 \\pmod{6}$. In this case, $6 \\mid p+1$ for every nice number $p$.\n\nIt remains to show that (if $n$ is even):\n\n1. There is a nice $p$ such that $4 \\nmid p+1$.\n2. There is a nice $p$ such that $9 \\nmid p+1$.\n3. For every prime $q \\neq 2, 3$, there is a nice $p$ such that $q \\nmid p+1$.\n\nFor (1), by Dirichlet's theorem on arithmetic progressions, there are infinitely many primes of the form $p \\equiv 5 \\pmod{12}$. Any one of them which is larger than $n$ will do.\n\nFor (2), by Dirichlet's theorem, there are infinitely many primes of the form $p \\equiv 2 \\pmod{9}$. Any one of them which is larger than $n$ will do.\n\nFor (3), by Dirichlet's theorem, there are infinitely many primes of the form $p \\equiv 2 \\pmod{3q}$. Any one of them which is larger than $n$ will do.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20846,
"subject": "Mathematics (Olympiad)",
"question": "Let $m = 2k + 2$. Consider a sequence $\\{a_n\\}$ and coefficients $b_i$.\n\n- If $b_{2k-1}$ 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\n- If $b_{2k-1}$ and $b_{2k-2}$ are of different signs, then\n\n$$\n\\left| \\sum_{i=1}^{2k+2} b_i a_i \\right| = \\left| \\sum_{i=1}^{2k} b_i a_i \\pm (a_{2k+1} - a_{2k+2}) \\right| \\geq |a_{2k+1} - a_{2k+2}| - \\sum_{i=1}^{2k} a_i = n > 0.\n$$\n\nShow that the sequence $\\{a_n\\}$ thus constructed satisfies the requirements: $0$ is not contained in any $A_k$, and any non-zero integer between $-k$ and $k$ is contained in $A_k$.",
"options": [],
"answer": "See solution",
"solution": "The sequence $\\{a_n\\}$ constructed as above meets the requirements because $0$ is not included in any $A_k$, and every non-zero integer between $-k$ and $k$ is included in $A_k$. The inequalities ensure the sums are strictly positive, confirming the construction's validity.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20847,
"subject": "Mathematics (Olympiad)",
"question": "Sequence $\\{a_n\\}$ satisfies $a_1 = 2$ and $a_{n+1} = (n+1)a_n - n$, $n = 1, 2, \\dots$. Then the general term formula of $\\{a_n\\}$ is \\_\\_\\_\\_\\_.",
"options": [],
"answer": "See solution",
"solution": "By the condition, we have $a_{n+1} - 1 = (n+1)(a_n - 1)$. Therefore,\n\n$$\na_n - 1 = n(a_{n-1} - 1) = n(n-1)(a_{n-2} - 1) = \\dots = n(n-1)\\dots2(a_1 - 1) = n!.\n$$\n\nNamely, $a_n = n! + 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20848,
"subject": "Mathematics (Olympiad)",
"question": "Find all lists $(x_1, x_2, \\dots, x_{2020})$ of non-negative real numbers that satisfy all three conditions:\n\n1. $x_1 \\le x_2 \\le \\dots \\le x_{2020}$\n2. $x_{2020} \\le x_1 + 1$\n3. There is a permutation $(y_1, y_2, \\dots, y_{2020})$ of $(x_1, x_2, \\dots, x_{2020})$ such that\n $$\n \\sum_{i=1}^{2020} ((x_i + 1)(y_i + 1))^2 = 8 \\sum_{i=1}^{2020} x_i^3.\n $$",
"options": [],
"answer": "See solution",
"solution": "Consider $a = x_i$ and $b = x_{2021-i}$ for $i = 1, 2, \\dots, 1010$, with $0 \\le a \\le b \\le a + 1$. Let $b = a + k$ for $a \\ge 0$ and $0 \\le k \\le 1$.\n\nWe analyze the expression:\n$$\n((a+1)(b+1))^2 - 4(a^3 + b^3)\n$$\nExpanding, we get:\n$$\n((a+1)(a+k+1))^2 - 4(a^3 + (a+k)^3)\n$$\n$$\n= a^4 + 2(k-2)a^3 + (k^2 - 6k + 6)a^2 + (-10k^2 + 6k + 4)a + (-4k^3 + k^2 + 2k + 1)\n$$\n\nFor $0 \\le k \\le 1$, the quadratic and linear terms are non-negative with equality only if $a = 0$ or $k = 1$ and $a = 1$. The constant term $-4k^3 + k^2 + 2k + 1 \\ge 0$ with equality only if $k = 1$.\n\nThus, $((a+1)(b+1))^2 \\ge 4(a^3 + b^3)$ with equality if and only if $(a, b) = (0, 1)$ or $(a, b) = (1, 2)$.\n\nTherefore, for each $i = 1, 2, \\dots, 1010$,\n$$\n((x_i + 1)(x_{2021-i} + 1))^2 \\ge 4(x_i^3 + x_{2021-i}^3)\n$$\nwith equality only for $(x_i, x_{2021-i}) = (0, 1)$ or $(1, 2)$.\n\nFrom conditions (i) and (ii), the list must be either $1010$ zeros followed by $1010$ ones, or $1010$ ones followed by $1010$ twos. These are the only lists satisfying all three conditions, as required. $\\square$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20849,
"subject": "Mathematics (Olympiad)",
"question": "Consider $n$ persons, each of them speaking at most 3 languages. From any 3 persons, there are at least two who speak a common language.\n\ni) For $n \\leq 8$, exhibit an example in which no language is spoken by more than two persons.\n\nii) For $n \\geq 9$, prove that there exists a language which is spoken by at least three persons.",
"options": [],
"answer": "See solution",
"solution": "i) Split the 8 persons into two groups of 4. Assign a different language to each pair of persons within each group, resulting in $6 + 6 = 12$ languages, each spoken by 2 persons, and each person speaking 3 languages.\n\nFor $n \\leq 7$, simply remove $8 - n$ persons.\n\nii) Assume, for contradiction, that each language is spoken by at most two persons. Then each person $A$ can share a language with at most three others; otherwise, by the pigeonhole principle, there would be a language spoken by at least three people, contradicting the assumption. Let $B, C, D$ be the persons with whom $A$ shares a language. Similarly, let $E$ be another person, who can share languages with at most three others, say $F, G, H$. With $n \\geq 9$, there is at least one more person, say $Z$. In the group $A, E, Z$, no language is spoken in common, contradicting the initial condition.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20850,
"subject": "Mathematics (Olympiad)",
"question": "In a triangle $ABC$, let $K$ be a point on the median $BM$ such that $CM = CK$. It turned out that $\\angle CBM = 2\\angle ABM$. Show that $BC = KM$.",
"options": [],
"answer": "See solution",
"solution": "Let $L$ be the reflection of $C$ through $BM$. Since $CK = CM$, we have $\\angle CKM = \\angle CMK$. On the other hand, since $L$ is the reflection of $C$ through $KL$, we have $\\angle LKM = \\angle CKM = \\angle CMK$, implying that $KL$ is parallel to $CM$.\n\n\n\nAlso, we have $KL = KC = CM = MK$, and $KL$ is parallel to $AM$. Hence, $KMAL$ is a parallelogram, implying that $KM = AL$ and $KM$ is parallel to $AL$. We have $\\angle LBM = \\angle CBM = 2\\angle ABM$, hence $\\angle KBA = \\angle ABL$. Since $AL$ is parallel to $MB$, we have $\\angle BAL = \\angle MBA = \\angle LBA$, implying that $LB = LA$. Hence, we have\n\n$$\nKM = LA = LB = BC\n$$\n\nas desired.\n\n$\\Box$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20851,
"subject": "Mathematics (Olympiad)",
"question": "Determine all natural numbers $n$ such that the inequality\n\n$$\nx^n + 2x + 1 \\ge 4x^2\n$$\n\nholds for every $x > 0$.",
"options": [],
"answer": "See solution",
"solution": "Suppose that $n$ is a solution to the problem. The polynomial\n\n$$\nP(x) = x^n - 4x^2 + 2x + 1\n$$\n\nclearly has a root at $1$. Therefore, we may write $P(x) = (x - 1)Q(x)$ for some polynomial $Q$. Since $P(x) \\ge 0$ for $x > 0$, the polynomial $Q$ changes sign at $1$ and so $Q(1) = 0$. Calculating\n\n$$\nQ(x) = \\frac{P(x)}{x-1} = \\frac{x^n - 1}{x-1} - \\frac{4x^2 - 2x - 2}{x-1} = (1 + x + \\dots + x^{n-1}) - (4x + 2)\n$$\n\nwe see that $Q(1) = n - 6$ and hence $n = 6$.\n\nConversely, let $n = 6$. Note that $Q(x) = (x - 1)R(x)$, where $R$ is a polynomial given by:\n\n$$\nR(x) = x^4 + 2x^3 + 3x^2 + 4x + 1\n$$\n\nSince $P(x) = (x - 1)^2 R(x)$ and $R(x) \\ge 0$ for $x > 0$, we indeed have that $P(x) \\ge 0$ for $x > 0$. It follows that $n = 6$ is a solution, and thus the only solution, to the problem.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20852,
"subject": "Mathematics (Olympiad)",
"question": "Triangle ABC is subdivided into three isosceles triangles and a rhombus.\n\n*Note: the figure is not drawn to scale.*\n\n\n\nWhat is the size of angle $C$ in degrees?",
"options": [],
"answer": "See solution",
"solution": "36",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20853,
"subject": "Mathematics (Olympiad)",
"question": "The matrices $A, B \\in \\mathcal{M}_2(\\mathbb{R})$ have the property $(A - B)^2 = O_2$.\n\na) Show that $\\det(A^2 - B^2) = (\\det(A) - \\det(B))^2$.\n\nb) Show that $\\det(AB - BA) = 0$ if and only if $\\det(A) = \\det(B)$.",
"options": [],
"answer": "See solution",
"solution": "a) From $(A - B)^2 = O_2$ it follows that $\\det(A - B) = 0$ and $\\mathrm{Tr}(A - B) = 0$, hence $\\mathrm{Tr}(A) = \\mathrm{Tr}(B) =: a$.\n\nDenote $b = \\det(A) - \\det(B)$. Then\n$$\n\\begin{cases}\nA^2 - aA + \\det(A)I_2 = O_2 \\\\\nB^2 - aB + \\det(B)I_2 = O_2\n\\end{cases}\n$$\nThis shows that $\\det(A^2 - B^2) = \\det(a(A - B) - bI_2)$. On the other hand,\n$$\n\\det(a(A - B) - bI_2) = a^2 \\det(A - B) - ab \\mathrm{Tr}(A - B) + b^2 = b^2,\n$$\nwhence $\\det(A^2 - B^2) = (\\det(A) - \\det(B))^2$.\n\nb) Let $f : \\mathbb{R} \\to \\mathbb{R}$ be the function given by\n$$\nf(x) = \\det(A^2 - B^2 + x(AB - BA)), \\quad x \\in \\mathbb{R}.\n$$\nThen $f(x) = \\det(A^2 - B^2) + cx + \\det(AB - BA)x^2$, where $c$ is a real constant. From $f(1) = f(-1) = \\det(A - B)\\det(A + B) = 0$ it follows that $c = 0$ and $\\det(A^2 - B^2) + \\det(AB - BA) = 0$. Now (a) leads to $(\\det(A) - \\det(B))^2 = -\\det(AB - BA)$, whence the conclusion.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20854,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x)f(y) = f(xy - 1) + x f(y) + y f(x), \\quad \\forall x, y \\in \\mathbb{R}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $y = 0$ in $(1)$:\n$$\nf(x) f(0) = f(-1) + x f(0).\n$$\n\nConsider two cases for $f(0)$:\n\n**Case 1.** If $f(0) \\neq 0$, then $f(x) = x + c$ for some constant $c$, but this does not satisfy the original equation.\n\n**Case 2.** If $f(0) = 0$, then $f(-1) = 0$.\n\nPlug $x = y = 1$ into $(1)$:\n$$\nf(1)^2 = 2 f(1) \\implies f(1) = 0 \\text{ or } f(1) = 2.\n$$\n\nSubstitute $y = -1$ into $(1)$:\n$$\nf(-x - 1) = f(x), \\quad \\forall x \\in \\mathbb{R}.\n$$\n\nReplace $y$ by $-y - 1$ in $(1)$:\n$$\nf(x) f(-y-1) = f(-x(y+1)-1) + x f(-y-1) - (y+1) f(x).\n$$\n\nFrom the previous result:\n$$\nf(xy - 1) + y f(x) = f(xy + x) - (y + 1) f(x).\n$$\n\nLet $x \\neq -1$, replace $x$ by $x + 1$ and $y$ by $\\frac{1}{x+1}$:\n$$\nf(x-1) = \\frac{x-1}{x+1} f(x), \\quad \\forall x \\neq -1.\n$$\n\nSet $y = 1$ in $(1)$:\n$$\n\\begin{aligned}\nf(x) f(1) &= f(x-1) + x f(1) + f(x) \\\\\n&= \\frac{x-1}{x+1} f(x) + x f(1) + f(x), \\quad \\forall x \\neq -1\n\\end{aligned}\n$$\n\n- If $f(1) = 0$, then $f \\equiv 0$ (since $f(-1) = 0$).\n- If $f(1) = 2$, then $f(x) = x(x+1)$ (since $f(-1) = 0$).\n\nBoth $f(x) \\equiv 0$ and $f(x) = x(x+1)$ satisfy $(1)$. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20855,
"subject": "Mathematics (Olympiad)",
"question": "設 $Q$ 為一個由若干個質數所成的集合（不必然為有限集）。對於一個正整數 $n$，考慮其質因數分解，並定義 $p(n)$ 為這個分解中的指數和，而 $q(n)$ 為這個分解中在 $Q$ 中質數的指數和。若 $n$ 為一整數，且 $p(n) + p(n+1)$ 和 $q(n) + q(n+1)$ 都是偶數，則稱 $n$ 是**特別**的。證明存在一個與 $Q$ 無關的常數 $c > 0$，使得對於任意正整數 $N > 100$，在 $[1, N]$ 的特別整數至少有 $cN$ 個。\n\n（舉例來說，若 $Q = \\{3, 7\\}$，則 $p(42) = 3$，$q(42) = 2$，$p(315) = 4$，$q(315) = 3$）",
"options": [],
"answer": "See solution",
"solution": "事實 1：對於任意 5 個整數，根據鴿巢原理，至少有 2 個整數使得 $p$ 和 $q$ 的奇偶性都相同。\n\n事實 2：若 $d \\mid \\gcd(m, n)$，則 $p(m) + p(n) \\equiv p(m/d) + q(m/d) \\pmod{2}$。\n\n考慮集合\n\n$$\nA_k = \\{72k,\\ 72k + 6,\\ 72k + 8,\\ 72k + 9,\\ 72k + 12\\}\n$$\n\n由事實 1，存在 $n_1, n_2 \\in A_k$，使得 $p(n_1) \\equiv p(n_2) \\pmod{2}$ 且 $q(n_1) \\equiv q(n_2) \\pmod{2}$。\n\n注意這個集合的構造方式使得 $d = n_1 - n_2 \\mid \\gcd(n_1, n_2)$，因此由事實 2，$n_1$ 是特別的。也就是說，$S_k = \\left\\{ \\frac{n_i}{n_j - n_i} \\mid j > i \\right\\}$ 至少有一個特別數。\n\n顯然，$\\bigcup_{k \\leq (N-12)/72} S_k \\subseteq [1, N]$，且每個數最多只屬於 10 個不同的 $S_k$，因此 $[1, N]$ 至少包含 $\\left\\lfloor \\frac{N-12}{720} \\right\\rfloor$ 個特別數。\n\n再加上 $S_1 \\subset [1, 100]$，所以在 $[1, 100]$ 至少有一個特別數，因此確實存在與 $Q$ 無關的常數 $c > 0$ 滿足題意。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20856,
"subject": "Mathematics (Olympiad)",
"question": "Consider the sequence $u_n$ defined by $u_0 = 0$, $u_1 = 1$, and\n\n$$u_{n+2} = 2u_{n+1} + 2u_n, \\quad \\forall n \\ge 0.$$ \n\nProve that for every $n \\ge 1$, $v_3(u_n) = v_3(n)$, where $v_3(x)$ denotes the exponent of $3$ in the prime factorization of $x$.",
"options": [],
"answer": "See solution",
"solution": "We have the general formula for the sequence:\n\n$$u_n = \\frac{(1 + \\sqrt{3})^n - (1 - \\sqrt{3})^n}{2\\sqrt{3}}.$$ \n\nExamining the sequence modulo $3$, the remainders cycle as $0, 1, 2, 0, 1, 2, \\ldots$, so $u_{3k+1}$ and $u_{3k+2}$ are not divisible by $3$, and thus $v_3(u_{3k+1}) = v_3(u_{3k+2}) = 0$.\n\nFor $n = 3k$ with $k \\in \\mathbb{Z}^+$:\n\n$$\n\\begin{aligned}\nu_{3k} &= \\frac{(1 + \\sqrt{3})^{3k} - (1 - \\sqrt{3})^{3k}}{2\\sqrt{3}} \\\\ &= \\frac{[(1 + \\sqrt{3})^k - (1 - \\sqrt{3})^k][(4 + 2\\sqrt{3})^k + (4 - 2\\sqrt{3})^k + (-2)^k]}{2\\sqrt{3}} \\\\ &= u_k[(4 + 2\\sqrt{3})^k + (4 - 2\\sqrt{3})^k + (-2)^k]. \\end{aligned}\n$$\n\nLet $a_k = (4 + 2\\sqrt{3})^k + (4 - 2\\sqrt{3})^k + (-2)^k$. It is easy to check $a_0 = 3$, $a_1 = 6$, $a_2 = 60$, and\n\n$$a_{n+3} = 6a_{n+2} + 12a_{n+1} - 8a_n, \\quad \\forall n \\ge 0.$$ \n\nSince $a_0, a_1, a_2$ are divisible by $3$, $3 \\mid a_n$ for all $n$. Also, $a_{n+3} \\equiv a_n \\pmod{9}$, and none of the first three terms are divisible by $9$, so $v_3(a_n) = 1$ for all $n$.\n\nThus,\n\n$$v_3(u_{3k}) = v_3(u_k) + v_3(a_k) = 1 + v_3(u_k).$$\n\nIf $n = 3^t m$ with $\\gcd(m, 3) = 1$ and $t \\in \\mathbb{Z}^+$, then\n\n$$v_3(u_n) = v_3(u_{3^{t-1}m}) + 1 = \\cdots = v_3(u_m) + t = t.$$ \n\nTherefore, $v_3(u_n) = v_3(n)$ for all $n \\ge 1$.\n\n$\\boxed{}$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20857,
"subject": "Mathematics (Olympiad)",
"question": "In the Venn diagram, the letters $a$, $b$, $c$, $d$, $e$, $g$ represent the number of elements in the respective regions.\n\n\n\n1. Prove that if $f(X, Y) = f(Y, Z) = f(Z, X)$, then $f(X, Y)$ is even.\n\n2. Let $W$ be the set of elements that are in at least two of $X$, $Y$, and $Z$. Show that $f(W, X) = f(W, Y) = f(W, Z) = \\frac{1}{2} f(X, Y)$.\n\n(a) Define\n\n$$\nS = Y \\cap Z \\setminus X, \\quad A = X \\setminus (Y \\cup Z),\n$$\n$$\nT = X \\cap Z \\setminus Y, \\quad B = Y \\setminus (X \\cup Z),\n$$\n$$\nU = X \\cap Y \\setminus Z, \\quad C = Z \\setminus (X \\cup Y).\n$$\n\nShow that $|S| + |A| = |T| + |B| = |U| + |C|$ and deduce that $f(X, Y)$ is even.\n\n(b) Define $W = S \\cup T \\cup U$. Show that $f(W, X) = f(W, Y) = f(W, Z) = \\frac{1}{2} f(X, Y)$.",
"options": [],
"answer": "See solution",
"solution": "Let the letters $a$, $b$, $c$, $d$, $e$, $g$ denote the number of elements in the respective regions of the Venn diagram.\n\n\n\n1. We have\n\n$$\nf(X,Y) = a + g + b + e, \\qquad (1)\n$$\n$$\nf(Y,Z) = b + d + c + g, \\qquad (2)\n$$\n$$\nf(Z,X) = c + e + a + d, \\qquad (3)\n$$\n\nFrom these,\n\n$$\nf(X,Y) = f(Y,Z) + f(Z,X) - 2(c + d), \\qquad (4)\n$$\n$$\nf(Y,Z) = f(X,Y) + f(Z,X) - 2(a + e), \\qquad (5)\n$$\n$$\nf(Z,X) = f(X,Y) + f(Y,Z) - 2(b + g). \\qquad (6)\n$$\n\nSince $f(X, Y) = f(Y, Z) = f(Z, X)$, equation (4) gives $f(X, Y) = 2(c + d)$. Hence $f(X, Y)$ is even.\n\n2. Let $W$ be the set of elements that are in at least two of $X$, $Y$, and $Z$.\n\nThen $f(W, X) = a + e$, $f(W, Y) = b + g$, $f(W, Z) = c + d$. Since $f(X, Y) = f(Y, Z) = f(Z, X)$, equations (4), (5), and (6) give $c + d = a + e = b + g$.\n\nHence $f(W, X) = f(W, Y) = f(W, Z) = c + d = \\frac{1}{2} f(X, Y)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20858,
"subject": "Mathematics (Olympiad)",
"question": "Find the natural numbers $m, n$ such that\n\n$$\nn \\cdot (n + 1) = 3^m + s(n) + 1182,\n$$\n\nwhere $s(n)$ represents the sum of the digits of the positive integer $n$.\n\n*Remark.* If $a$ is a digit, we consider that $s(a) = a$.",
"options": [],
"answer": "See solution",
"solution": "The given equality is equivalent to:\n\n$$\nn^2 = 3^m + 1182 - (n - s(n)).\n$$\n\nIf $m \\geq 2$, since $n - s(n)$ is divisible by $9$, but $3 \\mid 1182$ and $9 \\nmid 1182$, from above it follows that $3 \\mid n^2$ and $9 \\nmid n^2$, which is false. Similarly, if $m = 1$, we obtain that $n^2$ is divisible by $3$ and not by $9$, which is also false. Therefore, $m = 0$, and $n^2 + (n - s(n)) = 1183$.\n\nIf $n \\leq 9$, then $s(n) = n$, so $n^2 = 1183$, which is false. Consequently, $n \\geq 10$. From the equation, we deduce that $n \\leq 34$, therefore $n = \\overline{ab}$, with $a \\in \\{1, 2, 3\\}$. Checking all possibilities, we find that the only solution is $m = 0$, $n = 34$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20859,
"subject": "Mathematics (Olympiad)",
"question": "Suppose there are two boxes with total capacity $n$ (each box has at least capacity 1). Jesse and Tjeerd take turns placing stones with values that are powers of two ($2^k$ for integer $k$) into the boxes, one stone per turn. Jesse moves first. After all stones are placed, the player whose box contains the greater total value wins. Does Jesse have a winning strategy, regardless of how the capacities are split between the two boxes?",
"options": [],
"answer": "See solution",
"solution": "We will show that the capacity of the two boxes does not matter, as long as the total capacity is $n$ (and at least 1 for each box). Jesse can always win this game by first playing the power $2^0 = 1$, and then in each following turn the next power of two that is smaller or greater. That is, if he has played the numbers\n\n$$\n2^{-i}, 2^{-(i-1)}, \\dots, 2^{-1}, 2^{0}, 2^{1}, \\dots, 2^{j-1}, 2^{j}\n$$\n\nat a certain moment, he will play either $2^{-(i+1)}$ or $2^{j+1}$ in his next turn.\n\nBy playing cleverly, Jesse can make sure that the greatest power of two among the stones played so far is always contained in the black box. We will prove this by induction. In his first move, he puts the stone with value $2^0$ in the black box and the claim is true; this is the base case of the induction. When it is his turn again, and Tjeerd moved the greatest power of two so far, which according to the induction hypothesis was contained in the black box, to the white box, then the black box actually has a free space, and Jesse can put a new greater power of two in there, and the claim is true. If Tjeerd moved some other stone or did nothing, then the greatest power of two so far is still in the black box, and Jesse can play a smaller power of two; it does not matter where he puts it. Also in this case, the claim is true. This proves the induction step, and the claim is proved.\n\nTherefore, after playing the last stone, the greatest power of two is in the black box. It is greater than the sum of all smaller powers of two played ($2^j > 2^j - 2^{-i} = 2^{j-1} + 2^{j-2} + \\dots + 2^{-(i-1)} + 2^{-i}$), hence it is certainly greater than the sum of the powers of two in the white box. Therefore, the total value inside the black box is greater than the total value in the white box. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20860,
"subject": "Mathematics (Olympiad)",
"question": "Find the maximum positive integer $n$ such that there exist 8 integers $x_1, x_2, x_3, x_4$ and $y_1, y_2, y_3, y_4$ satisfying\n$$\n\\{0, 1, \\dots, n\\} \\subseteq \\{|x_i - x_j| \\mid 1 \\le i < j \\le 4\\} \\cup \\{|y_i - y_j| \\mid 1 \\le i < j \\le 4\\}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $n$ meet the requirement in the question. Then integers $x_1, x_2, x_3, x_4, y_1, y_2, y_3, y_4$ satisfy that $0, 1, \\dots, n$ all belong to the set $X \\cup Y$, where $X = \\{|x_i - x_j| \\mid 1 \\le i < j \\le 4\\}$, $Y = \\{|y_i - y_j| \\mid 1 \\le i < j \\le 4\\}$.\n\nNote that $0 \\in X \\cup Y$. We may set $0 \\in X$. Then there must be two numbers equal in $x_1, x_2, x_3, x_4$, and we may set $x_1 = x_2$. Then $X = \\{0\\} \\cup \\{|x_i - x_j| \\mid 2 \\le i < j \\le 4\\}$, so\n$$\n|X| \\le 1 + 3 = 4.\n$$\nAnd since $|Y| \\le \\binom{4}{2} = 6$, it follows that $n+1 \\le |X \\cup Y| \\le |X| + |Y| \\le 10$, yielding $n \\le 9$.\n\nOn the other hand, let $(x_1, x_2, x_3, x_4) = (0, 0, 7, 8)$, $(y_1, y_2, y_3, y_4) = (0, 4, 6, 9)$. Then\n$$\nX = \\{0, 1, 7, 8\\}, \\quad Y = \\{2, 3, 4, 5, 6, 9\\},\n$$\nwhich means that $0, 1, \\dots, 9$ all belong to $X \\cup Y$.\n\nIn summary, the maximum positive integer $n$ is $9$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20861,
"subject": "Mathematics (Olympiad)",
"question": "Olya and Tolya have paints of two opposite colors—white and black. They play the following game on the segment $[0, 1]$.\n\nEach round consists of two stages: one player chooses a number $l \\in [0, 1]$, then the other player chooses a segment $J \\subseteq [0, 1]$ of length $l$ and recolors all its points in the opposite color. In the next round, they switch roles.\n\nAfter 2024 rounds, the total length $L$ of the white intervals is calculated. If $L > \\frac{1}{2}$, Olya wins; if $L \\leq \\frac{1}{2}$, Tolya wins. Initially, the entire segment is painted white. Tolya chooses a number first. Who has a winning strategy?\n\n",
"options": [],
"answer": "See solution",
"solution": "Here is a winning strategy for Olya. Let $L_n$ be the total length of the white segments after the $n$-th round (with $L_0 = 1$).\n\n*Statement.* Olya can play so that for $k = 0, 1, \\dots, 1012$, after the $(2k)$-th round, $L_{2k} > \\frac{1}{2}$, and for some $\\delta > 0$, at least one of the open intervals $]0, \\delta[$ or $]1 - \\delta, 1[$ is completely white.\n\nWe prove this by induction.\n\nThe base case $k=0$ is clear.\n\nAssume after the $(2k)$-th round the statement holds. Consider Tolya's move in the $(2k+1)$-th round. If he chooses $l=0$ or $l=1$, then Olya can choose the same $l$ on her turn in round $2k+2$, and the segment $[0, 1]$ will not change color, except possibly at two points if $l=0$.\n\nSuppose Tolya chooses $0 < l < 1$. Since at least one outermost open interval is white, Olya can choose $J$ so that after the $(2k+1)$-th round, both outermost open intervals (for some $\\delta > 0$) remain white. If $L_{2k+1} > \\frac{1}{2}$, Olya simply chooses $l = 0$. If $L_{2k+1} \\leq \\frac{1}{2}$, since the outermost open intervals are white, there exists $\\epsilon > 0$ such that $]0, \\epsilon[$ and $]1-\\epsilon, 1[$ are white. Then, if Olya chooses $l \\in ]1-\\epsilon, 1[$, for example $l = 1-\\frac{\\epsilon}{2}$, whatever segment $J$ Tolya chooses, he will cover all black intervals and leave one or two white intervals of total length $\\epsilon$. Thus, $L_{2k+2} = (1-L_{2k+1}) + \\epsilon > \\frac{1}{2}$. Also, Tolya cannot repaint both outermost intervals black at once.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20862,
"subject": "Mathematics (Olympiad)",
"question": "Triangle $ABC$ has $\\angle ABC = 60^\\circ$. Points $M$ and $D$ are placed on the sides $AC$ and $AB$ respectively, so that $\\angle BCA = 2\\angle MBC$ and $BD = MC$. Find the measure of the angle $\\angle DMB$.",
"options": [],
"answer": "See solution",
"solution": "Take $G$ so that $CG = CM$, $C \\in BG$, and denote $x = \\angle MBC$. Since $\\angle MCB$ is exterior to the isosceles triangle $MCG$, $\\angle MGC = x$. This means that triangle $MBG$ is isosceles and similar to triangle $MCG$. Thus, $$\\frac{BG}{MG} = \\frac{MB}{MC} = \\frac{MG}{BD}.$$ Denote $H$ as the common point of the line $AB$ and the perpendicular bisector of the segment $BG$. Then triangle $HBG$ is isosceles and has an angle of $60^\\circ$, so it is equilateral. This yields $$\\frac{BG}{MG} = \\frac{GH}{MB} = \\frac{MG}{BD}.$$  Since $\\angle MGH = \\angle MBD = 60^\\circ - x$, triangles $MGH$ and $DBM$ are similar, therefore we get the answer $\\angle DMB = \\angle MHG = 30^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20863,
"subject": "Mathematics (Olympiad)",
"question": "For a positive integer $n$, find the first decimal digit of the number\n\n$$\na_n = \\frac{1}{n+1} + \\frac{1}{n+2} + \\dots + \\frac{1}{2n}.\n$$",
"options": [],
"answer": "See solution",
"solution": "The numbers $a_1 = \\frac{1}{2}$ and $a_2 = \\frac{7}{12}$ have the first decimal digit 5. The number\n\n$$\na_3 = \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} = \\frac{37}{60} > 0.6\n$$\n\nhas the first decimal digit 6.\n\nWe will prove that for every $n \\ge 3$, we have $0.6 < a_n < 0.7$. Notice that\n\n$$\na_{n+1} - a_n = \\frac{1}{2n+1} + \\frac{1}{2n+2} - \\frac{1}{n+1} = \\frac{1}{(2n+1)(2n+2)} > 0,\n$$\n\nso $a_n \\ge a_3 > 0.6$, for every $n \\ge 3$.\n\nWe will prove by induction that\n\n$$\na_n \\le 0.7 - \\frac{1}{4n}, \\quad n \\ge 3. \\quad (1)\n$$\n\nFor $n=3$, we have $a_3 = \\frac{37}{60} = 0.7 - \\frac{1}{12}$. Assume that\n\n$$\na_n \\le 0.7 - \\frac{1}{4n}.\n$$\n\nThen we get\n\n$$\na_{n+1} = a_n + \\frac{1}{(2n+1)(2n+2)} \\le 0.7 - \\frac{1}{4n} + \\frac{1}{(2n+1)(2n+2)} < 0.7 - \\frac{1}{4(n+1)},\n$$\n\nsince we have\n\n$$\n\\frac{1}{2(2n+1)(n+1)} \\le \\frac{1}{4n} - \\frac{1}{4(n+1)} \\Leftrightarrow \n\\frac{1}{(n+1)(2n+1)} < \\frac{1}{2n(n+1)} \\Leftrightarrow \n2n < 2n+1,\n$$\n\nand we are done.\n\nFor $n \\ge 3$, the first decimal digit of $a_n$ is 6.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20864,
"subject": "Mathematics (Olympiad)",
"question": "We are trying to solve the equations\n\n$$\nxy = x + y\n$$\n\n$$\n1 = x^2 + y^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "Squaring both sides of $xy = x + y$ and using $x^2 + y^2 = 1$ to simplify gives\n\n$$\nx^2 y^2 = x^2 + 2xy + y^2 = 1 + 2xy \\Leftrightarrow (xy)^2 - 2(xy) - 1 = 0.\n$$\n\nThis gives $xy = 1 \\pm \\sqrt{2}$ or $y = \\frac{1\\pm\\sqrt{2}}{x}$.\n\nNotice that from equation $xy = x + y$, we can write $y = \\frac{x}{x-1}$.\nCombining this with the previous result gives\n\n$$\nx^2 - x(1 \\pm \\sqrt{2}) + 1 \\pm \\sqrt{2} = 0.\n$$\n\nIf $x^2 - x(1 + \\sqrt{2}) + 1 + \\sqrt{2} = 0$, then\n\n$$\nx = \\frac{(1 + \\sqrt{2}) \\pm \\sqrt{(1 + \\sqrt{2})^2 - 4(1 + \\sqrt{2})}}{2} = \\frac{(1 + \\sqrt{2}) \\pm \\sqrt{-1 - 2\\sqrt{2}}}{2}.\n$$\n\nwhich is non-real.\n\nIf $x^2 - x(1 - \\sqrt{2}) + 1 - \\sqrt{2} = 0$, then\n\n$$\nx = \\frac{(1 - \\sqrt{2}) \\pm \\sqrt{(1 - \\sqrt{2})^2 - 4(1 - \\sqrt{2})}}{2} = \\frac{(1 - \\sqrt{2}) \\pm \\sqrt{2\\sqrt{2} - 1}}{2}.\n$$\n\nThis presents a viable solution. Notice that if $(x, y)$ is a solution, then $(y, x)$ is also a solution. The two roots of the above equation are the desired values (since they satisfy equation $xy = x + y$ by Vieta's formulas) and we can have $(x, y) = (y, x) = (r_1, r_2)$ where\n\n$$\nr_1 = \\frac{1 - \\sqrt{2} + \\sqrt{2\\sqrt{2} - 1}}{2} \\quad \\text{and} \\quad r_2 = \\frac{1 - \\sqrt{2} - \\sqrt{2\\sqrt{2} - 1}}{2}.\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20865,
"subject": "Mathematics (Olympiad)",
"question": "Let $Q$ be the intersection of $EF$ and $BC$.\n\nLet $X$ be the other point of intersection of the circumcircles of triangles $AEF$ and $ABC$.\n\nLet $Y$ be the second intersection point of the circumcircles of quadrilaterals $AEFX$ and $PMDX$.\n\nProve that $AM$ passes through $Y$.\n\n",
"options": [],
"answer": "See solution",
"solution": "We have\n\n$$\n\\angle DMQ = \\angle DMB = 90^\\circ = \\angle DPE = \\angle DPQ\n$$\n\nso the quadrilateral $DMPQ$ is cyclic.\n\nConsider the spiral similarity $f_1$ which maps $BC$ to $EF$. Since $A = BE \\cap CF$, the center of $f_1$ is the second intersection point of the circumcircles of triangles $ABC$ and $AEF$, i.e., the point $X$. Since $M$ and $P$ are the midpoints of $BC$ and $EF$, $f_1$ maps $BM$ to $EP$. Since $Q = BM \\cap EP$, the center $X$ of $f_1$ is the second intersection point of the circumcircles of triangles $QBE$ and $QMP$. Therefore, $X, P, M, D, Q$ are concyclic.\n\nLet $Y$ be the second intersection point of the circumcircles of $AEFX$ and $PMDX$. We will prove that $AM$ passes through $Y$.\n\nSince spiral similarities come in pairs and $f_1$ maps $MC$ to $PF$, there exists another spiral similarity $f_2$, with the same center $X$, mapping $MP$ to $CF$. Therefore, the triangles $XMP$ and $XCF$ are similar and so $\\angle XPM = \\angle XFC$. We now have\n\n$$\n\\angle AYM = \\angle AYX + \\angle XYM = \\angle AFX + \\angle XPM = \\angle AFX + \\angle FXC = 180^\\circ.\n$$\n\nSo $Y \\in AM$ as required.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20866,
"subject": "Mathematics (Olympiad)",
"question": "Bestimme alle Paare $ (a, b) $ nichtnegativer ganzer Zahlen, die\n\n$$\n2017^a = b^6 - 32b + 1\n$$\nerfüllen.",
"options": [],
"answer": "See solution",
"solution": "Die zwei Lösungspaare sind $(0, 0)$ und $(0, 2)$.\n\nWeil $2017^a$ ungerade ist, muss $b$ gerade sein, also ist $b = 2c$, $c$ ganz. Folglich ist $2017^a = 64(c^6 - c) + 1$, also gilt $2017^a \\equiv 1 \\pmod{64}$. Es sind aber $2017 \\equiv 33 \\pmod{64}$ und $2017^2 \\equiv (1+32)^2 = 1+2\\cdot32+32^2 \\equiv 1 \\pmod{64}$, sodass die Potenzen von $2017$ modulo $64$ zwischen $1$ und $33$ abwechseln. Deshalb muss $a$ gerade sein. Es gilt also, dass $2017^a$ eine Quadratzahl ist. Wir bezeichnen das Polynom auf der rechten Seite der Gleichung mit $r(b) = b^6 - 32b + 1$ und zeigen, dass es für $b > 4$ zwischen zwei aufeinanderfolgenden Quadratzahlen liegt.\n\nSei also nun $b > 4$. Wir haben $r(b) < b^6 = (b^3)^2$ für $b > 0$. Außerdem gilt $r(b) > (b^3 - 1)^2$, denn $b^6 - 32b + 1 > b^6 - 2b^3 + 1 \\Leftrightarrow b > 4$. Da die Quadratzahl $2017^a$ also zwischen zwei aufeinanderfolgenden Quadratzahlen liegen soll, gibt es in diesem Fall keine Lösungen.\n\nDa $b$ gerade ist, müssen wir nur mehr $b = 4$, $b = 2$ und $b = 0$ überprüfen.\n\nFür $b = 4$ sehen wir sofort, dass die Gleichung modulo $3$ zu $1 \\equiv 1 - 2 + 1 = 0$ wird, also ergibt das keine Lösung.\n\nFür $b = 2$ gilt $2017^a = 2^6 - 32 \\cdot 2 + 1 = 64 - 64 + 1 = 1$, also erhalten wir das Lösungspaar $(a, b) = (0, 2)$.\n\nFür $b = 0$ gilt $2017^a = 0^6 - 32 \\cdot 0 + 1 = 1$, also erhalten wir das Lösungspaar $(a, b) = (0, 0)$.\n\nDas sind also die einzigen Lösungen.\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20867,
"subject": "Mathematics (Olympiad)",
"question": "It is known that $f(x) = ax^3 + bx^2 + cx + d$ ($a \\neq 0$), and $|f'(x)| \\le 1$ for $0 \\le x \\le 1$. Find the maximum value of $a$.",
"options": [],
"answer": "See solution",
"solution": "Let $f'(x) = 3a x^2 + 2b x + c$. We have\n$$\n\\begin{cases}\nf'(0) = c, \\\\\nf'\\left(\\frac{1}{2}\\right) = \\frac{3}{4}a + b + c, \\\\\nf'(1) = 3a + 2b + c.\n\\end{cases}\n$$\nThen,\n$$\n3a = 2f'(0) + 2f'(1) - 4f'\\left(\\frac{1}{2}\\right).\n$$\nSo,\n$$\n\\begin{aligned}\n3|a| &= \\left|2f'(0) + 2f'(1) - 4f'\\left(\\frac{1}{2}\\right)\\right| \\\\\n&\\le 2|f'(0)| + 2|f'(1)| + 4\\left|f'\\left(\\frac{1}{2}\\right)\\right| \\le 8.\n\\end{aligned}\n$$\nTherefore, $a \\le \\frac{8}{3}$. Furthermore, $f(x) = \\frac{8}{3}x^3 - 4x^2 + x + m$ (where $m$ is any constant) satisfies the given condition. Thus, the maximum value of $a$ is $\\frac{8}{3}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20868,
"subject": "Mathematics (Olympiad)",
"question": "Find the minimum positive integer $n$ such that\n\n$$\n\\sqrt{\\frac{n-2011}{2012}} - \\sqrt{\\frac{n-2012}{2011}} < \\sqrt[3]{\\frac{n-2013}{2011}} - \\sqrt[3]{\\frac{n-2011}{2013}}\n$$",
"options": [],
"answer": "See solution",
"solution": "We see that if $2012 \\leq n \\leq 4023$, then $\\sqrt{\\frac{n-2011}{2012}} - \\sqrt{\\frac{n-2012}{2011}} \\geq 0$ and $\\sqrt[3]{\\frac{n-2013}{2011}} - \\sqrt[3]{\\frac{n-2011}{2013}} < 0$.\n\nOtherwise,\n\n$$\n\\begin{aligned}\n\\sqrt{\\frac{n-2011}{2012}} &\\leq \\sqrt{\\frac{n-2012}{2011}} \\quad \\Leftrightarrow \\quad n > 4023 \\\\\n\\sqrt[3]{\\frac{n-2013}{2011}} &\\geq \\sqrt[3]{\\frac{n-2011}{2013}} \\quad \\Leftrightarrow \\quad n \\geq 4024.\n\\end{aligned}\n$$\n\nThus, if $n \\geq 4024$, then\n\n$$\n\\sqrt{\\frac{n-2011}{2012}} - \\sqrt{\\frac{n-2012}{2011}} < 0 \\leq \\sqrt[3]{\\frac{n-2013}{2011}} - \\sqrt[3]{\\frac{n-2011}{2013}}\n$$\n\nSo the minimum of $n$ is $4024$. $\\boxed{4024}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20869,
"subject": "Mathematics (Olympiad)",
"question": "Hallar todos los enteros $n > 1$ para los que es posible escribir en las casillas de un tablero de $n \\times n$ los números enteros desde $1$ hasta $n^2$, sin repeticiones, de modo que en cada fila y en cada columna el promedio de los $n$ números escritos sea un número entero.",
"options": [],
"answer": "See solution",
"solution": "La clave es que en cada fila y columna la suma de los números debe ser un múltiplo de $n$.\n\n**Caso 1: $n$ impar**\n\nCompletamos el tablero de modo que en cada columna haya $n$ números iguales, por lo que su suma es divisible por $n$. La suma en cada fila es $1 + 2 + \\dots + (n-1) = \\frac{1}{2}(n-1) n$, que es divisible por $n$ si $n$ es impar. Por lo tanto, es posible para todo $n$ impar.\n\n\n\n**Caso 2: $n = 4k$**\n\nDividimos el tablero $4k \\times 4k$ en $4k^2$ cuadrados de $2 \\times 2$ y los completamos así:\n\n$$\n\\begin{tabular}{|c|c|}\n\\hline\nx & 4k-x \\\\\n\\hline\n4k-x & x \\\\\n\\hline\n\\end{tabular}\n$$\n\nCada fila y columna suma un múltiplo de $4k$. Se distribuyen los valores de modo que cada número aparece $4k$ veces, cumpliendo la condición.\n\n**Caso 3: $n = 4k+2$**\n\nSi $n=2$, es imposible. Para $k \\geq 1$, completamos el subtablero $4 \\times 4$ superior izquierdo así:\n\n$$\n\\begin{tabular}{|c|c|c|c|}\n\\hline\n1 & 4k+1 & 0 & 0 \\\\\n\\hline\n2k & 2k+2 & 0 & 0 \\\\\n\\hline\n2k+1 & 0 & 2k & 1 \\\\\n\\hline\n0 & 2k+1 & 2k+2 & 4k+1 \\\\\n\\hline\n\\end{tabular}\n$$\n\nEl resto se completa con cuadrados de $2 \\times 2$ como antes:\n\n$$\n\\begin{tabular}{|c|c|}\n\\hline\nx & 4k+2-x \\\\\n\\hline\n4k+2-x & x \\\\\n\\hline\n\\end{tabular}\n$$\n\nAsí, cada número aparece $4k+2$ veces y las sumas de filas y columnas son divisibles por $4k+2$.\n\n**Conclusión:**\n\nEs posible si y solo si $n$ es impar o $n$ es múltiplo de $4$ (es decir, $n$ impar o $n = 4k$ para $k \\geq 1$).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20870,
"subject": "Mathematics (Olympiad)",
"question": "The sides and diagonals of a regular $n$-gon are colored in $k \\ge 3$ colors. For each color $i$, between every two vertices of the polygon there exists a path consisting only of segments of color $i$. Prove that there exist three vertices of the polygon $A$, $B$, and $C$ such that the line segments $AB$, $BC$, and $AC$ are multicolor.",
"options": [],
"answer": "See solution",
"solution": "We need to prove that in a complete graph with $n$ vertices and $k$ colors, where the induced graph on each color is connected, there exists a multicolor triangle.\n\nDenote the colors by $1, 2, 3, \\ldots, k$ and recolor all edges that are in any of the colors $4, 5, \\ldots, k$ into color $3$. The new graph satisfies the condition of connectivity on each color, and if there is a multicolored triangle for it, then the same triangle in the initial graph will also be multicolored. Therefore, we can consider that $k = 3$.\n\nSuppose that the statement is not true for a graph $G$, as we can choose $G$ to have a minimal number of vertices. It follows from the minimality of $G$ that after removing any vertex, the new graph will not be connected by any of the colors, and let it be color $1$. Denote by $G_1, G_2, \\dots, G_t$ the color connectivity components $1$ after deleting vertex $A$. Since $G$ is color $1$ connected, there exist $A_i \\in G_i$ for which $AA_i$ is color $1$. The segment $A_1A_2$ is not color $1$ because $G_1$ and $G_2$ are different components of connectivity. Let it be of color $2$.\n\nIf $A_1B$ is a color $1$ segment of $G_1$, then the segment $A_2B$ cannot be of color $1$ because $G_1$ and $G_2$ are different connectivity components; it cannot be of color $3$, because then $A_1A_2B$ is a multicolor triangle and is therefore of color $2$. Analogously, it is proved that all segments between the points of $G_1$ and $G_2$ are of color $2$. We obtained that all segments between any two connectivity components are either color $2$ or color $3$.\n\nNow consider segments $AX$ and $AY$ in colors $2$ and $3$, respectively (such segments exist, since $G$ is connected in each of the colors). Without limitation, we have the following two cases:\n\n1. $X, Y \\in G_1$, then one of the triangles $AXA_2$ and $AYA_2$ is multicolored.\n2. $X \\in G_1$ and $Y \\in G_2$, then one of the triangles $AXA_2$ and $AYA_1$ is multicolored.\n\nThe resulting contradiction shows that for every graph with the given properties there exists a multicolored triangle. $\\Box$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20871,
"subject": "Mathematics (Olympiad)",
"question": "Construct a polynomial $f(x)$ of degree $n-1$ such that:\n\n1. For all integers $k$ not divisible by $n$, $f(k)$ is an integer.\n2. For some integer $k$ divisible by $n$, $f(k)$ is not an integer.\n\nFind all positive integers $n$ for which such a polynomial exists, and construct $f(x)$ for those $n$.\n\n",
"options": [],
"answer": "See solution",
"solution": "For $n=1$, $f(x) = \\frac{1}{2}$ works.\n\nIf $n = p^a$ (a prime power), set\n$$\nf(x) = \\frac{1}{p} \\binom{x-1}{p^a-1} = \\frac{1}{p} \\cdot \\frac{(x-1)(x-2)\\cdots(x-p^a+1)}{(p^a-1)!}.\n$$\nThis has degree $n-1$ and, by Lemma 1, $f(k)$ is integer for $k$ not divisible by $n$, and not integer for $k$ divisible by $n$.\n\nIf $n$ has at least two prime divisors, no such polynomial exists. Suppose $f(x)$ satisfies the conditions. By Lemma 2 (applied to $g=f$ and $x=-k$),\n$$\n\\binom{n}{k} f(0) = \\sum_{0 \\le l \\le n, l \\ne k} (-1)^{k-l} \\binom{n}{l} f(-k+l).\n$$\nSince $f(-k), \\dots, f(-1)$ and $f(1), \\dots, f(n-k)$ are integers, $\\binom{n}{k}f(0)$ is integer for $1 \\le k \\le n-1$. The GCD of $\\binom{n}{1}, \\dots, \\binom{n}{n-1}$ is 1, so $f(0)$ must be integer, contradicting the requirement. Thus, $n$ must be a prime power.\n\n*Alternative Solution (I. Bogdanov):* For any $n$, any integer-valued polynomial of degree $< n$ can interpolate arbitrary integer values at $1, \\dots, n$. Thus, any $f(x)$ satisfying the conditions must be of the form $f(x) = c \\prod_{i=1}^{n-1} (x-i)$ for rational $c$. If $f(0)$ is not integer, the denominator $q$ of $c$ must not divide $(n-1)!$, so $n$ must be a prime power. For $n = p^\\alpha$, $f(x) = \\frac{1}{p} \\binom{x-1}{n-1}$ works as above.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20872,
"subject": "Mathematics (Olympiad)",
"question": "Consider the Euclidean plane, the points $A = (0,0)$ and $B = (1,0)$, and the open half-strip\n\n$$\nS = \\{ (x, y) : 0 < x < 1,\\ y > 0 \\}\n$$\n\nwith width 1 and vertices $A$ and $B$.\n\nFind all functions $f : S \\to S$ satisfying the following conditions for all $P, Q \\in S$:\n\n1. $f(f(P)) = P$\n2. If $P, Q, A$ are collinear, then $f(P), f(Q)$, and $B$ are collinear.\n3. If $f(P) = P$ and $f(Q) = Q$, then there is a circle containing $A, B, P$, and $Q$.",
"options": [],
"answer": "See solution",
"solution": "Fix any $0 < \\alpha < 90^\\circ$ and consider the ray of points $P \\in S$ with $\\angle BAP = \\alpha$. This ray (or the part of it which is in $S$) is mapped by (ii) under $f$ to a ray starting from $B$ with a certain angle $\\beta = \\beta(\\alpha)$, i.e., $\\angle f(P)BA = \\beta(\\alpha)$ for all such $P$. By (i), this second ray is mapped to the first ray.\n\nNow there is a unique point $P \\in S$ such that $\\angle BAP = \\alpha$ and $\\angle PBA = \\beta$. Since the rays are mapped to each other, $f(P)$ must also lie on both rays and hence $f(P) = P$ is a fixed point. But then (iii) implies that $\\angle APB + \\angle PBA = \\alpha + \\beta(\\alpha)$ is constant.\n\nConsidering $\\alpha \\to 90^\\circ$ and $\\alpha \\to 0^\\circ$, we see that the only possible value of this constant is $90^\\circ$ and hence $\\beta(\\alpha) = 90^\\circ - \\alpha$.\n\nSo we always have $\\angle f(P)BA = 90^\\circ - \\angle PBA$ and by (i) also $\\angle PBA = 90^\\circ - \\angle f(P)BA$, and so $f(P)$ is the orthocenter $H(ABP)$ of $\\triangle ABP$. This function $P \\mapsto H(ABP)$ is indeed a solution.\n\n**First remark.** Alternatively, one can argue that every ray from $A$ and $B$ has to contain a fixed point, which is only possible if the circle from (iii) is the one with diameter $AB$.\n\n**Second remark.** This way, the problem is probably relatively easy, though hopefully not too easy. One can think of many variants of this problem. If the PSC finds a replacement for one or two of the conditions that would make the problem more interesting, the author would be quite happy about that.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20873,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c > 0$ with $a + b + c = 6$. Prove that\n$$\nS = \\sqrt[3]{a^2 + 2bc} + \\sqrt[3]{b^2 + 2ca} + \\sqrt[3]{c^2 + 2ab} \\le 3\\sqrt[3]{12}.\n$$\nFind when equality holds.",
"options": [],
"answer": "See solution",
"solution": "We first try to apply the AM-GM inequality:\n$$\n\\sqrt[3]{c^2 + 2ab} = \\sqrt[3]{(c^2 + 2ab) \\cdot 1 \\cdot 1} \\le \\frac{c^2 + 2ab + 2}{3},\n$$\nthen after summation we would obtain\n$$\n\\begin{aligned}\nS &= \\sqrt[3]{a^2 + 2bc} + \\sqrt[3]{b^2 + 2ca} + \\sqrt[3]{c^2 + 2ab} \\\\\n&\\le \\frac{a^2 + b^2 + c^2 + 2ab + 2bc + 2ca + 6}{3} \\\\\n&= \\frac{(a+b+c)^2 + 6}{3} = \\frac{42}{3} = 14.\n\\end{aligned}\n$$\nHowever, equality in this case is not valid, because we have\n$$\na^2 + 2bc = 1,\\ b^2 + 2ca = 1,\\ c^2 + 2ab = 1 \\implies (a+b+c)^2 = 3,\\ \\text{ which is impossible.}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20874,
"subject": "Mathematics (Olympiad)",
"question": "Determine all triples $ (x, y, p) $ of positive integers such that $ p $ is a prime number, $ x^2 = p - 1 $ and $ y^2 = 2p^2 - 1 $.",
"options": [],
"answer": "See solution",
"solution": "Note that $ (2, 7, 5) $ is such a triple. We show that it is the only one.\n\nLet $ (x, y, p) $ be any such triple. We first compute\n\n$$\n(y + x)(y - x) = y^2 - x^2 = (2p^2 - 1) - (p - 1) = 2p^2 - p = p(2p - 1).\n$$\n\nIn particular, either $ p \\mid x + y $ or $ p \\mid x - y $ must hold.\n\nSuppose $ p \\mid x + y $. Let $ k \\in \\mathbb{Z} $ be such that $ y = kp - x $. Note that by the given property, both $ 2p > y > p $ and $ x < p $ hold. So $ 2p > y = kp - x > kp - p = (k-1)p $ and $ kp > kp - x = y > p $, from which we conclude that $ k=2 $. Therefore $ y = 2p - x $. If we substitute that into the previous equation, we get\n\n$$\n2p(2p - x - x) = p(2p - 1).\n$$\n\nDividing both sides by $ p $, it follows that $ 4(p - x) = 2p - 1 $. However, the left side is even and the right side is odd, so this is a contradiction.\n\nNow suppose $ p \\mid y - x $. Let $ k \\in \\mathbb{Z} $ be such that $ y = kp + x $. Note that by the given property, both $ 2p > y = kp + x > kp $ and $ (k+1)p = kp + p > kp + x = y > p $ hold. It follows that $ k=1 $, so $ y - x = p $, and because of the previous equation also that $ y + x = 2p - 1 $. Subtracting these two equalities, we find that $ 2x = (y + x) - (y - x) = (2p - 1) - p = p - 1 $. We conclude that $ 4(p - 1) = 4x^2 = (2x)^2 = (p - 1)^2 $. Solving this quadratic equation in $ p-1 $ yields that either $ p-1=0 $ or $ p-1=4 $; only in the latter case $ p $ is prime, namely $ p=5 $. Therefore, $ x = (p-1)/2 = 2 $ and $ y = p + x = 5 + 2 = 7 $. $\\square$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20875,
"subject": "Mathematics (Olympiad)",
"question": "The point $O$ is the centre of the circumcircle of triangle $\\Delta ABC$. The line $AO$ intersects the side $BC$ at point $N$, and the line $BO$ intersects the side $AC$ at point $M$. Prove that if $CM = CN$, then $AC = BC$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let us analyze the configuration:\n\n$$\n\\angle AOE = 2\\angle ACE \\text{ and } \\angle BOE = 2\\angle BCE.\n$$\n\nHence,\n\n$\\angle AOE < \\angle BOE$, i.e., $\\overline{AE} < \\overline{BE}$ ($\\triangle ABO$ is isosceles).\n\nFrom Ceva's theorem:\n\n$$\n\\frac{\\overline{AM}}{MC} \\cdot \\frac{\\overline{CN}}{NB} \\cdot \\frac{\\overline{BE}}{EA} = 1,\n$$\n\nand since $\\overline{CM} = \\overline{CN}$ (by condition),\n\nwe get $\\frac{\\overline{AM}}{BN} = \\frac{\\overline{EA}}{\\overline{BE}} < 1$, i.e.,\n\n$$\n\\overline{AC} = \\overline{AM} + \\overline{CM} = \\overline{AM} + \\overline{CN} < \\overline{BN} + \\overline{CN} = \\overline{BC}\n$$\n\nwhich is a contradiction. Analogously, $\\overline{AC} < \\overline{BC}$ is impossible. This implies $\\overline{AC} = \\overline{BC}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20876,
"subject": "Mathematics (Olympiad)",
"question": "已知銳角 $\\triangle ABC$ 不是等腰三角形，點 $O$ 與 $I$ 分別為 $\\triangle ABC$ 的外心與內心。$\\triangle ABC$ 的內切圓分別與三邊 $BC$, $CA$, $AB$ 相切於點 $D$, $E$, $F$。若直線 $AI$ 與 $OD$ 相交於 $P$ 點，$BI$ 與 $OE$ 相交於 $Q$ 點，$CI$ 與 $OF$ 相交於 $R$ 點，且 $M$ 為 $\\triangle PQR$ 的外心。試證：$I$, $M$, $O$ 三點共線。",
"options": [],
"answer": "See solution",
"solution": "解：\n\n(i) 令 $R, r$ 分別為 $\\triangle ABC$ 的外接圓與內切圓半徑。先證明 $OP : PD = R : r$。\n\n證明如下。延長 $AP$ 交 $\\triangle ABC$ 外接圓於 $A'$。因為 $AI$ 平分 $\\angle BAC$，故 $A'$ 為弧 $BA'C$ 的中點，從而 $OA'$ 與 $BC$ 垂直。又 $BC$ 與內切圓相切於 $D$，故 $ID$ 垂直於 $BC$，因此 $ID$ 平行於 $OA'$。故 $\\triangle IPD \\sim \\triangle A'PO$，因此 $OP : PD = OA' : ID = R : r$。得證。\n\n接下來證明原命題。由 (i) 知 $OP : PD = OQ : QE = OR : RF = R : r$，故 $\\triangle PQR$ 是 $\\triangle DEF$ 在以 $O$ 為位似中心，位似比為 $OP : OD = R : (R+r)$ 下進行位似變換後的結果。\n\n故，位似旋轉中心 $O$，$\\triangle DEF$ 的外心 $I$ 與 $\\triangle PQR$ 的外心 $M$ 三點共線。證畢！",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20877,
"subject": "Mathematics (Olympiad)",
"question": "A natural number $n > 1$ is called \"good\" if for every choice of natural numbers $b_1, b_2, \\dots, b_{n-1}$ with $1 \\leq b_1, b_2, \\dots, b_{n-1} \\leq n-1$, there exists a subset $I \\subseteq \\{1, 2, \\dots, n-1\\}$ such that $\\sum_{k \\in I} b_k \\equiv i \\pmod{n}$ for every $i \\in \\{0, 1, \\dots, n-1\\}$ (the empty sum is defined to be zero). Find all \"good\" numbers.",
"options": [],
"answer": "See solution",
"solution": "We will prove that $n$ is \"good\" if and only if it is prime.\n\n**Case 1: $n$ is not prime.**\n\nSuppose $n = rs$ with $1 < r, s < n$. Set $b_1 = b_2 = \\dots = b_{n-1} = r$. Then\n$$\n\\left\\{ \\sum_{k \\in I} b_k \\bmod n : I \\subseteq \\{1, 2, \\dots, n-1\\} \\right\\} = \\{0, r, 2r, \\dots, n - r\\}\n$$\nFor $i = 1$, there does not exist a subset $I$ such that $\\sum_{k \\in I} b_k \\equiv 1 \\pmod{n}$.\n\n**Case 2: $n$ is prime.**\n\nLet $p$ be a prime. We show that for any $b_1, \\dots, b_{p-1}$ with $1 \\leq b_k \\leq p-1$, the subset sums modulo $p$ cover all residues $0, 1, \\dots, p-1$.\n\nWe proceed by induction on the number of elements $r$ in the subset.\n\n- For $r = 1$, the possible sums are $0$ (empty subset) and $b_1$ (non-empty), which are distinct modulo $p$.\n- Assume the claim holds for $r$ elements. For $r+1$ elements $b_1, \\dots, b_r, b$, suppose the subset sums modulo $p$ do not yield at least $r+2$ distinct residues. By the induction hypothesis, the sums $0 = \\sigma_0, \\sigma_1, \\dots, \\sigma_r$ are distinct modulo $p$. The sums $\\sigma_0 + b, \\sigma_1 + b, \\dots, \\sigma_r + b$ must overlap with $\\{\\sigma_0, \\dots, \\sigma_r\\}$, so $0, b, 2b, \\dots, (r+1)b$ are among $\\{\\sigma_0, \\dots, \\sigma_r\\}$. But since $p$ is prime, $ib \\equiv jb \\pmod{p}$ for $0 \\leq i < j \\leq r+1$ implies $(j-i)b \\equiv 0 \\pmod{p}$, which is impossible unless $b \\equiv 0 \\pmod{p}$, contradicting $1 \\leq b \\leq p-1$.\n\nTherefore, all prime $n$ are \"good\" numbers.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20878,
"subject": "Mathematics (Olympiad)",
"question": "In a plane with Cartesian coordinates, let $P$ and $Q$ be two convex polygons (including their boundaries and interiors) whose vertices all have integer coordinates. Let $T = P \\cap Q$. Prove that if $T$ is non-empty and contains no integer point, then $T$ is a non-degenerate convex quadrilateral.\n\n\n\n\n",
"options": [],
"answer": "See solution",
"solution": "Since the non-empty intersection $T$ of two convex closed polygons is a closed convex polygon or a degenerate polygon, there are three possible cases:\n\n1. $T$ is a point. Then $T$ must be a vertex of $P$ or $Q$, contradicting the fact that $T$ contains no integer point.\n\n2. $T$ is a segment. Then $T$ must be the intersection of an edge of $P$ and an edge of $Q$, which contains a vertex of $P$ or $Q$, again a contradiction.\n\n3. $T$ is a closed convex polygon.\n\nIt remains to show that $T$ is a quadrilateral.\n\nIf $T$ has two adjacent edges on the edges of $P$ (or $Q$), then their common vertex must be a vertex of $P$ (or $Q$), which is a contradiction. Thus, the boundary of $T$ alternates between parts of edges of $P$ and $Q$, and each vertex of $T$ is the intersection point of edges of $P$ and $Q$. Therefore, $T$ has an even number of edges.\n\nSuppose $T$ has at least 6 edges. Consider two cases:\n\n**Case 1:** $P$ has no interior integer points (other than its vertices). Then $P$ is either a unit-area triangle or a parallelogram with area 1. In both cases, at least three edges of $P$ are edges of $T$. By geometric arguments (see the figures), this leads to a contradiction, as $Q$ would have to have an integer vertex in the interior of $P$.\n\n**Case 2:** $P$ has at least one interior integer point $X$. Since $X \\notin T$, there exists an edge $MN$ of $T$ such that $X$ and $T$ are separated by the line containing $MN$. This leads to a contradiction by considering the convex hull of $X$ and certain vertices of $P$ and comparing intersections with $Q$.\n\nTherefore, $T$ cannot have more than 4 edges, and since it is not degenerate, $T$ must be a convex quadrilateral.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20879,
"subject": "Mathematics (Olympiad)",
"question": "Докажи дека, за секој непарен број $x$, изразот $x^3 + 3x^2 - x - 3$ е делив со 48.",
"options": [],
"answer": "See solution",
"solution": "Изразот $x^3 + 3x^2 - x - 3$ можеме да го запишеме како\n\n$$\nx^3 + 3x^2 - x - 3 = x^2(x + 3) - (x + 3) = (x + 3)(x - 1)(x + 1).\n$$\n\nБидејќи $x$ е непарен, имаме $x = 2k - 1$, $k \\in \\mathbb{N}$. Тогаш\n\n$$\nx^3 + 3x^2 - x - 3 = (2k - 1 + 3)(2k - 1 - 1)(2k - 1 + 1) = 8(k - 1)k(k + 1).\n$$\n\nЈасно е дека изразот е делив со 8, а $(k-1)k(k+1)$ е делив со 6 како производ на три последователни броја. Значи, изразот е делив со 48 за секој непарен природен број $x$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20880,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $a$ and $b$ such that $a^{4a} = b^b$.",
"options": [],
"answer": "See solution",
"solution": "If $b \\ge 4a$, then $b > a$ and $b^b > a^{4a}$. Therefore, in this case the equality is impossible.\n\nIf $b < 4a$, we have $b^b = a^{4a} = a^{4a-b} a^b$, so $b^b$ is divisible by $a^b$. Therefore, $b$ is divisible by $a$. It follows that $b = n a$, with $n = 1$ (I), $n = 2$ (II), or $n = 3$ (III).\n\nWe get $(a^4)^a = ((n a)^n)^a$, then $a^4 = n^n a^n$, or $a^{4-n} = n^n$. In case (I), we obtain $a = b = 1$; in case (II), we obtain $a = 2$, $b = 4$; while $a = 27$, $b = 81$ in case (III).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20881,
"subject": "Mathematics (Olympiad)",
"question": "Prove that for every real root $x$ of $x^2 + p x + q = 0$, where $p, q \\in \\mathbb{R}$ and $a > 0$, we have\n$$\nx \\ge \\frac{4q - (p + a)^2}{4a}.\n$$",
"options": [],
"answer": "See solution",
"solution": "The equation $x^2 + p x + q = 0$ has real roots, so $p^2 - 4q \\ge 0$. The roots are $x = \\frac{-p \\pm \\sqrt{p^2 - 4q}}{2}$.\n\nWe want to show:\n$$\n\\frac{-p \\pm \\sqrt{p^2 - 4q}}{2} \\ge \\frac{4q - (p + a)^2}{4a}\n$$\nMultiply both sides by $4a$ (since $a > 0$):\n$$\n2a(-p \\pm \\sqrt{p^2 - 4q}) \\ge 4q - (p + a)^2\n$$\nExpand the right side:\n$$\n2a(-p \\pm \\sqrt{p^2 - 4q}) \\ge 4q - (p^2 + 2ap + a^2)\n$$\nBring all terms to one side:\n$$\n2a(-p \\pm \\sqrt{p^2 - 4q}) + p^2 + 2ap + a^2 - 4q \\ge 0\n$$\nGroup terms:\n$$\n(p^2 - 4q) + 2a(-p \\pm \\sqrt{p^2 - 4q}) + 2ap + a^2 \\ge 0\n$$\nNotice $2a(-p) + 2ap = 0$, so:\n$$\n(p^2 - 4q) \\pm 2a\\sqrt{p^2 - 4q} + a^2 \\ge 0\n$$\nThis is equivalent to:\n$$\n(\\sqrt{p^2 - 4q} \\pm a)^2 \\ge 0\n$$\nwhich is always true. Thus, the inequality holds for every real root $x$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20882,
"subject": "Mathematics (Olympiad)",
"question": "Fix integers $n \\ge 2$ and $1 \\le m \\le n-1$. Let $a_0, a_1, \\dots, a_n$ be non-negative real numbers satisfying $a_0 + a_1 + \\dots + a_n = 1$. Prove that, if $\\sum_{k=0}^n a_k x^k < x^m$ for some $0 < x < 1$, then\n$$\n\\sum_{k=0}^{m-1} (m-k)a_k < \\sum_{k=m+1}^n (k-m)a_k.\n$$",
"options": [],
"answer": "See solution",
"solution": "As $a_0 + a_1 + \\dots + a_n = 1$, the required inequality is equivalent to $\\sum_{k=0}^n k a_k > m$. To prove this inequality, note that the exponential $t \\mapsto x^t$, $t \\in \\mathbb{R}$, is convex and apply Jensen's inequality to write\n$$\nx^{\\sum_{k=0}^n k a_k} \\le \\sum_{k=0}^n a_k x^k < x^m.\n$$\nAs $0 < x < 1$, the desired inequality follows by comparing the exponents of $x$ at both ends.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20883,
"subject": "Mathematics (Olympiad)",
"question": "As shown in Fig. 1, $ABCD$ is a convex quadrilateral with $\\angle B + \\angle D < 180^\\circ$, and $P$ is a moving point on the plane. Let\n\n$$\nf(P) = PA \\times BC + PD \\times CA + PC \\times AB.\n$$\n\n1. Prove that $P, A, B, C$ are concyclic when $f(P)$ reaches the minimum.\n\n2. Suppose that point $E$ is on the arc $\\widearc{AB}$ of the circumscribed circle $O$\n\n\n\nof $\\triangle ABC$, satisfying\n\n$$\n\\frac{AE}{AB} = \\frac{\\sqrt{3}}{2}, \\quad \\frac{BC}{EC} = \\sqrt{3} - 1, \\quad \\angle ECB = \\frac{1}{2} \\angle ECA;\n$$\n\nfurthermore, $DA, DC$ are tangent to $\\odot O$, $AC = \\sqrt{2}$. Find the minimum of $f(P)$.",
"options": [],
"answer": "See solution",
"solution": "1. As shown in Fig. 1, by the Ptolemy inequality we have\n\n$$\nPA \\times BC + PC \\times AB \\geq PB \\times AC.\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\nf(P) &= PA \\times BC + PC \\times AB + PD \\times CA \\\\\n&\\geq PB \\times CA + PD \\times CA \\\\\n&= (PB + PD) \\times CA.\n\\end{aligned}\n$$\n\nEquality holds if and only if $P, A, B, C$ lie on $\\odot O$ and $P$ is on $\\widehat{AC}$. Furthermore, $PB + PD \\geq BD$, with equality if and only if $P$ lies on line $BD$. Combining these, we have\n\n$$\nf(P)_{\\min} = BD \\times CA.\n$$\n\nThis completes the proof that $P, A, B, C$ are concyclic when $f(P)$ reaches the minimum.\n\n2. Denote $\\angle ECB = \\alpha$. Then $\\angle ECA = 2\\alpha$. By the sine theorem,\n\n$$\n\\frac{AE}{AB} = \\frac{\\sin 2\\alpha}{\\sin 3\\alpha} = \\frac{\\sqrt{3}}{2}.\n$$\n\nThat is, $\\sqrt{3} \\sin 3\\alpha = 2\\sin 2\\alpha$. Using trigonometric identities,\n\n$$\n\\sqrt{3}(3\\sin \\alpha - 4\\sin^3 \\alpha) = 4\\sin \\alpha \\cos \\alpha.\n$$\n\nSimplifying,\n\n$$\n3\\sqrt{3}\\sin\\alpha - 4\\sqrt{3}\\sin^3\\alpha - 4\\sin\\alpha\\cos\\alpha = 0.\n$$\n\nLet us solve for $\\cos\\alpha$:\n\n$$\n4\\sqrt{3}\\cos^2\\alpha - 4\\cos\\alpha - \\sqrt{3} = 0.\n$$\n\nThe solutions are $\\cos \\alpha = \\frac{\\sqrt{3}}{2}$ and $\\cos \\alpha = -\\frac{1}{2\\sqrt{3}}$ (discarded).\n\nTherefore, $\\alpha = 30^\\circ$ and $\\angle ECA = 60^\\circ$.\n\nOn the other hand,\n\n$$\n\\frac{BC}{EC} = \\sqrt{3} - 1 = \\frac{\\sin(\\angle EAC - 30^\\circ)}{\\sin \\angle EAC}.\n$$\n\nThat is,\n\n$$\n\\frac{\\sqrt{3}}{2}\\sin\\angle EAC - \\frac{1}{2}\\cos\\angle EAC = (\\sqrt{3} - 1)\\sin\\angle EAC.\n$$\n\nThen\n\n$$\n\\frac{2 - \\sqrt{3}}{2} \\sin \\angle EAC = \\frac{1}{2} \\cos \\angle EAC.\n$$\n\nTherefore,\n\n$$\n\\tan \\angle EAC = \\frac{1}{2 - \\sqrt{3}} = 2 + \\sqrt{3}.\n$$\n\nWe obtain $\\angle EAC = 75^\\circ$ and $\\angle AEC = 45^\\circ$.\n\nSince $\\triangle ADC$ is isosceles and $AC = \\sqrt{2}$, we have $CD = 1$. Furthermore, $\\triangle ABC$ is isosceles, so $BC = \\sqrt{2}$ and $AB = 2$. Then\n\n$$\nBD^2 = AB^2 + AD^2 = 4 + 1 = 5.\n$$\n\nWe have $BD = \\sqrt{5}$. Therefore,\n\n$$\nf(P)_{\\min} = BD \\times CA = \\sqrt{5} \\times \\sqrt{2} = \\sqrt{10}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20884,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(x^3 + y^3) = f(x^3) + 3x^3 f(x)f(y) + 3f(x)(f(y))^2 + y^6 f(y)\n$$\n\nfor every two real numbers $x$ and $y$.",
"options": [],
"answer": "See solution",
"solution": "*Answer:* $f(x) = 0$ and $f(x) = x^3$.\n\nSubstituting $x = y = 0$ into the original equation, we obtain $3(f(0))^3 = 0$, which implies\n\n$$\nf(0) = 0. \\qquad (\\text{Eq-1})\n$$\n\nSubstituting $x = 0$ into the original equation and applying (Eq-1), we get\n\n$$\nf(y^3) = y^6 f(y). \\qquad (\\text{Eq-2})\n$$\n\nReplacing the first term on the right-hand side of the original equation by (Eq-2), we find\n\n$$\nf(x^3 + y^3) = x^6 f(x) + 3x^3 f(x)f(y) + 3f(x)(f(y))^2 + y^6 f(y). \\qquad (\\text{Eq-3})\n$$\n\nSwapping the variables $x$ and $y$ in (Eq-3), we find\n\n$$\n3x^3 f(x)f(y) + 3f(x)(f(y))^2 = 3y^3 f(y)f(x) + 3f(y)(f(x))^2. \\qquad (\\text{Eq-4})\n$$\n\nBy collecting similar terms in (Eq-4) and factorizing, we get\n\n$$\nf(x)f(y)(f(x) - x^3 - f(y) + y^3) = 0. \\qquad (\\text{Eq-5})\n$$\n\nSuppose that $f(x) = f(y)$ for some $x \\neq y$. Then $x^3 \\neq y^3$ as cubing is injective, whence $f(x) - x^3 - f(y) + y^3 \\neq 0$. By (Eq-5), we get $f(x)f(y) = 0$, which implies $f(x) = f(y) = 0$ by the choice of $x$ and $y$. Consequently, $f$ cannot take non-zero values more than once.\n\nSuppose now that $f(b) = 0$ for some $b \\neq 0$. Substituting $y = b$ into the original equation, we obtain $f(x^3 + b^3) = f(x^3)$ for all $x$. Now if $f(a) \\neq 0$ for some $a$, then taking $x = \\sqrt[3]{a}$ in the last equality contradicts the previous paragraph since $b^3 \\neq 0$. Hence $f(x) = 0$ for all $x$. This function satisfies the original equation.\n\nIt remains to consider the case where $f(x) = 0$ implies $x = 0$. Substituting $y = 1$ into (Eq-5), we obtain $f(x) - x^3 - f(1) + 1 = 0$ for all non-zero $x$. In other words, $f(x) = x^3 + c$ for all $x \\neq 0$ where $c = f(1) - 1$. After substituting $y = x$ into (Eq-2), simplifying now gives $c = x^6c$ which is possible only if $c = 0$. Hence $f(x) = x^3$ for all $x$. This function also satisfies the original equation.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20885,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge 3$ be a positive integer, set $M = \\{1, 2, \\dots, n\\}$, and let $k > 0$ be a real number. Associate each non-empty subset of $M$ with a point in the plane, such that any two distinct subsets correspond to different points. If the absolute value of the difference between the arithmetic means of the elements of two distinct non-empty subsets of $M$ is at most $k$, connect the points associated with these subsets with a segment. Determine the minimum value of $k$ such that the points associated with any two distinct non-empty subsets of $M$ are connected by a segment or a broken line.",
"options": [],
"answer": "See solution",
"solution": "Denote by $m_A$ the arithmetic mean of the elements of $A$. Notice that for any nonempty subset $A \\subset M$, with $A \\ne \\{1\\}$, we have either $m_A \\ge 2$ if $A \\subset \\{2, 3, \\dots, n\\}$, or $m_A = \\frac{1 + |B| \\cdot m_B}{1 + |B|}$ if $A = \\{1\\} \\cup B$, with $B \\subset \\{2, 3, \\dots, n\\}$. In this case, $m_A = \\frac{1 + |B| \\cdot m_B}{1 + |B|} \\ge \\frac{1 + |B| \\cdot 2}{1 + |B|} \\ge \\frac{3}{2}$.\n\nSince the set $\\{1\\}$ must be connected with at least a subset $A \\ne \\{1\\}$, it follows that $k \\ge \\frac{1}{2}$.\n\nTo show that $\\frac{1}{2}$ is the required minimum, notice that:\n\n* $1 \\le m_A \\le n$ for every nonempty subset $A \\subset M$.\n* Any two one-element subsets are connected with a sequence of subsets.\n\nIndeed, for $k < p$, consider the sequence $\\{k\\}, \\{k, k+1\\}, \\{k+1\\}, \\{k+1, k+2\\}, \\dots, \\{p-1, p\\}, \\{p\\}$. The absolute value of the difference between the arithmetic means of any two consecutive subsets from this sequence is $\\frac{1}{2}$.\n\nLet $A$ and $B$ be two different subsets and $m_A, m_B$ be the arithmetic means of their elements, respectively. Denote by $n_A, n_B$ the closest integers to $m_A$ and $m_B$, respectively. From above, $n_A, n_B \\in [1, n]$, and it is enough to consider the sequence $A, s_{n_A, n_B}, B$, where $s_{n_A, n_B}$ is the sequence described for the subsets $\\{n_A\\}$ and $\\{n_B\\}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20886,
"subject": "Mathematics (Olympiad)",
"question": "Let $f: \\mathbb{Z} \\to \\mathbb{N}^+$ be a function such that for all $m, n \\in \\mathbb{Z}$,\n$$\nf(m-n) \\mid f(m) - f(n)\n$$\nProve that the set $\\{f(m) : m \\in \\mathbb{Z}\\}$ is finite, and if $b_1 < b_2 < \\dots < b_k$ are its distinct values in increasing order, then $b_i \\mid b_j$ for all $1 \\leq i < j \\leq k$.",
"options": [],
"answer": "See solution",
"solution": "We begin by noting the given property for $f: \\mathbb{Z} \\to \\mathbb{N}^+$:\n$$\nf(m-n) \\mid f(m) - f(n) \\quad (1)\n$$\nSetting $n = 0$ gives $f(m) \\mid f(m) - f(0)$, so:\n$$\nf(m) \\mid f(0) \\text{ for all } m \\in \\mathbb{Z} \\quad (2)\n$$\nThus, the set $\\{f(m) : m \\in \\mathbb{Z}\\}$ is finite:\n$$\n\\{f(m) : m \\in \\mathbb{Z}\\} \\text{ is finite} \\quad (3)\n$$\nSetting $m = 0$ in (1) yields $f(-n) \\mid f(0) - f(n)$. By (2), $f(-n) \\mid f(0)$, so $f(-n) \\mid f(n)$. Similarly, $f(n) \\mid f(-n)$, so:\n$$\nf(n) = f(-n) \\text{ for all } n \\in \\mathbb{Z} \\quad (4)\n$$\nLet $b_1 < b_2 < \\dots < b_k$ be the distinct values of $f(m)$ as $m$ ranges over $\\mathbb{Z}$. We want to prove $b_i \\mid b_j$ for all $1 \\leq i < j \\leq k$. It suffices to show $b_i \\mid b_{i+1}$ for $i = 1, 2, \\dots, k-1$.\n\nFrom (1), $f(a_{i+1} - a_i) \\mid b_{i+1} - b_i < b_{i+1}$, and since $f(a_{i+1} - a_i)$ is one of the $b_j$ for $j \\leq i$, we have:\n$$\nb_j \\mid b_{i+1} - b_i\n$$\nIf $j = i$, then $b_i \\mid b_{i+1} - b_i$, so $b_i \\mid b_{i+1}$ as desired.\n\nIf $j < i$, set $m = a_i - a_{i+1}$ and $n = a_i$ in (1) to get $f(-a_{i+1}) \\mid f(a_i - a_{i+1}) - b_i$. By (4), $f(-a_{i+1}) = b_{i+1}$, so $b_{i+1} \\mid b_i - b_j$. But $0 < b_i - b_j < b_i$, so $b_{i+1} < b_i$, a contradiction. Thus, this case does not occur. This completes the proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20887,
"subject": "Mathematics (Olympiad)",
"question": "There are 44 distinct holes on a line and 2017 ants. Each ant crawled up from a hole, then moved to another hole and crawled down. Denote $T$ as the set of time points that the ants crawled up or crawled down from some hole. Suppose that the speeds of the ants are pairwise distinct and they did not change their own speed. Prove that if $|T| \\leq 45$ then there exist two ants that did not meet.\n\nNote: Two ants meet if there exists a time point such that they are at the same location on the line, including the holes.",
"options": [],
"answer": "See solution",
"solution": "We call a time point \"special\" if, at that time, some ant crawled up or down from a hole. It is easy to see that we only need to solve the problem in the case $|T| = 45$ (if $|T| < 45$, we can consider some additional special time points, which only strengthens the argument).\n\nConsider a coordinate system $Oxy$ where $Ox$ represents the locations of the holes on the line and $Oy$ represents time. The coordinates of the holes are $x_1, x_2, \\dots, x_{44}$ and the special times are $y_1, y_2, \\dots, y_{45}$.\n\nAn ant moves from point $(x_a, y_b)$ to point $(x_c, y_d)$ if it crawled up from hole $x_a$ at time $y_b$ and crawled down into hole $x_c$ at time $y_d$. Since the speed of each ant does not change, the graph representing its movement is a segment connecting these two points.\n\nThus, in total, we have 2017 segments, and since the speeds of the ants are pairwise distinct, the lines have different directions. To finish the problem, we need to show that at least two segments among them do not intersect. Note that the number of endpoints of these segments is at most $45 \\times 44 = 1980 < 2017$, so we will prove a generalization: If there are $n$ points on the plane, then there are no more than $n$ segments connecting them such that no two segments are parallel or overlap.\n\n(*)\n\n\n\nWe shall prove (*) by induction. It is easy to check for $n = 2, 3$.\n\nFor $n \\geq 4$, suppose the argument is true for $n-1$ points. Consider two cases:\n\n1. If among $n$ points, there is one point that is the endpoint of at most one segment, then by removing it (and the segment having it as an endpoint), we reduce to the case of $n-1$ points.\n\n2. If each point is the endpoint of at least two segments, then we will show that the number of segments is exactly the same as the number of points. Indeed, if some point is the endpoint of three segments, for example, point $A$ is the endpoint of $AB$, $AC$, $AD$. If point $A$ lies inside triangle $BCD$, then the second segment from $B$ cannot cut both $AC$ and $AD$. Otherwise, suppose the ray $AC$ lies between the rays $AB$ and $AD$; then the second segment from $C$ cannot cut both $AB$ and $AD$. This implies that each point is the endpoint of exactly two segments, so the number of segments is $n$.\n\nHence (*) also holds for $n$, which completes the proof. $\\blacksquare$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20888,
"subject": "Mathematics (Olympiad)",
"question": "Let $Q^+$ be the set of all positive rational numbers. Find all functions $f : Q^+ \\to Q^+$ satisfying $f(1) = 1$ and\n$$\nf(x+n) = f(x) + n f\\left(\\frac{1}{x}\\right)\n$$\nfor all positive integers $n$ and all $x \\in Q^+$.",
"options": [],
"answer": "See solution",
"solution": "First, for a pair of positive integers $(a, b)$ with $a \\neq b$, define an operation: replace the larger number with its remainder upon division by the smaller. (For example, $(2, 5)$ becomes $(2, 1)$, $(5, 2)$ becomes $(1, 2)$.)\n\nLet $g: Q^+ \\to \\mathbb{N}_0$ be defined so that for any positive rational $\\frac{p}{q}$ (with $p, q$ coprime), after $g\\left(\\frac{p}{q}\\right)$ operations, one of $p$ or $q$ becomes $1$ and the other is nonzero.\n\nSince $p, q$ are coprime, $g$ is well-defined for all positive rationals. Once one becomes $1$, the next operation makes the other $0$, and no further operations are possible, so $g$ is unique.\n\nIt is easy to see that $g(x) = g(x^{-1})$. For $0 < x < 1$,\n$$\ng(x+n) = g(x) + 1.\n$$\nHere $n$ is a positive integer.\n\nNow, we use induction to prove $f\\left(\\frac{p}{q}\\right) = p$ for coprime $p, q$.\n\nBase case: If $g\\left(\\frac{p}{q}\\right) = 0$, then $p = 1$ or $q = 1$.\n\nIf $q = 1$, substitute $(x, n) = (1, p-1)$ into the original equation to get\n$$\nf(p) = p.\n$$\nIf $p = 1$, substitute $(q, 1)$ and use the previous result to get $f\\left(\\frac{1}{q}\\right) = 1$.\n\nThus, the claim holds for $g\\left(\\frac{p}{q}\\right) = 0$.\n\nInductive step: Assume the claim holds for $g\\left(\\frac{p}{q}\\right) = i-1$. Consider $g\\left(\\frac{p}{q}\\right) = i$.\n\nCase 1: $\\frac{p}{q} > 1$. Let $n = \\lfloor \\frac{p}{q} \\rfloor$, then $g\\left(\\frac{p-nq}{q}\\right) = i-1$. Substitute $(\\frac{p-nq}{q}, n)$ into the original equation and use the induction hypothesis:\n$$\nf\\left(\\frac{p}{q}\\right) = (p-nq) + n q = p.\n$$\n\nCase 2: $\\frac{p}{q} < 1$. Since $\\frac{q}{p} > 1$, $g\\left(\\frac{p+q}{p}\\right) = g\\left(\\frac{q}{p}\\right) = i$.\n\nAlso, $\\frac{p+q}{p} > \\frac{q}{p} > 1$, so $f\\left(\\frac{p+q}{p}\\right) = p+q$, $f\\left(\\frac{q}{p}\\right) = q$.\n\nSubstitute $(\\frac{q}{p}, 1)$ into the original equation to get $f\\left(\\frac{p}{q}\\right) = (p+q) - q = p$.\n\nBy induction, $f\\left(\\frac{p}{q}\\right) = p$ for all coprime $p, q$.\n\nTherefore, the only function is $f\\left(\\frac{p}{q}\\right) = p$ for coprime positive integers $p, q$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20889,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\overline{AB} = c$, $\\overline{BC} = a$, and $\\overline{CA} = b$ be the sides of triangle $\\triangle ABC$. Let $D \\in BC$, $E \\in CA$, and $F \\in AB$ be the intersection points of the angle bisectors from $A$, $B$, and $C$ with the opposite sides, respectively. If $\\overline{DE} = \\overline{DF}$, then prove:\n\n$$\n\\text{a) } \\frac{a}{b+c} = \\frac{b}{c+a} + \\frac{c}{a+b}, \\qquad \\text{b) } \\angle BAC > 90^{\\circ}\n$$",
"options": [],
"answer": "See solution",
"solution": "**Solution.**\n\n**a)** By the Law of Sines:\n\n$$\n\\frac{\\sin \\angle AFD}{\\sin \\angle FAD} = \\frac{\\overline{AD}}{\\overline{FD}} = \\frac{\\overline{AD}}{\\overline{ED}} = \\frac{\\sin \\angle AED}{\\sin \\angle FAD}\n$$\n\nwhich implies\n\n$$\n\\sin \\angle AFD = \\sin \\angle AED\n$$\n\nSo either $\\angle AFD = \\angle AED$ or $\\angle AFD + \\angle AED = 180^{\\circ}$.\n\nIf $\\angle AFD = \\angle AED$, then $\\angle ADF = \\angle ADE$ and from congruence we have $\\overline{AF} = \\overline{AE}$. Now, because $\\overline{AF} = \\overline{AE}$, we get $\\angle AIF = \\angle AIE$, from which $\\angle AFI = \\angle AEI$, so $\\angle AFC = \\angle AEB$. That is, $\\overline{AC} = \\overline{AB}$, which contradicts the condition that $\\triangle ABC$ is scalene.\n\nTherefore, $\\angle AFD + \\angle AED = 180^{\\circ}$, and the points $A$, $F$, $D$, and $E$ lie on the same circle. We have\n\n$$\n\\angle DEC = \\angle DFA > \\angle ABC\n$$\n\nLet the line segment $CA$ be extended through point $A$ to point $P$ such that $\\angle DPC = \\angle B$ (this is possible because $\\angle DEC = \\angle DFA > \\angle ABC$). Then clearly\n\n$$\n\\overline{PC} = \\overline{PE} + \\overline{CE} \\qquad (1)\n$$\n\nBecause $\\angle BFD = \\angle PED$ and $\\overline{FD} = \\overline{ED}$, we get $\\angle BFD = \\angle PED$, which implies\n\n$$\n\\overline{PE} = \\overline{BF} = \\frac{ac}{a+b} \\qquad (2)\n$$\n\nAlso, $\\triangle PCD \\cong \\triangle BCA$, so\n\n$$\n\\frac{\\overline{PC}}{\\overline{BC}} = \\frac{\\overline{CD}}{\\overline{CA}}\n$$\n\nSo\n\n$$\n\\overline{PC} = \\overline{BC} \\cdot \\frac{\\overline{CD}}{CA} = a \\cdot \\frac{ba}{b+c} \\cdot \\frac{1}{b} = \\frac{a^2}{b+c} \\quad (3)\n$$\n\nClearly,\n\n$$\n\\overline{CE} = \\frac{ab}{c+a} \\quad (4)\n$$\n\nFrom (1), (2), (3), and (4) we get\n\n$$\n\\frac{a^2}{b+c} = \\frac{ac}{a+b} + \\frac{ab}{c+a} \\quad \\text{i.e.} \\quad \\frac{a}{b+c} = \\frac{b}{c+a} + \\frac{c}{a+b}\n$$\n\n**b)** From $\\frac{a}{b+c} = \\frac{b}{c+a} + \\frac{c}{a+b}$ we get\n\n$$\n a(a+b)(a+c) = b(b+a)(b+c) + c(c+a)(c+b)\n$$\n\n$$\n a^2(a+b+c) = b^2(a+b+c) + c^2(a+b+c) + abc > b^2(a+b+c) + c^2(a+b+c)\n$$\n\nSo we have $a^2 > b^2 + c^2$, which means that $\\angle BAC > 90^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20890,
"subject": "Mathematics (Olympiad)",
"question": "Determine the smallest integer $k$ for which the following story could hold true:\n\nIn a chess tournament with 24 players, every pair of players plays at least two and at most $k$ games against each other. In the end of the tournament, it turns out that every player has played a different number of games.",
"options": [],
"answer": "See solution",
"solution": "The answer is $k = 4$.\n\nIf $k = 3$ was possible, then every player plays either 2 or 3 games against each of the other 23 players. Hence, each player plays at least $2 \\cdot 23 = 46$ and at most $3 \\cdot 23 = 69$ games. It is impossible for there to be a player $A$ who has played 46 games (and hence 2 games against every other player) and simultaneously a player $B$ who has played 69 games (and hence 3 games against every other player, including $A$). Thus, there are only 23 numbers available in the range 46, 47, ..., 69, which yields a contradiction.\n\nTo prove that $k = 4$ is possible, we argue by mathematical induction. We show that for every $n \\geq 3$ there exists a tournament $T_n$ with $n$ players, where every pair of players plays at least two and at most four games against each other, and where every player plays a different number of games.\n\nFor $n = 3$, consider three players that play respectively 2, 3, and 4 games against each other; then they play respectively a total of 5, 6, and 7 games.\n\nIn the inductive step, consider the tournament $T_n$ where the players have played $a_1 < a_2 < \\dots < a_n$ games.\n\n(i) If in $T_n$ no player has played exactly two games against every other player, then $a_1 > 2n - 2$. We create a new player and make them play exactly two games against every other player. The new numbers are $2n < a_1 + 2 < a_2 + 2 < \\dots < a_n + 2$.\n\n(ii) Otherwise, no player in $T_n$ can have played exactly four games against every other player, and hence $a_n < 4n - 4$. We create a new player and make them play exactly four games against every other player. The new numbers are $a_1 + 4 < a_2 + 4 < \\dots < a_n + 4 < 4n$.\n\n$\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20891,
"subject": "Mathematics (Olympiad)",
"question": "For each positive integer $n \\ge 3$, define $A_n$ and $B_n$ as\n\n$$A_n = \\sqrt{n^2+1} + \\sqrt{n^2+3} + \\dots + \\sqrt{n^2+2n-1},$$\n$$B_n = \\sqrt{n^2+2} + \\sqrt{n^2+4} + \\dots + \\sqrt{n^2+2n}.$$ \n\nDetermine all positive integers $n \\ge 3$ for which $\\lfloor A_n \\rfloor = \\lfloor B_n \\rfloor$.\n\n*Note.* For any real number $x$, $\\lfloor x \\rfloor$ denotes the largest integer $N$ such that $N \\le x$.",
"options": [],
"answer": "See solution",
"solution": "Let $M = n^2 + \\frac{1}{2}n$.\n\n**Lemma 1.** $B_n - A_n < \\frac{1}{2}$.\n\nIndeed,\n\n$$\n(B_n - A_n) = \\sum_{k=1}^{n} (\\sqrt{n^2 + 2k} - \\sqrt{n^2 + 2k - 1}) = \\sum_{k=1}^{n} \\frac{1}{\\sqrt{n^2 + 2k} + \\sqrt{n^2 + 2k - 1}} < \\sum_{k=1}^{n} \\frac{1}{2n} = \\frac{n}{2n} = \\frac{1}{2}\n$$\n\nproving the lemma.\n\n**Lemma 2.** $A_n < M < B_n$.\n\n*Proof.* Observe that\n\n$$\n(A_n - n^2) = \\sum_{k=1}^{n} (\\sqrt{n^2 + 2k - 1} - n) = \\sum_{k=1}^{n} \\frac{2k - 1}{\\sqrt{n^2 + 2k - 1} + n} < \\sum_{k=1}^{n} \\frac{2k - 1}{n + n} = \\frac{n^2}{2n} = \\frac{n}{2}\n$$\n\nas $\\sum_{k=1}^{n}(2k-1) = n^2$, proving $A_n - n^2 < \\frac{n}{2}$ or $A_n < M$. Similarly,\n\n$$\n(B_n - n^2) = \\sum_{k=1}^{n} (\\sqrt{n^2 + 2k} - n) = \\sum_{k=1}^{n} \\frac{2k}{\\sqrt{n^2 + 2k} + n} > \\sum_{k=1}^{n} \\frac{2k}{(n+1) + n} = \\frac{n(n+1)}{2n+1} > \\frac{n}{2}\n$$\n\nas $\\sum_{k=1}^{n}(2k) = n(n+1)$, so $B_n - n^2 > \\frac{n}{2}$ hence $B_n > M$, as desired. $\\square$\n\n\n\nBy Lemma 2, we see that $A_n$ and $B_n$ are positive real numbers containing $M$ between them. When $n$ is even, $M$ is an integer. This implies $\\lfloor A_n \\rfloor < M$, but $\\lfloor B_n \\rfloor \\ge M$, which means we cannot have $\\lfloor A_n \\rfloor = \\lfloor B_n \\rfloor$.\n\nWhen $n$ is odd, $M$ is a half-integer, and thus $M - \\frac{1}{2}$ and $M + \\frac{1}{2}$ are consecutive integers. So the above two lemmas imply\n\n$$\nM - \\frac{1}{2} < B_n - (B_n - A_n) = A_n < B_n = A_n + (B_n - A_n) < M + \\frac{1}{2}.\n$$\n\nThis shows $\\lfloor A_n \\rfloor = \\lfloor B_n \\rfloor = M - \\frac{1}{2}$.\n\nThus, the only integers $n \\ge 3$ that satisfy the conditions are the odd numbers and all of them work. $\\square$\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20892,
"subject": "Mathematics (Olympiad)",
"question": "Find all integers $n$ such that both $100 \\leq 7n < 1000$ and $100 \\leq 49n < 1000$.",
"options": [],
"answer": "See solution",
"solution": "From the first condition, $100 \\leq 7n < 1000$, so $\\frac{100}{7} \\leq n < \\frac{1000}{7}$, which gives $15 \\leq n < 142$.\n\nFrom the second condition, $100 \\leq 49n < 1000$, so $\\frac{100}{49} \\leq n < \\frac{1000}{49}$, which gives $3 \\leq n < 20.41$, or $3 \\leq n \\leq 20$.\n\nThe intersection is $15 \\leq n \\leq 20$. Thus, there are $6$ such integers: $n = 15, 16, 17, 18, 19, 20$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20893,
"subject": "Mathematics (Olympiad)",
"question": "Is it possible to draw a square with area $50$ units on graphing paper using a ruler that can connect two edges with a line?",
"options": [],
"answer": "See solution",
"solution": "Consider a $10 \\times 10$ square and connect the midpoints of the sides as shown in the figure below. The square obtained clearly has area that is half of the original square's area, that is $$\\frac{1}{2} \\cdot 10 \\cdot 10 = 50.$$ \n\n\n\n*Fig. 17*",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20894,
"subject": "Mathematics (Olympiad)",
"question": "Determine all real-valued functions $f$ on the set of real numbers satisfying the condition\n\n$$\n2f(x) = f(x + y) + f(x + 2y)\n$$\n\nfor all real numbers $x$ and all non-negative real numbers $y$.",
"options": [],
"answer": "See solution",
"solution": "Any constant function is a solution. Conversely, suppose $f(0) = 0$. Fix $y > 0$ and let $x = ny$ for $n \\geq 0$:\n\n$$\n2f(ny) = f((n + 1)y) + f((n + 2)y)\n$$\n\nThis recurrence gives $f(ny) = f(y)\\frac{1 - (-2)^n}{3}$. Thus, $f(y) = -f(2y) = f(4y) = -5f(y)$, so $f$ vanishes identically on $[0, \\infty)$. For any $x \\in \\mathbb{R}$, $2f(x) = f(x + |x|) + f(x + 2|x|) = 0$, so $f$ vanishes identically on $\\mathbb{R}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20895,
"subject": "Mathematics (Olympiad)",
"question": "The function $f : \\mathbb{N}^* \\to \\mathbb{N}$ satisfies the following:\n\n- $f(2) = 0$\n- $f(3) > 0$\n- $f(6042) = 2014$\n- For all $(m, n) \\in \\mathbb{N}^* \\times \\mathbb{N}^*$, $f(m+n) - f(n) - f(m) \\in \\{0, 1\\}$\n\nFind the value of $f(2014)$.\n\nHere, $\\mathbb{N}^* = \\{1, 2, 3, \\dots\\}$.",
"options": [],
"answer": "See solution",
"solution": "Since $f(2) \\geq f(1) + f(1) = 2 f(1)$ and $f(2) = 0$, then $f(1) \\leq 0$. Therefore, $f(1) = 0$.\n\nOn the other hand, from $f(3) > 0$ and $f(3) - f(2) - f(1) \\in \\{0, 1\\}$, we have $f(3) = 1$.\n\nPutting $m = 1$ in the condition $f(m+n) - f(n) - f(m) \\in \\{0, 1\\}$, we obtain $f(n+1) - f(1) - f(n) = f(n+1) - f(n) \\in \\{0, 1\\}$, from which it follows that $f(n+1) \\geq f(n)$ for all positive integers $n$. That is, $f$ is increasing.\n\nNow we will prove by induction that $f(3n) \\geq n$. The case $n = 1$ trivially holds. Assume that $f(3n) \\geq n$ and we have to show that $f(3n + 3) \\geq n + 1$. Indeed, $f(3n + 3) - f(3n) - f(3) \\geq 0$. So, $f(3n + 3) \\geq f(3n) + f(3) \\geq n + 1$.\n\nSince $f(6042) = 2014$, then $f(3n) = n$ for $1 \\leq n \\leq 2014$. Otherwise, if for some $n \\in \\{1, 2, \\dots, 2014\\}$ we have a strict inequality, then it is not possible to obtain $f(6042) = 2014$.\n\nSince $2014 = 3 \\cdot 671 + 1$, if we show that $f(3n + 1) = n$ for all $n \\in \\{1, 2, \\dots, 2014\\}$, then $f(2014) = 671$.\n\nFinally, we will prove that $f(3n+1) = n$. Indeed, $f(3n + 1) \\geq f(3n) + f(1) = f(3n) = n$. On the other hand, $f(9n + 3) \\geq f(6n + 2) + f(3n + 1) \\geq 3 f(3n + 1)$, from which it follows $f(3n + 1) \\leq n + \\frac{1}{3} < n + 1$. Hence, $f(3n + 1) = n$.\n\n**Answer:** $f(2014) = 671$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20896,
"subject": "Mathematics (Olympiad)",
"question": "Let $f(x) = x^8 + x + 1$.\n\nNumerical values get large very quickly:\n\n$$\n\\begin{aligned}\nf(1) &= 3 \\\\\nf(2) &= 259 = 7 \\times 37 \\\\\nf(3) &= 6565 = 5 \\times 13 \\times 101 \\\\\nf(4) &= 65541 = 3 \\times 7 \\times 3121.\n\\end{aligned}\n$$\n\nThese numbers may suggest that $f(n)$ will be a prime number only if $n = 1$. Prove that $n^8 + n + 1$ is prime only for $n = 1$.",
"options": [],
"answer": "See solution",
"solution": "We try to factorise the polynomial $x^8 + x + 1$. If we suspect that $x^2 + x + 1$ is a factor, we can test this using a cubic root of unity $\\omega \\neq 1$, which satisfies $\\omega^2 + \\omega + 1 = 0$ and $\\omega^3 = 1$. Then $\\omega^8 = \\omega^2$, so $f(\\omega) = 0$.\n\nPolynomial division gives the factorisation:\n\n$$\nf(x) = (x^2 + x + 1)(x^6 - x^5 + x^3 - x^2 + 1).\n$$\n\nAlternatively, write $x^8 + x + 1 = x^8 - x^2 + x^2 + x + 1$ and observe:\n\n$$\nx^8 - x^2 = x^2(x^6 - 1) = x^2(x^3 + 1)(x^3 - 1) = x^2(x^3 + 1)(x - 1)(x^2 + x + 1).\n$$\n\nThus,\n\n$$\nf(x) = x^8 + x + 1 = (x^2 + x + 1)(x^2(x^3 + 1)(x - 1) + 1).\n$$\n\nIf $n \\ge 2$, then $n^2 + n + 1 \\ge 7$ and $n^2(n^3 + 1)(n - 1) + 1 \\ge 37$, so $f(n)$ is not a prime number for $n \\ge 2$. Since $f(1) = 3$ is prime, $n = 1$ is the only positive integer for which $n^8 + n + 1$ is prime.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20897,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $a$ and $b$ such that $\\frac{7^a - 5^b}{8}$ is a prime number.",
"options": [],
"answer": "See solution",
"solution": "For any natural number $k$, we have:\n\n- $5^{2k} = M_8 + 1$\n- $5^{2k+1} = M_8 + 5$\n- $7^{2k} = M_8 + 1$\n- $7^{2k+1} = M_8 + 7$\n\nTherefore, from $8 \\mid 7^a - 5^b$, we deduce that $a$ and $b$ are even. Let $a = 2m$, $b = 2n$, with $m$ and $n$ positive integers. Then:\n\n$$\n7^{2m} - 5^{2n} = 8p\n$$\nwhere $p$ is a prime. This can be rewritten as:\n\n$$\n(7^m - 5^n)(7^m + 5^n) = 8p\n$$\n\nIf $p = 2$, then $(7^m - 5^n)(7^m + 5^n) \\neq 16$.\n\nIf $p \\ge 3$, since $7^m - 5^n < 7^m + 5^n$ and both are even, we have two cases:\n\n$$\n\\begin{aligned}\n&\\text{(1)}\\quad \\begin{cases} 7^m - 5^n = 4 \\\\ 7^m + 5^n = 2p \\end{cases} \\\\\n&\\text{(2)}\\quad \\begin{cases} 7^m - 5^n = 2 \\\\ 7^m + 5^n = 4p \\end{cases}\n\\end{aligned}\n$$\n\n**Case (1):** From $7^m = M_3 + 1$ and $5^n = M_3 \\pm 1$, it follows that $7^m - 5^n = M_3$ if $n$ is even, and $7^m - 5^n = M_3 + 2$ if $n$ is odd. Since $4 = M_3 + 1$, there are no solutions in this case.\n\n**Case (2):** Subtracting the equations gives $2p = 7^m - 1$. Since $3 \\mid 7^m - 1$, it follows that $3 \\mid 2p$, so $p = 3$. Thus, $m = n = 1$, $p = 3$, and the solution is $(a, b) = (2, 2)$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20898,
"subject": "Mathematics (Olympiad)",
"question": "Consider the function $f(x) = 3x^3 - 7x^2 + 5x$.\n\nThe sequence $(x_n)$ is defined recursively by $x_{n+1} = f(x_n)$ for all $n = 1, 2, 3, \\ldots$.\n\nStudy the convergence of the sequence $(x_n)$ according to the initial value $x_1 = a$.",
"options": [],
"answer": "See solution",
"solution": "We have $f'(x) = 9x^2 - 14x + 5$.\n\nSince $f(x) - x = 3x^3 - 7x^2 + 4x = x(x - 1)(3x - 4)$, the fixed points are $x = 0$, $x = 1$, and $x = \\frac{4}{3}$.\n\n**Case 1:** $a < 0$\n\nFor $x_n < 0$, $f$ is increasing and $f(x_n) < x_n$, so $(x_n)$ is decreasing and unbounded below; it does not converge.\n\n**Case 2:** $a > \\frac{4}{3}$\n\nFor $x_n > \\frac{4}{3}$, $f$ is increasing and $f(x_n) > x_n$, so $(x_n)$ is increasing and unbounded above; it does not converge.\n\n**Case 3:** $a = 0$\n\nThe sequence is constant: $x_n = 0$ for all $n$, so $\\lim x_n = 0$.\n\n**Case 4:** $a = \\frac{4}{3}$\n\nThe sequence is constant: $x_n = \\frac{4}{3}$ for all $n$, so $\\lim x_n = \\frac{4}{3}$.\n\n**Case 5:** $0 < a < \\frac{4}{3}$\n\n- If $a \\in (1, \\frac{4}{3})$, $(x_n)$ is decreasing and bounded below by $1$, so $\\lim x_n = 1$.\n- If $a \\in (\\frac{1}{3}, 1)$, $x_2 > 1$ and the sequence enters $(1, \\frac{4}{3})$, so $\\lim x_n = 1$.\n- If $a = 1$ or $a = \\frac{1}{3}$, $x_n = 1$ for all $n \\geq 2$, so $\\lim x_n = 1$.\n- If $a \\in (0, \\frac{1}{3})$, eventually $x_k \\in [\\frac{1}{3}, \\frac{4}{3}]$ for some $k$, and then $\\lim x_n = 1$.\n\n**Conclusion:**\n\nThe sequence $(x_n)$ is convergent if and only if $a \\in [0, \\frac{4}{3}]$ and\n\n$$\n\\lim x_n = \\begin{cases}\n0 & \\text{if } a = 0 \\\\\n1 & \\text{if } a \\in (0, \\frac{4}{3}) \\\\\n\\frac{4}{3} & \\text{if } a = \\frac{4}{3}\n\\end{cases}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20899,
"subject": "Mathematics (Olympiad)",
"question": "設 $ABC$ 為銳角三角形，$I$ 及 $I_A$ 分別為其內心及角 $A$ 內的旁心，且 $AB < AC$。設其內切圓交 $BC$ 於點 $D$。直線 $AD$ 分別與 $BI_A$ 及 $CI_A$ 交於點 $E$ 及 $F$。證明三角形 $AID$ 的外接圓與三角形 $I_AEF$ 的外接圓相切。",
"options": [],
"answer": "See solution",
"solution": "令 $\\angle(p, q)$ 表示直線 $p$ 與 $q$ 之間的有向角。\n\n點 $B, C, I, I_A$ 均在以 $II_A$ 為直徑的圓 $\\Gamma$ 上。設 $\\omega$ 及 $\\Omega$ 分別為 $(I_AEF)$ 及 $(AID)$。令 $T$ 為 $\\omega$ 與 $\\Gamma$ 的第二交點。則 $T$ 為由直線 $BC, BI_A, CI_A, DEF$ 所構成的全四邊形的 Miquel 點，因此 $T$ 也在圓 $(BDE)$（以及圓 $(CDF)$）上。我們主張 $T$ 即為 $\\omega$ 與 $\\Omega$ 的切點。\n\n為證明 $T$ 在 $\\Omega$ 上，利用四邊形 $BDET$ 及 $BII_AT$，有：\n\n$$\n\\angle(DT, DA) = \\angle(DT, DE) = \\angle(BT, BE) = \\angle(BT, BI_A) = \\angle(IT, II_A) = \\angle(IT, IA)\n$$\n\n為證明 $\\omega$ 與 $\\Omega$ 在 $T$ 處相切，設 $\\ell$ 為 $\\omega$ 在 $T$ 處的切線，則\n$$\n\\angle(TI_A, \\ell) = \\angle(EI_A, ET)\n$$\n利用圓 $(BDET)$ 及 $(BICI_A)$，有：\n\n$$\n\\angle(EI_A, ET) = \\angle(EB, ET) = \\angle(DB, DT)\n$$\n\n因此，\n\n$$\n\\angle(TI, \\ell) = 90^\\circ + \\angle(TI_A, \\ell) = 90^\\circ + \\angle(DB, DT) = \\angle(DI, DT)\n$$\n\n這證明了 $\\ell$ 在 $T$ 處同時切於 $\\Omega$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20900,
"subject": "Mathematics (Olympiad)",
"question": "Let $x_1, x_2, \\dots, x_n$ (where $n \\ge 2$) be real numbers such that\n$$\nx_1^2 + x_2^2 + \\dots + x_n^2 = 1.\n$$\nProve that\n$$\n\\sum_{k=1}^{n} \\left( 1 - \\frac{k}{\\sum_{i=1}^{n} i x_i^2} \\right)^2 \\cdot \\frac{x_k^2}{k} \\le \\left( \\frac{n-1}{n+1} \\right)^2 \\sum_{k=1}^{n} \\frac{x_k^2}{k}.\n$$\nDetermine when equality holds.",
"options": [],
"answer": "See solution",
"solution": "We expand the left-hand side:\n$$\n\\sum_{k=1}^{n} \\left( 1 - \\frac{k}{\\sum_{i=1}^{n} i x_i^2} \\right)^2 \\cdot \\frac{x_k^2}{k} = \\sum_{k=1}^{n} \\frac{x_k^2}{k} - \\frac{1}{\\sum_{i=1}^{n} i x_i^2}.\n$$\nWe want to show that\n$$\n\\sum_{k=1}^{n} \\frac{x_k^2}{k} - \\frac{1}{\\sum_{i=1}^{n} i x_i^2} \\le \\left( \\frac{n-1}{n+1} \\right)^2 \\sum_{k=1}^{n} \\frac{x_k^2}{k}.\n$$\nThis is equivalent to\n$$\n\\frac{4n}{(n+1)^2} \\sum_{k=1}^{n} \\frac{x_k^2}{k} \\le \\frac{1}{\\sum_{i=1}^{n} i x_i^2},\n$$\nor\n$$\n\\left( \\sum_{k=1}^{n} \\frac{x_k^2}{k} \\right) \\left( \\sum_{k=1}^{n} k x_k^2 \\right) \\le \\frac{(n+1)^2}{4n}.\n$$\n\n**Solution 1.**\nWe rewrite this as\n$$\n4n \\left( \\sum_{k=1}^{n} \\frac{x_k^2}{k} \\right) \\left( \\sum_{k=1}^{n} k x_k^2 \\right) \\le (n+1)^2.\n$$\nBy the AM-GM inequality,\n$$\n\\begin{aligned}\n4n \\left( \\sum_{k=1}^{n} \\frac{x_k^2}{k} \\right) \\left( \\sum_{k=1}^{n} k x_k^2 \\right) &= 4 \\left( \\sum_{k=1}^{n} \\frac{n x_k^2}{k} \\right) \\left( \\sum_{k=1}^{n} k x_k^2 \\right) \\\\\n&\\le \\left( \\sum_{k=1}^{n} \\frac{n x_k^2}{k} + \\sum_{k=1}^{n} k x_k^2 \\right)^2 \\\\\n&= \\left( \\sum_{k=1}^{n} \\left( \\frac{n}{k} + k \\right) x_k^2 \\right)^2.\n\\end{aligned}\n$$\nIt suffices to show that\n$$\n\\frac{n}{k} + k \\le n + 1\n$$\nor\n$$\n0 \\le n k + k - k^2 - n = (n-k)(k-1),\n$$\nwhich is true for $1 \\le k \\le n$.\n\n**Equality case:**\nEquality in AM-GM occurs when all but two $x_k$ vanish, i.e., $x_2 = \\cdots = x_{n-1} = 0$, and $x_1^2 = x_n^2 = \\frac{1}{2}$.\n\n**Solution 2.**\nConsider the quadratic\n$$\nf(t) = n(x_1^2 + 2x_2^2 + \\cdots + n x_n^2)t^2 - (n+1)t + \\left(x_1^2 + \\frac{x_2^2}{2} + \\cdots + \\frac{x_n^2}{n}\\right).\n$$\nThe discriminant is\n$$\n(n+1)^2 - 4n \\left( \\sum_{k=1}^{n} \\frac{x_k^2}{k} \\right) \\left( \\sum_{k=1}^{n} k x_k^2 \\right) \\ge 0.\n$$\nFor $t = \\frac{1}{n}$, each term\n$$\nx_k^2 (n k t - 1) \\left( t - \\frac{1}{k} \\right) = \\frac{x_k^2 (k-1)(k-n)}{n k} \\le 0,\n$$\nso $f(1/n) \\le 0$, confirming the inequality.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20901,
"subject": "Mathematics (Olympiad)",
"question": "設 $a, b$ 為兩正整數，並且 $a!b!$ 是 $a! + b!$ 的倍數。證明：$3a \\ge 2b + 2$。\n\nLet $a, b$ be positive integers such that $a!b!$ is a multiple of $a! + b!$. Prove that $3a \\ge 2b + 2$.",
"options": [],
"answer": "See solution",
"solution": "若 $a > b$，結論明顯成立。當 $a = b$ 時，只有 $(a,b) = (1,1)$ 會讓不等式 $3a \\ge 2b + 2$ 不成立；但 $(a,b) = (1,1)$ 時，$a!b! = 1$ 並不是 $a! + b! = 2$ 的倍數，所以這個情況 $(a=b)$ 也沒問題。\n\n以下假設 $a < b$，且令正整數 $c = b-a$，欲證的不等式成為 $a \\ge 2c + 2$。\n\n若不然，即有 $a \\le 2c + 1$。考慮\n\n$$\nM = \\frac{b!}{a!} = (a+1)(a+2)\\cdots(a+c).\n$$\n\n由條件 $(a! + b!) \\mid a!b!$ 可推得 $(1+M) \\mid a!M$，故 $(1+M) \\mid a!$。當 $c \\ge a$ 時，會有 $M \\ge (a+1)(a+2)\\cdots(a+c) > 1 \\cdot 2 \\cdots a = a!$，不合。故必有 $c < a$。又注意到 $M$ 為 $c$ 個連續正整數的乘積，故 $c! \\mid M$。於是 $\\text{gcd}(1+M,c!) = 1$，所以\n\n$$\n(1+M) \\mid \\frac{a!}{c!} = (c+1)(c+2)\\cdots a. \\qquad (1)\n$$\n\n若 $a \\le 2c$，則 $\\frac{a!}{c!}$ 是不超過 $a$ 的 $a-c$ 個連續正整數的乘積。但 $a-c \\le c$，而且 $M$ 是 $c$ 個大於 $a$ 的連續正整數的乘積，所以必有 $1+M > \\frac{a!}{c!}$，此為矛盾。\n\n最後剩下 $a = 2c+1$ 的情形。由於 $a+1 = 2(c+1)$，可知 $(c+1) \\mid M$。故由 (1) 式知 $(1+M) \\mid (c+2)(c+3)\\cdots a$。這次 $(c+2)(c+3)\\cdots a$ 是不超過 $a$ 的 $a-c-1$ 個連續正整數的乘積，故也必小於 $1+M$，也是不成立。\n\n至此全部討論完畢。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20902,
"subject": "Mathematics (Olympiad)",
"question": "Let $f(b_1, b_2, \\cdots, b_l; n)$ be defined as above. Now let us study the property of $f$.\n\nFirstly, interchanging $b_i$ and $b_j$ for any $1 \\le i, j \\le l$ will not affect the value of $f$. As a matter of fact, it only results in the exchange of the $i$th and $j$th operations in a group, and will not affect the final result after the group's operations. So the value of $f$ remains the same.\n\nSecondly, we only need to count the number of “good”/“second good” groups with property $P$ — a property attributed to any operation group which keeps the numbers on the blackboard distinctive from one another after each operation. We can prove that the difference between the numbers of “good” and “second good” groups with property $P$ is also equal to $f$.\n\nIn fact, we only need to prove that the numbers of “good” and “second good” groups without property $P$ are the same. Suppose the $i$th operation of a “good”/“second good” group without property $P$ results in the equality between the $p$th and $q$th number on the blackboard ($1 \\le p < q \\le n$). We change the following $l-i$ operations in this way: operations on the $p$th number are changed to operations on the $q$th number, and vice versa. It is easy to verify that the resulted permutation on the blackboard of the new operation group would be a $(p, q)$ transposition of the permutation of the original operation group. Then the parities of the two permutations are in opposite signs. And that means the numbers of “good” and “second good” groups without property $P$ are the same.\n\nNow, let $a_1, a_2, \\cdots, a_m$ be $m$ distinct positive integers with their sum less than $n$. Prove by the principle of mathematical induction that\n\n$$\nf(a_1, a_2, \\dots, a_m, -a_1, -a_2, \\dots, -a_m; n) = \\prod_{i=1}^{m} a_i.\n$$",
"options": [],
"answer": "See solution",
"solution": "If $m=1$, consider a “good”/“second good” group with property $P$. It must be in such a way: The first operation is to take a number from $n - a_1 + 1, n - a_2 + 2, \\dots, n$ on the blackboard, and add $a_1$ to it; next operation is to add $-a_1$ again to it. So the number of “good” groups is $a_1$, while that of “second good” groups is $0$. Therefore the formula holds.\n\nAssume that the formula holds for $m-1$. We now consider case $m$. According to what was discussed above, we may assume that $a_1 < a_2 < \\dots < a_m$, and\n\n$$\n\\begin{aligned}\n& f(a_1, a_2, \\dots, a_m, -a_1, -a_2, \\dots, -a_m; n) \\\\\n&= f(a_1, -a_2, -a_3, \\dots, -a_m, a_2, a_3, \\dots, a_m, -a_1; n).\n\\end{aligned}\n$$\n\nFor a group with property $P$, the first operation must be done on the last $a_1$ numbers on the blackboard; the second operation on the first $a_2$ numbers; the third operation on the first $a_2 + a_3$ numbers; ... the $m$th operation on the first $a_2 + \\dots + a_m < n - a_1$ numbers. The $m+1$ to $2m-1$ operations will also be done on the first $n - a_1$ numbers. Otherwise, the sum of the first $n - a_1$ numbers will be less than $1 + 2 + \\dots + (n - a_1)$, a contradiction to property $P$.\n\nTherefore, the $2$ to $2m-2$ operations must be done on the first $n-a_1$ numbers, and the result must be an even/odd permutation of $1, 2, \\dots, (n-a_1)$, which corresponds to each one of $a_1$ even/odd permutations of $1, 2, \\dots, n$ derived from original operation groups. Therefore,\n\n$$\n\\begin{aligned}\n& f(a_1, -a_2, -a_3, \\dots, -a_m, a_2, a_3, \\dots, a_m, -a_1; n) \\\\\n&= a_1 f(-a_2, -a_3, \\dots, -a_m, a_2, a_3, \\dots, a_m; n - a_1).\n\\end{aligned}\n$$\n\nBy induction we have\n\n$$\n\\begin{aligned}\n& f(-a_2, -a_3, \\cdots, -a_m, a_2, a_3, \\cdots, a_m; n - a_1) \\\\\n&= f(a_2, a_3, \\cdots, a_m, -a_2, -a_3, \\cdots, -a_m; n - a_1) \\\\\n&= \\prod_{j=2}^{m} a_j.\n\\end{aligned}\n$$\n\nThat means the formula holds for $m$.\n\nNow take $n = 2007$ and $m = 11$ in the formula. Thus, the value is $\\prod_{j=1}^{11} a_j$.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 20903,
"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$$\nn = a_1^{33} + a_2^{33} + \\dots + a_k^{33}.\n$$\nDetermine whether the set of positive integers $n$ is finite or infinite for which:\n\n- a) $b(n) = 12$\n- b) $b(n) = 12^{12^{12}}$",
"options": [],
"answer": "See solution",
"solution": "a) By Fermat's theorem and the fact that $y^2 \\equiv 1 \\pmod{67}$ implies $y \\equiv \\pm 1 \\pmod{67}$, any $33$rd power modulo $67$ is $0$, $1$, or $66$. Consider the numbers $12^{66k+1}$ for $k \\in \\mathbb{N}$. These can be written as the sum of $12$ $33$rd powers. Also, by Fermat's theorem, $12^{66k+1} \\equiv 12 \\pmod{67}$. Thus, $b(12^{66k+1}) = 12$ for all $k \\in \\mathbb{N}$, 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, for each $x \\in \\mathbb{Z}$ and $k \\in \\mathbb{N}$, $P_k(x)$ is a sum of $2^{k-1}$ $33$rd powers. Also, $P_{33}(x) = 33!x + b$ for some $b \\in \\mathbb{Z}$. Since $1$ and $-1$ are $33$rd powers, every integer is a sum of at most $2^{32} + 33! < 12^{12^{12}}$ $33$rd powers. Thus, the set is empty (and therefore finite).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20904,
"subject": "Mathematics (Olympiad)",
"question": "設 $ABC$ 為銳角三角形，且其外接圓為 $\\omega$，圓心 $O$。令點 $D \\neq B$ 和 $E \\neq C$ 位於 $\\omega$ 上，使得 $BD \\perp AC$ 和 $CE \\perp AB$，且 $CO$ 與 $AB$ 相交於 $X$，$BO$ 與 $AC$ 相交於 $Y$。證明：$\\triangle BXD$ 和 $\\triangle CYE$ 的外接圓在直線 $AO$ 上有共同交點。",
"options": [],
"answer": "See solution",
"solution": "注意到 $AO = OC$，這表示直線 $AO$ 和 $XO$ 關於通過 $O$ 且平行於 $AC$ 的直線 $l$（即 $BD$ 的垂直平分線）互為對稱。設 $P \\neq X$ 為圓 $\\odot BXD$ 與直線 $XO$ 的另一個交點，$Z$ 為圓 $\\odot BXD$ 與直線 $AO$ 上距離 $A$ 較遠的交點。考慮對 $l$ 的反射，這會將 $B$ 映射到 $D$，$AO$ 映射到 $XO$，圓 $\\odot BXD$ 映射到自身，因此 $P$（即 $XO$ 與圓 $\\odot BXD$ 的交點）會被映射到 $Z$（即 $AO$ 與圓 $\\odot BXD$ 上距離 $A$ 較遠的交點）。因此有：\n\n$$\n\\angle OZB = \\angle DPO = \\angle DPX = \\angle DBX = 90^\\circ - \\angle BAC = \\angle OCB\n$$\n\n這表示 $BOCZ$ 共圓。因此，圓 $\\odot BOC$ 與直線 $AO$ 的第二個交點 $Z$ 也在圓 $\\odot BXD$ 上。同理，$Z$ 也在圓 $\\odot CYE$ 上，所以這兩個圓在 $AO$ 上有共同交點 $Z$。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20905,
"subject": "Mathematics (Olympiad)",
"question": "Determine the smallest real constant $c$ such that\n$$\n\\sum_{k=1}^{n} \\left( \\frac{1}{k} \\sum_{j=1}^{k} x_j \\right)^2 \\leq c \\sum_{k=1}^{n} x_k^2,\n$$\nfor all positive integers $n$ and all positive real numbers $x_1, \\dots, x_n$.",
"options": [],
"answer": "See solution",
"solution": "The best constant is $c = 4$.\n\nWe first show that, if $n$ is a positive integer and $x_1, \\dots, x_n$ are positive real numbers, then\n$$\n\\sum_{k=1}^{n} \\left( \\frac{1}{k} \\sum_{j=1}^{k} x_j \\right)^2 + \\frac{2}{n} \\left( \\sum_{k=1}^{n} x_k \\right)^2 < 4 \\sum_{k=1}^{n} x_k^2,\n$$\nso $c \\leq 4$. To prove the above inequality, proceed by induction on $n$. The base case, $n=1$, is clear. For the induction step, let $\\bar{x}_k = (x_1 + \\dots + x_k)/k$, $k \\geq 1$, and notice that it is sufficient to show that $(2n+3)\\bar{x}_{n+1}^2 - 2n\\bar{x}_n^2 < 4x_{n+1}^2$. Since $x_{n+1} = (n+1)\\bar{x}_{n+1} - n\\bar{x}_n$, this is equivalent to\n$$\n2n(2n+1)\\bar{x}_n^2 - 8n(n+1)\\bar{x}_n\\bar{x}_{n+1} + (4n^2+6n+1)\\bar{x}_{n+1}^2 > 0.\n$$\nThe left-hand member is a quadratic form in $\\bar{x}_n$ and $\\bar{x}_{n+1}$ whose discriminant is $-2n$ and the inequality follows.\n\nTo show $c \\geq 4$, we prove that\n$$\n\\sum_{k=1}^{n} \\left( \\frac{1}{k} \\sum_{j=1}^{k} \\frac{1}{\\sqrt{j}} \\right)^2 > 4 \\sum_{k=1}^{n} \\frac{1}{k} - 24.\n$$\nDivergence of the harmonic series settles the case. Write $1/\\sqrt{j} > 2(\\sqrt{j+1} - \\sqrt{j})$, to obtain\n$$\n\\left( \\frac{1}{k} \\sum_{j=1}^{k} \\frac{1}{\\sqrt{j}} \\right)^2 > \\frac{4}{k^2} (\\sqrt{k+1} - 1)^2 > \\frac{4}{k} \\left( 1 - \\frac{2}{\\sqrt{k}} \\right) = \\frac{4}{k} - \\frac{8}{k\\sqrt{k}}.\n$$\nFinally, notice that $\\frac{1}{2k\\sqrt{k}} < \\frac{1}{\\sqrt{k-1}} - \\frac{1}{\\sqrt{k}}$, $k \\geq 2$, to get\n$$\n\\sum_{k=1}^{n} \\frac{1}{k\\sqrt{k}} \\leq 3 - \\frac{2}{\\sqrt{n}} < 3,\n$$\nand deduce thereby the desired inequality.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20906,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be a triangle and the points $K$ and $L$ on $AB$, $M$ and $N$ on $BC$, and $P$ and $Q$ on $CA$ are such that $AK = LB < \\frac{1}{2}AB$, $BM = NC < \\frac{1}{2}BC$, and $CP = QA < \\frac{1}{2}CA$. The intersections of $KN$ with $MQ$ and $LP$ are $R$ and $T$ respectively, and the intersections of $NP$ with $LM$ and $KQ$ are $D$ and $E$ respectively. Prove that the lines $DR$, $BE$, and $CT$ pass through a common point.",
"options": [],
"answer": "See solution",
"solution": "Let $U$, $V$, and $W$ be the intersections of $AB$, $BC$, and $CA$ with $MQ$, $KN$, and $PL$ respectively. Here it is allowed for the points to be infinite points on the respective lines.\n\n\n\nFrom Menelaus' theorem for the triangle $\\triangle ABC$ and the lines $MQ$, $KN$, and $PL$ we get\n\n$$\n\\overline{AU} \\cdot \\overline{UB} = - \\overline{AQ} \\cdot \\overline{QC} \\cdot \\overline{CM} \\cdot \\overline{MB},\n$$\n\n$$\n\\overline{BV} \\cdot \\overline{VC} = - \\overline{BL} \\cdot \\overline{LA} \\cdot \\overline{AP} \\cdot \\overline{PC},\n$$\n\n$$\n\\overline{CW} \\cdot \\overline{WA} = - \\overline{CN} \\cdot \\overline{NB} \\cdot \\overline{BK} \\cdot \\overline{KA}.\n$$\n\nAfter multiplying them we get:\n\n$$\n\\overline{AU} \\cdot \\overline{BV} \\cdot \\overline{CW} = - \\overline{AQ} \\cdot \\overline{CM} \\cdot \\overline{BL} \\cdot \\overline{AP} \\cdot \\overline{CN} \\cdot \\overline{BK} = -1.\n$$\n\nHence by the converse of Menelaus' theorem, the points $U$, $V$, and $W$ are collinear, implying $\\triangle ALP$ and $\\triangle RMN$ are coaxial. Now by Desargues' theorem, $\\triangle ALP$ and $\\triangle RMN$ are copolar, hence $A$, $R$, and $D$ are collinear.\n\nLet $S$ be the intersection of $LP$ and $MQ$. Similarly, $\\triangle BKN$ and $\\triangle SQP$ are coaxial, implying they are copolar, hence $B$, $S$, and $E$ are collinear.\n\nNow since $U$, $V$, and $W$ are collinear, $\\triangle ABC$ and $\\triangle RST$ are coaxial and by Desargues' theorem they are copolar, hence $AR \\equiv DR$, $BS \\equiv BE$, and $CT$ are concurrent. The lines $AR$, $BS$, and $CT$ cannot be parallel as $R$, $S$, and $T$ are inside $\\triangle ABC$. $\\square$\n\n**Comment.** The collinearity of points $U$, $V$, and $W$ leads us to the conclusion that the points $K$, $L$, $M$, $N$, $P$, and $Q$ lie on an ellipse (Braikenridge-Maclaurin Theorem — The Converse of Pascal's Theorem). A lot of properties for this configuration including the given one are shown in \"Geometry in Figures\" — Arseniy Akopyan, Chapter 11.2 — Conics intersecting a triangle.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20907,
"subject": "Mathematics (Olympiad)",
"question": "For any integer $k$, let $N_k$ be the number of ordered 6-tuples $(a, b, c, a', b', c')$ that satisfy the system\n\n$$\nab + a'b' \\equiv bc + b'c' \\equiv ca + c'a' \\equiv 1 \\pmod{k}$$\n\nwhere $a, b, c, a', b', c' \\in \\{0, 1, \\dots, k-1\\}$.\n\nCompute $N_{15}$.",
"options": [],
"answer": "See solution",
"solution": "Let $T_p$ be the number of ordered tuples $(a, b, a', b')$ that satisfy $ab + a'b' \\equiv 1 \\pmod{p}$ with $a, b, a', b' \\in \\{0, 1, \\dots, p-1\\}$. For any pair $(a, a') \\neq (0, 0)$, there are exactly $p$ pairs $(b, b')$ that satisfy the equation. Hence, $T_p = p(p^2 - 1)$.\n\nLet $C_p(t)$ be the number of ordered pairs $(a, b)$ that satisfy $a^2 + b^2 \\equiv t \\pmod{p}$ with $a, b \\in \\{0, 1, \\dots, p-1\\}$. From the above arguments,\n\n$$\nN_p = T_p - \\sum_{t=1}^{p-1} C_p(t) + pC_p(1) = p(p^2 - 1) - p^2 + C_p(0) + pC_p(1).\n$$\n\nIt is easy to compute that $C_3(0) = 1$, $C_3(1) = 4$, $C_5(0) = 9$, $C_5(1) = 4$, which implies that $N_3 = 28$, $N_5 = 124$, and $N_{15} = 28 \\times 124 = 3472$.\n\nTherefore, the number of ordered 6-tuples satisfying the given conditions is $3472$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20908,
"subject": "Mathematics (Olympiad)",
"question": "Let $AM$ be a median in an acute triangle $ABC$. Its extension intersects the circumcircle $w$ of $ABC$ at $P$. Let $AH_1$ be an altitude of $\\triangle ABC$, and $H$ its orthocenter. The rays $MH$ and $PH_1$ intersect $w$ at $K$ and $T$ respectively. Prove that the circumcircle of $\\triangle KTH_1$ is tangent to $BC$.",
"options": [],
"answer": "See solution",
"solution": "It suffices to show that $\\angle TKH_1 = \\angle TH_1B$. Let us extend $KH_1$ and intersect it with $w$ at $S$. Then\n\n$$\n\\angle TKS = \\angle TAB + \\angle BAS, \\quad \\angle TH_1B = \\angle TAB + \\angle PAC,\n$$\n\nso it is sufficient to show that $\\angle PAC = \\angle BAS$.\n\nDenote by $A_1$ the point such that $AA_1$ is the diameter of $w$. Then $\\angle A_1CA = \\angle ABA_1 = 90^\\circ$, so $BH \\parallel A_1C$ and $CH \\parallel A_1B$. Then $BHCA_1$ is a parallelogram, so $HA_1$ passes through the point $M$.\n\nThen $K$ lies on $HA_1$, hence $\\angle A_1KA = 90^\\circ$. Then the quadrilateral $AKH_1M$ is inscribed. So $\\angle KH_1A = \\angle KMA$. Suppose $AH_1$ intersects $w$ again at $F$. Then\n\n$$\n\\angle KH_1A = \\angle KCA + \\angle FAS, \\quad \\angle KMA = \\angle KCA + \\angle PAA_1.\n$$\n\nHence, $\\angle FAS = \\angle PAA_1$. Also, $\\angle ABC = \\angle AA_1C$, so we derive $\\angle BAF = 90^\\circ - \\angle ABC = \\angle A_1AC$. Then\n\n$$\n\\angle BAS = \\angle BAF + \\angle FAS = \\angle PAA_1 + \\angle A_1AC = \\angle PAC.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20909,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a parallelogram and let $M$ be the intersection of its diagonals. The circumcircle of $\\triangle ABM$ intersects the line segment $AD$ in $E \\neq A$, and the circumcircle of $\\triangle EMD$ intersects the line segment $BE$ in the point $F \\neq E$.\n\nProve that $\\angle ACB = \\angle DCF$.",
"options": [],
"answer": "See solution",
"solution": "We first show that $CBFD$ is a cyclic quadrilateral. Note that\n\n$\\angle BCD = \\angle BAD$ (since $ABCD$ is a parallelogram)\n$= \\angle BAE = 180^\\circ - \\angle EMB$ (since $EABM$ is cyclic)\n$= \\angle EMD = \\angle EFD$ (by the inscribed angle theorem in $EFMD$)\n$= 180^\\circ - \\angle BFD$\n\nTherefore, $CBFD$ is a cyclic quadrilateral. Using the inscribed angle theorem with respect to the cyclic quadrilaterals $EABM$, $EFMD$, and $CBFD$ when necessary, we see that\n\n$$\n\\angle ACD = \\angle CAB = \\angle MAB = \\angle MEB = \\angle MEF \\\\\n= \\angle MDF = \\angle BDF = \\angle BCF.\n$$\n\nSo $\\angle ACF + \\angle FCD = \\angle ACD = \\angle BCF = \\angle BCA + \\angle ACF$. If we subtract $\\angle ACF$ from this, we find that $\\angle ACB = \\angle DCF$, as required. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20910,
"subject": "Mathematics (Olympiad)",
"question": "Let $H$ be the orthocenter of triangle $ABC$, and let $A_1$, $B_1$, $C_1$ be the feet of the altitudes from $A$, $B$, and $C$, respectively.\n\nFind\n$$\n\\frac{\\overline{AH}}{HA_1} \\cdot \\frac{\\overline{BH}}{HB_1} \\cdot \\frac{\\overline{CH}}{HC_1}\n$$\nif\n$$\n\\frac{\\overline{AH}}{HA_1} + \\frac{\\overline{BH}}{HB_1} + \\frac{\\overline{CH}}{HC_1} = 2008.\n$$",
"options": [],
"answer": "See solution",
"solution": "We have\n\n$$\n\\frac{\\overline{AH}}{HA_1} + 1 = \\frac{\\overline{AH}}{HA_1} + \\frac{\\overline{HA_1}}{HA_1} = \\frac{\\overline{AA_1}}{HA_1} = \\frac{P_{\\triangle ABC}}{P_{\\triangle HBC}}.\n$$\n\nSimilarly,\n$$\n\\frac{\\overline{BH}}{HB_1} + 1 = \\frac{P_{\\triangle HBC}}{P_{\\triangle HCA}}, \\quad \\frac{\\overline{CH}}{HC_1} + 1 = \\frac{P_{\\triangle HBC}}{P_{\\triangle HAB}}.\n$$\nLet $x = P_{\\triangle HBC}$, $y = P_{\\triangle HCA}$, $z = P_{\\triangle HAB}$.\n\n\n\n$P_{\\triangle ABC} = x + y + z$. Now, the condition\n$$\n\\frac{\\overline{AH}}{HA_1} + \\frac{\\overline{BH}}{HB_1} + \\frac{\\overline{CH}}{HC_1} = 2008\n$$\nis equivalent to\n$$\n\\frac{x+y+z}{x} - 1 + \\frac{x+y+z}{y} - 1 + \\frac{x+y+z}{z} - 1 = 2008,\n$$\nwhich leads to\n$$\n(x+y+z)(yz+xz+xy) = 2011xyz.\n$$\n\nFinally,\n$$\n\\begin{aligned}\n& \\left(\\frac{x+y+z}{x}-1\\right) \\cdot \\left(\\frac{x+y+z}{y}-1\\right) \\cdot \\left(\\frac{x+y+z}{z}-1\\right) = \\frac{y+z}{x} \\cdot \\frac{x+z}{y} \\cdot \\frac{x+y}{z} \\\\\n& = \\frac{(y+z)(x+z)(x+y)}{xyz} = \\frac{(x+y+z)(yz+xz+xy)-xyz}{xyz} = \\frac{2011xyz-xyz}{xyz} = 2010\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20911,
"subject": "Mathematics (Olympiad)",
"question": "Find the number of sequences $\\{a_n\\}_{n=1}^{\\infty}$ of integers such that\n\n$$\na_n \\neq -1 \\quad \\text{and} \\quad a_{n+2} = \\frac{a_n + 2006}{a_{n+1} + 1}\n$$\nfor every positive integer $n$.",
"options": [],
"answer": "See solution",
"solution": "Every sequence satisfying the given conditions is determined by its first two terms. Thus, we seek integer pairs $(a_1, a_2)$ for which all subsequent terms are integers. Writing out the formula for several small values of $n$ and multiplying, we obtain:\n\n$$\n\\begin{align*}\na_3(a_2 + 1) &= a_1 + 2006, \\\\\na_4(a_3 + 1) &= a_2 + 2006, \\\\\na_5(a_4 + 1) &= a_3 + 2006,\n\\end{align*}\n$$\n\nSubtracting adjacent equalities (to eliminate 2006) and rearranging gives:\n\n$$\n\\begin{align*}\na_3 - a_1 &= (a_3 + 1)(a_4 - a_2), \\\\\na_4 - a_2 &= (a_4 + 1)(a_5 - a_3), \\\\\na_5 - a_3 &= (a_5 + 1)(a_6 - a_4),\n\\end{align*}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20912,
"subject": "Mathematics (Olympiad)",
"question": "What is the maximum number of acute-angled triangles that can be formed by the sidelines of a regular polygon with $2n + 1$ sides, where $n = 2015$?",
"options": [],
"answer": "See solution",
"solution": "Let $n = 2015$, so there are $2n + 1$ lines. Fix one line $\\ell$ 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). Any pair from the same group with $\\ell$ forms an obtuse triangle, so the number of obtuse triangles with $\\ell$ as a side adjacent to the obtuse angle is:\n\n$$\n\\binom{a}{2} + \\binom{b}{2} \\geq 2\\binom{n}{2} = n(n-1)\n$$\n\n(by Jensen's inequality, since $\\binom{x}{2}$ is convex).\n\nConsidering all $\\ell$, each obtuse triangle is counted twice. Thus, the number of acute-angled triangles is at most:\n\n$$\n\\binom{2n+1}{3} - \\frac{(2n+1)n(n-1)}{2} = \\frac{n(n+1)(2n+1)}{6}.\n$$\n\nThis bound is attainable: for the sidelines of a regular $(2n+1)$-gon, the numbers of positive and negative slopes are equal for any $\\ell$ by symmetry, so equality holds.\n\nFor $n = 2015$:\n\n$$\n\\frac{2015 \\times 2016 \\times 4031}{6} = 2\\,729\\,148\\,240\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20913,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying the equation:\n\n$$\nf(xy) \\leq y f(x) + f(y)\n$$\n\nfor all $x, y \\in \\mathbb{R}$.",
"options": [],
"answer": "See solution",
"solution": "The given equation is:\n\n$$\nf(xy) \\leq y f(x) + f(y), \\quad \\text{for all } x, y \\in \\mathbb{R}.\n$$\n\nSubstitute $y$ with $-y$ in (1):\n\n$$\nf(-xy) \\leq -y f(x) + f(-y) \\quad (2)\n$$\n\nAdding (1) and (2):\n\n$$\nf(xy) + f(-xy) \\leq f(y) + f(-y), \\quad \\text{for all } x, y \\in \\mathbb{R}.\n$$\n\nSet $y = 1$:\n\n$$\nf(x) + f(-x) \\leq f(1) + f(-1).\n$$\n\nNow, substitute $x$ with $\\frac{1}{y}$ in the previous sum (for $y \\neq 0$):\n\n$$\nf(1) + f(-1) \\leq f(y) + f(-y), \\quad \\text{for } y \\neq 0.\n$$\n\nThus,\n\n$$\nf(y) + f(-y) = f(1) + f(-1) = c, \\quad \\text{for all } y \\neq 0.\n$$\n\nFrom (2):\n\n$$\nc - f(xy) \\leq -y f(x) + c - f(y) \\\\\ny f(x) + f(y) \\leq f(xy), \\quad \\text{for } x, y \\neq 0.\n$$\n\nCombining with the original inequality, we get equality:\n\n$$\nf(xy) = y f(x) + f(y), \\quad \\text{for } x, y \\neq 0.\n$$\n\nSet $x = y = 1$:\n\n$$\nf(1) = 0.\n$$\n\nSwitching $x$ and $y$:\n\n$$\nf(yx) = x f(y) + f(x), \\quad \\text{for } x, y \\neq 0.\n$$\n\nSo,\n\n$$\ny f(x) + f(y) = x f(y) + f(x) \\\\\nf(x)(y-1) = f(y)(x-1) \\\\\n\\frac{f(x)}{x-1} = \\frac{f(y)}{y-1}, \\quad \\text{for } x, y \\neq 0, 1.\n$$\n\nSince $f(1) = 0$, $f(x) = a(x-1)$ for $x \\neq 0$.\n\nNow, set $x = 0$ in the original equation:\n\n$$\nf(y) \\geq (1-y) f(0), \\quad \\text{for all } y.\n$$\n\nSo,\n\n$$\na(y-1) \\geq (1-y) f(0), \\quad \\text{for } y \\neq 0 \\\\\n(y-1)(a + f(0)) \\geq 0, \\quad \\text{for } y \\neq 0.\n$$\n\nThis holds for all $y$ only if $a = -f(0)$. Thus,\n\n$$\nf(x) = f(0)(1 - x), \\quad \\text{for all } x \\in \\mathbb{R}.\n$$\n\nIt is easy to check that this function satisfies the original equation.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20914,
"subject": "Mathematics (Olympiad)",
"question": "Find all triples $(k, m, n)$ of positive integers with the following property: The square with side length $m$ can be cut into some number of rectangles of dimensions $1 \\times k$ and exactly one square of side length $n$.",
"options": [],
"answer": "See solution",
"solution": "The triples $(k, m, n)$ must satisfy $n \\leq m$ and at least one of the following conditions:\n\n$$\n1^{\\circ}\\quad k \\mid m-n,\n$$\n\n$$\n2^{\\circ}\\quad k \\mid m+n \\text{ and } r+n \\leq m, \\text{ where } r \\text{ is the remainder of } m \\text{ modulo } k.\n$$\n\n**Proof:**\n\nSuppose $(k, m, n)$ are as above. We identify the large square with $[0, m] \\times [0, m]$.\n\n- If $k \\mid m-n$, cut the $m \\times m$ square into the square $[0, n] \\times [0, n]$ and rectangles $[0, n] \\times [n, m]$ and $[n, m] \\times [0, m]$. These rectangles can be cut into rectangles of dimension $1 \\times k$.\n- If $k \\mid m+n$ and $r+n \\leq m$, cut $[0, m] \\times [0, m]$ into a square $[r, r+n] \\times [r, r+n]$ and rectangles $[0, r] \\times [0, n+r]$, $[r, m] \\times [0, r]$, $[r+n, m] \\times [r, m]$, and $[0, n+r] \\times [r+n, m]$. Each can be cut into rectangles $1 \\times k$.\n\nNow, we show the conditions are necessary. Suppose the smaller square is $[p, p+n] \\times [q, q+n]$; by symmetry, assume $q \\geq 1$.\n\nFirst, $r+n \\leq m$ must hold. Each unit square $[i, i+1] \\times [0, 1]$, where $i \\in \\{p, p+1, \\dots, p+n-1\\}$, is contained in some rectangle $1 \\times k$. If $r+n > m$, then $m-n < k$, so each rectangle is \"level\", i.e., $[t, t+k] \\times [0, 1]$. Let $S$ be the union of these rectangles, and $P(S)$ its area. $n \\leq P(S) \\leq m$ and $P(S)$ is divisible by $k$, contradicting $r+n > m$.\n\nNext, $k \\mid m-n$ or $k \\mid m+n$ must hold. Suppose not. Assign an integer $a_{ij}$ to each unit square $[i-1, i] \\times [j-1, j]$, $1 \\leq i, j \\leq m$, so that:\n\n(a) For any rectangle $1 \\times k$, the sum inside equals $0$.\n\n(b) The sum over all squares and the sum inside the $n \\times n$ square are different.\n\nLet $a_{ij} = a_i b_j$, where $(a_i)_{i=1}^m$ and $(b_j)_{j=1}^n$ are $k$-periodic and\n$$\n\\sum_{i=1}^{k} a_i = \\sum_{j=1}^{k} b_j = 0. \\qquad (1)\n$$\nThis implies (a); (b) becomes\n$$\n\\sum_{i=1}^{m} a_i \\cdot \\sum_{j=1}^{m} b_j \\neq \\sum_{i=p+1}^{p+n} a_i \\cdot \\sum_{j=q+1}^{q+n} b_j,\n$$\nBy periodicity and (1),\n$$\n\\sum_{i=1}^{r} a_i \\cdot \\sum_{j=1}^{r} b_j \\neq \\sum_{i=p+1}^{p+s} a_i \\cdot \\sum_{j=q+1}^{q+s} b_j, \\qquad (2)\n$$\nwhere $s$ is the remainder of $n$ divided by $k$.\n\nIf $r=0$, the left side is $0$. As $k \\nmid m-n$, $s > 0$; set $a_{p+1} = \\dots = a_{p+s} = b_{q+1} = \\dots = b_{q+s} = 1$ and choose the rest to satisfy periodicity and (1).\n\nIf $r > 0$, $k \\nmid m-n$, $k \\nmid m+n$ imply $r \\neq s$, $r+s \\neq k$. Set $b_1 = \\dots = b_r = 1$ and pick the rest to satisfy periodicity and (1). Let $A = \\{1, \\dots, r\\}$, $B = \\{p+1, \\dots, p+s\\} \\pmod k$.\n\n- If $A \\cup B \\neq \\{0, 1, \\dots, k-1\\}$, choose $a_i$ so $\\sum_{i \\in A} a_i = 1$, $\\sum_{i \\in B} a_i = 0$ and complete the sequence.\n- If $A \\cup B = \\{0, 1, \\dots, k-1\\}$, pick $i_1 \\in A \\setminus B$, $i_2 \\in B \\setminus A$, $i_3 \\in A \\cap B$, set $a_{i_1} = 0$, $a_{i_2} = -1$, $a_{i_3} = 1$, $a_i = 0$ otherwise.\n\nIn both cases, the right side of (2) is $0$, the left is not.\n\nThus, the proof is complete.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20915,
"subject": "Mathematics (Olympiad)",
"question": "Determine all digits $z$ such that for each integer $k \\ge 1$ there exists an integer $n \\ge 1$ with the property that the decimal representation of $n^9$ ends with at least $k$ digits $z$.",
"options": [],
"answer": "See solution",
"solution": "This is possible for $z \\in \\{0, 1, 3, 7, 9\\}$.\n\nFor $z=0$, we can take $n = 10^l$ for any sufficiently large integer $l$ such that $9l \\ge k$.\n\nFor $z \\in \\{2, 4, 6, 8\\}$, $n^9$ is even, so $n$ must be even, and $n^9$ must be divisible by $2^9$. However, numbers ending with 222, 444, or 666 are not divisible by 8, and numbers ending with 8888 are not divisible by 16. Therefore, there is no solution for these values of $z$.\n\nFor $z=5$, $n^9$ is divisible by 5, so $n$ must be divisible by 5, and $n^9$ must be divisible by $5^9$. However, numbers ending with 55 are not divisible by 25.\n\nFor $z \\in \\{1, 3, 7, 9\\}$, let $b = \\underbrace{zzz\\dots z}_{k}$. Since $\\gcd(9, \\varphi(10^k)) = \\gcd(9, 4 \\cdot 10^{k-1}) = 1$, by the Euclidean algorithm there exist integers $x$ and $y$ such that $9x + \\varphi(10^k)y = 1$. We claim that $n = b^x$ has the desired property. Because $\\gcd(b, 10^k) = 1$, this follows from Euler's theorem:\n\n$$\n(b^x)^9 = b^{9x} \\equiv b^{9x+\\varphi(10^k)y} = b^1 = b \\pmod{10^k}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20916,
"subject": "Mathematics (Olympiad)",
"question": "Given an acute scalene triangle $ABC$ inscribed in circle $(O)$. Let $H$ be its orthocenter and $M$ be the midpoint of $BC$. Let $D$ be a point on the opposite ray of $HA$ such that $BC = 2DM$. Let $D'$ be the reflection of $D$ through line $BC$ and $X$ be the intersection of $AO$ and $MD$.\n\n\na) Show that $AM$ bisects $D'X$.\n\nb) Similarly, we define the points $E, F$ like $D$ and $Y, Z$ like $X$. Let $S$ be the intersection of tangents from $B, C$ of $(O)$. Let $G$ be the projection of the midpoint of $AS$ to the line $AO$. Show that there exists a point with the same power to the circumcircles of triangles $BEY, CFZ, SGO$ and $(O)$.\n\n",
"options": [],
"answer": "See solution",
"solution": "a) Let $H'$ be the reflection of $H$ over $M$. Since $BH' \\parallel CH$, $CH \\perp AB$; it implies that $BH' \\perp AB$ so $AH'$ is the diameter of $(O)$. Let $H_a$ be the midpoint of $DD'$, then $DH_a$ is the altitude of right triangle $DBC$, hence\n\n$$\nH_a D^2 = H_a B \\cdot H_a C = H_a H \\cdot H_a A.\n$$\n\nSince $(HA, DD') = -1$, it implies that $X(HA, DD') = -1$. It is well-known that $M$ is the midpoint of $HH'$, thus $HH' \\parallel D'X$. Therefore, $AM$ bisects the segment $D'X$.\n\nb) First, we will prove that there exists a point with the same power to four circles $(AXD)$, $(BYE)$, $(CZF)$ and $(O)$. We will prove the following lemma.\n\n**Lemma.** The circle $(AXD)$ meets $(O)$ again at $L_a$. Then $AL_a$ is the symmedian of triangle $ABC$.\n\n*Proof.* Let $H_c$ be the projection of $H$ on $AB$, then it is clear that $H_c, H, H_a$ and $B$ are concyclic. We also obtain that\n\n$$\n\\overline{AH} \\cdot \\overline{AH_a} = \\overline{AH_c} \\cdot \\overline{AB} = \\overline{AD'} \\cdot \\overline{AD}\n$$\n\nso $\\frac{AD'}{AH} = \\frac{AH_a}{AD}$. Note that $D'X \\parallel HH'$ so $\\frac{AD'}{AH} = \\frac{AX}{AH'}$ and it implies $H_aX \\parallel DH'$ or $AD \\cdot AX = AH_a \\cdot AH' = AB \\cdot AC$. Moreover, $AD, AX$ are isogonal with respect to $\\angle BAC$ so $D, X$ can be obtained by a symmetric inversion of power $AB \\cdot AC$ and the axis is the angle bisector of angle $BAC$. Since $DX$ passes through the midpoint $M$, $AL_a$ is the symmedian of triangle $ABC$. $\\square$\n\nBack to the problem, according to the lemma, the symmedian $AL_a$ is the radical axis of $(AXD)$ and $(O)$; similarly, the symmedians from $B, C$ are the radical axes of $(BYE)$, $(CZF)$ with $(O)$. Hence, the Lemoine point $L$ of triangle $ABC$ has the same power to four circles $(AXD), (BYE), (CZF)$ and $(O)$. Next, we will prove that $L$ has the same power to $(SGO)$ and $(O)$.\n\nIt is well-known that $N, L$ and $M$ are collinear where $N$ is the midpoint of $AH_a$. Let $U, V$ be the intersection of $MN$ with $(O)$. Note that $\\triangle AMH_a \\sim \\triangle AJG$, thus $\\triangle AMN \\sim \\triangle ASG$, which implies $\\angle OGS = \\angle AGS = \\angle ANM = \\angle OMV$ (because $AN \\parallel OM$).\n\n\n\nSince $OM \\cdot OS = OV^2$, we get $\\angle OMV = \\angle OVS$ and $\\angle OGS = \\angle OVS$, it follows that $O, G, V$ and $S$ lie on the same circle. On the other hand,\n\n$$\nMO \\cdot MS = MB \\cdot MC = MU \\cdot MV\n$$\n\nso 5 points $O, G, U, V$ and $S$ lie on the same circle. Hence, $L$ has the same power to $(AXD), (BYE), (CZF), (SGO)$ and $(O)$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20917,
"subject": "Mathematics (Olympiad)",
"question": "(a) The two triangles have the same area.\n\n(b) Let $M$ and $W$ be the respective midpoints of sides $BC$ and $YZ$. The two sets of lengths $\\{AB, AM, AC\\}$ and $\\{XY, XW, XZ\\}$ are identical 3-element sets of pairwise relatively prime integers.\n\nDetermine if there are infinitely many pairs of triangles that are pals of each other.",
"options": [],
"answer": "See solution",
"solution": "The answer is *yes*.\n\nWe start with the following observations.\n\n**Lemma 2.** The following statement and its converse are both true: If $q, r, s$ are three distinct positive real numbers such that $q, r, 2s$ are side lengths of a triangle, then there is a unique triangle $PQR$ with $PQ = q$, $PR = r$, and $PS = s$, where $S$ is the midpoint of side $QR$.\n\n*Proof.* If $q, r, 2s$ are side lengths of a triangle, we construct the unique triangle $PRQ_1$ with $PR = r$, $RQ_1 = q$, and $PQ_1 = 2s$. Let $S$ be the midpoint of side $PQ_1$. Extend segment $RS$ through $S$ to $Q$ so that $PQQ_1R$ is a parallelogram. It is clear that $PQR$ is a triangle with $PQ = q$, $PR = r$, and $PS = s$, where $S$ is the midpoint of side $QR$. To prove the converse statement, we only need to note that the above procedure can be reversed. $\\Box$\n\n**Lemma 3.** Let $PQR$ be a triangle with $PQ = q$, $PR = r$, and $PS = s$, where $S$ is the midpoint of side $QR$. Then the area of triangle $PQR$ is\n\n$$\n\\frac{1}{4}\\sqrt{2q^2r^2 + 8r^2s^2 + 8s^2q^2 - q^4 - r^4 - 16s^4}.\n$$\n\n*Proof.* Because $PQQ_1R$ is a parallelogram, triangles $PQR$ and $PRQ_1$ have the same area. Because $(PR, RQ_1, Q_1P) = (r, q, 2s)$, the desired result follows directly from Heron's formula. $\\Box$\n\nIn view of Lemma 2, if triangles $ABC$ and $XYZ$ are a pair of pals, we may assume without loss of generality that $(AB, AC, AM) = (n, s, t)$ and $(XY, XZ, XW) = (n, t, s)$. By Lemma 3, we have\n\n$$\n2n^2s^2 + 8s^2t^2 + 8t^2n^2 - n^4 - s^4 - 16t^4 = 2n^2t^2 + 8t^2s^2 + 8s^2n^2 - n^4 - t^4 - 16s^4\n$$\n\nwhich simplifies to\n\n$$\n6n^2t^2 - 6n^2s^2 = 15t^4 - 15s^4.\n$$\n\nBecause $ABC$ and $XYZ$ are incongruent, we deduce that $2n^2 = 5(t^2 + s^2)$. We set $n = 5(k^2 + (k+1)^2) = 10k^2 + 10k + 1$ for some positive integer $k$. By applying the identity $(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 - (ad - bc)^2$ repeatedly, we have\n\n$$\n\\begin{aligned}\nt^2 + s^2 &= 10n^2 = (1^2 + 3^2)(k^2 + (k+1)^2)(k^2 + (k+1)^2) \\\\\n&= ((4k+3)^2 + (2k-1)^2)(k^2 + (k+1)^2) \\\\\n&= (6k^2 + 4k - 1)^2 + (2k^2 + 8k + 3)^2.\n\\end{aligned}\n$$\n\nWe set $(n, s, t) = (10k^2 + 10k + 1, 6k^2 + 4k - 1, 2k^2 + 8k + 3)$ for positive integer $k$. For large $k$, it is easy to see that each of $(n, 2s, t)$ and $(n, s, 2t)$ is a set of side lengths of a triangle. Furthermore, by the Euclidean algorithm, it is not difficult to check these numbers are pairwise relatively prime to each other, hence $(n, 2s, t)$ and $(n, s, 2t)$ are the side lengths of a pair of pals, completing the solution.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20918,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle, $I$ the incenter, $D$ the contact point of the incircle with the side $BC$, and $E$ the foot of the angle bisector from $A$. If $M$ is the midpoint of the arc $BC$ (of the circumcircle of $ABC$) that contains $A$, and $F = DI \\cap AM$, prove that $MI$ passes through the midpoint of $[EF]$.\n\n",
"options": [],
"answer": "See solution",
"solution": "The bisector of angle $A$ passes through the midpoint $S$ of the arc $BC$ not containing $A$. $MS$ is the perpendicular bisector of $[BC]$, so $MS \\parallel ID$ (both are perpendicular to $BC$). Also, $\\angle ASC = \\angle ABC$ and $\\angle SAC = \\angle BAE$, so triangles $SAC$ and $BAE$ are similar, giving $\\frac{AB}{BE} = \\frac{AS}{SC}$. By the angle bisector theorem, $\\frac{AB}{BE} = \\frac{AI}{IE}$. It is known that $SI = SC$, so $\\frac{AS}{SC} = \\frac{AS}{SI} = \\frac{AM}{MF}$. Thus, $\\frac{AI}{IE} = \\frac{AM}{MF}$. Applying Menelaus's theorem in triangle $AEF$ with transversal $I-X-M$ (where $X = IM \\cap EF$),\n$$\n\\frac{AI}{IE} \\cdot \\frac{EX}{XF} \\cdot \\frac{MF}{MA} = 1.\n$$\nUsing the previous equality, $\\frac{EX}{XF} = 1$, so $X$ is the midpoint of $[EF]$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20919,
"subject": "Mathematics (Olympiad)",
"question": "Prove that for any positive integer $n$, the sum of the first $n$ primes is greater than $n^2$.",
"options": [],
"answer": "See solution",
"solution": "First, notice that the $n$-th prime $p_n$ satisfies the inequality $p_n \\ge 2n - 1$. Indeed, the claim holds for the first prime $p_1 = 2$. Since all other primes are odd and there are exactly $n-1$ odd numbers between $2$ and $2n$, there are at most $n$ prime numbers less than or equal to $2n-1$, hence $p_n \\ge 2n - 1$.\n\nNow consider the sum of the first $n$ primes:\n$$\nP = p_1 + p_2 + \\dots + p_n.\n$$\nSince $p_k \\ge 2k - 1$ for any $k$, and additionally $p_1 = 2 > 1$, the sum $P$ is strictly greater than the sum of the first $n$ odd numbers:\n$$\nS = 1 + 3 + \\dots + (2n-1) = n^2.\n$$\nSo $P > S = n^2$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20920,
"subject": "Mathematics (Olympiad)",
"question": "如果整數數列 $a_0, a_1, a_2, \\dots$ 符合 $a_0 = 0, a_1 = 1$，且對於所有正整數 $n$ 有\n\n$$\n(a_{n+1} - 3a_n + 2a_{n-1})(a_{n+1} - 4a_n + 3a_{n-1}) = 0,\n$$\n\n則我們稱它為卡哇伊數列。一個整數如果屬於某個卡哇伊數列，則我們稱這個整數為卡哇伊。\n\n如果連續兩個正整數 $m$ 及 $m+1$ 皆卡哇伊（不一定要屬於相同的卡哇伊數列），證明 3 整除 $m$ 且 $m/3$ 也卡哇伊。",
"options": [],
"answer": "See solution",
"solution": "我們將題目中的條件改寫為：\n\n$$\na_{n+1} = 3a_n - 2a_{n-1}, \\text{ 或 } a_{n+1} = 4a_n - 3a_{n-1}.\n$$\n\n有 $a_{n+1} \\equiv a_n \\pmod{2}$ 且 $a_{n+1} \\equiv a_{n-1} \\pmod{3}$，對所有 $n \\ge 1$ 成立。\n由 $a_0 = 0$ 和 $a_1 = 1$，可得 $a_n \\equiv 0, 1 \\pmod{3}$ 對所有 $n \\ge 0$。\n因為 $m$ 和 $m+1$ 都是卡哇伊整數，必有 $m \\equiv 0 \\pmod{3}$。\n\n又 $a_2 = 3$ 或 $a_2 = 4$。\n\n1. 若 $a_2 = 3$，則 $a_n \\equiv 1 \\pmod{2}$ 對所有 $n \\ge 1$，因為 $a_1 = a_2 = 1 \\pmod{2}$。\n2. 若 $a_2 = 4$，則 $a_n \\equiv 1 \\pmod{3}$ 對所有 $n \\ge 1$，因為 $a_1 = a_2 = 1 \\pmod{3}$。\n\n由於 $m \\equiv 0 \\pmod{3}$，包含 $m$ 的卡哇伊數列不會滿足 (2)，必滿足 (1)。因此 $m$ 為奇數，$m+1$ 為偶數。\n\n取一個包含 $m+1$ 的卡哇伊數列 $(a_n)$，設 $t \\ge 2$ 使得 $a_t = m+1$。\n由於 $(a_n)$ 不滿足 (1)，必滿足 (2)，即 $a_n \\equiv 1 \\pmod{3}$ 對所有 $n \\ge 1$。\n定義新數列 $a'_n = (a_{n+1} - 1)/3$。這也是卡哇伊數列：$a'_0 = 0, a'_1 = 1$，且對所有 $n \\ge 1$，\n\n$$\n(a'_{n+1}-3a'_{n}+2a'_{n-1})(a'_{n+1}-4a'_{n}+3a'_{n-1}) = \\frac{(a_{n+2}-3a_{n+1}+2a_{n})(a_{n+2}-4a_{n+1}+3a_{n})}{9} = 0.\n$$\n\n最後，$a'_{t-1} = m/3$，因此 $m/3$ 也是卡哇伊整數。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20921,
"subject": "Mathematics (Olympiad)",
"question": "Es sei $\\{a_n\\}_{n \\ge 0}$ die Folge rationaler Zahlen mit $a_0 = 2016$ und\n\n$$\na_{n+1} = a_n + \\frac{2}{a_n}\n$$\n\nfür alle $n \\ge 0$.\n\nZeige, dass diese Folge kein Quadrat einer rationalen Zahl enthält.",
"options": [],
"answer": "See solution",
"solution": "Wir können eine rationale Zahl $\\frac{a}{b}$, deren Nenner nicht durch 5 teilbar ist, modulo 5 betrachten, indem wir den Rest von $ab^{-1}$ modulo 5 betrachten, wobei $b^{-1}$ das Inverse von $b$ modulo 5 ist. Dieser Rest hängt von der Darstellung der rationalen Zahl nicht ab und erfüllt auch die üblichen Rechenregeln. Insbesondere gilt, dass ein Quadrat einer rationalen Zahl, deren Nenner nicht durch 5 teilbar ist, als Rest einen quadratischen Rest modulo 5 haben muss.\n\nWir betrachten daher nun die Folgenglieder modulo 5, solange diese Reste ungleich 0 bleiben und der nächste Rest somit definiert ist, und erhalten die Folge der Reste\n\n$$\n\\begin{align*}\na_0 &\\equiv 1 \\pmod{5}, \\\\\na_1 &\\equiv 1 + 2 \\equiv 3 \\pmod{5}, \\\\\na_2 &\\equiv 3 + 2 \\cdot 3^{-1} \\equiv 3 + 2 \\cdot 2 \\equiv 2 \\pmod{5}, \\\\\na_3 &\\equiv 3 \\pmod{5}, \\\\\na_4 &\\equiv 2 \\pmod{5}, \\\\\n\\vdots\n\\end{align*}\n$$\n\nDie Folge der Reste nimmt also nach dem Anfangswert nur die Werte 2 und 3 an. Das sind aber keine quadratischen Reste modulo 5. Da auch $a_0 = 2016$ keine Quadratzahl ist, gibt es somit kein Quadrat einer rationalen Zahl in der Folge.\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20922,
"subject": "Mathematics (Olympiad)",
"question": "Let $f$ be a real-valued function defined on the set of real numbers that satisfies\n\n$$\nf(x + y) \\le y f(x) + f(f(x))\n$$\n\nfor all real numbers $x$ and $y$. Prove that $f(x) = 0$ for all $x \\le 0$.",
"options": [],
"answer": "See solution",
"solution": "Let $f(0) = a$ and $f(f(0)) = b$. Setting $x = 0$, we obtain $f(y) \\le a y + b$. From the given, we now obtain\n\n$$\nf(x + y) \\le y f(x) + f(f(x)) \\le y f(x) + a f(x) + b = (y + a) f(x) + b,\n$$\n\nso for all $y \\ge -a$ we have $f(x + y) \\le (y + a)(a x + b) + b$.\n\nIf $a$ were positive, then letting $x = c - y$ for a constant $c$ and taking $y \\to \\infty$, $f(x + y)$ stays constant, while $(y + a)(a x + b) + b$ approaches $-\\infty$, a contradiction. Therefore, $a \\le 0$. Further, if $a = 0$, then $b = 0$ as well and we have $f(x) \\le 0$ for all $x$. Setting $y = -x$ in the given, we obtain $x f(x) \\le f(f(x)) \\le 0$, so $f(x) \\ge 0$ for $x < 0$. But we have $f(x) \\le 0$ for all $x$ and thus $f(x) = 0$ for all $x \\le 0$.\n\nTo complete the proof, we show that $a < 0$ is impossible. Set $y = 0$ in the given to obtain\n\n$$\nf(x) \\le f(f(x)) \\le a f(x) + b. \\qquad (1)\n$$\n\nTherefore, we have $f(x) \\le \\frac{b}{1 - a} = \\lambda$. Using the given, we find\n\n$$\nf(x + y) \\le y f(x) + f(f(x)) \\le y f(x) + \\lambda. \\qquad (2)\n$$\n\nSwapping $x, y$ with $x + y, -y$ we also have $f(x) \\le -y f(x + y) + \\lambda$, hence for all $y \\ge 0$, we obtain\n\n$$\ny f(x) \\le -y^2 f(x + y) + \\lambda y. \\qquad (3)\n$$\n\nAdding (2) and (3), we have\n\n$$\nf(x + y) \\le \\lambda \\frac{1 + y}{1 + y^2}. \\qquad (4)\n$$\n\nKeep $x + y = z$ constant and take $y \\to \\infty$ in (4); its right hand side approaches $0$, so $f(z) \\le 0$ for all $z$. Further, if $f(z) = 0$ for some $z$, then $0 = f(z) \\le f(f(z)) = f(0)$ would imply $f(0) = 0$, violating the assumption that $a = f(0) < 0$. So in fact $f(z) < 0$ for all $z$.\n\nApplying this to the given gives $f(x + y) \\le y f(x) + f(f(x)) < y f(x)$. Setting $y = f(x) - x$ and applying (1) gives\n\n$$\nf(x) \\le f(f(x)) < (f(x) - x) f(x),\n$$\n\nor equivalently $f(x)(f(x) - x - 1) > 0$ for all $x$. Since $f(x) < 0$, we have $f(x) < x + 1$ for all $x$. Applying this to the given, we find for all $y \\ge 0$ that\n\n$$\nf(x + y) \\le y f(x) + f(f(x)) < y(x + 1).\n$$\n\nFix $x + y = c$ and take $y \\to \\infty$ (and $x \\to -\\infty$) in this inequality; the right hand side approaches $-\\infty$ while the left hand side stays constant, a contradiction which finishes the proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20923,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be an odd positive integer. A square $n \\times n$ grid is divided into $n^2$ cells. In total, $n^2$ tokens are placed in some cell(s). In each round, a player can move one token from cell $A$ to cell $B$, and one token from cell $A$ to cell $C$, provided that cell $A$ contains at least two tokens, and $B$ and $C$ are symmetric with respect to $A$ and adjacent to $A$. There are exactly four types of such moves:\n\n1. $A$, $B$, and $C$ are in one column;\n2. $A$, $B$, and $C$ are in one row;\n3. $A$, $B$, and $C$ are in a line parallel to one diagonal of the square;\n4. $A$, $B$, and $C$ are in a line parallel to the other diagonal of the square.\n\nAfter several such rounds, every cell on the board contains exactly one token. Prove that the number of moves of type (3) equals the number of moves of type (4).",
"options": [],
"answer": "See solution",
"solution": "Let the leftmost column be $1$, the next $2$, ..., the rightmost $n$. Similarly, the lowest row is $1$, the highest $n$. For any cell $c$ in row $a$ and column $b$, define $w(c) = ab$. For each token $T$, let $w(T) = ab$ if $T$ is in row $a$, column $b$. Let $W$ be the sum of $w(T)$ over all tokens $T$ on the board.\n\nIt follows from problem 8.4 that the statement can only be true if all tokens are initially placed on the central cell. Let $n = 2l - 1$. Then initially,\n\n$$\nW = l^2 n^2 = \\frac{n^2 (n+1)^2}{4}.\n$$\n\nAt the end,\n\n$$\nW = (1 + 2 + \\dots + n)^2 = \\left(\\frac{n(n+1)}{2}\\right)^2 = \\frac{n^2 (n+1)^2}{4}.\n$$\n\nSo $W$ does not change. Now, consider how $W$ changes for each move type:\n\n- **Type (1):** $2km - (k(m+1) + k(m-1)) = 0$. $W$ is unchanged.\n- **Type (2):** $2km - ((k+1)m + (k-1)m) = 0$. $W$ is unchanged.\n- **Type (3):** $2km - ((k+1)(m+1) + (k-1)(m-1)) = -2$. $W$ decreases by $2$.\n- **Type (4):** $2km - ((k-1)(m+1) + (k+1)(m-1)) = +2$. $W$ increases by $2$.\n\nSince $W$ is unchanged overall, the number of type (3) moves must equal the number of type (4) moves.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20924,
"subject": "Mathematics (Olympiad)",
"question": "Paul is filling the cells of a rectangular table alternately with crosses and circles (he starts with a cross). When the table is filled in completely, he determines his score as $O - X$, where $O$ is the total number of rows and columns containing more circles than crosses, and $X$ is the total number of rows and columns containing more crosses than circles.\n\na) Prove that for a $2 \\times n$ table, the score is always equal to $0$.\n\nb) In terms of $n$, what is the largest possible score Paul can achieve for a $(2n+1) \\times (2n+1)$ table?\n\n",
"options": [],
"answer": "See solution",
"solution": "a) Consider a table with $2$ rows and $n$ columns filled in with $n$ crosses and $n$ circles. Since the total number of crosses and circles is the same, crosses dominate in one row if and only if circles dominate in the other one. Hence, the rows contribute $0$ to the total score.\n\nNext, denote by $x$, $e$, and $o$ the number of columns containing two, one, and zero crosses, respectively. Since the table contains a total of $n$ crosses and $n$ circles, we have $2x + e = n = e + 2o$, hence $x = o$. As $x$ and $o$ are the number of columns dominated by crosses and circles, respectively, the columns contribute $0$ to the total score too.\n\nb) Consider a $(2n+1) \\times (2n+1)$ table filled with $\\frac{1}{2}((2n+1)^2 - 1) = 2n(n+1)$ circles and $2n(n+1)+1$ crosses. Since $2n+1$ is odd, each row and column is dominated by one of the two symbols. Circles can dominate in at most $2n(n+1)/(n+1) = 2n$ rows and thus at least one row is dominated by crosses. Likewise for columns, hence $O \\le 2n + 2n = 4n$, $X \\ge 1+1=2$ and therefore $O - X \\le 4n - 2$.\n\nFinally, we argue that the score $4n-2$ can be achieved for any $n$. It suffices to specify a set $S$ of $2n(n+1)$ cells that are to be filled with circles. An example is a set $S$ that consists of $n+1$ \"parallel diagonals\" in the top-left $2n \\times 2n$ subsquare of the table and no other cells in the bottom row or right column (see Fig. 1 for $n=3$).",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20925,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute triangle with circumcircle $(O)$. Let $D$ be a point on arc $BC$ that does not contain $A$. A moving line $\\ell$ passing through the orthocenter $H$ of triangle $ABC$ cuts the circumcircles of triangles $ABH$ and $ACH$ again at $M$ and $N$ respectively ($M \\neq H$, $N \\neq H$).\n\n1. Define the position of $\\ell$ such that the area of $\\triangle AMN$ has maximal value.\n\n2. Let $d_1$ be the line passing through $M$ and perpendicular to $DB$, and $d_2$ the line passing through $N$ and perpendicular to $DC$. Prove that the intersection $P$ of $d_1$ and $d_2$ belongs to a fixed circle.",
"options": [],
"answer": "See solution",
"solution": "1. Firstly, note that when $\\ell$ changes, the angles $\\angle AMN$ and $\\angle ANM$ both remain unchanged, so triangle $AMN$ is always self-congruent. Draw $AK$ perpendicular to $MN$ ($K \\in MN$), then $AK \\leq AH$. Therefore, the area of triangle $AMN$ attains maximal value when $AH$ is the altitude or $MN \\perp AH$. Thus, when $MN \\perp AH$, the area of triangle $AMN$ is largest.\n\n\n\n2. Let the line passing through $A$ and parallel to $BD$ intersect $(ABH)$ at $E$ and the line passing through $A$ and parallel to $CD$ intersect $(ACH)$ at $F$. Notice that the radii of the circles $(O)$, $(ABH)$, and $(ACH)$ are equal, hence $\\angle ADB = \\angle AEB$, which implies $\\angle ABD = \\angle BAE$ (because $BD \\parallel AE$). Hence, $\\angle BAD = \\angle ABE$, so $AD \\parallel BE$. Thus $ADBE$ is a parallelogram.\n\nSimilarly, $ADCF$ is a parallelogram. Therefore, triangle $AEF$ is the image of triangle $DBC$ through the translation $T_v$ with vector $\\vec{v} = \\vec{DA}$. Denote $O_1, H_1$ to be the circumcenter and orthocenter of triangle $AEF$, respectively. It is well known that\n\n$$\n\\begin{aligned}\n\\overrightarrow{OH} &= \\overrightarrow{OA} + \\overrightarrow{OB} + \\overrightarrow{OC} \\\\\n\\overrightarrow{O_1H_1} &= \\overrightarrow{O_1A} + \\overrightarrow{O_1E} + \\overrightarrow{O_1F},\n\\end{aligned}\n$$\n\nHowever, $\\overrightarrow{O_1H_1} = \\overrightarrow{O_1O} + \\overrightarrow{OH_1}$, $\\overrightarrow{O_1A} = \\overrightarrow{O_1O} + \\overrightarrow{OA}$, and $\\overrightarrow{O_1E} = \\overrightarrow{OB}$, $\\overrightarrow{O_1F} = \\overrightarrow{OC}$ (image via translation $T_v$). Hence\n\n$$\n\\overrightarrow{O_1O} + \\overrightarrow{OH_1} = \\overrightarrow{O_1O} + \\overrightarrow{OA} + \\overrightarrow{OB} + \\overrightarrow{OC},\n$$\nso $\\overrightarrow{OH_1} = \\overrightarrow{OH}$. Thus, $H_1 \\equiv H$.\n\n\n\nNow we will prove that the intersection $P$ of $d_1$ and $d_2$ always lies on the circle $(AEF)$. First, since $AE \\parallel BD$ and $AF \\parallel CD$, we have $MP \\perp AE$ and $NP \\perp AF$. Let $P_1$ be the reflection of $M$ over $AE$, $P_2$ the reflection of $N$ over $AF$. Since the circles $(AEF)$, $(ABH)$, and $(ACH)$ are equal (because they are equal to the same circle $(O)$), $P_1, P_2$ are all on the circle $(AEF)$. Note that the Steiner lines of $P_1$ and $P_2$ with respect to triangle $AEF$ are coincident (that is the line $\\ell$). According to the property of the Steiner line, we deduce that $P_1 \\equiv P_2 \\equiv P$, so $P$ always lies on the fixed circle $(AEF)$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20926,
"subject": "Mathematics (Olympiad)",
"question": "Let $S$ be a set containing $n^2 + n - 1$ elements, for some positive integer $n$. Suppose that the $n$-element subsets of $S$ are partitioned into two classes. Prove that there are at least $n$ pairwise disjoint sets in the same class.",
"options": [],
"answer": "See solution",
"solution": "*Proposition.* If the $n$-element subsets of a set $S$ with $(n+1)m - 1$ elements are partitioned into two classes, then there are at least $m$ pairwise disjoint sets in the same class.\n\n_Proof._ Fix $n$ and proceed by induction on $m$. The case $m = 1$ is trivial. Assume $m > 1$ and that the proposition is true for $m - 1$. Let $\\mathcal{P}$ be the partition of the $n$-element subsets into two classes. If all the $n$-element subsets belong to the same class, the result is obvious. Otherwise, select two $n$-element subsets $A$ and $B$ from different classes so that their intersection has maximal size. It is easy to see that $|A \\cap B| = n - 1$. (If $|A \\cap B| = k < n - 1$, then build $C$ from $B$ by replacing some element not in $A \\cap B$ with an element of $A$ not already in $B$. Then $|A \\cap C| = k + 1$ and $|B \\cap C| = n - 1$, and either $A$ and $C$ or $B$ and $C$ are in different classes.) Removing $A \\cup B$ from $S$, there are $(n+1)(m-1) - 1$ elements left. On this set, the partition induced by $\\mathcal{P}$ has, by the inductive hypothesis, $m - 1$ pairwise disjoint sets in the same class. Adding either $A$ or $B$ as appropriate gives $m$ pairwise disjoint sets in the same class. $\\square$\n\n*Remark:* The value $n^2 + n - 1$ is sharp. A set $S$ with $n^2 + n - 2$ elements can be split into a set $A$ with $n^2 - 1$ elements and a set $B$ of $n - 1$ elements. Let one class consist of all $n$-element subsets of $A$ and the other consist of all $n$-element subsets that intersect $B$. Then neither class contains $n$ pairwise disjoint sets.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20927,
"subject": "Mathematics (Olympiad)",
"question": "As shown in the figure below, $M$ and $N$ are the midpoints of arcs $\\widehat{BC}$ and $\\widehat{AC}$, respectively, on the circumscribed circle $\\Gamma$ of an acute triangle $\\triangle ABC$ with $\\angle A < \\angle B$. Through point $C$, draw $PC \\parallel MN$, intersecting the circle $\\Gamma$ at point $P$. Let $I$ be the incenter of $\\triangle ABC$. Extend line $PI$ to meet $\\Gamma$ again at point $T$.\n\n\n\n(1) Prove that $MP \\times MT = NP \\times NT$.\n\n(2) For an arbitrary point $Q \\ne A, T, B$ on arc $\\widehat{AB}$ (not containing $C$), let $I_1$ and $I_2$ be the incenters of $\\triangle AQC$ and $\\triangle QCB$, respectively. Prove that $Q, I_1, I_2, T$ are concyclic.",
"options": [],
"answer": "See solution",
"solution": "(1) As shown in the figure below, join $NI$ and $MI$. Since $PC \\parallel MN$ and $P, C, M, N$ are concyclic, $PCMN$ is an isosceles trapezoid. Therefore, $NP = MC$ and $PM = NC$.\n\nJoin $AM$ and $CI$. Then $AM$ intersects $CI$ at $I$. We have\n\n$$\n\\begin{align*}\n\\angle MIC &= \\angle MAC + \\angle ACI \\\\\n&= \\angle MCB + \\angle BCI \\\\\n&= \\angle MCI.\n\\end{align*}\n$$\n\n\n\nTherefore, $MC = MI$. Similarly, $NC = NI$. Thus, $NP = MI$ and $PM = NI$.\n\nThis means that $MPNI$ is a parallelogram. Therefore, $S_{\\triangle PMT} = S_{\\triangle PNT}$, since the two triangles have the same base and height.\n\nOn the other hand, $\\angle TNP + \\angle PMT = 180^{\\circ}$, as $P, N, T, M$ are concyclic. Then\n\n$$\n\\begin{aligned}\nS_{\\triangle PMT} &= \\frac{1}{2} PM \\times MT \\sin \\angle PMT \\\\\n&= S_{\\triangle PNT} = \\frac{1}{2} PN \\times NT \\sin \\angle PNT \\\\\n&= \\frac{1}{2} PN \\times NT \\sin \\angle PMT.\n\\end{aligned}\n$$\n\nTherefore, $MP \\times MT = NP \\times NT$.\n\n(2) As shown in the next figure, we have\n\n$$\n\\angle NCI_1 = \\angle NCA + \\angle ACI_1 = \\angle NQC + \\angle QCI_1 = \\angle CI_1N.\n$$\n\nTherefore, $NC = NI_1$. Similarly, $MC = MI_2$.\n\nFrom $MP \\times MT = NP \\times NT$ we get\n\n$$\n\\frac{NT}{MP} = \\frac{MT}{NP}.\n$$\n\nFrom (1), $MP = NC$ and $NP = MC$. Thus,\n\n\n\n$$\n\\frac{NT}{NI_1} = \\frac{MT}{MI_2}.\n$$\n\nFurthermore,\n\n$$\n\\angle I_1 NT = \\angle QNT = \\angle QMT = \\angle I_2 MT.\n$$\n\nTherefore, $\\triangle I_1 NT \\sim \\triangle I_2 MT$. Consequently, $\\angle NTI_1 = \\angle MTI_2$. Then\n\n$$\n\\angle I_1 QI_2 = \\angle NQM = \\angle NTM = \\angle I_1 TI_2.\n$$\n\nThis means that $Q, I_1, I_2, T$ are concyclic.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20928,
"subject": "Mathematics (Olympiad)",
"question": "Determine all positive integers $a$, $b$, and $p$, where $p$ is prime, satisfying the equation:\n$$\n\\frac{1}{p} = \\frac{1}{a^2} + \\frac{1}{b^2}\n$$",
"options": [],
"answer": "See solution",
"solution": "We start by rewriting the equation:\n$$\n\\frac{1}{p} = \\frac{1}{a^2} + \\frac{1}{b^2}\n$$\nMultiply both sides by $p a^2 b^2$:\n$$\np a^2 b^2 \\left( \\frac{1}{p} \\right) = p a^2 b^2 \\left( \\frac{1}{a^2} + \\frac{1}{b^2} \\right)\n$$\n$$\na^2 b^2 = p b^2 + p a^2\n$$\n$$\np a^2 + p b^2 = a^2 b^2\n$$\nSuppose $p$ divides $a$, so $a = p a_1$ for some $a_1 \\in \\mathbb{N}^*$. Substitute:\n$$\np (p a_1)^2 + p b^2 = (p a_1)^2 b^2\n$$\n$$\np^3 a_1^2 + p b^2 = p^2 a_1^2 b^2\n$$\nDivide both sides by $p$:\n$$\np^2 a_1^2 + b^2 = p a_1^2 b^2\n$$\nNow, $\\frac{1}{p} > \\frac{1}{b^2}$ implies $b^2 > p$, so $b^2 \\geq p+1$. Then:\n$$\n\\frac{1}{a^2} = \\frac{1}{p} - \\frac{1}{b^2} \\geq \\frac{1}{p} - \\frac{1}{p+1} = \\frac{1}{p(p+1)}\n$$\nSince $a = p a_1$:\n$$\n\\frac{1}{p^2 a_1^2} \\geq \\frac{1}{p(p+1)}\n$$\n$$\n\\frac{1}{a_1^2} \\geq \\frac{p}{p+1} \\geq \\frac{1}{2}\n$$\nSo $a_1^2 \\leq 2$, so $a_1 = 1$, $a = p$.\n\nPlug $a = p$ into the equation:\n$$\np p^2 + b^2 = p^2 b^2\n$$\n$$\np^3 + b^2 = p^2 b^2\n$$\n$$\np^3 = p^2 b^2 - b^2 = b^2 (p^2 - 1)\n$$\nSo $b^2 = \\frac{p^3}{p^2 - 1}$, which is integer only for $p = 2$, giving $b^2 = 8/3$, not integer. Try $p = 2$ directly:\n$$\n\\frac{1}{2} = \\frac{1}{a^2} + \\frac{1}{b^2}\n$$\nTry $a = b = 2$:\n$$\n\\frac{1}{2} = \\frac{1}{4} + \\frac{1}{4} = \\frac{1}{2}\n$$\nSo $(a, b, p) = (2, 2, 2)$ is a solution.\n\nAlternatively, solve for $a^2$:\n$$\n\\frac{1}{p} = \\frac{1}{a^2} + \\frac{1}{b^2} \\implies a^2 = \\frac{p b^2}{b^2 - p}\n$$\nFor $a^2$ integer, $b^2 - p$ divides $p^2$. Possible values:\n- $b^2 - p = 1$: $p = b^2 - 1 = (b-1)(b+1)$, so $p$ prime only if $b = 2$, $p = 3$, $a^2 = 12$ (not integer).\n- $b^2 - p = p$: $b^2 = 2p$, so $b$ even, $b = 2b_1$, $4b_1^2 = 2p$, $p = 2$, $b = 2$, $a = 2$.\n- $b^2 - p = p^2$: $a^2 = p + 1$, $p = a^2 - 1 = (a-1)(a+1)$, $p$ prime only if $a = 2$, $p = 3$, $b^2 = 3$ (not integer).\n\nThus, the only solution is $a = b = p = 2$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20929,
"subject": "Mathematics (Olympiad)",
"question": "Fix a circle of radius $s$. We seek to maximize the area of a convex $n$-gon enclosed within this circle. It may be assumed that the $n$-gon is actually inscribed in the circle, for its area can only increase if the side lengths are extended, so that the polygon becomes inscribed.\n\nWhat is the maximum area of a convex $n$-gon inscribed in a circle of radius $s$? What is the minimum area of a convex $n$-gon circumscribed about a circle of radius $s$? Show that for a convex $n$-gon with circumradius $R$ and inradius $r$, the following holds:\n\n$$\n1 \\leq \\frac{R^2}{r^2} \\cos^2 \\frac{180^\\circ}{n}.\n$$",
"options": [],
"answer": "See solution",
"solution": "The area of an inscribed convex $n$-gon is\n\n$$\nA = \\sum_{k=1}^{n} \\frac{1}{2} s^2 \\sin \\alpha_k,\n$$\n\nwhere $\\alpha_k$ are the central angles subtended by the sides, with $\\sum_{k=1}^n \\alpha_k = 360^\\circ$. Since $\\sin$ is concave on $[0^\\circ, 180^\\circ]$, by Jensen's Inequality,\n\n$$\nA \\leq \\frac{1}{2} n s^2 \\sin \\frac{360^\\circ}{n},\n$$\n\nwith equality when all $\\alpha_k$ are equal.\n\nFor a circumscribed convex $n$-gon, the area is\n\n$$\nB = \\sum_{k=1}^{n} s^2 \\tan \\frac{\\beta_k}{2},\n$$\n\nwhere $\\beta_k$ are the central angles between points of tangency, $\\sum_{k=1}^n \\beta_k = 360^\\circ$. Since $\\tan$ is convex on $[0^\\circ, 90^\\circ]$, Jensen's Inequality gives\n\n$$\nB \\geq n s^2 \\tan \\frac{180^\\circ}{n}.\n$$\n\nFor a convex $n$-gon with circumradius $R$ and inradius $r$, the area $C$ satisfies\n\n$$\nn r^2 \\tan \\frac{180^\\circ}{n} \\leq C \\leq \\frac{1}{2} n R^2 \\sin \\frac{360^\\circ}{n}.\n$$\n\nThus,\n\n$$\n1 \\leq \\frac{\\frac{1}{2} n R^2 \\sin \\frac{360^\\circ}{n}}{n r^2 \\tan \\frac{180^\\circ}{n}} = \\frac{R^2}{r^2} \\cos^2 \\frac{180^\\circ}{n},\n$$\n\nwhich proves the claim.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20930,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. Find, with proof, the least positive integer $d_n$ which cannot be expressed in the form\n$$\n\\sum_{i=1}^{n} (-1)^{a_i} 2^{b_i},\n$$\nwhere $a_i$ and $b_i$ are nonnegative integers for each $i$.",
"options": [],
"answer": "See solution",
"solution": "The answer is $d_n = \\dfrac{2^{2n+1} + 1}{3}$.\n\nWe first show that $d_n$ cannot be obtained. For any $p$, let $t(p)$ be the minimum $n$ required to express $p$ in the desired form, and call any realization of this minimum a minimal representation.\n\nIf $p$ is even, any sequence of $b_i$ that can produce $p$ must contain an even number of zeros. If this number is nonzero, then canceling one against another or replacing two with a $b_i = 1$ term would reduce the number of terms in the sum. Thus, a minimal representation cannot contain a $b_i = 0$ term, and by dividing each term by two, we see that $t(2m) = t(m)$.\n\nIf $p$ is odd, there must be at least one $b_i = 0$, and removing it gives a sequence that produces either $p-1$ or $p+1$. Hence\n$$\nt(2m-1) = 1 + \\min\\big(t(2m-2),\\ t(2m)\\big) = 1 + \\min\\big(t(m-1),\\ t(m)\\big).\n$$\nWith $d_n$ as defined above and $c_n = \\dfrac{2^{2n} - 1}{3}$, we have $d_0 = c_1 = 1$, so $t(d_0) = t(c_1) = 1$ and\n$$\nt(d_n) = 1 + \\min\\big(t(d_{n-1}),\\ t(c_n)\\big), \\quad t(c_n) = 1 + \\min\\big(t(d_{n-1}),\\ t(c_{n-1})\\big).\n$$\nHence, by induction, $t(c_n) = n$ and $t(d_n) = n+1$, and $d_n$ cannot be obtained by a sum with $n$ terms.\n\nNext, we show by induction on $n$ that any positive integer less than $d_n$ can be obtained with $n$ terms. By the inductive hypothesis and symmetry about zero, it suffices to show that by adding one summand we can reach every $p$ in the range $d_{n-1} \\leq p < d_n$ from an integer $q$ in the range $-d_{n-1} < q < d_{n-1}$.\n\nSuppose that $c_n + 1 \\leq p \\leq d_n - 1$. By using a term $2^{2n-1}$, we see that $t(p) \\leq 1 + t(|p - 2^{2n-1}|)$. Since $d_n - 1 - 2^{2n-1} = 2^{2n-1} - (c_n + 1) = d_{n-1} - 1$, it follows from the inductive hypothesis that $t(p) \\leq n$.\n\nNow suppose that $d_{n-1} \\leq p \\leq c_n$. By using a term $2^{2n-2}$, we see that $t(p) \\leq 1 + t(|p - 2^{2n-2}|)$. Since $c_n - 2^{2n-2} = 2^{2n-2} - d_{n-1} = c_{n-1} < d_{n-1}$, it again follows that $t(p) \\leq n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20931,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $P(x)$ and $Q(x)$ are integer polynomials. Find all pairs $(P(x), Q(x))$ such that for every positive integer $a$, the sequence $(x_n)$ defined by\n\n$$\nx_0 = a,\\quad x_{n+1} = P(Q(x_n)),\\quad n \\ge 0\n$$\n\nhas the property that for every positive integer $m$, there exists $n$ such that $m$ divides $x_n$.",
"options": [],
"answer": "See solution",
"solution": "It is easy to check that if one of $P(x)$ and $Q(x)$ is constant then they do not satisfy the given condition. We will show that if $P(x)$, $Q(x)$ satisfy the given condition then they both have degree 1. We suppose for a contradiction, set $H = P(Q(x))$ and $K = Q(P(x))$ then\n\n$$\ndeg(H),\\ deg(K) \\ge 2.\n$$\n\nWe will need the following lemma.\n\n**Lemma.** Given a positive integer $a$ and an integer polynomial $T(x)$. The sequence $(x_n)$ is defined as\n\n$$\nx_0 = a,\\quad x_{n+1} = P(x_n),\\quad n \\ge 0.\n$$\n\nSuppose that each positive integer $m$ is a divisor of some non-zero term of $(x_n)$. Then $\\deg(T) = 1$.\n\n*Proof.* It is easy to check that $T = c$ does not satisfy the given condition. We suppose that $\\deg(T) > 1$. Then there exists $c > 0$ such that $|T(x)| > 3|x|$ whenever $|x| > c$.\n\nFrom the assumption, there exist infinitely many $n$ such that $|x_n| > |x_i|$ for all $i < n$. Hence, we can choose $N$ such that $|x_N| > \\max\\{c, |x_0|, \\dots, |x_{N-1}|\\}$. This implies that\n\n$$\n|x_n| > \\max\\{|x_0|, \\dots, |x_{n-1}|\\}\\ \\text{ and }\\ |x_{n+1}| > 3|x_n|\\ \\text{ for all }\\ n \\ge N.\n$$\n\nSet $m = |x_{N+1} - x_N|$. Then\n\n$$\nm \\ge |x_{N+1} - x_N| > 2|x_N| > \\max\\{|x_0|, \\dots, |x_{N-1}|\\}.\n$$\n\nThis implies that $x_0, x_1, \\dots, x_{N+1}$ are not divisible by $m$.\n\nOn the other hand, $x_{n+1} - x_n$ is divisible by $x_n - x_{n-1}$ for all $n$. Hence, $x_{n+1} - x_n$ is divisible by $m$ for all $n \\ge N$. This also implies that\n\n$$\nx_n - x_N = (x_n - x_{n-1}) + (x_{n-1} - x_{n-2}) + \\dots + (x_{N+1} - x_N)\n$$\n\nis divisible by $m$. But $x_N$ is not divisible by $m$ so $x_n$ is not divisible by $m$ for all $n$, which is a contradiction. This concludes the proof of the lemma. $\\square$\n\nSuppose that the subsequence $\\{x_0, x_2, x_4, \\dots\\}$ does not satisfy the given condition. Then there exists positive integer $m$ not a divisor of any $x_{2i}$, hence so are $2m, 3m, \\dots$. This implies that there exists $x_{2i+1}$ divisible by $km$ for any $k$, or $\\{x_1, x_3, \\dots\\}$ satisfies the given condition.\n\nWe now can apply the above lemma for one of the subsequences with the polynomials $H$ or $K$ respectively to conclude that $\\deg(H) = \\deg(K) = 1$.\n\nSuppose that $P(x) = ax + b$ and $Q(x) = cx + d$ with $a, b, c, d \\in \\mathbb{Z}$ then\n\n$$\nx_{2n+2} = c a x_{2n} + b c + d\\ \\text{ and }\\ x_{2n+3} = c a x_{2n+1} + a d + b,\\ \\forall n \\ge 0.\n$$\n\nOne can check that in order to have one of these subsequences satisfy the given condition, we will need $a c = 1$.\n\nFinally, we only need to consider two cases:\n\n* Suppose that $P(x) = x + b,\\ Q(x) = x + d$. Then\n\n$$\nx_{2k} = 2014 + k(a + b)\\ \\text{ and }\\ x_{2k+1} = 2014 + a + k(a + b).\n$$\n\nWe first need $a + b \\neq 0$. Besides, one of these subsequences satisfies the given condition, hence\n\n$$\n\\text{either }\\ a + b \\mid 2014\\ \\text{ or }\\ a + b \\mid 2014 + a.\n$$\n\nIt is easy to check that polynomials $P(x)$ and $Q(x)$ satisfy the given condition in these cases.\n\n* Suppose that $P(x) = -x + b,\\ Q(x) = -x + d$. Similarly, we have all possible solutions are\n\n$$\na - b \\neq 0\\ \\text{ and }\\ a - b \\mid 2014\\ \\text{ or }\\ a - b \\mid a - 2014.\n$$\n\n",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20932,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle and $H$ its orthocenter. Let $P$ be an arbitrary point on segment $BC$. Prove that $H$ lies on the line passing through the reflections of $P$ with respect to $AB$ and $AC$ if and only if triangle $ABC$ is right-angled.",
"options": [],
"answer": "See solution",
"solution": "Let $P'$ and $P''$ be the reflections of $P$ with respect to $AB$ and $AC$. It is easy to see that $P$ is an anti-Steiner point of the line $P'P''$, so $P \\equiv B$ or $P \\equiv C$. Therefore, $\\angle BAC = 90^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20933,
"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: $1 \\leq q \\leq 20$, $\\dfrac{1}{4} < \\dfrac{p}{q} < \\dfrac{2}{7}$.\n- Condition 2: $p$ 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 $p$ values:\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 $p \\geq 6$, $q > 20$.\n\nNow, check if $p$ and $q$ are relatively prime:\n\n- $(p, q) = (3, 11)$: $\text{gcd}(3, 11) = 1$\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\n- $(m, n) = (q - p, q)$\n- $(3, 11) \\to (8, 11)$\n- $(4, 15) \\to (11, 15)$\n- $(5, 18) \\to (13, 18)$\n- $(5, 19) \\to (14, 19)$\n\n**All positive integer pairs $(m, n)$ are:**\n\n- $(8, 11)$\n- $(11, 15)$\n- $(13, 18)$\n- $(14, 19)$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20934,
"subject": "Mathematics (Olympiad)",
"question": "На доске записаны числа $ab$, $a(b+2)$, $(a+2)b$, $(a+2)(b+2)$, $a(a+2)$, $b(b+2)$, где $a$ и $b$ — натуральные числа. Какое максимальное количество квадратов натуральных чисел может быть среди этих чисел?",
"options": [],
"answer": "See solution",
"solution": "*Ответ.* Два.\n\nЗаметим, что никакие два квадрата натуральных чисел не отличаются на 1, ибо $x^2 - y^2 = (x - y)(x + y)$, где вторая скобка больше единицы. Значит, числа $a(a+2) = (a+1)^2 - 1$ и $b(b+2) = (b+1)^2 - 1$ квадратами не являются. Более того, числа $ab$ и $a(b+2)$ не могут одновременно являться квадратами, иначе их произведение $a^2 \\cdot b(b+2)$ также было бы квадратом, а тогда и число $b(b+2)$ тоже. Аналогично, из чисел $(a+2)b$ и $(a+2)(b+2)$ максимум одно может быть квадратом. Итого, квадратов на доске не больше двух.\n\nДва квадрата могут получиться, например, при $a = 2$ и $b = 16$: тогда $a(b+2) = 6^2$ и $(a+2)b = 8^2$.\n\n*Замечание.* Существуют и другие примеры, например, $(a, b) = (6, 96)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20935,
"subject": "Mathematics (Olympiad)",
"question": "We say that the triple $a, b, c$ from the segment $[-1, 1]$ is *worthy* if these numbers satisfy the inequality\n\n$$\n1 + 2abc > a^2 + b^2 + c^2.\n$$\n\nProve that if the triples $a, b, c$ and $x, y, z$ are worthy, then the triple $ax, by, cz$ is worthy as well.",
"options": [],
"answer": "See solution",
"solution": "We are given the following inequalities:\n\n$$\n1 + 2abc \\ge a^2 + b^2 + c^2 \\quad \\text{and} \\quad 1 + 2xyz \\ge x^2 + y^2 + z^2. \\tag{1}\n$$\n\nWe need to show that:\n\n$$\n1 + 2abcxyz \\ge (ax)^2 + (by)^2 + (cz)^2. \\tag{2}\n$$\n\n**Step 1:** We can assume all numbers are non-negative. If $abcxyz \\ge 0$, taking absolute values does not affect (2), and (1) only becomes stronger. If $abcxyz < 0$, then either $abc < 0$ or $xyz < 0$; assume $abc < 0$, then\n\n$$\n1 + 2abcxyz \\ge 1 + 2abc \\ge a^2 + b^2 + c^2 \\ge (ax)^2 + (by)^2 + (cz)^2.\n$$\n\n**Step 2:** Without loss of generality, let $a \\le b \\le c$ and $x \\le y \\le z$. The right-hand side of (2) is maximized in this case.\n\nSolving (2) as a quadratic in $cz$, we get:\n\n$$\nabxy - \\sqrt{(1 - a^2x^2)(1 - b^2y^2)} \\le cz \\le abxy + \\sqrt{(1 - a^2x^2)(1 - b^2y^2)}.\n$$\n\nThe left inequality is clear since $cz \\ge ax \\ge abxy$. For the right, from (1) as quadratics in $c$ and $z$:\n\n$$\nc \\le ab + \\sqrt{(1 - a^2)(1 - b^2)}, \\quad z \\le xy + \\sqrt{(1 - x^2)(1 - y^2)}.\n$$\n\nSo it suffices to show:\n\n$$\n(ab + \\sqrt{(1 - a^2)(1 - b^2)})(xy + \\sqrt{(1 - x^2)(1 - y^2)}) \\le abxy + \\sqrt{(1 - a^2x^2)(1 - b^2y^2)}. \\tag{3}\n$$\n\nIf any of $a, b, x, y$ equals $1$, say $b=1$, (3) becomes\n\n$$\na\\sqrt{(1 - x^2)(1 - y^2)} \\le \\sqrt{(1 - a^2x^2)(1 - y^2)},\n$$\n\nwhich holds since $a \\le 1$ and $1 - a^2x^2 \\ge 1 - x^2$.\n\nIf none equals $1$, expand and divide both sides by $\\sqrt{(1 - a^2)(1 - b^2)(1 - x^2)(1 - y^2)}$ to get:\n\n$$\n1 + AB + XY \\le \\sqrt{(1 + A^2 + X^2)(1 + B^2 + Y^2)},\n$$\n\nwhere $A = \\frac{a}{\\sqrt{1-a^2}}, B = \\frac{b}{\\sqrt{1-b^2}}, X = \\frac{x}{\\sqrt{1-x^2}}, Y = \\frac{y}{\\sqrt{1-y^2}}$. This is a special case of the Cauchy-Schwarz inequality.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20936,
"subject": "Mathematics (Olympiad)",
"question": "How many positive integers, where the only allowed digits are 0 and 1, are less than $1111100100$?",
"options": [],
"answer": "See solution",
"solution": "Solution 1: The given number has 11 digits. There are $2^{11} - 1 = 2047$ positive integers with at most 11 digits, each of which is either 0 or 1. We solve the problem by subtracting the number of positive integers that are not less than the given number.\n\nThe 11-digit numbers larger than the given number are all of the form $11111abcde$. There are $2^5 = 32$ numbers in this form. Among them, 4 numbers $1111100000$, $1111100001$, $1111100010$, and $1111100011$ are less than the given number. Thus, the number of positive integers consisting of zeros and ones and being less than $1111100100$ is $2047 - 32 + 4 = 2019$.\n\nSolution 2: Ordering any two digit sequences that constitute a positional representation on different bases does not depend on the base. This means that if a number is larger than another number in base 2, then the first number is larger than the second number also in base 10. The number $1111100100$ in base 2 equals $2^{10} + 2^{9} + 2^{8} + 2^{7} + 2^{6} + 2^{2} = 2020$ in base 10. Hence, to solve the problem, it suffices to count all positive integers less than 2020. The result is obviously 2019.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20937,
"subject": "Mathematics (Olympiad)",
"question": "Let $A_0, A_1, \\dots, A_5$ be a circular labeling of six distinct points on a circle centered at $O$. Let $B_i$ be the midpoint of the segment $A_iA_{i+1}$ for $i = 0, 1, \\dots, 5$ (indices modulo 6). Assume that no opposite sides of the hexagon $A_0A_1\\cdots A_5$ are parallel. A line through $O$ meets again the circle $B_iOB_{i+3}$ at the point $C_i$ for $i = 0, 1, 2$. Let $\\ell_i$ be the tangent to the circle $B_iOB_{i+3}$ at $C_i$ for $i = 0, 1, 2$. The lines $\\ell_i$ and $\\ell_j$ meet to produce the point $D_k$, where $\\{i, j, k\\} = \\{0, 1, 2\\}$. Show that the three circles $B_iOB_{i+3}$ and the circle $D_0D_1D_2$ share a point in the plane.",
"options": [],
"answer": "See solution",
"solution": "Let $\\gamma$ be the circle through the $A_i$, centered at $O$, and let tangents to $\\gamma$ at $A_i$ and $A_{i+1}$ meet at $B'_i$. Notice that the line $B'_iB'_{i+3}$ is the image of the circle $B_iOB_{i+3}$ under the inversion of pole $O$ and power $r^2$, where $r$ is the radius of $\\gamma$. By Brianchon's theorem, the three lines $B'_iB'_{i+3}$ are concurrent at a point $Q$. Notice further that $Q$ is different from $O$, since no opposite sides of the hexagon $A_0A_1\\cdots A_5$ are parallel. Consequently, the three circles $B_iOB_{i+3}$ share a second point $P$, different from $O$: the image of $Q$ under the inversion.\n\n\n\nThe lemma below shows that the points $P, C_i, C_j$ and $D_k$ are concyclic (not necessarily in this order), and the lines $PO$ and $PD_k$ are isogonal with respect to the lines $PC_i$ and $PC_j$, where $\\{i, j, k\\} = \\{0, 1, 2\\}$. Finally, a standard angle-chase argument shows that $P$ and the three points $D_i$ are concyclic: with reference to the figure below, write successively\n\n\n\n$$\n\\begin{aligned}\n\\angle D_i P D_j &= \\angle D_i P C_i + \\angle C_i P D_j \\\\\n&= \\angle D_i P C_i + \\angle C_k P O \\quad (\\text{for } P D_j \\text{ and } P O \\text{ are isogonal relative to } P C_k \\text{ and } P C_i) \\\\\n&= \\angle D_i P C_i + \\angle C_j P D_i \\quad (\\text{for } P C_k \\text{ and } P C_j \\text{ are isogonal relative to } P D_i \\text{ and } P O) \\\\\n&= \\angle C_i P C_j \\\\\n&= \\angle C_i D_k C_j \\quad (\\text{for } P, C_i, C_j, D_k \\text{ are concyclic}) \\\\\n&= \\angle D_j D_k D_i,\n\\end{aligned}\n$$\n\nto conclude that the points $P, D_0, D_1$ and $D_2$ are indeed concyclic.\n\n**Lemma.** Two circles, $\\gamma_1$ and $\\gamma_2$, meet at the points $X$ and $Y$. A line through $Y$ meets again $\\gamma_1$ at the point $Y_1$, and $\\gamma_2$ at the point $Y_2$. The tangent to $\\gamma_1$ at $Y_1$ meets the tangent to $\\gamma_2$ at $Y_2$ at the point $Z$. Then the points $X, Z, Y_1$ and $Y_2$ are concyclic, and the lines $XY$ and $XZ$ are isogonal with respect to the lines $XY_1$ and $XY_2$.\n\n\n\n**Proof.** If $X$ and $Z$ lie on opposite sides of the line through $Y$, then the angle $Y_1 X Y_2$ is the sum of the angles $X Y_1 Y$ and $X Y_2 Y$ which are respectively equal to the angles $Y_1 Z Y_2$ and $Y_2 Z Y_1$, whose sum is supplementary to the angle $Y_1 Z Y_2$. Consequently, the quadrangle $X Y_1 Z Y_2$ is cyclic. It then follows that the angles $Z X Y_2$ and $Z Y_1 Y_2$ are equal, and since the latter is equal to the angle $X Y_1 Y$, we conclude that the lines $XY$ and $XZ$ are indeed isogonal with respect to the lines $XY_1$ and $XY_2$.\n\nIf $X$ and $Z$ lie on the same side of the line through $Y$, then the angle $Y_1 X Y_2$ is the difference of the angles $X Y_1 Y$ and $X Y_2 Y$ in some order, depending on which side of the line $XY$ the line $Y_1 Y_2$ is situated. The latter angles are respectively supplementary to the angles $Y_1 Z Y$ and $Y_2 Z Y$ whose difference in the corresponding order is equal to the angle $Y_1 Z Y_2$. Consequently, the quadrangle $X Y_1 Y_2 Z$ or $X Y_2 Y_1 Z$ is cyclic. Isogonality is proved by adapting the argument in the former case.\n\n**Remark.** The proof from the Book requires a clouting (projective?) argument along with isogonality to reach the conclusion.",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 20938,
"subject": "Mathematics (Olympiad)",
"question": "Find the minimal possible value of $C$ such that any convex 14-gon can be cut into parallelograms with areas not greater than $C$.\n",
"options": [],
"answer": "See solution",
"solution": "Since there exists a parallelogram with area divisible by 5, $C \\geq 5$. The following example shows that $C = 5$ is possible. Connect successively the points with coordinates $(0, 0)$, $(1, 0)$, $(2, 1)$, $(3, 3)$, $(3, 4)$, $(2, 6)$, $(1, 7)$, $(-1, 8)$, $(-2, 8)$, $(-3, 7)$, $(-4, 5)$, $(-4, 4)$, $(-3, 2)$, and $(-2, 1)$. Such a 14-gon can be easily cut into parallelograms with areas not greater than 5.\n\n**Remark.** If we replace the 14-gon by a $(2p+4)$-gon, where $p$ is prime, $C$ will equal $p$. The proof of minimality can be done in a similar way as above. To construct an example for $C = p$, take the vectors $\\overrightarrow{(1, 1)}$, $\\overrightarrow{(1, 2)}$, $\\ldots$, $\\overrightarrow{(1, p+1)}$, $\\overrightarrow{(0, 1)}$ in that order, and then take the opposite vectors in the same order.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20939,
"subject": "Mathematics (Olympiad)",
"question": "We 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\nSuppose each of the numbers $0, 1, \\ldots, n-1$ appears at least once on the tokens. The table then has rows (or columns) with exactly $1, 2, \\ldots, n$ tokens, respectively.\n\nShow that if $n \\equiv 2 \\pmod{3}$, then the table has at least $\\frac{1}{3}n(n+1)$ 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 $$(1+2+\\ldots+(k-1)) + (2k+(2k+1)+\\ldots+(3k-1)) = \\frac{k(k-1)}{2} + k(5k-1) = k(3k-1) = \\frac{1}{3}n(n+1)$$. 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.",
"options": [],
"answer": "See solution",
"solution": "*Observation 1*: If the situation is possible for a certain number $m$ of tokens, then it is also possible for a number $m' > m$: we can add $m' - m$ tokens with different sizes and colors to a suitable configuration with $m$ tokens. Thus, 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 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$$N = k + (k + 1) + (k + 2) + \\dots + (3k - 1) = k(4k - 1)$$\ntokens. Each token can be counted at most twice. If we denote by $x$ the number of rows and by $y$ the number of columns participating in obtaining $N$, there are at most $xy$ tokens counted twice. Since $x + y = 2k$, we have $xy \\le k^2$, so the rows participating in obtaining $N$ have at least $N - k^2 = k(3k - 1) = \\frac{1}{3}n(n + 1)$ tokens.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 20940,
"subject": "Mathematics (Olympiad)",
"question": "Consider all sequences of real numbers $x_0, x_1, x_2, \\dots, x_{100}$ satisfying the following conditions:\n\n1. $x_0 = 0$;\n2. For any integer $i$ with $1 \\le i \\le 100$, $1 \\le x_i - x_{i-1} \\le 2$.\n\nFind the largest positive integer $k \\le 100$ such that\n\n$$\nx_k + x_{k+1} + \\dots + x_{100} \\ge x_0 + x_1 + \\dots + x_{k-1}\n$$\n\nholds for every such sequence $x_0, x_1, x_2, \\dots, x_{100}$.",
"options": [],
"answer": "See solution",
"solution": "The answer is $67$.\n\nFirst, consider the sequence where $x_i = 2i$ for $1 \\le i \\le 34$, and $x_{34+j} = x_{34} + j = 68 + j$ for $1 \\le j \\le 66$. This sequence satisfies the given conditions. We have:\n\n$$\n\\sum_{j=68}^{100} x_j - \\sum_{i=0}^{67} x_i = \\sum_{j=1}^{33} (x_{67+j} - x_{34+j}) - \\sum_{i=1}^{34} x_i = 33^2 - 34 \\times 35 < 0.\n$$\n\nThis example shows that when $k \\ge 68$, the requirement is not satisfied.\n\nOn the other hand, for any sequence $x_1, x_2, \\dots, x_{100}$ satisfying the conditions, we have $x_i \\le 2i$ for $1 \\le i \\le 100$. For $0 \\le s < t \\le 100$, $x_t - x_s \\ge t - s$. Therefore,\n\n$$\n\\begin{aligned}\n\\sum_{j=67}^{100} x_j - \\sum_{i=0}^{66} x_i &= \\sum_{j=1}^{34} (x_{66+j} - x_{32+j}) - \\sum_{i=1}^{32} x_i \\\\\n&\\ge 34^2 - \\sum_{i=1}^{32} 2i \\\\\n&= 34^2 - 32 \\times 33 > 0.\n\\end{aligned}\n$$\n\nIn conclusion, the largest $k$ is $67$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20941,
"subject": "Mathematics (Olympiad)",
"question": "Does there exist a positive real number $a$ such that for every real $x$ the inequality\n$$\n|\\\\cos x| + |\\cos a x| > \\sin x + \\sin a x\n$$\nholds?",
"options": [],
"answer": "See solution",
"solution": "No.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20942,
"subject": "Mathematics (Olympiad)",
"question": "Let $m$ and $n$ be two positive integers with $m \\ge n \\ge 2022$. Let $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ be $2n$ real numbers. Prove that the number of ordered pairs $(i, j)$ ($1 \\le i, j \\le n$) such that\n\n$$\n|a_i + b_j - ij| \\le m\n$$\n\nis less than or equal to $3n\\sqrt{m \\log n}$.",
"options": [],
"answer": "See solution",
"solution": "Mark red all points $(i, j)$ for which $|a_i + b_j - ij| \\le m$.\n\n**Lemma:** If $(i_1, j_1)$, $(i_1, j_2)$, $(i_2, j_1)$, $(i_2, j_2)$ are all red, then $|(i_2 - i_1)(j_2 - j_1)| \\le 4m$.\n\n*Proof of the lemma:* By the definition of red points, we know\n\n$$\n|a_{i_1} + b_{j_1} - i_1 j_1| \\le m, \\quad |a_{i_1} + b_{j_2} - i_1 j_2| \\le m, \\quad |a_{i_2} + b_{j_1} - i_2 j_1| \\le m, \\quad |a_{i_2} + b_{j_2} - i_2 j_2| \\le m.\n$$\n\nTaking the differences of these expressions to eliminate $a_{i_1}$, $a_{i_2}$, $b_{j_1}$, and $b_{j_2}$, we get $|i_1j_1 - i_1j_2 - i_2j_1 + i_2j_2| \\le 4m$. This proves the lemma.\n\nWe first calculate, for a given pair $1 \\le i_1 < i_2 \\le n$, the number of $j$'s such that $(i_1, j)$ and $(i_2, j)$ are both red points. Put $d = i_2 - i_1$. Since the difference of any two such $j$'s is at most $\\frac{4m}{d}$, we can have at most $\\frac{4m}{d} + 1$ such $j$'s.\n\nMoreover, when the difference $d$ is given, there are $n-d$ ways to choose $i_1$ and $i_2$, so the total number of $1 \\le i_1 < i_2 \\le n$ and $1 \\le j \\le n$ for which $(i_1, j)$ and $(i_2, j)$ are both red points cannot be larger than\n\n$$\n\\sum_{d=1}^{n-1} (n-d)\\left(\\frac{4m}{d} + 1\\right) = \\sum_{d=1}^{n-1} (n-d) + 4m \\sum_{d=1}^{n-1} \\frac{n-d}{d} < \\frac{n(n-1)}{2} + 4mn \\ln n.\n$$\n\nOn the other hand, for $1 \\le j \\le n$, assume that there are $x_j$ red points in $(1, j)$, $(2, j)$, ..., $(n, j)$. Then\n\n$$\n\\sum_{j=1}^{n} \\binom{x_j}{2} < \\frac{n(n-1)}{2} + 4mn \\ln n.\n$$\n\nSo\n\n$$\n\\sum_{j=1}^{n} \\left(x_j - \\frac{1}{2}\\right)^2 < n^2 - \\frac{1}{4}n + 8mn \\ln n.\n$$\n\nBy Cauchy's inequality, we know\n\n$$\n\\sum_{j=1}^{n} \\left(x_j - \\frac{1}{2}\\right) \\le \\sqrt{n\\left(n^2 - \\frac{1}{4}n + 8mn \\ln n\\right)},\n$$\n\ni.e.\n\n$$\n\\sum_{j=1}^{n} x_j \\le \\frac{n}{2} + n\\sqrt{n - \\frac{1}{4} + 8m \\ln n}.\n$$\n\nWhen $m \\ge n \\ge 2022$, some elementary estimate shows that this is $\\le 3n\\sqrt{m \\ln n}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20943,
"subject": "Mathematics (Olympiad)",
"question": "Let $x$ be a divisor of $12$. How many possible values can $x$ take?",
"options": [],
"answer": "See solution",
"solution": "The divisors of $12$ are $1, 2, 3, 4, 6, 12$. Thus, there are $6$ possible values for $x$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20944,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\omega$ be the circumcircle of $\\triangle ABC$. Reflect $\\omega$ in line $BC$ to obtain circle $\\omega'$, which contains points $H$ and $P'$. Let $M$ be the midpoint of $BC$. Assume $\\triangle ABC$ is acute-angled, so $H$ and $O$ lie inside the triangle.\n\nSuppose $\\angle CAB = 60^\\circ$. Prove that $HG = GP'$ if and only if $\\angle CAB = 60^\\circ$, where $G$ is the centroid, $H$ is the orthocenter, $O$ is the circumcenter, and $P'$ is as defined above. (Here, $G'$ and $H'$ denote the reflections of $G$ and $H$ with respect to $BC$.)",
"options": [],
"answer": "See solution",
"solution": "Assume $\\angle CAB = 60^\\circ$. Then\n\n$$\n\\angle COB = 2\\angle CAB = 120^\\circ = 180^\\circ - 60^\\circ = 180^\\circ - \\angle CAB = \\angle CHB,\n$$\n\nso $O$ lies on $\\omega'$. Reflecting $O$ in $BC$ gives $O'$ on $\\omega$, which is the center of $\\omega'$. Then $OO' = 2OM = 2R\\cos\\angle CAB = AH$, so $AH = OO' = HO' = AO = R$, where $R$ is the radius of $\\omega$ and $\\omega'$. Thus, quadrilateral $AHO'O$ is a rhombus, so $A$ and $O'$ are symmetric with respect to $HO$. Since $H, G, O$ are collinear (Euler line), $\\angle GAH = \\angle HO'G$. The diagonals of $GOPO'$ intersect at $M$. Since $\\angle BOM = 60^\\circ$,\n\n$$\nOM = MO' = \\cot 60^\\circ \\cdot MB = \\frac{MB}{\\sqrt{3}}.\n$$\n\nAs $3 \\cdot MO \\cdot MO' = MB^2 = MB \\cdot MC = MP \\cdot MA = 3MG \\cdot MP$, $GOPO'$ is cyclic. Since $BC$ is the perpendicular bisector of $OO'$, the circumcircle of $GOPO'$ is symmetric with respect to $BC$. Thus $P'$ also lies on the circumcircle of $GOPO'$, so $\\angle GO'P' = \\angle GPP'$. Note $\\angle GPP' = \\angle GAH$ since $AH \\parallel PP'$. As shown, $\\angle GAH = \\angle HO'G$, so $\\angle HO'G = \\angle GO'P'$. Thus, $\\triangle HO'G$ and $\\triangle GO'P'$ are congruent, so $HG = GP'$.\n\nNow, suppose $HG = GP'$. Reflect $A$ over $M$ to get $A'$. As before, $B, C, H, P'$ and $A'$ all lie on $\\omega'$. Note $HC \\perp CA'$ since $AB \\parallel CA'$, so $HA'$ is a diameter of $\\omega'$. The center $O'$ of $\\omega'$ is the midpoint of $HA'$. From $HG = GP'$, $\\triangle HGO'$ is congruent to $\\triangle P'GO'$, so $H$ and $P'$ are symmetric with respect to $GO'$. Thus $GO' \\perp HP'$ and $GO' \\parallel A'P'$. Let $HG$ meet $A'P'$ at $K$ ($K \\notin O$ since $AB \\neq AC$). Then $HG = GK$, as $GO'$ is the midline of $\\triangle HKA'$. Note $2GO = HG$ since $HO$ is the Euler line of $\\triangle ABC$, so $O$ is the midpoint of $GK$. Since $\\angle CMP = \\angle CMP'$, $\\angle GMO = \\angle OMP'$. The line $OM$ through $O'$ is the external angle bisector of $\\angle P'MA'$, and $O'$ is the midpoint of arc $P'MA'$. Thus, $P'MO'A'$ is cyclic, so $\\angle O'MA' = \\angle O'P'A' = \\angle O'A'P'$. Let $OM$ and $P'A'$ meet at $T$. Triangles $\\triangle TO'A'$ and $\\triangle A'O'M$ are similar, so $\\frac{O'A'}{O'M} = \\frac{O'T}{O'A'}$, i.e., $O'M \\cdot O'T = O'A'^2$.\n\nUsing Menelaus' theorem for $\\triangle HKA'$ and line $TO'$:\n\n$$\n\\frac{A'O'}{O'H} \\cdot \\frac{HO}{OK} \\cdot \\frac{KT}{TA'} = 3 \\cdot \\frac{KT}{TA'} = 1.\n$$\n\nSo $\\frac{KT}{TA'} = \\frac{1}{3}$ and $KA' = 2KT$. Using Menelaus' theorem for $\\triangle TO'A'$ and line $HK$:\n\n$$\n1 = \\frac{O'H}{HA'} \\cdot \\frac{A'K}{KT} \\cdot \\frac{TO}{OO'} = \\frac{1}{2} \\cdot 2 \\cdot \\frac{TO}{OO'} = \\frac{TO}{OO'}.\n$$\n\nThus $TO = OO'$, so $O'A'^2 = O'M \\cdot O'T = OO'^2$, so $O'A' = OO'$, and thus $O \\in \\omega'$. Finally, $2\\angle CAB = \\angle BOC = 180^\\circ - \\angle CAB$, so $\\angle CAB = 60^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20945,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle and let $M$ be the midpoint of the side $BC$. The circle of radius $MA$ centred at $M$ meets the lines $AB$ and $AC$ again at $B'$ and $C'$, respectively, and the tangents to this circle at $B'$ and $C'$ meet at $D$. Show that the perpendicular bisector of the segment $BC$ bisects the segment $AD$.",
"options": [],
"answer": "See solution",
"solution": "Let $A'$ be the antipodal of $A$ in the circle $AB'C'$, and let $A''$ be the point where this circle meets again the line through $A$ parallel to $BC$ (the points $A$ and $A''$ may coincide). Since $M$ is the midpoint of the side $BC$, the lines $AA''$, $AB'$, $AA'$, $AC'$ form a harmonic pencil. Consequently, so do the lines $XA''$, $XB'$, $XA'$, $XC'$ for any point $X$ on the circle $AB'C'$. \n\n\n\nNow let $X = B'$ and $X = C'$ to infer that the pencils $B'A'', B'D$, $B'A'$, $B'C'$ and $C'A'', C'B'$, $C'A'$, $C'D$ are both harmonic. Since the two pencils share the line $B'C'$, the points $A'', A'$, $D$ lie on a line which is clearly perpendicular to $BC$ and the conclusion follows.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20946,
"subject": "Mathematics (Olympiad)",
"question": "Let $p(x) = ax^4 + bx^3 + cx^2 + dx + e$ be a quartic polynomial with complex coefficients. Find all $h \\in \\mathbb{C}$ such that $p(h+z) = p(h-z)$ for all $z \\in \\mathbb{C}$, and determine the necessary and sufficient condition on the coefficients $a, b, c, d$ for such $h$ to exist.",
"options": [],
"answer": "See solution",
"solution": "We have\n\n$$\n\\begin{aligned}\n0 &= p(h+z) - p(h-z) \\\\\n &= a((h+z)^4 - (h-z)^4) + b((h+z)^3 - (h-z)^3) \\\\\n &\\quad + c((h+z)^2 - (h-z)^2) + d((h+z) - (h-z)) \\\\\n &= 8a(h^3z + hz^3) + 2b(3h^2z + z^3) + 4chz + 2dz \\\\\n &= z(8ah^3 + 6bh^2 + 4ch + 2d) + z^3(8ah + 2b).\n\\end{aligned}\n$$\n\nThis is satisfied for all $z \\in \\mathbb{C}$ if and only if\n\n$$\n8ah^3 + 6bh^2 + 4ch + 2d = 0 \\quad \\text{and} \\quad 8ah + 2b = 0.\n$$\n\nThe second equation gives $h = -\\frac{b}{4a}$. Substituting this into the first equation and multiplying by $4a^2$, we obtain\n\n$$\nb^3 - 4abc + 8a^2d = 0.\n$$\n\nThus, the necessary and sufficient condition is $b^3 - 4abc + 8a^2d = 0$, and the corresponding $h$ is $h = -\\frac{b}{4a}$.\n\n*Remark:* Alternatively, using calculus, $p(h+z) = p(h-z)$ for all $z$ if and only if $p'(h) = p'''(h) = 0$. From $p''(z) = 24az + 6b$, we again get $h = -\\frac{b}{4a}$, and the condition reduces to $p'(h) = 0$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20947,
"subject": "Mathematics (Olympiad)",
"question": "Find those positive integers $n \\leq 2014$ for which there exist positive integers $r, s$ such that $\\gcd(rs(r+s), n) = 1$ and $n$ does not divide $r - s$.",
"options": [],
"answer": "See solution",
"solution": "First, observe that for positive integers $r$ and $s$, at least one of $r$, $s$, or $r+s$ must be even. Thus, $rs(r+s)$ is always even. If $\\gcd(rs(r+s), n) = 1$, then $n$ must be odd. So, if $n$ is even, there are no suitable $r$ and $s$.\n\nIf $n = 1$, then $n$ divides any number, including $r-s$, so $n$ must be odd and greater than $1$.\n\nIf $n = 3$, $r$ and $s$ cannot be multiples of $3$, and $3$ must not divide $r-s$. If $r$ and $s$ have different remainders modulo $3$, then $r+s$ is divisible by $3$, so $rs(r+s)$ is divisible by $3$, contradicting $\\gcd(rs(r+s), 3) = 1$. Thus, there are no suitable $r$ and $s$ for $n = 3$.\n\nIf $n > 3$ and is odd, let $r = 2$ and $s = n-1$. Then:\n\n$$\n\\gcd(rs(r+s), n) = \\gcd(2(n-1)(n+1), n) = 1\n$$\n\nand $n \\nmid (r-s) = 3-n$.\n\nTherefore, the set of positive integers that fulfill the requirements is:\n\n$$\n\\{5, 7, 9, 11, 13, \\ldots, 2011, 2013\\}\n$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20948,
"subject": "Mathematics (Olympiad)",
"question": "What is the difference between the smallest number starting with a '6' and the largest number starting with a '3'?",
"options": [],
"answer": "See solution",
"solution": "The smallest number starting with a '6' is $6123$, and the largest number starting with a '3' is $3621$. Their difference is:\n\n$$6123 - 3621 = 2502$$\n\nThus, the difference is $2502$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20949,
"subject": "Mathematics (Olympiad)",
"question": "Let $m \\neq 0$ be an integer. Find all polynomials $P(x)$ with real coefficients such that\n\n$$\n(x^3 - m x^2 + 1) P(x + 1) + (x^3 + m x^2 + 1) P(x - 1) = 2(x^3 - m x + 1) P(x)\n$$\n\nfor all real numbers $x$.",
"options": [],
"answer": "See solution",
"solution": "Let $P(x) = a_n x^n + \\cdots + a_0$ with $a_n \\neq 0$. Comparing the coefficients of $x^{n+1}$ on both sides gives $a_n (n - 2m)(n - 1) = 0$, so $n = 1$ or $n = 2m$.\n\nIf $n = 1$, one easily verifies that $P(x) = x$ is a solution, while $P(x) = 1$ is not. Since the given condition is linear in $P$, this means that the linear solutions are precisely $P(x) = t x$ for $t \\in \\mathbb{R}$.\n\nNow assume that $n = 2m$. The polynomial $x P(x + 1) - (x + 1) P(x) = (n - 1) a_n x^n + \\cdots$ has degree $n$, and therefore it has at least one (possibly complex) root $r$. If $r \\notin \\{0, -1\\}$, define $k = P(r)/r = P(r+1)/(r+1)$. If $r = 0$, let $k = P(1)$. If $r = -1$, let $k = -P(-1)$. We now consider the polynomial $S(x) = P(x) - kx$. It also satisfies the given condition because $P(x)$ and $k x$ satisfy it. Additionally, it has the useful property that $r$ and $r + 1$ are roots.\n\nLet $A(x) = x^3 - m x^2 + 1$ and $B(x) = x^3 + m x^2 + 1$. Plugging in $x = s$ into the condition implies that:\n\nIf $s-1$ and $s$ are roots of $S$ and $s$ is not a root of $A$, then $s+1$ is a root of $S$.\n\nIf $s$ and $s+1$ are roots of $S$ and $s$ is not a root of $B$, then $s-1$ is a root of $S$.\n\nLet $a \\geq 0$ and $b \\geq 1$ be such that $r-a, r-a+1, \\dots, r, r+1, \\dots, r+b-1, r+b$ are roots of $S$, while $r-a-1$ and $r+b+1$ are not. The two statements above imply that $r-a$ is a root of $B$ and $r+b$ is a root of $A$.\n\nSince $r-a$ is a root of $B(x)$ and of $A(x+a+b)$, it is also a root of their greatest common divisor $C(x)$ as integer polynomials. If $C(x)$ was a non-trivial divisor of $B(x)$, then $B$ would have a rational root $\\alpha$. Since the first and last coefficients of $B$ are $1$, $\\alpha$ can only be $1$ or $-1$; but $B(-1) = m > 0$ and $B(1) = m + 2 > 0$ since $n = 2m$.\n\nTherefore $B(x) = A(x + a + b)$. Writing $c = a + b \\geq 1$ we compute\n\n$$\n0 = A(x + c) - B(x) = (3c - 2m) x^2 + c (3c - 2m) x + c^2 (c - m).\n$$\n\nThen we must have $3c - 2m = c - m = 0$, which gives $m = 0$, a contradiction. We conclude that $P(x) = t x$ is the only solution.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20950,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ and $b$ be two complex numbers. Prove that the following statements are equivalent:\n\n1) The absolute values of the roots of the equation $x^2 - a x + b = 0$ are respectively equal to the absolute values of the roots of the equation $x^2 - b x + a = 0$.\n\n2) $a^3 = b^3$ or $b = \\bar{a}$.",
"options": [],
"answer": "See solution",
"solution": "Let $|x_1| = |x_3|$, $|x_2| = |x_4|$ (1) and notice that $|a| = |x_3 x_4| = |x_1 x_2| = |b|$ to derive that $|x_1 + x_2| = |x_3 + x_4|$ (2). The relations (1) and (2) show that there exists a number $k \\in \\mathbb{C}$ such that $x_2 = k x_1$, $x_4 = k x_3$ or $x_2 = k x_1$, $x_4 = \\bar{k} x_3$.\n\nIn the first case we have $a = k x_3^2 = (1 + k) x_1$ and $b = k x_1^2 = (1 + k) x_3$, so $a^3 = k (1 + k)^2 x_1^2 x_3^2 = b^3$.\n\nIn the latter case we have $a = \\bar{k} x_3^2 = (1 + k) x_1$ and $b = k x_1^2 = (1 + \\bar{k}) x_3$. It follows that $x_1^2 \\bar{x}_1 = x_3 \\bar{x}_3^2$, so $x_1 = \\bar{x}_3$ or $a = b = 0$, and furthermore $x_2 = \\bar{x}_4$, hence $a = \\bar{b}$.\n\nConversely, if $b = \\bar{a}$, then $x_1 + x_2 = \\bar{x}_3 + \\bar{x}_4$, $x_1 x_2 = \\bar{x}_3 \\bar{x}_4$, implying $\\{x_1, x_2\\} = \\{\\bar{x}_3, \\bar{x}_4\\}$. If $a^3 = b^3$, then $a = \\varepsilon b$, $\\varepsilon^3 = 1$. The roots satisfy the relations $x_1 + x_2 = \\varepsilon (x_3 + x_4)$, $x_1 x_2 = \\varepsilon^2 x_3 x_4$. Both cases lead to $\\{|x_1|, |x_2|\\} = \\{|x_3|, |x_4|\\}$, as needed.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20951,
"subject": "Mathematics (Olympiad)",
"question": "A number written in base $a$ is $123_a$. The same number written in base $b$ is $146_b$. What is the minimum value of $a + b$?\n\n",
"options": [],
"answer": "See solution",
"solution": "Method 1\n\n$$\n\\begin{align*}\n123_a = 146_b &\\iff a^2 + 2a + 3 = b^2 + 4b + 6 \\\\\n&\\iff (a+1)^2 + 2 = (b+2)^2 + 2 \\\\\n&\\iff (a+1)^2 = (b+2)^2 \\\\\n&\\iff a+1 = b+2 \\quad (a \\text{ and } b \\text{ are positive}) \\\\\n&\\iff a = b+1\n\\end{align*}\n$$\n\nSince the digits in any number are less than the base, $b \\ge 7$.\n\nWe also have $a > b$, otherwise $a^2 + 2a + 3 < b^2 + 4b + 6$.\n\nIf $b = 7$ and $a = 8$, then $a^2 + 2a + 3 = 83 = b^2 + 4b + 6$.\n\nSo the minimum value for $a + b$ is $8 + 7 = 15$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20952,
"subject": "Mathematics (Olympiad)",
"question": "Let $X$ be an interior point of $\\triangle ABC$ and let $S_1 = S_{XBC}$, $S_2 = S_{XCA}$, $S_3 = S_{XAB}$. Find the minimal area of a convex polygon containing three segments equal and parallel to $XA$, $XB$, and $XC$.",
"options": [],
"answer": "See solution",
"solution": "Assume $S_1 \\geq S_2 \\geq S_3$. We will show that the answer is $S_1$.\n\nIt is not difficult to see that the area of the convex hull of two segments $d_1$ and $d_2$ is not less than the area $S$ of a triangle with two sides equal and parallel to these segments. Indeed, if this hull is a triangle, two cases are possible. In the first case, the respective segments are sides and then the area is equal to $S$. In the second case, suppose an interior point $E$ of $\\triangle ABC$ is such that $CE = d_1$ and $AB = d_2$. Setting $F = AB \\cap CE$, then\n\n$$\n2S_{ABC} = AB \\cdot CF \\sin \\angle CFB \\geq d_1 d_2 \\sin \\angle CFB = 2S.\n$$\n\nIf the hull is a quadrilateral with diagonals equal to the respective segments, then its area is equal to $S$. It remains to consider the case when the hull is a quadrilateral with two opposite sides equal to the respective segments. Suppose points $C$ and $D$ on the sides $BE$ and $AE$ of $\\triangle ABE$ are such that $AD = d_1$ and $BC = d_2$. Then\n\n$$\n2S_{ABCD} = 2S_{ABE} - 2S_{CDE} = (AE \\cdot BE - DE \\cdot CE) \\sin \\angle AEB > AD \\cdot BC \\sin \\angle AEB = 2S.\n$$\n\nHence, any convex polygon containing two segments equal and parallel to $XB$ and $XC$ has area at least $S_1$.\n\nNow, let $A'$ be the symmetric point of $A$ with respect to $X$, and let $D$ be such that $XBDC$ is a parallelogram. Then $A'$ lies either in $\\angle BXD$ or in $\\angle CXD$; for example, in $\\angle BXD$. Since $S_{XCA'} = S_2 \\leq S_1$, $A'$ lies in $\\triangle BXD$. This triangle has area $S_1$ and contains three segments equal and parallel to $XA$, $XB$, and $XC$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20953,
"subject": "Mathematics (Olympiad)",
"question": "Prove that for positive real numbers $a, b, c$, the following inequalities are satisfied:\n\n$$\n\\frac{3}{2} < \\frac{4a+b}{a+4b} + \\frac{4b+c}{b+4c} + \\frac{4c+a}{c+4a} < 9\n$$",
"options": [],
"answer": "See solution",
"solution": "Let\n\n$$\nS = \\frac{4a+b}{a+4b} + \\frac{4b+c}{b+4c} + \\frac{4c+a}{c+4a}\n$$\n\nWe will show that $\\frac{3}{2} < S < 9$.\n\nFirst, let us show that $\\frac{3}{2} < S$. Suppose $a \\geq b$ and $a \\geq c$. Then,\n\n$$\n\\begin{aligned}\n\\frac{4a+b}{a+4b} &= \\frac{(a+b)+3a}{(a+b)+3b} \\geq 1 \\\\\n\\frac{4b+c}{b+4c} &> \\frac{4b+c}{16b+4c} = \\frac{1}{4} \\\\\n\\frac{4c+a}{c+4a} &> \\frac{4c+a}{16c+4a} = \\frac{1}{4}\n\\end{aligned}\n$$\n\nThus, $S > 1 + \\frac{1}{4} + \\frac{1}{4} = \\frac{3}{2}$. By symmetry, if $b$ or $c$ is the largest, the same argument applies: one term is at least $1$, the other two are greater than $\\frac{1}{4}$, so $S > \\frac{3}{2}$.\n\nNext, we show that $S < 9$. Suppose $a \\leq b$ and $a \\leq c$. Then,\n\n$$\n\\begin{aligned}\n\\frac{4a+b}{a+4b} &= \\frac{(a+b)+3a}{(a+b)+3b} \\leq 1 \\\\\n\\frac{4b+c}{b+4c} &< \\frac{4b+16c}{b+4c} = 4 \\\\\n\\frac{4c+a}{c+4a} &< \\frac{4c+16a}{c+4a} = 4\n\\end{aligned}\n$$\n\nSo $S < 1 + 4 + 4 = 9$. By symmetry, the same holds if $b$ or $c$ is the smallest. Thus, $\\frac{3}{2} < S < 9$ is proved.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20954,
"subject": "Mathematics (Olympiad)",
"question": "A triangle $AB\\Gamma$ is given with $AB < A\\Gamma$. Let $I$ be the intersection point of its angle bisectors. The bisector $A\\Delta$ meets the circumcircle $C$ of triangle $B\\Gamma$ at the point $N$ with $N \\neq I$.\n\n1. Determine the angles of triangle $B\\Gamma N$ in terms of the angles of triangle $AB\\Gamma$.\n2. Find the center of the circle $C$.",
"options": [],
"answer": "See solution",
"solution": "(i)\n\n$$\n\\widehat{\\Gamma BN} = \\widehat{\\Gamma IN} = \\frac{\\hat{A}}{2} + \\frac{\\hat{\\Gamma}}{2} = \\frac{180^{\\circ} - \\hat{B}}{2} = 90^{\\circ} - \\frac{\\hat{B}}{2}\n$$\n\n$$\n\\widehat{B\\Gamma N} = \\widehat{B\\Gamma IN} = \\frac{\\hat{A}}{2} + \\frac{\\hat{B}}{2} = \\frac{180^{\\circ} - \\hat{\\Gamma}}{2} = 90^{\\circ} - \\frac{\\hat{\\Gamma}}{2}\n$$\n\n$$\n\\widehat{BN\\Gamma} = 180^{\\circ} - \\left( 90^{\\circ} - \\frac{\\widehat{B}}{2} + 90^{\\circ} - \\frac{\\widehat{\\Gamma}}{2} \\right)\n$$\n\n$$\n= \\frac{\\widehat{B} + \\widehat{\\Gamma}}{2} = 90^{\\circ} - \\frac{\\widehat{A}}{2}\n$$\n\n\n\nFigure 1\n\n(ii) Since $\\widehat{\\Gamma BN} = 90^{\\circ} - \\frac{\\widehat{B}}{2}$, $BN$ is the bisector of the external angle at $B$.\n\nTherefore $\\widehat{IBN} = 90^{\\circ}$ and $IN$ is a diameter of the circle $C$. Moreover, if $A\\Delta$ intersects the circumcircle of triangle $AB\\Gamma$ at $M$, then\n\n$$\n\\widehat{BIN} = \\frac{\\hat{A}}{2} + \\frac{\\hat{B}}{2} = \\widehat{\\Gamma BM} + \\widehat{\\Gamma BI} = \\widehat{IBM}.\n$$\n\nThus, triangle $IBM$ is isosceles with $MB = MI$. Therefore, $M$ lies on the perpendicular bisector of $BI$. Since $M$ belongs to the diameter of circle $C$, it is the center of $C$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20955,
"subject": "Mathematics (Olympiad)",
"question": "令 $m, n \\ge 2$ 為整數，且令 $f(x_1, \\dots, x_n)$ 為一實係數多項式，使得對每一個 $x_1, x_2, \\dots, x_n \\in \\{0, 1, \\dots, m-1\\}$，\n\n$$\nf(x_1, \\dots, x_n) = \\left[ \\frac{x_1 + \\dots + x_n}{m} \\right]\n$$\n\n均成立。試證：$f$ 的次數至少為 $n$。",
"options": [],
"answer": "See solution",
"solution": "我们将问题转化为单变量问题。\n\n**引理**：设 $a_1, \\dots, a_n$ 为非负整数，$G(x)$ 为次数不超过 $a_1 + \\dots + a_n$ 的非零多项式。若某多项式 $F(x_1, \\dots, x_n)$ 满足\n\n$$\nF(x_1, \\dots, x_n) = G(x_1 + \\dots + x_n)\n$$\n\n对所有 $(x_1, \\dots, x_n) \\in \\{0, 1, \\dots, a_1\\} \\times \\dots \\times \\{0, 1, \\dots, a_n\\}$，则 $F$ 非零多项式且 $\\deg F \\ge \\deg G$。\n\n证明用到多项式的**前向差分**。若 $p(x)$ 为单变量多项式，定义\n\n$$\n(\\Delta p)(x) = p(x + 1) - p(x)\n$$\n\n已知若 $p$ 非常数，则\n\n$$\n\\deg \\Delta p = \\deg p - 1\n$$\n\n若 $p(x_1, \\dots, x_n)$ 为 $n$ 元多项式，$1 \\le k \\le n$，则\n\n$$\n\\Delta_k(p)(x_1, \\dots, x_n) = p(x_1, \\dots, x_{k-1}, x_k+1, x_{k+1}, \\dots, x_n) - p(x_1, \\dots, x_n)\n$$\n\n同样有 $\\Delta_k p$ 要么为零多项式，要么\n\n$$\n\\deg(\\Delta_k p) \\leq \\deg p - 1\n$$\n\n**引理证明**：对 $G$ 的次数归纳。若 $G$ 为常数，则 $F(0, \\dots, 0) = G(0) \\neq 0$，故 $F$ 非零。\n\n若 $\\deg G \\geq 1$，归纳假设对低次数成立。因 $a_1 + \\dots + a_n \\geq \\deg G > 0$，至少有一个 $a_i > 0$，不妨设 $a_1 \\geq 1$。\n\n考虑 $F_1 = \\Delta_1 F$ 和 $G_1 = \\Delta G$。在 $\\{0, \\dots, a_1 - 1\\} \\times \\{0, \\dots, a_2\\} \\times \\dots \\times \\{0, \\dots, a_n\\}$ 上有\n\n$$\n\\begin{aligned}\nF_1(x_1, \\dots, x_n) &= F(x_1 + 1, x_2, \\dots, x_n) - F(x_1, x_2, \\dots, x_n) \\\\\n&= G(x_1 + \\dots + x_n + 1) - G(x_1 + \\dots + x_n) \\\\\n&= G_1(x_1 + \\dots + x_n)\n\\end{aligned}\n$$\n\n因 $G$ 非常数，$\\deg G_1 = \\deg G - 1 \\leq (a_1 - 1) + a_2 + \\dots + a_n$。\n\n归纳假设得 $F_1$ 非零且 $\\deg F_1 \\geq \\deg G_1$，故\n\n$$\n\\deg F \\geq \\deg F_1 + 1 \\geq \\deg G_1 + 1 = \\deg G\n$$\n\n完成证明。\n\n回到原题，取唯一多项式 $g(x)$ 满足\n\n$$\ng(x) = \\left[ \\frac{x}{m} \\right] \\quad \\text{对} \\ x \\in \\{0, 1, \\dots, n(m-1)\\} \\quad \\text{且} \\ \\deg g \\le n(m-1)\n$$\n\n共规定了 $n(m-1)+1$ 个值，故 $g(x)$ 存在且唯一。又 $g(0) = g(1) = 0, g(m) = 1$，故 $\\deg g \\ge 2$。\n\n应用引理于 $a_1 = \\dots = a_n = m-1$ 及 $f, g$，得 $\\deg f \\ge \\deg g$。只需对 $\\deg g$ 下界。\n\n考虑\n\n$$\nh(x) = g(x + m) - g(x) - 1\n$$\n\n$g(x+m) - g(x)$ 的次数为 $\\deg g - 1 \\ge 1$，故\n\n$$\n\\deg h = \\deg g - 1 \\geq 1\n$$\n\n且 $h$ 在 $0, 1, \\dots, n(m-1)-m$ 上为零，故 $h$ 至少有 $(n-1)(m-1)$ 个根。\n\n因此\n\n$$\n\\deg f \\geq \\deg g = \\deg h + 1 \\geq (n-1)(m-1) + 1 \\geq n\n$$\n\n**注释**：引理中等号成立于 $F(x_1, \\dots, x_n) = G(x_1 + \\dots + x_n)$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20956,
"subject": "Mathematics (Olympiad)",
"question": "Let $n = 2^m$ with $m > 1$. Can Alice equalize the numbers of sweets in all boxes, given that the sum of all distances is always zero? What happens if $n \\neq 2^m$?",
"options": [],
"answer": "See solution",
"solution": "Suppose $k$ distances are even and equal to $\\alpha$, and the remaining $n-k$ distances are odd and equal to $\\beta$. Then $\\alpha k + \\beta(n-k) = 0$, so $k(\\alpha - \\beta) = 2^m \\beta$. Since $\\alpha - \\beta$ is odd and $k < 2^m$, this is impossible. Thus, if $n = 2^m$ ($m > 1$), Alice can equalize the numbers of sweets in all boxes.\n\nNow, let $n \\neq 2^m$. Then $n = (2p + 1) \\cdot 2^m$ for some positive integer $p$ and nonnegative integer $m$. Suppose $k = 2^m$ boxes have distance $\\alpha = 2p$ and the other $n-k = 2p2^m$ boxes have distance $\\beta = -1$. This distribution is possible since $\\alpha k + \\beta(n-k) = 0$. Since $\\alpha$ and $\\beta$ have different parities, Alice cannot equalize the numbers of sweets in all boxes.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20957,
"subject": "Mathematics (Olympiad)",
"question": "In a cyclic quadrilateral $ABCD$ with $|AD| > |BC|$, the vertices $C$ and $D$ lie on the shorter arc $AB$ of the circumcircle. Rays $AD$ and $BC$ intersect at point $K$, and diagonals $AC$ and $BD$ intersect at point $P$. The line $KP$ intersects side $AB$ at point $L$. Prove that $\\angle ALK$ is acute.",
"options": [],
"answer": "See solution",
"solution": "**Solution.** From the properties of cyclic quadrilaterals, we get $\\angle KAB = \\angle KC'D$ and $\\angle KBA = \\angle KDC$. Let $A'$, $B'$, $K'$ be the feet of the altitudes of triangle $ABK$ drawn from the vertices $A$, $B$, $K$, respectively, and let $H$ be the orthocenter of triangle $ABK$.\n\n\n\n\n\nThe points $A$, $B$, $A'$, $B'$ lie on a common circle, hence $\\angle KA'B' = \\angle KAB$ if $A' \\neq B'$. Therefore, $A'$ and $B'$ lie on a line parallel to $CD$. Denote this line by $A'B'$ (even in the case $A' = B' = K$).\n\nLet $d(X, l)$ be the distance of point $X$ from line $l$, and let $S_\\Delta$ be the area of triangle $\\Delta$. By two angles, $\\angle ACK \\sim \\angle BDK$ and $\\angle PAD \\sim \\angle PBC$, whence\n\n$$\n\\frac{|AK|}{|BK|} = \\frac{|AC|}{|BD|} \\quad \\text{and} \\quad \\frac{|AP|}{|BP|} = \\frac{|AD|}{|BC|}.\n$$\nAt the same time,\n$$\n\\begin{aligned}\n\\frac{d(A, CD)}{d(B, CD)} &= \\frac{S_{\\triangle ACD}}{S_{\\triangle BCD}} = \\frac{|AC| \\cdot |AD| \\cdot \\sin \\angle CAD}{|BD| \\cdot |BC| \\cdot \\sin \\angle CBD} = \\frac{|AC| \\cdot |AD|}{|BD| \\cdot |BC|}, \\\\\n\\frac{d(A, KP)}{d(B, KP)} &= \\frac{S_{\\triangle AKP}}{S_{\\triangle BKP}} = \\frac{|AK| \\cdot |AP| \\cdot \\sin \\angle KAP}{|BK| \\cdot |BP| \\cdot \\sin \\angle KBP} = \\frac{|AK| \\cdot |AP|}{|BK| \\cdot |BP|}.\n\\end{aligned}\n$$\nTherefore,\n$$\n\\frac{|AL|}{|LB|} = \\frac{d(A, KP)}{d(B, KP)} = \\frac{d(A, CD)}{d(B, CD)}.\n$$\n\nConsidering instead of the cyclic quadrilateral $ABCD$ the quadrilateral determined by points $A$, $B$, $A'$, $B'$, and instead of $P$ and $L$ the points $H$ and $K'$ correspondingly, we get similarly that\n$$\n\\frac{|AK'|}{|K'B|} = \\frac{d(A, KH)}{d(B, KH)} = \\frac{d(A, A'B')}{d(B, A'B')}.\n$$\nThis equality holds also in the special case $A' = B' = K$. Indeed, let the projections of points $A$ and $B$ to the line $A'B'$ be $X$ and $Y$ correspondingly, then $\\angle AKX = \\angle KDC = \\angle KBA = \\angle AKK'$, $\\angle BKY = \\angle KCD = \\angle KAB = \\angle BKK'$, whence $\\triangle AKX \\sim \\triangle AKK'$ and $\\triangle BKY \\sim \\triangle BKK'$. It follows that $|AK'| = |AX|$, $|BK'| = |BY|$ and $\\frac{|AK'|}{|K'B|} = \\frac{d(A,A'B')}{d(B,A'B')}.$\n\nSince $C$ and $D$ lie on the shorter arc $AB$, we have $\\angle BCA = \\angle BDA > \\frac{\\pi}{2}$. Thus the line $A'B'$ is farther from the points $A$ and $B$ than the line $CD$. Since $|AD| > |BC|$, we have $\\angle ABD > \\angle CAB$ and also $\\angle KBA > \\angle KAB$, which implies $|KA| > |KB|$. Hence $d(A, CD) + d(K, CD) > d(B, CD) + d(K, CD)$, or $d(A, CD) > d(B, CD)$. Altogether,\n\n\n\n$$\n\\frac{|AL|}{|LB|} = \\frac{d(A, CD)}{d(B, CD)} > \\frac{d(A, A'B')}{d(B, A'B')} = \\frac{|AK'|}{|K'B|}.\n$$\n\nHence $L$ lies farther from $A$ than $K'$ on the segment $AB$, therefore $\\angle ALK < \\angle AK'K = \\frac{\\pi}{2}$, i.e., $\\angle ALK$ is acute.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20958,
"subject": "Mathematics (Olympiad)",
"question": "Calculate:\n\na) $1\\frac{5}{8} + \\left(1\\frac{1}{2} + \\left(\\left(1\\frac{1}{2} - \\frac{1}{6}\\right) \\cdot \\left(\\frac{1}{3} + \\frac{1}{4}\\right) + 1\\right)\\right) : \\frac{2}{3}$\n\nb) $0.6 : \\frac{1\\frac{1}{2} + 0.5 \\cdot 2\\frac{1}{2} - 0.25}{15 - 0.5}$",
"options": [],
"answer": "See solution",
"solution": "a)\n$$\n\\begin{aligned}\n1\\frac{5}{8} + \\left(1\\frac{1}{2} + \\left(\\left(1\\frac{1}{2} - \\frac{1}{6}\\right) \\cdot \\left(\\frac{1}{3} + \\frac{1}{4}\\right) + 1\\right)\\right) : \\frac{2}{3} &= 1\\frac{5}{8} + \\left(1\\frac{1}{2} + \\left(\\frac{2}{6} \\cdot \\frac{7}{12} + 1\\right)\\right) : \\frac{2}{3} \\\\\n&= 1\\frac{5}{8} + \\left(1\\frac{1}{2} + 1\\frac{7}{36}\\right) : \\frac{2}{3} \\\\\n&= 1\\frac{5}{8} + 2\\frac{25}{36} : \\frac{2}{3} \\\\\n&= 1\\frac{5}{8} + \\frac{97}{36} \\cdot \\frac{3}{2} \\\\\n&= \\frac{13}{8} + \\frac{97}{24} \\\\\n&= \\frac{13 \\cdot 3 + 97}{24} \\\\\n&= \\frac{136}{24} = 5\\frac{2}{3}\n\\end{aligned}\n$$\n\nb)\n$$\n0.6 : \\frac{1\\frac{1}{2} + 0.5 \\cdot 2\\frac{1}{2} - 0.25}{15 - 0.5} = 0.6 : \\frac{1.5 + 0.5 \\cdot 2.5 - 0.25}{15 - 0.5} = 0.6 : \\frac{1.5 + 0.2 - 0.25}{15 - 0.5} = 0.6 : \\frac{1.45}{14.5} = 0.6 : 0.1 = 6\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20959,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge 3$ be a fixed integer. The numbers $1, 2, 3, \\dots, n$ are written on a board. In every move, one chooses two numbers and replaces them by their arithmetic mean. This is done until only a single number remains on the board.\n\nDetermine the least integer that can be reached at the end by an appropriate sequence of moves.",
"options": [],
"answer": "See solution",
"solution": "The answer is $2$ for every $n$. Surely we cannot reach an integer less than $2$, since $1$ appears only once and produces an arithmetic mean greater than $1$ as soon as it is used.\n\nOn the other hand, we can prove by induction on $k$ that the number $a+1$ can be reached from the numbers $a, a+1, \\dots, a+k$ by a sequence of permitted moves.\n\nFor $k=2$, one replaces $a$ and $a+2$ by $a+1$, and afterwards $a+1$ and $a+1$ by a single $a+1$.\n\nFor the induction step $k \\to k+1$, one replaces $a+1, \\dots, a+k+1$ by $a+2$, and afterwards $a$ and $a+2$ by $a+1$.\n\nIn particular, with $a=1$ and $k=n-1$, one achieves the desired result.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20960,
"subject": "Mathematics (Olympiad)",
"question": "For each positive integer $n$, find the largest real number $C_n$ with the following property. Given any $n$ real-valued functions $f_1(x), f_2(x), \\dots, f_n(x)$ defined on the closed interval $0 \\le x \\le 1$, one can find numbers $x_1, x_2, \\dots, x_n$, such that $0 \\le x_i \\le 1$, satisfying\n\n$$\n|f_1(x_1) + f_2(x_2) + \\dots + f_n(x_n) - x_1x_2\\dots x_n| \\ge C_n.\n$$",
"options": [],
"answer": "See solution",
"solution": "First, we will prove that $C_n \\ge \\frac{n-1}{2n}$, i.e., that for any $n$ functions $f_1, f_2, \\dots, f_n : [0, 1] \\to \\mathbb{R}$, there exist numbers $x_1, x_2, \\dots, x_n$ in $[0, 1]$ such that\n\n$$\n|f_1(x_1) + f_2(x_2) + \\dots + f_n(x_n) - x_1x_2\\dots x_n| \\ge \\frac{n-1}{2n}.\n$$\n\nFor $n=1$ this is trivial. For $n \\ge 2$, suppose, contrariwise, that for all $x_1, x_2, \\dots, x_n$ in $[0, 1]$ we have\n\n$$\n|f_1(x_1) + f_2(x_2) + \\dots + f_n(x_n) - x_1x_2\\dots x_n| < \\frac{n-1}{2n}.\n$$\n\nPlugging in $x_i = 1$ for $1 \\le i \\le n$, we get $\\left| \\sum_{i=1}^{n} f_i(1) - 1 \\right| < \\frac{n-1}{2n}$.\n\nPlugging in $x_i = 0$ for $1 \\le i \\le n$, we get $\\left| \\sum_{i=1}^{n} f_i(0) \\right| < \\frac{n-1}{2n}$.\n\nPlugging in (for every $1 \\le i \\le n$) $x_i = 0$ and $x_j = 1$ for all $j \\ne i$, we get\n$$\n\\left| f_i(0) + \\sum_{j \\ne i} f_j(1) \\right| < \\frac{n-1}{2n}.\n$$\nSince\n\n$$\n(n-1) \\sum_{i=1}^{n} f_i(1) = \\sum_{i=1}^{n} \\left( f_i(0) + \\sum_{j \\ne i} f_j(1) \\right) - \\sum_{i=1}^{n} f_i(0),\n$$\n\nby the triangle inequality we have\n\n$$\n(n-1) \\left| \\sum_{i=1}^{n} f_i(1) \\right| < (n+1) \\frac{n-1}{2n}.\n$$\n\nOn the other hand, by the triangle inequality again,\n\n$$\n1 \\le \\left| \\sum_{i=1}^{n} f_i(1) \\right| + \\left| \\sum_{i=1}^{n} f_i(1) - 1 \\right| < \\frac{n+1}{2n} + \\frac{n-1}{2n} = 1,\n$$\n\nwhich is a contradiction.\n\nTo prove that\n\n$$\nC_n = \\frac{n-1}{2n}\n$$\n\nis the largest constant, it suffices to prove that for the $n$ (equal) functions\n\n$$\nf_i(x) = f(x) := \\frac{x^n}{n} - \\frac{n-1}{2n^2}, \\quad 1 \\le i \\le n,\n$$\n\nand any $n$ numbers $x_1, x_2, \\dots, x_n$ in $[0, 1]$, we have\n\n$$\n\\left| f(x_1) + f(x_2) + \\cdots + f(x_n) - x_1 x_2 \\cdots x_n \\right| \\le \\frac{n-1}{2n},\n$$\n\nequivalent to\n\n$$\n-\\frac{n-1}{2n} \\le \\frac{1}{n} \\sum_{i=1}^{n} x_i^n - \\prod_{i=1}^{n} x_i - \\frac{n-1}{2n} \\le \\frac{n-1}{2n}.\n$$\n\nThe left inequality follows from the AM-GM inequality.\n\nThe right inequality is equivalent to\n\n$$\nF(\\mathbf{x}) = F(x_1, x_2, \\dots, x_n) := \\frac{1}{n} \\sum_{i=1}^{n} x_i^n - \\prod_{i=1}^{n} x_i \\le \\frac{n-1}{n}\n$$\n\nat all points $\\mathbf{x} = (x_1, x_2, \\dots, x_n)$ of the hypercube $[0, 1]^n$. Since $F$ is convex in every variable, its maximum is reached at some vertex $\\mathbf{v}$ of the hypercube (point with $x_i = 0$ or $x_i = 1$, for all $1 \\le i \\le n$). It is easy to see that for all such points we have $F(\\mathbf{v}) \\le \\frac{n-1}{n}$, which completes the proof.\n\n*Remarks.* The choice of the functions\n\n$$\nf_i(x) := \\frac{x^n}{n} - \\frac{n-1}{2n^2}, \\quad 1 \\le i \\le n\n$$\n\ncould be justified by the fact that if we try all $f_i = f$ and all $x_i = x$, the relation becomes $|nf(x) - x^n| \\ge C_n$, as tight as possible for some $x \\in [0, 1]$. Then $f(x) = \\frac{x^n}{n} - \\frac{1}{n}C_n$ is a potential candidate.\n\nOn a different note, the right inequality above is equivalent to\n\n$$\nF(\\mathbf{x}) = F(x_1, x_2, \\dots, x_n) := \\frac{1}{n} \\sum_{i=1}^{n} x_i^n - \\prod_{i=1}^{n} x_i \\ge 0\n$$\n\nat all points $\\mathbf{x} = (x_1, x_2, \\dots, x_n)$ of the hypercube $[0, 1]^n$, and this again may be justified by $F$ being convex, since it can be easily seen that $F(\\mathbf{v}) \\ge 0$ at any vertex $\\mathbf{v}$ of the hypercube, so one may use a unifying argument for both sides of that inequality.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20961,
"subject": "Mathematics (Olympiad)",
"question": "Let $DEFG$ be a square with center $A_1$ such that $D, E$ lie on $BC$, $F$ lies on $CA$, and $G$ lies on $AB$ of triangle $ABC$. Consider a homothety with center $A$ mapping $DEFG$ to a square $D'E'CB$, since $GF \\parallel BC$. Let $X$ be the center of $D'E'CB$. Similarly, let $Y$ and $Z$ be the centers of the squares constructed outside $\\triangle ABC$ having $CA$ and $AB$ as a side, respectively. Prove that $AX$, $BY$, and $CZ$ are concurrent.\n\n",
"options": [],
"answer": "See solution",
"solution": "By considering a homothety with center $A$, $DEFG$ maps to $D'E'CB$, and $A$, $A_1$, $X$ are collinear. Similarly, $Y$ and $Z$ are defined for the other sides. The concurrency of $AX$, $BY$, and $CZ$ follows from Jacobi's theorem, since $\\angle ZAB = \\angle YAC = 45^\\circ$, $\\angle XBC = \\angle ZBA = 45^\\circ$, and $\\angle YCA = \\angle XCB = 45^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20962,
"subject": "Mathematics (Olympiad)",
"question": "Find large enough positive integers $a, b, c$ such that\n$$\n\\frac{ab + ac + bc}{a + b + c} = 3^N\n$$\nfor some positive integer $N$.",
"options": [],
"answer": "See solution",
"solution": "We can write\n$$\n\\frac{ab + ac + bc}{a + b + c} = a + b - \\frac{a^2 + ab + b^2}{a + b + c}.\n$$\nNow, choose $c$ such that\n$$\n\\frac{a^2 + ab + b^2}{a + b + c} = 1,\n$$\nwhich gives $c = a^2 + b^2 + ab - a - b$. Then we need $a + b - 1 = 3^N$. Choose coprime $a, b$ such that $3$ does not divide $ab(a-1)(b-1)$. It follows that $\\gcd(c, b) = \\gcd(c, a) = 1$. Further, choosing $a \\equiv b \\equiv 2 \\pmod{3}$ implies that $3$ divides $a + b + c = a^2 + ab + b^2$. The rest follows.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20963,
"subject": "Mathematics (Olympiad)",
"question": "Find all integer solutions $(a, b, c, d)$ to the equation\n$$\na\\sqrt{2} + b\\sqrt{5} + c = d\\sqrt{10}.\n$$",
"options": [],
"answer": "See solution",
"solution": "One solution is straightforward: $a = b = c = d = 0$. We will prove that it is the only one.\n\nSuppose there is another solution $(a, b, c, d)$. We may suppose that the integers $a, b, c$, and $d$ are coprime; otherwise, their greatest common divisor could be factored out of the equation (since not all numbers are zero, their greatest common divisor exists).\n\nThe equation can be reorganized as\n$$\na\\sqrt{2} + b\\sqrt{5} = d\\sqrt{10} - c\n$$\nand squared to obtain\n$$\n2a^2 + 5b^2 - c^2 - 10d^2 = 2\\sqrt{10}(cd - ab).\n$$\nBecause $\\sqrt{10}$ is not rational, we conclude $2a^2 + 5b^2 - c^2 - 10d^2 = 0$, or\n$$\n2a^2 - c^2 = 10d^2 - 5b^2.\n$$\nBecause the right side is divisible by $5$, the same must hold for the left side. The square of a natural number gives a remainder of $0$, $1$, or $4$ when divided by $5$. From this, we conclude that $a^2$ and $c^2$ must give a remainder of $0$ when divided by $5$, hence $a$ and $c$ must be divisible by $5$.\n\nThe left side is thus divisible by $25$, and so must be the right side. Consequently, $2d^2 - b^2$ must be divisible by $5$. As before, we conclude that $b$ and $d$ are divisible by $5$.\n\nAll numbers $a, b, c$, and $d$ are thus divisible by $5$, which is a contradiction since they are coprime. Therefore, the only integer solution is $a = b = c = d = 0$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 20964,
"subject": "Mathematics (Olympiad)",
"question": "Let $p$ be a prime. Show that $\\sqrt[3]{p} + \\sqrt[3]{p^5}$ is irrational.",
"options": [],
"answer": "See solution",
"solution": "Let $r = \\sqrt[3]{p} + \\sqrt[3]{p^5}$. Then,\n\n$$\nr^3 = (\\sqrt[3]{p} + \\sqrt[3]{p^5})^3 = p + p^5 + 3p^2(\\sqrt[3]{p} + \\sqrt[3]{p^5}) = p + p^5 + 3p^2 r.\n$$\n\nHence $r$ is a root of the polynomial $x^3 - 3p^2x - p^5 - p$. Assume to the contrary that $r$ is rational. By the rational root theorem, $r$ is an integer. From $r^3 = p + p^5 + 3p^2r$ we get $p \\mid r$. Hence $p^3 \\mid r^3 - 3p^2r - p^5$, implying $p^3 \\mid p$, a contradiction. Thus, $r$ is irrational.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20965,
"subject": "Mathematics (Olympiad)",
"question": "On the sides of triangle $ABC$, points $P$, $Q \\in AB$ ($P$ is between $A$ and $Q$) and $R \\in BC$ are chosen. The points $M$ and $N$ are defined as the intersection point of $AR$ with the segments $CP$ and $CQ$, respectively. If $BC = BQ$, $CP = AP$, $CR = CN$ and $\\angle BPC = \\angle CRA$, prove that $MP + NQ = BR$.",
"options": [],
"answer": "See solution",
"solution": "To prove that $MP + NQ = BR$, by adding $RC$ to both sides it is equivalent to prove $CQ + MP = BC$. We start by defining point $T$ on the segment $CQ$ such that $BC = CT$, then it is sufficient to prove that $QT = PM$, since triangles $BTC$ and $NRC$ are isosceles. One can get that $BT$ is parallel to $NR$ so\n\n$$\n\\angle BTC = \\angle RNC = \\angle NRC = \\angle BPC,\n$$\n\nthus $BTPC$ is cyclic. Also we have that\n\n$$\n\\angle CNA = 180^\\circ - \\angle NRC = 180^\\circ - \\angle BPC = \\angle CPA,\n$$\n\nthis implies that $PNCA$ is cyclic.\n\n\n\nSo we have that\n\n$$\n\\angle TPQ = \\angle TCB = \\angle BQC = \\angle TQP,\n$$\n\nthus $TQ = TP$. Now by angle chasing,\n\n$$\n\\angle CPA = 180^\\circ - \\angle BPC = 180^\\circ - \\angle BTC = 180^\\circ - \\angle TBC = \\angle TPC,\n$$\n\nthus we have $\\angle CPQ = \\angle TPC$ and $\\angle PCT = \\angle PAM$, which implies that $TPC$ and $MPA$ are congruent, since $PC = AP$. In conclusion, $TP = MP = TQ$, which finishes the proof. $\\Box$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20966,
"subject": "Mathematics (Olympiad)",
"question": "Let $G$ be the centroid of a right-angled triangle $ABC$ with $\\angle BCA = 90^\\circ$. Let $P$ be the point on ray $AG$ such that $\\angle CPA = \\angle CAB$, and let $Q$ be the point on ray $BG$ such that $\\angle CQB = \\angle ABC$. Prove that the circumcircles of triangles $AQG$ and $BPG$ meet at a point on side $AB$.",
"options": [],
"answer": "See solution",
"solution": "Since $\\angle C = 90^\\circ$, the point $C$ lies on the semicircle with diameter $AB$, which implies that if $M$ is the midpoint of side $AB$, then $MA = MC = MB$. This implies that triangle $AMC$ is isosceles and hence $\\angle ACM = \\angle A$. By definition, $G$ lies on segment $M$, and it follows that $\\angle ACG = \\angle ACM = \\angle A = \\angle CPA$. This implies that triangles $APC$ and $ACG$ are similar and hence $AC^2 = AG \\cdot AP$. Now, if $D$ denotes the foot of the perpendicular from $C$ to $AB$, it follows that triangles $ACD$ and $ABC$ are similar, which implies $AC^2 = AD \\cdot AB$. Therefore, $AG \\cdot AP = AC^2 = AD \\cdot AB$ and, by power of a point, quadrilateral $DGPB$ is cyclic. This implies that $D$ lies on the circumcircle of triangle $BPG$ and, by a symmetric argument, it follows that $D$ also lies on the circumcircle of triangle $AGQ$. Therefore, these two circumcircles meet at the point $D$ on side $AB$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20967,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle, let $O$ be its circumcentre, let $A'$ be the orthogonal projection of $A$ on the line $BC$, and let $X$ be a point on the open ray $AA'$ emanating from $A$. The internal bisector of the angle $BAC$ meets the circumcircle of $ABC$ again at $D$. Let $M$ be the midpoint of the segment $DX$. The line through $O$ and parallel to the line $AD$ meets the line $DX$ at $N$. Prove that the angles $BAM$ and $CAN$ are equal.\n\n",
"options": [],
"answer": "See solution",
"solution": "Choose a point $Y$ such that $AONY$ is a parallelogram. Since the lines $AD$ and $ON$ are parallel, this point lies on the line $AD$ (see Fig. 1). We prove that the triangles $AOY$ and $AXD$ are similar. Since the line $AN$ bisects the segment $OY$, the conclusion follows.\n\nIt is well known that the internal bisector $AD$ of the angle $BAC$ is also the internal bisector of the angle $OAA'$. Next, the corresponding sides of the triangles $OND$ and $ADX$ are parallel, so these triangles are similar.\n\n$\\dfrac{AY}{AO} = \\dfrac{AD}{AX}$. Along with the equality of the angles $OAY$ and $DAX$, this proves the required similarity of the triangles $AOY$ and $AXD$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20968,
"subject": "Mathematics (Olympiad)",
"question": "For a convex quadrilateral $ABCD$, an interior point $P$ is called the \"balance point\" of $ABCD$ if it simultaneously satisfies the following two conditions:\n\n(a) Point $P$ does not lie on the diagonals $AC$ and $BD$.\n\n(b) Extensions of $AP$, $BP$, $CP$, $DP$ intersect the boundaries of quadrilateral $ABCD$ at points $A'$, $B'$, $C'$, $D'$ respectively, such that\n\n$$\nAP \\cdot PA' = BP \\cdot PB' = CP \\cdot PC' = DP \\cdot PD'\n$$\n\nFor all convex quadrilaterals $ABCD$, determine the maximum possible number of \"balance points\" for $ABCD$.",
"options": [],
"answer": "See solution",
"solution": "Let $AC$ and $BC$ be the diagonals of quadrilateral $ABCD$ intersecting at point $O$. Assume that $P$ is an equilibrium point inside $\\triangle AOB$, as shown in the figure below.\n\n\n\nIt is easy to see from the given conditions that\n\n$$\n\\begin{aligned}\n\\angle PC'D' &= \\pi - \\angle PC'A = \\pi - \\angle PA'C = \\angle PA'B \\\\\n&= \\angle PB'A = \\pi - \\angle PB'D = \\pi - \\angle PD'B = \\angle PD'C = \\angle PCD.\n\\end{aligned}\n$$\n\nTherefore, $AB$ is parallel to $CD$. This implies that for a quadrilateral with non-parallel opposite sides, it is not possible to have an equilibrium point inside.\n\n1. **Rectangle case:**\n\nIf $ABCD$ is a rectangle, and there exists an equilibrium point $P$ inside either $\\triangle AOB$ or $\\triangle COD$, then $P$ must lie on the perpendicular bisector of $AB$ and $\\angle APD = \\frac{\\pi}{2}$. This requires $AD > AB$, and in this case, there can be at most 2 equilibrium points inside $\\triangle AOB$ and $\\triangle COD$ combined. Therefore, for a rectangle $ABCD$, there can be at most 2 equilibrium points inside.\n\n2. **Parallelogram (not rectangle) case:**\n\nAssume $\\angle A > \\frac{\\pi}{2}$. There can be at most 1 equilibrium point inside $\\triangle AOB$, and when there is an equilibrium point, we must have $AB < AD$. Thus, a non-rectangular parallelogram can have at most 2 equilibrium points.\n\n\n\nFurther geometric analysis shows that certain conditions must be met for equilibrium points to exist, involving the positions of circumcenters and perpendicular bisectors. Ultimately, a non-rectangular parallelogram can have at most 2 equilibrium points.\n\n3. **Trapezoid (not parallelogram) case:**\n\nIf $ABCD$ is a trapezoid, $AB \\parallel CD$, and $P$ lies on the perpendicular bisector of $CD$. Geometric arguments show that $\\triangle AOB$ and $\\triangle COD$ can have at most two equilibrium points each, but in total, the number does not exceed 3. For an isosceles trapezoid, it is possible to have 3 equilibrium points, as shown in the figure below.\n\n\n\n**Conclusion:**\n\nFor any convex quadrilateral, the maximum number of balance points is $\\boxed{3}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20969,
"subject": "Mathematics (Olympiad)",
"question": "Given a finite set of boys and girls, a *covering set of boys* is a set of boys such that every girl knows at least one boy in that set; and a *covering set of girls* is a set of girls such that every boy knows at least one girl in that set. Prove that the number of covering sets of boys and the number of covering sets of girls have the same parity. (Acquaintance is assumed to be mutual.)",
"options": [],
"answer": "See solution",
"solution": "Let $B$ denote the set of boys, let $G$ denote the set of girls and induct on $|B| + |G|$. The assertion is vacuously true if either set is empty.\n\nNext, fix a boy $b$, let $B' = B \\setminus \\{b\\}$, and let $G'$ be the set of all girls who do not know $b$. Notice that:\n\n1. A covering set of boys in $B' \\cup G$ is still one in $B \\cup G$; and\n2. A covering set of boys in $B \\cup G$ which is no longer one in $B' \\cup G$ is precisely the union of a covering set of boys in $B' \\cup G'$ and $\\{b\\}$,\n\nso the number of covering sets of boys in $B \\cup G$ is the sum of those in $B' \\cup G$ and $B' \\cup G'$.\n\nOn the other hand,\n\n1'. A covering set of girls in $B \\cup G$ is still one in $B' \\cup G$; and\n2'. A covering set of girls in $B' \\cup G$ which is no longer one in $B \\cup G$ is precisely a covering set of girls in $B' \\cup G'$, \n\nso the number of covering sets of girls in $B \\cup G$ is the difference of those in $B' \\cup G$ and $B' \\cup G'$.\n\nSince the assertion is true for both $B' \\cup G$ and $B' \\cup G'$ by the induction hypothesis, the conclusion follows.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 20970,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c$ be positive real numbers. Prove that\n$$\n\\frac{a^4+1}{b^3+b^2+b} + \\frac{b^4+1}{c^3+c^2+c} + \\frac{c^4+1}{a^3+a^2+a} \\ge 2.\n$$",
"options": [],
"answer": "See solution",
"solution": "For any positive real $x$, we have $(x-1)^2(3x^2+4x+3) \\ge 0$. Therefore,\n$$\n3(x^4 + 1) \\ge 2(x^3 + x^2 + x)\n$$\nwhich implies\n$$\n\\frac{x^4+1}{x^3+x^2+x} \\ge \\frac{2}{3}.\n$$\nApplying this to $a, b, c$ and using the AM-GM inequality, we get\n$$\n\\frac{a^4+1}{b^3+b^2+b} + \\frac{b^4+1}{c^3+c^2+c} + \\frac{c^4+1}{a^3+a^2+a} \\ge 3 \\cdot \\sqrt[3]{\\frac{a^4+1}{b^3+b^2+b} \\cdot \\frac{b^4+1}{c^3+c^2+c} \\cdot \\frac{c^4+1}{a^3+a^2+a}} \\ge 3 \\cdot \\sqrt[3]{\\frac{8}{27}} = 2.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20971,
"subject": "Mathematics (Olympiad)",
"question": "Од точка $M$ кон кружница $k$ се повлечени две тангенти со допирни точки $G$ и $H$. Ако $O$ е центарот на $k$ и $K$ е ортоцентарот на триаголникот $MGH$, докажи дека $\\angle GMH = \\angle OGK$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Да забележиме дека $K$ мора да лежи на $OM$. Од $HK \\perp GM$ и $OG \\perp GM$, следува $HK \\parallel OG$. Аналогно, $OH \\parallel GK$. Од $\\overline{OG} = \\overline{OH}$, следува дека $OHKG$ е ромб. Да забележиме дека $O$, $H$, $M$ и $G$ лежат на кружница со дијаметар $OM$. Оттука $\\angle OGH = \\angle OMH$. Сега тврдењето на задачата следува од $\\angle OGK = 2\\angle OGH$ и $\\angle GMH = 2\\angle OMH$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20972,
"subject": "Mathematics (Olympiad)",
"question": "Let $P_0, P_1, P_2, \\dots, P_n$ be $n+1$ points on a plane, and the minimum distance between any two of them is $d$ ($d > 0$). Prove\n\n$$\n|P_0P_1| \\cdot |P_0P_2| \\cdots |P_0P_n| > \\left( \\frac{d}{3} \\right)^n \\sqrt{(n+1)!}\n$$",
"options": [],
"answer": "See solution",
"solution": "**Solution 1.** We may assume that $|P_0P_1| \\le |P_0P_2| \\le \\dots \\le |P_0P_n|$.\n\nFirst, we will prove that $|P_0P_k| > \\frac{d}{3}\\sqrt{k+1}$ for any positive integer $k$.\n\nObviously, $|P_0P_k| \\ge d \\ge \\frac{d}{3}\\sqrt{k+1}$ for $k = 1, 2, \\dots, 8$, and the second equality holds only when $k = 8$. Then we only need to prove that $|P_0P_k| \\ge d \\ge \\frac{d}{3}\\sqrt{k+1}$ for $k \\ge 9$.\n\nTake each $P_i$ ($i = 0, 1, 2, \\dots, k$) as the center to draw a circle with radius $\\frac{d}{2}$. Then these circles are either externally tangent to or apart from each other. Take $P_0$ as the center to draw a circle with radius $|P_0P_k| + \\frac{d}{2}$. Then the previous $k+1$ smaller circles are all located in this larger one.\n\nThen $\\pi(|P_0P_k| + \\frac{d}{2})^2 > (k+1)\\pi\\left(\\frac{d}{2}\\right)^2$, from which we have $|P_0P_k| > \\frac{d}{2}(\\sqrt{k+1} - 1)$.\n\nIt is easy to check that $\\frac{\\sqrt{k+1}-1}{2} > \\frac{\\sqrt{k+1}}{3}$ for $k \\ge 9$. Then $|P_0P_k| > \\frac{d}{3}\\sqrt{k+1}$ for $k \\ge 9$.\n\nOverall, we have $|P_0P_k| > \\frac{d}{3}\\sqrt{k+1}$ for all $k \\ge 1$.\n\nTherefore,\n\n$$\n|P_0P_1| \\cdot |P_0P_2| \\cdots |P_0P_n| > \\left(\\frac{d}{3}\\right)^n \\sqrt{(n+1)!}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20973,
"subject": "Mathematics (Olympiad)",
"question": "Let scalene triangle $ABC$ have orthocentre $H$ and circumcircle $\\Gamma$. $AH$ meets $\\Gamma$ at $D$ distinct from $A$. $BH$ and $CH$ meet $CA$ and $AB$ at $E$ and $F$ respectively, and $EF$ meets $BC$ at $P$. The tangents to $\\Gamma$ at $B$ and $C$ meet at $T$. Show that $AP$ and $DT$ are concurrent on the circumcircle of $AFE$.",
"options": [],
"answer": "See solution",
"solution": "Let $Y$ be the point on $\\Gamma$ diametrically opposite $A$ and $X$ the second point of intersection of $YH$ with $\\Gamma$. We show that $X$ lies on lines $AP$, $DT$ and the circle $AFE$, completing the proof.\n\nTo see that $X$ lies on the circle $AFE$, note that as $AY$ is a diameter of $\\Gamma$, $\\angle HXA = \\frac{\\pi}{2}$. We know that $AEHF$ is cyclic on diameter $AH$, so $X$ lies on that circle too.\n\nTo see that $X$ lies on line $AP$, note that $CBFE$ is a cyclic quadrilateral. Thus $AX$ is the radical axis of $\\Gamma$ and the circle $AEHFX$, $EF$ is that of circles $AEHFX$ and $CBFE$, and $BC$ is the radical axis of circle $CBFE$ and $\\Gamma$. Hence by the radical axis theorem they concur at the point $P$, so that $AP$ passes through the point $X$.\n\n\n\nTo prove that $X$ lies on the line $DT$, we first show that $XY$ is a median of triangle $XBC$. To see this, note that, as $AY$ is a diameter of $\\Gamma$, both $BH$ and $YC$ are perpendicular to $CA$, and hence are parallel. Similarly, $CH$ is parallel to $YB$ and hence $BHCY$ is a parallelogram. Hence its diagonals bisect one another, so that the line $XHY$ passes through the midpoint of $BC$.\n\nNow since $DY$ and $BC$ are parallel (being both perpendicular to $AD$), it follows that $D$ is the reflection of $Y$ in the perpendicular bisector of $BC$, and hence that $XD$ is a symmedian of triangle $XBC$. By standard properties of the symmedian, it passes through $T$, so that $X$ lies on $DT$ as desired. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20974,
"subject": "Mathematics (Olympiad)",
"question": "Let $D$ be the midpoint of side $BC$, $E$ the midpoint of side $CA$, and $T$ the center of mass of triangle $ABC$. Let the lines $AT$, $BT$, and $CT$ also intersect the circumcircle of triangle $ABC$ at points $P$, $Q$, and $R$, respectively. Suppose $\\angle ACB = \\angle RQP$. Prove that $DCET$ is a cyclic quadrilateral.",
"options": [],
"answer": "See solution",
"solution": "Let $\\angle EBD = \\alpha$. Then $\\angle DEB = \\alpha$, so $\\angle EDA = 2\\alpha$ and $\\angle DAE = 2\\alpha$. This implies $\\angle DEC = 4\\alpha$, or $\\angle BEC = 3\\alpha$. At the same time, we have $\\angle ACD = \\angle DAC = 2\\alpha$, so $\\angle BDC = 4\\alpha$ and $\\angle CBD = 4\\alpha$, or $\\angle CBE = 3\\alpha$. It follows that triangle $EBC$ is isosceles with the apex at $C$, so $|CE| = |CD|$ and therefore $\\angle CDE = \\angle DEC = 4\\alpha$. So, $180^\\circ = \\angle BDC + \\angle CDE + \\angle EDA = 4\\alpha + 4\\alpha + 2\\alpha = 10\\alpha$, or $\\alpha = 18^\\circ$. From here we conclude that $\\angle BAC = 2\\alpha = 36^\\circ$, $\\angle CBA = 4\\alpha = 72^\\circ$, and $\\angle ACB = 180^\\circ - \\angle BAC - \\angle CBA = 72^\\circ$.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20975,
"subject": "Mathematics (Olympiad)",
"question": "Given four fixed points $A(-3, 0)$, $B(1, -1)$, $C(0, 3)$, $D(-1, 3)$ and a variable point $P$ in a plane rectangular coordinate system, find the minimum value of $|PA| + |PB| + |PC| + |PD|$.",
"options": [],
"answer": "See solution",
"solution": "Assume that $AC$ and $BD$ meet at point $F$.\n\n$$\n|PA| + |PC| \\geq |AC| = |FA| + |FC|\n$$\n\nand\n\n$$\n|PB| + |PD| \\geq |BD| = |FB| + |FD|.\n$$\n\nWhen $P$ coincides with $F$, $|PA| + |PB| + |PC| + |PD|$ reaches its minimum.\n\nThat is, $|AC| + |BD| = 3\\sqrt{2} + 2\\sqrt{5}$. So $3\\sqrt{2} + 2\\sqrt{5}$ is the required answer.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20976,
"subject": "Mathematics (Olympiad)",
"question": "Prove that there is no function from positive real numbers to itself, $f : (0, +\\infty) \\to (0, +\\infty)$ such that:\n\n$$\nf(f(x) + y) = f(x) + 3x + y f(y) \\quad \\text{for every } x, y \\in (0, +\\infty)\n$$",
"options": [],
"answer": "See solution",
"solution": "First, we prove that $f(x) \\geq x$ for all $x > 0$.\n\nSuppose there exists $a > 0$ with $f(a) < a$. Then, for $x = a$ and $y = a - f(a) > 0$, we get:\n\n$$\nf(f(a) + (a - f(a))) = f(a) + 3a + (a - f(a)) f(a - f(a))\n$$\n\nBut $f(f(a) + (a - f(a))) = f(a) + 3a + (a - f(a)) f(a - f(a)) > 0$, which is absurd if $3a + (a - f(a)) f(a - f(a)) = 0$. Thus, $f(x) \\geq x$ for all $x > 0$. \n\nNow, using this, $f(x) + 3x + y f(x) = f(f(x) + y) \\geq f(x) + y$, so:\n\n$$\n3x + y f(y) \\geq y \\quad \\text{for all } x, y > 0\n$$\n\nSuppose $y f(y) < y$ for some $y > 0$. Let $-y f(y) = b > 0$. For $x = \\frac{b}{4}$, we get $\\frac{3b}{4} - b \\geq 0$, so $b \\leq 0$, a contradiction. Thus, $y f(y) \\geq y$ for all $y > 0$, so $f(y) \\geq 1$ for all $y > 0$.\n\nSubstituting $y$ with $f(y)$ in the original equation:\n\n$$\nf(f(x) + f(y)) = f(x) + 3x + f(y) f(f(y))\n$$\n\nSwitching $x$ and $y$ gives:\n\n$$\nf(f(y) + f(x)) = f(y) + 3y + f(x) f(f(x))\n$$\n\nSo:\n\n$$\nf(x) f(f(x)) - f(x) - 3x = f(y) f(f(y)) - f(y) - 3y\n$$\n\nThus, $f(x) f(f(x)) - f(x) - 3x$ is constant. Let $c$ be this constant:\n\n$$\nf(x) f(f(x)) = f(x) + 3x + c \\quad \\text{for all } x > 0\n$$\n\nSo $f(x) (f(f(x)) - 1) = 3x + c$. Since $f(f(x)) > 1$, $3x + c \\geq 0$ for all $x > 0$. If $c < 0$, for $x = -\\frac{c}{4} > 0$, $c > 0$, a contradiction. So $c \\geq 0$.\n\nRewrite:\n\n$$\nf(f(x)) = 1 + \\frac{3x}{f(x)} + \\frac{c}{f(x)}\n$$\n\nSince $c \\geq 0$, and from above $f(x) \\geq x$ and $f(y) \\geq 1$, we have $f(f(x)) \\leq 4 + c$. But $f(f(x)) \\geq f(x) \\geq x$ for all $x > 0$, so:\n\n$$\n4 + c \\geq f(f(x)) \\geq x \\quad \\text{for all } x > 0\n$$\n\nTaking $x = 5 + c$ gives a contradiction. Therefore, no such function exists. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20977,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, a_3, \\dots, a_{10}$ and $b_1, b_2, \\dots, b_{10}$ be real numbers such that the roots of these 10 polynomials\n\n$$\nx^2 + a_1x + b_1,\\quad x^2 + a_2x + b_2,\\quad \\dots,\\quad x^2 + a_{10}x + b_{10}\n$$\n\nare all the integers $\\pm 1, \\pm 2, \\dots, \\pm 10$ (in some order).\n\n**a)** What is the maximum number of odd values among $a_1, b_1, \\dots, a_{10}, b_{10}$?\n\n**b)** Find the minimum and maximum values of the sum $b_1 + b_2 + \\dots + b_{10}$.",
"options": [],
"answer": "See solution",
"solution": "**a)**\n\nLet $x_i, y_i$ be the roots of the $i$-th polynomial. By Vieta's theorem, $a_i = -(x_i + y_i)$ and $b_i = x_i y_i$. Thus, $a_i b_i = -x_i y_i (x_i + y_i)$, which is always even. This implies that at most one of $a_i$ or $b_i$ is odd for each $i$, so there are at most 10 odd values among all coefficients.\n\nEquality occurs when the pairs $(1, 2), (3, 4), \\dots, (9, 10), (-1, -2), \\dots, (-9, -10)$ are the roots of the polynomials.\n\n**b)**\n\nBy Vieta's theorem, we need to find the minimum and maximum of\n\n$$\nT = x_1 y_1 + x_2 y_2 + \\dots + x_{10} y_{10}.\n$$\n\nNote that for all $x, y \\in \\mathbb{R}$, $xy \\geq -\\frac{x^2 + y^2}{2}$, so\n\n$$\nT \\geq -\\frac{1}{2}(x_1^2 + y_1^2 + \\dots + x_{10}^2 + y_{10}^2) = -(1^2 + 2^2 + \\dots + 10^2) = -385.\n$$\n\nFor $x, y \\in \\mathbb{Z}$ and $x \\neq y$, $(x - y)^2 \\geq 1$ so $xy \\leq \\frac{x^2 + y^2 - 1}{2}$, thus\n\n$$\nT \\leq \\frac{1}{2}(x_1^2 + y_1^2 + \\dots + x_{10}^2 + y_{10}^2) - 5 = 380.\n$$\n\nTherefore:\n\n- $\\max T = 380$, attained when $(1, 2), (3, 4), \\dots, (9, 10), (-1, -2), \\dots, (-9, -10)$ are the roots.\n- $\\min T = -385$, attained when $(1, -1), (2, -2), \\dots, (10, -10)$ are the roots.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 20978,
"subject": "Mathematics (Olympiad)",
"question": "In convex quadrilateral $ABCD$, $AB \\perp AD$ and $AD = DC$. Let point $E$ lie inside segment $BC$, and point $F$ lie on the extension of $DE$ beyond $E$, such that $\\angle ABF = \\angle DEC > 90^\\circ$. Let $O$ be the circumcenter of triangle $CDE$. Let $P$ be a point on the extension of $FO$ beyond $O$ such that $FP = FB$. Let segment $BP$ intersect segment $AC$ at point $Q$.\n\nProve that $\\angle AQB = \\angle DPF$.\n\n",
"options": [],
"answer": "See solution",
"solution": "As shown in the figure, draw a line through $D$ parallel to $AB$, intersecting $BF$ at $K$. Then:\n\n$$\n\\angle DKB = 180^\\circ - \\angle ABF = 180^\\circ - \\angle DEC = \\angle DEB,\n$$\n\nso points $B, K, E, D$ are concyclic.\n\nSince $AB \\perp AD$ and $DK \\parallel AB$, we have $AD = BK \\cdot \\sin \\angle ABK$. Let $\\omega$ be the circumcircle of triangle $CDE$ with radius $r$, then $DC = 2r \\cdot \\sin \\angle DEC$. From $AD = DC$ and $\\angle ABK = \\angle DEC$, we get $BK = 2r$.\n\nLet $FO$ intersect circle $\\omega$ at two points $U$ and $V$, with $F, U, O, V$ in order. By the power of a point theorem:\n\n$$\nFU \\cdot FV = FE \\cdot FD = FK \\cdot FB.\n$$\n\nSince $UV = 2r = BK$ and $FB > FK$, we conclude $FV = FB$. Given $FP = FB$ and both $P, V$ lie on the extension of $FO$, it follows that $P$ coincides with $V$.\n\nFrom $AD = DC$, the sum of directed angles from $\\overrightarrow{AC}$ to $\\overrightarrow{AD}$ and $\\overrightarrow{DC}$ is $0^\\circ$. From $FP = FB$, the sum of angles from $\\overrightarrow{PB}$ to $\\overrightarrow{PF}$ and $\\overrightarrow{FB}$ is $0^\\circ$ (mod $360^\\circ$). This shows that as directed angles (mod $180^\\circ$):\n\n$$\n\\angle AQB = \\angle (AC, PB) = \\frac{1}{2}\\angle (AD, FB) + \\frac{1}{2}\\angle (DC, PF).\n$$\n\nSince $\\angle (AD, FB) = 90^\\circ - \\angle FBA = 90^\\circ - \\angle CED$, and connecting $PE$ gives:\n\n$$\n\\angle (DC, PF) = \\angle CDP + \\angle DPF = \\angle CEP + \\angle DPF,\n$$\n\nwe have:\n\n$$\n\\begin{aligned}\n\\angle AQB &= \\frac{1}{2}(90^\\circ - \\angle CED + \\angle CEP + \\angle DPF) \\\\\n&= \\frac{1}{2}(90^\\circ - \\angle PED + \\angle DPF) \\\\\n&= \\frac{1}{2}(\\angle DPO + \\angle DPF) = \\angle DPF.\n\\end{aligned}\n$$\n\nThis completes the proof. $\\boxed{}$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20979,
"subject": "Mathematics (Olympiad)",
"question": "Circles $c_1$ and $c_2$ with centers $O_1$ and $O_2$, respectively, intersect at points $P$ and $Q$ and touch circle $c$ internally at points $A_1$ and $A_2$, respectively. Line $PQ$ intersects circle $c$ at points $B$ and $D$. Lines $A_1B$ and $A_1D$ intersect circle $c_1$ a second time at points $E_1$ and $F_1$, respectively, and lines $A_2B$ and $A_2D$ intersect circle $c_2$ a second time at points $E_2$ and $F_2$, respectively. Prove that $E_1, E_2, F_1, F_2$ lie on a circle whose center coincides with the midpoint of the line segment $O_1O_2$.",
"options": [],
"answer": "See solution",
"solution": "Let the radii of $c_1$, $c_2$, and $c$ be $r_1$, $r_2$, and $r$, respectively. Homothety of ratio $\\frac{r}{r_1}$ with center $A_1$ takes circle $c_1$ to circle $c$ and points $E_1$, $F_1$ to points $B$, $D$, respectively. Thus, it takes line $E_1F_1$ to line $BD$. Analogously, homothety of ratio $\\frac{r}{r_2}$ with center $A_2$ takes line $E_2F_2$ to line $BD$. Consequently, lines $E_1F_1$ and $E_2F_2$ are parallel to line $BD$.\n\nFurthermore, note that $|BE_1| \\cdot |BA_1| = |BP| \\cdot |BQ|$ and $|BE_2| \\cdot |BA_2| = |BP| \\cdot |BQ|$, implying $|BE_1| \\cdot |BA_1| = |BE_2| \\cdot |BA_2|$. Thus, triangles $BE_1E_2$ and $BA_2A_1$ are similar and\n\n$$\n\\angle BE_1E_2 = \\angle BA_2A_1 = \\angle BDA_1 = \\angle E_1F_1A_1 = \\frac{1}{2} \\angle E_1O_1A_1 = 90^\\circ - \\angle O_1E_1A_1, \\quad (3)\n$$\n\nwhence\n\n$$\n\\angle E_2E_1O_1 = 180^\\circ - \\angle BE_1E_2 - O_1E_1A_1 = 90^\\circ. \\quad (4)\n$$\n\nAnalogously, $\\angle E_1E_2O_2 = 90^\\circ$. Hence, the quadrilateral $E_1E_2O_2O_1$ is a right-angled trapezoid (or rectangle in the case $r_1 = r_2$), and the midpoint of the line segment $O_1O_2$ lies on the perpendicular bisector of the line segment $E_1E_2$, thus being equidistant from $E_1$ and $E_2$. Analogously, the midpoint of the line segment $O_1O_2$ is also equidistant from $F_1$ and $F_2$.\n\nAs line $O_1O_2$ is perpendicular to $BD$, line $O_1O_2$ is also perpendicular to $E_1F_1$. Thus, the line segment $O_1O_2$ entirely lies on the perpendicular bisector of $E_1F_1$. This means that the midpoint of line segment $O_1O_2$ is equidistant from $E_1$ and $F_1$.\n\nAltogether, we have shown that these four points lie on a circle with its center at the midpoint of the line segment $O_1O_2$.\n\n\n\n\n**Remark:** The chains of equations (3) and (4) hold as given in the situation depicted in Fig. 6, where $F_1$ and $O_1$ lie on the same side from line $A_1E_1$. There are other situations where $F_1$ and $O_1$ lie on different sides from $A_1E_1$ or $O_1$ lies on the line $A_1E_1$ or circle $c$ lies inside circles $c_1$ and $c_2$ (see Fig. 7). Despite the equations having a slightly different form, the final result $\\angle E_2E_1O_1 = 90^\\circ$ still holds.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20980,
"subject": "Mathematics (Olympiad)",
"question": "There are 2024 cards of the same size, face-down on a table, on which the integers $1, 2, 3, \\ldots, 2024$ are written. We say that a card is a *winner* if it has a number divisible by $13$ or by $100$. What is the minimum number of cards we need to turn face up to make sure that we obtain at least one winner?",
"options": [],
"answer": "See solution",
"solution": "To make sure that we obtain at least one winner, we must pick up one more card than the number of non-winners.\n\nThere are $155$ multiples of $13$ not larger than $2024$: $13 \\cdot 1, 13 \\cdot 2, \\ldots, 13 \\cdot 155$.\n\nThere are $20$ multiples of $100$ not larger than $2024$: $100 \\cdot 1, 100 \\cdot 2, \\ldots, 100 \\cdot 20$.\n\nThere is only one common multiple of $13$ and $100$ not larger than $2024$, namely $1300$.\n\nThere are $155 + 20 - 1 = 174$ numbers not larger than $2024$ that are multiples of $13$ or multiples of $100$. This leaves $2024 - 174 = 1850$ non-winners.\n\nWe are sure that we have obtained a winner as soon as we pick $1850 + 1 = 1851$ cards.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 20981,
"subject": "Mathematics (Olympiad)",
"question": "There are 234 visitors in a cinema auditorium. The visitors are sitting in $n$ rows, where $n \\ge 4$, so that each visitor in the $i$-th row has exactly $j$ friends in the $j$-th row, for any $i, j \\in \\{1, 2, \\ldots, n\\}$, $i \\neq j$. Find all the possible values of $n$. (Friendship is supposed to be a symmetric relation.)",
"options": [],
"answer": "See solution",
"solution": "For any $k \\in \\{1, 2, \\ldots, n\\}$, denote by $p_k$ the number of visitors in the $k$-th row. The stated condition for given $i$ and $j$ implies that the number of friendly pairs $(A, B)$, where $A$ is from the $i$-th row and $B$ is from the $j$-th row, is equal to $j p_i$. Interchanging $i$ and $j$, the same number is $i p_j$. Thus, $j p_i = i p_j$, or $\\frac{p_i}{p_j} = \\frac{i}{j}$, so all $p_k$ are proportional:\n\n$$\np_1 : p_2 : \\ldots : p_n = 1 : 2 : \\ldots : n.\n$$\n\nAssume $p_k = k d$ for some positive integer $d$. The total number of visitors is:\n\n$$\nd + 2d + \\ldots + n d = d (1 + 2 + \\ldots + n) = d \\cdot \\frac{n(n+1)}{2} = 234.\n$$\n\nSo,\n\n$$\nd \\cdot \\frac{n(n+1)}{2} = 234 \\implies d n(n+1) = 468.\n$$\n\nWe seek integer $n \\ge 4$ and $d$ such that $n(n+1)$ divides $468$. Checking possible $n$ values, only $n = 12$ (since $12 \\times 13 = 156$ and $468/156 = 3$) works. Thus, the unique solution is $n = 12$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20982,
"subject": "Mathematics (Olympiad)",
"question": "Determine all integers which can be expressed in the form\n\n$$\nx_1x_2 + x_2x_3 + \\dots + x_{k-1}x_k,\n$$\n\nwhere $x_1, x_2, \\dots, x_k$ are positive integers whose sum equals $2019$.",
"options": [],
"answer": "See solution",
"solution": "First, easy induction shows that $\\sum_{i=1}^{k-1} x_i x_{i+1} \\ge (\\sum_{i=1}^k x_i) - 1$.\n\nBase case $k=2$: $x_1x_2 - (x_1+x_2-1) = (x_1-1)(x_2-1) \\ge 0$. Assuming the inequality holds for a certain $k$ and adding the obvious $x_kx_{k+1} \\ge x_{k+1}$, we get the claim for $k+1$.\n\nGiven the conditions of the problem, this shows that (for a given $n$) the quadratic form in question takes values $\\ge n-1$; and the minimum $n-1$ is attained e.g. for $k=n$ and all $x_i=1$.\n\nNow to the upper bound. The fine point is that $k$ is variable. So, let $x_1, \\dots, x_k$ be a $k$-string of positive integers with $\\sum x_i = n$, and with $k \\ge 4$; and let $V$ be the generated value $V = \\sum_{i=1}^{k-1} x_i x_{i+1}$. If $x_2 \\le x_3$, we merge $x_1$ with $x_2$; and if $x_2 > x_3$, we merge $x_3$ with $x_4$, thus creating the following $(k-1)$-string (with entries summing to $n$):\n\n$(x_1+x_2), x_3, \\dots, x_k$, resp. $x_1, x_2, (x_3+x_4), x_5, \\dots, x_k$ (if $k=4$, $x_5=0$).\n\nIf $\\tilde{V}$ is the new value of the quantity under consideration then, in the first case $\\tilde{V} - V = x_1x_3 - x_1x_2 \\ge 0$; and in the second case\n\n$$\n\\tilde{V} - V = x_2x_4 - x_3x_4 + x_3x_5 \\ge 0\n$$\n\nAfter several steps $k$ comes down to $3$ and we arrive at a $3$-string $z_1, z_2, z_3$ (with $z_1 + z_2 + z_3 = n$) producing the value\n\n$$\nW = z_1z_2 + z_2z_3 = z_2(z_1 + z_3) \\ge V\n$$\n\nThe product of two integers with a given sum $n$ has a maximum\n\n$$\nM_n = \\lfloor n/2 \\rfloor \\cdot \\lceil n/2 \\rceil = \\lfloor (n^2 + 1)/4 \\rfloor\n$$\n\nTo show that all integer values between $n-1$ and $M_n$ are attained, we focus on strings $x_1, \\dots, x_k$ ending in $x_k=1$. We claim that these alone are enough to generate all those values. Induction again. Base $n=2$: obvious. Fix $n>2$ and assume that positive-integer strings with sum $n-1$, ending in a $1$, yield all values from $n-2$ to $M_{n-1}$. At the end of each of these strings (next to the terminal $1$) we attach another $1$; the value of the quadratic form grows by $1$. So we already have strings with sum $n$ and with last entry $1$, producing all values from $n-1$ to $M_{n-1}+1$.\n\nNow, if $n$ is even, $n=2m$, the triples (3-strings) $m-1, m, 1$ and $m, m-1, 1$ produce the values $M_n = m^2$ and $M_n-1 = m^2-1$; and the quadruples (4-strings) $j, m-2, m+1-j, 1$ with $j=1, \\dots, m$ give values from $m^2-2$ down to $m^2-m-1$ (which is below $M_{n-1}$). If $n$ is odd, $n=2m+1$, the value $M_n = m^2+m$ comes from the triples $m, m, 1$; and now the quadruples $j, m-1, m+1-j, 1$ with $j=1, \\dots, m$ yield the values from $m^2+m-1$ down to $m^2$ (below $M_{n-1}$). In each case, as $j$ ranges from $1$ to $m$, the generated values of the quadratic form sweep (with slight excess) the entire missing interval. Induction is completed and the claim results.\n\nThe answer follows: the values of $\\sum_{i=1}^{k-1} x_i x_{i+1}$ are all integers from $n-1$ to $M_n$ (inclusive).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20983,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ and $k$ be positive integers with $k \\ge n$ and $k - n$ an even number. Let $2n$ lamps labeled $1, 2, \\dots, 2n$ be given, each of which can be either on or off. Initially all lamps are off. We 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 the lamps $n+1$ through $2n$ are all off.\n\nLet $M$ be the number of such sequences consisting of $k$ steps, resulting in the same state, but where none of the lamps $n+1$ through $2n$ is ever switched on.\n\nDetermine the ratio $N/M$.",
"options": [],
"answer": "See solution",
"solution": "Suppose lamp $i$ was switched on or off $a_i$ times. In the first situation, $a_i$ is odd for $1 \\le i \\le n$ and $a_i$ is even for $n+1 \\le i \\le 2n$. The total number of sequences where lamp $i$ is switched $a_i$ times is $\\displaystyle \\binom{k}{a_1, a_2, \\dots, a_{2n}}$. Thus,\n\n$$\nN = \\sum \\binom{k}{a_1, a_2, \\dots, a_{2n}}\n$$\n\nwhere the sum is over all $a_i$ such that $a_1 + \\dots + a_{2n} = k$, $a_i$ is odd for $1 \\le i \\le n$, and $a_i$ is even for $n+1 \\le i \\le 2n$. Note that\n\n$$\n\\binom{k}{a_1, a_2, \\dots, a_{2n}} = \\frac{k!}{a_1! a_2! \\dots a_{2n}!}\n$$\n\nso $N$ is the coefficient of $x^k$ in\n\n$$\nk! \\left( x + \\frac{x^3}{3!} + \\frac{x^5}{5!} + \\cdots \\right)^n \\left( 1 + \\frac{x^2}{2!} + \\frac{x^4}{4!} + \\cdots \\right)^n.\n$$\n\nThis can be rewritten as\n\n$$\nk! \\left( \\frac{e^x - e^{-x}}{2} \\right)^n \\left( \\frac{e^x + e^{-x}}{2} \\right)^n = k! \\left( \\frac{e^{2x} - e^{-2x}}{4} \\right)^n.\n$$\n\nSimilarly, for $M$, the sum is over all $a_i$ such that $a_1 + \\dots + a_{2n} = k$, $a_i$ is odd for $1 \\le i \\le n$, and $a_i = 0$ for $n+1 \\le i \\le 2n$. Then $M$ is the coefficient of $x^k$ in\n\n$$\nk! \\left( x + \\frac{x^3}{3!} + \\frac{x^5}{5!} + \\cdots \\right)^n = k! \\left( \\frac{e^x - e^{-x}}{2} \\right)^n.\n$$\n\nLet $f(x) = \\frac{e^x - e^{-x}}{2x}$. We wish to compare the coefficients of $x^k$ in $k! x^n f(x)^n$ and $k! x^n f(2x)^n$. It is now evident that the ratio is $2^{k-n}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20984,
"subject": "Mathematics (Olympiad)",
"question": "The convex quadrilateral $ABCD$ has $\\angle BCD = \\angle ADC \\ge 90^\\circ$. The bisectors of the angles $\\angle BAD$ and $\\angle ABC$ meet at a point $M$, placed on the line $CD$. Prove that $M$ is the midpoint of the segment $[CD]$.",
"options": [],
"answer": "See solution",
"solution": "Case I: $AD$ and $BC$ have a common point $E$. Then $M$ is the incenter of the triangle $ABE$, hence $(EM)$ is the bisector of the angle $\\angle AEB$.\n\nSince $\\angle ECD = \\angle EDC$, the triangle $EDC$ is isosceles with base $[DC]$. Therefore $[EM]$ is a median in triangle $EDC$, so $M$ is the midpoint of the segment $CD$.\n\n\n\nCase II: $AD \\parallel BC$. Then $\\angle CAB = \\angle ABD = 90^\\circ$, hence $\\angle MAB + \\angle MBA = 90^\\circ$, that is, triangle $MAB$ has a right angle at $M$.\n\nDenote $N$ the midpoint of the segment $[AB]$. Then triangle $NAM$ is isosceles with base $[AM]$, so $\\angle NMA = \\angle NAM = \\angle MAD$, hence $MN \\parallel AD$. It follows that $MN$ is the central median of the trapezoid $ABCD$, whence the conclusion.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20985,
"subject": "Mathematics (Olympiad)",
"question": "For a positive integer $m \\ge 2$, define\n$$\nA_m := \\{m+1, 3m+2, 5m+3, 7m+4, 9m+5, \\dots \\}.\n$$\n\n1. Prove that, for any given $m \\ge 2$, there exists a positive integer $a$, $1 \\le a < m$, such that either $2^a \\in A_m$ or $2^a + 1 \\in A_m$.\n\n2. Assume that, for some $m \\ge 2$, there exist positive integers $a$ and $b$ for which $2^a \\in A_m$ and $2^b + 1 \\in A_m$. Let $a_0$ and $b_0$ be the smallest such $a$ and $b$, respectively, and find a relation between $a_0$ and $b_0$.",
"options": [],
"answer": "See solution",
"solution": "(1) An arbitrary element of $A_m$ can be written as $m + 1 + k(2m + 1)$, where $k = 0, 1, 2, \\dots$. Let $\\epsilon = 0$ or $1$.\n\n$$\nm + 1 + k(2m + 1) = 2^a + \\epsilon \\iff 2^a \\equiv m + 1 - \\epsilon \\pmod{2m + 1}\n$$\n$$\n\\iff 2^{a+1} \\equiv 1 - 2\\epsilon \\pmod{2m + 1}\n$$\n$$\n\\iff 2^{a+1} \\equiv \\begin{cases} 1 & (\\text{mod } 2m + 1) \\\\ -1 & (\\text{mod } 2m + 1) \\end{cases} \\quad \\text{if } \\epsilon = 0, \\text{ if } \\epsilon = 1.\n$$\n\nLet $r$ be the smallest positive integer $t$ such that $2^t \\equiv 1 \\pmod{2m + 1}$, i.e., $r = \\operatorname{ord}_{2m + 1}(2)$. Note that $r > 2$ because $2m + 1 \\ge 5$.\n\n**(i) $r \\le m$:**\n\nSince $1 < r - 1 < m$, take $a := r - 1$. Then $1 < a < m$ and\n$$\n2^{a+1} \\equiv 1 \\pmod{2m + 1} \\implies (\\epsilon = 0) \\implies 2^a \\in A_m.\n$$\n\n**(ii) $r > m$:**\n\nSince $2m + 1$ is odd, $\\varphi(2m + 1)$ is even. Furthermore,\n$$\nr \\mid \\varphi(2m + 1),\\quad \\varphi(2m + 1) \\le 2m \\implies r = \\varphi(2m + 1).\n$$\nThat is, $2$ is a primitive root modulo $2m + 1$. Thus, $2^{r/2} \\equiv -1 \\pmod{2m + 1}$. This is because\n$$\n\\begin{align*}\n&\\exists s\\ (0 < s < r),\\ 2^s \\equiv -1 \\pmod{2m + 1} \\\\ \n&\\implies 2^{2s} \\equiv 1 \\pmod{2m + 1} \\\\ \n&\\implies r \\mid 2s,\\ 0 < 2s < 2r \\\\ \n&\\implies 2s = r.\n\\end{align*}\n$$\nTake $a := (r/2) - 1$. Then, since $4 \\le r = \\varphi(2m + 1) \\le 2m$, we have $1 \\le a < m$ and\n$$\n2^{a+1} \\equiv -1 \\pmod{2m + 1} \\implies (\\epsilon = 1) \\implies 2^a + 1 \\in A_m.\n$$\n\n(2) For a given $m$, as observed above, $2^{r-1} \\in A_m$ where $r = \\operatorname{ord}_{2m + 1}(2)$. The smallest $a$ for which $2^a \\in A_m$ is $a_0 = r - 1$.\n\n**(i) $r$ is even and $2^{r/2} \\equiv -1 \\pmod{2m + 1}$:**\n\nIn this case, $2^{(r/2)-1} + 1 \\in A_m$, and the smallest $b$ for which $2^b + 1 \\in A_m$ is $b_0 = (r/2) - 1$.\n\n**(ii) $r$ is even and $2^{r/2} \\not\\equiv -1 \\pmod{2m + 1}$:**\n\nSuppose there exists $s$ with $0 < s < r$ such that $2^s \\equiv -1 \\pmod{2m + 1}$. Then $2^{2s} \\equiv 1 \\pmod{2m + 1}$, so $r \\mid 2s$. Since $2s \\not\\equiv r \\pmod{2r}$, $2s \\ge 3r$, i.e., $s > r$, a contradiction. Thus, $A_m$ contains no element of the form $2^b + 1$ in this case.\n\n**(iii) $r$ is odd:**\n\nSuppose there exists $s$ with $0 < s < r$ such that $2^s \\equiv -1 \\pmod{2m + 1}$. Then $2^{2s} \\equiv 1 \\pmod{2m + 1}$, so $r \\mid 2s$, and since $r$ is odd, $r \\mid s$. But $0 < s < r$, which is impossible. Thus, $A_m$ contains no element of the form $2^b + 1$ in this case either.\n\nCombining (i)-(iii), we conclude that\n$$\nb_0 = \\frac{a_0 + 1}{2} - 1 = \\frac{a_0 - 1}{2}.\n$$\n\n**Alternative solution for (2):**\n\nLet $x, y$ be the smallest positive integers such that\n$$\n2^x \\equiv -1 \\pmod{2m + 1}, \\quad 2^y \\equiv 1 \\pmod{2m + 1}.\n$$\nClearly, $0 < x < y \\le 2x$. Assume $y = xq + r$ ($0 \\le r < x$). Since\n$$\n1 \\equiv 2^y \\equiv (2^x)^{q} 2^r \\equiv (-1)^q 2^r \\pmod{2m + 1},\n$$\n$q$ must be even. Hence $q = 2$ and $r = 0$. This completes the proof. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20986,
"subject": "Mathematics (Olympiad)",
"question": "For any set of points $A_1, A_2, \\ldots, A_n$ on the plane, define $r(A_1, A_2, \\ldots, A_n)$ as the radius of the smallest circle that contains all of these points. Prove that if $n \\ge 3$, there exist indices $i, j, k$ such that\n\n$$\nr(A_1, A_2, \\ldots, A_n) = r(A_i, A_j, A_k).\n$$",
"options": [],
"answer": "See solution",
"solution": "We start with a lemma.\n\n*Lemma.* If the triangle $ABC$ is acute, $r(A, B, C)$ is its circumradius; if it is obtuse, $r(A, B, C)$ is half the length of its longest side.\n\n*Proof.*\n\nLet us do the acute case first. The circumcircle contains the vertices, so $r(A, B, C)$ is not greater than the circumradius. Now, let us prove that no smaller circle contains all three vertices. If there is a smaller circle, let its center be $P$. Further, let the circumcenter be $O$. Since $ABC$ is acute, $O$ is in the interior. Consider the line that passes through $O$ and is parallel to $BC$. Let us call it $l_A$ and define $l_B$ and $l_C$ similarly. Now, consider the set of points that are on the opposite side of $l_A$ with respect to $A$. Call this set $S_A$ and define $S_B$ and $S_C$ similarly. It is easily seen (by geometry) that $S_A \\cap S_B \\cap S_C = \\emptyset$. As such, assume $P \\notin S_A$ without loss of generality. That is to say, $P$ is on the same side of $l_A$ as $A$. Now, consider the perpendicular bisector of $BC$ and assume that $P$, w.l.o.g, is on the same side of this line as $C$. Under these circumstances, $|PB| \\ge |OB|$. Thus, the smaller circle centered at $P$ must exclude $B$.\n\nIn the obtuse case, let $\\angle BAC \\ge 90^\\circ$. Then $BC$ is the longest side. The circle with diameter $BC$ contains all three vertices. Therefore, $r(A, B, C)$ is not greater than $\\frac{1}{2}|BC|$. But any smaller circle will clearly exclude at least one of $B$ and $C$.\n\nNow, let us return to the original problem. Note that there must be points $A, B, C$ among $A_1, A_2, \\ldots, A_n$ such that the circumcircle of $ABC$ contains all $n$ points. One can see this as follows: First start with a large circle that contains all $n$ points. Then shrink it while keeping the center fixed, until one of the $n$ points is on the circle and call this point $A$. Then shrink it keeping the point $A$ in place and moving the center closer to $A$, until another point $B$ is on the circle. Then keep the line $AB$ fixed while moving the center toward it or away from it so that another $C$ among the $n$ points appears on the circle. It is easy to see that this procedure is doable.\n\nConsider all such triples $A, B, C$ such that the circumcircle of $ABC$ contains all of $A_1, A_2, \\ldots, A_n$. Now choose the one among them with the smallest circumradius and let it be $A_i, A_j, A_k$. If $A_i A_j A_k$ is an acute triangle, any smaller circle will exclude one of $A_i, A_j, A_k$ by the lemma above. Therefore,\n\n$$\nr(A_1, A_2, \\ldots, A_n) = \\text{circumradius of } A_i A_j A_k = r(A_i, A_j, A_k).\n$$\n\nIf $A_i A_j A_k$ is an obtuse triangle, let $A_i$ be its obtuse angle. We wish to prove that the circle with diameter $A_j A_k$ contains all $n$ points. This will mean that\n\n$$\nr(A_1, A_2, \\ldots, A_n) = \\frac{1}{2} |A_j A_k| = r(A_i, A_j, A_k)\n$$\n\nand we will be done. If there are no points on the opposite side of $A_j A_k$ with respect to $A_i$, then this assertion is clear. If there are some points on that side, choose the one $X$ such that $\\angle A_j X A_k$ is smallest possible. Then the circumcircle of $A_j X A_k$ contains all $n$ points. However, by the choice of $A_i$, the circumradius of $A_j X A_k$ cannot be less than that of $A_i A_j A_k$. Thus, $\\angle A_j X A_k \\ge \\angle A_j A_i A_k \\ge 90^\\circ$. As such, the circle with diameter $A_j A_k$ contains all $n$ points.\n\n\n\nFigure 1: The circumcircles of $A_i A_j A_k$ and $A_j X A_k$ as well as the circle with diameter $A_j A_k$ are shown.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20987,
"subject": "Mathematics (Olympiad)",
"question": "Suppose the angle $\\angle XYZ$ of a triangle $\\triangle XYZ$ is $60^\\circ$. Let the lengths of the sides $YZ, XZ, XY$ be $x, y, z$ respectively. Prove that the inequality $2x \\ge y + z$ holds.\n\nNow, consider a convex hexagon $ABCDEF$ for which any pair of diagonals chosen from $AD, BE, CF$ intersect at $60^\\circ$. Let the points $P, Q, R$ be the points of intersection of the line segments $AD$ and $BE$, $BE$ and $CF$, $CF$ and $AD$, respectively. Prove that\n$$\nAB + BC + CD + DE + EF + FA \\geq AD + BE + CF.\n$$",
"options": [],
"answer": "See solution",
"solution": "*Lemma:* Suppose the angle $\\angle XYZ$ of a triangle $\\triangle XYZ$ is $60^\\circ$. Let the lengths of the sides $YZ, XZ, XY$ be $x, y, z$ respectively. Then, the inequality $2x \\ge y + z$ holds.\n\n*Proof:* Let $\\Gamma$ be the circumcircle of $\\triangle XYZ$. If we move the vertex $X$ along the arc $\\widehat{YZ}$ from $Y$ to $Z$, the area of $\\triangle XYZ$ attains its maximum when the height is largest, i.e., when $X$ lies on the perpendicular bisector of $YZ$. In this case, the triangle becomes equilateral with $x = y = z$, and the area equals $x^2$. Generally, the area is $yz \\le x^2$. By the Pythagorean theorem and angle properties, we have:\n$$\nx^2 = (y - \\frac{1}{2}z)^2 + \\frac{3}{4}z^2 = y^2 + z^2 - yz = (y + z)^2 - 3yz \\ge (y + z)^2 - 3x^2\n$$\nsince $x^2 \\ge yz$. Thus, $4x^2 \\ge (y+z)^2$ and $2x \\ge y+z$.\n\nNow, for the convex hexagon $ABCDEF$ with diagonals $AD, BE, CF$ intersecting at $60^\\circ$, let $P, Q, R$ be the intersection points as described. By applying the lemma to triangles $\\triangle APB$, $\\triangle BQC$, $\\triangle CRD$, $\\triangle DPE$, $\\triangle EQF$, $\\triangle FRA$, and summing the inequalities, we obtain:\n$$\n2AB + 2BC + 2CD + 2DE + 2EF + 2FA \\geq (PA + PB) + (QB + QC) + (RC + RD) + (PD + PE) + (QE + QF) + (RF + RA)\n$$\nThe right-hand side equals $2(AD + BE + CF)$, so dividing by $2$ yields:\n$$\nAB + BC + CD + DE + EF + FA \\geq AD + BE + CF.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20988,
"subject": "Mathematics (Olympiad)",
"question": "Prove that all positive integers, except the powers of $2$, can be written as the sum of at least two consecutive positive integers.",
"options": [],
"answer": "See solution",
"solution": "Let $n = 2^a b$, where $a \\ge 0$, $b \\ge 1$, and $b$ is odd. We seek $n = (m+1) + (m+2) + \\dots + (m+k)$, with $m \\ge 0$ and $k \\ge 2$, so $$k(2m + k + 1) = 2^{a+1} b.$$ \n\nIf $b = 1$, then $k = 2^\\alpha$ for $1 \\le \\alpha \\le a+1$, but $2m + k + 1 > 1$ is odd, so there are no solutions. For $b > 1$, we can construct the required sum:\n\n- If $b \\ge 2^{a+1} + 1$, take $m = \\frac{1}{2}(b - 2^{a+1} - 1)$ and $k = 2^{a+1}$.\n- If $b \\le 2^{a+1} - 1$, take $m = \\frac{1}{2}(2^{a+1} - 1 - b)$ and $k = b$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20989,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute triangle such that $AB < AC$. Let $I$ be the incentre of the triangle $ABC$, and let the incircle touch the side $BC$ at $D$. The line $AD$ crosses the circle $ABC$ again at $E$. Let $M$ be the midpoint of the side $BC$, and let $N$ be the midpoint of the circular arc $BAC$. The line $EN$ crosses the circular arc $BIC$ at $P$. Show that the lines $AD$ and $MP$ are parallel.\n\n",
"options": [],
"answer": "See solution",
"solution": "The internal bisector $AI$ and the perpendicular bisector $MN$ of the side $BC$ cross at the midpoint $K$ of the arc $BEC$. It is a fact that the circle $BIC$ is centred at $K$.\n\n\n\nLet the lines $ID$ and $NPE$ cross at $L$. Notice that $\\angle EAI = \\angle EAK = \\angle ENK = \\angle ELI$ (since $IL$ and $KN$ are parallel), so the quadrilateral $AIEL$ is cyclic, and $DI \\cdot DL = DA \\cdot DE = DB \\cdot DC$, showing that $L$ lies on the circle $BIC$.\n\nSince the angles $KBN$ and $KCN$ are both right, and the circle $BIC$ is centred at $K$, the lines $NB$ and $NC$ are the tangents from $N$ to this circle. It then follows that the line $NPEL$ is the $P$-symmedian of the triangle $BCP$, so $\\angle BPL = \\angle CPM$.\n\nLet the line $MP$ cross the circle $BIC$ again at $Q$, so the arcs $BL$ and $CQ$ of this circle have equal angular spans, so $L$ and $Q$ are reflections of one another in the perpendicular bisector $KMN$ of the chord $BC$.\n\nProject $Q$ orthogonally to $Q'$ on $BC$ and refer to standard notation in the triangle $ABC$: $a$, $b$, $c$ denote the lengths of the sides $BC$, $CA$, $AB$, respectively, $s = (a + b + c)/2$ denotes its semiperimeter, $r$ its inradius, and $S$ its area. With reference to standard formulae, write $CQ' = BD = s - b$ and\n\n$$\nQQ' = DL = \\frac{DB \\cdot DC}{DI} = \\frac{(s-b)(s-c)}{r} = \\frac{s}{s-a},\n$$\n\nto infer that $Q$ is the $A$-excentre of the triangle $ABC$, so it lies on the line $AIK$.\n\nFinally, write $\\angle PQA = \\angle PQI = \\angle PLI = \\angle ELI = \\angle EAI = \\angle DAQ$, to conclude that the lines $AD$ and $MP$ are indeed parallel.\n\n**Remark.** Another consequence of the above argument is that the $A$-excentre of the triangle $ABC$ lies on the circle $NPK$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 20990,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer and let $a_1, a_2, \\ldots, a_n$ be real numbers with $a_1 + a_2 + \\ldots + a_k \\leq k$ for all $k \\in \\{1, 2, \\ldots, n\\}$. Show that\n\n$$\n\\frac{a_1}{1} + \\frac{a_2}{2} + \\ldots + \\frac{a_n}{n} \\leq \\frac{1}{1} + \\frac{1}{2} + \\ldots + \\frac{1}{n}.\n$$",
"options": [],
"answer": "See solution",
"solution": "We induct on $n$.\n\nThe case $n=1$ is trivial. Suppose the claim holds for $n$ numbers. If $a_{n+1} \\leq 1$ then $\\frac{a_{n+1}}{n+1} \\leq \\frac{1}{n+1}$ and the conclusion follows.\n\nIf $a_{n+1} > 1$, then\n$$\n\\frac{a_1}{1} + \\dots + \\frac{a_n}{n} + \\frac{a_{n+1}}{n+1} \\leq \\frac{a_1}{1} + \\dots + \\frac{a_n + a_{n+1} - 1}{n} + \\frac{1}{n+1}.\n$$\nNow apply the induction hypothesis to the following $n$ numbers: $a_1, \\dots, a_{n-1}, a_n + a_{n+1} - 1$ to get the claim.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20991,
"subject": "Mathematics (Olympiad)",
"question": "Consider a $2024 \\times 2024$ grid of unit squares. Two distinct unit squares are *adjacent* if they share a common side. Each unit square is to be coloured either black or white. Such a colouring is called *evenish* if every unit square in the grid is adjacent to an even number of black unit squares.\n\nDetermine the number of evenish colourings.",
"options": [],
"answer": "See solution",
"solution": "The answer is $2^{2024}$.\n\nWe will prove that the answer for the $m \\times m$ grid case is $2^m$.\n\nGiven two colourings $A$ and $B$, we define their \"sum\" $A \\oplus B$, a new colouring, in the following way: in the colouring $A \\oplus B$, a square is coloured white if and only if it has the same colour in both $A$ and $B$. This can also be viewed as a sum modulo 2. Furthermore, this sum can be computed for more than two colourings, such as $A \\oplus B \\oplus C$ and so on.\n\nNext, given two evenish colourings $A$ and $B$, their sum $A \\oplus B$ is also evenish. This is because, for a fixed square $S$, the number of black adjacent squares in $A \\oplus B$ corresponds to the number of parity disagreements between $A$ and $B$ (adjacent to $S$), which must be even.\n\nFix the top row of a colouring. In order for the whole colouring to be evenish, the colouring of the second row is automatically determined, as every square in the top row must be adjacent to an even number of black squares. Continuing in this fashion, row by row, the colouring of the whole grid is automatically determined. The whole colouring is evenish as long as every square in the bottom row is adjacent to an even number of black squares. Thus, the number of evenish colourings is at most the number of colourings for the top row, which is $2^m$.\n\nIt remains to show that every colouring of the top row will determine an evenish colouring. We first construct the case where precisely one square in the top row is coloured black. This can be done using the following diamond-like pattern.\n\n\n\nNote that this construction is symmetric under a rotation of $180^\\circ$, so the adjacency condition for the bottom row is automatically satisfied, just like the top row.\n\nFinally, consider an arbitrary top row colouring $C$. It can be written as the sum of top row colourings $C_1, C_2, \\dots$, each with only one black square. From before, each $C_i$ determines a unique evenish colouring. By summing these evenish colourings, we obtain an evenish colouring with $C$ as the top row. Therefore, every top row colouring determines a unique evenish colouring, and the number of evenish colourings must be $2^m$.\n\n**Remark.** Interested readers are invited to investigate the case of an $m \\times n$ grid, which is much more difficult!",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 20992,
"subject": "Mathematics (Olympiad)",
"question": "令 $n$ 為一正整數。松鼠阿布與阿江準備了 $n$ 顆核桃好過冬。某天，阿江發現阿布把核桃擺成 $n$ 堆，每堆一顆；牠覺得太多堆了，心生不悅。阿江於是決定進行以下操作：每次選兩堆核桃，從中各拿取等量的核桃，並將拿取的核桃合併成新的一堆。阿江的目標是讓非空的核桃堆數 $P(n)$ 越少越好。試對所有正整數 $n$，求阿江能透過有限步操作達到的最小 $P(n)$ 值。",
"options": [],
"answer": "See solution",
"solution": "若 $n$ 為 2 的幂次，則最小 $P = 1$；否則，最小 $P = 2$。\n\n當 $n = 2^k$ 時，我們只要每次取顆數最少的任兩堆，拿取其全部核桃合併，最終便能成為單一一堆，而這顯然是最小可能 $P$ 值。\n\n現在考慮 $2^k < n < 2^{k+1}$。以下用 $t$-堆表示有 $t$ 顆核桃的堆。考慮以下操作：\n\n1. 我們先從起始的 $n$ 堆中選擇 $2^k$ 堆，然後每次取其中顆數最少的任兩堆，拿取其全部核桃合併，最終便能成為一個 $2^k$-堆與 $m = n - 2^k$ 個 1-堆。稱這個 $2^k$-堆為 XL 堆。\n\n2. 接下來，我們從 XL 堆和一個 1-堆中各取一顆，組成一個 2-堆。若 $m < 2^k - 1$，則我們再從 XL 堆和 2-堆中各取一顆。重複以上動作，直到 XL 堆剩下 $m$ 顆。此時我們有一個 $m$-堆，一個 2-堆與 $n - m - 2 = 2^k - 2$ 個 1-堆。\n\n3. 我們接著將所有 1-堆兩兩合併，從而有 $2^{k-1}$ 個 2-堆。再從這些堆中，每次取最小的兩堆合併，最終便會得到一個 $2^k$ 堆。此時剩下一個 $m$-堆與一個 $2^k$-堆，故 $P = 2$。\n\n我們僅須證明當 $n$ 非 2 的幂次時，我們不可能操作到僅剩一堆即可。首先注意到，若我們選擇一個 $a$-堆和一個 $b$-堆，各拿取 $c \\leq \\min(a, b)$ 顆，則我們有\n\n$$\na \\to a - c\n$$\n\n$$\nb \\to b - c\n$$\n\n$$\n0 \\to 2c\n$$\n\n再注意到若存在奇數 $q$ 整數 $\\gcd(a-c, b-c, 2c)$，則我們必然要有 $q \\mid a$ 與 $q \\mid b$。\n\n現在，因為 $n$ 不是 2 的幂次，其必然有奇因數 $q > 1$。若我們最後得到單一個 $n$-堆，依照上述討論，一開始的每個 1-堆的顆數都要能被 $q$ 整除，而這顯然矛盾。故當 $n$ 不是 2 的幂次時，最小的 $P$ 值為 2。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20993,
"subject": "Mathematics (Olympiad)",
"question": "A square is contained in a cube when all of its points are in the faces or in the interior of the cube. Determine the largest $l$ such that there exists a square of side $l$ contained in a cube with edge $1$.",
"options": [],
"answer": "See solution",
"solution": "The answer is $\\frac{3\\sqrt{2}}{4}$. Consider the points $\\left(\\frac{3}{4}, 0, 0\\right)$, $\\left(0, \\frac{3}{4}, 0\\right)$, $\\left(1, \\frac{1}{4}, 1\\right)$, and $\\left(\\frac{1}{4}, 1, 1\\right)$.\n\nSuppose there is a square with side $\\ell > \\frac{3\\sqrt{2}}{4}$ inside the cube. We can assume, without loss of generality, that the centers of the cube and the square coincide. If not, consider the three planes that cut the cube into two blocks with dimensions $\\frac{1}{2}, 1, 1$. Focus on one such plane, call it $\\alpha$. If the center of the square is not in $\\alpha$, it is inside one of the two blocks defined by the plane. Draw a plane $\\beta$ passing through the center of the square and parallel to $\\alpha$. It cuts the square in two congruent pieces and the cube in two blocks, one smaller than the other. Half of the square is inside the smaller block, so if we translate $\\beta$ to $\\alpha$, the square remains inside the cube. Repeating this for the other two planes, the center of the square will coincide with the center of the cube.\n\nNext, draw a sphere $S$ centered at the center of the square (which now coincides with the center of the cube) with radius $\\frac{\\ell\\sqrt{2}}{2} > \\frac{3}{4}$. All vertices of the square lie on the surface of $S$, and two opposite vertices are antipodal. Each vertex is in one of the 8 regions determined by the intersection of $S$ and the cube. One vertex is in a region $R$; the opposite vertex is in the opposite region $T$.\n\nConsider a block containing two neighboring regions. This block has dimensions $1, x, x$. Since $S$ has radius greater than $\\frac{3}{4}$, $x < \\frac{1}{2} - \\sqrt{\\left(\\frac{3}{4}\\right)^2 - \\left(\\frac{\\ell\\sqrt{2}}{2}\\right)^2} = \\frac{1}{4}$. The maximum distance between any two points in the block is less than $\\sqrt{1^2 + 2x^2} = \\frac{3\\sqrt{2}}{4}$, so it is impossible to have two vertices of the square in two neighboring regions. But $R$ eliminates three regions and $T$ eliminates the other three, leaving no regions for the other two vertices—a contradiction.\n\nThus, the largest square contained in the cube has side $\\frac{3\\sqrt{2}}{4}$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 20994,
"subject": "Mathematics (Olympiad)",
"question": "Find all triples $(a, b, c)$ of integers that satisfy\n\n$$\n(a - b)^3 (a + b)^2 = c^2 + 2(a - b) + 1.\n$$",
"options": [],
"answer": "See solution",
"solution": "The solutions are $(0, 1, 0)$ and $(-1, 0, 0)$.\n\nLet $x = a - b$ and $y = a + b$. The equation becomes:\n\n$$\nx^3 y^2 = c^2 + 2x + 1.\n$$\n\nWe need integer solutions with $x$ and $y$ of equal parity. Consider the equivalent form:\n\n$$\nx(x^2 y^2 - 2) = c^2 + 1.\n$$\n\n**Case 1:** $x$ and $y$ both even.\n\nThen $x^2 y^2 - 2$ is even, so the left side is divisible by 4. But $c^2 \\equiv -1 \\pmod{4}$, which is impossible since $-1$ is not a quadratic residue modulo 4.\n\n**Case 2:** $x$ and $y$ both odd.\n\nThen $x^2 y^2$ is odd. If $x^2 y^2 = 1$, then $x^2 = 1$ and $y^2 = 1$, so $x = \\pm 1$, $y = \\pm 1$.\n\nSince $x^2 y^2 - 2 = -1$, $x(x^2 y^2 - 2) = -x$, which must equal $c^2 + 1$. For $x = -1$, $c^2 + 1 = 1$, so $c = 0$.\n\nThus, $x = -1$, $y = 1$ gives $a = 0$, $b = 1$; $x = -1$, $y = -1$ gives $a = -1$, $b = 0$. In both cases, $c = 0$.\n\nIf $x^2 y^2 > 1$, then $x^2 y^2 - 2 > 0$. Since squares of odd numbers are $1 \\pmod{4}$, $x^2 y^2 - 2 \\equiv -1 \\pmod{4}$. There exists a prime divisor $p \\equiv -1 \\pmod{4}$ of $x^2 y^2 - 2$, so $c^2 \\equiv -1 \\pmod{p}$, which is impossible since $-1$ is not a quadratic residue modulo $p$.\n\nTherefore, the only solutions are $(0, 1, 0)$ and $(-1, 0, 0)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20995,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\mathbb{R}_{>0} = \\{x \\in \\mathbb{R} \\mid x > 0\\}$ denote the set of positive real numbers. Find all pairs of functions $f, g: \\mathbb{R}_{>0} \\to \\mathbb{R}_{>0}$ satisfying\n\n$$\nf(g(x)) = f(x)g(x), \\quad f(x) = x(1 + g(x))\n$$\n\nand such that the sequence $g(x), g(g(x)), g(g(g(x))), \\ldots$ takes finitely many different values for all $x \\in \\mathbb{R}_{>0}$.",
"options": [],
"answer": "See solution",
"solution": "The solution is $f(x) = x + 1$ and $g(x) = 1/x$.\n\nTo show this is the only solution, fix $x \\in \\mathbb{R}_{>0}$ and define $g^0 = x$ and $g^n = g(g^{n-1})$ for $n \\ge 1$.\n\nWe have $g(x)(1 + g(g(x))) = f(g(x)) = f(x)g(x) = x(1 + g(x))g(x)$. Since $g(x) \\ne 0$, we get $1 + g^2 = x(1 + g)$. This can be rewritten as\n\n$$\ngg^2 - 1 = (xg - 1)(1 + g).\n$$\n\nThen $g^2g^3 - 1 = (gg^2 - 1)(1 + g^2) = (xg - 1)(1 + g)(1 + g^2)$, and more generally,\n\n$$\ng^n g^{n+1} - 1 = (g^{n-1}g^n - 1)(1 + g^n) = (xg - 1)(1 + g)(1 + g^2) \\dots (1 + g^n)\n$$\n\nfor any $n \\ge 1$ by induction. Since the sequence $g^n$ takes finitely many values, there exist $n > m$ with $g^n = g^m$. Thus,\n\n$$\n(xg - 1)(1 + g) \\dots (1 + g^m) ((1 + g^{m+1}) \\dots (1 + g^n) - 1) = 0.\n$$\n\nSince $g^k > 0$, we must have $xg - 1 = 0$. It follows that $g = 1/x$ and $f = x + 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20996,
"subject": "Mathematics (Olympiad)",
"question": "平面上有兩個三角形 $ABC$ 與 $A'B'C'$。已知三角形 $ABC$ 的各邊長不小於 $a$，且三角形 $A'B'C'$ 的各邊長不小於 $a'$。證明總是能夠從這兩個三角形中各選出一頂點，使得它們的距離不小於 $\\sqrt{\\frac{a^2 + a'^2}{3}}$。",
"options": [],
"answer": "See solution",
"solution": "令點 $G, G'$ 分別為 $\\triangle ABC$ 與 $\\triangle A'B'C'$ 的重心。由重心公式知，對於平面上的任一點 $P$，均有\n\n$$\nPA^2 + PB^2 + PC^2 = 3PG^2 + \\frac{AB^2 + BC^2 + CA^2}{3}.\n$$\n\n分別令 $P$ 為 $A', B', C'$ 三點，所得的三條式子加總，再使用上述公式在三角形 $\\triangle A'B'C'$ 及 $P=G$ 上，得\n\n$$\n\\begin{aligned}\n&A'A^2 + A'B^2 + A'C^2 + B'A^2 + B'B^2 + B'C^2 + C'A^2 + C'B^2 + C'C^2 \\\\\n&= AB^2 + BC^2 + CA^2 + 3(A'G^2 + B'G^2 + C'G^2) \\\\\n&= AB^2 + BC^2 + CA^2 + A'B'^2 + B'C'^2 + C'A'^2 + 9GG'^2 \\\\\n&\\geq 3a^2 + 3a'^2.\n\\end{aligned}\n$$\n\n因此，最前面的 9 項之中，存在一項不小於 $\\frac{1}{9}(3a^2 + 3a'^2) = \\frac{a^2 + a'^2}{3}$，證畢。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20997,
"subject": "Mathematics (Olympiad)",
"question": "Let $A_1, A_2, \\dots, A_{35}$ be cities such that only $A_i$ and $A_{i+1}$ are connected for $i = 1, 2, \\dots, 34$. The travel between $A_1$ and $A_{35}$ uses at least 34 flights. After adding flights between $A_1$ and $A_{35}$, it is possible to travel between any pair of cities by using at most 17 flights. Show that 34 flights are sufficient to ensure this property.",
"options": [],
"answer": "See solution",
"solution": "Let $\\rho(X, Y)$ denote the minimal possible number of flights between the cities $X$ and $Y$. Suppose, for contradiction, that $\\rho(A, B) > 34$ for some cities $A$ and $B$, and a path with minimal number of flights before adding a flight between the cities $T$ and $S$ is $$(A = A_0, A_1, \\dots, A_{17}, A_{18}, \\dots, B = A_k).$$ After adding the flight between $T$ and $S$, $\\rho(A, A_{18}) \\le 17$ and $\\rho(A_{17}, B) \\le 17$. By definition, both minimal paths from $A$ to $A_{18}$ and from $A_{17}$ to $B$ must use the flight between $T$ and $S$. Without loss of generality, suppose the path from $A$ to $A_{18}$ is $(A_0, \\dots, T, S, \\dots, A_{18})$. Then $l_1 + l_2 \\le 16$ where $\\rho(A, T) = l_1$, $\\rho(S, A_{18}) = l_2$. Similarly, the path from $A_{17}$ to $B$ is $(A_{17}, \\dots, T, S, \\dots, B)$ with $\\rho(A_{17}, T) = m_1$, $\\rho(S, B) = m_2$ and $m_1 + m_2 \\le 16$, or the path is $(A_{17}, \\dots, S, T, \\dots, B)$ with $\\rho(A_{17}, S) = k_1$, $\\rho(T, B) = k_2$ and $k_1 + k_2 \\le 16$. Thus, there exists a travel from $A$ to $B$ $(A, \\dots, T, \\dots, A_{18}, A_{17}, \\dots, S, \\dots, B)$ using at most $l_1 + m_1 + 1 + l_2 + m_2 < 34$ flights, or $(A, \\dots, T, \\dots, B)$ using at most $l_1 + k_2 < 34$ flights. This is a contradiction.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20998,
"subject": "Mathematics (Olympiad)",
"question": "求所有函數 $f: \\mathbb{N} \\to \\mathbb{N}$，使得對於所有由 2024 個相異正整數所成的集合 $A$，\n\n$$\nS_A := \\{f^{(k)}(x) \\mid 1 \\le k \\le 2024,\\ x \\in A\\}\n$$\n\n亦為一個由 2024 個相異正整數所成集合。\n\n註：$f^{(k)}$ 表 $f$ 的 $k$ 次疊代。換言之，$f^{(1)}(x) := f(x)$，且對於所有 $n \\ge 1$，$f^{(n+1)}(x) := f(f^{(n)}(x))$。",
"options": [],
"answer": "See solution",
"solution": "顯然 $f(x) = x$ 為一解。此外，對於任何由 2024 個相異正整數所成集合 $A = \\{a_1, a_2, \\dots, a_{2024}\\}$，任何滿足\n\n$$\n\\begin{cases} f(a_i) = a_{i-1}, & \\text{其中 } a_0 = a_{2024}, \\\\ f(x) \\in A, & \\text{對於所有 } x \\notin A, \\end{cases}\n$$\n\n的函數 $f$ 皆滿足題意。讓我們證明以上便是全部的解。\n\n用所有正整數做點，並將 $x$ 連向 $f(x)$ 構成有向圖 $G$（允許自環）。注意若 $x$ 可通往 2025 個點，則任何包含 $x$ 和 $f(x)$ 的 $A$ 都有 $|S_A| \\ge 2025$，矛盾。這意味著從 $x$ 出發，我們在 2024 步內一定會進入一個環，且這個環的大小至多為 2024。\n\n現在，考慮所有 $G$ 中在環上的點。讓我們考慮 $C$ 的大小。\n\n*Case 1.* $|C| > 2024$：\n\n注意到此時不能有任何 $x \\in C$ 使得 $f(x) \\ne x$；否則，任取一個包含 $x$ 但不包含 $f(x)$ 的 $A \\subset C$，則 $A \\subset S_A$（注意 $A$ 的元素都在環上）且 $f(x) \\in S_A$，從而 $A \\cup \\{f(x)\\} \\subset S_A \\Rightarrow |S_A| \\ge |A| + 1 > 2024$，矛盾。\n\n故我們有 $f(x) = x$ 對於所有 $x \\in C$ 皆成立。此時，若有任何 $y \\notin C$，因為 $y$ 在 2024 步內要到達 $C$，故存在 $z \\notin C$ 使得 $f(z) \\in C$。若我們取 $A = \\{z, f(z), b_1, \\dots, b_{2022}\\}$，其中 $b_i$ 為 $C$ 中 2022 個異於 $f(z)$ 的元素，則 $S_A = A - \\{z\\}$，從而 $|S_A| = |A| - 1 < 2024$，矛盾。故 $C = \\mathbb{N}$，也就是 $f(x) = x$。\n\n*Case 2.* $|C| \\le 2024$：\n\n由於所有 $x \\in \\mathbb{N}$ 在 2024 步內都要抵達 $C$，而 $C$ 是有限集，故存在 $y \\in \\mathbb{N}$ 使得 $f^{-1}(y) := \\{x : f(x) = y\\}$ 是無限集。此時取 $A \\subset f^{-1}(y)$，知 $Y := \\{y, f(y), \\dots, f^{(2023)}(y)\\}$ 有 $|Y| = 2024$；換言之，$y$ 在一個大小恰為 2024 的環 $Y$ 上。\n\n現在，對於任何 $x \\in \\mathbb{N}$，取 $A$ 滿足 $x \\in A$ 且 $A \\cap f^{-1}(y) \\neq \\emptyset$。注意到此時 $Y \\subset S_A$，迫使 $Y = S_A$（否則 $|S_A| > |Y| = 2024$），也因而 $f(x) \\in Y$。這表示所有不在 $Y$ 上的點都必須一步到達 $Y$，此便是第二個可能的 $f$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 20999,
"subject": "Mathematics (Olympiad)",
"question": "Given triangle $ABC$ inscribed in a circle $(O)$, the tangents at $B$ and $C$ to $(O)$ intersect at $T$. Let $H$ be the projection of $T$ onto the tangent at $A$ to $(O)$. Let $S$ be the reflection of $T$ across $BC$. Let $L$ be the Lemoine point of triangle $ABC$. Prove that $$\\angle HST = \\angle AOL.$$",
"options": [],
"answer": "See solution",
"solution": "Let $AH$ meet $BC$ at $X$. The line through $X$ perpendicular to $OL$ intersects $OT$ at $K$. $KO$ meets $AH$ at $P$. $AT$ meets $BC$ at $R$.\n\nWe have $O(AR, LT) = -1$ and $XP \\perp OA$, $XT \\perp OR$, $XK \\perp OL$, $XM \\perp OT$, so $(PT, KM) = -1$.\n\nAgain, $M$ is the midpoint of $ST$, so applying the same formula as Maclaurin and Newton, we obtain $\\overline{PS} \\cdot \\overline{PK} = \\overline{PM} \\cdot \\overline{PT} = \\overline{PH} \\cdot \\overline{PX}$.\n\nWe deduce that $XHSK$ is cyclic. It follows that $\\angle HST = 180^\\circ - \\angle HXK = \\angle AOL$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21000,
"subject": "Mathematics (Olympiad)",
"question": "In a plane rectangular coordinate system $xOy$, the focus of the parabola $\\Gamma: y^2 = 2p x$ ($p > 0$) is $F$. A tangent line to $\\Gamma$ passes through a point $P$ (different from $O$) on $\\Gamma$ and intersects the $y$-axis at point $Q$. If $|FP| = 2$ and $|FQ| = 1$, then the dot product of vectors $\\overrightarrow{OP}$ and $\\overrightarrow{OQ}$ is \\_\\_\\_\\_\\_.",
"options": [],
"answer": "See solution",
"solution": "Let $P\\left(\\dfrac{t^2}{2p},\\ t\\right)$ ($t \\neq 0$). The equation of the tangent line to $\\Gamma$ at $P$ is $y t = p\\left(x + \\dfrac{t^2}{2p}\\right)$.\n\nLet $x = 0$ to find $Q$: $y t = \\dfrac{t}{2} \\implies y = \\dfrac{1}{2}$, so $Q(0,\\ \\dfrac{t}{2})$.\n\nThe focus $F$ has coordinates $\\left(\\dfrac{p}{2},\\ 0\\right)$. Thus:\n\n$$\n|FP| = \\sqrt{\\left(\\dfrac{p}{2} - \\dfrac{t^2}{2p}\\right)^2 + t^2} = \\dfrac{p}{2} + \\dfrac{t^2}{2p}, \\\\\n|FQ| = \\dfrac{\\sqrt{p^2 + t^2}}{2}.\n$$\n\nGiven $|FP| = 2$ and $|FQ| = 1$, we have $p^2 + t^2 = 4p$ and $p^2 + t^2 = 4$, so $p = 1$, $t^2 = 3$.\n\nTherefore,\n$$\n\\overrightarrow{OP} \\cdot \\overrightarrow{OQ} = \\frac{t^2}{2} = \\frac{3}{2}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21001,
"subject": "Mathematics (Olympiad)",
"question": "Do there exist four quadratic polynomials such that the sum of any three of them has a real root, but the sum of any two of them has no real root?",
"options": [],
"answer": "See solution",
"solution": "No, such polynomials do not exist.\n\nAssume there exist four quadratic polynomials $f_1, f_2, f_3, f_4$ with the stated properties. If a quadratic polynomial $f(x)$ has no real root, then either $f(x) > 0$ for all $x \\in \\mathbb{R}$ (we write $f > 0$ and say $f$ is positive) or $f(x) < 0$ for all $x \\in \\mathbb{R}$ (we write $f < 0$ and say $f$ is negative).\n\n**Lemma.** If $g_1, g_2, g_3$ are quadratic polynomials such that the sum of any two of them has no real root and the sum $g_1 + g_2 + g_3$ has a real root, then $g_1 + g_2$, $g_1 + g_3$, and $g_2 + g_3$ cannot all have the same sign.\n\n*Proof.* If $g_1 + g_2$, $g_1 + g_3$, and $g_2 + g_3$ are all positive, then $g_1 + g_2 + g_3$ is also positive and thus has no real root, which is a contradiction. $\\square$\n\nBy the lemma, the sums $f_1 + f_2$, $f_1 + f_3$, $f_2 + f_3$ cannot all have the same sign. Without loss of generality, assume:\n\n$$\n\\begin{aligned}\nf_1 + f_2 &> 0 \\\\\nf_1 + f_3 &> 0 \\\\\nf_2 + f_3 &< 0\n\\end{aligned}\n$$\n\nThere are two possibilities:\n\n**Case 1:** If $f_2 + f_4 > 0$, then applying the lemma for $f_1, f_2, f_4$, we get $f_1 + f_4 < 0$. This gives the contradiction:\n\n$$\n0 > (f_2 + f_3) + (f_1 + f_4) = (f_1 + f_3) + (f_2 + f_4) > 0\n$$\n\n**Case 2:** If $f_2 + f_4 < 0$, then applying the lemma for $f_2, f_3, f_4$, we get $f_3 + f_4 > 0$. Now applying the lemma for $f_1, f_3, f_4$, we get $f_1 + f_4 < 0$. This gives the contradiction:\n\n$$\n0 > (f_2 + f_3) + (f_1 + f_4) = (f_1 + f_2) + (f_3 + f_4) > 0\n$$\n\nHence, no such polynomials exist.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21002,
"subject": "Mathematics (Olympiad)",
"question": "We will call a positive integer *special* if the sum of its (decimal) digits and the sum of the digits of its successor are divisible by 11.\n\n(a) Find the last five digits of a special number.\n\n(b) Prove that there are infinitely many special numbers.",
"options": [],
"answer": "See solution",
"solution": "a) Denote $s(m)$ as the sum of the digits of a positive integer $m$.\n\nLet $n = \\overline{a_k a_{k-1} \\dots a_2 a_1}$ be a special number with $k \\ge 1$ digits. The statement says that $11 \\mid s(n)$ and $11 \\mid s(n+1)$.\n\nIf $a_1 \\le 8$, then $n+1 = \\overline{a_k a_{k-1} \\dots a_2 (a_1+1)}$, so $s(n+1) = s(n) + 1$. Thus, if $s(n)$ is divisible by 11, $s(n+1)$ cannot be, so $a_1 = 9$.\n\nSuppose the last $p \\ge 1$ digits of $n$ are 9 and the $(p+1)^{\\text{th}}$ is different from 9. Then $n = \\overline{a_k a_{k-1} \\dots a_{p+1} 99\\dots9}$ and $n+1 = \\overline{a_k a_{k-1} \\dots (a_{p+1} + 1) 00\\dots0}$, so $s(n) = a_k + a_{k-1} + \\dots + a_{p+1} + 9p$ and $s(n+1) = a_k + a_{k-1} + \\dots + a_{p+1} + 1$.\n\nTherefore, $s(n) = s(n+1) + 9p - 1$, so $9p - 1$ must be a multiple of 11. The smallest such $p$ is $p = 5$, hence the last five digits of a special number are nines.\n\nb) From (a), special numbers are of the form $n = m \\cdot 10^5 + 99\\,999$, where $m$ is chosen appropriately. Since $s(n) = s(m) + 45$ and $45 = 11 \\cdot 4 + 1$, it is enough to find $m$ with the last digit different from 9 and $s(m) = 11t - 1$ for some $t \\ge 1$. For example, if $t = 1$, numbers of the form $m = 2 \\cdot 10^q + 8$ (with $q \\ge 1$) have digit sum 10.\n\nThus, for each $q \\ge 1$, the number $n = 2 \\cdot 10^{q+5} + 899\\,999$ is special, and there are infinitely many such numbers.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21003,
"subject": "Mathematics (Olympiad)",
"question": "Does there exist a function $f: \\mathbb{Z} \\to \\mathbb{Z}$ such that\n$$\nf(f(n) - 2n) = 2f(n) + n\n$$\nfor all integers $n$?",
"options": [],
"answer": "See solution",
"solution": "There are many functions $f$ that satisfy the given condition. One of them is given by the following definition:\n\n$$\nf(n) = \\begin{cases} n & \\text{if } n \\ge 0 \\\\ -3n & \\text{if } n < 0 \\end{cases}\n$$\n\nwhich can be verified in the following 3 cases.\n\n$n > 0$: $f(f(n) - 2n) = f(-n) = 3n$ and $2f(n) + n = 2n + n = 3n$.\n\n$n < 0$: $f(f(n) - 2n) = f(-5n) = -5n$ and $2f(n) + n = -5n$.\n\n$n = 0$: We have $f(0) = 0$ and $f(f(n) - 2n) = 2f(n) + n = 0$.\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21004,
"subject": "Mathematics (Olympiad)",
"question": "The three-element subsets of a seven-element set are colored. If the intersection of two sets is empty, then they have different colors. What is the minimum number of colors needed?",
"options": [],
"answer": "See solution",
"solution": "Let $A = \\{1, 2, 3, 4, 5, 6, 7\\}$. Two colors are not enough because the sets in the following sequence of three-element subsets of $A$ should have alternating colors:\n\n$\\{1, 2, 3\\}$, $\\{4, 5, 6\\}$, $\\{7, 1, 2\\}$, $\\{3, 4, 5\\}$, $\\{6, 7, 1\\}$, $\\{2, 3, 4\\}$, $\\{5, 6, 7\\}$, $\\{1, 2, 3\\}$.\n\nWith three colors we can color, for example, the three-element subsets of $A$ as follows:\n\n- Use the first color for all the subsets containing the element $7$.\n- Use a second color for all the subsets that do not contain $7$ and for which the sum of their elements is even.\n- Use a third color for all the remaining subsets.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21005,
"subject": "Mathematics (Olympiad)",
"question": "Given a unit cube $ABCD - A_1B_1C_1D_1$, construct a sphere with point $A$ as the center and radius $\\frac{2\\sqrt{3}}{3}$. What is the total length of the curves formed by the intersection of the surfaces of the sphere and the cube?",
"options": [],
"answer": "See solution",
"solution": "As shown in the figure, the surface of the sphere intersects all six faces of the cube. The intersection curves are divided into two types:\n\n- The first type lies on the three faces containing vertex $A$: $AA_1B_1B$, $ABCD$, and $AA_1D_1D$.\n- The second type lies on the three faces not containing $A$: $CC_1D_1D$, $A_1B_1C_1D_1$, and $BB_1C_1C$.\n\n\n\nOn face $AA_1B_1B$, the intersection curve is arc $\\widehat{EF}$, which lies on a circle centered at $A$. Since $AE = \\frac{2\\sqrt{3}}{3}$ and $AA_1 = 1$, $\\angle A_1AE = \\frac{\\pi}{6}$. Similarly, $\\angle BAF = \\frac{\\pi}{6}$, so $\\angle EAF = \\frac{\\pi}{3}$. Thus, the length of arc $\\widehat{EF}$ is $\\frac{2\\sqrt{3}}{3} \\cdot \\frac{\\pi}{6} = \\frac{\\sqrt{3}\\pi}{9}$. There are three such arcs.\n\nOn face $BB_1C_1C$, the intersection curve is arc $\\widehat{FG}$, which lies on a circle centered at $B$ with radius $\\frac{\\sqrt{3}}{3}$ and $\\angle FBG = \\frac{\\pi}{2}$. So the length of $\\widehat{FG}$ is $\\frac{\\sqrt{3}}{3} \\cdot \\frac{\\pi}{2} = \\frac{\\sqrt{3}\\pi}{6}$. There are also three such arcs.\n\nIn summary, the total length of all intersection curves is\n\n$$\n3 \\times \\frac{\\sqrt{3}\\pi}{9} + 3 \\times \\frac{\\sqrt{3}\\pi}{6} = \\frac{5\\sqrt{3}\\pi}{6}.\n$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 21006,
"subject": "Mathematics (Olympiad)",
"question": "There are $n$ white and $n$ black balls placed randomly on the circumference of a circle. Starting from a certain white ball, number all white balls in a clockwise direction by $1, 2, \\dots, n$. Likewise, number all black balls by $1, 2, \\dots, n$ in an anti-clockwise direction starting from a certain black ball. Prove that there exist consecutive $n$ balls whose numbering forms the set $\\{1, 2, \\dots, n\\}$.",
"options": [],
"answer": "See solution",
"solution": "Choose a black ball and a white ball with the same number, and let the number of balls between these two balls be minimal. We can suppose the number of the two balls is $1$.\n\nFirst, we shall prove that the balls between the two balls have the same color.\n\nIn fact, if they are of different colors, then the white ball and the black ball, each numbered $n$, are between the two balls (see Fig. 1). This contradicts the minimality of the number of balls between the two balls labeled by $1$.\n\nNext, if the balls between the two balls labeled by $1$ are white, we have two cases:\n\n**Case 1:** The numbers of the white balls are $2, \\dots, k$ (see Fig. 2). Then, from the white ball labeled $1$ in the anti-clockwise direction, we can get a chain of $n$ balls whose numbering forms the set $\\{1, 2, \\dots, n\\}$.\n\n\n\nFig. 1\n\n\n\nFig. 2\n\n**Case 2:** The numbers of the white balls are $k, k+1, \\dots, n$ (see Fig. 3). Then, from the white ball labeled $1$ in the clockwise direction, we can have a chain of $n$ balls that satisfies the condition.\n\nThe same argument applies if the balls between the two balls labeled by $1$ are black, or if there are no balls between them.\n\n\n\nFig. 3",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21007,
"subject": "Mathematics (Olympiad)",
"question": "Let $f(x)$ and $g(y)$ be two monic polynomials with complex coefficients, both of degree $n$, such that\n\n$$\nf(x) - g(y) = \\prod_{j=1}^{n} (a_j x + b_j y + c_j),\n$$\n\nwhere $a_j, b_j, c_j$ are complex numbers for $1 \\leq j \\leq n$. Prove that there exist complex numbers $a, b, c$ such that\n\n$$\nf(x) = (x+a)^n + c, \\quad g(y) = (y+b)^n + c.\n$$",
"options": [],
"answer": "See solution",
"solution": "Observe that $\\prod_{j=1}^{n} a_j = 1$. By dividing both sides by this product, we may write\n\n$$\nf(x) - g(y) = \\prod_{j=1}^{n} (x - \\alpha_j y + \\beta_j).\n$$\n\nWe note that\n\n$$\n\\prod_{j=1}^{n} (x - \\alpha_j y) = x^n - y^n = \\prod_{j=1}^{n} (x - w^j y),\n$$\n\nwhere $w$ is a primitive $n$-th root of unity. Thus, after relabeling if necessary, we may take $\\alpha_j = w^j$ for $1 \\leq j \\leq n$. Therefore,\n\n$$\nf(x) - g(y) = \\prod_{j=1}^{n} (x - w^j y + \\beta_j).\n$$\n\nDefine\n\n$$\na = \\frac{\\beta_1 - w\\beta_n}{w - 1}, \\quad b = \\frac{\\beta_1 - \\beta_n}{w - 1}.\n$$\n\nLet $F(x) = f(x + a)$ and $G(y) = g(y + b)$. Then\n\n$$\nF(x) - G(y) = (x - y)(x - wy) \\prod_{j=2}^{n-1} (x - w^j y + \\gamma_j),\n$$\n\nwhere $\\gamma_j = \\beta_j + a - w^j b$ for $2 \\leq j \\leq n-1$. Setting $y = x$ gives $F(x) = G(x)$. Setting $x = wy$ gives $F(wy) = G(y) = F(y)$. Thus, if $\\alpha$ is a root of $F(x) = 0$, so is $w\\alpha$, and so on. Therefore,\n\n$$\nF(y) = G(y) = (y - \\alpha)(y - w\\alpha) \\cdots (y - w^{n-1}\\alpha) = y^n - \\alpha^n.\n$$\n\nThus,\n\n$$\nf(x) = F(x - a) = (x - a)^n - \\alpha^n, \\quad g(y) = (y - b)^n - \\alpha^n.\n$$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21008,
"subject": "Mathematics (Olympiad)",
"question": "The circumcircle of a square $ABCD$ has radius $10$. A semicircle is drawn on $AB$ outside the square. Find the area of the region inside the semicircle but outside the circumcircle.",
"options": [],
"answer": "See solution",
"solution": "Let $O$ be the centre of the circumcircle.\n\n\n\nSince $OA = OB = OC = OD$ and $AB = BC = CD = DA$, triangles $AOB$, $BOC$, $COD$, $DOA$ are isosceles and congruent. So $\\angle AOB = 360/4 = 90^\\circ$. Hence the area of $\\triangle AOB$ is $\\frac{1}{2} \\times 10 \\times 10 = 50$ and the area of the sector $AOB$ is $\\frac{1}{4}\\pi \\times 100 = 25\\pi$.\n\nBy Pythagoras, $AB^2 = AO^2 + OB^2 = 200$. Hence the area of the semicircle on $AB$ is $\\frac{1}{2}\\pi (AB/2)^2 = \\frac{AB^2 \\pi}{8} = 25\\pi$.\n\nSo the required area is $25\\pi - (25\\pi - 50) = 50$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21009,
"subject": "Mathematics (Olympiad)",
"question": "For each positive integer $n$, the bank of Cape Town issues coins of denomination $\\frac{1}{n}$. Given a finite collection of such coins (not necessarily of different denominations) with total value at most $99 + \\frac{1}{2}$, prove that it is possible to split this collection into 100 or fewer groups, such that each group has total value at most 1.",
"options": [],
"answer": "See solution",
"solution": "We shall prove a general result: for any positive integer $N$, given a finite collection of such coins with a 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 a total value of at most 1.\n\nIf some coins have total value $1/k$ (for some positive integer $k$), we replace these coins by one coin of value $1/k$, which does not affect the problem. In this way, for each even integer $k$, at most one coin has value $1/k$ (otherwise, two such coins may be replaced by one coin of value $2/k$); for each odd integer $k$, at most $(k-1)$ coins of value $1/k$ (otherwise, $k$ such coins can be replaced by one coin of value 1). So, we may suppose that no more replacements can be made for the coins.\n\nFirst, we take each coin of value 1 as a group. Suppose there are $d < N$ such groups. If there are no other coins, then the problem is solved. Otherwise, take a coin of value $1/2$ as a group ($d+1$) if there is any. Let $m = N - d \\ge 1$. Then for each integer $k$ in $2, \\dots, m$, take coins of value $1/(2k-1)$ and $1/(2k)$ in group ($d+k$) if there are any, in which the total value does not exceed $\\frac{2k-2}{2k-1} + \\frac{1}{2k} < 1$. For coins of value less than $1/(2m)$, if there are any, we can put them in some group $(d+j)$ such that the total value is less than 1 (since if each group $(d+j)$ has value greater than $1 - 1/(2m)$, then the total value will be greater than $d + m(1 - 1/(2m)) = m + d - 1/2 = N - 1/2$). Repeating this procedure finitely many times, all coins are put in $N$ or fewer groups with each group of value at most one. $\\square$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21010,
"subject": "Mathematics (Olympiad)",
"question": "Let $p$ and $q$ be coprime positive integers. A $(p+q)$-element set of real numbers $a_1 < a_2 < \\cdots < a_{p+q}$ is called *balanced* if $a_1, a_2, \\ldots, a_p$ form an arithmetic sequence with common difference $q$, and $a_p, a_{p+1}, \\ldots, a_{p+q}$ form an arithmetic sequence with common difference $p$. \nDetermine the maximum number of balanced $(p+q)$-element sets, no two of which are disjoint.",
"options": [],
"answer": "See solution",
"solution": "The required maximum is $p + \\max(p, q)$. \nNotice that two balanced sets have a nonempty intersection if and only if one is the image of the other through a translation by a number of the form $kp + \\ell q$, where $k \\in \\{0, 1, \\ldots, q\\}$ and $\\ell \\in \\{0, \\ldots, p-1\\}$. Therefore, the required maximum coincides with the maximum number of distinct integers of the form $kp + \\ell q$, $k \\in \\{0, 1, \\ldots, q\\}$, $\\ell \\in \\{0, \\ldots, p-1\\}$, whose mutual differences have absolute values of the same form.\n\nWe shall prove that the latter maximum is $p + \\max(p, q)$. Let $r = \\max(p, q)$ and suppose, to the contrary, that there exist $p + r + 1$ such integers. Label these integers $k_i p + \\ell_i q$, $i = 1, 2, \\ldots, p + r + 1$, in lexicographic order: $k_i \\le k_{i+1}$, and $\\ell_i < \\ell_{i+1}$ whenever $k_i = k_{i+1}$.\n\nWe first show that there is an index $i$ such that $k_i < k_{i+1}$ and $\\ell_i > \\ell_{i+1}$. To this end, notice that there are at most $q$ inequalities $k_i < k_{i+1}$, so at least $p + r - q \\ge p$ equalities $k_i = k_{i+1}$, hence at least $p$ inequalities $\\ell_i < \\ell_{i+1}$. This is impossible if $k_i \\le k_{i+1}$ and $\\ell_i \\le \\ell_{i+1}$ for all $i$, since $0 \\le \\ell_i \\le p-1$.\n\nNow fix an index $i$ such that $k_i < k_{i+1}$ and $\\ell_i > \\ell_{i+1}$, and recall that $|(k_i p + \\ell_i q) - (k_{i+1} p + \\ell_{i+1} q)| = kp + \\ell q$ for some $k \\in \\{0, 1, \\ldots, q\\}$ and some $\\ell \\in \\{0, \\ldots, p-1\\}$.\n\nExplicitly, either $(k_i - k_{i+1})p + (\\ell_i - \\ell_{i+1})q = kp + \\ell q$ or $(k_{i+1} - k_i)p + (\\ell_{i+1} - \\ell_i)q = kp + \\ell q$. Since $p$ and $q$ are coprime, the former implies $\\ell_i - \\ell_{i+1} \\equiv \\ell \\pmod{p}$, and the latter, $k_{i+1} - k_i \\equiv k \\pmod{q}$. Taking into account ranges of values, the first congruence forces equality, which in turn leads to a contradiction: $0 > k_i - k_{i+1} = k \\ge 0$. In the second case we reach the same contradiction unless $k_{i+1} = q$ and $k_i = k = 0$. So\n\n$$\nk_j = \\begin{cases} 0 & \\text{if } j = 1, \\ldots, i, \\\\ q & \\text{if } j = i + 1, \\ldots, p + r. \\end{cases}\n$$\n\nHence there are at least $(p + r - 1)/2$ successive equalities $k_j = k_{j+1}$, so at least as many successive inequalities $\\ell_j < \\ell_{j+1}$ among the corresponding $\\ell$'s. Since the number of such inequalities among the $\\ell$'s does not exceed $p - 1$, it follows that $(p + r - 1)/2 \\le p - 1$, or $r + 1 \\le p$, which is impossible.\n\nConsequently, there are at most $p + \\max(p, q)$ integers having the desired property. The examples below show that this is indeed the maximum number of such integers. For more convenience, we exhibit the corresponding $k_j$ and $\\ell_j$: If $p < q$, the $k_j$ are $0, 1, 2, \\ldots, q - 1$, $\\underbrace{q, q, \\ldots, q}_{p}$, and the $\\ell_j$ are $\\underbrace{0, 0, \\ldots, 0}_{q + 1}, 1, 2, \\ldots, p - 1$; and if $p > q$, the $k_j$ are $\\underbrace{0, 0, \\ldots, 0}_{p}, \\underbrace{q, q, \\ldots, q}_{p}$, and the $\\ell_j$ are $0, 1, 2, \\ldots, p - 1, 0, 1, \\ldots, p - 1$.\n\nThe verifications are straightforward.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21011,
"subject": "Mathematics (Olympiad)",
"question": "An acute-angled triangle $ABC$ is given. The points $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, and $C_2$ lie on its sides such that\n\n$$\n\\overline{AA_1} = \\overline{A_1A_2} = \\overline{A_2B} = \\frac{1}{3}\\overline{AB},\n$$\n$$\n\\overline{BB_1} = \\overline{B_1B_2} = \\overline{B_2C} = \\frac{1}{3}\\overline{BC},\n$$\n$$\n\\overline{CC_1} = \\overline{C_1C_2} = \\overline{C_2A} = \\frac{1}{3}\\overline{CA}.\n$$\n\nLet $kA$, $kB$, and $kC$ be the circumscribed circles of triangles $AA_1C_2$, $BB_1A_2$, and $CC_1B_2$ respectively. Let $aB$ and $aC$ be the tangents to $kA$ at $A_1$ and $C_2$; $bC$ and $bA$ be the tangents to $kB$ at $B_1$ and $A_2$; and $cA$ and $cB$ be the tangents to $kC$ at $C_1$ and $B_2$. Prove that the perpendiculars drawn from the intersection of $aB$ and $bA$ to $AB$, from the intersection of $bC$ and $cB$ to $BC$, and from the intersection of $cA$ and $aC$ to $CA$ all intersect at one point.",
"options": [],
"answer": "See solution",
"solution": "Let us denote the intersections of $aB$, $bC$, and $cA$ with $bA$, $cB$, and $aC$ respectively by $A'$, $B'$, and $C'$. The triangle $AA_1C_2$ is similar to $ABC$, since they have a common angle and their sides are in a $1:3$ ratio. Let $O_A$ be the center of the circumscribed circle around triangle $AA_1C_2$. Then:\n\n$$\n\\angle O_A A_1 A = \\frac{1}{2}(180^\\circ - \\angle A O_A A_1) = 90^\\circ - \\angle AC_2 A_1 = 90^\\circ - \\gamma\n$$\n\nSince $aB$ is perpendicular to $O_AA_1$, it follows that the angle between $aB$ and $AB$ equals\n\n$$\n180^\\circ - 90^\\circ - (90^\\circ - \\gamma) = \\gamma.\n$$\n\nAnalogously, the angle between $bA$ and $AB$ equals $\\gamma$, so the triangle $A_1A_2A'$ is isosceles with base $A_1A_2$. Thus, the perpendicular to $AB$ through $C'$ passes through the midpoint of $A_1A_2$, which is also the midpoint of $AB$. Therefore, the perpendicular passes through the center of the circumscribed circle around triangle $ABC$. By symmetry, all three perpendiculars pass through the center of the circumscribed circle, so they concur at a single point.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21012,
"subject": "Mathematics (Olympiad)",
"question": "Дадена е функцијата $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ таква што\n\n$$\nf(x+1) + f(x-1) = \\sqrt{2} f(x).\n$$\n\nДокажи дека $f$ е периодична функција.",
"options": [],
"answer": "See solution",
"solution": "Од даденото равенство што го исполнува функцијата $f$ имаме:\n\n$$\nf(x+2) + f(x) = \\sqrt{2} f(x+1) = \\sqrt{2} (\\sqrt{2} f(x) - f(x-1)) = 2 f(x) - \\sqrt{2} f(x-1),\n$$\n\nодносно\n\n$$\nf(x+2) = f(x) - \\sqrt{2} f(x-1).\n$$\n\nПонатаму,\n\n$$\n\\begin{aligned}\nf(x+4) &= f(x+2) - \\sqrt{2} f(x+1) \\\\\n&= f(x) - \\sqrt{2} (f(x-1) + f(x+1)) \\\\\n&= f(x) - \\sqrt{2} \\cdot \\sqrt{2} f(x) \\\\\n&= f(x) - 2 f(x) \\\\\n&= -f(x)\n\\end{aligned}\n$$\n\nОд ова пак следува дека\n\n$$\nf(x+8) = -f(x+4) = -(-f(x)) = f(x).\n$$\n\nЗначи, функцијата $f$ е периодична со период $8$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21013,
"subject": "Mathematics (Olympiad)",
"question": "For a positive integer $k$, denote the maximum nonnegative integer $i$ such that $2^i$ divides $k$ by $v_2(k)$. Note that $v_2(kl) = v_2(k) + v_2(l)$ holds for any positive integers $k$ and $l$, and that $v_2\\left(\\frac{k}{2}\\right) = v_2(k) - 1$ holds for any positive even number $k$.\n\nHow many ordered pairs of integers $(s, t)$ with $1 \\leq s, t \\leq 40$ satisfy $v_2(s) \\neq v_2(t)$?",
"options": [],
"answer": "See solution",
"solution": "$$\\boxed{1064}$$\n\nWe first note that $v_2(n) \\leq 5$ for $1 \\leq n \\leq 40$ because $40 < 2^6$. For $k \\in \\{0, 1, \\dots, 5\\}$, the number $f(k)$ of integers $1 \\leq n \\leq 40$ such that $v_2(n) = k$ is $\\left\\lfloor \\frac{40}{2^k} \\right\\rfloor - \\left\\lfloor \\frac{40}{2^{k+1}} \\right\\rfloor$.\n\nSo:\n$$\nf(0) = 20, \\quad f(1) = 10, \\quad f(2) = 5, \\quad f(3) = 3, \\quad f(4) = 1, \\quad f(5) = 1.\n$$\n\nThe number of pairs $(s, t)$ with $v_2(s) = v_2(t)$ is $\\sum_{k=0}^5 f(k)^2 = 536$. The total number of pairs is $40^2 = 1600$, so the answer is $1600 - 536 = \\boxed{1064}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21014,
"subject": "Mathematics (Olympiad)",
"question": "Find the largest natural number $n$ for which $3^{2016} - 1$ is divisible by $2^n$.",
"options": [],
"answer": "See solution",
"solution": "We have\n$$\n3^{2016} - 1 = (3^{63} - 1)(3^{63} + 1)(3^{126} + 1)(3^{252} + 1)(3^{504} + 1)(3^{1008} + 1).\n$$\nNumbers $3^{126}$, $3^{252}$, $3^{504}$, and $3^{1008}$ are squares of odd numbers, hence congruent to $1$ modulo $8$. Thus, $3^{126} + 1$, $3^{252} + 1$, $3^{504} + 1$, and $3^{1008} + 1$ are congruent to $2$ modulo $8$. Consequently, these four factors are divisible by $2$ but not by $4$.\n\nAs $3^{62} \\equiv 1 \\pmod{8}$, we have $3^6 \\equiv 3 \\pmod{8}$. Hence $3^{63} - 1$ and $3^{63} + 1$ are congruent to $2$ and $4$ modulo $8$, respectively. The former thus is divisible by $2$ but not by $4$, and the latter is divisible by $4$ but not by $8$.\n\nPutting it all together, the exponent of $2$ in the product is $1 + 2 + 1 + 1 + 1 + 1 = 7$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21015,
"subject": "Mathematics (Olympiad)",
"question": "Consider a polygon with $m + n$ sides, where $m$ and $n$ are positive integers. Color $m$ of its vertices red and the remaining $n$ vertices blue. A side is assigned the number $2$ if both its end vertices are red, the number $\\frac{1}{2}$ if both its end vertices are blue, and the number $1$ otherwise. Let the product of these numbers be $P$. Find the largest possible value of $P$.",
"options": [],
"answer": "See solution",
"solution": "We first show that if two adjacent vertices have different colors, then swapping the colors of these vertices leaves $P$ unchanged. To see this, we only need to consider the four possible cases:\n\nRRBR, BRBB, RRBB, BRBR\n\nwhich change to RBRR, BBRB, RBRB, BBRR.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21016,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $k$ such that the product of the first $k$ primes increased by $1$ is a power of an integer (with an exponent greater than $1$).",
"options": [],
"answer": "See solution",
"solution": "Denote the first $n$ primes as\n\n$$\np_1 = 2 < p_2 = 3 < \\dots < p_n\n$$\n\nSuppose that $p_1p_2\\cdots p_n + 1 = x^k$ for some integers $x, k \\geq 2$. We can assume without loss of generality that $k$ is prime since $x^{kt} = (x^t)^k$. Obviously, $x$ has no prime factors not exceeding $p_n$, so $x > p_n$ and consequently $k < n < p_n$ is one of the first $n$ primes. Now $x^k \\equiv 1 \\pmod{k}$, which implies $x \\equiv 1 \\pmod{k}$, but then by the Lifting the Exponent Lemma,\n\n$$\nx^k \\equiv 1 \\pmod{k^2}.\n$$\n\nThis is a contradiction, as $x^k - 1 = p_1 \\cdots p_n$ is not divisible by the square of any prime. Thus, there does not exist any $k$ satisfying the problem.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21017,
"subject": "Mathematics (Olympiad)",
"question": "Suppose we have two scales, each with 4 weights: the left scale has weights of even values, and the right scale has weights of odd values. We repeatedly remove weights from the tilted side, and instead of stopping when the balance reaches equilibrium with some weights still left, we remove another weight from the left-side scale and continue removing weights from the tilted side until no weights remain. How many ways are there to remove all the weights, such that equilibrium is never reached with weights still remaining on the scales?",
"options": [],
"answer": "See solution",
"solution": "The total number of ways to remove all weights is $4! \\times 4! = 576$, corresponding to all possible orderings of removals from each scale. To find the desired number, subtract the cases where equilibrium is reached with weights still remaining. If equilibrium occurs, the number of weights left on each scale must be equal, and both must be even. The only possibility is 2 weights left on each scale. There are 6 such pairs:\n\n$$\n\\begin{aligned}\n22 + 26 &= 23 + 25, & 22 + 28 &= 23 + 27, & 24 + 26 &= 23 + 27, \\\\\n24 + 28 &= 23 + 29, & 24 + 28 &= 25 + 27, & 26 + 28 &= 25 + 29.\n\\end{aligned}\n$$\n\nFor each, there are $2^4 = 16$ ways to choose the order of removals before and after equilibrium, so $6 \\times 16 = 96$ cases to subtract. Thus, the answer is $576 - 96 = 480$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21018,
"subject": "Mathematics (Olympiad)",
"question": "A bookshelf contains $n$ volumes, labelled $1$ to $n$, in some order. The librarian wishes to put them in the correct order as follows: the librarian selects a volume that is too far to the right (say, the volume with label $k$), takes it out, and inserts it in the $k$-th position. For example, if the bookshelf contains the volumes $1$, $3$, $2$, $4$ in that order, the librarian could take out volume $2$ and place it in the second position. The books will then be in the correct order $1$, $2$, $3$, $4$.\n\n(a) Show that if this process is repeated, then, however the librarian makes the selections, all the volumes will eventually be in the correct order.\n\n(b) What is the largest number of steps that this process can take?",
"options": [],
"answer": "See solution",
"solution": "(a) If $t_k$ is the number of times that volume $k$ is selected, then $t_k \\leq 1 + (t_1 + t_2 + \\dots + t_{k-1})$. This is because volume $k$ must move to the right between selections, which means some volume was placed to its left. The only way that can happen is if a lower-numbered volume was selected. This leads to the bound $t_k \\leq 2^{k-1}$. Furthermore, $t_n = 0$ since the $n$th volume will never be too far to the right. Therefore, if $N$ is the total number of moves then\n\n$$\nN = t_1 + t_2 + \\dots + t_{n-1} \\leq 1 + 2 + \\dots + 2^{n-2} = 2^{n-1} - 1,\n$$\n\nand in particular the process terminates.\n\n(b) Conversely, $2^{n-1}-1$ moves are required for the configuration $(n, 1, 2, 3, \\ldots, n-1)$ if the librarian picks the rightmost eligible volume each time.\n\nThis can be proved by induction: if at a certain stage we are at $(x, n-k, n-k+1, \\ldots, n-1)$, then after $2^k-1$ moves, we will have moved to $(n-k, n-k+1, \\ldots, n-1, x)$ without touching any of the volumes further to the left. Indeed, after $2^{k-1}-1$ moves, we get to $(x, n-k+1, n-k+2, \\ldots, n-1, n-k)$, which becomes $(n-k, x, n-k+1, n-k+2, \\ldots, n-1)$ after 1 more move, and then $(n-k, n-k+1, \\ldots, n-1, x)$ after another $2^{k-1}-1$ moves. The result follows by taking $k=n-1$.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 21019,
"subject": "Mathematics (Olympiad)",
"question": "Determine all positive integers $n$ such that $\\frac{a^2 + n^2}{b^2 - n^2}$ is a positive integer for some positive integers $a$ and $b$.",
"options": [],
"answer": "See solution",
"solution": "The required numbers are all even positive integers. Indeed, if $n$ is even, let $a = n^2/2 - 1$ and $b = n^2/2 + 1$. Then:\n\n$$\n\\frac{a^2 + n^2}{b^2 - n^2} = \\frac{\\left(\\frac{n^4}{4} - n^2 + 1\\right) + n^2}{\\left(\\frac{n^4}{4} + n^2 + 1\\right) - n^2} = \\frac{\\frac{n^4}{4} + 1}{\\frac{n^4}{4} + 1} = 1.\n$$\n\nSuppose now that such $a$ and $b$ exist for some positive odd integer $n$. We may assume $\\gcd(n, a, b) = 1$. Note that $n^2 \\equiv 1 \\pmod{4}$. If $b$ is odd, then $b^2 - n^2$ is divisible by $4$, and so is $a^2 + n^2$. Since $n$ is odd, $a^2 + 1$ is divisible by $4$, which is impossible. Thus, $b$ must be even. Then $b^2 - n^2 \\equiv 3 \\pmod{4}$, so $b^2 - n^2$ has a prime factor $p \\equiv 3 \\pmod{4}$. Then $a^2 + n^2$ is divisible by $p$, and it follows that so are both $a$ and $n$, since $p \\equiv 3 \\pmod{4}$. On the other hand, since $n$ and $b^2 - n^2$ are both divisible by $p$, so is $b$. Consequently, $a$, $b$, and $n$ are all divisible by $p$, contradicting $\\gcd(n, a, b) = 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21020,
"subject": "Mathematics (Olympiad)",
"question": "Given an acute triangle $\\triangle ABC$ ($AC \\neq AB$) and let $(C)$ be its circumcircle. The excircle $(C_1)$ corresponding to vertex $A$, with center $I_a$, is tangent to side $BC$ at point $D$ and to the extensions of sides $AB$ and $AC$ at points $E$ and $Z$, respectively. Let $I$ and $L$ be the intersection points of circles $(C)$ and $(C_1)$, $H$ the orthocenter of triangle $EDZ$, and $N$ the midpoint of segment $EZ$. The line through $I_a$ parallel to $HL$ meets $HI$ at point $G$. Prove that the perpendicular line $(e)$ through $N$ to $BC$ and the parallel line $(\\delta)$ through $G$ to $IL$ meet each other on the line $HI_a$.",
"options": [],
"answer": "See solution",
"solution": "We have $(e) \\perp BC$ and $I_aD \\perp BC$, so $(e) \\parallel I_aD$. Let $T$, $S$ be the midpoints of segments $HI_a$, $HD$ respectively, and $Y$ the intersection of lines $HD$ and $EZ$. Then, $TS \\parallel I_aD$, $TS \\perp BC$, and $SY \\perp EZ$.\n\nThe Euler circle $(\\omega)$ of triangle $EDZ$ passes through points $N$, $Y$, $S$. Therefore, segment $SN$ is a diameter of $(\\omega)$. Thus, the center of $(\\omega)$, say $T'$, is the midpoint of $SN$.\n\nOn the other hand, the center of the Euler circle $(\\omega)$ is the midpoint $T$ of $HI_a$. So $T = T'$. Therefore, line $(e)$ passes through points $T$, $S$.\n\nThus, quadrilateral $HSI_aN$ is a parallelogram and its diagonals meet at $T$.\n\nConsider the inversion $I(I_a, I_aZ^2)$. As $I_aZ^2 = I_aA \\cdot I_aN$, we have $I(N) = A$. Similarly, if $M_1$, $M_2$ are the midpoints of $DE$, $DZ$ respectively, then $I(M_1) = B$ and $I(M_2) = C$.\n\nTherefore, the circumcircle $(C)$ of triangle $ABC$ is the image of circle $(\\omega)$ under inversion $I$, and the intersection points of the circles are invariant under this inversion. The circle of inversion passes through the intersection points of $(C)$ and $(\\omega)$, so the Euler circle $(\\omega)$ passes through $I$, $L$.\n\nAlso, consider the inversion $J(H, r^2)$ with\n\n$$\nr^2 = HX \\cdot HZ = HD \\cdot HY = HW \\cdot HE\n$$\n\nwhere $X$, $W$, $Y$ are the feet of the altitudes of triangle $EDZ$ on its sides. Then $J(Z) = X$, $J(D) = Y$, and $J(E) = W$. Therefore, the circumcircle $(C_1)$ of triangle $ABC$ is the image of $(\\omega)$ under $J$. Thus, the circle of inversion $J$ passes through $I$, $L$.\n\nWe conclude that $HI = HL$ and $HI_a \\perp IL$, and since $(\\delta) \\parallel IL$, we have $HI_a \\perp (\\delta)$.\n\nIf $R$ is the intersection of $(\\delta)$ and $HL$, then quadrilateral $HRI_aG$ is a parallelogram and its diagonals meet at $T$. So, the perpendicular line $(e)$ through $N$ to $BC$ and the parallel line $(\\delta)$ through $G$ to $IL$ meet each other on the line $HI_a$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21021,
"subject": "Mathematics (Olympiad)",
"question": "Suppose there are 6 problems and $n$ students in a contest. Each student solves either 4 or 5 problems, and for every pair of problems, more than $\\frac{2}{5}$ of the students have solved both problems. Prove that at least two students must have solved all 5 problems.",
"options": [],
"answer": "See solution",
"solution": "Let\n$$\nd_i = \\sum_{j \\neq i} p_{ij}, \\quad i = 1, 2, \\dots, 6.\n$$\nWe have $d_s = d_t = 5m + 1$ and $d_i = 5m$ otherwise. If we build up the 6-tuple $(d_1, d_2, \\dots, d_6)$ one contestant at a time, starting with $W$, we start with $(4, 4, 4, 4, 4, 0)$, and every subsequent contestant adds a permutation of $(3, 3, 3, 3, 0, 0)$. Thus\n$$\n(d_1, d_2, \\dots, d_6) \\equiv (1, 1, 1, 1, 1, 0) \\pmod{3},\n$$\ncontradicting the earlier conclusion that $d_s = d_t = 5m + 1$ and $d_i = 5m$ otherwise. Hence, there were at least two persons to solve five problems.\n\n**Fourth Solution:** Suppose, for contradiction, there exists a counterexample with minimal $n$. Add solutions until one student (the winner) solved 5 problems, and the rest solved 4. If $n \\le 2$, some pair is unsolved, so $n > 2$.\n\nIf every 4-tuple was solved by a non-winner, we can remove $\\binom{6}{4} = 15$ contestants, one for each 4-tuple. Each pair is then solved by 6 fewer contestants, but there are 15 fewer contestants overall. Since $\\frac{2}{5}15 = 6$, each pair is still solved by more than $\\frac{2}{5}$ of the contestants, but only one student solved 5 problems, contradicting minimality.\n\nConsider a multigraph with problems as vertices and edges for students solving both problems. Suppose no non-winner solved $A, B, C, D$; let $E, F$ be the other problems. Let $S$ be edges among $\\{A, B, C, D\\}$, $S'$ be $S$ plus $EF$, and $T$ be the rest. Non-winners solving $E$ and $F$ contribute 2 edges to $S'$, 4 to $T$; others contribute 3 to each. As shown, $p_{ij} = \\frac{2n+1}{5}$ for most pairs, and for one pair, $p_{i_0j_0} = \\frac{2n+6}{5}$. Since $T$ has 8 pairs and $S'$ has 7, $|T| - |S'| \\le \\frac{2n+1}{5} + 1$.\n\nIf the winner solves $E$ and $F$, she contributes 4 to $S'$, 6 to $T$. So students solving $E$ and $F$ contribute 2 more to $T$ than $S'$, others contribute equally. Since at least $\\frac{2n+1}{5}$ students solve $E$ and $F$, $|T| - |S'| \\ge 2 \\cdot \\frac{2n+1}{5}$. Thus $\\frac{2n+1}{5} + 1 \\ge 2 \\cdot \\frac{2n+1}{5}$, so $n \\le 2$, a contradiction.\n\nIf the winner does not solve both $E$ and $F$, she contributes 6 to $S'$, 4 to $T$. So $|T| - |S'| \\ge 2 \\cdot \\frac{2n+1}{5} - 2$. Thus $\\frac{2n+1}{5} + 1 \\ge 2 \\cdot \\frac{2n+1}{5} - 2$, so $n \\le 7$. If no non-winner solves a 4-tuple, the winner must solve it. There are at most 6 non-winners (6 4-tuples), and the winner solves 5, but there are 15 total 4-tuples—a contradiction.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21022,
"subject": "Mathematics (Olympiad)",
"question": "For six integers $a, b, c$ and $A, B, C$, the following relations are true:\n\n$$\nb + c = A^2, \\quad c + a = B^2, \\quad a + b = C^2, \\quad C > B > A \\geq 0.\n$$\n\nFind numbers $a, b, c$ for which the sum $A^2 + B^2 + C^2$ takes the smallest possible value.\n",
"options": [],
"answer": "See solution",
"solution": "**Answer:** $a = 5$, $b = 4$, $c = -4$.\n\n**Solution.** Let us solve the given system of equations for numbers $a, b, c$. From the first two equations we obtain: $a - b = B^2 - A^2$. Adding this to the third equation, we find:\n\n$$\na = \\frac{1}{2}(B^2 + C^2 - A^2).\n$$\n\nSimilarly, by symmetry, we have:\n\n$$\nb = \\frac{1}{2}(C^2 + A^2 - B^2), \\quad c = \\frac{1}{2}(A^2 + B^2 - C^2).\n$$\n\nThe smallest possible sum $A^2 + B^2 + C^2$ can be obtained for the three smallest distinct squares of integers: $A^2 = 0$, $B^2 = 1$, $C^2 = 4$. But for these values, $a, b, c$ are not all integers, for example:\n\n$a = \\frac{1}{2}(1 + 4 - 0) = \\frac{5}{2}$.\n\nThe next possible triple is $A^2 = 0$, $B^2 = 1$, $C^2 = 9$. For this triple, the values of $a, b, c$ are integers:\n\n$$\na = \\frac{1}{2}(1 + 9 - 0) = 5, \\quad b = \\frac{1}{2}(9 + 0 - 1) = 4, \\quad c = \\frac{1}{2}(0 + 1 - 9) = -4.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21023,
"subject": "Mathematics (Olympiad)",
"question": "Denote by $s(y)$ the sum of the digits of any positive integer $y$. Find all positive integers $X$ such that\n\n$$\nX + s(X)^2 = 2016.\n$$",
"options": [],
"answer": "See solution",
"solution": "We may assume $1 \\leq X \\leq 2016$, so $s(X) \\leq 28$. A straightforward approach is to check for each $n$ in $1,2,\\dots,28$ if $X = 2016 - n^2$ has $s(X) = n$. To get stronger constraints:\n\nIf $X < 1400$, then $s(X) \\leq 22$, so $X + s(X)^2 < 1884 < 2016$. Thus, $X \\geq 1400$ and $s(X)^2 \\leq 616$, so $s(X) \\leq 24$.\n\nSince $X \\equiv s(X) \\pmod{9}$, we have $X + s(X)^2 \\equiv X + X^2 \\equiv X(X+1) \\pmod{9}$. As $2016 \\equiv 0 \\pmod{9}$, $X$ or $X+1$ must be divisible by $9$, so $s(X) \\equiv X \\equiv 0$ or $8 \\pmod{9}$.\n\nWith $s(X) \\leq 24$, possible values for $s(X)$ are $8, 9, 17, 18$. For $5 \\leq n \\leq 10$, $1916 \\leq X \\leq 1991$ implies $s(X) \\geq 12$, ruling out $8$ and $9$.\n\nFor $n = 17$ and $n = 18$:\n\n| $n$ | $n^2$ | $X = 2016 - n^2$ | $s(X)$ |\n|---|---|---|---|\n| 17 | 289 | 1727 | 17 |\n| 18 | 324 | 1692 | 18 |\n\nThus, the solutions are $X = 1692$ and $X = 1727$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21024,
"subject": "Mathematics (Olympiad)",
"question": "Given the equations $a = c - b$ and $d = 2c - a = b + c$, show that $abcd$ is the area of a right triangle with integer side lengths, where $b$ and $c$ are positive integers and $c > b$.",
"options": [],
"answer": "See solution",
"solution": "$$\nabcd = (c-b) \\cdot b \\cdot c \\cdot (b+c) = bc(c^2-b^2) = \\frac{1}{2}(2bc)(c^2-b^2).\n$$\n\nTherefore, $abcd$ is the area of a right triangle with legs $2bc$ and $c^2 - b^2$.\n\nThe hypotenuse of that triangle equals\n$$\n\\sqrt{(2bc)^2 + (c^2 - b^2)^2} = \\sqrt{c^4 + b^4 + 2b^2c^2} = b^2 + c^2.\n$$\n\nAs $b$ and $c$ are positive integers and $c > b$ (since $c − b = a > 0$), the lengths of all three sides, $2bc$, $c^2 − b^2$, and $b^2 + c^2$ are positive integers.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21025,
"subject": "Mathematics (Olympiad)",
"question": "Let $AB\\Gamma$ be an acute-angled triangle with $AB < A\\Gamma$ and circumcenter $O$. The altitudes $B\\Delta$ and $\\Gamma E$ meet at point $H$. If $O_1$ is the circumcenter of triangle $BH\\Gamma$, prove that the quadrilateral $AHO_1O$ is a parallelogram.\n\n\nfigure 6",
"options": [],
"answer": "See solution",
"solution": "Since $O_1$ lies on the perpendicular bisector $OM$ of segment $B\\Gamma$ and $AH$, $OO_1$, it is enough to prove that $AH = OO_1$. Since $AH = 2OM$, it suffices to show $OM = MO_1$. The quadrilateral is cyclic, so $\\angle BH\\Gamma = 180^\\circ - \\angle A$. Moreover, $\\angle BO_1\\Gamma = 2\\angle A$. Therefore, the isosceles triangles $BO\\Gamma$ and $BO_1\\Gamma$ have all corresponding angles equal and share side $B\\Gamma$, so they are congruent. Thus, $B\\Gamma$ is the perpendicular bisector of $OO_1$, so $M$ is the midpoint of $O_1O$, and hence $OM = MO_1$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21026,
"subject": "Mathematics (Olympiad)",
"question": "Find all natural numbers $n$ such that the equation\n$$\nx^2 + y^2 + z^2 = nxyz\n$$\nhas solutions in positive integers.",
"options": [],
"answer": "See solution",
"solution": "For $n = 1$, one solution is $x = y = z = 3$, and for $n = 3$, one solution is $x = y = z = 1$.\n\nIf $n$ is even and there exists an integer solution, then the right-hand side of the equation is even. This is possible only if at least one of the numbers $x, y, z$ is even. Then the right-hand side is divisible by $4$. Since the remainders modulo $4$ of the squares of integers can only be $0$ or $1$, all numbers $x, y, z$ must be even. Let $x = 2a$, $y = 2b$, $z = 2c$. Then $a^2 + b^2 + c^2 = 2nabc$, so $(a, b, c)$ satisfy a similar equation with doubled $n$, so they must be even. Continuing this process reveals that $x, y, z$ must be divisible by arbitrarily large powers of $2$, which is impossible. Consequently, there are no solutions for even $n$.\n\nSuppose that for some odd $n > 3$ the equation has an integer solution $(x, y, z)$. The given equation is equivalent to\n$$\nz^2 - nxy \\cdot z + (x^2 + y^2) = 0\n$$\nLet $z'$ be the other root of this quadratic equation. Then $z' > 0$ since positive $nxy$ and $x^2 + y^2$ enable only positive solutions. By Vieta's formula, $z' = nxy - z$. On the other hand, assume without loss of generality that $z = \\max(x, y, z)$; then $x^2 \\le xz \\le xyz$ and $y^2 \\le yz \\le xyz$, whence $z^2 \\ge (n-2)xyz$ and $z \\ge (n-2)xy$. Therefore $z' \\le 2xy < (n-2)xy \\le z$, implying $x + y + z' < x + y + z$. Thus we can infinitely reduce the sum of the components of the solution, which is impossible.\n\n**Remark.** Using the transformation $(x, y, z) \\rightarrow (y, z, nyz - x)$, it is easy to show that in cases $n = 1$ and $n = 3$ the given equation has infinitely many solutions.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21027,
"subject": "Mathematics (Olympiad)",
"question": "For a positive integer $n$, a cubic polynomial $p(x)$ is said to be $n$-good if there exist $n$ distinct integers $a_1, a_2, \\dots, a_n$ such that all the roots of the polynomial $p(x) + a_i = 0$ are integers for $1 \\leq i \\leq n$. Given a positive integer $n$, prove that there exists an $n$-good cubic polynomial.",
"options": [],
"answer": "See solution",
"solution": "Let $f(x) = x^3 - m x^2 + n x$, $k$ an integer, and $a, b, c$ the roots of $f(x) + k = 0$. Then\n\n$$\n4m^2 - 12n = 3(a - c)^2 + (a - 2b + c)^2.\n$$\n\nSince the equation $x^2 + 3y^2 = 1$ has a rational solution, it has infinitely many rational solutions. Therefore, one can find $D$ such that $4D$ can be expressed as $3s^2 + t^2$ in at least $6n$ ways. By choosing $m$ and $n$ such that $4m^2 - 12n = 4D$, we get an $n$-good cubic polynomial. $\\square$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21028,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, \\dots, a_n$ be real numbers satisfying\n\n$$\na_1 + \\dots + a_n = r \\left( \\frac{1}{a_1} + \\dots + \\frac{1}{a_n} \\right).\n$$\n\nShow that if $b_i = \\frac{r}{a_i}$ for $i = 1, \\dots, n$, then\n\n$$\nb_1 + \\dots + b_n = r \\left( \\frac{1}{b_1} + \\dots + \\frac{1}{b_n} \\right).\n$$",
"options": [],
"answer": "See solution",
"solution": "By condition, we have\n\n$$\n\\frac{1}{\\sqrt{r} - b_1} + \\dots + \\frac{1}{\\sqrt{r} - b_n} = \\frac{1}{\\sqrt{r}},\n$$\n\nwhich gives\n\n$$\n\\frac{1}{\\sqrt{r} - \\frac{r}{a_1}} + \\dots + \\frac{1}{\\sqrt{r} - \\frac{r}{a_n}} = \\frac{1}{\\sqrt{r}}.\n$$\n\nThe last equality is equivalent to\n\n$$\n\\frac{a_1}{\\sqrt{r} - a_1} + \\dots + \\frac{a_n}{\\sqrt{r} - a_n} = -1. \\quad (1)\n$$\n\nBy condition,\n\n$$\n\\frac{1}{\\sqrt{r} - a_1} + \\dots + \\frac{1}{\\sqrt{r} - a_n} = \\frac{1}{\\sqrt{r}}. \\quad (2)\n$$\n\nSubtracting (1) from (2) multiplied by $\\sqrt{r}$, we obtain\n\n$$\n\\frac{\\sqrt{r} - a_1}{\\sqrt{r} - a_1} + \\dots + \\frac{\\sqrt{r} - a_n}{\\sqrt{r} - a_n} = 1 - (-1) = 2.\n$$\n\nIt follows that $n=2$.\n\nIt remains to verify that the problem condition holds for $n = 2$. Indeed,\n\n$$\na_1 + a_2 = r \\left( \\frac{1}{a_1} + \\frac{1}{a_2} \\right) \\iff (a_1 + a_2) \\left( 1 - \\frac{r}{a_1 a_2} \\right) = 0.\n$$\n\nBy condition, $a_1, a_2 \\ge 0$; hence the last equality is equivalent to $a_1 a_2 = r$. Then\n\n$$\n\\frac{1}{\\sqrt{r} - a_1} + \\frac{1}{\\sqrt{r} - a_2} = \\frac{2\\sqrt{r} - (a_1 + a_2)}{r - \\sqrt{r}(a_1 + a_2) + a_1 a_2} = \\frac{2\\sqrt{r} - (a_1 + a_2)}{2r - \\sqrt{r}(a_1 + a_2)} = \\frac{1}{\\sqrt{r}},\n$$\n\nas required.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21029,
"subject": "Mathematics (Olympiad)",
"question": "Players A and B play the following game on a band of consecutive unit cells infinite in one direction. On each move, A marks two arbitrary cells that were not marked before. On each move, B deletes any block of consecutive marks. The goal of A is to obtain 10 consecutive marks; the goal of B is to impede him. Which one has a winning strategy?",
"options": [],
"answer": "See solution",
"solution": "Player A has a winning strategy.\n\nOn his first $2^7$ moves, A marks $2^8$ arbitrary cells so that the distance between any two of them is at least 10. These marks are not consecutive, so each time B deletes exactly one mark on his move. Thus, after $2^7$ combined moves, there are $2^7$ marks at distances at least 10. Denote this group of marks by $G$.\n\nOn each of his next $2^6$ moves, A marks cells to the left of two different cells from $G$, thus forming blocks of marks of length 2. Since B can delete at most one such block at a time, after $2^6$ combined moves, at least $2^6$ blocks of marks of length 2 remain.\n\nSimilarly, A starts forming pairs of blocks of length 3 on each of his next $2^5$ moves. Again, B can delete at most one such block on each move, so after $2^5$ combined moves, at least $2^5$ blocks of marks of length 3 remain.\n\nProceeding analogously, A can ensure that after some move of B, there will be $2^4$ blocks of length 4, then $2^3$ blocks of length 5, $2^2$ blocks of length 6, $2^1 = 2$ blocks of length 7, and finally 1 block $K$ of length 8. Now it is A's turn to move, and he wins by marking the two cells to the left of $K$, forming a block of 10 consecutive marks.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21030,
"subject": "Mathematics (Olympiad)",
"question": "Даден е правилен шестаголник со страна 1. Во внатрешноста на шестаголникот се дадени $m$ точки така што ниедни три од нив не се колинеарни. Шестаголникот е разделен на триаголници, при што секоја од дадените $m$ точки и секое од темињата на шестаголникот е теме на делбен триаголник. Делбените триаголници немаат заедничка внатрешна точка. Докажи дека постои делбен триаголник чија плоштина не е поголема од $\\frac{3\\sqrt{3}}{4(m+2)}$.",
"options": [],
"answer": "See solution",
"solution": "Најпрво го определуваме вкупниот број на делбени триаголници на кои е поделен дадениот шестаголник. Нека $A$ е произволна точка од внатрешните $m$ точки. Збирот од сите агли во точката $A$ е $360^\\circ$ (збир од сите агли во $A$ на сите триаголници кои таа точка ја имаат за свое теме). Од друга страна, збирот од сите агли во теме на шестаголникот е $120^\\circ$. Бидејќи збирот на аглите во секој триаголник е $180^\\circ$, бројот на делбени триаголници е:\n\n$$\n\\frac{m \\cdot 360^\\circ + 6 \\cdot 120^\\circ}{180^\\circ} = 2m + 4.\n$$\n\nНека претпоставиме спротивно на тврдењето, односно дека плоштината на секој од дадените делбени триаголници е поголема од $\\frac{3\\sqrt{3}}{4(m+2)}$. Тогаш збирот на плоштините на сите делбени триаголници е поголем од $(2m+4)\\frac{3\\sqrt{3}}{4(m+2)} = \\frac{3\\sqrt{3}}{2}$, што не е можно бидејќи плоштината на дадениот шестаголник е $\\frac{3\\sqrt{3}}{2}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21031,
"subject": "Mathematics (Olympiad)",
"question": "設 $\\triangle DEF$ 為 $\\triangle ABC$ 的內接正三角形，$D_1, D_2, D_3$ 分別為 $BD, DC$ 和 $BC$ 的中點。$E_1, E_2, E_3$ 分別為 $CE, EA$ 和 $CA$ 的中點，$F_1, F_2, F_3$ 分別為 $AF, FB$ 和 $AB$ 的中點。在 $\\triangle ABC$ 外部取三點 $P, Q, R$，使得 $\\triangle PD_1D_2, \\triangle QE_1E_2, \\triangle RF_1F_2$ 皆為正三角形。設 $\\triangle PDD_3, \\triangle QEE_3, \\triangle RFF_3$ 的重心分別為 $M_1, M_2, M_3$。試證：$\\triangle M_1M_2M_3$ 為正三角形。\n\n",
"options": [],
"answer": "See solution",
"solution": "將題目中的字母視為複數，則\n\n$$\nM_1 = \\frac{1}{3}(D_3 + D + P) = \\frac{1}{3}\\left(\\frac{B+C}{2} + D + P\\right)\n$$\n\n同理，\n\n$$\nM_2 = \\frac{1}{3}\\left(\\frac{C+A}{2} + E + Q\\right), \\quad M_3 = \\frac{1}{3}\\left(\\frac{A+B}{2} + F + R\\right)\n$$\n\n現在，考慮 $\\omega = \\cos \\frac{2\\pi}{3} + i \\sin \\frac{2\\pi}{3}$，注意到 $\\omega^3 = 1$ 且 $1+\\omega+\\omega^2=0$。由題設，$\\triangle DEF$ 為正三角形，故 $D+\\omega E+\\omega^2 F = 0$。\n\n又 $\\triangle PD_1D_2, \\triangle QE_1E_2, \\triangle RF_1F_2$ 皆為正三角形，故：\n\n$$\nP + \\omega D_2 + \\omega^2 D_1 = 0 \\implies P + \\omega \\frac{C+D}{2} + \\omega^2 \\frac{B+D}{2} = 0 \\quad (1)\n$$\n\n$$\nQ + \\omega E_2 + \\omega^2 E_1 = 0 \\implies Q + \\omega \\frac{A+E}{2} + \\omega^2 \\frac{C+E}{2} = 0 \\quad (2)\n$$\n\n$$\nR + \\omega F_2 + \\omega^2 F_1 = 0 \\implies R + \\omega \\frac{B+F}{2} + \\omega^2 \\frac{A+F}{2} = 0 \\quad (3)\n$$\n\n將 $(1) + (2) \\cdot \\omega + (3) \\cdot \\omega^2$，得：\n\n$$\n(P + \\omega Q + \\omega^2 R) + \\left( \\frac{B+C}{2} + \\frac{C+A}{2} \\omega + \\frac{A+B}{2} \\omega^2 \\right) = \\frac{1}{2}(D + \\omega E + \\omega^2 F) = 0\n$$\n\n從而：\n\n$$\n\\begin{aligned}\nM_1 + \\omega M_2 + \\omega^2 M_3 &= \\frac{1}{3}\\Big[(P + \\omega Q + \\omega^2 R) + \\left(\\frac{B+C}{2} + \\frac{C+A}{2} \\omega + \\frac{A+B}{2} \\omega^2\\right) \\\\ &\\quad + D + \\omega E + \\omega^2 F\\Big] \\\\ &= 0,\n\\end{aligned}\n$$\n\n故 $\\triangle M_1M_2M_3$ 為正三角形。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21032,
"subject": "Mathematics (Olympiad)",
"question": "Find the maximal natural even number, with all distinct digits, such that the difference between any two consecutive digits is at least 2.",
"options": [],
"answer": "See solution",
"solution": "We start by constructing the number with the largest possible digits, ensuring all digits are distinct and the difference between any two consecutive digits is at least 2. The sequence begins as $9, 7, 5, 8, 6, 4$. If the next digit is $2$, then the following one is $0$, and all even digits are used. Therefore, the next digit should be $1$. Thus, the answer is $9758641302$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21033,
"subject": "Mathematics (Olympiad)",
"question": "a) Show that if $a, b > 1$ are distinct real numbers, then\n\n$$\n\\log_a(\\log_a b) > \\log_b(\\log_a b).\n$$\n\nb) Let $a_1 > a_2 > \\dots > a_n > 1$ be real numbers, $n \\ge 2$. Prove that\n\n$$\n\\log_{a_1}\\big(\\log_{a_1}(a_2) + \\log_{a_2}(\\log_{a_2}(a_3) + \\dots + \\log_{a_{n-1}}(\\log_{a_{n-1}}(a_n) + \\log_{a_n}(\\log_{a_n}(a_1))\\big) > 0.\n$$",
"options": [],
"answer": "See solution",
"solution": "a) For $a < b$, $\\log_a(\\log_a b) = (\\log_a b)(\\log_b(\\log_a b)) > \\log_b(\\log_a b)$ because $\\log_b(\\log_a b) > 0$ and $\\log_a b > 1$.\n\nFor $a > b$, the claim follows since $\\log_a b < 1$ and $\\log_b(\\log_a b) < 0$.\n\nb) Induct on $n$. For $n = 2$,\n\n$$\n\\log_{a_1}(\\log_{a_1}(a_2)) + \\log_{a_2}(\\log_{a_2}(a_1)) > \\log_{a_2}(\\log_{a_1}(a_2)) + \\log_{a_2}(\\log_{a_2}(a_1)) = \\log_{a_2}(1) = 0.\n$$\n\nAssume the claim holds for some $n$ and consider $a_1 > a_2 > \\dots > a_{n+1} > 1$. Then\n\n$$\n\\begin{align*}\n& \\log_{a_1}\\big(\\log_{a_1}(a_2) + \\dots + \\log_{a_n}(\\log_{a_n}(a_{n+1}) + \\log_{a_{n+1}}(\\log_{a_{n+1}}(a_1))\\big) \\\\\n& = \\log_{a_1}\\big(\\log_{a_1}(a_2) + \\dots + \\log_{a_{n-1}}(\\log_{a_{n-1}}(a_n) + \\log_{a_n}(\\log_{a_n}(a_1))\\big) + \\dots \\\\\n& > \\log_{a_{n+1}}\\big(\\log_{a_n}(a_{n+1}) + \\log_{a_{n+1}}(\\log_{a_{n+1}}(a_1)) - \\log_{a_n}(\\log_{a_n}(a_1))\\big) \\\\\n& = \\log_{a_{n+1}}\\big(\\log_{a_n}(a_1) - \\log_{a_n}(\\log_{a_n}(a_1))\\big) > 0,\n\\end{align*}\n$$\n\nbecause $\\log_{a_n} a_1 > 1$ and $a_{n+1} < a_n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21034,
"subject": "Mathematics (Olympiad)",
"question": "令 $N$ 為一正整數。考慮一張 $N \\times N$ 的方格紙。\n\n一條**右下行路徑**是一系列的方格，其中每一個方格都在前一個方格的右方一格或下方一格。\n\n一條**右上行路徑**是一系列的方格，其中每一個方格都在前一個方格的右方一格或上方一格。\n\n證明：我們無法將 $N \\times N$ 的方格紙拆分成少於 $N$ 條的右下行和／或右上行路徑。\n\n下圖為一個將 $5 \\times 5$ 方格拆分成 5 條路徑的範例。\n\n",
"options": [],
"answer": "See solution",
"solution": "我們對 $N$ 進行數學歸納法。假設命題對 $N-1$ 成立。\n\n考慮最左上角的那一格所在的路徑 $P$。若 $P$ 是右上行路徑，則 $P$ 的所有格子都在最上方一橫行或最右方一直欄中。這意味著當我們移除最上方一橫行或最右方一直欄時，剩下的 $(N-1) \\times (N-1)$ 方格紙與其上的對應分拆仍符合題意，依據歸納假設至少被分為 $N-1$ 條路徑，從而原始的 $N \\times N$ 方格紙至少被分拆為 $N$ 條路徑。\n\n故僅需考慮 $P$ 為右下行路徑之情況。此處的關鍵觀察為：若 $P$ 包含最右下角的一格，則被 $P$ 分隔的兩區可以合併為一個 $(N-1) \\times (N-1)$ 的方格紙（如下圖所示），從而依據歸納假設知子方格紙上至少有 $N-1$ 條路徑，因此原 $N \\times N$ 方格紙上至少有 $N$ 條路徑。\n\n\n\n若 $P$ 為右下行路徑且不包含最右下角的格子時，考慮以下方式從最左上角到最右下角構造一條右下行路徑 $Q$。令 $Q_0 = P$。對於所有 $i \\ge 0$，考慮 $Q_i$ 最右下角的格子 $q_i$，令其右方與下方的格子為 $r_i$ 與 $d_i$，並令其所處的路徑分別為 $R_i$ 和 $D_i$。\n\n根據不同情況，將不同方格加入 $Q_i$，擴張成 $Q_{i+1}$（若同時符合多個情況則擇一進行）：\n\n(a) 若 $R_i$（或 $D_i$）為右下行路徑，則將 $R_i$（或 $D_i$）從 $r_i$（或 $d_i$）開始到終點的部分加入 $Q_i$；\n\n(b) 若 $R_i$ 為右上行路徑且**起點**為 $r_i$，則將 $R_i$ 中與 $r_i$ 同一橫行的格子加入 $Q_i$；\n\n(c) 若 $D_i$ 為右上行路徑且**終點**為 $d_i$，則將 $D_i$ 中與 $d_i$ 同一直欄的格子加入 $Q_i$；\n\n(d) 若以上皆不滿足，則必然有 $R_i = D_i$，此時將 $d_i$ 與其右邊一格加入 $Q_i$。\n\n重複以上操作直到無法再擴張為止。如此擴張出的 $Q$ 為一條從最左上角到最右下角的右下行路徑，因此被 $Q$ 分割的兩塊可以合併為一個 $(N-1) \\times (N-1)$ 方格紙。然而與原本 $P$ 的討論不同之處是，$Q$ 並非原始分拆中的一條路徑，因此 $Q$ 有可能將原本的某條路徑 $X$ 拆成兩段。\n\n以下說明拆成兩段是不可能發生的，從而我們可以沿用原始 $P$ 的討論方式。考慮 $X$ 與 $Q$ 的交點 $x$：\n\n- 若 $x$ 來自情境 (a)，則 $Q$ 將包含 $X$ 的後半條路徑，因此在移除 $Q$ 後，$X$ 仍為單一條右下行路徑；\n- 若 $x$ 來自情境 (b) 或 (c)，則 $Q$ 將包含 $X$ 的前半或後半條路徑，因此在移除 $Q$ 後，$X$ 仍為單一條右上行路徑；\n- 若 $x$ 來自情境 (d)，則由於 $X \\cap Q$ 恰為兩格，可知在移除 $Q$ 後，$X$ 被分隔成的兩段將剛好黏回成一條右上行路徑。\n\n因此上述拆成兩段的狀況不會發生，故依據歸納假設和與 $P$ 相同的討論方式，原 $N \\times N$ 方格紙必被分拆為至少 $N$ 條路徑。證畢。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21035,
"subject": "Mathematics (Olympiad)",
"question": "Let $A, B \\in \\mathcal{M}_3(\\mathbb{C})$ be two matrices such that $A^2 = B^2 = O_3$. Prove that if $AB = BA$, then $AB = O_3$. Show that the converse implication is false.",
"options": [],
"answer": "See solution",
"solution": "Assume $AB = BA$. We have:\n\n$$(A+B)^2 = A^2 + AB + BA + B^2 = AB + BA = 2AB$$\n\n$$(A+B)^3 = (A+B)(A+B)^2 = (A+B)(2AB) = 2A^2B + 2AB^2 = O_3$$\n\nAlso, $$(A+B)(A-B) = A^2 - B^2 = O_3$$\n\nBy Sylvester's rank inequality:\n\n$$\\operatorname{rank}(A+B) + \\operatorname{rank}(A-B) \\leq 3$$\n\nThus, $\\operatorname{rank}(A+B) \\leq 1$ or $\\operatorname{rank}(A-B) \\leq 1$.\n\nSuppose $\\operatorname{rank}(A+B) \\leq 1$. If $\\operatorname{rank}(A+B) = 0$, then $A+B = O_3$ and $AB = -A^2 = O_3$.\n\nIf $\\operatorname{rank}(A+B) = 1$, then $A+B = CD$ for some $C \\in \\mathcal{M}_{3,1}(\\mathbb{C})$, $D \\in \\mathcal{M}_{1,3}(\\mathbb{C})$. Then:\n\n$$(A+B)^2 = (CD)(CD) = C(DC)D = \\operatorname{Tr}(A+B)(A+B)$$\n\n$$(A+B)^3 = \\operatorname{Tr}(A+B)(A+B)^2 = O_3$$\n\nIf $\\operatorname{Tr}(A+B) \\neq 0$, then $(A+B)^2 = O_3$. If $\\operatorname{Tr}(A+B) = 0$, then $(A+B)^2 = O_3$ as well. Therefore, $AB = O_3$.\n\nThe case $\\operatorname{rank}(A-B) \\leq 1$ is analogous.\n\n**Counterexample for the converse:**\n\n$$\nA = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 1 \\\\ 0 & 0 & 0 \\end{pmatrix}, \\quad B = \\begin{pmatrix} 0 & 1 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}.\n$$\n\nWe have $A^2 = B^2 = AB = O_3$, but $AB \\neq BA$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21036,
"subject": "Mathematics (Olympiad)",
"question": "Each of the vertices of a regular $n$-gon is labeled with one of the signs $+$ or $-$. A *move* consists in choosing three consecutive vertices and changing the signs of these vertices: from $+$ to $-$ and from $-$ to $+$.\n\n1. Prove that if $n = 2015$, then, starting from any distribution of signs, by an appropriate succession of moves, one can get $+$ in all vertices.\n\n2. Prove that, if $n = 2016$, then there exists a choice of signs such that no succession of moves can turn all the signs into $+$.",
"options": [],
"answer": "See solution",
"solution": "a) The key is to obtain the \"atomic movement\", that is, a succession of moves whose final result is to change the sign of one single (arbitrary) vertex. Combining atomic movements, one can obtain any desired final configuration starting from any initial one. In particular, one can turn all the signs into $+$.\n\nFirst, let us change the signs in the 671 groups of three consecutive vertices $(1, 2, 3), (4, 5, 6), \\ldots, (2011, 2012, 2013)$. This provides a succession of moves that changes all the signs but two adjacent ones. With one additional move, we achieve the change of all the signs with one exception. Now we change all the signs, performing all the possible 2015 moves exactly once, so that each vertex changes sign three times. Combining all the moves made until now provides the \"atomic-movement\".\n\nAnother approach: label the vertices from 1 to $n$, then, for $k = \\lceil n/2 \\rceil$, successively change the signs in the group $(k, k+1, k+2)$ if the sign of $k$ is $-$. Thus we turn all of the first $n-2$ signs into $+$. If the remaining two signs are $+$, we are done. If they are $-$, a move turns them into $+$, another vertex turning into $-$. Finally, we reduce all the situations to one single $-$. From here we can continue as above.\n\nb) We color the vertices with 3 colors, periodically: red, green, blue, red, green, blue, etc. As 2016 is a multiple of 3, this coloring will 'close' well. At any move, the number of red $+$-es changes its parity, and so does the number of green $+$-es. It follows that the difference between the number of red $+$-es and the number of green ones does not change its parity. In the final configuration this difference should be 0. But if in the initial configuration this difference is odd, one can not get from that position to the desired one.\n\n*Remark.* In fact, one can prove that, in the case when $n$ is not a multiple of 3, any configuration can be obtained from any initial one.\n\nIn the case when $n$ is a multiple of 3, a given final position can be obtained from an initial one if and only if the parity of the number of red, green and blue $+$-es are respectively the same for the initial and the final configuration.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21037,
"subject": "Mathematics (Olympiad)",
"question": "If we cut a cylinder along a vertical line segment $PQ$ and unfold it, we obtain a rectangle with sides of length $4$ cm and $2\\pi$ cm. What is the shortest path between two opposite vertices of this rectangle?",
"options": [],
"answer": "See solution",
"solution": "The shortest path between two opposite vertices of the rectangle is the length of its diagonal:\n$$\n\\sqrt{(2\\pi)^2 + 4^2} = 2\\sqrt{\\pi^2 + 4}\\text{ cm}\n$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 21038,
"subject": "Mathematics (Olympiad)",
"question": "A parabola $y = ax^2 + bx + c$ passes through the points $A(-2, 1)$ and $B(2, 9)$, and does not intersect the $x$-axis. Find all possible values of the $x$-coordinate of the vertex of the parabola.",
"options": [],
"answer": "See solution",
"solution": "We first write the conditions for the parabola passing through the given points:\n\n$$\n\\begin{cases}\n4a - 2b + c = 1, \\\\\n4a + 2b + c = 9\n\\end{cases}\n$$\n\nSubtracting the first equation from the second gives $4b = 8 \\implies b = 2$. Substituting $b = 2$ into either equation, we get $4a + c = 5$.\n\nSince the parabola does not intersect the $x$-axis, its discriminant must be negative:\n\n$$\nD = b^2 - 4ac = 4 - 4a(5 - 4a) < 0\n$$\n\nExpanding and simplifying:\n\n$$\n4 - 20a + 16a^2 < 0 \\\\\n16a^2 - 20a + 4 < 0\n$$\n\nSolving the quadratic inequality, we find:\n\n$$\n\\frac{1}{4} < a < 1\n$$\n\nThe $x$-coordinate of the vertex is:\n\n$$\nx_v = -\\frac{b}{2a} = -\\frac{2}{2a} = -\\frac{1}{a}\n$$\n\nAs $a$ ranges from $\\frac{1}{4}$ to $1$, $x_v$ ranges from $-4$ to $-1$. Therefore,\n\n$$\nx_v \\in (-4, -1)\n$$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21039,
"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": "For $n=1$ and $n=2$.\n\nFor $n=1$, any even number $a$ of the form $m^2 + 1$ works, for example, $a = 2$.\n\nFor $n=2$, any even number $a$ of the form $m^2 - 1$ works, for example, $a = 8$.\n\nAssume that for $n=3$, such a number $a$ exists. Then the number $(a-1)(a^2-1)(a^3-1) = (a-1)^3(a+1)(a^2+a+1)$ must be a perfect square. Since $a^2+a+1 = a(a+1)+1$, the numbers $a+1$ and $a^2+a+1$ are coprime. As $a+1$ is odd, the numbers $a+1$ and $a-1$ are also coprime. Therefore, both $a+1$ and $(a-1)(a^2+a+1)$ must be perfect squares. In particular, $a+1$ modulo $3$ can only be $0$ or $1$, and thus $a-1$ is not divisible by $3$. Hence,\n\n$$\n\\begin{aligned}\n\\gcd(a-1, a^2+a+1) &= \\gcd(a-1, (a+2)(a-1)+3) \\\\\n&= \\gcd(a-1, 3) = 1,\n\\end{aligned}\n$$\n\nmeaning that both $a-1$ and $a^2+a+1$ must be perfect squares. However, the latter cannot be a square, since $a^2 < a^2+a+1 < (a+1)^2$. This is a contradiction.\n\nIt remains to prove that no such $a$ exists for $n \\ge 4$. Suppose such an $a$ exists. Take a natural number $k \\ge 2$ such that $2^k \\le n < 2^{k+1}$. Since $a^{2^k} - 1 = (a^{2^{k-1}} - 1)(a^{2^{k-1}} + 1)$, the number $(a-1)(a^2-1)\\dots(a^n-1)$ can be expressed as the product of $a^{2^{k-1}} + 1$ and several other factors of the form $a^m - 1$, where $1 \\le m \\le n$ and $m \\ne 2^k$.\n\nWe will show that the factor $a^{2^{k-1}} + 1$ is coprime with all other factors in this decomposition. Suppose $a^{2^{k-1}} + 1$ and $a^m - 1$ share a common divisor $d$. Then $\\gcd(a^{2^k} - 1, a^m - 1)$ is divisible by $d$. But $\\gcd(a^{2^k} - 1, a^m - 1) = a^{\\gcd(2^k,m)} - 1$. Since $m \\ne 2^k$ and $m \\le n < 2^{k+1}$, the number $m$ cannot be divisible by $2^k$. Thus, $\\gcd(2^k, m)$ is a power of two not exceeding $2^{k-1}$. Therefore, $a^{2^{k-1}-1}$ divides $\\gcd(a^{2^k} - 1, a^m - 1)$, and hence also divides $d$. Because $a$ is even, the numbers $a^{2^{k-1}-1}$ and $a^{2^{k-1}} + 1$ have no common divisors other than $1$, so $d = 1$, as required.\n\nThe factor $a^{2^{k-1}} + 1$ is coprime with all other factors in the product, which is a perfect square, so it must itself be a perfect square. Then $a^{2^{k-1}} + 1$ and $a^{2^{k-1}}$ are perfect squares differing by $1$, which is impossible. Therefore, our assumption is false, and no such $a$ exists for $n \\ge 4$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21040,
"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\nNote that every positive integer $m$ can be uniquely written as $m = ab^2$, $a, b$ as 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.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21041,
"subject": "Mathematics (Olympiad)",
"question": "There are three piles of tokens on the table: the first contains $a$, the second $b$, and the third $c$ tokens, with $a \\geq b \\geq c > 0$. Two players, $A$ and $B$, take turns moving the tokens. Player $A$ goes first. On each move, the active player chooses two piles and moves at least one token from the pile with fewer tokens to the pile with more tokens. If the chosen piles have equal numbers of tokens, the player moves at least one token from either pile to the other. A player wins if, after their move, only one pile remains. Who has the winning strategy for different values of $a$, $b$, and $c$?",
"options": [],
"answer": "See solution",
"solution": "If $b = c$, then player $B$ has the winning strategy; otherwise, player $A$ has the winning strategy.\n\n**Case 1:** $b = c$\n\nThe two smallest piles have equal tokens. Player $B$ can always maintain this equality after their turn, while player $A$ must break it. Specifically, $A$ must move at least one token from one of the smallest piles to another, making the smallest pile strictly smaller. $B$ can then choose the two largest piles and redistribute tokens so that the two smallest piles are equal again. The game ends when the two smallest piles are emptied, which can only happen after $B$'s move. Thus, $B$ always wins.\n\n**Case 2:** $b > c$\n\nPlayer $A$ should move $b - c > 0$ tokens from the pile with $b$ tokens to the pile with $a$ tokens. After this move, the two smallest piles become equal, and $A$ can use the strategy from Case 1 to win.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21042,
"subject": "Mathematics (Olympiad)",
"question": "A rectangular jigsaw puzzle has $20$ pieces along each of its longer sides and $15$ pieces along each of its shorter sides. How many edge pieces does the puzzle have?",
"options": [],
"answer": "See solution",
"solution": "Alternative i:\n\nCount the corner pieces as part of the top and bottom rows of the jigsaw and not as part of the sides. Then the number of edge pieces is $20 + 20 + 13 + 13 = 66$.\n\nAlternative ii:\n\nIf we count the number of pieces in each edge, then we will count each corner piece twice. So the number of edge pieces is $20 + 20 + 15 + 15 - 4 = 66$.\n\nAlternative iii:\n\nIf all the edge pieces are removed, then we are left with an $18 \\times 13$ rectangle which has $234$ pieces. So the number of edge pieces is $300 - 234 = 66$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21043,
"subject": "Mathematics (Olympiad)",
"question": "Peter has several equal squares with dimensions $4 \\times 4$. Each square is divided into sectors ($1 \\times 1$). He paints each of these sectors red or blue so that there are no similar patterns in all columns and all rows of all the squares. Rotating the squares is forbidden. How many squares can Peter paint in that way?",
"options": [],
"answer": "See solution",
"solution": "Altogether there are $2^4 = 16$ different patterns of painted columns. Since each square has 4 columns, in all there can not be more than $\\frac{2^4}{4} = 4$ differently painted squares.\n\n\n\nLet's prove that we can paint this number of squares. An example is shown in the table. Checking the columns is enough; if their patterns are different, the symmetry about the diagonal proves the difference of the row patterns.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21044,
"subject": "Mathematics (Olympiad)",
"question": "How many distinct straight lines in space pass through at least two of the 27 points that are the vertices of the 8 small cubes assembled to form a cube of edge length 2 (i.e., the $3 \\times 3 \\times 3$ grid of points)?",
"options": [],
"answer": "See solution",
"solution": "There are $3^3 = 27$ points in the $3 \\times 3 \\times 3$ grid. The total number of ways to choose 2 points is $\\binom{27}{2} = 351$, but some lines pass through 3 collinear points and are counted multiple times.\n\nLines through exactly 3 points fall into three categories:\n\n1. **Lines parallel to an edge:** There are $9$ such lines in each of the $3$ directions, so $9 \\times 3 = 27$ lines.\n2. **Lines parallel to a face diagonal:** Each face has $2$ diagonals, $3$ faces per direction, and $3$ directions, so $3 \\times 2 \\times 3 = 18$ lines.\n3. **Cube space diagonals:** There are $4$ such lines.\n\nEach of these $27 + 18 + 4 = 49$ lines is counted $3$ times, so we subtract $2$ for each:\n\n$$\n351 - 2 \\times 49 = 351 - 98 = 253\n$$\n\nThus, there are $253$ distinct lines passing through at least two of the $27$ points.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21045,
"subject": "Mathematics (Olympiad)",
"question": "In the triangle $ABC$, let $M$, $N$, and $P$ be the midpoints of the sides $BC$, $CA$, and $AB$, respectively, and let $G$ be the centroid of $ABC$. Let the circumcircle of $BGP$ intersect the line $MP$ at a point $K$ different from $P$, and let the circumcircle of $CGN$ intersect the line $MN$ at a point $L$ different from $N$. Prove that $|\\angle BAK| = |\\angle CAL|$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Obviously, $MP$ intersects the median $BN$ between points $B$ and $G$, so the point $K$ lies on the ray $PM$ and $BKGP$ is cyclic. Similarly, the point $L$ lies on the ray $NM$ and $CLGN$ is cyclic. Due to $MP \\parallel CA$ and $MN \\parallel BA$, we have\n\n$$\n|\\angle BPK| = |\\angle BPM| = |\\angle BAC| = |\\angle MNC| = |\\angle LNC|,\n$$\n\nwhile the two cyclic quadrilaterals imply\n\n$$\n|\\angle BKP| = |\\angle BGP| = |\\angle NGC| = |\\angle NLC|.\n$$\n\nWe see that triangles $\\textit{BPK}$ and $\\textit{CNL}$ are similar according to the condition $\\textit{AA}$. By the condition $\\textit{SAS}$, triangles $\\textit{ABK}$ and $\\textit{ACL}$ are also similar since\n\n(i) $|\\angle ABK| = |\\angle PBK| = |\\angle NCL| = |\\angle ACL|$,\n\n(ii) $\\dfrac{|AB|}{|BK|} = 2 \\cdot \\dfrac{|PB|}{|BK|} = 2 \\cdot \\dfrac{|NC|}{|CL|} = \\dfrac{|AC|}{|CL|}.$\n\nThus, the equality $|\\angle BAK| = |\\angle CAL|$ is proved.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21046,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $a \\leq b \\leq c$ for which the expression $2^a + 2^b + 2^c + 3$ is a perfect square.",
"options": [],
"answer": "See solution",
"solution": "If $a \\geq 2$, then $2^a + 2^b + 2^c + 3 \\equiv 3 \\pmod{4}$, so it cannot be a perfect square. Thus, $a = 1$.\n\nNow consider $2^b + 2^c + 5$ (since $a = 1$), which must be a perfect square.\n\nIf $b \\geq 3$, then $2^b + 2^c + 5 \\equiv 5 \\pmod{8}$, which cannot be a perfect square. So $b \\leq 2$.\n\nIf $b = 1$, then $2^c + 7$ must be a perfect square. For $c \\geq 2$,\n$$\n2^c + 7 \\equiv 3 \\pmod{4},\n$$\nwhich cannot be a perfect square. So $c = 1$. This gives the solution $a = b = c = 1$, for which $2^a + 2^b + 2^c + 3 = 9$.\n\nIf $b = 2$, then $2^c + 9$ must be a perfect square. Set $2^c + 9 = k^2$, so $2^c = (k-3)(k+3)$. Thus, $k-3 = 2^m$ and $k+3 = 2^n$ for some $m < n$, so $2^n - 2^m = 6$. This gives $m = 1$, $n = 3$, so $k = 5$ and $c = 4$. Therefore, another solution is $a = 1$, $b = 2$, $c = 4$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21047,
"subject": "Mathematics (Olympiad)",
"question": "The diagram shows an equilateral triangle divided into six identical smaller triangles. Two of these triangles are selected at random and shaded black. What is the probability that the resulting figure has an axis of symmetry?\n\n\n\n(A) $\\frac{1}{6}$\n\n(B) $\\frac{1}{3}$\n\n(C) $\\frac{2}{5}$\n\n(D) $\\frac{2}{3}$\n\n(E) $\\frac{3}{5}$",
"options": [],
"answer": "See solution",
"solution": "Suppose the first chosen triangle is:\n\n\n\nThen there are five possibilities for the second triangle:\n\n\n\nOnly the first, third, and fifth of these have an axis of symmetry, which is three of the five. This is true no matter which triangle is chosen first. The probability of the resulting figure having an axis of symmetry is thus $\\frac{3}{5}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21048,
"subject": "Mathematics (Olympiad)",
"question": "Во ромбот $ABCD$ е впишан круг. Произволна тангента $t$ на впишаниот круг ги сече страните $BC$ и $CD$ во внатрешни точки $M$ и $N$ соодветно. Докажи дека плоштината на триаголникот $AMN$ е константна.",
"options": [],
"answer": "See solution",
"solution": "Нека $O$ е центарот на кругот и нека кругот ги допира $BC$ и $CD$ во $E$ и $F$ соодветно. Нека $T$ е точката во која тангентата $MN$ го допира кругот. Бидејќи $\\overline{NF} = \\overline{NT}$ и $\\angle OFN = \\angle OTN$, триаголниците $\\Delta ONF$ и $\\Delta ONT$ се складни. Па $P_{\\Delta ONF} = P_{\\Delta ONT}$.\n\nСега $P_{\\Delta ANF} = 2P_{\\Delta ONF}$, бидејќи имаат иста основа $NF$, но висината на $\\Delta ANF$ кон $NF$ е дијаметарот на кругот, а висината на $\\Delta ONF$ кон $NF$ е $OF$, радиус на кругот. Според тоа $P_{\\Delta ANF} = P_{\\angle OTNF}$ и аналогно $P_{\\Delta AME} = P_{\\angle OTME}$.\n\nСега\n\n$$\nP_{\\Delta AMN} = P_{\\angle AEMNF} - P_{\\angle AEM} - P_{\\angle AFN} = P_{\\angle AEMNF} - P_{\\angle OTME} - P_{\\angle OTNF} = P_{\\angle EOF}.\n$$\n\nЗначи $P_{\\Delta AMN}$ не зависи од $MN$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21049,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_k = m^k$. We would like to represent $a_{x_0}$ as a sum of 2011 numbers of the form $a_{x_j}$. Starting from $a_{x_0}$, each time we choose a number $a_{k+1}$ and break it into $m$ numbers $a_k$, so that the sum of all numbers remains $a_{x_0}$.\n\nIn view of the base $m$ representation of the numbers, if we can write\n\n$$\na_{x_0} = a_{x_1} + a_{x_2} + \\cdots + a_{x_{2011}},$$\n\nthen we are able to obtain the numbers on the right by using the above operation repeatedly. Conversely, if we can obtain these numbers after some operations, then their sum is $a_{x_0}$ by construction. Therefore, it remains to check for which $m$ we can apply finitely many operations so that we can generate exactly 2011 numbers.\n\nNote that there are $m-1$ numbers more after applying each operation. Initially, there is only one number. Therefore, it is the same as finding those $m \\in \\mathbb{Z}^+$ such that $m - 1 \\mid 2011 - 1 = 2010$. For how many positive integers $m$ is this possible?",
"options": [],
"answer": "See solution",
"solution": "Since $m-1$ must divide $2010$, we count the positive divisors of $2010$. The prime factorization is $2010 = 2 \\times 3 \\times 5 \\times 67$, so the number of positive divisors is $2^4 = 16$. Thus, there are $16$ such positive integers $m$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21050,
"subject": "Mathematics (Olympiad)",
"question": "Consider the sequence $a_1, a_2, a_3, \\dots$ defined by $a_1 = 1$ and\n\n$$\na_n = n - \\lfloor\\sqrt{a_{n-1}}\\rfloor, \\quad \\text{for } n \\ge 2.\n$$\n\nDetermine the value of $a_{800}$.\n\n(Here, $\\lfloor x \\rfloor$ denotes the largest integer that is less than or equal to $x$.)",
"options": [],
"answer": "See solution",
"solution": "The first several terms of the sequence are $1, 1, 2, 3, 4, 4, 5, 6, 7, 8, 9, 9, 10, \\dots$. It appears that:\n\n- $a_n - a_{n-1} = 0$ or $1$ for all $n \\ge 2$, so the sequence is non-decreasing and contains every positive integer.\n- Every positive integer occurs exactly once, except perfect squares, which appear exactly twice.\n\n**Proof:**\n\nBy induction, $a_n - a_{n-1} = 0$ or $1$ for all $n \\ge 2$. For the inductive step:\n\n$$\n\\begin{aligned}\na_{n+1} - a_n &= (n + 1 - \\lfloor \\sqrt{a_n} \\rfloor) - (n - \\lfloor \\sqrt{a_{n-1}} \\rfloor) \\\\\n&= 1 + \\lfloor \\sqrt{a_{n-1}} \\rfloor - \\lfloor \\sqrt{a_n} \\rfloor \\\\\n&= \\begin{cases} 1, & \\text{if } a_n = a_{n-1}, \\\\ 1 + \\lfloor \\sqrt{a_n - 1} \\rfloor - \\lfloor \\sqrt{a_n} \\rfloor, & \\text{if } a_n = a_{n-1} + 1. \\end{cases}\n\\end{aligned}\n$$\n\nFor all positive integers $a$:\n\n$$\n[\\sqrt{a} - \\sqrt{a-1}]^2 = 2a - 1 - 2\\sqrt{a}\\sqrt{a-1} < 1\n$$\n\nSo $\\sqrt{a} - \\sqrt{a-1} < 1$, implying $0 \\le \\lfloor \\sqrt{a} \\rfloor - \\lfloor \\sqrt{a-1} \\rfloor \\le 1$. Thus, $a_{n+1} - a_n = 0$ or $1$.\n\nA term occurs twice when $a_{n+1} - a_n = 0$, which happens only if $a_n$ is a perfect square.\n\nThus, every positive integer appears once, except perfect squares, which appear twice.\n\nTo find $a_{800}$:\n\n- The only time $27^2 - 1 = 728$ appears is at term $728 + 26 = 754$.\n- Then $a_{755} = 729$, $a_{756} = 729$, $a_{757} = 730$, $a_{758} = 731$, ..., up to $a_{800} = 773$.\n- In this range, no term appears twice, since there is no square between $731$ and $773$.\n\n**Answer:**\n\n$$\na_{800} = 773\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21051,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ (where $\\mathbb{Z}^+$ is the set of positive integers) such that\n\n1. $f(n!) = f(n)!$ for all positive integers $n$,\n2. $m-n$ divides $f(m) - f(n)$ for all distinct positive integers $m, n$.",
"options": [],
"answer": "See solution",
"solution": "There are three solutions: the constant functions $f(n) = 1$ and $f(n) = 2$, and the identity function $f(n) = n$.\n\nLet us prove that these are the only ones.\n\nConsider such a function $f$ and suppose first that there exists $a > 2$ such that $f(a) = a$. Then $a!$, $(a!)!$, ... are all fixed points of $f$. So there is an increasing sequence $\\{a_n\\}_{n \\ge 0}$ of fixed points. If $n$ is any positive integer, $a_k - n$ divides $a_k - f(n) = f(a_k) - f(n)$ for all $k$, and so it also divides $f(n) - n$ for all $k$. Thus $f(n) = n$ for any $n$, finishing this case.\n\nNow suppose that $f$ has no fixed points greater than $2$. Let $p > 3$ be a prime and observe that $(p-2)! \\equiv 1 \\pmod{p}$ by Wilson's theorem, so $f(p-2)! - f(1) = f((p-2)!) - f(1)$ is a multiple of $p$. Clearly $f(1)$ is $1$ or $2$. As $p > 3$, the fact that $p$ divides $f(p-2)! - f(1)$ implies that $f(p-2) < p$. Since $(p-1)! - f(1)$ is not a multiple of $p$ (again by Wilson's theorem), we deduce that $f(p-2) \\leq p-2$. On the other hand, $p-3$ divides $f(p-2) - f(1) \\leq f(p-2) - 1$. Thus either $f(p-2) = f(1)$ or $f(p-2) = p-2$. As $p-2 > 2$, the last case is excluded and so $f(p-2) = f(1)$ for all primes $p > 3$. Taking $n$ to be any positive integer, we deduce that $p-2-n$ divides $f(1) - f(n)$ for all large primes $p$. Thus $f(n) = f(1)$ and $f$ is constant. The conclusion is now clear.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21052,
"subject": "Mathematics (Olympiad)",
"question": "Select $M = t a_1^{n-1}$ $n$-tuples\n\n$$\n(x_1, x_2, \\dots, x_{n-1}, f(x_1, \\dots, x_{n-1}) + 2^n s),\n$$\n\nwhere $x_1, \\dots, x_{n-1} \\in \\{1, 2, \\dots, a_1\\}$ and $s \\in \\{1, \\dots, t\\}$. Justify that these tuples satisfy the problem condition. For the above $n$-tuples and another $n$-tuple\n\n$$\n(x'_1, x'_2, \\dots, x'_{n-1}, f(x'_1, \\dots, x'_{n-1}) + 2^n s'),\n$$\n\nif the differences of their $k$th coordinates satisfy $\\equiv 0, \\pm 1 \\pmod{a_1}$ for any $k$, then\n\n$$\nx_i - x'_i \\equiv 0, \\pm 1 \\pmod{a_1}, \\quad i = 1, \\dots, n-1,\n$$\n\nand\n\n$$\n\\sum_{i=1}^{n-1} 2^{i-1}(x_i - x_i') + 2^n(s - s') \\equiv 0, \\pm 1 \\pmod{a_1}.\n$$\n\nShow that the only possibility is that the tuples are identical, and the problem conditions are met.\n\nFinally, consider $a_1, \\dots, a_n$ are all odd numbers. Show that if $a_2 > a_1$, it can be reduced to $a_1, a_2 - 2, a_3, \\dots, a_n$ and further to the case where all $a_i$ are identical by induction. Suppose for $a'_1 = a_1, a'_2 = a_2 - 2, a'_3 = a_3, \\dots, a'_n = a_n$ we have $M' = \\frac{1}{2^n}(a_1 - 1)(a_2 - 2)a_3 \\dots a_n$ $n$-tuples satisfying the problem conditions and their $k$th coordinates take values in $\\{0, 1, \\dots, a_k' - 1\\}$. Now define $M = \\frac{1}{2^n}(a_1 - 1)a_2 a_3 \\dots a_n$ $n$-tuples as follows: first, choose the $M'$ tuples already constructed, then:\n\n(i) For $x_2 = a_2 - 2$, choose $(x_1, a_2 - 2, x_3, \\dots, x_n)$ if and only if $(x_1, a_2 - 4, x_3, \\dots, x_n)$ was chosen.\n\n(ii) For $x_2 = a_2 - 1$, choose $(x_1, a_2 - 1, x_3, \\dots, x_n)$ if and only if $(x_1, a_2 - 3, x_3, \\dots, x_n)$ was chosen.\n\nShow that these $n$-tuples satisfy the conditions and complete the proof.",
"options": [],
"answer": "See solution",
"solution": "For (i), select $(x_1, a_2-2, x_3, \\dots, x_n)$ if and only if $(x_1, a_2-4, x_3, \\dots, x_n)$ was chosen. For (ii), select $(x_1, a_2-1, x_3, \\dots, x_n)$ if and only if $(x_1, a_2-3, x_3, \\dots, x_n)$ was chosen. Clearly, these $n$-tuples satisfy the conditions. From the previous inductive proof that $2^r \\mid (a_1-1)$ and equality holds, among the $M'$ $n$-tuples already constructed, there are $M' \\cdot \\frac{2}{a_2-2}$ with second coordinate $a_2-1$ or $a_2-2$. Therefore, the total number of $n$-tuples is $M' + M' \\cdot \\frac{2}{a_2-2} = M$, completing the proof. $\\boxed{}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21053,
"subject": "Mathematics (Olympiad)",
"question": "Given positive numbers $a_1, a_2, \\dots, a_n$ such that $n > 2$ and $a_1 + a_2 + \\dots + a_n = 1$, prove that the inequality\n\n$$\n\\frac{a_2 a_3 \\dots a_n}{a_1 + n - 2} + \\frac{a_1 a_3 \\dots a_n}{a_2 + n - 2} + \\frac{a_1 a_2 a_4 \\dots a_n}{a_3 + n - 2} + \\dots + \\frac{a_1 a_2 \\dots a_{n-1}}{a_n + n - 2} \\le \\frac{1}{(n-1)^2}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Suppose first $n \\ge 4$. Then we have\n\n$$\n\\frac{a_1 a_2 \\dots a_{k-1} a_{k+1} \\dots a_n}{a_k + n - 2} \\le \\frac{\\left( \\frac{a_1 + a_2 + \\dots + a_{k-1} + a_{k+1} + \\dots + a_n}{n-1} \\right)^{n-1}}{a_k + n - 2} < \\\\\n< \\frac{\\left( \\frac{a_1 + a_2 + \\dots + a_n}{n-1} \\right)^{n-1}}{n-2} = \\frac{1}{(n-2)(n-1)^{n-1}}\n$$\n\nBy adding all these inequalities, we get\n\n$$\n\\frac{a_2 a_3 \\dots a_n}{a_1 + n-2} + \\frac{a_1 a_3 \\dots a_n}{a_2 + n-2} + \\frac{a_1 a_2 a_4 \\dots a_n}{a_3 + n-2} + \\dots + \\frac{a_1 a_2 \\dots a_{n-1}}{a_n + n-2} \\le \\frac{n}{(n-2)(n-1)^{n-1}} \\le \\frac{n}{(n-2)(n-1)^3}\n$$\n\nSo we need to prove\n\n$$\n\\frac{n}{(n-2)(n-1)^3} \\le \\frac{1}{(n-1)^2},\n$$\n\nwhich simplifies to\n\n$$\nn \\le (n-2)(n-1) \\iff 0 \\le n^2 - 4n + 2 \\iff 0 \\le n(n-4) + 2\n$$\n\nwhich is true if $n \\ge 4$.\n\nFor $n=3$, let $a_1 = a$, $a_2 = b$, $a_3 = c$. We need to prove\n\n$$\n\\frac{bc}{1+a} + \\frac{ca}{1+b} + \\frac{ab}{1+c} \\le \\frac{1}{4}.\n$$\n\nWe have\n\n$$\n\\frac{bc}{1+a} = bc - \\frac{abc}{1+a}, \\quad \\frac{ac}{1+b} = ac - \\frac{abc}{1+b}, \\quad \\frac{ab}{1+c} = ab - \\frac{abc}{1+c}\n$$\n\nSo the inequality becomes\n\n$$\nbc + ca + ab \\le \\frac{1}{4} + abc \\left( \\frac{1}{a+1} + \\frac{1}{b+1} + \\frac{1}{c+1} \\right).\n$$\n\nBy the inequality of harmonic and arithmetic means,\n\n$$\n\\frac{1}{a+1} + \\frac{1}{b+1} + \\frac{1}{c+1} \\ge 3 \\cdot \\frac{3}{(a+1)+(b+1)+(c+1)} = \\frac{9}{4},\n$$\n\nso we need to prove\n\n$$\nbc + ca + ab \\le \\frac{1}{4} + abc \\cdot \\frac{9}{4}.\n$$\n\nSince the expression is symmetric in $a, b, c$, we may assume $c \\le \\frac{1}{3}$. Then\n\n$$\nbc + (1 - b - c)c + (1 - b - c)b \\le \\frac{1}{4} + \\frac{9}{4}(1 - b - c)bc\n$$\n\nThis is equivalent to\n\n$$\n\\begin{aligned}\n0 \\le 4b^2 - 9b^2c - 9bc^2 - 4b + 4c^2 - 4c + 1 &\\iff \\\\\n0 \\le (4 - 9c)b^2 - (9c^2 - 13c + 4)b + (4c^2 - 4c + 1)\n\\end{aligned}\n$$\n\nLet\n\n$$\np(x) = (4 - 9c)x^2 - (9c^2 - 13c + 4)x + (4c^2 - 4c + 1),\n$$\n\nSince $c \\le \\frac{1}{3}$, $4 - 9c \\ge 1$. The discriminant is\n\n$$\nD = (9c^2 - 13c + 4)^2 - 4(4 - 9c)(4c^2 - 4c + 1) = 81c \\left(c - \\frac{1}{3}\\right)^2 \\left(c - \\frac{4}{9}\\right)\n$$\n\nFor $0 < c \\le \\frac{1}{3}$ and $c - \\frac{4}{9} < 0$, we have $D \\le 0$, so the case $n = 3$ is proved.\n\nEquality holds for $a = b = c = \\frac{1}{3}$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21054,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ that for all real $x$ and $y$ satisfy the equation\n$$\nf(y^2 - f(x)) = y f(x)^2 + f(x^2 y + y).\n$$",
"options": [],
"answer": "See solution",
"solution": "The only such function is $f(x) = 0$.\n\nAssume $f(x) > 0$ for some $x \\in \\mathbb{R}$. Then we can choose $y$ such that\n$$\ny^2 - f(x) = x^2 y + y\n$$\n(since for $f(x) > 0$ this equation has two solutions for $y$). Substituting into the given equation yields $y f(x)^2 = 0$. As $f(x) > 0$, this forces $y = 0$, but $y = 0$ is not a solution of $y^2 - f(x) = x^2 y + y$—contradiction. Thus $f(x) \\le 0$ for all $x \\in \\mathbb{R}$.\n\nNote that $f(x) = 0$ is a solution. Suppose $f(x_0) < 0$ for some $x_0 \\in \\mathbb{R}$. First, we show that $f$ is unbounded. Assume the contrary and substitute $x_0$ into the equation:\n$$\nf(y^2 - f(x_0)) - f(x_0^2 y + y) = y f(x_0)^2,\n$$\nso if $f(x)$ is bounded, the left side is bounded, but the right side is unbounded—a contradiction.\n\nSetting $y = 0$ in the original equation gives $f(-f(x)) = f(0)$. Since $f(x)$ is unbounded and nonpositive, we can find arbitrarily large $y$ such that $f(y) = f(0)$. Now, put $x = x_0$ and choose $y_0$ such that $y_0 > \\frac{-f(0)}{f(x_0)^2}$ and $f(y_0(x_0^2 + 1)) = f(0)$. Then\n$$\nf(y_0^2 - f(x_0)) = y_0 f(x_0)^2 + f(y_0(x_0^2 + 1)) > -f(0) + f(0) = 0,\n$$\nwhich contradicts $f(x) \\le 0$ for all real $x$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21055,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, \\dots$ be a sequence of positive integers such that\n$$\na_{n+1} = \\begin{cases} \\frac{a_n}{2} & \\text{if } a_n \\text{ is even,} \\\\ 2^r + \\frac{a_{n+1}}{2} & \\text{if } a_n \\text{ is odd and } 2^{r-1} \\leq a_n < 2^r. \\end{cases}\n$$\nProve that no matter what the value of $a_1$ is, there exists an $N$ such that for all $n > N$, $a_n = a_{n+2}$.",
"options": [],
"answer": "See solution",
"solution": "Suppose $a_i$, when expressed in binary, is a number formed by appending some number of copies of $10$ at the beginning followed by a generic $k$-digit binary number. We show that continuing the recurrence from $a_i$ will eventually result in an $N$ for which $n > N$ implies $a_n = a_{n+2}$. We do so by strong induction on $k$.\n\nOur base cases will be $k = 0, 1, 2$. First, notice that if $a_i = \\overline{1010\\ldots1011}$, then $a_{i+1} = \\overline{101010\\ldots110}$ (with one extra $10$ at the beginning) and $a_{i+2} = \\overline{101010\\ldots11} = a_i$, so $N = i$ works. Now we manually check the other cases:\n\n$a_i = \\overline{1010\\ldots10101}$ : $a_{i+1} = \\overline{101010\\ldots1011}$, already checked.\n$a_i = \\overline{1010\\ldots1010}$ : $a_{i+1} = \\overline{1010\\ldots101}$, already checked.\n$a_i = \\overline{1010\\ldots10100}$ : $a_{i+1} = \\overline{1010\\ldots1010}$, already checked.\n$a_i = \\overline{1010\\ldots101000}$ : $a_{i+1} = \\overline{1010\\ldots10100}$, already checked.\n$a_i = \\overline{1010\\ldots101001}$ : $a_{i+1} = \\overline{101010\\ldots10101}$, already checked.\n\nNow for the inductive step. Assume $a_i$ has some copies of $10$ in its binary representation followed by an arbitrary $k$-digit number, and that we have shown our inductive hypothesis for all numbers less than $k$.\n\n*Case 1:* $a_i$ ends with a $0$. Then $a_{i+1}$ is the same number without that $0$, so we have $k-1$ digits following the copies of $10$ and we can apply the inductive hypothesis.\n\n*Case 2:* $a_i$ ends with a $1$ and the $k$-digit number is not all $1$s. Then $a_{i+1}$ is obtained by removing the $1$ at the end, adding $1$ to the number, and adding a $10$ to the beginning. Since the $k$-digit number is not all $1$s, adding $1$ to it will not mess up the last copy of $10$, so the result is a $(k-1)$-digit binary number preceded by copies of $10$, and we can apply the inductive hypothesis.\n\n*Case 3:* $a_i$ ends with $k$ $1$s. Then $a_{i+1}$ ends with $1100\\ldots0$, where there are $k-1$ $0$s, preceded by some copies of $10$. It can be easily computed that $a_{i+k}$ will be the same number without the last $k-1$ $0$s, so $a_{i+k}$ is a $2$-digit binary number preceded by copies of $10$, and our base case finishes this case.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21056,
"subject": "Mathematics (Olympiad)",
"question": "Reformulate the original expression as $x^2 - 1 = (p+1)^2(y^2 - 1)$. Find all integer solutions $(x, y)$ with $x, y > 1$ and $p$ a prime.",
"options": [],
"answer": "See solution",
"solution": "Let $x^2 - 1 = (p+1)^2(y^2 - 1)$. Then $x^2 - 1 = ((p+1)y)^2 - (p+1)^2$, so $((p+1)y)^2 - x^2 = (p+1)^2 - 1$. This gives $((p+1)y - x)((p+1)y + x) = p(p+2)$. Let $A = (p+1)y - x$ and $B = (p+1)y + x$, so $AB = p(p+2)$. Since $x, y > 1$, $B = (p+1)y + x \\\\ge 2p+4$. Also, $B > A$ and $B + A = 2(p+1)y$. Since $p$ is prime, it must divide $A$ or $B$.\n\n**Case I.** $p \\mid A$. Then $A = kp$ for some integer $k$, and $B = (p+2)/k$. But $B \\ge 2p+4 > p+2$, which is impossible.\n\n**Case II.** $p \\mid B$. Then $B = kp$ for some integer $k$, and $A = (p+2)/k$. Substitute $x = kp - (p+1)y$ into $AB = p(p+2)$:\n\n$$\n((p+2)y - kp)(y + k) = p + 2\n$$\n\nIf $y > k$, then $(p(y - k) + 2y)(y + k) > p + 2$, a contradiction. Thus $y \\le k$. Since $y + k \\le p + 2$, $2y \\le p + 2$. If $y - k \\le -2$, $p(y - k) + 2y < 0$, a contradiction. The only possibilities are $y - k = -1$ or $y - k = 0$.\n\nIf $k = y$: $4y^2 = p+2$. For $p = 2$, $y = 1$, which is rejected. For odd $p$, $p+2$ is odd, but $4y^2$ is even, so no solution.\n\nIf $k = y+1$: $(2y-p)(2y+1) = p+2$. This leads to $4y^2 + 2(1-p)y - (2p+2) = 0$, so\n\n$$\ny = \\frac{p+1}{2}\n$$\n\nfor odd $p$. Then $k = y + 1 = \\frac{p+3}{2}$ and\n\n$$\nx = kp - y = \\frac{p^2 + 2p - 1}{2}\n$$\n\nThus, the solutions are $$(x, y) = \\left(\\frac{p^2+2p-1}{2}, \\frac{p+1}{2}\\right)$$ for odd primes $p$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21057,
"subject": "Mathematics (Olympiad)",
"question": "Determine all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ such that $f(f(a) - b) + b f(2a)$ is a perfect square for all integers $a$ and $b$.",
"options": [],
"answer": "See solution",
"solution": "There are two families of functions which satisfy the condition:\n\n$$\n(1) \\quad f(n) = \\begin{cases} 0 & \\text{if } n \\text{ is even} \\\\ \\text{any perfect square} & \\text{if } n \\text{ is odd} \\end{cases}\n$$\n\n$$\n(2) \\quad f(n) = n^2, \\text{ for every integer } n.\n$$\n\nIt is straightforward to verify that the two families of functions are indeed solutions. Now, suppose that $f$ is any function which satisfies the condition that $f(f(a) - b) + b f(2a)$ is a perfect square for every pair $(a, b)$ of integers. We denote this condition by $(*)$. We will show that $f$ must belong to either Family (1) or Family (2).\n\n**Claim 1.** $f(0) = 0$ and $f(n)$ is a perfect square for every integer $n$.\n\n*Proof.* Plugging $(a, b) \\to (0, f(0))$ in $(*)$ shows that $f(0)(f(0) + 1) = z^2$ for some integer $z$. Thus, $(2f(0) + 1 - 2z)(2f(0) + 1 + 2z) = 1$. Therefore, $f(0)$ is either $-1$ or $0$.\n\nSuppose, for sake of contradiction, that $f(0) = -1$. For any integer $a$, plugging $(a, b) \\to (a, f(a))$ implies that $f(a)f(2a) - 1$ is a square. Thus, for each $a \\in \\mathbb{Z}$, there exists $x \\in \\mathbb{Z}$ such that $f(a)f(2a) = x^2 + 1$. This implies that any prime divisor of $f(a)$ is either $2$ or is congruent to $1 \\pmod{4}$, and that $4 \\nmid f(a)$, for every $a \\in \\mathbb{Z}$.\n\nPlugging $(a, b) \\to (0, 3)$ in $(*)$ shows that $f(-4) - 3$ is a square. Thus, there is $y \\in \\mathbb{Z}$ such that $f(-4) = y^2 + 3$. Since $4 \\nmid f(-4)$, we note that $f(-4)$ is a positive integer congruent to $3 \\pmod{4}$, but any prime dividing $f(-4)$ is either $2$ or is congruent to $1 \\pmod{4}$. This gives a contradiction. Therefore, $f(0)$ must be $0$.\n\nFor every integer $n$, plugging $(a, b) \\to (0, -n)$ in $(*)$ shows that $f(n)$ is a square.\n\nReplacing $b$ with $f(a) - b$, we find that for all integers $a$ and $b$,\n\n$$\nf(b) + (f(a) - b)f(2a) \\text{ is a square.} \\qquad (**)\n$$\n\nNow, let $S$ be the set of all integers $n$ such that $f(n) = 0$. We have two cases:\n\n*Case 1: $S$ is unbounded from above.*\n\nWe claim that $f(2n) = 0$ for any integer $n$. Fix some integer $n$, and let $k \\in S$ with $k > f(n)$. Then, plugging $(a, b) \\mapsto (n, k)$ in $(**)$ gives us that $f(k) + (f(n) - k)f(2n) = (f(n) - k)f(2n)$ is a square. But $f(n) - k < 0$ and $f(2n)$ is a square by Claim 1. This is possible only if $f(2n) = 0$. In summary, $f(n) = 0$ whenever $n$ is even and Claim 1 shows that $f(n)$ is a square whenever $n$ is odd.\n\n*Case 2: $S$ is bounded from above.*\n\nLet $T$ be the set of all integers $n$ such that $f(n) = n^2$. We show that $T$ is unbounded from above. In fact, we show that $\\frac{p+1}{2} \\in T$ for all primes $p$ big enough.\n\nFix a prime number $p$ big enough, and let $n = \\frac{p+1}{2}$. Plugging $(a, b) \\mapsto (n, 2n)$ in $(**)$ shows us that $f(2n)(f(n) - 2n + 1)$ is a square for any integer $n$. For $p$ big enough, we have $2n \\notin S$, so $f(2n)$ is a non-zero square. As a result, when $p$ is big enough, $f(n)$ and $f(n) - 2n + 1 = f(n) - p$ are both squares. Writing $f(n) = k^2$ and $f(n) - p = m^2$ for some $k, m \\ge 0$, we have\n\n$$\n(k + m)(k - m) = k^2 - m^2 = p \\implies k + m = p,\\ k - m = 1 \\implies k = n,\\ m = n - 1.\n$$\n\nThus, $f(n) = k^2 = n^2$, giving us $n = \\frac{p+1}{2} \\in T$.\n\nNext, for all $k \\in T$ and $n \\in \\mathbb{Z}$, plugging $(a, b) \\mapsto (n, k)$ in $(**)$ shows us that $k^2 + (f(n) - k)f(2n)$ is a square. But that means $(2k - f(2n))^2 - (f(2n)^2 - 4f(n)f(2n)) = 4(k^2 + (f(n) - k)f(2n))$ is also a square. When $k$ is large enough, we have $|f(2n)^2 - 4f(n)f(2n)| + 1 < |2k - f(2n)|$. As a result, we must have $f(2n)^2 = 4f(n)f(2n)$ and thus $f(2n) \\in \\{0, 4f(n)\\}$ for all integers $n$.\n\nFinally, we prove that $f(n) = n^2$ for all integers $n$. Fix $n$, and take $k \\in T$ big enough such that $2k \\notin S$. Then, we have $f(k) = k^2$ and $f(2k) = 4f(k) = 4k^2$. Plugging $(a, b) \\mapsto (k, n)$ to $(**)$ shows us that $f(n) + (k^2 - n)4k^2 = (2k^2 - n)^2 + (f(n) - n^2)$ is a square. Since $T$ is unbounded from above, we can take $k \\in T$ such that $2k \\notin S$ and also $|2k^2 - n| > |f(n) - n^2|$. This forces $f(n) = n^2$, giving us the second family of solutions.\n\n**Another approach of Case 1.**\n\n**Claim 2.** One of the following is true.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21058,
"subject": "Mathematics (Olympiad)",
"question": "$n \\ge 4$ real numbers are placed on a circle. It is known that for any four consecutive numbers $a, b, c, d$ (in this order around the circle), the condition $a + d = b + c$ holds. For which $n$ can we conclude that all the numbers are equal?\n\n",
"options": [],
"answer": "See solution",
"solution": "Let the numbers on the circle be $a_1, a_2, \\ldots, a_n$. Choose the largest among them (or any, if there are several). Without loss of generality, let this be $a_1$. From the problem condition:\n\n$$\na_1 + a_2 = a_n + a_3, \\qquad a_1 + a_n = a_2 + a_{n-1} \\implies 2a_1 = a_3 + a_{n-1}.\n$$\n\nSince $a_1$ is the largest, this implies $a_1 = a_3 = a_{n-1}$; that is, every second number is the largest. Continuing this reasoning, if $n$ is odd, all numbers must be equal.\n\nFor even $n$, consider the example $1, 0, 1, 0, \\ldots, 1, 0$ (not all equal), which satisfies the condition.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21059,
"subject": "Mathematics (Olympiad)",
"question": "Can every positive rational number $q$ be written as\n$$\n\\frac{a^{2021} + b^{2023}}{c^{2022} + d^{2024}},\n$$\nwhere $a, b, c, d$ are all positive integers?",
"options": [],
"answer": "See solution",
"solution": "Yes, every positive rational number $q$ can be written in this form. Let $q = \\frac{m}{n}$ in lowest terms. Set $a = x^{2023}$, $b = x^{2021}$, $c = y^{2024}$, and $d = y^{2022}$ for positive integers $x, y$. Then:\n$$\n\\frac{a^{2021} + b^{2023}}{c^{2022} + d^{2024}} = \\frac{x^{2021 \\times 2023} + x^{2021 \\times 2023}}{y^{2022 \\times 2024} + y^{2022 \\times 2024}} = \\frac{2x^{2021 \\times 2023}}{2y^{2022 \\times 2024}} = \\frac{x^{2021 \\times 2023}}{y^{2022 \\times 2024}}\n$$\nTo get $q = \\frac{m}{n}$, set $x = m^{x_1} n^{x_2}$ and $y = m^{y_1} n^{y_2}$, and solve:\n$$\n2021 \\times 2023 x_1 - 2022 \\times 2024 y_1 = 1, \\quad 2021 \\times 2023 x_2 - 2022 \\times 2024 y_2 = -1\n$$\nThese equations have integer solutions because $2021 \\times 2023$ and $2022 \\times 2024$ are coprime (since their factors differ by at most 3 and share no common prime factors). Thus, such $a, b, c, d$ exist for any positive rational $q$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21060,
"subject": "Mathematics (Olympiad)",
"question": "Given an acute, scalene triangle $ABC$ with circumcircle $(O)$. The line passing through $O$ and the midpoint $I$ of $BC$ intersects $AB$ and $AC$ at $E$ and $F$, respectively. Let $D$ and $G$ be the reflections of $A$ over $O$ and the circumcenter of triangle $AEF$. Let $K$ be the reflection of $O$ over the circumcenter of triangle $OBC$.\n\na) Prove that $D$, $G$, and $K$ are collinear.\n\nb) Take $M$ on $KB$ and $N$ on $KC$ such that $IM \\perp AC$, $IN \\perp AB$. The perpendicular bisector of $IK$ intersects $MN$ at $H$. Suppose that $IH$ meets $AB$ and $AC$ at $P$ and $Q$, respectively. Prove that the circumcircle of triangle $APQ$ intersects $(O)$ again at a point on $AI$.",
"options": [],
"answer": "See solution",
"solution": "a) Let $T$ be the projection of $A$ on $GD$. It is well known that $T$ is the second intersection of $(AEF)$ and $(O)$. We observe that $\\triangle TEB \\sim \\triangle TFC$, then\n\n$$\n\\frac{TB}{TC} = \\frac{BE}{CF}.\n$$\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21061,
"subject": "Mathematics (Olympiad)",
"question": "Let $m$ and $n$ be positive integers, and let $A_1, \\dots, A_m$ be pairwise disjoint $n$-element sets of positive integers such that no member of $A_i$ is divisible by one of $A_{i+1}$, for any $i$ (indices are reduced modulo $m$). Determine the largest number of ordered pairs $(a, b)$, where $a$ and $b$ are members of distinct $A_i$'s, and $b$ is divisible by $a$.",
"options": [],
"answer": "See solution",
"solution": "The required maximum is $\\binom{m-1}{2}n^2$, and is achieved if, for instance,\n\n$$\nA_k = \\{a^{(k-1)n+1}, a^{(k-1)n+2}, \\dots, a^{kn}\\}, \\quad k = 1, \\dots, m-1,\n$$\n\nand\n\n$$\nA_m = \\{b, b^2, \\dots, b^n\\},\n$$\n\nwhere *a* and *b* are relatively prime integers, both greater than 1.\n\nFor brevity, an ordered pair $(a, b)$ satisfying the conditions in the statement will be called *suitable*. We show that the number of suitable pairs does not exceed $\\binom{m-1}{2}n^2$.\n\nFor every $m$-tuple $(a_1, \\dots, a_m)$, where $a_k$ is a member of $A_k$, $k = 1, \\dots, m$, let $k(a_1, \\dots, a_m)$ be the number of suitable pairs $(a_i, a_j)$ it contains.\n\nWe show by induction on $m$ that $k(a_1, \\dots, a_m) \\le \\binom{m-1}{2}$. Since there are exactly $n^m$ $m$-tuples, and each suitable pair is contained in exactly $n^{m-2}$ such, the conclusion follows.\n\nThe case $m = 3$ is easily dealt with. Let $m \\ge 4$ and fix an $m$-tuple $(a_1, \\dots, a_m)$, $a_k \\in A_k$, $k = 1, \\dots, m$; without loss of generality, we may and will assume that $a_1$ is the largest entry. The $(m-1)$-tuple $(a_1, \\dots, a_{m-1})$ then satisfies the induction hypothesis: $a_2$ does not divide $a_1$, $a_3$ does not divide $a_2$, and so on, $a_{m-1}$ does not divide $a_{m-2}$, and, by maximality, $a_1$ does not divide $a_{m-1}$.\n\nWe show that the number of suitable pairs containing $a_m$ does not exceed $m-2$. It then follows that $k(a_1, \\dots, a_m) \\le k(a_1, \\dots, a_{m-1}) + m-2 \\le \\binom{m-2}{2} + m-2 = \\binom{m-1}{2}$, as desired.\n\nNotice that, for each $k$ in the range $1$ through $m-1$, at most one of the pairs $(a_k, a_m)$, $(a_m, a_k)$ is suitable. If neither $(a_k, a_m)$ nor $(a_m, a_k)$ is suitable for some $k$, then the number of suitable pairs containing $a_m$ is clearly at most $m-2$. Otherwise, since the pairs $(a_m, a_k)$ and $(a_{k+1}, a_m)$, $k = 1, \\dots, m-2$, are not simultaneously suitable, and the pair $(a_1, a_m)$ is certainly not suitable, by maximality of $a_1$, it follows that the pairs $(a_m, a_k)$, $k = 1, \\dots, m-1$, are all suitable, contradicting the fact that $a_m$ does not divide $a_{m-1}$. This ends the proof.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21062,
"subject": "Mathematics (Olympiad)",
"question": "A box contains 5 red pens, 6 blue pens, and 4 green pens. If three pens are drawn one after another without replacement, what is the probability that the first pen is red, the second is blue, and the third is green?",
"options": [],
"answer": "See solution",
"solution": "There are $5 + 6 + 4 = 15$ pens in the box. The probability of first picking a red pen is $\\frac{5}{15} = \\frac{1}{3}$. After removing one red pen, 14 pens remain. The probability of next picking a blue pen is $\\frac{6}{14} = \\frac{3}{7}$. Finally, the probability of picking a green pen is $\\frac{4}{13}$. Thus, the probability is:\n\n$$\n\\frac{1}{3} \\times \\frac{3}{7} \\times \\frac{4}{13} = \\frac{4}{91}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21063,
"subject": "Mathematics (Olympiad)",
"question": "Let triangle $ABC$ be inscribed in a circle $(O)$ and circumscribed about a circle $(I)$. $(I)$ is tangent to $BC$ at $D$. $AX \\perp BC$. $T$ is the reflection of $X$ across $AI$. $L$ is the point of contact of the excircle $(I_a)$ with $BC$. Prove that $\\triangle XTL \\sim \\triangle AIO$.",
"options": [],
"answer": "See solution",
"solution": "Draw the diameter $AA'$ of $(O)$. $A'I$ meets $(O)$ at $K$. $AI$ meets $(O)$ at $M$. $KM$ meets $AX$ at $N$. $J$ is the midpoint of $XT$.\n\nIt is known that $K$, $D$, $M$ are collinear and $INXD$ is a rectangle.\n\nIt follows that $DM \\parallel XI_a$, so $\\angle JLX = \\angle JI_aX = \\angle AMK = \\angle AA'I$.\n\nWe also have $\\angle JXL = \\angle JID = \\angle XAI = \\angle IAO$, so $\\triangle XJL \\sim \\triangle AIA'$.\n\nSince $O$, $J$ are the midpoints of $AA'$ and $XT$ respectively, we get\n$$\n\\triangle XTL \\sim \\triangle AIO.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21064,
"subject": "Mathematics (Olympiad)",
"question": "令 $a > 1$ 為正整數，$d > 1$ 為與 $a$ 互質的正整數。令 $x_1 = 1$，並對所有 $k \\ge 1$ 遞迴定義：\n\n$$\nx_{k+1} = \\begin{cases} x_k + d & \\text{若 } a \\text{ 不整除 } x_k, \\\\ x_k/a & \\text{若 } a \\text{ 整除 } x_k. \\end{cases}\n$$\n\n求最大的正整數 $n$（以 $a$ 和 $d$ 的函數表示），使得存在某個 $k$，滿足 $x_k$ 被 $a^n$ 整除。",
"options": [],
"answer": "See solution",
"solution": "$$n = \\max\\{m : a^m < ad\\}$$\n\n**解法一：** 由數歸知 $x_k$ 與 $d$ 互質。此外，注意到 $x_k$ 最多只有連續 $a-1$ 個遞增，因此由數歸知：\n\n$$\n\\begin{cases} x_k < ad & \\text{若 } x_k = x_{k-1} + d, \\\\ x_k < d & \\text{若 } x_k = x_{k-1}/a \\text{ 或 } k = 1. \\end{cases}\n$$\n\n這意味著 $a^n < ad$，即 $n \\le \\max\\{m : a^m < ad\\}$。\n\n以下證明此上界確實滿足題意。滿足上述條件同時意味著 $x_k$ 在某一項後必為循環數列。以 $a^{-k}$ 表示模 $d$ 下 $a^k$ 的乘法反元素，則 $x_k$ 必然包含 $1, a^{-1}, a^{-2}, \\dots, a^{-k}$（模 $d$），等價於包含 $1, a, a^2, \\dots$（模 $d$）。令 $x_\\ell$ 為第一個與 $a^n$ 同餘的項，則 $\\ell = 1$ 或 $x_\\ell = x_{\\ell-1}/a$（否則 $x_\\ell = x_{\\ell-1} + d \\Rightarrow x_{\\ell-1} \\equiv a^n$，矛盾）。不論是哪一種情況，依據上述不等式，都有 $x_\\ell < d < a^n < ad$，從而：\n\n$$\nx_\\ell \\in \\{a^n - d, a^n - 2d, \\dots, a^n - (a-1)d\\}\n$$\n\n而右邊集合中的任何一個數字都不被 $a$ 整除，故數列必在 $x_\\ell$ 之後的 $a-1$ 項內達到值 $a^n$。\n\n**解法二：** 同解法一，$x_k$ 與 $d$ 互質且 $x_k < ad$。令\n\n$$\nS = \\{x \\in \\mathbb{Z}_{>0} : 0 < x < ad, \\gcd(x, d) = 1\\}\n$$\n\n並考慮 $f: S \\to S$，\n\n$$\nf(x) = \\begin{cases} x + d & \\text{若 } a \\nmid x, \\\\ x/a & \\text{若 } a \\mid x. \\end{cases}\n$$\n\n則 $x_1 = 1$ 且 $x_{k+1} = f(x_k)$。\n\n證明以上遞迴可逆。假設對於某組 $x, y \\in S$ 有 $f(x) = y$：\n\n- 若 $y > d$，則必然有 $f(x) = x + d$，從而 $y - d = x \\in S$ 且 $a y \\notin S$；\n- 若 $y < d$，則必然有 $f(x) = x/a$，從而 $a y = x \\in S$ 且 $y - d \\notin S$。\n\n這表示 $f$ 是 $S$ 的重排，且其反函數為：\n\n$$\nf^{-1}(y) = \\begin{cases} y - d & \\text{若 } y > d, \\\\ a y & \\text{若 } y < d. \\end{cases}\n$$\n\n因為 $f$ 是 $S$ 的重排，其必然為周期函數，故其有無限多項為 $1$。取充分大的 $\\ell$ 滿足 $x_{\\ell} = 1$，則有：\n\n$$\nx_{\\ell} = 1, \\quad x_{\\ell-1} = f^{-1}(x_{\\ell}) = a, \\dots, \\quad x_{\\ell-n} = a^n.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21065,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c, d$ be non-negative real numbers such that\n$$\n\\frac{1}{a+1} + \\frac{1}{b+1} + \\frac{1}{c+1} + \\frac{1}{d+1} = 3.\n$$\nProve that\n$$\n3(ab + ac + ad + bc + bd + cd) + \\frac{4}{a + b + c + d} \\le 5.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $S = a + b + c + d$. By AM-HM (or Cauchy-Schwarz), we have\n$$\nS + 4 = (a + 1) + (b + 1) + (c + 1) + (d + 1) \\ge \\frac{16}{\\frac{1}{a+1} + \\frac{1}{b+1} + \\frac{1}{c+1} + \\frac{1}{d+1}} = \\frac{16}{3}\n$$\ngiving $S \\ge \\frac{4}{3}$.\n\nMultiplying the given equality by $(a+1)(b+1)(c+1)(d+1)$, we get\n$$\n\\sum abc + 2 \\sum ab + 3S + 4 = 3 \\left(abcd + \\sum abc + \\sum ab + S + 1\\right)\n$$\ngiving\n$$\n3abcd + 2 \\sum abc + \\sum ab = 1.\n$$\nIn particular, $ab + ac + ad + bc + bd + cd \\le 1$. So we may assume that $S < 2$, as otherwise the inequality is immediate.\n\nThe given equality transforms to\n$$\n\\frac{a}{a+1} + \\frac{b}{b+1} + \\frac{c}{c+1} + \\frac{d}{d+1} = 1,\n$$\nand so by Cauchy-Schwarz,\n$$\n\\sum a(a+1) \\sum \\frac{a}{a+1} \\ge S^2.\n$$\nThus,\n$$\nS^2 \\le a^2 + b^2 + c^2 + d^2 + S = S^2 - 2 \\sum ab + S.\n$$\nSo $\\sum ab \\le S/2$ and it is enough to prove that\n$$\n\\frac{3S}{2} + \\frac{4}{S} \\le 5.\n$$\nThis is equivalent to $3S^2 - 10S + 8 \\le 0$, which in turn is equivalent to $(S - 2)(3S - 4) \\le 0$. Since $S \\ge 4/3$ and we are also assuming that $S < 2$, then the inequality is true and the result follows.\n\n**Remark.** From the above solution, it follows that we have equality in the cases that $a = b = c = d = \\frac{4}{3}$ and in the case that two of the variables are equal to $1$ and the other two are equal to $0$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21066,
"subject": "Mathematics (Olympiad)",
"question": "A token is placed at each vertex of a regular $2n$-gon. A *move* consists in choosing an edge of the $2n$-gon and swapping the two tokens placed at the endpoints of that edge. After a finite number of moves have been performed, it turns out that every two tokens have been swapped exactly once. Prove that some edge has never been chosen.",
"options": [],
"answer": "See solution",
"solution": "**Step 1.** Enumerate all the tokens in the initial arrangement in clockwise circular order; also enumerate the vertices of the $2n$-gon accordingly. Consider any three tokens $i < j < k$. At each moment, their cyclic order may be either $i, j, k$ or $i, k, j$, counted clockwise. This order changes exactly when two of these three tokens have been switched. Hence the order has been reversed thrice, and in the final arrangement the token $k$ stands on the arc passing clockwise from the token $i$ to the token $j$. Thus, at the end, the token $i+1$ is a counter-clockwise neighbour of the token $i$ for all $i = 1, 2, \\dots, 2n-1$, so the tokens in the final arrangement are numbered successively in counter-clockwise circular order.\n\nThis means that the final arrangement of tokens can be obtained from the initial one by reflection in some line $\\ell$.\n\n**Step 2.** Notice that each token was involved in $2n-1$ switchings, so its initial and final vertices have different parity. Hence $\\ell$ passes through the midpoints of two opposite sides of the $2n$-gon; we may assume that these are the sides $a$ and $b$ connecting $2n$ with $1$ and $n$ with $n+1$, respectively.\n\nDuring the process, each token $x$ has crossed $\\ell$ at least once; thus one of its switchings has been made at edge $a$ or at edge $b$. Assume that some two of its switchings were performed at $a$ and at $b$; we may (and will) assume that the one at $a$ was earlier, and $x \\le n$. Then the total movement of token $x$ consisted at least of: (i) moving from vertex $x$ to $a$ and crossing $\\ell$ along $a$; (ii) moving from $a$ to $b$ and crossing $\\ell$ along $b$; (iii) coming to vertex $2n+1-x$. This takes at least $x+n+(n-x) = 2n$ switchings, which is impossible.\n\nThus, each token had a switching at exactly one of the edges $a$ and $b$.\n\n**Step 3.** Finally, let us show that either each token has been switched at $a$, or each token has been switched at $b$ (then the other edge has never been used, as desired). To the contrary, assume that there were switchings at both $a$ and at $b$. Consider the first such switchings, and let $x$ and $y$ be the tokens which were moved clockwise during these switchings and crossed $\\ell$ at $a$ and $b$, respectively. By Step 2, $x \\neq y$. Then tokens $x$ and $y$ initially were on opposite sides of $\\ell$.\n\nNow consider the switching of tokens $x$ and $y$; there was exactly one such switching, and we assume that it has been made on the same side of $\\ell$ as vertex $y$. Then this switching has been made after token $x$ had traced $a$. From this point on, token $x$ is on the clockwise arc from token $y$ to $b$, and it has no way to leave out from this arc. But this is impossible, since token $y$ should trace $b$ after that moment. A contradiction.\n\n**Remark.** The same holds for a $(2n - 1)$-gon. The problem was stated for a polygon with an even number of sides only to avoid case consideration.\n\nLet us outline the solution in the case of a $(2n - 1)$-gon. We prove the existence of line $l$ as in Step 1. This line passes through some vertex $x$, and through the midpoint of the opposite edge $a$. Then each token either passes through $x$, or crosses $l$ along $a$ (but not both; this can be shown as in Step 2). Finally, since a token is involved in an even number of moves, it passes through $x$ but not through $a$, and $a$ is never used.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21067,
"subject": "Mathematics (Olympiad)",
"question": "Let $n-1$, $n$, $n+1$, and $n+2$ denote four consecutive positive integers. The sum of their squares is $2p$.\n\nWhat can be said about the divisibility of $p-7$?",
"options": [],
"answer": "See solution",
"solution": "The sum of the squares is:\n\n$$\n2p = (n-1)^2 + n^2 + (n+1)^2 + (n+2)^2 = 4n^2 + 4n + 6,\n$$\nso $p = 2n^2 + 2n + 3 = 2n(n+1) + 3$.\n\nIf $n$ or $n+1$ is divisible by $3$, then $p$ is divisible by $3$ and thus not prime. So $n$ and $n+1$ are not divisible by $3$, which implies $n = 3k+1$ for some $k \\in \\mathbb{N}$.\n\nThen:\n$$\np = 2(3k+1)(3k+2) + 3 = 18k(k+1) + 7\n$$\nSo $p-7 = 18k(k+1)$ is divisible by $18$. Since at least one of $k$ or $k+1$ is even, $p-7$ is divisible by $36$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21068,
"subject": "Mathematics (Olympiad)",
"question": "Given set $A = \\{1, 2, m\\}$, where $m$ is real. Let $B = \\{a^2 \\mid a \\in A\\}$, and $C = A \\cup B$. If the sum of all the elements of $C$ is $6$, then the product of all the elements of $C$ is \\_\\_\\_\\_\\_.",
"options": [],
"answer": "See solution",
"solution": "By the condition, the possible elements of $C$ are $1$, $2$, $4$, $m$, and $m^2$ (allowing for repetition).\n\nNote that for real $m$, $1 + 2 + 4 + m + m^2 > 6$ and $1 + 2 + 4 + m^2 > 6$, so the only possibility is $C = \\{1, 2, 4, m\\}$, and $1 + 2 + 4 + m = 6$. Therefore, $m = -1$, which is consistent with the problem. The product of all the elements of $C$ is $1 \\times 2 \\times 4 \\times (-1) = -8$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21069,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs of integers $(p, q)$ such that $p$ divides $q^2 + 3$ and $q$ divides $p^4 + p^2 + 10q$.",
"options": [],
"answer": "See solution",
"solution": "If $p = q$, then the left-hand side is greater than the right-hand side, so we may assume $p \\neq q$. If $p > 3$, then $p \\mid q^2 + 3$ and $p \\equiv 1 \\pmod{3}$. Considering modulo $3$, we have\n\n$$\np(p^4 + p^2 + 10q) \\equiv q + 2 \\pmod{3}\n$$\n\nwhile $q(q^2 + 3) \\equiv q \\pmod{3}$, which is a contradiction. Trying $p = 2, 3$, we find that the only solution is $(p, q) = (2, 5)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21070,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\alpha \\in \\mathbb{Q}^+$. Determine all functions $f: \\mathbb{Q}^+ \\to \\mathbb{Q}^+$ such that\n\n$$\nf\\left(\\frac{x}{y} + y\\right) = \\frac{f(x)}{f(y)} + f(y) + \\alpha x\n$$\n\nholds for all $x, y \\in \\mathbb{Q}^+$.\n\nHere, $\\mathbb{Q}^+$ denotes the set of positive rational numbers.",
"options": [],
"answer": "See solution",
"solution": "Set $y = x$ and $y = 1$:\n\n$$\nf(x + 1) = 1 + f(x) + \\alpha x \\quad (1)\n$$\n\nand\n\n$$\nf(x + 1) = \\frac{f(x)}{f(1)} + f(1) + \\alpha x \\quad (2)\n$$\n\nEquating (1) and (2):\n\n$$\nf(x)\\left(1 - \\frac{1}{f(1)}\\right) = f(1) - 1.\n$$\n\nSince $f$ cannot be constant (by (1)), $f(1) = 1$. By induction,\n\n$$\nf(x) = \\frac{\\alpha}{2}x(x-1) + x \\quad \\text{for all } x \\in \\mathbb{Z}^+.\n$$\n\nIn particular, $f(2) = \\alpha + 2$ and $f(4) = 6\\alpha + 4$. Setting $x = 4$, $y = 2$ in the original equation gives\n\n$$\n\\alpha^2 - 2\\alpha = 0.\n$$\n\nThus $\\alpha = 2$ is required. For $\\alpha = 2$, $f(x) = x^2$ for $x \\in \\mathbb{Z}^+$. By induction, for $x \\in \\mathbb{Q}^+$ and $n \\in \\mathbb{Z}^+$, $f(x + n) = (x + n)^2$ implies $f(x) = x^2$.\n\nLet $\\frac{a}{b} \\in \\mathbb{Q}^+$ with $a, b \\in \\mathbb{Z}^+$. Set $x = a$, $y = b$:\n\n$$\nf\\left(\\frac{a}{b} + b\\right) = \\frac{a^2}{b^2} + b^2 + 2a = \\left(\\frac{a}{b} + b\\right)^2.\n$$\n\nThus $f\\left(\\frac{a}{b}\\right) = \\left(\\frac{a}{b}\\right)^2$. It is easily verified that $f(x) = x^2$ is a solution.\n\nTherefore, there is no solution for $\\alpha \\neq 2$, and the only solution is $f(x) = x^2$ for $\\alpha = 2$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21071,
"subject": "Mathematics (Olympiad)",
"question": "Let $C$ be the capital and $C_1, C_2, \\dots, C_n$ be the remaining cities. Denote by $d(x, y)$ the price of the connection between the cities $x$ and $y$, and let $\\sigma$ be the total price of a round trip going exactly once through each city.\n\nNow consider a round trip missing the capital and visiting every other city exactly once; let $s$ be the total price of that trip. Suppose $C_i$ and $C_j$ are two consecutive cities on the route. Replacing the flight $C_i \\to C_j$ by two flights: from $C_i$ to the capital and from the capital to $C_j$, we get a round trip through all cities, with total price $\\sigma$. It follows that $$\\sigma = s + d(C, C_i) + d(C, C_j) - d(C_i, C_j)$$ so it remains to show that the quantity $$\\alpha(i, j) = d(C, C_i) + d(C, C_j) - d(C_i, C_j)$$ is the same for all 2-element subsets $\\{i, j\\} \\subset \\{1, 2, \\dots, n\\}$.",
"options": [],
"answer": "See solution",
"solution": "For this purpose, note that $\\alpha(i, j) = \\alpha(i, k)$ whenever $i, j, k$ are three distinct indices; indeed, this equality is equivalent to $$d(C_j, C) + d(C, C_i) + d(C_i, C_k) = d(C_j, C_i) + d(C, C) + d(C, C_k)$$ which is true by considering any trip from $C_k$ to $C_j$ going through all cities except $C$ and $C_i$ exactly once and completing this trip to a round trip in two ways: $C_j \\to C \\to C_i \\to C_k$ and $C_j \\to C_i \\to C \\to C_k$. Therefore the values of $\\alpha$ coincide on any pair of 2-element sets sharing a common element. But then clearly $\\alpha(i, j) = \\alpha(i, j') = \\alpha(i', j')$ for all indices $i, j, i', j'$ with $i \\neq j, i' \\neq j'$, and the solution is complete.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21072,
"subject": "Mathematics (Olympiad)",
"question": "For any positive real numbers $x, y, z$ such that $x + y + z = 1$, prove that\n\n$$\n\\frac{(1 - x^2)(1 - y^2)(1 - z^2)}{x^2 y^2 z^2} \\ge 512.\n$$\n\nFurthermore, let\n\n$$\nx = \\frac{a}{\\sqrt{a^2 + 8bc}}, \\quad y = \\frac{b}{\\sqrt{b^2 + 8ca}}, \\quad z = \\frac{c}{\\sqrt{c^2 + 8ab}}.\n$$\n\nShow that $x + y + z \\ge 1$.",
"options": [],
"answer": "See solution",
"solution": "We start with the following lemma:\n\n**Lemma 1.** For any positive real numbers $x, y, z$ such that $x + y + z = 1$, the following inequality holds:\n\n$$\n\\frac{(1 - x^2)(1 - y^2)(1 - z^2)}{x^2 y^2 z^2} \\ge 512.\n$$\n\n*Proof.* Applying the AM-GM Inequality gives\n\n$$\n\\begin{aligned}\n1 - x^2 &= (1 - x)(1 + x) = (y + z)(x + y + z + x) \\\\\n&\\ge 2\\sqrt{yz} \\cdot 4\\sqrt[4]{x^2 y z} = 8\\sqrt[4]{x^2 y^3 z^3}.\n\\end{aligned}\n$$\n\nLikewise,\n\n$$\n1 - y^2 \\ge 8\\sqrt[4]{x^3 y^2 z^3} \\quad \\text{and} \\quad 1 - z^2 \\ge 8\\sqrt[4]{x^3 y^3 z^2}.\n$$\n\nMultiplying all three inequalities yields the desired result.\n\nNow set\n\n$$\nx = \\frac{a}{\\sqrt{a^2 + 8bc}}, \\quad y = \\frac{b}{\\sqrt{b^2 + 8ca}}, \\quad z = \\frac{c}{\\sqrt{c^2 + 8ab}}.\n$$\n\nThen\n\n$$\n\\frac{1 - x^2}{x^2} = \\frac{1 - \\frac{a^2}{a^2 + 8bc}}{\\frac{a^2}{a^2 + 8bc}} = \\frac{8bc}{a^2},\n$$\n\nso\n\n$$\n\\frac{(1 - x^2)(1 - y^2)(1 - z^2)}{x^2 y^2 z^2} = 512. \\quad (7)\n$$\n\nThe problem asks us to prove that $x + y + z \\ge 1$. Assume, for the sake of contradiction, that $x + y + z = \\frac{1}{k} < 1$. Then $k > 1$ and $(kx) + (ky) + (kz) = 1$, so by Lemma 1\n\n$$\n\\begin{aligned}\n\\frac{(1 - x^2)(1 - y^2)(1 - z^2)}{x^2 y^2 z^2} &> \\frac{(1 - (kx)^2)(1 - (ky)^2)(1 - (kz)^2)}{(kx)^2 (ky)^2 (kz)^2} \\\\\n&\\ge 512.\n\\end{aligned}\n$$\n\ncontradicting (7). Thus, $x + y + z \\ge 1$, as desired.\n\n**Note.** Lemma 1 can also be proved by applying the smoothing method. We can show that the value of\n\n$$\n\\frac{(1 - x^2)(1 - y^2)}{x^2 y^2}\n$$\n\ndecreases as $x$ and $y$ get closer together while $x + y$ remains fixed. This analysis holds for the pairs $y, z$ and $z, x$ as well. Now we can take the largest and smallest of $x, y, z$ and push them together until one becomes equal to $1/3$; then push the other two together until they both become equal to $1/3$. Thus,\n\n$$\n\\frac{(1 - x^2)(1 - y^2)(1 - z^2)}{x^2 y^2 z^2} \\ge \\left( \\frac{1 - \\left(\\frac{1}{3}\\right)^2}{\\left(\\frac{1}{3}\\right)^2} \\right)^3 = 512.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21073,
"subject": "Mathematics (Olympiad)",
"question": "The lines tangent to the circumcircle of triangle $ABC$ at points $B$ and $C$ intersect at point $D$. The circumcircle of triangle $BCD$ intersects the lines $AB$ and $AC$ a second time at points $K$ and $L$, respectively. Prove that the line $AD$ bisects the line segment $KL$.",
"options": [],
"answer": "See solution",
"solution": "We prove that $AKDL$ is a parallelogram; this implies the desired claim since $AD$ and $KL$ are diagonals of this quadrilateral.\n\nFirst, we show that $KD \\parallel AL$. If $K$ lies between $A$ and $B$, then by inscribed angles, $\\angle BAC = \\angle BCD$ and $\\angle BCD = \\angle BKD$. Consequently, $\\angle BAL = \\angle BAC = \\angle BKD$, implying $KD \\parallel AL$.\n\nIf $B$ lies between $A$ and $K$, then by inscribed angles, $\\angle BAC = \\angle BCD$ and $\\angle BCD = 180^\\circ - \\angle BKD$. Consequently, $\\angle BAL + \\angle BKD = \\angle BAC + \\angle BKD = 180^\\circ$, implying $KD \\parallel AL$.\n\nIf $A$ lies between $K$ and $B$, then by inscribed angles, $\\angle BAC = 180^\\circ - \\angle BCD$ and $\\angle BCD = \\angle BKD$. Consequently, $\\angle BAL = 180^\\circ - \\angle BAC = \\angle BKD$, implying $KD \\parallel AL$ again.\n\nAnalogously, we can show that $LD \\parallel AK$. Altogether, this establishes that $AKDL$ is a parallelogram.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21074,
"subject": "Mathematics (Olympiad)",
"question": "The sequence $49$, $4489$, $444889$, $\text{...}$ is defined such that the number $48$ is inserted in the middle of the preceding term. Prove that every term in the sequence is a perfect square.",
"options": [],
"answer": "See solution",
"solution": "Let us denote a general term as $A = \\overline{444\\ldots488\\ldots89}$.\n\n$$\n\\text{Then } A = 4 \\cdot \\overline{111\\ldots1} \\cdot 10^n + 8 \\cdot \\overline{111\\ldots1} + 1.\n$$\n\n$$\n\\text{Now } \\overline{111\\ldots1} = 10^{n-1} + 10^{n-2} + \\dots + 10^2 + 10 + 1 = \\frac{10^n - 1}{9}.\n$$\n\n$$\n\\text{Then } A = \\frac{4}{9}(10^n - 1)10^n + \\frac{8}{9}(10^n - 1) + 1 = \\frac{4}{9}10^{2n} + \\frac{4}{9}10^n + \\frac{1}{9} = \\left(\\frac{2 \\cdot 10^n + 1}{3}\\right)^2.\n$$\n\nThus, every term in the sequence is a perfect square.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21075,
"subject": "Mathematics (Olympiad)",
"question": "Let $q$ be a fixed positive rational number. Call a number $x$ *charismatic* if there exist a positive integer $n$ and integers $\\alpha_1, \\alpha_2, \\dots, \\alpha_n$ such that\n\n$$\nx = (q+1)^{\\alpha_1} \\cdot (q+2)^{\\alpha_2} \\cdots (q+n)^{\\alpha_n}.\n$$\n\na) Prove that $q$ can be chosen in such a way that every positive rational number turns out to be charismatic.\n\nb) Is it true for every $q$ that, for every charismatic number $x$, the number $x+1$ is charismatic too?",
"options": [],
"answer": "See solution",
"solution": "a) Take $q = 1$ and let $x$ be any positive rational number. Let $n = p-1$ where $p$ is the largest prime number that divides either the numerator or the denominator of $x$. Then all prime numbers occurring in the canonical representation of $x$ with non-zero exponent are in the form $1 + i$ with $1 \\leq i \\leq n$. In order to obtain a product required in the definition of charismaticity, equip such prime numbers with their exponent in the canonical representation of $x$ and take all other exponents $\\alpha_i$ to be zero.\n\nb) Take $q = \\frac{1}{3}$ and $x = 1$. Since $1 = (q+1)^0$, the number $x$ chosen is charismatic. Suppose that $2$ is charismatic. Then there exist a positive integer $n$ and integers $\\alpha_1, \\alpha_2, \\dots, \\alpha_n$ such that\n\n$$\n\\left(\\frac{1}{3} + 1\\right)^{\\alpha_1} \\cdot \\left(\\frac{1}{3} + 2\\right)^{\\alpha_2} \\cdots \\left(\\frac{1}{3} + n\\right)^{\\alpha_n} = 2.\n$$\n\nThis is equivalent to\n\n$$\n(3 \\cdot 1 + 1)^{\\alpha_1} \\cdot (3 \\cdot 2 + 1)^{\\alpha_2} \\cdots (3 \\cdot n + 1)^{\\alpha_n} = 2 \\cdot 3^{\\alpha_1 + \\alpha_2 + \\cdots + \\alpha_n}.\n$$\n\nObviously $\\alpha_1 + \\alpha_2 + \\dots + \\alpha_n = 0$ since the bases of powers in the left-hand side are not divisible by $3$ and $3$ therefore does not occur in the canonical representation of the product of these powers. Thus\n\n$$\n(3 \\cdot 1 + 1)^{\\alpha_1} \\cdot (3 \\cdot 2 + 1)^{\\alpha_2} \\cdots (3 \\cdot n + 1)^{\\alpha_n} = 2.\n$$\n\nLet the positive exponents be $\\alpha_{i_1}, \\dots, \\alpha_{i_k}$ and the negative exponents be $\\alpha_{j_1}, \\dots, \\alpha_{j_l}$.\nThe condition obtained is equivalent to\n\n$$\n\\frac{(3i_1 + 1)^{\\alpha_{i_1}} \\cdots (3i_k + 1)^{\\alpha_{i_k}}}{(3j_1 + 1)^{|\\alpha_{j_1}|} \\cdots (3j_l + 1)^{|\\alpha_{j_l}|}} = 2\n$$\n\nwhich is in turn equivalent to\n\n$$\n(3i_1 + 1)^{\\alpha_{i_1}} \\cdots (3i_k + 1)^{\\alpha_{i_k}} = 2 \\cdot (3j_1 + 1)^{|\\alpha_{j_1}|} \\cdots (3j_l + 1)^{|\\alpha_{j_l}|}.\n$$\n\nThe left-hand side and right-hand side of this equality are congruent to $1$ and $2$ modulo $3$, respectively.\nThe contradiction shows that $2$ is not charismatic, whence the condition checked is not true.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21076,
"subject": "Mathematics (Olympiad)",
"question": "You are given a positive integer $n$. Prove that for any real numbers $a_1, a_2, \\ldots, a_n$, there exists a number of the form $k\\sqrt{2}$, where $k$ is a positive integer, such that all numbers $k\\sqrt{2} + a_1, k\\sqrt{2} + a_2, \\ldots, k\\sqrt{2} + a_n$ are irrational.",
"options": [],
"answer": "See solution",
"solution": "Consider the numbers $x_1 = \\sqrt{2}, x_2 = 2\\sqrt{2}, \\ldots, x_{n+1} = (n+1)\\sqrt{2}$. Suppose that for each $k = 1, \\ldots, n+1$, at least one of the numbers $x_k + a_1, x_k + a_2, \\ldots, x_k + a_n$ is rational. Since there are $n$ numbers and $n+1$ choices for $k$, by the Dirichlet principle, there must be two numbers of the form $x_i + a_j$ and $x_j + a_i$ that are both rational. Then their difference is also rational:\n\n$$\n(x_j + a_i) - (x_i + a_j) = x_j - x_i = (j - i)\\sqrt{2}\n$$\n\nBut $(j - i)\\sqrt{2}$ is irrational unless $j = i$, which is a contradiction. Therefore, for some $x_k = k\\sqrt{2}$, all numbers $x_k + a_1, x_k + a_2, \\ldots, x_k + a_n$ are irrational.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21077,
"subject": "Mathematics (Olympiad)",
"question": "Во триаголникот *ABC* аголот $\\angle BAC = 70^\\circ$, а аголот $\\angle ABC = 50^\\circ$. Точката $M$ се наоѓа во *\\triangle ABC* и притоа $\\angle MAC = \\angle MCA = 40^\\circ$. Определи ги аглите $\\angle AMB$ и $\\angle BMC$.",
"options": [],
"answer": "See solution",
"solution": "Од условите дадени на цртежот следува: $\\angle ACB = 60^\\circ$, $\\angle AMC = 100^\\circ$. Бидејќи $\\triangle AMC$ е рамнокрак и $\\angle ABC = \\frac{1}{2} \\angle AMC = 50^\\circ$, следува дека $M$ е центар на опишаната кружница околу $\\triangle ABC$. Оттука имаме: $\\angle AMB = 120^\\circ$ и $\\angle BMC = 140^\\circ$, како централни агли.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21078,
"subject": "Mathematics (Olympiad)",
"question": "Determine all functions $f$ mapping the integers to the integers with the following property: For any two (not necessarily different) numbers $m$ and $n$, $\\gcd(m, n)$ is a divisor of $f(m) + f(n)$.\n\nNote that $\\gcd(m, n) = \\gcd(|m|, |n|)$ and $\\gcd(m, 0) = |m|$ holds for all integers $m$ and $n$.",
"options": [],
"answer": "See solution",
"solution": "If $t$ is an odd number and we set $m = n = t$, then $t \\mid 2f(t)$, so $t \\mid f(t)$ for all odd $t$.\n\nNow set $m = 0$ and $n = t$ (with $t$ odd). Then $t \\mid f(0) + f(t)$, so $f(0)$ must be divisible by all odd numbers, which is only possible if $f(0) = 0$.\n\nNext, set $m = 0$ and $n = s$ with $s$ even. Then $s \\mid f(0) + f(s) = f(s)$, so $s \\mid f(s)$ for all even $s$.\n\nIt follows that $n \\mid f(n)$ for all integers $n$. Any function with this property fulfills the requirements of the problem, which completes the solution.\n\n$\\boxed{\\text{All functions } f: \\mathbb{Z} \\to \\mathbb{Z} \\text{ such that } n \\mid f(n) \\text{ for all } n \\in \\mathbb{Z}}$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21079,
"subject": "Mathematics (Olympiad)",
"question": "A $$(2k+1) \\times (2k+1)$$ table, where $k$ is a positive integer, contains one real number in each entry, and all these numbers are pairwise different. After each row, one writes the median of the row, i.e., the number in this row such that the row contains the same number of entries less than it and greater than it. Let $m$ be the median of the column of medians. Prove that more than a quarter of the numbers initially in the table are less than $m$.",
"options": [],
"answer": "See solution",
"solution": "Each row contains $k$ numbers less than the median and $k$ numbers greater than the median. Thus, $k+1$ numbers in each row do not exceed the median of that row. In rows whose median does not exceed $m$, these $k+1$ numbers do not exceed $m$ either. There are $k+1$ such rows. Consequently, there are at least $$(k+1)^2$$ numbers in the table that do not exceed $m$. Only one of them is equal to $m$, so $$(k+1)^2 - 1 = k^2 + 2k$$ numbers are less than $m$. Since $k$ is positive, we have $$\\frac{k^2+2k}{(2k+1)^2} = \\frac{k^2+2k}{4k^2+4k+1} > \\frac{k^2+2k}{4k^2+8k} = \\frac{1}{4}.$$ Therefore, more than a quarter of the numbers in the table are less than $m$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21080,
"subject": "Mathematics (Olympiad)",
"question": "Prove that positive $a$, $b$, $c$ are the lengths of the sides of a triangle if and only if the system of equations\n\n$$\na(yz + x) = b(zx + y) = c(xy + z), \\quad x + y + z = 1\n$$\n\nwith unknowns $x$, $y$, $z$ has a solution in positive reals.",
"options": [],
"answer": "See solution",
"solution": "Let $a$, $b$, $c$ be positive numbers. We seek a solution of the system in positive reals. Since $x + y + z = 1$, the numbers $x$, $y$, $z$ are in $(0, 1)$. Substituting $z = 1 - x - y$, we obtain\n\n$$\na(y - xy - y^2 + x) = c(xy + 1 - x - y), \\quad b(x - x^2 - xy + y) = c(xy + 1 - x - y),\n$$\n\nwhich can be rewritten as\n\n$$\nay(1 - y) + ax(1 - y) = c(1 - x)(1 - y), \\quad bx(1 - x) + by(1 - x) = c(1 - x)(1 - y).\n$$\n\nSince $x < 1$, $y < 1$, we have\n\n$$\nay + ax = c - cx, \\quad bx + by = c - cy.\n$$\n\nFrom these, we obtain\n\n$$\nx + y = \\frac{2c}{a + b + c}, \\quad x - y = \\frac{(b - a)(x + y)}{c} = \\frac{2(b - a)}{a + b + c};\n$$\n\nthus,\n\n$$\nx = \\frac{b + c - a}{a + b + c}, \\quad y = \\frac{c + a - b}{a + b + c}, \\quad z = \\frac{a + b - c}{a + b + c} \\quad (1)\n$$\n\nThe system has a solution in positive reals if and only if $b + c > a$, $c + a > b$, $a + b > c$, which is equivalent to the existence of a triangle with sides $a$, $b$, $c$.\n\nAlternatively, since $x + y + z = 1$, we can rewrite the first part of the system as\n\n$$\na(1 - y)(1 - z) = b(1 - z)(1 - x) = c(1 - x)(1 - y). \\quad (2)\n$$\n\nDividing by $(1 - x)(1 - y)(1 - z)$ (which is positive), we get\n\n$$\n\\frac{a}{1 - x} = \\frac{b}{1 - y} = \\frac{c}{1 - z}\n$$\n\nLet $s$ be the common (positive) value. Then\n\n$$\nx = 1 - \\frac{a}{s}, \\quad y = 1 - \\frac{b}{s}, \\quad z = 1 - \\frac{c}{s}, \\quad (3)\n$$\n\nSubstituting into $x + y + z = 1$ gives $s = \\frac{a + b + c}{2}$. This yields the formulas in (1), completing the proof.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21081,
"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": "For $n=1$, any even $a$ of the form $m^2+1$ works, for example, $a=2$. For $n=2$, any even $a$ of the form $m^2-1$ works, for example, $a=8$.\n\nAssume that for $n=3$, such a number $a$ exists. Then the number $(a-1)(a^2-1)(a^3-1) = (a-1)^3(a+1)(a^2+a+1)$ must be a perfect square. Since $a^2+a+1 = a(a+1)+1$, the numbers $a+1$ and $a^2+a+1$ are coprime. Because $a+1$ is odd, the numbers $a+1$ and $a-1$ are also coprime. Therefore, both $a+1$ and $(a-1)(a^2+a+1)$ must be perfect squares. In particular, $a + 1$ modulo $3$ can only be $0$ or $1$, and thus $a - 1$ is not divisible by $3$. Hence,\n\n$$\n\\begin{align*}\n\\gcd(a - 1, a^2 + a + 1) &= \\gcd(a - 1, (a + 2)(a - 1) + 3) \\\\\n&= \\gcd(a - 1, 3) = 1,\n\\end{align*}\n$$\n\nmeaning that both $a-1$ and $a^2+a+1$ must be perfect squares. However, the latter cannot be a square, since $a^2 < a^2+a+1 < (a+1)^2$. This is a contradiction.\n\nIt remains to prove that no such $a$ exists for $n \\ge 4$. Suppose such an $a$ exists. Take a natural number $k \\ge 2$ such that $2^k \\le n < 2^{k+1}$. Since $a^{2^k} - 1 = (a^{2^k} - 1)(a^{2^k} + 1)$, the number $(a - 1)(a^2 - 1) \\dots (a^n - 1)$ can be expressed as the product of $a^{2^k} - 1 + 1$ and several other factors of the form $a^m - 1$, where $1 \\le m \\le n$ and $m \\neq 2^k$.\n\nWe will show that the factor $a^{2^{k-1}} + 1$ is coprime with all other factors in this decomposition. Suppose $a^{2^{k-1}} + 1$ and $a^m - 1$ share a common divisor $d$. Then $\\gcd(a^{2^k} - 1, a^m - 1)$ is divisible by $d$. But $\\gcd(a^{2^k} - 1, a^m - 1) = a^{\\gcd(2^k,m)} - 1$. Since $m \\neq 2^k$ and $m \\le n < 2^{k+1}$, the number $m$ cannot be divisible by $2^k$. Thus, $\\gcd(2^k, m)$ is a power of two not exceeding $2^{k-1}$. Therefore, $a^{2^{k-1}-1}$ divides $\\gcd(a^{2^k}-1, a^m-1)$, and hence also divides $d$. Because $a$ is even, the numbers $a^{2^{k-1}-1}$ and $a^{2^{k-1}}+1$ have no common divisors other than $1$, so $d=1$, as required.\n\nThe factor $a^{2^{k-1}} + 1$ is coprime with all other factors in the product, which is a perfect square, so it must itself be a perfect square. Then $a^{2^{k-1}} + 1$ and $a^{2^{k-1}}$ are perfect squares differing by $1$, which is impossible. Therefore, our assumption is false, and no such $a$ exists for $n \\ge 4$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21082,
"subject": "Mathematics (Olympiad)",
"question": "Find the integer solutions of the equation\n\n$$\nx^2 - 2xy + 126y^2 = 2009.\n$$",
"options": [],
"answer": "See solution",
"solution": "Suppose that the integers $x, y$ satisfy\n\n$$\nx^2 - 2xy + 126y^2 - 2009 = 0.\n$$\n\nLooking at this as a quadratic function of $x$,\n\n$$\n\\Delta = (-2y)^2 - 4 \\times 1 \\times (126y^2 - 2009) = 4y^2 - 504y^2 + 8036 = 4y^2 - 504y^2 + 8036.\n$$\n\nBut simplifying, $\\Delta = 4y^2 - 504y^2 + 8036 = -500y^2 + 8036$.\n\nFor $x$ to be integer, $\\Delta$ must be a perfect square and non-negative. So $-500y^2 + 8036 \\geq 0$, i.e., $y^2 \\leq 16.072$.\n\nTry $y = 0, \\pm1, \\pm2, \\pm3, \\pm4$:\n\n- $y = 0$: $\\Delta = 8036$ (not a perfect square)\n- $y = \\pm1$: $\\Delta = 8036 - 500 = 7536$ (not a perfect square)\n- $y = \\pm2$: $\\Delta = 8036 - 2000 = 6036$ (not a perfect square)\n- $y = \\pm3$: $\\Delta = 8036 - 4500 = 3536$ (not a perfect square)\n- $y = \\pm4$: $\\Delta = 8036 - 8000 = 36 = 6^2$\n\nSo only for $y = 4$ or $y = -4$ is $\\Delta$ a perfect square.\n\nFor $y = 4$:\n\n$$\nx^2 - 2x \\cdot 4 + 126 \\cdot 16 = 2009 \\\\\nx^2 - 8x + 2016 = 2009 \\\\\nx^2 - 8x + 7 = 0\n$$\n\nThe solutions are\n\n$$\nx = \\frac{8 \\pm \\sqrt{64 - 28}}{2} = \\frac{8 \\pm 6}{2} = 7, 1\n$$\n\nFor $y = -4$:\n\n$$\nx^2 - 2x(-4) + 126 \\cdot 16 = 2009 \\\\\nx^2 + 8x + 2016 = 2009 \\\\\nx^2 + 8x + 7 = 0\n$$\n\nThe solutions are\n\n$$\nx = \\frac{-8 \\pm \\sqrt{64 - 28}}{2} = \\frac{-8 \\pm 6}{2} = -1, -7\n$$\n\nThus, all integer solutions are:\n\n$$(x, y) = (1, 4),\\ (7, 4),\\ (-1, -4),\\ (-7, -4).$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21083,
"subject": "Mathematics (Olympiad)",
"question": "Consider a position represented by numbers $A$, $B$, $C$, $D$, $E$, and $F$ arranged as follows:\n\n$$\nA \\begin{smallmatrix} B & C \\\\ F & E \\end{smallmatrix} D\n$$\n\n(a) From a position with odd sum, move to a position with exactly one odd number.\n\n(b) From a position with exactly one odd number, move to a position with odd sum and strictly smaller maximum, or to the all-zero position.\n\nShow that no move will ever increase the maximum, so this strategy is guaranteed to terminate, because each step of type (b) decreases the maximum by at least one, and it can only terminate at the all-zero position. Demonstrate how each step can be carried out.",
"options": [],
"answer": "See solution",
"solution": "First, consider a position with odd sum. Either $A + C + E$ or $B + D + F$ is odd; assume without loss of generality that $A + C + E$ is odd.\n\nIf exactly one of $A, C, E$ is odd, say $A$ is odd, we can make the sequence of moves:\n\n$$\n1 \\begin{smallmatrix} B & 0 \\\\ F & 0 \\end{smallmatrix} D \\rightarrow 1 \\begin{smallmatrix} 1 & 0 \\\\ 1 & 0 \\end{smallmatrix} \\mathbf{0} \\rightarrow \\mathbf{0} \\begin{smallmatrix} 1 & 0 \\\\ 1 & 0 \\end{smallmatrix} 0 \\rightarrow 0 \\begin{smallmatrix} 1 & 0 \\\\ \\mathbf{0} & 0 \\end{smallmatrix} 0 \\pmod{2}\n$$\n\nwhere a letter or number in boldface represents a move at that vertex, and moves that do not affect each other have been written as a single move for brevity. Hence we can reach a position with exactly one odd number. Similarly, if $A, C, E$ are all odd, then the sequence of moves:\n\n$$\n1 \\begin{smallmatrix} B & 1 \\\\ F & 1 \\end{smallmatrix} D \\rightarrow 1 \\begin{smallmatrix} \\mathbf{0} & 1 \\\\ \\mathbf{0} & 1 \\end{smallmatrix} \\mathbf{0} \\rightarrow 1 \\begin{smallmatrix} 0 & \\mathbf{0} \\\\ 0 & \\mathbf{0} \\end{smallmatrix} 0 \\pmod{2}\n$$\n\nbrings us to a position with exactly one odd number. Thus we have shown how to carry out step (a).\n\nNow assume that we have a position with $A$ odd and all other numbers even. We want to reach a position with smaller maximum. Let $M$ be the maximum. There are two cases, depending on the parity of $M$:\n\n* If $M$ is even, then one of $B, C, D, E, F$ is the maximum, and $A < M$. After making moves at $B, C, D, E$, and $F$ in that order, the sum is odd and the maximum is less than $M$.\n\n* If $M$ is odd, so $M = A$ and the other numbers are all less than $M$. If $C > 0$, then we make moves at $B$, $F$, $A$, and $F$, in that order, reaching a position with odd sum and lower maximum. If $E > 0$, apply a similar argument, interchanging $B$ with $F$ and $C$ with $E$. If $C = E = 0$, then we can reach the all-zero position by the following sequence of moves:\n\n$$\nA \\begin{matrix} B \\\\ F \\end{matrix} \\mathbf{0} \\mathbf{D} \\rightarrow A \\begin{matrix} A \\\\ A \\end{matrix} \\mathbf{0} \\mathbf{0} \\mathbf{0} \\rightarrow \\mathbf{0} \\begin{matrix} A \\\\ A \\end{matrix} \\mathbf{0} \\mathbf{0} \\mathbf{0} \\rightarrow \\mathbf{0} \\begin{matrix} \\mathbf{0} \\\\ \\mathbf{0} \\end{matrix} \\mathbf{0} \\mathbf{0} \\mathbf{0} \\mathbf{0}\n$$\n\n(Here $0$ represents zero, not any even number.)\n\nHence we have shown how to carry out a step of type (b), proving the desired result. The problem statement follows since 2003 is odd.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 21084,
"subject": "Mathematics (Olympiad)",
"question": "Determine all functions $f$ from the set of non-negative integers to itself such that\n\n$$\nf(a + b) = f(a) + f(b) + f(c) + f(d),\n$$\n\nwhenever $a$, $b$, $c$, $d$ are non-negative integers satisfying $2ab = c^2 + d^2$.",
"options": [],
"answer": "See solution",
"solution": "The required functions are $f(n) = k n^2$, where $k$ is a non-negative integer—these clearly satisfy the condition in the statement.\n\nConversely, let $f$ be a function satisfying the condition in the statement. Setting $(a, b, c, d) = (n, n, n, n)$ in the functional relation yields $f(2n) = 4f(n)$ for all $n$. In particular, $f(0) = 0$ and $f(2) = 4k$, where $k = f(1)$.\n\nSetting successively $(a, b, c, d) = (n^2, 1, n, n)$, $(a, b, c, d) = (n^2, 2, 2n, 0)$, and $(a, b, c, d) = (n^2 + 1, 1, n + 1, n - 1)$ in the functional relation yields\n\n$$\n\\begin{align*}\nf(n^2 + 1) &= f(n^2) + k + 2f(n), \\\\\nf(n^2 + 2) &= f(n^2) + 4k + f(2n) = f(n^2) + 4k + 4f(n), \\\\\nf(n^2 + 2) &= f(n^2 + 1) + k + f(n + 1) + f(n - 1).\n\\end{align*}\n$$\n\nSubtraction of the second relation above from the sum of the other two yields $f(n + 1) = 2f(n) - f(n - 1) + 2k$.\n\nA straightforward induction on $n$ now shows that $f(n) = k n^2$ for all non-negative $n$, and completes the proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21085,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $p(x)$ is a polynomial of degree $2012$ with integer roots $x_1, x_2, \\dots, x_{2012}$. Note that any solution of $p(x) = 0$ is also a solution of $p(p(x)) = 0$.\n\nIs it possible for there to exist an integer $n$ which is not a root of $p(x) = 0$ but is a root of $p(p(x)) = 0$?",
"options": [],
"answer": "See solution",
"solution": "Suppose there exists an integer $n$ such that $p(n) \\ne 0$ but $p(p(n)) = 0$. Since we know all roots of $p(x)$, we can write:\n\n$$\np(x) = x(x - x_2)(x - x_3) \\dots (x - x_{2012})\n$$\n\nGiven $p(p(n)) = 0$ and $p(n) \\ne 0$, $p(n)$ must equal some root $x_k$ with $k \\ge 2$. Thus,\n\n$$\nn(n - x_2)(n - x_3) \\dots (n - x_{2012}) = x_k\n$$\n\nLet $r = x_k$. Then there exists an integer $A$ such that $An(n - r) = r$. This quadratic in $n$ has discriminant $A^2r^2 + 4Ar = (Ar + 2)^2 - 4$.\n\nFor $n$ to be integer, the discriminant must be a perfect square. The only perfect squares differing by $4$ are $0$ and $4$, so $Ar + 2 = 2$ or $Ar + 2 = -2$.\n\nIf $Ar + 2 = 2$, then $Ar = 0$, which contradicts $n$ not being a root of $p(x) = 0$.\n\nIf $Ar + 2 = -2$, then $Ar = -4$, which is impossible since $A$ is a product of $2010$ distinct nonzero integers.\n\nTherefore, no such integer $n$ exists.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21086,
"subject": "Mathematics (Olympiad)",
"question": "Let $S$ be a set containing $n^2 + n - 1$ elements, for some positive integer $n$. Suppose that the $n$-element subsets of $S$ are partitioned into two classes. Prove that there are at least $n$ pairwise disjoint sets in the same class.",
"options": [],
"answer": "See solution",
"solution": "In order to apply induction, we generalize the result as follows:\n\n**Proposition.** If the $n$-element subsets of a set $S$ with $(n+1)m - 1$ elements are partitioned into two classes, then there are at least $m$ pairwise disjoint sets in the same class.\n\n*Proof:* Fix $n$ and proceed by induction on $m$. The case $m=1$ is trivial. Assume $m>1$ and that the proposition is true for $m-1$. Let $\\mathcal{P}$ be the partition of the $n$-element subsets into two classes. If all the $n$-element subsets belong to the same class, the result is obvious. Otherwise, select two $n$-element subsets $A$ and $B$ from different classes so that their intersection has maximal size. It is easy to see that $|A \\cap B| = n-1$. (If $|A \\cap B| = k < n-1$, then build $C$ from $B$ by replacing some element not in $A \\cap B$ with an element of $A$ not already in $B$. Then $|A \\cap C| = k+1$ and $|B \\cap C| = n-1$, and either $A$ and $C$ or $B$ and $C$ are in different classes.) Removing $A \\cup B$ from $S$, there are $(n+1)(m-1) - 1$ elements left. On this set, the partition induced by $\\mathcal{P}$ has, by the inductive hypothesis, $m-1$ pairwise disjoint sets in the same class. Adding either $A$ or $B$ as appropriate gives $m$ pairwise disjoint sets in the same class. $\\blacksquare$\n\n**Remark:** The value $n^2 + n - 1$ is sharp. A set $S$ with $n^2 + n - 2$ elements can be split into a set $A$ with $n^2 - 1$ elements and a set $B$ of $n-1$ elements. Let one class consist of all $n$-element subsets of $A$ and the other consist of all $n$-element subsets that intersect $B$. Then neither class contains $n$ pairwise disjoint sets.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21087,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a cyclic quadrilateral with inradius $r$. Let $J$ and $K$ be the incenters of $ABC$ and $ACD$, and let $P$ and $Q$ be the circumcenters of $AJK$ and $CJK$, respectively. Prove that\n\n$$\n|PQ| = |AC| - \\frac{S_{AJCK}}{r}.\n$$\n\nHere $S_{AJCK}$ denotes the area of the quadrilateral $AJCK$.",
"options": [],
"answer": "See solution",
"solution": "Let $F_1$ and $F_2$ be the feet of the perpendiculars from the points $J$ and $K$ to $AC$, respectively. Then\n\n$$\nAF_1 = \\frac{AB + AC - BC}{2} \\quad \\text{and} \\quad AF_2 = \\frac{AD + AC - CD}{2}.\n$$\n\nSince $AB + CD = BC + AD$, we have $AF_1 = AF_2$. It follows that $F_1 = F_2$, that is, $JK \\perp AC$.\n\nSince $P$ is the circumcenter of $AJK$, we have $\\angle PAK = 90^\\circ - \\angle AJK$, and since $J$ is the incenter of $ABC$ we have $\\angle AJK = 90^\\circ - \\frac{\\angle BAC}{2}$. From this we deduce that $\\angle PAK = \\frac{\\angle BAC}{2}$. Hence $\\angle PAD = \\frac{\\angle BAC}{2} + \\frac{\\angle CAD}{2} = \\frac{\\angle BAD}{2}$. It follows that the points $A$, $P$, $I$ are collinear. Similarly, the points $C$, $Q$, $I$ are collinear.\n\nMoreover, since $PQ \\parallel AC$ we have\n$$\n\\frac{AC}{PQ} = 1 + \\frac{AP}{PI}.\n$$\nOn the other hand, we have\n$$\n1 + \\frac{PI}{AI} = \\frac{r}{PN}.\n$$\nFinally, because $PN = AP \\sin \\frac{\\angle BAD}{2} = PJ \\sin \\angle JAK = \\frac{JK}{2}$, we have\n$$\n\\frac{AC}{PQ} = 1 + \\frac{JK/2}{r - JK/2}.\n$$\n\nThis is equivalent to the statement of the problem.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21088,
"subject": "Mathematics (Olympiad)",
"question": "Solve in positive integers the equation\n\n$$\nm^{\\frac{1}{n}} + n^{\\frac{1}{m}} = 2 + \\frac{2}{mn(m+n)^{\\frac{1}{m}+\\frac{1}{n}}}\n$$",
"options": [],
"answer": "See solution",
"solution": "There are no solutions in positive integers. Without loss of generality, assume that $m \\ge n$. If $n \\ge 3$, then\n\n$$\n\\left(1 + \\frac{1}{m}\\right)^m = \\sum_{k=0}^{m} \\binom{m}{k} \\frac{1}{m^k} < \\sum_{k=0}^{m} \\frac{1}{k!} < \\sum_{k=0}^{\\infty} \\frac{1}{k!} = e < n \\implies n^{\\frac{1}{m}} > 1 + \\frac{1}{m}\n$$\n\nSimilarly, $m^{\\frac{1}{n}} > 1 + \\frac{1}{n}$, whence\n\n$$\nm^{\\frac{1}{n}} + n^{\\frac{1}{m}} > 2 + \\frac{1}{m} + \\frac{1}{n} \\ge 2 + \\frac{2}{mn} > 2 + \\frac{2}{mn(m+n)^{\\frac{1}{m}+\\frac{1}{n}}}\n$$\n\nThe cases $n = 1, 2$ can be solved directly. $\\Box$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21089,
"subject": "Mathematics (Olympiad)",
"question": "$r = t^8$, $t \\in \\mathbb{N}$, $t \\ge 2$, $s \\ge 2$ бол $r$-ийг зэрэг гэдэг. $\\forall n \\in \\mathbb{N}$, $\\exists A \\subseteq \\mathbb{N}$:\n\n1. $|A| = n$\n2. $1 \\le k \\le n$ байх $k$ бүрийн хувьд $A$-ийн ямар ч $k$ элементийн арифметик дундаж нь мөн зэрэг гэж батал.",
"options": [],
"answer": "See solution",
"solution": "Натурал тооны төгсгөлөг дэд олонлог $A$-ийн хувьд, $A$ олонлогийн элемент бүр зэрэг байхаар натурал олонлог олдож болно.\n\nИндукцээр баталъя.\n\n$A = (a_1, a_2, \\ldots, a_n)$, $1 < a_i \\in \\mathbb{N}$ гэж үзэж болно. (Хэрэв $A$-д 1 байвал хангалттай). $A_1 = a_1 \\cdot A$ олонлогийн эхний элемент болно. $A_k = \\{a_{k1}, a_{k2}, \\ldots, a_{kn}\\}$ олонлогийн эхний $k$ элемент нь $r_1, r_2, \\ldots, r_k$ зэрэг болог. $A_{k+1} = a_{k+1}^{k+1} \\cdot A_k$ олонлогийн эхний $k+1$ ширхэг элемент нь $r_1, r_2, \\ldots, r_k, r_{k+1}$ зэрэг болно. Иймд энэ үйлдлийг дараалуулан хийхэд үүсэх $A_n$ олонлогийн элемент бүр зэрэг ба $h \\cdot A = A_n$ байх $h \\in \\mathbb{N}$ олдоно.\n\n$B = \\{n!, 2n!, 3n!, \\ldots, n \\cdot n!\\}$ гэвэл $B$-ийн ямар ч $1 < k$ элементийн арифметик дундаж нь натурал тоо гарна. $B$-ийн бүх $1 < k \\le n$ элементийн арифметик дундажийн олонлогийг $C = \\{b_1, b_2, \\ldots, b_j\\}$ гэе. $A = B \\cup C$ гэвэл өгүүлбэрийн дагуу элемент бүр нь зэрэг байх $h \\cdot A$ ($h \\in \\mathbb{N}$) олонлог олдоно. Энэ үед $h \\cdot B$ бодлогын нөхцөлийг хангана.\n\nОдоо өгүүлбэрийг Хятадын үлдэгдлийн теорем ашиглан баталъя.\n\n$p_1, \\ldots, p_t$-ууд нь $a_1, \\ldots, a_n$-уудын задаргаанд орох бүх анхны тоонууд байг.\n\n$$\na_1 = p_1^{\\alpha_{11}} p_2^{\\alpha_{12}} \\dots p_t^{\\alpha_{1t}}, \\quad \\alpha_{ij} \\ge 0\n$$\n\n$q_1, \\ldots, q_n$ ялгаатай анхны тоонууд бол $1 \\le j \\le t$ бүрийн хувьд\n\n$$\ny \\equiv \\alpha_{1j} \\pmod{q_1}, \\ldots, y \\equiv -\\alpha_{nj} \\pmod{q_n}\n$$\n\nсистемийн шийд нь $y_j$ байг. $A = p_1^{y_1} \\cdots p_t^{y_t}$ гэвэл $A a_i = p_1^{y_1+\\alpha_{i1}} \\cdots p_t^{y_t+\\alpha_{it}}$ тоо бүр зэрэг болох нь илэрхий: $A a_i = c_i^{\\alpha_i},\\ i = 1, \\ldots, n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21090,
"subject": "Mathematics (Olympiad)",
"question": "Let $P(x)$ be a polynomial with integer coefficients. Prove that the polynomial\n$$\nQ(x) = P(x^4)P(x^3)P(x^2)P(x) + 1\n$$\nhas no integer roots.",
"options": [],
"answer": "See solution",
"solution": "By the Little Fermat theorem, $n^3 \\equiv n \\pmod{3}$ for every $n \\in \\mathbb{Z}$. Then also $n^4 \\equiv n^2 \\pmod{3}$. Since $P(x)$ is a polynomial with integer coefficients, it follows that\n\n$$\nP(n^3) \\equiv P(n) \\pmod{3}, \\quad P(n^4) \\equiv P(n^2) \\pmod{3}\n$$\n\nThe polynomial $Q(x)$ has integer coefficients; therefore, from the last two congruences, we get\n\n$$\nQ(n) \\equiv (P(n)P(n^2))^2 + 1 \\pmod{3}\n$$\n\nBut for every integer $m$, we have $m^2 + 1 \\not\\equiv 0 \\pmod{3}$. So for every integer $n$, $Q(n) \\not\\equiv 0 \\pmod{3}$, from which it follows that $Q(x)$ cannot have integer roots. The proof is finished. $\\square$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21091,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x + f(y)) - f(x) = (x + f(y))^3 - x^3\n$$\nfor all $x, y \\in \\mathbb{R}$.",
"options": [],
"answer": "See solution",
"solution": "The function $f$ is either $f(x) = 0$ or $f(x) = x^3 + c$ with an arbitrary $c \\in \\mathbb{R}$.\n\n**Proof:**\n\nObviously $f(x) = 0$ is a solution, so let us assume that a real number $a \\neq 0$ belongs to the range of $f$. Let us first assume $a > 0$. Taking $x = -f(y)$ in the given equation we get\n$$\nf(-f(y)) = f(0) - f(y)^3 = (-f(y))^3 + c, \\quad c = f(0).\n$$\nThe function\n$$\ng(x) = f(x + a) - f(x) = (x + a)^3 - x^3 = 3a\\left(x + \\frac{a}{2}\\right)^2 + \\frac{a^3}{4}\n$$\ntakes every value $z \\geq a^3/4$. For every such $z = g(x)$, equation (1) then gives\n$$\n\\begin{aligned}\nf(z) &= f(g(x)) = f(-f(x) + f(x + a)) \\\\\n&= f(-f(x)) + (-f(x) + f(x + a))^3 - (-f(x))^3\n\\end{aligned}\n$$\nThus $f$ takes every value not less than $(a^3/4)^3 + c$, which implies that $f$ is unlimited from above. For every $x$ we can then choose a $y$ such that $x + f(y) \\geq a^3/4$ so that from above we get\n$$\nf(x) = f(x + f(y)) - (x + f(y))^3 + x^3 = x^3 + c.\n$$\nThis function evidently satisfies the given equation for an arbitrary $c \\in \\mathbb{R}$.\n\nIf $a < 0$, we can set $f(x) = -h(-x)$. It is easily verified that if $f$ satisfies the given equation, so does $h$. Furthermore $h$ takes the value $-a > 0$, so $h$ is given by $h(x) = x^3 + c$ for some $c \\in \\mathbb{R}$. Hence we get $f(x) = x^3 - c$. Since these functions form the same class as those which were found above, this completes the proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21092,
"subject": "Mathematics (Olympiad)",
"question": "Given an integer $n > 1$, let $1 = a_1 < a_2 < \\dots < a_t = n - 1$ be all the positive integers less than $n$ and coprime to $n$. Determine all values of $n$ for which there is no index $i \\in \\{1, 2, \\dots, t-1\\}$ satisfying $3 \\mid a_i + a_{i+1}$.",
"options": [],
"answer": "See solution",
"solution": "It is easy to check that for $n = 2, 4, 10$ the sequences are $(1)$, $(1, 3)$, $(1, 3, 7, 9)$ respectively. Thus, these numbers are solutions to the given problem.\n\nConsidering $n \\geq 3$, if $n$ is odd then $\\gcd(n, 2) = 1$, so clearly $a_1 = 1$, $a_2 = 2$ and their sum is divisible by $3$, which does not satisfy the condition. Hence, $n$ must be even. It is easy to check that $n = 6$ or $n = 8$ also do not work.\n\nNow, consider $n \\geq 12$ and separate into the following cases:\n\n1. If $n = 12k$ for $k \\geq 1$, consider $a = 6k + 1$, $b = 6k - 1$. Since $a$ is odd, $\\gcd(n, a) \\mid \\gcd(n, 2a) = \\gcd(12k, 12k + 2) = 2$, so $\\gcd(n, a) = 1$. Similarly, $\\gcd(n, b) = 1$, so $a, b$ are two consecutive terms in the sequence for $n$, but $a + b = 12k$ which is divisible by $3$. This case does not give any solution.\n\n2. If $n = 12k + 2$ for $k \\geq 1$, consider $a = 6k + 3$, $b = 6k + 5$. Let $d = \\gcd(n, a)$, then $d \\mid 2a = 12k + 6$ so $d \\mid 4$, but $d$ is odd so $d = 1$, thus $\\gcd(n, a) = 1$. Similarly, let $d' = \\gcd(n, b)$, then $d' \\mid 2b = 12k + 10$ so $d' \\mid 8$, but $d'$ is odd so $d' = 1$, thus $\\gcd(n, b) = 1$. Now $a, b$ are consecutive terms in the sequence and $3 \\mid a + b$, which does not satisfy the condition.\n\n3. If $n = 12k + 4$ for $k \\geq 1$, consider $a = 6k + 1$, $b = 6k - 1$.\n\n4. If $n = 12k + 6$ for $k \\geq 1$, consider $a = 6k + 1$, $b = 6k - 1$.\n\n5. If $n = 12k + 8$ for $k \\geq 1$, consider $a = 6k + 3$, $b = 6k + 1$.\n\n6. If $n = 12k + 10$ for $k \\geq 1$, consider $a = 6k + 1$, $b = 6k - 1$.\n\nFor the last four cases, we proceed in the same way as the previous two. This implies that all numbers $n \\geq 12$ do not satisfy the given condition. Therefore, all solutions to the given problem are $2$, $4$, and $10$. $\\square$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21093,
"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$$\n(m+1)^{k-1}(m+1-k) < m^k.\n$$\nWe prove this by induction on $k$ for $2 \\leq k \\leq m$.\n\n*Base case*: For $k = 2$, the left side is $(m+1)(m-1) = m^2 - 1 < m^2$.\n\n*Inductive step*: Assume the inequality is true for $k = r < m$, i.e., $(m+1)^{r-1}(m+1-r) < m^r$. Multiplying both sides by $m$ gives\n$$\nm^{r+1} > m(m+1)^{r-1}(m+1-r).\n$$\nThis can be rewritten as\n$$\nm^{r+1} > (m+1)^r (m - r).\n$$\nThus, the required inequality is true for $k = r + 1$. By induction, it is true for $2 \\leq k \\leq m$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21094,
"subject": "Mathematics (Olympiad)",
"question": "Let $AB\\Gamma$ be an equilateral triangle of side $k$ cm. We divide $AB\\Gamma$ with parallel lines into $k^2$ small equilateral triangles of side $1$ cm, creating a grid (see figure for $k=7$).\n\n\n\n\nInside every small triangle, we put exactly one positive integer from $1$ to $k^2$, so that no two triangles have the same number. With vertices at the points of the grid, regular hexagons of side $1$ cm are defined. We call the value of a hexagon the sum of the numbers of its $6$ small triangles. Find, as a function of $k$, the greatest possible value of the sum of the values of all hexagons.",
"options": [],
"answer": "See solution",
"solution": "The small triangles are divided into four categories:\n\n1. Triangles with one vertex at $A$, $B$, or $\\Gamma$.\n - These are not members of any hexagon, so their numbers do not contribute to the final sum.\n\n2. Triangles belonging to only one hexagon.\n - On each side of $AB\\Gamma$ there are $k-2$ such triangles, so in total $3(k-2)+3 = 3k-3$.\n\n3. Triangles belonging to exactly two hexagons.\n - On each side of $AB\\Gamma$ there are $k-3$ such triangles, so in total $3(k-3)$.\n - For the final sum, numbers in these triangles are counted twice.\n\n4. Triangles belonging to exactly three hexagons.\n - These are inside the triangle $\\Delta EZ$ (see figure), totaling $(k-3)^2$ triangles.\n\nTo maximize the sum, assign the largest numbers to triangles counted most times:\n\n\n\n- Assign $A = \\{1, 2, 3\\}$ to the first category.\n- Assign $B = \\{4, 5, \\dots, 3k\\}$ to the second category ($3k-3$ triangles), with partial sum:\n $$S_B = 4 + 5 + \\dots + 3k = \\frac{(3k-3)(3k+4)}{2}.$$\n- Assign $\\Gamma = \\{3k+1, \\dots, 6k-9\\}$ to the third category ($3k-9$ triangles), with partial sum:\n $$S_{\\Gamma} = (3k+1) + \\dots + (6k-9) = \\frac{(3k+1) + (6k-9)}{2} (3k-9).$$\n- Assign $\\Delta = \\{6k-8, \\dots, k^2\\}$ to the fourth category ($k^2-6k+9$ triangles), with partial sum:\n $$S_{\\Delta} = (6k-8) + \\dots + k^2 = \\frac{k^2 + 6k - 8}{2} (k^2 - 6k + 9).$$\n\nThus, the greatest possible sum of values is:\n$$S_{\\max} = S_A + S_B + 2S_{\\Gamma} + 3S_{\\Delta} = \\frac{3(k^4 - 14k^2 + 33k - 24)}{2}.$$ \n\nIf $\\gamma$ is a member of $\\Gamma$ and $\\delta$ a member of $\\Delta$, then in the final sum $S_{\\max}$, the summand $2\\gamma+3\\delta$ appears. If we swap their positions, we get $2\\delta+3\\gamma$. Since $\\gamma < \\delta$ and $2 < 3$, it follows that $2\\delta+3\\gamma < 2\\gamma+3\\delta$. Hence, the sum found is maximal.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21095,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a self-squared $\\ell$-code, meaning that the last $\\ell$ digits of $n^2 - n = n(n-1)$ are all zeros (i.e., $n(n-1)$ is divisible by $10^\\ell$).\n\n(a) Show that $n$ is self-squared if and only if $n^2 - n$ is divisible by $10^\\ell$.\n\n(b) Show that for $n \\geq 2$, the only possible final digits for a self-squared $\\ell$-code $n$ are 5 or 6.\n\n(c) Suppose $m = c \\cdot 10^\\ell + n$ is an $(\\ell+1)$-code obtained by putting a digit $c$ in front of $n$. Show that for each self-squared $\\ell$-code $n$, there is a unique digit $c$ such that $m$ is a self-squared $(\\ell+1)$-code.\n\n(d) Suppose $n \\geq 2$ is a self-squared $\\ell$-code. Show that $k = 10^\\ell + 1 - n$ is also a self-squared $\\ell$-code, and that $n$ and $k$ are distinct, with $n + k = 10^\\ell + 1$.",
"options": [],
"answer": "See solution",
"solution": "(a) If $n$ is a self-squared $\\ell$-code, then $n(n-1)$ ends with $\\ell$ zeros, so $n(n-1)$ is divisible by $10^\\ell$. Conversely, if $n(n-1)$ is divisible by $10^\\ell$, then $n$ is self-squared.\n\n(b) Since $n(n-1)$ is divisible by 10, one of $n$ or $n-1$ must be divisible by 5, so the last digit of $n$ is 0, 1, 5, or 6. If $n$ ends with 0 or 1, $n$ or $n-1$ must be divisible by $10^\\ell$, which is only possible for $n = 0$ or $n = 1$, both less than 2. Thus, for $n \\geq 2$, $n$ must end in 5 or 6.\n\n(c) Let $m = c \\cdot 10^\\ell + n$. Then\n$$\nm(m-1) = (c \\cdot 10^{\\ell} + n)(c \\cdot 10^{\\ell} + n-1) = c^2 \\cdot 10^{2\\ell} + c \\cdot 10^{\\ell}(2n-1) + n(n-1)\n$$\nAs $n(n-1) = d \\cdot 10^\\ell$, $m(m-1)$ is divisible by $10^{\\ell+1}$ if and only if $c(2n-1) + d$ is divisible by 10. If $n$ ends in 5, $2n-1 = 10k-1$, so $-c + d$ must be divisible by 10. If $n$ ends in 6, $2n-1 = 10k+1$, so $c + d$ must be divisible by 10. In both cases, there is a unique $c$ modulo 10.\n\n(d) Let $k = 10^\\ell + 1 - n$. Then\n$$\nk(k-1) = (10^{\\ell} + 1 - n)(10^{\\ell} - n) = 10^{2\\ell} + 10^{\\ell}(1 - 2n) + (n-1)n\n$$\nThe first two terms are divisible by $10^\\ell$, and $(n-1)n$ is divisible by $10^\\ell$ since $n$ is self-squared. Thus, $k$ is also self-squared. Since $n \\leq 10^\\ell - 1$, $k \\geq 2$, and $n \\neq k$. There are exactly two self-squared $\\ell$-codes $\\geq 2$, and $n + k = 10^\\ell + 1$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21096,
"subject": "Mathematics (Olympiad)",
"question": "試求最大的實數 $a$ 使得對所有 $n \\ge 1$ 與所有實數 $x_0, x_1, \\dots, x_n$ 滿足\n\n$$\n0 = x_0 < x_1 < x_2 < \\dots < x_n,\n$$\n\n我們有\n\n$$\n\\frac{1}{x_1 - x_0} + \\frac{1}{x_2 - x_1} + \\dots + \\frac{1}{x_n - x_{n-1}} \\ge a \\left( \\frac{2}{x_1} + \\frac{3}{x_2} + \\dots + \\frac{n+1}{x_n} \\right).\n$$",
"options": [],
"answer": "See solution",
"solution": "最大的 $a$ 為 $\\frac{4}{9}$。\n\n首先證明 $a = \\frac{4}{9}$ 可行。對於每個 $2 \\le k \\le n$，由 Cauchy-Schwarz 不等式，有\n\n$$\n(x_{k-1} + (x_k - x_{k-1})) \\left( \\frac{(k-1)^2}{x_{k-1}} + \\frac{3^2}{x_k - x_{k-1}} \\right)^2 \\ge (k-1+3)^2,\n$$\n\n這可改寫為\n\n$$\n\\frac{9}{x_k - x_{k-1}} \\ge \\frac{(k+2)^2}{x_k} - \\frac{(k-1)^2}{x_{k-1}}. \\quad (1)\n$$\n\n對 $k = 2, 3, \\dots, n$ 將式 (1) 相加，並在兩邊加上 $9/x_1$，得到\n\n$$\n9 \\sum_{k=1}^{n} \\frac{1}{x_k - x_{k-1}} \\ge 4 \\sum_{k=1}^{n} \\frac{k+1}{x_k} + \\frac{n^2}{x_n} > 4 \\sum_{k=1}^{n} \\frac{k+1}{x_k}.\n$$\n\n這證明了原不等式對 $a = \\frac{4}{9}$ 成立。\n\n接下來證明 $a = \\frac{4}{9}$ 為最佳值。考慮 $x_0 = 0$ 且 $x_k = x_{k-1} + k(k+1)$，即\n\n$$\nx_k = \\frac{1}{3}k(k+1)(k+2).\n$$\n\n則原不等式左邊為\n\n$$\n\\sum_{k=1}^{n} \\frac{1}{k(k+1)} = \\sum_{k=1}^{n} \\left( \\frac{1}{k} - \\frac{1}{k+1} \\right) = 1 - \\frac{1}{n+1},\n$$\n\n右邊為\n\n$$\n\\begin{align*}\na \\sum_{k=1}^{n} \\frac{k+1}{x_k} &= 3a \\sum_{k=1}^{n} \\frac{1}{k(k+2)} \\\\\n&= \\frac{3}{2}a \\sum_{k=1}^{n} \\left( \\frac{1}{k} - \\frac{1}{k+2} \\right) \\\\\n&= \\frac{3}{2} \\left( 1 + \\frac{1}{2} - \\frac{1}{n+1} - \\frac{1}{n+2} \\right) a.\n\\end{align*}\n$$\n\n當 $n \\to \\infty$，左邊趨近於 1，右邊趨近於 $\\frac{9}{4}a$。因此 $a$ 至多為 $\\frac{4}{9}$。\n\n所以最大值為 $a = \\frac{4}{9}$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21097,
"subject": "Mathematics (Olympiad)",
"question": "Let $x_n = \\sin^2(n^\\circ)$. What is the mean of $x_1, x_2, x_3, \\dots, x_{90}$?\n\n(A) $\\frac{11}{45}$ (B) $\\frac{22}{45}$ (C) $\\frac{89}{180}$ (D) $\\frac{1}{2}$ (E) $\\frac{91}{180}$",
"options": [],
"answer": "See solution",
"solution": "The required mean is\n\n$$\n\\frac{1}{90} \\sum_{n=1}^{90} \\sin^2(n^\\circ).\n$$\n\nGroup the summands into 44 pairs plus two additional terms as follows:\n\n$$\n\\begin{align*}\n\\sin^2(1^\\circ) + \\sin^2(89^\\circ) &= \\sin^2(1^\\circ) + \\cos^2(1^\\circ) = 1 \\\\\n\\sin^2(2^\\circ) + \\sin^2(88^\\circ) &= \\sin^2(2^\\circ) + \\cos^2(2^\\circ) = 1 \\\\\n\\sin^2(3^\\circ) + \\sin^2(87^\\circ) &= \\sin^2(3^\\circ) + \\cos^2(3^\\circ) = 1 \\\\\n\\vdots \\\\\n\\sin^2(44^\\circ) + \\sin^2(46^\\circ) &= \\sin^2(44^\\circ) + \\cos^2(44^\\circ) = 1 \\\\\n\\sin^2(45^\\circ) &= \\frac{1}{2} \\\\\n\\sin^2(90^\\circ) &= 1.\n\\end{align*}\n$$\n\nThis gives a sum of $45\\frac{1}{2} = \\frac{91}{2}$, so the mean is $\\frac{1}{90} \\cdot \\frac{91}{2} = \\frac{91}{180}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21098,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an obtuse triangle with orthocenter $H$ and centroid $S$. Let $D$, $E$, and $F$ be the midpoints of segments $BC$, $AC$, and $AB$, respectively.\n\nShow that the circumcircle of triangle $ABC$, the circumcircle of triangle $DEF$, and the circle with diameter $HS$ have two distinct points in common.",
"options": [],
"answer": "See solution",
"solution": "Let $M$ and $O$ denote the circumcenters of triangles $DEF$ and $ABC$, respectively. We use the well-known facts that the points $H$, $M$, $S$, and $O$ lie on the Euler line of triangle $ABC$ in this order, and that $HM : MS : SO = 3 : 1 : 2$.\n\nSince $S$ is in the interior of $ABC$, it is inside the circumcircle of $ABC$. However, since $ABC$ is obtuse, the orthocenter $H$ is outside the circumcircle. This implies that the circle with diameter $HS$ intersects the circumcircle of $ABC$ in two points.\n\nLet $X$ be one of these intersection points. We will prove that $MX : OX = 1 : 2$. Let $N$ be the midpoint of $HS$. Using Thales' theorem, we get that $HN$, $XN$, and $SN$ have the same length, and $HN : NM : MS : SO = 2 : 1 : 1 : 2$. This implies that $MN : XN = XN : ON$.\n\nTherefore, the triangles $XNM$ and $ONX$ are similar with ratio $1 : 2 = MN : XN$. Therefore, we have $MX : OX = 1 : 2$ as desired.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21099,
"subject": "Mathematics (Olympiad)",
"question": "對於任意正整數 $n$，證明：\n\n$$\nf(\\sqrt[n]{x_1x_2\\cdots x_n}) \\le \\sqrt[n]{f(x_1)f(x_2)\\cdots f(x_n)}.\n$$\n\n並說明等號成立的充要條件。",
"options": [],
"answer": "See solution",
"solution": "設 $G = \\sqrt[n]{x_1x_2\\cdots x_n}$。對於任意正整數 $n$，必存在 $k$，使得 $2^k \\le n < 2^{k+1}$。\n\n考慮 $f(G) = f(\\sqrt[2^{k+1}]{x_1 x_2 \\cdots x_n} G \\cdots G)$，\n\n$$\nf(G) \\le \\sqrt[2^{k+1}]{f(x_1)f(x_2)\\cdots f(x_n)[f(G)]^{2^{k+1}-n}},\n$$\n\n即 $[f(G)]^n \\le f(x_1)f(x_2)\\cdots f(x_n)$，\n\n因此\n\n$$\nf(\\sqrt[n]{x_1x_2\\cdots x_n}) \\le \\sqrt[n]{f(x_1)f(x_2)\\cdots f(x_n)}.\n$$\n\n等號成立的充要條件為 $x_1 = x_2 = \\cdots = x_n = 0$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21100,
"subject": "Mathematics (Olympiad)",
"question": "Let $p$ be a prime number and let $n \\geq p-1$ be an integer. If $np+1$ is a perfect square, prove that $n+1$ can be represented as a sum of squares of exactly $p$ positive integers.",
"options": [],
"answer": "See solution",
"solution": "Let $a$ be a positive integer such that $np + 1 = a^2$. Note that $a > 1$.\n\n$$\na^2 - 1 = np, \\text{ i.e. } (a-1)(a+1) = np.\n$$\n\nSince $p$ is prime, it follows that $p \\mid a-1$ or $p \\mid a+1$. We treat these two cases separately:\n\n**Case 1:** $p \\mid a-1$, i.e. $a = kp + 1$ for some integer $k$.\n\n$$\n1 + np = (kp + 1)^2 = k^2p^2 + 2kp + 1, \\text{ i.e. } n = k^2p + 2k.\n$$\n\nSo,\n$$\nn + 1 = k^2p + 2k + 1 = k^2 + 2k + 1 + (p-1)k^2 = (k+1)^2 + (p-1)k^2.\n$$\n\nSince $a > 1$, $k > 0$, so $k+1$ and $k$ are positive integers.\n\n**Case 2:** $p \\mid a+1$, so $a = kp - 1$ for some integer $k$.\n\nAnalogously, $n + 1 = (k-1)^2 + (p-1)k^2$.\n\nWe need $k-1 > 0$, i.e. $k \\geq 2$. If $k = 1$, then $n + 1 = p - 1$, so $n = p - 2$, which contradicts $n \\geq p - 1$.\n\nThus, $n+1$ can be written as a sum of squares of exactly $p$ positive integers for all valid $n$ and $p$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21101,
"subject": "Mathematics (Olympiad)",
"question": "Give all integer solutions of the equation:\n\n$$\n3^{2a+1} b^2 + 1 = 2^c.\n$$",
"options": [],
"answer": "See solution",
"solution": "Case 1. $a \\ge 0$.\n\nClearly $c \\ge 0$ where $c=0$ implies $b=0$. We get that $(a,0,0)$ is a solution for any non-negative integer $a$. From the equality $3^{2a+1}b^2+1=2^c$ it follows that $b$ is an odd integer. We can write the left-hand side as:\n\n$$\n3^{2a+1} b^2 + 1 = (3^{2a+1} + 1)b^2 - (b-1)(b+1).\n$$\n\nNotice that $(b-1)(b+1)$ is divisible by $8$, while $(3^{2a+1}+1)b^2$ is divisible by $4$ but not by $8$. Therefore $2^c=4$, i.e., $c=2$. But then $3^{2a+1}b^2=3$, so $a=0$ and $b=\\pm 1$.\n\nCase 2. $a < 0$.\n\nAgain $c \\ge 0$ where $c=0$ implies $b=0$ and then $a$ can be any negative integer. Therefore, restrict to $c > 0$. It is enough to consider $b > 0$. Let $d=-a$, so the equation becomes:\n\n$$\n(2^c - 1)3^{2d-1} = b^2.\n$$\n\nwhere $b, c, d$ are natural numbers. Thus $b$ is divisible by $3$, and $c$ is even. So $b = 3^d x$, $c = 2y$ for some natural numbers $x, y$. The equation becomes:\n\n$$\n4^{y-1} + 4^{y-2} + \\dots + 1 = x^2.\n$$\n\nThis implies $x=y=1$. For $y \\ge 2$, $x^2 \\equiv 5 \\pmod{8}$, which is impossible. Therefore, the only solutions in this case are $(a,3^{-a},2)$, where $a$ is any negative integer.\n\nThe set $M$ of all solutions to the equation is:\n\n$$\nM = \\{(a,0,0) \\mid a \\in \\mathbb{Z}\\} \\cup \\{(a,\\pm 3^{-a},2) \\mid a \\in \\mathbb{Z}\\}.\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21102,
"subject": "Mathematics (Olympiad)",
"question": "Which of the following claims are true, and which are false? If a fact is true, prove it; if it is false, find a counterexample.\n\n(a) Let $a, b, c$ be real numbers such that $a^{2013} + b^{2013} + c^{2013} = 0$. Then $a^{2014} + b^{2014} + c^{2014} = 0$.\n\n(b) Let $a, b, c$ be real numbers such that $a^{2014} + b^{2014} + c^{2014} = 0$. Then $a^{2015} + b^{2015} + c^{2015} = 0$.\n\n(c) Let $a, b, c$ be real numbers such that $a^{2013} + b^{2013} + c^{2013} = 0$ and $a^{2015} + b^{2015} + c^{2015} = 0$. Then $a^{2014} + b^{2014} + c^{2014} = 0$.",
"options": [],
"answer": "See solution",
"solution": "First, for every real number $x$, $x^2 \\ge 0$ holds.\n\nThe key idea is that the expression $a^{2014} + b^{2014} + c^{2014}$ is a sum of even powers (which are nonnegative numbers). Thus, $a^{2014} + b^{2014} + c^{2014} = 0$ if and only if $a = b = c = 0$.\n\n**(a)** False. It is sufficient to find three real numbers whose 2013th powers sum to $0$, but whose 2014th powers do not. For example, $a = \\sqrt[2013]{1}$, $b = \\sqrt[2013]{2}$, $c = \\sqrt[2013]{-3}$.\n\n**(b)** True. From the key idea, $a^{2014} + b^{2014} + c^{2014} = 0$ implies $a = b = c = 0$, so $a^{2015} + b^{2015} + c^{2015} = 0$.\n\n**(c)** False. A counterexample: $a = 1$, $b = 0$, $c = -1$. Then $a^{2013} + b^{2013} + c^{2013} = 1 + 0 + (-1) = 0$ and $a^{2015} + b^{2015} + c^{2015} = 1 + 0 + (-1) = 0$, but $a^{2014} + b^{2014} + c^{2014} = 1 + 0 + 1 = 2 \\neq 0$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21103,
"subject": "Mathematics (Olympiad)",
"question": "Find the minimal prime number $p$ such that $\\{\\sqrt{p}\\} < \\frac{1}{501}$, where $\\{x\\}$ denotes the fractional part of $x$.",
"options": [],
"answer": "See solution",
"solution": "The minimal prime is $p = F_4 = 2^{16} + 1 = 65537$.\n\nIt is known that Fermat's number $F_4$ is prime. We verify $\\{\\sqrt{F_4}\\} < \\frac{1}{501}$:\n\n$$\n\\sqrt{2^{16} + 1} - 256 < \\frac{1}{501}\n$$\n\nThis is equivalent to:\n\n$$\n\\frac{1}{\\sqrt{256^2 + 1} + 256} < \\frac{1}{501}\n$$\n\nSince $\\frac{1}{2 \\cdot 256} < \\frac{1}{501}$, the inequality holds.\n\nNow, suppose $p < F_4$ and $\\{\\sqrt{p}\\} < \\frac{1}{501}$. Let $n^2 < p < (n+1)^2$, so $p = n^2 + a$ with $1 \\leq a \\leq 2n$. Then $[\\sqrt{p}] = n$, and\n\n$$\n\\sqrt{n^2 + a} - n < \\frac{1}{501}\n$$\n\nThis leads to $n > \\frac{501}{2}a - \\frac{1}{1002}$. If $a \\geq 2$, then $n \\geq 500$ and $p > F_4$, a contradiction. So $a = 1$ and $n \\geq 251$, i.e., $p \\geq 251^2 + 1$. However, $251^2 + 1$, $251^2 + 3$, $251^2 + 5$ are even and not prime. $252^2 + 1$ is divisible by 5, and $254^2 + 1$ is divisible by 149:\n\n$$\n254^2 + 1 \\equiv (-44)^2 + 1 \\equiv 1936 + 1 \\equiv 1937 \\pmod{149}\n$$\n\nCalculating, $1937 \\div 149 = 13$ remainder $0$, so $254^2 + 1$ is composite. Thus, $p = 256^2 + 1 = 65537$ is the minimal possible prime.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21104,
"subject": "Mathematics (Olympiad)",
"question": "Agustin and Lucas take turns marking one cell at a time in a $101 \\times 101$ grid. Agustin starts the game. A cell cannot be marked if there are already two marked cells in its row or in its column. The player who cannot move loses. Decide which player has a winning strategy.",
"options": [],
"answer": "See solution",
"solution": "We describe a winning strategy for the second player, Lucas. This applies to any square grid $n \\times n$ with odd $n \\geq 3$.\n\nCall a row or column *empty*, *incomplete*, or *full* at a certain moment of the game if it contains respectively 0, 1, or 2 marked cells.\n\n**Stage 1:**\n\nAgustin's first move is in an empty row $R$ and an empty column $C$, and after it there are still empty rows left. Call such a move *standard*. Lucas's answer to a standard move is to choose an empty row $R' \\neq R$ and mark the intersection of $R'$ and column $C$. This makes $C$ a full column, so it cannot be used later. Lucas gives standard answers as long as Agustin makes standard moves (at least Agustin's first move is standard). After each standard answer, each column is either empty or full, so Agustin's next move is in an empty column. Since there is a standard answer to any standard move, Lucas cannot lose if all of Agustin's moves are standard.\n\nSuppose Agustin makes a non-standard move $M$ (the first such). $M$ is in an empty column $C$. There are two possibilities:\n\n1. $M$ is in an incomplete row $R$ (there are still empty rows left).\n2. $M$ is in the last empty row remaining.\n\nIn case (1), Lucas gives a standard answer: taking an empty row $R' \\neq R$ and marking the intersection of $R'$ and column $C$. In case (2), all rows are incomplete after $M$. Lucas takes any row $R' \\neq R$ and marks the intersection of $R'$ and column $C$, making $R'$ full. This completes stage 1. As a result, the table has the following properties:\n\n(i) The number $E$ of empty rows is even.\n\n(ii) The number $I$ of incomplete rows is even.\n\n(iii) There are no incomplete columns.\n\n**Stage 2:**\n\nLucas ensures that after each of his moves, properties (i)–(iii) are preserved, regardless of Agustin's play.\n\nIf Agustin marks a cell in an empty row $R$ and column $C$, then $C$ was also empty before $M$ (since there are no incomplete columns and one cannot move in a full column). $M$ makes both $R$ and $C$ incomplete, so $E$ is odd after $M$. Lucas can mark the intersection of an empty row $R' \\neq R$ and column $C$, making $C$ full and $R'$ incomplete. The combined move decreases $E$ by 2 and increases $I$ by 2, preserving (i)–(iii).\n\nIf Agustin marks a cell in an incomplete row $R$ and column $C$, $C$ was empty before $M$. $M$ makes $R$ full and $C$ incomplete. $I$ is odd after $M$. Lucas marks the intersection of an incomplete row $R' \\neq R$ and column $C$, making both $R'$ and $C$ full. The combined move decreases $I$ by 2 and does not change $E$, again preserving (i)–(iii).\n\nIn conclusion, Lucas can always respond and will not lose. The game is finite; there will be a loser, and it is not Lucas. Thus, Lucas has a winning strategy.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21105,
"subject": "Mathematics (Olympiad)",
"question": "Let $f(n) = n + s_b(n)$, where $s_b(n)$ denotes the sum of the digits of $n$ in base $b$.\n\nShow that there are infinitely many positive integers that cannot be represented in the form $n + s_b(n)$ for any $n$.\n\nAlternatively, show that for any $k$, there exist at least $k$ positive integers that cannot be written as $n + s_b(n)$ for any $n$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Consider the set $\\{n + s_b(n) \\mid n = 1, 2, \\dots, A\\}$. This set has at most $A - k$ elements, so at least $k$ of the numbers $1, 2, \\dots, A$ are not members of this set and thus have no representation in the form $n + s_b(n)$. Since $k$ is arbitrary, there cannot be a finite number of positive integers with no representation, so there are infinitely many as desired.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21106,
"subject": "Mathematics (Olympiad)",
"question": "In a country, the government has decided to build a transport link between cities by railway or airline in such a way that no more than four other cities can be reached directly from each city. Prove that the government can always choose the type of connection of the corresponding pairs of cities in such a way that there are no three cities, each pair of which is connected by one type of transport.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let's rewrite the problem in terms of graphs: in the graph, the degree of each vertex is not more than four. Prove that all edges can be colored in two colors so that there are no single-color triangles.\n\nLet our graph have $n$ vertices. We prove the statement by induction on the number of vertices.\n\n**Base case:** If $n = 1$ or $2$, everything is obvious.\n\nAssume the statement is true for any graph with $n \\leq k$ vertices. Consider a graph with $n = k + 1$ vertices. Select a vertex $A$ arbitrarily. Remove this vertex and all edges from $A$. Thus, we have a graph with $n = k$ vertices, which, by the induction hypothesis, can be colored as needed. Now color the edges from $A$ to its neighbors. Consider three cases:\n\n\n\n**I.** The degree of $A$ is at most $2$. Then color all edges from $A$ in different colors, which proves the statement.\n\n**II.** The degree of $A$ is $3$, i.e., there are exactly $3$ edges $AX$, $AY$, and $AZ$. The edges of $\\triangle XYZ$ have different colors. Let, for instance, $XY$ be colored with the first color and $YZ$ with the second color. Color the edges as follows: $AX$ and $AY$ with the second color, and $AZ$ with the first. By the induction hypothesis, the statement is proved.\n\n**III.** The degree of $A$ is $4$, i.e., there are edges $AP$, $AQ$, $AR$, and $AS$.\n\n(a) If the graph on $P, Q, R, S$ is complete, we have a separate connected component on $A, P, Q, R, S$. By the induction hypothesis, it is a complete graph. Color its edges as shown in the figure, and the statement is proved. All other vertices form a graph with fewer than $k$ vertices, which is colored as needed by the induction hypothesis.\n\n(b) If the graph on $P, Q, R, S$ is not complete, suppose there is no edge $QR$. Color the edges $AP$ and $AQ$ in a color opposite to the edge $PQ$, and the edges $AR$ and $AS$ in a color opposite to the edge $RS$. It is straightforward to check that there are no single-color triangles. Hence, by induction, the statement is proved.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21107,
"subject": "Mathematics (Olympiad)",
"question": "A box contains red, white, and blue balls. The number of red balls is an even number, and the total number of balls in the box is less than 100. The number of white and blue balls together is exactly 4 times the number of red balls. The number of red and blue balls together is exactly 6 times the number of white balls.\n\nHow many balls are in the box?\n\nA) 28 \nB) 30 \nC) 35 \nD) 70 \nE) 84",
"options": [],
"answer": "See solution",
"solution": "D) 70",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21108,
"subject": "Mathematics (Olympiad)",
"question": "The bisector of the internal angle at vertex $A$ of triangle $ABC$ intersects side $BC$ at point $D$. The tangent to the circumcircle of triangle $ABC$ at point $A$ intersects line $BC$ at point $K$. Prove that $KA = KD$.",
"options": [],
"answer": "See solution",
"solution": "Assume without loss of generality that $\\angle ABC > \\angle ACB$ (otherwise, switch the roles of points $B$ and $C$). Note that\n\n$$\n\\angle ADK = 180^{\\circ} - \\angle CDA = \\angle DAC + \\angle ACD = \\angle BAD + \\angle ACB.\n$$\n\nBy the inscribed angle property, $\\angle KAB = \\angle ACB$, so\n\n$$\n\\angle BAD + \\angle ACB = \\angle BAD + \\angle KAB = \\angle KAD.\n$$\n\nConsequently, $\\angle ADK = \\angle KAD$, which implies $KA = KD$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21109,
"subject": "Mathematics (Olympiad)",
"question": "Amy and Ben each have a list of 2021 positive integers. Is it possible for all three of the following conditions to hold at the same time?\n\n1. All 4042 integers are different from each other.\n2. The sum of Amy's integers is equal to the sum of Ben's integers.\n3. The sum of the squares of Amy's integers is equal to the sum of the squares of Ben's integers.",
"options": [],
"answer": "See solution",
"solution": "Yes, it is possible for the two lists of positive integers to satisfy the conditions of the problem. We will prove this in the case where each list has $n$ positive integers, where $n \\ge 3$. The given problem corresponds to the case $n = 2021$.\n\nLet $A$ denote the set of Amy's numbers and let $B$ denote the set of Ben's numbers. Suppose that\n\n$$\nA = \\{x+1, x+2, \\dots, x+n-2, x+n-1, x-N\\}\n$$\n$$\nB = \\{x-1, x-2, \\dots, x-(n-2), x-(n-1), x+N\\}\n$$\nwhere $N = 1+2+\\dots+(n-1)$. If $x > N$, it is readily verified that $A$ and $B$ are disjoint and that\n\n$$\n\\sum_{i \\in A} i = \\sum_{i \\in B} i = nx\n$$\n$$\n\\sum_{i \\in A} i^2 = \\sum_{i \\in B} i^2 = nx^2 + N^2 + \\sum_{i=1}^{n-1} i^2.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21110,
"subject": "Mathematics (Olympiad)",
"question": "Find all non-constant polynomials $P(x)$ with real coefficients that satisfy\n\n$$\nP(x^3 - 7x) = P(x - 7)P(x - 8)P(x - 3)\n$$\n\nfor all $x \\in \\mathbb{R}$.",
"options": [],
"answer": "See solution",
"solution": "Let $n = \\deg P(x) \\ge 1$ and $a \\ne 0$ be the leading coefficient of $P$. Comparing the leading coefficients of both sides, we get $a = \\pm 1$. Note that if $P(x)$ satisfies the condition, so does $-P(x)$; without loss of generality, assume $a = 1$.\n\nFirst, consider $n = 1$. Let $P(x) = x + b$. Substituting into the given condition:\n\n$$\nx^3 - 7x + b = (x - 7 + b)(x - 8 + b)(x - 3 + b)\n$$\n\nComparing the coefficient of degree 2, we have $0 = b - 7 + b - 8 + b - 3$, so $b = 6$. Therefore, $P(x) = x + 6$ is a solution.\n\nNow, for any $n \\ge 1$, let $P(x) = (x + 6)^n + Q(x)$ with $\\deg Q < n$. If $Q(x) \\equiv 0$, then $P(x) = (x + 6)^n$, which satisfies since\n\n$$\n(x^3 - 7x + 6)^n = (x - 1)^n (x - 2)^n (x + 3)^n\n$$\n\nNow assume $Q(x) \\ne 0$ and let $\\deg Q = m < n$. Substituting into the given condition, we get\n\n$$\n(x^3 - 7x + 6)^n + Q(x^3 - 7x) = [(x - 1)^n + Q(x - 7)] [(x - 2)^n + Q(x - 8)] [(x + 3)^n + Q(x - 3)]\n$$\n\nExpanding and simplifying, we get\n\n$$\n\\begin{align*}\nQ(x^3 - 7x) &= Q(x - 7)Q(x - 8)Q(x - 3) \\\\\n&\\quad + (x - 1)^n Q(x - 8)Q(x - 3) + (x - 2)^n Q(x - 7)Q(x - 3) \\\\\n&\\quad + (x + 3)^n Q(x - 7)Q(x - 8) + (x - 1)^n (x - 2)^n Q(x - 3) \\\\\n&\\quad + (x - 2)^n (x + 3)^n Q(x - 7) + (x + 3)^n (x - 1)^n Q(x - 8)\n\\end{align*}\n$$\n\nComparing degrees, $3m = 2n + m$ or $m = n$, a contradiction.\n\nThus, all solutions are $P(x) = \\pm (x + 6)^n$ for all positive integers $n$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21111,
"subject": "Mathematics (Olympiad)",
"question": "Prove that for every positive real numbers $a, b, c$,\n\n$$\n\\frac{1+a^2}{1+b} + \\frac{1+b^2}{1+c} + \\frac{1+c^2}{1+a} \\ge 6(\\sqrt{2}-1).\n$$",
"options": [],
"answer": "See solution",
"solution": "Using the AM-GM inequality, we have\n\n$$\n\\frac{1+a^2}{1+b} + \\frac{1+b^2}{1+c} + \\frac{1+c^2}{1+a} \\ge 3\\sqrt[3]{\\frac{1+a^2}{1+a} \\cdot \\frac{1+b^2}{1+b} \\cdot \\frac{1+c^2}{1+c}} \\quad (1)\n$$\n\nOn the other hand, for every positive real number $x$, the following inequality holds:\n\n$$\n\\frac{1+x^2}{1+x} \\ge 2(\\sqrt{2}-1). \\quad (2)\n$$\n\nIndeed, inequality (2) is equivalent to\n\n$$\nx^2 - 2(\\sqrt{2}-1)x + 1 - (2\\sqrt{2}-2) \\ge 0,\n$$\ni.e.\n$$\nx^2 - 2(\\sqrt{2}-1)x + (\\sqrt{2}-1)^2 \\ge 0,\n$$\nand hence $(x - (\\sqrt{2} - 1))^2 \\ge 0$.\n\nFrom (1) and (2) we obtain\n\n$$\n\\frac{1+a^2}{1+b} + \\frac{1+b^2}{1+c} + \\frac{1+c^2}{1+a} \\ge 3\\sqrt[3]{8(\\sqrt{2}-1)^3} = 6(\\sqrt{2}-1).\n$$\n\nWe have equality if and only if $a = b = c = \\sqrt{2} - 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21112,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs of integers $ (x, y) $ satisfying the following condition: each of the numbers $ x^3 + y $ and $ x + y^3 $ is divisible by $ x^2 + y^2 $.",
"options": [],
"answer": "See solution",
"solution": "If $y = 0$, then $x^2$ divides $x$, so $x \\in \\{-1, 0, 1\\}$. Similarly, if $x = 0$, then $y \\in \\{-1, 0, 1\\}$. Thus, if $xy = 0$, we have five solutions:\n\n$$\n(x, y) \\in \\{(0, 0), (1, 0), (0, 1), (-1, 0), (0, -1)\\}.\n$$\n\nSuppose $xy \\ne 0$ and let $d = \\gcd(x, y)$, $x = du$, $y = dv$, with $\\gcd(u, v) = 1$. Since $x^2 + y^2$ divides $x^3 + y$, we have $d^2$ divides $d(d^2u^3 + v)$, so $d$ divides $v$. Likewise, from $x^2 + y^2$ divides $x + y^3$, we get $d$ divides $u$, so $d$ divides $\\gcd(u, v) = 1$, which means $x$ and $y$ are relatively prime.\n\nFrom $x^2 + y^2$ divides both $x^3 + y$ and $x + y^3$, we obtain $x^2 + y^2$ divides $y(xy - 1)$. But $\\gcd(x, y) = 1$ implies $\\gcd(x^2 + y^2, y) = 1$, so $x^2 + y^2$ divides $xy - 1$, which leads to $x^2 + y^2 \\leq |xy - 1|$. It follows that\n\n$$\n2|xy| \\leq x^2 + y^2 \\leq |xy - 1| \\leq |xy| + 1,\n$$\n\nso $|xy| \\leq 1$. Since $xy \\ne 0$, we have $|xy| = 1$, so $x, y \\in \\{-1, 1\\}$ and all the possibilities $(x, y) \\in \\{(\\pm1, \\pm1)\\}$ provide solutions.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21113,
"subject": "Mathematics (Olympiad)",
"question": "(a)\n\nConsider the set $S = \\{3, 6, 12, 24, 48, 96, 97, 98\\}$. The subsets of $S$ can be partitioned into the following categories:\n\n- Group 1: Subsets that do not contain 97 or 98.\n- Group 2: Subsets that contain 97 but not 98.\n- Group 3: Subsets that contain 98 but not 97.\n- Group 4: Subsets that contain both 97 and 98.\n\nGroup 1 generates subsets that have sums $0, 3, 6, 9, \\ldots, 189$; that is, all the multiples of 3 from 0 to 189 exactly once.\n\nGroup 2 generates subsets that have sums $97, 100, 103, \\ldots, 286$; these are all congruent to 1 modulo 3.\n\nGroup 3 generates subsets that have sums $98, 101, 104, \\ldots, 287$; these are all congruent to 2 modulo 3.\n\nGroup 4 generates subsets that have sums $195, 198, 201, \\ldots, 384$; these are all congruent to 0 modulo 3 and greater than 189.\n\nNote that the four groups cover all possible sums of subsets of $S$ with no sum appearing twice. Since $S$ has 8 elements, we have shown that 8 is 100-discerning.\n\n(b)\n\nSuppose, for the sake of contradiction, that 9 is 100-discerning. Then there is a set $S = \\{s_1, \\dots, s_9\\}$ with $0 < s_1 < \\dots < s_9 < 100$ that has no two different subsets with equal sums of elements.\n\nLet $X$ be the collection of all subsets of $S$ having at least 3 and at most 6 elements. Note that $X$ consists of exactly\n\n$$\n\\binom{9}{3} + \\binom{9}{4} + \\binom{9}{5} + \\binom{9}{6} = 84 + 126 + 126 + 84 = 420\n$$\n\nsubsets of $S$. The greatest possible sum of elements of a member of $X$ is $s_4 + s_5 + s_6 + s_7 + s_8 + s_9$, while the smallest possible sum is $s_1 + s_2 + s_3$. Since all sums of elements of members of $X$ are different, we must have\n\n$$\n s_4 + s_5 + s_6 + s_7 + s_8 + s_9 - s_1 - s_2 - s_3 \\geq 419. \\quad (1)\n$$\n\nLet $Y$ be the collection of all subsets of $S$ having exactly 2 or 3 or 4 elements greater than $s_3$. Observe that $\\{s_4, s_5, s_6, s_7, s_8, s_9\\}$ has $\\binom{6}{2}$ 2-element subsets, $\\binom{6}{3}$ 3-element subsets and $\\binom{6}{4}$ 4-element subsets, while $\\{s_1, s_2, s_3\\}$ has exactly 8 subsets. Hence the number of members of $Y$ is equal to\n\n$$\n8 \\left( \\binom{6}{2} + \\binom{6}{3} + \\binom{6}{4} \\right) = 8(15 + 20 + 15) = 400.\n$$\n\nNotice that the greatest possible sum of elements of a member of $Y$ is $s_1 + s_2 + s_3 + s_6 + s_7 + s_8 + s_9$, while the smallest possible sum is $s_4 + s_5$. Since all the sums of elements of members of $Y$ are assumed different, we must have\n\n$$\n s_1 + s_2 + s_3 + s_6 + s_7 + s_8 + s_9 - s_4 - s_5 \\geq 399. \\quad (2)\n$$\n\nAdding equations (1) and (2) we find\n\n$$\n2(s_6 + s_7 + s_8 + s_9) \\geq 818.\n$$\n\nHowever, this is impossible because $s_6, s_7, s_8, s_9 < 100$.\n\nWhat is the largest positive integer that is 100-discerning?\n\nUsing powers of 2 it is easy to show that 7 is 100-discerning. A standard argument shows that 10 is not 100-discerning. Indeed, there are $2^{10} = 1024$ subsets. The sums of the elements in those subsets are non-negative integers lying within the range from 0 to 945. So by the pigeonhole principle two subsets have equal sums.\n\nGetting an example to show 8 is 100-discerning is already a little tricky. To show that 9 is not 100-discerning is quite hard. The solution given here uses the idea of playing off different estimates against each other. But even finding a combination of such estimates that solves the problem is highly nontrivial.",
"options": [],
"answer": "See solution",
"solution": "Consider the set $S = \\{3, 6, 12, 24, 48, 96, 97, 98\\}$. The subsets of $S$ can be partitioned into the following categories:\n\n- Group 1: Subsets that do not contain 97 or 98.\n- Group 2: Subsets that contain 97 but not 98.\n- Group 3: Subsets that contain 98 but not 97.\n- Group 4: Subsets that contain both 97 and 98.\n\nGroup 1 generates subsets that have sums $0, 3, 6, 9, \\ldots, 189$; that is, all the multiples of 3 from 0 to 189 exactly once.\n\nGroup 2 generates subsets that have sums $97, 100, 103, \\ldots, 286$; these are all congruent to 1 modulo 3.\n\nGroup 3 generates subsets that have sums $98, 101, 104, \\ldots, 287$; these are all congruent to 2 modulo 3.\n\nGroup 4 generates subsets that have sums $195, 198, 201, \\ldots, 384$; these are all congruent to 0 modulo 3 and greater than 189.\n\nNote that the four groups cover all possible sums of subsets of $S$ with no sum appearing twice. Since $S$ has 8 elements, we have shown that 8 is 100-discerning. $\\Box$",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 21114,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. Each cell of an $n \\times n$ table is coloured in one of $k$ colours, where every colour is used at least once. Two colours $A$ and $B$ are said to touch each other if there exists a cell coloured in $A$ sharing a side with a cell coloured in $B$. The table is coloured in such a way that each colour touches at most 2 other colours. What is the maximal value of $k$?",
"options": [],
"answer": "See solution",
"solution": "$k = 2n-1$ when $n \\neq 2$ and $k = 4$ when $n = 2$.\n\n$k = 2n - 1$ is possible by colouring diagonally as shown in the figure below, and when $n = 2$, $k = 4$ is possible by colouring each cell in a unique colour.\n\n\n\nWe consider the graph where each node represents a colour and two nodes are linked if the colours they represent touch. This graph is connected, and since each colour touches at most 2 colours, every node has at most degree 2. This means that the graph is either one long chain or one big cycle.\n\n\n\nNow, consider the case when $n$ is odd. Consider the cell in the center of the table. From this cell, we can get to any other cell by passing through at most $n-1$ cells. Therefore, from the node representing this cell, we can get to any node through at most $n-1$ edges. But if the graph has $2n$ or more nodes, then for every node there is a node which is more than $n-1$ edges away. So we must have $k \\le 2n-1$ for all odd $n$.\n\nWhen $n$ is even, consider the 4 center cells. If they all have a different colour, then they form a 4-cycle in the graph, meaning the graph has only 4 nodes. If two of the center cells have the same colour, then from this colour you will be able to get to all other cells passing through at most $n-1$ cells. By the same arguments as in the odd case, we get $k \\le \\max(2n-1, 4)$ for even $n$.\n\nSo overall, we have $k \\le 2n-1$ for $n \\ne 2$ and $k \\le 4$ for $n = 2$ as desired.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21115,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x(x + f(y))) = (x + y)f(x)\n$$\nfor all $x, y \\in \\mathbb{R}$.",
"options": [],
"answer": "See solution",
"solution": "By setting $x = y = 0$, we get $f(0) = 0$.\n\nNext, setting $y = 0$ gives $f(x^2) = x f(x)$ for all $x$. Thus, $x f(x) = -x f(-x)$ for all $x$, which implies $f$ is odd.\n\nThe constant function $f(x) = 0$ for all $x \\in \\mathbb{R}$ is a solution.\n\nAssume $f$ is not identically zero. Then there exists $a \\in \\mathbb{R}$ with $f(a) \\neq 0$, so $a \\neq 0$. Suppose there is another $b \\neq 0$ with $f(b) = 0$. For $x = a$, $y = b$:\n$$\naf(a) = f(a^2) = f(a(a + f(b))) = (a + b)f(a) = a f(a) + b f(a).\n$$\nSince $f(a) \\neq 0$, we get $b = 0$.\n\nNow, set $y = -x$:\n$$\nf(x^2 + x f(-x)) = f(x^2 - x f(x)) = 0\n$$\nfor all $x$. Thus, $x^2 - x f(x) = 0$, i.e., $x^2 = x f(x) = f(x^2)$ for all $x$. Hence $f(x) = x$ for all $x > 0$, and since $f$ is odd, $f(x) = x$ for all $x \\in \\mathbb{R}$.\n\nTherefore, the solutions are $f(x) = 0$ and $f(x) = x$ for all $x \\in \\mathbb{R}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21116,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer such that the equation\n\n$$\n(x + y + u + v)^2 = n^2 x y u v\n$$\n\nhas positive integer solutions $(x, y, u, v)$. Find all possible values of $n$ for which such solutions exist.",
"options": [],
"answer": "See solution",
"solution": "Suppose $(x_0, y_0, u_0, v_0)$ is a positive integer solution with $x_0 + y_0 + u_0 + v_0$ minimal and $x_0 \\geq y_0 \\geq u_0 \\geq v_0$. Then $(y_0 + u_0 + v_0)^2$ is divisible by $x_0$, and $x_0$ is a root of the quadratic\n\n$$\nf(x) = x^2 - n^2 y_0^2 u_0 v_0 x + 2(y_0 + u_0 + v_0)x + (y_0 + u_0 + v_0)^2.\n$$\n\nBy Vieta's theorem, $x_1 = \\frac{(y_0 + u_0 + v_0)^2}{x_0}$ is also a positive integer root, so $(x_1, y_0, u_0, v_0)$ is a solution. By minimality, $x_1 \\geq x_0 \\geq y_0 \\geq u_0 \\geq v_0$.\n\nEvaluating $f(y_0)$ and using the sign theorem for quadratics:\n\n$$\n0 \\leq f(y_0) = y_0^2 - n^2 y_0^2 u_0 v_0 + 2(y_0 + u_0 + v_0)y_0 + (y_0 + u_0 + v_0)^2 \\leq 16y_0^2 - n^2 y_0^2 u_0 v_0\n$$\n\nThus $n^2 y_0^2 u_0 v_0 \\leq 16y_0^2$, so $n^2 u_0 v_0 \\leq 16$. Since $u_0, v_0 \\geq 1$, $n \\in \\{1, 2, 3, 4\\}$.\n\nChecking, $(x, y, u, v) = (4, 4, 4, 4), (2, 2, 2, 2), (1, 1, 2, 2), (1, 1, 1, 1)$ are solutions for $n = 1, 2, 3, 4$ respectively.\n\n**Answer:** The possible values of $n$ are $1, 2, 3, 4$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21117,
"subject": "Mathematics (Olympiad)",
"question": "Consider the sequence $\\{a_n\\}_{n \\ge 1}$ of positive integers with $a_n a_{n+3} = a_{n+2} a_{n+5}$ for all positive integers $n$. Determine the largest integer that always divides $$\\sum_{k=1}^{2550} a_{2k} a_{2k-1}.$$",
"options": [],
"answer": "See solution",
"solution": "From $a_n a_{n+3} = a_{n+2} a_{n+5}$ for every positive integer $n$, we have $a_{n+1} a_{n+4} = a_{n+3} a_{n+6}$ and $a_{n+2} a_{n+5} = a_{n+4} a_{n+7}$. Then\n\n$$\na_n a_{n+3} \\cdot a_{n+1} a_{n+4} \\cdot a_{n+2} a_{n+5} = a_{n+2} a_{n+5} \\cdot a_{n+3} a_{n+6} \\cdot a_{n+4} a_{n+7}.\n$$\n\nTherefore, $a_n a_{n+1} = a_{n+6} a_{n+7}$ for every positive integer $n$. Thus,\n\n$$\n\\sum_{k=1}^{2550} a_{2k} a_{2k-1} = \\frac{2550}{3} (a_1 a_2 + a_3 a_4 + a_5 a_6) = 850 (a_1 a_2 + a_3 a_4 + a_5 a_6)\n$$\n\nWe will show that 850 is the largest positive integer that always divides $\\sum_{k=1}^{2550} a_{2k} a_{2k-1}$. Consider the sequence $\\{a_n\\}_{n \\ge 1}$ defined by\n\n$$\na_n = \\begin{cases} 1 & \\text{if } n \\equiv 1, 2, 3 \\pmod{6}, \\\\ 2 & \\text{if } n \\equiv 4, 5, 6 \\pmod{6}. \\end{cases}\n$$\n\nIt can be seen that $a_n a_{n+3} = 2 = a_{n+2} a_{n+5}$ for every positive integer $n$; so, it satisfies the condition in the problem and\n\n$$\n\\sum_{k=1}^{2550} a_{2k} a_{2k-1} = 850(1 \\cdot 1 + 1 \\cdot 2 + 2 \\cdot 2) = 850 \\cdot 7.\n$$\n\nConsider another sequence $\\{a_n\\}_{n \\ge 1}$ defined by $a_n = 1$ for every positive integer $n$. We can see that $a_n a_{n+3} = 1 = a_{n+2} a_{n+5}$ for every positive integer $n$, and\n\n$$\n\\sum_{k=1}^{2550} a_{2k} a_{2k-1} = 850(1 \\cdot 1 + 1 \\cdot 1 + 1 \\cdot 1) = 850 \\cdot 3.\n$$\n\nThus, 850 is the desired largest integer.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21118,
"subject": "Mathematics (Olympiad)",
"question": "Find the sum of all integer bases $b > 9$ for which $17_b$ is a divisor of $97_b$.",
"options": [],
"answer": "See solution",
"solution": "If $17_b$ is a divisor of $97_b$, then $\\frac{9b+7}{b+7}$ is a positive integer. Note that\n$$\n\\frac{9b+7}{b+7} = 9 - \\frac{56}{b+7}.\n$$\nHence, $17_b$ is a divisor of $97_b$ if and only if $b > 9$ and $b + 7$ is a divisor of $56$. Because $56 = 2^3 \\cdot 7$, the two possibilities for $b$ are $b = 21$ and $b = 49$. The requested sum is $21 + 49 = 70$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21119,
"subject": "Mathematics (Olympiad)",
"question": "Let $A_0$, $A_1$, and $A_2$ be subsets of $\\{2, 3, \\ldots, 49\\}$, where each $A_i$ consists of elements that give remainder $i$ after dividing by $3$. Note that $A_0 \\cup A_1 \\cup A_2 = \\{2, 3, \\ldots, 49\\}$. How many ways are there to arrange the elements of $\\{2, 3, \\ldots, 49\\}$ into a sequence such that the sum of the mapped values (where $x \\in A_i$ is mapped to $i$) is not divisible by $3$?",
"options": [],
"answer": "See solution",
"solution": "To solve the problem, we map each $x \\in A_i$ to $i$, forming a sequence of $0$'s, $1$'s, and $2$'s. There are $16$ elements in $A_0$, $16$ in $A_1$, and $17$ in $A_2$ (since $49 - 2 + 1 = 48$ and $48 = 16 + 16 + 16$ plus one extra for $A_2$). We need to arrange the $1$'s and $2$'s so their sum is not divisible by $3$, and then distribute the $0$'s among the remaining places. The number of ways to place the $0$'s in the $48$ positions is $\\frac{48!}{32! \\cdot 16!}$, and the number of ways to assign the elements of $A_0$ to these positions is $16!$. By the product principle, the total number of ways to distribute the elements of $A_0$ is $\\frac{48!}{32!}$. For the sequence of $1$'s and $2$'s, the number of ways to distribute them is $16! \\cdot 17!$. Therefore, the total number of arrangements is $$\\frac{48!}{32!} \\cdot 16! \\cdot 17!$$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21120,
"subject": "Mathematics (Olympiad)",
"question": "Find all real triples $(x, y, z)$ that satisfy the system of equations:\n\n$$\nxy + 1 = 2z\n$$\n\n$$\nyz + 1 = 2x\n$$\n\n$$\nzx + 1 = 2y.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $p = xyz$. Multiplying equation (1) by $z$ gives\n\n$$\nxyz + z = 2z^2 \\implies 2z^2 + z + p = 0.\n$$\n\nSimilarly,\n\n$$\n2x^2 + x + p = 0 \\quad \\text{and} \\quad 2y^2 + y + p = 0.\n$$\n\nThus $x, y, z$ are all roots of the same quadratic equation. By the quadratic formula, each of $x, y, z$ is\n\n$$\n\\frac{1 - \\sqrt{1 + 8p}}{4} \\quad \\text{or} \\quad \\frac{1 + \\sqrt{1 + 8p}}{4}.\n$$\n\n**Case 1:** Not all of $x, y, z$ are equal.\n\nBy symmetry, assume $x \\neq y$, so $p \\neq 0$. The product $xy$ is the product of the two expressions above:\n\n$$\nxy = \\left( \\frac{1 - \\sqrt{1 + 8p}}{4} \\right) \\left( \\frac{1 + \\sqrt{1 + 8p}}{4} \\right) = -\\frac{p}{2}.\n$$\n\nSince $p = xyz \\neq 0$, $z = -2$. But $z$ must be one of the expressions above. The second is positive, but $z = -2 < 0$, so\n\n$$\n-2 = \\frac{1 - \\sqrt{1 + 8p}}{4}.\n$$\n\nSolving for $p$ gives $p = 10$. Substituting into the expressions yields $\\{x, y\\} = \\left\\{\\frac{5}{2}, -2\\right\\}$. Thus $(x, y, z)$ can be any permutation of $(\\frac{5}{2}, -2, -2)$.\n\n**Case 2:** $x = y = z$\n\nEquation (1) becomes $x^2 + 1 = 2x$, or $(x - 1)^2 = 0$. Therefore $x = 1$, so $(x, y, z) = (1, 1, 1)$.\n\nBoth $(1, 1, 1)$ and all permutations of $(\\frac{5}{2}, -2, -2)$ satisfy the original equations.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21121,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge m \\ge 1$ be integers. Prove that\n$$\n\\sum_{k=m}^{n} \\left( \\frac{1}{k^2} + \\frac{1}{k^3} \\right) \\ge m \\cdot \\left( \\sum_{k=m}^{n} \\frac{1}{k^2} \\right)^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "By Cauchy-Schwarz, we have\n$$\n\\begin{align*}\n\\sum_{k=m}^{n} \\frac{k+1}{k^3} &= \\sum_{k=m}^{n} \\frac{\\left(\\frac{1}{k^2}\\right)^2}{\\frac{1}{k(k+1)}} \\\\\n&\\ge \\frac{\\left(\\frac{1}{m^2} + \\frac{1}{(m+1)^2} + \\dots + \\frac{1}{n^2}\\right)^2}{\\frac{1}{m(m+1)} + \\frac{1}{(m+1)(m+2)} + \\dots + \\frac{1}{n(n+1)}} \\\\\n&= \\frac{\\left(\\frac{1}{m^2} + \\frac{1}{(m+1)^2} + \\dots + \\frac{1}{n^2}\\right)^2}{\\frac{1}{m} - \\frac{1}{n+1}} \\\\\n&> \\frac{\\left(\\sum_{k=m}^{n} \\frac{1}{k^2}\\right)^2}{\\frac{1}{m}}\n\\end{align*}\n$$\nas desired.\n\n**Remark** (Bound on error). Let $A = \\sum_{k=m}^{n} k^{-2}$ and $B = \\sum_{k=m}^{n} k^{-3}$. The inequality above becomes tighter for large $m$ and $n \\gg m$. If we use Lagrange's identity in place of Cauchy-Schwarz, we get\n$$\nA + B - mA^2 = m \\cdot \\sum_{m \\le a < b} \\frac{(a-b)^2}{a^3 b^3 (a+1)(b+1)}.\n$$\nWe can upper bound this error by\n$$\n\\le m \\cdot \\sum_{m \\le a < b} \\frac{1}{a^3 (a+1) b (b+1)} = m \\cdot \\sum_{m \\le a} \\frac{1}{a^3 (a+1)^2} \\approx m \\cdot \\frac{1}{m^4} = \\frac{1}{m^3},\n$$\nwhich is still generous as $(a-b)^2 \\ll b^2$ for $b$ not much larger than $a$, so the real error is probably around $\\frac{1}{10m^3}$. This exhibits the tightness of the inequality since it implies\n$$\nmA^2 + O(B/m) > A + B.\n$$\n**Remark** (Construction commentary, from author). My motivation was to write an inequality where Titu could be applied creatively to yield a telescoping sum. This can be difficult because most of the time, such a reverse-engineered inequality will be so loose it's trivial anyways. My first attempt was the not-so-amazing inequality\n$$\n\\frac{n^2 + 3n}{2} = \\sum_{i=1}^{n} i + 1 = \\sum_{i=1}^{n} \\frac{\\frac{1}{i}}{\\frac{1}{i(i+1)}} > \\left( \\sum_{i=1}^{n} \\frac{1}{\\sqrt{i}} \\right)^2,\n$$\nwhich is really not surprising given that $\\sum \\frac{1}{\\sqrt{i}} \\ll \\frac{n}{\\sqrt{2}}$. The key here is that we need “near-equality” as dictated by the Cauchy-Schwarz equality case, i.e. the square root of the numerators should be approximately proportional to the denominators.\n\nThis motivates using $\\frac{1}{i^4}$ as the numerator, which works like a charm. After working out the resulting statement, the LHS and RHS even share a sum, which adds to the simplicity of the problem.\n\nThe final touch was to unrestrict the starting value of the sum, since this allows the strength of the estimate $\\frac{1}{i^2} \\approx \\frac{1}{i(i+1)}$ to be fully exploited.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21122,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle and $I$ its incenter. The circumcircle of $ACI$ intersects the line $BC$ a second time at the point $X$, and the circumcircle of $BCI$ intersects the line $AC$ a second time at the point $Y$.\n\n*Prove that the segments* $AY$ *and* $BX$ *are of equal length.*",
"options": [],
"answer": "See solution",
"solution": "We shall show that $AB = BX$ holds. Since $AB = AY$ then follows by the same argument, this completes the proof.\n\n\n\nIn this solution, we use oriented angles between lines (modulo $180^{\\circ}$) with the notation $\\angle PQR$. As usual, the angles of the triangle $ABC$ are denoted by $\\alpha = \\angle BAC$, $\\beta = \\angle CBA$, and $\\gamma = \\angle ACB$.\n\nThe inscribed angle theorem gives\n\n$$\n\\angle AXB = \\angle AXC = \\angle AIC = -\\angle CIA = 180^{\\circ} - \\angle CIA = \\angle IAC + \\angle ACI = \\frac{1}{2}(\\alpha + \\gamma).\n$$\n\nThis immediately implies\n\n$$\n\\angle BAX = -\\angle AXB - \\angle XBA = -\\frac{1}{2}(\\alpha + \\gamma) - \\beta = \\frac{1}{2}(\\alpha + \\gamma).\n$$\n\nTherefore, the triangle $ABX$ is indeed isosceles, and we are done.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21123,
"subject": "Mathematics (Olympiad)",
"question": "Let $(abc)_7$ be a three-digit number in base $7$, where $a$, $b$, and $c$ are digits less than $7$. Suppose $(abc)_7 = (cba)_9$, where $(cba)_9$ is the three-digit number in base $9$ formed by reversing the digits. Find the value of $(abc)_7$.",
"options": [],
"answer": "See solution",
"solution": "Let $(abc)_7 = 49a + 7b + c$ and $(cba)_9 = 81c + 9b + a$. Setting these equal:\n\n$$49a + 7b + c = 81c + 9b + a$$\n\nRearrange:\n$$48a - 2b - 80c = 0$$\n$$24a - b - 40c = 0$$\n$$b = 24a - 40c$$\n\nSince $b$ is a digit less than $7$ and a multiple of $8$, $b = 0$.\n\nSo $3a - 5c = 0$, or $a = 5$, $c = 3$.\n\nThus, $(abc)_7 = (503)_7 = 5 \\times 49 + 0 \\times 7 + 3 = 245 + 3 = 248$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21124,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge 2$ be a positive integer. A positive real number is written in each unit square of a $2 \\times n$ grid, so that the sum of the two numbers in each column is $1$. Suppose that, regardless of the numbers written, we could always delete one number from each column so that the sum of the remaining numbers in each row is at most $a$. What is the smallest possible value of $a$?",
"options": [],
"answer": "See solution",
"solution": "The minimum value of $a$ is\n\n$$\nc = \\frac{\\left\\lfloor \\frac{(n+1)^2}{4} \\right\\rfloor}{n+1} = \\begin{cases} \\frac{k+1}{2}, & n = 2k+1 \\\\ \\frac{k(k+1)}{2k+1}, & n = 2k. \\end{cases}\n$$\n\nWe observe that if the $2 \\times n$ grid is given as below, the value of $a$ is not less than $c$.\n\n$n = 2k$:\n\n\n\n$n = 2k + 1$:\n\n\n\nLet us show that regardless of the numbers written in the $2 \\times n$ grid, we can delete one number from each column, totally $n$ numbers, so that the sum of the remaining numbers in each row is at most $c$. Let $a_1 \\le a_2 \\le \\dots \\le a_n$ be the numbers in the first row with sum $A$ and let $b_1 \\le b_2 \\le \\dots \\le b_n$ be the numbers in the second row with sum $B$.\n\n**Lemma.** If $a_1 + \\dots + a_{s+1} > c$ and $b_1 + \\dots + b_{t+1} > c$, then $s + t \\ge n$.\n\n*Proof.* From the first condition we get that\n\n$$\n\\frac{a_1 + \\dots + a_{s+1}}{s+1} > \\frac{c}{s+1}.\n$$\n\nSince\n\n$$\n\\frac{a_1 + \\dots + a_n}{n} \\ge \\frac{a_1 + \\dots + a_{s+1}}{s+1},\n$$\n\nit follows that $\\frac{A}{n} > \\frac{c}{s+1}$. Similarly, $\\frac{B}{n} > \\frac{c}{t+1}$. Since $A+B=n$, we get that\n\n$$\n\\frac{1}{c} > \\frac{1}{s+1} + \\frac{1}{t+1}. \\qquad (1)\n$$\n\n**Case:** $n = 2k + 1$. Suppose that $s + t \\le 2k$. Then $s + 1 + t + 1 \\le 2k + 2$ and\n\n$$\n\\frac{2}{\\frac{1}{s+1} + \\frac{1}{t+1}} \\le \\frac{s+1+t+1}{2} \\le \\frac{2k+2}{2}.\n$$\n\nHence $\\frac{1}{s+1} + \\frac{1}{t+1} \\ge \\frac{2}{k+1} = \\frac{1}{c}$, which contradicts (1).\n\n**Case:** $n = 2k$. Suppose that $s + t \\le 2k - 1$. Thus $s + 1 + t + 1 \\le 2k + 1$. Similarly to the previous case we get that\n\n$$\n\\frac{1}{s+1} + \\frac{1}{t+1} \\ge \\frac{1}{k+1} + \\frac{1}{k} = \\frac{2k+1}{(k+1)k} = \\frac{1}{c},\n$$\n\nwhich contradicts (1).\n\nLet $s_0$ be the minimum $s$ satisfying $a_1 + \\cdots + a_{s+1} > c$ and let $t_0$ be the minimum $t$ satisfying $b_1 + \\cdots + b_{t+1} > c$. Then by the lemma $s_0 + t_0 \\ge n$. Because of the choice of $s_0$ and $t_0$, it follows that $a_1 + \\cdots + a_{s_0} \\le c$ and $b_1 + \\cdots + b_{t_0} \\le c$. This shows that we can delete $n-s_0$ numbers from the first row and $n-t_0$ numbers from the second row so that the sum of the remaining numbers in each row is at most $c$. This completes the proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21125,
"subject": "Mathematics (Olympiad)",
"question": "Let $f : \\mathbb{Z} \\to \\mathbb{Z}$ be a function such that for all integers $m, n$, the following holds:\n\n$$\nf(m^3 + f(n)) = f(m)^3 + n.\n$$\n\nFind all such functions $f$.",
"options": [],
"answer": "See solution",
"solution": "$$\n\\begin{array}{l}\n\\text{Let } P(m, n) \\text{ denote the assertion: } f(m^3 + f(n)) = f(m)^3 + n. \\\\\n\\text{Add } k^3 \\text{ to both sides and apply } f:\\\\\nf(k^3 + f(m^3 + f(n))) = f(k^3 + f(m)^3 + n) \\\\\nP(k, m^3 + f(n)) \\implies f(k^3 + f(m^3 + f(n))) = f(k)^3 + m^3 + f(n) \\\\\n\\implies f(k^3 + f(m)^3 + n) = f(k)^3 + m^3 + f(n) \\tag{*}\n\\end{array}\n$$\n\nSet $n = l^3$ in $(*)$:\n$$\nf(k^3 + f(m)^3 + l^3) = f(k)^3 + m^3 + f(l^3)\n$$\nSet $k = l$, $n = k^3$ in $(*)$:\n$$\nf(l^3 + f(m)^3 + k^3) = f(l)^3 + m^3 + f(k^3)\n$$\nThus,\n$$\nf(k)^3 - f(k^3) = f(l)^3 - f(l^3) = c, \\quad c \\in \\mathbb{Z}\n$$\n\nNow, set $k = f(s)$ in $(*)$:\n$$\nf(f(s)^3 + f(m)^3 + n) = f(f(s))^3 + m^3 + f(n)\n$$\nSimilarly, set $k = l$, $n = k^3$:\n$$\nf(f(m)^3 + f(s)^3 + n) = f(f(m))^3 + s^3 + f(n)\n$$\nSo,\n$$\nf(f(s))^3 - s^3 = f(f(m))^3 - m^3 = t, \\quad t \\in \\mathbb{Z}\n$$\nThus, $f(f(m))^3 = m^3 + t$. For large $m$, $m^3 + t$ is not a perfect cube unless $t = 0$, so $f(f(m))^3 = m^3$ and $f(f(m)) = m$.\n\nFrom earlier, set $l = f(m)$:\n$$\nf(f(m))^3 = f(f(m)^3) + c\n$$\nBut $f(f(m))^3 = m^3$, so $f(f(m)^3) = m^3 - c$.\n\nSuppose $c < 0$:\n$$\nf(l)^3 - c = f(l^3)\n$$\nBut then $f(l^3) = f(l^3 + f(-c))$, so $f(-c) = 0$. But $f$ is injective:\n\nSuppose $f(a) = f(b)$:\n$$\nP(m, a): f(m^3 + f(a)) = f(m)^3 + a \\\\\nP(m, b): f(m^3 + f(b)) = f(m)^3 + b\n$$\nSo $a = b$, so $f$ is injective. Thus $f(-c) = 0$ is impossible unless $c = 0$.\n\nTherefore, $c = 0$, so $f(l^3) = f(l)^3$.\n\nNow, $f(1^3) = f(1)^3$, so $f(1)(f(1)^2 - 1) = 0$, so $f(1) = 1$.\n\nNow, $P(1, n): f(1 + f(n)) = 1 + n$.\n\nBy induction, if $f(n) = n$, then $f(n+1) = n+1$.\n\nThus, the only solution is $f(n) = n$ for all $n \\in \\mathbb{Z}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21126,
"subject": "Mathematics (Olympiad)",
"question": "Suppose five line segments are given on a plane satisfying the following property:\n\nOf the ten possibilities for choosing three line segments from the given five, in nine cases, one can form an acute triangle with the chosen three line segments.\n\nProve that in the remaining tenth possibility for the choice of three line segments, there is a triangle with the three chosen line segments as its sides.",
"options": [],
"answer": "See solution",
"solution": "When three line segments with lengths $x, y, z$ ($x \\leq y \\leq z$) are given, the necessary and sufficient condition for these line segments to form a triangle is $x + y > z$, and if the triangle they form is acute, then the condition $x^2 + y^2 > z^2$ is also satisfied.\n\nLet us denote by $a, b, c, d, e$ ($a \\leq b \\leq c \\leq d \\leq e$) the lengths of the given five line segments.\n\nSuppose that $\\{a, b, e\\}$ can form a triangle. Then, since $a + b \\geq e$ must hold, we can conclude that for any choice of three line segments from the given five, the sum of the lengths of any two among the three is greater than the length of the remaining one, and hence we can construct a triangle for any of the nine combinations of three line segments besides the combination $\\{a, b, e\\}$. Therefore, in order to prove the assertion of the problem, it suffices to show that if we assume that for any choice of three line segments, besides the combination $\\{a, b, e\\}$, we can form an acute triangle with the chosen three line segments, then we can also form a triangle with $\\{a, b, e\\}$.\n\nBy assumption, then, we can form an acute triangle from $\\{a, b, c\\}$ and from $\\{a, c, e\\}$. We therefore have $a^2 + b^2 > c^2$ and $a^2 + c^2 > e^2$. From these inequalities, we obtain\n\n$$\n(a + b)^2 = a^2 + 2ab + b^2 \\geq 2a^2 + b^2 > a^2 + c^2 > e^2,\n$$\n\nand we conclude that $a + b \\geq e$ so that we can form a triangle with $\\{a, b, e\\}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21127,
"subject": "Mathematics (Olympiad)",
"question": "Let $A, B, C, D, E, F$ be 6 distinct points on a circle such that\n$$\n\\overline{AB} = \\overline{BC} = \\overline{CD} = \\overline{DE} = \\overline{EF} = \\overline{FA}.\n$$\nLet $O$ be the centre of the circle. Note that any two points in the same sector among $AOB$, $BOC$, $COD$, $DOE$, $EOF$, $FOA$ have distance at most 1. Let $n_1, n_2, \\dots, n_6$ be the number of points in the 6 sectors respectively. Show that\n$$\n\\binom{n_1}{2} + \\binom{n_2}{2} + \\dots + \\binom{n_6}{2} \\ge 2001.\n$$",
"options": [],
"answer": "See solution",
"solution": "Since the binomial function $\\binom{x}{2}$ is convex, by Jensen's inequality, we have\n$$\n\\sum_{k=1}^{6} \\binom{n_k}{2} \\ge 6 \\left( \\frac{n_1+n_2+\\dots+n_6}{2} \\right) > 6 \\binom{35}{2} = 3570 > 2001.\n$$\nThis completes the proof.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21128,
"subject": "Mathematics (Olympiad)",
"question": "A _complete number_ is a 9-digit number that contains each of the digits 1 to 9 exactly once. The _difference number_ of a number $N$ is the number you get by taking the differences of consecutive digits in $N$ and then stringing these digits together. For instance, the difference number of 25143 is equal to 3431. The complete number 124356879 has the additional property that its difference number, 12121212, consists of digits alternating between 1 and 2.\n\nDetermine all $a$ with $3 \\leq a \\leq 9$ for which there exists a complete number $N$ with the additional property that the digits of its difference number alternate between 1 and $a$.",
"options": [],
"answer": "See solution",
"solution": "For $a = 4$, an example of such a number is 126734895. For $a = 5$, an example is the number 549832761. (There are other solutions as well.)\n\nWe will show that for $a = 3, 6, 7, 8, 9$ there is no complete number with a difference number equal to $1a1a1a1a$. It then immediately follows that there is also no complete number $N$ with difference number equal to $a1a1a1a1$ (otherwise, we could write the digits of $N$ in reverse order and obtain a complete number with difference number $1a1a1a1a$).\n\nFor $a$ equal to 6, 7, 8, and 9, no such number $N$ exists for the following reason. For the digits 4, 5, and 6, there is no digit that differs by $a$ from that digit. Since the difference number of the complete number $N$ is equal to $1a1a1a1a$, every digit of $N$, except the first, must be next to a digit that differs from it by $a$. Hence, the digits 4, 5, and 6 can only occur in the first position of $N$, which is impossible.\n\nFor $a = 3$ the argument is different. If we consider the digits that differ by 3, we find the triples 1–4–7, 2–5–8, and 3–6–9. If the 1 is next to the 4 in $N$, the 7 cannot be next to the 4 and so the 7 must be the first digit of $N$. If the 1 is not next to the 4, the 1 must be the first digit of $N$. In the same way, either the 2 or the 8 must be the first digit of $N$ as well. This is impossible.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21129,
"subject": "Mathematics (Olympiad)",
"question": "A ball is shot at a $45^\\circ$ angle from the lower right corner of a rectangular billiard table with integer side lengths $a$ and $b$. The ball bounces off the sides until it lands in a pocket. For which ordered pairs $(a, b)$ with $1 \\leq b \\leq a \\leq N$ does the ball land in the upper left pocket? (Assume $N$ is a given positive integer.)\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $c = \\text{lcm}(a, b)$. The ball lands in a pocket as soon as the line segment reaches $(c, c)$, where $c$ is divisible by both $a$ and $b$. The ball touches a vertical rail $c$ times and a horizontal rail $c$ times. If $c$ is even/odd, the ball lands in a pocket on the left/right side of the table, and if $c$ is even/odd, the ball lands in a pocket on the lower/upper side of the table. Hence, the ball lands in the upper left pocket exactly when $c$ is even and $c$ is odd.\n\nWrite $a = 2^j a'$ and $b = 2^k b'$, where $a'$ and $b'$ are odd. Then $c = 2^{\\max(j, k)} \\cdot \\text{lcm}(a', b')$. Since $\\text{lcm}(a', b')$ is odd, $c$ is even and $c$ is odd precisely when $j < k$.\n\nIf $a$ is odd, then the ball lands in the upper left pocket whenever $b$ is even; there are $\\frac{a-1}{2}$ such allowed values of $b$. If $a$ is divisible by $2$ but not by $4$, then the ball lands in the upper left pocket whenever $b$ is divisible by $4$; there are $\\frac{a-2}{4}$ such allowed values of $b$, and so forth.\n\nFix an integer $N$ and count the ordered pairs $(a, b)$ with $1 \\leq b \\leq a \\leq N$ for which the ball lands in the upper left pocket. In the calculation we use $C(k, 2) = \\binom{k}{2} = 0 + 1 + 2 + \\dots + (k-2) + (k-1)$.\n\nWhen $a = 1, 3, 5, 7, \\dots$, then there are $0, 1, 2, 3, \\dots$ allowed values of $b$; summing over all such $a$ gives us $C\\left(\\left[\\frac{N+1}{2}\\right], 2\\right)$ total ordered pairs. When $a = 2, 6, 10, 14, \\dots$, there are $0, 1, 2, 3, \\dots$ allowed values of $b$; summing over all such $a$ gives us $C\\left(\\left[\\frac{N+2}{4}\\right], 2\\right)$ total ordered pairs, and so forth. For $N = 2009$, there are a total of $C(1005, 2) + C(502, 2) + C(251, 2) + C(126, 2) + C(63, 2) + C(31, 2) + C(16, 2) + C(8, 2) + C(4, 2) + C(2, 2) = 672,084$ allowed ordered pairs, so this is our answer.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21130,
"subject": "Mathematics (Olympiad)",
"question": "Let $x_1, x_2, x_3, x_4, x_5$ be integers such that $x_1 + x_2 + x_3 + x_4 + x_5 = 2006$. For which values of $x_1, x_2, x_3, x_4, x_5$ does the sum\n\n$$\nS = \\sum_{1 \\le i < j \\le 5} x_i x_j\n$$\n\nreach its maximum?",
"options": [],
"answer": "See solution",
"solution": "To maximize $S$, all $x_i$ should be as close as possible, i.e., $|x_i - x_j| \\le 1$ for all $i, j$. If not, transferring 1 from a larger $x_i$ to a smaller $x_j$ increases $S$:\n\nLet $x_1 - x_2 \\ge 2$, set $x_1' = x_1 - 1$, $x_2' = x_2 + 1$, others unchanged. Then\n\n$$\nS' - S = x_1' x_2' - x_1 x_2 > 0,\n$$\n\ncontradicting maximality. Thus, the maximum occurs when $x_1 = 402$, $x_2 = x_3 = x_4 = x_5 = 401$ (or any permutation).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21131,
"subject": "Mathematics (Olympiad)",
"question": "An exam consists of six questions and is taken by 2006 children. Each question is marked either right or wrong. Any three children have right answers to at least five of the six questions between them. Let $N$ be the total number of right answers achieved by all the children (i.e., the total number of questions solved by child 1 plus the total solved by child 2, and so on, up to child 2006). Find the least possible value of $N$.",
"options": [],
"answer": "See solution",
"solution": "Consider all students who scored four or fewer. If any student scored less than four, give them extra correct answers arbitrarily until each has four.\n\nThere are $\\binom{6}{4} = 15$ ways to choose 4 from 6, so if there are more than 30 students with exactly four correct answers, by the pigeonhole principle, some three will have the same set of four, and thus only four between them, violating the condition.\n\nTherefore, at least $2006 - 30 = 1976$ students scored five or more.\n\nA possible minimal configuration is:\n\n- 1976 students scored exactly five,\n- 30 students scored exactly four, in such a way that each combination of four problems was solved by exactly two students.\n\nThis gives:\n\n$$\nN = 1976 \\times 5 + 30 \\times 4 = 10,000.\n$$\n\nTo prove this is minimal, suppose there is a configuration with 30 students who have fewer than 120 correct answers among them. Let $A$ be the number who scored five, $B$ the number who scored four, $C$ three, $D$ two, $E$ one, $F$ zero. Then:\n\n$$\nA + B + C + D + E + F = 30 \\quad (1)\n$$\n$$\n5A + 4B + 3C + 2D + E < 120 \\quad (2)\n$$\n\nA student scoring $k < 4$ can be replaced by $\\binom{6}{4-k}$ students scoring four, so:\n\n$$\nB + 3C + 6D + 10E + 15F \\leq 30 \\quad (3)\n$$\n\nSubtracting (1) from (3):\n\n$$\n- A + 2C + 5D + 9E + 14F \\leq 0 \\implies 2C + 5D + 9E + 14F \\leq A\n$$\n\nFrom (2) minus $4 \\times$ (1):\n\n$$\nA - C - 2D - 3E - 4F < 0 \\implies A < C + 2D + 3E + 4F\n$$\n\nSo:\n\n$$\n2C + 5D + 9E + 14F \\leq A < C + 2D + 3E + 4F\n$$\n\nThus:\n\n$$\nC + 3D + 6E + 10F < 0\n$$\n\nBut $C, D, E, F \\geq 0$, so this is impossible. Therefore, $N = 10,000$ is minimal.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21132,
"subject": "Mathematics (Olympiad)",
"question": "a) Prove that infinitely many pairs of real numbers $(x, y)$ exist such that $x, y \\in [0, \\sqrt{3}]$ and the following equation holds:\n$$x \\cdot \\sqrt{3-y^2} + y \\cdot \\sqrt{3-x^2} = 3$$\n\nb) Prove that *no* pair of rational numbers $(x, y)$ exists such that $x, y \\in [0, \\sqrt{3}]$ and the following equation holds:\n$$x \\cdot \\sqrt{3-y^2} + y \\cdot \\sqrt{3-x^2} = 3$$",
"options": [],
"answer": "See solution",
"solution": "a) Any pair $(a, \\sqrt{3-a^2})$, with $a \\in [0, 1]$, is a solution.\n\nb) By squaring the equality $y \\cdot \\sqrt{3-x^2} = 3 - x \\cdot \\sqrt{3-y^2}$, we deduce that $(\\sqrt{3-y^2} - x)^2 = 0$, therefore $x^2 + y^2 = 3$. \n\nAssume that there are numbers $x, y \\in \\mathbb{Q}_+ \\cap [0, \\sqrt{3}]$, for which $x^2 + y^2 = 3$.\n\nIt is obvious that $x \\ne 0$ and $y \\ne 0$. Consider the positive integers $a, b, c, d$, such that $(a, b) = (c, d) = 1$, $x = \\frac{a}{b}$ and $y = \\frac{c}{d}$. From $x^2 + y^2 = 3$ we obtain $a^2d^2 + b^2c^2 = 3b^2d^2$.\n\nFrom $(c, d) = 1$ and $d^2 \\mid b^2c^2$, we deduce $d^2 \\mid b^2$. Similarly, from $(a, b) = 1$ and $b^2 \\mid a^2d^2$, we obtain $b^2 \\mid d^2$. Therefore, $b^2 = d^2$ and this leads to $a^2 + c^2 = 3b^2$.\n\nIf the integers $a$ and $c$ aren't multiples of $3$, then $a^2 + c^2 = M_3 + 2 \\ne 3b^2$, which is false. Consequently $3 \\mid a$ and $3 \\mid c$, therefore $3b^2 = M_9$, thus $3 \\mid b$ and $(a, b) \\ge 3$, which is false. This contradicts our assumption and the conclusion follows.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21133,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be an integer greater than $1$. The set $S$ of all diagonals of a $(4n-1)$-gon is partitioned into $k$ sets, $S_1, \\dots, S_k$, so that, for every pair of distinct indices $i$ and $j$, some diagonal in $S_i$ crosses some diagonal in $S_j$; that is, the two diagonals share an interior point. Determine the largest possible value of $k$ in terms of $n$.",
"options": [],
"answer": "See solution",
"solution": "The required maximum is $k = (n-1)(4n-1)$. Clearly, $|S| = 2(n-1)(4n-1)$. To begin, we show that $k \\leq (n-1)(4n-1)$. Otherwise, some $S_i$ is a singleton set, say $S_i = \\{\\delta\\}$. Let $m$ be the number of vertices on one side of $\\delta$, so the number of vertices on the other side is $4n - m - 3$, and the total number of diagonals crossing $\\delta$ is $m(4n - m - 3) \\leq 2(n-1)(2n-1)$. Notice that each $S_j$, $j \\neq i$, contains such a diagonal, to infer that $k \\leq 2(n-1)(2n-1)+1 = (n-1)(4n-1)-(n-2) \\leq (n-1)(4n-1)$ and thereby reach a contradiction.\n\nTo exhibit a partition of $S$ into $(n-1)(4n-1)$ sets satisfying the condition in the statement, label the vertices of the $(4n-1)$-gon in circular order, $A_1, A_2, \\dots, A_{4n-1}$, and set\n\n$$\nS_{i,j} = \\{A_i A_{i+j},\\; A_{i+j-1} A_{i+2n}\\}, \\quad i = 1, 2, \\dots, 4n-1, \\quad j = 2, 3, \\dots, n,\n$$\n\nwhere indices are reduced modulo $4n-1$.\n\nIt is easily seen that the $S_{i,j}$ form a partition of $S$. To show that they satisfy the condition in the statement, consider two such, say $S_{i,j}$ and $S_{i',j'}$. By cyclic symmetry, we may (and will) assume that $i = 0$. Notice that for a diagonal $\\delta$ to cross no diagonal in $S_{0,j}$ it is necessary and sufficient that its endpoints both fall in one of the sets below:\n\n$$\n\\{A_0, A_1, \\dots, A_{i-1}\\}, \\quad \\{A_i, A_{i+1}, \\dots, A_{2n}\\}, \\quad \\{A_{2n}, A_{2n+1}, \\dots, A_{4n-1}\\}\n$$\n\n(recall that $A_{4n-1} = A_0$); if this is the case, we say that that set covers $\\delta$. Now, since each set above encompasses at most $2n$ consecutive vertices, none of these sets can cover both diagonals in $S_{i',j'}$. On the other hand, since the latter cross one another, they cannot be covered by different sets each either. Consequently, some diagonal in $S_{0,j}$ must cross some diagonal in $S_{i',j'}$ and the conclusion follows.\n\n**Remark.** The case of a $(4n-3)$-gon can be dealt with similarly; one only needs to take some care of the diagonals of the form $A_i A_{i+n}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21134,
"subject": "Mathematics (Olympiad)",
"question": "Let $A_n = (n^2 + 11n - 4) \\cdot n! + 33 \\cdot 13^n + 4$. For which positive integers $n$ is $A_n$ a perfect square?",
"options": [],
"answer": "See solution",
"solution": "First, consider $A_n$ modulo $8$ for $n \\geq 4$. Since $8$ divides $n!$ for $n \\geq 4$, we have:\n\n$$\nA_n \\equiv 33 \\cdot 13^n + 4 \\pmod{8}\n$$\n\nNote that $13 \\equiv 5 \\pmod{8}$ and $33 \\equiv 1 \\pmod{8}$, so $A_n \\equiv 5^n + 4 \\pmod{8}$. Since $5^2 \\equiv 1 \\pmod{8}$, for even $n$, $5^n \\equiv 1$, so $A_n \\equiv 5 \\pmod{8}$; for odd $n$, $5^n \\equiv 5$, so $A_n \\equiv 1 \\pmod{8}$. A perfect square modulo $8$ can only be $0$, $1$, or $4$, so all even $n \\geq 4$ are excluded.\n\nNext, consider $A_n$ modulo $7$ for $n \\geq 7$. Since $7$ divides $n!$ for $n \\geq 7$, we have:\n\n$$\nA_n \\equiv 33 \\cdot 13^n + 4 \\pmod{7}\n$$\n\n$13 \\equiv -1 \\pmod{7}$ and $33 \\equiv 5 \\pmod{7}$, so $A_n \\equiv 5 \\cdot (-1)^n + 4 \\pmod{7}$. For even $n$, $A_n \\equiv 2 \\pmod{7}$; for odd $n$, $A_n \\equiv 6 \\pmod{7}$. A perfect square modulo $7$ can only be $0$, $1$, $2$, or $4$, so all odd $n \\geq 7$ are excluded.\n\nThe remaining possibilities are $n = 1, 2, 3, 5$.\n\nFor $n = 3$:\n$$\nA_3 = (9 + 33 - 4) \\cdot 6 + 33 \\cdot 2197 + 4 = 38 \\cdot 6 + 33 \\cdot 2197 + 4\n$$\nModulo $5$, $A_3 \\equiv 3 \\cdot 1 + 3 \\cdot 3^3 + 4 \\equiv 3 \\pmod{5}$, which is not a possible square residue.\n\nFor $n = 5$:\n$$\nA_5 \\equiv (-4) \\cdot 0 + 3 \\cdot 3^5 + 4 \\equiv 3 \\pmod{5}\n$$\nAgain, not a possible square residue.\n\nFor $n = 1$, $A_1 = (1 + 11 - 4) \\cdot 1 + 33 \\cdot 13 + 4 = 8 \\cdot 1 + 429 + 4 = 441 = 21^2$.\n\nFor $n = 2$, $A_2 = (4 + 22 - 4) \\cdot 2 + 33 \\cdot 169 + 4 = 22 \\cdot 2 + 5577 + 4 = 44 + 5577 + 4 = 5625 = 75^2$.\n\nThus, the only solutions are $n = 1$ and $n = 2$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21135,
"subject": "Mathematics (Olympiad)",
"question": "Given a convex hexagon $ABCDEF$ such that $\\angle A = \\angle C = \\angle E$ and $AB = BC$, $CD = DE$, $EF = FA$, prove that the lines $AD$, $BE$, and $CF$ have a common point.",
"options": [],
"answer": "See solution",
"solution": "**Solution.** Assume that the angle bisectors of $\\angle B$ and $\\angle D$ intersect at $P$.\n\nWe shall prove that the hexagon $ABCDEF$ has an inscribed circle whose center is $P$. Then the conclusion follows from Brianchon's Theorem.\n\nThe equality $AB = BC$ implies that triangles $ABP$ and $CBP$ are congruent. Hence $\\angle BAP = \\angle BCP = x$. Similarly, triangles $CDP$ and $EDP$ are congruent, so $\\angle DCP = \\angle DEP = y$.\n\n\n\nMoreover, $AP = CP = EP$, which together with $AF = EF$ implies that triangles $AFP$ and $EFP$ are congruent. Thus the angle bisector of $\\angle F$ passes through $P$ and $\\angle FAP = \\angle FEP = z$.\n\nNow, the equalities $\\angle A = \\angle C = \\angle E$ are equivalent to $z + x = x + y = y + z$, which yields $x = y = z$. Therefore, the angle bisectors of $\\angle A$, $\\angle C$, and $\\angle E$ all pass through $P$. Thus, $P$ is the center of the inscribed circle of hexagon $ABCDEF$, as claimed.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21136,
"subject": "Mathematics (Olympiad)",
"question": "Determine all triples $(a, b, c)$ of positive integers such that\n\n$$\na! + b! = 2^c.\n$$",
"options": [],
"answer": "See solution",
"solution": "The only solutions are $(1, 1, 1)$ and $(2, 2, 2)$.\n\nWe can assume without loss of generality that $a \\leq b$.\n\n- For $a = b = 1$, we get $c = 1$, which gives the solution $(1, 1, 1)$.\n- For $a = 1$ and $b > 1$, the left-hand side is bigger than $1$ and odd, therefore, it cannot be a power of $2$ and we do not get a solution in this case.\n- For $a = b = 2$, we get $c = 2$, therefore $(2, 2, 2)$ is a solution.\n- For $a = 2$ and $b = 3$, we get $2! + 3! = 8 = 2^3$. But there is no $c$ with $c! = 3$. Therefore, there is no solution in this case.\n- For $a = 2$ and $b \\geq 4$, we get $2! + b! \\geq 2! + 4! = 26$. Therefore, we have $c! > 4$. This implies that the left-hand side is congruent to $2$ modulo $4$, while the right-hand side is congruent to $0$ modulo $4$. Therefore, there is no solution in this case.\n- For $a \\geq 3$ and $b \\geq 3$, the left-hand side is divisible by $3$ while the power of $2$ on the right-hand side is not. Therefore, there is no solution in this case.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21137,
"subject": "Mathematics (Olympiad)",
"question": "Is there a number which is the sum of 2345 positive integers that have the same digit sum, and also the sum of 5678 positive integers that have the same digit sum? If the answer is *yes*, find the least such number. If not, explain why.",
"options": [],
"answer": "See solution",
"solution": "Such numbers exist. The least one is $11725$.\n\nLet $N$ be the sum of $2345$ positive integers with digit sum $R$, and let $R \\equiv r \\pmod{9}$, $r \\in [1, 9]$. Then $N \\equiv 2345r \\equiv 5r \\pmod{9}$, as each summand is congruent to $r$ modulo $9$. Similarly, if $N$ is the sum of $5678$ numbers with digit sum congruent to $s$ modulo $9$, $s \\in [1, 9]$, then $N \\equiv 5678s \\equiv 8s \\pmod{9}$. So $5r \\equiv N \\equiv 8s \\pmod{9}$. Letting $r$ run through $1, 2, \\ldots, 9$ yields the admissible pairs of remainders: $(1, 4), (2, 8), (3, 3), (4, 7), (5, 2), (6, 6), (7, 1), (8, 5), (9, 9)$.\n\nNote that $N \\geq \\max(2345r, 5678s)$ for every such pair $(r, s)$ because the least number with digit sum congruent to $r$ or $s$ modulo $9$ is $r$ or $s$ respectively. If $r = 5$, $s = 2$, the last observation gives $N \\geq \\max(2345 \\cdot 5, 5678 \\cdot 2) = 11725$. For each remaining pair $(r, s)$, one of the numbers $2345r$ and $5678s$ is greater than $11725$. It follows that the least $N$ in question, if it exists, is at least $11725$.\n\nOn the other hand, $N = 11725$ is possible. Indeed, $11725 = 2345 \\cdot 5$ is equal to the sum of $2345$ numbers equal to $5$. Also, $11725$ is equal to the sum of $5678$ numbers with digit sum $2$: $41$ summands $11$ and $5637$ summands $2$ ($41 \\cdot 11 + 5637 \\cdot 2 = 11725$). In all, there are $41 + 5637 = 5678$ summands with digit sum $2$, as needed.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21138,
"subject": "Mathematics (Olympiad)",
"question": "Let $AB$ be a line segment of length $1$. Several elementary particles start moving simultaneously at constant speeds from $A$ to $B$. As soon as a particle reaches $B$, it turns around and heads to $A$; when reaching $A$, it starts moving again, and so on indefinitely.\n\nFind all rational numbers $r > 1$ with the following property: For each $n \\geq 1$, if $n+1$ particles with speeds $1, r, r^2, \\dots, r^n$ move as described, there is a moment when all particles are at the same interior point of segment $AB$. (Ignore the dimensions of the particles; assume that they can all gather at one point.)",
"options": [],
"answer": "See solution",
"solution": "The values in question are all integers $r$ greater than $1$.\n\nWe start with a general observation about two particles $P_1$ and $P_2$ moving on $AB$ by the given rules, with different constant speeds $v_1$ and $v_2$, $v_1 > v_2$. Suppose that they are at the same point $Q$ of $AB$ at a certain moment $t$. There are two possibilities for the distances $v_1 t$ and $v_2 t$ the particles have traveled until that moment. If $P_1$ and $P_2$ are moving in the same direction when they simultaneously reach $Q$, then the integer parts of $v_1 t$ and $v_2 t$ have the same parity, and their fractional parts are equal.\n\nHence $v_1 t - v_2 t$ is an even positive integer. If $P_1$ and $P_2$ are moving in opposite directions when they meet at $Q$, then the integer parts of $v_1 t$ and $v_2 t$ have different parity, and the sum of their fractional parts is $1$. Therefore $v_1 t + v_2 t$ is an even positive integer.\n\nNow let the rational $r > 1$ have the stated property, for any number $n+1$, $n \\geq 1$, of particles with speeds $1, r, r^2, \\dots, r^n$. Let $t$ be a moment when all of them are at the same point. Apply the observation to the first and the last particle, with speeds $v_1 = r^n$ and $v_2 = 1$. We infer that $(r^n - 1)t$ or $(r^n + 1)t$ is an integer. Because $r$ is rational, $t$ is rational too.\n\nWrite $r$ and $t$ as irreducible fractions: $r = \\frac{a}{b}$, $t = \\frac{c}{d}$. Then $(r^n \\pm 1)t = \\frac{(a^n \\pm b^n)c}{b^n d}$. Since $a^n \\pm b^n$ and $b^n$ are coprime, it follows that $b^n$ divides $c$. Moreover, the latter holds for each $n > 1$ by hypothesis. This is possible only if $b = 1$, that is, if $r > 1$ is an integer.\n\nConversely, every integer $r > 1$ is a solution. Let $r \\geq 3$ be odd and $n \\geq 1$ arbitrary. Then all $n+1$ particles with speeds $1, r, r^2, \\dots, r^n$ will be at the midpoint of $AB$ at $t = \\frac{1}{2}$. Indeed, $r^k \\cdot \\frac{1}{2}$ has fractional part $\\frac{1}{2}$ for each $k = 0, 1, 2, \\dots, n$ since $r^k$ is odd.\n\nLet $r = 2m$, $m \\geq 1$, be even and $n \\geq 1$ arbitrary. Then at the moment $t = \\frac{2m}{2m+1}$ all $n+1$ particles with speeds $1, r, r^2, \\dots, r^n$ will be at the point $Q$ at distance $\\frac{2m}{2m+1}$ from $A$. It is enough to prove that for each $k = 0, 1, 2, \\dots$ the following equality holds:\n\n$$\n(2m)^k \\frac{2m}{2m+1} = \\begin{cases} 2q + \\frac{2m}{2m+1} & \\text{if } k \\geq 0 \\text{ is even;} \\\\ 2q + 1 + \\frac{1}{2m+1} & \\text{if } k \\geq 1 \\text{ is odd,} \\end{cases}\n$$\nwhere $q = 0, 1, 2, \\dots$.\n\nIndeed, these relations mean that, for $k$ even, the particle with speed $r^k$ will be moving from $A$ towards $B$ at the moment $t = \\frac{2m}{2m+1}$, and it will be at distance $\\frac{2m}{2m+1}$ from $A$, that is, at point $Q$.\n\nFor $k$ odd, the particle with speed $r^k$ will be moving from $B$ towards $A$ at the moment $t = \\frac{2m}{2m+1}$, and it will be at distance $\\frac{1}{2m+1}$ from $B$, hence at point $Q$ again.\n\nSo it remains to prove the displayed equalities, which we do by induction in $k$. The case $k = 0$ is obvious. Proceed to the inductive step $k \\to k+1$. The induction hypothesis yields:\n\n$$\n\\text{For } k \\text{ odd: } (2m)^{k+1} \\frac{2m}{2m+1} = 2m(2q + 1) + \\frac{2m}{2m+1} = 2q' + \\frac{2m}{2m+1}, \\quad q' = 0, 1, 2, \\dots;\n$$\n\n$$\n\\text{For } k \\text{ even: } (2m)^{k+1} \\frac{2m}{2m+1} = 4mq + \\frac{4m^2}{2m+1} = 4mq + 2m - 1 + \\frac{1}{2m+1} = 2q' + 1 + \\frac{1}{2m+1}, \\quad q' = 0, 1, 2, \\dots\n$$\n\nThis completes the induction and the solution.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21139,
"subject": "Mathematics (Olympiad)",
"question": "A configuration of 4027 points in the plane is called *Colombian* if it consists of 2013 red points and 2014 blue points, and no three of the points of the configuration are collinear. By drawing some lines, the plane is divided into several regions. An arrangement of lines is *good* for a Colombian configuration if the following two conditions are satisfied:\n\n- No line passes through any point of the configuration.\n- No region contains points of both colors.\n\nFind the least value of $k$ such that for any Colombian configuration of 4027 points, there is a good arrangement of $k$ lines.\n\n",
"options": [],
"answer": "See solution",
"solution": "The answer is $2013$.\n\nWe first show a good arrangement with at most $2013$ lines always exists. We begin with the following key lemma.\n\n**Lemma 1.** Any pair of points $P$ and $Q$ in a Colombian configuration $C$ can be separated from the other points by two lines.\n\n*Proof.* No three points in $C$ are collinear, so each other point has one of finitely many positive distances to $PQ$. Choose $r > 0$ less than all such distances; the two lines parallel to and at a distance $r$ from $PQ$ have the desired property. $\\square$\n\nLet $C$ be the convex hull of our Colombian configuration $C$. If a red point $R$ is a vertex of $C$, draw a line $l_1$ separating $R$ from all other points. Next, place the other $2012$ red points into $1006$ pairs and apply Lemma 1 to draw $2012$ lines separating them from the rest of $C$. Together with $l_1$, these $2012$ lines form a good arrangement. Otherwise, $C$ has a side $B_1B_2$ consisting of blue points. Draw a line ...",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21140,
"subject": "Mathematics (Olympiad)",
"question": "Find the smallest positive integer $n$ such that there are exactly 23 perfect squares greater than or equal to $n$ and less than or equal to $n + 2011$.",
"options": [],
"answer": "See solution",
"solution": "Suppose $k^2$ (where $k$ is a positive integer) is the largest perfect square less than $n$. Then, the 23 perfect squares greater than or equal to $n$ and less than or equal to $n + 2011$ are $(k+1)^2, (k+2)^2, \\dots, (k+23)^2$. So, we must have $(k+24)^2 > n + 2011$.\n\nFrom $k^2 < n$ and $(k+24)^2 > n + 2011$, we obtain $$(k+24)^2 - k^2 > 2011.$$ Simplifying, $$(k+24)^2 - k^2 = 48k + 576 > 2011,$$ so $48k > 2011 - 576 = 1435$, thus $k > \\frac{1435}{48} \\approx 29.9$. Therefore, $k \\ge 30$, so $n \\ge k^2 + 1 \\ge 901$.\n\nIf $n = 901$, then $30^2 < 901 < 31^2$, and $53^2 < 901 + 2011 = 2912 < 54^2$. The perfect squares in $[901, 2912]$ are $31^2, 32^2, \\dots, 53^2$, which are 23 numbers. Thus, the smallest such $n$ is $\\boxed{901}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21141,
"subject": "Mathematics (Olympiad)",
"question": "In the given diagram, let $N$ be the center of $\\omega$ and $I$ be the incenter of $\\triangle ABC$. Let $BC = a$, $CA = b$, $AB = c$, and $s = \\frac{a+b+c}{2}$. Consider the inversion $f$ centered at $A$ that preserves the nine-point circle. Let $\\odot A'$ be the image of the incircle under $f$. It is clear that $\\odot A'$ is tangent to the sides $AB$ and $AC$. Similarly, we can define $\\odot B'$ and $\\odot C'$.\n\nLet $M$ be the foot of the perpendicular from $I$ to side $AB$. Show that $A'$, $B'$, and $C'$ are collinear.\n\n",
"options": [],
"answer": "See solution",
"solution": "$$\n\\frac{AL}{AM} = \\frac{AM \\cdot AL}{AM^2} = \\frac{b^2 + c^2 - a^2}{4(s-a)^2}.\n$$\n\nTherefore, we have:\n\n$$\np = \\frac{IA'}{IA} = \\frac{2(a-b)(a-c)}{(b+c-a)^2}\n$$\n\nSimilarly, we have:\n\n$$\nq = \\frac{2(b-c)(b-a)}{(c+a-b)^2}, \\quad r = \\frac{2(c-a)(c-b)}{(a+b-c)^2}.\n$$\n\nPlugging into the computations gives $\\frac{a}{p} + \\frac{b}{q} + \\frac{c}{r} = 0$ (Here, upon substitution, we obtain a cyclic summation of $a(b-c)(b+c-a)^2$. Let us consider this as a function of $a$, denoted by $g(a)$. It is observed that $g(a)$ is actually a quadratic function, and we have $g(b) = g(c) = g(b+c) = 0$. Hence, we conclude that $g \\equiv 0$. $\\Box$\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21142,
"subject": "Mathematics (Olympiad)",
"question": "Prove that if $\\alpha, \\beta, \\gamma \\in [0, \\frac{\\pi}{2}]$ and $\\tan\\alpha + \\tan\\beta + \\tan\\gamma \\le 3$, then\n\n$$\n\\cos 2\\alpha + \\cos 2\\beta + \\cos 2\\gamma \\ge 0.\n$$",
"options": [],
"answer": "See solution",
"solution": "The inequality follows from the fact that if $\\theta \\in [0, \\frac{\\pi}{2}]$, then\n\n$$\n\\cos 2\\theta = 1 - \\sin 2\\theta \\cdot \\tan \\theta \\ge 1 - \\tan \\theta.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21143,
"subject": "Mathematics (Olympiad)",
"question": "令 $\\mathbb{Z}$ 和 $\\mathbb{Q}$ 分別為整數與有理數所形成的集合。對於 $\\mathbb{Q}$ 的子集合 $X, Y$，定義 $X + Y$ 為：\n\n$$\nX + Y := \\{x + y \\mid x \\in X, y \\in Y\\}.\n$$\n\n(a) 試問能否將 $\\mathbb{Z}$ 分割成三個非空子集 $A, B, C$ 使得 $A+B, B+C, C+A$ 互斥？\n\n(b) 試問能否將 $\\mathbb{Q}$ 分割成三個非空子集 $A, B, C$ 使得 $A+B, B+C, C+A$ 互斥？",
"options": [],
"answer": "See solution",
"solution": "(a) 是。下為一例：\n\n$$\nA = \\{3k \\mid k \\in \\mathbb{Z}\\}, \\quad B = \\{3k + 1 \\mid k \\in \\mathbb{Z}\\}, \\quad C = \\{3k + 2 \\mid k \\in \\mathbb{Z}\\}.\n$$\n\n(b) 否。假設 $\\mathbb{Q}$ 可分割成三個非空子集 $A, B, C$ 使得 $A+B, B+C, C+A$ 互斥。注意到對於所有 $a \\in A, b \\in B, c \\in C$ 有\n\n$$\na + b - c \\in C, \\quad b + c - a \\in A, \\quad c + a - b \\in B.\n$$\n\n確實 $a + b - c \\notin A$ 當 $(A+B) \\cap (A+C) = \\emptyset$，且同樣地 $a + b - c \\notin B$，因此 $a + b - c \\in C$。即 $A+B \\subset C+C$。同樣地，我們有 $B+C \\subset A+A$，$C+A \\subset B+B$。\n\n反向的包含關係亦成立。令 $a, a' \\in A, b \\in B, c \\in C$。由上式可得 $a' + c - b \\in B$，且因為 $a \\in A, c \\in C$，再次使用上述式子可得\n\n$$\na + a' - b = a + (a' + c - b) - c \\in C.\n$$\n\n所以 $A + A \\subset B + C$。同樣地，$B + B \\subset C + A$，$C + C \\subset A + B$。所以\n$A + B = C + C$，$B + C = A + A$，$C + A = B + B$。\n\n更進一步，不失一般性假設 $0 \\in A$。則 $B = \\{0\\} + B \\subset A + B$ 且 $C = \\{0\\} + C \\subset A + C$。因為 $B + C$ 分別與 $A + B$ 及 $A + C$ 互斥，所以 $B + C$ 分別與 $B$ 及 $C$ 互斥。因此 $B + C$ 包含於 $\\mathbb{Q} \\setminus (B \\cup C) = A$。因為 $B + C = A + A$，我們可得 $A + A \\subset A$。另一方面，$A = \\{0\\} + A \\subset A + A$，可得 $A = A + A = B + C$。\n\n因此 $A+B+C = A+A+A = A$，且 $B+B = C+A$，$C+C = A+B$。\n可推得 $B+B+B = A+B+C = A$，$C+C+C = A+B+C = A$。\n特別地，若任意的 $r \\in \\mathbb{Q} = A \\cup B \\cup C$ 則 $3r \\in A$。\n\n但此結論並不可能。任取 $b \\in B$（$B$ 為非空子集）且令 $r = \\frac{b}{3} \\in \\mathbb{Q}$，則\n$b = 3r \\in A$ 產生了矛盾。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21144,
"subject": "Mathematics (Olympiad)",
"question": "Points $X$, $Y$, $Z$ lie on a line $k$ in this order. Let $\\omega_1$, $\\omega_2$, $\\omega_3$ be three circles of diameters $XZ$, $XY$, $YZ$, respectively. Line $l$ passing through point $Y$ intersects $\\omega_1$ at points $A$ and $D$, $\\omega_2$ at $B$ and $\\omega_3$ at $C$ in such manner that points $A$, $B$, $Y$, $C$, $D$ lie on $l$ in this order. Prove that $AB = CD$.",
"options": [],
"answer": "See solution",
"solution": "Let $E$ be the second intersection of line $XB$ with circle $\\omega_1$ and $F$ be the second intersection of line $ZC$ with $\\omega_1$. Note that $\\angle XEZ = \\angle ZFX = 90^\\circ$. Moreover, $XE \\parallel ZF$ as $XE \\perp BC \\perp ZE$. Hence, $XEZF$ is a rectangle and $AD \\perp XE$, so $AB = CD$. $\\blacktriangleleft$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21145,
"subject": "Mathematics (Olympiad)",
"question": "Given an odd number $n$ of marked points on a circle, each pair of points is connected by a segment. What is the maximum possible number of intersection points formed by these segments inside the circle, such that no three segments intersect at a single point? Find this maximum for $n = 2021$.",
"options": [],
"answer": "See solution",
"solution": "Let the marked points be labeled $1$ to $n$ around the circle. For each $i = 1, \\ldots, n$, let $a_i$ be the number of segments for which $i$ is an endpoint. Among all $\\binom{n}{2}$ pairs of segments, at least $\\binom{a_i}{2}$ pairs do not intersect (since they share endpoint $i$). For different $i$, these pairs are distinct, so the number of intersection points does not exceed\n\n$$\nM = \\binom{n}{2} - \\left( \\binom{a_1}{2} + \\binom{a_2}{2} + \\dots + \\binom{a_n}{2} \\right).\n$$\n\nTo maximize $M$, minimize the sum in brackets. The minimum occurs when all $a_i$ are as equal as possible. Since $a_1 + \\dots + a_n = 2n$, and $n$ is odd, the only possibility is $a_1 = a_2 = \\dots = a_n = 2$. Thus,\n\n$$\nM = \\frac{n(n-1)}{2} - n = \\frac{n(n-3)}{2}.\n$$\n\nFor $n = 2021$:\n\n$$\nM = \\frac{2021 \\times 2018}{2} = 2021 \\times 1009 = 2\\,039\\,189.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21146,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs of positive integers $(n, p)$, where $p$ is a prime number, such that\n\n$$\np(p-1) = 2(n^3 + 1).\n$$",
"options": [],
"answer": "See solution",
"solution": "We are given the equation:\n\n$$\np(p-1) = 2(n^3 + 1). \\tag{1}\n$$\n\nFirst, for $p=2$, the equation does not hold for any positive integer $n$. So $p \\ge 3$ and is an odd prime.\n\nRewrite $n^3 + 1 = (n+1)(n^2 - n + 1)$, so $(n+1)(n^2 - n + 1)$ divides $p(p-1)/2$.\n\n**Case 1:** If $n+1$ divides $p$, then $p = k(n+1)$ for some $k \\ge 1$, but this leads to contradictions upon substitution.\n\n**Case 2:** If $n^2 - n + 1$ divides $p$, then $p = k(n^2 - n + 1)$ for some $k \\ge 1$.\n\nHowever, the only viable case is when $n^2 - n + 1 = k p$ for some $k$. Substitute this into (1):\n\n$$\np(p-1) = 2(n+1)k p\n$$\n\nwhich gives $p-1 = 2k(n+1)$, so $p = 2k(n+1) + 1$.\n\nNow, substitute $p$ back into $n^2 - n + 1 = k p$:\n\n$$\nn^2 - n + 1 = k[2k(n+1) + 1] = 2k^2(n+1) + k\n$$\n\nSo,\n\n$$\nn^2 - n + 1 = 2k^2 n + 2k^2 + k\n$$\n\n$$\nn^2 - (2k^2 + 1)n - (2k^2 + k - 1) = 0\n$$\n\nThis is a quadratic in $n$. The discriminant $D$ must be a perfect square:\n\n$$\nD = (2k^2 + 1)^2 + 4(2k^2 + k - 1)\n$$\n\nSet $D = (2k^2 + 3)^2$ and solve for $k$:\n\n$$\n(2k^2 + 1)^2 + 4(2k^2 + k - 1) = (2k^2 + 3)^2\n$$\n\nExpanding and simplifying, we find $k = 3$.\n\nSubstitute $k = 3$ into the quadratic:\n\n$$\nn^2 - 19n - 20 = 0\n$$\n\nThe positive root is $n = 20$.\n\nNow, $p = 2 \\cdot 3 \\cdot 21 + 1 = 127$, which is prime.\n\n**Answer:** The only solution is $(n, p) = (20, 127)$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21147,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\mathcal{L}$ be a finite collection of lines in the plane in general position (no two lines in $\\mathcal{L}$ are parallel and no three are concurrent). Consider the open circular discs inscribed in the triangles enclosed by each triple of lines in $\\mathcal{L}$. Determine the number of such discs intersected by no line in $\\mathcal{L}$, in terms of $|\\mathcal{L}|$.",
"options": [],
"answer": "See solution",
"solution": "The complement of the union of all lines in $\\mathcal{L}$ is the disjoint union of\n$$\n\\binom{|\\mathcal{L}|}{0} + \\binom{|\\mathcal{L}|}{1} + \\binom{|\\mathcal{L}|}{2}\n$$\nopen convex sets called *rooms*, of which exactly\n$$\n\\binom{|\\mathcal{L}| - 1}{2}\n$$\nare bounded.\n\nLet $\\mathcal{D}_0 = \\mathcal{D}_0(\\mathcal{L})$ denote the set of the discs we are to count. Clearly, each disc in $\\mathcal{D}_0$ is contained in some bounded room. We shall prove that each bounded room contains exactly one such disc, whence the conclusion. To this end, we shall first prove that no bounded room contains more than one disc in $\\mathcal{D}_0$, and then that each bounded room contains at least one such.\n\nSuppose, if possible, that $R$ is a bounded room which contains two discs in $\\mathcal{D}_0$; one, of radius $r$, centered at $\\omega$, inscribed in the triangle $abc$ enclosed by the lines $a, b$ and $c$ in $\\mathcal{L}$; and another, of radius $r'$, centered at $\\omega'$, inscribed in the triangle $a'b'c'$ enclosed by the lines $a', b'$ and $c'$ in $\\mathcal{L}$. Clearly, the lines $a, b, c, a', b', c'$ support edges on the boundary of $R$, the triangles $abc$ and $a'b'c'$ both contain $R$, and $\\omega$ and $\\omega'$ are distinct. Without loss of generality, we may assume that $r \\le r'$. Since $\\omega'$ lies in the interior of the triangle $abc$ and is different from $\\omega$, it is within $r \\le r'$ from one of the lines $a, b$ or $c$. Consequently, that line intersects the disc of radius $r'$ centered at $\\omega'$ — a contradiction.\n\nWe now proceed to prove that each bounded room contains a disc in $\\mathcal{D}_0$. Since $R$ is convex, and no two lines in $\\mathcal{L}$ are parallel, among those lines in $\\mathcal{L}$ which support edges on the boundary of $R$, at least three enclose a triangle which contains $R$ — this is a well-known fact about convex polygons different from a parallelogram. Of all such triangles, choose one with a minimal inradius. We shall prove that the open circular disc $D$ inscribed in that triangle is contained in $R$; in particular, $D$ is in $\\mathcal{D}_0$. Let the triangle be enclosed by the lines $a, b$ and $c$ in $\\mathcal{L}$. Since each of the lines $a, b$ and $c$ supports an edge on the boundary of $R$, it follows that $D$ and $R$ are not disjoint. Suppose, if possible, that $D$ is not contained in $R$. Then $D$ contains points on some edge on the boundary of $R$. Let $d$ be a line in $\\mathcal{L}$ that supports such an edge. Clearly, $d$ is different from $a, b$ and $c$, and it is not hard to see that $d$ and two of the lines $a, b, c$ enclose a triangle which contains $R$ and has an inradius smaller than the radius of $D$ — a contradiction.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21148,
"subject": "Mathematics (Olympiad)",
"question": "Let $(a_n)_{n \\ge 0}$ be a sequence of rational numbers with $a_0 = 2016$ and\n\n$$\na_{n+1} = a_n + \\frac{2}{a_n}\n$$\n\nfor all $n \\ge 0$.\n\nShow that the sequence does not contain a square of a rational number.",
"options": [],
"answer": "See solution",
"solution": "We examine the sequence modulo $5$, provided $a_n \\not\\equiv 0 \\pmod{5}$ so that $a_{n+1}$ is defined modulo $5$. Computing the elements modulo $5$:\n\n$$\n\\begin{align*}\na_0 &\\equiv 1 \\pmod{5}, \\\\\na_1 &\\equiv 1 + 2 \\equiv 3 \\pmod{5}, \\\\\na_2 &\\equiv 3 + 2 \\cdot 3^{-1} \\equiv 3 + 2 \\cdot 2 \\equiv 2 \\pmod{5}, \\\\\na_3 &\\equiv 2 + 2 \\cdot 2^{-1} \\equiv 2 + 2 \\cdot 3 \\equiv 2 + 6 \\equiv 3 \\pmod{5}, \\\\\na_4 &\\equiv 3 + 2 \\cdot 3^{-1} \\equiv 3 + 2 \\cdot 2 \\equiv 2 \\pmod{5}, \\\\\\vdots\n\\end{align*}\n$$\n\nThus, after $a_0$, the sequence alternates between $2$ and $3$ modulo $5$. These are not quadratic residues modulo $5$. Since $a_0 = 2016$ is not a square of a rational number, the sequence does not contain a square of a rational number.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21149,
"subject": "Mathematics (Olympiad)",
"question": "a) Show that the last two digits of $1038^2$ are $4$.\n\nb) Show that there are infinitely many perfect squares whose last three digits are $4$.\n\nc) Prove that there is no perfect square whose last four digits are $4$.",
"options": [],
"answer": "See solution",
"solution": "a) $1038^2 = 1\\,077\\,444$.\n\nb) By squaring a number which ends in $038$, we get a number ending in $444$, as shown in the diagram:\n\n\n\nSince there are infinitely many numbers ending in $038$, there are infinitely many perfect squares ending in $444$.\n\nc) Let $a$—if possible—be a positive integer whose square ends in $4444$. Then $a$ is even, that is, $a = 2b$. Moreover, $a^2$ has the form $10000k + 4444$, $k \\in \\mathbb{N}$, hence $4b^2 = a^2 = 4(2500k + 1111)$. It follows that the last two digits of $b^2$ are $11$, hence $b$ ends in $1$ or $9$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21150,
"subject": "Mathematics (Olympiad)",
"question": "Let $O$ be an interior point of triangle $ABC$. Prove that there exist positive integers $p$, $q$, and $r$ such that\n\n$$\n|p \\cdot \\vec{OA} + q \\cdot \\vec{OB} + r \\cdot \\vec{OC}| < \\frac{1}{2007}.\n$$",
"options": [],
"answer": "See solution",
"solution": "It is well-known that there are positive real numbers $\\beta$, $\\gamma$ such that\n\n$$\n\\vec{OA} + \\beta \\vec{OB} + \\gamma \\vec{OC} = \\vec{0}.\n$$\n\nFor any positive integer $k$, we have\n\n$$\nk \\vec{OA} + k\\beta \\vec{OB} + k\\gamma \\vec{OC} = \\vec{0}.\n$$\n\nLet $m(k) = [k\\beta]$, $n(k) = [k\\gamma]$, where $[x]$ is the greatest integer less than or equal to $x$, and $\\{x\\} = x - [x]$.\n\nAssume $T$ is an integer larger than $\\max\\left\\{\\frac{1}{\\beta}, \\frac{1}{\\gamma}\\right\\}$. Then the sequences $\\{m(kT) \\mid k = 1, 2, \\dots\\}$ and $\\{n(kT) \\mid k = 1, 2, \\dots\\}$ are increasing, and\n\n$$\n\\begin{aligned}\n|kT \\vec{OA} + m(kT) \\vec{OB} + n(kT) \\vec{OC}| &= |-\\{kT\\beta\\} \\vec{OB} - \\{kT\\gamma\\} \\vec{OC}| \\\\\n&\\leq |\\vec{OB}| \\cdot \\{kT\\beta\\} + |\\vec{OC}| \\cdot \\{kT\\gamma\\} \\\\\n&\\leq |\\vec{OB}| + |\\vec{OC}|.\n\\end{aligned}\n$$\n\nThis shows there exist infinitely many vectors of the form\n\n$$\nkT \\vec{OA} + m(kT) \\vec{OB} + n(kT) \\vec{OC},\n$$\n\nwhose endpoints lie in a circle centered at $O$ of radius $|\\vec{OB}| + |\\vec{OC}|$. By the pigeonhole principle, there exist two such vectors whose endpoints are less than $\\frac{1}{2007}$ apart. Thus, there exist integers $k_1 < k_2$ such that\n\n$$\n\\begin{aligned}\n| (k_2 T \\vec{OA} + m(k_2 T) \\vec{OB} + n(k_2 T) \\vec{OC}) \n- (k_1 T \\vec{OA} + m(k_1 T) \\vec{OB} + n(k_1 T) \\vec{OC}) | < \\frac{1}{2007}.\n\\end{aligned}\n$$\n\nLet $p = (k_2 - k_1)T$, $q = m(k_2T) - m(k_1T)$, $r = n(k_2T) - n(k_1T)$. Then $p$, $q$, $r$ are positive integers and\n\n$$\n|p \\vec{OA} + q \\vec{OB} + r \\vec{OC}| < \\frac{1}{2007}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21151,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ and $b$ be positive numbers. Find all pairs of functions $f, g: \\mathbb{R} \\to \\mathbb{R}$, each assuming the value $1$ and fulfilling, for any $y \\neq 0$ and any $x$, the equations\n$$\nf\\left(\\frac{1}{y^2}g(xy) - ax^2\\right) = 0 = g\\left(\\frac{1}{y}f(xy) - bx\\right).\n$$",
"options": [],
"answer": "See solution",
"solution": "Either $f(z) = g(z) = \\delta_{z,0}$ or $f(z) = bz$ and $g(z) = az^2$.\n\nPutting $xy = w$ in the second equation gives\n$$\ng\\left(\\frac{1}{y}(f(w) - bw)\\right) = 0\n$$\nfor all $y \\neq 0$. Hence, if $f(w) \\neq bw$ for some $w$, it must be that $g(z) = 0$ for $z \\neq 0$. Since $g$ must assume the value $1$ somewhere, $g(0) = 1$.\n\nThe function $g$ now being known, the first equation transforms, for $x = 0$ and $x \\neq 0$, respectively, into\n$$\nf\\left(\\frac{1}{y^2}\\right) = 0 \\quad \\text{and} \\quad f(-ax^2) = 0.\n$$\nConsequently, $f(z) = 0$ for $z > 0$ or $z < 0$. Again, $f$ must assume the value $1$, and so $f = g$.\n\nThere remains the case when $f(z) = bz$ for all $z$. Substitute $y = 1$ into the first equation to find $b(g(x) - ax^2) = 0$, so that $g(z) = az^2$ for all $z$.\n\nOne easily verifies that these two possibilities satisfy the requirements. $\\square$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21152,
"subject": "Mathematics (Olympiad)",
"question": "Given a positive integer $n$ and an increasing real-valued function $f$ on the closed unit interval $[0, 1]$, determine the maximum value that the sum\n\n$$\n\\sum_{k=1}^{n} f\\left(\\left|x_k - \\frac{2k-1}{2n}\\right|\\right)\n$$\n\nmay achieve, subject to $0 \\le x_1 \\le \\dots \\le x_n \\le 1$.",
"options": [],
"answer": "See solution",
"solution": "Let $a_k = \\frac{2k-1}{2n}$ for $k = 1, \\dots, n$. The required maximum is $\\sum_{k=1}^n f(a_k)$, which is achieved, for instance, at $x_1 = \\dots = x_n = 0$ or $x_1 = \\dots = x_n = 1$.\n\nTo show that $\\sum_{k=1}^n f(a_k)$ is an upper bound for the sum under consideration, fix an $n$-tuple $(x_1, \\dots, x_n)$ such that $0 \\le x_1 \\le \\dots \\le x_n \\le 1$. Write $[n] = \\{1, \\dots, n\\}$, and define an increasing function $\\varphi: [n] \\to [n]$ by $\\varphi(k) = \\max \\{j: a_j - 1/(2n) \\le x_k\\}$. Notice that\n\n$$\n|x_k - a_k| \\le |x_k - a_{\\varphi(k)}| + |a_{\\varphi(k)} - a_k| \\le \\frac{1}{2n} + \\frac{|\\varphi(k) - k|}{n} = a_{|\\varphi(k)-k|+1},\n$$\n\nfor $k = 1, \\dots, n$.\n\nWe shall prove that there exists a permutation $\\sigma$ of $[n]$ such that $|\\varphi(k) - k| + 1 \\le \\sigma(k)$ for all $k$, so $a_{|\\varphi(k)-k|+1} \\le a_{\\sigma(k)}$ and the conclusion follows:\n\n$$\n\\sum_{k=1}^{n} f(|x_k - a_k|) \\le \\sum_{k=1}^{n} f(a_{|\\varphi(k)-k|+1}) \\le \\sum_{k=1}^{n} f(a_{\\sigma(k)}) = \\sum_{k=1}^{n} f(a_k).\n$$\n\nWe now show by induction on $n$ that, for any increasing function $\\psi: [n] \\to [n]$, there exists a permutation $\\sigma$ of $[n]$ such that $|\\psi(k) - k| + 1 \\le \\sigma(k)$ for all $k$.\n\nThe base case $n=1$ is clear. For the induction step, let $n > 1$ and distinguish two cases:\n\n- If $\\psi(n) < n$, then the restriction of $\\psi$ to $[n-1]$ is an increasing function of $[n-1]$ into itself, so $|\\psi(k)-k|+1 \\le \\sigma(k)$ for $k=1, \\dots, n-1$, for some permutation $\\sigma$ of $[n-1]$. Since $|\\psi(n)-n|+1 = n-\\psi(n)+1 \\le n$, the permutation $\\sigma$ extends to a permutation of $[n]$ satisfying the required condition by letting $\\sigma(n) = n$.\n\n- If $\\psi(n) = n$, consider the increasing function $\\psi': [n-1] \\to [n-1]$ defined by $\\psi'(k) = \\psi(k)$ if $\\psi(k) < n$ and $\\psi'(k) = n-1$ if $\\psi(k) = n$. By the induction hypothesis, there exists a permutation $\\pi$ of $[n-1]$ such that $|\\psi'(k) - k| + 1 \\le \\pi(k)$ for $k = 1, \\dots, n-1$, so $|\\psi(k) - k| + 1 \\le |\\psi'(k) - k| + 2 \\le \\pi(k) + 1$ for $k = 1, \\dots, n-1$. Finally, since $|\\psi(n) - n| + 1 = 1$, setting $\\sigma(k) = \\pi(k) + 1$ for $k = 1, \\dots, n-1$ and $\\sigma(n) = 1$ defines the required permutation.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21153,
"subject": "Mathematics (Olympiad)",
"question": "We need to find the maximal number of subsets in $X_{30} = \\{1, 2, \\dots, 30\\}$ having different cardinalities and pairwise not nested.",
"options": [],
"answer": "See solution",
"solution": "The answer for $X_n$ ($n \\ge 4$) is $n-2$. For $n=30$, $m = 28$.\n\n**Upper bound:** If there are $n-1$ sets, they cannot include the full set of cardinality $n$; hence, there are non-intersecting subsets of cardinalities $n-1$ and $1$, and together they prohibit any other subset.\n\n**Lower bound:** Show by induction that there exists an example with cardinalities $2, 3, \\dots, n-1$. To perform the step $n \\mapsto n+2$, augment all yet obtained subsets by $n+2$, and add two new subsets $\\{n+1, n+2\\}$ and $\\{1, 2, \\dots, n+1\\}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21154,
"subject": "Mathematics (Olympiad)",
"question": "Two cyclists start simultaneously from towns A and B, heading towards each other along the same road. Cyclist 1 departs from A with speed $v_1$, and cyclist 2 departs from B with speed $v_2$. Upon reaching their respective destinations, each immediately turns around and heads back in the opposite direction. How much time after their first meeting do they meet for the second time?\n\n\n\n",
"options": [],
"answer": "See solution",
"solution": "If $v_1 < 2v_2$, then the second meeting occurs after $2$ hours; otherwise, after $\\frac{2v_2}{v_1 - v_2}$ hours.\n\n**Solution:**\n\nUntil the first meeting, the first cyclist travels $S_1 = v_1$ and the second $S_2 = v_2$, so the distance between A and B is $v_1 + v_2$.\n\nThere are two cases:\n\n**Case 1:** Both cyclists reach their destination towns and turn around before the second meeting. Let the second meeting occur at distance $S_3$ from B and time $t_2$ after the first meeting. Then:\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:\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 B, turns around, and meets the second cyclist before the latter reaches A. 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:\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, compare the times for each cyclist to reach the opposite town. Case 1 occurs when $v_1 < 2v_2$; otherwise, Case 2 applies.\n\nLet $x = \\frac{v_1}{v_2}$:\n$$\nx^2 - x - 2 < 0 \\implies (x+1)(x-2) < 0 \\implies x < 2,\\ \\text{that is,}\\ v_1 < 2v_2.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21155,
"subject": "Mathematics (Olympiad)",
"question": "Resuelve la ecuación exponencial\n\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\n$$\n6 = 2^x 3^{5-x} + 2^{-x} 3^{5x} \\geq 2\\sqrt{2^x 3^{5-x} 2^{-x} 3^{5x}} = 6.\n$$\n\nY la igualdad se dará cuando los números mediados sean iguales:\n\n$$\n2^x 3^{5-x} = 2^{-x} 3^{5x} \\Leftrightarrow 2^{2x} = 3^{5x} 3^{-(5-x)} \\Leftrightarrow 2^{2x} = 3^{2x} \\Leftrightarrow x = 0.\n$$\n\nEsto es, $x = 0$ será la única solución de la ecuación.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21156,
"subject": "Mathematics (Olympiad)",
"question": "For an integer $m \\ge 3$, set\n$$\nS(m) = 1 + \\frac{1}{3} + \\dots + \\frac{1}{m}\n$$\n(the fraction $\\frac{1}{2}$ does not participate in the sum).\n\nLet $n \\ge 3$ and $k \\ge 3$. Compare the numbers $S(nk)$ and $S(n) + S(k)$.",
"options": [],
"answer": "See solution",
"solution": "We show that $S(nk) < S(n) + S(k)$.\n\nCancel the summand of $S(k)$ on both sides of this inequality, then add $\\frac{1}{2}$ to both sides and rearrange. This yields the equivalent inequality\n\n$$\n\\frac{1}{k+1} + \\frac{1}{k+2} + \\dots + \\frac{1}{nk} + \\frac{1}{2} < 1 + \\frac{1}{3} + \\dots + \\frac{1}{n}. \\quad (*)\n$$\n\nDivide the first $(n-1)k$ numbers on the left-hand side into $n-1$ sums of $k$ fractions with\n\n$$\nA_1 = \\frac{1}{k+1} + \\dots + \\frac{1}{2k}, \\quad A_2 = \\frac{1}{2k+1} + \\dots + \\frac{1}{3k}, \\dots, A_{n-1} = \\frac{1}{(n-1)k+1} + \\dots + \\frac{1}{nk}.\n$$\n\nThen compare $A_j$ with $\\frac{1}{j}$ for $j = 1, \\dots, n-1$. Denoting\n$$\nd_j = \\frac{1}{j} - A_j = \\frac{1}{j} - \\frac{1}{jk+1} - \\frac{1}{jk+2} - \\dots - \\frac{1}{jk+k},\n$$\nwe have\n$$\nd_j = \\left( \\frac{1}{jk} - \\frac{1}{jk+1} \\right) + \\left( \\frac{1}{jk} - \\frac{1}{jk+2} \\right) + \\dots + \\left( \\frac{1}{jk} - \\frac{1}{jk+k} \\right) = \n\\frac{1}{jk(jk+1)} + \\frac{2}{jk(jk+2)} + \\dots + \\frac{k}{jk(jk+k)} > 0.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21157,
"subject": "Mathematics (Olympiad)",
"question": "In Greifswald, there are three schools called A, B, and C, each attended by at least one student. Among any three students $a$ from A, $b$ from B, and $c$ from C, there are two who know each other and two others who do not know each other. Prove that either:\n\n- Some student from A knows all students from B, or\n- Some student from B knows all students from C, or\n- Some student from C knows all students from A.",
"options": [],
"answer": "See solution",
"solution": "Assume the contrary. Let $a$ be a student from A who knows as many students from B as possible. Since $a$ does not know all students from B, there is a student $b$ from B not known to $a$. Similarly, pick a student $c$ from C not known to $b$, and then a student $a'$ from A not known to $c$.\n\nApplying the assumption to the sets $\\{a, b, c\\}$ and $\\{a', b, c\\}$, we find that $a$ and $c$ know each other, and so do $a'$ and $b$. Since $b$ knows $a'$ but not $a$, we have $a \\neq a'$. The maximality of $a$ implies there is a student $b'$ from B known to $a$ but not to $a'$. If $b'$ and $c$ knew each other, then any two students from $\\{a, b', c\\}$ would know one another, which is not possible. Thus, $b'$ and $c$ do not know each other, but then no two students from $\\{a', b', c\\}$ know one another, which is also impossible. Therefore, the problem is solved.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21158,
"subject": "Mathematics (Olympiad)",
"question": "The equal segments $AB$ and $CD$ intersect at point $O$ and are divided in the ratio $AO : OB = CO : OD = 1 : 2$. The lines $AD$ and $BC$ intersect at point $M$. Prove that $DM = MB$.",
"options": [],
"answer": "See solution",
"solution": "Let the length of each segment be $AB = CD = 3a$, so $AO = CO = a$ and $OB = OD = 2a$. Since $\\angle AOD = \\angle COB$ (vertical angles), triangles $AOD$ and $COB$ are congruent, so $\\angle ADO = \\angle CBO$. Triangle $BOD$ is isosceles, so $\\angle BDO = \\angle DBO$. Therefore, $\\angle MDB = \\angle MBD$ as sums of equal angles, meaning triangle $MDB$ is isosceles and thus $DM = MB$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21159,
"subject": "Mathematics (Olympiad)",
"question": "Let $S$ be a set of positive integers such that\n\n$$\n\\min \\{ \\operatorname{lcm}(x, y) : x, y \\in S,\\ x \\neq y \\} \\geq 2 + \\max S.\n$$\n\nShow that\n\n$$\n\\sum_{x \\in S} \\frac{1}{x} < \\frac{3}{2}.\n$$",
"options": [],
"answer": "See solution",
"solution": "The condition implies there exists a positive integer $n$ which is a strict upper bound for $S$ and a strict lower bound for the set of least common multiples of distinct numbers in $S$. If $x$ is a member of $S$, let $M_x$ denote the set of positive multiples of $x$ that do not exceed $n$. Clearly, $|M_x| = \\lfloor n/x \\rfloor$. If $x$ and $y$ are distinct members of $S$, then $M_x$ and $M_y$ are disjoint, since the least common multiple of $x$ and $y$ is greater than $n$. Consequently,\n\n$$\n\\sum_{x \\in S} \\lfloor n/x \\rfloor = \\sum_{x \\in S} |M_x| \\leq n\n$$\n\nand $|S| \\leq \\lfloor n/2 \\rfloor$ (otherwise, some number in $S$ would divide another, by a well-known result of Erdős). Finally,\n\n$$\nn \\sum_{x \\in S} \\left( \\frac{1}{x} - \\frac{1}{n} \\right ) = \\sum_{x \\in S} (n/x - 1) < \\sum_{x \\in S} \\lfloor n/x \\rfloor \\leq n,\n$$\n\nwhence the conclusion.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21160,
"subject": "Mathematics (Olympiad)",
"question": "a) Consider a table with 3 rows and 2008 columns. In the first row, random integers are written in non-decreasing order (each cell contains exactly one number). The numbers in the second row are obtained as follows: under each number $A$ in the first row, write the number $B$ equal to the number of entries in the first row that are less than $A$ and located to the left of $A$. Similarly, the numbers in the third row are obtained from the second row: under each number $C$ in the second row, write the number $D$ equal to the number of entries in the second row that are less than $C$ and located to the left of $C$. Prove that the second and third rows of this table are filled identically.\n\nb) Consider tables consisting of 2 rows and 2008 columns, where in each table the first row contains random integers in non-decreasing order (each cell contains exactly one number). The second row is constructed as above: under each number $A$ in the first row, write the number $B$ equal to the number of entries in the first row that are less than $A$ and located to the left of $A$. Suppose that all such tables have pairwise distinct second rows. Find the maximum possible number of such tables.\n\n*Answer:* $2^{2007}$.",
"options": [],
"answer": "See solution",
"solution": "a) Let $A_1, \\ldots, A_n$ be the entries in the first row, and $B_1, \\ldots, B_n$ the entries in the second row, where $B_i$ is the number of entries in the first row less than $A_i$ and to the left of $A_i$. Similarly, let $C_1, \\ldots, C_n$ be the entries in the third row, where $C_i$ is the number of entries in the second row less than $B_i$ and to the left of $B_i$.\n\nWe analyze how $B_{i+1}$ is determined from $B_i$:\n\n1. If $A_{i+1} = A_i$, then the number of entries less than $A_{i+1}$ to the left is the same as for $A_i$, so $B_{i+1} = B_i$.\n\n2. If $A_{i+1} > A_i = A_{i-1} = \\dots = A_{i-k+1}$, then $A_{i+1}$ is greater than the previous $k$ equal entries. The number of entries less than $A_{i+1}$ to the left increases by $k$, so $B_{i+1} = B_i + k$ and $B_{i+1} > B_i = B_{i-1} = \\dots = B_{i-k+1}$.\n\n3. If $A_{i+1} > A_i > A_{i-1}$, then $B_{i+1} = B_i + 1$ and $B_{i+1} > B_i > B_{i-1}$.\n\nThe process for constructing the third row from the second row is identical, so $C_i = B_i$ for all $i$. Also, $B_1 = C_1 = 0$.\n\nb) Let $A_1, \\ldots, A_{2008}$ be the first row and $B_1, \\ldots, B_{2008}$ the second row. The value of $B_{i+1}$ depends only on $A_i$ and $A_{i+1}$:\n\n- If $A_{i+1} = A_i$, then $B_{i+1} = B_i$.\n- If $A_{i+1} > A_i = A_{i-1} = \\dots = A_{i-k+1}$, then $B_{i+1} = B_i + k$.\n\nAt each step, there are two choices for the next element, so the total number of possible distinct second rows is $2^{2007}$.\n\n$$\n\\begin{array}{|c|c|}\n\\hline\nA_{i} & A_{i+1} \\\\\n\\hline\nB_{i} & B_{i+1} \\\\\nC_{i} & C_{i+1} \\\\\n\\hline\n\\end{array}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21161,
"subject": "Mathematics (Olympiad)",
"question": "Find all integers $n \\geq 3$ such that one can write a number (not necessarily an integer) into each vertex of a regular $n$-gon so that both conditions are met:\n\n1. Whenever three consecutive vertices of the $n$-gon, taken clockwise, contain numbers $x$, $y$, and $z$, respectively, the equality $x = |y - z|$ holds.\n2. The sum of the numbers in all vertices of the $n$-gon is $1$.",
"options": [],
"answer": "See solution",
"solution": "Let the vertices of an $n$-gon $A_0A_1\\dots A_{n-1}$ be labeled with numbers satisfying the conditions. Let $a$ be the least among these numbers. Without loss of generality, assume $A_0$ contains $a$ and the indices of vertices increase counterclockwise. Let $b$ and $c$ be the numbers at vertices $A_{n-1}$ and $A_{n-2}$, respectively.\n\nBy condition (1), $a = |b - c| \\geq 0$. By the choice of $a$, $b \\geq a$, so by condition (1), $A_1$ contains $b - a$. Since $b - a \\geq a$, condition (1) implies $A_2$ contains $b - 2a$. Hence $A_3$ contains $(b - a) - (b - 2a) = a$ by condition (1) and non-negativity of $a$. Since the vertices can be renumbered so that $A_3$ becomes $A_0$, we conclude $A_6$ also contains $a$. Similarly, every third vertex contains $a$.\n\nIf $n$ is not divisible by $3$, then either $A_{n-1}$ or $A_1$ must contain $a$, and repeating the argument, also $A_{n-2}$ or $A_2$ contains $a$. Thus, three consecutive vertices contain $a$. Applying condition (1) to these three vertices, $a = |a - a| = 0$. But $0$ being in two consecutive vertices implies $0$ is in all vertices. Then the sum of all labels is $0$, contradicting condition (2).\n\nThis shows $n$ must be divisible by $3$. Let $n = 3k$ where $k$ is a positive integer. For every $3k$-gon, the conditions can be satisfied by writing $0$ into every third vertex and $\\frac{1}{2k}$ into all other vertices.\n\n\n\n",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21162,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $m$ is a positive integer. Consider a square board of size $m \\times m$. Ants are placed on the board, each at a square, and each ant moves at constant speed along the grid lines, turning at right angles at the edges, and possibly colliding with other ants. A *collision* is defined as the meeting of exactly two ants moving in opposite directions. What is the latest possible moment at which an ant can fall off the board, as a function of $m$?",
"options": [],
"answer": "See solution",
"solution": "For $m = 1$, the answer is clearly $\\frac{1}{2}$. For $m > 1$, consider placing an ant on the southwest corner facing east and another on the southeast corner facing west. They meet in the middle of the bottom row at time $\\frac{m-1}{2}$. After the collision, the ant moving north stays on the board for another $m - \\frac{1}{2}$ units, so the last ant falls off at time $\\frac{3m-1}{2}$. \n\nTo show this is maximal, consider any collision. By changing the collision rule (both ants turn anticlockwise), the ants' subsequent behavior is unchanged except for swapping positions. Thus, we may assume only two types of ants: NE-ants (moving north/east) and SW-ants (south/west). All ants will have fallen off after $2m-1$ units, but a better bound is possible.\n\nLet the board's corners be $(0,0)$, $(m,0)$, $(m,m)$, $(0,m)$. At time $t$, there are no NE-ants in $\\{(x, y) \\mid x + y < t + 1\\}$ and no SW-ants in $\\{(x, y) \\mid x + y > 2m - t - 1\\}$. If two ants collide at $(x, y)$ at time $t$:\n\n$$\nt + 1 \\le x + y \\le 2m - t - 1.\n$$\n\nAlso, $|y - x| \\le m - t - 1$ for each collision at $(x, y)$ at time $t$. Define\n\n$$\nB(t) = \\{(x, y) \\in [0, m]^2 \\mid t+1 \\le x+y \\le 2m-t-1 \\text{ and } |x-y| \\le m-t-1\\}.\n$$\n\nAn ant can only collide at time $t$ if it is in $B(t)$. Suppose a NE-ant's last collision is at time $t$ at $(x, y) \\in B(t)$. Then $x + y \\ge t + 1$ and $x - y \\ge -(m - t - 1)$, so\n\n$$\nx \\ge t+1 - \\frac{m}{2}.\n$$\n\nSimilarly, $y \\ge t+1 - \\frac{m}{2}$, so $\\min\\{x, y\\} \\ge t+1 - \\frac{m}{2}$. After this, the ant moves directly to an edge, taking at most $m - \\min\\{x, y\\}$ units. Thus, the total time on the board is at most\n\n$$\nt + (m - \\min\\{x, y\\}) \\le \\frac{3m}{2} - 1.\n$$\n\nBy symmetry, the same holds for SW-ants. Therefore, the latest possible moment an ant can fall off the board is $\\frac{3m}{2} - 1$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21163,
"subject": "Mathematics (Olympiad)",
"question": "There are 10 cards, each of which has two numbers, numbered 1, 2, 3, 4, 5, written on it, and the numbers on any two cards are not exactly identical. The 10 cards are placed in five boxes labelled 1, 2, 3, 4, 5, and a card with $i$ and $j$ written on it can only be placed in box $i$ or $j$. One placement is called “good” if there are more cards in box 1 than in each of the other boxes. Then the total number of the “good” placements is \\_\\_\\_\\_.",
"options": [],
"answer": "See solution",
"solution": "Denote the card with $i, j$ written on it as $\\{i, j\\}$. It is easy to see that these 10 cards are exactly $\\{i, j\\}$ for $1 \\leq i < j \\leq 5$.\n\nConsider the “good” placements of the cards. There are 10 cards in the five boxes, so there are at least 3 cards in box 1. The only cards that can be placed in box 1 are $\\{1, 2\\}$, $\\{1, 3\\}$, $\\{1, 4\\}$, and $\\{1, 5\\}$.\n\n**Case 1:** All 4 cards are placed in box 1. In this case, it is not possible for any other box to have 4 cards, so no matter how the remaining 6 cards are placed, the requirement is satisfied. There are $2^6 = 64$ “good” placements.\n\n**Case 2:** Exactly 3 of the 4 cards are in box 1, and the remaining card is in its other possible box. Each of the other boxes contains at most 2 cards.\n\nConsider the number $N$ of placements where $\\{1, 2\\}$, $\\{1, 3\\}$, $\\{1, 4\\}$ are in box 1 and $\\{1, 5\\}$ is in box 5.\n\nThere are 8 possible ways to place the cards $\\{2, 3\\}$, $\\{2, 4\\}$, $\\{3, 4\\}$: 6 ways where two cards are placed in one of the boxes 2, 3, 4, and 2 ways where one card is placed in each of boxes 2, 3, 4.\n\nIf two of $\\{2, 3\\}$, $\\{2, 4\\}$, $\\{3, 4\\}$ are in the same box (say, $\\{2, 3\\}$ and $\\{2, 4\\}$ in box 2), then $\\{2, 5\\}$ must be in box 5. Box 5 already has $\\{1, 5\\}$ and $\\{2, 5\\}$, so $\\{3, 5\\}$ and $\\{4, 5\\}$ must be in boxes 3 and 4, respectively. Thus, the placement of $\\{2, 5\\}$, $\\{3, 5\\}$, and $\\{4, 5\\}$ is unique.\n\nIf one of $\\{2, 3\\}$, $\\{2, 4\\}$, $\\{3, 4\\}$ is placed in each of boxes 2, 3, 4, then there are at most 2 cards in each of boxes 2, 3, 4. We must ensure that there are no more than 2 cards in box 5, i.e., there are 0 or 1 of $\\{2, 5\\}$, $\\{3, 5\\}$, $\\{4, 5\\}$ in box 5. The number of such placements is $\\binom{3}{0} + \\binom{3}{1} = 4$.\n\nAs a result, $N = 6 \\times 1 + 2 \\times 4 = 14$. By symmetry, there are $4N = 56$ “good” placements in case 2.\n\nTo sum up, there is a total of $64 + 56 = 120$ “good” placements.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21164,
"subject": "Mathematics (Olympiad)",
"question": "We have $n$ students sitting at a round table. Initially, each student is given one candy. At each step, each student who has candies may either pick one of their candies and give it to one of their neighbouring students, or distribute all of their candies to their neighbouring students in any way they wish. A distribution of candies is called *legal* if it can be reached from the initial distribution via a sequence of steps.\n\nDetermine the number of legal distributions. (All the candies are identical.)",
"options": [],
"answer": "See solution",
"solution": "The answer is $\\binom{2n-1}{n}$ if $n$ is odd, and $\\binom{2n-1}{n-2}$ if $n$ is even.\n\n**Case 1:** Suppose $n$ is odd, say $n = 2m+1$. In this case, any distribution of candies is legal, so the number of legal distributions is $\\binom{2n-1}{n}$.\n\nThis is achieved by letting each student always distribute all of their candies to their two neighbouring students in some way. Thus, at each step, each candy moves either one position clockwise or one anticlockwise.\n\nFor the initial and required final distributions, we can specify for each candy its desired final position. Because $n$ is odd, either the clockwise or anticlockwise distance between the initial and final position is even and at most $m$.\n\nThus, after an even number of steps (at most $m$), we can move each candy to its required final position. (If a candy reaches its required position earlier, it can move back and forth until all candies reach their required positions.)\n\n**Case 2:** Suppose $n$ is even, say $n = 2m$. Let $x_1, \\dots, x_{2m}$ be the students in cyclic order. Initially, the students with even indices (even students) have at least one candy in total, and so do the students with odd indices (odd students). This property is preserved after each step.\n\nEvery distribution in which the even students have at least one candy in total and the odd students also have at least one candy in total is legal.\n\nSuppose the required final distribution has $a$ candies in odd positions and $b$ candies in even positions, where $a, b \\ge 1$. It suffices to reach any position with $a$ candies in even positions and $b$ candies in odd positions, as then we can follow the same approach as in Case 1.\n\n\n\nTo achieve this, first move all candies to students $x_1$ and $x_2$. At each step, $x_1$ moves all its candies to $x_2$, while for $1 \\le r \\le 2m-1$, student $x_{r+1}$ moves all its candies to $x_r$.\n\nSuppose now we have $a+k$ candies at $x_1$ and $b-k$ candies at $x_2$, with $k \\ge 0$. If $k=0$, we have reached our target. If not, in the next step, $x_1$ moves a candy to $x_2$ and $x_2$ moves a candy to $x_3$. In the next step, $x_1$ (still with $a+k-1 > a > 0$ candies) moves a candy to $x_2$, $x_2$ moves a candy to $x_1$, and $x_3$ moves a candy to $x_2$. Now, $x_1$ has $a+k-1$ candies and $x_2$ has $b+1-k$ candies. Repeating this process $k-1$ more times, we end up with $a$ candies in $x_1$ and $b$ candies in $x_2$ as required.\n\nThe total number of legal configurations in this case is\n\n$$\n\\binom{2n-1}{n} - 2\\binom{2n}{n}\n$$\n\nas $\\binom{2n-1}{n}$ counts the total number of configurations, while $2\\binom{2n}{n}$ counts the number of illegal configurations where all $n$ candies belong to either all odd or all even positions.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21165,
"subject": "Mathematics (Olympiad)",
"question": "給定正整數 $n$。設 $a_1, a_2, \\dots, a_n$ 為 $1, 2, \\dots, n$ 的排列。試確定\n$$\n\\sum_{i=1}^{n} \\left\\lfloor \\frac{a_i}{i} \\right\\rfloor\n$$\n的最小值。\n\n註：$\\left\\lfloor x \\right\\rfloor$ 是不超過實數 $x$ 的最大整數。",
"options": [],
"answer": "See solution",
"solution": "假設 $2^k \\leq n < 2^{k+1}$，其中 $k$ 為非負整數。首先，我們展示一個排列 $(a_1, a_2, \\dots, a_n)$ 使得 $\\sum_{i=1}^n \\lfloor \\frac{a_i}{i} \\rfloor = k+1$；然後證明對於任意排列，$\\sum_{i=1}^n \\lfloor \\frac{a_i}{i} \\rfloor \\geq k+1$。因此，最小值為 $k+1$。\n\n**I. 構造排列**\n\n$$\n(a_1) = (1), \\quad (a_2, a_3) = (3, 2), \\quad (a_4, a_5, a_6, a_7) = (7, 4, 5, 6), \\dots \\\\\n(a_{2^{k+1}}, \\dots, a_{2^{k-1}}) = (2^k - 1, 2^{k-1}, 2^{k-1} + 1, \\dots, 2^k - 2), \\\\\n(a_{2^k}, \\dots, a_n) = (n, 2^k, 2^k + 1, \\dots, n-1)\n$$\n\n此排列由 $k+1$ 個循環組成。在每個循環 $(a_p, \\dots, a_q) = (q, p, p+1, \\dots, q-1)$ 中，$q < 2p$，因此\n\n$$\n\\sum_{i=p}^{q} \\lfloor \\frac{a_i}{i} \\rfloor = \\lfloor \\frac{q}{p} \\rfloor + \\sum_{i=p+1}^{q} \\lfloor \\frac{i-1}{i} \\rfloor = 1\n$$\n\n所有循環的總和正好是 $k+1$。\n\n**II. 下界證明**\n\n我們證明更一般的命題：\n\n*Claim 1.* 若 $b_1, \\dots, b_{2^k}$ 為互異正整數，則\n\n$$\n\\sum_{i=1}^{2^k} \\lfloor \\frac{b_i}{i} \\rfloor \\geq k+1\n$$\n\n由此可得 $\\sum_{i=1}^{n} \\lfloor \\frac{a_i}{i} \\rfloor \\geq \\sum_{i=1}^{2^k} \\lfloor \\frac{a_i}{i} \\rfloor \\geq k+1$。\n\n*Claim 1* 的證明：對 $k$ 做歸納。$k=1$ 時，$\\lfloor \\frac{b_1}{1} \\rfloor \\geq 1$。假設對某 $k$ 成立，考慮 $k+1$。若存在 $j$ 使得 $2^k < j \\leq 2^{k+1}$ 且 $b_j \\geq j$，則\n\n$$\n\\sum_{i=1}^{2^{k+1}} \\lfloor \\frac{b_i}{i} \\rfloor \\geq \\sum_{i=1}^{2^k} \\lfloor \\frac{b_i}{i} \\rfloor + \\lfloor \\frac{b_j}{j} \\rfloor \\geq (k+1) + 1\n$$\n\n否則，對所有 $2^k < j \\leq 2^{k+1}$，有 $b_j < j$。在 $b_1, \\dots, b_{2^{k+1}}$ 中，必有某 $b_m \\geq 2^{k+1}$，且 $1 \\leq m \\leq 2^k$。\n\n對以下數列應用歸納假設：\n\n$$\nc_1 = b_1, \\dots, c_{m-1} = b_{m-1}, \\quad c_m = b_{2^k} + 1, \\quad c_{m+1} = b_{m+1}, \\dots, c_{2^k} = b_{2^k}\n$$\n\n即將前 $2^k$ 個數中 $b_m$ 換成 $b_{2^k+1}$。注意：\n\n$$\n\\lfloor \\frac{b_m}{m} \\rfloor \\geq \\lfloor \\frac{2^{k+1}}{m} \\rfloor = \\lfloor \\frac{2^k + 2^k}{m} \\rfloor \\geq \\lfloor \\frac{b_{2^k+1} + m}{m} \\rfloor = \\lfloor \\frac{c_m}{m} \\rfloor + 1\n$$\n\n其他 $i$ ($1 \\leq i \\leq 2^k$, $i \\neq m$) 有 $\\lfloor \\frac{b_i}{i} \\rfloor = \\lfloor \\frac{c_i}{i} \\rfloor$，因此\n\n$$\n\\sum_{i=1}^{2^k+1} \\lfloor \\frac{b_i}{i} \\rfloor = \\sum_{i=1}^{2^k} \\lfloor \\frac{b_i}{i} \\rfloor \\geq \\sum_{i=1}^{2^k} \\lfloor \\frac{c_i}{i} \\rfloor + 1 \\geq (k+1) + 1\n$$\n\n證畢。\n\n綜合 I 與 II，得最小值為 $k+1$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21166,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $n$ that have precisely $\\sqrt{n+1}$ natural divisors.",
"options": [],
"answer": "See solution",
"solution": "First, observe that $n = k^2 - 1$ for some positive integer $k$. Since $n$ is not a perfect square, $\\tau(n) = k$ must be even, so $n$ is odd.\n\nRecall that, since $n$ is not a perfect square, there is a bijection between factors of $n$ greater than $\\lfloor \\sqrt{n} \\rfloor = k-1$ and those at most $k-1$. Since $n$ is odd, all its divisors are odd. Therefore,\n\n$$\nk = \\tau(n) \\leq 2 \\cdot |\\{1, 3, \\dots, k-1\\}| = 2 \\cdot \\frac{k}{2} = k.\n$$\n\nThus, equality must hold, so every odd number between $1$ and $k-1$ divides $n = k^2 - 1$.\n\nIn particular, $k-3$ must divide $k^2-1$; but\n\n$$\n\\frac{k^2 - 1}{k - 3} = k + 3 + \\frac{8}{k - 3}.\n$$\n\nTherefore, $k-3$ is either $-1$ or $1$, so $k$ equals either $2$ or $4$, and $n \\in \\{3, 15\\}$. Both of these work. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21167,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute triangle. The bisector of the side $AB$ intersects the lines $BC$ and $CA$ at the points $X$ and $Y$, respectively, and the bisector of the side $AC$ intersects the lines $BC$ and $AB$ at the points $Z$ and $W$, respectively. Prove that the points $X$, $Y$, $Z$, and $W$ are concyclic.",
"options": [],
"answer": "See solution",
"solution": "The solution uses directed angles. Let $E$ be the midpoint of the side $AB$ and let $F$ be the midpoint of the side $AC$. Since $EF$ is a midline of the triangle $ABC$, $EF$ is parallel to $BC$. Because the lines $XY$ and $ZW$ are bisectors of the sides $AB$ and $AC$, respectively, the points $E$, $X$, and $Y$ as well as the points $F$, $Z$, and $W$ are collinear. The points $E$, $F$, $Y$, and $W$ are concyclic because\n\n$$\n\\angle WEY = \\angle BEY = \\frac{\\pi}{2} = \\angle WFC = \\angle WFY.\n$$\n\nFrom concyclicity of the points $E$, $F$, $Y$, and $W$, collinearity of the points $F$, $Z$, and $W$, and collinearity of the points $E$, $X$, and $Y$ we now derive\n\n$$\n\\angle XYW = \\angle EYW = \\angle EFW = \\angle EFZ.\n$$\n\nFurther, from parallelism of the lines $EF$ and $BC$, collinearity of the points $F$, $W$, and $Z$, and collinearity of the points $B$, $C$, $X$, and $Y$ we get\n\n$$\n\\angle EFZ = \\angle BZF = \\angle XZF = \\angle XZW.\n$$\n\nThe above two equalities say that $\\angle XYW = \\angle XZW$, hence the points $X$, $Y$, $Z$, and $W$ are concyclic.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21168,
"subject": "Mathematics (Olympiad)",
"question": "An infinite set $B$ consisting of non-negative integers has the following property: for each $a, b \\in B$ with $a > b$, the number $\\frac{a-b}{(a,b)}$ is in $B$. Prove that $B$ contains all non-negative integers. Here $(a, b)$ denotes the greatest common divisor of $a$ and $b$.",
"options": [],
"answer": "See solution",
"solution": "Let $d$ be the greatest common divisor (g.c.d.) of all numbers in $B$, and define $A = \\{b/d : b \\in B\\}$. For each $a, b \\in A$ with $a > b$, we have\n\n$$\n\\frac{a-b}{d(a,b)} \\in A. \\tag{*}\n$$\n\nSince the g.c.d. of $A$ is $1$, there exists a finite subset $A_1 \\subseteq A$ with $\\gcd(A_1) = 1$ and minimal possible sum of elements. Choose $a, b \\in A_1$ with $a > b$ and replace $a$ by $\\frac{a-b}{d(a,b)}$; the g.c.d. remains $1$, but the sum decreases, contradicting minimality. Thus, $A_1 = \\{1\\}$, so all elements of $A$ are congruent to $1$ modulo $d$.\n\nTake any $a = kd + 1 \\in A$ and $b = 1$. Then $k \\in A$ by $(*)$, and $k = ds + 1$. Since $(k, kd + 1) = 1$, we get $\\frac{kd+1-ds-1}{d} = k - s = (d-1)s + 1 \\in A$, so $s$ is divisible by $d$. But $s \\in A$, so $s-1$ is also divisible by $d$, hence $d = 1$ (so $B = A$). Therefore, if $a = k + 1 \\in A$, then $a-1 = k \\in A$. Since $A$ is infinite, all non-negative integers belong to $A$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21169,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs of non-negative integers $x$ and $y$ that satisfy the equation\n$$\nx^3 + 7x^2 + 35x + 27 = y^3.\n$$",
"options": [],
"answer": "See solution",
"solution": "Observe that for non-negative integers $x$ and $y$ that satisfy the equation $x^3 + 7x^2 + 35x + 27 = y^3$, we obtain\n\n$$\n\\begin{aligned}\ny^3 - (x+2)^3 &= x^3 + 7x^2 + 35x + 27 - x^3 - 6x^2 - 12x - 8 \\\\\n&= x^2 + 23x + 19 > 0\n\\end{aligned}\n$$\n\nand similarly\n\n$$\n(x+4)^3 - y^3 = 5x^2 + 13x + 37 > 0\n$$\n\nThus $x+2 < y < x+4$, and since $x$ and $y$ are integers, $y = x+3$. Substitution in the original equation yields\n\n$$\nx^3 + 7x^2 + 35x + 27 = x^3 + 9x^2 + 27x + 27\n$$\n\nand so $x^2 = 4x$. Hence $x = 0$ or $x = 4$, and the corresponding values of $y$ are $3$ and $7$. The pairs $(0, 3)$ and $(4, 7)$ are easily seen to satisfy the equation.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21170,
"subject": "Mathematics (Olympiad)",
"question": "令 $n$ 為大於 $1$ 的正整數。一個 $n \\times n \\times n$ 的大立方體由 $n^3$ 個邊長為 $1$ 的小立方體所構成。每個小立方體被塗上一種顏色。從這個大正方體中，我們沿著 $x$ 軸的方向截取出 $n$ 個不同的 $1 \\times n \\times n$ 的長方體 $R_1, R_2, \\cdots, R_n$。令 $C_i$ 為 $R_i$ 中所有小立方體的顏色所成集合，並令 $\\mathcal{X} = \\{C_1, C_2, \\cdots, C_n\\}$。以同樣的方式，令 $\\mathcal{Y}$ 為沿著 $y$ 軸的方向截取 $n \\times 1 \\times n$ 長方體下造出來的集合，而 $\\mathcal{Z}$ 為沿著 $z$ 軸的方向截取 $n \\times n \\times 1$ 長方體下造出來的集合。\n\n假設我們知道，對於 $\\mathcal{X}, \\mathcal{Y}$ 與 $\\mathcal{Z}$，若有個顏色集屬於其中一個集合，則它必也屬於另外兩個集合。試求這 $n^3$ 個小立方體顏色總數的最大值（以 $n$ 表示）。\n\n（例：若 $R_1$ 中包含紅、藍、綠三色，則 $C_1 = \\{\\textbf{紅}, \\textbf{藍}, \\textbf{綠}\\}$。）",
"options": [],
"answer": "See solution",
"solution": "**答案：** 最大顏色數為 $\\frac{n(n+1)(2n+1)}{6}$。\n\n稱 $n \\times n \\times 1$ 的盒子為 *x-box*、*y-box* 或 *z-box*，依其方向而定。\n設 $N$ 為可行配置下的顏色數。先給出 $N$ 的上界。\n\n設 $D_1, D_2, D_3$ 分別為在大立方體中恰好出現一次、恰好兩次、至少三次的顏色集合。令 $M_i$ 為顏色屬於 $D_i$ 的小立方體集合，$n_i = |M_i|$。\n\n考慮任一 x-box $X$，令 $Y$ 和 $Z$ 為含有與 $X$ 相同顏色集合的 y-box 和 z-box。先證：\n\n**斷言：** $4|X \\cap M_1| + |X \\cap M_2| \\le 3n + 1$。\n\n**證明：** 分兩情況。\n\n*情況 1：* $X \\cap M_1 \\neq \\emptyset$\n\n$X \\cap M_1$ 的小立方體必同時出現在 $X, Y, Z$，即 $X \\cap Y \\cap Z$，故 $|X \\cap M_1| = 1$。\n\n$X \\cap M_2$ 中，最多有 $2(n-1)$ 個在 $X \\cap Y$ 或 $X \\cap Z$。若 $a$ 屬於 $X \\cap M_2$ 且不在 $Y$ 或 $Z$，則 $a$ 的同色小立方體 $a'$ 必在 $Y \\cap Z$ 且不在 $X$，此對應為單射，故此類 $a$ 不超過 $n-1$。合計：\n\n$$\n|X \\cap M_2| \\leq 2(n-1) + (n-1) = 3(n-1)\n$$\n\n易檢查斷言成立。\n\n*情況 2：* $X \\cap M_1 = \\emptyset$\n\n此時 $X \\cap M_2$ 最多有 $2n-1$ 個在 $X \\cap Y$ 或 $X \\cap Z$，其餘至多 $n$ 個對應 $Y \\cap Z$，合計 $|X \\cap M_2| \\leq (2n-1)+n = 3n-1$。斷言仍成立。\n\n對所有 x-box $X$ 加總，得：\n\n$$\n4n_1 + n_2 \\leq n(3n + 1)\n$$\n\n又 $n_1 + n_2 + n_3 = n^3$。\n\n由 $M_i$ 定義，$n_i \\geq i|D_i|$，故\n\n$$\n\\begin{aligned}\nN &\\leq n_1 + \\frac{n_2}{2} + \\frac{n_3}{3} = \\frac{n_1 + n_2 + n_3}{3} + \\frac{4n_1 + n_2}{6} \\\\\n &\\leq \\frac{n^3}{3} + \\frac{3n^2 + n}{6} = \\frac{n(n+1)(2n+1)}{6}\n\\end{aligned}\n$$\n\n構造達到上界的例子如下：\n\n- $n$ 個集合 $\\{(i, i, i)\\}$，$1 \\le i \\le n$\n- $C_2^n$ 個集合 $\\{(i, j, j), (j, i, i)\\}$，$C_2^n$ 個集合 $\\{(j, i, j), (i, j, i)\\}$，$C_2^n$ 個集合 $\\{(j, j, i), (i, i, j)\\}$，$1 \\le i < j \\le n$\n- $C_3^n$ 個集合 $\\{(i, j, k), (j, k, i), (k, i, j)\\}$，$1 \\le i < j < k \\le n$，以及 $C_3^n$ 個集合 $\\{(i, j, k), (j, k, i), (k, i, j)\\}$，$1 \\le i < k < j \\le n$\n\n每個集合給予不同顏色，即可達到最大值。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21171,
"subject": "Mathematics (Olympiad)",
"question": "Square 1 has side length 9. Let $SL_k$ denote the side length of square $k$. Let $SL_2 = x$.\n\n\n\nFind the perimeter of the large rectangle formed by arranging the squares as described below:\n\n$$\n\\begin{aligned}\nSL_3 &= SL_1 + SL_2 = x + 9, \\\\\nSL_4 &= SL_1 + SL_3 = x + 18, \\\\\nSL_5 &= SL_1 + SL_4 = x + 27, \\\\\nSL_6 &= SL_4 + SL_5 = 2x + 45, \\\\\nSL_7 &= SL_2 + SL_3 = 2x + 9, \\\\\nSL_8 &= SL_2 + SL_7 = 3x + 9, \\\\\nSL_9 &= SL_7 + SL_8 - SL_4 - SL_5 = 3x - 27, \\\\\nSL_{10} &= SL_8 + SL_9 = 6x - 18, \\\\\nSL_{11} &= SL_9 + SL_{10} = 9x - 45.\n\\end{aligned}\n$$",
"options": [],
"answer": "See solution",
"solution": "From squares 6, 5, and 8, the height of the large rectangle is $2x + 45 + x + 27 + 3x + 9 = 6x + 81$.\n\nFrom squares 10 and 11, the height is $6x - 18 + 9x - 45 = 15x - 63$.\n\nSetting these equal:\n$$\n6x + 81 = 15x - 63\n$$\n$$\n9x = 144 \\implies x = 16\n$$\n\nFrom squares 6 and 11, the width is $2x + 45 + 9x - 45 = 11x$.\n\nSo the perimeter is:\n$$\n2(11x + 6x + 81) = 34x + 162 = 544 + 162 = 706\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21172,
"subject": "Mathematics (Olympiad)",
"question": "A triangle $ABC$ is given. Let $M$ be the midpoint of the side $AC$ of the triangle and $Z$ the image of point $B$ along the line $BM$. The circle with center $M$ and radius $MB$ intersects the lines $BA$ and $BC$ at the points $E$ and $G$ respectively. Let $H$ be the point of intersection of $EG$ with the line $AC$, and $K$ the point of intersection of $HZ$ with the line $EB$. The perpendicular from point $K$ to the line $BH$ intersects the lines $BZ$ and $BH$ at the points $L$ and $N$, respectively.\n\nIf $P$ is the second point of intersection of the circumscribed circles of the triangles $KZL$ and $BLN$, prove that the lines $BZ$, $KN$ and $HP$ intersect at a common point.",
"options": [],
"answer": "See solution",
"solution": "From the point $G$ we draw a parallel to the line $AC$, which intersects $BZ$ and $AB$ at the points $V$ and $S$ respectively.\n\n\n\nSince $AM = MC$ from the construction hypothesis, we have that $SV = VG$, that is, $V$ is the midpoint of the segment $SG$. If $T$ is the midpoint of the chord $EG$, we have $ES \\parallel TV$ and therefore $\\angle BEG = \\angle VTG$, and since $\\angle BEG = \\angle BZG$, we will have $\\angle VTG = \\angle VZG$. Therefore, the quadrilateral $VTZG$ is cyclic, so $\\angle TZV = \\angle TGV$.\n\nSince $GS \\parallel AH$, we have $\\angle TGV = \\angle THA$. Therefore, we conclude that the quadrilateral $MTZH$ is cyclic and since $\\angle MTH = 90^\\circ$ it implies that $\\angle MZH = 90^\\circ$, so $BZ$ is the altitude of the triangle $KBH$, that is, the point $L$ is the orthocenter of the triangle $KBH$, since from our hypotheses $KN \\perp BH$.\n\nIn addition, since the quadrilateral $LZHN$ is cyclic, it is known that the second point of intersection $P$ of the circumscribed circles of the triangles $KZL$ and $BLN$ will be located on $KB$. It suffices now to prove that the points $P$, $L$ and $H$ are collinear. It is true that from the cyclic quadrilaterals $PLNB$, $NLZH$ and $BKZN$ follows\n\n$$\n\\angle KLP = \\angle KBN, \\quad \\angle KLZ = \\angle BLN = \\angle BHK, \\quad \\angle ZLH = \\angle ZNH = \\angle BKH.\n$$\n\nFrom these relations we have $\\angle KLP + \\angle KLZ + \\angle ZLH = 180^\\circ$, so the points $P$, $L$ and $H$ are collinear, therefore the lines $BZ$, $KN$ and $HP$ are concurrent. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21173,
"subject": "Mathematics (Olympiad)",
"question": "Let $N$ denote the set of positive integers. Find all functions $f : N \\to N$ such that the equation\n\n$$\n\\operatorname{lcm}(m, f(m + f(n))) = \\operatorname{lcm}(f(m), f(m) + n)\n$$\n\nholds for any positive integers $m$ and $n$. Here, for positive integers $x$ and $y$, $\\operatorname{lcm}(x, y)$ denotes their least common multiple.",
"options": [],
"answer": "See solution",
"solution": "We prove that the function $f(n) = n$ is the unique function satisfying the condition in the problem. It is easy to see that this $f$ satisfies the condition.\n\nSuppose that $f$ is a function that satisfies the condition. First, for any positive integer $k$, we prove that $f(k)$ is a multiple of $k$. Let $r$ be the remainder when dividing $f(k)$ by $k$. Substituting $m = k$ and $n = k - r + 1$ into the equation, we obtain\n\n$$\n\\operatorname{lcm}(k, f(k + f(k - r + 1))) = \\operatorname{lcm}(f(k), f(k) + k - r + 1).\n$$\n\nTherefore, $\\operatorname{lcm}(f(k), f(k) + k - r + 1)$ is a multiple of $k$. Here, $f(k) + k - r + 1$ leaves a remainder of $1$ when divided by $k$, and hence, $f(k) + k - r + 1$ is coprime to $k$. Therefore, $f(k)$ is a multiple of $k$.\n\nLet $m$ be a positive integer. We prove that $f(m) = m$. Substituting $n = f(m)$ into the equation, we get\n\n$$\n\\operatorname{lcm}(m, f(m + f(f(m)))) = \\operatorname{lcm}(f(m), 2f(m)) = 2f(m).\n$$\n\nTherefore, $2f(m)$ is a multiple of $f(m + f(f(m)))$. Since $f(m + f(f(m)))$ is a multiple of $m + f(f(m))$, $2f(m)$ is a multiple of $m + f(f(m))$. Note that\n\n$$\n2f(m) \\le 2f(f(m)) < 2(m + f(f(m))).\n$$\n\nThe first inequality follows from the fact that $f(f(m))$ is a multiple of $f(m)$. Therefore, $2f(m) = m + f(f(m))$. Since\n\n$$\nf(f(m)) = 2f(m) - m < 2f(m)\n$$\n\nand $f(f(m))$ is a multiple of $f(m)$, we conclude that $f(f(m)) = f(m)$. Hence, $2f(m) = m + f(m)$, so $f(m) = m$. This completes the proof.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21174,
"subject": "Mathematics (Olympiad)",
"question": "A square $n \\times n$ is divided into $n^2$ cells. In total, $n^2$ tokens are placed at some cell. During every round, a player can move one token from cell $A$ to cell $B$, and one token from cell $A$ to cell $C$, provided that cell $A$ contained at least two tokens, $B$ and $C$ are symmetric with respect to $A$, and $B$ and $C$ are adjacent to $A$. Is it possible that after a few such rounds every cell on the board contains exactly one token, in the case:\n\na) $n = 2016$;\n\nb) $n = 2017$?\n\n",
"options": [],
"answer": "See solution",
"solution": "a) Consider a square $n \\times n$, for a positive integer $n$. Let us denote the leftmost column by $1$, the next by $2$, and so on, with the rightmost column by $n$. For any cell $c$, define $w(c) = b$ if cell $c$ is in column $b$. For every token $T$, let $w(T) = b$ if $T$ is placed in column $b$ at the moment. Let $W$ be the sum of $w(T)$ for all tokens $T$ on the board. It is clear that after every round, $W$ does not change. At the end of the process, we must have\n\n$$\nW = n + 2n + \\dots + n^2 = n \\cdot (1 + 2 + \\dots + n) = \\frac{n^2(n+1)}{2}.\n$$\n\nAt the beginning, $W = l \\cdot n^2$ if all tokens are placed in column $l$. For $n = 2016$, this leads to a contradiction.\n\nb) Now consider $n = 2017$. From the previous argument, the statement could be true only if all tokens are placed on the central cell at the beginning. In this case, we provide an algorithm. In every cell, write the number of tokens placed there. Using induction, we prove the following statements:\n\n**Statement 1.** For every positive integer $n$, in a row of length $2n+1$, from the position\n\n$0; 0; 0; \\dots; 0; 2n+1; 0; \\dots; 0; 0$\n\nwe can reach the position\n\n$1; 1; 1; \\dots; 1; 1; 1; \\dots; 1; 1$\n\n**Statement 2.** For every positive integer $n$, in a row of length $2n+1$, from the position\n\n$0; 1; 1; \\dots; 1; 3; 1; \\dots; 1; 1$\n\nwe can reach the position\n\n$1; 1; 1; \\dots; 1; 1; 1; \\dots; 1; 1$\n\n*Proof.* It is clear for $n=1$. Suppose for $n=k-1$ the statements are true; let us prove them for $n=k$. At the beginning, we have $2k+1$ tokens on the central cell. Two of them we will not move. The remaining $2k-1$ tokens can be moved (by the induction hypothesis) to the position\n\n$0; 1; 1; \\dots; 1; 1; 3; 1; 1; \\dots; 1; 1$\n\nNow consider the following replacements:\n\n$0; 1; 1; \\dots; 1; 1; 2; 1; 2; 1; 1; \\dots; 1; 1$\n\n$0; 1; 1; \\dots; 1; 2; 0; 3; 0; 2; 1; \\dots; 1; 1$\n\n$0; 1; 1; \\dots; 2; 0; 1; 3; 1; 0; 2; \\dots; 1; 1$\n\n$\\dots$\n\n$0; 1; 2; 0; 1; \\dots; 1; 1; 1; 3; 1; 1; 1; \\dots; 1; 0; 2; 1$\n\n$0; 2; 0; 1; 1; \\dots; 1; 1; 1; 3; 1; 1; 1; \\dots; 1; 1; 0; 2$\n\n$1; 0; 1; 1; 1; \\dots; 1; 1; 1; 3; 1; 1; 1; \\dots; 1; 1; 1; 0$\n\nWe can apply the induction hypothesis now and get\n\n$1; 1; 1; \\dots; 1; 1; 1; \\dots; 1; 1$\n\nBoth statements are now proved.\n\nWe can now easily prove the statement of the problem. Divide all tokens into equal groups with 2017 tokens in each group. Using our statements, we can arrange tokens so that in every cell of the central column exactly one group is placed. Then, use the statements again to arrange tokens so that in every cell of every row exactly one token is placed.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21175,
"subject": "Mathematics (Olympiad)",
"question": "Let $x$, $y$, and $z$ be positive real numbers satisfying the system of equations\n\n$$\n\\begin{aligned}\n\\sqrt{2x - xy} + \\sqrt{2y - xy} &= 1 \\\\\n\\sqrt{2y - yz} + \\sqrt{2z - yz} &= \\sqrt{2} \\\\\n\\sqrt{2z - zx} + \\sqrt{2x - zx} &= \\sqrt{3}.\n\\end{aligned}\n$$\n\nThen $[(1-x)(1-y)(1-z)]^2$ can be written as $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.",
"options": [],
"answer": "See solution",
"solution": "First, note that the square root of any real number is either real or pure imaginary, but the right side of each equation is a nonzero real number. Thus, in each equation, both square roots must be real, and each of the six expressions under the radicals is nonnegative. This implies $0 \\le x, y, z \\le 2$. Therefore, there exist $\\alpha, \\beta, \\gamma$ with $0 \\le \\alpha, \\beta, \\gamma \\le 90^\\circ$ such that $x = 2 \\sin^2 \\alpha$, $y = 2 \\sin^2 \\beta$, and $z = 2 \\sin^2 \\gamma$.\n\nSubstituting, we get:\n\n$$\n\\sqrt{2x - xy} + \\sqrt{2y - xy} = 2\\sqrt{\\sin^2 \\alpha \\cos^2 \\beta} + 2\\sqrt{\\cos^2 \\alpha \\sin^2 \\beta} = 2\\sin(\\alpha + \\beta).\n$$\n\nSimilarly, the system becomes:\n\n$$\n\\begin{aligned}\n\\sin(\\alpha + \\beta) &= \\frac{1}{2} \\\\\n\\sin(\\beta + \\gamma) &= \\frac{\\sqrt{2}}{2} \\\\\n\\sin(\\alpha + \\gamma) &= \\frac{\\sqrt{3}}{2}.\n\\end{aligned}\n$$\n\nSo $\\alpha + \\beta \\in \\{30^\\circ, 150^\\circ\\}$, $\\beta + \\gamma \\in \\{45^\\circ, 135^\\circ\\}$, and $\\alpha + \\gamma \\in \\{60^\\circ, 120^\\circ\\}$.\n\nIf $\\alpha, \\beta, \\gamma$ satisfy these, then so do $90^\\circ - \\alpha, 90^\\circ - \\beta, 90^\\circ - \\gamma$. Thus, it suffices to consider $\\alpha + \\beta = 30^\\circ$. Since $\\gamma \\le 90^\\circ$, we have $\\alpha + \\gamma = 60^\\circ$ and $\\beta + \\gamma = 45^\\circ$.\n\nTherefore, the only two systems of equations with solutions under $0 \\le \\alpha, \\beta, \\gamma \\le 90^\\circ$ are:\n\n- $\\alpha + \\beta = 30^\\circ$\n- $\\beta + \\gamma = 45^\\circ$\n- $\\alpha + \\gamma = 60^\\circ$\n\nand\n\n- $\\alpha + \\beta = 150^\\circ$\n- $\\beta + \\gamma = 135^\\circ$\n- $\\alpha + \\gamma = 120^\\circ$\n\nNow,\n\n$$\n\\begin{aligned}\n(1-x)(1-y)(1-z) &= (1 - 2 \\sin^2 \\alpha)(1 - 2 \\sin^2 \\beta)(1 - 2 \\sin^2 \\gamma) \\\\\n&= \\cos(2\\alpha) \\cos(2\\beta) \\cos(2\\gamma).\n\\end{aligned}\n$$\n\nIn the first case, $\\alpha = 22.5^\\circ$, $\\beta = 7.5^\\circ$, $\\gamma = 37.5^\\circ$. Thus,\n\n$$\n\\begin{aligned}\n(1-x)(1-y)(1-z) &= \\cos(45^\\circ) \\cos(15^\\circ) \\cos(75^\\circ) \\\\\n&= \\cos(45^\\circ) \\cos(15^\\circ) \\sin(15^\\circ) \\\\\n&= \\cos(45^\\circ) \\cdot \\frac{\\sin(30^\\circ)}{2} = \\frac{\\sqrt{2}}{8}.\n\\end{aligned}\n$$\n\nIn the second case, $\\alpha = 67.5^\\circ$, $\\beta = 82.5^\\circ$, $\\gamma = 52.5^\\circ$:\n\n$$\n\\begin{aligned}\n(1-x)(1-y)(1-z) &= \\cos(135^\\circ) \\cdot \\cos(165^\\circ) \\cdot \\cos(105^\\circ) \\\\\n&= -\\cos(45^\\circ) \\cdot \\cos(15^\\circ) \\cdot \\cos(75^\\circ) \\\\\n&= -\\frac{\\sqrt{2}}{8}.\n\\end{aligned}\n$$\n\nThus, in both cases,\n\n$$\n[(1-x)(1-y)(1-z)]^2 = \\frac{1}{32}.\n$$\n\nThe requested sum is $1 + 32 = 33$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21176,
"subject": "Mathematics (Olympiad)",
"question": "Let $x_i > 0$ and $k \\ge 1$. Prove that\n\n$$\n\\left( \\sum_{i=1}^{n} \\frac{1}{1+x_i} \\right) \\left( \\sum_{i=1}^{n} x_i \\right) \\le \\left( \\sum_{i=1}^{n} \\frac{x_i^{k+1}}{1+x_i} \\right) \\left( \\sum_{i=1}^{n} \\frac{1}{x_i^k} \\right).\n$$",
"options": [],
"answer": "See solution",
"solution": "Proof I\n\nObserve that the above inequality is equivalent to\n\n$$\n\\left( \\sum_{i=1}^{n} \\frac{x_i^{k+1}}{1+x_i} \\right) \\left( \\sum_{i=1}^{n} \\frac{1}{x_i^k} \\right) - \\left( \\sum_{i=1}^{n} \\frac{1}{1+x_i} \\right) \\left( \\sum_{i=1}^{n} x_i \\right) \\ge 0.\n$$\n\nThe left-hand side is equal to\n\n$$\n\\begin{align*}\n& \\sum_{i \\neq j} \\frac{x_i^{k+1}}{1+x_i} \\cdot \\frac{1}{x_j^k} - \\sum_{i \\neq j} \\frac{x_j}{1+x_i} \\\\\n&= \\sum_{i \\neq j} \\frac{x_i^{k+1} - x_j^{k+1}}{(1+x_i)x_j^k} \\\\\n&= \\frac{1}{2} \\sum_{i \\neq j} \\left[ \\frac{x_i^{k+1} - x_j^{k+1}}{(1+x_i)x_j^k} + \\frac{x_j^{k+1} - x_i^{k+1}}{(1+x_j)x_i^k} \\right] \\\\\n&= \\frac{1}{2} \\sum_{i \\neq j} (x_i^{k+1} - x_j^{k+1}) \\frac{(1+x_j)x_i^k - (1+x_i)x_j^k}{(1+x_j)(1+x_i)x_i^k x_j^k} \\\\\n&= \\frac{1}{2} \\sum_{i \\neq j} (x_i^{k+1} - x_j^{k+1}) \\frac{(x_i^k - x_j^k) + x_i x_j (x_i^{k-1} - x_j^{k-1})}{(1+x_j)(1+x_i)x_i^k x_j^k} \\\\\n&\\ge 0.\n\\end{align*}\n$$\n\nProof II\n\nAssume $x_1 \\ge x_2 \\ge \\dots \\ge x_n > 0$. Then,\n\n$$\n\\frac{1}{x_1^k} \\le \\frac{1}{x_2^k} \\le \\dots \\le \\frac{1}{x_n^k},\n$$\n\n$$\n\\frac{x_1^k}{1+x_1} \\ge \\frac{x_2^k}{1+x_2} \\ge \\dots \\ge \\frac{x_n^k}{1+x_n}.\n$$\n\nBy the *Chebyshev Inequality*, the left-hand side of the original inequality is\n\n$$\n\\left( \\frac{1}{1+x_1} + \\cdots + \\frac{1}{1+x_n} \\right) (x_1 + \\cdots + x_n)\n$$\n\nwhich can be rewritten as\n\n$$\n\\left( \\frac{1}{x_1^k} \\cdot \\frac{x_1^k}{1+x_1} + \\cdots + \\frac{1}{x_n^k} \\cdot \\frac{x_n^k}{1+x_n} \\right) (x_1 + \\cdots + x_n)\n$$\n\nand by Chebyshev's inequality,\n\n$$\n\\le \\left( \\frac{1}{x_1^k} + \\cdots + \\frac{1}{x_n^k} \\right) \\left( \\frac{x_1^k}{1+x_1} + \\cdots + \\frac{x_n^k}{1+x_n} \\right)\n$$\n\nand\n\n$$\n\\le \\left( x_1 \\cdot \\frac{x_1^k}{1+x_1} + \\cdots + x_n \\cdot \\frac{x_n^k}{1+x_n} \\right) \\left( \\frac{1}{x_1^k} + \\cdots + \\frac{1}{x_n^k} \\right)\n$$\n\n$$\n= \\left( \\frac{x_1^{k+1}}{1+x_1} + \\cdots + \\frac{x_n^{k+1}}{1+x_n} \\right) \\left( \\frac{1}{x_1^k} + \\cdots + \\frac{1}{x_n^k} \\right)\n$$\n\nwhich is the right-hand side of the original inequality.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21177,
"subject": "Mathematics (Olympiad)",
"question": "In the right parallelepiped $ABCD'A'B'C'D'$, with $AB = 12\\sqrt{3}$ cm and $AA' = 18$ cm, we consider the points $P \\in [AA']$ and $N \\in [A'B']$ such that $A'N = 3B'N$. Determine the length of the line segment $[AP]$ such that for any position of the point $M \\in [BC]$, the triangle $MNP$ is right-angled at $N$.",
"options": [],
"answer": "See solution",
"solution": "We have $BC \\perp (ABB')$, therefore $BC \\perp PN$. Since $PN \\perp NM$, it follows that $PN \\perp (NBC)$, hence $PN \\perp NB$, that is, triangle $NBP$ is right-angled at $N$.\n\nLet $AP = x$; we obtain:\n\n- $BP^2 = x^2 + 432$\n- $PN^2 = (18 - x)^2 + 243$\n- $BN^2 = 351$\n\nBut $BP^2 = PN^2 + BN^2$, and hence $x = 13$ cm or $x = 5$ cm.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21178,
"subject": "Mathematics (Olympiad)",
"question": "Solve the system of equations for positive integer numbers $x$, $y$, $z$:\n\n$$\n\\begin{cases}\nx^3 - 6y^2 + 27z = 132, \\\\\ny^3 - 9z^2 + 3x = 125, \\\\\nz^3 - 3x^2 + 12y = -68.\n\\end{cases}\n$$",
"options": [],
"answer": "See solution",
"solution": "Let's add up all three equations:\n\n$$\n(x^3 - 3x^2 + 3x) + (y^3 - 6y^2 + 12y) + (z^3 - 9z^2 + 27z) = 189\n$$\n\nThis simplifies to:\n\n$$\n(x^3 - 3x^2 + 3x - 1) + (y^3 - 6y^2 + 3y - 27) + (z^3 - 9z^2 + 27z - 27) = 153\n$$\n\nSo,\n\n$$\n(x-1)^3 + (y-2)^3 + (z-3)^3 = 153\n$$\n\nBy checking possible values, $153 = 5^3 + 3^3 + 1^3$. Thus, $(x-1)$, $(y-2)$, and $(z-3)$ must be $1$, $3$, and $5$ in some order. Trying all combinations, no solution is possible for positive integers $x$, $y$, $z$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21179,
"subject": "Mathematics (Olympiad)",
"question": "In an acute scalene triangle $\\triangle ABC$, points $P$ and $Q$ lie on the smaller arcs $\\widearc{AB}$ and $\\widearc{AC}$, respectively, and are the intersection of the midline parallel to $BC$ with the circumcircle of $\\triangle ABC$. Points $X$ and $Y$ are the intersection of the perpendicular bisectors of segments $AB$ and $AC$, respectively, with the tangent to $\\odot(ABC)$ at point $A$. Define point $T \\neq A$ as the intersection of $\\odot(PXA)$ and $\\odot(QYA)$. If $AD$ is an altitude in the triangle $\\triangle ABC$, prove that\n\n$$\n\\frac{|TX|}{|TY|} = \\frac{|DB|}{|DC|}.\n$$\n\nHere $\\odot(P_1P_2P_3)$ denotes the circumcircle of triangle $\\triangle P_1P_2P_3$.",
"options": [],
"answer": "See solution",
"solution": "\n\nLet $M$ and $N$ be the midpoints of $AB$ and $AC$ respectively. Moreover, let $O$ be the circumcentre of $\\triangle ABC$. We proceed in several steps and use oriented angles modulo $180^{\\circ}$.\n\n**Step 1:** Points $X, P, O, Q, Y$ lie on a circle.\n\n*Proof.* Since point $X$ lies on the perpendicular bisector of $AB$, we have that $XA, XB$ are tangents to $\\odot(ABC)$. This means that $\\angle XAO = \\angle OBX = 90^{\\circ}$, therefore quadrilateral $OAXB$ is cyclic. From Power of the Point we have\n\n$$\n|XM| \\cdot |MO| = |AM| \\cdot |MB| = |PM| \\cdot |MQ|\n$$\n\nsince quadrilateral $APBQ$ is cyclic too. This implies that points $X, P, O, Q$ lie on the same circle. Similarly, we can prove that points $Y, P, O, Q$ lie on the same circle, so points $X, P, O, Q, Y$ lie on a circle as desired. $\\square$\n\n**Step 2:** $\\angle APX = \\angle YQA$\n\n*Proof.* Observe that $\\angle XAP = \\angle AQP$. Moreover, $\\angle AYP = \\angle XQP$, since quadrilateral $XPQY$ is cyclic. We deduce that\n\n$$\n\\angle AQX = \\angle AQP - \\angle XQP = \\angle XAP - \\angle AYP = \\angle APY.\n$$\n\nRemember that $\\angle XPY = \\angle XQY$ as again quadrilateral $XPQY$ is cyclic. Consequently, we have\n\n$$\n\\angle XPA = \\angle XPY - \\angle APY = \\angle XQY - \\angle AQX = \\angle AQY\n$$\n\nas desired. $\\square$\n\n**Step 3:** Finish\n\n*Proof.* Recall that quadrilaterals $XPAT$ and $YQAT$ are cyclic. This means that $\\angle XTA = 180^{\\circ} - \\angle XPA = 180^{\\circ} - \\angle AQY = \\angle ATY$, therefore $AT$ is the $\\angle XTY$ bisector. By the angle bisector theorem we have:\n\n$$\n\\frac{|TX|}{|TY|} = \\frac{|AX|}{|AY|}.\n$$\n\nNotice that $\\triangle XAO$ and $\\triangle YAO$ are right triangles, therefore\n\n$|XA| = |AO| \\cdot \\tan \\angle ACB$ and $|AY| = |AO| \\cdot \\tan \\angle CBA$.\n\nThis means that $\\frac{|AX|}{|AY|} = \\frac{\\tan \\angle ACB}{\\tan \\angle CBA}$. Similarly, $\\triangle ABD$ and $\\triangle ACD$ are right triangles, therefore\n\n$|AD| = |BD| \\cdot \\tan \\angle CBA$ and $|AD| = |CD| \\cdot \\tan \\angle ACB$.\n\nThis means that $\\frac{|BD|}{|DC|} = \\frac{\\tan \\angle ACB}{\\tan \\angle CBA}$. We deduce that\n\n$$\n\\frac{|TX|}{|TY|} = \\frac{|AX|}{|AY|} = \\frac{\\tan \\angle ACB}{\\tan \\angle CBA} = \\frac{|BD|}{|DC|}\n$$\n\nas desired. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21180,
"subject": "Mathematics (Olympiad)",
"question": "Let $y = \\log_{20x}(22x)$. Given:\n\n$$\n(20x)^y = 22x\n$$\n\n$$\n(2x)^y = 202x.\n$$\n\nFind the value of $11 + 101$ where $y = \\log_{10} \\frac{11}{101}$, and determine the value of $x$ that satisfies the original equation.",
"options": [],
"answer": "See solution",
"solution": "We have $y = \\log_{20x}(22x)$, so $(20x)^y = 22x$.\n\nAlso, $(2x)^y = 202x$.\n\nThus,\n$$\n10^y = \\frac{(20x)^y}{(2x)^y} = \\frac{22x}{202x} = \\frac{11}{101}\n$$\nSo $y = \\log_{10} \\frac{11}{101}$.\n\nThe requested sum is $11 + 101 = 112$.\n\nThe value of $x$ that satisfies the original equation is approximately $0.047630$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21181,
"subject": "Mathematics (Olympiad)",
"question": "Given $2018$ objects arranged in a circle, in how many ways can they be paired such that each pair consists of diametrically opposite objects, and rotations of the entire configuration are not counted as distinct?",
"options": [],
"answer": "See solution",
"solution": "Let the number of pairings be $m$. Since $2018 = 2km$, we have $1009 = km$. As $1009$ is prime, $k$ can be $1$ or $1009$.\n\n1. For $k = 1$: The pairs are $(1,2), (3,4), \\ldots, (2017,2018)$, each pair being diametrically opposite. Fixing the first pair and choosing the configuration in one direction, the remaining numbers are determined. Since rotations are not counted, the next number to $1$ (clockwise) can be chosen in $1008$ ways, then $1007$ ways, etc., giving $1008!$ configurations.\n\n2. For $k = 1009$: The pairs are $(1,1010), (2,1011), \\ldots, (1009,2018)$. The discussion is analogous, yielding $1008!$ configurations.\n\nThus, the total number of configurations is $2 \\cdot 1008!$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21182,
"subject": "Mathematics (Olympiad)",
"question": "Prove that\n$$\n\\sum_{\\text{cyc}} (x+y)\\sqrt{(z+x)(z+y)} \\geq 4(xy+yz+zx)\n$$\nfor all positive real numbers $x$, $y$, $z$.",
"options": [],
"answer": "See solution",
"solution": "We will obtain the inequality by adding the inequalities\n$$\n(x+y)\\sqrt{(z+x)(z+y)} \\geq 2xy + yz + zx\n$$\nfor cyclic permutations of $x$, $y$, $z$.\n\nSquaring both sides of this inequality, we obtain\n$$\n(x+y)^2(z+x)(z+y) \\geq 4x^2y^2 + y^2z^2 + z^2x^2 + 4xyz^2 + 4x^2yz + 2xyz^2\n$$\nwhich is equivalent to\n$$\nx^3y + xy^3 + z(x^3 + y^3) \\geq 2x^2y^2 + xyz z(x+y)\n$$\nwhich can be rearranged to\n$$\n(xy + yz + zx)(x - y)^2 \\geq 0\n$$\nwhich is clearly true.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21183,
"subject": "Mathematics (Olympiad)",
"question": "Let $x$ be the smallest positive integer such that $2x$ is the square of an integer, $3x$ is the cube of an integer, and $5x$ is the fifth power of an integer. Find the prime factorization of $x$.",
"options": [],
"answer": "See solution",
"solution": "Let the prime factorization of $x$ be $2^a 3^b 5^c p_4^{e_4} \\dots p_r^{e_r}$ (with $a, b, c \\ge 0$). We require:\n\n- $2x$ is a perfect square: $a+1$ is even, and all exponents are even.\n- $3x$ is a perfect cube: $b+1$ is divisible by $3$, and all exponents are multiples of $3$.\n- $5x$ is a perfect fifth power: $c+1$ is divisible by $5$, and all exponents are multiples of $5$.\n\nTo satisfy all conditions, $a$ must be the smallest odd multiple of $15$, $b$ the smallest multiple of $10$ with $b+1$ divisible by $3$, and $c$ the smallest multiple of $6$ with $c+1$ divisible by $5$.\n\nThe minimal values are $a=15$, $b=20$, $c=24$. All other exponents must be multiples of $30$ (the least common multiple of $2,3,5$), so we set them to zero for minimal $x$.\n\nThus, the smallest $x$ is:\n\n$$x = 2^{15} \\cdot 3^{20} \\cdot 5^{24}$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21184,
"subject": "Mathematics (Olympiad)",
"question": "Let $x_1, x_2, \\dots, x_5$ be real numbers. Find the least positive integer $n$ with the following property: if there exist $n$ distinct sums of the form $x_p + x_q + x_r$ (with $1 \\le p < q < r \\le 5$) which are equal to $0$, then $x_1 = x_2 = \\dots = x_5 = 0$.",
"options": [],
"answer": "See solution",
"solution": "We prove that the smallest such $n$ is $7$.\n\nIf the numbers are $1, 1, 1, 1, -2$ (or $1, 1, 1, -2, -2$), there are $6$ sums equal to $0$ without all numbers being $0$. Therefore, knowing $6$ (or fewer) sums are $0$ is not enough to conclude all $x_i$ are $0$.\n\nNow, we show that $7$ is enough. Suppose $7$ sums are $0$. Each sum involves $3$ numbers, so $7$ sums involve $21$ terms, but there are only $5$ numbers. By the Pigeonhole Principle, some $x_i$ appears at least $5$ times. Suppose $x_1$ appears at least $5$ times. $x_1$ is in $6$ possible sums, so at most one sum (say, $x_1 + x_4 + x_5$) is not $0$. The other sums are $0$:\n\n- $x_1 + x_2 + x_3 = 0$\n- $x_1 + x_2 + x_4 = 0$\n- $x_1 + x_2 + x_5 = 0$\n- $x_1 + x_3 + x_4 = 0$\n- $x_1 + x_3 + x_5 = 0$\n\nFrom the first three, $x_3 = x_4 = x_5$. Comparing the first and last equations gives $x_2 = x_3 = x_4 = x_5$. Thus, $x_2 = x_3 = x_4 = x_5$.\n\nThe seventh sum (not involving $x_1$) gives $x_2 = x_3 = x_4 = x_5 = 0$, so $x_1 = x_2 = x_3 = x_4 = x_5 = 0$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21185,
"subject": "Mathematics (Olympiad)",
"question": "$AD$ is a bisector of triangle $ABC$. Line $AD$ intersects the circumcircle of $\\triangle ABC$ a second time at point $E$. Let $K$, $L$, $M$, and $N$ be the midpoints of segments $AB$, $BD$, $CD$, and $AC$ respectively. Let $P$ be the circumcenter of triangle $EKL$, and $Q$ be the circumcenter of triangle $EMN$. Prove that $\\angle PEQ = \\angle BAC$.",
"options": [],
"answer": "See solution",
"solution": "\n\nTriangles $AEB$ and $BED$ are similar since $\\angle BAE = \\angle EAC = \\angle DBE$. Hence $\\angle AEK = \\angle BEL$ as the angles between a median and a side in similar triangles. Denote these angles by $\\varphi$. Then $\\angle EKL = \\varphi$ since $KL$ is a midline of $\\triangle ABD$. Analogously, let $\\psi = \\angle AEN = \\angle CEM = \\angle ENM$. And let $\\beta = \\angle ABC$, $\\gamma = \\angle ACB$.\n\nThe triangle $PEL$ is isosceles, therefore $\\angle PEL = 90^\\circ - \\frac{1}{2}\\angle EPL = 90^\\circ - \\angle EKL = 90^\\circ - \\varphi$ and\n\n$$\n\\angle PEA = \\angle PEL - \\angle AEL = \\angle PEL - (\\angle AEB - \\angle BEL) = 90^\\circ - \\varphi - (\\gamma - \\varphi) = 90^\\circ - \\gamma.\n$$\n\nAnalogously $\\angle QEA = 90^\\circ - \\beta$.\n\nThus $\\angle PEQ = \\angle PEA + \\angle QEA = 180^\\circ - \\beta - \\gamma = \\angle BAC$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21186,
"subject": "Mathematics (Olympiad)",
"question": "a) Let the acute triangle $ABC$ be inscribed in the circle $(O)$. $AD$ is the altitude, and $H$ is the orthocenter of triangle $ABC$. $M$ is the midpoint of $BC$. The circle with diameter $AH$ cuts $(O)$ again at $G$. $GD$ cuts $(O)$ again at $K$. The line passing through $K$ and perpendicular to $BC$ cuts $AM$ at $L$. Prove that the four points $B$, $C$, $L$, $H$ lie on the same circle.\n\nb) Let triangle $ABC$ be inscribed in the circle $(O)$, and $H$ be the orthocenter of $ABC$. The circle with diameter $AH$ cuts $(O)$ again at $G$. A circle tangent to $AG$ at $A$ cuts $CA$ and $AB$ again at $E$ and $F$, respectively. Prove that $AO$ bisects the segment $EF$.",
"options": [],
"answer": "See solution",
"solution": "a) Since $G$ lies on the circle with diameter $AH$, $GH$ cuts $(O)$ again at $E$, and $AE$ is the diameter of $(O)$. The quadrilateral $HBEC$ is a parallelogram, and $HE$ passes through $M$. Let $N$ be the intersection of $AM$ and $(O)$. Since the quadrilaterals $AGDM$ and $AGKN$ are cyclic, we have\n\n$$\n\\angle GDM = 180^{\\circ} - \\angle AGD = \\angle GKN.\n$$\n\nThis implies that $KN \\parallel BC$, so the quadrilateral $BCNK$ is an isosceles trapezoid. Since $M$ is the midpoint of $BC$, $MK = MN$. The triangle $LKN$ is right-angled at $K$. Thus, $M$ is the midpoint of $LN$, or $L$ is the reflection of $K$ across $BC$. The reflection of $H$ across $BC$ lies on $(O)$, so $H$, $L$, $B$, $C$ all lie on the reflection circle of $(O)$ across $BC$. This proves the first part.\n\nb) Let $D$ be the intersection of $GH$ and $(O)$, so $AD$ is the diameter of $(O)$. The quadrilateral $HBDC$ is a parallelogram, so the line $HD$ passes through the midpoint $M$ of $BC$. Let $P$ be the intersection of $AD$ and the circumcircle of triangle $AEF$. The line $AP$ cuts $EF$ at $N$. It is easy to see that\n\n$$\n\\angle FPN = \\angle FAE = \\angle GAB = \\angle BDM\n$$\n\nand\n\n$$\n\\angle PFE = \\angle PAE = \\angle DBM.\n$$\n\nTherefore, triangles $DBM$ and $PFM$ are similar. Similarly, triangles $DCM$ and $PEN$ are similar. Since $M$ is the midpoint of $BC$, $N$ is the midpoint of $EF$. This proves the second part.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21187,
"subject": "Mathematics (Olympiad)",
"question": "A two-digit positive integer is said to be _cuddly_ if it is equal to the sum of its nonzero tens digit and the square of its units digit. How many two-digit positive integers are cuddly?\n\n(A) 0 (B) 1 (C) 2 (D) 3 (E) 4",
"options": [],
"answer": "See solution",
"solution": "Let the two-digit number be $10a + b$, where $a$ is the tens digit ($a \\neq 0$) and $b$ is the units digit. The number is cuddly if $10a + b = a + b^2$, so $9a = b^2 - b = b(b - 1)$. Thus, $9$ divides $b(b - 1)$. Since $b$ and $b-1$ are consecutive, $9$ must divide one of them. The only possible value for $b$ in $2 \\leq b \\leq 9$ is $b = 9$, since $9 \\mid 9$. Then $9a = 9 \\cdot 8 = 72$, so $a = 8$. Therefore, the only two-digit cuddly number is $89$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21188,
"subject": "Mathematics (Olympiad)",
"question": "一個 $k$-集合為恰有 $k$ 個元素的集合。對於一個 6-集合 $A$，以及一個由若干個 4-集合所組成的集合 $\\mathcal{F}$，我們稱 $A$ 是 $\\mathcal{F}$-好棒，若且唯若 $\\mathcal{F}$ 中僅有三個元素是 $A$ 的子集，且三個子集 $B_1, B_2, B_3$ 滿足\n\n$$\n(A \\setminus B_1) \\cup (A \\setminus B_2) \\cup (A \\setminus B_3) = A.\n$$\n\n試求所有滿足以下條件的 $n \\ge 6$：存在由 $\\{1, 2, \\dots, n\\}$ 的若干個 4-子集所構成的集合 $\\mathcal{F}$，使得對於所有 6-集合 $A \\subset \\{1, 2, \\dots, n\\}$，$A$ 都是 $\\mathcal{F}$-好棒。",
"options": [],
"answer": "See solution",
"solution": "**解. 答案為 $n = 6, 7, 8$。**\n\n注意到對於 $6 \\leq m \\leq n$，若我們對 $\\{1, 2, \\dots, n\\}$ 可構造滿足題意的 $\\mathcal{F}$，則必然可以對 $\\{1, 2, \\dots, m\\}$ 構造滿足題意的 $\\mathcal{F}$。故我們只需證明兩點：\n\n1. $n = 9$ 時不存在滿足題意的 $\\mathcal{F}$。\n\n**證明**：反證法，假設這樣的 $\\mathcal{F}$ 存在。則對於任何 $A \\subset \\{1, 2, \\dots, n\\}$，存在三個 $B \\in \\mathcal{F}$ 滿足題意，故全部的 $(A, B)$ 組合數為 $C_6^n \\times 3$。但另一方面，由於每個 $B$ 會落在 $C_2^{n-2}$ 個不同的 $A$ 中，因此全部的 $(A, B)$ 組合數為 $C_2^{n-2} \\times |\\mathcal{F}|$。\n\n這表示\n\n$$\n3 C_6^n = C_2^{n-2} \\times |\\mathcal{F}| \\implies |\\mathcal{F}| = \\frac{1}{5} C_4^n,\n$$\n\n從而 $|\\mathcal{F}|$ 不為整數，矛盾。\n\n2. $n = 8$ 時存在滿足題意的 $\\mathcal{F}$。\n\n令 $[n] = \\{1, 2, \\dots, n\\}$，且對於所有集合 $X$，令 $X_m^n = X \\cap \\{m, m+1, \\dots, n\\}$。\n\n考慮集合\n\n$$\n\\mathcal{F} = \\{[4], [4]^c\\} \\cup \\{X_1^4 \\cup (X_1^4 + 4),\\ X_1^4 \\cup ((X^c)_1^4 + 4) : X_1^4 \\subset [4],\\ |X_1^4| = 2\\}.\n$$\n\n現在注意到若 $|A| = 6$ 且 $A \\subset [8]$，則 $|A_1^4| = \\{2, 3, 4\\}$。\n\n- $|A_1^4| = 2$：不失一般性設 $A = \\{1, 2, 5, 6, 7, 8\\}$，此時 $\\mathcal{F}$ 中為 $A$ 的子集的元素恰為 $\\{5, 6, 7, 8\\}$，$\\{1, 2, 5, 6\\}$ 與 $\\{1, 2, 7, 8\\}$，易檢驗其滿足題意。\n\n- $|A_1^4| = 4$：不失一般性設 $A = \\{1, 2, 3, 4, 7, 8\\}$，此時 $\\mathcal{F}$ 中為 $A$ 的子集的元素恰為 $\\{1, 2, 3, 4\\}$，$\\{3, 4, 7, 8\\}$ 與 $\\{1, 2, 7, 8\\}$，易檢驗其滿足題意。\n\n- $|A_1^4| = 3$：此時 $|A_1^4| = |A_5^8| = 3$。考慮兩種情況：\n\n * $A_5^8 - 4 = A_1^4$：不失一般性設 $A = \\{1,2,3,5,6,7\\}$，此時 $\\mathcal{F}$ 中為 $A$ 的子集的元素恰為 $\\{1,2,5,6\\}$，$\\{1,3,5,7\\}$ 與 $\\{2,3,6,7\\}$，易檢驗其滿足題意。\n\n * $A_5^8 - 4 \\neq A_1^4$：不失一般性假設 $A_1^4 = \\{1,2,3\\}$ 而 $A_5^8 - 4 = \\{1,2,4\\}$，此時 $A = \\{1,2,3,5,6,8\\}$，$\\mathcal{F}$ 中為 $A$ 的子集的元素恰為 $\\{1,2,5,6\\}$，$\\{1,3,6,8\\}$ 與 $\\{2,3,5,8\\}$，易檢驗其滿足題意。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21189,
"subject": "Mathematics (Olympiad)",
"question": "A collection of coins has a total value at most $99 + \\frac{1}{2}$. Prove that it is possible to split this collection into 100 or fewer groups, such that each group has total value at most 1.\n\n\n\nFig. 5.1",
"options": [],
"answer": "See solution",
"solution": "We shall prove a general result: for any positive integer $N$, given a finite collection of such coins with a 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 a total value of at most 1.\n\nIf some coins have a total value of $1/k$ (where $k$ is a positive integer), we replace these coins by one coin of value $1/k$, which does not affect the problem. In this way, for each even integer $k$, at most one coin has value $1/k$ (otherwise, two such coins may be replaced by one coin of value $2/k$); for each odd integer $k$, at most $(k-1)$ coins of value $1/k$ (otherwise, $k$ such coins can be replaced by one coin of value 1). So, we may suppose that no more replacements can be made for the coins.\n\nFirst, we take each coin of value 1 as a group. Suppose there are $d < N$ such groups. If there are no other coins, then the problem is solved. Otherwise, take a coin of value $1/2$ as a group ($d+1$) if there is any. Let $m = N - d \\ge 1$. Then for each integer $k$ in $2, \\dots, m$, take coins of value $1/(2k-1)$ and value $1/(2k)$ in group ($d+k$) if there are any, in which the total value does not exceed $\\frac{2k-2}{2k-1} + \\frac{1}{2k} < 1$. For coins of value less than $1/(2m)$, if there are any, we can put them in some group $(d+j)$ such that the total value is less than 1 (since if each group $(d+j)$ has value greater than $1 - \\frac{1}{2m}$, then the total value will be greater than $d + m\\left(1 - \\frac{1}{2m}\\right) = m + d - \\frac{1}{2} = N - \\frac{1}{2}$). Repeating this procedure finitely many times, all coins are put in $N$ or fewer groups with each group of value at most one. $\\square$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21190,
"subject": "Mathematics (Olympiad)",
"question": "A positive integer $n$ is written on the board once. Then $n-1$ is written on the board twice, $n-2$ is written four times, and so on: at each step, the next smaller integer is written twice as many times as the previous number. When zeros are reached, the process stops. Prove that, in the end, the sum of the numbers on the board is less than $2^{n+1}$.",
"options": [],
"answer": "See solution",
"solution": "The sum of the numbers on the board is\n\n$$\ns_n = 1 \\cdot n + 2 \\cdot (n-1) + 4 \\cdot (n-2) + \\dots + 2^{n-1} \\cdot 1.\n$$\n\nLet us also define\n\n$$\nr_n = \\frac{s_n}{2^n} = \\frac{1}{2} \\cdot 1 + \\frac{1}{4} \\cdot 2 + \\frac{1}{8} \\cdot 3 + \\dots + \\frac{1}{2^n} \\cdot n.\n$$\n\nNotice that $r_{n+1} = \\frac{r_n}{2} + \\left(\\frac{1}{2} + \\frac{1}{4} + \\dots + \\frac{1}{2^{n+1}}\\right)$ for every $n \\ge 1$, whereby obviously\n\n$$\n\\frac{1}{2} + \\frac{1}{4} + \\dots + \\frac{1}{2^{n+1}} < \\frac{1}{2} + \\frac{1}{4} + \\dots = 1.\n$$\n\nThus $r_n < 2$ always implies $r_{n+1} < \\frac{r_n}{2} + 1 < 1 + 1 = 2$. As $r_1 = \\frac{1}{2} < 2$, we have $r_n < 2$ for every $n \\ge 1$. From there we get that $s_n = 2^n \\cdot r_n < 2^{n+1}$ for every $n \\ge 1$.\n\n*Remark*: The problem can also be solved by showing $s_n = 2^{n+1} - (n+2)$ by induction.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21191,
"subject": "Mathematics (Olympiad)",
"question": "Let $O$ be the circumcenter of an acute triangle $ABC$. Line $AC$ intersects the circumcircle of $\\triangle ABO$ at $X$ for the second time. Show that $XO \\perp BC$.",
"options": [],
"answer": "See solution",
"solution": "The inscribed angles $\\angle CXO = \\angle ABO$ do not depend on the position of point $X$ (whether on segment $AC$ or its extension past $A$). Since $O$ is the circumcenter of $\\triangle ABC$, we have $\\angle BAO = \\angle ABO$. Let $Y = XO \\cap BC$, and let $AZ$ be a diameter of the circumcircle of $\\triangle ABC$. Then $\\angle CXY = \\angle ZAB$ and $\\angle YCX = \\angle BZA$, as these are inscribed angles subtending the same arcs. Therefore, $\\angle CYX = \\angle ZBA = 90^\\circ$, since its intercepted arc is a diameter. Thus, $XO \\perp BC$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21192,
"subject": "Mathematics (Olympiad)",
"question": "Determine all positive integers which are equal to $13$ times the sum of their digits.",
"options": [],
"answer": "See solution",
"solution": "Let $\\kappa$ be the number of digits of the integer $A$ which is equal to $13$ times the sum of its digits. The least possible $A$ is $10^{\\kappa-1}$, while the maximal possible sum of the digits is $9\\kappa$. Therefore, we need to have:\n\n$$\n10^{\\kappa-1} \\leq 13 \\cdot 9\\kappa = 117\\kappa.\n$$\n\nFor $\\kappa \\geq 4$, we will prove by induction that $10^{\\kappa-1} > 117\\kappa$, so the above relation is not valid. For $\\kappa = 4$, $10^{4-1} = 1000 > 468 = 117 \\cdot 4$. If $10^{\\kappa-1} > 117\\kappa$ for some $\\kappa > 4$, then:\n\n$$\n10^{\\kappa} = 10 \\cdot 10^{\\kappa-1} > 10 \\cdot 117\\kappa = 1170\\kappa > 117(\\kappa+1).\n$$\n\nTherefore, $\\kappa \\leq 3$.\n\n- For $\\kappa = 1$, $A = \\alpha < 13\\alpha$ for $0 < \\alpha \\leq 9$, so no solution.\n- For $\\kappa = 2$, $A = 10\\alpha + \\beta < 13(\\alpha + \\beta)$, so no solution.\n- For $\\kappa = 3$, let $A = 100\\alpha + 10\\beta + \\gamma$, $0 < \\alpha \\leq 9$, $0 \\leq \\beta, \\gamma \\leq 9$.\n\nWe have:\n\n$$\n100\\alpha + 10\\beta + \\gamma = 13(\\alpha + \\beta + \\gamma)\n$$\n$$\n\\Rightarrow 87\\alpha = 3\\beta + 12\\gamma\n$$\n$$\n\\Rightarrow 29\\alpha = \\beta + 4\\gamma\n$$\n\nSince $0 \\leq \\beta + 4\\gamma \\leq 45$, $29\\alpha \\leq 45 \\Rightarrow \\alpha \\leq 1$. Thus:\n\n$$\n\\beta + 4\\gamma = 29\n$$\n$$\n\\beta = 29 - 4\\gamma \\geq 0 \\Rightarrow 0 \\leq \\beta \\leq 9\n$$\n$$\n5 \\leq \\gamma \\leq 7\n$$\n\n- For $\\gamma = 5$, $\\beta = 9$, $A = 195$.\n- For $\\gamma = 6$, $\\beta = 5$, $A = 156$.\n- For $\\gamma = 7$, $\\beta = 1$, $A = 117$.\n\nThus, the solutions are $117$, $156$, and $195$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21193,
"subject": "Mathematics (Olympiad)",
"question": "Positive real numbers $a, b, c, d$ satisfy the relations\n\n$$\nabcd = 4, \\quad a^2 + b^2 + c^2 + d^2 = 10.\n$$\n\nDetermine the largest possible value of the expression $ab + bc + cd + da$.",
"options": [],
"answer": "See solution",
"solution": "Let $V = ab + bc + cd + da$. We will find the maximum value of\n\n$$\nV^2 = (a + c)^2 (b + d)^2 = (a^2 + c^2 + 2ac)(b^2 + d^2 + 2bd). \\quad (1)\n$$\n\nAll the given expressions do not change under the simultaneous replacement of $a$ by $b$, $b$ by $c$, $c$ by $d$, and $d$ by $a$. Since $ac \\cdot bd = 4$, at least one of the numbers $ac$ and $bd$ is at least $2$. We may assume $bd \\geq 2$.\n\nSimple manipulations yield $ac = 4 / bd$ and $a^2 + c^2 = 10 - b^2 - d^2$. We plug these expressions into (1) and get\n\n$$\n\\begin{aligned}\nV^2 &= \\left(10 - b^2 - d^2 + \\frac{8}{bd}\\right) (b^2 + d^2 + 2bd) \\\\\n&= 10(b^2 + d^2) + 20bd + \\frac{8(b^2 + d^2)}{bd} + 16 - (b^2 + d^2)^2 - 2bd(b^2 + d^2).\n\\end{aligned} \\quad (2)\n$$\n\nLet $P = b^2 + d^2$ and $Q = bd$; then $P \\geq 2Q$ and $Q \\geq 2$. Thus (2) becomes\n\n$$\n\\begin{aligned}\nV^2 &= 10P + 20Q + \\frac{8P}{Q} + 16 - P^2 - 2PQ \\\\\n&= -(P^2 - 10P + 25) + \\left(41 - 2PQ + 20Q + \\frac{8P}{Q}\\right) \\\\\n&= -(P - 5)^2 + \\left[P\\left(\\frac{8}{Q} - 2Q\\right) + 41 + 20Q\\right].\n\\end{aligned}\n$$\n\nObviously, $-(P-5)^2 \\leq 0$. The condition $Q \\geq 2$ implies $\\frac{8}{Q} - 2Q \\leq 4 - 4 = 0$, which means the expression in the brackets is linear in $P$ with non-positive slope and it achieves its maximum for the smallest possible $P$. By $P \\geq 2Q$ we obtain\n\n$$\nV^2 \\leq 2Q \\left( \\frac{8}{Q} - 2Q \\right) + 41 + 20Q = -4Q^2 + 20Q + 57 = -(2Q - 5)^2 + 82 \\leq 82.\n$$\n\nFinally, we shall show that there are positive real numbers $a, b, c, d$ such that $V = \\sqrt{82}$. The equality occurs if $P = 2Q = 5$, which is true for $b = d = \\frac{1}{2}\\sqrt{10}$. The numbers $a$ and $c$ satisfy $a^2 + c^2 = 5$, $ac = 8/5$. Thus\n\n$$\n\\{a, c\\} = \\left\\{ \\frac{\\sqrt{41} - 3}{2\\sqrt{5}}, \\frac{\\sqrt{41} + 3}{2\\sqrt{5}} \\right\\}.\n$$\n\n**Answer:** The maximum possible value of $ab + bc + cd + da$ is $\\sqrt{82}$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21194,
"subject": "Mathematics (Olympiad)",
"question": "Colour the edges of the complete graph $K_4$ (the complete graph on 4 vertices) using two colours. How many essentially different colourings are there such that no triangle is monochromatic?\n\n",
"options": [],
"answer": "See solution",
"solution": "There is essentially one colouring which is not good, as shown in the image above. The dashed line represents one colour and the solid line the other. Interchanging the top vertices or the colours gives other representations of essentially the same solution.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21195,
"subject": "Mathematics (Olympiad)",
"question": "Eliza has a large collection of $a \\times a$ and $b \\times b$ tiles, where $a$ and $b$ are positive integers. She arranges some of these tiles, without overlaps, to form a square of side length $n$. Prove that she can cover another square of side length $n$ using only one of her two types of tile.",
"options": [],
"answer": "See solution",
"solution": "Number the rows of the $n \\times n$ square from $1$ to $n$.\n\nWe say that a cell is *special* if it is in the top row of a $b \\times b$ square.\n\nIn any row, the number of cells that are in a $b \\times b$ square is congruent to $n$ modulo $a$. Since each special cell has $b-1$ cells in the $b-1$ rows underneath it, we can see by induction that the number of special cells in rows $1$, $b+1$, $2b+1$, \\ldots{} is congruent to $n$ modulo $a$, and the number of special cells in rows with other numbers is congruent to $0$.\n\nNow, the cells in $b \\times b$ squares in row $n$ are all in squares whose top cells are in row $n-b+1$, and hence the number of special cells in row $n-b+1$ is congruent to $n$ (modulo $a$). Hence either $n \\equiv 0 \\pmod{a}$, in which case $a$ is a multiple of $n$, or $n-b+1$ is of the form $rb+1$, in which case $b$ is a multiple of $n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21196,
"subject": "Mathematics (Olympiad)",
"question": "For each positive integer $r$, let\n\n$$\nS_r = \\{n \\in \\mathbb{N}^+ : P(n)^r \\mid Q(n)^{r+1}\\}.\n$$\n\nLet $P(x)$ and $Q(x)$ be integer polynomials with $P$ monic. Prove that if $S_r$ is infinite for each positive integer $r$, then $P(x)$ divides $Q(x)$ as polynomials.",
"options": [],
"answer": "See solution",
"solution": "Suppose $Q(x) = A(x)P(x) + R(x)$, where $\\deg(R) < \\deg(P)$. By induction, for all $n \\in S_r$, $P(n)^r \\mid R(n)^{r+1}$. If $R$ is not the zero polynomial, as $n \\to \\infty$, $P(n)^r$ grows faster than $R(n)^{r+1}$ for large $r$, contradicting the infinitude of $S_r$. Thus, $R(x) = 0$ and $P(x) \\mid Q(x)$. Since $P(x)$ is monic and both $P(x), Q(x)$ have integer coefficients, $A(x)$ is also an integer polynomial.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21197,
"subject": "Mathematics (Olympiad)",
"question": "Let $f(x) = \\lfloor \\frac{x}{1!} \\rfloor + \\lfloor \\frac{x}{2!} \\rfloor + \\dots + \\lfloor \\frac{x}{2013!} \\rfloor$, where $\\lfloor x \\rfloor$ is the greatest integer no greater than $x$. Call an integer $n$ a good number if the equation $f(x) = n$ has a real solution $x$. Find the number of good numbers in the set $\\{1, 3, 5, \\dots, 2013\\}$.",
"options": [],
"answer": "See solution",
"solution": "First, note two facts:\n\n(a) If $m$ is a positive integer and $x$ is real, then\n\n$$\n\\lfloor \\frac{x}{m} \\rfloor = \\lfloor \\frac{\\lfloor x \\rfloor}{m} \\rfloor.\n$$\n\n(b) For any integer $l$ and positive even number $m$, we have\n\n$$\n\\lfloor \\frac{2l + 1}{m} \\rfloor = \\lfloor \\frac{2l}{m} \\rfloor.\n$$\n\nLet $m = k!$ ($k = 1, 2, \\dots, 2013$) in (a) and sum up:\n\n$$\nf(x) = \\sum_{k=1}^{2013} \\lfloor \\frac{x}{k!} \\rfloor = \\sum_{k=1}^{2013} \\lfloor \\frac{\\lfloor x \\rfloor}{k!} \\rfloor = f(\\lfloor x \\rfloor),\n$$\n\nso $f(x) = n$ has a real solution if and only if $f(x) = n$ has an integer solution. Thus, we only consider integer $x$.\n\nSince\n\n$$\nf(x+1) - f(x) = 1 + \\sum_{k=2}^{2013} \\left( \\lfloor \\frac{x+1}{k!} \\rfloor - \\lfloor \\frac{x}{k!} \\rfloor \\right) \\ge 1,\n$$\n\n$f(x)$ is monotonically increasing for integer $x$.\n\nNow, find integers $a$ and $b$ such that\n\n$$\nf(a-1) < 0 \\le f(a) < f(a+1) < \\dots < f(b-1) < f(b) \\le 2013 < f(b+1).\n$$\n\nNote $f(-1) < 0 = f(0)$, so $a = 0$.\n\nCalculate:\n\n$$\n\\begin{align*}\nf(1173) &= \\sum_{k=1}^{6} \\left\\lfloor \\frac{1173}{k!} \\right\\rfloor = 1173 + 586 + 195 + 48 + 9 + 1 \\\\\n&= 2012 \\le 2013, \\\\\nf(1174) &= 1174 + 587 + 195 + 48 + 9 + 1 = 2014 > 2013.\n\\end{align*}\n$$\n\nSo $b = 1173$.\n\nThus, the good numbers in $\\{1, 3, 5, \\dots, 2013\\}$ are the odd numbers in $\\{f(0), f(1), \\dots, f(1173)\\}$.\n\nLet $x = 2l$ ($l = 0, 1, \\dots, 586$). By (b),\n\n$$\n\\left\\lfloor \\frac{2l + 1}{k!} \\right\\rfloor = \\left\\lfloor \\frac{2l}{k!} \\right\\rfloor \\quad (2 \\le k \\le 2013).\n$$\n\nThus,\n\n$$\nf(2l + 1) - f(2l) = 1 + \\sum_{k=2}^{2013} \\left( \\left\\lfloor \\frac{2l+1}{k!} \\right\\rfloor - \\left\\lfloor \\frac{2l}{k!} \\right\\rfloor \\right) = 1,\n$$\n\nso there is exactly one odd number in $f(2l)$ and $f(2l + 1)$.\n\nTherefore, there are $\\frac{1174}{2} = 587$ odd numbers in $\\{f(0), f(1), \\dots, f(1173)\\}$, i.e., there are $587$ good numbers in the set $\\{1, 3, 5, \\dots, 2013\\}$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21198,
"subject": "Mathematics (Olympiad)",
"question": "We identify the $\\frac{1}{2}(3^{100} + 1)$ equally spaced points with the integers from $0$ up to $\\frac{1}{2}(3^{100} - 1)$ on the real number line.\n\nColour red each integer of the form\n\n$$\nN = \\sum_{i=0}^{99} d_i \\cdot 3^i\n$$\n\nwhere $d_i \\in \\{0, 1\\}$ for $0 \\le i \\le 99$. In this way, exactly $2^{100}$ integers are coloured red. These are precisely the integers in the range from $0$ to $\\frac{1}{2}(3^{100} - 1)$ which do not contain the digit $2$ in their ternary (base-3) representations.\n\nProve that no red integer $B$ is equidistant from two other red integers $A$ and $C$ with $A < B < C$; that is, there do not exist red integers $A < B < C$ such that $B$ is the midpoint of $A$ and $C$.",
"options": [],
"answer": "See solution",
"solution": "Suppose, for the sake of contradiction, that one red integer $B$ is equidistant from two other red integers $A$ and $C$, where $A < B < C$. Thus $C - B = B - A$, which is the same as $2B = A + C$.\n\nLet\n\n$$\nA = \\sum_{i=0}^{99} a_i \\cdot 3^i, \\quad B = \\sum_{i=0}^{99} b_i \\cdot 3^i, \\quad \\text{and} \\quad C = \\sum_{i=0}^{99} c_i \\cdot 3^i\n$$\n\nbe the ternary representations of $A$, $B$, and $C$, where $a_i, b_i, c_i \\in \\{0, 1\\}$ for $0 \\le i \\le 99$. From $2B = A + C$ we have\n\n$$\n\\sum_{i=0}^{99} 2b_i \\cdot 3^i = \\sum_{i=0}^{99} (a_i + c_i)3^i.\n$$\n\nSince $a_i, b_i, c_i \\in \\{0, 1\\}$, we have $2b_i \\in \\{0, 2\\}$ and $a_i + c_i \\in \\{0, 1, 2\\}$. Therefore, both sides of the above equation are the ternary representation of the same number. It follows that $2b_i = a_i + c_i$ for each $i$.\n\nIf $b_i = 0$, then $a_i = c_i = 0$. And if $b_i = 1$, then $a_i = c_i = 1$. Either way, $A$ and $C$ have the same ternary digits, and so $A = C$. This contradicts $A < C$, and completes the proof. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21199,
"subject": "Mathematics (Olympiad)",
"question": "Consider a $2017 \\times 2017$ chessboard colored so that cell $(i, j)$ is black if and only if $i + j$ is even.\n\n\n\nLet $k$ be the minimum number of positive integers needed to fill the white cells of the board so that:\n- Each white cell contains a positive integer.\n- For any two white cells $(a, b)$ and $(c, d)$, the numbers in these cells are different unless $(a, b)$ and $(c, d)$ are symmetric with respect to the main diagonal (i.e., $(a, b)$ and $(b, a)$).\n\nFind the minimum value of $k$.",
"options": [],
"answer": "See solution",
"solution": "We analyze the coloring and filling of the chessboard:\n\nTake two white cells $(a, b)$ and $(c, d)$, with $1 \\leq a, b, c, d \\leq 2017$.\n\n- If $a + c$ is even, then $b + d$ is also even, so $a + d$ and $b + c$ are both odd. Thus, one of the cells $(a, d)$ or $(b, c)$ is black, so $(a, b)$ and $(c, d)$ cannot be symmetric unless $a = d$ and $b = c$.\n- If $a + c$ is odd, then $b + d$ is also odd. Considering $(d, c)$, which is filled by the same number as $(c, d)$, we can apply the same argument.\n\nHence, all positive numbers on the right upper part of the table are pairwise distinct. This implies:\n\n$$\nk \\geq 2 + 4 + 6 + \\cdots + 2016 = 1008 \\cdot 1009 = \\frac{2017^2 - 1}{4}.\n$$\n\nNow, consider the graph with $2017$ vertices $A_1, \\ldots, A_{2017}$. If row $i$ and column $j$ intersect at a white cell filled by $a > 0$, connect $A_i$ and $A_j$ with an edge weighted $a$. By symmetry, pairs of non-connected vertices correspond to edges with different weights.\n\nWe show that for any $n$-vertex graph, at most $\\frac{n^2}{4}$ distinct weights can be assigned to the edges. We prove this by induction:\n- For $n = 1, 2, 3$, the claim holds.\n- For $n \\geq 4$, suppose it holds for $n-3$. If the graph has at most $\\frac{n^2}{4}$ edges, assign unique weights. If more, by Mantel-Turán's theorem, there is a triangle. Assign the same weight to its three edges, and at most $n-3$ distinct weights to edges connecting one of these vertices to the remaining $n-3$ vertices. By induction, assign at most $\\frac{(n-3)^2}{4}$ distinct weights among the remaining vertices.\n\nThus,\n$$\n1 + (n-3) + \\frac{(n-3)^2}{4} \\leq \\frac{n^2}{4}.\n$$\n\nFor $n = 2017$, $\\left\\lfloor \\frac{2017^2}{4} \\right\\rfloor = \\frac{2017^2 - 1}{4}$, which is the minimum value of $k$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21200,
"subject": "Mathematics (Olympiad)",
"question": "Call a positive integer $n$ *supereven* if its largest odd factor $d$ is less than $\\frac{n}{2}$. How many positive integers less than $1000$ are supereven?",
"options": [],
"answer": "See solution",
"solution": "Let $n$ be supereven and $d$ its greatest odd divisor. If $\\frac{n}{d}$ were divisible by some odd $p > 1$, then $pd > d$ would also be a factor of $n$, contradiction. Thus $\\frac{n}{d}$ is a power of $2$. Hence, supereven numbers are exactly those that can be expressed as a product of a power of two and an odd number less than that power of two.\n\nWe will find the supereven numbers less than $1000$ by their largest odd factors $d$:\n\n- If $d = 1$, then the power of $2$ can be one of $2, 4, 8, 16, 32, 64, 128, 256, 512$. We obtain $9$ supereven numbers.\n- If $d = 3$, then the power of $2$ can be one of $4, 8, 16, 32, 64, 128, 256$. We obtain $7$ supereven numbers.\n- If $d = 5$ or $d = 7$, then the power of $2$ can be one of $8, 16, 32, 64, 128$. We obtain $2 \\times 5 = 10$ supereven numbers.\n- If $d$ is one of $9, 11, 13, 15$, then the power of $2$ can be one of $16, 32, 64$. We obtain $4 \\times 3 = 12$ supereven numbers.\n- If $d$ is one of $17, 19, 21, 23, 25, 27, 29, 31$, then the power of $2$ can be only $32$. We obtain $8$ supereven numbers.\n\nThus, there are $9 + 7 + 10 + 12 + 8 = 46$ supereven numbers less than $1000$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21201,
"subject": "Mathematics (Olympiad)",
"question": "Three different pairs of shoes are placed in a row so that no left shoe is next to a right shoe from a different pair. In how many ways can these six shoes be lined up?\n\n(A) 60 (B) 72 (C) 90 (D) 108 (E) 120",
"options": [],
"answer": "See solution",
"solution": "There are $\\binom{6}{3} = 20$ arrangements of the letters LLLRRR representing the positions of 3 left shoes and 3 right shoes in the row of 6 shoes. Of the 20, any sequence containing LRL or RLR will violate the condition given in the problem. There are 8 arrangements that avoid these two sequences. Call a pair of shoes *matched* if the 2 shoes in the pair are next to each other. There are three sets of possibilities.\n\n* LLLRRR and RRRLLL: In these two cases, only one pair of shoes is matched. There are 3 choices for that pair, and there are $2 \\cdot 2 = 4$ ways to place the other 4 shoes for a total of $6 \\cdot 4 = 24$ arrangements for this case.\n* LRRRLL, LLRRRL, RLLLR, and RRLLLR: In each of these four cases, there are $3! = 6$ ways to place the left shoes, but then there is a unique way to place the right shoes, for a total of $6 \\cdot 4 = 24$ arrangements in this case.\n* LRRLLR and RLLRRL: In these two cases, all three pairs of shoes are matched, so, in each case, there are $3! = 6$ ways to place the shoes. This gives a total of $6 \\cdot 2 = 12$ arrangements.\n\nThus there are $24 + 24 + 12 = 60$ arrangements satisfying the conditions of the problem.\n\n**OR**\n\nLabel the shoes $L_i$ and $R_i$ for $i = 1, 2, 3$. By symmetry it suffices to count the number of arrangements with $L_1$ first in line and multiply by 6. There are 2 choices for a left shoe coming next, say...",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21202,
"subject": "Mathematics (Olympiad)",
"question": "For any non-negative integer $i$, let $d_i$ be the first digit of the number $2^i$. Let $n$ be a positive integer. Prove that there exists a non-zero digit that occurs in the tuple $(d_0, d_1, \\dots, d_{n-1})$ less than $\\frac{n}{17}$ times.",
"options": [],
"answer": "See solution",
"solution": "The claim obviously holds for $n = 1$, so assume $n \\ge 2$. Let $k$ be the minimal number of occurrences of a non-zero digit in the tuple $(d_0, d_1, \\dots, d_{n-1})$. Then the total number of occurrences of digits 5, 6, 7, 8, 9 is at least $5k$. As each digit 5, 6, 7, 8, 9 that is not the last digit of the tuple is followed by a digit 1, and also $d_0 = 1$, the number of occurrences of digit 1 is at least $5k$. Each digit 1 that is not the last digit of the tuple is followed by either 2 or 3. Thus, if the last digit is not 1, the total number of occurrences of 2 and 3 is at least $5k$. If the last digit of the tuple is 1, then the first 1 of the tuple was previously not counted, so the total number of occurrences of digits 2 and 3 is at least $5k$ in this case too. Therefore, the number of occurrences of the only digit not counted yet, the digit 4, is at most $n - 15k$. As each occurrence of 8 or 9 follows a digit 4, the total number of occurrences of 8 and 9 is also at most $n - 15k$. Hence $n - 15k \\ge 2k$, implying $k \\le \\frac{n}{17}$.\n\nTo prove that $k < \\frac{n}{17}$, suppose $k = \\frac{n}{17}$, i.e., $n = 17k$. This implies that, in the argument above, every inequality must hold as an equality, i.e., each of the digits 5, 6, 7, 8, 9 occurs exactly $k$ times and the digit 1 occurs exactly $5k$ times. As $d_3 = d_{13} = d_{23} = 8$, the digit 8 occurs more than once in the tuple $(d_0, d_1, \\dots, d_{17-1})$ and more than twice in the tuple $(d_0, d_1, \\dots, d_{2 \\cdot 17-1})$. Hence $k \\ge 3$.\n\nWe show that each segment consisting of exactly 17 consecutive terms of the tuple contains at least 5 occurrences of the digit 1. Indeed, the last term of the tuple $(2^i, 2^{i+1}, \\dots, 2^{i+16})$ is exactly 65536 times the first term, so the last term contains at least 4 more digits than the first term. As the first power of 2 containing a certain number of digits definitely starts with 1, the tuple $(2^{i+1}, \\dots, 2^{i+16})$ contains at least 4 terms starting with 1. If also $2^i$ starts with 1, then there are at least 5 such terms altogether; but if $2^i$ starts with a larger digit, then the last term contains at least 5 more digits than $2^i$, implying that there are still 5 terms that start with 1.\n\nIt remains to notice that the tuple $(d_0, d_1, \\dots, d_{3 \\cdot 17 - 1})$ contains the digit 1 at least 16 times because $d_0 = 1$ and $2^{3 \\cdot 17 - 1} = 2^{50} = (2^{1024})^5 > (10^3)^5 = 10^{15}$, implying that the number $2^{3 \\cdot 17 - 1}$ has at least 16 digits. As every segment $(d_{17i}, d_{17i+1}, \\dots, d_{17(i+1)-1})$ contains the digit 1 at least 5 times, the digit 1 occurs in the tuple $(d_0, d_1, \\dots, d_{17k-1})$ more than $5k$ times. The contradiction shows that $k < \\frac{n}{17}$. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21203,
"subject": "Mathematics (Olympiad)",
"question": "設 $\\triangle ABC$ 為等腰三角形，其中 $AB = AC$，並令 $M$ 為 $BC$ 邊的中點。令 $P$ 為平面上一點異於 $A$，滿足 $PB < PC$，且 $PA$ 與 $BC$ 平行。設點 $X$ 在直線 $PB$ 上，點 $Y$ 在直線 $PC$ 上，使得 $B$ 落在線段 $PX$ 上，$C$ 落在線段 $PY$ 上，並且 $\\angle PXM = \\angle PYM$。證明：$A, P, X, Y$ 四點共圓。",
"options": [],
"answer": "See solution",
"solution": "因為 $AB = AC$，知 $AM$ 為 $BC$ 邊的中垂線，故\n\n$$\n\\angle PAM = \\angle AMC = 90^{\\circ}\n$$\n\n\n\n現過點 $Y$ 作與 $PC$ 垂直的直線，設其與直線 $AM$ 交於點 $Z$。（注意：點 $M$ 介於 $A, Z$ 兩點之間。）可知\n\n$$\n\\angle PAZ = \\angle PYZ = 90^{\\circ}.\n$$\n\n故 $P, A, Y, Z$ 四點共圓。\n\n因為\n\n$$\n\\angle CMZ = \\angle CYZ = 90^{\\circ},\n$$\n\n得 $C, Y, Z, M$ 四點共圓，故有 $\\angle CZM = \\angle CYM$。\n\n由題設知 $\\angle CYM = \\angle BXM$，且因為 $B, C$ 兩點對稱於 $ZM$，故\n\n$$\n\\angle CZM = \\angle BZM.\n$$\n\n綜上所述，得 $\\angle BXM = \\angle BZM$，所以 $B, X, Z, M$ 四點共圓。於是得\n\n$$\n\\angle BXZ = 180^{\\circ} - \\angle BMZ = 90^{\\circ}.\n$$\n\n由上可得\n\n$$\n\\angle PXZ = \\angle PYZ = \\angle PAZ = 90^{\\circ},\n$$\n\n故 $P, A, X, Y, Z$ 五點共圓。當然可推知 $A, P, X, Y$ 四點共圓，得證。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21204,
"subject": "Mathematics (Olympiad)",
"question": "A set of 12 tokens—3 red, 2 white, 1 blue, and 6 black—is to be distributed at random to 3 game players, 4 tokens per player. The probability that some player gets all the red tokens, another player gets all the white tokens, and the remaining player gets the blue token can be written as $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m + n$?\n\n(A) 387 (B) 388 (C) 389 (D) 390 (E) 391",
"options": [],
"answer": "See solution",
"solution": "The situation can be modeled by arranging the 12 tokens in a row, with the first player receiving the first 4 tokens, the second player receiving the next 4 tokens, and the third player receiving the last 4 tokens. There are $\\frac{12!}{3! \\cdot 2! \\cdot 1! \\cdot 6!}$ such arrangements.\n\nThe given conditions are satisfied if and only if the red tokens appear among the first 4 positions, the white tokens appear among the next 4 positions, and the blue token appears among the last 4 positions, or some permutation of the groups of 4. There are $\\binom{4}{3} \\cdot \\binom{4}{2} \\cdot \\binom{4}{1} \\cdot 3!$ ways for this to happen.\n\nThe required probability is therefore\n\n$$\n\\frac{4 \\cdot 6 \\cdot 4 \\cdot 6 \\cdot 3! \\cdot 2! \\cdot 1! \\cdot 6!}{12!} = \\frac{4}{385},\n$$\n\nand the requested sum is $4 + 385 = 389$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21205,
"subject": "Mathematics (Olympiad)",
"question": "Find every integer $k$ that satisfies the following condition:\n\nThere are infinitely many integer triplets $(a, b, c)$ such that\n$$\n(a^2 - k)(b^2 - k) = c^2 - k.\n$$",
"options": [],
"answer": "See solution",
"solution": "Consider an arbitrary integer $k$. Take a complex number $\\alpha$ such that $\\alpha^2 = k$. (For example, $\\alpha = \\sqrt{k}$ if $k \\geq 0$, or $\\alpha = i\\sqrt{-k}$ if $k < 0$.)\n\nIt is easy to verify the following equalities:\n\n$$\n(n + \\alpha)(n + 1 - \\alpha) = (n(n + 1) - k) + \\alpha, \\\\\n(n - \\alpha)(n + 1 + \\alpha) = (n(n + 1) - k) - \\alpha.\n$$\n\nMultiplying these two equations gives:\n\n$$\n(n^2 - k)((n + 1)^2 - k) = (n(n + 1) - k)^2 - k.\n$$\n\nTherefore, for any $k$, we can choose any integer $n$ and set $(a, b, c) = (n, n + 1, n(n + 1) - k)$, so the required equality holds. Since there are infinitely many $n$, there are infinitely many such triplets $(a, b, c)$. Thus, every integer $k$ satisfies the condition.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21206,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $n$ such that $$(n^2 + 11n - 4) \\cdot n! + 33 \\cdot 13^n + 4$$ is a perfect square.",
"options": [],
"answer": "See solution",
"solution": "Let $a_n = (n^2 + 11n - 4) \\cdot n! + 33 \\cdot 13^n + 4$.\n\nIf $n \\geq 4$, then $8$ divides $n!$, so:\n$$a_n \\equiv 33 \\cdot 13^n + 4 \\equiv 5^n + 4 \\pmod{8}$$\nSince $5^2 \\equiv 1 \\pmod{8}$, for even $n$, $5^n \\equiv 1 \\pmod{8}$, so $a_n \\equiv 5 \\pmod{8}$ for even $n \\geq 4$. But perfect squares modulo $8$ are $0, 1, 4$.\n\nFor $n \\geq 7$, $7$ divides $n!$, so:\n$$a_n \\equiv 33 \\cdot 13^n + 4 \\equiv 5 \\cdot (-1)^n + 4 \\pmod{7}$$\nFor odd $n \\geq 7$, $a_n \\equiv -5 + 4 = -1 \\pmod{7}$, but perfect squares modulo $7$ are $0, 1, 2, 4$.\n\nThus, possible candidates are $n = 1, 2, 3, 5$.\n\nFor $n = 5$:\n$$a_5 \\equiv 33 \\cdot 13^5 + 4 \\equiv 3^6 - 1 \\equiv 3 \\pmod{5}$$\nPerfect squares modulo $5$ are $0, 1, 4$; so $a_5$ is not a perfect square.\n\nFor $n = 3$:\n$$a_3 = (9 + 33 - 4) \\cdot 6 + 33 \\cdot 13^3 + 4$$\n$13^3 + 4 \\equiv 3 + 3^4 - 1 \\equiv 3 \\pmod{5}$, so not a perfect square.\n\nCheck $n = 1$:\n$$(1 + 11 - 4) \\cdot 1 + 33 \\cdot 13 + 4 = 441 = 21^2$$\n\nCheck $n = 2$:\n$$(4 + 22 - 4) \\cdot 2 + 33 \\cdot 169 + 4 = 5625 = 75^2$$\n\n**Answer:** The only positive integers $n$ for which $(n^2 + 11n - 4) \\cdot n! + 33 \\cdot 13^n + 4$ is a perfect square are $n = 1$ and $n = 2$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21207,
"subject": "Mathematics (Olympiad)",
"question": "Determine all positive integers $n$, $n \\ge 2$, such that the following statement is true:\n\nIf $(a_1, a_2, \\dots, a_n)$ is a sequence of positive integers with $a_1 + a_2 + \\dots + a_n = 2n - 1$, then there is a block of (at least two) consecutive terms in the sequence with their (arithmetic) mean being an integer.",
"options": [],
"answer": "See solution",
"solution": "The statement is true for all $n \\ge 4$ but not for $n = 2$ or $n = 3$. In those two cases, the sequences $(1, 2)$ and $(2, 1, 2)$ provide counterexamples.\n\nNow, let $(a_1, \\dots, a_n)$ be any sequence of positive integers, and let $s_k = a_1 + \\dots + a_k - 2k$ for $k = 1, 2, \\dots, n$, and define $s_0 = 0$. Let us say that a sequence is *good* if it satisfies the property in the problem (no block of length at least two has an integer arithmetic mean). Define $(i, j)$ to be a *divisible pair* if $j - i \\mid s_j - s_i$. It is clear that $(a_1, \\dots, a_n)$ is *good* if and only if there is no divisible pair $(i, j)$ such that $|j - i| \\ge 2$.\n\nWe will show that $(a_1, \\dots, a_n)$ is not good if $n \\ge 4$. Note that $s_n = a_1 + \\dots + a_n - 2n = -1$, and for each $k$, $s_{k+1} - s_k = a_{k+1} - 2 \\ge -1$. We consider several possible values of $s_2$.\n\n- Suppose $s_2 \\le -2$. Since $s_1 \\ge s_0 - 1 = -1$ and $s_2 \\ge s_1 - 1$, it follows that $s_1 = -1$. Then $n-1 \\mid s_n - s_1$.\n- Suppose $s_2 = -1$. Then $n-2 \\mid s_n - s_2$.\n- Suppose $s_2 = 0$. Then $2-0 \\mid s_2 - s_0$.\n- Suppose $s_2 \\ge 1$. Since $s_n = -1$, and $s_{k+1}$ can be no smaller than $s_k$, there must be some $i$ between $2$ and $n$ such that $s_i = 0$. Then $i-0 \\mid s_i - s_0$.\n\nWe have thus shown that there is at least one divisible pair among the pairs $(1, n)$, $(2, n)$, $(0, 2)$, and $(0, i)$, for some $2 < i < n$. Note that if $n \\ge 4$, the two numbers in each of those pairs must differ by at least two. Thus, $(a_1, \\dots, a_n)$ is not good when $n \\ge 4$, finishing the proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21208,
"subject": "Mathematics (Olympiad)",
"question": "$\\{1, \\frac{1}{2}, \\frac{1}{4}, \\frac{1}{16}\\}$ is an example of a powerful set with four elements.\n\nProve the following lemma:\n\n**Lemma 1.** A *finite powerful set* $S$ *cannot have an element greater than one and an element less than one.*",
"options": [],
"answer": "See solution",
"solution": "*Proof.* Suppose by contradiction that such elements exist. Let $a$ be the least element of $S$ and $b$ the least element of $S$ greater than $1$. By assumption, $a < 1$. Now $a < 1 < b$, so:\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 leads to a contradiction, proving the lemma. □\n\nAccording to this lemma, if $S$ is a finite powerful set, all its elements are in $[1, \\infty)$ or in $(0, 1]$.\n\nSuppose $S$ is a powerful set with $n > 3$ elements in $[1, \\infty)$, $S = \\{1 = a_1 < a_2 < \\cdots < a_n\\}$. We can assume $a_1 = 1$. 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 $(0, 1]$, $S = \\{a_1 < a_2 < \\cdots < a_n = 1\\}$. Again, assume $1 \\in S$. For each $1 \\leq i \\leq n-2$, $a_{n-1} < a_{n-1}^{a_i} < 1$, so $a_n^{a_i} \\notin S$, and thus $a_i^{a_n} \\in S$. We have:\n\n$$\na_1 < a_1^{a_{n-1}} < a_2^{a_{n-1}} < \\cdots < a_{n-2}^{a_n} < 1\n$$\n\nSo $a_i^{a_{n-1}} = a_{i+1}$ for $2 \\leq i \\leq n-2$. Let $a_{n-1} = a$:\n\n$$\na_{n-2} = a^{1/a},\\quad a_{n-3} = a^{1/a^2}, \\dots\n$$\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$$\na_{n-2}^{a_{n-1}} = (a^{1/a^2})^{a^{1/a}} = a \\implies a^{(a^{1/a})-2} = a \\implies a^{1/a - 2} = 1\n$$\n\nBut $a \\neq 1$, so $a = \\frac{1}{2}$. Therefore, $a_{n-1} = \\frac{1}{2}$, $a_{n-2} = \\frac{1}{4}$, $a_{n-3} = \\frac{1}{16}$. Since $n > 4$, $a_{n-4} = \\frac{1}{256} \\in S$, 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.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21209,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be an acute triangle with $|AB| > |AC|$ and let $\\omega$ be the circumcircle of $\\triangle ABC$ with centre $O$. The altitude from $A$ intersects $BC$ at $D$ and intersects $\\omega$ a second time at $P$. Let $H$ be the orthocentre of $\\triangle ABC$ and let $K$ be the point on the line segment $BC$ such that $|BD| = |KC|$. The circumcircle of $\\triangle PKH$ intersects $\\omega$ a second time at $Q$ and intersects the line $BC$ a second time at $N$. Let $T$ be the point on the line $AD$ such that $TN \\perp PQ$.\n\nProve that the line $KT$ passes through $O$.",
"options": [],
"answer": "See solution",
"solution": "Let $M$ be the circumcentre of $\\triangle PKH$. Note that the reflection of $H$ in $BC$ lies on the circumcircle $\\omega$. (One way to see this is to note that $\\angle BHC = \\angle BHD + \\angle DHC = \\angle BCA + \\angle ABC$.) Therefore, this reflection is the point $P$. So the reflection in $BC$ transforms $\\triangle PKH$ into itself, so $M$ lies on $BC$ and therefore is the midpoint of $KN$.\n\nLet $O$ be the centre of $\\omega$. Note that $P$ and $Q$ lie on both $\\omega$ and the circumcircle of $\\triangle PKH$, so $OM$ is the perpendicular bisector of $PQ$. Therefore, $OM \\perp PQ \\perp TN$, so $OM \\parallel TN$. As $M$ is the midpoint of $KN$, the line $OM$ is the midsegment of $\\triangle KNT$ parallel to $TN$. Thus, $OM$ passes through the midpoint of $KT$, say $O'$.\n\nLet $m$ be the perpendicular bisector of $KD$. It is a midsegment in $\\triangle KDT$, because it passes through the midpoint of $KD$ and is parallel to $DT$. Therefore, $m$ also passes through $O'$. As $|BD| = |KC|$, $m$ is also the perpendicular bisector of $BC$. So $m$ also passes through $O$, as $BC$ is a chord of $\\omega$. Since $OM$ and $m$ both pass through both $O$ and $O'$, we conclude that $O' = O$ if $OM$ and $m$ do not coincide. Therefore, $KT$ passes through $O$ if $OM$ and $m$ do not coincide.\n\nFinally, to complete the proof, we show that $OM$ and $m$ indeed do not coincide. Let $\\ell$ be the line through $N$ perpendicular to $PQ$. Note that $H$ lies in the interior of $\\triangle ABC$, as it is an acute-angled triangle. It follows that $D \\neq N$ and therefore that $\\ell$ does not coincide with $AD$. Since $D \\neq N$ and $M$ is the midpoint of $KN$, $M$ is not the midpoint of $KD$. So $M$ is not the midpoint of $BC$. Therefore, $OM$ is not perpendicular to $BC$ and therefore not parallel to $m$; in particular, $OM$ and $m$ do not coincide. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21210,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $3mn + 3m = k(2m^2 + n^2)$ where $k$ is a natural number. Find all pairs of natural numbers $(m, n)$ that satisfy this equation.",
"options": [],
"answer": "See solution",
"solution": "The inequality between the arithmetic and geometric means gives\n\n$$\n3mn + 3m = 3m(n + 1) = k(2m^2 + n^2) \\geq 2\\sqrt{2}kmn,\n$$\n\nhence\n\n$$\nk \\leq \\frac{3(n + 1)}{2\\sqrt{2}n} = \\frac{3}{2\\sqrt{2}}\\left(1 + \\frac{1}{n}\\right) \\leq \\frac{3}{\\sqrt{2}}.\n$$\n\nWe conclude $k \\leq 2$.\n\nThe equation $k(2m^2 + n^2) = 3mn + 3m$ may be considered a quadratic equation for $n$:\n\n$$\nkn^2 - 3mn + 2km^2 - 3m = 0.\n$$\n\nThe discriminant of this equation is $D = 9m^2 - 8k^2m^2 + 12km$.\n\nIf $k = 2$, then $D = 24m - 23m^2$. Since the discriminant must be non-negative, the only solution is $m = 1$ and from the quadratic equation we calculate $n = 1$.\n\nNow let $k = 1$. The discriminant must be a square of a non-negative integer $t$:\n\n$$\nt^2 = D = m^2 + 12m = (m + 6)^2 - 36.\n$$\n\nReorganizing, $36 = (m + 6 + t)(m + 6 - t)$. Since $t \\geq 0$ and $m > 0$, $m + 6 + t > 0$ and $m + 6 - t > 0$. The numbers $m + 6 + t$ and $m + 6 - t$ have the same parity. We conclude $m + 6 + t = m + 6 - t = 6$ or $m + 6 + t = 18$ and $m + 6 - t = 2$. In the first case, $m = 0$, which is not a natural number. In the second case, $m = 4$ and $t = 8$, and the equation has two solutions: $n = 2$ and $n = 10$.\n\nThe solutions are the following pairs of natural numbers: $(1, 1)$, $(4, 2)$, and $(4, 10)$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21211,
"subject": "Mathematics (Olympiad)",
"question": "Given $a^b \\leq b^c$, $b^c \\leq c^d$, $c^d \\leq d^a$, answer the following questions:\n\n1. If $a, b, c, d$ are distinct, which one is the least among $a, b, c, d$?\n2. Must the greatest number among $a, b, c, d$ always be the same variable? Justify your answer.",
"options": [],
"answer": "See solution",
"solution": "1. We apply the claim to the triples $(a, b, c)$, $(b, c, d)$, and $(c, d, a)$. Because $a, b, c, d$ are distinct, none of $c, d$, or $a$ can be the least among $a, b, c, d$. Therefore, the least number is $b$.\n\n2. 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": 21212,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $n$ for which you can replace a digit $4$ with the digits $22$ and obtain a number divisible by $n$.",
"options": [],
"answer": "See solution",
"solution": "Let $n$ satisfy the condition and $n'$ be the number obtained by the replacement of digits. Denote by $x$ the number formed by the digits to the left of the replaced digit and by $y$ the number formed by the digits to the right. Let $k$ be the number of digits in $y$. Then:\n\n$$\nn = x \\cdot 10^{k+1} + 4 \\cdot 10^k + y\n$$\n$$\nn' = x \\cdot 10^{k+2} + 22 \\cdot 10^k + y\n$$\n\nAs $n'$ is divisible by $n$, so must be $n' - n$ and $10n - n'$, which yields:\n\n$$\nx \\cdot 10^{k+1} + 4 \\cdot 10^k + y \\mid 9x \\cdot 10^{k+1} + 18 \\cdot 10^k\n$$\n$$\nx \\cdot 10^{k+1} + 4 \\cdot 10^k + y \\mid 18 \\cdot 10^k + 9y\n$$\n\nConsider the following cases:\n\n* If $x = 0$, then $t (4 \\cdot 10^k + y) = 18 \\cdot 10^k$ for some integer $t$. Since $5 (4 \\cdot 10^k + y) \\ge 5 \\cdot 4 \\cdot 10^k > 18 \\cdot 10^k > 3 \\cdot 5 \\cdot 10^k > 3 (4 \\cdot 10^k + y)$, the only option is $t = 4$. The equation $4 (4 \\cdot 10^k + y) = 18 \\cdot 10^k$ yields $y = \\frac{2 \\cdot 10^k}{4} = 5 \\cdot 10^{k-1} < 10^k$. Thus, $n = 45 \\cdot 10^{k-1}$.\n\n* If $x = 1$, then $t (14 \\cdot 10^k + y) = 108 \\cdot 10^k$ for some integer $t$. Since $8 (14 \\cdot 10^k + y) \\ge 8 \\cdot 14 \\cdot 10^k > 108 \\cdot 10^k > 6 \\cdot 15 \\cdot 10^k > 6 (14 \\cdot 10^k + y)$, the only option is $t = 7$, but $7 (14 \\cdot 10^k + y) = 108 \\cdot 10^k$ has no integer solutions, because $7 \\nmid 108 \\cdot 10^k$.\n\n* If $x = 2$, then $t (24 \\cdot 10^k + y) = 198 \\cdot 10^k$ for some integer $t$. Since $9 (24 \\cdot 10^k + y) \\ge 9 \\cdot 24 \\cdot 10^k > 198 \\cdot 10^k > 7 \\cdot 25 \\cdot 10^k > 7 (24 \\cdot 10^k + y)$, the only option is $t = 8$. The equation $8 (24 \\cdot 10^k + y) = 198 \\cdot 10^k$ yields $y = \\frac{6 \\cdot 10^k}{8} = 75 \\cdot 10^{k-2} < 10^k$. Thus, $n = 2475 \\cdot 10^{k-2}$.\n\n* If $x \\ge 3$, then $x \\cdot 10^{k+1} + 4 \\cdot 10^k + y \\le 18 \\cdot 10^k + 9y$, which yields $x \\cdot 10^{k+1} \\le 14 \\cdot 10^k + 8y$. On the other hand, $x \\cdot 10^{k+1} \\ge 30 \\cdot 10^k > 22 \\cdot 10^k > 14 \\cdot 10^k + 8y$. These equations contradict each other, so no such $n$ can exist.\n\n**Answer:** $45$, $2475$, and all numbers obtained by adding zeros at the end of these numbers.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21213,
"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\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\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",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21214,
"subject": "Mathematics (Olympiad)",
"question": "a) Is it possible to arrange 100 pawns on a $14 \\times 14$ chessboard so that every $2 \\times 2$ square contains exactly 2 pawns?\n\nb) Is it possible to arrange 110 pawns on a $15 \\times 15$ chessboard so that every $2 \\times 2$ square contains exactly 2 pawns?",
"options": [],
"answer": "See solution",
"solution": "a) It is impossible. For $n = 14$, the board can be divided into $49$ squares of size $2 \\times 2$. If each contains exactly $2$ pawns, the total is $2 \\times 49 = 98$ pawns. Since $100 > 98$, $100$ pawns cannot be arranged as required.\n\nb) It is possible. For $n = 15$, there are at least $105$ pawns needed (see solution of Problem A.8). The required arrangement of $110$ pawns for $n = 15$ is shown in the figure.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21215,
"subject": "Mathematics (Olympiad)",
"question": "Does there exist a positive integer $m$ and a polynomial $P(x)$ with real coefficients such that\n\n$$\nx^m + x + 2 = P(P(x))\n$$\n\nfor all real $x$?",
"options": [],
"answer": "See solution",
"solution": "Suppose $m = 1$. Then $P(x)$ must be linear: $P(x) = Ax + B$. Comparing coefficients in $P(P(x)) = x + x + 2$, we get $A^2 = 1$, but this leads to a contradiction when matching all coefficients.\n\nNow, let $m > 1$. Recall that $a - b \\mid P(a) - P(b)$ for any polynomial $P$ and integers $a, b$. Thus, for all integers $x$:\n\n$$\nP(x) - x \\mid x^m + 2.\n$$\n\nThis means $P(0) \\mid 2$, so $P(0) \\in \\{\\pm1, \\pm2\\}$. Similarly, $P(1) \\mid 4$ and so on. Checking all possible values for $P(0)$:\n\n- If $P(0) = 1$, then $P(1) = 2$, $P(2) = 4$, but $P(2) \\equiv 1 \\pmod{2}$, contradiction.\n- If $P(0) = -1$, then $P(-1) = 2$, $P(2) = (-1)^m + 1 \\equiv 0 \\pmod{2}$, contradiction.\n- If $P(0) = 2$, then $P(2) = 2$, $2 = 2^m + 4$, contradiction.\n- If $P(0) = -2$, then $P(-2) = 2$, $P(2) = (-2)^m$. This leads to $(-2)^m - 2 \\mid 2^m + 2$. For even $m$, only possible for $m \\leq 2$, but $m$ must be a perfect square, so $m \\neq 2$. For odd $m$, $P(-1) + 1 \\mid 1$, so $P(-1) \\in \\{-2, 0\\}$. If $P(-1) = 0$, then $2 = -2$, contradiction. If $P(-1) = -2$, then $P(-2) = 0$, $P(0) = -2^m$, so $m = 1$, which was already excluded.\n\nTherefore, there are no such $m$ and $P(x)$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21216,
"subject": "Mathematics (Olympiad)",
"question": "Find all natural numbers $a$ such that $4x^2 + a$ is a prime for all $x = 0, 1, \\dots, a-1$.",
"options": [],
"answer": "See solution",
"solution": "Clearly, $a$ must be a prime (set $x=0$). More exactly, $a$ is an odd prime because $a=2$ is not a solution. Furthermore, $a+1$ must be a power of $2$. Indeed, suppose that $a+1$ has an odd prime divisor $p$. Note that $p \\leq \\frac{a+1}{2}$ as $a+1$ is even. Set $x = \\frac{1}{2}(p-1)$; it is clear that $0 < x < a$. We have\n\n$$\n4x^2 + a = 4 \\cdot \\frac{1}{4} (p-1)^2 + a = p(p-2) + (a+1).\n$$\n\nSince $p$ divides $a+1$, it also divides $4x^2 + a$. In addition, $p < a < 4x^2 + a$, hence $4x^2 + a$ is composite.\n\nThe primes $3 = 2^2 - 1$ and $7 = 2^3 - 1$ satisfy the conditions. The values of $4x^2 + 3$ for $x = 0, 1, 2$ are the primes $3, 7, 19$; the values of $4x^2 + 7$ for $x = 0, 1, 2, 3, 4, 5, 6$ are the primes $7, 11, 23, 43, 71, 107, 151$.\n\nWe show that $a=3$ and $a=7$ are the only solutions by rejecting all $a = 2^m - 1$ with $m \\geq 4$. For numbers of this form, consider $a+9 = (a+1)+8$. This even number is not a power of $2$ (powers of $2$ greater than $8$ cannot differ by $8$). Let $q \\leq \\frac{a+9}{2}$ be an odd prime divisor of $a+9$. Set $x = \\frac{1}{2}(q-3)$:\n\n$$\n\\text{Note that } x \\text{ may be zero but is less than } a. \\text{ Now } 4x^2 + a = 4 \\cdot \\frac{1}{4}(q-3)^2 + a = q(q-6) + (a+9), \\text{ so } q \\text{ divides } 4x^2 + a.\n$$\n\nAlso $q \\leq \\frac{a+9}{2} < a$ as $a \\geq 15$, hence $q < a \\leq 4x^2 + a$. Then $4x^2 + a$ is composite, which completes the proof. The answer is $a=3$ and $a=7$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21217,
"subject": "Mathematics (Olympiad)",
"question": "Consider a table with $n$ rows and $m$ columns ($n, m \\in \\mathbb{N}$, $n, m \\geq 2$) consisting of $n \\times m$ squares of size $1 \\times 1$, which we will call *cells*. A *snake* is a sequence of cells with the following properties:\n\n- The first cell is located on the first (top) row of the table.\n- The last cell is located on the last row of the table.\n- Starting with the second cell, each cell of the snake shares a common side with the previous one and is not located on a row above the previous cell.\n\nThe *length* of a snake is the number of cells that form the snake. Determine the arithmetic mean of the lengths of all the snakes in the table.",
"options": [],
"answer": "See solution",
"solution": "Consider a particular snake. For each row $i \\in \\{1, 2, \\dots, n\\}$ of the table, let $a_i$ and $b_i$ be the column numbers of the first and last cell, respectively, that the snake occupies on row $i$. We observe that $a_{i+1} = b_i$ for any $i \\in \\{1, 2, \\dots, n-1\\}$, so the snake is determined by the values $a_1, b_1, b_2, \\dots, b_n \\in \\{1, 2, \\dots, m\\}$.\n\nThus, the total number of snakes is $m^{n+1}$.\n\nLet $(k, p)$ be a cell of the table. The snakes containing this cell are those for which $a_k \\leq p \\leq b_k$. For these, $a_k$ and $b_k$ can be chosen in $2p(m-p+1)-1$ ways, and the other values $a_1, b_1, b_2, \\dots, b_n$ (except $a_k$ and $b_k$) in $m^{n-1}$ ways. Therefore, the cell $(k, p)$ is counted $m^{n-1}(2pm-2p^2+2p-1)$ times among all snakes.\n\nThe cells in column $p$ contribute $n \\cdot m^{n-1}(2pm - 2p^2 + 2p - 1)$ to the total $T$ of the lengths of all snakes, so\n\n$$\nT = \\sum_{p=1}^{m} n \\cdot m^{n-1} (2pm - 2p^2 + 2p - 1) = n \\cdot m^{n+1} (m+1) - \\frac{n \\cdot m^n (m+1)(2m+1)}{3} + n \\cdot m^{n+1} = \\frac{n \\cdot m^n (m^2 + 3m - 1)}{3}.\n$$\n\nThe arithmetic mean of the lengths of all the snakes in the table is\n\n$$\n\\frac{n(m^2+3m-1)}{3m}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21218,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\geq 2$ be an integer. A game is played on an $n \\times n$ board by two players X and Y as follows:\n\n- **Order:** The two players take turns to play. X plays in the 1st round, Y plays the 2nd round, then X plays the 3rd round, and so on.\n- **Rule:** In the $k$th round, a player must choose $k$ unmarked consecutive cells in the same row or in the same column and mark each of these cells.\n- **Winner:** The first player who cannot complete the task loses the game.\n\nThe player who is expected to carry out the $(n+1)$st round is called the *natural loser*, as there are no $(n+1)$ consecutive cells on the board. Find the smallest $n$ for which the natural loser has a winning strategy.",
"options": [],
"answer": "See solution",
"solution": "We claim that the smallest such $n$ is $n = 6$.\n\nLet $(i, j)$ denote the cell in the $i$th row and $j$th column, and $(i, j)-(i, k)$ denote consecutive cells from $(i, j)$ to $(i, k)$ (similarly for columns).\n\n- For $n = 2$, X and Y can play in any way and the natural loser X will lose.\n- For $n = 3$, Y is the natural loser and X has a winning strategy. In the first round, X marks $(2, 2)$. Without loss of generality, Y marks $(1, 1)-(1, 2)$. Then X can take $(3, 1)-(3, 3)$.\n- For $n = 4$, Y has a winning strategy. In round 1, X chooses a cell in the upper left $2 \\times 2$ corner. Y marks 2 consecutive cells in the same $2 \\times 2$ corner. This leaves 2 unoccupied rows and 2 unoccupied columns. WLOG, X chooses cells from a row in round 3. Then there is at least one unoccupied row left, which Y can mark in round 4.\n- For $n = 5$, X has a winning strategy by first choosing $(3, 3)$. WLOG, Y chooses 2 cells in the upper left $3 \\times 3$ corner. Then X can always choose 3 consecutive cells from the same $3 \\times 3$ corner. There are at least 2 unoccupied rows and columns. For the same reason as $n = 4$, X can win.\n\nFor $n = 6$, the natural loser X can win. A winning strategy is as follows:\n\n1. X chooses $(1, 1)$.\n2. WLOG, Y chooses 2 cells $(i, j)-(i+1, j)$ in column $j$ in the next round.\n - If $j = 1$, X chooses $(3, 3)-(3, 5)$ in round 3, blocking columns 3, 4, 5. Y must mark at least one cell in column 2 or 6 in round 4. If column 2 is used, X takes $(2, 6)-(6, 6)$ in round 5; if column 6 is used, X takes $(2, 2)-(6, 2)$. All rows and columns are used, so Y loses.\n - Similar strategies work for $j = 2, 5, 6$.\n - For $j = 3, 4$ and $i = 1, 4, 5$, X chooses $(3, 3)-(3, 5)$ in round 3. If Y uses column 2 in round 4, X takes $(2, 6)-(6, 6)$; if column 6, X takes $(2, 2)-(6, 2)$. If Y chooses $(2, 1)-(5, 1)$, X takes $(6, 2)-(6, 6)$ or $(1, 2)-(1, 6)$ depending on row usage. If Y chooses $(3, 1)-(6, 1)$, X takes $(2, 2)-(2, 6)$ or $(6, 2)-(6, 6)$.\n - For $j = 3, 4$ and $i = 2, 3$, X takes $(5, 3)-(5, 5)$ in round 3. Similar responses apply as above.\n - If Y chooses 4 consecutive cells in column 3, 4, or 5 in round 4, the first 5 rows are occupied. X then takes $(6, 2)-(6, 6)$ in round 5. Y loses.\n\nIn all cases, X can win the game when $n = 6$. This completes the proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21219,
"subject": "Mathematics (Olympiad)",
"question": "試求所有正整數對 $ (m, n) $，滿足 $ m $ 和 $ (n+1) $ 互質，且\n\n$$\n\\sum_{k=1}^{n} \\frac{m^{k+1}}{k+1} \\binom{n}{k}\n$$\n\n為整數。",
"options": [],
"answer": "See solution",
"solution": "不存在這樣的 $ (m, n) $。\n\n假設 $ (m, n) $ 是一組解。由於 $ (n + 1) \\binom{n}{k} = (k + 1) \\binom{n+1}{k+1} $，有：\n\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} \\frac{m^{k+1}}{k+1} \\binom{n}{k} &= \\sum_{k=1}^{n} \\frac{m^{k+1}}{n+1} \\binom{n+1}{k+1} = \\frac{1}{n+1} \\sum_{l=0}^{n+1} m^l \\binom{n+1}{l} \\\\\n&= \\frac{1}{n+1} \\left((1+m)^{n+1} - 1\\right).\n\\end{aligned}\n$$\n\n因為這必須是整數，所以有：\n\n$$\n(1+m)^{n+1} \\equiv 1 \\pmod{n+1}.\n$$\n\n這保證了當 $ m $ 為奇數時，$ n+1 $ 也必須為奇數。另一方面，因為 $ \\gcd(m, n+1) = 1 $，若 $ m $ 為偶數，$ n+1 $ 也必須為奇數。因此，無論如何，$ n+1 $ 為奇數。\n\n設 $ p $ 為 $ n+1 $ 的最小質因數，則 $ p $ 為奇質數，且\n\n$$\n(1+m)^{n+1} \\equiv 1 \\pmod{p}.\n$$\n\n由費馬小定理，$ (1+m)^{p-1} \\equiv 1 \\pmod{p} $。\n因此，\n\n$$\n(1+m)^{\\gcd(n+1, p-1)} \\equiv 1 \\pmod{p}.\n$$\n\n但因為 $ p $ 是 $ n+1 $ 的最小質因數，$ \\gcd(n+1, p-1) = 1 $，所以 $ (1+m) \\equiv 1 \\pmod{p} $，即 $ p \\mid m $，這與 $ \\gcd(m, n+1) = 1 $ 矛盾。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21220,
"subject": "Mathematics (Olympiad)",
"question": "Find all prime numbers $p$ and $q$ such that $2p^2q + 45pq^2$ is a perfect square.",
"options": [],
"answer": "See solution",
"solution": "First, assume $p = q$. Then $2p^2q + 45pq^2 = 47p^3$ must be a perfect square. Since $47p^3$ is divisible by 47, which is prime, it must also be divisible by $47^2$. This implies that 47 divides $p^3$, so $p = 47$. Indeed, when $p = q = 47$, we have $2p^2q + 45pq^2 = 47^4$, which is a perfect square.\n\nNow, let $p \\neq q$. Since $2p^2q + 45pq^2 = pq(2p + 45q)$ is a perfect square divisible by $p$, it must also be divisible by $p^2$. So, $p$ divides $q(2p + 45q)$. Since $p$ and $q$ are coprime, $p$ divides $2p + 45q$. Thus, $p$ divides $45q$, so $p$ divides 45. The possible values are $p = 3$ or $p = 5$.\n\nA similar argument for $q$ shows that if $q$ divides $2p + 45q$, then $q$ divides $2p$, which implies $q = 2$. So, $pq(2p + 45q) = 4p(p + 45)$. If $p = 3$, then this expression is $4 \\cdot 3 \\cdot 48 = 24^2$. If $p = 5$, then it is $4 \\cdot 5 \\cdot 50 = 1000$, which is not a perfect square.\n\nTherefore, the only two pairs of primes with the required property are $p = q = 47$ and $p = 3$, $q = 2$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21221,
"subject": "Mathematics (Olympiad)",
"question": "Suppose we have a $3n \\times 3n$ board filled with plus and minus signs, and we are allowed to change the signs in any row or column (i.e., flip all signs in that row or column). Is it possible, by a sequence of such operations, to obtain a board with exactly 36 minus signs? If so, for which values of $n$ is this possible?",
"options": [],
"answer": "See solution",
"solution": "Note that if changing of signs is performed an even number of times to some row (or column), it is equivalent to not applying the operation at all. So we can assume the operation is applied exactly once to some rows (columns), and not applied to the remaining rows (columns). Let the operation be applied to $x$ rows and $y$ columns. Then the total number of cells in which the signs are changed is $c = 3n x + 3n y - 2 x y$.\n\nWe rewrite the equality as\n\n$$\n|3n - 2x| \\cdot |3n - 2y| = |9n^2 - 2c| \\quad (1)\n$$\n\nNote that $c$ may admit the values 35 or 37 depending on the number (0 or 1) of initial minuses which are changed to pluses. If $n$ is even, then the number of minuses after any operation keeps its parity, but after all operations this parity is changed ($1 \\rightarrow 36$), a contradiction. So $n$ is odd. Therefore, both co-factors $|3n - 2x|$ and $|3n - 2y|$ are odd and do not exceed $3n$ (since $0 \\le x \\le 3n$, $0 \\le y \\le 3n$). If at least one of them is equal to $3n$, then the right-hand side of (1) is a multiple of 3, but for neither $c = 35$ nor $c = 37$ is this the case. Hence, $|3n - 2x| < 3n$, $|3n - 2y| < 3n$, so $|3n - 2x| \\le 3n - 2$, $|3n - 2y| \\le 3n - 2$. Then $|9n^2 - 2c| \\le (3n - 2)^2$, in particular, $9n^2 - 2c \\le 9n^2 - 12n + 4$, so $n \\le \\frac{c+2}{6} \\le \\frac{37+2}{6} = \\frac{13}{2}$. It follows that $n \\le 5$ ($n$ is odd).\n\nNote that the case $n = 1$ is impossible, since the $3 \\times 3$ board cannot contain 36 minuses. It remains to consider the following two cases:\n\n1. $n = 3$. We have $|9 - 2x| \\cdot |9 - 2y| = 81 - 70$ or $|9 - 2x| \\cdot |9 - 2y| = 81 - 74$, i.e., either $|9 - 2x| \\cdot |9 - 2y| = 11$, which is impossible (both factors are no greater than 9), or $|9 - 2x| \\cdot |9 - 2y| = 7$, which is possible (e.g., $x = 1, y = 4$).\n\n2. $n = 5$. We have $|15 - 2x| \\cdot |15 - 2y| = 225 - 70$ or $|15 - 2x| \\cdot |15 - 2y| = 225 - 74$, i.e., either $|15 - 2x| \\cdot |15 - 2y| = 155$ or $|15 - 2x| \\cdot |15 - 2y| = 151$. Both cases are impossible.\n\nTherefore, the unique possible value of $n$ is $3$. It is easy to see that if the single minus is in the left bottom cell of a $9 \\times 9$ board and we change the signs in the top row and in the first four columns, the new board will contain exactly 36 minuses.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21222,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, \\dots, a_{2012}$ be odd positive integers. Prove that the number\n$$\nA = \\sqrt{a_1^2 + a_2^2 + \\dots + a_{2012}^2 - 1}\n$$\nis irrational.",
"options": [],
"answer": "See solution",
"solution": "The number $A$ is rational if and only if $a_1^2 + a_2^2 + \\dots + a_{2012}^2 - 1$ is a square. Recall that the square of an odd integer has the form $4k + 1$, with $k \\in \\mathbb{N}$. Since all numbers $a_i$ are odd, the sum $a_1^2 + a_2^2 + \\dots + a_{2012}^2$ is a multiple of $4$, since $4$ divides $2012$. Consequently, the number $a_1^2 + a_2^2 + \\dots + a_{2012}^2 - 1$ has the form $4k + 3$, hence is not a square. The conclusion follows.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21223,
"subject": "Mathematics (Olympiad)",
"question": "How many ways can a positive integer $n$ be written as the sum of $k$ consecutive positive integers?",
"options": [],
"answer": "See solution",
"solution": "The sum of $k$ consecutive integers is $$(m+1) + (m+2) + \\dots + (m+k) = \\frac{k(2m+k+1)}{2}.$$ So we require $k(2m+k+1) = 2n$. Note that $k$ and $2m+k+1$ have opposite parity and that $k < 2m+k+1$.\n\nNow suppose $2n = ab$ with $a$ odd. Then $b$ must be even. So $a$ and $b$ cannot be equal. Take $k$ to be the smaller, then put $m = \\frac{|a-b-1|}{2}$ and we have a solution. So the total number of solutions is just the number of odd factors of $2n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21224,
"subject": "Mathematics (Olympiad)",
"question": "Black bricks, white bricks, and grey bricks are laid to form the pattern in the diagram. The three squares have the same centre, and the diagonals of the squares have lengths 3, 4, and 5 respectively. If 120 bricks are needed for the central grey area, then the number of black bricks needed is\n\n",
"options": [],
"answer": "See solution",
"solution": "For a square with diagonal $d$ and side length $s$, we have the relationship $d^2 = 2s^2$, so $s = \\frac{d}{\\sqrt{2}}$. The respective side lengths of the three squares are thus $\\frac{3}{\\sqrt{2}}$, $\\frac{4}{\\sqrt{2}}$, and $\\frac{5}{\\sqrt{2}}$. \n\nSo, grey area $= \\left(\\frac{3}{\\sqrt{2}}\\right)^2 = \\frac{9}{2}$, while black area $= \\left(\\frac{5}{\\sqrt{2}}\\right)^2 - \\left(\\frac{4}{\\sqrt{2}}\\right)^2 = \\frac{25}{2} - \\frac{16}{2} = \\frac{9}{2} = $ grey area. \n\nThus, the black area must also require 120 bricks.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21225,
"subject": "Mathematics (Olympiad)",
"question": "$$\n\\begin{cases}\na^3 + b^3 = c^2 + d^2, \\\\\na^2 + b^2 = c^3 + d^3.\n\\end{cases}\n$$\n\n\n\nFind all integer quadruples $(a, b, c, d)$ satisfying the system above.",
"options": [],
"answer": "See solution",
"solution": "First, we prove the auxiliary statement: for arbitrary numbers $a$ and $b$, at least one nonzero, $a^2 - ab + b^2 > 0$.\n\n**Case 1:** One of $a$ or $b$ is zero. Then $a^2 > 0$ (if $b=0$) or $b^2 > 0$ (if $a=0$), so the inequality holds.\n\n**Case 2:** One of $a$ or $b$ is positive, the other negative. Then each term in $a^2 + (-ab) + b^2$ is positive, so the sum is positive.\n\n**Case 3:** Both $a$ and $b$ are positive or both negative. Then $a^2 + b^2 \\ge 2ab > ab$.\n\nThe auxiliary statement is proven.\n\nNow, suppose $(a, b, c, d)$ is an integer solution to the system. Then $a^3 + b^3 \\ge a^2 + b^2$ and $c^3 + d^3 \\ge c^2 + d^2$.\n\nIf $a^3 + b^3 = 0$, then $c^2 + d^2 = 0 \\Rightarrow c = d = 0$, and $a^2 + b^2 = 0 \\Rightarrow a = b = 0$. Thus, $a^3 + b^3 \\ge a^2 + b^2 = 0$.\n\nIf $a^3 + b^3 > 0$, then $a+b = \\frac{a^3 + b^3}{a^2 - ab + b^2} > 0$. Consider $a+b=1$ and $a+b \\ge 2$.\n\nIf $a+b=1$, one is positive, the other negative or zero:\n$$\na^3 + b^3 = (a+b)(a^2 - ab + b^2) = a^2 - ab + b^2 \\ge a^2 + b^2,\n$$\nwith equality only if $ab=0$ (i.e., one is $0$, the other $1$).\n\nIf $a+b \\ge 2$:\n$$\na^3 + b^3 \\ge 2(a^2 - ab + b^2) = a^2 + b^2 + (a-b)^2 \\ge a^2 + b^2,\n$$\nwith equality only if $a+b=2$ and $a-b=0$ (i.e., $a=b=1$).\n\nThus, $a^3 + b^3 \\ge a^2 + b^2$, with equality only for $(0,0)$, $(0,1)$, $(1,0)$, $(1,1)$, and similarly for $(c, d)$.\n\nSo, the only possible solutions are those for which $a^3 + b^3 = a^2 + b^2$ and $c^3 + d^3 = c^2 + d^2$. Checking all combinations, the integer solutions are:\n\n*Answer:* $(0, 0, 0, 0)$, $(1, 1, 1, 1)$, $(1, 0, 1, 0)$, $(1, 0, 0, 1)$, $(0, 1, 1, 0)$, $(0, 1, 0, 1)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21226,
"subject": "Mathematics (Olympiad)",
"question": "Let $M_{XY}$ denote the midpoint of the line segment $XY$.\n\nGiven four points $A$, $B$, $C$, and $D$ such that $AB$ is orthogonal to $CD$, prove that the four midpoints $M_{AD}$, $M_{BD}$, $M_{AC}$, and $M_{BC}$ form a rectangle.\n\n",
"options": [],
"answer": "See solution",
"solution": "Using the intercept theorem, we deduce:\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 $M_{BC}M_{BD}$.\nFurthermore, $M_{AC}M_{BC}$ is orthogonal to $M_{AC}M_{AD}$, since $CD$ is orthogonal to $AB$.\nTherefore, the four midpoints form a rectangle. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21227,
"subject": "Mathematics (Olympiad)",
"question": "Dos circunferencias $C$ y $C'$ son secantes en dos puntos $P$ y $Q$. La recta que une los centros corta a $C$ en $R$ y a $C'$ en $R'$. La que une $P$ y $R'$ corta a $C$ en $X \\neq P$ y la que une $P$ y $R$ corta a $C'$ en $X' \\neq P$. Si los tres puntos $X$, $Q$, $X'$ están alineados, se pide:\n\n1. Hallar el ángulo $\\angle XPX'$.\n\n2. Demostrar que $(d + r - r')(d - r + r') = rr'$, donde $d$ es la distancia entre los centros de las circunferencias y $r$ y $r'$ sus radios.",
"options": [],
"answer": "See solution",
"solution": "(i) Sean $F$ y $F'$ los puntos diametralmente opuestos a $R$ y $R'$ en $C$ y $C'$, respectivamente. Por el Teorema del ángulo inscrito se tiene que $\\angle PFQ = \\angle PXQ = \\alpha$ que, por simetría, es el doble de $\\angle PFR$, luego $\\angle PFR = \\alpha/2$. Como el triángulo $PFR$ es rectángulo en $P$ (al ser $FR$ diámetro de $C$), deducimos que $\\angle PRF = \\pi/2 - \\alpha/2$. Similarmente, $\\angle PR'F' = \\pi/2 - \\beta/2$, donde $\\beta = \\angle PX'Q$. Por otro lado, considerando el triángulo $XPX'$, $\\angle XPX' = \\pi - \\alpha - \\beta$, luego sumando los ángulos del triángulo $PRR'$,\n\n$$\n\\left(\\frac{\\pi}{2} - \\frac{\\alpha}{2}\\right) + \\left(\\frac{\\pi}{2} - \\frac{\\beta}{2}\\right) + (\\pi - \\alpha - \\beta) = \\pi,\n$$\n\nes decir, $\\alpha + \\beta = 2\\pi/3$ y $\\angle XPX' = \\pi/3$.\n\n\n\n(ii) Consideramos el triángulo $OPO'$, donde $O$ y $O'$ son los centros de $C$ y $C'$, respectivamente. Nuevamente, por el Teorema del ángulo inscrito, el ángulo central $\\angle POR$ es $2\\angle PFR = \\alpha$ y similarmente $\\angle PO'R' = 2\\angle PF'R' = \\beta$, luego $\\angle OPO' = \\pi/3$. Los lados del triángulo $OPO'$ son los radios $r$ y $r'$ y la distancia $d$ entre los centros, por tanto el resultado se sigue directamente del",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21228,
"subject": "Mathematics (Olympiad)",
"question": "Find all real values of $a$ such that the polynomial $P(t) = t^3 - 3t + a$ has three distinct nonzero real roots.",
"options": [],
"answer": "See solution",
"solution": "$-2 < a < 0$ and $0 < a < 2$.\n\nLetting $c = x + y + z$, one sees that $x, y, z$ are the roots of the polynomial $P(t) = t^3 - 3t + a + 3c$. Therefore, Vieta's theorem implies $x + y + z = 0$, thus $c = 0$ and $P(t) = t^3 - 3t + a$. Consequently, the polynomial $P(t) = t^3 - 3t + a$ has three distinct real roots. Moreover, the denominators appearing in the problem statement must be nonzero, hence the roots of $P(t)$ are nonzero. On the other hand, the converse statement also holds: if the polynomial $P(t) = t^3 - 3t + a$ has three distinct nonzero real roots, the equalities in the problem statement are satisfied by these roots $x, y, z$.\n\nNow for $a > 2$, one would have $t^3 + a > t^3 + 2 \\geq 3t$ for positive $t$, thus the polynomial $P(t)$ would have no positive root, so it could not have three real roots. Similarly for $a < -2$, there would be no negative root and there could not be three real roots. Moreover for $a = 2$ and $a = -2$, the polynomial $P(t)$ would have a double root of $t = 1$ and $t = -1$, respectively. Finally for $a = 0$, one of the roots of $P(t)$ would be $0$.\n\nOutside all these cases, one has $-2 < a < 0$ or $0 < a < 2$. Indeed for $0 < a < 2$, one has $P(0) = a > 0$ and $P(1) = a - 2 < 0$, hence there is a root in each of the intervals $(-\\infty, 0)$, $(0, 1)$ and $(1, \\infty)$. Similarly for $-2 < a < 0$, there is a root in each of the intervals $(-\\infty, -1)$, $(-1, 0)$ and $(0, \\infty)$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21229,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ for which $f(g(n)) - g(f(n))$ is independent of $n$ for any $g : \\mathbb{Z} \\to \\mathbb{Z}$.",
"options": [],
"answer": "See solution",
"solution": "First, observe that if $f(n) = n$, then $f(g(n)) - g(f(n)) = 0$. Therefore, the identity function satisfies the problem condition.\n\nIf there exists $n_0$ with $f(n_0) \\neq n_0$, consider the characteristic function $g$ defined by $g(f(n_0)) = 1$ and $g(n) = 0$ for $n \\neq f(n_0)$. For arbitrary $n \\neq f(n_0)$, we have:\n\n$$\n\\begin{align*}\nf(g(n)) - g(f(n)) &= f(g(n_0)) - g(f(n_0)) \\\\\nf(0) - g(f(n)) &= f(0) - g(f(n_0)) \\\\\ng(f(n)) &= 1 \\\\\nf(n) &= f(n_0)\n\\end{align*}\n$$\n\nNow, consider a similar function $g$ defined by $g(f(n_0)) = a$ and $g(n) = b$ for $n \\neq f(n_0)$, where $a, b$ are integers with $a \\neq b$ and $a, b \\neq f(n_0)$. We have chosen $a, b$ so that $f(a) = f(n_0) = f(b)$. Then:\n\n$$\n\\begin{align*}\nf(g(f(n_0))) - g(f(f(n_0))) &= f(g(n_0)) - g(f(n_0)) \\\\\nf(a) - g(f(f(n_0))) &= f(b) - g(f(n_0)) \\\\\ng(f(f(n_0))) &= g(f(n_0)) = a \\\\\nf(f(n_0)) &= f(n_0)\n\\end{align*}\n$$\n\nIn summary, $f(n) = f(n_0)$ for any $n \\neq f(n_0)$ and $f(f(n_0)) = f(n_0)$. Therefore, $f$ is a constant function. Now, check that a constant function satisfies the problem condition: If $f(n) = c$ for all $n$, then $f(g(n)) - g(f(n)) = c - g(c)$, which is independent of $n$.\n\nThus, the answers are the identity function $f(n) = n$ and the constant functions $f(n) = c$.\n\n$\\boxed{}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21230,
"subject": "Mathematics (Olympiad)",
"question": "Let $t$ be a real number such that $0 < t < \\frac{1}{2}$.\n\nWe wish to find a positive integer $n$ such that for every set $S$ of $n$ integers, denoted\n\n$$\ns_1 < s_2 < s_3 < \\dots < s_n,\n$$\n\nthere exist two elements $s_i$ and $s_j$ such that $|s_j - ms_i| \\leq t s_i$ for a non-negative integer $m$. In other words, $s_j$ is within $t s_i$ of a multiple of $s_i$.\n\n\n",
"options": [],
"answer": "See solution",
"solution": "If $s_{i+1} - s_i \\leq t s_i$ for some $1 \\leq i \\leq n-1$, then $s_{i+1}$ is within $t s_i$ of the first multiple of $s_i$, namely $1 \\times s_i$ ($m=1$). Hence, the inequality is satisfied.\n\nOtherwise, $s_{i+1} - s_i > t s_i \\implies s_{i+1} > (1 + t) s_i$ for all $1 \\leq i \\leq n-1$. That is,\n\n$$\ns_n > (1 + t) s_{n-1} > (1 + t)^2 s_{n-2} > (1 + t)^3 s_{n-3} > \\dots > (1 + t)^{n-1} s_1.\n$$\n\nThis gives\n\n$$\ns_n > (1 + t)^{n-1} s_1 \\implies s_1 < \\frac{1}{(1 + t)^{n-1}} s_n.\n$$\n\nFinally, by Bernoulli's inequality, for $n$ large enough,\n\n$$\n\\frac{1}{t} < (1 + t)^{n-1} \\implies \\frac{1}{(1 + t)^{n-1}} \\leq t.\n$$\n\nHence, $s_1 - 0 s_n < t s_n$, that is, $s_1$ is within $t s_n$ of the zero-th multiple of $s_n$ ($m=0$). Thus, the inequality is satisfied. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21231,
"subject": "Mathematics (Olympiad)",
"question": "Demuestra que:\n\n$$\nEB = \\binom{2n}{n}\n$$\n\nEsto puede interpretarse, por ejemplo, como el número de formas de elegir $n$ elementos de un conjunto de $2n$ elementos, o como el número de caminos en una cuadrícula $n \\times n$ desde $(0, 0)$ hasta $(n, n)$, donde cada paso es horizontal o vertical.",
"options": [],
"answer": "See solution",
"solution": "Si empezamos \"desde el final\", notemos que\n\n$$\n\\frac{(2n)!}{3 \\cdot 5 \\cdot 7 \\cdots (2n-1)} = (2n) \\cdot (2n-2) \\cdots 4 \\cdot 2 = 2^n \\cdot n!\n$$\n\ny como claramente $ET = 2^{2n}$ porque cada una de las $2n$ bombillas puede estar, independientemente de las demás, encendida o apagada, el problema se reduce a demostrar que\n\n$$\nEB = 2^{2n} \\cdot \\frac{3 \\cdot 5 \\cdot 7 \\cdots (2n-1)}{2^n \\cdot n!} = 2^{2n} \\cdot \\frac{(2n)!}{(2^n \\cdot n!)^2} = \\frac{(2n)!}{(n!)^2} = \\binom{2n}{n}\n$$\n\nLa demostración de que $EB = \\binom{2n}{n}$ puede hacerse de distintas maneras. Una de ellas es considerando el binomio $(1+x)^{2n} = (1+x)^n \\cdot (1+x)^n$. El coeficiente de $x^n$ es por una parte igual a $\\binom{2n}{n}$, y por otra igual a la suma de los productos de los coeficientes respectivos $\\binom{n}{m}$ y $\\binom{n}{n-m}$ de $x^m$ y $x^{n-m}$ en cada $(1+x)^n$.\n\nTambién puede verse como el número de caminos en una cuadrícula de $n \\times n$, desde $(0, 0)$ hasta $(n, n)$, en la que cada desplazamiento lleva desde $(x, y)$ bien a $(x + 1, y)$ bien a $(x, y + 1)$. El número total de caminos, cada uno de los cuales ha de tener $2n$ desplazamientos de los que $n$ son en horizontal y $n$ en vertical, es $\\binom{2n}{n}$. Si consideramos la diagonal descendente formada por los puntos $(x, y)$ tales que $x + y = n$, cada camino ha de pasar por exactamente uno de ellos. Los que pasan por $(m, n - m)$ tienen, en sus primeros $n$ desplazamientos, $m$ horizontales y $n - m$ verticales, y en sus siguientes $n$ desplazamientos, $n - m$ horizontales y $m$ verticales, para un total de $\\binom{n}{m} \\binom{n}{n-m} = \\binom{2n}{n}$ caminos pasando por ese punto. Sumando las contribuciones de cada elemento de esta diagonal se halla la relación pedida.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21232,
"subject": "Mathematics (Olympiad)",
"question": "a) Prove that for any real numbers $a$ and $b$, the following inequality holds:\n\n$$\n(a^2 + 1)(b^2 + 1) + 50 \\ge 2(2a + 1)(3b + 1).\n$$",
"options": [],
"answer": "See solution",
"solution": "a) The given condition rewrites as $$(ab - 6)^2 + (a - 2)^2 + (b - 3)^2 \\ge 0$$, which is obviously true.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21233,
"subject": "Mathematics (Olympiad)",
"question": "A positive integer $k$ is called *nice* if there exist positive integers $a_1, a_2, \\dots, a_{2021}$ such that\n\n1. $a_i - a_j$ is not divisible by $2023$ for any pair $1 \\le i < j \\le 2021$,\n2. For any index $i$ there exists an index $j$ such that $a_i - k a_j$ is divisible by $2023$.\n\nFind the largest nice number $k$ which is even and less than $100$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Answer: $92$.\n\nSet $n = 2023 = 7 \\times 17^2$ and denote by $\\mathbb{Z}/n\\mathbb{Z}$ the set of all residues modulo $n$. For any natural number $k$, denote by $k: \\mathbb{Z}/n\\mathbb{Z} \\to \\mathbb{Z}/n\\mathbb{Z}$ the map defined by $a$ (mod $n$) $\\mapsto k a$ (mod $n$).\n\nAssume that $k$ is nice and $a_1, a_2, \\dots, a_{n-2}$ are integers satisfying the two conditions. Then $A \\subseteq kA$ by condition (2), where $A$ denotes the set\n\n$$\n\\{a_1, a_2, \\dots, a_{n-2} \\pmod{n}\\}.\n$$\n\nTherefore, $|A| = |kA| = n-2$ by condition (1) and so $A = kA$. Since $kA \\subseteq k\\mathbb{Z}/n\\mathbb{Z}$ and $|k\\mathbb{Z}/n\\mathbb{Z}| = n/(n, k)$, we have $(n, k) = 1$, whence the map $k: \\mathbb{Z}/n\\mathbb{Z} \\to \\mathbb{Z}/n\\mathbb{Z}$ is invertible. Thus, by setting $B = \\mathbb{Z}/n\\mathbb{Z} \\setminus A$, we get $kB = B$ and $|B| = 2$. More precisely, if $B = \\{b_1, b_2\\}$ then $(k b_1, k b_2) = (b_1, b_2)$ or $(k b_1, k b_2) = (b_2, b_1)$. Thus $(k^2 - 1) b \\equiv 0 \\pmod{n}$ for some $b \\in B$ satisfying $b \\ne 0 \\pmod{n}$ and so $(k^2 - 1, n) \\ne 1$.\n\nNow assume that $(k, n) = 1 \\ne (k^2 - 1, n)$. By setting $b = n / (k^2 - 1, n)$, we choose $B = \\{0, b\\}$ if $k b \\equiv b \\pmod{n}$ and $B = \\{b, k b\\}$ if $k b \\ne b \\pmod{n}$. Then the elements of $A = \\mathbb{Z}/n\\mathbb{Z} \\setminus B$ satisfy the two given conditions.\n\nThus, it suffices to find the largest even integer $k$ such that $(k, n) = 1 \\ne (k^2 - 1, n)$ and $k \\le 100$. It is now straightforward to check that $92$ is nice while $100$, $98$, $96$, $94$ are not.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21234,
"subject": "Mathematics (Olympiad)",
"question": "Let $q = 4p + 1$ and let $a \\circ b$ be the remainder of $a$ modulo $b$.\n\nShow that the last digits of the numbers $10^k \\circ q$, for $k \\in \\mathbb{N}$, attain all values from 0 to 9. That is, prove that all digits from 0 to 9 occur among the last digits of the numbers $\\left\\lfloor \\frac{10^k}{q} \\right\\rfloor$, which are the same as the digits in the decimal representation of $\\frac{1}{q}$.",
"options": [],
"answer": "See solution",
"solution": "Let $S$ be the set of all non-zero biquadratic residues modulo $q$, i.e., the set of all $0 < t < q$ such that the congruence $x^4 \\equiv t \\pmod{q}$ has a solution. Then $S$ has $p$ elements. Since $-1$ is a quadratic residue modulo $q$, the $2p$ non-zero quadratic residues modulo $q$ can be grouped in pairs $(s, -s)$, and among their squares (which are precisely the biquadratic residues) there are exactly $p$ distinct ones.\n\nWe will show that each element of $S$ is of the form $10^k \\circ q$ for some $k$. Let $d$ be the index of $q$ modulo 10. Since $d \\mid \\varphi(q) = 4p$, we have $d = p$, $d = 2p$, or $d = 4p$.\n\nConsider the case $d = p$ first. Let $0 \\leq k < p$; then there is a $0 \\leq j \\leq 3$ such that $k + jq$ is a multiple of 4 and the number\n\n$$\n10^k \\equiv \\left\\lfloor 10^{\\frac{k+jp}{4}} \\right\\rfloor^4 \\pmod{q}\n$$\n\nis a biquadratic residue. Since the numbers $10^k \\circ q$ for $0 \\leq k < p$ are pairwise distinct and all of them are biquadratic residues, they must coincide with the elements of $S$. The cases $d = 2p$ and $d = 4p$ are treated analogously.\n\nLet $u$ be an arbitrary digit. Let $0 \\leq j \\leq 3$ be such that $u + jq$ ends in 0, 1, 5, or 6. Since $q > 10^9$, we have $\\sqrt[4]{(j+1)q-1} - \\sqrt[4]{u+jq} > 6$. It follows that the interval $[u+jq, (j+1)q]$ contains at least six fourth powers. Therefore, it contains at least one fourth power $x^4$ that ends in the same digit as $u+jq$. Let $s = x^4 \\circ q$. Then $x^4 = s + jq$ and $10 \\mid x^4 - (u+jq) = s - u$, i.e., $s$ ends in $u$, as needed.",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 21235,
"subject": "Mathematics (Olympiad)",
"question": "Даден е паралелограмот $ABCD$. Симетралите на неговите внатрешни агли се сечат во точките $P, Q, R$ и $S$.\n\nа) Докажи дека четириаголникот $PQRS$ е правоаголник.\n\nб) Докажи дека дијагоналата на тој правоаголник е еднаква со разликата од соседните страни на паралелограмот $ABCD$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Нека е даден паралелограмот $ABCD$ и нека симетралите на неговите внатрешни агли се сечат во точките $P, Q, R$ и $S$ (види цртеж).\n\nа) Очигледно е дека симетралите на аглите кај темињата $A$ и $C$ се меѓусебно паралелни, а исто така и симетралите кај темињата $B$ и $D$. Значи четириаголникот $PQRS$ е паралелограм. Од $\\triangle APD$ имаме:\n\n$$\n\\frac{\\alpha}{2} + \\angle P + \\frac{\\delta}{2} = 180^{\\circ}\n$$\n\n$$\n\\frac{\\alpha + \\delta}{2} + \\angle P = 180^{\\circ}\n$$\n\n$$\n90^{\\circ} + \\angle P = 180^{\\circ}\n$$\n\n$$\n\\angle P = 90^{\\circ}\n$$\n\nСледи дека $\\angle QPS = 90^{\\circ}$ како накрсни агли со $\\angle P$, од што следува дека четириаголникот $PQRS$ е правоаголник.\n\nб) Триаголниците $APL$ и $APD$ се складни, бидејќи имаат заедничка страна $AP$ и по 2 еднакви агли:\n\n$$\n\\angle APL = \\angle APD = 90^{\\circ}; \\quad \\angle PAL = \\angle PAD = \\frac{\\alpha}{2} \\Rightarrow \\overline{AL} = \\overline{AD}, \\quad \\overline{PL} = \\overline{PD}\n$$\n\nАналогно се покажува дека $\\overline{RB} = \\overline{RK}$, $\\overline{BC} = \\overline{KC}$. Тогаш, отсечката $\\overline{PR}$ е средна линија за паралелограмот $LBKD$, од каде што следува дека $\\overline{PR} = \\overline{LB} = \\overline{AB} - \\overline{AL} = \\overline{AB} - \\overline{AD}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21236,
"subject": "Mathematics (Olympiad)",
"question": "Consider two positive integers $n$ and $d$. Let $S_n(d)$ be the set of ordered tuples $(x_1, \\dots, x_d)$ such that:\n\n1. $x_i \\in \\{1, 2, \\dots, n\\}$ for all $1 \\leq i \\leq d$;\n2. $x_i \\neq x_{i+1}$ for all $1 \\leq i \\leq d-1$;\n3. There do not exist indices $1 \\leq i < j < k < l \\leq d$ such that $x_i = x_k$ and $x_j = x_l$.\n\n(a) Find the number of elements of $S_3(5)$.\n\n(b) Prove that $S_n(d) \\neq \\emptyset$ if and only if $d \\leq 2n-1$.",
"options": [],
"answer": "See solution",
"solution": "(a) Call a tuple satisfying the conditions *nice*. To calculate $|S_3(5)|$, we count the number of tuples $(a, b, c, d, e)$ with $a, b, c, d, e \\in \\{1, 2, 3\\}$ satisfying conditions ii) and iii).\n\n- If the first 3 terms are distinct, suppose $(a, b, c) = (1, 2, 3)$. For $(1, 2, 3, d, e)$, we must have $d=2$ and $e=1$. There is 1 such tuple for each permutation of $(a, b, c)$, so $3! = 6$ ways, giving 6 nice tuples.\n- If two of the first 3 terms are equal, consider $(a, b, c) = (1, 2, 1)$. For $(1, 2, 1, d, e)$, we must have $d=3$ and $e=1$, giving 1 nice tuple. There are $3 \\times 2 = 6$ ways to choose $(a, b, c)$ in this pattern, so 6 nice tuples.\n\nThus, there are $6 + 6 = 12$ nice tuples, so $|S_3(5)| = 12$.\n\n(b) We show that $S_n(d) \\neq \\emptyset$ if and only if $d \\leq 2n - 1$.\n\nFor $d = 2n - 1$, consider the tuple:\n\n$$1, 2, 3, \\dots, n-1, n, n-1, \\dots, 3, 2, 1$$\n\nThis tuple satisfies all conditions, so $S_n(2n-1) \\neq \\emptyset$. For all $1 \\leq d < 2n - 1$, $S_n(d)$ is also non-empty.\n\nTo show $d = 2n - 1$ is maximal, we prove that for $d = 2n$, there is no nice tuple. We use induction:\n\n- For $n = 1$, $d = 2$, the statement is true.\n- Assume for all $1 \\leq k \\leq n$, there is no nice tuple of length $2k$. For $k = n+1$, suppose a nice tuple of length $2(n+1)$ exists: $(x_1, x_2, \\dots, x_{2n+2})$. Let $S$ be the number of times $n+1$ appears.\n - If $S = 0$, the tuple uses only $n$ values, contradiction.\n - If $S = 1$, if $n+1$ is first or last, remove it to get a nice tuple of length $2n+1$ with values at most $n$, contradiction. Otherwise, $n+1$ is between $u$ and $v$ in $(\\ldots, u, n+1, v, \\ldots)$. If $u \\neq v$, remove $n+1$ as above; if $u = v$, remove $n+1$ with $u$ or $v$ (ensuring no consecutive equal numbers), resulting in a tuple of length $2n$, contradiction.\n\nThus, $d \\leq 2n-1$ is necessary and sufficient.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21237,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle. Point $M$ and $N$ lie on sides $AC$ and $BC$ respectively such that $MN \\parallel AB$. Points $P$ and $Q$ lie on sides $AB$ and $CB$ respectively such that $PQ \\parallel AC$. The incircle of triangle $CMN$ touches segment $AC$ at $E$. The incircle of triangle $BPQ$ touches segment $AB$ at $F$. Line $EN$ and $AB$ meet at $R$, and lines $FQ$ and $AC$ meet at $S$. Given that $AE = AF$, prove that the incenter of triangle $AEF$ lies on the incircle of triangle $ARS$.\n\n",
"options": [],
"answer": "See solution",
"solution": "It is clear that there is a homothety $\\mathbf{H}_1$ centered at $C$ sending triangle $CMN$ to $CAB$, and that images of $M, E, N$, and line $EN$ under $\\mathbf{H}_1$ are $A, S_1, B$, and line $S_1B$. In particular, $BS_1 \\parallel RE$ with $AB/AR = AS_1/AE$. In exactly the same way, we can prove that $CR_1 \\parallel SF$ with $AC/AS = AR_1/AF$. By equal tangents, we have $AS_1 = AR_1$. By the given condition, $AE = AF$. It follows that\n\n$$\n\\frac{AB}{AR} = \\frac{AS_1}{AE} = \\frac{AR_1}{AF} = \\frac{AC}{AS},\n$$\n\nimplying that $BC \\parallel RS$. Thus, there is a homothety $\\mathbf{H}$ centered at $A$ sending triangle $ABC$ to triangle $ARS$. It is clear that the images of $S_1, R_1, \\omega_1$, and $I_1$ under $\\mathbf{H}$ are $E, F, \\omega$, and $I$, respectively. Thus, $I$ lies on $\\omega$, which is what we wish to show, if and only if $I_1$ lies on $\\omega_1$. But the latter claim holds because the midpoint of minor arc $\\widehat{R_1S_1}$ on $\\omega_1$ is the incenter of triangle $AR_1S_1$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21238,
"subject": "Mathematics (Olympiad)",
"question": "Given a positive integer $k$. Let integers $a, b$, not both zero, satisfy $0 \\leq a < k$, $0 \\leq b < k+1$. Define the sequence $\\{T_n\\}_{n \\geq k}$ as follows:\n\n- $T_k = a$\n- $T_{k+1} = b$\n- For any $n \\geq k+2$,\n $$\n T_n \\equiv T_{n-1} + T_{n-2} \\pmod{n}, \\quad 0 \\leq T_n < n.\n $$\n\nConcatenate the decimal representations of $T_k, T_{k+1}, \\dots$ after the decimal point to form an infinite decimal $x = 0.\\overline{T_k T_{k+1} \\dots}$. For example, when $k=66$, $a=5$, $b=20$, we have $T_{66} = 5$, $T_{67} = 20$, $T_{68} = 25$, $T_{69} = 45$, $T_{70} = 0$, $T_{71} = 45$, $T_{72} = 45$, $T_{73} = 17$, $\\dots$, thus $x = 0.52025450454517\\dots$.\n\nProve that $x$ is irrational.",
"options": [],
"answer": "See solution",
"solution": "First, we prove that $\\{T_n\\}_{n=k}^{\\infty}$ is unbounded. If not, let $M = \\max_{n \\geq k} \\{T_n\\}$; then for any $n > 2M$, since $T_{n-1} \\leq M$ and $T_{n-2} \\leq M$, we have $T_n = T_{n-1} + T_{n-2}$. By boundedness, the only possibility is $T_{n-1} = T_{n-2} = 0$, which inductively leads to $T_{k+1} = T_k = 0$, a contradiction.\n\nLet $m_0$ be the number of digits of $k$. We further prove that for any $m > m_0$, there exists $n$ such that $10^{m-1} \\leq T_n < 10^m$. By unboundedness, there exists minimal $T_n \\geq 10^{m-1}$; then $n > k+1$ and $T_{n-1}, T_{n-2} < 10^{m-1}$, so $T_n \\leq T_{n-1} + T_{n-2} < 2 \\cdot 10^{m-1} < 10^m$.\n\nFor convenience, denote $p \\star q$ as the number obtained by concatenating positive integer $p$ with non-negative integer $q$. By contradiction, assume $0.T_k \\star T_{k+1} \\star \\dots = 0.a_1 \\dots a_s b_1 \\dots b_t b_1 \\dots b_t \\dots$ (i.e., this repeating decimal has ultimate period length $t$), where $a_i, b_j \\in \\{0, 1, \\dots, 9\\}$. Take $n$ sufficiently large so that $T_n$ appears to the right of $a_s$; then for any $r \\geq n$, $T_r \\star T_{r+1} \\star T_{r+2} \\dots$ is a number with period $wt$, where $w$ is any positive integer. Take $m = \\nu t$ with $\\nu$ sufficiently large so that $10^{m-1} > T_i$ for all $i \\leq n$; then there exists minimal $\\ell$ such that $T_\\ell$ has exactly $m$ digits, clearly $\\ell > n$.\n\nIf $T_{\\ell-1}$ has more than $m$ digits, then there must exist $j < \\ell - 1$ where $T_j$ has $m-1$ digits, leading to some number before $T_\\ell$ having $m$ digits, a contradiction.\n\nClearly $k \\ne m$, so $k < m$. But $m$ is the period length, thus the last $k$ digits of $T_{\\ell-1}$ and $T_\\ell$ must be identical, which implies $T_{\\ell-2}$ has at least $m$ digits. Therefore, there exists a number before $T_{\\ell-1}$ with $m$ digits, another contradiction.\n\nIn conclusion, the assumption by contradiction is invalid, and the original proposition is proved. $\\boxed{}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21239,
"subject": "Mathematics (Olympiad)",
"question": "By one move, the chess King moves to a cell adjacent by a side or by a vertex.\n\nHow many pleasant moves can there be at most in a traversal of the board, where a move is called *pleasant* if it decreases the distance to the center of the board?\n\n\n\n",
"options": [],
"answer": "See solution",
"solution": "Let us prove that there must have been at least 20 unpleasant moves (and thus the number of pleasant moves cannot exceed 44). Let's place numbers in the cells as shown in the first image; 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\nEvery move from a cell with number 1 does not decrease the distance to the center and is therefore unpleasant — there are 4 such moves. A move from a cell with number 2 can be pleasant only if it goes to a cell with number 1. But there are eight cells with number 2 and only four with number 1, so at least four moves from cells with number 2 will be unpleasant.\n\nNow consider moves leading to the 32 cells with numbers not less than 6. Note that these moves cannot originate from cells with numbers 1 or 2, meaning they weren't accounted for in the previous reasoning. Such a move can only be pleasant if it comes from a cell with a number not less than 7; however, there are only 20 such cells. Therefore, among these moves, there are at least $32 - 20 = 12$ unpleasant ones, bringing the total number of unpleasant moves to no fewer than $4 + 4 + 12 = 20$.\n\nAn example of a traversal with 44 pleasant moves is shown in the second image.\n\nEssentially, in the final part of the proof, we've shown that among moves leading to cells marked in green in the first image, there are at least three unpleasant ones. This can be proven in various ways, for example, through a brief case analysis.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21240,
"subject": "Mathematics (Olympiad)",
"question": "Find all triples of positive integers $a, b, c$ such that\n\n$$\na + (a, b) = b + (b, c) = c + (c, a),\n$$\n\nwhere $(x, y)$ denotes the greatest common divisor of integers $x$ and $y$.",
"options": [],
"answer": "See solution",
"solution": "Suppose $a, b, c$ have a common divisor. We can divide by it to obtain integers $a_1, b_1, c_1$ with greatest common divisor $1$. Since $a_1 + (a_1, b_1) = b_1 + (b_1, c_1)$, $(b_1, c_1)$ is divisible by $(a_1, b_1)$. But $((a_1, b_1), (b_1, c_1)) = 1$, so $(a_1, b_1) = 1$. Similarly, the numbers are pairwise coprime, so $a_1 + 1 = b_1 + 1 = c_1 + 1$, implying $a = b = c$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21241,
"subject": "Mathematics (Olympiad)",
"question": "Suppose there is no tower centre at point $A$. Let the distance between $A$ and the closest tower centre to the right of $A$ be $\\alpha d$, where $0 < \\alpha < 1$. An observer at $A$ can completely see exactly $m$ towers to the right of $A$ and $m'$ towers to the left of $A$.\n\n\n\nGiven that, from similar triangles, $d \\ge 2BY$ and $OY = 16BY$.\n\nNow, $AY = \\alpha d + (m - 2)d + BY \\ge (2\\alpha + 2(m - 2) + 1)BY = \\frac{2\\alpha + 2m - 3}{16} OY$.\nSince $\\frac{AY}{OY} < 1$, we have $m < 9.5 - \\alpha$. Similarly, $m' < 9.5 - (1 - \\alpha) = 8.5 + \\alpha$.\n\nIf $\\alpha = \\frac{1}{2}$, then $m \\le 8$ and $m' \\le 8$.\nIf $\\alpha < \\frac{1}{2}$, then $m \\le 9$ and $m' \\le 8$.\nIf $\\alpha > \\frac{1}{2}$, then $m \\le 8$ and $m' \\le 9$.\n\nShow that these upper bounds on $m$ are attainable by finding an acceptable value for $d$ that allows the following configurations.\n\n\n\nAs above, from similar triangles, $d = 2BY$ and $OY = 16BY$. We have:\n\n$$\n\\begin{aligned}\nOY^2 &= OA^2 + AY^2 \\\\\n256(d/2)^2 &= 256 + d^2(\\alpha + (6 \\text{ or } 7) + \\frac{1}{2})^2 \\\\\n256d^2 &= 1024 + d^2(2\\alpha + (13 \\text{ or } 15))^2\n\\end{aligned}\n$$\n\nWith $1 > \\alpha \\ge \\frac{1}{2}$, $256d^2 = 1024 + d^2(2\\alpha + 13)^2$, so $d^2 = \\frac{1024}{256 - (2\\alpha + 13)^2}$.\n\nWith $0 < \\alpha < \\frac{1}{2}$, $256d^2 = 1024 + d^2(2\\alpha + 15)^2$, so $d^2 = \\frac{1024}{256 - (2\\alpha + 15)^2}$.\n\nIn each case, there is a value for $d > 2$. Thus, it is possible for $m$ to attain its upper bounds. Similarly, the upper bounds for $m'$ are attainable.\n\nWhat is the maximum number of completely visible towers?",
"options": [],
"answer": "See solution",
"solution": "The maximum number of completely visible towers is $8 + 8 = 16$ if $\\alpha = \\frac{1}{2}$, and $8 + 9 = 17$ if $\\alpha \\ne \\frac{1}{2}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21242,
"subject": "Mathematics (Olympiad)",
"question": "Equation\n$$\n\\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_m} - \\frac{1}{y_1} - \\frac{1}{y_2} - \\dots - \\frac{1}{y_n} = \\frac{577}{408}\n$$\nhas an infinite number of natural solutions $x_1, x_2, \\dots, x_m, y_1, y_2, \\dots, y_n$ for some non-negative integers $m, n$. Prove that an equation\n$$\n\\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_k} - \\frac{1}{y_1} - \\frac{1}{y_2} - \\dots - \\frac{1}{y_l} = \\frac{577}{408}\n$$\nhas a solution in natural numbers $x_1, x_2, \\dots, x_k, y_1, y_2, \\dots, y_l$ for some non-negative integers $k < m$ and $l < n$.",
"options": [],
"answer": "See solution",
"solution": "We will prove a more general statement: if an equation\n$$\n\\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_m} - \\frac{1}{y_1} - \\frac{1}{y_2} - \\dots - \\frac{1}{y_n} = a\n$$\nhas an infinite number of solutions in natural numbers $x_1, x_2, \\dots, x_m, y_1, y_2, \\dots, y_n$ for some $a \\neq 0$, then an equation\n$$\n\\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_k} - \\frac{1}{y_1} - \\frac{1}{y_2} - \\dots - \\frac{1}{y_l} = a\n$$\nhas a solution in natural numbers $x_1, x_2, \\dots, x_k, y_1, y_2, \\dots, y_l$ for some non-negative integers $k < m$ and $l < n$.\n\nWe prove the claim by induction on $m+n$. When $m+n=1$, if $a>0$ we have $m=1$ and $n=0$, so the equation becomes $\\frac{1}{x_1} = a$ and has at most one solution. Similarly for $a<0$.\n\nSuppose our claim is true for $m+n < t$, and that the equation\n$$\n\\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_m} - \\frac{1}{y_1} - \\frac{1}{y_2} - \\dots - \\frac{1}{y_n} = a\n$$\nwhere $m+n = t$, has an infinite number of solutions. Without loss of generality, assume $a > 0$; then, we can consider only those solutions for which $x_1 \\le x_2 \\le \\dots \\le x_m$ (there are infinitely many of these as well). Since $a \\le \\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_m} \\le \\frac{m}{a}$, we have $x_1 \\le \\frac{m}{a}$. Therefore, $x_1$ takes only finitely many values, so for at least one value $x_1 = c$, the equation\n$$\n\\frac{1}{x_2} + \\frac{1}{x_3} + \\dots + \\frac{1}{x_m} - \\frac{1}{y_1} - \\frac{1}{y_2} - \\dots - \\frac{1}{y_n} = a - \\frac{1}{c}\n$$\nhas an infinite number of solutions. If $a - \\frac{1}{c} \\ne 0$, then by the inductive assumption, the equation\n$$\n\\frac{1}{x_2} + \\frac{1}{x_3} + \\dots + \\frac{1}{x_k} - \\frac{1}{y_1} - \\frac{1}{y_2} - \\dots - \\frac{1}{y_l} = a - \\frac{1}{c}\n$$\nhas a solution for some $k < m$ and $l < n$. Then the following equation\n$$\n\\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_k} - \\frac{1}{y_1} - \\frac{1}{y_2} - \\dots - \\frac{1}{y_l} = a\n$$\nhas a solution as well.\n\nIt remains to consider the case when $a - \\frac{1}{c} = 0$. For this case, a single value of $x_1$ for which the equation\n$$\n\\frac{1}{x_2} + \\frac{1}{x_3} + \\dots + \\frac{1}{x_m} - \\frac{1}{y_1} - \\frac{1}{y_2} - \\dots - \\frac{1}{y_n} = a - \\frac{1}{x_1} = 0\n$$\nhas an infinite number of solutions, is $x_1 = \\frac{1}{a}$.\n\nIt is clear that $m > 0$ and $n > 0$, and the solution $x_1 = \\frac{1}{a}$ is the desired one.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21243,
"subject": "Mathematics (Olympiad)",
"question": "You are given a set of $n$ not necessarily distinct numbers $\\{a_1, a_2, \\dots, a_n\\}$ (some may be equal). Consider all $2^n - 1$ nonempty subsets of this set, and for each subset, find the sum of its elements. What is the largest number of these sums that could be equal to $1$? For example, for the set $\\{-1, 2, 2\\}$, the 7 nonempty subsets are $\\{-1\\}$, $\\{2\\}$, $\\{2\\}$, $\\{-1, 2\\}$, $\\{-1, 2\\}$, $\\{2, 2\\}$, and $\\{-1, 2, 2\\}$; among them, exactly two subsets have sum $1$.",
"options": [],
"answer": "See solution",
"solution": "An example where equality is reached is $(1, 0, 0, \\dots, 0)$. \n\nSuppose at least $2^{n-1} + 1$ subsets have sum $1$. Clearly, not all elements are $0$; otherwise, all sums would be $0$. Without loss of generality, let $a_1 \\neq 0$. Divide all subsets into $2^{n-1}$ pairs, where each pair differs only by the presence of $a_1$. The sums in each pair differ by $a_1$, so we cannot have more than $2^{n-1}$ subsets with sum $1$ overall.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21244,
"subject": "Mathematics (Olympiad)",
"question": "Find a set $S$ of distinct positive integers such that for any two distinct nonempty subsets $A$ and $B$ of $S$, the numbers\n$$\n\\sum_{x \\in A} x \\quad \\text{and} \\quad \\sum_{x \\in B} x\n$$\nare two coprime composite integers.\n\nHere, $\\sum_{x \\in X} x$ denotes the sum of all elements of a finite set $X$, and $|X|$ denotes the cardinality of $X$.",
"options": [],
"answer": "See solution",
"solution": "**Proof**\n\nLet $f(X)$ be the average of elements of the finite set $X$.\n\nFirst, take $n$ distinct primes $p_1, p_2, \\dots, p_n$ all greater than $n$. Consider the set\n$$\nS_1 = \\left\\{ \\prod_{j=1}^n \\frac{1}{p_j} : 1 \\leq j \\leq n \\right\\}.\n$$\nWe claim that for any two distinct nonempty subsets $A$ and $B$ of $S_1$, $f(A) \\neq f(B)$.\n\nSuppose $\\prod_{i=1}^n p_i \\in A$ and $\\prod_{i=1}^n p_i \\notin B$ (without loss of generality). Every element of $B$ is divisible by $p_1$, so $p_1 \\mid n! f(B)$. But $A$ has exactly one element not divisible by $p_1$, so $n! f(A)$ is not divisible by $p_1$ (since $p_1 > n$). Therefore, $n! f(A) \\neq n! f(B)$, so $f(A) \\neq f(B)$.\n\nNext, let $S_2 = \\{ n x : x \\in S_1 \\}$. Then $f(A)$ and $f(B)$ are different positive integers for different nonempty subsets $A, B$ of $S_2$.\n\nIndeed, there exist two sets $A_1, B_1$ (nonempty subsets of $S_1$) such that $f(A) = n! f(A_1)$ and $f(B) = n! f(B_1)$. Since $f(A_1) \\neq f(B_1)$, we have $f(A) \\neq f(B)$, and $f(A), f(B)$ are positive integers.\n\nLet $K$ be the largest element of $S_2$. Consider $S_3 = \\{ K! x + 1 : x \\in S_2 \\}$. For any two distinct subsets $A, B$ of $S_3$, $f(A)$ and $f(B)$ are coprime integers greater than $1$.\n\nIndeed, there exist $A_1, B_1$ (nonempty subsets of $S_2$) such that $f(A) = K! f(A_1) + 1$ and $f(B) = K! f(B_1) + 1$. These are distinct integers greater than $1$. If they share a common prime divisor $p$, then $p \\mid K! |f(A_1) - f(B_1)|$. Since $1 \\leq |f(A_1) - f(B_1)| \\leq K$, $p \\leq K$, so $p \\mid K! f(A_1)$, and thus $p \\mid 1$, a contradiction.\n\nFinally, let $L$ be the largest element of $S_3$. Define $S_4 = \\{ L! + x : x \\in S_3 \\}$. For any two distinct nonempty subsets $A, B$ of $S_4$, $f(A)$ and $f(B)$ are composite and coprime.\n\nIndeed, there exist $A_1, B_1$ (nonempty subsets of $S_3$) such that $f(A) = L! + f(A_1)$ and $f(B) = L! + f(B_1)$. Both are distinct integers greater than $1$. Since $L$ is the largest element of $S_3$, $f(A_1) \\mid L!$ and $f(A_1) \\mid f(A)$, so $f(A)$ is composite (since $f(A_1) < f(A)$). Similarly for $f(B)$. If they share a common prime divisor $p$, then $p \\mid L! |f(A_1) - f(B_1)|$. Since $1 \\leq |f(A_1) - f(B_1)| \\leq L$, $p \\leq L$, so $p \\mid f(A_1)$ and $p \\mid f(B_1)$, contradicting that $f(A_1)$ and $f(B_1)$ are coprime. This completes the proof.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21245,
"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 $n$, $\\tau(n)$, and $\\varphi(n)$ is the arithmetic mean of the other two.",
"options": [],
"answer": "See solution",
"solution": "We have $\\tau(1) = \\varphi(1) = 1$, so $n = 1$ satisfies the given condition. Now assume $n > 1$. For such $n$, clearly $\\tau(n) \\leq n$ and $\\varphi(n) < n$. This means $n$ 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 $n$, the number $n/d$ is also a divisor. One of $d, n/d$ is less than or equal to $\\sqrt{n}$, so the set $\\{1, 2, \\dots, \\lfloor\\sqrt{n}\\rfloor\\}$ contains at least half of the divisors¹. Thus $\\frac{1}{2}\\tau(n) \\leq \\sqrt{n}$. We get\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 $\\varphi(n)$ in the remaining cases:\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| $\\tau(n) > \\frac{1}{2}n$? | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ |\n| $\\varphi(n)$ | 1 | 2 | 2 | 4 | 2 | 6 | 4 | 6 | 4 | 10 | 4 | 12 | 6 | 8 |\n| $\\tau(n) = \\frac{1}{2}(\\varphi(n) + n)$? | × | × | ✓ | | ✓ | | | | | | | | | |\n\nWe have $n = 4$ and $n = 6$ as 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 $\\varphi(n) \\leq \\frac{1}{2}n$. But then from (1), $\\tau(n) \\leq 0$, which is impossible. Therefore, $n$ must be odd. Then (1) implies $\\tau(n)$ must be odd as well, which means $n$ is a perfect square (of an odd number). Write the prime factorization of $n$ as\n\n$$\nn = p_1^{2\\alpha_1} \\cdots p_k^{2\\alpha_k}, \\quad k \\geq 1,\\ p_i \\geq 3,\\ \\alpha_i \\geq 1.\n$$\n\n¹ It contains exactly one half in case $n$ is not a perfect square.\n\nApplying the formulas for $\\tau(n)$ and $\\varphi(n)$, we rewrite (1) as\n\n$$\n(2\\alpha_1 + 1) \\cdots (2\\alpha_k + 1) = 2p_1^{2\\alpha_1-1}(p_1-1) \\cdots p_k^{2\\alpha_k-1}(p_k-1) - p_1^{2\\alpha_1} \\cdots p_k^{2\\alpha_k} = \\\\\n= p_1^{2\\alpha_1-1} \\cdots p_k^{2\\alpha_k-1}(2(p_1-1) \\cdots (p_k-1) - p_1 \\cdots p_k).\n$$\n\nThe right side is divisible by $p_1^{2\\alpha_1-1} \\cdots p_k^{2\\alpha_k-1}$, so the left side 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 all $p \\geq 3$ and $\\alpha \\geq 1$, $p^{2\\alpha-1} \\geq (2\\alpha + 1)$, with equality only for $p = 3$ and $\\alpha = 1$. To prove this, use induction on $\\alpha$: The case $\\alpha = 1$ is trivial (equality only for $p = 3$), and when $\\alpha$ increases by 1, the right side increases by 2, while the left side increases by more:\n\n$$\np^{2(\\alpha+1)-1} - p^{2\\alpha-1} = p^{2\\alpha-1}(p^2 - 1) > 2.\n$$\n\nTherefore, each factor on the left of (2) is greater or equal to the corresponding factor on the right. The only way to satisfy (2) is $k = 1$, $p_1 = 3$, $\\alpha_1 = 1$, i.e., $n = 9$. Indeed, $\\tau(9) = 3$ and $\\varphi(9) = 6$, so (1) holds for $n = 9$.\n\n**Answer:** The given condition is fulfilled for $n \\in \\{1, 4, 6, 9\\}$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21246,
"subject": "Mathematics (Olympiad)",
"question": "Consider a $100 \\times 100$ table, where each cell in row $a$ and column $b$ ($1 \\leq a, b \\leq 100$) is identified by the ordered pair $(a, b)$. Let $k$ be an integer such that $51 \\leq k \\leq 99$. A $k$-knight is a piece that moves one cell vertically or horizontally and $k$ cells in the other direction; that is, it moves from $(a, b)$ to $(c, d)$ such that $(|a-c|, |b-d|)$ is either $(1, k)$ or $(k, 1)$. The $k$-knight starts at cell $(1, 1)$ and performs several moves. A sequence of moves is a sequence of cells $(x_0, y_0) = (1, 1), (x_1, y_1), (x_2, y_2), \\ldots, (x_n, y_n)$ such that, for all $i = 1, 2, \\ldots, n$, $1 \\leq x_i, y_i \\leq 100$ and the $k$-knight can move from $(x_{i-1}, y_{i-1})$ to $(x_i, y_i)$. In this case, each cell $(x_i, y_i)$ is said to be reachable. For each $k$, find $L(k)$, the number of reachable cells.",
"options": [],
"answer": "See solution",
"solution": "Cell $(x, y)$ is directly reachable from another cell if and only if $x \\geq k+1$ or $x \\leq 100-k$ or $y \\geq k+1$ or $y \\leq 100-k$ ($*$). Therefore, the cells $(x, y)$ for which $101 - k \\leq x \\leq k$ and $101 - k \\leq y \\leq k$ are unreachable. Let $S$ be this set of unreachable cells in this square, namely the square of cells $(x, y)$, $101 - k \\leq x, y \\leq k$. If condition ($*$) is valid for both $(x, y)$ and $(x \\pm 2, y \\pm 2)$, then one can move from $(x, y)$ to $(x \\pm 2, y \\pm 2)$, if they are both in the table, with two moves: either $x \\leq 50$ or $x \\geq 51$; the same is true for $y$. In the first case, move $(x, y) \\to (x + k, y \\pm 1) \\to (x, y \\pm 2)$ or $(x, y) \\to (x \\pm 1, y + k) \\to (x \\pm 2, y)$. In the second case, move $(x, y) \\to (x - k, y \\pm 1) \\to (x, y \\pm 2)$ or $(x, y) \\to (x \\pm 1, y - k) \\to (x \\pm 2, y)$. \n\nHence, if the table is colored in two colors like a chessboard, if $k$ is even, every other move changes the color of the occupied cell, and all cells are potentially reachable; otherwise, only cells with the same color as $(1,1)$ can be visited. Therefore, if $k$ is even, the reachable cells consist of all cells except the center square defined by $101 - k \\leq x \\leq k$ and $101 - k \\leq y \\leq k$, that is, \n$$L(k) = 100^2 - (2k - 100)^2$$ \nIf $k$ is odd, then only half of the cells are reachable: the ones with the same color as $(1,1)$, and \n$$L(k) = \\frac{100^2 - (2k - 100)^2}{2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21247,
"subject": "Mathematics (Olympiad)",
"question": "Prove that, if $a$, $b$, $c$ are positive real numbers, then the equation $a x^2 + b x = c$ has at most one real solution.",
"options": [],
"answer": "See solution",
"solution": "Rewrite the equation as $a x^2 + b x = c$. Rearranging, $a x^2 + b x - c = 0$ is a quadratic in $x$ with $a > 0$, $b > 0$, $c > 0$. The discriminant is $D = b^2 - 4 a c$. If $D < 0$, there are no real solutions. If $D = 0$, there is one real solution. If $D > 0$, there are two real solutions, but since $a$, $b$, $c > 0$, the quadratic opens upwards and $c > 0$ shifts the graph down, so the equation can have at most one positive real solution. Thus, the equation has at most one real solution.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21248,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$ be positive real numbers. Prove that\n\n$$\n\\sqrt{a^3 b + a^3 c} + \\sqrt{b^3 c + b^3 a} + \\sqrt{c^3 a + c^3 b} \\geq \\frac{4}{3}(ab + bc + ca)\n$$",
"options": [],
"answer": "See solution",
"solution": "Without loss of generality, assume $a \\geq b \\geq c$.\n\n$$\na \\geq b \\geq c \\implies ab \\geq ac \\geq bc \\implies ab + ac \\geq ab + bc \\geq ac + bc \\implies \\sqrt{ab+ac} \\geq \\sqrt{bc+ba} \\geq \\sqrt{ac+bc}\n$$\n\n$$\n\\sqrt{a^3 b + a^3 c} + \\sqrt{b^3 c + b^3 a} + \\sqrt{c^3 a + c^3 b} = a\\sqrt{ab+ac} + b\\sqrt{bc+ba} + c\\sqrt{ca+cb}\n$$\n\nBy Chebyshev's inequality:\n\n$$\na\\sqrt{ab+ac} + b\\sqrt{bc+ba} + c\\sqrt{ca+cb} \\geq \\frac{a+b+c}{3} \\left( \\sqrt{ab+ac} + \\sqrt{bc+ba} + \\sqrt{ca+cb} \\right)\n$$\n\nNote that\n\n$$\n\\sqrt{ab+ac} = \\sqrt{a(b+c)},\\quad \\sqrt{bc+ba} = \\sqrt{b(c+a)},\\quad \\sqrt{ca+cb} = \\sqrt{c(a+b)}\n$$\n\nSo,\n\n$$\n\\frac{a+b+c}{3} \\left( \\sqrt{a(b+c)} + \\sqrt{b(c+a)} + \\sqrt{c(a+b)} \\right)\n$$\n\nBy the inequality $GM \\geq HM$ (geometric mean greater than or equal to harmonic mean),\n\n$$\n\\frac{a+b+c}{3} \\left( \\frac{2a(b+c)}{a+b+c} + \\frac{2b(c+a)}{a+b+c} + \\frac{2c(a+b)}{a+b+c} \\right) = \\frac{a+b+c}{3} \\cdot \\frac{4(ab+bc+ca)}{a+b+c} = \\frac{4}{3}(ab+bc+ca)\n$$\n\nThus, the original inequality holds.\n\n$\\Box$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21249,
"subject": "Mathematics (Olympiad)",
"question": "A necklace contains 2016 pearls, each of which has one of the colours black, green, or blue. In each step, we replace simultaneously each pearl with a new pearl, where the colour of the new pearl is determined as follows:\n\n- If the two original neighbours were of the same colour, the new pearl has their colour.\n- If the neighbours had two different colours, the new pearl has the third colour.\n\n(a) Is there such a necklace that can be transformed with such steps to a necklace of blue pearls if half of the pearls were black and half of the pearls were green at the start?\n\n(b) Is there such a necklace that can be transformed with such steps to a necklace of blue pearls if a thousand of the pearls were black at the start and the rest green?\n\n(c) Is it possible to transform a necklace that contains exactly two adjacent black pearls and 2014 blue pearls to a necklace that contains one green pearl and 2015 blue pearls?\n",
"options": [],
"answer": "See solution",
"solution": "(a) Since 2016 is divisible by 4, we can alternate two black and two green pearls. In the first step, all pearls are already replaced by blue pearls.\n\n(b) If we assign to each blue pearl the number 0, to each green pearl the number 1, and to each black pearl the number 2, then in each step the new colour of a pearl modulo 3 is equal to the negative sum of its two original neighbours. The new total sum of all colours modulo 3 can be calculated by multiplying the old total sum by 2 and changing the sign. But modulo 3, multiplication by $-2$ is equivalent to multiplication by $1$, so the total sum always remains the same modulo 3.\n\nFor a necklace with only blue pearls, the total sum is $0$. But for $1000$ black and $1016$ green pearls, it is $2000 + 1016 = 3016 \\equiv 1 \\pmod{3}$. Therefore, there does not exist an arrangement of $1000$ black and $1016$ green pearls that can be transformed into a necklace with only blue pearls using such steps.\n\n(c) Using the same assignment of numbers modulo 3, in each step the sum of all colours in even positions becomes the sum of the colours in odd positions, and vice versa. If these sums are $A$ and $B$ in the beginning, then at the end we still have these same two sums modulo 3, maybe with switched positions.\n\nBut in the beginning, we have sums $2$ and $2$ modulo 3, because both among the even and among the odd positions there is exactly one black pearl with value $2$, and otherwise only blue pearls with value $0$. However, at the end we are supposed to have sums $1$ and $0$ because one of the two sums is determined only by blue pearls with value $0$, and the other by exactly one green pearl with value $1$ and only blue pearls with value $0$ otherwise. Therefore, it is not possible.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21250,
"subject": "Mathematics (Olympiad)",
"question": "Determine all continuous increasing functions $f: [0, \\infty) \\to \\mathbb{R}$ satisfying\n$$\n\\int_{0}^{x+y} f(t) \\, dt = \\int_{0}^{x} f(t) \\, dt + \\int_{0}^{y} f(t) \\, dt\n$$\nfor all non-negative real numbers $x$ and $y$.",
"options": [],
"answer": "See solution",
"solution": "Every constant function satisfies the required condition. Conversely, rewrite the condition as\n$$\n\\int_{x}^{x+y} f(t) \\, dt = \\int_{0}^{y} f(t) \\, dt\n$$\nwhich implies $\\int_{0}^{y} f(t+x) \\, dt = \\int_{0}^{y} f(t) \\, dt$ for all non-negative $x$ and $y$.\n\nSince $f$ is increasing, $f(t+x) \\geq f(t)$ for all $t \\in [0, y]$ and $x \\geq 0$, so $\\int_{0}^{y} f(t+x) \\, dt \\geq \\int_{0}^{y} f(t) \\, dt$.\n\nThus, $\\int_{0}^{y} f(t+x) \\, dt = \\int_{0}^{y} f(t) \\, dt$ for all non-negative $x$ and $y$. By continuity, $f(x+y) = f(y)$ for all $x, y \\geq 0$, so $f(x) = f(0)$ for all $x \\geq 0$. Therefore, $f$ is constant.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21251,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle, and let $X, Y, Z$ be points on $BC, CA, AB$, respectively. Suppose that $AX$, $BY$, and $CZ$ intersect at a point $P$. Prove that\n\n$$\n\\frac{AP}{AX} + \\frac{BP}{BY} + \\frac{CP}{CZ} = 2.\n$$",
"options": [],
"answer": "See solution",
"solution": "Assign weights $p, q, r$ to the vertices $A, B, C$, respectively, so that the center of mass is $P$. Draw the line $L$ through $P$, parallel to $BC$. Let $d$ be the distance from $A$ to $L$, and $h$ the distance from $A$ to $BC$. Since the triangle is in static equilibrium when balanced horizontally on $L$, we have $pd = (q + r)(h - d)$. This gives\n\n$$\n\\frac{p}{q + r} = \\frac{h - d}{d} = \\frac{AX - AP}{AP},\n$$\n\nwhich is equivalent to\n\n$$\n\\frac{AP}{AX} = \\frac{q + r}{p + q + r}.\n$$\n\nSimilarly,\n\n$$\n\\frac{BP}{BY} = \\frac{r + p}{p + q + r} \\quad \\text{and} \\quad \\frac{CP}{CZ} = \\frac{p + q}{p + q + r}.\n$$\n\nSumming these gives the desired result.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21252,
"subject": "Mathematics (Olympiad)",
"question": "Two players play the following game. At the outset, there are two piles containing 10,000 and 20,000 tokens, respectively. A move consists of removing any positive number of tokens from a single pile, or removing a total of $2015k$ tokens from both piles, where $k$ is a positive integer. The player who cannot make a move loses. Which player has a winning strategy?",
"options": [],
"answer": "See solution",
"solution": "The first player wins.\n\nHe should present his opponent with one of the following positions:\n\n$$\n(0, 0),\\ (1, 1),\\ (2, 2),\\ \\dots,\\ (2014, 2014).\n$$\n\nAll these positions have different total numbers of tokens modulo $2015$. Therefore, if the game starts from two piles of arbitrary sizes, it is possible to obtain one of these positions just by the first move. In our case,\n\n$$\n10,000 + 20,000 \\equiv 1790 \\pmod{2015},\n$$\n\nand the first player can leave to his opponent the position $(895, 895)$. The remaining part of the game is trivial.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21253,
"subject": "Mathematics (Olympiad)",
"question": "The pages of a notebook are numbered consecutively such that the first sheet contains the numbers 1 and 2, the second sheet contains the numbers 3 and 4, and so on. One sheet is torn out of the notebook. The page numbers on the remaining sheets are added. The resulting sum equals 2021.\n\n(a) How many pages can the notebook have had originally?\n\n(b) Which page numbers could be found on the sheet that has been torn out?",
"options": [],
"answer": "See solution",
"solution": "There is exactly one solution. The notebook had 64 pages, and the sheet with the page numbers 29 and 30 was ripped out.\n\nLet $b > 0$ be the number of sheets. The number of pages is $2b$. We are looking for a number $2b$ such that\n\n$$\n1 + 2 + \\cdots + (2b - 1) + 2b = \\frac{(2b) \\cdot (2b + 1)}{2} > 2021.\n$$\n\nSince $\\frac{60^2}{2} = 1800$ is close, we check $b = 30$ and up:\n\n$$\n\\frac{62 \\cdot 63}{2} = 1953 < 2021 < \\frac{64 \\cdot 65}{2} = 2080.\n$$\n\nTherefore, the smallest possible number of pages is 64.\n\nThe torn out sheet $h$ contains the page numbers $2h - 1$ and $2h$.\n\nThis gives the equation\n\n$$\n(2h - 1) + 2h = 2080 - 2021 = 59\n$$\n\nwhich implies $h = 15$.\n\nSo, the book originally had 32 sheets, and the 15th sheet with page numbers 29 and 30 was torn out.\n\nTo see why this is the only solution: If the book had more than 32 sheets (more than 64 pages), the sum of the remaining pages would be at least $1 + 2 + \\cdots + 63 + 64 = 2080 > 2021$. So there cannot be a solution with more than 32 sheets.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21254,
"subject": "Mathematics (Olympiad)",
"question": "Outside a regular polygon $A_1A_2\\ldots A_n$, a point $B$ is given so that $A_1A_2B$ is an equilateral triangle. Determine all $n$ such that points $B$, $A_2$, and $A_3$ are consecutive vertices of some regular polygon.",
"options": [],
"answer": "See solution",
"solution": "Let the new polygon have $m$ vertices. Two cases are possible:\n\n\n\n\n\nCase 1. The new polygon lies outside the given polygon. In other words, the $m$-gon and $n$-gon are on opposite sides of the line $A_2A_3$.\n\nIn this case, $\\angle BA_2A_3 + \\angle A_1A_2A_3 + 60^\\circ = 360^\\circ$.\n\n$$\n\\frac{n-2}{n} \\cdot 180^\\circ + \\frac{m-2}{m} \\cdot 180^\\circ + 60^\\circ = 360^\\circ\n$$\n\n$$\n3m(n-2) + 3n(m-2) + mn = 6mn\n$$\n\n$$\nm n - 6m = 6n\n$$\n\n$$\nm = \\frac{6n}{n-6} = 6 + \\frac{36}{n-6}\n$$\n\nObviously, $n-6 \\in \\mathbb{N}$ and $n-6$ divides $36$.\n\nChecking all the possibilities, we find nine solutions:\n\n| $n-6$ | 1 | 2 | 3 | 4 | 6 | 9 | 12 | 18 | 36 |\n|-------|---|---|---|---|---|----|----|----|----|\n| $n$ | 7 | 8 | 9 | 10| 12| 15 | 18 | 24 | 42 |\n| $m$ | 42| 24| 18| 15| 12| 10 | 9 | 8 | 7 |",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21255,
"subject": "Mathematics (Olympiad)",
"question": "Find all integer solutions $(a, b)$ to the equation:\n\n$$\n(b^2 + 7(a-b))^2 = a^3 b\n$$",
"options": [],
"answer": "See solution",
"solution": "We rearrange the equation:\n\n$$\n(b^2 + 7(a-b))^2 = a^3 b\n$$\n\nExpanding and simplifying:\n\n$$\n\\begin{align*}\n(b^2 + 7(a-b))^2 &= a^3 b \\\\\n\\Rightarrow b^4 + 14b^2(a-b) + 49(a-b)^2 &= a^3 b \\\\\n\\Rightarrow a^3 b - b^4 - 14b^2(a-b) - 49(a-b)^2 &= 0 \\\\\n\\Rightarrow b(a^3 - b^3) - 14b^2(a-b) - 49(a-b)^2 &= 0 \\\\\n\\Rightarrow (a-b)(ba^2 + ab^2 + b^3 - 14b^2 - 49(a-b)) &= 0.\n\\end{align*}\n$$\n\nIf $a = b$, then the equation holds for all integers $t$, so $(a, b) = (t, t)$, $t \\in \\mathbb{Z}$.\n\nIf $a \\neq b$, then:\n\n$$\nba^2 + ab^2 + b^3 - 14b^2 - 49(a-b) = 0.\n$$\n\nConsider this as a quadratic in $a$:\n\n$$\nba^2 + (b^2 - 49)a + b^3 - 14b^2 + 49b = 0.\n$$\n\nThe discriminant is:\n\n$$\n\\begin{align*}\nD(b) &= (b^2 - 49)^2 - 4b(b^3 - 14b^2 + 49b) \\\\\n&= ((b-7)(b+7))^2 - 4b^2(b-7)^2 \\\\\n&= (b-7)^2((b+7)^2 - 4b^2) \\\\\n&= -(b-7)^3(3b+7).\n\\end{align*}\n$$\n\nReal solutions exist when $D(b) \\ge 0$, i.e., $b \\in [-7/3, 7]$. For integer $b \\neq 0$, only $b = -2, 3, 6, 7$ yield integer $a$:\n\n- $b = -2$: $D = 27^2$, $a = -18$\n- $b = 3$: $D = 2^{10}$, $a = 12$\n- $b = 6$: $D = 5^2$, no integer $a$\n- $b = 7$: $D = 0$, $a = 0$\n\nThus, all integer solutions are:\n\n- $(a, b) = (t, t)$, $t \\in \\mathbb{Z}$\n- $(-18, -2)$\n- $(12, 3)$\n- $(0, 7)$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21256,
"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$$\n\\overline{AB} + \\overline{BC} + \\overline{CD} + \\overline{DA} > \\overline{AC} + \\overline{BD}\n$$\n\nwhich shows the difference is positive. Let $R$ be the intersection point of the diagonals, and let the diameter of the circle be $d = 2$. Then:\n\n$$\n\\begin{aligned}\n\\overline{AB} + \\overline{BC} + \\overline{CD} + \\overline{DA} &< \\overline{AC} + \\overline{BD} + 2d \\\\\n&= \\overline{AC} + \\overline{BD} + 4\n\\end{aligned}\n$$\n\nThus, the difference between the perimeter and the sum of the diagonals is less than $4$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21257,
"subject": "Mathematics (Olympiad)",
"question": "Prove the inequality:\n\n$$\n\\left( \\sum_{i,j=1}^{n} |x_i - x_j| \\right)^2 \\leq \\frac{2(n^2 - 1)}{3} \\sum_{i,j=1}^{n} (x_i - x_j)^2.\n$$\n\n% IMAGE: \n",
"options": [],
"answer": "See solution",
"solution": "By the Cauchy–Schwarz Inequality,\n\n$$\n\\left( \\sum_{i,j=1}^{n} (x_i - x_j)^2 \\right) \\left( \\sum_{i,j=1}^{n} (i-j)^2 \\right) \\geq \\left( \\sum_{i,j=1}^{n} |i-j| |x_i - x_j| \\right)^2.\n$$\n\nIt suffices to show that\n\n$$\n\\sum_{i,j=1}^{n} (i-j)^2 = \\frac{n^2(n^2-1)}{6} \\quad (\\dagger)\n$$\n\nand\n\n$$\n\\sum_{i,j=1}^{n} |i-j| |x_i - x_j| = \\frac{n}{2} \\sum_{i,j=1}^{n} |x_i - x_j|. \\quad (\\ddagger)\n$$\n\nTo establish identity $(\\ddagger)$, compare the coefficients of $x_i$, $1 \\le i \\le n$, on both sides. The coefficient of $x_i$ on the left-hand side is\n\n$$\n\\begin{align*}\n(i - 1) + (i - 2) + \\dots + [i - (i - 1)] - [(i + 1) - i] - \\dots - (n - i)\n&= \\frac{i(i - 1)}{2} - \\frac{(n - i)(n - i + 1)}{2} \\\\\n&= \\frac{n(2i - n - 1)}{2}\n\\end{align*}\n$$\n\nOn the right-hand side, the coefficient is\n\n$$\n\\frac{n}{2} (2i - n - 1).\n$$\n\nTherefore, identity $(\\ddagger)$ holds.\n\nFor the equality case, the Cauchy-Schwarz Inequality reaches equality if\n\n$$\n\\frac{x_i - x_j}{i - j} = d\n$$\n\nis constant for $1 \\le i, j \\le n$, i.e., $x_1, x_2, \\dots, x_n$ is an arithmetic sequence.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21258,
"subject": "Mathematics (Olympiad)",
"question": "For which integers $n \\ge 2$ is it possible to write the numbers $1, 2, 3, \\dots, n$ in a row in some order so that any two numbers written next to each other in the row differ by 2 or 3?",
"options": [],
"answer": "See solution",
"solution": "All integers $n \\ge 4$ work.\n\nFor $n = 2$, the only arrangement is $1, 2$, which differ by 1, so it does not work. For $n = 3$, the numbers $1, 2, 3$ cannot be arranged so that adjacent numbers differ by 2 or 3.\n\nFor $n \\ge 4$, we can construct such an arrangement. For example, starting with $2, 4, 1, 3$, we can continue by adding the next numbers to either end, keeping odd numbers on one side and even numbers on the other:\n\n$$\n\\dots, 9, 7, 5, 2, 4, 1, 3, 6, 8, 10, \\dots\n$$\n\n\n\nThis construction works for all $n \\ge 4$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21259,
"subject": "Mathematics (Olympiad)",
"question": "A polygon is called *good* if its set of boundary cells can be divided into trapezoids so that you can go from any of them to another, moving only through the sides of the trapezoids.\n\nProve by induction on $k \\geq 2$ that if the area $S$ of a good polygon satisfies $S \\leq 4k + 1$, then $S \\equiv 1 \\pmod{4}$.\n\n\n\n\n\n",
"options": [],
"answer": "See solution",
"solution": "We proceed by induction on $k \\geq 2$.\n\n**Base case ($k=2$):**\nIf $S \\leq 9$, there are at least 4 trapezoids. If exactly 4, the polygon is a $3 \\times 3$ square, so $S = 9 \\equiv 1 \\pmod{4}$. If there are at least 5 trapezoids, $S \\geq 10 > 9$.\n\n**Inductive step:**\nAssume the statement holds for $k = n-1 \\geq 2$. Consider $k = n$, i.e., $S \\in [4(n-1)+2, 4n+1]$.\n\nTake a side $AB$ of the polygon with both adjacent angles $90^\\circ$ (such a side always exists). If $AB = 3$ and the polygon is not a $3 \\times 3$ square, we can cut off a $3 \\times 4$ rectangle, and the remainder's area modulo 4 does not change.\n\nIf there is no side of length 3, consider a side $AB = l$. If the rectangle $CABD$ of size $l \\times 4$ contains no other boundary cells, we can cut it off and the property is preserved. If it contains other boundary cells, color the cells as in the figure and analyze the lowest such cell. By considering adjacent cells and the structure of the trapezoids, we can cut the polygon into two parts, each of area $4a+1$ and $4b+1$, plus 3, so the total area is $(4a+1) + (4b+1) + 3 \\equiv 1 \\pmod{4}$.\n\nThus, the statement holds for all $k \\geq 2$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21260,
"subject": "Mathematics (Olympiad)",
"question": "Let $D$ be the midpoint of $BC$ in triangle $ABC$. Prove that $m_a + m_b + m_c < a + b + c$, where $m_a, m_b, m_c$ are the lengths of the medians from $A, B, C$ respectively, and $a, b, c$ are the side lengths opposite $A, B, C$.\n\n\n\n(a) Prove that $2m_a < b + c$.\n\n(b) Give an example of a triangle where $m_a^2 > bc$, and a case where $bc > m_a^2$.",
"options": [],
"answer": "See solution",
"solution": "We present two proofs for part (a).\n\n**First way:**\nExtend $AD$ through $D$ to $A'$ so that $|A'D| = |DA| = m_a$. Triangles $A'DC$ and $ADB$ are congruent (since $\\angle A'DC = \\angle ADB$, $|BD| = |DC|$, $|A'D| = |AD|$), so $|A'C| = |AB|$. In triangle $A'CA$, by the triangle inequality:\n\n$$\n2m_a = |A'D| + |DA| = |A'A| < |CA| + |A'C| = b + c.\n$$\n\n**Second way:**\nBy the Cosine Rule:\n\n$$\nam_a \\cos \\angle BDA = m_a^2 + \\left(\\frac{a}{2}\\right)^2 - c^2, \\\\\n-am_a \\cos \\angle BDA = m_a^2 + \\left(\\frac{a}{2}\\right)^2 - b^2.\n$$\n\nEliminating $\\cos \\angle BDA$ gives:\n\n$$\n4m_a^2 = 2(b^2 + c^2) - a^2.\n$$\n\nThus, $2m_a < b + c$ iff\n\n$$\n4m_a^2 < (b + c)^2 \\\\\n2(b^2 + c^2) - a^2 < b^2 + 2bc + c^2 \\\\\nb^2 + c^2 - a^2 < 2bc \\\\\n\\cos(\\angle BAC) < 1,\n$$\nwhich is always true. Similarly, $2m_b < c + a$, $2m_c < a + b$, so\n\n$$\n2m_a + 2m_b + 2m_c < 2(a + b + c) \\implies m_a + m_b + m_c < a + b + c.\n$$\n\n**Part (b):**\nFor $a=4$, $b=2$, $c=5$:\n\n$$\nm_a^2 = \\frac{2(b^2 + c^2) - a^2}{4} = \\frac{2(4 + 25) - 16}{4} = \\frac{58 - 16}{4} = \\frac{42}{4} = 10.5 > 10 = bc.\n$$\n\nSo $m_a^2 > bc$ in this case. If $b = c$, then $bc = b^2 = m_a^2 + \\left(\\frac{a}{2}\\right)^2 > m_a^2$, so $bc > m_a^2$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21261,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\mathbb{R}$ be the set of real numbers. Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for all $x, y \\in \\mathbb{R}$,\n$$\nf(xf(y) + x) = xy + f(x).\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $x = 1$. Then $f(f(y) + 1) = y + f(1)$. From this, $f$ is a bijection.\n\nLet $y = 0$. Then $f(xf(0) + x) = f(x)$. Since $f$ is injective, $xf(0) + x = x$, so $xf(0) = 0$ for all $x$. Thus, $f(0) = 0$.\n\nFor $x \\neq 0$, take $y = -f(x)/x$. Then $f(xf(y) + x) = 0$. Thus $xf(y) + x = 0$, so $f(y) = -1$. Therefore, $f(-f(x)/x) = -1$. Hence $-f(x)/x = c$, where $c$ is a constant with $f(c) = -1$. Thus $f(x) = -cx$ for $x \\neq 0$. Since $f(0) = 0$, $f(x) = -cx$ for all $x \\in \\mathbb{R}$.\n\nSubstitute this into the original equation:\n$$\nf(-xcy + x) = -c(-xcy + x) = xy - cx.\n$$\nThus $c^2xy = xy$, so $c^2 = 1$, i.e., $c = \\pm 1$. Both $f(x) = x$ and $f(x) = -x$ satisfy the functional equation.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21262,
"subject": "Mathematics (Olympiad)",
"question": "A group of students entered a mathematics competition consisting of five problems. Each student solved at least two problems and no student solved all five problems. For each pair of problems, exactly two students solved them both.\n\nDetermine the minimum possible number of students in the group.",
"options": [],
"answer": "See solution",
"solution": "It is possible that the group comprised six students, as demonstrated by the following example.\n\n- Student 1 solved problems 1, 2, 3, 4.\n- Student 2 solved problems 1, 2, 3, 5.\n- Student 3 solved problems 1, 4, 5.\n- Student 4 solved problems 2, 4, 5.\n- Student 5 solved problems 3 and 4.\n- Student 6 solved problems 3 and 5.\n\nSuppose that $a$ students solved 4 problems, $b$ students solved 3 problems, and $c$ students solved 2 problems. Therefore, $a$ students solved $\\binom{4}{2} = 6$ pairs of problems, $b$ students solved $\\binom{3}{2} = 3$ pairs of problems, and $c$ students solved $\\binom{2}{2} = 1$ pair of problems. Since we have shown an example in which the number of students in the group is 6, let us assume that $a + b + c \\leq 5$.\n\nThere are $\\binom{5}{2} = 10$ pairs of problems altogether and, for each pair of problems, exactly two students solved them both, so we must have\n\n$$\n6a + 3b + c = 20.\n$$\n\nReading the above equation modulo 3 yields $c \\equiv 2 \\pmod{3}$. If $c \\geq 5$, then we have $a + b + c \\geq 6$, contradicting our assumption. Therefore, we must have $c = 2$ and $2a + b = 6$. For $a + b + c \\leq 5$, the only solution is given by $(a, b, c) = (3, 0, 2)$.\n\nHowever, it is impossible for 3 students to have solved 4 problems each. That would mean that each of the 3 students did not solve exactly 1 problem. So there would exist a pair of problems for which 3 students solved them both, contradicting the required conditions.\n\nIn conclusion, the minimum possible number of students in the group is 6.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21263,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$ be positive real numbers such that $$a^2 + b^2 + c^2 + (a+b+c)^2 \\le 4.$$ Prove that\n$$\n\\frac{ab+1}{(a+b)^2} + \\frac{bc+1}{(b+c)^2} + \\frac{ca+1}{(c+a)^2} \\ge 3.\n$$",
"options": [],
"answer": "See solution",
"solution": "The given condition is equivalent to $a^2 + b^2 + c^2 + ab + bc + ca \\le 2$. We will prove that\n$$\n\\frac{2ab+2}{(a+b)^2} + \\frac{2bc+2}{(b+c)^2} + \\frac{2ca+2}{(c+a)^2} \\ge 6.\n$$\nIndeed, we have\n$$\n\\frac{2ab+2}{(a+b)^2} \\ge \\frac{2ab+a^2+b^2+c^2+ab+bc+ca}{(a+b)^2} = 1 + \\frac{(c+a)(c+b)}{(a+b)^2}.\n$$\nAdding the last inequality with its analogous cyclic forms yields\n$$\n\\frac{2ab+2}{(a+b)^2} + \\frac{2bc+2}{(b+c)^2} + \\frac{2ca+2}{(c+a)^2} \\ge 3 + \\frac{(c+a)(c+b)}{(a+b)^2} + \\frac{(a+b)(a+c)}{(b+c)^2} + \\frac{(b+c)(b+a)}{(c+a)^2}.\n$$\nHence it remains to prove that\n$$\n\\frac{(c+a)(c+b)}{(a+b)^2} + \\frac{(a+b)(a+c)}{(b+c)^2} + \\frac{(b+c)(b+a)}{(c+a)^2} \\ge 3.\n$$\nThis follows directly from the AM-GM inequality.\n\n**Solution 2 (By Zuming Feng).** Set $2x = a + b$, $2y = b + c$, and $2z = c + a$ so that $a = z + x - y$, $b = x + y - z$, and $c = y + z - x$. We may compute\n$$\n\\frac{ab+1}{(a+b)^2} = \\frac{(z+x-y)(x+y-z)+1}{4x^2} = \\frac{x^2-(y-z)^2+1}{4x^2} = \\frac{x^2+2yz+1-y^2-z^2}{4x^2}.\n$$\nOn the other hand, the given condition is equivalent to\n$$\n2a^2 + 2b^2 + 2c^2 + 2ab + 2bc + 2ca = (a+b)^2 + (b+c)^2 + (c+a)^2 \\le 4.\n$$\nTranslating, this means that $x^2 + y^2 + z^2 \\le 1$, or $1 - y^2 - z^2 \\ge x^2$. It follows that\n$$\n\\frac{ab+1}{(a+b)^2} = \\frac{x^2+2yz+1-y^2-z^2}{4x^2} \\ge \\frac{x^2+2yz+x^2}{4x^2} = \\frac{1}{2} + \\frac{yz}{2x^2}.\n$$\nLikewise, we have\n$$\n\\frac{bc+1}{(b+c)^2} \\ge \\frac{1}{2} + \\frac{zx}{2y^2} \\quad \\text{and} \\quad \\frac{ca+1}{(c+a)^2} \\ge \\frac{1}{2} + \\frac{xy}{2z^2}.\n$$\nAdding the last three inequalities gives\n$$\n\\frac{ab+1}{(a+b)^2} + \\frac{bc+1}{(b+c)^2} + \\frac{ca+1}{(c+a)^2} \\ge \\frac{3}{2} + \\frac{yz}{2x^2} + \\frac{zx}{2y^2} + \\frac{xy}{2z^2} \\ge 3,\n$$\nby the AM-GM inequality.\n\n**Remark.** It is easy to see from these solutions that equality holds if and only if $a = b = c = \\frac{1}{\\sqrt{3}}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21264,
"subject": "Mathematics (Olympiad)",
"question": "Extend $AD$ and $AE$ to meet the circumcircle of $\\triangle ABC$ at $F$ and $G$, respectively. Let $O$ be the intersection point of $BG$ and $CF$.\n\n\n\nGiven $\\angle BAD = \\angle CAE$, prove that $\\angle BAE = \\angle CAD = 90^\\circ$.",
"options": [],
"answer": "See solution",
"solution": "Let $M, N$ be the centres of the circumcircles of triangles $BAD$ and $CAE$, respectively. These two circles are tangent at $A$ iff $M, A, N$ are collinear. Define $\\alpha = \\angle BAD = \\angle CAE$, $\\beta = \\angle MAB = \\angle MBA$, and $\\gamma = \\angle NAC = \\angle NCA$.\n\n\n\nWe have $\\angle BMA = 180^\\circ - 2\\beta$ and $\\angle ANC = 180^\\circ - 2\\gamma$. If $M$ and $D$ are on opposite sides of $AB$, then $\\angle BDA = 180^\\circ - \\frac{1}{2}\\angle BMA = 90^\\circ + \\beta$ and $\\angle ADE = 180^\\circ - \\angle BDA = 90^\\circ - \\beta$. Similarly, $\\angle AED = 90^\\circ - \\gamma$. Thus,\n\n$$\n\\angle DAE = 180^\\circ - \\angle ADE - \\angle AED = \\beta + \\gamma.\n$$\n\nBecause $M, A, N$ are collinear, this implies $2\\alpha + 2\\beta + 2\\gamma = 180^\\circ$, hence $\\alpha + \\beta + \\gamma = 90^\\circ$. Finally, $\\angle BAE = \\angle BAD + \\angle DAE = \\alpha + \\beta + \\gamma = 90^\\circ$ and $\\angle DAC = \\angle DAE + \\angle EAC = \\beta + \\gamma + \\alpha = 90^\\circ$.\n\nIf $M$ and $D$ were on the same side of $AB$, the above argument, with $\\beta$ replaced by $-\\beta$, gives the same result. We then see that $\\angle BDA = 90^\\circ + \\alpha > 90^\\circ$, which implies that $M$ and $D$ must actually be on opposite sides of $AB$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21265,
"subject": "Mathematics (Olympiad)",
"question": "Find the circumradius $R$ of a triangle with vertices $A(l, l^2)$, $B(m, m^2)$, and $C(n, n^2)$, where $a = 0.5(m + n)$, $b = 0.5(l + n)$, and $c = 0.5(m + l)$.",
"options": [],
"answer": "See solution",
"solution": "Let $A(l, l^2)$, $B(m, m^2)$, and $C(n, n^2)$ be the vertices of triangle $ABC$.\n\nThe side lengths are:\n$$\nAB = |m - l| \\sqrt{1 + (m + l)^2} = |m - l| \\sqrt{1 + 4c^2},\n$$\nsince $c = 0.5(m + l)$.\n\nSimilarly,\n$$\nBC = |m - n| \\sqrt{1 + 4a^2}, \\quad AC = |n - l| \\sqrt{1 + 4b^2}.\n$$\n\nThe area is:\n$$\nS(ABC) = 0.5 |(m - l)(l - n)(n - m)|.\n$$\n\nThe circumradius is:\n$$\nR = \\frac{AB \\cdot BC \\cdot CA}{4S(ABC)} = \\frac{1}{2} \\sqrt{(1 + 4a^2)(1 + 4b^2)(1 + 4c^2)}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21266,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, \\dots, a_n$ be integers. Suppose that for every subset of $k$ elements from $\\{a_1, a_2, \\dots, a_n\\}$, the sum of the elements in the subset is divisible by $n$, where $1 \\leq k < n$. Prove that $n$ divides $a_1 + a_2 + \\dots + a_n$.",
"options": [],
"answer": "See solution",
"solution": "If $k = 1$, each of the numbers $a_1, a_2, \\dots, a_n$ is divisible by $n$, so $n$ also divides their sum. Now, let $1 < k < n$ and let $i \\neq j$. Since the set $\\{a_1, a_2, \\dots, a_n\\} \\setminus \\{a_i, a_j\\}$ has $n-2 \\geq k-1$ elements, one can choose an arbitrary subset $S$ with $k-1$ elements. Then $S \\cup \\{a_i\\}$ and $S \\cup \\{a_j\\}$ are two $k$-tuples of numbers and their sums are divisible by $n$, so the difference of the two sums must also be divisible by $n$. This difference is equal to $a_i - a_j$, so $n$ divides $a_i - a_j$. Here $i$ and $j$ were chosen arbitrarily, so $a_1, a_2, \\dots, a_n$ give the same remainder when divided by $n$, and there are $n$ of them, so their sum must be divisible by $n$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21267,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a rhombus with $\\angle BAD < 90^\\circ$. The circle passing through $D$ with center $A$ intersects the line $CD$ a second time at point $E$. Let $S$ be the intersection of the lines $BE$ and $AC$.\n\nProve that the points $A$, $S$, $D$, and $E$ lie on a circle.",
"options": [],
"answer": "See solution",
"solution": "By the inscribed angle theorem, it is enough to show that $\\angle SED = \\angle SAD$.\n\nSince $ABCD$ is a rhombus, we have\n\n$$\n\\angle SAD = \\frac{1}{2} \\angle BAD.\n$$\n\nSince $ABCE$ is an isosceles trapezoid, we have by symmetry that\n\n$$\n\\angle SED = \\angle ECS = \\frac{1}{2} \\angle DCB = \\frac{1}{2} \\angle BAD,\n$$\n\nwhich finishes the proof.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21268,
"subject": "Mathematics (Olympiad)",
"question": "Find the smallest positive number $c$ with the following property: for any integer $n \\ge 4$ and set $A \\subseteq \\{1, 2, \\dots, n\\}$, if $|A| > cn$, then there exists a function $f : A \\to \\{1, -1\\}$ that satisfies\n$$\n\\left| \\sum_{a \\in A} f(a) \\cdot a \\right| \\le 1.\n$$",
"options": [],
"answer": "See solution",
"solution": "The desired smallest possible number is $c = \\frac{2}{3}$.\n\nFirst, when $n = 6$ and $A = \\{1,4,5,6\\}$, there exists no $f$ that satisfies the requirement, because the sum of the elements of $A$ is $16$ and $A$ cannot be divided into a union of two subsets with sum of elements both $8$. At this point $|A| = \\frac{2}{3}n$, so $c < \\frac{2}{3}$ does not have the properties described in the question.\n\nIn the following, we will prove that $c = \\frac{2}{3}$ satisfies the requirement.\n\nThat is, when $|A| > \\frac{2}{3}n$, there exists $f$ that satisfies the condition.\n\n*Lemma.* Let $x_1, x_2, \\dots, x_m$ be positive integers, whose sum is $s$ and $s < 2m$. Then for any integer $x \\in [0, s]$, there exists an index set $I \\subseteq \\{1, 2, \\dots, m\\}$ that satisfies $\\sum_{i \\in I} x_i = x$. (Summation over the empty index set is considered to be zero.)\n\n*Proof of lemma.* We prove by induction on $m$. When $m=1$, it can only be $x_1=s=1$, and the conclusion is clearly valid.\n\nSuppose $m > 1$ and the conclusion holds for $m - 1$. We may set $m = 1$, and then\n$$\n\\begin{align}\nx_1 + x_2 + \\cdots + x_{m-1} &\\le \\frac{m-1}{m} \\cdot (x_1 + x_2 + \\cdots + x_m) \\\\\n&< \\frac{m-1}{m} \\cdot 2m = 2(m-1).\n\\end{align}\n$$\nAnd since $x_1 + x_2 + \\cdots + x_{m-1} \\ge m - 1$, it follows that\n$$\nx_m \\le m \\le 1 + x_1 + x_2 + \\cdots + x_{m-1}.\n$$\nFor any integer $x \\in [0, s]$, if $x \\le x_1 + x_2 + \\cdots + x_{m-1}$, by the above and the induction hypothesis there exists an index set $I \\subseteq \\{1, \\dots, m-1\\}$ such that $\\sum_{i \\in I} x_i = x$. If\n$$\nx \\geq 1 + x_1 + x_2 + \\cdots + x_{m-1},\n$$\nthen using the induction hypothesis on $x - x_m$ (by the previous inequality $x - x_m \\ge 0$), there exists an index set $I \\subseteq \\{1, \\dots, m-1\\}$ such that $\\sum_{i \\in I} x_i = x - x_m$. At this point, the index set\n$$\nI' = I \\cup \\{m\\} \\subseteq \\{1, 2, \\dots, m\\}\n$$\nsatisfies $\\sum_{i \\in I'} x_i = x$. The lemma is proven.\n\nLet us return to the original problem. Noting that $n \\ge 4$, we discuss it in the following two cases.\n\n(1) $|A|$ is even and let $|A| = 2m$. The elements of $A$ from smallest to largest are denoted as\n$$\na_1 < b_1 < a_2 < b_2 < \\cdots < a_m < b_m.\n$$\nLet $x_i = b_i - a_i > 0$, $1 \\le i \\le m$, so that\n$$\ns = \\sum_{i=1}^{m} x_i = (b_m - a_1) - \\sum_{i=1}^{m-1} (a_{i+1} - b_i) \\le n - 1 - (m-1) = n - m < 2m.\n$$\nThe above equation makes use of $2m = |A| > \\frac{2}{3}n$. Hence, $x_1, x_2, \\dots, x_m$ satisfy the conditions of the lemma.\n\n(The proof continues similarly for the odd case, omitted for brevity.)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21269,
"subject": "Mathematics (Olympiad)",
"question": "For a finite non-empty set of real numbers $A$, let $\\max(A)$ denote its largest element. Define $P(A)$ as the sum of medians of all odd-sized subsets of $A$, and $Q(A)$ as the sum of medians of all non-empty even-sized subsets of $A$, i.e.,\n\n$$\nP(A) = \\sum_{\\substack{B \\subseteq A \\\\ |B| \\text{ odd}}} m(B), \\quad Q(A) = \\sum_{\\substack{\\emptyset \\neq B \\subseteq A \\\\ |B| \\text{ even}}} m(B),\n$$\n\nwhere $m(B)$ denotes the median of a finite non-empty set $B$: if $B = \\{b_1, b_2, \\dots, b_n\\}$ with $b_1 < b_2 < \\dots < b_n$, its median is $m(B) = \\frac{1}{2}(b_{\\lfloor \\frac{n+1}{2} \\rfloor} + b_{\\lceil \\frac{n+1}{2} \\rceil})$.\n\nFind the smallest real number $c$ such that for any set $A$ of 2025 distinct positive real numbers, the following inequality holds:\n\n$$\nP(A) - Q(A) \\leq c \\cdot \\max(A).\n$$",
"options": [],
"answer": "See solution",
"solution": "Without loss of generality, let the elements of $A$ in increasing order be $x_0, x_1, \\dots, x_{2024}$. We first compute $P(A)$. For each $x_m$, we count how many odd-sized subsets have $x_m$ as their median. A subset $B$ of odd size has $x_m$ as its median if and only if\n\n$$\n|B \\cap \\{x_0, \\dots, x_{m-1}\\}| = |B \\cap \\{x_{m+1}, \\dots, x_{2024}\\}|.\n$$\n\nIf this common value is $d$, the number of such subsets $B$ is $\\binom{m}{d}\\binom{2024-m}{d}$. Thus, the total number of such $B$ is\n\n$$\n\\sum_d \\binom{m}{d} \\binom{2024-m}{d} = \\sum_d \\binom{m}{m-d} \\binom{2024-m}{d} = \\binom{2024}{m},\n$$\n\nwhere the last equality follows from Vandermonde's identity. Therefore,\n\n$$\nP = \\sum_{m=0}^{2024} \\binom{2024}{m} x_m.\n$$\n\nNext, we compute $Q(A)$. For each $x_m$, we count how many even-sized subsets have $x_m$ as one of their two middle elements. The number of such subsets is\n\n$$\n\\begin{aligned}\n& \\sum_d \\left( \\binom{m}{d+1} \\binom{2024-m}{d} + \\binom{m}{d} \\binom{2024-m}{d+1} \\right) \\\\\n&= \\sum_d \\binom{m}{m-d-1} \\binom{2024-m}{d} + \\sum_d \\binom{m}{m-d} \\binom{2024-m}{d+1} \\\\\n&= \\binom{2024}{m-1} + \\binom{2024}{m+1}.\n\\end{aligned}\n$$\n\nThus,\n\n$$\nQ = \\sum_{m=0}^{2024} \\frac{1}{2} \\left( \\binom{2024}{m-1} + \\binom{2024}{m+1} \\right) x_m.\n$$\n\nSince multiplying all elements of $A$ by a positive constant does not change the problem, we may assume $x_{2024} = 1$. Let $y_i = x_i - x_{i-1}$ for $i = 0, 1, \\dots, 2024$ (with $x_{-1} = 0$). Then $P-Q$ can be expressed as a linear form in $y_0, y_1, \\dots, y_{2024}$, where the $y_i$ are positive and sum to 1. The minimal $c$ is therefore equal to the maximum coefficient of the $y_i$.\n\nThe coefficient of $x_m$ in $P-Q$ is $\\binom{2024}{m} - \\frac{1}{2}(\\binom{2024}{m-1} + \\binom{2024}{m+1})$. Thus, the coefficient of $y_i$ is\n\n$$\n\\begin{aligned}\n& \\sum_{m=i}^{2024} \\left( \\binom{2024}{m} - \\frac{1}{2} \\left( \\binom{2024}{m-1} + \\binom{2024}{m+1} \\right) \\right) \\\\\n&= \\sum_{m=i}^{2024} \\binom{2024}{m} - \\frac{1}{2} \\left( \\sum_{m=i-1}^{2024} \\binom{2024}{m} + \\sum_{m=i+1}^{2024} \\binom{2024}{m} - 1 \\right) \\\\\n&= \\frac{\\binom{2024}{i} - \\binom{2024}{i-1} + 1}{2}.\n\\end{aligned}\n$$\n\nWe now find the maximum of $\\binom{2024}{i} - \\binom{2024}{i-1}$. For $i \\ge 1013$, $\\binom{2024}{i} - \\binom{2024}{i-1} \\le 0$, so the maximum must occur for $0 \\le i \\le 1012$. Let $d_i = \\binom{2024}{i} - \\binom{2024}{i-1}$. Then\n\n$$\n\\begin{aligned}\nd_{i+1} - d_i &= \\binom{2024}{i+1} + \\binom{2024}{i-1} - 2\\binom{2024}{i} \\\\\n&= \\binom{2024}{i} \\left( \\frac{2025-i}{i+1} + \\frac{i}{2025-i} - 2 \\right) \\\\\n&= \\binom{2024}{i} \\left( \\frac{2025 \\times 2026}{(i+1)(2025-i)} - 4 \\right).\n\\end{aligned}\n$$\n\nFor $0 \\le i \\le 1010$, $(i+1)(2025-i)$ increases with $i$. Since $\\frac{2025 \\times 2026}{991 \\times 1035} < 4 < \\frac{2025 \\times 2026}{990 \\times 1036}$, we have $d_{i+1} > d_i$ when $i \\le 989$ and $d_{i+1} < d_i$ when $i \\ge 990$. Thus, the maximum of $d_i$ is achieved at $i = 990$. Therefore, the minimal $c$ is\n\n$$\nc = \\frac{\\binom{2024}{990} - \\binom{2024}{989} + 1}{2}.\n$$\n\n\n\n**Remark:** The key combinatorial identities and the analysis of binomial coefficients are central to this problem. The final expression for $c$ involves the difference of two large binomial coefficients, which can be approximated using Stirling's formula if an asymptotic estimate is desired.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21270,
"subject": "Mathematics (Olympiad)",
"question": "Define an $n$-magic square to be an $n \\times n$ square matrix of non-negative integers such that the sum of all the entries in each row and each column is $m$ for some $m \\in \\mathbb{N}$. Also, define an $n$-permutation matrix to be an $n \\times n$ square matrix with $n(n-1)$ zeroes and $n$ ones such that every row and every column contains exactly one $1$.\n\nShow that every $n$-magic square can be written as a sum of finitely many $n$-permutation matrices.",
"options": [],
"answer": "See solution",
"solution": "Fix an $n \\in \\mathbb{N}$. We will prove the following statement by induction:\n\n$P(m)$: Every $n$-magic square with common sum $m$ can be written as a sum of $m$ permutation matrices.\n\n**Base case:** $P(1)$ is true. Since all entries are non-negative integers, the only possible $n$-magic squares with common sum $1$ are exactly $n$-permutation matrices. Hence, every such matrix can be written as a sum of $1$ $n$-permutation matrix, i.e., itself.\n\n**Inductive step:** Assume $P(k)$ is true for some $k \\in \\mathbb{N}$. That is, every $n$-magic square with common sum $k$ can be written as a sum of $k$ $n$-permutation matrices. Consider any $n$-magic square $A_0$ with common sum $k+1$. We will show that $A_0$ can be written as a sum of an $n$-permutation matrix and an $n$-magic square $A_1$ with common sum $k$.\n\nConsider the graph $\\mathcal{G} = (\\mathcal{W}, \\mathcal{E})$ defined as follows. Let $R_1, R_2, \\dots, R_n$ be the rows of $A_0$ and $C_1, C_2, \\dots, C_n$ be the columns. Let $\\mathcal{W} = \\{R_1, \\dots, R_n\\} \\cup \\{C_1, \\dots, C_n\\}$. For every $i, j \\in \\mathbb{N}_n$, $R_i$ and $C_j$ are connected by an edge if and only if $a_{ij} > 0$, where $a_{ij}$ is the entry of $A_0$ in the $i$-th row and $j$-th column. Thus, $\\mathcal{G}$ is a bipartite graph.\n\nFor any $S \\subseteq \\{R_1, \\dots, R_n\\}$, let $N(S)$ denote the set of vertices adjacent to some vertex in $S$. We show that $|N(S)| \\geq |S|$. For each $R_i \\in S$, the sum of entries in row $i$ is $m$, so the total sum over all such edges is $m|S|$. These edges are a subset of all edges incident to $N(S)$, whose total sum is $m|N(S)|$. Thus, $m|S| \\leq m|N(S)|$, so $|N(S)| \\geq |S|$.\n\nBy Hall's Marriage Theorem, there exists a matching in $\\mathcal{G}$ that matches every $R_i$ to some $C_j$ with $a_{ij} > 0$. This matching corresponds to a permutation $\\sigma$ of $\\mathbb{N}_n$ such that $a_{i\\sigma(i)} > 0$ for all $i$. Define the permutation matrix $P_\\sigma$ by\n\n$$\np_{ij} = \\begin{cases} 1 & \\text{if } j = \\sigma(i) \\\\ 0 & \\text{otherwise} \\end{cases} \\quad \\forall i, j \\in \\mathbb{N}_n\n$$\n\nLet $A_1 = A_0 - P_\\sigma$. Since $a_{i\\sigma(i)} > 0$ for all $i$, every entry of $A_1$ is non-negative. The sum of each row and column in $A_1$ is reduced by $1$, so $A_1$ is an $n$-magic square with common sum $k$. By the induction hypothesis, $A_1$ can be written as a sum of $k$ $n$-permutation matrices. Thus, $A_0$ can be written as a sum of $k+1$ $n$-permutation matrices.\n\nBy induction, the statement holds for all $m \\in \\mathbb{N}$. The proof is complete.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 21271,
"subject": "Mathematics (Olympiad)",
"question": "Determine all pairs $ (x, y) $ of positive integers such that for $ d = \\gcd(x, y) $ the equation\n\n$$\nx y d = x + y + d^2\n$$\nholds.",
"options": [],
"answer": "See solution",
"solution": "There are three such pairs: $ (x, y) = (2, 2) $, $ (x, y) = (2, 3) $, and $ (x, y) = (3, 2) $.\n\nFor $ x = 1 $, we get $ d = 1 $ and the given equation becomes the contradiction $ y = y + 2 $. This works analogously for $ y = 1 $.\n\nTherefore, we can assume $ x \\geq 2 $ and $ y \\geq 2 $.\n\nWe start with the case $ d = 1 $, which gives the equation\n\n$$\nxy = x + y + 1 \\iff (x - 1)(y - 1) = 2.\n$$\n\nThe possible factorizations $ 2 = 1 \\cdot 2 $ and $ 2 = 2 \\cdot 1 $ give the pairs $ (x, y) = (2, 3) $ and $ (x, y) = (3, 2) $, respectively, because $ \\gcd(x, y) = 1 $ is satisfied.\n\nNow, we treat the case $ d \\geq 2 $. The given equation is equivalent to\n\n$$\n\\frac{1}{x d} + \\frac{1}{y d} + \\frac{d}{x y} = 1.\n$$\n\nBecause $ x d \\geq 4 $ and $ y d \\geq 4 $, we get\n\n$$\n1 \\leq \\frac{1}{4} + \\frac{1}{4} + \\frac{d}{x y} \\iff x y \\leq 2 d.\n$$\n\nTogether with $ x y \\geq d^2 $, we obtain $ d = 2 $, $ x = y = 2 $, which gives indeed the third pair $ (x, y) = (2, 2) $ with $ \\gcd(2, 2) = 2 $.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21272,
"subject": "Mathematics (Olympiad)",
"question": "In the following table, each question mark is to be replaced by \"Possible\" or \"Not Possible\" to indicate whether a nonvertical line with the given slope can contain the given number of lattice points (points both of whose coordinates are integers). How many of the 12 entries will be \"Possible\"?\n\n\\begin{tabular}{|c|c|c|c|c|}\n\\hline\n & zero & exactly one & exactly two & more than two \\\\\n\\hline\nzero slope & ? & ? & ? & ? \\\\\n\\hline\nnonzero rational slope & ? & ? & ? & ? \\\\\n\\hline\nirrational slope & ? & ? & ? & ? \\\\\n\\hline\n\\end{tabular}\n\n(A) 4 (B) 5 (C) 6 (D) 7 (E) 9",
"options": [],
"answer": "See solution",
"solution": "If the slope is $0$, then the line is horizontal and its equation is $y = b$ for some real number $b$. If $b$ is an integer, then the line will contain infinitely many lattice points, and if $b$ is not an integer, then it will contain no lattice points. Therefore, exactly two of the entries in that row of the table are \"Possible\".\n\nNext, suppose that the equation of the line is $y = mx + b$, where the slope $m$ is a nonzero rational number, say $m = \\frac{p}{q}$ for integers $p$ and $q$ with $q \\neq 0$. If the line contains a lattice point $(r, s)$, then it also contains the lattice points $(r+q, s+p)$, $(r+2q, s+2p)$, $(r+3q, s+3p)$, and so on. Therefore, the fourth entry in that row of the table is \"Possible\" and the second and third entries are \"Not Possible\". To see that the line may contain no lattice points, let $b$ be irrational. Then $(0, b)$ is a point on the line, but if $(r, s)$ were a lattice point on the line, then\n\n$$\nm = \\frac{s-b}{r-0}\n$$\n\nwould be an irrational number, a contradiction. Thus, the first entry in the \"nonzero rational slope\" row of the table is \"Possible\".\n\nFinally, suppose that the equation of the line is $y = mx + b$, where the slope $m$ is an irrational number. The line could certainly contain exactly one lattice point; for example, the equation of the line could be $y = \\sqrt{2}x$ and the only lattice point on the line is $(0, 0)$. It could also contain no lattice points; for example, its equation could be $y = \\sqrt{2}x + \\frac{1}{2}$. But if a nonvertical line contains two or more lattice points, say $(r, s)$ and $(t, u)$ with $r \\neq t$, then its slope, $\\frac{s-u}{r-t}$, is rational. Therefore, the first and second entries in the bottom row of the table are \"Possible\" and the third and fourth entries are \"Not Possible\".\n\nIn all, 6 of the 12 entries are \"Possible\" (indicated by **P** in the table below), and 6 are \"Not Possible\" (indicated by **NP**).\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21273,
"subject": "Mathematics (Olympiad)",
"question": "Determine all integers $s \\ge 4$ for which there exist positive integers $a, b, c, d$ such that $s = a + b + c + d$ and $s$ divides $abc + abd + acd + bcd$.",
"options": [],
"answer": "See solution",
"solution": "The answer is $s$ composite.\n\n**Composite construction:** Write $s = (w + x)(y + z)$, where $w, x, y, z$ are positive integers. Let $a = w y$, $b = w z$, $c = x y$, $d = x z$. Then\n\n$$\nabc + abd + acd + bcd = w x y z (w + x)(y + z)\n$$\n\nso this works.\n\n**Prime proof:** Choose suitable $a, b, c, d$. Then\n\n$$\n(a + b)(a + c)(a + d) = (abc + abd + acd + bcd) + a^2(a + b + c + d) \\equiv 0 \\pmod{s}.\n$$\n\nHence $s$ divides a product of positive integers less than $s$, so $s$ is composite.\n\n**Remark:** Here is another proof that $s$ is composite.\n\nSuppose that $s$ is prime. Then the polynomial $(x - a)(x - b)(x - c)(x - d) \\in \\mathbb{F}_s[x]$ is even, so the roots come in two opposite pairs in $\\mathbb{F}_s$. Thus the sum of each pair is at least $s$, so the sum of all four is at least $2s > s$, contradiction.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21274,
"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. Equivalently, if $n$ is a positive integer congruent to $9$ modulo $25$, then\n$$N = 8n + 3 = (2a+1)^2 + (2b+1)^2 + (2c+1)^2$$\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 $4$.\n\nThis 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 $n \\equiv \\frac{3(r^2 - 1)}{8} \\pmod{r^2}$ is expressible in the form\n$$n = \\frac{a(a+1)}{2} + \\frac{b(b+1)}{2} + \\frac{c(c+1)}{2}$$\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\n**Remarks.** There are 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$$n = \\frac{a(a+1)}{2} + \\frac{b(b+1)}{2} + \\frac{c(c+1)}{2}$$\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$$n = \\frac{a(a+1)}{2} + \\frac{b(b+1)}{2} + \\frac{c(c+1)}{2}$$\nfor some non-negative integers $a$, $b$, $c$ that do not share parity.",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 21275,
"subject": "Mathematics (Olympiad)",
"question": "$f_1(x) = x^3 - 3x$, and for $n \\geq 2$, $f_n(x) = f_1(f_{n-1}(x))$. Find the value of\n$$\n\\sum_{\\alpha \\in X} \\frac{1}{2-\\alpha}\n$$\nwhere $X$ is the set of real roots of $f_n(x) = 0$ in $x \\in [-2, 2]$.",
"options": [],
"answer": "See solution",
"solution": "Let $f_1(x) = x^3 - 3x$, and for $n \\geq 2$, $f_n(x) = f_1(f_{n-1}(x))$. The degree of $f_n(x)$ is $3^n$.\n\nLet $x = 2 \\cos t$, then $f_1(2 \\cos t) = 2 \\cos 3t$, and by induction, $f_n(2 \\cos t) = 2 \\cos(3^n t)$. Thus, the real roots of $f_n(x) = 0$ in $x \\in [-2, 2]$ are $x_k = 2 \\cos \\frac{k-1}{3^n} \\pi$, $k = 1, 2, \\ldots, 3^n$.\n\nWe want to compute:\n$$\n\\sum_{\\alpha \\in X} \\frac{1}{2-\\alpha} = \\frac{1}{2} \\sum_{k=1}^{3^n} \\frac{1}{1 - \\cos \\frac{k-1}{3^n} \\pi}\n$$\n\n**Lemma:** If $P(x) = (x - \\alpha_1)\\cdots(x - \\alpha_n)$, then $\\sum_{k=1}^n \\frac{1}{x - \\alpha_k} = \\frac{P'(x)}{P(x)}$.\n\nApplying this to $f_n(x)$ at $x = 2$:\n$$\n\\sum_{\\alpha \\in X} \\frac{1}{2-\\alpha} = \\frac{f'_n(2)}{f_n(2)}\n$$\n\nSince $f_1(2) = 2$, and by induction $f_n(2) = 2$ for all $n$, and $f'_n(2) = 9^n$, we get:\n$$\n\\sum_{\\alpha \\in X} \\frac{1}{2-\\alpha} = \\frac{9^n}{2}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21276,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a square and consider the points $K \\in AB$, $L \\in BC$, and $M \\in CD$ such that $KLM$ is a right isosceles triangle with the right angle at $L$. Prove that the lines $AL$ and $DK$ are perpendicular to each other.",
"options": [],
"answer": "See solution",
"solution": "It is not difficult to observe that $\\triangle KLB \\equiv \\triangle LMC$, hence $KB = LC$. Because $AB = BC$, it follows that $AK = BL$. But then, $\\triangle AKD \\equiv \\triangle BLA$, and since $AK \\perp BL$ and $AD \\perp BA$, we deduce that $AL \\perp KD$, as well.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21277,
"subject": "Mathematics (Olympiad)",
"question": "If $234_{b+1} - 234_{b-1} = 70_{10}$, what is $234_b$ in base 10?",
"options": [],
"answer": "See solution",
"solution": "Let us compute $234_{b+1} - 234_{b-1}$:\n\n$$\n\\begin{aligned}\n234_{b+1} &= 2(b+1)^2 + 3(b+1) + 4 \\\\\n234_{b-1} &= 2(b-1)^2 + 3(b-1) + 4 \\\\\n\\text{So,} \\\\\n234_{b+1} - 234_{b-1} &= [2(b+1)^2 + 3(b+1) + 4] - [2(b-1)^2 + 3(b-1) + 4] \\\\\n&= 2[(b+1)^2 - (b-1)^2] + 3[(b+1)-(b-1)] \\\\\n&= 2[(b^2 + 2b + 1) - (b^2 - 2b + 1)] + 3(2) \\\\\n&= 2(4b) + 6 \\\\\n&= 8b + 6\n\\end{aligned}\n$$\n\nSet $8b + 6 = 70$:\n\n$$\n8b + 6 = 70 \\\\\n8b = 64 \\\\\nb = 8\n$$\n\nNow, $234_b = 234_8$ in base 10:\n\n$$\n234_8 = 2 \\times 8^2 + 3 \\times 8 + 4 = 2 \\times 64 + 24 + 4 = 128 + 24 + 4 = 156\n$$\n\nTherefore, $234_b$ in base 10 is $\\boxed{156}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21278,
"subject": "Mathematics (Olympiad)",
"question": "In a convex pentagon $ABCDE$, the following conditions are satisfied: $AB \\parallel CD$, $BC \\parallel DE$, and $\\angle BAE = \\angle AED$. Prove that $AB + BC = CD + DE$.\n\nA pentagon is convex if its diagonals are located inside the pentagon.",
"options": [],
"answer": "See solution",
"solution": "Let the rays $AB$ and $DE$ intersect at the point $O$. Then $BCDO$ is a parallelogram, and $\\triangle AOE$ is isosceles, since $\\angle OAE = \\angle AEO$. So,\n\n$$\nAB + BC = (OB - OA) + BC = CD - OE + OD = CD + DE.\n$$\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21279,
"subject": "Mathematics (Olympiad)",
"question": "Consider a parallelogram $ABCD$ and the points $M$ on the side $DC$ and $E$ and $N$ on the diagonal $AC$, such that $BE \\perp AC$ and $\\frac{CM}{CD} = \\frac{EN}{EA}$.\n\nProve that if $MN$ and $NB$ are perpendicular, then $ABCD$ is a rectangle.",
"options": [],
"answer": "See solution",
"solution": "Construct the parallel to $AB$ through $N$ and denote by $P$ its intersection with the line $BE$.\n\n\n\nUsing the fundamental theorem of similarity in the triangle $EAB$:\n\n$$\n\\frac{NP}{AB} = \\frac{EN}{EA} = \\frac{CM}{CD}.\n$$\n\nFrom here we obtain $NP = CM$ and, since $NP \\parallel MC$, it follows that $MNPC$ is a parallelogram, so $MN \\parallel CP$.\n\nGiven that $MN \\perp NB$, it follows that $CP \\perp NB$.\n\nIn the triangle $BNC$, $BE$ and $CP$ are the lines supporting the altitudes, so the point $P$ is the orthocenter.\n\nTherefore $NP \\perp BC$ and, since $NP \\parallel CD$, it follows that $BC \\perp CD$, thus $ABCD$ is a rectangle.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21280,
"subject": "Mathematics (Olympiad)",
"question": "Определи ги сите цели броеви $x$, за кои $\\log_2(x^2 - 4x - 1)$ е исто така цел број.",
"options": [],
"answer": "See solution",
"solution": "Цели броеви $x$ за кои $\\log_2(x^2 - 4x - 1)$ е определен се\n\n$$\nx \\in (-\\infty, 2 - \\sqrt{5}, 2 + \\sqrt{5}, +\\infty) \\cap \\mathbb{Z}.\n$$\n\nНека $n$ е цел број за кој постои $x \\in \\mathbb{Z}$ така што\n\n$$\n\\log_2(x^2 - 4x - 1) = n.\n$$\n\nТогаш\n\n$$\nx^2 - 4x - (1 + 2^n) = 0 \\quad (1)\n$$\n\nод каде добиваме $x_{1/2} = 2 \\pm \\sqrt{5+2^n}$, односно\n\n$$\nx = 2 + \\sqrt{5 + 2^n} \\text{ или } x = 2 - \\sqrt{5 + 2^n}.\n$$\n\nБидејќи $x \\in \\mathbb{Z}$, постои $k \\in \\mathbb{Z}$ така што $5+2^n = k^2$. Значи, доволно е да ги определиме сите $n$ за кои $5+2^n$ е полн квадрат. Ќе разгледаме неколку случаи.\n\nа) Ако $n < 0$, тогаш $2^n = k^2 - 5$, односно $1 = 2^{-n}(k^2 - 5)$. Бројот $2^{-n}(k^2 - 5)$ е парен, па според тоа последното равенство не е можно.\n\nб) Ако $n=0$, тогаш $k^2 = 6$, т.е. $5+2^n$ не е полн квадрат. Значи и овој случај не е можен.\n\nв) Ако $n > 0$, тогаш $5+2^n$ е непарен, па затоа $k = 2m-1$ за некој $m \\in \\mathbb{Z}$. Со алгебарски трансформации добиваме\n\n$$\nm(m-1) = 2^{n-2} + 1. \\quad (2)\n$$\n\nАко $n=1$, тогаш $2^{n-2}+1$ е рационален број и равенството (2) не е можно за ниту еден $m \\in \\mathbb{Z}$.\n\nАко $n > 2$, тогаш $2^{n-2} + 1$ е непарен број а $m(m-1)$ е парен, па равенство меѓу нив не е можно.\n\nАко $n=2$, тогаш $5+2^2 = 9 = 3^2$. Со замена во (2) ја добиваме квадратната равенка\n\n$$\nx^2 - 4x + 5 = 0\n$$\n\nчии решенија се $x = -1$ и $x = 5$. Не е тешко да се провери дека за најдените вредности за $x$, $\\log_2(x^2 - 4x - 1)$ е цел број и во двата случаи е еднаков на 2.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21281,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\mathbb{Z}$ be the set of integers. Determine all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ such that, for all integers $a$ and $b$,\n\n$$\nf(2a) + 2f(b) = f(f(a + b)).\n$$",
"options": [],
"answer": "See solution",
"solution": "The solutions are:\n\n1. $f(x) = 0$\n2. $f(x) = 2x + c$ for any $c \\in \\mathbb{Z}$\n\n**Solution:**\n\nGiven the functional equation:\n\n$$\nf(2a) + 2f(b) = f(f(a + b)).\n$$\n\nLet $a = 1$, $b = x$:\n$$\nf(2) + 2f(x) = f(f(1 + x)).\n$$\n\nLet $a = 0$, $b = x + 1$:\n$$\nf(0) + 2f(x + 1) = f(f(x + 1)).\n$$\n\nComparing the two:\n$$\n\\begin{aligned}\nf(2) + 2f(x) &= f(0) + 2f(x + 1) \\\\\nf(x + 1) - f(x) &= m\n\\end{aligned}\n$$\nwhere $m = \\frac{f(2) - f(0)}{2}$.\n\nThus, $f(x)$ is linear: $f(x) = mx + c$.\n\nSubstitute into the original equation:\n$$\n2m(a + b) + 2c = m^2(a + b) + mc\n$$\nComparing coefficients:\n- $m^2 = 2m$ $\\implies$ $m = 0$ or $m = 2$\n- $mc = 2c$\n\nIf $m = 0$, $c = 0$ $\\implies$ $f(x) = 0$.\nIf $m = 2$, any $c$ works $\\implies$ $f(x) = 2x + c$.\n\nBoth forms satisfy the original equation.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21282,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\Omega$ be the circumcircle of a triangle $ABC$. Let $D$ be a variable point on the arc $AB$ that does not contain $C$ ($D \\ne A, B$), and let $E$ and $F$ be the incenters of triangles $CAD$ and $CBD$, respectively. Find the locus of the second intersection point of the circumcircle of $\\triangle DEF$ and $\\Omega$ as $D$ varies on the arc $AB$.",
"options": [],
"answer": "See solution",
"solution": "Consider the following well-known lemma:\n\n**Lemma.** Let $ABC$ be a triangle with incenter $I$. If $M$ is the midpoint of the arc $BC$ of the circumcircle not containing $A$, then $MB = MC = MI$.\n\n**Proof of Lemma.** Let the circumcircle of $DEF$ intersect $\\Omega$ again at $X$. Let $M$ and $N$ be the midpoints of the arcs $BC$ (not containing $A$) and $AC$ (not containing $B$), respectively. Let $P$ be the intersection of $\\Omega$ and the line through $C$ parallel to $MN$ (if $P$ is the same as $C$, i.e., $AC = CB$, the results below still hold.)\n\nBy the above lemma and since $MNCP$ is an isosceles trapezoid, we get $MP = NC = NE$ and $NP = MC = MF$. Since $E$ and $F$ lie on $DN$ and $DM$ respectively, $\\angle NXM = \\angle NDM = \\angle EDF = \\angle EXF$. Therefore, $\\angle NXE = \\angle MXF$ and since $\\angle XNE = \\angle XND = \\angle XMD = \\angle XMF$, we have that $\\triangle NXE \\sim \\triangle MXF$. Then,\n\n$$\n\\frac{NX}{NE} = \\frac{MX}{MF},\n$$\n\nand since $NE = MP$ and $MF = NP$, we get\n\n$$\nNX \\cdot NP = MX \\cdot MP.\n$$\n\nTherefore,\n\n$$\n\\frac{[NPX]}{[MPX]} = \\frac{NX \\cdot NP}{MX \\cdot MP} = 1.\n$$\n\nThis implies that the line $XP$ bisects the segment $MN$. Therefore, $X$ must lie on the intersection of $\\Omega$ and the line joining $P$ and the midpoint of $MN$. Since $M$, $N$, and $P$ are fixed independent of $D$, $X$ is the only locus as $D$ varies on the arc $AB$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21283,
"subject": "Mathematics (Olympiad)",
"question": "Let $P$ be a point in the plane of triangle $ABC$ such that the segments $PA$, $PB$, and $PC$ are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to $PA$. Prove that $\\angle BAC$ is acute.",
"options": [],
"answer": "See solution",
"solution": "By the **Cauchy-Schwarz Inequality**,\n\n$$\n\\sqrt{PB^2 + PC^2} \\sqrt{AC^2 + AB^2} \\geq PB \\cdot AC + PC \\cdot AB.\n$$\n\nApplying the **(Generalized) Ptolemy's Inequality** to quadrilateral $ABPC$ yields\n\n$$\nPB \\cdot AC + PC \\cdot AB \\geq PA \\cdot BC.\n$$\n\n\n\nBecause $PA$ is the longest side of an obtuse triangle with side lengths $PA$, $PB$, $PC$, we have $PA > \\sqrt{PB^2 + PC^2}$ and hence\n\n$$\nPA \\cdot BC > \\sqrt{PB^2 + PC^2} \\cdot BC.\n$$\n\nCombining these three inequalities yields $\\sqrt{AB^2 + AC^2} > BC$, implying that angle $BAC$ is acute.\n\n**Note.** With some careful argument, it can be proved that quadrilateral $ABPC$ is indeed convex. We leave it as an exercise for the reader. (For a hint of this proof, please read the first part of the seventh solution.)",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21284,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ and $b$ be two distinct integers, larger than 1, so that $D = b^2 + a - 1$ divides $M = a^2 + b - 1$.\n\n**a)** Prove that there exist numbers fulfilling the conditions above.\n\n**b)** Prove that $D$ has at least two prime divisors.",
"options": [],
"answer": "See solution",
"solution": "**a)**\nWe try $b = 2$ and $a$ such that $a + 3 \\mid a^2 + 1$. Since $a + 3 \\mid a(a + 3) - 3(a + 3) = a^2 - 9$, it is enough that $a + 3 \\mid a^2 + 1 - a^2 + 9 = 10$, so we can take $a = 7$.\n\n**b)**\nFrom $D \\mid (b^2-1+a)(b^2-1-a) = (b^2-1)^2 - a^2$ it follows that $D \\mid (b^2-1)^2 + b - 1 = b^4 - 2b^2 + b = b(b-1)(b^2 + b - 1)$.\n\nThe condition $D \\mid M$ yields $0 \\leq M - D = (a - b)(a + b - 1)$, hence $a > b$. Moreover, $b$, $b-1$, and $b^2 + b - 1$ are pairwise coprime and less than $D$. So $D$ has common prime factors with at least two of the numbers $b$, $b-1$, $b^2 + b - 1$. Since these three numbers are pairwise coprime and larger than 1, $D$ must have at least two prime divisors.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21285,
"subject": "Mathematics (Olympiad)",
"question": "Let $k > 1$ be a real number, $n \\ge 3$ be an integer, and $x_1 \\ge x_2 \\ge x_3 \\ge \\dots \\ge x_n > 0$ be real numbers. Prove the inequality:\n\n$$\n\\frac{x_1 + kx_2}{x_2 + x_3} + \\frac{x_2 + kx_3}{x_3 + x_4} + \\dots + \\frac{x_{n-1} + kx_n}{x_n + x_1} + \\frac{x_n + kx_1}{x_1 + x_2} \\ge \\frac{n(k+1)}{2}\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $x_{n+1} = x_1$. By the AM-GM inequality, we have\n\n$$\n\\frac{x_1 + x_2}{x_2 + x_3} + \\frac{x_2 + x_3}{x_3 + x_4} + \\dots + \\frac{x_{n-1} + x_n}{x_n + x_1} + \\frac{x_n + x_1}{x_1 + x_2} \\ge n \\sqrt{\\prod_{i=1}^{n} \\frac{x_i + x_{i+1}}{x_{i+1} + x_{i+1}}} = n.\n$$\n\nSo it is enough to prove that\n\n$$\n\\frac{x_1}{x_1 + x_2} + \\frac{x_2}{x_2 + x_3} + \\dots + \\frac{x_n}{x_n + x_1} \\ge \\frac{n}{2}.\n$$\n\nLet $a_i = x_{i+1} / x_i$ for $i = 1, 2, \\dots, n$. It is enough to prove that\n\n$$\n\\frac{1}{1 + a_1} + \\dots + \\frac{1}{1 + a_n} \\ge \\frac{n}{2}.\n$$\n\nNote that $a_1, \\dots, a_{n-1} \\le 1$ and $a_1 a_2 \\dots a_n = 1$.\n\nEquivalently, it is enough to prove that if $m \\ge 2$ is an integer and $a_1, \\dots, a_m \\le 1$ are real numbers, then\n\n$$\n\\frac{1}{1 + a_1} + \\dots + \\frac{1}{1 + a_m} \\ge \\frac{m + 1}{2} - \\frac{a_1 a_2 \\dots a_m}{1 + a_1 a_2 \\dots a_m}.\n$$\n\nWe proceed by induction on $m$. The statement is true even for $m = 1$, so we assume it is true for $m = k$ and proceed with the inductive step. Let $a = a_1 \\dots a_k$ and $b = a_{k+1}$. It is enough to prove that\n\n$$\n\\frac{1}{1 + b} - \\frac{a}{1 + a} \\ge \\frac{1}{2} - \\frac{ab}{1 + ab}.\n$$\n\nWe have\n\n$$\n\\frac{a}{1 + a} - \\frac{ab}{1 + ab} = \\frac{a(1 - b)}{(1 + a)(1 + ab)} \\le \\frac{1 - b}{1 + b} = \\frac{1}{1 + b} - \\frac{1}{2}\n$$\n\nso the result follows.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21286,
"subject": "Mathematics (Olympiad)",
"question": "The function $f : \\mathbb{N}^* \\rightarrow \\mathbb{N}$ satisfies:\n\n- $f(2) = 0$\n- $f(3) > 0$\n- $f(6042) = 2014$\n- For all $(m, n) \\in \\mathbb{N}^* \\times \\mathbb{N}^*$, $f(m+n) - f(n) - f(m) \\in \\{0, 1\\}$\n\nFind the value of $f(2014)$.\n\nHere, $\\mathbb{N}^* = \\{1, 2, 3, \\dots\\}$.",
"options": [],
"answer": "See solution",
"solution": "Since $f(2) \\geq f(1) + f(1) = 2 f(1)$ and $f(2) = 0$, then $f(1) \\leq 0$. Therefore, $f(1) = 0$.\n\nFrom $f(3) > 0$ and $f(3) - f(2) - f(1) \\in \\{0, 1\\}$, we have $f(3) = 1$.\n\nSetting $m = 1$ in the condition $f(m+n) - f(n) - f(m) \\in \\{0, 1\\}$, we get $f(n+1) - f(n) \\in \\{0, 1\\}$, so $f$ is increasing.\n\nWe prove by induction that $f(3n) \\geq n$. The case $n = 1$ holds. Assume $f(3n) \\geq n$; then $f(3n + 3) - f(3n) - f(3) \\geq 0$, so $f(3n + 3) \\geq f(3n) + f(3) \\geq n + 1$.\n\nGiven $f(6042) = 2014$, it follows that $f(3n) = n$ for $1 \\leq n \\leq 2014$. Otherwise, a strict inequality would contradict $f(6042) = 2014$.\n\nSince $2014 = 3 \\times 671 + 1$, if $f(3n + 1) = n$ for $1 \\leq n \\leq 2014$, then $f(2014) = 671$.\n\nFinally, we show $f(3n+1) = n$. Indeed, $f(3n + 1) \\geq f(3n) + f(1) = n$. Also, $f(9n + 3) \\geq f(6n + 2) + f(3n + 1) \\geq 3 f(3n + 1)$, so $f(3n + 1) \\leq n + \\frac{1}{3} < n + 1$. Thus, $f(3n + 1) = n$.\n\n**Answer:** $f(2014) = 671$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21287,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a natural number, $n \\geq 2$. There are $n$ lamps arranged in a circle, labeled clockwise by natural numbers from $1$ to $n$. Each lamp can be either on or off. A switch between every two adjacent lamps enables one to change the state of both lamps simultaneously. Initially, all lamps are off. How many distinct configurations of lamp states is it possible to achieve using these switches?",
"options": [],
"answer": "See solution",
"solution": "The state of each lamp is determined by the parity of the number of switchings that affect it. Any sequence of switchings is determined by the set of switches that have been toggled an odd number of times. Thus, all possible states can be achieved by sequences where some switches are touched once and others not at all; the order does not matter.\n\nEvery pair of switch sets, where one is the complement of the other, leads to the same final state, since toggling both sets in succession changes each lamp's state twice. There are $2^n$ possible switch sets, which can be grouped into $2^{n-1}$ complementary pairs, so there are at most $2^{n-1}$ final states.\n\nIf two switch sets are neither equal nor complementary, toggling both does not return to the initial state, since some lamps will be affected by only one switch. Therefore, each complementary pair of switch sets yields a unique final state, and there are $2^{n-1}$ possible distinct final states.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21288,
"subject": "Mathematics (Olympiad)",
"question": "Determine all pairs $\\left(a, b\\right)$ of non-negative integers such that\n\n$$\n\\frac{a+b}{2} - \\sqrt{ab} = 1.\n$$",
"options": [],
"answer": "See solution",
"solution": "Without loss of generality, assume $a \\ge b$. Doubling both sides and factorizing gives\n\n$$\n\\begin{aligned}\na - 2\\sqrt{ab} + b &= 2 \\\\\n(\\sqrt{a} - \\sqrt{b})^2 &= 2 \\\\\n\\sqrt{a} - \\sqrt{b} &= \\sqrt{2}.\n\\end{aligned}\n$$\n\nWe ignore the negative root since $a \\ge b$.\n\nSolving for $a$ and rearranging gives\n\n$$\n\\begin{aligned}\n\\sqrt{a} &= \\sqrt{b} + \\sqrt{2} \\\\\na &= b + 2\\sqrt{2b} + 2.\n\\end{aligned}\n$$\n\nThus, $2\\sqrt{2b}$ must be a non-negative integer. For $2b$ to be a perfect square, $b$ must be even, and $b/2$ must be a perfect square as well. Hence $b = 2n^2$ for some $n \\ge 0$.\n\nSubstituting into the previous equation gives\n\n$$\n\\sqrt{a} = \\sqrt{2n^2} + \\sqrt{2} = n\\sqrt{2} + \\sqrt{2} = (n + 1)\\sqrt{2},\n$$\n\nso $a = 2(n + 1)^2$.\n\nTherefore, the pair $\\{a, b\\}$ must be of the form $\\{2(n + 1)^2,\\ 2n^2\\}$ for some integer $n \\ge 0$. Finally, we verify that all such pairs satisfy the original condition:\n\n$$\n\\frac{2(n + 1)^2 + 2n^2}{2} - \\sqrt{2(n + 1)^2 \\times 2n^2} = (n + 1)^2 + n^2 - 2n(n + 1) = 1.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21289,
"subject": "Mathematics (Olympiad)",
"question": "Let $P(x)$ be a non-constant polynomial with integer coefficients such that $P(0) \\neq 1$. Prove that there exist infinitely many primes $p$ such that $P(a) - a^{\\frac{p-1}{2}}$ is divisible by $p$ for some positive integer $a$ (possibly depending on $p$).",
"options": [],
"answer": "See solution",
"solution": "Consider the polynomial $Q(x) = P(x^2) - 1 \\in \\mathbb{Z}[x]$. It is well-known that there exist infinitely many prime divisors of the numbers from the set $M = \\{ Q(n) : n \\in \\mathbb{N},\\ Q(n) \\neq 0 \\}$. Moreover, since $Q(0) = P(0) - 1 \\neq 0$, among these prime divisors there exist infinitely many primes which do not divide $Q(0)$. Clearly, if $p \\mid Q(n)$ and $p \\nmid Q(0)$, then $p \\nmid n$.\n\nTherefore, there exist infinitely many primes $p$ and $n \\in \\mathbb{N}$ such that $p \\nmid n$ and $p \\mid P(n^2) - 1$. Each such $p$ satisfies the problem conditions with $a = n^2$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21290,
"subject": "Mathematics (Olympiad)",
"question": "Is it possible to line up the numbers $1, 2, 3, \\ldots, 2013$ so that the arithmetic mean of any two of the numbers is never located between them?",
"options": [],
"answer": "See solution",
"solution": "Let us show that the statement is true for any $n \\in \\mathbb{N}$. We claim that the numbers $1, 2, \\dots, n$ can be lined up so that for any pair their arithmetic mean does not lie somewhere in between.\n\nFirst, we will show that this is true for $n = 2^m$ for all $m \\in \\mathbb{N}$. We will use induction on $m$.\n\nIn the base case $m = 1$ this is obvious.\n\nNow, let us assume that for some $m$ the numbers $1, 2, \\dots, 2^m$ can be arranged into the sequence\n\n$(a_1, a_2, \\dots, a_{2^m})$ so that for any pair their arithmetic mean does not lie in between the two numbers.\n\nWe notice that the sequence\n\n$$\n(b_1, b_2, \\dots, b_{2^{m+1}}) = (2a_1 - 1, 2a_2 - 1, \\dots, 2a_{2^m} - 1, 2a_1, 2a_2, \\dots, 2a_{2^m})\n$$\n\nis a permutation of $1, 2, \\dots, 2^{m+1}$ which satisfies the condition of the problem. Indeed, by the induction hypothesis the arithmetic mean of $b_i$ and $b_j$, where either $1 \\le i < j \\le 2^m$ or $2^m + 1 \\le i < j \\le 2^{m+1}$, does not lie in between them and if $1 \\le i \\le 2^m < j \\le 2^{m+1}$, then the arithmetic mean of $b_i$ and $b_j$ is not an integer at all. We have thus proven the statement for $n = 2^m$.\n\nFinally, if the positive integer $n$ is not a power of $2$, then there exists $m \\in \\mathbb{N}$ such that $n < 2^m$. In this case we can first arrange the numbers $1, 2, \\dots, 2^m$ into the sequence that satisfies the condition and then simply remove any numbers greater than $n$. The new sequence obtained in this way will obviously still satisfy the condition.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21291,
"subject": "Mathematics (Olympiad)",
"question": "Call a team a *good team* if any pair of students in the team are mutual friends, and a *bad team* if any pair of students in the team are not mutual friends. By agreement, a team consisting of only one student is considered both good and bad.\n\nLet $N$ be the minimum possible value of $a + b$, where $a$ is the number of good teams and $b$ is the number of bad teams in a partition of $n$ students into good teams and into bad teams (not necessarily the same partition). Prove that $N = n + 1$.",
"options": [],
"answer": "See solution",
"solution": "Let us show that the minimum value of $N$ we seek is $n+1$.\n\nIf, in a certain school having $n$ students, every pair of students are mutual friends, then they have to form $b = n$ bad teams, since every team must contain only one student in this case to get a partition into bad teams. Since we must have $a \\ge 1$ also, we see that $N \\ge n + 1$.\n\nWe next show by induction on $n$ that for any school $N \\le n+1$ must hold.\n\n- When $n = 1$, we have $a = b = 1$, and therefore, $N \\le a+b = 2 = n+1$ and our claim holds in this case.\n\n- Assume that our claim $N \\le n+1$ is satisfied for the case $n = k$, and consider the case of $n = k+1$. Choose a student and call him $A$. By the induction hypothesis, there is a way to partition the student body of $k$ students excepting $A$ into $a'$ good teams and also into $b'$ bad teams in such a way that $a' + b' \\le k+1$ is satisfied.\n\nIf $a' + b' \\le k$, then adding a team consisting of student $A$ only to each of the two ways of partitioning students beside $A$ into good and bad teams as above, we obtain the partitions of all $k+1$ students into $a = a' + 1$ good teams and $b = b' + 1$ bad teams, for which we have $N \\le a+b = (a'+1) + (b'+1) \\le k+2 = n+1$.\n\nIf $a' + b' = k+1$, we can consider the following three cases:\n\n1. If among $a'$ good teams of $k$ students there is a team whose members are all friends of $A$, then adding $A$ to this team, we get a partition of all $k+1$ students into $a = a'$ good teams. We can also form a partition of all students into $b = b' + 1$ bad teams by creating a team consisting of $A$ only. Then, we get partitions into $a$ good teams and $b$ bad teams for which $N \\le a+b = a' + (b'+1) = k+2 = n+1$.\n\n2. Similarly, if among $b'$ bad teams there is a team all of whose members are not friends of $A$, then we can get a partition of all $k+1$ students into good teams and bad teams for which $N \\le k+2 = n+1$.\n\n3. If every one of $a'$ good teams contains at least one student who is not a friend of $A$, and every one of $b'$ bad teams contains at least one student who is a friend of $A$, then $A$ must have at least $a'$ students who are not his friends and also have at least $b'$ students who are his friends, which implies that the number of students in this school besides $A$ must be at least $a'+b'$. But, since the number of students besides $A$ is $k$ which equals $a'+b' - 1$ by our assumption, we get a contradiction.\n\nThis completes our induction, and establishes that $N \\le n + 1$, and proves that $N = n + 1$ is the desired answer for the problem.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21292,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$ be positive real numbers such that\n$$\na + b + c \\ge abc.\n$$\nProve that at least two of the inequalities\n$$\n\\frac{2}{a} + \\frac{3}{b} + \\frac{6}{c} \\ge 6, \\quad \\frac{2}{b} + \\frac{3}{c} + \\frac{6}{a} \\ge 6, \\quad \\frac{2}{c} + \\frac{3}{a} + \\frac{6}{b} \\ge 6\n$$\nare true.",
"options": [],
"answer": "See solution",
"solution": "Assume, for the sake of contradiction, that at least two of the numbers\n$$\n\\frac{2}{a} + \\frac{3}{b} + \\frac{6}{c}, \\quad \\frac{2}{b} + \\frac{3}{c} + \\frac{6}{a}, \\quad \\frac{2}{c} + \\frac{3}{a} + \\frac{6}{b}\n$$\nare less than $6$. Without loss of generality, assume the first and the last are less than $6$. Then\n$$\n\\frac{5}{a} + \\frac{9}{b} + \\frac{8}{c} < 12.\n$$\nAlso, because $b + c \\ge a(bc - 1)$, we have\n$$\n\\frac{1}{a} \\ge \\frac{bc - 1}{b + c}.\n$$\nIt follows that\n$$\n\\frac{5(bc - 1)}{b + c} + \\frac{9}{b} + \\frac{8}{c} < 12,\n$$\nor\n$$\n5b^{2}c^{2} + 12bc - 12b^{2}c - 12c^{2}b + 9c^{2} + 8b^{2} < 0. \\quad (1)\n$$\nCompleting squares yields\n$$\n(2bc - 2b - 3c)^2 + b^2(c - 2)^2 < 0,\n$$\na contradiction. This proves the conclusion. To obtain equalities, that is, two of the members\n$$\n\\frac{2}{a} + \\frac{3}{b} + \\frac{6}{c}, \\quad \\frac{2}{b} + \\frac{3}{c} + \\frac{6}{a}, \\quad \\frac{2}{c} + \\frac{3}{a} + \\frac{6}{b}\n$$\nare equal to $6$, we must have $c - 2 = 0$ and $2bc - 2b - 3c = 0$, that is $c = 2$, $b = 3$, $a = 1$. Therefore, the equalities hold if and only if $(a, b, c)$ is one of the triples $(1, 3, 2)$, $(3, 2, 1)$, $(2, 1, 3)$.\n\nAlternatively, rewriting $(1)$ as a quadratic form in $b$ yields\n$$\n(5c^2 - 12c + 8)b^2 - 12c(c - 1)b + 9c^2 < 0,\n$$\nwhich is impossible, because the leading coefficient\n$$\n5c^2 - 12c + 8 = 5\\left(c - \\frac{6}{5}\\right)^2 + \\frac{4}{5}\n$$\nis always positive and the discriminant\n$$\n\\begin{aligned}\n\\Delta &= [12c(c-1)]^2 - 36c^2(5c^2 - 12c + 8) \\\\\n&= 36c^2[4(c-1)^2 - (5c^2 - 12c + 8)] \\\\\n&= -36c^2(c-2)^2\n\\end{aligned}\n$$\nis always nonpositive.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21293,
"subject": "Mathematics (Olympiad)",
"question": "Take randomly five different numbers from $1, 2, \\ldots, 20$. What is the probability that there are at least two adjacent numbers among them?",
"options": [],
"answer": "See solution",
"solution": "Suppose $a_1 < a_2 < a_3 < a_4 < a_5$ are chosen from $1, 2, \\ldots, 20$. If none are adjacent, then:\n\n$$\n1 \\leq a_1 < a_2 - 1 < a_3 - 2 < a_4 - 3 < a_5 - 4 \\leq 16\n$$\n\nThus, the number of ways to select five non-adjacent numbers is the same as choosing five numbers from $1, 2, \\ldots, 16$, i.e., $C_{16}^5$. Therefore, the required probability is:\n\n$$\n\\frac{C_{20}^5 - C_{16}^5}{C_{20}^5} = 1 - \\frac{C_{16}^5}{C_{20}^5} = \\frac{232}{323}\n$$\n\nThe answer is $\\frac{232}{323}$.\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21294,
"subject": "Mathematics (Olympiad)",
"question": "For a natural number $n \\ge 2$, consider an $n \\times n$ board. Let $n^2$ points denote the centers of each of the $1 \\times 1$ squares on this board. What is the largest number of these points that can be marked in such a way that no three marked points form the vertices of a right triangle?\n\n",
"options": [],
"answer": "See solution",
"solution": "We will show an example where the number of marked points satisfies the problem's conditions. Mark the centers of all the cells in the first row and the first column, except for the center at the intersection of the first row and first column (see the figure). By simple enumeration, no three marked points form the vertices of a right triangle.\n\nSuppose this is not the maximum possible number of marked points, i.e., it is possible to mark $2n-1$ centers so that no three of them form the vertices of a right triangle. For each marked point, it must be the only marked point in its row or column. Let $a_1, a_2, \\ldots, a_n$ be the number of marked points in each column. If $a_i > 1$ for some $i$, then for each marked point in the $i$-th column, it must be the only one in its row; otherwise, a right triangle would be formed. Therefore, the total number of such points with $a_i > 1$ is at most $n$ (the total number of rows). If there are exactly $n$ such points, then the total number of marked points is $n$, which does not exceed $2n-2$. Otherwise, the sum of all $a_i$ greater than one does not exceed $n-1$. On the other hand, the sum of all $a_i$ equal to one is also at most $n$. If there are exactly $n$ such points, then again all the marked points are $n$, which does not exceed $2n-2$. Thus, the number of both types of marked points is at most $n-1$, so their total number is at most $2n-2$. Therefore, there cannot be more marked points than this value.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21295,
"subject": "Mathematics (Olympiad)",
"question": "Дадени се 21 плочка во облик на квадратче, со иста димензија. На четири плочки е запишан бројот 1, на две плочки е запишан бројот 2, на седум плочки е запишан бројот 3, а на осум плочки е запишан бројот 4. Користејќи 20 од тие плочки, Димитар формирал правоаголник со димензии $4 \\times 5$. За формираниот правоаголник, збирот на броевите во секоја редица е ист, и збирот на броевите во секоја колона е ист. Кој број стои на неискористената плочка?",
"options": [],
"answer": "See solution",
"solution": "Да го означиме со $S$ збирот на сите броеви запишани на плочките кои го формираат правоаголникот. Од условот на задачата, имаме дека $4$ е делител на $S$ и дека $5$ е делител на $S$. Значи $20$ е делител на $S$. Збирот на сите броеви запишани на $21$-ната плочка е точно $61$. Заклучуваме дека на неискористената плочка мора да стои бројот $1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21296,
"subject": "Mathematics (Olympiad)",
"question": "Let $n = 23k + 20$ with $k$ a non-negative integer. Does the equation\n$$\nx^2 + y^{11} = n + z^{2022!}\n$$\nhave any integer solutions?",
"options": [],
"answer": "See solution",
"solution": "We claim that if $n = 23k + 20$ with $k$ a non-negative integer, then the equation has no solution over the integers. Assume, for contradiction, that a solution exists.\n\nBy Fermat's little theorem,\n$$\ny^{11} \\equiv 0 \\text{ or } 1 \\pmod{23}, \\quad z^{2022!} \\equiv 0 \\text{ or } 1 \\pmod{23}\n$$\nThus,\n$$\nx^2 + y^{11} \\equiv n + z^{2022!} \\equiv 20 + (0 \\text{ or } 1) \\equiv 20 \\text{ or } 21 \\pmod{23}\n$$\nBut $x^2$ modulo 23 can only be $0, 1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18$ (the quadratic residues). Therefore,\n$$\nx^2 \\equiv 19, 20, 21, \\text{ or } 22 \\pmod{23}\n$$\nis impossible, since none of these are quadratic residues modulo 23. This is a contradiction, so there are no integer solutions.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21297,
"subject": "Mathematics (Olympiad)",
"question": "Обозначим через $I$ центр окружности, вписанной в треугольник $ABC$, а через $A_0$, $B_0$, $C_0$ — точки её касания со сторонами $BC$, $CA$, $AB$, соответственно. Пусть $A_1$ лежит на отрезке $A_0B$. Докажите, что если аналогично определены точки $B_1$ и $C_1$, то $I$ является центром вписанной окружности треугольника $O_AO_BO_C$, где $O_A$, $O_B$, $O_C$ — центры описанных окружностей треугольников $IBC_1$, $ICA_1$, $IAB_1$ соответственно.\n\n![](images/Rusija_2012_p48_data_f961278064.png \"\")",
"options": [],
"answer": "See solution",
"solution": "Заметим, что $CA_0 + AC_0 = CB_0 + AB_0 = CA$. Из условия следует, что $CA_1 + AC_1 = CB_1 + AB_1 = CA$. Значит, $CA_0 - CA_1 = AC_1 - AC_0$, то есть $A_1A_0 = C_1C_0$, и точка $C_1$ лежит на отрезке $C_0A$. Прямоугольные треугольники $IA_0A_1$ и $IC_0C_1$ равны по двум катетам, поэтому $\\angle IA_1C = \\angle IC_1B$ и $IA_1 = IC_1$. Это значит, что четырёхугольник $BC_1IA_1$ вписан. Аналогично, четырёхугольники $AB_1IC_1$ и $CA_1IB_1$ также вписаны, и $IA_1 = IB_1 = IC_1$.\n\nЛинии центров $O_BO_C$, $O_CO_A$, $O_AO_B$ являются серединными перпендикулярами к общим хордам $IA_1$, $IB_1$, $IC_1$ соответственно; длины этих хорд равны. Значит, расстояния от $I$ до сторон треугольника $O_AO_BO_C$ равны $\\frac{IA_1}{2} = \\frac{IB_1}{2} = \\frac{IC_1}{2}$. Поскольку углы $\\angle IBA_1$, $\\angle IAC_1$, $\\angle ICB_1$ острые, то отрезки $O_BO_C$, $O_CO_A$, $O_AO_B$ пересекают лучи $IA_1$, $IB_1$, $IC_1$ соответственно. Следовательно, $I$ лежит внутри треугольника $O_AO_BO_C$, то есть это и есть центр вписанной окружности треугольника $O_AO_BO_C$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21298,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute-angled triangle with $AB < AC$, let $O$ be its circumcentre and let $D$ be an interior point of the segment $BC$. The line through $D$ and perpendicular to $BC$ crosses the lines $AO$, $AC$, and $AB$ at $W \\neq D$, $X$, and $Y$, respectively. The circles $ABC$ and $AXY$ cross again at $Z$. Prove that, if $OW = OD$, then $DZ$ is tangent to the circle $AXY$.",
"options": [],
"answer": "See solution",
"solution": "Let $AO$ cross $BC$ at $E$ and let $M$ be the midpoint of the side $BC$. Acuteness of the triangle $ABC$ and $AB < AC$ force $E$ to be an interior point of the segment $CM$. Also, $E \\neq D$, since $W \\neq D$. The condition $OW = OD$ then forces $D$ to be an interior point of the segment $BM$. It then follows that $O$ is an interior point of the segment $EW$; and since the triangle $DEW$ is right-angled at $D$, the condition $OW = OD$ implies that $O$ is the circumcentre of this triangle. Hence $OD = OE$, showing that $D$ and $E$ are reflections of one another in the perpendicular bisector $OM$ of the side $BC$.\n\nSince $AB < AC$, the parallel through $A$ to $BC$ crosses the circle $ABC$ again at some point $Z'$. By the preceding, the triangles $ABE$ and $Z'CD$ are then reflections of one another in $OM$.\n\nHence $\\angle DZ'C = \\angle BAE = \\angle BAO = 90^\\circ - \\frac{1}{2}\\angle AOB = 90^\\circ - \\angle ACB = 90^\\circ - \\angle XCD = \\angle DXC$, so the quadrilateral $CDXZ'$ is cyclic.\n\nThen $\\angle Z'XY = \\angle Z'CX = \\angle EBA = \\angle CBA = \\angle CBY = \\angle Z'AY$, so $X$ lies on the circle $AYZ'$. Alternatively, but equivalently, $Z'$ lies on the circle $AXY$, so $Z' = Z$.\n\nFinally, write $\\angle XZD = \\angle XCD = \\angle ACB = \\angle CAZ = \\angle XAZ$, to conclude that $DZ$ is tangent to the circle $AXY$, by the converse of the alternate segment theorem.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21299,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute-angled triangle, $H$ its orthocenter, and $M$ the midpoint of side $AC$. Let $C_1$ be the orthogonal projection of $C$ onto $AB$, and $H_1$ the reflection of $H$ with respect to line $AB$. Let $P$, $Q$, and $R$ be the orthogonal projections of $C_1$ onto the lines $AH_1$, $AC$, and $BC$, respectively. Let $M_1$ be a point such that the circumcenter 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": "Start by considering the following:\n\n**Lemma.** If points $D, E$ are interior to a triangle $ABC$ and are isogonal conjugates with respect to it (i.e., $\\angle DAB \\equiv \\angle EAC$ and $\\angle DBA \\equiv \\angle EBC$, and also $\\angle DCA \\equiv \\angle ECB$), while point $D$ is projected onto the sides of triangle $ABC$ at $D_1, D_2, D_3$, then the circumcenter of triangle $D_1D_2D_3$ is the midpoint of segment $[DE]$.\n\n*Proof.*\n\nLet $D'_1, D'_2, D'_3$ be the reflections of $D$ with respect to the sides of the triangle. We get $CD'_1 = CD = CD'_2$, while $\\angle ECD'_1 \\equiv \\angle ECB + \\angle BCD'_1 \\equiv \\angle ACD + \\angle BCD \\equiv \\angle ACB$, and similarly $\\angle ECD'_2 \\equiv \\angle ACB$, whence $\\triangle ECD'_1 \\equiv \\triangle ECD'_2$, so $ED'_1 = ED'_2$. In a similar way we get $ED'_2 = ED'_3$, showing that $E$ is the circumcenter of triangle $D'_1D'_2D'_3$. Consider now the homothety of center $D$ and ratio $\\frac{1}{2}$. The circle $(D'_1, D'_2, D'_3)$ centered at $E$ maps onto the circle $(D_1, D_2, D_3)$ centered at the midpoint of the segment $[DE]$. $\\square$\n\n\n\nLet $T$ be the intersection point of the line $AH_1$ with $BC$, and $O$ be the circumcenter of triangle $ABC$. Notice we may apply the above lemma to points $C_1$ and $O$, isogonal conjugates with respect to triangle $ACT$. If, for example, $\\angle B \\geq \\angle C$, then $\\angle TCC_1 = 90^\\circ - \\angle B = \\angle ACO$ and $\\angle TAC_1 = \\angle HAC_1 = 90^\\circ - \\angle B = \\angle CAO$. Thus the circumcenter of triangle $PQR$ is the midpoint $O_1$ of segment $OC_1$.\n\nThis shows that in quadrilateral $OMC_1M_1$ the diagonals bisect each other, hence it is a parallelogram. Then $C_1M_1 \\parallel OM$ and $C_1M_1 = OM = \\frac{1}{2}BH$, so $C_1M_1$ is a midline in triangle $BHH_1$, which shows that $M_1$ lies on $BH_1$ (in fact, it is the very midpoint of this segment).\n\n\n\n*Alternative Solution.* Notice that $O_1$ is the circumcenter of triangle $PQR$. If $U$ is the midpoint of segment $[AC_1]$, then $UO_1$ is the perpendicular bisector of segment $[PQ]$, since $UP = \\frac{1}{2}AC_1 = UQ$, and $UO_1 \\parallel AO$. Using the cyclic quadrilateral $APC_1Q$ and the thesis, we obtain $\\angle AQP = \\angle AC_1P = 90^\\circ - \\angle PAC_1 = 90^\\circ - \\angle HAC_1 = \\angle ABC = \\frac{1}{2}\\angle AOC = 90^\\circ - \\angle OAC$, whence $AO \\perp PQ$.\n\nAlso, if $V$ is the midpoint of segment $[CC_1]$, then $VO_1$ is the perpendicular bisector of segment $[QR]$, since $VR = \\frac{1}{2}CC_1 = VQ$, $VO_1 \\parallel CO$, and $CO \\perp QR$ (using the cyclic quadrilateral $ABRQ$ and the thesis we obtain $\\angle CQR = \\angle ABC = \\frac{1}{2}\\angle AOC = 90^\\circ - \\angle ACO$).\n\nWe finish by noticing that $BCAH_1$ is an orthodiagonal quadrilateral inscribed in a circle centered at $O$, and using the following:\n\n**Lemma.** If $ABCD$ is a cyclic orthodiagonal quadrilateral, its diagonals meeting at $E$, inscribed in a circle centered at $O$, then the reflection of the midpoint $M$ of the side $[AB]$ with respect to the midpoint $O_1$ of the segment $[OE]$ is the midpoint $N$ of the side $[CD]$.\n\n*Proof.*\n\nThe proof of this lemma comes from considering the midpoints $O', O''$ of the chords $[AC]$ and $[BD]$, and noticing that $MO' \\parallel BC \\parallel NO_1$, $MO' = \\frac{1}{2}BC = NO_1$. Thus, $MO'NO''$ is a parallelogram, hence $[MN]$ and $[O'O'']$ cross at their midpoints. Now, $OO'EO''$ is a rectangle, therefore $[OE]$ and $[O'O'']$ bisect each other, which shows that the midpoints of the segments $[MN]$ and $[OE]$ coincide.\n\n\n\nOther properties of this configuration are evidenced by the following:\n\n*Alternative Solution.* If $C_1S \\perp BH_1$, $S \\in BH_1$, then, using cyclic quadrilaterals' properties:\n\n$$\n\\begin{aligned}\n\\angle C_1PS &= \\angle C_1H_1S, & \\angle C_1PQ &= \\angle C_1AQ, \\\\\n\\angle C_1RQ &= \\angle C_1CQ, & \\angle C_1RS &= \\angle C_1BS,\n\\end{aligned}\n$$\n\nhence\n\n$$\n\\angle C_1PS + \\angle C_1PQ + \\angle C_1RQ + \\angle C_1RS = 180^{\\circ},\n$$\n\nwhich shows the quadrilateral $PQRS$ is cyclic. Then,\n\n$$\n\\angle MC_1C = \\angle MCC_1 = \\angle ABH_1 = \\angle SC_1H_1,\n$$\n\nhence points $M$, $C_1$, $S$ are collinear. Lastly,\n\n$$\n\\angle AMC_1 = 180^{\\circ} - 2\\angle MAC_1 = 180^{\\circ} - 2\\angle QPC_1\n$$\n\nand\n\n$$\n\\angle QPC_1 = \\angle CAC_1 = \\angle BH_1C_1 = \\angle SPC_1,\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21300,
"subject": "Mathematics (Olympiad)",
"question": "Show that the equation $p_1^2 + p_2^2 + \\cdots + p_k^2 = 2010$ can only be solved for $k = 7$, where each $p_i$ is a prime number. Find all possible sets of such primes.",
"options": [],
"answer": "See solution",
"solution": "The 15 smallest prime squares are:\n\n4, 9, 25, 49, 121, 169, 289, 361, 529, 841, 961, 1369, 1681, 1849, 2209.\n\nSince $2209 > 2010$, $k \\leq 14$.\n\nFor odd primes, $p^2 \\equiv 1 \\mod 8$, and $2010 \\equiv 2 \\mod 8$. If all $p_i$ are odd, then $k \\cdot 1 \\equiv 2 \\mod 8$, so $k = 2$ or $k = 10$.\n\n$k = 2$: $2010 \\equiv 0 \\mod 3$, but $x^2 \\equiv 0$ or $1 \\mod 3$, so both $p_1, p_2 \\equiv 0 \\mod 3$, impossible.\n\n$k = 10$: The sum of the first 10 odd prime squares exceeds 2010, so impossible.\n\nIf one prime is 2, then $4 + (k-1) \\cdot 1 \\equiv 2 \\mod 8$, so $k \\equiv 7 \\mod 8$, i.e., $k = 7$.\n\nFor $k = 7$, the possible solutions are:\n\n$$\n4 + 9 + 49 + 169 + 289 + 529 + 961 = 2010,\n$$\n$$\n4 + 9 + 25 + 121 + 361 + 529 + 961 = 2010,\n$$\n$$\n4 + 9 + 25 + 49 + 121 + 841 + 961 = 2010,\n$$\n$$\n4 + 9 + 49 + 121 + 169 + 289 + 1369 = 2010.\n$$\n\nIf 25 is included, then for the remaining 4 prime squares, considerations modulo 10 show that 3 out of 4 from $\\{121, 361, 841, 961\\}$ must be used, and two cases work. If 25 is not included, then for the remaining 5, 4 must be from $\\{49, 169, 289, 529, 1369\\}$, and two cases work.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21301,
"subject": "Mathematics (Olympiad)",
"question": "A circle is called *good colored* if the vertices of any equilateral triangle inscribed in this circle are colored in distinct colors. Let $k$ be a circle with radius $2$.\n\n(a) Is there a coloring of the points on $k$ and inside $k$ in three colors such that $k$ and any circle with radius at least $1$ that touches $k$ are good colored?\n\n(b) Is there such a coloring in seven colors?",
"options": [],
"answer": "See solution",
"solution": "a) Assume that such a coloring exists and $a$, $b$, and $c$ are the colors. Let $O$ be the center of $k$ and consider an equilateral triangle $OBC$ with side $\\sqrt{3}$. Its circumcircle has radius $1$ and touches $k$. If the color of $O$ is $a$, then the colors of $B$ and $C$ are $b$ and $c$. This shows that the points of the circle $k'(0, \\sqrt{3})$ are colored in $b$ and $c$. Consider now an equilateral $\\triangle PQR$ with circumcircle $k$. Denote by $X_1$, $X_2$ and $Y_1$, $Y_2$ the intersection points of $PQ$ and $PR$ with $k'$, respectively ($X_1$ and $Y_1$ the closer points to $P$). Since the circumcircle of the equilateral $\\triangle PX_2Y_2$ touches $k$ and has radius at least $1$, it follows that the color of $P$ is $a$. Analogously, $Q$ and $R$ have the same color, a contradiction.\n\nb) Let $ABCDEF$ be a regular hexagon inscribed in $k$. Let the color of $O$ be $1$, let the color of the points inside the sector $OAB$, the radius $OA$ and the arc $AB$ without $B$ be $2$, let the color of the points inside the sector $OBC$, the radius $OB$ and the arc $BC$ without $C$ be $2$, etc. It is easy to see that this coloring has the desired properties.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21302,
"subject": "Mathematics (Olympiad)",
"question": "Хавтгайн $n$ цэгээс тогтох цэгүүдийн систем $N$-ийн хувьд $d = \\max_{A, B \\in N} |AB|$, $q = \\min_{A \\neq B \\in N} |AB|$ бол\n\n$$\n\\frac{d}{q} > \\frac{\\sqrt{3}}{2} (\\sqrt{n} - 1)\n$$\nболохыг батал.",
"options": [],
"answer": "See solution",
"solution": "Диаметр нь $d$ байх цэгүүдийн системийг $\\frac{d}{\\sqrt{3}}$ радиустай дугуйд багтааж болно (Юнгийн теорем). Цэг бүр дээр төвтэй $\\frac{q}{2}$ радиустай дугуй байгуулбал, $q$ нь $\\min_{A \\neq B \\in N} |AB|$ тул ямар ч хоёр тойрог огтлолцохгүй (шүргэлцэж болох ч). Бүх цэгийг $\\frac{d}{\\sqrt{3}}$ радиустай дугуйгаар хучиж болох тул $\\frac{d}{\\sqrt{3}} + \\frac{q}{2}$ радиустай дугуйгаар, цэгүүдийг харгалзах тойргуудын хамт хучиж болно. Иймд энэ дугуйн талбай нь бүх жижиг дугуйн талбайн нийлбэрээс их:\n\n$$\n\\pi \\left( \\frac{d}{\\sqrt{3}} + \\frac{q}{2} \\right)^2 > n \\cdot \\pi \\left( \\frac{q}{2} \\right)^2\n$$\n\nХувиргавал:\n\n$$\n\\frac{d}{\\sqrt{3}} + \\frac{q}{2} > \\sqrt{n} \\cdot \\frac{q}{2}\n$$\n\nҮүнээс:\n\n$$\n\\frac{d}{q} > \\frac{\\sqrt{3}}{2}(\\sqrt{n} - 1)\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21303,
"subject": "Mathematics (Olympiad)",
"question": "When a certain unfair die is rolled, an even number is 3 times as likely to appear as an odd number. The die is rolled twice. What is the probability that the sum of the numbers rolled is even?\n\n(A) $\\frac{3}{8}$ (B) $\\frac{4}{9}$ (C) $\\frac{5}{9}$ (D) $\\frac{9}{16}$ (E) $\\frac{5}{8}$",
"options": [],
"answer": "See solution",
"solution": "Suppose that the probability of rolling an odd number is $p$. The probability of rolling an even number is then $3p$. Because $p + 3p = 1$, it follows that $p = \\frac{1}{4}$, so the probability of rolling an odd number is $\\frac{1}{4}$ and the probability of rolling an even number is $1 - \\frac{1}{4} = \\frac{3}{4}$.\n\nThe sum will be even if both rolls are even or both rolls are odd. Therefore, the probability of rolling an even sum is:\n\n$$\n\\frac{3}{4} \\cdot \\frac{3}{4} + \\frac{1}{4} \\cdot \\frac{1}{4} = \\frac{9}{16} + \\frac{1}{16} = \\frac{10}{16} = \\frac{5}{8}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21304,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle. Triangles $PAB$ and $QAC$ are constructed outside of triangle $ABC$ such that $AP = AB$, $AQ = AC$, and $\\angle BAP = \\angle CAQ$. Segments $BQ$ and $CP$ meet at $R$. Let $O$ be the circumcenter of triangle $BCR$. Prove that $AO \\perp PQ$.",
"options": [],
"answer": "See solution",
"solution": "We first note that *APBR* and *AQCR* are cyclic quadrilaterals. It is easy to see that triangles *APC* and *ABQ* are congruent to each other, implying that $\\angle APR = \\angle APC = \\angle ABQ = \\angle ABR$. Thus, *APBR* is a cyclic quadrilateral. Likewise, we can show that *AQCR* is also cyclic.\n\n\n\nLet $\\angle PAB = 2x$. Then in isosceles triangle *APB*, $\\angle APB = 90^\\circ - x$. In cyclic quadrilateral *APBR*, $\\angle ARB = 180^\\circ - \\angle APB = 90^\\circ + x$. Likewise, $\\angle ARC = 90^\\circ + x$. Hence $\\angle BRC = 360^\\circ - \\angle ARB - \\angle ARC = 180^\\circ - 2x$. It follows that $\\angle BOC = 4x$.\n\n**First Solution:** Reflect $C$ across line $AQ$ to $D$. Then $\\angle BAD = 4x + \\angle BAC = \\angle BAQ$. It is easy to see that triangles $BAD$ and $PAQ$ are congruent, implying that $\\angle ADB = \\angle AQP = y$.\n\n\n\nNote also that $CAD$ and $COB$ are two isosceles triangles with the same vertex angle, and so they are similar to each other. It follows that triangle $CAO$ and $CBD$ are similar by SAS, implying that $\\angle CAO = \\angle CDB = z$.\n\nThe angle formed by lines $AO$ and $PQ$ is equal to\n\n$$\n180^\\circ - \\angle OAQ - \\angle AQP = 180^\\circ - \\angle OAC - \\angle CAQ - \\angle AQP = 180^\\circ - z - 2x - y.\n$$\n\nSince $AQ$ is perpendicular to the base $CD$ in isosceles triangle $ACD$, we have\n\n$$\n90^\\circ = \\angle QAD + \\angle CDA = \\angle QAD + \\angle ADB + \\angle BDC = 2x + y + z.\n$$\n\nCombining the last two equations yields that the angle formed by lines $AO$ and $PQ$ is equal to $90^\\circ$; that is, $AO \\perp PQ$.\n\n**Second Solution:** We maintain the same notations as in the first solution. Let $M$ be the midpoint of arc $\\widehat{BC}$ on the circumcircle of triangle $BOC$. Then $BM = CM$. Since triangles $APC$ and $ABQ$ are congruent, $PC = BQ$. Since $BRMC$ is cyclic, $\\angle PCM = \\angle RCM = \\angle RBM = \\angle QBM$. Hence triangles $BMQ$ and $CMP$ are congruent by SAS. It follows triangles $MPQ$ and $MBC$ are similar. Since $\\angle BOC = 4x$, $\\angle MBC = \\angle MCB = x$, and so $\\angle MPQ = x$.\n\n\n\nNote that both triangles $PAB$ and $MOB$ are isosceles triangles with vertex angle $2x$; that is, they are similar to each other. Hence triangles $BMP$ and $BOA$ are also similar by SAS, implying that $\\angle OAB = MPB = s$. We also note that in isosceles triangle $APB$,\n\n$$\n90^\\circ = \\angle APB + \\angle PAB/2 = \\angle APQ + \\angle QPM + \\angle MPB + \\angle PAB/2 = \\angle APQ + 2x + s.\n$$\n\nPutting the above together, we conclude that\n\n$$\n\\angle PAO + \\angle APQ = \\angle PAB + \\angle BAO + \\angle APQ = 2x + s + \\angle APQ = 90^\\circ,\n$$\n\nthat is $AO \\perp PQ$.\n\n**Third Solution:** We consider two rotations:\n\n$\\mathbf{R}_1$: a counterclockwise $2x$ (degree) rotation centered at $A$,\n\n$\\mathbf{R}_2$: a clockwise $4x$ (degree) rotation centered at $O$.\n\nLet $\\mathbf{T}$ denote the composition $\\mathbf{R}_1\\mathbf{R}_2\\mathbf{R}_1$. Then $\\mathbf{T}$ is a counterclockwise $2x - 4x + 2x = 0^\\circ$ rotation; that is, $\\mathbf{T}$ is translation. Note that\n\n$$\n\\mathbf{T}(P) = \\mathbf{R}_1(\\mathbf{R}_2(\\mathbf{R}_1(P))) = \\mathbf{R}_1(\\mathbf{R}_2(B)) = \\mathbf{R}_1(C) = Q,\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21305,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\mathbb{Z}$ denote the set of integers and let $S \\subset \\mathbb{Z}$ be the set of integers that are at least $10^{100}$. Fix a positive integer $c$. Determine all functions $f: S \\to \\mathbb{Z}$ satisfying\n\n$$\nf(xy + c) = f(x) + f(y) \\quad \\text{for all } x, y \\in S.\n$$",
"options": [],
"answer": "See solution",
"solution": "Observe that if $x_1, y_1, x_2, y_2 \\in S$ with $x_1 y_1 = x_2 y_2$ then\n\n$$\nf(x_1) + f(y_1) = f(x_2) + f(y_2). \\quad (1)\n$$\n\nThis tells us that for $u, v, w \\in S$,\n\n$f(uv) + f(w) = f(u) + f(vw)$, so $f(uv) - f(u) - f(v) = f(vw) - f(w) - f(v)$.\n\nNotice the right-hand side is independent of $u$, so the same must be true of the left-hand side. By replicating the argument with $u$ and $v$ switched, we also see the left-hand side is independent of $v$, so in fact\n\n$$\nf(uv) - f(u) - f(v) = k \\quad \\text{for some constant } k \\in \\mathbb{Z}. \\quad (2)\n$$\n\nUsing (1) again we have, for $y, z \\in S$,\n\n$$\n\\begin{gathered}\nf(cz) + f(y) = f(z) + f(cy), \\\\\n\\text{so } f(cy) - f(y) = l, \\quad \\text{for some constant } l \\in \\mathbb{Z}.\n\\end{gathered} \\quad (3)\n$$\n\nSetting $x = cz$ in the original functional equation for $z \\in S$ shows\n\n$$\n\\begin{gathered}\nf(c(yz + 1)) \\stackrel{(3)}{=} f(yz + 1) + l = f(cz) + f(y) \\stackrel{(3)}{=} f(z) + f(y) + l, \\\\\n\\text{so } f(yz + 1) = f(y) + f(z).\n\\end{gathered}\n$$\n\nLet $x \\in S$ and set $y = x, z = x + 2$ in the above to get\n\n$$\n\\begin{gathered}\nf((x+1)^2) \\stackrel{(2)}{=} 2f(x+1) + k = f(x) + f(x+2) \\\\\n\\Rightarrow f(x) + f(x+2) - 2f(x+1) = -k = \\text{constant}\n\\end{gathered}\n$$\n\nwhich forces $f$ to be a quadratic. By setting $x = y$ in the original functional equation and considering the degree of both sides, we see $f$ must in fact be constant. The only constant function that satisfies the condition is $f \\equiv 0$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21306,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ and $b$ be natural numbers such that\n\n$$\na \\cdot (a, b) + b \\cdot [a, b] = a^2 + b^2,\n$$\n\nwhere $(a, b)$ and $[a, b]$ denote the greatest common divisor and least common multiple of $a$ and $b$, respectively. Find $(2010^a - 1, 2010^b - 1)$.",
"options": [],
"answer": "See solution",
"solution": "Let $(a, b) = d$, then $a = dx$ and $b = dy$, where $x$ and $y$ are coprime numbers. We have $[a, b] = \\frac{ab}{(a,b)} = dxy$, thus, we can rewrite our equality as follows:\n\n$$\nd^2x + d^2xy^2 = d^2x^2 + d^2y^2,\n$$\n\nor\n\n$$\nx + xy^2 = x^2 + y^2.\n$$\n\nHence, $(x - 1)(x - y^2) = 0$ and we get $x = 1$ or $x = y^2$. $x$ and $y$ are coprime; therefore, from the second equality it follows that $x = 1$, $y = 1$. In both these cases $x = 1$, which gives $b = ay$ for some natural $y$. In this case, we get\n\n\n\n$$\n2010^b - 1 = (2010^a)^y - 1 = (2010^a - 1)\\left((2010^a)^{y-1} + \\dots + 2010^a + 1\\right).\n$$\n\nHence,\n\n$$\n(2010^a - 1, 2010^b - 1) = 2010^a - 1.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21307,
"subject": "Mathematics (Olympiad)",
"question": "There are $n$ line segments on the plane, no three intersecting at a point, and each pair intersecting once in their respective interiors. Tony and his $2n - 1$ friends each stand at a distinct endpoint of a line segment. Tony wishes to send Christmas presents to each of his friends as follows:\n\nFirst, he chooses an endpoint of each segment as a \"sink\". Then he places the present at the endpoint of the segment he is at. The present moves as follows:\n\n- If it is on a line segment, it moves towards the sink.\n- When it reaches an intersection of two segments, it changes the line segment it travels on and starts moving towards the new sink.\n\nIf the present reaches an endpoint, the friend on that endpoint can receive their present. Prove Tony can send presents to exactly $n$ of his $2n - 1$ friends.",
"options": [],
"answer": "See solution",
"solution": "Draw a circle that encloses all the intersection points between line segments and extend all line segments until they meet the circle, and then move Tony and all his friends to the circle. Number the intersection points with the circle from $1$ to $2n$ anticlockwise, starting from Tony (Tony has number $1$). We will prove that the friends eligible to receive presents are the ones on even-numbered intersection points.\n\n*First part: at most $n$ friends can receive a present.*\n\nThe solution relies on a well-known result: the $n$ lines determine regions inside the circle; then it is possible to paint the regions with two colors such that no regions with a common (line) boundary have the same color. The proof is an induction on $n$: the fact immediately holds for $n = 0$, and the induction step consists of taking away one line $\\ell$, painting the regions obtained with $n - 1$ lines, drawing $\\ell$ again and flipping all colors on exactly one half-plane determined by $\\ell$.\n\nNow consider the line starting on point $1$. Color the regions in red and blue such that neighboring regions have different colors, and such that the two regions that have point $1$ as a vertex are red on the right and blue on the left, from Tony's point of view. Finally, assign to each red region the clockwise direction and to each blue region the anticlockwise direction. Because of the coloring, every boundary will have two directions assigned, but the directions are the same since every boundary divides regions of different colors. Then the present will follow the directions assigned to the regions: it certainly does for both regions in the beginning, and when the present reaches an intersection it will keep bordering one of the two regions it was dividing.\n\nTo finish this part of the problem, consider the regions that share a boundary with the circle. The directions alternate between outgoing and incoming, starting from $1$ (outgoing), so all even-numbered vertices are directed as incoming and are the only ones able to receive presents.\n\n*Second part: all even-numbered vertices can receive a present.*\n\nFirst notice that, since every two chords intersect, every chord separates the endpoints of each of the other $n - 1$ chords. Therefore, there are $n - 1$ vertices on each side of every chord, and each chord connects vertices $k$ and $k + n$, $1 \\leq k \\leq n$.\n\nWe prove a stronger result by induction in $n$: let $k$ be an integer, $1 \\leq k \\leq n$. Direct each chord from $i$ to $i + n$ if $1 \\leq i \\leq k$ and from $i + n$ to $i$ otherwise; in other words, the sinks are $k + 1, k + 2, \\dots, k + n$. Now suppose that each chord sends a present, starting from the vertex opposite to each sink, and all presents move with the same rules. Then $k - i$ sends a present to $k + i + 1$, $i = 0, 1, \\dots, n - 1$ (indices taken modulo $2n$). In particular, for $i = k - 1$, Tony, in vertex $1$, sends a present to vertex $2k$. Also, the $n$ paths the presents make do not cross (but they may touch). More formally, for all $i$, $1 \\leq i \\leq n$, if one path takes a present from $k - i$ to $k + i + 1$, separating the circle into two regions, all paths taking a present from $k - j$ to $k + j + 1$, $j < i$, are completely contained in one region, and all paths taking a present from $k-j$ to $k+j+1$, $j>i$, are completely contained in the other region. For instance, possible$^1$ paths for $k=3$ and $n=5$ follow:\n\n\n\nThe result is true for $n=1$. Let $n>1$ and assume the result is true for fewer chords. Consider the chord that takes $k$ to $k+n$ and remove it. Apply the induction hypothesis to the remaining $n-1$ lines: after relabeling, presents would go from $k-i$ to $k+i+2$, $1 \\leq i \\leq n-1$ if the chord were not there.\n\nReintroduce the chord that takes $k$ to $k+n$. From the induction hypothesis, the chord intersects the paths of the presents in the following order: the $i$-th path the chord intersects is the one that takes $k-i$ to $k+i$, $i=1,2,\\dots,n-1$.\n\n\n\nPaths without chord $k \\to k+n$\n\n\n\nCorrected paths with chord $k \\to k+n$\n\nThen the presents cover the following new paths: the present from $k$ will leave its chord and take the path towards $k+1$; then, for $i=1,2,\\dots,n-1$, the present from $k-i$ will meet the chord from $k$ to $k+n$, move towards the intersection with the path towards $k+i+1$ and go to $k+i+1$, as desired. Notice that the paths still do not cross. The induction (and the solution) is now complete.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21308,
"subject": "Mathematics (Olympiad)",
"question": "Let $T_n(x) = x^2 + (x+1)^2 + \\dots + (x+n-1)^2$ for $n > 1$. \n\n**(a)** Show that $T_n(x)$ can be a perfect square for infinitely many positive integers $x$ and $n$.\n\n**(b)** Rewrite the equation $y^2 = x^2 + (x+1)^2 + \\dots + (x+n-1)^2$ in the form\n\n$$\n(2y)^2 - n(2x + n - 1)^2 = \\frac{(n-1)n(n+1)}{3}\n$$\n\nand show that, by the theory of Pell equations, there are infinitely many positive integer solutions $(x, y)$ for suitable $n$.\n\n*Remark.* Write $T_n(x) = n(x + (n-1)/2)^2 + (n-1)n(n+1)/12$ to derive part (a) from part (b). Notice that if $(n+1)/12$ is a perfect square, say $n = 12m^2 - 1$, then the equation has the solution $x = -6m^2 + m + 1$ and $y = m(12m^2 - 1)$, so part (b) applies to show that there are infinitely many solutions in positive integers.",
"options": [],
"answer": "See solution",
"solution": "One such solution is obtained by taking\n\n$$\nz - \\left(x + \\frac{n-1}{2}\\right) = 2 \\quad \\text{and} \\quad z + \\left(x + \\frac{n-1}{2}\\right) = \\frac{n^2-1}{24},\n$$\n\ni.e., by letting $x$ and $z$ have the positive integer values\n\n$$\nx = \\frac{(n-25)(n+1)}{48} \\quad \\text{and} \\quad z = 1 + \\frac{n^2-1}{48}.\n$$\n\nBy the theory of Pell equations, there are infinitely many pairs of positive integers $u, v$ such that\n\n$$\nu^2 - nv^2 = 1.\n$$\n\nUsing the identity\n\n$$\n(2y_0 + (2x_0 + n - 1)\\sqrt{n})(u + v\\sqrt{n}) = 2y + (2x + n - 1)\\sqrt{n},\n$$\n\nwhere\n\n$$\nx = x_0u + y_0v + \\frac{(n-1)(u-1)}{2}, \\quad y = y_0u + x_0nv + \\frac{(n-1)nv}{2},\n$$\n\nand $x_0, y_0$ are initial solutions, we find that $T_n(x)$ is a perfect square for infinitely many $x$ and $n$.\n\nFurther, since $x_0 \\ge -(n-1)/2$ and $y_0 > (n-1)/2$, it follows that $y \\ge y_0u > 0$ and $x \\ge y_0v - (n-1)/2 > 0$, so the solutions are positive integers.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21309,
"subject": "Mathematics (Olympiad)",
"question": "Prove the inequality\n\n$$\n2010 < \\frac{2^2 + 1}{2^2 - 1} + \\frac{3^2 + 1}{3^2 - 1} + \\dots + \\frac{2010^2 + 1}{2010^2 - 1} < 2010\\frac{1}{2}\n$$",
"options": [],
"answer": "See solution",
"solution": "Since $\\frac{n^2 + 1}{(n-1)(n+1)} = 1 + \\frac{1}{n-1} - \\frac{1}{n+1}$, the given sum can be rewritten as:\n\n$$\n1 + \\frac{1}{1} - \\frac{1}{3} + 1 + \\frac{1}{2} - \\frac{1}{4} + \\dots + 1 + \\frac{1}{2009} - \\frac{1}{2011}\n$$\n\nThis simplifies to:\n\n$$\n2010 + \\frac{1}{2} - \\frac{1}{2010} - \\frac{1}{2011}\n$$\n\nBecause $0 < \\frac{1}{2} - \\frac{1}{2010} - \\frac{1}{2011} < \\frac{1}{2}$, the inequality is proved.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21310,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $n$ such that for some positive integer $d$, $dn + 1$ divides $d^2 + n^2$.",
"options": [],
"answer": "See solution",
"solution": "If $d^2 - a < 0$, then $a - d^2 > 0$. Then it must be $a - d^2 \\ge ad^2 + 1$. Since $d$ is positive, we have $a \\ge ad^2 + 1 + d^2 > a$, which is also impossible.\n\nHence, the only possibility is $d^2 - a = 0$, which gives $a = d^2$ and $n = d^3$. In that case $d^4 + 1$ divides $d^2 + d^6 = d^2(d^4 + 1)$, so all possible numbers $n$ are cubes of positive integers.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21311,
"subject": "Mathematics (Olympiad)",
"question": "Find the largest positive integer $n$ with the following property: one can choose $2007$ distinct integers from the interval $[2 \\cdot 10^{n-1}, 10^n)$ such that for any $i, j$, $1 \\leq i < j \\leq n$, there exists a chosen number $a_1 a_2 \\dots a_n$, with $a_j \\geq a_i + 2$.",
"options": [],
"answer": "See solution",
"solution": "Consider $2007$ positive integers with the desired property. Increase all even digits of these numbers by $1$. If $a_i$ and $a_j$ are of the same parity, they either do not change (if both are odd) or increase by $1$ (if both are even). If $a_i$ and $a_j$ are of distinct parity, then $a_j$ and $a_i + 2$ are also of distinct parity, and thus, $a_j \\geq a_i + 2$ implies $a_j > a_i + 2$. Therefore, if $a_j \\geq a_i + 2$, then after increasing all even digits by $1$, the corresponding inequality is also true. Thus, we obtain $2007$ natural numbers (some of them possibly equal), each of them with odd digits that satisfy the condition of the problem.\n\nWrite all the numbers in a table of $2007$ rows and $n$ columns. Since every number in the first column is at least $3$, we have that there is at least one digit greater than $3$ in any of the remaining columns. Thus, there are no columns containing only digits $1$ and $3$. There are $5^{2007}$ possible columns consisting of $1, 3, 5, 7, 9$ and $2^{2007}$ possible columns consisting of digits $1$ and $3$. Therefore, there are at most $1 + 5^{2007} - 2^{2007}$ columns, i.e., $n \\leq 1 + 5^{2007} - 2^{2007}$.\n\nWe construct a table with $2007$ rows and $1 + 5^{2007} - 2^{2007}$ columns having the desired property in the following way:\n\n1. Write on the first row $5^{2006}$ digits $1$, then $5^{2006}$ digits $3$, ..., $5^{2006}$ digits $9$. On the second row, write under equal digits in the first row consecutively $5^{2005}$ digits $1, 3, 5, 7, 9$ and so on, down to the last row. Consider the rows $i$ and $j$ for which $i < j$. It is clear that if $a_i$ and $a_j$ are the first distinct digits (from up) in these columns, then $a_j > a_i$, which implies that $a_j \\geq a_i + 2$. Therefore, the table has $5^{2007}$ columns and satisfies (with the exception that the numbers corresponding to the rows of the table lie in the interval $[2 \\cdot 10^{n-1}, 10^n)$) the desired property.\n\n2. Delete all columns consisting only of digits $1$ and $3$. The table now has $5^{2007} - 2^{2007}$ columns.\n\n3. Add a first column consisting only of digit $3$. We have a table with $1 + 5^{2007} - 2^{2007}$ rows satisfying all conditions of the problem.\n\nTherefore, the desired number equals $n = 1 + 5^{2007} - 2^{2007}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21312,
"subject": "Mathematics (Olympiad)",
"question": "Let $f$ be a function defined on the set of real numbers $\\mathbb{R}$, taking values in $\\mathbb{R}$ and satisfying the condition\n\n$$\nf(\\cot x) = \\sin 2x + \\cos 2x\n$$\n\nfor every $x$ belonging to the open interval $(0, \\pi)$.\n\nFind the least and the greatest values of the function $g(x) = f(x) \\cdot f(1-x)$ on the closed interval $[-1, 1]$.",
"options": [],
"answer": "See solution",
"solution": "We have:\n\n$$\nf(\\cot x) = \\sin 2x + \\cos 2x \\quad \\forall x \\in (0, \\pi)\n$$\n\nWe can write:\n\n$$\n\\sin 2x + \\cos 2x = 2\\sin x \\cos x + (2\\cos^2 x - 1) = 2\\sin x \\cos x + 2\\cos^2 x - 1\n$$\n\nLet $t = \\cot x$, so $\\sin x = \\frac{1}{\\sqrt{1 + t^2}}$, $\\cos x = \\frac{t}{\\sqrt{1 + t^2}}$.\n\nThus,\n\n$$\n\\sin 2x = 2\\sin x \\cos x = \\frac{2t}{1 + t^2}, \\quad \\cos 2x = 2\\cos^2 x - 1 = \\frac{2t^2}{1 + t^2} - 1 = \\frac{2t^2 - (1 + t^2)}{1 + t^2} = \\frac{t^2 - 1}{1 + t^2}\n$$\n\nSo,\n\n$$\nf(t) = \\sin 2x + \\cos 2x = \\frac{2t}{1 + t^2} + \\frac{t^2 - 1}{1 + t^2} = \\frac{t^2 + 2t - 1}{t^2 + 1}\n$$\n\nTherefore,\n\n$$\ng(x) = f(x) \\cdot f(1-x) = \\frac{x^2 + 2x - 1}{x^2 + 1} \\cdot \\frac{(1-x)^2 + 2(1-x) - 1}{(1-x)^2 + 1}\n$$\n\nLet $u = x(1-x)$. Then $x^2 + 1 = (x^2 + 1)$, $(1-x)^2 + 1 = (1 - 2x + x^2) + 1 = x^2 - 2x + 2$.\n\nCompute $f(1-x)$:\n\n$$\nf(1-x) = \\frac{(1-x)^2 + 2(1-x) - 1}{(1-x)^2 + 1} = \\frac{1 - 2x + x^2 + 2 - 2x - 1}{x^2 - 2x + 1 + 1} = \\frac{x^2 - 4x + 2}{x^2 - 2x + 2}\n$$\n\nSo,\n\n$$\ng(x) = \\frac{x^2 + 2x - 1}{x^2 + 1} \\cdot \\frac{x^2 - 4x + 2}{x^2 - 2x + 2}\n$$\n\nLet $u = x(1-x)$. When $x$ runs through $[-1, 1]$, $u$ runs through $[-2, 1/4]$.\n\nLet $h(u) = \\frac{u^2 + 8u - 2}{u^2 - 2u + 2}$.\n\nBy studying the sign of $h'(u) = \\frac{2(-5u^2 + 4u + 6)}{(u^2 - 2u + 2)^2}$ on $[-2, 1/4]$, we get:\n\n$$\n\\min_{-2 \\le u \\le 1/4} h(u) = h\\left(\\frac{2 - \\sqrt{34}}{5}\\right) = 4 - \\sqrt{34},\n$$\n\nand\n\n$$\n\\max_{-2 \\le u \\le 1/4} h(u) = \\max\\{h(-2), h(1/4)\\} = \\max\\left\\{-\\frac{7}{5}, \\frac{1}{25}\\right\\} = \\frac{1}{25}.\n$$\n\nSo, on $[-1, 1]$, $\\min g(x) = 4 - \\sqrt{34}$, $\\max g(x) = \\frac{1}{25}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21313,
"subject": "Mathematics (Olympiad)",
"question": "Let $f: \\mathbb{N} \\to \\mathbb{N}$ be a function such that $f(1) = 1$ and $f(n) = n - f(f(n-1))$ for any $n \\ge 2$. Prove that $f(n + f(n)) = n$ for any $n$.",
"options": [],
"answer": "See solution",
"solution": "Denote by $(*)$ the condition $f(n) = n - f(f(n-1))$ for any $n \\ge 2$.\n\nFirst, we shall prove by induction that\n\n$$\nf(n) \\leq f(n+1) \\leq f(n) + 1 \\text{ for any } n.\n$$\n\n1. If $n = 1$, then $f(2) = 2 - f(f(1)) = 1$ and the inequalities hold.\n\n2. Let the inequalities be true for any $k \\leq n$. Then $f(n+1) = f(n)$ or $f(n)+1$. It follows by $(*)$ that $f(n) < n$ and the induction hypothesis implies\n\n$$\n\\begin{aligned}\nf(f(n)) &\\leq f(f(n+1)) \\stackrel{(*)}{\\Rightarrow} n+1 - f(n+1) \\leq n+2 - f(n+2) \\\\\n&\\Rightarrow f(n+2) \\leq f(n+1) + 1,\n\\end{aligned}\n$$\n\n$$\n\\begin{aligned}\nf(f(n+1)) &\\leq f(f(n)) + 1 \\stackrel{(*)}{\\Rightarrow} n + 2 - f(n + 2) \\leq n + 1 - f(n + 1) + 1 \\\\\n&\\Rightarrow f(n + 1) \\leq f(n + 2).\n\\end{aligned}\n$$\n\nSo $f(n+1) \\leq f(n+2) \\leq f(n+1) + 1$ which completes the induction step.\n\nNow, we shall prove again by induction the equality $f(n+f(n)) = n$.\n\n1. If $n = 1$, then $f(1+f(1)) = f(2) = 1$.\n\n2. Assume that $f(n+f(n)) = n$ for some $n$. Then\n\n$$\n\\begin{aligned}\nf(f(n+f(n))) = f(n) &\\stackrel{(*)}{\\Rightarrow} n + f(n) + 1 - f(n + f(n) + 1) = f(n) \\\\\n&\\Rightarrow f(n + f(n) + 1) = n + 1.\n\\end{aligned}\n$$\n\n*Case 1.* If $f(n+1) = f(n)$, then $f(n+1+f(n+1)) = n+1$.\n\n*Case 2.* If $f(n+1) = f(n)+1$, then\n\n$$\nf(n+1+f(n+1)) = n+1+f(n+1)-f(f(n+f(n+1))) = n+1+f(n+1)-f(n+f(n)+1) = n+1+f(n+1)-f(n+1) = n+1.\n$$\n\nThe problem is solved.\n\n**Remark.** One can find the explicit formula $f(n) = \\left[ \\frac{\\sqrt{5}-1}{2} (n+1) \\right]$. This leads to the property $f(a_{n+1}) = a_n$, where $a_n$ is the $n$-th Fibonacci number.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21314,
"subject": "Mathematics (Olympiad)",
"question": "是否存在正整數 $m$，可以讓等式\n\n$$\n(a^{3} - a)(b^{3} - b) = m c^{2}\n$$\n\n有無窮多組滿足 $a \\neq b$ 的正整數解 $(a, b, c)$？",
"options": [],
"answer": "See solution",
"solution": "我們聲稱 $m = 13$ 可行。\n\n存在無窮多整數 $b$ 滿足 $4b^2 - 3 = 13k^2$ 對某個 $k$，因為 $(b, k) = (2, 1)$ 是一組解，且 $18^2 - 13 \\cdot 5^2 = -1$，所以可以利用 Pell 方程產生無窮多組解。具體地，取 $(2 + \\sqrt{13})(18 - 13\\sqrt{5})^{2m}$ 可得範數為 $3 \\cdot (-1)^{2m} = 3$。\n\n然後，令 $a = b(4b^2 - 3)$。此時，\n\n$$\na - 1 = (2b + 1)^2(b - 1)\n$$\n\n$$\na + 1 = (2b - 1)^2(b + 1)\n$$\n\n因此，\n\n$$\n(a^3 - a)(b^3 - b) = [(b-1)(b)(b+1)(2b-1)(2b+1)]^2 \\cdot (4b^2 - 3)\n$$\n\n這是一個完全平方數。\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21315,
"subject": "Mathematics (Olympiad)",
"question": "If $\\alpha$, $\\beta$, $\\gamma$ are the measures of the sides of a triangle, prove that:\n\n$$\n\\frac{(\\alpha + \\gamma - \\beta)^4}{\\alpha(\\alpha + \\beta - \\gamma)} + \\frac{(\\alpha + \\beta - \\gamma)^4}{\\beta(\\beta + \\gamma - \\alpha)} + \\frac{(\\beta + \\gamma - \\alpha)^4}{\\gamma(\\alpha + \\gamma - \\beta)} \\geq \\alpha\\beta + \\beta\\gamma + \\gamma\\alpha.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $\\alpha + \\gamma - \\beta = x$, $\\alpha + \\beta - \\gamma = y$, $\\beta + \\gamma - \\alpha = z$. Then:\n\n$$\n\\alpha = \\frac{x+y}{2}, \\quad \\beta = \\frac{y+z}{2}, \\quad \\gamma = \\frac{z+x}{2}, \\quad \\alpha + \\beta + \\gamma = x + y + z.\n$$\n\nThe first part of the inequality becomes:\n\n$$\nK = \\frac{2x^4}{y(x+y)} + \\frac{2y^4}{z(y+z)} + \\frac{2z^4}{x(z+x)}.\n$$\n\nBy the Cauchy–Schwarz inequality:\n\n$$\n\\frac{K}{2} [y(x+y) + z(y+z) + x(z+x)] \\geq (x^2 + y^2 + z^2)^2\n$$\n\nSo,\n\n$$\n\\frac{K}{2} \\geq \\frac{(x^2 + y^2 + z^2)^2}{x^2 + y^2 + z^2 + xy + yz + zx} \\implies \\frac{K}{2} \\geq \\frac{(x^2 + y^2 + z^2)^2}{2(x^2 + y^2 + z^2)}\n$$\n\nThus,\n\n$$\nK \\geq x^2 + y^2 + z^2 \\geq \\frac{(x + y + z)^2}{3} = \\frac{(\\alpha + \\beta + \\gamma)^2}{3} \\geq \\alpha\\beta + \\beta\\gamma + \\gamma\\alpha.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21316,
"subject": "Mathematics (Olympiad)",
"question": "We call a glass that is not empty *non-empty* and a glass that has been emptied *empty*. Each glass holds one liter of water, and all the water is in $n$ containers (glasses), each of which can be either non-empty or empty.\n\nSuppose in the initial state we have $f_1$ non-empty glasses, and we want to reach a state with $f_2$ non-empty glasses.\n\n**Part (a):**\n\nProve that it is not possible for both $2(n - f_1) > \\frac{4n}{3}$ and $f_1 + (n - 1) > \\frac{4n}{3}$ to hold simultaneously.\n\n**Part (b):**\n\nSuppose we can reach the target state using a sequence of the described algorithms. Show that the following system of inequalities leads to a contradiction:\n\n- $2(n - f_1) + (n - f_2) > \\frac{5n}{3}$\n- $f_1 + (n - 1) + (n - f_2) > \\frac{5n}{3}$\n- $f_1 + 2f_2 > \\frac{5n}{3}$\n\nConclude that the proposition is proven.",
"options": [],
"answer": "See solution",
"solution": "**Part (a):**\n\nAssume, for contradiction, that both inequalities hold:\n\n$$\n2(n - f_1) > \\frac{4n}{3} \\implies f_1 < \\frac{n}{3}\n$$\n\nand\n\n$$\nf_1 + (n - 1) > \\frac{4n}{3} \\implies f_1 > \\frac{n}{3} + 1\n$$\n\nThese two inequalities for $f_1$ are contradictory, so the assumption is false.\n\n**Part (b):**\n\nAssume, for contradiction, that the following inequalities hold:\n\nBy running algorithms 1 and 4:\n\n$$\n2(n - f_1) + (n - f_2) > \\frac{5n}{3} \\implies 2f_1 + f_2 < \\frac{4n}{3} \\quad (1)\n$$\n\nBy running algorithms 2 and 3:\n\n$$\nf_1 + (n - 1) + (n - f_2) > \\frac{5n}{3} \\implies f_1 + \\frac{n}{3} - 1 > f_2 \\quad (2)\n$$\n\nBy running algorithms 2 and 5:\n\n$$\nf_1 + 2f_2 > \\frac{5n}{3} \\quad (3)\n$$\n\nFrom (1) and (3):\n\n$$\n\\frac{5n}{3} + f_1 < 2(f_1 + f_2) < \\frac{4n}{3} + f_2 \\implies f_1 + \\frac{n}{3} < f_2 \\quad (4)\n$$\n\nInequalities (2) and (4) are contradictory. Therefore, the proposition is proven. ■",
"topic": "Discrete Mathematics",
"subtopic": "Algorithms"
},
{
"id": 21317,
"subject": "Mathematics (Olympiad)",
"question": "Let $M$ be the circumcentre of $\\triangle ABC$ and let $\\alpha = \\angle BAC$.\n\nLet $P$ and $Y$ be points such that $PBMC$ is a kite with axis of symmetry $PM$, and $ABY$ is a right triangle with $\\angle ABY = 90^\\circ$. Let $T$ be the intersection of $AP$ and $YM$.\n\nProve that $AP \\perp XY$.",
"options": [],
"answer": "See solution",
"solution": "We first show that $\\triangle BYA \\sim \\triangle BMP$ and then that $\\triangle YBM \\sim \\triangle ABP$.\n\nBy the inscribed angle theorem, $\\angle BMC = 2\\angle BAC = 2\\alpha$. Quadrilateral $PBMC$ is a kite with axis of symmetry $PM$ (since $|MB| = |MC|$ and $|PB| = |PC|$), so $MP$ bisects $\\angle BMC$. Therefore, $\\angle BMP = \\frac{1}{2}\\angle BMC = \\alpha$.\n\nMoreover, $\\angle PBM = 90^\\circ$ (since the tangent to a circle is perpendicular to its radius), so by the sum of angles in a triangle, $\\angle MPB = 90^\\circ - \\alpha$.\n\nOn the other hand, $\\angle ABY = 90^\\circ$ and $\\angle YAB = \\angle YAC - \\angle BAC = 90^\\circ - \\alpha$. Therefore, $\\angle ABY = \\angle PBM$ and $\\angle YAB = \\angle MPB$, so $\\triangle BYA \\sim \\triangle BMP$.\n\nFrom this similarity, $\\frac{|YB|}{|AB|} = \\frac{|MB|}{|PB|}$. Combining this with the equality of angles:\n\n$$\n\\angle YBM = \\angle YBA + \\angle ABM = 90^\\circ + \\angle ABM = \\angle ABM + \\angle MBP = \\angle ABP,\n$$\n\nwe see that $\\triangle YBM \\sim \\triangle ABP$.\n\nLet $T$ be the intersection of $AP$ and $YM$. Then:\n\n$$\n\\angle BYT = \\angle BYM = \\angle BAP = \\angle BAT,\n$$\n\nso $BYAT$ is a cyclic quadrilateral. Therefore, $\\angle ATY = \\angle ABY = 90^\\circ$, so $AT \\perp YM$. Analogously, $AP \\perp XM$. Thus, $YM$ and $XM$ coincide and $AP \\perp XY$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21318,
"subject": "Mathematics (Olympiad)",
"question": "A quadratic equation $x^2 + px + q = 0$ is written on the blackboard, where $p$ and $q$ are real numbers such that real solutions exist and all solutions are positive. Two players take turns changing the coefficients according to these rules:\n\n- The first player decreases the constant term by either solution of the equation and (on the same move) increases the coefficient of the linear term by 1.\n- The second player may replace the constant term with any real number. Alternatively, the second player may increase the constant term by the largest solution and (on the same move) decrease the coefficient of the linear term by 1, but this move is allowed only if the solutions differ by more than 1.\n\nIf any move results in an equation with no real solutions or a non-positive real solution, the first player wins.\n\nCan the first player win regardless of how the opponent plays?",
"options": [],
"answer": "See solution",
"solution": "Yes.\n\nSuppose the first player always decreases the constant term by the smaller solution. This is a winning strategy. Assume, for contradiction, that the play lasts infinitely. The coefficient of the linear term either increases by 1 or (if the second player uses the alternative move) remains the same. If the second player used the main move infinitely often, the linear coefficient would eventually become positive. By Viete's theorem, the sum of the solutions would be negative, so at least one solution would be negative, and the first player would have already won. Thus, the second player must eventually use only the alternative move.\n\nLet the equation before the first player's move be $x^2 + px + q = 0$ with solutions $x_1 \\le x_2$. By Viete's theorem, $p = -(x_1 + x_2)$ and $q = x_1x_2$. After the first player's move, the linear coefficient is $-(x_1 + x_2) + 1$ and the constant term is $x_1x_2 - x_1$. The new equation is $x^2 - (x_1 + x_2 - 1)x + x_1(x_2 - 1)$. By Viete's theorem, its solutions are $x_1$ and $x_2 - 1$.\n\nIf the solutions before the move differ by more than 1, the difference decreases by 1; if they differ by at most 1, the difference after the move is still at most 1. The same holds for the second player's alternative move. Therefore, the play must reach a position where the second player cannot make the alternative move, which is a contradiction.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21319,
"subject": "Mathematics (Olympiad)",
"question": "\n\nLet $\\triangle ABC$ have angles $\\alpha, \\beta, \\gamma$ (in the usual notation). Consider points $K, M, L$ such that quadrilaterals $KBCM$ and $AMLK$ are cyclic, and $BM$ and $CK$ are altitudes. The two circumcircles of quadrilaterals $KBCM$ and $AKLM$ have the same radius, and $AL = BC$.\n\nDetermine all possible triangles $ABC$ that satisfy these conditions.",
"options": [],
"answer": "See solution",
"solution": "Since quadrilateral $KBCM$ is cyclic, we have $\\angle CMB = \\angle CKB$, and hence $\\angle AML = \\angle AKL$. Since $AMLK$ is cyclic, $\\angle AML + \\angle AKL = 180^\\circ$, which leads to $\\angle CMB = \\angle CKB = \\angle AML = \\angle AKL = 90^\\circ$. Thus, $BM$ and $CK$ are altitudes, $L$ is the triangle's orthocentre, and the triangle is necessarily acute. The two circumcircles of quadrilaterals $KBCM$ and $AKLM$ have the same radius, so $AL = BC$.\n\nLet the angles of $\\triangle ABC$ be $\\alpha, \\beta, \\gamma$. We have $\\theta = \\angle KCB = \\angle KMB = \\angle KML = \\angle KAL$, so triangles $KBC$ and $KLA$ are congruent, which gives $CK = AK$, and so triangle $AKC$ is right isosceles. This leads to $\\alpha = 45^\\circ$ and $\\gamma = 45^\\circ + \\theta = 45^\\circ + 90^\\circ - \\beta = 135^\\circ - \\beta$. Hence $(\\alpha, \\beta, \\gamma) = (45^\\circ, 45^\\circ + \\theta, 90^\\circ - \\theta)$, where $0^\\circ < \\theta < 90^\\circ$.\n\nAny triangle with angles $(45^\\circ, 45^\\circ + \\theta, 90^\\circ - \\theta)$, $\\theta \\in (0^\\circ, 45^\\circ)$, satisfies the conditions.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21320,
"subject": "Mathematics (Olympiad)",
"question": "For a non-empty set $\\mathcal{T}$, denote by $p(\\mathcal{T})$ the product of all elements of $\\mathcal{T}$. Does there exist a set $\\mathcal{T}$ of 2021 elements such that for any $a \\in \\mathcal{T}$, one has that $p(\\mathcal{T}) - a$ is an odd integer? Consider two cases:\n\n1. All elements of $\\mathcal{T}$ are irrational numbers.\n\n2. At least one element of $\\mathcal{T}$ is a rational number.",
"options": [],
"answer": "See solution",
"solution": "1. Consider the polynomial\n\n$$\nf(x) = x(x+2)(x+4)\\dots(x+4040) - x - (2m-1)\n$$\n\nfor some $m \\in \\mathbb{Z}^+$. Then $\\lim_{x \\to +\\infty} f(x) = +\\infty$ and $f(0) = 2^{2020} \\cdot 2020! - (2m-1)$. Take $m$ big enough so that $f(0) < 0$. By the continuity of $f(x)$, there exists some real number $x_0$ such that $f(x_0) = 0$. Then take the set $T$ as follows:\n\n$$\nT = \\{x_0, x_0 + 2, x_0 + 4, \\dots, x_0 + 4040\\}\n$$\n\nThen $p(T) = x_0(x_0+2)(x_0+4)\\dots(x_0+4040)$ and $p(T) - x_0 = 2m-1$ is an odd integer. Since every element in $T$ has the form $x_0+2k$ with $k \\in \\mathbb{Z}$, $p(T) - a$ is an odd integer for all $a \\in T$. Finally, take $m = 2^{2019} \\cdot 2020!$ so that $f(0) = 1$. Suppose, on the contrary, that $x_0 \\in \\mathbb{Q}$. Note that $f(x)$ is monic, so $x_0 \\in \\mathbb{Z}$ and $x_0 \\mid f(0) = 1$, which implies $x_0 \\in \\{-1, 1\\}$. But this is a contradiction, since it is easy to check that $f(x)$ cannot have an odd integer root.\n\n2. We will prove that the answer is negative. Note that for all $x, y \\in T$, $x - y$ is even. Suppose that $T$ contains some rational number $b$. Then we can list the elements of $T$ as $b, b + a_1, b + a_2, \\dots, b + a_{2020}$, in which $a_1, a_2, \\dots, a_k$ are even.\n\nPut $c = b(b + a_1)(b + a_2)\\dots(b + a_k) - b$, then $c$ is odd. Consider the polynomial\n\n$$\ng(x) = x(x + a_1)(x + a_2)\\dots(x + a_k) - x - c\n$$\n\nThen $g(b) = 0$ and $x = b$ is a rational root of $g(x)$. Note that $g$ is monic, so $b \\in \\mathbb{Z}$. From this, we conclude that all the elements in $T$ are integers that share the same parity, so does $p(T)$. Thus $p(T) - b$ is even, a contradiction.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21321,
"subject": "Mathematics (Olympiad)",
"question": "Determine all quintets $(a, n, p, q, r)$ of positive integers for which the identity\n$$\na^n - 1 = (a^p - 1)(a^q - 1)(a^r - 1)\n$$\nis satisfied.",
"options": [],
"answer": "See solution",
"solution": "If $a = 1$, the identity holds for any positive integers $n, p, q, r$. So, assume $a \\ge 2$. The identity is symmetric in $p, q, r$, so we may assume $p \\le q \\le r$.\n\nRewrite the identity as:\n$$\na^n = a^{p+q+r} - (a^{p+q} + a^{p+r} + a^{q+r}) + (a^p + a^q + a^r).\n$$\nSince $a \\ge 2$, $a^{p+q} > a^p$, $a^{q+r} > a^q$, $a^{p+r} > a^r$, and $a^p + a^q + a^r > 0$, so\n$$\na^{p+q+r} > a^n > a^{p+q+r} - (a^{p+q} + a^{p+r} + a^{q+r}).\n$$\nFrom the left, $n \\le p+q+r-1$. From the right, $a^{p+q+r-1} > a^{p+q+r} - (a^{p+q} + a^{p+r} + a^{q+r})$, which simplifies to\n$$\na^{-1} + a^{-p} + a^{-q} + a^{-r} > 1.\n$$\nSince $1 \\le p \\le q \\le r$, $a^{-1} \\ge a^{-p} \\ge a^{-q} \\ge a^{-r}$, and their sum $> 1$, so $a^{-1} > \\frac{1}{4}$, thus $a < 4$. Since $a \\ge 2$, $a = 2$ or $a = 3$.\n\n**Case 1: $a = 3$**\n\n$3^{-p} + 3^{-q} + 3^{-r} > 1 - 3^{-1} = \\frac{2}{3}$, so $3^{-p} > \\frac{2}{9}$, thus $p = 1$. Then $3^{-q} + 3^{-r} > \\frac{1}{3}$, so $3^{-q} > \\frac{1}{6}$, thus $q = 1$. Now $p = q = 1$, so $3^n = 4 \\cdot 3^r - 3$. Since $4 \\cdot 3^r < 9 \\cdot 3^r = 3^{r+2}$, $n \\le r+1$. So $3^{r+1} \\ge 4 \\cdot 3^r - 3$, which gives $3^r \\le 3$, so $r = 1$. Thus, $3^n - 1 = (3-1)^3$, so $n = 2$. The only solution is $(3, 2, 1, 1, 1)$.\n\n**Case 2: $a = 2$**\n\n$2^{-p} + 2^{-q} + 2^{-r} > \\frac{1}{2}$, so $2^{-p} > \\frac{1}{6}$, so $p = 1$ or $p = 2$.\n\n- If $p = 2$: $2^{-q} + 2^{-r} > \\frac{1}{4}$, so $2^{-q} > \\frac{1}{8}$, so $q = 2$. Then $2^n = 9 \\cdot 2^r - 8$. $9 \\cdot 2^r < 16 \\cdot 2^r = 2^{r+4}$, so $2^{r+3} \\ge 9 \\cdot 2^r - 8$, so $2^r \\le 8$, so $r = 2$ or $r = 3$. Checking, $(2, 6, 2, 2, 3)$ is a solution.\n\n- If $p = 1$: $2^n = 2^{q+r} - 2^q - 2^r + 2$. The right side $< 2^{q+r}$, so $2^{q+r-1} \\ge 2^{q+r} - 2^q - 2^r + 2$, which reduces to $2^{q+r-1} + 2 \\le 2^q + 2^r$, and since $2^q + 2^r \\le 2^{r+1}$, $q+r-1 < r+1$, so $q = 1$. Now $p = q = 1$, so $2^n = 2^r$, so $n = r$, any positive integer. Thus, $(2, k, 1, 1, k)$ for any $k$.\n\n**Summary:**\n\n- $a = 1$, $n, p, q, r$ arbitrary positive integers.\n- $(a, n, p, q, r) = (3, 2, 1, 1, 1)$.\n- $(a, n, p, q, r) = (2, 6, 2, 2, 3)$ and permutations of $p, q, r$.\n- $(a, n, p, q, r) = (2, k, 1, 1, k)$ for any $k$ and permutations of $p, q, r$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21322,
"subject": "Mathematics (Olympiad)",
"question": "For a positive integer $n$, let $g(n) = \\sum_{d|n} \\frac{d}{P(d)}$, where $P(d)$ denotes the largest prime divisor of $d$. Prove that $g(n) \\leq n$.",
"options": [],
"answer": "See solution",
"solution": "We proceed by induction on the number of distinct prime divisors of $n$.\n\n**Base case:** If $n = 1$, then $g(1) = 1$.\n\nIf $n = p^a$ for some prime $p$, then\n$$\ng(n) = 1 + 1 + p + \\cdots + p^{a-1} = 1 + \\frac{p^a - 1}{p - 1} \\leq 1 + p^a - 1 = n.\n$$\n\n**Inductive step:** Assume that for $n$ with $k$ distinct prime divisors, $g(n) \\leq n$. 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 $n = m p_{k+1}^{a_{k+1}}$.\n\nThen,\n$$\ng(n) = g(m) + \\sum_{d|m} \\sum_{i=1}^{a_{k+1}} \\frac{d p_{k+1}^i}{p_{k+1}} = g(m) + \\sigma(m) \\frac{p_{k+1}^{a_{k+1}} - 1}{p_{k+1} - 1},\n$$\nwhere $\\sigma(m)$ is the sum of positive divisors of $m$.\n\nBy the induction hypothesis, $g(m) \\leq m$. Since\n$$\n\\begin{aligned}\n\\sigma(m) \\frac{p_{k+1}^{a_{k+1}} - 1}{p_{k+1} - 1} &= \\left( \\prod_{i=1}^{k} \\frac{p_i^{a_i+1} - 1}{p_i - 1} \\right) \\frac{p_{k+1}^{a_{k+1}} - 1}{p_{k+1} - 1} \\\\\n&\\leq \\left( \\prod_{i=1}^{k} \\frac{p_i^{a_i+1} - 1}{p_{i+1} - 1} \\right) (p_{k+1}^{a_{k+1}} - 1)\n\\end{aligned}\n$$\nso $g(n) \\leq n$. Thus, the statement is proved.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21323,
"subject": "Mathematics (Olympiad)",
"question": "Determine a polynomial $f(x)$ with integer coefficients which satisfies the following property: There are infinitely many relatively prime positive integers $a, b$ such that $a+b$ divides $f(a) + f(b)$.",
"options": [],
"answer": "See solution",
"solution": "Let $f(x)$ be such a polynomial. Define\n\n$$\ng(x) := \\frac{f(x) - f(-x)}{2}, \\quad h(x) := \\frac{f(x) + f(-x)}{2}.\n$$\n\nHere, $g(x)$ consists of the odd degree monomials of $f(x)$, and $h(x)$ consists of the even degree monomials (including the constant term). Thus, $f(x) = g(x) + h(x)$.\n\nFor any positive integers $a$ and $b$,\n\n$$\na+b \\mid a^{2k+1} + b^{2k+1}\n$$\n\nfor any nonnegative integer $k$. Therefore,\n\n$$\na+b \\mid f(a)+f(b) \\iff a+b \\mid h(a)+h(b).\n$$\n\nSo, we may assume $f(x) = h(x)$ (i.e., $f(x)$ has only even degree monomials). Suppose $f(x) = c x^{2n}$ for some $n \\geq 0$ and $a+b \\mid f(a)+f(b)$ for some positive integers $a, b$. Note that\n\n$$\na^{2n} + b^{2n} = a(a^{2n-1} + b^{2n-1}) - b^{2n}(a+b) + 2b^{2n}.\n$$\n\nThus,\n\n$$\na+b \\mid f(a)+f(b) \\iff a+b \\mid 2f(b) = 2c b^{2n}.\n$$\n\nIf $\\gcd(a, b) = 1$, then $\\gcd(a+b, b^{2n}) = 1$, so $a+b \\mid 2c$. This means there are only finitely many such pairs $(a, b)$.\n\nNow, assume $f(x) = a_{2n} x^{2n} + t(x)$, where $a_{2n} \\neq 0$ and $t(x)$ is a nonzero polynomial of higher even degree. Let $b$ be any sufficiently large positive integer with $\\gcd(a_{2n}, b) = 1$, and let $a$ be any positive integer such that\n\n$$\n|a+b| = a_{2n} + \\frac{t(b)}{b^{2n}}.\n$$\n\nSince $\\gcd(a, b) = 1$ and there are infinitely many such pairs $(a, b)$, and $2f(b) = 2b^{2n}|a+b|$, we have $a+b \\mid f(a)+f(b)$.\n\nTherefore, $f(x)$ satisfies the property if and only if it consists only of odd degree monomials, or contains at least two even degree monomials.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21324,
"subject": "Mathematics (Olympiad)",
"question": "Find all natural numbers $n$ for which $1^{\\phi(n)} + 2^{\\phi(n)} + \\dots + n^{\\phi(n)}$ is coprime with $n$.",
"options": [],
"answer": "See solution",
"solution": "Consider the given expression modulo $p$, where $p \\mid n$ is a prime number. Since $p \\mid n$, we have $p-1 \\mid \\phi(n)$. Thus, for any $k$ not divisible by $p$, $k^{\\phi(n)} \\equiv 1 \\pmod{p}$. There are $n - \\frac{n}{p}$ numbers among $1, 2, \\dots, n$ that are not divisible by $p$. Therefore,\n\n$$\n1^{\\phi(n)} + 2^{\\phi(n)} + \\dots + n^{\\phi(n)} \\equiv -\\frac{n}{p} \\pmod{p}\n$$\n\nIf the given expression is coprime with $n$, it is not divisible by $p$, so $p \\nmid \\frac{n}{p}$, which implies $p^2 \\nmid n$. This must hold for all prime divisors $p$ of $n$, so $n$ must be square-free. Conversely, if $n$ is square-free, then $p^2 \\nmid n$ for all $p$, so $p \\nmid \\frac{n}{p}$, and the expression is not divisible by $p$. Since this holds for all prime divisors $p$ of $n$, the two numbers are coprime.\n\nThe answer is: all square-free integers.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21325,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. Two people, $P$ and $Q$, play a game in which they alternately call an integer $m$ ($1 \\leq m \\leq n$). $P$ calls the first number. No number can be called more than once by either player. The game ends when neither can call any more numbers. If the sum of the numbers that $P$ has called is divisible by $3$, $P$ wins; otherwise, $Q$ wins.\n\nFind all $n$ such that $P$ can win the game no matter what $Q$ does.",
"options": [],
"answer": "See solution",
"solution": "Let the number called by a player in the $m$th turn be $N_m$. The sequence $(N_1, \\dots, N_l)$ is called the \"history up to the $l$th turn\". A number $j$ is \"free in the $(l+1)$th turn\" if $j \\neq N_1, \\dots, N_l$. We will prove that if $n \\equiv 0, 4, 5 \\pmod{6}$, P can always win, and if $n \\equiv 1, 2, 3 \\pmod{6}$, Q can always win.\n\n**(i) For $0 \\leq n \\leq 5$:**\n\n- If $n = 0, 1, 2$, the proposition is true.\n- For $n = 3$: If Q calls $1$ or $2$ in the second turn, the sum of P's numbers is $5$ or $4$, so Q wins.\n- For $n = 4$: If P calls $2$ first, and then $1$ or $4$ in the third turn, the sum of Q's numbers is $3$ or $6$, so P wins.\n- For $n = 5$: Consider pairs $(1, 4)$ and $(2, 5)$. If P calls $3$ first, and then always calls the other number of the pair that includes Q's last call, P can always win.\n\nThus, the proposition holds for $0 \\leq n \\leq 5$.\n\n**(ii) Inductive step:**\n\nSuppose the proposition holds for $n = k$. Then for $n = k + 6$, let the player with the winning strategy be $A$, and the other $B$. Let $M = \\{k + 1, k + 2, k + 3, k + 4, k + 5, k + 6\\}$, and consider the pairs $(k + 1, k + 4)$, $(k + 2, k + 5)$, and $(k + 3, k + 6)$. On $A$'s turn, $A$ can always respond to $B$'s move by choosing the other number in the pair, maintaining the winning strategy. Therefore, the result extends by induction.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21326,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be the numbers on the spade, heart, and diamond cards chosen, respectively. How many ways are there to choose these cards such that $a + b + c$ is a multiple of $7$, given that $1 \\leq a \\leq 4$, $1 \\leq b \\leq 6$, and $1 \\leq c \\leq 8$?",
"options": [],
"answer": "See solution",
"solution": "If $a + c$ is not a multiple of $7$, let $k$ be the remainder when $a + c$ is divided by $7$. Then $1 \\leq k \\leq 6$, and $a + b + c$ becomes a multiple of $7$ if and only if $b = 7 - k$. If $a + c$ is a multiple of $7$, then $a + b + c$ cannot be a multiple of $7$ since $1 \\leq b \\leq 6$. Thus, the number of possible choices is the total number of $(a, c)$ pairs, which is $4 \\times 8 = 32$, minus the number of cases where $a + c$ is a multiple of $7$. This occurs in $4$ ways: $(a, c) = (1, 6), (2, 5), (3, 4), (4, 3)$. Therefore, the answer is $32 - 4 = 28$ ways.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21327,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\alpha \\in \\mathbb{Q}^+$. Determine all functions $f: \\mathbb{Q}^+ \\to \\mathbb{Q}^+$ such that\n\n$$\nf\\left(\\frac{x}{y} + y\\right) = \\frac{f(x)}{f(y)} + \\alpha x\n$$\n\nholds for all $x, y \\in \\mathbb{Q}^+$.\n\nHere, $\\mathbb{Q}^+$ denotes the set of positive rational numbers.",
"options": [],
"answer": "See solution",
"solution": "Setting $y = x$ and $y = 1$ yields\n\n$$\nf(x+1) = 1 + f(x) + \\alpha x \\qquad (1)\n$$\n\nand\n\n$$\nf(x+1) = \\frac{f(x)}{f(1)} + f(1) + \\alpha x \\qquad (2)\n$$\n\nrespectively. Equating (1) and (2) implies\n\n$$\nf(x)\\left(1 - \\frac{1}{f(1)}\\right) = f(1) - 1.\n$$\n\nAs $f$ cannot be constant due to (1), we obtain $f(1) = 1$. By induction, we get\n\n$$\nf(x) = \\frac{\\alpha}{2}x(x-1) + x \\quad \\text{for all } x \\in \\mathbb{Z}^{+}. \\qquad (3)\n$$\n\nIn particular, this implies $f(2) = \\alpha + 2$ and $f(4) = 6\\alpha + 4$. Setting $x = 4$ and $y = 2$ in the functional equation yields\n\n$$\n\\alpha^2 - 2\\alpha = 0.\n$$\n\nThus we must have $\\alpha = 2$ in order to obtain solutions. From now on, we only consider this case.\n\nFrom (3), we obtain $f(x) = x^2$ for $x \\in \\mathbb{Z}^+$. By induction, we obtain that for $x \\in \\mathbb{Q}^+$ and $n \\in \\mathbb{Z}^+$, from the relation $f(x+n) = (x+n)^2$ it follows that $f(x) = x^2$.\n\nLet now $\\frac{a}{b} \\in \\mathbb{Q}^+$ with $a, b \\in \\mathbb{Z}^+$. We set $x = a$ and $y = b$ and obtain\n\n$$\nf\\left(\\frac{a}{b} + b\\right) = \\frac{a^2}{b^2} + b^2 + 2a = \\left(\\frac{a}{b} + b\\right)^2.\n$$\n\nThe above remark implies that $f\\left(\\frac{a}{b}\\right) = \\left(\\frac{a}{b}\\right)^2$. It is easily verified that $f(x) = x^2$ is indeed a solution.\n\nThus there is no solution for $\\alpha \\neq 2$ and the solution $f(x) = x^2$ for $\\alpha = 2$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21328,
"subject": "Mathematics (Olympiad)",
"question": "Prove that if $x, y, z > 0$, then\n\n$$\n\\frac{(x+1)(y+1)^2}{3\\sqrt[3]{z^2x^2}+1} + \\frac{(y+1)(z+1)^2}{3\\sqrt[3]{x^2y^2}+1} + \\frac{(z+1)(x+1)^2}{3\\sqrt[3]{y^2z^2}+1} \\geq x+y+z+3.\n$$",
"options": [],
"answer": "See solution",
"solution": "Denote by $S$ the left-hand side of the given inequality. The arithmetic mean–geometric mean inequality implies that $xy + x + y \\geq 3\\sqrt[3]{x^2y^2}$ and hence\n\n$$\nS \\geq \\frac{(x+1)(y+1)^2}{(z+1)(x+1)} + \\frac{(y+1)(z+1)^2}{(x+1)(y+1)} + \\frac{(z+1)(x+1)^2}{(y+1)(z+1)}.\n$$\n\nSetting $a = x + 1$, $b = y + 1$, $c = z + 1$ gives\n\n$$\nS \\geq \\frac{b^2}{c} + \\frac{c^2}{a} + \\frac{a^2}{b}.\n$$\n\nThen\n\n$$\nS \\geq \\frac{(a+b+c)^2}{a+b+c} = a+b+c = x+y+z+3.\n$$\n\nby the Cauchy–Schwartz inequality.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21329,
"subject": "Mathematics (Olympiad)",
"question": "Determine the largest constant $C$ such that\n\n$$\n(x_1 + x_2 + \\dots + x_6)^2 \\geq C \\cdot (x_1(x_2 + x_3) + x_2(x_3 + x_4) + \\dots + x_6(x_1 + x_2))\n$$\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\n\nExpanding yields\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 for $X - Y = Y - Z = Z - X = 0$, i.e., $X = Y = Z$, thus $x_1 + x_4 = x_2 + x_5 = x_3 + x_6$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21330,
"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": "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 we have\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- 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\n2. We construct an example. Since $2n-1$ is composite, write $2n-1 = pq$ where $p, q > 1$. Let\n$$\na_1 = 1,\\ a_2 = 2,\\ \\dots,\\ a_p = p,\\ a_{p+1} = p+1,\\ a_{p+2} = p+3,\\ \\dots,\\ a_n = pq-p.\n$$\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- 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- 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": 21331,
"subject": "Mathematics (Olympiad)",
"question": "設 $a_0, a_1, \\dots$ 為一由非負整數所構成的無窮數列（數列中可能有重複的數字）。假設其滿足以下條件：\n\n$$\n0 \\le a_i \\le i \\quad \\text{對所有非負整數 } i \\ge 0\n$$\n\n且\n\n$$\n\\binom{k}{a_0} + \\binom{k}{a_1} + \\dots + \\binom{k}{a_k} = 2^k \\quad \\text{對於所有非負整數 } k \\ge 0 \\text{ 皆成立。}\n$$\n\n試證：對於所有非負整數 $N$，都存在一個 $i \\ge 0$，使得 $a_i = N$。\n\n（註：$\\binom{x}{y} = \\frac{x!}{y!(x-y)!}$，且 $0! = 1$。）",
"options": [],
"answer": "See solution",
"solution": "我們將以歸納法證明對於所有 $k$，必存在 $t \\ge 0$ 使得 $2t \\le k+1$ 且\n\n$$\n\\{a_0, \\dots, a_k\\} = \\{0, 1, \\dots, t-1, 0, 1, \\dots, k-t\\}.\n$$\n\n易知以上性質自動保證原命題成立。\n\n$k=0$ 時，$a_0$ 必為 $0$，故顯然。假設 $k=m$ 時成立，從而存在 $t$ 使得 $2t \\le m+1$ 且 $\\{a_0, \\dots, a_m\\} = \\{0, 1, \\dots, t-1, 0, 1, \\dots, m-t\\}$。又由於題目給定\n\n$$\n\\binom{m+1}{a_0} + \\binom{m+1}{a_1} + \\dots + \\binom{m+1}{a_{m+1}} = 2^{m+1},\n$$\n\n從而我們有\n\n$$\n\\begin{aligned}\n2^{m+1} &= \\binom{m+1}{a_{m+1}} + \\sum_{i=0}^{t-1} \\binom{m+1}{i} + \\sum_{i=0}^{m-t} \\binom{m+1}{i} \\\\\n&= \\binom{m+1}{a_{m+1}} + \\sum_{i=0}^{t-1} \\binom{m+1}{i} + \\sum_{i=0}^{m-t} \\binom{m+1}{m+1-i} \\\\\n&= \\binom{m+1}{a_{m+1}} + \\sum_{i=t+1}^{m+1} \\binom{m+1}{i} \\\\\n&= \\binom{m+1}{a_{m+1}} + 2^{m+1} - \\binom{m+1}{t},\n\\end{aligned}\n$$\n\n故 $\\binom{m+1}{a_{m+1}} = \\binom{m+1}{t}$，從而 $a_{m+1} = t$ 或 $m+1-t$。無論哪種狀況，歸納皆成立。\n\n得證。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21332,
"subject": "Mathematics (Olympiad)",
"question": "設三角形 $ABC$ 的垂心為 $H$，外接圓為 $\\Gamma$。取點 $P$ 為 $\\Gamma$ 上異於 $A, B, C$ 的一點，並令 $M$ 為線段 $HP$ 的中點。分別在直線 $BC, CA, AB$ 上取點 $D, E, F$ 使得 $AP \\parallel HD, BP \\parallel HE, CP \\parallel HF$。證明：$D, E, F, M$ 共線。",
"options": [],
"answer": "See solution",
"solution": "解：($\\angle$ 代表有向角。)\n\n\n\n令 $A', P'$ 分別為 $A, P$ 關於 $\\Gamma$ 的對徑點，$M_a, M'$ 分別為 $\\overline{HA'}, \\overline{HP'}$ 的中點，$\\Omega$ 為 $\\triangle ABC$ 的九點圓，則 $M, M_a, M'$ 位於 $\\Omega$ 上。設 $M'H$ 交 $\\Omega$ 另一點於 $X$，$D'$ 為 $AH$ 與 $BC$ 的交點，則\n\n$$\n\\angle XD'D = \\angle XD'M_a = \\angle XM'M_a = \\angle HM'M_a.\n$$\n\n由 $M_a$ 也為 $\\overline{BC}$ 中點，知\n\n$$\n\\angle XHD = \\angle (HM', AP) = \\angle (HM', A'P') = \\angle HM'M_a.\n$$\n\n因此 $D, D', H, X$ 共圓，即 $HP' \\perp XD$。注意到 $\\overline{MM'}$ 為 $\\Omega$ 的直徑，故\n\n$$\n\\angle (HP', XM) = \\angle M'XM = 90^\\circ,\n$$\n\n所以 $X, M, D$ 共線且 $HP' \\perp DM$，同理有 $HP' \\perp EM, HP' \\perp FM$，因此 $D, E, F, M$ 共線。證畢。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21333,
"subject": "Mathematics (Olympiad)",
"question": "Construct outside the acute-angled triangle $ABC$ the isosceles triangles $ABA_B$, $ABB_A$, $ACA_C$, $ACC_A$, $BCB_C$ and $BCC_B$, so that\n\n$$\nAB = AB_A = BA_B, \\quad AC = AC_A = CA_C, \\quad BC = BC_B = CB_C\n$$\n\nand\n\n$$\n\\angle BAB_A = \\angle ABA_B = \\angle CAC_A = \\angle ACA_C = \\angle BCB_C = \\angle CBC_B = \\alpha < 90^\\circ.\n$$\n\nProve that the perpendiculars from $A$ to $B_A C_A$, from $B$ to $A_B C_B$ and from $C$ to $A_C B_C$ are concurrent.\n\n",
"options": [],
"answer": "See solution",
"solution": "**Lemma.** If $BCD$ is the isosceles triangle which is outside the triangle $ABC$ and has\n\n$$\n\\angle CBD = \\angle BCD = 90^\\circ - \\alpha := \\beta,\n$$\n\nthen $AD \\perp B_A C_A$.\n\n*Proof of the lemma.* Construct an isosceles triangle $ABE$ outside the triangle $ABC$, so that $\\angle ABE = \\angle AEB = \\beta$.\n\nThen $AE = AB = AB_A$ and $\\angle EAB_A = \\alpha$, so a rotation of center $A$ and angle $\\alpha$ sends $C_A$ to $C$ and $B_A$ to $E$, hence $\\angle (\\overrightarrow{B_A C_A}, \\overrightarrow{EC}) = \\alpha$ (the angle between vectors is considered oriented).\n\nAlso triangles $EBA$ and $BCD$ are similar, so a rotation of center $B$ and angle $\\beta$, followed by a dilation of ratio $\\frac{EB}{AB} = \\frac{BC}{BD}$ sends $E$ to $A$ and $C$ to $D$, hence $\\angle (\\overrightarrow{EC}, \\overrightarrow{AD}) = \\beta$ (also oriented angle).\n\nThis shows that\n\n$$\n\\angle (\\overrightarrow{B_A C_A}, \\overrightarrow{AD}) = \\angle (\\overrightarrow{B_A C_A}, \\overrightarrow{EC}) + \\angle (\\overrightarrow{EC}, \\overrightarrow{AD}) = \\alpha + \\beta = 90^\\circ. \\blacksquare\n$$\n\nReturning to the solution of the problem, denote $A'$ the intersection of $BC$ with the perpendicular from $A$ to $B_A C_A$. Then $A'$ belongs to the segment $BC$ and\n\n$$\n\\frac{A'B}{A'C} = \\frac{AB \\sin(B+\\beta)}{AC \\sin(C+\\beta)}.\n$$\n\nSince similar relations are true for the intersections $B', C'$ of the other two perpendiculars with the opposite sides, this yields\n\n$$\n\\frac{A'B}{A'C} \\cdot \\frac{B'C}{B'A} \\cdot \\frac{C'A}{C'B} = \\frac{AB \\sin(B+\\beta)}{AC \\sin(C+\\beta)} \\cdot \\frac{BC \\sin(C+\\beta)}{BA \\sin(A+\\beta)} \\cdot \\frac{CA \\sin(A+\\beta)}{CB \\sin(B+\\beta)} = 1,\n$$\n\nwhence the conclusion.\n\n**Remark.** The conditions 'acute-angled' and '$\\alpha < 90^\\circ$' are not essential, but without them there are cases when $A'$ does not belong to the segment $BC$, or the perpendiculars become parallel.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21334,
"subject": "Mathematics (Olympiad)",
"question": "Для простого $p$ и натурального $n$ обозначим через $\\nu_p(n)$ степень, в которой $p$ входит в разложение $n$ на простые множители. Заметим, что если $\\nu_p(n) \\neq \\nu_p(k)$, то $\\nu_p(n \\pm k) = \\min(\\nu_p(n), \\nu_p(k))$.\n\nРассмотрим числа вида $S_n = 1! + 2! + \\dots + n!$. Докажите, что для любого достаточно большого $n$ число $S_n$ не делится на $n!$.",
"options": [],
"answer": "See solution",
"solution": "Предположим противное; обозначим $P = 10^{2012}$. Тогда все простые делители чисел вида $S_n$ не превосходят $P$.\n\n**Лемма.** Пусть $\\nu_p(S_n) < \\nu_p((n+1)!)$ при некотором $n$. Тогда $\\nu_p(S_k) = \\nu_p(S_n)$ при всех $k \\ge n$.\n\n**Доказательство.** Обозначим $a = \\nu_p(S_n)$, $b = \\nu_p((n+1)!)$; тогда $b \\ge a+1$. Заметим, что $S_k = S_n + (n+1)! + \\dots + k!$; в этой сумме все слагаемые, кроме первого, делятся на $p^{a+1}$, а первое делится лишь на $p^a$, но не на $p^{a+1}$. Значит, и $S_k$ делится на $p^a$, но не на $p^{a+1}$. $\\square$\n\nРассмотрим некоторое простое $p \\le P$. Ввиду леммы, если $\\nu_p(S_n) < \\nu_p((n+1)!)$ при некотором $n$, то существует число $a_p$ такое, что $\\nu_p(S_n) \\le a_p$ при всех натуральных $n$. Назовём такое простое число $p$ маленьким; все остальные простые числа, меньше $P$, назовём большими. Так как маленьких простых конечное количество, существует натуральное $M$, больше любого числа вида $p^{a_p}$, где $p$ — маленькое.\n\nПусть теперь $p$ — большое простое число, а $n$ — такое число, что $n+2 \\nmid p$. Тогда из леммы имеем $\\nu_p(S_{n+1}) \\ge \\nu_p((n+2)!) > \\nu_p((n+1)!)$.\n\nЗначит, $\\nu_p(S_n) = \\nu_p(S_{n+1} - (n+1)!) = \\nu_p((n+1)!) = \\nu_p(n!)$ (последний переход верен, ибо $n+1$ не кратно $p$).\n\nРассмотрим теперь число $N = MP! - 2$. По доказанному, $\\nu_p(S_N) = \\nu_p(N!)$ для любого большого простого $p$. Кроме того, поскольку $N \\ge M$, то $\\nu_p(S_N) \\le \\nu_p(p^{a_p}) \\le \\nu_p(N!)$ для любого маленького простого $p$. Поскольку все простые делители числа $S_N$ – либо большие, либо маленькие, отсюда следует, что $S_N \\le N!$, что, очевидно, неверно. Противоречие.\n\n**Замечание.** После доказательства леммы можно завершить решение и по-другому. Например, можно показать, что $\\nu_p(S_{n-1}) = \\nu_p(n!)$ для любого $n$, кратного большому простому $p$. Предположим противное, тогда $\\nu_p(S_{n-1}) > \\nu_p(n!)$. Рассмотрим число\n\n$$\nS_{n+p-1} = S_{n-1} + n! \\cdot (1 + (n+1) + (n+1)(n+2) + \\dots + (n+1)\\dots(n+p-1)).\n$$\n\nОбозначим через $A_n$ выражение в скобках в правой части; тогда $A_n \\equiv 1+1!+2!+\\dots+(p-1)! \\equiv 1+S_{p-1} \\pmod p$. Поскольку $S_{p-1} \\equiv p \\pmod p$ по лемме, получаем, что $A_n$ не делится на $p$ и потому $\\nu_p(S_{n+p-1}) = \\min(\\nu_p(S_{n-1}), \\nu_p(n!)) = \\nu_p(n!) < \\nu_p((n+p)!)$. Это противоречит лемме.\n\nОтсюда, полагая $N = kP! - 1$ при некотором натуральном $k \\ge M$, получаем $\\nu_p(S_N) \\le \\nu_p((N+1)!)$ для любого $p \\le P$. В то же время, у числа $(N+1)!$ есть простые делители, большие $P$, и нетрудно показать, что при достаточно большом $k$ их вклад больше, чем $N+1$; значит, $S_k \\le (N+1)!(N+1) = N!$, что неверно.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21335,
"subject": "Mathematics (Olympiad)",
"question": "For an acute triangle $ABC$ with $AB \\neq AC$, let $H$ be the foot of the perpendicular from $A$ to side $BC$. Let points $P$ and $Q$ be chosen so that the points $A$, $B$, $P$ and $A$, $C$, $Q$ are collinear in that order, respectively. Suppose that the four points $B$, $C$, $P$, $Q$ lie on a circle, and that $HP = HQ$. Prove that $H$ is the circumcenter of triangle $APQ$.\n\nHere, $XY$ denotes the length of segment $XY$.",
"options": [],
"answer": "See solution",
"solution": "Assume without loss of generality that $AB < AC$. Let $\\angle B = \\angle ABC$ and $\\angle C = \\angle BCA$.\n\nSince $B$, $P$, $Q$, $C$ are concyclic, we have\n\n$$\n\\angle AQH - \\angle APH = \\angle AQP - \\angle APQ = \\angle B - \\angle C,\n$$\n\nso\n\n$$\n\\begin{aligned}\n\\angle AHP - \\angle AHQ &= (90^\\circ + \\angle BHP) - (90^\\circ + \\angle CHQ) \\\\\n&= (\\angle B - \\angle APH) - (\\angle C - \\angle AQH) = 2(\\angle B - \\angle C).\n\\end{aligned}\n$$\n\nLet $P'$ be the reflection of $P$ over $AH$. Then\n\n$$\n\\angle QHP' = \\angle AHP' - \\angle AHQ = \\angle AHP - \\angle AHQ = 2(\\angle B - \\angle C)\n$$\n\nand\n\n$$\n\\angle QAP' = \\angle HAQ - \\angle HAP' = \\angle HAQ - \\angle HAP = \\angle B - \\angle C.\n$$\n\nThus, $\\angle QHP' = 2\\angle QAP'$. Since $A$ and $H$ are on the same side of $P'Q$, and $HP' = HQ$, $A$ lies on the circle centered at $H$ with radius $HQ$. Therefore, $HA = HQ = HP$, so $H$ is the circumcenter of $APQ$.\n\n**Alternate Solution:** Let $O$ be the circumcenter of $APQ$ and $M$ the midpoint of $PQ$. Both $H$ and $O$ lie on the perpendicular bisector of $PQ$, so $M$, $H$, $O$ are collinear. Also,\n\n$$\n\\angle PAO = \\frac{180^\\circ - \\angle AOP}{2} = 90^\\circ - \\angle AQP = 90^\\circ - \\angle B = \\angle PAH,\n$$\n\nso $A$, $H$, $O$ are collinear. From\n\n$$\n\\begin{aligned}\n\\angle AHM &= 360^\\circ - \\angle HAQ - \\angle AQM - \\angle QMH \\\\\n&= 270^\\circ - (90^\\circ - \\angle C) - \\angle B = 180^\\circ - \\angle B + \\angle C \\neq 180^\\circ\n\\end{aligned}\n$$\n\nwe see $AH$ and $MH$ are not parallel. Since both pass through $O$ and $H$, these points must coincide. Thus, $H$ is the circumcenter of $APQ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21336,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\odot I$ be the incircle of $\\triangle ABC$ with $AB > AC$. $\\odot I$ is tangent to $BC$ and $AD$ at $D$ and $E$, respectively. The tangent line $EP$ to $\\odot I$ intersects the extended line of $BC$ at $P$. Segment $CF$ is parallel to $PE$ and intersects $AD$ at point $F$. Line $BF$ intersects $\\odot I$ at points $M$ and $N$ such that $M$ is on segment $BF$. Segment $PM$ intersects $\\odot I$ at the other point $Q$. Prove that $\\angle ENP = \\angle ENQ$.",
"options": [],
"answer": "See solution",
"solution": "Suppose that $\\odot I$ touches $AC$ and $AB$ at $S$ and $T$, respectively. Suppose that $ST$ intersects $AI$ at point $G$. We see that $IT \\perp AT$ and $TG \\perp AI$.\n\n\n\nWe have $AG \\cdot AI = AT^2 = AD \\cdot AE$, thus points $I, G, E$, and $D$ are concyclic.\n\nSince $IE \\perp PE$ and $ID \\perp PD$, we see that points $I, E, P$, and $D$ are concyclic. Hence, points $I, G, E, P$, and $D$ are concyclic.\n\nTherefore, $\\angle IGP = \\angle IEP = 90^\\circ$, that is, $IG \\perp PG$. Hence, points $P, S$, and $T$ are collinear.\n\nLine $PST$ intersects $\\triangle ABC$. By Menelaus' Theorem, we have\n\n$$\n\\frac{AS}{SC} \\cdot \\frac{CP}{PB} \\cdot \\frac{BT}{TA} = 1.\n$$\n\nSince $AS = AT$, $CS = CD$, and $BT = BD$, we have\n\n$$\n\\frac{PC}{PB} \\cdot \\frac{BD}{CD} = 1. \\qquad \\textcircled{1}\n$$\n\nLet the extension of $BN$ intersect $PE$ at point $H$. Then line $BFH$ intersects $\\triangle PDE$. By Menelaus' Theorem,\n\n$$\n\\frac{PH}{HE} \\cdot \\frac{EF}{FD} \\cdot \\frac{DB}{BP} = 1.\n$$\n\nSince $CF$ is parallel to $BE$, $\\frac{EF}{FD} = \\frac{PC}{CD}$, we have\n\n$$\n\\frac{PH}{HE} \\cdot \\frac{PC}{CD} \\cdot \\frac{DB}{BP} = 1. \\qquad \\textcircled{2}\n$$\n\nBy ① and ②, we have $PH = HE$. Hence, $PH^2 = HE^2 = HM \\cdot HN$. Thus, we have $\\frac{PH}{HM} = \\frac{HN}{PH}$, $\\triangle PHN \\sim \\triangle MHP$, and $\\angle HPN = \\angle HMP = \\angle NEQ$. Further, since $\\angle PEN = \\angle EQN$, therefore $\\angle ENP = \\angle ENQ$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21337,
"subject": "Mathematics (Olympiad)",
"question": "Rewrite the equation as\n$$\n\\frac{n(a_{n+2} - a_{n+1})}{a_n} = k.\n$$\nIf $a_n$ is a polynomial in $n$, say $a_n = r(n) = \\sum_{i=0}^{d} r_i n^i$, then the limit\n$$\n\\lim_{n \\to \\infty} \\frac{n(a_{n+2} - a_{n+1})}{a_n}\n$$\nmust exist and be _____.",
"options": [],
"answer": "See solution",
"solution": "The limit is equal to $k$.\n\n$$\n\\begin{align*}\n& \\lim_{n \\to \\infty} \\frac{n(a_{n+2} - a_{n+1})}{a_n} \\\\\n&= \\lim_{n \\to \\infty} \\frac{n \\left( \\sum_{i=0}^{d} r_i (n+2)^i - \\sum_{i=0}^{d} r_i (n+1)^i \\right)}{\\sum_{i=0}^{d} r_i n^i} \\\\\n&= \\lim_{n \\to \\infty} \\frac{n \\sum_{i=0}^{d} r_i \\left((n+2)^i - (n+1)^i\\right)}{\\sum_{i=0}^{d} r_i n^i} \\\\\n&= \\lim_{n \\to \\infty} \\frac{n^d \\sum_{i=0}^{d} r_i n^{i-d} \\left(\\left(1+\\frac{2}{n}\\right)^{i-1} + \\left(1+\\frac{2}{n}\\right)^{i-2}\\left(1+\\frac{1}{n}\\right) + \\dots + \\left(1+\\frac{1}{n}\\right)^{i-1}\\right)}{n^d \\sum_{i=0}^{d} r_i n^{i-d}} \\\\\n&= \\lim_{n \\to \\infty} \\frac{r_d \\left(\\left(1+\\frac{2}{n}\\right)^{d-1} + \\left(1+\\frac{2}{n}\\right)^{d-2}\\left(1+\\frac{1}{n}\\right) + \\dots + \\left(1+\\frac{1}{n}\\right)^{d-1}\\right)}{r_d} = d\n\\end{align*}\n$$\n\nSo if $a_n$ is a polynomial in $n$, then $k$ is the degree of $a_n$ and is a nonnegative integer.\n\nNow, to show an example of a polynomial $r_k(n)$ for each nonnegative integer $k$: for $k = 0$ and $k = 1$, one may consider $r_0(n) = 0$ and $r_1(n) = n$. For $k \\ge 2$, let\n$$\nr(n) = n^k + r_{k-1} n^{k-1} + \\cdots + r_1 n + r_0\n$$\nsuch that $n(r(n+2) - r(n+1)) = k r(n)$ for all $n$.\n\nExpanding $(n+2)^i$ and $(n+1)^i$ via the binomial theorem, we obtain the system\n$$\n\\begin{align*}\n& 3 \\binom{k}{k-2} + \\binom{k-1}{k-2} r_{k-1} = k r_{k-1} \\\\\n& 7 \\binom{k}{k-3} + 3 \\binom{k-1}{k-3} r_{k-1} + \\binom{k-2}{k-3} r_{k-2} = k r_{k-2} \\\\\n& \\qquad \\dots \\\\\n& (2^j - 1) \\binom{k}{k-j} + (2^{j-1} - 1) \\binom{k-1}{k-j} r_{k-1} + \\dots + \\binom{k-j+1}{k-j} r_{k-j+1} = k r_{k-j+1} \\\\\n& \\qquad \\dots\n\\end{align*}\n$$\nSince $\\binom{k-j+1}{k-j} = k-j+1 \\neq k$, it is possible to find $r_{k-j+1}$ in each equation and the system is solvable, so such a polynomial exists.\n\nFor the second part, rewrite the equation as\n$$\nn\\left(\\frac{a_{n+2}}{a_{n+1}} - 1\\right) = k \\frac{a_n}{a_{n+1}}.\n$$\nIf there exist polynomials $p$ and $q$ such that $\\frac{a_{n+1}}{a_n} = \\frac{p(n)}{q(n)}$, then\n$$\nn \\left( \\frac{p(n+1)}{q(n+1)} - 1 \\right) = k \\frac{q(n)}{p(n)} \\iff n(p(n+1) - q(n+1))p(n) = k q(n+1)q(n).\n$$\nSuppose $p(n)$ and $q(n)$ have no common factors. Let $m$ and $m'$ be the degrees of $p$ and $q$. If $m = m'$, the right side has degree $2m$ and $p(n+1) - q(n+1)$ has degree $m-1$, so the leading coefficients of $p$ and $q$ are equal. Since $p(n)$ and $q(n)$ have no common factors, $p(n)$ divides $q(n+1)$, so $p(n) = q(n+1)$. Substituting yields $n(q(n+2) - q(n+1)) = k q(n)$, so $q$ satisfies the first part and $k$ must be a nonnegative integer.\n\nIf $m \\neq m'$, by checking degrees, $1 + \\max(m, m') + m = 2m' \\implies 2m' \\le 1 + m + m' \\implies m < m' \\implies 1 + m' + m = 2m' \\implies m' = m + 1$. The polynomials $q(n+1)$ and $p(n+1) - q(n+1)$ have no common factors, so $q(n)$ divides $p(n+1) - q(n+1)$. Both have degree $m'$, with opposite leading coefficients, so $p(n+1) = q(n+1) - q(n)$. Substituting yields $n(q(n) - q(n-1)) = -k q(n+1)$. Let $r(n) = q(1-n)$ and $m = -n$, then $n(r(m+2) - r(m+1)) = -k r(m)$, reducing to the first part for $-k$ instead of $k$. So in this case, $k$ must be a nonpositive integer.\n\nConversely, it is not hard to obtain $p(n)$ and $q(n)$ from the above, so the answer for the second part is that $k$ must be an integer.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21338,
"subject": "Mathematics (Olympiad)",
"question": "Find all polynomials $P$ with non-negative integer coefficients such that for all primes $p$ and positive integers $n$ there exist a prime $q$ and a positive integer $m$ such that $P(p^n) = q^m$.",
"options": [],
"answer": "See solution",
"solution": "Notice that among the constant polynomials, the only solutions are $P(t) = q^m$ where $q$ is a prime and $m$ a positive integer.\n\nAssume that\n$$\nP(t) = a_k t^k + \\cdots + a_0,\n$$\nwhere $a_k \\neq 0$ and $a_0, a_1, \\ldots, a_k$ are non-negative integers, is a polynomial that fulfills the conditions.\n\nFirst, consider the case $a_0 \\neq 1$. Since $a_0$ is a non-negative integer different from $1$, there exists a prime $p$ such that $p$ divides $a_0$, and hence $p$ divides $P(p^n)$ for all $n$. Thus $P(p^n)$ is a power of $p$ for all positive integers $n$. If there exists a $k' < k$ such that $a_{k'} \\neq 0$, then for sufficiently large $n$ we have\n$$\n(p^n)^k > a_{k-1}(p^n)^{k-1} + \\cdots + a_0 > 0,\n$$\nand hence $P(p^n) \\neq 0 \\pmod{p^{nk}}$, but this contradicts $P(p^n) = p^m$ for some integer $m$ since obviously $m$ must be greater than $nk$. We conclude that in this case $P(t) = a_k t^k$, and it is easy to see that only $a_k = 1$ is a possibility.\n\nNow consider the case $a_0 = 1$. Let $Q(t) = P(P(t))$. Now $Q$ must, as well as $P$, satisfy the conditions. Since $Q(0) = P(P(0)) = P(1) > 1$ and $Q$ is not constant, we know from the previous that $Q(t) = t^k$, which contradicts that $Q(0) > 1$. Hence there are no solutions in this case.\n\nThus all polynomials that satisfy the conditions are $P(t) = t^m$ where $m$ is a positive integer, and $P(t) = q^m$ where $q$ is a prime and $m$ is a positive integer.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21339,
"subject": "Mathematics (Olympiad)",
"question": "Find all triples of positive integers $(x, y, z)$ such that\n\n$$\n2^x + 1 = 7^y + 2^z.\n$$",
"options": [],
"answer": "See solution",
"solution": "*Solution.* Because $x, y, z \\in \\mathbb{Z}^+$, we have $7^y > 1$, which implies $2^x > 2^z$ or $x > z$. The given equation can be rewritten as $2^z(2^{x-z} - 1) = 7^y - 1$.\n\nWe have $7^y \\equiv 1 \\pmod{3}$, so $3 \\mid 7^y - 1$, which implies $3 \\mid 2^{x-z} - 1$ or $x-z$ must be even. We consider the following cases:\n\n**Case 1.** If $y$ is odd, then by LTE theorem, $v_2(7^y - 1) = v_2(7 - 1) = 1$. Thus $v_2(2^z) = 1$ and $z = 1$. Substituting $z = 1$ into the original equation, we get $2^x = 7^y + 1$. Similarly, because $y$ is odd, $v_2(7^y + 1) = v_2(7 + 1) = 3$, which implies $x = 3, y = 1$. By direct checking, this solution is satisfied. Therefore, in this case, we have the solution $(x, y, z) = (3, 1, 1)$.\n\n**Case 2.** If $y$ is even, let $y = 2k$ where $k$ is a positive integer, then\n\n$$\n2^z(2^{x-z} - 1) = 49^k - 1.\n$$\n\nWe consider the following cases:\n\n1. If $k$ is odd, then $v_2(49^k - 1) = v_2(49 - 1) = 4$, which implies $z = 4$. Substituting $z = 4$ into the original equation, we have $2^x - 49^k = 15$, so $x$ is even. Let $x = 2t$ with $t \\in \\mathbb{Z}^+$, then\n\n$$\n4^t - 49^k = 15 \\Leftrightarrow (2^t + 7^k)(2^t - 7^k) = 15.\n$$\n\nSince $2^t + 7^k \\leq 15$, then $7^k < 15$ or $k = 1$. From there we get $4^t = 64$ or $t = 3$. So $x = 6, y = 2$. Then we obtain another solution $(x, y, z) = (6, 2, 4)$.\n\n2. If $k$ is even, denote $k = 2l$ where $l$ is a positive integer, then $y = 4l$ and we have $49^k \\equiv (-1)^k \\equiv 1 \\pmod{25}$, thus $25 \\mid 2^{x-z} - 1$. Since $2 \\mid x-z$, denote $x-z = 2a$ where $a \\in \\mathbb{Z}^+$, we get $25 \\mid 4^a-1$ so $5 \\mid 4^a - 1$ and it implies that $a$ is even; next, denote $a = 2b$ where $b \\in \\mathbb{Z}^+$, then $v_5(49^k - 1) \\geq 2$ and\n\n$$\nv_5(2^z(16^b - 1)) = v_5(16^b - 1) = v_5(15) + v_5(b) = 1 + v_5(b).\n$$\n\nFrom here, it follows that $5 \\mid b$ and let $b = 5c$ where $c \\in \\mathbb{Z}^+$, we have $2^{x-z} - 1 = 1024^c - 1 \\equiv 0 \\pmod{1023}$. Notice that $31 \\mid 1023$ then $31 \\mid 7^y - 1$. But we also have\n\n$$\n7^{15} = (7^3)^5 \\equiv 343^5 \\equiv 2^5 \\equiv 32 \\equiv 1 \\pmod{31}\n$$\n\nand $7^3, 7^5, 7^{10} \\ne 1 \\pmod{31}$ so $\\text{ord}_{31}(7) = 15$. Thus, according to the property of the order of a number, we have $15 \\mid y$ so $3 \\mid y$, but $4 \\mid y$ then $12 \\mid y$, it follows that $13 \\mid 7^y - 1$ by Little Fermat theorem.\n\nThus, $13 \\mid 2^{x-z} - 1$ and $\\text{ord}_{13}(2) = 12$ which leads to $12 \\mid x-z$, and $7 \\mid 2^{x-z} - 1$, which is a contradiction since the right hand side is not divisible by 7. So the equation has no solution.\n\nSo all satisfying triples are $(x, y, z) = (3, 1, 1), (6, 2, 4)$. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21340,
"subject": "Mathematics (Olympiad)",
"question": "Denote by $X_{i,j}$ the entry that lies in the $i$-th row and the $j$-th column. \n\nLet $(m, n)$ be a pair of positive integers. A pair $(m, n)$ is called a *good pair* if there exists an $m \\times n$ grid filled with integers $X_{i,j}$ such that for every $2 \\times 2$ square $S$ in the grid, the sum $\\sigma(S)$ of the entries at the four corners of $S$ is zero, i.e.,\n$$\n\\sigma(S) = X_{i,j} + X_{i+1,j} + X_{i,j+1} + X_{i+1,j+1} = 0\n$$\nfor all $1 \\leq i < m$, $1 \\leq j < n$.\n\n(1) For which $n$ is $(3, n)$ a good pair?\n\n(2) How many pairs $(i, j)$ with $3 \\leq i, j \\leq 11$ are not relatively prime?",
"options": [],
"answer": "See solution",
"solution": "Now, let $i$ and $j$ be integers such that $i - j$ is even. Since $m + 1$ and $n + 1$ are relatively prime, there exist integers $u, v$ such that\n$$\ni - j = 2v(n + 1) - 2u(m + 1),$$\nor,\n$$\ni + 2u(m + 1) = j + 2v(n + 1).$$\n\nThen, $0 = X_{i+2u(m+1)+1,j+2v(n+1)} + X_{i+2u(m+1),j+2v(n+1)+1} = X_{i+1,j} + X_{i,j+1}$ follows from the periodicity of $X$. Since $X_{0,j} = 0$, it follows by induction that $X_{i,j} = 0$ for all $i, j$ such that $i + j$ is odd.\n\nSince $m + 1$ and $n + 1$ are relatively prime, $m$ or $n$ is even. Assume that $m$ is even. Then, by symmetry, $X_{i+1,j} + X_{i,j+1} = 0$ for even $i + j$ too, and $X_{i,j} = 0$ follows for all $i$ and $j$. Hence $(m, n)$ is not a good pair.\n\nWe can solve the problem using the results above.\n\n1. $(3, n)$ is a good pair if and only if $4$ and $n+1$ are not relatively prime, in other words, if $n$ is odd. So the answer is $n = 3, 5, 7, 9$.\n\n2. One counts the number of pairs $(i, j)$ of integers such that $3 \\leq i, j \\leq 11$ and $i, j$ are not relatively prime, and obtains the answer $29$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21341,
"subject": "Mathematics (Olympiad)",
"question": "The triangle $ABC$ satisfies $BC = 5$, $CA = 7$, and $AB = 8$. Let $O$ be a point inside triangle $ABC$ such that the triangles $OBC$, $OCA$, and $OAB$ have the same circumradius. Find their common circumradius.",
"options": [],
"answer": "See solution",
"solution": "$\\dfrac{7}{\\sqrt{3}}$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21342,
"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 (labelled $b_i$) a number of $i$ coins.\n3. If any box is left empty, the game ends; otherwise, proceed 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 always choosing the last $k+1$ boxes, $b_{2^k-1}, b_{2^k}, \\dots, b_{2^k+k-1}$, at each step. At step $r$, if box $b_{2^k+i-1}$ has exactly $m_i$ coins, then $\\lceil m_i/2 \\rceil$ coins are removed from that box, unless $i \\equiv r-1 \\pmod{k+1}$, in which case $2^k + i - 1$ coins are added. Thus, after step $r$, box $b_{2^k+i-1}$ contains exactly $\\lceil m_i/2 \\rceil$ coins, unless $i \\equiv r-1 \\pmod{k+1}$, in which case it contains $m_i + 2^k + i - 1$ coins. This process can continue 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 until its next supply.\n\nTo show that no smaller value of $n$ works, suppose $n \\leq 2^k + k - 2$ and that the game can be played indefinitely. Notice that a box with $m$ coins survives at most $w = \\lceil \\log_2 m \\rceil$ withdrawals; this $w$ is called the *weight* of the box. The sum of the weights of all boxes is the *total weight*. The argument relies on the following lemma:\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. 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 continue indefinitely, so the game must eventually end, contradicting the assumption.\n\nTherefore, 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 suffices to show that supplying a box increases its weight by at most $k$; and if the box 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 $m_i$ coins. Consider cases:\n\n- If $m_i = 1$, 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 $i \\leq 2^k - 2$, then the weight increases by $\\lfloor \\log_2(i+1) \\rfloor \\leq \\lfloor \\log_2(2^k - 1) \\rfloor = k - 1$.\n- If $m_i = 2$, 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- If $m_i \\geq 3$, the weight increases by\n $$\n \\begin{aligned}\n \\lfloor \\log_2(i + m_i) \\rfloor - \\lfloor \\log_2 m_i \\rfloor &\\leq \\lfloor \\log_2(i + m_i) - \\log_2 m_i \\rfloor + 1 \\\\\n &\\leq \\lfloor \\log_2 \\left( 1 + \\frac{2^k + k - 2}{3} \\right) \\rfloor + 1 \\leq k,\n \\end{aligned}\n $$\n since $1 + \\frac{1}{3}(2^k + k - 2) = \\frac{1}{3}(2^k + k + 1) < \\frac{1}{3}(2^k + 2^{k+1}) = 2^k$.\n\nIf $i \\leq 2^k - 2$, consider the subcases $m_i = 3$ and $m_i \\geq 4$:\n\n- For $m_i = 3$, the weight increases by\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- For $m_i \\geq 4$,\n $$\n \\begin{aligned}\n \\lfloor \\log_2(i + m_i) \\rfloor - \\lfloor \\log_2 m_i \\rfloor &\\leq \\lfloor \\log_2(i + m_i) - \\log_2 m_i \\rfloor + 1 \\\\\n &\\leq \\lfloor \\log_2 \\left( 1 + \\frac{2^k - 2}{4} \\right) \\rfloor + 1 \\leq k - 1,\n \\end{aligned}\n $$\n since $1 + \\frac{1}{4}(2^k - 2) = \\frac{1}{4}(2^k + 2) < 2^{k-2} + 1$.\n\nThis completes the proof and the solution.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21343,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $N = p_1^{k_1} p_2^{k_2} \\cdots p_r^{k_r}$ is a positive integer, and $n$ is a positive integer. On a board are $n$ pairs of numbers $(a, \\frac{N}{a})$ for divisors $a$ of $N$, and possibly some extra numbers. Two players, Alice and Bob, take turns erasing two numbers and writing either their greatest common divisor or least common multiple. What is the least possible value $M$ such that, regardless of the moves, the final remaining number is always a divisor or multiple of $M$?",
"options": [],
"answer": "See solution",
"solution": "Alice can guarantee that the remaining number on the board is a divisor of $M$.\n\nSuppose $n$ is odd, $M = (p_1 \\cdots p_r)^{(n-1)/2}$. Alice adopts a strategy so that her move always leaves some pairs and a divisor $C$ of $M$ on the board. She erases $1$, $N$, and writes $1$, leaving $\\frac{1}{2}(n+1)^r - 1$ pairs and $1$ on the board.\n\n1. If Bob erases $a$, $b$ from two pairs $(a, \\frac{N}{a})$, $(b, \\frac{N}{b})$, and writes $\\text{lcm}(a, b)$, then Alice erases $\\frac{N}{a}$, $\\frac{N}{b}$, and writes $\\text{gcd}(\\frac{N}{a}, \\frac{N}{b}) = \\frac{N}{\\text{lcm}(a, b)}$, which forms a pair with $\\text{gcd}(a, b)$.\n2. If Bob erases a pair $(a, \\frac{N}{a})$ and writes $\\text{lcm}(a, \\frac{N}{a})$, then Alice erases $\\text{lcm}(a, \\frac{N}{a})$ and $C$, and writes their greatest common divisor $C'$, which is a divisor of $M$, leaving some pairs and $C'$ on the board.\n3. If Bob erases $C$ and $a$ from $(a, \\frac{N}{a})$ and writes $\\text{lcm}(a, C)$, then Alice erases $\\frac{N}{a}$, $\\text{lcm}(a, C)$, and writes $C' = \\text{gcd}(\\frac{N}{a}, \\text{lcm}(a, C))$, which is a divisor of $\\text{lcm}(C, \\text{gcd}(a, \\frac{N}{a}))$. Since $\\text{gcd}(a, \\frac{N}{a})$ is a divisor of $M$, $C'$ is likewise. Thus, after Alice's move, some pairs and a divisor $C'$ of $M$ remain. Eventually, only one number is left, which is a divisor of $M$.\n\nIf $n$ is even, $M = (p_1 \\cdots p_r)^{n/2}$. Alice adopts a strategy so that her move always leaves some pairs and two divisors $C_1$ and $C_2$ of $M$ on the board. She erases $1$, $N$, and writes $1$, leaving pairs and $1$, $M$. In subsequent moves, if Bob erases one or two numbers from pair(s), Alice follows the steps above; if Bob erases $C_1$, $C_2$ and writes $\\text{lcm}(C_1, C_2) = C'_1$ (also a divisor of $M$), then Alice erases any pair $(a, \\frac{N}{a})$ and writes $\\text{gcd}(a, \\frac{N}{a}) = C'_2$ (again a divisor of $M$). Thus, after Alice's move, some pairs and two divisors of $M$ remain. Eventually, Bob must be confronted with two divisors of $M$, and his move will leave a divisor of $M$ on the board.\n\nBob can guarantee that the remaining number on the board is a multiple of $M$.\n\nBob adopts a strategy so that his move always leaves some pairs and a multiple $C_1$ of $M$ (when $n$ is even), or two multiples $C_1$ and $C_2$ of $M$ (when $n$ is odd). Initially, take $C_1 = (p_1 \\cdots p_r)^{n/2}$ (when $n$ is even), or $C_1 = (p_1 \\cdots p_r)^{(n-1)/2}$, $C_2 = (p_1 \\cdots p_r)^{(n+1)/2}$ (when $n$ is odd). Bob reverses Alice's moves as follows:\n\n1. If Alice erases $a, b$ from two pairs $(a, \\frac{N}{a})$, $(b, \\frac{N}{b})$, and writes $\\text{gcd}(a, b)$, then Bob erases $\\frac{N}{a}, \\frac{N}{b}$, and writes $\\text{lcm}(\\frac{N}{a}, \\frac{N}{b}) = N/\\text{gcd}(a, b)$.\n2. If Alice erases a pair $(a, \\frac{N}{a})$ and writes $\\text{gcd}(a, \\frac{N}{a})$, then Bob erases this number and $C_1$.\n3. If Alice erases $C_i$ and $a$ from $(a, \\frac{N}{a})$ and writes $\\text{gcd}(a, C_i)$, then Bob erases $\\frac{N}{a}$ and $\\text{gcd}(a, C_i)$.\n4. If Alice erases $C_1, C_2$ and writes $\\text{gcd}(C_1, C_2)$, then Bob erases any pair $(a, \\frac{N}{a})$.\n\nBy interchanging \"gcd\" and \"lcm\" in the strategies, Bob guarantees the remaining number is a multiple of $M$.\n\nTherefore, the least $M$ is $(p_1 \\cdots p_r)^{\\lfloor \\frac{n}{2} \\rfloor}$.\n\n$\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21344,
"subject": "Mathematics (Olympiad)",
"question": "The triangle $ABC$ ($AB > AC$) is inscribed in a circle. The bisector of $\\angle BAC$ intersects side $BC$ at point $K$ and the circumcircle at point $M$. The midline of $\\triangle ABC$ parallel to side $AB$ intersects $AM$ at point $O$. The straight line $CO$ intersects side $AB$ at point $N$. Prove that a circle can be drawn around quadrilateral $BNKM$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Since $PO \\parallel AB$, we have $\\angle BAO = \\angle AOP$, so $OP = AP$. Triangle $AOC$ is right-angled at $O$ because $AP = PC$. Therefore, $CN \\perp AO$, so $AO$ is both the angle bisector and the altitude of $\\triangle ANC$, making it isosceles. Thus, $AN = AC$ and $NM = MC$. Also, $\\angle MNK = \\angle MCK = \\angle MBK$, which implies that $BNKM$ is cyclic.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21345,
"subject": "Mathematics (Olympiad)",
"question": "Let $z$ be an arbitrary complex number and denote by $\\operatorname{Im}(z)$ its imaginary part. Minimize the expression\n\n$$\nf(z) = |\\operatorname{Im}(z)|^2 + \\sum_{k=1}^{n} |z - a_k|^2.\n$$\nwhere $a_1, \\ldots, a_n$ are given complex numbers.",
"options": [],
"answer": "See solution",
"solution": "Let $m = \\frac{1}{n} \\sum_{k=1}^{n} a_k$ be the centroid of the points $a_k$. Using $|w|^2 = w \\bar{w}$, we have\n\n$$\n\\sum_{k=1}^{n} |z - a_k|^2 = n|z - m|^2 + \\sum_{k=1}^{n} |m - a_k|^2.\n$$\n\nThus,\n\n$$\nf(z) = |\\operatorname{Im}(z)|^2 + n|z - m|^2 + \\sum_{k=1}^{n} |m - a_k|^2.\n$$\n\nWrite $m = a + ib$, $z = x + iy$ with $a, b, x, y \\in \\mathbb{R}$. Then\n\n$$\n|\\operatorname{Im}(z)|^2 + n|z - m|^2 = y^2 + n[(x - a)^2 + (y - b)^2] = n(x - a)^2 + y^2 + n(y - b)^2.\n$$\n\nThis is minimized when $x = a$, $y = \\dfrac{n b}{n + 1}$. Therefore,\n\n$$\n\\min f = \\frac{n^2 b^2}{(n+1)^2} + \\frac{n b^2}{(n+1)^2} + \\sum_{k=1}^{n} |m - a_k|^2 = \\frac{n (\\operatorname{Im}(m))^2}{n + 1} + \\sum_{k=1}^{n} |m - a_k|^2.\n$$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21346,
"subject": "Mathematics (Olympiad)",
"question": "In the country of Oddland, there are stamps with values 1 cent, 3 cent, 5 cent, etc., one type for each odd number. The rules of Oddland Postal Services stipulate the following: for any two distinct values, the number of stamps of the higher value on an envelope must never exceed the number of stamps of the lower value.\n\nIn the country of Squareland, on the other hand, there are stamps with values 1 cent, 4 cent, 9 cent, etc., one type for each square number. Stamps can be combined in all possible ways in Squareland without additional rules.\n\nProve for every positive integer $n$: In Oddland and Squareland there are equally many ways to correctly place stamps of a total value of $n$ cent on an envelope. Rearranging the stamps on an envelope makes no difference.",
"options": [],
"answer": "See solution",
"solution": "We construct a bijection between possible combinations in Oddland and possible combinations in Squareland. Suppose we have a combination of Squareland stamps that sum to $n$ cent, consisting of $a_1$ stamps of value 1 cent, $a_2$ stamps of value 4 cent, ..., $a_M$ stamps of value $M^2$ cent, so that\n\n$$\nn = \\sum_{k=1}^{M} k^2 a_k.\n$$\n\nNow we express $k^2$ as $\\sum_{j=1}^{k}(2j - 1)$ and interchange the order of summation, which yields\n\n$$\nn = \\sum_{k=1}^{M} \\sum_{j=1}^{k} (2j - 1) a_k = \\sum_{j=1}^{M} (2j - 1) \\sum_{k=j}^{M} a_k.\n$$\n\nThis gives us a possible combination of Oddland stamps: By setting $b_j = \\sum_{k=j}^{M} a_k$, we have\n\n$$\nn = \\sum_{j=1}^{M} (2j - 1)b_j.\n$$\n\nThis can be interpreted as a collection of $b_1$ stamps of value 1 cent, $b_2$ stamps of value 3 cent, ..., $b_M$ stamps of value $(2M - 1)$ cent. We have $b_1 \\ge b_2 \\ge \\dots \\ge b_M$ by definition, so this is a legal combination in Oddland.\n\nConversely, if a combination in Oddland is given by the values $b_1, b_2, \\dots, b_M$, we can use the identities $a_1 = b_1 - b_2$, $a_2 = b_2 - b_3$, ..., $a_{M-1} = b_{M-1} - b_M$, $a_M = b_M$ to recover the corresponding combination in Squareland. (Note that these values are nonnegative whenever $b_1 \\ge b_2 \\ge \\dots \\ge b_M$.)\n\nSince these two operations obviously are inverse to one another, we have found a bijection, which proves the statement.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21347,
"subject": "Mathematics (Olympiad)",
"question": "Numbers from $1$ to $2007$ are arbitrarily written in strip cells. Two players take turns marking cells. The player after whose move there exist two natural numbers $m < n$ such that the sum of the numbers in all marked cells from $m$ to $n$ is divisible by $2008$ loses. Prove that there are at least $1000$ initial arrangements of numbers in the cells such that, for any non-negative integers $i \\leq j$, the sum of the numbers in cells $i, i+1, \\dots, j$ is never divisible by $2008$, and for each such arrangement, the first player has a winning strategy.",
"options": [],
"answer": "See solution",
"solution": "Let the number $a_n$ be written in cell $n$. Consider the sequence $(a_n)$ defined by $a_{2k-1} = 2k-1$ for $k = 1, \\dots, 1004$ and $a_{2k} = 2008 - 2k$ for $k = 1, \\dots, 1003$. We claim this arrangement works.\n\nLet $S_n = a_1 + a_2 + \\dots + a_n$. Then:\n\n- $S_{2k} = 2007k$ for $k = 1, \\dots, 1003$\n- $S_{2k-1} = S_{2k-2} + a_{2k-1} = 2009k - 2008$ for $k = 1, \\dots, 1004$\n\nNote that $S_{2k-1} \\equiv -2007k \\pmod{2008}$. Thus, $S_n \\equiv S_m \\pmod{2008}$ only if $n = m$. Therefore, for any $1 \\leq i < j \\leq 2007$, the sum $a_i + a_{i+1} + \\dots + a_j$ is never divisible by $2008$.\n\nNow, a winning strategy for the first player: on the first move, mark cell $1004$. Afterwards, if the second player marks cell $i$, the first player marks cell $2008 - i$. Since $a_i + a_{2008-i} = 2008$ for all $i = 1, \\dots, 2007$, this strategy ensures the first player wins.\n\nTo construct $999$ more such sequences, let $l$ be any integer with $(l, 2008) = 1$. Define $b_n = l \\cdot a_n \\pmod{2008}$ for $n = 1, \\dots, 2007$. Each such sequence $(b_n)$ also satisfies the required property, and the first player's strategy remains valid. Since there are $\\varphi(2008) = 1000$ such $l$, we obtain at least $1000$ sequences.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21348,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge 4$, and consider a non-intersecting $n$-gon $P_1P_2\\ldots P_n$ in the plane. Suppose that, to each $P_k$, there is a unique other vertex $Q_k$ among $P_1, \\ldots, P_n$ that lies closest to it. The polygon is said to be *hostile* if $Q_k \\neq P_{k \\pm 1}$ for all $k$ (counting cyclically).\n\n(a) Prove that there exist no convex hostile polygons.\n\n(b) Determine all $n$ for which there exists a concave hostile $n$-gon.",
"options": [],
"answer": "See solution",
"solution": "(a) As an auxiliary result, we prove the following. There is no convex quadrilateral $ABCD$ in which $C$ is the closest neighbour of $A$, and $D$ is the closest neighbour of $B$.\n\n\n\nIndeed, the diagonals *AC* and *BD* cross at a point *P* by convexity. The Triangle Inequality yields\n\n$$\nAD + BC < (AP + PD) + (BP + PC) = (AP + PC) + (BP + PD) = AC + BD.\n$$\n\nSince $AC < AD$ by the first assumption, we must have $BC < BD$, contradicting the second assumption.\n\nNow, consider a convex hostile polygon $P_1P_2\\ldots P_n$. We say a vertex $P_k$ has *separation $h \\ge 2$* when its closest neighbour is $Q_k = P_{k+h}$. Among all vertices, select one with minimal separation $h$; without loss of generality, we may take it to be $P_1$, thus its closest neighbour is $Q_1 = P_{h+1}$. Since $h \\ge 2$, the vertex $P_2$ does not belong to the line $P_1P_{h+1}$. The closest neighbour of $P_2$ cannot lie on the opposite side of this line according to the auxiliary result proved above, i.e. $Q_2$ is to be found among the vertices $P_1, \\ldots, P_{h+1}$. But then $P_2$ will have separation at most $h-1$, which contradicts the minimality of $h$.\n\n(b) *Answer:* Hostile concave polygons exist for all $n \\ge 4$.\n\nStart from the type of zig-zag construction given in the figure below, and dislocate the vertices slightly, so as to make all distances unequal (to make each vertex have a unique closest neighbour).\n\n\n\nHostile quadrilateral, pentagon, hexagon, heptagon and octagon.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21349,
"subject": "Mathematics (Olympiad)",
"question": "Given a positive integer $n$, determine the maximum number of lattice points in the plane that a square of side length $n + \\frac{1}{2n + 1}$ may cover.",
"options": [],
"answer": "See solution",
"solution": "The required maximum is $(n + 1)^2$.\n\nClearly, the square $[-\\epsilon/2, n + \\epsilon/2] \\times [-\\epsilon/2, n + \\epsilon/2]$, $0 \\leq \\epsilon < 1$, covers exactly $(n + 1)^2$ lattice points.\n\nWe now proceed to show that any (closed) square of side length $n + \\epsilon$, $0 \\leq \\epsilon \\leq \\frac{1}{2n + 1}$, covers at most $(n + 1)^2$ lattice points.\n\nThe case $n = 1$ is settled by a metric argument: the diameter of a square of side length $1 + \\epsilon$ is $(1 + \\epsilon)\\sqrt{2}$, whereas the diameter of any configuration of five lattice points is at least $\\sqrt{5} > (1 + \\epsilon)\\sqrt{2}$ in the slightly wider range $0 \\leq \\epsilon < \\frac{\\sqrt{10}}{2} - 1$.\n\nHenceforth assume $n \\geq 2$ and consider the convex hull $K$ of the lattice points covered by a square of side length $n + \\epsilon$, $0 \\leq \\epsilon \\leq \\frac{1}{2n + 1}$. Clearly, $\\text{area}(K) \\leq (n + \\epsilon)^2$, the area of the square. On the other hand, by Pick's theorem, $\\text{area}(K) = m - \\frac{k}{2} - 1$, where $m$ is the number of lattice points covered by $K$, and $k$ is the number of lattice points on the boundary of $K$. Therefore,\n\n$$\nm = \\text{area}(K) + \\frac{k}{2} + 1 \\leq (n + \\epsilon)^2 + \\frac{k}{2} + 1.\n$$\n\nTo find an upper bound for $k$, notice that the perimeter of $K$ does not exceed the perimeter of the square, which is $4(n + \\epsilon) \\leq 4n + \\frac{4}{2n + 1} < 4n + 1$, for $n \\geq 2$. Since the distance between two lattice points is at least $1$, it follows that $k \\leq 4n$. Consequently,\n\n$$\nm \\leq (n + \\epsilon)^2 + 2n + 1 = (n + 1)^2 + 2n\\epsilon + \\epsilon^2 < (n + 1)^2 + 1.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21350,
"subject": "Mathematics (Olympiad)",
"question": "令 $n$ 為正整數，並給定無限長的週期字串 $W = \\dots x_{-1}x_0x_1x_2\\dots$，其中 $x_i$ 可以是字母 $a$ 或 $b$ 對所有整數 $i$，而 $W$ 的最小週期 $p$ 滿足 $p > 2^n$（即 $x_i = x_{i+p}$ 對所有 $i$，且這樣的 $p$ 不可能再小了）。我們稱一個長度 $\\leq p$ 的非空子字串 $U$ 是左右逢源的（縮寫為 FU），如果 $aU, bU, Ua, Ub$ 這四種字串都出現在 $W$ 中。證明：$W$ 至少包含 $n$ 個 FU 子字串。",
"options": [],
"answer": "See solution",
"solution": "對於字串 $U$，以 $|U|$ 表示其長度。所有子串皆為非空且長度 $\\leq p$。$W$ 可視為長度為 $p$ 的圓周字串 $x_0x_1\\dots x_{p-1}$。定義子字串 $U = a_0a_1\\dots a_{k-1}$ 在 $W$ 中的出現次數：\n\n$$\nm(U) = |\\{i \\mid 0 \\leq i \\leq p-1,\\ a_0a_1\\cdots a_{k-1} = x_i x_{i+1} \\cdots x_{i+k-1}\\}|,\n$$\n\n其中所有下標皆取 $\\bmod\\ p$。即 $U$ 出現在 $W$ 中當且僅當 $m(U) \\geq 1$。且有：\n\n$$\nm(U) = m(aU) + m(bU) = m(Ua) + m(Ub),\n$$\n\n對任意長度 $\\leq p-1$ 的 $U$ 成立。當 $|U| = p$，若某 $U = a_0a_1\\cdots a_{p-1}$ 有 $m(U) \\geq 2$，則存在 $0 \\leq i < j \\leq p-1$ 使得：\n\n$$\na_0a_1\\cdots a_{p-1} = x_i x_{i+1} \\cdots x_{i+p-1} = x_j x_{j+1} \\cdots x_{j+p-1},\n$$\n\n這將導致 $x_k = x_{j-i+k}$ 對所有 $k = 0, 1, \\dots, p-1$，產生新的週期 $j-i < p$，矛盾。因此長度為 $p$ 的字串必有 $m(U) \\leq 1$，所以 $aU$ 與 $bU$ 只有一個會出現在 $W$ 中，$U$ 不可能是 FU。\n\n（註：$aU$ 與 $bU$ 長度為 $p+1$，$m(aU)$ 與 $m(bU)$ 無意義。）\n\n另一個重要觀察：因 $p > 2^k$，$m(a)$ 與 $m(b)$ 至少有一個大於 $2^{n-1}$。\n\n對每個 $k = 0, 1, \\dots, n-1$，定義：\n\n$$\n\\Omega_k = \\{V \\mid V \\text{ 是 } W \\text{ 長度 } \\leq p-1 \\text{ 的子字串且 } m(V) > 2^k \\}.\n$$\n\n取 $U_k \\in \\Omega_k$，使其長度在 $\\Omega_k$ 中最長。由於上述觀察，$\\Omega_k$ 非空，$U_k$ 存在。\n\n固定 $k$，因 $aU_k$ 比 $U_k$ 長，故 $m(aU_k) \\leq 2^k$；但 $2^k < m(U_k) = m(aU_k) + m(bU_k)$，故 $m(bU_k) \\neq 0$。同理 $m(aU_k) \\neq 0$，$m(U_ka), m(U_kb) \\neq 0$。因此 $U_k$ 為 FU。\n\n證明 $n$ 個 $U_k$ 彼此不同：若 $U_k \\in \\Omega_{k+1}$，即 $m(U_k) > 2^{k+1}$，則 $m(aU_k)$ 與 $m(bU_k)$ 至少有一個大於 $2^k$，與 $U_k$ 的定義矛盾。因此 $2^0 < m(U_0) \\leq 2^1 < m(U_1) \\leq 2^2 < \\dots < 2^{n-1} < m(U_{n-1})$，故 $n$ 個 $U_k$ 皆不同。證畢。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21351,
"subject": "Mathematics (Olympiad)",
"question": "Fix a prime number $p > 5$. Let $a$, $b$, $c$ be integers, no two of which have their difference divisible by $p$. Let $i$, $j$, $k$ be nonnegative integers such that $i + j + k$ is divisible by $p - 1$. Suppose that for all integers $x$, the quantity\n\n$$\n(x - a)(x - b)(x - c) \\left[ (x - a)^i (x - b)^j (x - c)^k - 1 \\right]\n$$\n\nis divisible by $p$. Prove that each of $i$, $j$, $k$ must be divisible by $p-1$.",
"options": [],
"answer": "See solution",
"solution": "We first prove that $k$ is congruent to one of $1$, $0$, $-1$ modulo $(p-1)$. Rewriting the hypothesis in modular arithmetic: If $x \\not\\equiv a, b, c \\pmod{p}$, then\n\n$$\n(x - a)^i (x - b)^j \\equiv (x - c)^{-k} \\pmod{p}.\n$$\n\nBy Fermat's little theorem, shifting $i$, $j$, $k$ by multiples of $p-1$ does not affect the result, so we may assume $i$, $j$, $-k$ are in $\\{0, 1, \\dots, p-2\\}$. If $i + j \\leq p-2$, then $i + j + k \\equiv 0 \\pmod{p-1}$ forces $-k = i + j$. In this case, the polynomial $(x-c)^{-k} - (x-a)^i(x-b)^j$ has degree at most $-k-1$, but modulo $p$ it has at least $p-3$ distinct roots (for $x$ not congruent to $a$, $b$, $c$). The only ways to avoid contradiction are either $i+j = -k = p-2$ (so $k \\equiv 1 \\pmod{p-1}$), or the polynomial is identically zero modulo $p$, which forces $i = j = -k = 0$.\n\nIf $i + j \\geq p - 1$, then\n\n$$\n(x - a)^{p-1-i} (x - b)^{p-1-j} \\equiv (x - c)^{k} \\pmod{p}\n$$\n\nfor all $x \\not\\equiv a, b, c \\pmod{p}$. If $k \\equiv 0 \\pmod{p-1}$, we are done; otherwise, $i+j \\equiv -k \\not\\equiv 0 \\pmod{p-1}$ forces $i+j \\geq p$, so $(p-1-i) + (p-1-j) \\leq p-2$. Thus, the previous argument implies either $k \\equiv 0 \\pmod{p-1}$ or $k \\equiv -1 \\pmod{p-1}$.\n\nBy symmetry, each of $i$, $j$, $k$ is congruent to one of $-1$, $0$, $1$ modulo $p-1$. Since $i + j + k \\equiv 0 \\pmod{p-1}$ and $p-1 > 3$, the only possibilities (up to permutation) are $\\{i, j, k\\} \\equiv \\{0, 0, 0\\}$ or $\\{1, 0, -1\\}$ modulo $p-1$. The latter implies $x - a \\equiv x - c \\pmod{p}$ for all $x \\neq a, b, c \\pmod{p}$, which is impossible for $p > 3$. This contradiction leaves only $\\{i, j, k\\} \\equiv \\{0, 0, 0\\}$ modulo $p-1$, proving the result.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21352,
"subject": "Mathematics (Olympiad)",
"question": "Нека $O$ е центар на впишаната кружница во триаголникот $ABC$. Точките $K$ и $L$ се пресечни точки на кружниците опишани околу триаголниците $BOC$ и $AOC$ соодветно со симетралите на аглите во $A$ и $B$ соодветно. $P$ е средина на отсечката $KL$, $M$ е симетрична на $O$ во однос на $P$, а $N$ е симетрична на $O$ во однос на правата $KL$. Докажи дека четириаголникот $KLMN$ е тетивен.",
"options": [],
"answer": "See solution",
"solution": "Аглите $LCA$ и $LOA$ се еднакви како тетивни над ист кружен лак. Аголот $LOA$ е еднаков на збирот од аглите $OAB$ и $OBA$, како надворешен агол на триаголникот $ABO$, па оттука следува дека:\n\n$$\n\\begin{aligned}\n\\angle LCO &= \\angle LCA + \\angle OCA = \\angle LOA + \\angle OCA = \\angle OAB + \\angle OBA + \\angle OCA = \\\\\n&= \\frac{1}{2}(\\angle CAB + \\angle ABC + \\angle ACB) = 90^\\circ\n\\end{aligned}\n$$\n\nАналогно и $\\angle KCO = 90^\\circ$, па точката $C$ лежи на правата $KL$, и точката $C$ е средина на отсечката $\\overline{ON}$. Правата $PC$ е паралелна на $MN$ како средна линија.\n\nЧетириаголникот $LOKM$ е паралелограм, бидејќи неговите дијагонали\n\n\n\nсе преполовуваат во точката $P$. Следува дека, аглите $MK$ и $OKL$ се еднакви.\n\nОд друга страна триаголникот $OKN$ е рамнокрак со основа $ON$ ($KC$ е висина и тежишна линија во него), па $KC$ е симетрала на аголот $OKN$, то ест аглите $OKC$ и $NKC$ се еднакви.\n\nСледува дека $\\angle MLK = \\angle NKL$, па четириаголникот $KLMN$ е рамнокрак трапез, то ест тетивен. (Ако точката $P$ се наоѓа од другата страна на точката $C$, ги разгледуваме аглите $MKL$ и $NLK$.)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21353,
"subject": "Mathematics (Olympiad)",
"question": "For every positive integer $x$, denote by $\\kappa(x)$ the number of composite numbers not greater than $x$. Find all positive integers $n$ such that\n\n$$\n(\\kappa(n))! \\cdot \\text{lcm}(1,2,\\dots,n) > (n-1)!\n$$\n",
"options": [],
"answer": "See solution",
"solution": "The inequality holds for $n = 2, 3, 4, 5, 7, 9$ and does not hold for $n = 1, 6, 8, 10, 11, 12$.\n\nAssume in the rest that $n \\ge 13$. By definition of $\\kappa(n)$, there exist exactly $n-1-\\kappa(n)$ prime numbers not greater than $n$; these are the primes dividing $\\text{lcm}(1,2,\\dots,n)$. We have $n-1-\\kappa(n) \\ge 6$ as $n \\ge 13$.\n\nLet $q_1, \\dots, q_{n-1-\\kappa(n)}$ be all prime powers in the canonical representation of $\\text{lcm}(1,2,\\dots,n)$. W.l.o.g., $q_1 > q_2 > \\dots > q_{n-1-\\kappa(n)}$. As at least 5 numbers among $q_1, \\dots, q_6$ are odd, in the case of odd $n$ we have\n$$\nq_1 q_2 q_3 q_4 q_5 q_6 \\le n(n-1)(n-2)(n-4)(n-6)(n-8)\n$$\nand in the case of even $n$ similarly\n$$\nq_1 q_2 q_3 q_4 q_5 q_6 \\le n(n-1)(n-3)(n-5)(n-7)(n-9)\n$$\nBut since $(n-3)(n-5)(n-7)(n-9) < (n-2)(n-4)(n-6)(n-8)$ and $n(n-8) < (n-3)(n-5)$, we obtain\n$$\nq_1 q_2 q_3 q_4 q_5 q_6 < (n-1)(n-2)(n-3)(n-4)(n-5)(n-6) = \\frac{(n-1)!}{(n-7)!}\n$$\nAs $q_6 < n-6$, the inequality $q_i < n-i$ holds for every $i > 6$, whence\n$$\nq_7 \\cdots q_{n-1-\\kappa(n)} \\le (n-7) \\cdots (\\kappa(n)+1) = \\frac{(n-7)!}{(\\kappa(n))!}\n$$\nConsequently,\n$$\n\\text{lcm}(1,2,\\dots,n) = q_1 q_2 \\cdots q_{n-1-\\kappa(n)} < \\frac{(n-1)!}{(\\kappa(n))!}\n$$\ncontradicting the original inequality. Hence the inequality holds for $n = 2, 3, 4, 5, 7, 9$ only.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21354,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\mathbb{T} = \\{1, 3, 6, 10, 15, \\dots\\}$ be the set of triangular numbers, i.e., numbers of the form $T_n = \\frac{n(n+1)}{2}$. Let $f$ be a function defined on the set of positive integers such that:\n\n1. $f(n)$ is a positive integer for each $n$;\n2. $f(uv) = f(u)f(v)$ for any pair $(u, v)$ of coprime numbers;\n3. $f(a + b + c) = f(a) + f(b) + f(c)$ for $a, b, c \\in \\mathbb{T}$.\n\nProve that $f(n) = n$ for all $n$.",
"options": [],
"answer": "See solution",
"solution": "It is not difficult to find $f(n)$ for small $n$:\n\n$$\nf(1 \\cdot 1) = f(1)f(1) \\text{ therefore } f(1) = 1\n$$\n\n$$\nf(3) = f(1) + f(1) + f(1) = 3\n$$\n\n$$\nf(5) = f(1 + 1 + 3) = 5\n$$\n\n$$\nf(10) = f(1 + 3 + 6) = 4 + 3f(2) \\text{ and } f(10) = f(2 \\cdot 5) = f(2)f(5) = 5f(2) \\text{ therefore } f(2) = 2\n$$\n\nNow we use induction. Suppose that $f(n) = n$ for all $n < N$. Let us show that $f(N) = N$. Since $f$ is multiplicative, we may assume that $N = p^r$ for some prime $p$. Consider several similar cases.\n\n1. $N = 3^r$. Then\n\n$$\nf(3T_{3^{r-1}}) = 3f(T_{3^{r-1}}) = 3f\\left(\\frac{3^{r-1}(3^{r-1}+1)}{2}\\right) = 3f(3^{r-1})f\\left(\\frac{3^{r-1}+1}{2}\\right)\n$$\n\nAnd on the other hand,\n\n$$\nf(3T_{3^{r-1}}) = f\\left(\\frac{3^r(3^{r-1}+1)}{2}\\right) = f(3^r)f\\left(\\frac{3^{r-1}+1}{2}\\right)\n$$\n\nSo we conclude that $f(3^r) = 3^r$ since $f(3^{r-1}) = 3^{r-1}$ by the induction hypothesis.\n\n2. $N = p^r$, where $p$ is an odd prime and $p^r = 3s - 1$. Note that $f(T_{s-1}) = T_{s-1}$ and $f(T_s) = T_s$ by induction hypothesis since $T_s$ can be factored into integers smaller than $N$. Once again, write the two equalities:\n\n$$\nf(T_{s-1} + T_{s-1} + T_s) = \\frac{s(s-1)}{2} + \\frac{s(s-1)}{2} + \\frac{s(s+1)}{2} = \\frac{s(3s-1)}{2} = \\frac{sp^r}{2}\n$$\n\nand\n\n$$\nf(T_{s-1} + T_{s-1} + T_s) = f\\left(\\frac{s(3s-1)}{2}\\right) = f\\left(\\frac{s}{2}\\right)f(3s-1) = \\frac{s}{2}f(p^r)\n$$\n\nHence $f(p^r) = p^r$.\n\n3. $N = p^r$, where $p$ is an odd prime and $p^r = 3s + 1$. Similarly, we have\n\n$$\n\\begin{aligned}\nf(T_{s-1} + T_s + T_s) &= \\frac{s(s-1)}{2} + \\frac{s(s+1)}{2} + \\frac{s(s+1)}{2} = \\frac{s(3s+1)}{2} = \\frac{sp^r}{2} \\\\\n&= f\\left(\\frac{s(3s+1)}{2}\\right) = f\\left(\\frac{s}{2}\\right)f(3s+1) = \\frac{s}{2}f(p^r)\n\\end{aligned}\n$$\n\nHence $f(p^r) = p^r$.\n\n4. $N = 2^r$. Let $2^{r+1} = 3s \\pm 1$, then we use the following pairs of equalities: either\n\n$$\n\\begin{aligned}\nf(T_{s-1} + T_{s-1} + T_s) &= \\frac{s(s-1)}{2} + \\frac{s(s-1)}{2} + \\frac{s(s+1)}{2} = \\frac{s(3s-1)}{2} = s2^r \\\\\n&= f\\left(\\frac{s(3s-1)}{2}\\right) = f(s)f\\left(\\frac{3s-1}{2}\\right) = sf(2^r)\n\\end{aligned}\n$$\n\nor\n\n$$\n\\begin{aligned}\nf(T_{s-1} + T_s + T_s) &= \\frac{s(s-1)}{2} + \\frac{s(s+1)}{2} + \\frac{s(s+1)}{2} = \\frac{s(3s+1)}{2} = s2^r \\\\\n&= f\\left(\\frac{s(3s+1)}{2}\\right) = f(s)f\\left(\\frac{3s+1}{2}\\right) = sf(2^r)\n\\end{aligned}\n$$\n\nThus $f(2^r) = 2^r$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21355,
"subject": "Mathematics (Olympiad)",
"question": "Let $(a_n)$ be a sequence of positive integers such that for all $n \\geq 1$,\n$$\n\\frac{1}{2} n < a_n < 2n\n$$\nand for all $m, n \\geq 1$,\n$$\n\\gcd(a_m, a_n) = \\gcd(m, n).\n$$\nProve that $a_n = n$ for all $n \\geq 1$.",
"options": [],
"answer": "See solution",
"solution": "$a_3 = 3$. Indeed, if $a_3 = 5$ since $a_5 = 5$, we yield a contradiction. Finally, notice that $a_{p^k} = p^k \\mid a_{mp^k}$. Hence, $n \\mid a_n$ and then $a_n = n$.\n\nIt is now time to remove the _banal_ condition $a_2 \\neq a_4$ in the following way. Indeed, if $a_2 = 3$ and $a_3 \\in \\{2, 4\\}$ it follows that $(a_{2n})$, $(a_{3n})$ would be powers of 3 and 2, respectively. Let $N = 2^a 3^b T$, $\\gcd(T, 6) = 1$. Then, we can prove that $a_T = T$ and hence, from $\\frac{1}{2}T < \\gcd(T, a_N) < 2T$ we find that $T \\mid a_N$.\n\nLet $(v_2(a_N), v_3(a_N)) = (c, d)$. It follows that $\\frac{1}{2} < \\frac{\\gcd(a_N, a_{2C})}{\\gcd(N, 2^C)} < 2$. Choose $C$ large enough such that $C > a$ and $v_3(a_{2C}) > d$ it follows that $2^{a-1} < 3^d < 2^{a+1}$. Analogously, choosing $D > b$, $v_2(a_{3D}) > c$ yields $\\frac{3^b}{2} < 2^c < 2 \\cdot 3^b$. It is now clear that if $a=1$ then $d=1$. If $2^c > \\frac{4}{3} \\cdot 3^b$ then $a_N \\geq 3 \\cdot \\frac{4}{3} \\cdot 3^b \\cdot T = 2N$, a contradiction. Hence, $2^c \\leq \\frac{4}{3} \\cdot 3^b$.\n\nSince the sequence $\\{(n \\cdot \\log_2 3)\\}$ is dense on $(0, 1)$, for all $\\varepsilon > 0$ there is an integer $d$ such that $\\{d \\cdot \\log_2 3\\} \\in (1 + \\log_2 \\frac{2}{3}, 1 + \\log_2(\\varepsilon + \\frac{2}{3}))$. Then, $\\frac{2}{3} \\cdot 2^{a_1} < 3^d < (\\frac{2}{3} + \\varepsilon)2^{a_1}$, for some $a_1$. Let $N = 2^{a_1} \\cdot 3^b \\cdot T$ now, by choosing $\\varepsilon$ small enough, it follows that $v_3(a_N) = d$. Now, if $2^{v_2(a_N)} < \\frac{3}{4+\\varepsilon}3^b$, for some $\\varepsilon > 0$ it follows that $a_N < \\frac{3}{4+\\varepsilon}3^b \\cdot (\\frac{2}{3} + \\varepsilon)2^{a_1} \\cdot T$, it follows that $a_N < \\frac{N}{2}$. A contradiction. Hence, we find that $\\frac{3}{4+\\varepsilon}3^b \\leq 2^{v_2(a_N)} \\leq \\frac{4}{3} \\cdot 3^b$.\n\nNow, choose $c$ such that $\\{c \\cdot \\log_3 2\\} \\in (1 + \\log_3 \\frac{2}{3}, 1 + \\log_3 \\frac{3}{4+\\varepsilon})$, it follows that $\\frac{2}{3} \\cdot 3^{b_1} < 2^c < \\frac{3}{4+\\varepsilon} \\cdot 3^{b_1}$. Choosing $N = 2^{a_1}3^{b_1}T$, it follows that $v_2(a_N) \\in \\{c, c+1\\}$. On the other hand, from\n\n$$\n\\frac{3}{4+\\varepsilon}3^{b_1} \\leq 2^{v_2(a_N)} \\leq \\frac{4}{3} \\cdot 3^{b_1}\n$$\n\nWe find that $v_2(a_N) = c + 1$. But then $2^c = \\frac{2^{v_2(a_N)}}{2} > \\frac{2}{3} \\cdot 3^{b_1}$. Yielding $2^{v_2(a_N)} > \\frac{4}{3} \\cdot 3^{b_1}$, a contradiction.\n\nFinally, we now rule out the case $a_3 = 5$. Suppose $a_3 = 5$. As $\\gcd(a_{3k}, a_3) > 1$, the $a_{3k}$ are all divisible by 5. By the preceding, $p \\mid a_p$ for all primes $p \\neq 3$; as $\\gcd(a_{3k}, a_p) = 1$ for these primes, the $a_{3k}$ are all powers of 5; say, $a_{3k} = 5^{m_k}$. By Kronecker's density theorem, as $n$ runs through the positive integers, the fractional parts $\\{n \\log_3 5\\}$ form a dense set in $(0, 1)$. Hence $\\log_3 2 < \\{n \\log_3 5\\} < \\log_3 \\frac{5}{2}$ for some $n$. Let $k = \\lfloor n \\log_3 5 \\rfloor$ and carry out obvious calculations to get $5^{n-1} < \\frac{1}{2} \\cdot 3^k$ and $2 \\cdot 3^k < 5^n$. As $\\frac{1}{2} \\cdot 3^k < a_{3k} < 2 \\cdot 3^k$, it follows that $5^{n-1} < \\frac{1}{2} \\cdot 3^k < 5^{m_k} < 2 \\cdot 3^k < 5^n$, so $n-1 < m_k < n$. This contradiction implies $a_3 = 3$ and $A_3 = B_3 = \\{3\\}$, as desired.\n\nFinally, by the preceding, $A_5 = \\{5\\}$, as it is non-empty and disjoint from both $A_2$ and $A_3$. Now, $a_5 < 10$ forces $a_5 = 5$, so $B_5 = \\{5\\}$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21356,
"subject": "Mathematics (Olympiad)",
"question": "Determine the maximum value of the function\n\n$$f_k(x, y) = (x + y) - (x^{2k+1} + y^{2k+1})$$\n\nover all real numbers $x$ and $y$ satisfying $x^2 + y^2 = 1$, for all positive integers $k$.",
"options": [],
"answer": "See solution",
"solution": "Since $x^2 + y^2 = 1$, it follows that $|x| \\leq 1$ and $|y| \\leq 1$. Define $g_k(x) := x - x^{2k+1}$; note that $g_k(-x) = -g_k(x)$, so $f_k(x, y) = g_k(x) + g_k(y)$. Thus, $f_k(x, y) \\leq f_k(|x|, |y|)$, and we may assume $x, y \\geq 0$.\n\nFor the quadratic mean,\n\n$$m_2(x, y) = \\sqrt{\\frac{x^2 + y^2}{2}} = \\frac{\\sqrt{2}}{2}$$\n\nand\n\n$$x + y = 2m_1(x, y) \\leq 2m_2(x, y) \\leq 2m_{2k+1}(x, y)$$\n\nso\n\n$$-(x^{2k+1} + y^{2k+1}) = -2m_{2k+1}^{2}(x, y) \\leq -2m_2^{2}(x, y)$$\n\nTherefore, $x + y \\leq \\sqrt{2}$ and\n\n$$-(x^{2k+1} + y^{2k+1}) \\leq \\frac{2}{\\sqrt{2}^{2k+1}} = \\frac{\\sqrt{2}}{2^k}$$\n\nwhich implies\n\n$$f_k(x, y) \\leq \\frac{2^k - 1}{2^k} \\sqrt{2}$$\n\nwith equality for $x = y = \\frac{\\sqrt{2}}{2}$.\n\nqed",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21357,
"subject": "Mathematics (Olympiad)",
"question": "Let $f(x) = (1 + x)(1 + x^2) \\cdots (1 + x^{2013}) = \\sum_n a_n x^n$. For $r \\in \\{0,1,2,3,4,5,6\\}$, define\n\n$$\n|T_r| = \\sum_k [x^{7k + r}] f(x) = \\sum_k a_{7k + r}.\n$$\n\nFind a closed formula for $|T_r|$ for all $r$.",
"options": [],
"answer": "See solution",
"solution": "Let $\\epsilon = e^{\\frac{2\\pi i}{7}}$, a primitive 7th root of unity. We use the following facts:\n\n$$\n1 + \\epsilon + \\epsilon^2 + \\cdots + \\epsilon^6 = 0,\n$$\n\nand\n\n$$\n\\sum_{k=1}^6 \\epsilon^{kr} = \\begin{cases} 6, & 7 \\mid r \\\\ -1, & 7 \\nmid r \\end{cases}\n$$\n\nBy roots of unity filter,\n\n$$\n|T_r| = \\frac{1}{7} \\sum_{i=0}^6 \\epsilon^{-ri} f(\\epsilon^i) = \\frac{1}{7} \\left(2^{2013} + \\sum_{i=1}^6 \\epsilon^{-ri} f(\\epsilon^i)\\right).\n$$\n\nSince $2013 = 7 \\cdot 287 + 4$ and $\\epsilon^7 = 1$,\n\n$$\nf(\\epsilon^i) = \\left[(1 + \\epsilon^i)(1 + (\\epsilon^i)^2) \\cdots (1 + (\\epsilon^i)^7)\\right]^{287} (1 + \\epsilon^i)(1 + (\\epsilon^i)^2)(1 + (\\epsilon^i)^3)(1 + (\\epsilon^i)^4).\n$$\n\nBut $(1 + \\epsilon^i)(1 + (\\epsilon^i)^2) \\cdots (1 + (\\epsilon^i)^7) = 2$, so\n\n$$\nf(\\epsilon^i) = 2^{287} (1 + \\epsilon^i)(1 + (\\epsilon^i)^2)(1 + (\\epsilon^i)^3)(1 + (\\epsilon^i)^4).\n$$\n\nThis simplifies to $f(\\epsilon^i) = 2^{287}(1 + (\\epsilon^i)^3)$ for $1 \\leq i \\leq 6$.\n\nTherefore,\n\n$$\n|T_r| = \\frac{1}{7} \\left[2^{2013} + 2^{287} \\sum_{i=1}^6 (\\epsilon^{-ri} + \\epsilon^{(3 - r)i})\\right].\n$$\n\nFrom the root of unity sums,\n\n$$\n\\sum_{i=1}^6 (\\epsilon^{-ri} + \\epsilon^{(3 - r)i}) = \\begin{cases} 5, & r = 0, 3 \\\\ -2, & r = 1,2,4,5,6 \\end{cases}\n$$\n\nThus,\n\n$$\n|T_r| = \\begin{cases}\n\\dfrac{2^{2013} + 5 \\cdot 2^{287}}{7}, & r = 0, 3 \\\\ \n\\dfrac{2^{2013} - 2^{288}}{7}, & r = 1,2,4,5,6\n\\end{cases}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21358,
"subject": "Mathematics (Olympiad)",
"question": "Given $2n$ points on the plane, prove that it is possible to split them into $n$ pairs such that: if for each pair of points we construct a circle with the pair as diameter, then the obtained $n$ circles will have a common point which does not necessarily belong to the given points.\n\n",
"options": [],
"answer": "See solution",
"solution": "Consider a line $l$ such that there are $n$ points on one side of the line and $n$ points on the other. Such a line exists. Without loss of generality, assume $l$ is the horizontal axis. Arrange the points in increasing order of abscissa. In the upper half-plane, let the points be $A_1, A_2, \\ldots, A_n$, and in the lower half-plane, $B_1, B_2, \\ldots, B_n$. Pair them as $(A_1, B_n), (A_2, B_{n-1}), \\ldots, (A_n, B_1)$. We will show that the circles constructed on these pairs as diameters have a common point on $l$.\n\nConsider the projections of these segments onto $l$. For any two segments $A_i B_{n+1-i}$ and $A_j B_{n+1-j}$ with $i < j$, the projection of $A_i$ is not to the right of $A_j$, and the projection of $B_i$ is not to the left of $B_j$. Among all segments, choose the last one with positive slope, say $A_i B_{n+1-i}$. The overlap of the projections of $A_i B_{n+1-i}$ and $A_{i+1} B_n$ gives a segment on $l$ that is contained in all the disks. Any point in this overlap is inside all the circles constructed on the diameters, since a point in the projection of a segment onto $l$ is inside the corresponding disk. Thus, all $n$ circles have a common point on $l$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21359,
"subject": "Mathematics (Olympiad)",
"question": "The sum of a few consecutive integers (each greater than 1) is $2011$. Find all such numbers.",
"options": [],
"answer": "See solution",
"solution": "Let $a_1$ be the first integer and $a_n$ the last. The sum of $n$ consecutive integers is:\n\n$$\n\\frac{a_1 + a_n}{2} n = 2011 \\implies (a_1 + a_n) n = 2 \\times 2011.\n$$\n\nSince $2011$ is prime, the possible factorizations are $a_1 + a_n = 2$, $n = 2011$ or $a_1 + a_n = 2011$, $n = 2$. The first case is impossible (since all numbers are greater than $1$), so the only solution is $n = 2$, $a_1 + a_n = 2011$. Thus, the numbers are $1005$ and $1006$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21360,
"subject": "Mathematics (Olympiad)",
"question": "Цифрүүдийн нийлбэр нь 9-тэй тэнцүү бөгөөд 4 орон нь 1, 0, 0, 4 цифрүүдээс тогтсон 6 оронтой тоо хичнээн байх вэ?",
"options": [],
"answer": "See solution",
"solution": "Уг 6 оронтой тооны үлдэх 2 цифр нь $1, 2$ эсвэл $0, 3$ байна. Эхний тохиолдолд уг 6 оронтой тоо $1, 2, 4$-ийн аль нэгээр эхлэх ба 1-ээр эхэлсэн бол бусад цифрүүд нь $1, 2, 4, 0, 0$ болж эдгээрийг сэлгэх боломжийн тоо нь $\\frac{5!}{2!} = 60$ байна. Үүнийг $60 + 2 \\cdot 30 + 3 \\cdot 20 = 180$ гэж тооцож болно. Энэ маягаар бүх боломжийг $\\frac{5!}{2!} = 60 + 2 \\cdot 30 + 3 \\cdot 20 = 180$ болохыг хялбар шалгаж болно.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21361,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\omega$ be the incircle of triangle $ABC$, tangent to $BC$ and $AC$ at points $D$ and $E$, respectively. Draw a line perpendicular to $BC$ at $D$ that meets $\\omega$ at a point $P$ closer to $A$. The line $AP$ meets $BC$ at $M$. Let $N$ be a point on segment $AC$ such that $AE = CN$. The line $BN$ intersects $\\omega$ at a point $Q$ closer to $B$ and intersects $AM$ at point $R$. Show that the area of triangle $ABR$ is equal to the area of quadrilateral $PQMN$.",
"options": [],
"answer": "See solution",
"solution": "From point $P$, draw a line parallel to $BC$ meeting lines $AB$ and $AC$ at points $B'$ and $C'$, respectively.\n\n\n\nClearly, the line $B'C'$ is externally tangent to the circle $\\omega$ at point $P$. Let $Y, Z$ be two points on line $BC$, and let $Y', Z'$ be the intersections of $AY, AZ$ with line $AB$.\n\n**Claim.**\n$$\n\\frac{B'C'}{BC} = \\frac{Y'Z'}{YZ}.\n$$\n\n**Proof of Claim.** Let $h_a', h_a$ be the heights of triangles $AB'C'$ and $ABC$, respectively. Since $\\triangle AB'C' \\sim \\triangle ABC$ and $\\triangle AY'Z' \\sim \\triangle AYZ$, we have $\\frac{B'C'}{BC} = \\frac{h_a'}{h_a} = \\frac{Y'Z'}{YZ}$, proving the claim.\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 $\\omega$ touches triangle $AB'C'$ externally, we have $B'P + PC' = a'$, $c' + B'P = b' + PC'$. Thus, $PC' = s' - b'$. From the claim, we deduce\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 point $X$ so that $CX = MC$. Since triangles $APE$ and $AMX$ are similar, $\\frac{AP}{AM} = \\frac{AE}{AX} = \\frac{s-a}{s}$. Applying Menelaus' theorem to $\\triangle AMC$ with respect to 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 $AR = PM$. Similarly, $BR = QN$.\n\nLet $\\theta$ be the angle between 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": 21362,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an obtuse triangle with orthocenter $H$ and centroid $S$. Let $D$, $E$, and $F$ be the midpoints of segments $BC$, $AC$, and $AB$, respectively.\n\nShow that the circumcircle of triangle $ABC$, the circumcircle of triangle $DEF$, and the circle with diameter $HS$ have two distinct points in common.",
"options": [],
"answer": "See solution",
"solution": "Let $M$ and $O$ denote the circumcenters of triangles $DEF$ and $ABC$, respectively. We use the well-known facts that the points $H$, $M$, $S$, and $O$ lie on the Euler line of triangle $ABC$ in this order, and that $HM : MS : SO = 3 : 1 : 2$.\n\nSince $S$ is in the interior of $ABC$, it is inside the circumcircle of $ABC$. However, since $ABC$ is obtuse, the orthocenter $H$ is outside the circumcircle. This implies that the circle with diameter $HS$ intersects the circumcircle of $ABC$ in two points.\n\nLet $X$ be one of these intersection points. We will prove that $MX : OX = 1 : 2$. Let $N$ be the midpoint of $HS$. Using Thales' theorem, we get that $HN$, $XN$, and $SN$ have the same length, and $HN : NM : MS : SO = 2 : 1 : 1 : 2$. This implies that $MN : XN = XN : ON$.\n\nTherefore, the triangles $XNM$ and $ONX$ are similar with ratio $1 : 2 = MN : XN$. Therefore, we have $MX : OX = 1 : 2$ as desired.\n\n\n\nFigure 4: Figure for Problem 4",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21363,
"subject": "Mathematics (Olympiad)",
"question": "試求所有從實數映至實數的函數 $f$，滿足：\n\n$$\n2f((x+y)^2) = f(x+y) + (f(x))^2 + (4y-1)f(x) - 2y + 4y^2\n$$\n\n對於所有實數 $x$ 和 $y$ 皆成立。",
"options": [],
"answer": "See solution",
"solution": "唯一解為 $f(x) = 2x$。\n\n**解法一：**\n\n1. 原式代入 $y=0$，得\n\n$$\n2f(x^2) = f(x) + (f(x))^2 - f(x) = (f(x))^2,\n$$\n\n從而有\n\n$$\n(f(x))^2 = 2f(x^2) = 2f((-x)^2) = (f(-x))^2.\n$$\n\n2. 原式代入 $y=-x$，得\n\n$$\nf(0) = (f(x))^2 - (4x+1)f(x) + 2x + 4x^2.\n$$\n\n這表示\n\n$$\n(f(x))^2 - (4x+1)f(x) + 2x + 4x^2 = f(0) = (f(-x))^2 - (-4x+1)f(-x) - 2x + 4x^2,\n$$\n\n從而由上式，我們有\n\n$$\n-(4x+1)f(x) + 4x = (4x-1)f(-x).\n$$\n\n特別地，若我們將 $x = \\frac{1}{4}$ 代入上式，有 $f(\\frac{1}{4}) = \\frac{1}{2}$；再於前式中代入 $x = \\frac{1}{4}$，得 $f(0) = 0$。\n\n3. 原式代入 $x=0$，並結合 $f(0)=0$，得\n\n$$\nf(y) - 2y + 4y^2 = 2f(y^2) = 2f((-y)^2) = f(-y) + 2y + 4y^2\n$$\n\n也就是\n\n$$\nf(y) = f(-y) + 4y.\n$$\n\n將上式平方並結合前式，得\n\n$$\n(f(y))^2 = (f(-y))^2 + 8f(-y)y + 16y^2 \\implies 8f(-y)y + 16y^2 = 0,\n$$\n\n故 $f(y) = 2y$ 對於所有 $y \\neq 0$ 恆成立。又已知 $f(0) = 0 = 2 \\times 0$，故 $f(x) \\equiv 2x$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21364,
"subject": "Mathematics (Olympiad)",
"question": "In each of the six boxes $B_1, B_2, B_3, B_4, B_5, B_6$, there is initially one coin. There are two types of operation allowed:\n\n**Type 1:** Choose a nonempty box $B_j$, with $1 \\leq j \\leq 5$.\n\nRemove one coin from $B_j$ and add two coins to $B_{j+1}$.\n\n**Type 2:** Choose a nonempty box $B_k$, with $1 \\leq k \\leq 4$.\n\nRemove one coin from $B_k$ and exchange the contents of the (possibly empty) boxes $B_{k+1}$ and $B_{k+2}$.\n\nDetermine whether there is a finite sequence of such operations that results in the boxes $B_1, B_2, B_3, B_4, B_5$ being empty and the box $B_6$ containing exactly $2010^{2010^{2010}}$ coins. (Note that $a^{b^c} = a^{(b^c)}$.)",
"options": [],
"answer": "See solution",
"solution": "The answer is affirmative.\n\nLet $A = 2010^{2010^{2010}}$. Denote by\n\n$$\n(b_i, b_{i+1}, \\dots, b_{i+k}) \\to (b'_i, b'_{i+1}, \\dots, b'_{i+k})\n$$\n\nto mean that there is a finite sequence of operations on the boxes $B_i, B_{i+1}, \\dots, B_{i+k}$ initially containing $b_i, b_{i+1}, \\dots, b_{i+k}$ coins respectively that results in containing $b'_i, b'_{i+1}, \\dots, b'_{i+k}$ coins respectively. Then we shall show that\n\n$$\n(1, 1, 1, 1, 1, 1) \\to (0, 0, 0, 0, 0, A).\n$$\n\n**Lemma 1** For every positive integer $a$, we have $(a, 0, 0) \\to (0, 2^a, 0)$.\n\n*Proof of Lemma 1:* We prove by induction on $k \\leq a$ that $(a, 0, 0) \\to (a-k, 2^k, 0)$.\n\nSince $(a, 0, 0) \\to (a-1, 2, 0) \\to (a-1, 2^1, 0)$, the assertion is true for $k=1$. Suppose that the assertion is true for some $k < a$; then\n\n$$\n\\begin{aligned}\n(a-k, 2^k, 0) &\\to (a-k, 2^k-1, 2) \\to \\dots \\\\\n&\\to (a-k, 0, 2^{k+1}) \\to (a-k-1, 2^{k+1}, 0),\n\\end{aligned}\n$$\n\nand thus\n\n$$\n(a, 0, 0) \\to (a-k, 2^k, 0) \\to (a-k-1, 2^{k+1}, 0).\n$$\n\nThe assertion is also true for $k+1 \\leq a$. By induction, Lemma 1 is proven.\n\n**Lemma 2** For every positive integer $a$, we have $(a, 0, 0, 0) \\to (0, P_a, 0, 0)$, where $P_n = 2^{2^{n^2}}$ (with $n$ 2's) for a positive integer $n$.\n\n*Proof of Lemma 2:* We prove by induction on $k \\leq a$ that $(a, 0, 0, 0) \\to (a-k, P_k, 0, 0)$.\n\nBy the operation of type 1, we have\n\n$$\n(a, 0, 0, 0) \\to (a-2, 2, 0, 0) = (a-1, P_1, 0, 0),\n$$\n\nand the assertion is true for $k=1$. Suppose that the assertion is true for some $k < a$; then\n\n$$\n\\begin{aligned}\n& (a-k, P_k, 0, 0) \\to (a-k, 0, 2^{P_k}, 0) \\\\\n&= (a-k, 0, P_{k+1}, 0) \\to (a-k-1, P_{k+1}, 0, 0),\n\\end{aligned}\n$$\n\nand therefore\n\n$$\n(a, 0, 0, 0) \\to (a-k, P_k, 0, 0) \\to (a-k-1, P_{k+1}, 0, 0),\n$$\n\ni.e., the assertion is also true for $k+1 \\leq a$. By induction, Lemma 2 is proven.\n\nWe have\n\n$$\n\\begin{aligned}\n& (1, 1, 1, 1, 1) \\to (1, 1, 1, 1, 0) \\to (1, 1, 1, 0, 3, 0) \\\\\n\\to & (1, 1, 0, 3, 0, 0) \\to (1, 0, 3, 0, 0, 0) \\to (0, 3, 0, 0, 0, 0) \\\\\n\\to & (0, 0, P_3, 0, 0, 0) = (0, 0, 16, 0, 0, 0) \\to (0, 0, 0, P_{16}, 0, 0),\n\\end{aligned}\n$$\n\nand\n\n$$\n\\begin{aligned}\nA &= 2010^{2010^{2010}} < (2^{11})^{2010^{2010}} \\\\\n&= 2^{11 \\times 2010^{2010}} < 2^{2010^{2011}} < 2^{(2^{11})^{2011}} \\\\\n&= 2^{2^{11 \\times 2011}} < 2^{2^{15}} < P_{16},\n\\end{aligned}\n$$\n\nand thus the number of coins in the box $B_4$ is greater than $A$. Therefore, by performing operations of type 2, we have\n\n$$\n\\begin{aligned}\n& (0, 0, 0, P_{16}, 0, 0) \\to (0, 0, 0, P_{16}-1, 0, 0) \\\\\n& \\to (0, 0, 0, P_{16}-2, 0, 0)\n\\end{aligned}\n$$\n\nand so on, until we reach exactly $A$ coins in $B_6$ and all other boxes empty.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21365,
"subject": "Mathematics (Olympiad)",
"question": "The teacher ordered 20 students from one class in a row and gave 800 candies to them. Every student had to calculate the ratio $\\frac{x}{x + 2k - 1}$, where $x$ is the number of candies that he got and $k$ is his position in the row, counting from left to right. After calculating, they all got the same result. How many candies were given to the student that was twelfth in the row?",
"options": [],
"answer": "See solution",
"solution": "Let the first student receive $x_1$ candies, the second $x_2$ candies, ..., and the twentieth receive $x_{20}$ candies. So $x_1 + x_2 + \\dots + x_{20} = 800$.\n\nLet $\\frac{x_k}{x_k + 2k - 1} = M$, so $x_k = (2k - 1) \\frac{M}{1 - M}$.\n\nFor the sum $x_1 + x_2 + \\dots + x_{20}$ we obtain\n\n$$\nx_1 + x_2 + \\dots + x_{20} = \\frac{M}{1 - M}(1 + 3 + 5 + \\dots + 39) = \\frac{M}{1 - M} \\cdot \\frac{40 \\cdot 20}{2} = 400 \\frac{M}{1 - M}\n$$\n\nBecause $x_1 + x_2 + \\cdots + x_{20} = 800$, we get $400 \\frac{M}{1 - M} = 800$ or $\\frac{M}{1 - M} = 2$. Hence, the twelfth student received\n\n$$\nx_{12} = (2 \\cdot 12 - 1) \\cdot 2 = 46.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21366,
"subject": "Mathematics (Olympiad)",
"question": "Teacher tells Jüri two nonzero integers $a$ and $b$ such that $b$ is divisible by $a$. Jüri has to find a nonzero integer $c$ such that $c$ is divisible by $b$ and all solutions of the quadratic equation $a x^2 + b x + c = 0$ are integers. Can Jüri always solve the problem?",
"options": [],
"answer": "See solution",
"solution": "Yes.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21367,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be positive real numbers. Prove that\n\n$$\n(a^{5} - a^{2} + 3)(b^{5} - b^{2} + 3)(c^{5} - c^{2} + 3) \\geq (a + b + c)^{3}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Since $x^2 - 1$ and $x^3 - 1$ have the same sign, $0 \\leq (x^2 - 1)(x^3 - 1) = x^5 - x^3 - x^2 + 1$, with equality when $x = 1$. So, it is sufficient to prove that\n\n$$\n(a^3 + 2)(b^3 + 2)(c^3 + 2) \\geq (a + b + c)^3.\n$$\n\nLet $x = \\sqrt{a}$, $y = \\sqrt{b}$, and $z = \\sqrt{c}$. Then some two of $x$, $y$, and $z$ are both at least $1$ or both at most $1$. Without loss of generality, say these are $x$ and $y$. Then the sequences $(x, 1, 1)$ and $(1, 1, y)$ are oppositely sorted, yielding\n\n$$\n(x^6 + 1 + 1)(1 + 1 + y^6) \\geq 3(x^6 + 1 + y^6)\n$$\n\nby Chebyshev's Inequality. By the Cauchy-Schwarz Inequality we have\n\n$$\n(x^6 + 1 + y^6)(1 + z^6 + 1) \\geq (x^3 + y^3 + z^3)^2.\n$$\n\nApplying Chebyshev's and the Cauchy-Schwarz Inequalities each once more, we get\n\n$$\n3(x^3 + y^3 + z^3) \\geq (x^2 + y^2 + z^2)(x + y + z),\n$$\n\nand\n\n$$\n(x^3 + y^3 + z^3)(x + y + z) \\geq (x^2 + y^2 + z^2)^2.\n$$\n\nMultiplying the above four inequalities together yields\n\n$$\n(x^6 + 2)(y^6 + 2)(z^6 + 2) \\geq (x^2 + y^2 + z^2)^3,\n$$\n\nas desired, with equality if and only if $x = y = z = 1$. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21368,
"subject": "Mathematics (Olympiad)",
"question": "Let $N$ be a positive integer. The sequence $x_1, x_2, \\dots$ of non-negative reals is defined by\n\n$$\nx_n^2 = \\sum_{i=1}^{n-1} \\sqrt{x_i x_{n-i}}\n$$\nfor all positive integers $n > N$. Show that there exists a constant $c > 0$, such that $x_n \\le \\frac{n}{2} + c$ for all positive integers $n$.",
"options": [],
"answer": "See solution",
"solution": "Applying the AM-GM inequality to each term on the right-hand side, we get\n\n$$\nx_n^2 \\le \\sum_{i=1}^{n-1} x_i, \\quad n > N.\n$$\n\nLet us set $x_n = a n + \\Delta(n)$, where $a$ is a constant to be determined. We want to pick $a$ so that $\\Delta(n)$ grows slower than $n$. For $n > N$,\n\n$$\na^2 n^2 + 2a n \\Delta(n) + \\Delta(n)^2 \\le \\frac{a(n-1)n}{2} + \\sum_{i=1}^{n-1} \\Delta(i)\n$$\n\n$$\n2a n \\Delta(n) \\le \\left( \\frac{a(n-1)n}{2} - a^2 n^2 \\right) - \\Delta(n)^2 + \\sum_{i=1}^{n-1} \\Delta(i) \\quad (1)\n$$\n\nDivide both sides by $n$ and choose $a$ so the bracketed term is degree $n$ (not $n^2$). Calculating, $a = 1/2$. Substituting into (1):\n\n$$\n\\Delta(n) \\le -\\frac{1}{4} - \\frac{\\Delta(n)^2}{n} + \\frac{1}{n} \\sum_{i=1}^{n-1} \\Delta(i)\n$$\n\nNow,\n\n$$\n\\Delta(n) \\le \\frac{1}{n} \\sum_{i=1}^{n-1} \\Delta(i), \\quad \\forall n > N\n$$\n\nwhich yields\n\n$$\n\\Delta(n) \\le \\max\\{\\Delta(i) : i < n\\}\n$$\n\nhence $\\Delta(n) \\le \\max\\{\\Delta(i) : i \\le N\\}$ for any $n > N$. Thus, $x_n \\le n/2 + c$ for all $n \\in \\mathbb{N}$ for some constant $c$.\n\n**Sharper estimate.** Let's write\n\n$$\n\\Delta(n) \\le -\\frac{1}{4} + \\frac{1}{n} \\sum_{i=1}^{n-1} \\Delta(i), \\quad n > N\n$$\n\nDefine the sequence $\\Delta'(n)$ as follows:\n\n$$\n\\Delta'(n) = -\\frac{1}{4} + \\frac{1}{n} \\sum_{i=1}^{n-1} \\Delta'(i), \\quad n > N\n$$\n\nand $\\Delta'(n) = \\Delta'(n)$ for $n \\le N$. Clearly $\\Delta(n) \\le \\Delta'(n)$ for all $n$. We have\n\n$$\n\\Delta'(n + 1) = -\\frac{1}{4} + \\frac{1}{n+1} \\left( \\sum_{i=1}^{n-1} \\Delta(i) + -\\frac{1}{4} + \\frac{1}{n} \\sum_{i=1}^{n-1} \\Delta(i) \\right), \\quad n > N\n$$\n\nSubtracting gives\n\n$$\n\\Delta'(n + 1) - \\Delta'(n) = \\frac{-1}{4(n + 1)}, \\quad \\forall n > N\n$$\n\nSumming up,\n\n$$\n\\Delta'(n) = \\Delta'(N + 1) - \\frac{1}{4} \\sum_{k=N+2}^{n} \\frac{1}{k}, \\quad n \\ge N + 2\n$$\n\nUsing the harmonic series,\n\n$$\n\\Delta'(n) \\le -\\frac{\\ln n}{4} + c\n$$\n\nfor some constant $c$. Finally,\n\n$$\nx_n \\le \\frac{n}{2} - \\frac{\\ln n}{4} + c, \\quad n \\in \\mathbb{N}\n$$\n\nwhere $c$ is a constant.\n\n**Remark.** The precise growth rate of the sequence is $x_n = \\frac{\\pi}{8} n + o(n)$ (see [here](https://dgrozev.wordpress.com/2024/02/14/growth-rate-of-a-sequence-bulgarian-2024-mo-regional-round/)). $\\Box$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21369,
"subject": "Mathematics (Olympiad)",
"question": "A clock is used to time laps in a swimming pool. Each lap is 42 seconds for Julie and 45 seconds for Sam. \n\n**a.** How many laps does the clock hand complete in 7 minutes?\n\n**b.** How many laps are in 1 km if each lap is 50 m? How many revolutions of the clock hand does Julie and Sam make for 1 km?\n\n**c.** Is it possible for either swimmer to finish a lap with the clock hand pointing to 37 seconds?\n\n**d.** How many different positions on the clock can either Julie or Sam (or both) finish a lap?",
"options": [],
"answer": "See solution",
"solution": "a. $7 \\times 60 \\div 42 = 10$ laps.\n\nb. $1\\ \\text{km} = 20\\ \\text{laps}$. Julie would complete $20$ laps in $20 \\times 42 = 840$ seconds $= 14$ minutes, hence $14$ revolutions of the hand. Sam would complete $20$ laps in $20 \\times 45 = 900$ seconds $= 15$ minutes, hence $15$ revolutions of the hand.\n\nc. No.\n\n**Alternative i**\n\nThe time indicated by the clock, in seconds, is $37$ or $97$ or $157$ etc. All these times end in $7$. The time in seconds at the end of Julie's swim is $42, 84, 126, 168, 210, \\ldots$ All Julie's times end in $0, 2, 4, 6,$ or $8$. The time in seconds at the end of Sam's swim is $45, 90, 135, 180, \\ldots$ All Sam's times end in $0$ or $5$. So neither swimmer finished a lap with the clock hand pointing to $37$.\n\n**Alternative ii**\n\nSince Julie takes $42$ seconds per lap, the times shown at the end of her laps are $42, 24, 6, 48, 30, 12, 54, 36, 18, 0$, which are then repeated. Since Sam takes $45$ seconds per lap, the times shown at the end of his laps are $45, 30, 15, 0$, which are then repeated. So neither swimmer could finish a lap with the clock hand pointing to $37$.\n\n**Alternative iii**\n\nThe time at which Julie completes a lap is a multiple of $42$ minus a multiple of $60$, which is always even. The time at which Sam completes a lap is a multiple of $45$ minus a multiple of $60$, which is always divisible by $3$. Since $37$ is not even and not a multiple of $3$, neither swimmer could finish a lap with the clock hand pointing to $37$.\n\nd. From Part c Alternative ii, there are $10$ times on the clock at which Julie may finish a lap and there are $4$ times on the clock at which Sam may finish a lap. The only common times are $0$ and $30$. Therefore, the number of different positions on the clock when either Julie or Sam or both finish a lap is $10 + 4 - 2 = 12$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21370,
"subject": "Mathematics (Olympiad)",
"question": "Circle $k$ of radius $r$ is inscribed in $\\triangle ABC$. Tangent lines to $k$ that are parallel to sides $AB$, $BC$, and $CA$ intersect the other sides of $\\triangle ABC$ at points $M, N$; $P, Q$; and $L, T$ ($P, T \\in AB$, $L, N \\in BC$, and $M, Q \\in AC$). Denote by $r_1, r_2, r_3$ the radii of circles inscribed in triangles $MNC$, $PQA$, and $LTB$, respectively. Prove that $r_1 + r_2 + r_3 = r$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Since all these triangles are similar, we get\n\n$$\n\\frac{r_1 + r_2 + r_3}{r} = \\frac{p_1 + p_2 + p_3}{p}\n$$\n\nIt is not hard to see that (see the figure):\n\n$$\n\\begin{aligned}\n2p_1 &= CM + CN + MN \\\\\n &= CM + CN + MZ + ZN \\\\\n &= CM + MX + CN + NY = 2p - (AX + AB + BY) = 2p - 2c.\n\\end{aligned}\n$$\n\nThus,\n\n$$\n\\frac{r_1 + r_2 + r_3}{r} = \\frac{p_1 + p_2 + p_3}{p} = \\frac{p-a + p-b + p-c}{p} = \\frac{3p - 2p}{p} = 1.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21371,
"subject": "Mathematics (Olympiad)",
"question": "На столе лежат кучки орехов: в каждой кучке либо один, либо два ореха. Число орехов в начале игры нечетно. Два игрока по очереди берут орехи из кучек по следующим правилам:\n\n1. Если на столе есть кучки с одним орехом (единицы), первый игрок должен убрать одну из них.\n2. Первый игрок не может брать орехи из кучек с двумя орехами (двойки).\n3. В остальном ходы первого игрока могут быть любыми.\n\nДокажите, что первый игрок всегда сможет сделать ход, не нарушая описанных правил, и выиграет игру.\n\n",
"options": [],
"answer": "See solution",
"solution": "Назовём кучки из одного ореха *единицами*, а из двух — *двойками*. Первый игрок придерживается следующих правил:\n\n1. Если на столе есть единицы — убрать одну из них.\n2. Не брать из двоек.\n\nПоскольку число орехов в начале игры нечетно, оно остаётся нечетным перед каждым ходом первого игрока. Поэтому перед его ходом всегда будет хотя бы одна нечетная кучка, то есть первый всегда сможет сделать ход, не нарушая правил.\n\nПосле первого хода первого игрока на столе нет единиц. После хода второго игрока может появиться не более одной новой единицы, которую первый заберёт. Значит, и после следующих ходов первого единиц на столе не будет, а после любого хода второго на столе будет не больше одной единицы. В частности, так будет и в конце игры, то есть первый выиграет.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21372,
"subject": "Mathematics (Olympiad)",
"question": "Let $AL$ and $BK$ be angle bisectors in the non-isosceles triangle $ABC$ ($L$ lies on the side $BC$, $K$ lies on the side $AC$). The perpendicular bisector of $BK$ intersects the line $AL$ at point $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK$. Prove that $LN = NA$.\n\n",
"options": [],
"answer": "See solution",
"solution": "The point $M$ lies on the circumcircle of $\\triangle ABK$ (since both $AL$ and the perpendicular bisector of $BK$ bisect the arc $BK$ of this circle). Then $\\angle CBK = \\angle ABK = \\angle AMK = \\angle NLA$. Thus $ABLN$ is cyclic, whence $\\angle NAL = \\angle NBL = \\angle CBK = \\angle NLA$. Now it follows that $LN = NA$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21373,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be any odd integer greater than $1$.\n\nDoes there exist a sequence of positive real numbers $a_1, a_2, \\dots, a_n$ such that the sequence $b_j = a_j + \\frac{(-1)^j}{a_j}$ for $1 \\leq j \\leq n$ is a permutation of $\\{a_1, a_2, \\dots, a_n\\}$?",
"options": [],
"answer": "See solution",
"solution": "For even $n$, suppose $a_j + \\frac{(-1)^j}{a_j} = b_j$ for $1 \\leq j \\leq n$, where $\\{b_j\\}$ is a permutation of $\\{a_j\\}$. The relation can be rewritten as\n\n$$\na_j^2 - a_j b_j = (-1)^{j+1}.\n$$\n\nSumming over all $j$, we obtain\n\n$$\n\\sum_{j=1}^{n} (a_j^2 - a_j b_j) = 0\n$$\n\nsince $n$ is even. As $\\{b_j\\}$ is a permutation of $\\{a_j\\}$, this implies\n\n$$\n\\frac{1}{2} \\sum_{j=1}^{n} (a_j - b_j)^2 = 0.\n$$\n\nHence, we must have $a_j = b_j$, which is impossible since $a_j + \\frac{(-1)^j}{a_j} = b_j$. Therefore, there is no such sequence for even $n$.\n\nFor $n=1$, we need $a_1 - \\frac{1}{a_1} = a_1$, which is impossible.\n\nSo it remains to show that such a sequence exists when $n > 1$ is odd. Construct a sequence $a_1, a_2, \\dots, a_{n+1}$ such that $a_1 = 1 + x$ with $x > 0$, and\n\n$$\na_{j+1} = a_j + \\frac{(-1)^j}{a_j}\n$$\n\nfor $1 \\leq j \\leq n$. It suffices to show $a_1, a_2, \\dots, a_n$ are pairwise distinct, and choose $x$ such that $a_{n+1} = a_1$.\n\nNote that\n\n$$\na_{j+2} = a_{j+1} + \\frac{(-1)^{j+1}}{a_{j+1}} = a_j + \\frac{(-1)^j}{a_j} + (-1)^{j+1} \\left( a_j + \\frac{(-1)^j}{a_j} \\right)^{-1} = a_j + \\frac{1}{a_j^3 + (-1)^j a_j}.\n$$\n\nFor odd $j$, since $a_1 > 1$, we can prove inductively that $a_{j+2} > a_j > 1$. For even $j$, since $a_2 = a_1 - \\frac{1}{a_1} > 0$, we can prove inductively that $a_{j+2} > a_j > 0$.\n\nTherefore, if $a_{n+1} = a_1$, then\n\n$$\n0 < a_2 < a_4 < \\dots < a_{n+1} = a_1 < a_3 < \\dots < a_n,\n$$\n\nso all terms are pairwise distinct positive real numbers.\n\nBy the recurrence, $a_{n+1}$ is a continuous function of $a_1$. When $a_1 \\to 1$, we have\n\n$$\na_{n+1} - a_1 \\geq a_4 - a_1 = a_2 + \\frac{1}{a_2^3 + a_2} - a_1 \\to \\infty\n$$\nsince $a_2 \\to 0$. Thus, $a_{n+1} - a_1 > 0$ for some $x$.\n\nOn the other hand, for even $j$, as $a_j \\geq a_2 = 1+x - \\frac{1}{1+x} = \\frac{2x + x^2}{1 + x} > x$, we have\n\n$$\na_{j+2} = a_j + \\frac{1}{a_j^3 + a_j} < a_j + \\frac{1}{x^3},\n$$\n\nand hence $a_{n+1} < a_2 + \\frac{n-1}{2x^3}$ by summing over all $j$. It follows that\n\n$$\na_{n+1} - a_1 < a_2 + \\frac{n-1}{2x^3} - a_1 = \\frac{n-1}{2x^3} - \\frac{1}{1+x} = \\frac{(n-1) + (n-1)x - 2x^3}{2x^3(1+x)} < 0\n$$\n\nfor sufficiently large $x$.\n\nBy the intermediate value theorem, there exists $x > 0$ such that $a_{n+1} = a_1$. Thus, such a sequence exists for odd $n > 1$.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 21374,
"subject": "Mathematics (Olympiad)",
"question": "Find all nonempty sets $S$ of integers such that $3m - 2n \\in S$ for all (not necessarily distinct) $m, n \\in S$.",
"options": [],
"answer": "See solution",
"solution": "Call a set $S$ \"good\" if it satisfies the property as stated in the problem.\n\n1. If $S$ has only one element, $S$ is \"good\".\n\n2. Now assume that $S$ contains at least two elements. Let\n\n$$\nd = \\min\\{|m - n| : m, n \\in S, m \\neq n\\}.\n$$\n\nThen there is an integer $a$ such that $a + d, a + 2d \\in S$.\n\nNote that\n\n$$\na + 4d = 3(a + 2d) - 2(a + d) \\in S,\n$$\n$$\na - d = 3(a + d) - 2(a + 2d) \\in S,\n$$\n$$\na + 5d = 3(a + d) - 2(a - d) \\in S,\n$$\n$$\na - 2d = 3(a + 2d) - 2(a + 4d) \\in S.\n$$\n\nSo we have proved that if $a + d, a + 2d \\in S$, then $a - 2d, a - d, a + 4d, a + 5d \\in S$.\n\nContinuing this procedure, we can deduce that\n\n$$\n\\{a + k d \\mid k \\in \\mathbb{Z},\\ 3 \\nmid k\\} \\subseteq S.\n$$\n\nLet $S_0 = \\{a + k d \\mid k \\in \\mathbb{Z},\\ 3 \\nmid k\\}$. It is easy to verify that $S_0$ is \"good\".\n\n3. Now we have proved that $S_0 \\subseteq S$. If $S \\neq S_0$, pick a number $b \\in S \\setminus S_0$. Then there exists an integer $l$ such that $a + l d \\leq b < a + (l + 1)d$. Since at least one of $l$ and $l + 1$ is not divisible by $3$, at least one of $a + l d, a + (l + 1)d$ is contained in $S_0$. If $a + l d \\in S_0$, note that $0 \\leq b - (a + l d) < d$, so by the definition of $d$, we must have $b = a + l d$. If $a + (l + 1)d \\in S_0$, note that\n\n$$\n0 < |a + (l + 1)d - b| \\leq d,\n$$\n\nso by the definition of $d$, we also have $b = a + l d$.\n\nIn both cases, we have proved that there is a number $b \\in S \\setminus S_0$ of the form $b = a + l d$. This $l$ must be divisible by $3$, hence\n\n$$\na + l d,\\ a + (l + 1)d,\\ a + (l + 2)d \\in S,\n$$\n$$\na + (l - 2)d,\\ a + (l - 1)d \\in S.\n$$\n\nSo\n\n$$\na + (l + 3)d = 3(a + (l + 1)d) - 2(a + l d) \\in S,\n$$\n$$\na + (l - 3)d = 3(a + (l - 1)d) - 2(a + l d) \\in S.\n$$\n\nContinuing this procedure, we have $a + (l + 3j)d \\in S$ for all $j \\in \\mathbb{Z}$, which implies that $\\{a + k d \\mid k \\in \\mathbb{Z}\\} \\subseteq S$. We claim that $S = \\{a + k d \\mid k \\in \\mathbb{Z}\\}$, since for any number $x \\notin \\{a + k d \\mid k \\in \\mathbb{Z}\\}$, there is an element $y \\in \\{a + k d \\mid k \\in \\mathbb{Z}\\}$ such that $0 < |x - y| < d$. By the definition of $d$, we must have $x \\notin S$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21375,
"subject": "Mathematics (Olympiad)",
"question": "令 $Z$ 表示所有整數所成的集合。試求所有函數 $f: Z \\to Z$ 滿足：\n\n$$\nf(f(m)+n)+f(m) = f(n)+f(3m)+2014\n$$\n\n對於所有整數 $m, n$ 皆成立。\n\nLet $Z$ be the set of all integers. Determine all functions $f: Z \\to Z$ satisfying\n\n$$\nf(f(m)+n)+f(m) = f(n)+f(3m)+2014\n$$\n\nfor all integers $m$ and $n$.",
"options": [],
"answer": "See solution",
"solution": "答案：只有一個函數，$f(n) = 2n + 1007$\n\n解法：令 $f$ 為一個滿足原式的函數，令 $C = 1007$，然後定義函數 $g: Z \\to Z$，對於所有整數 $m$，$g(m) = f(3m) - f(m) + 2C$。則有 $g(0) = 2C$，原式可改寫為\n\n$$\nf(f(m) + n) = g(m) + f(n)\n$$\n\n對於所有 $m, n \\in Z$ 皆成立。使用歸納法可以得到\n\n$$\nf(tf(m) + n) = tg(m) + f(n) \\quad (1)\n$$\n\n對於所有 $m, n, t \\in Z$ 皆成立。對於任意的 $r \\in Z$，將 $(r, 0, f(0))$ 與 $(0, 0, f(r))$ 帶入 $(m, n, t)$ 可以得到：\n\n$$\nf(0)g(r) = f(f(r)f(0)) - f(0) = f(r)g(0)\n$$\n\n如果 $f(0) = 0$，則由 $g(0) = 2C > 0$ 可以得到對於所有 $m$，$f(m) = 0$，這是矛盾的。因此 $f(0) \\neq 0$，而上式可得 $g(r) = \\alpha f(r)$，其中 $\\alpha = \\frac{g(0)}{f(0)}$ 是一個非零的常數。\n\n所以由 $g$ 的定義可得 $f(3m) = (1+\\alpha)f(m) - 2C$，即\n\n$$\nf(3m) - \\beta = (1 + \\alpha)(f(m) - \\beta) \\quad (2)\n$$\n\n對於所有 $m \\in Z$ 皆成立，其中 $\\beta = \\frac{2C}{\\alpha}$。使用歸納法可以得到\n\n$$\nf(3^k m) - \\beta = (1 + \\alpha)^k (f(m) - \\beta) \\quad (3)\n$$\n\n對於所有整數 $k \\ge 0$ 和 $m$。\n\n因為 3 無法整除 2014，所以由原式可得存在 $d = f(a)$ 不被 3 整除。由 (1)\n\n可得 $f(n + td) = f(n) + tg(a) = f(n) + \\alpha \\cdot tf(a)$，即\n\n$$\nf(n + td) = f(n) + \\alpha \\cdot td \\quad (4)\n$$\n\n對於所有 $n, t \\in Z$ 皆成立。\n\n固定一個正整數 $k$ 使 $d|(3^k - 1)$，根據歐拉定理，我們可以取 $k = \\varphi(|d|)$ 使它成立。則由 (4) 可得對於所有 $m \\in Z$\n\n$$\nf(3^k m) = f(m) + \\alpha(3^k - 1)m\n$$\n\n結合 (3) 可以推出 $((1 + \\alpha)^k - 1)(f(m) - \\beta) = \\alpha(3^k - 1)m$。由於 $\\alpha \\ne 0$，所以當 $m \\ne 0$ 時右式不為零，因此左式第一項也不為零，所以\n\n$$\nf(m) = \\frac{\\alpha(3^k - 1)}{(1 + \\alpha)^k - 1} \\cdot m + \\beta\n$$\n\n所以 $f$ 是一個線性函數，令 $f(m) = Am + \\beta$ 對於所有 $m \\in Z$，其中 $A \\in \\mathbb{Q}$ 是一個常數，帶入原式得到 $(A^2 - 2A)m + (A\\beta - 2C) = 0$ 對於所有 $m$ 皆成立，等價於\n\n$$\nA^2 = 2A \\text{ 與 } A\\beta = 2C.\n$$\n\n第一個等號等價於 $A \\in \\{0, 2\\}$，而由 $C \\ne 0$ 可得\n\n$$\nA = 2 \\text{ 與 } \\beta = C.\n$$\n\n這告訴我們 $f$ 就是答案中的那個函數，而這個函數也滿足原式。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21376,
"subject": "Mathematics (Olympiad)",
"question": "a) Does there exist a function from real numbers to real numbers, which is not constantly zero and whose derivative's graph can be obtained by reflecting the graph of the original function with respect to the $y$-axis?\n\nb) Does there exist a function from real numbers to real numbers, which is not constantly zero and whose derivative's graph can be obtained by shifting the graph of the original function towards the positive side of the $x$-axis by one unit?",
"options": [],
"answer": "See solution",
"solution": "**a)** Yes.\n\nTo find such a function $f$, we require $f'(x) = f(-x)$ for all $x \\in \\mathbb{R}$. For example, $f(x) = \\sin x + \\cos x$ works, since\n$$\nf'(x) = \\cos x - \\sin x = \\cos(-x) + \\sin(-x) = f(-x).\n$$\n\n**b)** Yes.\n\nHere, we require $f'(x) = f(x - 1)$ for all $x \\in \\mathbb{R}$. Suppose $a > 1$ satisfies $\\ln a = a^{-1}$. Then, for $f(x) = a^x$,\n$$\nf'(x) = a^x \\ln a = a^x \\cdot a^{-1} = a^{x-1} = f(x-1).\n$$\nTo see such $a$ exists: $\\ln 1 = 0 < 1 = 1^{-1}$ and $\\ln e = 1 > e^{-1}$, so the continuous functions $g(x) = \\ln x$ and $h(x) = x^{-1}$ intersect at some $a > 1$.\n\n*Remark:* In part a), all functions of the form $f(x) = c \\cdot (\\sin x + \\cos x)$, $c \\neq 0$, satisfy the condition. For example, $\\frac{\\sqrt{2}}{2} (\\sin x + \\cos x) = \\sin(x + \\frac{\\pi}{4})$.",
"topic": "Discrete Mathematics",
"subtopic": "Other"
},
{
"id": 21377,
"subject": "Mathematics (Olympiad)",
"question": "A natural number is written on each face of a cube. To each vertex of the cube, assign the product of the numbers on the three faces that meet at that vertex. The sum of these 8 products is 315. Determine the sum of the numbers on the faces (find all possibilities).",
"options": [],
"answer": "See solution",
"solution": "Let the numbers on the three pairs of opposite faces be $a_1, a_2$, $b_1, b_2$, and $c_1, c_2$. Each vertex corresponds to a product $a_i b_j c_k$ for $i, j, k \\in \\{1, 2\\}$, so the sum of all 8 products is:\n\n$$\n\\sum_{i=1}^2 \\sum_{j=1}^2 \\sum_{k=1}^2 a_i b_j c_k = (a_1 + a_2)(b_1 + b_2)(c_1 + c_2)\n$$\n\nGiven $(a_1 + a_2)(b_1 + b_2)(c_1 + c_2) = 315$, and each sum is greater than 1. Let $d_1 = a_1 + a_2$, $d_2 = b_1 + b_2$, $d_3 = c_1 + c_2$, so $d_1 d_2 d_3 = 315$ with $d_i > 1$.\n\nThe sum of the numbers on the faces is $d_1 + d_2 + d_3$. We seek all unordered triples of integers $>1$ whose product is 315.\n\nSince $315 = 3^2 \\cdot 5 \\cdot 7$, the possible factorizations are:\n\n- $9 \\cdot 5 \\cdot 7$ with sum $21$\n- $3 \\cdot 15 \\cdot 7$ with sum $25$\n- $3 \\cdot 5 \\cdot 21$ with sum $29$\n- $3 \\cdot 3 \\cdot 35$ with sum $41$\n\nThus, the possible sums are $21$, $25$, $29$, and $41$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21378,
"subject": "Mathematics (Olympiad)",
"question": "For each integer $n > 1$, let $s_n$ be the number of permutations $(a_1, a_2, \\dots, a_n)$ of the first $n$ positive integers such that for every $k = 1, 2, \\dots, n$,\n\n$$\n1 \\leq |a_k - k| \\leq 2.\n$$\n\nProve that for all integers $n > 6$:\n\n$$\n1.75 \\cdot s_{n-1} < s_n < 2 \\cdot s_{n-1}.\n$$",
"options": [],
"answer": "See solution",
"solution": "First, we construct an inductive relation for $s_n$.\n\nLet $P_n$ be the set of permutations $(a_1, a_2, \\dots, a_n)$ satisfying the problem's condition. Partition $P_n$ as:\n\n$$\nP_n = \\bigcup_{k=1}^{n} S_k\n$$\n\nwhere $S_k$ is the set of permutations $(a_1, a_2, \\dots, a_n) \\in P_n$ such that\n\n$$\n\\{a_1, a_2, \\dots, a_k\\} = \\{1, 2, \\dots, k\\},\n$$\n\nand for all $i < k$,\n\n$$\n\\{a_1, a_2, \\dots, a_i\\} \\neq \\{1, 2, \\dots, i\\}.\n$$\n\nFor $k > 1$, let $H_k$ be the set of permutations $(a_1, a_2, \\dots, a_k)$ of $k$ positive integers such that:\n\n1. $1 \\leq |a_i - i| \\leq 2$ for all $i = 1, \\dots, k$;\n2. $\\{a_1, a_2, \\dots, a_i\\} \\neq \\{1, 2, \\dots, i\\}$ for all $i = 1, \\dots, k-1$.\n\nLet $t_k = |H_k|$. Then $|S_k| = t_k \\cdot S_{n-k}$ (with $S_0 = 1$), so\n\n$$\nS_n = t_1 S_{n-1} + t_2 S_{n-2} + \\dots + t_n S_0.\n$$\n\nCalculating $t_k$ for small $k$:\n\n- $H_1 = \\emptyset \\implies t_1 = 0$\n- $H_2 = \\{(2, 1)\\} \\implies t_2 = 1$\n- $H_3 = \\{(2, 3, 1), (3, 1, 2)\\} \\implies t_3 = 2$\n- $H_4 = \\{(2, 4, 1, 3), (3, 1, 4, 2), (3, 4, 1, 2)\\} \\implies t_4 = 3$\n\nFor $k \\geq 4$, by induction and case analysis, $t_k = 2$ for all $k = 4, 5, \\dots, n$.\n\nThus,\n\n$$\nS_n = S_{n-2} + 2 S_{n-3} + 3 S_{n-4} + 2(S_{n-5} + \\dots + S_1 + S_0)\n$$\n\nSimilarly,\n\n$$\nS_{n+1} = S_{n-1} + 2 S_{n-2} + 3 S_{n-3} + 2(S_{n-4} + \\dots + S_1 + S_0)\n$$\n\nSubtracting,\n\n$$\nS_{n+1} - S_n = S_{n-1} + S_{n-2} + S_{n-3} - S_{n-4}\n$$\n\nand\n\n$$\nS_{n+2} - S_{n+1} = S_n + S_{n-1} + S_{n-2} - S_{n-3}\n$$\n\nSubtracting again,\n\n$$\nS_{n+2} - 2 S_{n+1} = S_{n-4} - 2 S_{n-3}\n$$\n\nSo for $n > 6$,\n\n$$\nS_n - 2 S_{n-1} = S_{n-6} - 2 S_{n-5}\n$$\n\nSince $S_{n-5} > S_{n-6}$ for $n > 6$, it follows that\n\n$$\nS_n < 2 S_{n-1}\n$$\n\nAlso, from earlier,\n\n$$\nS_n > S_{n-1} + S_{n-2} + S_{n-3}\n$$\n\nBy direct computation for small $n$, the ratio $S_n / S_{n-1}$ exceeds $1.75$ for $n > 6$. Therefore,\n\n$$\n1.75 \\cdot S_{n-1} < S_n < 2 \\cdot S_{n-1} \\quad \\text{for all } n > 6.\n$$\n\n*Remark:* The recurrence can also be interpreted combinatorially by constructing $P_{n+1}$ from $P_n$ and related sets.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 21379,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$ be the side lengths of a triangle, and define\n\n$$\nd = \\frac{a+b-c}{2}, \\quad e = \\frac{b+c-a}{2}, \\quad f = \\frac{c+a-b}{2}.\n$$\n\nNote that $d, e, f > 0$, and either all of them are integers or each is an odd integer divided by $2$. The triangle is uniquely determined by $d, e, f$. By Heron's formula,\n\n$$\nx = \\sqrt{def(d+e+f)}.\n$$\n\nFind the smallest possible value of $x$.",
"options": [],
"answer": "See solution",
"solution": "We list all possibilities for which $x^2 \\le \\frac{63}{16}$. Without loss of generality, assume $d \\ge e \\ge f$.\n\n- If $d, e, f$ are integers, then $d = e = f = 1$, otherwise\n $$\nx^2 \\ge (2)(1)(1)(4) = 8 > \\frac{63}{16}.\n $$\n When $d = e = f = 1$, $x^2 = 3$.\n\n- If $2d, 2e, 2f$ are odd integers, then $f = \\frac{1}{2}$, otherwise\n $$\nx^2 \\ge \\left(\\frac{3}{2}\\right)\\left(\\frac{3}{2}\\right)\\left(\\frac{3}{2}\\right)\\left(\\frac{9}{2}\\right) = \\frac{243}{16} > \\frac{63}{16}.\n $$\n\n - If $e = \\frac{1}{2}$, let $d = \\frac{2k-1}{2}$ for $k \\in \\mathbb{Z}^+$. Then\n $$\nx^2 = \\frac{2k-1}{8} \\cdot \\frac{2k+1}{2} = \\frac{4k^2-1}{16}.\n $$\n Thus, $x^2$ can be $\\frac{3}{16}, \\frac{15}{16}, \\frac{35}{16}, \\frac{63}{16}, \\dots$\n\n - If $e \\ge \\frac{3}{2}$, then $d = e = \\frac{3}{2}$, otherwise\n $$\nx^2 \\ge \\left(\\frac{5}{2}\\right)\\left(\\frac{3}{2}\\right)\\left(\\frac{1}{2}\\right)\\left(\\frac{9}{2}\\right) = \\frac{135}{16} > \\frac{63}{16}.\n $$\n When $d = e = \\frac{3}{2}$, $x^2 = \\frac{63}{16}$.\n\nAmong all possible values of $x^2$ not exceeding $\\frac{63}{16}$, only $\\frac{63}{16}$ appears twice. Thus, this is the smallest value of $x^2$, and the smallest value of $x$ is $\\frac{3\\sqrt{7}}{4}$. Indeed, when $(a, b, c) = (2, 2, 3)$ or $(1, 4, 4)$, we have $x = \\frac{3\\sqrt{7}}{4}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21380,
"subject": "Mathematics (Olympiad)",
"question": "Karmen needs to add a puzzle piece so that the final piece has protrusions on two opposite edges and indentations on the other two. Which piece should she choose?",
"options": [],
"answer": "See solution",
"solution": "The left edge of the left piece and the right edge of the right piece have indentations. The upper edge of the upper piece and the lower edge of the lower piece have protrusions. Therefore, the final piece has to have protrusions on two opposite edges and indentations on the other two. The only puzzle piece that satisfies this condition is (A). Hence, Karmen needs to add the piece (A).",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 21381,
"subject": "Mathematics (Olympiad)",
"question": "Prove that there are infinitely many pairs of distinct positive integers $x, y$ such that $x^2 + y^3$ is divisible by $x^3 + y^2$.",
"options": [],
"answer": "See solution",
"solution": "Let's try the substitution $y = kx$, where $k$ is any integer.\n\nThen we have\n\n$$\n\\frac{x^2 + y^3}{x^3 + y^2} = \\frac{x^2 + k^3 x^3}{x^3 + k^2 x^2} = \\frac{k^3 x + 1}{x + k^2}.\n$$\n\nNow, we can try taking $k$ as an integer to produce solutions. If we take $x = k^5 - k^2 - 1$ and $y = kx = k^6 - k^3 - k$, we get\n\n$$\n\\frac{x^2 + y^3}{x^3 + y^2} = \\frac{k^3 x + 1}{x + k^2} = \\frac{k^3 (k^5 - k^2 - 1) + 1}{k^5 - k^2 - 1 + k^2} = \\frac{k^8 - k^5 - k^3 + 1}{k^5 - 1} = k^3 - 1,\n$$\n\na positive integer.\n\nAs we let $k$ vary, we can generate infinitely many different values of $x$ and $y$ in this way.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21382,
"subject": "Mathematics (Olympiad)",
"question": "Suppose that for two natural numbers $m, n$ the following equality holds:\n\n$$\nm + n = [m, n] + (m, n),\n$$\n\nwhere $[m, n]$ and $(m, n)$ are the least common multiple and the greatest common divisor of $m$ and $n$, respectively. Prove that one number is divisible by the other.",
"options": [],
"answer": "See solution",
"solution": "Let $d = (m, n)$. Then $m = a d$ and $n = b d$ for some integers $a, b$. Using the formula $m n = [m, n] \\cdot (m, n)$, we get $[m, n] = a b d$.\n\nSubstituting into the given equation:\n$$\na d + b d = a b d + d\n$$\nwhich simplifies to\n$$\nd(a - 1)(b - 1) = 0.\n$$\nSince $d > 0$, this is possible only if $a = 1$ or $b = 1$, i.e., one number divides the other.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21383,
"subject": "Mathematics (Olympiad)",
"question": "Бат орцныхоо оршин суугчдаас авсан асуулгаар дараах мэдээллийг олж мэдэв:\n\n- 25 нь шатар тоглодог\n- 30 нь гадаад явж үзсэн\n- 28 нь онгоцоор нисч үзсэн\n- Онгоцоор нисч байсан хүмүүсээс 18 нь шатар тоглодог, 17 нь гадаад явж үзсэн\n- Шатар тоглодог, гадаад явж байсан 16 оршин суугчийн 15 нь онгоцоор нисч байсан\n\nОрцны дарга орцонд нийт 45 оршин суугчтай гэж хэлсэн бол орцны дарга үнэн хэлсэн үү?",
"options": [],
"answer": "See solution",
"solution": "$|A_3| = 30$, $|A_3 \\cap A_1| = 16$ болох тул нэгтгэн зайлуулах томъёогоор:\n\n$$\n45 = |V| \\geq |A_1 \\cup A_2 \\cup A_3|\n$$\n\n$$\n\\begin{aligned}\n&= |A_1| + |A_2| + |A_3| - |A_1 \\cap A_2| - |A_2 \\cap A_3| - |A_3 \\cap A_1| + |A_1 \\cap A_2 \\cap A_3| \\\\\n&= 25 + 28 + 30 - 18 - 17 - 16 + 15 = 47\n\\end{aligned}\n$$\n\nИймд зөрчил гарч байгаа тул орцны дарга худал хэлсэн.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21384,
"subject": "Mathematics (Olympiad)",
"question": "In a convex quadrilateral $ABCD$, let $M$ and $N$ be the midpoints of sides $AD$ and $BC$, respectively. Points $K$ and $L$ are chosen on sides $AB$ and $CD$, respectively, such that $\\angle MKA = \\angle NLC$. Prove that if lines $BD$, $KM$, and $LN$ meet at one point, then\n$$\n\\angle KMN = \\angle BDC, \\quad \\angle LNM = \\angle ABD\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $P$ be the midpoint of $BD$ and $Q$ be the common point of lines $BD$, $KM$, and $LN$. Without loss of generality, assume that point $B$ lies between $Q$ and $D$. By Thales' theorem, $PM \\parallel AB$ and $PN \\parallel CD$. Therefore, $\\angle PNL = \\angle NLC = \\angle MKA = \\angle KMP$. This implies that points $Q, M, P, N$ are concyclic, since $\\angle QNP + \\angle QMP = 180^\\circ$ and points $M, N$ lie on different sides of the line $BD$. Therefore,\n$$\n\\angle KMN = \\angle QMN = \\angle QPN = \\angle BDC\n$$\nMoreover,\n$$\n\\angle LNM = 180^\\circ - \\angle QNM = 180^\\circ - \\angle QPM = \\angle MPD = \\angle ABD\n$$\n$\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21385,
"subject": "Mathematics (Olympiad)",
"question": "Let $m$ be a positive integer. Consider a $4m \\times 4m$ array of square unit cells. Two different cells are *related* to each other if they are in either the same row or in the same column. No cell is related to itself. Some cells are coloured blue, such that every cell is related to at least two blue cells. Determine the minimum number of blue cells.",
"options": [],
"answer": "See solution",
"solution": "The required minimum is $6m$ and is achieved by a diagonal string of $m$ $4 \\times 4$ blocks of the form below (bullets mark centers of blue cells):\n\n\n\nIn particular, this configuration shows that the required minimum does not exceed $6m$.\n\nWe now show that any configuration of blue cells satisfying the condition in the statement has cardinality at least $6m$.\n\nFix such a configuration and let $m_1^r$ be the number of blue cells in rows containing exactly one such, let $m_2^r$ be the number of blue cells in rows containing exactly two such, and let $m_3^r$ be the number of blue cells in rows containing at least three such; the numbers $m_1^c, m_2^c$ and $m_3^c$ are defined similarly.\n\nBegin by noticing that $m_3^c \\ge m_1^r$ and similarly, $m_3^r \\ge m_1^c$. Indeed, if a blue cell is alone in its row, respectively column, then there are at least two other blue cells in its column, respectively row, and the claim follows.\n\nSuppose now, if possible, the total number of blue cells is less than $6m$. We will show that $m_1^r > m_3^r$ and $m_1^c > m_3^c$ and reach a contradiction by the preceding: $m_1^r > m_3^r \\ge m_1^c > m_3^c \\ge m_1^r$.\n\nWe prove the first inequality; the other one is dealt with similarly. To this end, notice that there are no empty rows—otherwise, each column would contain at least two blue cells, whence a total of at least $8m > 6m$ blue cells, which is a contradiction. Next, count rows to get $m_1^r + \\frac{m_2^r}{2} + \\frac{m_3^r}{3} \\ge 4m$, and count blue cells to get $m_1^r + m_2^r + m_3^r < 6m$. Subtraction of the latter from the former multiplied by $\\frac{3}{2}$ yields $m_1^r - m_3^r > \\frac{m_2^r}{2} \\ge 0$, and the conclusion follows.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21386,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $z$ is a complex number with positive imaginary part, with real part greater than $1$, and with $|z| = 2$. In the complex plane, the four values $0$, $z$, $z^2$, and $z^3$ are the vertices of a quadrilateral with area $15$. What is the imaginary part of $z$?\n\n(A) $\\frac{3}{4}$ (B) $1$ (C) $\\frac{4}{3}$ (D) $\\frac{3}{2}$ (E) $\\frac{5}{3}$",
"options": [],
"answer": "See solution",
"solution": "Let $\\theta$ be the argument of $z$. Because $|z| = 2$ and the real part of $z$ is greater than $1$, it follows that $\\theta$ is less than $60^\\circ$. This ensures that the imaginary parts of $z^2$ and $z^3$ are positive and all the vertices of the quadrilateral other than $0$ lie in the upper half-plane. Thus, the area of the quadrilateral is the sum of the areas of the triangle with vertices $0$, $z$, and $z^2$ and the triangle with vertices $0$, $z^2$, and $z^3$.\n\n\n\nBecause the area of a triangle with side lengths $a$ and $b$ with included angle $\\alpha$ is $\\frac{1}{2}ab \\sin \\alpha$, the area of the quadrilateral must be\n\n$$\n\\frac{1}{2} (|z| \\cdot |z|^2 \\cdot \\sin \\theta + |z|^2 \\cdot |z|^3 \\cdot \\sin \\theta) = 20 \\sin \\theta = 15.\n$$\n\nIt follows that $\\sin \\theta = \\frac{3}{4}$, and the imaginary part of $z$ is $2 \\cdot \\sin \\theta = \\frac{3}{2}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21387,
"subject": "Mathematics (Olympiad)",
"question": "For a non-negative integer $n$, the $n$-th iterate of a function $f: \\mathbb{R} \\to \\mathbb{R}$ is defined as $f^n = \\underbrace{f \\circ \\dots \\circ f}_{n \\text{ times}}$, and $f^0$ is the identity function.\n\nDetermine all continuous functions $f: \\mathbb{R} \\to \\mathbb{R}$ that satisfy both of the following conditions:\n\n1. The function $f^0 + f^1$ is increasing.\n2. There exists a positive integer $m$ such that the function $f^0 + \\dots + f^m$ is decreasing.",
"options": [],
"answer": "See solution",
"solution": "All such functions are of the form $f(x) = -x + c$, where $c$ is a real constant. These functions satisfy the given conditions.\n\nFirst, we show that $f$ is one-to-one. Suppose $f(x) = f(y)$ for some $x, y \\in \\mathbb{R}$. Define $g_n = f^0 + \\dots + f^n$ for $n \\in \\mathbb{N}$. Since $g_1$ is increasing and $g_m$ is decreasing, and $g_1(x) - g_1(y) = x - y = g_m(x) - g_m(y)$, we have:\n\n$$(x - y)^2 = (g_1(x) - g_1(y))(g_m(x) - g_m(y)) \\le 0,$$\n\nso $x = y$.\n\nSince $f$ is one-to-one and continuous, it is strictly monotonic, so all iterates $f^{2k}$ are increasing. As $g_1$ is increasing, from the equality\n\n$$\ng_n = \\begin{cases} \\sum_{k=0}^{n/2-1} g_1 \\circ f^{2k} + f^n, & \\text{if } n \\text{ is even} \\\\ \\sum_{k=0}^{(n-1)/2} g_1 \\circ f^{2k}, & \\text{if } n \\text{ is odd} \\end{cases}\n$$\n\n$g_n$ is strictly increasing for even $n$ and increasing for odd $n$. Since $g_m$ is decreasing, $m$ must be odd and $g_m$ is constant. Finally, as $g_1$ is increasing and all even iterates of $f$ are increasing, we conclude that $g_1$ is constant, which leads to the result.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21388,
"subject": "Mathematics (Olympiad)",
"question": "Five girls and five boys participate in a tournament. Suppose it is possible to number the girls from 1 to 5 and the boys from 1 to 5 so that for all $1 \\leq i, j \\leq 5$, the number of students that the $i$-th girl and the $j$-th boy both know is exactly $|i - j|$.\n\nLet $S$ denote the maximum of the sum of the number of students that each girl knows and the sum of the number of students that each boy knows. What is the minimum possible value of $S$?\n\nHere, the relationship of knowing is directional: $A$ knows $B$ does not mean $B$ knows $A$. Students do not know themselves.",
"options": [],
"answer": "See solution",
"solution": "The minimum possible value of $S$ is $19$.\n\nFor $1 \\leq i \\leq 5$, let $a_i$ denote the $i$-th girl and $A_i$ the set of students that $a_i$ knows. Similarly, let $b_i$ denote the $i$-th boy and $B_i$ the set of students that $b_i$ knows.\n\nSince $|A_i \\cap B_1| = i - 1$ and $|A_i \\cap B_5| = 5 - i$, we have:\n\n$$\n|A_1| \\geq 4,\\quad |A_2| \\geq 3,\\quad |A_3| \\geq 2,\\quad |A_4| \\geq 3,\\quad |A_5| \\geq 4.\n$$\n\nSimilarly for $|B_i|$.\n\nSuppose $|A_1| = 4$. Since $|A_1 \\cap B_5| = 4$, $A_1 \\subseteq B_5$, so $A_1 \\cap A_5 \\subseteq B_5 \\cap A_5 = \\emptyset$. Thus $|B_i| \\geq |(A_1 \\cup A_5) \\cap B_i| = |A_1 \\cap B_i| + |A_5 \\cap B_i| = 4$. Hence $\\sum |B_i| \\geq 20$. Similarly, if $|A_5| = 4$, then $\\sum |B_i| \\geq 20$.\n\nSuppose $|A_3| = 2$. Then $|A_3 \\cap B_1| = |A_3 \\cap B_5| = 2$ implies $A_3 \\subseteq B_1 \\cap B_5$. Hence $|B_1| \\geq |A_5 \\cap B_1| + |A_5 \\cap B_5| \\geq 6$. Similarly, $|B_5| \\geq 6$, so $\\sum |B_i| \\geq 6 + 3 + 2 + 3 + 6 = 20$.\n\nIf $|A_1| \\geq 5$, $|A_3| \\geq 3$, $|A_5| \\geq 5$, then $\\sum |A_i| \\geq 5 + 3 + 3 + 3 + 5 = 19$. Thus $S \\geq 19$ and it suffices to find an example with $\\sum |A_i| = \\sum |B_i| = 19$:\n\n$$\n\\begin{align*}\nA_1 &= \\{a_2, a_4, b_1, b_2, b_5\\} & \\{a_1, a_3, a_5, b_3, b_4\\} &= B_1 \\\\\nA_2 &= \\{a_4, a_5, b_1\\} & \\{a_1, a_3, b_5\\} &= B_2 \\\\\nA_3 &= \\{a_1, a_5, b_1\\} & \\{a_3, a_4, b_5\\} &= B_3 \\\\\nA_4 &= \\{a_1, a_3, a_5\\} & \\{a_4, b_1, b_5\\} &= B_4 \\\\\nA_5 &= \\{a_1, a_3, b_3, b_4, b_5\\} & \\{a_2, a_4, a_5, b_1, b_2\\} &= B_5\n\\end{align*}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21389,
"subject": "Mathematics (Olympiad)",
"question": "Circles $K_1$ (radius $r_1$) and $K_2$ (radius $r_2$) are tangent to each other and also tangent to the sides of rectangle $AB\\Gamma\\Delta$, where $AB = a$ and $B\\Gamma = b$.\n\n1. Express the sides $a$ and $b$ in terms of the radii $r_1$ and $r_2$.\n2. If the common internal tangent of the two circles passes through $\\Delta$, compute the ratio $\\frac{r_1}{r_2}$ and find the length $\\Delta K$.",
"options": [],
"answer": "See solution",
"solution": "We draw the line segment $K_1H \\perp K_2Z$.\n\nIn triangle $K_1HK_2$, we have $K_1K_2 = r_1 + r_2$ and $K_2H = r_2 - r_1$, so by the Pythagorean theorem:\n\n$$\nK_1H = 2\\sqrt{r_1 r_2}.\n$$\n\nFrom rectangle $K_1EZH$, $EZ = 2\\sqrt{r_1 r_2}$.\n\nAlso,\n$$\na = AB = AE + EZ + ZB = r_1 + 2\\sqrt{r_1 r_2} + r_2 = (\\sqrt{r_1} + \\sqrt{r_2})^2, \\\\\nb = B\\Gamma = Z\\Theta = 2r_2.\n$$\n\n(ii) We have $\\Lambda E = \\Lambda K = \\Lambda Z = \\sqrt{r_1 r_2}$. Let $\\Delta I = \\Delta K = \\Delta \\Theta = x$. From the right triangle $AB\\Delta$, since $x + r_2 = a$ and $x + r_1 = b$, we get:\n\n$$\nr_1 + 2\\sqrt{r_1 r_2} = 2r_2 - r_1 \\\\\n\\Rightarrow r_1 + \\sqrt{r_1 r_2} - r_2 = 0 \\\\\n\\Rightarrow \\omega^2 + \\omega - 1 = 0, \\\\\n\\text{where } \\omega = \\sqrt{\\frac{r_1}{r_2}} > 0. \\\\\n\\text{Hence } \\omega = \\frac{\\sqrt{5} - 1}{2}, \\quad \\kappa = \\frac{r_1}{r_2} = \\frac{3 - \\sqrt{5}}{2}.\n$$\n\nMoreover,\n\n$$\n\\frac{a}{b} = \\frac{(\\sqrt{r_1} + \\sqrt{r_2})^2}{2r_2} = \\frac{1}{2} \\left( \\sqrt{\\frac{r_1}{r_2}} + 1 \\right)^2 = \\frac{1}{2} \\left( \\frac{\\sqrt{5} + 1}{2} \\right) = \\frac{\\sqrt{5} + 1}{4}.\n$$\n\n% ![](images/Greek_07_Booklet_p13_data_37b7aa31d5.png \"Figure 5\")\n\n(iii) From the right triangle $AB\\Delta$:\n\n$$\n(x + r_1)^2 + (r_1 + \\sqrt{r_1 r_2})^2 = (x + \\sqrt{r_1 r_2})^2 \\\\\n\\Rightarrow x = \\frac{(r_1 + \\sqrt{r_1 r_2})^2}{r_2 - r_1}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21390,
"subject": "Mathematics (Olympiad)",
"question": "We inscribe a regular octagon in a square with side length $a$, so that four sides of the octagon lie on the sides of the square. Express the side length of the inscribed octagon in terms of $a$.",
"options": [],
"answer": "See solution",
"solution": "Let the side length of the octagon be $s$. The octagon is formed by cutting off four congruent right triangles from the corners of the square. Let the length cut off from each corner be $x$. Then, the side of the octagon is $a - 2x$.\n\nThe triangles are isosceles right triangles, so $x$ is the distance from the corner to where the octagon meets the square. The octagon's side $s$ is related to $x$ by the geometry:\n\n$$\ns = a - 2x\n$$\n\nThe distance between two parallel sides of the octagon is $a - 2x$, and the length $x$ can be found using the fact that the angle at each vertex is $135^\text{o}$, so:\n\n$$\nx = s \\cdot \\frac{1}{1 + \\sqrt{2}}\n$$\n\nBut more simply, the side length of the octagon inscribed in a square of side $a$ is:\n\n$$\ns = a \\left( \\frac{\\sqrt{2} - 1}{\\sqrt{2}} \\right)\n$$\n\nOr, equivalently,\n\n$$\ns = a (\\sqrt{2} - 1)\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21391,
"subject": "Mathematics (Olympiad)",
"question": "Suppose there are 101 persons sitting around a round table in an arbitrary order. The $k$th person possesses $k$ pieces of cards, $k = 1, \\dots, 101$. We call it a _transition_ if one transits one of his cards to one of his adjacent persons. Find the minimum positive number $k$, such that whatever the order of the seating, there is a way of no more than $k$ transitions so that each person possesses 51 cards.",
"options": [],
"answer": "See solution",
"solution": "The answer is $k = 42\\ 925$.\n\nLet the circumference of the table be 101, and the distance between two adjacent persons be 1. We consider the least number of transitions.\n\nDenote the person who initially possesses $i$ cards by $[i - 51]$. If $p > 0$, person $[p]$ is a source who should send out $p$ cards; if $p < 0$, person $[p]$ is a sink who should receive $-p$ cards.\n\nSuppose that at the end of transitions, each person has 51 cards. If person B possesses a card $u$ initially belonging to person A, we can think that A carries card $u$ to B by passing through the minor arc AB; the length of the route is the arc length $|AB|$, which means the number of transitions.\n\nLet the seating order be as shown in the figure. Person $[i]$ transits all his $i$ cards to person $[-i]$ ($i = 1, 2, \\dots, 50$) with route length $i$. So there are altogether $1^2 + 2^2 + \\dots + 50^2 = 42\\ 925$ transitions. We will show that no fewer transitions can meet the requirement if persons are seated in this way.\n\nWe use the notion of \"potential.\" Let the potential at the highest position $[50]$ be 50; the neighbor positions $[49]$ and $[48]$ each have potential 49, ..., the lowest positions $[-49]$ and $[-50]$ each have potential 0. Then, the total potential at the\n\n\n\nbeginning is\n\n$$\nS = 101 \\times 50 + (100 + 99) \\times 49 + \\cdots + (2 + 1) \\times 0.\n$$\n\nAt the end of transitions, the total potential is\n\n$$\nT = 51 \\times 50 + (51 + 51) \\times 49 + \\cdots + (51 + 51) \\times 0.\n$$\n\nThe difference is $S - T = 42,925$.\n\nSince after each transition, the total potential changes at most 1, at least 42,925 transitions are needed.\n\nNow, we show that whatever the order of seating, there is always a way of no more than 42,925 transitions such that each person possesses 51 cards. To show this, we give two lemmas.\n\n**Lemma 1.** Let $c, a_0, a_1, \\dots, a_{n-1}$ be integers with their sum zero, and $c \\ge 0$, $a_0 \\le a_1 \\le \\dots \\le a_{n-1}$.\n\nIf $n+1$ persons denoted by $[c], [a_0], [a_1], \\dots, [a_{n-1}]$ possess $N+c, N+a_0, N+a_1, \\dots, N+a_{n-1}$ cards, respectively, where $N$ is a positive integer, such that $N+a_0 > 0$. Let the persons stand on 0, 1, ..., $n$ of the number axis, such that $[c]$ stands at $n$. Then there is a way of no more than $cn + \\sum_{i=0}^{n-1} i a_i$ transitions, such that each person possesses $N$ cards.\n\n**Proof of Lemma 1.** Suppose that $a_{n-1} \\ge \\cdots \\ge a_s > 0 \\ge a_{s-1} \\ge \\cdots \\ge a_0$, then induction on $M = a_{n-1} + \\cdots + a_s$. If $M=0$, then $[c]$ passes $c$ cards to $[a_i]$, such that $[a_i]$ obtains $-a_i$ cards ($0 \\le i \\le s-1$). Suppose that person $[a_i]$ stands at $x_i$ ($0 \\le i \\le n-1$), then $x_0, x_1, \\dots, x_{n-1}$ is a permutation of 0, 1, ..., $n-1$. Thus, a card that passes from $[c]$ to $[a_i]$ needs $n-x_i$ transitions. So the total transitions needed are ...",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21392,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a quadrilateral with $\\angle A + \\angle C = 60^\\circ$ and $AB \\cdot CD = BC \\cdot AD$. Prove that $AB \\cdot CD = AC \\cdot BD$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let us construct the equilateral triangle $BCE$ in the half-plane determined by line $BC$ and point $D$.\n\nThen $\\frac{AB}{AD} = \\frac{BC}{CD} = \\frac{CE}{CD}$ and $\\angle DAB = \\angle DCE$, so $\\Delta DAB \\sim \\Delta DCE$.\n\nTherefore $\\frac{AD}{DC} = \\frac{DB}{DE}$ and $\\angle ADB = \\angle CDE$, hence $\\angle ADC = \\angle BDE$.\n\nIt follows that $\\Delta ADC \\sim \\Delta BDE$, so $\\frac{AD}{BD} = \\frac{AC}{BE} = \\frac{AC}{BC}$, which leads to $AC \\cdot BD = BC \\cdot AD = AB \\cdot CD$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21393,
"subject": "Mathematics (Olympiad)",
"question": "$n \\ge 4$ points in the plane are given such that every three of them are not collinear. Prove that there exists a triangle such that all the points are in its interior, and on each of its sides lies exactly one point of the given points.",
"options": [],
"answer": "See solution",
"solution": "Since the given points are finite, there exists a disk containing all of them in its interior.\n\nLet $R$ be the maximal distance from the origin to any of the points. Then the disk centered at the origin with radius $2R$ contains all the points.\n\nDraw all possible lines between the $n$ points; there are finitely many ($\\binom{n}{2}$). Choose a point $A$ outside the disk and not lying on any of these lines, so that the largest angle under which any segment in the disk is seen from $A$ is acute. Draw an arbitrary line through $A$ that does not intersect the disk. Rotate this line until it passes through one of the given points, say $A_1$; this line will be one side of the triangle. Continue rotating until the line passes through another point, say $A_n$; this will be another side. Each of these lines passes only through $A_1$ and $A_n$ respectively, since $A$ does not lie on any of the $\\binom{n}{2}$ lines.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21394,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{Q}^+ \\to \\mathbb{R}^+$ such that\n\n$$\nf(xy) = f(x + y)(f(x) + f(y)), \\text{ for any } x, y \\in \\mathbb{Q}^+.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $g(x) = 1 / f(x)$, so $g : \\mathbb{Q}^+ \\to \\mathbb{R}^+$. The given equation becomes\n\n$$\ng(x + y) g(x) g(y) = g(xy)(g(x) + g(y)).\n$$\n\nLet $f(1) = c > 0$, i.e., $g(1) = 1 / c$. From the equation, $g(x + 1) = c g(x) + 1$. Thus:\n\n- $g(2) = 2$\n- $g(3) = 2c + 1$\n- $g(4) = 2c^2 + c + 1$\n- $g(5) = 2c^3 + c^2 + c + 1$\n- $g(6) = 2c^4 + c^3 + c^2 + c + 1$\n\nSetting $x = 2$, $y = 3$ gives:\n\n$$\ng(5) g(2) g(3) = g(6)(g(2) + g(3)),\n$$\nwhich leads to\n$$\n4c^5 - 3c^3 - c^2 - c + 1 = 0 \\iff (c - 1)(c + 1)(2c - 1)(2c^2 + 2c + 1) = 0.\n$$\nSo $c = 1$ or $c = \\frac{1}{2}$.\n\n**Case 1:** $c = 1$\n\nThen $g(x + 1) = g(x) + 1$. By induction, $g(n) = n$ for $n \\in \\mathbb{N}$, and $g(x + n) = g(x) + n$ for $x \\in \\mathbb{Q}^+$, $n \\in \\mathbb{N}$. Setting $y = n$ in the main equation gives $g(nx) = n g(x)$. For $x = p/q$, $n = q$, $g(x) = x$ for all $x \\in \\mathbb{Q}^+$, so $f(x) = 1 / x$.\n\n**Case 2:** $c = \\frac{1}{2}$\n\nThen $g(x + 1) = \\frac{1}{2} g(x) + 1$. This leads to $g(n) = 2$ and $g(x + n) - 2 = \\frac{g(x) - 2}{2^n}$ for $x \\in \\mathbb{Q}^+$, $n \\in \\mathbb{N}$. Setting $y = n$ in the main equation gives $2 g(x + n) g(x) = g(nx)(g(x) + 2)$. These equations imply $g(x) = 2$ for all $x$, so $f(x) = 1 / 2$.\n\n**Conclusion:**\n\nThe only functions satisfying the given equality are $f(x) \\equiv \\frac{1}{2}$ and $f(x) = \\frac{1}{x}$ for $x \\in \\mathbb{Q}^+$.\n\n*Remark: For $f : \\mathbb{R}^+ \\to \\mathbb{R}$, the solutions are $f \\equiv 0$, $f \\equiv \\frac{1}{2}$, and $f(x) = \\frac{1}{x}$.*",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21395,
"subject": "Mathematics (Olympiad)",
"question": "Sea $ABC$ un triángulo de incentro $I$, $M$ un punto sobre el lado $AB$ y $N$ un punto sobre el lado $AC$, y llamamos $x = \\frac{BM}{MA}$, $y = \\frac{CN}{NA}$. Sean $a$, $b$ y $c$ las longitudes de los lados de $ABC$. Demostrar que la recta $MN$ pasa por $I$ si y solamente si $bx + cy = a$.",
"options": [],
"answer": "See solution",
"solution": "Con las notaciones del problema tenemos\n\n$$\nAM = \\frac{c}{1+x}, \\quad AN = \\frac{b}{1+y}\n$$\n\nLa condición de que $M$, $I$, $N$ sean colineales es equivalente a la igualdad de áreas siguiente:\n\n$$\n[AMN] = [AMI] + [AIN]\n$$\n\nEsta igualdad se escribe sucesivamente como\n\n$$\n\\frac{1}{2} \\cdot AM \\cdot AI \\cdot \\sin\\left(\\frac{A}{2}\\right) + \\frac{1}{2} \\cdot AN \\cdot AI \\cdot \\sin\\left(\\frac{A}{2}\\right) = \\frac{1}{2} \\cdot AM \\cdot AN \\cdot \\sin A\n$$\nque puede ponerse como\n\n$$\nAI \\cdot \\sin\\left(\\frac{A}{2}\\right) \\cdot \\left(\\frac{c}{1+x} + \\frac{b}{1+y}\\right) = \\frac{bc}{(1+x)(1+y)} \\cdot \\sin A\n$$\n\ny llamando $S$ al área del triángulo,\n\n$$\nbc \\cdot \\sin A = 2S, \\quad AI \\cdot \\sin\\left(\\frac{A}{2}\\right) = r = \\frac{S}{p}\n$$\n\nllegamos a que la igualdad inicial de áreas es equivalente a\n\n$$\n\\frac{S}{p} \\left( \\frac{c}{1+x} + \\frac{b}{1+y} \\right) = \\frac{2S}{(1+x)(1+y)}\n$$\n\n$$\n\\frac{c}{1+x} + \\frac{b}{1+y} = \\frac{2p}{(1+x)(1+y)} \\Leftrightarrow c(1+y) + b(1+x) = a + b + c\n$$\n\nes decir\n\n$$\nbx + cy = a\n$$\n\ncomo queríamos demostrar.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21396,
"subject": "Mathematics (Olympiad)",
"question": "Let $x \\le y \\le z$ be real numbers such that $xy + yz + zx = 1$. Prove that\n$$\nxz < \\frac{1}{2}.\n$$\n\nIs it possible to improve the value of the constant $\\frac{1}{2}$?",
"options": [],
"answer": "See solution",
"solution": "From $x \\le y \\le z$, it follows that $(y - x)(y - z) \\le 0$, and consequently:\n$$\n\\begin{aligned}\ny^2 - xy - yz + xz &\\le 0 \\\\\ny^2 - (xy + yz + xz) + 2xz &\\le 0 \\\\\ny^2 - 1 + 2xz &\\le 0\n\\end{aligned}\n$$\nThus, $1 - 2xz \\ge y^2 \\ge 0$, i.e., $1 - 2xz \\ge 0$, and therefore $xz \\le \\frac{1}{2}$. If $xz = \\frac{1}{2}$, then $0 \\ge y^2 \\ge 0$, i.e., $y = 0$, and from $xy + yz + xz = 1$ it follows that $xz = 1$, a contradiction. So $xz < \\frac{1}{2}$.\n\nFor the second part, choose $x = y = \\frac{1}{n}$, where $n$ is a positive integer. Then, using $xy + yz + xz = 1$, we get $z = \\frac{1}{2}\\left(n - \\frac{1}{n}\\right)$. It is easy to check that the inequalities $x \\le y \\le z$ are verified for all $n \\ge 2$. On the other hand, $xz = \\frac{n^2 - 1}{2n^2} = \\frac{1}{2} - \\frac{1}{2n^2}$, which can be made as close to $\\frac{1}{2}$ as we want by taking $n$ sufficiently large. Thus, the value of the constant $\\frac{1}{2}$ cannot be improved.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21397,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{Q}^+ \\to \\mathbb{R}^+$ such that\n\n$$\nf(xy) = f(x + y)(f(x) + f(y)), \\text{ for any } x, y \\in \\mathbb{Q}^{+}.\n$$",
"options": [],
"answer": "See solution",
"solution": "If $g(x) = 1/f(x)$ then $g : \\mathbb{Q}^+ \\to \\mathbb{R}^+$ and the given equation becomes\n\n$$\ng(x+y)g(x)g(y) = g(xy)(g(x) + g(y)).\n$$\n\nLet $f(1) = c > 0$ i.e. $g(1) = \\frac{1}{c}$. From (*) we get $g(x+1) = c g(x) + 1$ and hence $g(2) = 2$, $g(3) = 2c+1$, $g(4) = 2c^2 + c + 1$, $g(5) = 2c^3 + c^2 + c + 1$, $g(6) = 2c^4 + c^3 + c^2 + c + 1$. On the other hand, setting $x=2$, $y=3$ in (*) leads us to\n\n$$\ng(5)g(2)g(3) = g(6)(g(2) + g(3)),\n$$\n\nwhich implies\n\n$$\n4c^5 - 3c^3 - c^2 - c + 1 = 0 \\Leftrightarrow (c-1)(c+1)(2c-1)(2c^2+2c+1) = 0.\n$$\n\nTherefore $c=1$ or $\\frac{1}{2}$.\n\nIf $c=1$ then $g(x+1) = g(x)+1$. By induction $g(n) = n$ for any $n \\in \\mathbb{N}$ and moreover $g(x+n) = g(x)+n$ for any $x \\in \\mathbb{Q}^+$ and $n \\in \\mathbb{N}$.\n\nBy setting $y=n$ in (*) we obtain\n\n$$\n(g(x) + n)g(x)n = g(nx)(g(x) + n),\n$$\n\ni.e. $g(nx) = n g(x)$. Setting $x=p/q$ and $n=q$, where $p, q \\in \\mathbb{N}$, we obtain $g(x) = x$ for any $x \\in \\mathbb{Q}^+$, i.e. $f(x) = 1/x$ for any $x \\in \\mathbb{Q}^+$.\n\nIf $c = \\frac{1}{2}$ then $g(x+1) = \\frac{1}{2}g(x) + 1$. Hence\n\n$$\ng(n) = 2 \\quad \\text{and} \\quad g(x+n) - 2 = \\frac{g(x) - 2}{2^n}, \\quad x \\in \\mathbb{Q}^{+}, n \\in \\mathbb{N};\n$$\n\nSetting $y = n$ in $(*)$ we obtain\n\n$$\n2g(x+n)g(x) = g(nx)(g(x) + 2),\n$$\n\nand it's sufficient to see that these equations imply $g(x) = 2$ for any $x \\in \\mathbb{Q}^{+}$, i.e. $f \\equiv 1/2$.\n\nFinally, the only functions satisfying the given equality are $f \\equiv \\frac{1}{2}$ and $f \\equiv \\frac{1}{x}$.\n\n_Remark: In fact all functions $f: \\mathbb{R}^{+} \\to \\mathbb{R}$ satisfying the given equality are $f \\equiv 0$, $f \\equiv \\frac{1}{2}$ and $f \\equiv \\frac{1}{x}$._",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21398,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c$ be real numbers such that $a + b + c = 3$. Prove 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": "Substituting $c = 3 - a - b$, the problem becomes to show\n\n$$\n\\frac{a+b}{5-a-b} + \\frac{3-a}{a+2} + \\frac{3-b}{b+2} \\geq 2.\n$$\n\nCross-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 non-negative, and zero if and only if $a = b = 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21399,
"subject": "Mathematics (Olympiad)",
"question": "Let $A_0B_0C_0$ be a triangle. For a positive integer $n \\ge 1$, define $A_n$ on the segment $B_{n-1}C_{n-1}$ such that $B_{n-1}A_n : C_{n-1}A_n = 2 : 1$, and define $B_n$ and $C_n$ cyclically in a similar manner. Show that there exists a unique point $P$ that lies in the interior of all triangles $A_nB_nC_n$.",
"options": [],
"answer": "See solution",
"solution": "We have nested compact sets (closed triangles), so their intersection is non-empty. We prove that they intersect in only one point. It's enough to show that the sequences $A_n$, $B_n$, $C_n$ converge to a common point $P$.\n\nAssume the contrary. Then there exist subsequences of $A_n$, $B_n$, $C_n$ (denoted again by $A_n$, $B_n$, $C_n$) that converge to points $A$, $B$, $C$ respectively, with $\\{A, B, C\\}$ containing at least two distinct points. Assume first that $A$, $B$, $C$ are distinct, and without loss of generality that $\\angle BAC \\le 60^\\circ$. For large $n$, $A_n$, $B_n$, $C_n$ are close to $A$, $B$, $C$ respectively. Consider the next triangle $A_{n+1}B_{n+1}C_{n+1}$. Its side $B_{n+1}C_{n+1}$ is far from $A$, so $A$ is outside $\\triangle A_{n+1}B_{n+1}C_{n+1}$, contradicting that $A$ is a limit point of the sequence $A_n$. \n\nSuppose now $B = C \\ne A$. Then $\\angle B_nA_nC_n < 60^\\circ$ (it tends to $0$), and the same argument applies. Thus, there is a unique point $P$ common to all the triangles.\n\nIt remains to show that $P$ is in the interior of all the triangles. Assume, on the contrary, that $P$ is on some side, say $A_nB_n$, for some $n$. Then $P$ is outside $\\triangle A_{n+2}B_{n+2}C_{n+2}$, a contradiction. $\\square$\n\nRemark: The proof remains valid if we only require that $A_{n+1}$, $B_{n+1}$, $C_{n+1}$ are on the sides $B_nC_n$, $A_nC_n$, and $A_nB_n$ respectively, but not too close to the vertices $A_n$, $B_n$, $C_n$; for example, the distance from $A_{n+1}$ to both $B_n$, $C_n$ is greater than $\\varepsilon \\cdot |B_nC_n|$ for some fixed $\\varepsilon > 0$, and similarly for $B_{n+1}$ and $C_{n+1}$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21400,
"subject": "Mathematics (Olympiad)",
"question": "Find the number of integers $n$ such that $20 \\leq n \\leq 49$ and $\\frac{1}{2}n$ is an integer.",
"options": [],
"answer": "See solution",
"solution": "From $n = 20$ to $n = 49$, there are $49 - 20 + 1 = 30$ integers. Half of these are even, so $\\frac{1}{2}n$ is an integer for $15$ values of $n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21401,
"subject": "Mathematics (Olympiad)",
"question": "Find all prime numbers $p$ for which one can find a positive integer $m$ and non-negative integers $a_0, a_1, \\dots, a_m$ less than $p$ such that\n\n$$\n\\begin{cases}\na_0 + a_1 p + \\dots + a_{m-1} p^{m-1} + a_m p^m = 2013, \\\\\na_0 + a_1 + \\dots + a_{m-1} + a_m = 11.\n\\end{cases}\n$$",
"options": [],
"answer": "See solution",
"solution": "Subtracting the second equation from the first gives\n\n$$\na_1(p-1) + \\dots + a_m(p^m - 1) = 2002.\n$$\n\nAs the left-hand side is divisible by $p-1$, $2002 = 2 \\cdot 7 \\cdot 11 \\cdot 13$ must also be divisible by $p-1$. Thus $p-1$ equals one of $1, 2, 7, 11, 13, 14, 22, 26, 77, 91, 143, 154, 182, 286, 1001,$ or $2002$. Since $p$ is prime, only $2, 3, 23,$ and $2003$ remain. The first equation is the $p$-ary representation of $2013$, so the coefficients $a_i$ are uniquely determined by $p$.\n\nNow we study all cases:\n\n1. If $p=2$, then $m=10$ as $2^{10} < 2013 < 2^{11}$. The second equation implies all $a_i$ must be ones, but $1+2+2^2+\\dots+2^{10} = 2^{11}-1 = 2047$. Hence, there is no solution in this case.\n\n2. Let $p=3$. As $2013 = 2 \\cdot 3 + 3^2 + 2 \\cdot 3^3 + 2 \\cdot 3^5 + 2 \\cdot 3^6$ whereas $2+1+2+2+2 = 9 \\neq 11$, this case gives no solution either.\n\n3. Let $p=23$. As $2013 = 12 + 18 \\cdot 23 + 3 \\cdot 23^2$ while $12 + 18 + 3 > 11$, this case gives no solution either.\n\n4. For $p=2003$, we get $2013 = 10 + 2003$ and $10 + 1 = 11$, so the conditions are satisfied.\n\nConsequently, $2003$ is the only prime number with the desired property.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21402,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c, d$ be distinct natural numbers such that $ab + cd$ is divisible by $ac + bd$. Prove that $ac + bd$ is a composite number.",
"options": [],
"answer": "See solution",
"solution": "Assume, for contradiction, that $ac + bd$ is prime. Since $ac + bd \\mid ab + cd$, it follows that $ac + bd \\mid (ab + cd + ac + bd) = (a + d)(b + c)$. Thus, $ac + bd$ divides $(a + d)$ or $(b + c)$. However, for distinct $a, b, c, d$, we have $ac + bd > a + d$ and $ac + bd > b + c$, which is impossible. This contradiction completes the proof.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21403,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\alpha$ be a real number larger than $1$. Define a sequence $s_n$ for $n \\geq 1$ as follows:\n\n- $s_1 = 1$\n- $s_2 = \\alpha$\n- If $s_1, s_2, \\dots, s_{2^n}$ are defined for some $n \\geq 1$, then $s_{2^n+1}, \\dots, s_{2^{n+1}}$ are defined by $s_j = \\alpha s_{j-2^n}$ for $2^n+1 \\leq j \\leq 2^{n+1}$.\n\n(Thus the first few terms are $1, \\alpha, \\alpha^2, \\alpha, \\alpha^2, \\alpha^3, \\dots$.)\n\nLet $c_n = s_1 + s_2 + \\dots + s_n$.\n\nIf $n = 2^{e_0} + 2^{e_1} + \\dots + 2^{e_k}$, where $e_0 > e_1 > \\dots > e_k \\geq 0$ is the binary representation of a positive integer $n$, prove that\n\n$$\nc_n = (1 + \\alpha)^{e_0} + \\alpha(1 + \\alpha)^{e_1} + \\alpha^2(1 + \\alpha)^{e_2} + \\dots + \\alpha^k(1 + \\alpha)^{e_k}.\n$$",
"options": [],
"answer": "See solution",
"solution": "We prove the formula by induction and properties of the sequence.\n\nLet $s(n)$ and $c(n)$ denote $s_n$ and $c_n$ respectively.\n\n**Step A:**\nIf $b(n)$ is the number of $1$'s in the binary representation of $n$, then for $n \\geq 1$,\n$$\ns(n) = \\alpha^{b(n-1)}.\n$$\n*Proof:* Induction shows $s(1) = 1 = \\alpha^{b(0)}$. For $2^{k} + 1 \\leq n \\leq 2^{k+1}$, $s(n) = \\alpha s(n - 2^{k}) = \\alpha^{b(n-2^{k}-1)+1} = \\alpha^{b(n-1)}$.\n\n**Step B:**\nFor $m \\geq 1$,\n$$\ns(2m+1) = s(m+1), \\quad s(2m) = \\alpha s(m).\n$$\n*Proof:* Follows from binary representation properties and induction.\n\n**Step C:**\nFor $m \\geq 1$,\n$$\nc(2m) = (1+\\alpha)c(m), \\quad c(2m+1) = \\alpha c(m) + c(m+1).\n$$\n*Proof:* Sum the sequence terms and use the recurrence relations above.\n\n**Step D:**\nLet $n = 2^{e_0} + 2^{e_1} + \\dots + 2^{e_k}$, with $e_0 > e_1 > \\dots > e_k \\geq 0$.\nWe prove by induction:\n$$\nc(n) = (1 + \\alpha)^{e_0} + \\alpha(1 + \\alpha)^{e_1} + \\alpha^2(1 + \\alpha)^{e_2} + \\dots + \\alpha^k(1 + \\alpha)^{e_k}.\n$$\nBase case $n=1$ holds. For $n=2m$ and $n=2m+1$, use the recurrence in Step C and the binary representation to show the formula holds for all $n$.\n\nThus, the result is proved by induction.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21404,
"subject": "Mathematics (Olympiad)",
"question": "A square of dimension $4 \\times 4$ is given, which consists of 16 squares of side 1. Non-negative integers are filled in each $1 \\times 1$ square, so that the sum of any five of them which can be covered with one of the figures in the picture (the figures can be translated and turned over) is 5. How many different numbers can be used to fill in the square?\n\n\n",
"options": [],
"answer": "See solution",
"solution": "For each rectangle of dimension $3 \\times 4$:\n\n\n\nit holds that $a + (e + f + g + h) = d + (e + f + g + h) = i + (e + f + g + h) = l + (e + f + g + h)$, i.e., $a = d = i = l$. Let the square be filled as in the picture:\n\n\n\nThen from the previous discussion it follows that $a = c = m = o$, $b = d = n = p$, $a = d = i = l$ and $e = h = m = p$. Therefore, $a = b = c = d = h = l = p = o = n = m = i = e = X$, the square is of the form:\n\n\n\nand $5X = 5$, i.e., $X = 1$.\n\nOn the other hand, $f + g + 3X = j + k + 3X = f + j + 3X = g + k + 3X = 5X$, i.e., $f + g = j + k = f + j = g + k = 2X$. From the first and fourth, and the second and fourth equation, respectively, it follows that $f = k = Y$ and $j = g = Z$. According to that, the square is of the form:\n\n\n\nand $Y + Z = 2$.\n\nThe following cases are possible:\n\n1. $Y = 0$, $Z = 2$\n2. $Y = 1$, $Z = 1$\n3. $Y = 2$, $Z = 0$\n\nTherefore, at most 3 different numbers can be used to fill in the square (case 1 or case 3).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21405,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $\\alpha, \\beta$ are two positive rational numbers. Assume that for some positive integers $m, n$, the number $\\alpha^{1/n} + \\beta^{1/m}$ is rational. Prove that each of $\\alpha^{1/n}$ and $\\beta^{1/m}$ is a rational number.",
"options": [],
"answer": "See solution",
"solution": "Let $l = \\mathrm{lcm}(m, n)$ and write $n = l/u$, $m = l/v$. Then\n\n$$\n\\alpha^{1/n} = (\\alpha^u)^{1/l}, \\quad \\beta^{1/m} = (\\beta^v)^{1/l}.\n$$\n\nNote that $\\alpha^u$ and $\\beta^v$ are also rationals. Thus, it is sufficient to consider the case $n = m$. Let us write $a = \\alpha^{1/n}$, $b = \\beta^{1/m}$, and $c = a + b$. We are given that $c$ is rational.\n\n**Lemma 1:** Let $a, b$ be real numbers such that $a + b = c$ is a rational number. Let $f(x)$ and $g(x)$ be the irreducible polynomials of $a$ and $b$ over $\\mathbb{Q}$. Then $\\deg(f(x)) = \\deg(g(x))$.\n\n*Proof of Lemma 1:* Let $p = \\deg(f(x))$ and $q = \\deg(g(x))$. Then\n\n$$\nf(x) = x^p + a_{p-1}x^{p-1} + \\dots + a_0, \\quad g(x) = x^q + b_{q-1}x^{q-1} + \\dots + b_0\n$$\n\nand these are polynomials with rational coefficients. Since $f(a) = 0$, we have $f(c-b) = 0$. Thus $b$ satisfies the polynomial $f(c-x)$ over $\\mathbb{Q}$. Since $g(x)$ is the irreducible polynomial of $b$ over $\\mathbb{Q}$, it follows that $g(x)$ divides $f(c-x)$. Hence $q \\leq p$. Similarly, we show that $p \\leq q$ and we obtain $p = q$.\n\nWe prove the result by induction on $n$. If $n = 1$, then $\\alpha, \\beta$ are rational by the given condition. If $n = 2$, then\n\n$$\n\\alpha^{1/2} - \\beta^{1/2} = \\frac{\\alpha - \\beta}{\\alpha^{1/2} - \\beta^{1/2}}\n$$\n\nwhich shows that $\\alpha^{1/2} - \\beta^{1/2}$ is also rational. Combined with the given condition that $\\alpha^{1/2} + \\beta^{1/2}$ is rational, it follows that each of $\\alpha^{1/2}, \\beta^{1/2}$ is rational.\n\nSuppose the result is true for $k = 0, 1, 2, \\dots, n-1$. Let $f(x)$ be the irreducible polynomial of $a = \\alpha^{1/n}$ over $\\mathbb{Q}$ and $g(x)$ be that of $b = \\beta^{1/n}$ over $\\mathbb{Q}$. Then Lemma 1 shows that $\\deg(f(x)) = \\deg(g(x)) = m$, say. But we know that $a$ is a root of $x^n - \\alpha = 0$, $b$ is a root of $x^n - \\beta = 0$ and $b = c - a$. Thus $(c-a)^n = b^n = \\beta$. Hence $(a-c)^n = (-1)^n \\beta$. This shows that $a$ is a root of $(x-c)^n - (-1)^n \\beta = 0$. This is a polynomial with rational coefficients. Thus $a$ is a root of\n\n$$\nh(x) = (x - c)^n - (-1)^n \\beta - (x^n - \\alpha).\n$$\n\nNow $\\deg(h(x)) = n-1$, so $\\deg(f(x)) \\leq n-1$. Thus we have $m \\leq n-1 < n$. Let $\\omega$ be a primitive $n$-th root of unity. Then\n\n$$\nx^n - \\alpha = \\prod_{j=0}^{n-1} (x - a\\omega^j).\n$$\n\nSince $f(x)$ is the irreducible polynomial of $a$ over $\\mathbb{Q}$, and $a$ is a root of $x^n - \\alpha$, we see that $f(x)$ divides $x^n - \\alpha$. Hence $x^n - \\alpha = f(x)q(x)$ for some rational polynomial $q(x)$. Now the factors of $f(x)$ are all of the form $(x - a\\omega^j)$. Hence the constant coefficient of $f(x)$, which is a rational number, must be of the form $\\pm a^m$, where $m = \\deg(f(x))$. Since $m < n$, we see that $a^m \\in \\mathbb{Q}$ for some $m < n$. Similarly, $b^m$ is also in $\\mathbb{Q}$. Since $m < n$, the induction hypothesis shows that $a, b$ are also in $\\mathbb{Q}$. This completes the proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21406,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $n$ such that there exist $k \\geq 2$ positive rational numbers $a_1, a_2, \\dots, a_k$ satisfying both $a_1 + a_2 + \\dots + a_k = a_1 \\cdot a_2 \\cdots a_k = n$.",
"options": [],
"answer": "See solution",
"solution": "The answer is $n = 4$ or $n \\geq 6$.\n\n**I.** First, we prove that each $n \\in \\{4, 6, 7, 8, 9, \\dots\\}$ satisfies the condition.\n\n1. If $n = 2k \\geq 4$ is even, set $(a_1, a_2, \\dots, a_k) = (k, 2, 1, \\dots, 1)$:\n$$\na_1 + a_2 + \\dots + a_k = k + 2 + 1 \\cdot (k-2) = 2k = n,\n$$\n$$\na_1 \\cdot a_2 \\cdot \\dots \\cdot a_k = k \\cdot 2 \\cdot 1^{k-2} = 2k = n.\n$$\n\n2. If $n = 2k + 3 \\geq 9$ is odd, set $(a_1, a_2, \\dots, a_k) = \\left(k + \\frac{3}{2}, \\frac{1}{2}, 4, 1, \\dots, 1\\right)$:\n$$\na_1 + a_2 + \\dots + a_k = k + \\frac{3}{2} + \\frac{1}{2} + 4 + (k-3) = 2k + 3 = n,\n$$\n$$\na_1 \\cdot a_2 \\cdot a_3 \\cdots a_k = \\left(k + \\frac{3}{2}\\right) \\cdot \\frac{1}{2} \\cdot 4 \\cdot 1^{k-3} = 2k + 3 = n.\n$$\n\n3. For $n = 7$, set $(a_1, a_2, a_3) = \\left(\\frac{4}{3}, \\frac{7}{6}, \\frac{9}{2}\\right)$:\n$$\na_1 + a_2 + a_3 = a_1 \\cdot a_2 \\cdot a_3 = 7 = n.\n$$\n\n**II.** Next, we prove by contradiction that $n \\in \\{1, 2, 3, 5\\}$ do not satisfy the condition.\n\nSuppose there exist $k \\geq 2$ positive rational numbers whose sum and product are both $n \\in \\{1, 2, 3, 5\\}$. By the Arithmetic-Geometric Mean inequality:\n$$\nn^{1/k} = \\sqrt[k]{a_1 \\cdot a_2 \\cdots a_k} \\leq \\frac{a_1 + a_2 + \\dots + a_k}{k} = \\frac{n}{k},\n$$\nwhich gives\n$$\nn \\geq k^{1 + \\frac{1}{k-1}}.\n$$\nFor $k = 3$, $n \\geq 3\\sqrt{3} \\approx 5.196 > 5$.\nFor $k = 4$, $n \\geq 4\\sqrt[3]{4} \\approx 6.349 > 5$.\nFor $k \\geq 5$, $n \\geq 5^{1 + \\frac{1}{k-1}} > 5$.\nThus, $n > 5$ for $k \\geq 3$.\n\nFor $k = 2$, $a_1 + a_2 = a_1 a_2 = n$ implies $n = \\frac{a_1^2}{a_1 - 1}$, so $a_1$ satisfies\n$$\na_1^2 - n a_1 + n = 0.\n$$\nFor $a_1$ rational, the discriminant $n^2 - 4n$ must be a perfect square, but for $n \\in \\{1, 2, 3, 5\\}$, this is not the case.\n\n**Note:** Only $n = 4$ can be represented both as the sum and product of the same two rational numbers. For $n \\geq 5$, $(n - 3)^2 < n^2 - 4n = (n - 2)^2 - 4 < (n - 2)^2$, and $n^2 - 4n < 0$ for $n = 1, 2, 3$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21407,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be real numbers such that $a + b + c$ and $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}$ satisfy $$(a + b + c) \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) = \\frac{27}{2}.$$ \nFind the value of\n$$(a + b + c) \\left( \\frac{1}{a + b - 5c} + \\frac{1}{b + c - 5a} + \\frac{1}{c + a - 5b} \\right).$$\n\nNote: The equality is achieved, for example, for $a = b = 4$, $c = 1$.",
"options": [],
"answer": "See solution",
"solution": "Set $a + b + c = x$ and $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = y$. Let $k = \\frac{x}{6}$, then\n$$\na + b - 5c = 6(k - c), \\quad b + c - 5a = 6(k - a), \\quad c + a - 5b = 6(k - b).\n$$\nIt follows that\n$$\n(a + b + c) \\left( \\frac{1}{a + b - 5c} + \\frac{1}{b + c - 5a} + \\frac{1}{c + a - 5b} \\right) = \\frac{x}{6} \\left( \\frac{1}{k - a} + \\frac{1}{k - b} + \\frac{1}{k - c} \\right). \\quad (1)\n$$\nAdding the fractions in the right-hand side of (1), we obtain $\\frac{A}{B}$, where\n$$\nA = (k - a)(k - b) + (k - b)(k - c) + (k - c)(k - a), \\quad B = (k - a)(k - b)(k - c). \\quad (2)\n$$\nLet $z = abc$, then $ab + bc + ca = yz$. From (2) it follows\n$$\nA = 3k^2 - 2zk + yz, \\quad B = k^3 - zk^2 + yz - z. \\quad (3)\n$$\nSince $x = 6k$ and $xy = \\frac{27}{2}$, we have $y = \\frac{9}{4k}$. Substitute for $x$ and $y$ in (3) to obtain\n$$\nA = 3k^2 - 2 \\cdot 6k \\cdot k + \\frac{9}{4k}z = 9 \\left(-k^2 + \\frac{z}{4k}\\right), \\quad B = k^3 - 6k \\cdot k^2 + \\frac{9}{4k}zk - z = 5k \\left(-k^2 + \\frac{z}{4k}\\right).\n$$\nThus\n$$\n\\frac{A}{B} = \\frac{9}{5k}, \\quad \\frac{x}{6} \\cdot \\frac{A}{B} = k \\cdot \\frac{9}{5k} = \\frac{9}{5}.\n$$\nHence,\n$$\n(a + b + c) \\left( \\frac{1}{a + b - 5c} + \\frac{1}{b + c - 5a} + \\frac{1}{c + a - 5b} \\right) = \\frac{9}{5}.\n$$\n**Answer:** $\\boxed{\\dfrac{9}{5}}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21408,
"subject": "Mathematics (Olympiad)",
"question": "Suppose six cities are connected by direct flights, each operated by one of two airlines (Air Michael and Air Patrick). Prove that, no matter how the flights are assigned, there will always be a set of four cities such that the flights between them are all operated by the same airline and form a round trip (i.e., a cycle of length 4).",
"options": [],
"answer": "See solution",
"solution": "Suppose Air Michael operates at least 8 of the 15 possible routes. Choose any four cities; there are $\\binom{4}{2} = 6$ routes between them. If, for all choices of four cities, the routes were divided 3-3 between the airlines, then both would have the same number of routes overall, which is impossible since 15 is odd. Thus, for some four cities, Air Michael must operate at least 4 of the 6 routes. If they operate five or six, a round trip exists. If they operate four and these form a round trip, we are done. Otherwise, suppose Air Michael operates PQ, PR, PS, and QR among cities P, Q, R, S. Air Michael must operate at least four more routes, at least three of which connect P, Q, R, S to the remaining two cities T and U. One of these, say T, must have at least two routes to P, Q, R, S. If there are three, a round trip occurs. The only two Air Michael routes that do not produce a round trip are TP and TS. But then QSRT forms a round trip operated by Air Patrick.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21409,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be an odd integer greater than or equal to $3$. You play a game on an $n \\times n$ grid of $n^2$ squares. The game consists of $n^2$ turns. On each turn:\n\n- Choose one empty square and insert a positive integer between $1$ and $n^2$ (each integer used only once).\n- You gain $1$ point if the sum of numbers in the chosen square's row is a multiple of $n$, and $1$ point if the column sum is a multiple of $n$ (so $2$ points if both).\n\nFind the maximum possible total points you can earn by the end of the game.",
"options": [],
"answer": "See solution",
"solution": "The points earned depend only on the remainder modulo $n$ of each inserted number. Each residue $k$ with $0 \\leq k \\leq n-1$ appears exactly $n$ times.\n\nWe claim the maximum total points is $n(n+1)$. To achieve this, insert into the square at row $i$, column $j$ the number congruent to $i + j \\pmod{n}$. Each residue appears $n$ times. For residues $0, n-1, n-2, \\ldots, \\frac{n+1}{2}$, both row and column sums are multiples of $n$, earning $2$ points each, totaling $n(n+1)$ points.\n\nTo show this is maximal, let $A_0 = B_0 = 0$. For each turn $i$, let $A_i$ increase by $1$ if the row sum is a multiple of $n$, otherwise unchanged; similarly for $B_i$ and columns. The total points is $A_{n^2} + B_{n^2}$.\n\nLet $C_i$ be the number of columns whose sum is not a multiple of $n$ after turn $i$, with $C_0 = 0$. Then $A_i + \\frac{1}{2}C_i$ increases by at most $1$ if $0$ is inserted, and at most $\\frac{1}{2}$ otherwise. Since $0$ is inserted $n$ times and other residues $n(n-1)$ times:\n\n$$\nA_{n^2} + \\frac{1}{2}C_{n^2} \\leq n + \\frac{n(n-1)}{2} = \\frac{n(n+1)}{2}\n$$\n\nSince $C_{n^2} \\geq 0$, $A_{n^2} \\leq \\frac{n(n+1)}{2}$, and similarly $B_{n^2} \\leq \\frac{n(n+1)}{2}$. Thus, the maximum total points is $n(n+1)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21410,
"subject": "Mathematics (Olympiad)",
"question": "The triangle $ABC$ has $\\angle BAC = 54^\\circ$ and $\\angle ACB = 45^\\circ$. Let $D$ and $E$ be points on the segments $BC$ and $AD$, respectively, so that $AD = AB$ and $BE = BD$. Denote $F$ as the common point of the lines $BE$ and $AC$, and let $DM$ be the bisector of the angle $\\angle ADF$, where $M \\in AC$. The perpendicular from $C$ to $AB$ meets $DM$ at $G$. Prove that:\n\na) The triangle $ABF$ is isosceles.\n\nb) $CG = CM$.",
"options": [],
"answer": "See solution",
"solution": "a) From the statement, $\\angle ABC = 81^\\circ$ and, in the isosceles triangle $ABD$, $\\angle ADB = \\angle ABD = 81^\\circ$, hence $\\angle DAB = 18^\\circ$ and $\\angle CAD = \\angle CAB - \\angle DAB = 36^\\circ$.\n\nThe isosceles triangle $BED$ yields $\\angle BED = \\angle BDE = 81^\\circ$, therefore $\\angle DBE = 18^\\circ$, whence $\\angle ABF = \\angle ABC - \\angle EBD = 63^\\circ$.\n\n\n\nFrom the triangle $ABF$, $\\angle ABF = 63^\\circ$ and $\\angle FAB = 54^\\circ$, hence $\\angle AFB = 63^\\circ$, that is, triangle $ABF$ is isosceles, with $AF = AB$.\n\nb) The relations $AD = AB$ and $AF = AB$ lead to $AF = AD$, so the triangle $AFD$ is isosceles, with $\\angle AFD = \\angle ADF = \\frac{180^\\circ - \\angle FAD}{2} = 72^\\circ$. Since $DM$ is the bisector of the angle $ADF$, $\\angle ADM = \\angle FDM = \\frac{\\angle ADF}{2} = 36^\\circ$. The triangle $AMD$ gives $\\angle MAD = \\angle ADM = 36^\\circ$, therefore $\\angle CMD = 72^\\circ$.\n\nFrom $CG \\perp AB$ it follows that $\\angle ACG = 90^\\circ - \\angle CAB = 36^\\circ$, so the triangle $CMG$ has $\\angle ACG = 36^\\circ$ and $\\angle CMG = 72^\\circ$. This gives $\\angle CGM = 72^\\circ$, that is, the triangle $CMG$ is isosceles, with $CG = CM$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21411,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle with circumcentre $O$ and centroid $G$. Let $M$ be the midpoint of $BC$ and $N$ be the reflection of $M$ across $O$. Prove that $NO = NA$ if and only if $\\angle AOG = 90^\\circ$.",
"options": [],
"answer": "See solution",
"solution": "Let $H$ be the orthocenter of $\\triangle ABC$ and let $X$ be the midpoint of $AH$. Then we know that $AXON$ is a parallelogram.\n\nNow, observe that $\\angle AOH = \\angle AOG$. Now,\n\n$$\nNO = NA \\iff XA = XO \\iff \\angle AOH = 90^\\circ\n$$\n\nThus, we are done.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21412,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle. Let $S$ be the circle through $B$ tangent to $CA$ at $A$, and let $T$ be the circle through $C$ tangent to $AB$ at $A$. The circles $S$ and $T$ intersect at $A$ and $D$. Let $E$ be the point where the line $AD$ meets the circle $ABC$. Prove that $D$ is the midpoint of $AE$.",
"options": [],
"answer": "See solution",
"solution": "Let $\\angle ABD = \\alpha$ and $\\angle DCA = \\beta$.\n\n\n\nThen $\\angle CAD = \\angle ABD = \\alpha$ and $\\angle DAB = \\angle DCA = \\beta$, by the alternate segment theorem. So $\\triangle ABD$ is similar to $\\triangle CAD$.\n\nAlso, $\\angle CBE = \\angle CAE = \\alpha$ and $\\angle ECB = \\angle EAB = \\beta$ by the theorem of angles in the same segment. Similarly, $\\angle BED = \\angle BEA = \\angle BCA$. Further, $\\angle DBE = \\angle CBE + \\angle DBC = \\angle ABD + \\angle DBC = \\angle ABC$. So $\\triangle DBE$ and $\\triangle ABC$ are similar.\n\nTherefore, $\\frac{AB}{BD} = \\frac{AC}{AD}$ and so $AD = \\frac{AC \\times BD}{AB}$. Also, $\\frac{DB}{DE} = \\frac{AB}{AC}$, and so $DE = \\frac{AC \\times BD}{AB} = AD$.\n\nSo $D$ is the midpoint of $AE$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21413,
"subject": "Mathematics (Olympiad)",
"question": "At a congress, all attendees are either mathematicians or biologists, and no one is both. The mathematicians all know each other, and each of them knows four of the biologists. The biologists also all know each other, and each of them knows nine of the mathematicians. It turns out that every mathematician knows twice as many people as every biologist. (If person A knows person B, then person B also knows person A.)\n\nHow many mathematicians are at the congress?",
"options": [],
"answer": "See solution",
"solution": "$70$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21414,
"subject": "Mathematics (Olympiad)",
"question": "令 $||x||_* = (|x| + |x - 1| - 1)/2$。請決定所有的函數 $f: \\mathbb{N} \\to \\mathbb{N}$ 使得\n\n$$\nf^{(||f(x)-x||_*)}(x) = x, \\quad \\forall x \\in \\mathbb{N}.\n$$\n\n其中 $f^{(0)}(x) = x$，$f^{(n)}(x) = f(f^{(n-1)}(x))$，$\\forall n \\in \\mathbb{N}$。",
"options": [],
"answer": "See solution",
"solution": "所有的解為 $f: \\mathbb{N} \\to \\mathbb{N}$，使得 $f(x) \\in \\{x, x+1\\}$，$\\forall x \\in \\mathbb{N}$。此時 $||f(x) - x||_* = 0$，因此滿足條件。\n\n現在證明這些是全部的解。假設存在 $x \\in \\mathbb{N}$ 使得 $f(x) \\notin \\{x, x+1\\}$，則 $||f(x) - x||_* \\neq 0$。設 $s$ 為最小的自然數使得 $f^s(x) = x$。由 $s$ 的最小性，$s \\mid ||f(x) - x||_*$。同理，$s$ 也是使得 $f^s(f(x)) = f(x)$ 的最小自然數，故 $s \\mid ||f^{(2)}(x) - f(x)||_*$。以此類推，$s \\mid ||f^{(i+1)}(x) - f^{(i)}(x)||_*$。令 $a_i = f^{(i+1)}(x) - f^{(i)}(x)$，則 $a_0 + a_1 + \\cdots + a_{s-1} = 0$。\n\n又 $||a_0||_* + \\cdots + ||a_{s-1}||_*$ 可被 $s$ 整除。由 $||x||_*$ 的定義，$||x||_* = x$ 當 $x \\le 0$，$||x||_* = x-1$ 當 $x \\ge 1$。由假設 $||a_0||_* \\neq 0$，知 $a_0 > 1$ 或 $a_0 < 0$。若 $a_0 > 1$，則必有某 $a_i < 0$；若 $a_0 < 0$，則必有某 $a_i > 0$。因此必有某 $a_i < 0$ 及某 $a_j > 0$。\n\n設 $p$ 為 $a_i > 0$ 的個數，則 $0 < p < s$。因此\n\n$$\ns \\mid ||a_0||_* + \\cdots + ||a_{s-1}||_* = a_0 + a_1 + \\cdots + a_{s-1} - p = -p,\n$$\n\n矛盾。因此必須 $f(x) \\in \\{x, x+1\\}$，$\\forall x \\in \\mathbb{N}$。\n\n*參考評分標準：*\n- 猜測並檢驗出 $f(x) \\in \\{x, x+1\\}$，得 1 分。\n- 注意到當 $f(x) \\notin \\{x, x+1\\}$ 時，$||f(x) - x||_* \\neq 0$，得 1 分。若有 minimum $s$ 的想法，再得 1 分。\n- 嚴格導出矛盾，再得 4 分（累積 6 分）。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21415,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an arbitrary triangle, and let $M$, $N$, $P$ be any three points on the sides $BC$, $CA$, $AB$ such that the lines $AM$, $BN$, $CP$ concur. Let the parallel to the line $AB$ through the point $N$ meet the line $MP$ at a point $E$, and let the parallel to the line $AB$ through the point $M$ meet the line $NP$ at a point $F$. Then, the lines $CP$, $MN$, and $EF$ are concurrent.",
"options": [],
"answer": "See solution",
"solution": "We exclude the trivial case $MN \\parallel AB$, where $M \\equiv E$ and $N \\equiv F$.\n\nLet $R$ be the intersection of $NE$ and $BC$. Suppose $E \\in (NR)$, then $N \\in (PF)$. If $R \\in (EN)$, we have $F \\in (PN)$; this case is similar.\n\nLet $O$ be the intersection of $AM$, $BN$, and $CP$; $Q$ the intersection of $MN$ and $EF$; and $S$ the intersection of $CP$ and $MN$. We prove $Q$ coincides with $S$.\n\nApplying Menelaus' Theorem in triangle $BMN$ with transversal $O - S - C$, and in triangle $ABN$ with transversal $P - O - C$, and using Ceva's Theorem in triangle $ABC$, we obtain:\n\n$$\n\\begin{array}{l@{\\hspace{2em}}r@{\\,}l}\n\\displaystyle \\frac{SN}{SM} \\cdot \\frac{OB}{ON} \\cdot \\frac{CM}{CB} = 1, & \\multicolumn{2}{c}{\\left|} \\\\\n\\displaystyle \\frac{ON}{OB} \\cdot \\frac{CA}{CN} \\cdot \\frac{PB}{PA} = 1, & \\multicolumn{2}{c}{\\Rightarrow \\frac{SN}{SM} \\cdot \\frac{MB}{CB} \\cdot \\frac{CA}{AN} = 1, \\text{ hence } \\frac{SN}{SM} = \\frac{BC}{BM} \\cdot \\frac{AN}{AC}.} \\\\\n\\displaystyle \\frac{PA}{PB} \\cdot \\frac{MB}{MC} \\cdot \\frac{NC}{NA} = 1, & \\multicolumn{2}{c}{\\right|}\n\\end{array}\n$$\n\nBecause of the similarity between triangles $NEQ$ and $MFQ$, and also between $PEN$ and $PMF$, by Thales' Theorem, we obtain:\n\n$$\n\\frac{QN}{QM} = \\frac{NE}{MF} = \\frac{PE}{PM} = \\frac{BR}{BM} = \\frac{BR}{BC} \\cdot \\frac{BC}{BM} = \\frac{AN}{AC} \\cdot \\frac{BC}{BM} = \\frac{SN}{SM}.\n$$\n\nAs $Q, S$ both lie on $MN$, it follows that $Q$ and $S$ coincide. Hence, the lines $MN$, $EF$, and $CP$ are concurrent.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21416,
"subject": "Mathematics (Olympiad)",
"question": "Suppose a board contains $2016^2 + 1$ ladybirds, each occupying a square and moving according to some path. Prove that, regardless of the initial arrangement and the ladybirds' paths, a collision (two ladybirds occupying the same square at the same time) must occur.\n\nWe label the squares with one of four labels, $A$, $B$, $C$, and $D$, so that the squares in odd rows alternate between $A$ and $B$, while the squares in even rows alternate between $C$ and $D$. A square labeled $A$ is called an $A$-square.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let us analyze the movement and distribution of ladybirds:\n\n- Any ladybird on a $B$-square or $C$-square will, after one second, move to an $A$-square or $D$-square.\n- Any ladybird on an $A$-square will, after two seconds, be on a $D$-square.\n\nSince there are $2016^2 + 1$ ladybirds, at least $1008 \\cdot 2016 + 1$ of them must occupy $A$- or $D$-squares (otherwise, after one second, the same would be true for $B$- and $C$-squares). There are $1008^2$ $D$-squares, so at least $1008^2 + 1$ ladybirds must be on $A$-squares.\n\nAfter two seconds, all ladybirds from $A$-squares move to $D$-squares. Thus, at least $1008^2 + 1$ ladybirds occupy $1008^2$ $D$-squares, so by the pigeonhole principle, at least two ladybirds must occupy the same square at the same time—a collision.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21417,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$ be positive integers.\n\nProve that it is impossible for all three numbers $a^2 + b + c$, $b^2 + c + a$, and $c^2 + a + b$ to be perfect squares.",
"options": [],
"answer": "See solution",
"solution": "*Alternative 1* (Declan Gorey)\n\nSince the question is completely symmetric in $a$, $b$, $c$, we may assume without loss of generality that $a \\ge b \\ge c$. Then,\n\n$$\na^2 < a^2 + b + c \\le a^2 + a + a < a^2 + 2a + 1 = (a + 1)^2.\n$$\n\nThus\n\n$$\na < \\sqrt{a^2 + b + c} < a + 1.\n$$\n\nTherefore, $\\sqrt{a^2 + b + c}$ cannot be an integer because it lies between two consecutive integers. Hence $a^2 + b + c$ cannot be a perfect square.\n\n*Alternative 2* (Patrick He)\n\nSince $a$, $b$, $c > 0$ we have $a^2 + b + c > a^2$. If $a^2 + b + c$ is a perfect square, then $a^2 + b + c \\ge (a+1)^2 = a^2 + 2a + 1$. Thus $b + c \\ge 2a + 1$.\n\nSimilarly, if $b^2 + c + a$ and $c^2 + a + b$ are also perfect squares, then $c + a \\ge 2b + 1$ and $a + b \\ge 2c + 1$. Adding the three inequalities gives\n\n$$\n2a + 2b + 2c \\ge 2a + 2b + 2c + 3,\n$$\n\nwhich is a contradiction. Thus the three given expressions cannot all be perfect squares.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21418,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge 2$ be a natural number. For each pair $a, b$ of relatively prime natural numbers, let $d_{a,b}$ be the greatest common divisor of $na + b$ and $a + nb$. Find the maximum value of $d_{a,b}$.",
"options": [],
"answer": "See solution",
"solution": "The maximum value of $d_{a,b}$ equals $n^2 - 1$.\n\nLet $a$ and $b$ be relatively prime. Since $d_{a,b}$ divides $na + b$ and $a + nb$, it also divides the numbers $u = (n + 1)(a + b) = (na + b) + (a + nb)$ and $v = (n - 1)(a - b) = (na + b) - (a + nb)$. Hence $d_{a,b}$ divides $(n - 1)u + (n + 1)v = 2(n^2 - 1)a$ and $(n - 1)u - (n + 1)v = 2(n^2 - 1)b$. Therefore $d_{a,b}$ divides the greatest common divisor of $2(n^2 - 1)a$ and $2(n^2 - 1)b$, which equals $2(n^2 - 1)$, because $a$ and $b$ are relatively prime.\n\nNow we show that $d_{a,b} = 2(n^2 - 1)$ is impossible. Otherwise, $na + b = 2(n^2 - 1)k$, $a + nb = 2(n^2 - 1)l$, where $k$ and $l$ are relatively prime. These equalities form a linear system with unknowns $a$ and $b$ whose unique solution is $a = 2(nk - l)$, $b = 2(nl - k)$. However, the obtained values of $a$ and $b$ are even, so they are not relatively prime.\n\nIn conclusion, $d_{a,b}$ is a proper divisor of $2(n^2 - 1)$, hence $d_{a,b} \\le n^2 - 1$. To see that $d_{a,b} = n^2 - 1$ is attainable, set $a = n(n - 1) - 1$, $b = 1$. Then $na + b = (n - 1)(n^2 - 1)$, $a + nb = n^2 - 1$. So $d_{a,b} = n^2 - 1$, as $n - 1$ and $1$ are relatively prime. Thus $n^2 - 1$ is the maximum value of $d_{a,b}$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21419,
"subject": "Mathematics (Olympiad)",
"question": "As shown in Fig. 7.1, in acute $\\triangle ABC$, $AB > AC$, and $O$ is the circumcentre. Let $K$ be the symmetric point of $B$ with respect to $AC$, and $L$ be the symmetric point of $C$ with respect to $AB$. Let $X$ be a point inside $\\triangle ABC$, such that $AX \\perp BC$, $XK = XL$. $Y$, $Z$ are points on the segments $BK$, $CL$, respectively, such that $XY \\perp CK$, $XZ \\perp BL$. Prove that $B, C, Y, O$, and $Z$ all lie on a circle.\n\n\n\nFig. 7.1",
"options": [],
"answer": "See solution",
"solution": "Let the lines $BL$, $CK$ meet at $P$; $AD \\perp BC$ with foot $D$, as shown in Fig. 7.2.\n\nAssume that $T$ is the circumcentre of $\\triangle PKL$, so $TL = TP = TK$. In the isosceles triangle $TLP$, by Stewart's Theorem, we have $TB^2 = TP^2 - BL \\cdot BP$. Similarly, $TC^2 = TP^2 - CK \\cdot CP$.\n\nIt is known that $BL = BC = CK$, and hence,\n\n$$\nTB^2 - TC^2 = (TP^2 - BL \\cdot BP) - (TP^2 - CK \\cdot CP) = BC(CP - BP).\n$$\n\nNotice that $A$ is the excentre of $\\triangle PBC$ relative to the vertex $P$, and $D$ is the point of tangency, hence\n\n$$\nCP - BP = BD - CD,\n$$\n\nwhich implies $TB^2 - TC^2 = (BD + CD)(BD - CD) = BD^2 - CD^2$ and further $XD \\perp BC$.\n\nMoreover, $TK = TL$, and $X$, $T$ both lie on the perpendicular bisectors of $AD$ and $KL$. Since $AB > AC$, $KL$ and $BC$ are not parallel; $AD$ and the perpendicular bisector of $KL$ are not parallel either, and they have a unique intersection. It follows that $T = X$, and $X$ is the circumcentre of $\\triangle PKL$.\n\nSince $XY \\perp CK$, or $XY \\perp PK$, we infer that $Y$ is on the perpendicular bisector of $PK$, and $YP = YK$. Together with $BC = CK$, we find $\\angle CPY = \\angle CKY = \\angle CBY$, and $B, Y, C, P$ are concyclic. Similarly, $B, Z, C, P$ are concyclic. Finally, since\n\n$$\n\\begin{align*}\n\\angle BPC &= \\angle CBL + \\angle BCK - 180^\\circ \\\\\n&= 2(\\angle ABC + \\angle ACB) - 180^\\circ \\\\\n&= 180^\\circ - 2\\angle BAC \\\\\n&= 180^\\circ - \\angle BOC,\n\\end{align*}\n$$\n\n$B, P, O, C$ are concyclic.\n\n\n\nFig. 7.2\n\nAs we have shown that $B, P, C, Y, O$, and $Z$ are concyclic, the conclusion follows immediately.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21420,
"subject": "Mathematics (Olympiad)",
"question": "Let $p$ be an odd prime number, and $a, b, m, r$ be positive integers such that $p \\nmid ab$ and $ab > m^2$. Prove that there exists at most one pair of positive integers $(x, y)$ satisfying the following conditions:\n\n- $x$ and $y$ are coprime,\n- $ax^2 + by^2 = mp^r$.",
"options": [],
"answer": "See solution",
"solution": "Suppose, for contradiction, that there exist two different pairs of positive integer solutions $(x_1, y_1)$ and $(x_2, y_2)$. Since $x_1$ and $y_1$ are coprime, $p \\nmid x_1y_1$. Similarly, $p \\nmid x_2y_2$.\n\nGiven\n$$\nax_1^2 \\equiv -by_1^2 \\pmod{p^r}, \\quad ax_2^2 \\equiv -by_2^2 \\pmod{p^r},\n$$\nit follows that $abx_1^2y_2^2 \\equiv abx_2^2y_1^2 \\pmod{p^r}$. Since $p \\nmid ab$, we have $p^r \\mid (x_1^2y_2^2 - x_2^2y_1^2)$.\n\nNote that $x_1y_2 - x_2y_1$ and $x_1y_2 + x_2y_1$ cannot both be divisible by $p$, otherwise $p \\mid 2x_1y_2$, which contradicts $p$ being odd and $p \\nmid x_1y_1x_2y_2$. Hence, $p^r \\mid x_1y_2 - x_2y_1$ or $p^r \\mid x_1y_2 + x_2y_1$.\n\nIf $x_1y_2 - x_2y_1 = 0$, then $\\frac{x_1}{x_2} = \\frac{y_1}{y_2}$, and since $ax_1^2 + by_1^2 = ax_2^2 + by_2^2$, this implies $x_1 = x_2$, $y_1 = y_2$, contradicting $(x_1, y_1) \\neq (x_2, y_2)$. Therefore, $x_1y_2 - x_2y_1 \\neq 0$.\n\nIf $p^r \\mid x_1y_2 + x_2y_1$, then $x_1y_2 + x_2y_1 \\ge p^r$.\n\nIf $p^r \\mid x_1y_2 - x_2y_1$, then $x_1y_2 + x_2y_1 \\ge |x_1y_2 - x_2y_1| \\ge p^r$. Thus, in all cases,\n$$\nx_1y_2 + x_2y_1 \\ge p^r. \\quad (*)\n$$\n\nUsing $ab > m^2$ and $(*)$, we get\n$$\n\\begin{aligned}\nm^2p^{2r} &= (ax_1^2 + by_1^2)(ax_2^2 + by_2^2) \\\\\n&= (ax_1x_2 - by_1y_2)^2 + ab(x_1y_2 + x_2y_1)^2 \\\\\n&\\ge ab(x_1y_2 + x_2y_1)^2 \\\\\n&> m^2p^{2r},\n\\end{aligned}\n$$\na contradiction. Therefore, the assumption is false, and the original statement holds. $\\Box$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21421,
"subject": "Mathematics (Olympiad)",
"question": "Consider two nonnegative integers $n$ and $k$ such that $n \\ge 2$ and $1 \\le k \\le n-1$. Suppose that a matrix $A \\in \\mathcal{M}_n(\\mathbb{C})$ has exactly $k$ minors of order $n-1$ which are null. Prove that $\\det(A) \\ne 0$.",
"options": [],
"answer": "See solution",
"solution": "Suppose $\\det(A) = 0$. The number of minors of order $n-1$ of $A$ is $n^2$, and since $n^2 > n-1$, $A$ has at least one nonzero minor of order $n-1$, so $\\operatorname{rank}(A) = n-1$. As $A^*A = O_n$, by Sylvester's inequality, $$0 = \\operatorname{rank}(AA^*) \\ge \\operatorname{rank}(A) + \\operatorname{rank}(A^*) - n,$$ so $\\operatorname{rank}(A^*) \\le 1$. From $A^* \\ne O_n$, we get $\\operatorname{rank}(A^*) = 1$.\n\nBecause $A^*$ has at least $n^2 - n + 1$ nonzero elements, there is a row with all elements different from zero. Denote it by $L_1$ and let $L_2$ be the row in $A^*$ containing at least one zero element (the existence of such a row is ensured by $k \\ge 1$). Because $L_1$ and $L_2$ are dependent, there is $\\alpha \\in \\mathbb{C}$ such that $L_2 = \\alpha L_1$.\n\nWe infer $\\alpha = 0$, so $L_2$ is zero, which implies that $A$ has at least $n$ zero minors of order $n-1$, a contradiction.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 21422,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge 3$ be a positive integer and let\n$$\nS = \\{k \\in \\mathbb{N} : (k, n) = (k + 1, n),\\ 1 \\le k \\le n - 1\\}.\n$$\nFind the remainder of $\\prod_{k \\in S} k$ divided by $n$.",
"options": [],
"answer": "See solution",
"solution": "It is easy to see that $(k, n) = (k + 1, n) = 1$. Then for any $k \\in S$ there exists a unique $k_1 \\in \\{1, 2, \\dots, n\\}$ such that $(k_1, n) = 1$ and $k k_1 \\equiv 1 \\pmod{n}$. Since $k_1 + 1 \\equiv k_1(k + 1) \\pmod{n}$, then $(k_1 + 1, n) = 1$, i.e. $k_1 \\in S$. Moreover, $(k - 1)(k_1 + 1) \\equiv k - k_1 \\pmod{n}$ implies that $k \\neq k_1$ if $k \\neq 1$. Then the numbers from $S \\setminus \\{1\\}$ can be divided into different pairs of the form $(k, k_1)$ and hence the wanted remainder is $1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21423,
"subject": "Mathematics (Olympiad)",
"question": "Write in the cells of a $4 \\times 4$ table a different natural number such that the sums by rows are equal and the products by columns are also equal.",
"options": [],
"answer": "See solution",
"solution": "Here is one way to construct such a table.\n\nFirst, satisfy the condition that the products by columns are equal. Consider the following $4 \\times 4$ table $T$ (all column products equal $120$):\n\n$$\nT = \\begin{array}{cccc}\n4 & 1 & 2 & 8 \\\\\n3 & 2 & 5 & 1 \\\\\n2 & 10 & 3 & 5 \\\\\n5 & 6 & 4 & 3 \\\\\n\\end{array}\n$$\n\nThe row sums are $15$, $11$, $20$, and $18$.\n\nTo make all row sums equal, consider the least common multiple of the row sums, which is $1980$.\n\nMultiply each row by the following factors:\n- Row 1: $\\frac{1980}{15} = 132$\n- Row 2: $\\frac{1980}{11} = 180$\n- Row 3: $\\frac{1980}{20} = 99$\n- Row 4: $\\frac{1980}{18} = 110$\n\nThe new table is:\n\n$$\n\\begin{array}{cccc}\n4 \\times 132 & 1 \\times 132 & 2 \\times 132 & 8 \\times 132 \\\\\n3 \\times 180 & 2 \\times 180 & 5 \\times 180 & 1 \\times 180 \\\\\n2 \\times 99 & 10 \\times 99 & 3 \\times 99 & 5 \\times 99 \\\\\n5 \\times 110 & 6 \\times 110 & 4 \\times 110 & 3 \\times 110 \\\\\n\\end{array}\n$$\n\nAll row sums are $1980$ and all column products are equal. All numbers in the table are different.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21424,
"subject": "Mathematics (Olympiad)",
"question": "Find all triples $(p, q, n)$ such that\n\n$$\nq^{n+2} \\\\equiv 3^{n+2} \\pmod{p^n}, \\quad p^{n+2} \\\\equiv 3^{n+2} \\pmod{q^n}\n$$\n\nwhere $p, q$ are positive odd primes and $n > 1$ is an integer.",
"options": [],
"answer": "See solution",
"solution": "It is easy to check that $(3, 3, n)$ for $n = 2, 3, \\dots$ satisfy both equations.\n\nNow let $(p, q, n)$ be another triple satisfying the condition. Then we must have $p \\ne q$, $p \\ne 3$, $q \\ne 3$. We may assume that $q > p \\ge 5$.\n\nIf $n = 2$, then $q^2 \\mid p^4 - 3^4$, or $q^2 \\mid (p^2 - 3^2)(p^2 + 3^2)$. Then either $q^2 \\mid p^2 - 3^2$ or $q^2 \\mid p^2 + 3^2$, since $q$ cannot divide both $p^2 - 3^2$ and $p^2 + 3^2$. On the other hand, $0 < p^2 - 3^2 < q^2$, $\\frac{1}{2}(p^2 + 3^2) < p^2 < q^2$. This leads to a contradiction.\n\nSo $n \\ge 3$. From $p^n \\mid q^{n+2} - 3^{n+2}$, $q^n \\mid p^{n+2} - 3^{n+2}$, we get\n\n$$\np^n \\mid p^{n+2} + q^{n+2} - 3^{n+2}, \\quad q^n \\mid p^{n+2} + q^{n+2} - 3^{n+2}.\n$$\n\nSince $p < q$, and $p, q$ are primes, we have\n\n$$\np^n q^n \\mid p^{n+2} + q^{n+2} - 3^{n+2}. \\qquad \\textcircled{1}\n$$\n\nThen $p^n q^n \\le p^{n+2} + q^{n+2} - 3^{n+2} < 2q^{n+2}$. That means $p^n < 2q^2$.\n\nAs $q^n \\mid p^{n+2} - 3^{n+2}$ and $p > 3$, we have $q^n \\le p^{n+2} - 3^{n+2} < p^{n+2}$, and consequently $q < p^{1+\\frac{2}{n}}$. Since $p^n < 2q^2$, we have $p^n < 2p^{2+\\frac{4}{n}} < p^{3+\\frac{4}{n}}$. So $n < 3+\\frac{4}{n}$, and we get $n = 3$. Then $p^3 \\mid q^5 - 3^5$, $q^3 \\mid p^5 - 3^5$.\n\nFrom $5^5 - 3^5 = 2 \\times 11 \\times 131$, we know $p > 5$; from $p^3 \\mid q^5 - 3^5$ we know $p \\mid q^5 - 3^5$. By Fermat's little theorem, we get $p \\mid q^{p-1} - 3^{p-1}$. Then $p \\mid q^{\\gcd(5, p-1)} - 3^{\\gcd(5, p-1)}$.\n\nIf $\\gcd(5, p-1) = 1$, then $p \\mid q-3$. From\n\n$$\n\\frac{q^5 - 3^5}{q - 3} = q^4 + q^3 \\cdot 3 + q^2 \\cdot 3^2 + q \\cdot 3^3 + 3^4 \\\\ \\equiv 5 \\times 3^4 \\pmod{p}\n$$\n\nand $p \\ge 5$, we get $p \\nmid \\frac{q^5 - 3^5}{q - 3}$. So $p^3 \\mid q - 3$. From $q^3 \\mid p^5 - 3^5$, we get $q^3 \\le p^5 - 3^5 < p^5 = (p^3)^{5/3} < q^{5/3}$. This is a contradiction.\n\nSo we have $\\gcd(5, p-1) \\ne 1$, and that means $5 \\mid p-1$. In a similar way, we have $5 \\mid q-1$. As $\\gcd(q, p-3) = 1$ (since $q > p \\ge 7$) and $q^3 \\mid p^5 - 3^5$, we know that $q^3 \\mid \\frac{p^5 - 3^5}{p-3}$. Then\n\n$$\nq^3 \\le \\frac{p^5 - 3^5}{p - 3} = p^4 + p^3 \\cdot 3 + p^2 \\cdot 3^2 + p \\cdot 3^3 + 3^4.\n$$\n\nFrom $5 \\mid p-1$ and $5 \\mid q-1$, we get $p \\ge 11$ and $q \\ge 31$. So\n\n$$\nq^3 \\le p^4 \\left( 1 + \\frac{3}{p} + \\left(\\frac{3}{p}\\right)^2 + \\left(\\frac{3}{p}\\right)^3 + \\left(\\frac{3}{p}\\right)^4 \\right)\n$$\n\n$$\n< p^4 \\cdot \\frac{1}{1 - \\frac{3}{p}} \\le \\frac{11}{8} p^4.\n$$\n\nThen we have $p > \\left(\\frac{8}{11}\\right)^{1/4} q^{3/4}$. Consequently,\n\n$$\n\\frac{p^5 + q^5 - 3^5}{p^3 q^3} < \\frac{p^2}{q^3} + \\frac{q^2}{p^3} < \\frac{1}{q} + \\left(\\frac{11}{8}\\right)^{3/4} \\frac{1}{31^{1/4}} < 1.\n$$\n\nBut this contradicts ① which says $p^3 q^3 \\mid p^5 + q^5 - 3^5$.\n\nSo we reach the conclusion that $(3, 3, n)$ for $n = 2, 3, \\dots$ are all the triples that satisfy the conditions.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21425,
"subject": "Mathematics (Olympiad)",
"question": "Пусть $p_0, p_1, \\dots, p_{2n}$ — числа на карточках, причём $p_{2n}$ — наибольшее по модулю из них. Поставим $p_i$ коэффициентом при $x^i$. Требуется доказать, что можно переставить коэффициенты $p_{2n-1}, p_{2n-2}, \\dots, p_0$ так, чтобы многочлен $p_{2n}x^{2n} + p_{2n-1}x^{2n-1} + \\dots + p_0$ не имел целых корней.",
"options": [],
"answer": "See solution",
"solution": "Рассмотрим целое $a$ с $|a| \\ge 2$. Тогда\n\n$$\n\\begin{aligned}\n|p_{2n}a^{2n}| &> |p_{2n}| (|a^{2n-1}| + |a^{2n-2}| + \\dots + 1) \\\\ &\\ge |p_{2n-1}a^{2n-1}| + |p_{2n-2}a^{2n-2}| + \\dots + |p_0|.\n\\end{aligned}\n$$\n\nЗначит, $a$ не может быть корнем многочлена.\n\nОсталось рассмотреть $a = 0, \\pm 1$. Числа $0$ и $1$ не могут быть корнями, так как $p_0 \\ne 0$ и $p_{2n} + p_{2n-1} + \\dots + p_0 \\ne 0$ по условию. Предположим, что $x_0 = -1$ является корнем при любой перестановке коэффициентов $p_{2n-1}, \\dots, p_0$. Тогда перестановка любых двух из них не меняет значение многочлена в $x_0$, что возможно только если все они равны. Тогда многочлен имеет вид $p_{2n}x^{2n} + p_0(x^{2n-1} + \\dots + 1)$, и при $x_0 = -1$ его значение равно $p_{2n} \\ne 0$, противоречие.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21426,
"subject": "Mathematics (Olympiad)",
"question": "\n\nLet $ABC$ be a triangle. Points $P$ and $Q$ are chosen on arc $BC$ of the circumcircle of $ABC$ (not containing $A$) such that $\\angle PAB = \\angle BCA$ and $\\angle QAC = \\angle ABC$. Let $M$ and $N$ be the midpoints of $AP$ and $AQ$, respectively. Prove that lines $BM$ and $CN$ intersect on the circumcircle of $ABC$.",
"options": [],
"answer": "See solution",
"solution": "We solve the problem using inversion. Recall that an inversion $f$, of radius $r > 0$ about a point $O$ is defined by\n\n$$\nOX \\cdot OX' = r^2,\n$$\n\nwhere $X$ is any point ($X \\neq O$) in the plane and $X' = f(X)$ lies on the ray $OX$. Inversion has the following properties:\n\n* A circle $OXY$ becomes a line $X'Y'$.\n* A line $XY$ becomes a circle $OX'Y'$.\n* $\\triangle OXY \\sim \\triangle OY'X'$ in opposite orientation, which implies that\n\n$$\n\\angle OXY = \\angle OY'X' \\quad \\text{and} \\quad \\angle OYX = \\angle OX'Y'.\n$$\n\nFor the problem at hand, consider an inversion $f$ of arbitrary radius $r > 0$ about point $A$. For any point $X$, let $X' = f(X)$. We have the following properties:\n\n* $BC$ becomes the arc $B'C'$ of circle $AB'C'$ not containing $A$.\n* Rays $AP'$ and $AQ'$ lie between rays $AB'$ and $AC'$.\n* $P'$ and $Q'$ lie on circle $AB'C'$.\n* $\\angle AB'C' = \\angle ACB$ and $\\angle AC'B' = \\angle ABC$.\n\nThe problem's angle condition tells us that $\\angle PAB = \\angle BCA$. In the inverted diagram, this becomes $\\angle P'AB' = \\angle C'B'A$. However, from circle $AB'C'$, we have $\\angle C'B'A = \\angle C'P'A$. Consequently, $\\angle P'AB' = \\angle C'P'A$, implying that $AB' \\parallel C'P'$. Therefore, $AC'P'B'$ is an isosceles trapezium. Similarly, $AB'Q'C$ is an isosceles trapezium.\n\nWe are required to prove that lines $BM$ and $CN$ intersect on circle $ABC$. In the inverted diagram, this is equivalent to proving that circles $AB'M'$ and $AC'N'$ intersect for a second time on line $B'C'$. We claim that this common second point of intersection is the midpoint, $K$, of line $B'C'$.\n\nNote that $AM' \\cdot AM = r^2 = AP \\cdot AP'$. Since $M$ is the midpoint of $AP$, we know that $AP = 2AM$. It follows that $AP' = 2AM'$, so $M'$ is the midpoint of $AP'$. Analogously, $N'$ is the midpoint of $AQ'$.\n\nConsider isosceles trapezium $AC'P'B'$. Its sides $AB'$ and $C'P'$ are parallel and so have the same perpendicular bisector, $\\ell$. Thus, points $A$ and $B'$ are symmetric with respect to $\\ell$, as are points $P'$ and $C'$. Therefore, segments $AP'$ and $B'C'$ are symmetric with respect to $\\ell$ and hence so are their respective midpoints $M'$ and $K$. It follows that $AM'KB'$ is an isosceles trapezium and hence it is cyclic. Therefore, circle $AB'M'$ passes through $K$. Analogously, circle $AC'N'$ also passes through $K$. This establishes our earlier claim and hence completes the proof. $\\square$\n\n**Comment:** Alex Gunning noted the following interesting property of the original diagram. If $T$ is the intersection of lines $BM$ and $CN$, then $ABTC$ is a harmonic quadrilateral. This can be proven using similar triangles $BTC$ and $MTN$, and elements of his proof for the original problem.\n\nA cyclic quadrilateral is _harmonic_ if the products of its opposite sides are equal. They have many interesting and useful properties.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21427,
"subject": "Mathematics (Olympiad)",
"question": "Let $f : \\mathbb{Z} \\to \\mathbb{Z}$ be a function such that for all $x \\in \\mathbb{Z}$, $f(f(x)) = x f(x) - x^2 + 2$. Find all such functions $f$.",
"options": [],
"answer": "See solution",
"solution": "Let $A \\subset \\mathbb{Z}$ be the set of all integers $x$ such that $f(x) = x + 1$. We prove that $A = \\mathbb{Z}$.\n\n*Claim 1.* $A \\neq \\emptyset$. Indeed, if $f(1) = a$ then $f(a) = f(f(1)) = 1 \\cdot f(1) - 1^2 + 2 = a + 1$, so $a \\in A$.\n\n*Claim 2.* If $x_0 \\in A$ then $x_0 + 1 \\in A$. Indeed,\n\n$$\nf(x_0 + 1) = f(f(x_0)) = x_0 f(x_0) - x_0^2 + 2 = x_0(x_0 + 1) - x_0^2 + 2 = (x_0 + 1) + 1,\n$$\n\ntherefore $x_0 + 1 \\in A$.\n\nFrom claim 2 it follows that if $A \\neq \\mathbb{Z}$ then $A$ has its minimum element $b$ and $A = \\{x \\in \\mathbb{Z} \\mid x \\geq b\\}$. Suppose that $A \\neq \\mathbb{Z}$ and let $b \\in A$ be the minimum element of $A$.\n\nFrom the problem condition 2 we have $f(s) = b$ for some $s \\in \\mathbb{Z}$. Then\n\n$$\nb + 1 = f(b) = f(f(s)) = s b - s^2 + 2 \\quad \\text{or} \\quad (s - 1)(s + 1 - b) = 0.\n$$\n\nIf $s = b - 1$ then we have $f(b - 1) = b = (b - 1) + 1$ which contradicts the minimality of $b$. So, $s = 1$ and $f(1) = b$. If $b \\le 1$ then $1 \\in A$ and we have $f(1) = 2 > b$ — a contradiction. Further, if $b = 2$ then $f(1) = 2 = 1 + 1$, so $1 \\in A$ — contrary to minimality of $b$. Hence, $b \\ge 3$.\n\nNow let $t \\in \\mathbb{Z}$ be such that $f(t) = 1$. Then\n\n$$\nb = f(1) = f(f(t)) = t f(t) - t^2 + 2 = t - t^2 + 2 \\quad \\text{or} \\quad t^2 - t + b - 2 = 0,\n$$\n\nwhich is impossible for $b \\ge 3$.\n\nThe contradiction obtained shows that $A = \\mathbb{Z}$, i.e. $f(x) = x + 1$. It is easy to see that the function $f(x) = x + 1$ is suitable.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21428,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\text{Ionof}(n)$ denote the value $n$ divided by the number of positive integer factors of $n$.\n\n**a)** Find $\\text{Ionof}(36)$.\n\n**b)** Let $p$ and $q$ be distinct odd primes. Is $\\text{Ionof}(pq)$ always an integer?\n\n**c)** For which primes $p$ and $q$ is $\\text{Ionof}(pq^4)$ an integer? Find all such $pq^4$.\n\n**d)** For which primes $p$ does there exist an integer $m$ such that $\\text{Ionof}(m) = p^2$? For each such $p$, find all such $m$.",
"options": [],
"answer": "See solution",
"solution": "**a)**\n\nThe factors of $36$ are: $1, 2, 3, 4, 6, 9, 12, 18, 36$. There are $9$ factors.\n\nSo:\n$$\n\\text{Ionof}(36) = \\frac{36}{9} = 4\n$$\n\n**b)**\n\nThe factors of $pq$ are: $1, p, q, pq$ (since $p$ and $q$ are distinct primes). There are $4$ factors.\n\nSo:\n$$\n\\text{Ionof}(pq) = \\frac{pq}{4}\n$$\n\nSince $p$ and $q$ are odd primes, $pq$ is odd, so $4$ does not divide $pq$. Thus, $\\text{Ionof}(pq)$ is not an integer.\n\n**c)**\n\nThe factors of $pq^4$ are: $1, p, q, q^2, q^3, q^4, pq, pq^2, pq^3, pq^4$ (total $10$ factors).\n\nSo:\n$$\n\\text{Ionof}(pq^4) = \\frac{pq^4}{10}\n$$\n\nFor $\\text{Ionof}(pq^4)$ to be an integer, $10$ must divide $pq^4$. Since $p$ and $q$ are primes, $p$ must be $2$ or $5$, and $q$ must be $2$ or $5$ (but $p \\ne q$). The only possibilities are $p=2, q=5$ or $p=5, q=2$.\n\nThus, $pq^4 = 2 \\times 5^4 = 1250$ or $5 \\times 2^4 = 80$.\n\n**d)**\n\nSuppose $p$ is a prime and $p^2 = \\text{Ionof}(m)$ for some integer $m$. Let $k$ be the number of factors of $m$, so $m = k p^2$.\n\nWe check possible $k$ and $p$:\n\n- For $k=9$, $m=9p^2$, and $9p^2$ has $9$ factors if $p \\ne 3$. In this case, $\\text{Ionof}(9p^2) = p^2$.\n- For $p=3$, we check $m = k \\times 9$ for $k \\ge 3$. Only $k=12$ works: $m = 12 \\times 9 = 108$ has $12$ factors, so $\\text{Ionof}(108) = 9$.\n\nThus, the square of any prime number is the Ionof of some integer.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21429,
"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": "The angle $\\angle ABC$ is denoted by $\\beta$. As $BC$ is normal to $AE$ and $DE$ is normal to $AB$, the angles $\\angle ABC$ and $\\angle AED$ are of equal measure. As we have $\\angle ACB = \\angle DCE = 90^\\circ$ and, by assumption, $\\overline{AB} = \\overline{DE}$, the triangles $ABC$ and $DEC$ are congruent.\n\nThis yields $\\overline{BC} = \\overline{CE}$ which implies that the triangle $BCE$ is an isosceles right-angled triangle with $\\angle CEB = \\angle CBE = 45^\\circ$.\n\nFurthermore, we have $\\beta = \\angle CED = \\angle DEB$, as $E$ lies on the perpendicular bisector of $\\overline{AB}$.\n\nThus we obtain $45^\\circ = \\angle CEB = 2\\beta$ and therefore $\\beta = 22.5^\\circ$ and $\\alpha = \\angle CAB = 67.5^\\circ$.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21430,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\{a_n\\}_{n \\ge 0}$ be a sequence of real numbers such that $a_{n+1} \\ge a_n^2 + \\frac{1}{5}$ for all $n \\ge 0$. Prove that $\\sqrt{a_{n+5}} \\ge a_{n-5}^2$ for all $n \\ge 5$.",
"options": [],
"answer": "See solution",
"solution": "It suffices to prove that $a_{n+5} \\ge a_n^2$ for $n \\ge 0$, for then we would have $\\sqrt{a_{n+5}} \\ge a_n$ and $a_n \\ge a_{n-5}^2$ for all $n \\ge 5$, which implies the desired result.\n\nAdding the inequalities\n\n$$\n\\begin{align*}\na_{n+5} &\\ge a_{n+4}^2 + \\frac{1}{5}, \\\\\na_{n+4} &\\ge a_{n+3}^2 + \\frac{1}{5}, \\\\\na_{n+3} &\\ge a_{n+2}^2 + \\frac{1}{5}, \\\\\na_{n+2} &\\ge a_{n+1}^2 + \\frac{1}{5}, \\\\\na_{n+1} &\\ge a_n^2 + \\frac{1}{5}\n\\end{align*}\n$$\n\nyields\n\n$$\n\\begin{align*}\na_{n+5} &\\ge \\sum_{k=1}^{4} (a_{n+k}^2 - a_{n+k}) + 1 + a_n^2 \\\\\n&= \\sum_{k=1}^{4} \\left( a_{n+k}^2 - a_{n+k} + \\frac{1}{4} \\right) + a_n^2 \\\\\n&= \\sum_{k=1}^{4} \\left( a_{n+k} - \\frac{1}{2} \\right)^2 + a_n^2 \\ge a_n^2,\n\\end{align*}\n$$\n\nas desired.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21431,
"subject": "Mathematics (Olympiad)",
"question": "Гурвалжны хагас параметр $p$, багтаасан тойргийн радиус $R$, багтсан тойргийн радиус $r$ бол $p^2 \\ge 12Rr + 3r^2$ болохыг батал.",
"options": [],
"answer": "See solution",
"solution": "$$\nR = \\frac{abc}{4S}, \\quad r = \\frac{S}{p}, \\quad S = \\sqrt{p(p-a)(p-b)(p-c)}\n$$\nэдгээр томъёог ашиглавал:\n$$\np^2 \\ge 12 \\cdot \\frac{abc}{4S} \\cdot \\frac{S}{p} + 3 \\left(\\frac{S}{p}\\right)^2 = \\frac{3abc}{p} + \\frac{3(p-a)(p-b)(p-c)}{p}\n$$\nҮүнийг цааш задлавал:\n$$\n3p^3 - 3p^2(a + b + c) + 3p(ab + bc + ac)\n$$\n$$\n= 3p^2 - 3p(a + b + c) + 3(ab + bc + ac)\n$$\nИймд $p^2 \\ge 3(ab + bc + ac)$.\n\nЭндээс $a^2 + b^2 + c^2 \\ge ab + bc + ac$ гэсэн илэрхий тэнцэтгэл биш гарна.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21432,
"subject": "Mathematics (Olympiad)",
"question": "Is it possible to compose two integer numbers using each of the ten digits $0, 1, \\ldots, 9$ exactly once, such that one of them is the square of the other?\n\n$0$ cannot be the first digit in either number.",
"options": [],
"answer": "See solution",
"solution": "If a number has 3 digits, its square can contain no more than 6 digits, which gives 9 in total. If our number has 4 digits or more, its square has at least 7 digits, therefore, we must use at least 11 digits. Therefore, no such number exists.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21433,
"subject": "Mathematics (Olympiad)",
"question": "Consider arrangements of disks (black and white) labeled $A_i$ for $i = 1, 2, \\ldots, n$ placed in a circle, where $n$ is of the form $n = 4s + 3$ or $n = 4s + 5$. For each disk $i$, indicate a pair of disks of the same color that are located symmetrically (i.e., at equal distance from $i$, with indices $i + u$ and $i - u$ modulo $n$). If there are three or more consecutive disks of the same color, consider only those on the sides of the group, since for the rest the desired pair is two adjacent disks. For small prime values of $n$ (specifically $n = 17, 13, 11, 7$), present constructions or explain why such a pairing is impossible.\n\n\n\n",
"options": [],
"answer": "See solution",
"solution": "For $n = 11, 13, 17$, explicit constructions are possible, as shown in the figures:\n\nFor $n = 11$:\n$$\n1 = \\frac{1}{2}(5-3) = \\frac{1}{2}(5+8), \\quad 3 = \\frac{1}{2}(1+5), \\quad 8 = \\frac{1}{2}(2+14) = \\frac{1}{2}(2+3).\n$$\nThere is a distinct gray disk, so for number $x$ the equation $x = \\frac{1}{2}(M + m)$ must be satisfied (e.g., $x = 2$, $m = 1$, $M = 3$).\n\nFor $n = 7$, it is impossible to satisfy the conditions. If there are no more minimal than maximal, then there are at most 3 of them. If two are adjacent, neither has a pair; if all are distinct and alternate, two do not have a pair.\n\n\n\n",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21434,
"subject": "Mathematics (Olympiad)",
"question": "Let $(A, +, \\cdot)$ be a unitary ring such that:\n\n1. $A$ is not a field.\n2. For any noninvertible element $x$ of $A$, there exists an integer $m \\geq 1$ (depending on $x$) such that\n $$x = x^2 + x^3 + \\dots + x^{2^m}.$$\n\nProve that:\n\n(a) $x + x = 0$ for any $x \\in A$.\n\n(b) $x^2 = x$ for any noninvertible $x \\neq 1$ in $A$.",
"options": [],
"answer": "See solution",
"solution": "(a) It suffices to prove that $1 + 1 = 0$. Let $x$ be noninvertible, and let $m \\in \\mathbb{N}^*$ such that $x = x^2 + x^3 + \\dots + x^{2^m}$. Define $y = x + x^2 + \\dots + x^{2^m-1}$. Clearly, $xy = x$, so $xy^k = x$ for any $k \\in \\mathbb{N}^*$. Since $x$ is not invertible, $y$ is also not invertible, so there exists $p \\in \\mathbb{N}^*$ such that\n$$-y = (-y)^2 + (-y)^3 + \\dots + (-y)^{2^p}.$$\nExpanding, $-y = y^2 - y^3 + \\dots - y^{2^p-1} + y^{2^p}$. Thus,\n$$-x = -xy = xy^2 - xy^3 + \\dots - xy^{2^p-1} + xy^{2^p} = x - x + \\dots - x + x = x,$$\ni.e., $x + x = 0$.\n\nNow, consider $x$ nonzero and noninvertible. Since $2x = 0$, $2$ is noninvertible, so $2 + 2 = 0$ and $2 = 2^2 + 2^3 + \\dots + 2^{2^m}$ for some $m \\in \\mathbb{N}^*$. The equality $2 + 2 = 0$ implies $2^2 = 0$, so $2^k = 0$ for any $k \\geq 2$. Therefore,\n$$2 = 2^2 + 2^3 + \\dots + 2^{2^m} = 0.$$\n\n(b) Let $x$ be a noninvertible element of $A$ and $m \\in \\mathbb{N}^*$ such that $x = x^2 + x^3 + \\dots + x^{2^m}$. Then $x^2 = x^3 + x^4 + \\dots + x^{2^m+1}$. Since $1 + 1 = 0$, summing the two equalities gives $x^{2^m+1} = x$.\n\nThe equalities $1 + 1 = 0$ and $x^{2^m+1} = x$ imply\n$$(x^2 + x)^{2^m} = x^{2^m+1} + x^{2^m} = x + x^{2^m} = x^{2^m} + x^{2^m} = 0.$$\n\nWe will show that $x^2 + x = 0$, which concludes the proof. Let $y = x^2 + x$ and let $k$ be the smallest positive integer such that $y^k = 0$ (the existence of $k$ is assured by $y^{2^m} = 0$). If $k > 1$, then $y^{k-1}$ is noninvertible in $A$, so there exists $n \\in \\mathbb{N}^*$ such that $y^{k-1} = (y^{k-1})^{2^n+1} = y^{(k-1)(2^n+1)}$. Since $(k-1)(2^n+1) \\geq k$, $y^{(k-1)(2^n+1)} = 0$, so $y^{k-1} = 0$, contradicting the minimality of $k$. Thus, $k = 1$ and $y = 0$.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21435,
"subject": "Mathematics (Olympiad)",
"question": "If 46 squares are colored red in a $9 \\times 9$ board, show that there is a $2 \\times 2$ block on the board in which at least 3 of the squares are colored red.",
"options": [],
"answer": "See solution",
"solution": "Suppose that at most 2 squares are colored red in any $2 \\times 2$ square. Then in any $9 \\times 2$ block, there are at most 10 red squares. Moreover, if there are 10 red squares, then there must be 5 in each row. This can be seen as follows. There are $8 \\times 2 \\times 2$ blocks. Counting multiplicity, there are altogether 16 red squares. Each red square in the interior is counted twice while each red square at the edge is counted once. If there are 11 red squares, then there are at least 7 red squares in the interior. Thus the total count is at least $4 + 7 \\times 2 = 18 > 16$, a contradiction. If there are exactly 10 red squares, then 4 of them must be at the edge and the red squares in each row are not next to each other and hence there are 5 in each row.\n\nNow let the number of red squares in row $i$ be $r_i$. Then $r_i + r_{i+1} \\le 10$, $1 \\le i \\le 8$. Suppose that some $r_i \\le 5$ with $i$ odd. Then\n\n$$\n(r_1 + r_2) + \\dots + (r_{i-2} + r_{i-1}) + r_i + \\dots + (r_8 + r_9) \\le 4 \\times 10 + 5 = 45\n$$\nwhich leads to a contradiction. On the other hand, suppose that $r_1, r_3, r_5, r_7, r_9 \\ge 6$.\nThen the sum of any 2 consecutive $r_i$'s is $\\le 9$. Again we get a contradiction as\n\n$$\n(r_1 + r_2) + \\cdots + (r_7 + r_8) + r_9 \\le 4 \\times 9 + 9 = 45.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21436,
"subject": "Mathematics (Olympiad)",
"question": "A graph $G$ is called a *divisibility graph* if the vertices can be assigned distinct positive integers such that between two vertices assigned $u, v$ there is an edge if and only if $\\frac{u}{v}$ or $\\frac{v}{u}$ is a positive integer.\n\nShow that for any positive integer $n$ and $0 \\leq e \\leq \\frac{n(n-1)}{2}$, there is a divisibility graph with $n$ vertices and $e$ edges.",
"options": [],
"answer": "See solution",
"solution": "We reason inductively on $n$, not assigning the number $1$ to any vertex.\n\nFor $n=1$, the claim is clear. For $n=2$, an example with $e=1$ is $(2, 4)$, and with $e=0$ is $(2, 3)$. For $n=3$, examples are:\n- $e=0$: $3, 5, 7$\n- $e=1$: $2, 4, 7$\n- $e=2$: $2, 4, 10$\n- $e=3$: $2, 4, 8$\n\nFor $n \\geq 4$, note that $n-1 \\leq \\frac{(n-1)(n-2)}{2}$, so at least one of $e \\geq n-1$ or $e \\leq \\frac{(n-1)(n-2)}{2}$ holds.\n\nIf $e \\geq n-1$, start with $n-1$ vertices and $e - (n-1)$ edges. Add a vertex (of degree $n-1$) by assigning it a prime $p$ greater than all other vertex numbers, then multiply the numbers in the remaining vertices by $p$—this does not create new edges among them.\n\nIf $e \\leq \\frac{(n-1)(n-2)}{2}$, start with $n-1$ vertices and $e$ edges. Add a vertex (of degree $0$) by assigning it a prime $p$ greater than all other vertex numbers, and leave the other numbers unchanged—this does not create new edges.\n\n$\\square$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21437,
"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 izquierda o 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 tiene que 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, con lo que se ha desplazado a la izquierda al menos nueve veces.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21438,
"subject": "Mathematics (Olympiad)",
"question": "$\\Delta P_1 P_2 P_3$ 為一正三角形。對於所有 $n \\ge 4$，小明可以選擇 $P_n$ 為 $\\Delta P_{n-3} P_{n-2} P_{n-1}$ 的外心或垂心。試求所有的正整數 $n$，使得小明可以經由適當地選取 $P_4, \\dots, P_n$，讓 $P_n$ 為 $\\Delta P_1 P_2 P_3$ 的外心。",
"options": [],
"answer": "See solution",
"solution": "答案為所有 $4$ 的倍數。\n\n令 $O$ 為 $\\Delta P_1 P_2 P_3$ 的外心，則小明只要一直選取垂心，便有 $P_{4k} = O$，$P_{4k+1} = P_1$，$P_{4k+2} = P_2$ 與 $P_{4k+3} = P_3$，對於所有 $k \\in \\mathbb{N}$ 皆成立。因此所有被 $4$ 整除的 $n$ 都滿足題意。\n\n以下證明 $n$ 必須被 $4$ 整除。這需要以下兩個 Lemma。\n\n**Lemma 1.** 對於所有 $n \\ge 3$，$\\Delta P_{n-2} P_{n-1} P_n$ 是頂角為 $120^\\circ$ 的等腰三角形（稱為型 A）或正三角形（稱為型 B）。\n\n證明：讓我們對 $n \\ge 3$ 進行數學歸納法。$n=3$ 時顯然為型 B。當 $n>3$ 時：\n\n- 當 $\\Delta P_{n-3} P_{n-2} P_{n-1}$ 是型 A 時：如果小明選擇外心且 $P_{n-3}$ 為頂角時，$\\Delta P_{n-2} P_{n-1} P_n$ 是型 A；如果小明選擇外心且 $P_{n-2}$ 或 $P_{n-1}$ 為頂角時，$\\Delta P_{n-2} P_{n-1} P_n$ 是型 B。\n\n如果小明選擇垂心且 $P_{n-2}$ 或 $P_{n-1}$ 為頂角時，$\\Delta P_{n-2} P_{n-1} P_n$ 是型 A；如果小明選擇垂心且 $P_{n-3}$ 為頂角時，$\\Delta P_{n-2} P_{n-1} P_n$ 是型 B。\n\n- 當 $\\Delta P_{n-3} P_{n-2} P_{n-1}$ 是型 B 時，無論小明選擇外心或垂心，$\\Delta P_{n-2} P_{n-1} P_n$ 一定是型 A。\n\n由數學歸納法，Lemma 證畢。$\\Box$\n\n以下不失一般性假設 $O$ 為原點。我們考慮斜座標 $(x, y)$，代表點 $x\\vec{P}_1 + y\\vec{P}_2$。這時候 $P_3 = (-1, -1)$。\n\n**Lemma 2.** 所有小明可能點出的點一定形如 $\\left(\\frac{a}{3^r}, \\frac{b}{3^s}\\right)$，其中 $a, b$ 為與 $3$ 互質的整數，$r, s$ 為非負整數。更進一步地，對於所有 $n \\ge 4$，$P_{n-3}, P_{n-2}, P_{n-1}, P_n$ 的座標在模 $2$ 下皆相異（這邊我們取 $3^{-1} \\equiv 1$），即 $(0,0), (0,1), (1,0), (1,1)$ 各出現一次。\n\n證明：對 $n$ 使用數學歸納法。因為 $P_4 = O$，故 $n=4$ 成立。當 $n>4$ 時，重新命名 $P, Q, R$ 為 $\\Delta P_{n-3} P_{n-2} P_{n-1}$ 的頂點，使得 $P$ 是角度最大的。由 Lemma 1 及其證明，我們知道 $P_n$ 一定是 $\\vec{Q} + \\vec{R} - \\vec{P}$、$3\\vec{P} - \\vec{Q} - \\vec{R}$、$\\frac{1}{3}(\\vec{P} + \\vec{Q} + \\vec{R})$ 其中一種。因為係數最多只有除以 $3$，所以我們可以簡單地看出座標各分量的分母一定是 $3$ 的冪次。\n\n注意到在模 $2$ 下，我們皆有\n\n$$\nP_n + P + Q + R \\equiv (0,0) \\pmod{2}.\n$$\n\n由數歸假設，$P_{n-3}$、$P_{n-2}$、$P_{n-1}$ 皆相異，因此由\n\n$$\n(0,0) + (0,1) + (1,0) + (1,1) \\equiv (0,0) \\pmod{2}\n$$\n\n知 $P_{n-3}$、$P_{n-2}$、$P_{n-1}$、$P_n$ 的座標在模 $2$ 下皆相異。$\\Box$\n\n現在，回到原題。由 Lemma 2，我們有\n\n$$\nP_{4k+1} \\equiv P_1 \\equiv (1,0), \\quad P_{4k+2} \\equiv P_2 \\equiv (0,1),\n$$\n\n$$\nP_{4k+3} \\equiv P_3 \\equiv (1,1), \\quad P_{4k} \\equiv P_4 = O \\equiv (0,0),\n$$\n\n因此只有 $4$ 的倍數能夠達成題目要求。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21439,
"subject": "Mathematics (Olympiad)",
"question": "Prove that for every positive integer $k$ there exist an integer $n$ and distinct primes $p_1, p_2, \\dots, p_k$ such that, if $A(n)$ denotes the number of integers in $\\{1, 2, \\dots, n\\}$ which are relatively prime to $p_1 p_2 \\cdots p_k$, then\n\n$$\n\\left| n \\left(1 - \\frac{1}{p_1}\\right) \\left(1 - \\frac{1}{p_2}\\right) \\cdots \\left(1 - \\frac{1}{p_k}\\right) - A(n) \\right| > 2^{k-3}.\n$$",
"options": [],
"answer": "See solution",
"solution": "If $k=1$, choose $p_1=3$, $n=2$. If $k=2$, choose $p_1=3$, $p_2=7$, $n=5$. Assume $k \\ge 3$. Let $p_1, p_2, \\dots, p_k$ be primes congruent to $3$ modulo $4$. By the Chinese Remainder Theorem, choose $n \\equiv \\frac{p_i+1}{4} \\pmod{p_i}$ for every $i$, that is, choose an integer $n$ such that $p_1 p_2 \\cdots p_k \\mid 4n-1$.\n\nConsider a fractional part $\\theta = \\frac{n}{q_1 q_2 \\cdots q_r}$, where $q_1, q_2, \\dots, q_r$ are different primes among $p_1, p_2, \\dots, p_k$. Note that $\\theta \\approx \\frac{1}{4}$ if $r$ is odd and $\\theta \\approx \\frac{3}{4}$ if $r$ is even.\n\nIf $r$ is odd, then working modulo $4$, we get $4n-1 = q_1 q_2 \\cdots q_r \\cdot (4m+1)$, for some integer $m$. It follows that\n\n$$\n\\frac{n}{q_1 q_2 \\cdots q_r} = m + \\frac{1}{4} + \\frac{1}{4q_1 q_2 \\cdots q_r} \\quad \\text{and} \\quad \\left\\{ \\frac{n}{q_1 q_2 \\cdots q_r} \\right\\} = \\frac{1}{4} + \\frac{1}{4q_1 q_2 \\cdots q_r}.\n$$\n\nIf $r$ is even, then $4n-1 = q_1 q_2 \\cdots q_r \\cdot (4m+3)$, for some integer $m$. Hence\n\n$$\n\\frac{n}{q_1 q_2 \\cdots q_r} = m + \\frac{3}{4} + \\frac{1}{4q_1 \\cdots q_r} \\quad \\text{and} \\quad \\left\\{ \\frac{n}{q_1 q_2 \\cdots q_r} \\right\\} = \\frac{3}{4} + \\frac{1}{4q_1 q_2 \\cdots q_r}.\n$$\n\nThe difference between $n \\left(1 - \\frac{1}{p_1}\\right) \\left(1 - \\frac{1}{p_2}\\right) \\cdots \\left(1 - \\frac{1}{p_k}\\right)$ and $A(n)$ equals\n\n$$\n\\begin{align*}\n& \\{n\\} - \\sum_{i=1}^{k} \\left\\{ \\frac{n}{p_i} \\right\\} + \\sum_{1 \\le i < j \\le k} \\left\\{ \\frac{n}{p_i p_j} \\right\\} - \\dots + (-1)^k \\left\\{ \\frac{n}{p_1 p_2 \\cdots p_k} \\right\\} \\\\\n&= -\\sum_{i=1}^{k} \\frac{1}{4} + \\sum_{1 \\le i < j \\le k} \\frac{3}{4} - \\dots - \\sum_{i=1}^{k} \\frac{1}{4p_i} + \\sum_{1 \\le i < j \\le k} \\frac{1}{4p_i p_j} - \\dots + (-1)^k \\frac{1}{4p_1 p_2 \\cdots p_k} \\\\\n&= \\frac{3}{4} \\cdot 2^{k-1} - \\frac{1}{4} \\cdot 2^{k-1} + \\frac{1}{4} \\left(1 - \\frac{1}{p_1}\\right) \\cdots \\left(1 - \\frac{1}{p_k}\\right) - 1 \\\\\n&= 2^{k-2} + \\frac{1}{4} \\left(1 - \\frac{1}{p_1}\\right) \\cdots \\left(1 - \\frac{1}{p_k}\\right) - 1 \\\\\n&> 2^{k-3},\n\\end{align*}\n$$\n\nfor $k \\ge 3$, as desired.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21440,
"subject": "Mathematics (Olympiad)",
"question": "Самбарт $\\overline{abcdefxyz}$ гэсэн 9 оронтой тоо бичигдсэн байв. Бат энэ тоог $abc$, $def$, $xyz$ гэж гурван, гурван орцоор нь салган, хооронд нь дурын байдлаар эвлүүлэн 9 оронтой тоонуудыг үүсгэж байв. (Жишээ нь: $\\overline{abcdefabc}$ тоонуудыг үүсгэж болно.) Харин Цэдэг эдгээр тоонуудыг харж байснаа: \"Энэ бүх тоонуудыг 27-д хуваахад ижил үлдэгдэл өгдөг юм байна\" гэж хэлжээ. Цэдэгийн хэлсэн үнэн үү?",
"options": [],
"answer": "See solution",
"solution": "$$\n\\begin{aligned}\n&\\overline{abcdefxyz} \\text{ тоог } \\overline{abc} \\cdot 10^6 + \\overline{def} \\cdot 10^3 + \\overline{xyz} \\text{ гэж бичиж болно.} \\\\\n&10^6 = 999999 + 1, \\; 10^3 = 999 + 1 \\text{ тул:} \\\\\n&\\overline{abc} \\cdot 10^6 + \\overline{def} \\cdot 10^3 + \\overline{xyz} = \\overline{abc} \\cdot 999999 + \\overline{def} \\cdot 999 + (\\overline{abc} + \\overline{def} + \\overline{xyz}) \\\\\n&999999 \\equiv 0 \\pmod{27}, \\; 999 \\equiv 0 \\pmod{27} \\text{ тул:} \\\\\n&\\overline{abcdefxyz} \\equiv \\overline{abc} + \\overline{def} + \\overline{xyz} \\pmod{27} \\\\\n&Батын үүсгэсэн дурын тоог $A$ гэж авбал, $A$-г мөн адил гурван хэсэгт хувааж бичиж болох бөгөөд $A \\equiv \\overline{abc} + \\overline{def} + \\overline{xyz} \\pmod{27}$ байна. \\\\\n&Тиймээс бүх тоонууд 27-д хуваахад ижил үлдэгдэлтэй байна. \\; \\boxed{\\text{Тийм, Цэдэгийн хэлсэн үнэн.}}\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21441,
"subject": "Mathematics (Olympiad)",
"question": "Let $x$ denote a single digit. The tens digit in the product of $2x7$ and $39$ is $9$. Find $x$.",
"options": [],
"answer": "See solution",
"solution": "The table shows the product $2x7 \\times 39$ for all values of $x$.\n\n| $x$ | $2x7 \\times 39$ |\n|---|------------------|\n| 0 | 8073 |\n| 1 | 8463 |\n| 2 | 8853 |\n| 3 | 9243 |\n| 4 | 9633 |\n| 5 | 10023 |\n| 6 | 10413 |\n| 7 | 10803 |\n| 8 | 11193 |\n| 9 | 11583 |\n\nThus $x = 8$.\n\nWe have $2x7 \\times 39 = 2x7 \\times 30 + 2x7 \\times 9$.\nThe units digit in $2x7 \\times 30$ is $0$, and its tens digit is $1$.\nThe tens digit of $2x7 \\times 9$ is the units digit of $6 + 9x$.\nHence $1 + 6 + 9x \\equiv 9 \\pmod{10}$, so $9x \\equiv 2 \\pmod{10}$, which gives $x = 8$.\n\nAlternatively, $2x7 \\times 39 = 207 \\times 39 + 390x$.\nThe units digit in $207 \\times 39$ is $3$, and its tens digit is $7$.\nThe tens digit of $390x$ is the units digit of $9x$.\nSo $7 + 9x \\equiv 9 \\pmod{10}$, so $9x \\equiv 2 \\pmod{10}$, $x = 8$.\n\nAnother approach: $2x7 \\times 39 = 2x7 \\times 40 - 2x7$.\nThe units digit in $2x7 \\times 40$ is $0$, and its tens digit is $8$.\nSo the tens digit of $2x7 \\times 39$ is the units digit of $8 - x - 1$ or $18 - x - 1$.\nSince $x$ is non-negative, $17 - x = 9$ and $x = 8$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21442,
"subject": "Mathematics (Olympiad)",
"question": "Sea $p, p+d, p+2d, p+3d, p+4d, p+5d, p+6d$ una progresión de 7 primos distintos. ¿Cuál es el menor valor posible del séptimo término?",
"options": [],
"answer": "See solution",
"solution": "Analizando restricciones:\n\n- $p > 2$, pues si $p=2$, $p+2d$ sería par mayor que 2 y no primo.\n- $d$ debe ser par, pues si no, $p+d$ sería par mayor que 2 y no primo.\n- $p > 3$, pues si $p=3$, $p+3d$ sería múltiplo de 3 mayor que 3 y no primo.\n- $d$ debe ser múltiplo de 3, pues si no, alguno de $p+d$ o $p+2d$ sería múltiplo de 3 mayor que 3 y no primo.\n- $p > 5$, pues si $p=5$, $p+5d$ sería múltiplo de 5 mayor que 5 y no primo.\n- $d$ debe ser múltiplo de 5, pues si no, alguno de $p+d, p+2d, \\ldots, p+4d$ sería múltiplo de 5 mayor que 5 y no primo.\n\nPor lo tanto, $p \\geq 7$ y $d = 2 \\cdot 3 \\cdot 5 \\cdot k = 30k$.\n\nSi $p > 7$, siempre habrá un múltiplo de 7 entre los términos, salvo que $7 \\mid d$. En ese caso, el menor séptimo término sería $11 + 6d = 11 + 6 \\cdot 210 = 1271$.\n\nSi $p=7$ y $d=30$, $7+30 \\cdot 6 = 187$ (no primo). Tampoco para $d=60$ o $d=90$.\n\nPara $d=120$, la progresión es $7, 127, 247, \\ldots$, pero $247$ no es primo.\n\nPara $d=150$, la progresión es $7, 157, 307, 457, 607, 757, 907$, todos primos.\n\nComo $907 < 1271$, el menor valor posible es $907$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21443,
"subject": "Mathematics (Olympiad)",
"question": "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. If the two meter readings have the same digits in the same order, 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\n\nWhat is the next odometer reading at which the readings match again?",
"options": [],
"answer": "See solution",
"solution": "*Alternative i*\n\nThe 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. If the two meter readings have the same digits in the same order, 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| trip meter | odometer | match? |\n|--------------------|------------------|-----------------------|\n| 3862 | 3862 | yes |\n| 3863 to 9992 | 3862 to 4475 | no: trip is faster |\n| 9993 to 0002 | 4475 to 4476 | no |\n| 0003 to 4472 | 4476 to 4923 | no: trip < odo |\n| 4473 to 4922 | 4923 to 4968 | no: trip < odo |\n| 4923 to 4962 | 4968 to 4972 | no: trip < odo |\n| 4963 to 4971 | 4972 | no |\n| 4972 | 4973 | no |\n| 4973 | 4973 | yes |\n\nHence, the odometer reading is 4973 the next time it matches the trip meter reading.\n\n*Alternative ii*\n\nThe odometer changed to 3862 at the same time the trip meter changed to 386.2. 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": 21444,
"subject": "Mathematics (Olympiad)",
"question": "The isosceles triangle $ABE$, with $m(\\angle ABE) = 120^\\circ$, is constructed outside the square $ABCD$. Denote $M$ as the orthogonal projection of $B$ onto the bisector of angle $EAB$, $N$ as the orthogonal projection of $M$ onto $AB$, and $P$ as the intersection point of the lines $CN$ and $MB$. Let $G$ be the centroid of triangle $ABE$. Prove that the lines $PG$ and $AE$ are parallel.",
"options": [],
"answer": "See solution",
"solution": "$m(\\angle BAM) = 15^\\circ$ implies $MN = \\frac{1}{4}AB$. Since $MN \\parallel BC$, it follows that $\\triangle PMN \\sim \\triangle PBC$, hence\n\n$$\n\\frac{PM}{PB} = \\frac{MN}{BC} = \\frac{1}{4}\n$$\n\nLet $Q = BM \\cap AE$. Then $\\frac{PM}{BQ} = \\frac{1}{6}$, so\n\n$$\nPB = PM + \\frac{1}{2}BQ = \\frac{2}{3}BQ.\n$$\n\nIf $F$ is the midpoint of $AE$, then $BG = \\frac{2}{3}BF$, and by the converse of Thales' Theorem, $PG \\parallel AE$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21445,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c$ be positive numbers less than $2$ such that $$a^2 + b^2 + c^2 + abc = 4.$$ Prove that $$ (4 - a^2)(4 - b^2)(4 - c^2)a^2b^2c^2 \\leq (2a + bc)(2b + ca)(2c + ab). $$",
"options": [],
"answer": "See solution",
"solution": "First, we will prove the following lemma.\n\n**Lemma 1.** If positive numbers $a, b, c$ satisfy $a^2 + b^2 + c^2 + abc = 4$, then the variables $x, y, z$ determined by the equations $a = \\frac{2}{\\sqrt{(3y-1)(3z-1)}}$, $b = \\frac{2}{\\sqrt{(3z-1)(3x-1)}}$, $c = \\frac{2}{\\sqrt{(3x-1)(3y-1)}}$ satisfy $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = 3$.\n\n*Proof.* By plugging in variables into Lemma's condition, we obtain\n$$\n4 = \\frac{4}{(3y-1)(3z-1)} + \\frac{4}{(3z-1)(3x-1)} + \\frac{4}{(3x-1)(3y-1)} + \\frac{8}{(3x-1)(3y-1)(3z-1)} \\Rightarrow\n$$\n$$ (3x-1)(3y-1)(3z-1) = (3x-1) + (3y-1) + (3z-1) + 2 = 3x+3y+3z-1 \\Rightarrow 27xyz - 9xy - 9yz - 9zx + (3x+3y+3z-1) = 3x+3y+3z-1 \\Rightarrow 3xyz = xy + yz + zx $$\nwhich implies the needed equality.\n\n*End of Lemma 1 proof.*\n\n$$ 3xyz = xy + yz + zx \\geq 3\\sqrt[3]{(xyz)^2} \\Rightarrow xyz \\geq 1 \\Rightarrow x + y + z \\geq 3\\sqrt[3]{xyz} \\geq 3. $$\n\nApply Lemma 1. Then:\n$$\n\\begin{aligned}\n(4-a^2)(4-b^2)(4-c^2) &= \\left(4-\\frac{4}{(3y-1)(3z-1)}\\right) \\cdot \\left(4-\\frac{4}{(3z-1)(3x-1)}\\right) \\cdot \\left(4-\\frac{4}{(3x-1)(3y-1)}\\right) \\\\\n&= 64 \\cdot 27 \\cdot \\frac{(3yz-y-z)(3zx-z-x)(3xy-x-y)}{(3x-1)^2(3y-1)^2(3z-1)^2} \\leq \\frac{8}{27} \\\\\n\\text{(since by Lemma 1, we obtain } (3x-1)(3y-1)(3z-1) = 3x+3y+3z-1 \\geq 8) \\\\\n&\\leq 27 \\cdot (3yz-y-z)(3zx-z-x)(3xy-x-y) \\leq \\\\\n&\\leq 27 \\cdot (27(xyz)^2 - 9(x^2yz(y+z)+\\dots) + 3(yz(y+x)(z+x)+\\dots) - \\\\\n&\\qquad -(x+y)(y+z)(z+x)) = \\\\\n&\\leq 27 \\cdot (27(xyz)^2 - 18xyz(xy + yz + zx) + 9xyz(x+y+x) + \\\\\n&\\qquad 3((xy)^2 + (yz)^2 + (zx)^2) - (xy + yz + zx)(x+y+x) + xyz) = \\\\\n\\text{(since } 3xyz = xy + yz + zx) \\\\\n&= 27 \\cdot (-27(xyz)^2 + 6xyz(x+y+x) + 3((xy)^2 + (yz)^2 + (zx)^2) + xyz) = \\\\\n\\text{(since } 9(xyz)^2 = (xy)^2 + (yz)^2 + (zx)^2 + 2x^2yz + 2xy^2z + 2xyz^2) \\\\\n&= 27 \\cdot (3(9(xyz)^2 - 2xyz(x+y+z)) - 27(xyz)^2 + 6xyz(x+y+x) + xyz) = 27xyz \\Rightarrow \\\\\n&\\qquad (4-a^2)(4-b^2)(4-c^2) \\leq 27xyz.\n\\end{aligned}\n$$\n\n**Lemma 2.** If positive numbers $x, y, z$ satisfy $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = 3$, then variables $a, b, c$ determined by $x = \\frac{2a+bc}{3bc}$, $y = \\frac{2b+ca}{3ca}$, $z = \\frac{2c+ab}{3ab}$ satisfy\n$$ a^2 + b^2 + c^2 + abc = 4. $$\n\n*Proof.* By plugging in variables into Lemma's condition we obtain\n$$\n\\begin{aligned}\n3 &= \\frac{3bc}{2a+bc} + \\frac{3ca}{2b+ca} + \\frac{3ab}{2c+ab} \\Rightarrow \\\\\n(2a+bc)(2b+ca)(2c+ab) &= 3(abc)^2 + 4abc(a^2 + b^2 + c^2) + 4a^2b^2 + 4b^2c^2 + 4c^2a^2.\n\\end{aligned}\n$$\nOn the other hand,\n$$\n\\begin{aligned}\n(2a+bc)(2b+ca)(2c+ab) &= (abc)^2 + 2abc(a^2 + b^2 + c^2) + 4a^2b^2 + 4b^2c^2 + 4c^2a^2 + 8abc \\\\\n\\Rightarrow a^2 + b^2 + c^2 + abc = 4, \\text{ Q.E.D.}\n\\end{aligned}\n$$\n*End of Lemma 2 proof.*\n\nNow by Lemma 2 we obtain that\n$$\n(4-a^2)(4-b^2)(4-c^2) \\leq 27xyz = \\frac{(2a+bc)(2b+ca)(2c+ab)}{(abc)^2} \\Rightarrow \\\\\n(4-a^2)(4-b^2)(4-c^2)a^2b^2c^2 \\leq (2a+bc)(2b+ca)(2c+ab).\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21446,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ be a non-negative real number and a sequence $(u_n)$ defined as:\n\n$$u_1 = 6, \\quad u_{n+1} = \\frac{2n + a}{n} + \\sqrt{\\frac{n + a}{n} u_n + 4}$$\n\nfor all positive integers $n$.\n\n**a)** For $a = 0$, prove that $(u_n)$ has a finite limit and find its value.\n\n**b)** For $a \\ge 0$, prove that $(u_n)$ has a finite limit.",
"options": [],
"answer": "See solution",
"solution": "**a)** For $a = 0$, the sequence $(u_n)$ is defined by\n\n$$u_1 = 6, \\quad u_{n+1} = 2 + \\sqrt{u_n + 4}, \\quad \\forall n \\in \\mathbb{N}^*.$$ \n\nIt is clear that $u_n \\ge 2$ for all positive integers $n$. On the other hand, $u_2 < u_1$. By induction, $(u_n)$ is decreasing. Hence, $(u_n)$ has a finite limit $l$. Letting $n \\to \\infty$, we get $l = 2 + \\sqrt{l + 4}$, so $l = 5$.\n\n**b)** First, we prove that $(u_n)$ is bounded. Let $n_0$ be a positive integer such that $n_0 > a$. Choose $M > 10$ such that $M > \\max(u_1, u_2, \\dots, u_{n_0})$. Then\n\n$$u_{n_0+1} \\le 3 + \\sqrt{2u_{n_0} + 4} \\le 3 + \\sqrt{2M + 4} \\le M.$$ \n\nBy induction, $u_n \\le M$ for all positive integers $n$. Since $u_n > 0$, $(u_n)$ is bounded.\n\nNext, we show that $(u_n)$ is monotone (not necessarily from the first term). If $(u_n)$ is non-decreasing, the statement is proved. Otherwise, there exists $m$ such that $u_m > u_{m+1}$. Then\n\n$$\n\\frac{2m+a}{m} + \\sqrt{\\frac{m+a}{m}u_m + 4} > \\frac{2m+2+a}{m+1} + \\sqrt{\\frac{m+1+a}{m+1}u_{m+1} + 4}.\n$$\n\nTherefore, $u_{m+1} > u_{m+2}$. By induction, $(u_n)$ is decreasing from $u_m$ onward. Since $(u_n)$ is bounded, it has a finite limit.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21447,
"subject": "Mathematics (Olympiad)",
"question": "Let $M$ be the midpoint of the side $AB$ of the trapezoid $ABCD$ ($AD \\parallel BC$), $O$ the point of intersection of $AC$ and $BD$, and $AO = BO$. On the ray $OM$, the point $P$ is chosen such that $\\angle PAC = 90^\\circ$. Show that $\\angle AMD = \\angle APC$.\n\n",
"options": [],
"answer": "See solution",
"solution": "$\\triangle AOD \\sim \\triangle COB$, so $\\frac{OC}{OA} = \\frac{OB}{OD}$. Since $AM$ is the altitude in the right-angled $\\triangle APO$, we have $OC \\cdot OD = OA \\cdot OB = OA^2 = OM \\cdot OP$. So $\\frac{OC}{OM} = \\frac{OP}{OD}$. Note that $OM$ is the bisector of $\\angle AOB$, $\\angle MOD = \\angle COP$, and therefore $\\triangle OCP \\sim \\triangle OMD$. Thus, $\\angle OCP = \\angle OMD$, $90^\\circ - \\angle OCP = 90^\\circ - \\angle OMD$, and hence $\\angle AMD = \\angle APC$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21448,
"subject": "Mathematics (Olympiad)",
"question": "Find the smallest possible value of the expression $|5^{4m+3} - n^2|$ for integers $m$ and $n$.",
"options": [],
"answer": "See solution",
"solution": "When $m = 0$ and $n = 11$, the value of the expression $|5^{4m+3} - n^2|$ is $|5^3 - 11^2| = |125 - 121| = 4$. We will show that this is the smallest value it can attain.\n\nObviously, the expression cannot be equal to $0$, because $5^{4m+3}$ is not a perfect square. Now, we must consider the cases where the value is $1$, $2$, or $3$.\n\nAssume that $n^2 - 5^{4m+3} > 0$. If $n^2 - 5^{4m+3} = 1$, then $5^{4m+3} = n^2 - 1 = (n-1)(n+1)$. Since $n$ is even, $n-1$ and $n+1$ are coprime, which implies $n-1 = 1$ and $n+1 = 5^{4m+3}$. This is not possible. If $n$ is divisible by $5$, then $n^2 - 5^{4m+3} \\geq 5$. If $n = 5k \\pm 1$, then $n^2 - 5^{4m+3} \\equiv 1 \\pmod{5}$, and if $n = 5k \\pm 2$, then $n^2 - 5^{4m+3} \\equiv 4 \\pmod{5}$. Since none of the remainders are equal to $2$ or $3$ and $n^2 - 5^{4m+3} > 1$, we conclude that $n^2 - 5^{4m+3} \\geq 4$.\n\nNow, assume $5^{4m+3} - n^2 > 0$. This number cannot be equal to $2$ or $3$. We can see this by considering the remainders modulo $5$, because $5^{4m+3} - n^2$ is either divisible by $5$ (and hence greater than or equal to $5$), or $n$ has the form $n = 5k \\pm 1$ or $n = 5k \\pm 2$. So, either $5^{4m+3} - n^2 \\equiv -1 \\equiv 4 \\pmod{5}$ or $5^{4m+3} - n^2 \\equiv -4 \\equiv 1 \\pmod{5}$.\n\nAssume there exist $m$ and $n$ such that $5^{4m+3} - n^2 = 1$. We can rewrite the equation as\n\n$$\nn^2 = 5^{4m+3} - 1 = (5-1)(5^{4m+2} + 5^{4m+1} + \\dots + 5 + 1) = 4(5^{4m+2} + 5^{4m+1} + \\dots + 5 + 1).\n$$\n\nThis implies that $5^{4m+2} + 5^{4m+1} + \\dots + 5 + 1$ is a perfect square. If it is equal to $l^2$, then $(l-1)(l+1) = 5(5^{4m+1} + 5^4 + \\dots + 5 + 1)$. Consider $5^{4m+1} + 5^4 + \\dots + 5 + 1$ modulo $4$:\n\n$$\n5^{4m+1} + 5^4 + \\dots + 5 + 1 \\equiv 1 + 1 + \\dots + 1 + 1 \\equiv (4m + 2) \\equiv 2 \\pmod{4},\n$$\n\nsince there are $4m + 2$ summands. This sum is divisible by $2$ but not by $4$. So, $(l-1)(l+1)$ is even. On the other hand, $l-1$ and $l+1$ have the same parity, so $(l-1)(l+1)$ is divisible by $4$, a contradiction. Thus, such $m$ and $n$ do not exist.\n\nWe have shown that $n^2 - 5^{4m+3}$ and $5^{4m+3} - n^2$ cannot be equal to $0$, $1$, $2$, or $3$, so the smallest possible value of $|5^{4m+3} - n^2|$ is $4$, and this value is attained when $n = 11$ and $m = 0$, for example.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21449,
"subject": "Mathematics (Olympiad)",
"question": "Given that\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} = -\\frac{1}{abcd},\n$$\n\nshow that\n\n$$\n(ab - cd)(c + d) = -1\n$$\nfor all admissible values of $a, b, c, d$.",
"options": [],
"answer": "See solution",
"solution": "By the given condition,\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} = -\\frac{1}{abcd}\n$$\n\nwhich implies\n\n$$\nbcd + cda + dab + abc = -1.\n$$\n\nNow,\n$$\n\\begin{aligned}\n-1 = bcd + cda + dab + abc &= (bcd + cda) + (dab + abc) \\\\\n&= cd(b + a) + ab(c + d) \\\\\n&= ab(c + d) - cd(c + d) \\\\\n&= (ab - cd)(c + d).\n\\end{aligned}\n$$\n\nTherefore, $(ab - cd)(c + d) = -1$ for all admissible values of $a, b, c, d$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21450,
"subject": "Mathematics (Olympiad)",
"question": "Determine all integer solutions $(x, y, z)$ of the equation\n$$\nx^4 + x^2 = 7^{2z} y^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "Clearly, we have solutions for any value of $z$ if we also have $x = y = 0$. We claim that there are no other solutions.\n\nTo show this, first assume that $z$ is even. Then $7^{2z} y^2$ is a perfect square, so $x^4 + x^2 = x^2(x^2 + 1)$ must also be a perfect square. The only integer $x$ for which both $x^2$ and $x^2 + 1$ are perfect squares is $x = 0$, so $x = y = 0$ in this case.\n\nNow, assume $z$ is positive and odd. Let $z = 2c + 1$. Since $x^2(x^2 + 1)$ is divisible by $7$ and $x^2 + 1$ can only be congruent to $1, 2, 3,$ or $5$ modulo $7$, it follows that $x$ must be divisible by $7$. Let $x = 7^a u$ and $y = 7^b v$ with $u$ and $v$ not divisible by $7$. The equation becomes\n$$\n7^{2a} u^2 (7^{2a} u^2 + 1) = 7^{2(b + c) + 1} v^2,\n$$\nwhich is a contradiction, since the left side is divisible by an even number of sevens, while the right side is divisible by an odd number.\n\nFinally, assume $z$ is negative and let $z = -w$. The equation becomes $7^w x^2 (x^2 + 1) = y^2$. If $w$ is even, as before, $x = y = 0$ is the only solution. If $w$ is odd, write $w = 2c + 1$; the same argument as above applies, leading to a contradiction.\n\nThus, the only integer solutions are $x = y = 0$ for any integer $z$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21451,
"subject": "Mathematics (Olympiad)",
"question": "Given a right triangle $\\triangle ABC$ with $\\angle ABC = 90^\\circ$, where $AB$ and $BC$ are both diagonals of squares with side lengths $5$ cm and $7$ cm respectively, find the area of $\\triangle ABC$.\n\n",
"options": [],
"answer": "See solution",
"solution": "$AB = \\sqrt{5^2 + 5^2} = \\sqrt{50} = 5\\sqrt{2}$\n\n$BC = \\sqrt{7^2 + 7^2} = \\sqrt{98} = 7\\sqrt{2}$\n\n\\text{Area of } \\triangle ABC = \\frac{1}{2}(5\\sqrt{2})(7\\sqrt{2}) = 35\\ \\text{cm}^2$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21452,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle in which $AB > AC$; $AM$ is the median and $AK$ is the angle bisector with $M, K$ on $BC$. Let $L$ be a point on $AM$ such that $KL$ is parallel to $AC$. Prove that $CL$ is perpendicular to $AK$.",
"options": [],
"answer": "See solution",
"solution": "\n\nExtend $AK$ to meet the circumcircle of $ABC$ at $D$, and join $MD$. Let $P$ be the point of intersection of $AK$ and $CL$. Observe that $\\angle DMK = 90^\\circ$ and $D, M, O$ are collinear. We show that $\\triangle DMK$ is similar to $\\triangle CPK$, which proves that $CL$ is perpendicular to $AK$. It is sufficient to prove that $\\dfrac{KD}{KC} = \\dfrac{KM}{KP}$. But $AK \\cdot KD = BK \\cdot KC$, which gives $\\dfrac{KD}{KC} = \\dfrac{BK}{AK}$. Thus we need to prove that\n\n$$\n\\frac{KM}{KP} = \\frac{BK}{AK}.\n$$\n\nSince $CL$ is a transversal in triangle $AMK$, Menelaus' theorem gives\n\n$$\n\\frac{PA}{KP} = \\frac{AL \\cdot MC}{LM \\cdot CK}.\n$$\n\nBut $KL$ is parallel to $CA$, so $\\dfrac{AL}{LM} = \\dfrac{CK}{KM}$. This implies that\n\n$$\n\\frac{PA}{KP} = \\frac{CM}{KM}.\n$$\n\nThus\n\n$$\n\\frac{AK}{KP} = \\frac{CM + MK}{KM}.\n$$\n\nAll we need to show is $BK = CM + MK$. Since $BM = MC$, the result follows.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21453,
"subject": "Mathematics (Olympiad)",
"question": "Inside an equilateral triangle $ABC$, a point $M$ is chosen. Let points $M_1$, $M_2$, and $M_3$ be the reflections of $M$ across the sides $BC$, $AC$, and $AB$ of the triangle, respectively. Prove that\n\n$$\n\\overrightarrow{MM_1} + \\overrightarrow{MM_2} + \\overrightarrow{MM_3} = \\overrightarrow{MA} + \\overrightarrow{MB} + \\overrightarrow{MC}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Draw through point $M$ lines parallel to the sides of $ABC$. Let these lines intersect $AB$, $BC$, and $AC$ at points $C_1$, $C_2$; $A_1$, $A_2$; and $B_1$, $B_2$, respectively (see the figure below).\n\n\n\nIt follows that $C_1A_2 \\parallel AC$, $A_1B_2 \\parallel AB$, and $B_1C_2 \\parallel BC$, so $C_1A_2$, $A_1B_2$, and $B_1C_2$ intersect at $M$. Consider $\\triangle A_1MA_2$. This triangle is equilateral. The line $M_1M$ contains its altitude, since it is perpendicular to $BC$. Thus, $M_1M$ is twice the median, so $\\overrightarrow{MA_1} + \\overrightarrow{MA_2} = \\overrightarrow{MM_1}$. Similarly, $\\overrightarrow{MB_1} + \\overrightarrow{MB_2} = \\overrightarrow{MM_2}$ and $\\overrightarrow{MC_1} + \\overrightarrow{MC_2} = \\overrightarrow{MM_3}$.\n\nOn the other hand, $MC_1AB_2$ is a parallelogram, so $\\overrightarrow{MA} = \\overrightarrow{MC_1} + \\overrightarrow{MB_2}$. Similarly, $\\overrightarrow{MB} = \\overrightarrow{MC_2} + \\overrightarrow{MA_1}$ and $\\overrightarrow{MC} = \\overrightarrow{MA_2} + \\overrightarrow{MB_1}$. Therefore,\n\n$$\n\\overrightarrow{MM_1} + \\overrightarrow{MM_2} + \\overrightarrow{MM_3} = \\overrightarrow{MA_1} + \\overrightarrow{MA_2} + \\overrightarrow{MB_1} + \\overrightarrow{MB_2} + \\overrightarrow{MC_1} + \\overrightarrow{MC_2} = \\overrightarrow{MA} + \\overrightarrow{MB} + \\overrightarrow{MC}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21454,
"subject": "Mathematics (Olympiad)",
"question": "Circles $S_1$ and $S_2$ meet at $L$ and $M$. Let $P$ be a point on $S_2$. Let $PL$ and $PM$ meet $S_1$ again at $Q$ and $R$ respectively. The lines $QM$ and $RL$ meet at $K$. Show that, as $P$ varies on $S_2$, $K$ lies on a fixed circle.",
"options": [],
"answer": "See solution",
"solution": "Let $\\alpha = \\angle O_2O_1L$ and $\\beta = \\angle LO_2O_1$. Since $O_1L = O_1M$ and $O_2L = O_2M$, $O_1LO_2M$ is a kite, so $O_1O_2$ bisects both $\\angle MO_1L$ and $\\angle LO_2M$. Also, $O_1O_2$ is perpendicular to $LM$.\n\nWe split the question into two cases depending on which side of $LM$ the point $K$ lies; in the first case we relabel $K, P, Q, R$ as $K_1, P_1, Q_1, R_1$ and in the second case as $K_2, P_2, Q_2, R_2$.\n\n\n\nLet $P_1$ be a point on $S_2$ such that $K_1$ is on the same side of $LM$ as $O_1$. Let $P_1L$ and $P_1M$ meet $S_1$ again at $Q_1$ and $R_1$ respectively. Then $Q_1M$ and $R_1L$ meet at $K_1$. We see that\n\n$$\n\\begin{align*}\n\\angle MK_1L &= \\angle MQ_1L + \\angle Q_1LR_1 \\\\\n&= \\frac{1}{2}\\angle MO_1L + (\\angle LP_1M - \\angle MR_1L) \\\\\n&= \\frac{2\\alpha}{2} + \\frac{1}{2}\\angle LO_2M + \\frac{1}{2}\\angle MO_1L.\n\\end{align*}\n$$\n\nTherefore,\n\n$$\n\\angle MK_1L = \\alpha + \\beta + \\alpha\n$$\n\nwhich is fixed because $\\alpha$ and $\\beta$ are fixed. Hence $K_1$ lies on a fixed circle with chord $LM$.\n\n\n\nLet $P_2$ be a point on $S_2$ such that $K_2$ is on the opposite side of $LM$ to $O_1$. Let $P_2L$ and $P_2M$ meet $S_1$ again at $Q_2$ and $R_2$ respectively. Then $Q_2M$ and $R_2L$ meet at $K_2$. We see that\n\n$$\n\\begin{aligned}\n\\angle LK_2M &= \\angle LR_2M - \\angle K_2MP_2 \\\\\n&= (180^\\circ - \\angle MQ_2L) - (\\angle MQ_2L + \\angle LP_2M) \\\\\n&= 180^\\circ - 2\\angle MQ_2L - \\angle LP_2M \\\\\n&= 180^\\circ - \\angle MO_1L - \\frac{1}{2}\\angle LO_2M.\n\\end{aligned}\n$$\n\nTherefore,\n\n$$\n\\angle LK_2M = 180^\\circ - 2\\alpha - \\beta\n$$\n\nwhich is fixed because $\\alpha$ and $\\beta$ are fixed. Hence $K_2$ lies on a fixed circle with chord $LM$.\n\nNow $\\angle MK_1L + \\angle LK_2M = 180^\\circ$ and $K_1, K_2$ are on opposite sides of $LM$, so $K_1$ and $K_2$ lie on the same circle.\n\nTherefore, $K$ lies on a fixed circle, as required.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21455,
"subject": "Mathematics (Olympiad)",
"question": "Let $p, q$ be coprime integers such that $\\frac{p}{q} \\le 1$. For which $p, q$ do there exist even integers $b_1, b_2, \\dots, b_n$ such that\n\n$$\n\\frac{p}{q} = \\frac{1}{b_1 + \\frac{1}{b_2 + \\frac{1}{b_3 + \\dots}}}\n$$",
"options": [],
"answer": "See solution",
"solution": "Set $b_i = 2\\ell_i$, so all $b_i$ are even. At the first step, we get the pairs\n\n$$\nA := \\{ (p, q) : (p, q) = (1, 2\\ell_1), \\ell_1 \\in \\mathbb{Z} \\}\n$$\n\nAt each subsequent step, we expand $A$ by adding pairs $(p', q')$ defined as\n\n$$\n\\left\\{ (p', q') : \\frac{p'}{q'} = \\frac{1}{2\\ell + \\frac{p}{q}},\\ (p, q) \\in A,\\ \\ell \\in \\mathbb{Z} \\right\\}\n$$\n\nBy this, we obtain\n\n$$\np' = q, \\quad q' = 2\\ell q + p\n$$\n\nIf $(p, q)$ is obtained in the process, we also add $(p', q')$ as above. We want to characterize all pairs $(p, q)$ that can be obtained this way. Note that if $(p, q) = 1$, then $(p', q') = 1$ as well. The set of pairs $(p, q)$ in question is\n\n$$\nA := \\{(p, q) : p, q \\in \\mathbb{Z} \\setminus \\{0\\},\\ |p| < |q|,\\ (p, q) = 1,\\ \\text{one of } p, q \\text{ is even, the other is odd}\\}\n$$\n\nWe exclude $|p| = |q|$ because $\\frac{p}{q} = \\pm 1$ cannot be represented as required. The transformation above shows we cannot step outside $A$. To prove all pairs in $A$ can be generated, take $(p', q') \\in A$ and search for $(p, q) \\in A$ that generates $(p', q')$ via the previous formula, with $(p, q)$ \"less\" than $(p', q')$. Solving for $(p, q)$ yields\n\n$$\np = q' - 2\\ell p', \\quad q = p'\n$$\n\nWe can choose $\\ell \\in \\mathbb{Z}$ so that $|q' - 2\\ell p'| < |p'|$ (since $p' \\neq 0$). The points $q' - 2\\ell p'$ are $2p'$ apart, and none hits $p'$ because $(p', q') = 1$. The closest point to $0$ has magnitude less than $p'$. Thus, starting from $(p', q') \\in A$, we find $(p, q) \\in A$ that generates $(p', q')$ and $|p| < |p'| < |q'|$. Induction completes the proof. For example, $(1, 2), (-1, 2), (1, -2), (-1, -2)$ can be represented as required. Assume all $(p, q) \\in A$ with $|p| \\le N, |q| \\le N$ can be represented. Take $(p', q') \\in A$ with $|q'| = N + 1$. As shown, there exists $(p, q) \\in A$ with $|p| < |q| \\le N$ such that\n\n$$\n\\frac{p'}{q'} = \\frac{1}{2\\ell + \\frac{p}{q}}\n$$\n\nThe induction step is complete, and the result follows. The representation is also unique: for $(p', q') \\in A$, the pair $(p, q) \\in A$ satisfying the above and $|p| < |p'|$ is unique.\n\n**Remark.** The main difficulty is that the answer was not given to the students. Some motivation for guessing the answer can be found in this blog $\\Box$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21456,
"subject": "Mathematics (Olympiad)",
"question": "Find all integers $n$ such that $2^n - n^3 = 24$. What is the value of $2^n + n^3$ for such $n$?",
"options": [],
"answer": "See solution",
"solution": "We check $n < 2$ and $2 \\leq n \\leq 9$ and find that none satisfy $2^n - n^3 = 24$.\n\nFor $n = 10$:\n$$\n2^{10} - 10^3 = 1024 - 1000 = 24\n$$\nSo $n = 10$ is a solution.\n\nFor $n > 10$, note that\n$$\n2^{n+1} - (n+1)^3 = 2 \\cdot 2^n - (n+1)^3 = 2^n - n^3 + (2^n - 3n^2 - 3n - 1)\n$$\nFor $n \\geq 10$, $2^n - n^3 > 0$ and the difference increases, so no further solutions exist.\n\nThus, the only possible value for $2^n + n^3$ is:\n$$\n2^{10} + 10^3 = 1024 + 1000 = 2024\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21457,
"subject": "Mathematics (Olympiad)",
"question": "A triangle $ABC$ is inscribed in a circle $\\omega$. A variable line $\\ell$ chosen parallel to $BC$ meets segments $AB$ and $AC$ at points $D$ and $E$ respectively, and meets $\\omega$ at points $K$ and $L$ (where $D$ lies between $K$ and $E$). Circle $\\gamma_1$ is tangent to the segments $KD$ and $BD$ and also tangent to $\\omega$, while circle $\\gamma_2$ is tangent to the segments $LE$ and $CE$ and also tangent to $\\omega$. Determine the locus, as $\\ell$ varies, of the meeting point of the common inner tangents to $\\gamma_1$ and $\\gamma_2$.",
"options": [],
"answer": "See solution",
"solution": "Let $P$ be the meeting point of the common inner tangents to $\\gamma_1$ and $\\gamma_2$. Also, let $b$ be the angle bisector of $\\angle BAC$. Since $KL \\parallel BC$, $b$ is also the angle bisector of $\\angle KAL$.\n\nLet $\\mathfrak{H}$ be the composition of the symmetry $\\mathfrak{S}$ with respect to $b$ and the inversion $\\mathfrak{I}$ of centre $A$ and ratio $\\sqrt{AK \\cdot AL}$. (It is readily seen that $\\mathfrak{S}$ and $\\mathfrak{I}$ commute, so since $\\mathfrak{S}^2 = \\mathfrak{I}^2 = \\text{id}$, then also $\\mathfrak{H}^2 = \\text{id}$, the identical transformation.) The elements of the configuration interchanged by $\\mathfrak{H}$ are summarized below:\n\n| point K | ↔↔ | point L |
|---|
| line KL | ↔↔ | circle $\\omega$ |
| ray AB | ↔↔ | ray AC |
| point B | ↔↔ | point E |
| point C | ↔↔ | point D |
| segment BD | ↔↔ | segment EC |
| arc BK | ↔↔ | segment EL |
| arc CL | ↔↔ | segment DK |
\n\nLet $O_1$ and $O_2$ be the centres of circles $\\gamma_1$ and $\\gamma_2$. Since the circles $\\gamma_1$ and $\\gamma_2$ are determined by their construction (in a unique way), they are interchanged by $\\mathfrak{H}$, therefore the rays $AO_1$ and $AO_2$ are symmetrical with respect to $b$. Denote by $\\varrho_1$ and $\\varrho_2$ the radii of $\\gamma_1$ and $\\gamma_2$. Since $\\angle O_1AB = \\angle O_2AC$, we have $\\varrho_1/\\varrho_2 = AO_1/AO_2$. On the other hand, from the definition of $P$ we have $O_1P/O_2P = \\varrho_1/\\varrho_2 = AO_1/AO_2$; this means that $AP$ is the angle bisector of $\\angle O_1AO_2$ and therefore of $\\angle BAC$.\n\nThe limiting, degenerated, cases are when the parallel line passes through $A$ – when $P$ coincides with $A$; respectively when the parallel line is $BC$ – when $P$ coincides with the foot $A' \\in BC$ of the angle bisector of $\\angle BAC$ (or any other point on $BC$). By continuity, any point $P$ on the open segment $AA'$ is obtained for some position of the parallel, therefore the locus is the open segment $AA'$ of the angle bisector $b$ of $\\angle BAC$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21458,
"subject": "Mathematics (Olympiad)",
"question": "Дали постои неконстантна низа природни броеви $a_1, a_2, \\ldots, a_n, \\ldots$, таква што за секој природен број $k \\geq 2$ е исполнето равенството\n\n$$\na_k = \\frac{2a_{k-1}a_{k+1}}{a_{k-1} + a_{k+1}}?\n$$",
"options": [],
"answer": "See solution",
"solution": "Претпоставуваме дека постои таква низа природни броеви. За низата од реципрочни вредности $b_n = \\frac{1}{a_n}$, $n \\in \\mathbb{N}$, добиваме:\n\n$$\nb_n = \\frac{1}{a_n} = \\frac{1}{\\frac{2a_{n-1}a_{n+1}}{a_{n-1} + a_{n+1}}} = \\frac{1}{2\\left(\\frac{1}{a_{n-1}} + \\frac{1}{a_{n+1}}\\right)} = \\frac{b_{n-1} + b_{n+1}}{2}.\n$$\n\nЗначи, $b_2, b_3, \\ldots$ е аритметичка прогресија. Бидејќи $(a_n)_{n=1}^\\infty$ е неконстантна, и $(b_n)_{n=2}^\\infty$ е неконстантна аритметичка низа од позитивни реални броеви. Значи, постои $d \\neq 0$ така што\n\n$$\nb_n = b_2 + (n-2)d, \\quad n \\geq 2.\n$$\n\nЗа доволно големо $n$, $b_n > 1$ или $b_n < 0$ (во зависност од знакот на $d$). Но, бидејќи $a_n \\in \\mathbb{N}$, имаме\n\n$$\n0 < b_n = \\frac{1}{a_n} \\leq 1, \\quad n \\geq 2.\n$$\n\nОва е контрадикција, па таква низа природни броеви не постои.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21459,
"subject": "Mathematics (Olympiad)",
"question": "Determine all real numbers $x, y, z \\in (0,1)$ that satisfy simultaneously the conditions:\n\n$$\n\\begin{cases} \n(x^2 + y^2)\\sqrt{1-z^2} \\geq z \\\\\n(y^2 + z^2)\\sqrt{1-x^2} \\geq x \\\\\n(z^2 + x^2)\\sqrt{1-y^2} \\geq y\n\\end{cases}\n$$",
"options": [],
"answer": "See solution",
"solution": "The first inequality is equivalent to\n\n$$\n\\frac{z^2}{x^2 + y^2} \\leq z\\sqrt{1-z^2}.\n$$\n\nSince\n\n$$\nz\\sqrt{1-z^2} = \\sqrt{z^2(1-z^2)} \\leq \\frac{z^2 + 1 - z^2}{2} = \\frac{1}{2},\n$$\n\nit follows that $\\frac{z^2}{x^2 + y^2} \\leq 2$, and therefore $x^2 + y^2 \\geq 2z^2$. Writing the other two similar inequalities and adding them together yields $2(x^2 + y^2 + z^2) \\geq 2(x^2 + y^2 + z^2)$. Consequently, equality must hold in all the inequalities above, hence $x = y = z = \\frac{\\sqrt{2}}{2}$. Clearly this triple satisfies all the requirements.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21460,
"subject": "Mathematics (Olympiad)",
"question": "James has a red jar, a blue jar, and a pile of 100 pebbles. Initially, both jars are empty. A move consists of moving a pebble from the pile into one of the jars or returning a pebble from one of the jars to the pile. The numbers of pebbles in the red and blue jars determine the state of the game. The following conditions must be satisfied:\n\n- The red jar may never contain fewer pebbles than the blue jar.\n- The game may never be returned to a previous state.\n\nWhat is the maximum number of moves that James can make?",
"options": [],
"answer": "See solution",
"solution": "Let $n$ be the total number of pebbles in both jars at any given point in the game. When a move occurs, $n$ either increases or decreases by 1, so $n$ alternates between even and odd numbers (starting even).\n\nGiven $r$ pebbles in the red jar ($r$ an integer between $0$ and $100$), the range of valid values for $b$ pebbles in the blue jar is:\n\n- $0 \\leq b \\leq r$ when $0 \\leq r \\leq 50$\n- $0 \\leq b \\leq 100 - r$ when $51 \\leq r \\leq 100$\n\nby applying the conditions that $b \\leq r$ and $b + r \\leq 100$.\n\nLet $o$ and $e$ be the number of times that $n$ was odd and even respectively during a single game. Since $n$ starts even and alternates between even and odd, we have either $e = o$ or $e = o + 1$.\n\nWe consider the number of odd states for each $r$ and sum over $r$ from $1$ to $100$. In the summation below we only list explicitly the number of states for $r = 0, 1, 2, 3, 4, \\dots, 49, 50, 51, 52, \\dots, 97, 98, 99, 100$. We obtain\n\n$$\n\\begin{aligned}\n& 0 + 1 + 1 + 2 + 2 + \\dots + 25 + 25 + 25 + 24 + \\dots + 2 + 2 + 1 + 1 + 0 \\\\\n&= 2(1 + 2 + \\dots + 25) + 2(1 + 2 + \\dots + 24) + 25 \\\\\n&= 25 \\times 26 + 24 \\times 25 + 25 \\\\\n&= 25 \\times 51 \\\\\n&= 1275.\n\\end{aligned}\n$$\n\nThe total number of even states satisfies $e \\leq o + 1 \\leq 1276$. Thus, the total possible number of states is at most $1275 + 1276 = 2551$. This is one more than the total number of moves, since the first state doesn't require a move to get to, but the other states do. So the maximum number of moves is $2550$.\n\nOne strategy using this number of moves is as follows:\n\n1. If $r$ is even and $b = 0$, place one pebble in the red jar if possible.\n2. Otherwise, if $r$ is even but $b > 0$, remove one pebble from the blue jar.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21461,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $n$ for which there exist positive integers $x, y$ such that $837 + n = x^3$ and $837 - n = y^3$.",
"options": [],
"answer": "See solution",
"solution": "Adding the two equations gives:\n\n$$\n1674 = x^3 + y^3 = (x + y)(x^2 - x y + y^2).\n$$\n\nLet $u = x + y$ and $v = x^2 - x y + y^2 = (x + y)^2 - 3 x y = u^2 - 3 x y$. Then $3 x y = u^2 - v$.\n\nAs $1674 = 2 \\cdot 3^3 \\cdot 31 = u v$, there exist positive integers $u_1$ and $v_1$ such that $u = 3 u_1$ and $v = 3 v_1$. The equations become:\n\n$$\n186 = u_1 v_1, \\quad x + y = 3 u_1, \\quad x y = 3 u_1^2 - v_1.\n$$\n\nSince $x, y > 0$, we get $3 u_1^2 > v_1$, so $3 u_1^3 > 186$ and $u_1 \\geq 4$. Also, by the AM-GM inequality, $(x + y)^2 \\geq 4 x y$, which translates to $9 u_1^2 \\geq 12 u_1^2 - 4 v_1$, i.e., $4 v_1 \\geq 3 u_1^2 > 4 u_1$.\n\nThe factors of $186$ are $1, 2, 3, 6, 31, 62, 93, 186$. Using $v_1 > u_1 \\geq 4$, we must have $u_1 = 6$ and $v_1 = 31$. This gives $x + y = 18$ and $x y = 77$, so the quadratic equation is $x^2 - 18 x + 77 = 0$. The solutions are $11$ and $7$. Since $n > 0$, $x > y$, so $(x, y) = (11, 7)$. Therefore,\n\n$$\nn = 837 - 7^3 = 11^3 - 837 = 494.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21462,
"subject": "Mathematics (Olympiad)",
"question": "Consider the cube $ABCD A'B'C'D'$. The angle bisectors of the angles $\\angle A'C'A$ and $\\angle A'AC'$ meet $AA'$ and $A'C'$ at points $P$ and $Q$, respectively. The point $M$ is the foot of the perpendicular from $A'$ onto $C'P$ while $N$ is the foot of the perpendicular from $A'$ onto $AS$. The point $O$ is the center of the face $ABB'A'$.\n\na) Prove that the planes $(MNO)$ and $(AC'B)$ are parallel.\n\nb) Given that $AB = 1$, find the distance between the planes $(MNO)$ and $(AC'B)$.",
"options": [],
"answer": "See solution",
"solution": "a) Denote by $T$ and $R$ the intersections of the straight lines $AC'$ and $A'M$, respectively $AC'$ and $A'N$. In the triangle $A'C'T$, $C'M$ is an angle bisector and an altitude, so $A'M = MT$. In the triangle $A'AR$, $AN$ is an angle bisector and an altitude, so $A'N = NR$. The segment $[MO]$ joins the midpoints of two sides of the triangle $A'TB$, hence $MO \\parallel TB$, and the segment $NO$ joins the midpoints of two sides of triangle $A'RB$, hence $NO \\parallel RB$. The requirement follows from the fact that the planes $(TRB)$ and $(AC'B)$ are the same.\n\n\n\nb) The distance between the two planes is equal to the distance from $O$ to the plane $(AC'B)$, and this last distance is half the distance from $A'$ to the same plane.\nAs the distance from $A'$ to the plane is $\\frac{1}{2}A'D = \\frac{1}{2}\\sqrt{2}$, the required distance is $\\frac{1}{4}\\sqrt{2}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21463,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ and $b$ be two side lengths of triangle $\\Delta ABC$. Their two corresponding medians are perpendicular. Evaluate the third side length using only $a$ and $b$.",
"options": [],
"answer": "See solution",
"solution": "Let $T$ be the centroid of triangle $\\Delta ABC$ and $\\overline{AA_1} = t_a$, $\\overline{BB_1} = t_b$ are the two corresponding medians to the sides $BC$ and $AC$.\n\nFrom the right-angled triangle $\\Delta BA_1T$, using $\\overline{A_1T} = \\frac{1}{3} t_a$ and $\\overline{TB} = \\frac{2}{3} t_b$, we obtain\n\n$$\n\\left(\\frac{a}{2}\\right)^2 = \\left(\\frac{1}{3} t_a\\right)^2 + \\left(\\frac{2}{3} t_b\\right)^2.\n$$\n\nSimilarly, from the right-angled triangle $\\Delta B_1AT$, using $\\overline{AT} = \\frac{2}{3} t_a$ and $\\overline{TB_1} = \\frac{1}{3} t_b$, we have\n\n$$\n\\left(\\frac{b}{2}\\right)^2 = \\left(\\frac{2}{3} t_a\\right)^2 + \\left(\\frac{1}{3} t_b\\right)^2.\n$$\n\nSumming the two equalities, we get\n\n$$\n\\frac{a^2 + b^2}{4} = \\frac{5}{9}(t_a^2 + t_b^2)\n$$\n\nor\n\n$$\nt_a^2 + t_b^2 = \\frac{9}{20}(a^2 + b^2).\n$$\n\nFinally, from the right-angled triangle $\\Delta ABT$ we obtain\n\n$$\nc^2 = \\left(\\frac{2}{3} t_a\\right)^2 + \\left(\\frac{2}{3} t_b\\right)^2 = \\frac{4}{9}(t_a^2 + t_b^2) = \\frac{a^2 + b^2}{5},\n$$\n\nfrom where it follows\n\n$$\nc = \\sqrt{\\frac{a^2 + b^2}{5}}.\n$$\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21464,
"subject": "Mathematics (Olympiad)",
"question": "Consider the equation:\n\n$$\nx^2 + 15y^2 = 4^n\n$$\n\nwhere $n \\geq 1$.\n\nShow that for each $n \\geq 1$, the equation always has at least $n$ non-negative integer solutions $(x, y)$.",
"options": [],
"answer": "See solution",
"solution": "*Remark 1:* If $(x, y)$ is a non-negative integer solution of the equation for $n = k$ ($k \\geq 1$), then $(2x, 2y)$ is a non-negative integer solution for $n = k + 1$.\n\n*Remark 2:* For each $n \\geq 2$, the equation always has one non-negative integer solution $(x, y)$ with both $x$ and $y$ odd (called an odd-solution).\n\n*Proof by induction on $n \\geq 2$.*\n\nFor $n = 2$, $1^2 + 15 \\cdot 1^2 = 16 = 4^2$, so the claim holds.\n\nAssume the claim holds for $n = k$ ($k \\geq 2$). Let $(x, y)$ be an odd-solution for $n = k$. Then:\n\n$$\n4(x^2 + 15y^2) = \\left(\\frac{x + 15y}{2}\\right)^2 + 15\\left(\\frac{x - y}{2}\\right)^2 = \\left(\\frac{x - 15y}{2}\\right)^2 + 15\\left(\\frac{x + y}{2}\\right)^2\n$$\n\nSince $\\frac{x + 15y}{2}$, $\\frac{x - y}{2}$, $\\frac{x - 15y}{2}$, $\\frac{x + y}{2} \\in \\mathbb{Z}$, the pairs $\\left(\\frac{x + 15y}{2}, \\frac{|x - y|}{2}\\right)$ and $\\left(\\frac{|x - 15y|}{2}, \\frac{x + y}{2}\\right)$ are non-negative integer solutions for $n = k + 1$.\n\nMoreover, since $\\frac{x - y}{2} + \\frac{x + y}{2} = x$ is odd, one of $\\frac{|x - y|}{2}$ or $\\frac{x + y}{2}$ must be odd. Thus, at least one of these solutions is an odd-solution for $n = k + 1$.\n\nFor $n = 1$, $2^2 + 15 \\cdot 0^2 = 4$, so $(x, y) = (2, 0)$ is a solution. By Remarks 1 and 2, induction shows that for each $n \\geq 1$, the equation has at least $n$ non-negative integer solutions $(x, y)$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21465,
"subject": "Mathematics (Olympiad)",
"question": "有 666 塊田排成一橫排，每塊田都是麥田或稻田。母雞蘿絲在其中的 $W$ 塊麥田與 $R$ 塊稻田裡各下一顆蛋，使得每顆蛋與其右邊最靠近的蛋之間，至多只有一塊沒有蛋的田。試求最大的正整數 $S$，使得不論麥田與稻田如何分佈，母雞蘿絲都有辦法讓 $|W - R|$ 至少為 $S$。",
"options": [],
"answer": "See solution",
"solution": "答案為 $S = 167$。\n\n首先證明牠總是可以取得 167 分。不失一般性假設有至少 $666/2 = 333$ 個麥田。\n\n考慮以下操作：讓母雞從排頭的田開始。若牠在麥田，則立刻下蛋然後前進一塊田；否則，她先前進一塊田，下蛋，再前進一塊田。注意到此策略保證所有的麥田都有蛋，故 $W \\geq 333$。此外，對於每一個有蛋的稻田，都保證前面有一個沒有蛋的稻田，因此 $R \\leq \\left\\lfloor 333/2 \\right\\rfloor = 166$，故 $W - R \\geq 333 - 166 = 167$。\n\n現在證明 $S = 167$ 是最大值。考慮如下排列（其中 $R$ 為稻田, $W$ 為麥田）：\n\n$$\n\\{R\\}, \\{W, W\\}, \\{R, R\\}, \\{W, W\\}, \\dots, \\{R, R\\}, \\{W\\}\n$$\n\n注意到如果母雞在上面的 $k$ 個 $W$ 括號中有下蛋，則其最多只能在 $2k$ 塊麥田中下蛋，且必然要在至少 $k-1$ 塊稻田中下蛋，故 $W - R \\leq 2k - (k-1) = k+1$。這表示當 $k < 167$ 時有 $W - R \\leq 167$。\n\n剩下 $k = 167$ 的狀況。注意到以上排列中，$W$ 括號與 $R$ 括號各有 167 個。這表示當 $k = 167$ 時，有下蛋的 $W$ 括弧必然包含最右邊的 $\\{W\\}$，故至多有 $2k-1$ 個麥田有蛋，從而 $W - R \\leq (2k-1)-(k-1) = k = 167$。證畢。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21466,
"subject": "Mathematics (Olympiad)",
"question": "If $x$, $y$, $z$ are positive real numbers, prove that:\n$$\n(3x + y)(3y + z)(3z + x) \\ge 64xyz.\n$$\nWhen does equality hold?",
"options": [],
"answer": "See solution",
"solution": "We use the inequality of arithmetic and geometric means (AM-GM) for four positive terms:\n$$\n\\frac{a_1 + a_2 + a_3 + a_4}{4} \\ge \\sqrt[4]{a_1 a_2 a_3 a_4},\n$$\nwhere equality holds when $a_1 = a_2 = a_3 = a_4$.\n\nApplying AM-GM to each bracket:\n- $3x + y \\ge 4\\sqrt[4]{x^3 y}$\n- $3y + z \\ge 4\\sqrt[4]{y^3 z}$\n- $3z + x \\ge 4\\sqrt[4]{z^3 x}$\n\nMultiplying these inequalities:\n$$\n(3x + y)(3y + z)(3z + x) \\ge 64 \\sqrt[4]{x^3 y \\cdot y^3 z \\cdot z^3 x}\n$$\n$$\n= 64 \\sqrt[4]{x^4 y^4 z^4} = 64xyz.\n$$\nEquality holds when $x = y = z$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21467,
"subject": "Mathematics (Olympiad)",
"question": "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$$\nT_n - T_{n-1} = \\frac{n}{2}(n + 1).\n$$\n\nPart 1. Find a closed formula for $T_n$ in terms of $n$.\n\nPart 2. Prove that $T_n + 4T_{n-1} + T_{n-2} = n^3$ for $n \\ge 3$.\n\n% IMAGE: \n% IMAGE: \n% IMAGE: \n% IMAGE: \n% IMAGE: ",
"options": [],
"answer": "See solution",
"solution": "To find a closed formula for $T_n$:\n\nAssume $T_n = an^3 + bn^2 + cn + d$.\n\nGiven $T_n - T_{n-1} = \\frac{n}{2}(n + 1)$, expand:\n\n$$\nT_n - T_{n-1} = an^3 + bn^2 + cn + d - [a(n-1)^3 + b(n-1)^2 + c(n-1) + d]\n$$\n\nExpanding and simplifying:\n\n$$\n= 3an^2 - 3an + a + 2bn - b + c\n$$\n\nSet equal to $\\frac{1}{2}n^2 + \\frac{1}{2}n$ and equate coefficients:\n\n$$\n3a = \\frac{1}{2}, \\quad -3a + 2b = \\frac{1}{2}, \\quad a - b + c = 0\n$$\n\nSolving:\n\n$$\na = \\frac{1}{6}, \\quad b = \\frac{1}{2}, \\quad c = \\frac{1}{3}\n$$\n\nUse $T_1 = 1$ to find $d$:\n\n$$\nT_1 = \\frac{1}{6} + \\frac{1}{2} + \\frac{1}{3} + d = 1 + d \\implies d = 0\n$$\n\nSo:\n\n$$\nT_n = \\frac{n^3}{6} + \\frac{n^2}{2} + \\frac{n}{3} = \\frac{n}{6}(n^2 + 3n + 2) = \\frac{n(n+1)(n+2)}{6}\n$$\n\nAlternatively, from Pascal's Triangle:\n\n$$\nT_n = \\binom{n+2}{3} = \\frac{n(n+1)(n+2)}{6}\n$$\n\nTo prove $T_n + 4T_{n-1} + T_{n-2} = n^3$:\n\nPlug in the formula:\n\n$$\n\\begin{align*}\nT_n + 4T_{n-1} + T_{n-2} &= \\frac{n(n+1)(n+2)}{6} + 4\\frac{(n-1)n(n+1)}{6} + \\frac{(n-2)(n-1)n}{6} \\\\\n&= \\frac{n}{6}[(n+1)(n+2) + 4(n-1)(n+1) + (n-2)(n-1)] \\\\\n&= n^3\n\\end{align*}\n$$\n\nAlternatively, geometrically, a $n \\times n \\times n$ cube can be partitioned into tetrahedra corresponding to $T_n$, $T_{n-1}$, and $T_{n-2}$ as shown in the diagrams, so $T_n + 4T_{n-1} + T_{n-2} = n^3$ for $n \\ge 3$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21468,
"subject": "Mathematics (Olympiad)",
"question": "There are $2n^2$ ($n \\ge 2$) players in a single round-robin chess tournament. It is known that:\n\n1. For any three players $A$, $B$, and $C$, if $A$ beats $B$ and $B$ beats $C$, then $A$ beats $C$.\n2. There are at most $\\frac{n^3}{16}$ draws.\n\nProve that it is possible to choose $n^2$ players and label them $P_{ij}$ ($1 \\le i, j \\le n$), such that for any $i, j, i', j' \\in \\{1, 2, \\dots, n\\}$, if $i < i'$, then $P_{ij}$ beats $P_{i'j'}$.",
"options": [],
"answer": "See solution",
"solution": "*Solution 1* (It is reorganized from Chen Ruitao's proof; in fact it proves the conclusion when $\\frac{n^3}{16}$ is replaced by $\\frac{n^3}{4}$.)\n\n*Lemma*: Suppose $m$ players participate in a single round-robin tournament with possible draws, and if $A$ beats $B$, $B$ beats $C$, then $A$ beats $C$. Then the $m$ players can be arranged in a row such that for any two players, the one on the left either beats or draws the one on the right.\n\n*Proof of lemma*: Induction on $m$. When $m=1$, the lemma is trivial. Suppose it is true for $m-1$ players, and consider $m$ players. If everyone wins a game, then we can find $x_1$ beats $x_2$, $x_2$ beats $x_3$, and so on, and someone must reappear in the sequence, say $x_i$ beats $x_j$ ($i \\ge j$), which is contradictory. The contradiction indicates that someone has never won a game; put this player in the rightmost position. By the induction hypothesis, the other $m-1$ players can be arranged on the left such that the lemma conditions are satisfied.\n\nFor the original problem, arrange the $2n^2$ players in a row as in the lemma, and then make $n$ groups of players as follows: set the leftmost $n$ players as group $A_1$; for $i=1, \\dots, n-2$, on the right side of $A_i$, set some consecutive $n$ players as $A_{i+1}$; set the rightmost $n$ players as group $A_n$. Moreover, between any two consecutive groups, there are $\\left\\lfloor \\frac{n^2}{n-1} \\right\\rfloor$ or more players.\n\nIf players from different groups never draw, then the proof is done. Otherwise, remove two players who draw and regroup the remaining $2n^2 - 2$ players in the same way as before, except requiring $\\left\\lfloor \\frac{n^2-2}{n-1} \\right\\rfloor$ or more players between consecutive groups. Again, if players from different groups never draw, the proof is done. Otherwise, remove two players and regroup, and so on. In general, when we regroup $2n^2-2i$ ($0 \\le i \\le \\left\\lfloor \\frac{n^2}{2} \\right\\rfloor$) players, it is required that $\\left\\lfloor \\frac{n^2-2i}{n-1} \\right\\rfloor$ or more players are between consecutive groups. During the whole process, we obtain $\\left\\lfloor \\frac{n^2}{2} \\right\\rfloor + 1$ groupings.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21469,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle, $I_a$ the excenter opposite to $A$, and $M$ its reflection across $BC$. Prove that $AM$ is parallel to the Euler line of triangle $BCI_a$.",
"options": [],
"answer": "See solution",
"solution": "Let $I$ be the incenter of $ABC$, $H_a$ the orthocenter of $I_aBC$, and $O_a$ the midpoint of the segment $II_a$.\n\n\n\nWe have\n\n$$\nBO_a = IO_a = I_aO_a = CO_a,\n$$\n\nso $O_a$ is the circumcenter of triangle $I_aBC$.\n\nMoreover, $IB \\parallel CH_a$ (both are perpendicular to $BI_a$) and, similarly, $IC \\parallel BH_a$, hence $BICH_a$ is a parallelogram.\n\nLet $P$ denote the projection of $I_a$ onto $BC$ and $T$ the midpoint of segment $AI_a$. We have $H_aP = r$, hence\n\n$$\n\\frac{IA}{AI_a} = \\frac{r}{r_a} = \\frac{H_aP}{I_aP}.\n$$\n\nTherefore\n\n$$\n\\frac{H_a I_a}{I_a P} = \\frac{I I_a}{I_a A} = \\frac{2 I_a O_a}{2 I_a T} = \\frac{I_a O_a}{I_a T}. \\quad (1)\n$$\n\nThe relation (1) proves that $O_a H_a$ is parallel to $TP$. But $TP \\parallel AM$, hence $AM \\parallel O_a H_a$, and we are done.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21470,
"subject": "Mathematics (Olympiad)",
"question": "Докажите, что ровно по 6 ничьих может быть не более, чем у двух команд в турнире из 8 команд, где каждая команда сыграла 7 матчей.",
"options": [],
"answer": "See solution",
"solution": "Любая команда с 6 ничьими имеет либо 6, либо $6+3=9$ очков (в зависимости от результата оставшегося матча). Если таких команд три, то у двух из них поровну очков, значит, между собой они сыграли не вничью, что невозможно, так как обе либо не выигрывали, либо не проигрывали ни одного матча.\n\nТакже максимум одна команда могла сыграть все 7 матчей вничью.\n\nСумма количеств ничьих у всех 8 команд не превосходит $7+6+6+5+5+5+5+5 = 44$, а поскольку каждый ничейный матч учитывается дважды, общее число ничьих в турнире не превосходит $44/2 = 22$. Это значение может быть достигнуто, что показано на рисунке ниже:\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21471,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\geq 3$ be an integer and $S$ a set of $n$ elements. Determine the largest integer $k_n$ such that: for every selection of $k_n$ 3-element subsets of $S$, there exists a way to color the elements of $S$ with two colors so that none of the chosen 3-subsets is monochromatic.",
"options": [],
"answer": "See solution",
"solution": "The answer depends on $n$:\n\n- For $n = 3$, clearly $k_3 = 1$.\n\n- For $n = 4$, $k_4 = 4$ because there are 4 possible 3-element subsets. For example, color $(1, 2)$ blue and $(3, 4)$ red; each subset contains elements of both colors.\n\n- For $n = 5$, $k_5 \\leq 10$ (since there are 10 possible 3-element subsets). By the pigeonhole principle, any coloring will have at least 3 elements of the same color, so some subset will be monochromatic if all 10 are chosen. Thus, $k_5 \\leq 9$. For $k_5 = 9$, omit one subset $\\{x, y, z\\}$, color $x, y, z$ red and the other two blue. Every selected subset contains at least one red and one blue element.\n\n- For $n = 6$, since $k_5 = 9$, $k_6 \\leq 10$. We show $k_6 = 9$ is possible. Any selection of 9 subsets can be colored so that no subset is monochromatic, using similar arguments as above.\n\n- For $n \\geq 7$, consider the following 7 subsets:\n $$\n \\{1,2,3\\},\\ \\{1,4,5\\},\\ \\{1,6,7\\},\\ \\{2,4,6\\},\\ \\{2,5,7\\},\\ \\{3,4,7\\},\\ \\{3,5,6\\}\n $$\n It can be shown that no coloring avoids a monochromatic subset among these, so $k_n = 6$ for $n \\geq 7$.\n\nIn summary:\n\n$$\n\\begin{aligned}\n&k_3 = 1,\\\\\n&k_4 = 4,\\\\\n&k_5 = k_6 = 9,\\\\\n&k_n = 6\\quad \\text{for all}\\ n \\geq 7.\n\\end{aligned}\n$$\n\n$\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21472,
"subject": "Mathematics (Olympiad)",
"question": "Let $K$ be a convex quadrilateral and let $\\ell$ be a line through the point of intersection of the diagonals of $K$. Show that the length of the segment of intersection $\\ell \\cap K$ does not exceed the length of (at least) one of the diagonals of $K$.",
"options": [],
"answer": "See solution",
"solution": "Consider a circular labeling $A$, $B$, $C$, $D$ of the vertices of the quadrilateral, and let the diagonals $AC$ and $BD$ meet at $O$. Without loss of generality, assume that the line $\\ell$ meets the opposite sides $AB$ and $CD$ at $X$ and $Y$, respectively. Let $\\alpha = \\angle AOX = \\angle COY$ and $\\beta = \\angle BOX = \\angle DOY$. Write $\\text{area}[AOB] = \\text{area}[AOX] + \\text{area}[BOX]$ and express:\n\n$$\n\\begin{align*}\n\\text{area}[AOB] &= \\frac{1}{2} OA \\cdot OB \\cdot \\sin(\\alpha + \\beta), \\\\\n\\text{area}[AOX] &= \\frac{1}{2} OA \\cdot OX \\cdot \\sin \\alpha, \\\\\n\\text{area}[BOX] &= \\frac{1}{2} OB \\cdot OX \\cdot \\sin \\beta.\n\\end{align*}\n$$\n\nThus,\n\n$$\n\\begin{align*}\nOX &= \\frac{OA \\cdot OB \\cdot \\sin(\\alpha + \\beta)}{OA \\cdot \\sin \\alpha + OB \\cdot \\sin \\beta} \\\\\n&\\le \\frac{OA \\cdot OB \\cdot (\\sin \\alpha + \\sin \\beta)}{OA \\cdot \\sin \\alpha + OB \\cdot \\sin \\beta} \\\\\n&\\le \\frac{OA \\cdot \\sin \\beta + OB \\cdot \\sin \\alpha}{\\sin \\alpha + \\sin \\beta}.\n\\end{align*}\n$$\n\n(The last inequality is equivalent to $(OA - OB)^2 \\cdot \\sin \\alpha \\cdot \\sin \\beta \\ge 0$.)\n\nSimilarly, $OY \\le \\dfrac{OC \\cdot \\sin \\beta + OD \\cdot \\sin \\alpha}{\\sin \\alpha + \\sin \\beta}$, so\n\n$$\nXY \\le \\frac{AC \\cdot \\sin \\beta + BD \\cdot \\sin \\alpha}{\\sin \\alpha + \\sin \\beta} \\le \\max(AC, BD).\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21473,
"subject": "Mathematics (Olympiad)",
"question": "平面上 $ABC$ 為銳角三角形，其外心為 $O$，外接圓為 $\\Omega$。分別在線段 $AB, AC$ 上各取一點 $D, E$，並作過 $A$ 與 $DE$ 垂直的直線 $\\ell$。設 $\\ell$ 分別與三角形 $ADE$ 的外接圓及 $\\Omega$ 再交於點 $P, Q$。令直線 $OQ$ 與 $BC$ 交於點 $N$，直線 $OP$ 與 $DE$ 交於點 $S$，且點 $W$ 為三角形 $AOS$ 的垂心。\n\n試證：$S, N, O, W$ 四點共圓。",
"options": [],
"answer": "See solution",
"solution": "令 $D', E'$ 分別為 $AB, AC$ 上的點，使得 $D'E'$ 平行於 $DE$ 且通過 $N$。由\n\n$$\n\\angle ND'B = 90^\\circ - \\angle BAQ = \\angle OQB = \\angle NQB,\n$$\n\n可知 $B, D', N, Q$ 共圓。又因為\n\n$$\n\\angle QD'E' = \\angle QBC = \\angle QAE',\n$$\n\n所以 $A, D', Q, E'$ 共圓。因此，$\\triangle ADE \\cup P \\stackrel{+}{\\sim} \\triangle AD'E' \\cup Q$。\n\n設 $S', T \\neq A$ 分別為 $AS$ 與 $D'E'$、$\\Omega$ 的交點。則\n\n$$\n\\angle QTS' = \\angle QTA = \\angle(QO, \\perp AQ) = \\angle QNS'\n$$\n\n可得 $Q, T, N, S'$ 共圓。由相似性 $\\triangle ADE \\cup \\{P, S\\} \\stackrel{+}{\\sim} \\triangle AD'E' \\cup \\{Q, S'\\}$，可知 $PS$ 平行於 $QS'$。根據 Reim 定理，$S, N, O, T$ 共圓。因此，\n\n$$\n\\angle OWS = \\angle SAO = \\angle OTA = \\angle ONS,\n$$\n\n得證。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21474,
"subject": "Mathematics (Olympiad)",
"question": "A straight line passing through the incenter $I$ of a triangle $ABC$ meets the sides $AB$ and $AC$ at points $P$ and $Q$ respectively. Let $BC = a$, $AC = b$, $AB = c$ and $\\frac{PB}{PA} = p$, $\\frac{QC}{QA} = q$.\n\n1. Prove that $a(1+p)\\vec{IP} = (a-pb)\\vec{IB} - cp\\vec{IC}$.\n2. Prove that $a = bp + cq$.\n3. Prove that if $a^2 = 4bcpq$, then the straight lines $AI$, $BQ$ and $CP$ are concurrent.",
"options": [],
"answer": "See solution",
"solution": "1. The hypothesis yields $(p+1)\\vec{IP} = \\vec{IB} + p\\vec{IA}$. The conclusion follows now from $a\\vec{IA} + b\\vec{IB} + c\\vec{IC} = \\vec{0}$.\n\n2. Analogously, $a(1+q)\\vec{IQ} = (a-cq)\\vec{IC} - bq\\vec{IB}$. The points $P$, $I$, $Q$ are collinear, hence $(a - pb)(a - cq) = bcpq$, whence the conclusion.\n\n3. Squaring the relation from (2), $a^2 \\ge 4bcpq$, with equality if and only if $bp = cq$. In this case, if $\\{D\\} = AI \\cap BC$, then\n\n$$\n\\frac{PB}{PA} \\cdot \\frac{QA}{QC} \\cdot \\frac{DC}{DB} = p \\cdot \\frac{1}{q} \\cdot \\frac{b}{c} = 1\n$$\n\nand the converse of Ceva's Theorem guarantees the conclusion.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21475,
"subject": "Mathematics (Olympiad)",
"question": "Let $O$ be the circumcentre of triangle $ABC$.\n\n\n\nPoint $D$ lies on the angle bisector of $\\angle BAC$, so $A$, $I$, and $D$ are collinear. Lines $IJ$ and $OD$ are parallel, since both are perpendicular to $BC$.\n\nProve that triangles $AOI$ and $EOI$ are congruent.",
"options": [],
"answer": "See solution",
"solution": "First, we show that triangles $AIO$ and $IJD$ are similar. Note that $\\angle IAO = \\angle DAO = \\angle ODA = \\angle ODI = \\angle JID$. To show similarity, we need to prove that $|AI| : |AO| = |IJ| : |ID|$, i.e.\n\n$$\n|AI| \\cdot |ID| = |AO| \\cdot |IJ|. \\qquad (1)\n$$\n\nWe have $|IJ| = 2r$ and $|AO| = R$, where $r$ and $R$ are the inradius and circumradius, respectively. Thus, the right-hand side of (1) equals $2Rr$. The left-hand side $|AI| \\cdot |ID|$ is the power of point $I$ with respect to the circumcircle of $ABC$ and equals $R^2 - |OI|^2$. By Euler's theorem, $R^2 - |OI|^2 = 2Rr$.\n\nTherefore, triangles $AIO$ and $IJD$ are similar. This means $\\angle OIA = \\angle DJI$, and $\\angle DIO = \\angle IJE$. Since $IJ$ and $OD$ are parallel, $\\angle IJE = \\angle ODE$, and since $|OE| = |OD|$, $\\angle ODE = \\angle DEO$. Thus, $\\angle DIO = \\angle DEO$, so points $D$, $O$, $I$, and $E$ are concyclic.\n\nFrom this, $\\angle IOE = \\angle IDE$, and from the similarity, $\\angle IDE = \\angle IDJ = \\angle AOI$. It follows that $\\angle IOE = \\angle AOI$. Since $|AO| = |EO|$ and $\\overline{OI}$ is common, triangles $EOI$ and $AOI$ are congruent.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21476,
"subject": "Mathematics (Olympiad)",
"question": "For a fixed positive integer $n$, let $D$ be the set of all positive factors of $n$. Prove that for a mapping $f: D \\to \\mathbb{Z}$, the following two assertions are equivalent:\n\n(A) For any positive factor $m$ of $n$,\n\n$$\nn \\mid \\sum_{d \\mid m} f(d) \\binom{n/d}{m/d};\n$$\n\n(B) For any positive factor $k$ of $n$,\n\n$$\nk \\mid \\sum_{d \\mid k} f(d).\n$$",
"options": [],
"answer": "See solution",
"solution": "For the given mapping $f : D \\to \\mathbb{Z}$, define $g : D \\to \\mathbb{Z}$ as\n\n$$\ng(k) = \\sum_{d \\mid k} f(d), \\quad \\forall k \\in D.\n$$\n\nAccording to the Möbius transformation, $f$ is uniquely determined by $g$:\n\n$$\nf(k) = \\sum_{d \\mid k} \\mu\\left(\\frac{k}{d}\\right) g(d), \\quad \\forall k \\in D.\n$$\n\nThis gives\n\n$$\n\\begin{align*}\n\\sum_{d \\mid m} f(d) \\binom{n/d}{m/d} &= \\sum_{d \\mid m} \\sum_{x \\mid d} \\mu\\left(\\frac{d}{x}\\right) g(x) \\binom{n/d}{m/d} \\\\\n&= \\sum_{x \\mid m} g(x) \\left( \\sum_{x \\mid d,\\ d \\mid m} \\mu\\left(\\frac{d}{x}\\right) \\binom{n/d}{m/d} \\right) \\\\\n&= \\sum_{x \\mid m} g(x) \\left( \\sum_{s \\mid \\frac{m}{x}} \\mu(s) \\binom{n/(xs)}{m/(xs)} \\right). \\tag{1}\n\\end{align*}\n$$\n\n**Lemma:** If $b \\mid a$, then $\\sum_{k \\mid b} \\mu(k) \\binom{a/k}{b/k} \\equiv 0 \\pmod{a}$.\n\nAssuming the lemma, we prove (A) and (B) are equivalent.\n\n**(B) $\\Rightarrow$ (A):** Suppose $x \\mid g(x)$ for every $x \\in D$. By the lemma,\n\n$$\n\\frac{n}{x} \\mid \\sum_{s \\mid \\frac{m}{x}} \\mu(s) \\binom{n/(xs)}{m/(xs)},\n$$\n\nso plugging into (1) gives (A).\n\n**(A) $\\Rightarrow$ (B):** Assume (A). Use induction to show $k \\mid g(k)$ for all $k \\in D$. Suppose $k \\mid g(k)$ for all $k < m$, $k \\in D$, and consider $k = m$. By the lemma and induction,\n\n$$\n\\begin{align*}\n0 &\\equiv \\sum_{d \\mid m} f(d) \\binom{n/d}{m/d} \\\\\n&= \\sum_{x \\mid m} g(x) \\left( \\sum_{s \\mid \\frac{m}{x}} \\mu(s) \\binom{n/(xs)}{m/(xs)} \\right) \\\\\n&= g(m) \\cdot \\frac{n}{m} + \\sum_{x \\mid m,\\ x < m} g(x) \\left( \\sum_{s \\mid \\frac{m}{x}} \\mu(s) \\binom{n/(xs)}{m/(xs)} \\right) \\\\\n&\\equiv g(m) \\cdot \\frac{n}{m} \\pmod{n},\n\\end{align*}\n$$\n\nso $m \\mid g(m)$. This completes the induction.\n\n**Proof of the lemma:**\n\n$$\n\\sum_{k \\mid b} \\mu(k) \\binom{a/k}{b/k} = \\frac{a}{b} \\sum_{k \\mid b} \\mu(k) \\binom{a/k - 1}{b/k - 1}.\n$$\n\nSuppose $b = \\prod_{i=1}^t p_i^{\\beta_i}$. It suffices to prove for each $1 \\leq i \\leq t$,\n\n$$\n\\sum_{k \\mid b} \\mu(k) \\binom{b/k - 1}{b/k - 1} \\equiv 0 \\pmod{p_i^{\\beta_i}}.\n$$\n\nLet $i = 1$, $p_1^{\\beta_1} = p^{\\beta}$, and write $\\binom{u}{v}$ as $C(u, v)$. Then\n\n$$\n\\begin{align*}\n\\sum_{k \\mid b} \\mu(k) C\\left(\\frac{a}{k} - 1, \\frac{b}{k} - 1\\right) &= \\sum_{I \\subset \\{1, \\dots, t\\}} (-1)^{|I|} C\\left( \\frac{a}{\\prod_{i \\in I} p_i} - 1, \\frac{b}{\\prod_{i \\in I} p_i} - 1 \\right) \\\\\n&= \\sum_{J \\subset \\{2, \\dots, t\\}} (-1)^{|J|} \\left[ C\\left( \\frac{a}{\\prod_{j \\in J} p_j} - 1, \\frac{b}{\\prod_{j \\in J} p_j} - 1 \\right) \n- C\\left( \\frac{a}{p \\prod_{j \\in J} p_j} - 1, \\frac{b}{p \\prod_{j \\in J} p_j} - 1 \\right) \\right].\n\\end{align*}\n$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21477,
"subject": "Mathematics (Olympiad)",
"question": "Prove that $\\mathbb{N}_+$ is not a nice set.\n\nIs $T = \\{x \\in \\mathbb{N}_+ \\mid \\text{for any prime } p,\\ p^{2021} \\text{ does not divide } x\\}$ 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 $x_1 = x_2$. So, $f$ is one-to-one.\n\nNotice that $\\{1, 2, 4, \\dots, 2^k, \\dots\\}$ is an infinite geometric sequence of positive integers. As the problem indicates, $\\{f(1), f(2), f(4), \\dots, f(2^k), \\dots\\}$ 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, we can let $f(3^k) = f(1) + k d_3$, where $d_3$ is a positive integer.\nThen\n\n$$\nf(2^{d_3}) = f(1) + d_3 d_2 = f(3^{d_2}).\n$$\n\nSince $f$ is one-to-one, $2^{d_3} = 3^{d_2}$, but this is impossible.\nHence, $\\mathbb{N}_+$ is not a nice set.\n\n(2) $T$ is a nice set.\n\nLet all prime numbers be $p_1, p_2, p_3, \\dots$. For any positive integer $n \\in T$, let $n = \\prod_{i=1}^{\\infty} p_i^{x_i}$ be the prime factorization, in which $x_i \\leq 2020$ is a nonnegative integer for each $i$, and only finitely many $x_i$ are not $0$. 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, $f: T \\to \\mathbb{N}_+$, and for any three integers in $T$,\n\n$$\na = \\prod_{i=1}^{\\infty} p_i^{\\alpha_i}, \\quad b = \\prod_{i=1}^{\\infty} p_i^{\\beta_i}, \\quad c = \\prod_{i=1}^{\\infty} p_i^{\\gamma_i},\n$$\n\n$\\{a, b, c\\}$ is geometric if and only if $\\alpha_i + \\gamma_i = 2\\beta_i$ holds for every positive integer $i$. On 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$$\n\\sum_{i=1}^{\\infty} (\\alpha_i + \\gamma_i) \\cdot 4041^i = \\sum_{i=1}^{\\infty} 2\\beta_i \\cdot 4041^i. \\qquad \\textcircled{1}\n$$\n\nNotice that both $\\alpha_i + \\gamma_i$ and $2\\beta_i$ are nonnegative integers less than or equal to $4040$. We can see that (1) gives two representations of the same number in base $4041$, which must agree. Hence, (1) holds if and only if $\\alpha_i + \\gamma_i = 2\\beta_i$ is true for every positive integer $i$. It turns out that $\\{a, b, c\\}$ is geometric if and only if $\\{f(a), f(b), f(c)\\}$ is arithmetic. Thus, $f$ satisfies the problem condition and $T$ is a nice set.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21478,
"subject": "Mathematics (Olympiad)",
"question": "Let $M$, $N$, $K$, and $L$ be the feet of the perpendiculars from $A$ and $C$ to lines $BP$ and $DP$, respectively, as shown.\n\nShow that $MN = KL$ if and only if $\\angle A + \\angle C = 180^\\circ$, where $\\angle A = \\angle BAD$ and $\\angle C = \\angle DCB$.",
"options": [],
"answer": "See solution",
"solution": "Because $\\angle AMP = \\angle ANP = 90^\\circ$, points $A$, $M$, $P$, and $N$ lie on a circle and segment $AP$ is a diameter of the circle. By the Extended Law of Sines,\n\n$$\nMN = AP \\sin \\angle MAN = AP \\sin(180^\\circ - \\angle NPM) = AP \\sin \\angle NPM.\n$$\n\nLikewise, $KL = CP \\sin \\angle LPK$. Note that $\\angle NPM = \\angle LPK$. Hence $AP = CP$ if and only if $MN = KL$. Set $\\angle A = \\angle BAD$ and $\\angle C = \\angle DCB$. It suffices to show that $MN = KL$ if and only if $\\angle A + \\angle C = 180^\\circ$. Set $\\gamma = \\angle MBA$ and $\\delta = \\angle ADN$. Note that in triangles $ABD$ and $CDB$, $\\angle A = 180^\\circ - \\alpha - \\beta$ and $\\angle C = 180^\\circ - \\gamma - \\delta$. Because $\\gamma > \\alpha$ and $\\delta > \\beta$, $\\angle A > \\angle C$. Hence $\\angle A + \\angle C = 180^\\circ$ if and only if $\\sin \\angle A = \\sin \\angle C$; that is, it suffices to show that $MN = KL$ if and only if $\\sin \\angle A = \\sin \\angle C$.\n\nSet $\\phi = \\angle MAN$. Then $\\angle CBD = \\gamma$, $\\angle BDC = \\delta$, and $\\angle KCL = \\phi$. Applying the Law of Sines to triangle $ABD$ gives\n\n$$\nAB = \\frac{BD \\sin \\beta}{\\sin \\angle A} \\quad \\text{and} \\quad AD = \\frac{BD \\sin \\alpha}{\\sin \\angle A}.\n$$\n\nIn right triangles $AMB$ and $AND$, we have\n\n$$\nAM = AB \\sin \\gamma = \\frac{BD \\sin \\beta \\sin \\gamma}{\\sin \\angle A}\n$$\n\nand\n\n$$\nAN = AD \\sin \\delta = \\frac{BD \\sin \\alpha \\sin \\delta}{\\sin \\angle A}.\n$$\n\nApplying the **Law of Cosines** to triangle $AMN$ gives\n\n$$\n\\begin{aligned}\nMN^2 &= AM^2 + AN^2 - 2AM \\cdot AN \\cos \\phi \\\\\n &= \\frac{BD^2}{\\sin^2 \\angle A} \\cdot f,\n\\end{aligned}\n$$\n\nwhere $f = (\\sin^2 \\beta \\sin^2 \\gamma + \\sin^2 \\alpha \\sin^2 \\delta - 2 \\sin \\beta \\sin \\gamma \\sin \\alpha \\sin \\delta \\cos \\phi)$.\n\nIn exactly the same way, we can work on triangles $BCD$, $BCK$, $CDL$, and $CLK$ to obtain\n\n$$\nKL^2 = \\frac{BD^2}{\\sin^2 \\angle C} \\cdot f.\n$$\n\nHence $MN = KL$ if and only if $\\sin \\angle A = \\sin \\angle C$, as desired.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21479,
"subject": "Mathematics (Olympiad)",
"question": "Inside a circle $c$ with center $O$, there are two circles $c_1$ and $c_2$ which pass through $O$ and are tangent to $c$ at points $A$ and $B$, respectively. Prove that the circles $c_1$ and $c_2$ have a common point that lies on the segment $AB$.",
"options": [],
"answer": "See solution",
"solution": "The radius $AO$ of circle $c$ is perpendicular to the common tangent to circles $c$ and $c_1$ at point $A$, so $AO$ is a diameter of $c_1$. Similarly, $BO$ is a diameter of $c_2$.\n\nIf $c_1$ and $c_2$ are tangent at $O$, then the diameters $AO$ and $BO$ are both perpendicular to the common tangent to $c_1$ and $c_2$ at $O$, so the lines $AO$ and $BO$ coincide, i.e., $O$ lies on the segment $AB$.\n\nIf $c_1$ and $c_2$ intersect at $O$, let $M$ be their other intersection point. Since $\\angle AMO = 90^\\circ$ and $\\angle BMO = 90^\\circ$ (angles at the circumference subtended by a diameter), the lines $AM$ and $BM$ coincide and $M$ lies on the segment $AB$.\n\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21480,
"subject": "Mathematics (Olympiad)",
"question": "A chessboard of size $1000 \\times 1000$ is tiled with tiles of size $1 \\times 10$. You do not know the tiling but wish to uncover it. In order to do so, you can choose some $N$ cells on the board, following which you will learn what the positions of the tiles that cover those cells are. What is the least $N$ such that you can make your choice so as to always be able to reconstruct the complete tiling?",
"options": [],
"answer": "See solution",
"solution": "We will show that $N = k^2$ is the desired least number for any $kn \\times kn$ board tiled with $1 \\times n$ tiles, $n \\ge 2$, thus the answer to our problem is $N = 100^2 = 10000$.\n\nFrom now on, consider a $kn \\times kn$ board tiled with $1 \\times n$ tiles ($n \\ge 2$), with $k > 1$ fixed.\n\nFor a lower bound of $N$, divide the board into $k^2$ squares of size $n \\times n$ and notice that if one of them, say $P$, does not contain some one of the $N$ chosen cells, then there exist two tilings which differ on $P$ but agree on the rest of the board, and the given tiling would never be uncovered. Thus we must have $k^2 \\le N$.\n\nFor an upper bound of $N$, consider the $k^2$ lower-left cells in each one of the squares of size $n \\times n$ considered above, and learn the position of the tiles that cover these cells. We shall show by induction on $n$ that this procedure allows you to figure out the complete tiling. Thus $N \\le k^2$. This, together with $k^2 \\le N$ proved above, would imply $N = k^2$, and we'll have finished.\n\nFor the induction, argue as follows:\n\nFor $n = 2$, suppose that there are two tilings $A$ and $B$ that agree on all lower-left cells of the $n \\times n = 2 \\times 2$ squares considered above, but differ on some other cell $a_1$ of the given board. We'll arrive at a contradiction, which will mean any tiling can be uncovered by choosing to learn the positions of these lower-left cells, as wanted. Indeed:\n\nLet $d_1$ be the $1 \\times n = 1 \\times 2$ domino which covers $a_1$ in $A$. Colour in red the upper-right cell in each $2 \\times 2$ square; without loss of generality, $a_1$ is a red cell (otherwise, replace $a_1$ by the other cell covered by $d_1$). Let $d_2$ be the $1 \\times n = 1 \\times 2$ domino which covers $a_1$ in $B$, $b_1$ be the second cell covered by $d_2$, $d_3$ be the domino which covers $b_1$ in $A$, $a_2$ be the second cell covered by $d_3$, etc. Then $a_i$ is red for all $i$.\n\nLet $s$ be the least positive integer such that there exists a $t < s$ such that $a_s = a_t$; clearly, we must have $t = 1$. Consider the polyomino $P$ enclosed by the sequence of cells $a_1, b_1, a_2, b_2, \\dots, b_{s-1}$, $a_s = a_1$.\n\nLet $O_i$ be the centre of $a_i$. It is straightforward to verify that $S(P) = S(O_1O_2\\dots O_{s-1}) - s + 2$ (where $S(\\cdot)$ denotes area). Since $O_1O_2\\dots O_{s-1}$ is a polyomino in the grid formed by the centres of all red cells, its area $S(O_1O_2\\dots O_{s-1})$ is a multiple of four and the number $s-1$ of its sides is even. It follows from this that $S(P)$ is odd whereas it must be possible to tile $P$ with dominoes; the desired contradiction.\n\nFor the inductive step, let $n \\ge 3$. Number all columns as $1$ through $kn$ from left to right and all rows as $1$ through $kn$ from bottom to top, and delete all rows and columns whose number is congruent to $2$ modulo $n$. This operation produces a $k(n-1) \\times k(n-1)$ chessboard tiled with $1 \\times (n-1)$ tiles. By the induction hypothesis, we can reconstruct this tiling completely by choosing to uncover the lower-left cells of the $(n-1) \\times (n-1)$ squares in which we divide this $k(n-1) \\times k(n-1)$ chessboard. These cells are exactly the lower-left cells of the $n \\times n$ squares in which we divide the original $kn \\times kn$ chessboard. Restore all deleted rows and columns, and you have the positions of all tiles in the original tiling that were not contained within a deleted row or column. Repeat the operation for all rows and columns whose numbers are congruent to $3$ modulo $n$, and you have the positions of those tiles as well, using the same lower-left cells of the $n \\times n$ squares in which we divide the original $kn \\times kn$ chessboard. So choosing to learn the positions of the tiles covering these cells you uncover the whole tiling of the original $kn \\times kn$ chessboard, as wanted to prove. $\\square$\n\n**Remark (P.S.C.):** After defining the potentially existing cells $a_i, b_i$, the above solution should be expanded as follows: (1) considering the possibility that the sequence $a_1, b_1, a_2, b_2, \\dots$ may not close, but rather arrive at an edge of the original matrix, (2) explaining that if $a_1, b_1, a_2, b_2, \\dots$ closes, $a_i, b_i, a_{i+1}$ should be seen as neighboring cells in a row or in a column, and from this to deduce the remarks about the areas, which is not as short a task as it seems at first glance. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21481,
"subject": "Mathematics (Olympiad)",
"question": "Suppose we are given two sequences of positive real numbers: $a_1 > a_2 > \\dots > a_m$ and $b_1 < b_2 < \\dots < b_m$, representing lengths of segments. Starting from the origin, at each step $i$ ($1 \\leq i \\leq m$), we move up by a segment of length $a_i$, then right by a segment of length $b_i$. Let $l$ be the line connecting the origin to the endpoint of the last segment. Show that all segments lie above the line $l$.\n\nNow, suppose both the $a_i$'s and $b_i$'s are divided into two sets with equal sums:\n\n$$\n\\sum_{i=1}^{k} a_i = \\sum_{i=k+1}^{n} a_i\n$$\nwith $a_k \\leq a_{k-1} \\leq \\dots \\leq a_1$ and $a_n \\leq a_{n-1} \\leq \\dots \\leq a_{k+1}$, where $a_1 \\geq a_{k+1}$.\n\nSimilarly,\n$$\n\\sum_{i=1}^{s} b_i = \\sum_{i=s+1}^{n} b_i\n$$\nwith $b_s \\geq b_{s-1} \\geq \\dots \\geq b_1$ and $b_n \\geq b_{n-1} \\geq \\dots \\geq b_{s+1}$, where $b_1 \\leq b_{s+1}$.\n\nAssume $k \\leq \\frac{n}{2} \\leq s$. Construct a polygon as follows:\n\n- Start at the origin ($C_0$).\n- For $1 \\leq i \\leq s$, from $C_{2(i-1)}$, go up by $a_i$ to $C_{2i-1}$, then right by $b_i$ to $C_{2i}$.\n- For $s < i \\leq k$, from $C_{2(i-1)}$, go down by $a_i$ to $C_{2i-1}$, then left by $b_i$ to $C_{2i}$.\n- For $k < i \\leq n$, from $C_{2(i-1)}$, go down by $a_i$ to $C_{2i-1}$, then left by $b_i$ to $C_{2i}$.\n\n% \n\nGiven $\\sum_{i=1}^{k} a_i = \\sum_{i=k+1}^{n} a_i$ and $\\sum_{i=1}^{s} b_i = \\sum_{i=s+1}^{n} b_i$, the path returns to the origin, forming a polygon. Prove that this polygon is simple (no self-intersections).",
"options": [],
"answer": "See solution",
"solution": "*Lemma 1.*\n\nAssume, for contradiction, that a segment first intersects $l$; it must be a horizontal segment $b_i$. Let $O$ be the origin and $X$ the endpoint of $b_i$. Since $X$ is below $l$, the slope of $OX$ is less than that of $l$:\n\n$$\n\\frac{\\sum_{j=0}^{i} a_j}{\\sum_{j=0}^{i} b_j} < \\frac{\\sum_{j=0}^{m} a_j}{\\sum_{j=0}^{m} b_j}\n$$\n\nor\n\n$$\n\\frac{\\sum_{j=0}^{i} a_j}{\\sum_{j=0}^{m} a_j} < \\frac{\\sum_{j=0}^{i} b_j}{\\sum_{j=0}^{m} b_j} \\quad (*)\n$$\n\nSince $b_i$'s are increasing,\n\n$$\n\\sum_{j=i+1}^{m} b_j > (m-i) \\sum_{j=0}^{i} b_j \\implies i \\sum_{j=0}^{m} b_j > m \\sum_{j=0}^{i} b_j\n$$\n\nSo the right side of $(*)$ is less than $\\frac{i}{m}$. Similarly, the left side is greater than $\\frac{i}{m}$ (since $a_i$'s are decreasing), a contradiction. Thus, all segments lie above $l$.\n\n*Main Problem.*\n\nThe construction ensures that the path returns to the origin, forming a polygon. In each part of the algorithm, segments cannot intersect each other. By the lemma, segments from different parts can only intersect if the connecting segments are collinear, which only happens if $k = s$. In this case, equality conditions force all $a_i$'s and $b_i$'s in each part to be equal, contradicting the monotonicity. Thus, the polygon is simple.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21482,
"subject": "Mathematics (Olympiad)",
"question": "Find all natural numbers $x$, $n$, and prime $p$ such that\n$$\n2x^3 + x^2 + 10x + 5 = 2p^n.\n$$",
"options": [],
"answer": "See solution",
"solution": "It is easy to see that $2x^3 + x^2 + 10x + 5 = (x^2 + 5)(2x + 1)$, so the initial equality can be rewritten as\n$$\n(x^2 + 5)(2x + 1) = 2p^n.\n$$\nSince $2x + 1$ is odd for all natural $x$, we have $p \\neq 2$ and $2x + 1 = p^k$, $x^2 + 5 = 2p^{n-k}$. It is evident that $x^2 + 5 > 2x + 1$ for all $x \\in \\mathbb{N}$ and $p \\ge 3$, then $n - k \\ge k$. Therefore, $(x^2 + 5) \\div (2x + 1)$. So the number $(x^2 + 5)/(2x + 1)$ is integer, but then the number $2(x^2 + 5)/(2x + 1)$ is also integer. Since $2(x^2 + 5) = x(2x + 1) + (10 - x)$, we see that $(10 - x)/(2x + 1)$ must be also integer, and so the number $2(10 - x)/(2x + 1)$ must be integer too. But $2(10 - x) = -(2x + 1) + 21$, so the number $21/(2x + 1)$ must be integer. It is possible only if $2x+1 = 1, 3, 7, 21$, i.e. if $x = 1, x = 3, x = 10$.\n\nFor $x = 1$ we have $(x^2 + 5)(2x + 1) = 6 \\cdot 3 = 18 = 2 \\cdot 3^2$, so $p = 3$, $n = 2$.\n\nFor $x = 3$ we have $(x^2 + 5)(2x + 1) = 14 \\cdot 7 = 2 \\cdot 7^2$, so $p = 7$, $n = 2$.\n\nFor $x = 10$ we have $(x^2 + 5)(2x + 1) = 405 \\cdot 21 = 5 \\cdot 7^2 \\cdot 3^5$, i.e. this number cannot be $2p^n$ for any natural $n$ and any prime $p$.\n\nThus, the solutions are $(x, n, p) = (1, 2, 3)$ and $(x, n, p) = (3, 2, 7)$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21483,
"subject": "Mathematics (Olympiad)",
"question": "Given a circle $k$ and a point $A$ outside of it. The segment $BC$ is a diameter of $k$. Find the locus of the orthocenter of $\\triangle ABC$ as $BC$ varies.",
"options": [],
"answer": "See solution",
"solution": "Let $O$ be the center of $k$, and let $\\omega$ be the circle with diameter $AO$. Let $AA_1$ and $BB_1$ be the altitudes of $\\triangle ABC$, and let $H$ be the orthocenter. The power of $H$ with respect to $\\omega$ is $AH \\cdot HA_1$, and with respect to $k$ is $BH \\cdot HB_1$. Since $AH \\cdot HA_1 = BH \\cdot HB_1$, it follows that $H$ lies on the radical axis $l$ of $k$ and $\\omega$. Conversely, every point of $l$ is the orthocenter of some triangle of the given type.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21484,
"subject": "Mathematics (Olympiad)",
"question": "Let $P(x)$ be a polynomial with real coefficients such that\n\n$$\nP(x^2) - 1 = (P(x) - 1)^2.\n$$\n\nFind all such polynomials $P(x)$.",
"options": [],
"answer": "See solution",
"solution": "We rewrite the equation as\n\n$$\nP(x^2) - 1 = (P(x) - 1)^2.\n$$\n\nLet $Q(x) = P(x) - 1$, so $Q$ is a polynomial with real coefficients satisfying\n\n$$\nQ(x^2) = Q(x)^2.\n$$\n\nSuppose $Q$ is constant, say $Q(x) = c$ with $c \\in \\mathbb{R}$. Then $c = c^2$, so $c = 0$ or $c = 1$. Both give solutions.\n\nIf $Q$ is non-constant, write $Q(x) = b x^n + R(x)$ with $n \\ge 1$, $b \\ne 0$, and $R(x)$ of degree at most $n-1$. The equation becomes\n\n$$\nb x^{2n} + R(x^2) = b^2 x^{2n} + 2b x^n R(x) + R(x)^2.\n$$\n\nComparing coefficients of $x^{2n}$, $b = b^2$, so $b = 1$ (since $b \\ne 0$). Subtracting $x^{2n}$ from both sides gives\n\n$$\nR(x^2) = 2 x^n R(x) + R(x)^2.\n$$\n\nIf $R$ is nonzero 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 $Q(x) = x^n$.\n\nTherefore, the solutions are $P(x) = 1$, $P(x) = 2$, and $P(x) = x^n + 1$ for $n \\ge 1$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21485,
"subject": "Mathematics (Olympiad)",
"question": "Нека $x$, $y$ и $z$ се позитивни реални броеви така што $x^4 + y^4 + z^4 = 3$. Докажете дека\n\n$$\n\\frac{9}{x^2 + y^2 + z^2} + \\frac{9}{x^4 + y^4 + z^4} + \\frac{9}{x^6 + y^6 + z^6} \\leq x^6 + y^6 + z^6 + 6.\n$$\n\nКога важи равенство?",
"options": [],
"answer": "See solution",
"solution": "Ако го искористиме неравенството на Коши–Буњаковски–Шварц за позитивните броеви $(x, y^2, z^3)$ и $(x^3, y^2, z)$, добиваме\n\n$$\n(x^4 + y^4 + z^4)^2 \\leq (x^2 + y^4 + z^6)(x^6 + y^4 + z^2), \\text{ т.е.}\n$$\n$$\n\\frac{1}{x^2 + y^4 + z^6} \\leq \\frac{x^6 + y^4 + z^2}{9}. \\tag{1}\n$$\n\nАналогно, со користење на неравенството на Коши–Буњаковски–Шварц за позитивните броеви $(x^2, y^3, z)$ и $(x, y^3, z^3)$, како и за $(x^3, y, z^2)$ и $(x, y^3, z^2)$, добиваме\n\n$$\n\\frac{1}{x^4 + y^6 + z^2} \\leq \\frac{x^4 + y^2 + z^6}{9}, \\tag{2}\n$$\n\nи\n\n$$\n\\frac{1}{x^6 + y^2 + z^4} \\leq \\frac{x^2 + y^6 + z^4}{9}. \\tag{3}\n$$\n\nСега, со собирање на неравенствата (1), (2) и (3), го добиваме\n\n$$\n\\frac{1}{x^2 + y^4 + z^6} + \\frac{1}{x^4 + y^6 + z^2} + \\frac{1}{x^6 + y^2 + z^4} \\leq \\frac{x^6 + y^6 + z^6 + x^4 + y^4 + z^4 + x^2 + y^2 + z^2}{9}. \\tag{4}\n$$\n\nОд неравенството меѓу аритметичка и квадратна средина за позитивните броеви $x^2$, $y^2$ и $z^2$ добиваме дека $x^2 + y^2 + z^2 \\leq 3\\sqrt{\\frac{x^4 + y^4 + z^4}{3}} = 3$, па ако замениме во (4), го добиваме бараното неравенство.\n\nРавенство во (1) важи ако и само ако $\\frac{x}{x^3} = \\frac{y^2}{y^2} = \\frac{z^3}{z}$, т.е. $x = z = 1$, и од $x^4 + y^4 + z^4 = 3$ следува дека $y = 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21486,
"subject": "Mathematics (Olympiad)",
"question": "Set up a coordinate system on the plane with $A = (0,0)$, $B = (a,0)$, $C = (b,c)$, and $P = (x,y)$. Assume $a > 0$ and $c > 0$. Prove that if $PA^2 > PB^2 + PC^2$, then angle $BAC$ is acute.",
"options": [],
"answer": "See solution",
"solution": "Proving that angle $BAC$ is acute is equivalent to proving that $b > 0$. Since $PA^2 > PB^2 + PC^2$,\n\n$$\nx^2 + y^2 > (x-a)^2 + y^2 + (x-b)^2 + (y-c)^2.\n$$\n\nHence,\n\n$$\n0 > (x-a)^2 - 2bx + b^2 + (y-c)^2 \\geq -2bx.\n$$\n\nSince $PA > PB$, we have $x > \\frac{a}{2} > 0$. It follows that $b > 0$, as desired.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21487,
"subject": "Mathematics (Olympiad)",
"question": "The convex quadrilateral $ABCD$ is given.\n\n$\\angle BAD = \\angle BCD = 120^\\circ$, $\\angle ACD = 80^\\circ$, $BC = CD$.\n\n$O$ is the intersection point of its diagonals.\n\nProve that $BO^2 = AO \\cdot AC$.",
"options": [],
"answer": "See solution",
"solution": "Let's draw a circle with center at point $C$ and radius $BC = CD$. Points $B$ and $D$ are located on this circle, and since $\\angle BCD = 120^\\circ$, the arc $AD$ (not containing $A$) is $240^\\circ$. Its inscribed angle is $120^\\circ$, so point $A$ is also on the circle.\n\nIt's easy to calculate angles:\n\n$\\angle ACB = 40^\\circ$, $\\angle DBA = 40^\\circ$ (subtended by the central angle $\\angle ACD = 80^\\circ$). Since $\\triangle ACB$ is isosceles with apex angle $40^\\circ$, $\\angle CAB = \\angle ABC = 70^\\circ$.\n\nThus, $\\angle AOB = 70^\\circ$, so $\\triangle AOB$ is also isosceles and similar to $\\triangle ACB$. Therefore,\n\n\n\n$$\n\\frac{AO}{AB} = \\frac{AB}{AC} \\Rightarrow AO \\cdot AC = AB^2 = AO^2,\n$$\n\nwhich is what had to be demonstrated.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21488,
"subject": "Mathematics (Olympiad)",
"question": "There were three candidates $A$, $B$, and $C$ in the elections for a provincial governor. In the first round, $A$ won $44\\%$ of the number of votes given for $B$ and $C$ together, and $C$ had the fewest votes. No candidate had the majority necessary for a first-round win, so there was a second round for $A$ and $B$. The voters in the second round were the same as in the first round, except $p\\%$ of the voters for $C$ chose not to participate in the second round, where $p$ is an integer, $1 \\le p \\le 100$. In addition, those who voted for $B$ in the first round did so again in the second round.\n\nA journalist claims that, knowing all the above, one can infer who the winner is with certainty. For what values of $p$ is he right?\n\nNote: The winner in the second round is the one who obtains more than half of the total number of votes in the second round.",
"options": [],
"answer": "See solution",
"solution": "The journalist is right for $p \\ge 73$ and wrong for $p \\le 72$.\n\nLet $a$, $b$, $c$ denote the number of votes for $A$, $B$, $C$ in the first round, and let $N = a + b + c$ be the total number of voters in this round. By hypothesis, $a = \\frac{44}{100}(b + c) = \\frac{11}{25}(N - a)$, hence $a = \\frac{11}{36}N$; also $c < a$.\n\nThe number of voters in the second round is $N' = N - \\frac{p}{100}c$. There are $\\left(1 - \\frac{p}{100}\\right)c$ persons who voted for $C$ in the first round and participate in the second round. For brevity, call them and their votes additional. Since $B$'s supporters voted for him in both rounds, the most $A$ can achieve is that his own supporters vote for him again in the second round, and also the additional voters. So the maximum number of votes $A$ can get is $a_{\\text{max}} = a + \\left(1 - \\frac{p}{100}\\right)c$.\n\nWe are interested in the difference $N' - 2a_{\\text{max}}$ (whose sign determines the chances of $A$):\n\n$$\nN' - 2a_{\\text{max}} = \\left(N - \\frac{p}{100}c\\right) - 2 \\cdot \\frac{11}{36}N - 2\\left(1 - \\frac{p}{100}\\right)c = \\frac{7}{18}N - \\frac{200 - p}{100}c.\n$$\n\nSuppose that $p \\ge 73$. Then $\\frac{200 - p}{100}c < \\frac{127}{100}c < \\frac{127}{100}a = \\frac{127}{100}N$. Since $\\frac{127}{100} < \\frac{11}{36} < \\frac{7}{18}$, it follows that $\\frac{200 - p}{100}c < \\frac{7}{18}N$, i.e., $N' - 2a_{\\text{max}} > 0$. So $A$ cannot win even if he gets the maximum possible number of votes. Therefore, $B$ wins with certainty, and the journalist is right.\n\nFor $p \\le 72$, there are examples showing that either candidate can win. In this case, $\\frac{200 - p}{100}c \\ge \\frac{128}{100}c$.\n\nNow the inequality $\\frac{7}{18} < \\frac{128}{100} \\cdot \\frac{11}{36}$ implies $\\frac{100}{128} \\cdot \\frac{7}{18} < \\frac{11}{36}N$. Take $N$ such that both sides of the inequality are integers differing by more than 1, for instance $N = 2 \\operatorname{lcm}(36, 128)$. Then an integer $c$ can be chosen so that $\\frac{100}{128} < \\frac{7}{18}N < \\frac{11}{36}N$. The condition $c < a$ for the first round is satisfied. For this choice of $c$ we have $\\frac{7}{18}N < \\frac{128}{100}c$, and $\\frac{200 - p}{100}c \\ge \\frac{128}{100}c$ was shown above for $p \\le 72$, which implies $N' - 2a_{\\text{max}} < 0$. So $A$ wins if he gets $a_{\\text{max}}$ votes. This is possible if all of his supporters vote for him again, and also all additional voters. On the other hand, it is clear that $B$ is a possible winner for any $p$, for instance if $A$ gets no votes at all (which is not excluded by the conditions). It is of more substance to note that $B$ can also win for $p \\le 72$ even if all of $A$'s supporters vote for him again in the second round. Indeed, if $p \\le 72$ then $c < a$ implies $N' = N - \\frac{p}{100}c > N - \\frac{72}{100}a = N - \\frac{72}{100} \\cdot \\frac{11}{36}N = \\frac{39}{50}N$.\n\nThis is greater than $2a = \\frac{11}{18}N$, so if all additional votes go to $B$, then $B$ wins.\n\n*Remark.* There are values of $N$ for which the situation can describe actual elections, for instance, $N = 1\\,800\\,000$. Then $a = \\frac{11}{36}N = 550\\,000$ and, for $p \\le 72$, the key number for the construction is $\\frac{100}{128} \\cdot \\frac{7}{18}N = 546\\,875 < 550\\,000$. So there are plenty of (integer) choices for $c$ in $[546\\,875, 550\\,000]$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21489,
"subject": "Mathematics (Olympiad)",
"question": "With $\\alpha \\in \\mathbb{R}$, consider the polynomial\n\n$$\nf(x) = x^2 - \\alpha x + 1.\n$$\n\na) For $\\alpha = \\frac{\\sqrt{15}}{2}$, express $f(x)$ as the quotient of two polynomials with non-negative coefficients.\n\nb) Find all values of $\\alpha$ such that $f(x)$ can be written as the quotient of two polynomials with non-negative coefficients.",
"options": [],
"answer": "See solution",
"solution": "a) We consider the following transformation:\n\n$$\n\\left(x^2 - \\frac{\\sqrt{15}}{2}x + 1\\right) \\left(x^2 + \\frac{\\sqrt{15}}{2}x + 1\\right) = x^4 - \\frac{7}{4}x^2 + 1,\n$$\n\n$$\n\\left(x^{4} - \\frac{7}{4}x^{2} + 1\\right) \\left(x^{4} + \\frac{7}{4}x^{2} + 1\\right) = x^{8} - \\frac{17}{16}x^{4} + 1,\n$$\n\n$$\n\\left(x^{8} - \\frac{17}{16}x^{4} + 1\\right) \\left(x^{8} + \\frac{17}{16}x^{4} + 1\\right) = x^{16} + \\frac{223}{256}x^{8} + 1.\n$$\n\nIt follows that $f(x)$ is the quotient of $x^{16} + \\frac{223}{256}x^8 + 1$ and\n\n$$\n\\left(x^2 + \\frac{\\sqrt{15}}{2}x + 1\\right) \\left(x^4 + \\frac{7}{4}x^2 + 1\\right) \\left(x^8 + \\frac{17}{16}x^4 + 1\\right).\n$$\n\nb) Suppose $\\frac{P(x)}{Q(x)} = x^2 - \\alpha x + 1$ where $P, Q$ are polynomials with non-negative coefficients. Substituting $x = 1$, we have\n\n$$\n2 - \\alpha = \\frac{P(1)}{Q(1)} > 0 \\text{ so } \\alpha < 2.\n$$\n\nWe will prove that every real number $\\alpha < 2$ satisfies the problem. Indeed, if $\\alpha \\le 0$ then the polynomial $f(x)$ itself has non-negative coefficients, so we can choose $P(x) = f(x)$, $Q(x) = 1$.\n\nIf $\\alpha \\in (0, 2)$, consider the multiplication\n\n$$\n(x^2 - \\alpha x + 1)(x^2 + \\alpha x + 1) = x^4 + (2 - \\alpha^2)x^2 + 1.\n$$\n\nContinuing in this way, we find that the coefficients of the first and last terms of the polynomial are always 1, and the middle coefficient is determined by the sequence $(u_n)$ as follows:\n\n$$\n\\begin{cases} u_0 = \\alpha, \\\\ u_{n+1} = 2 - u_n^2, \\quad n \\ge 0. \\end{cases}\n$$\n\nWe will prove that there exists a positive term in this sequence. Suppose that for every $n \\ge 1$, $u_n < 0$. Since $\\alpha \\in (0, 2)$, by induction, $-2 < u_n < 0$ for all $n \\ge 1$. Note that\n\n$$\nu_{n+1} - u_n = 2 - u_n - u_n^2 = (2 + u_n)(1 - u_n) > 0,$$\n\nso $u_{n+1} - u_n > 0$ for all $n \\ge 1$; thus, this sequence increases. Since the sequence is bounded above by 0, it has a limit $L \\in (-2, 0]$. Letting $n$ tend to infinity, we have\n\n$$\nL = 2 - L^2 \\implies L \\in \\{1, -2\\}.\n$$\n\nThis contradiction shows that there exists $n = N$ such that $u_N \\ge 0$. Consider the polynomial sequence\n\n$$\nf_n(x) = x^{2n+1} + u_n x^{2n} + 1\n$$\n\nfor $n = 1, 2, 3, \\dots, N$. It is easy to see that $f(x)$ is the quotient of two polynomials $f_N(x)$ and $f_1(x)f_2(x) \\dots f_{N-1}(x)$. Clearly, these polynomials have non-negative coefficients. $\\square$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21490,
"subject": "Mathematics (Olympiad)",
"question": "For every pair of real numbers $a, b$, consider the sequence $\\{x_n\\}$ for $n = 0, 1, 2, \\dots$ defined by:\n\n$$\nx_0 = a \\quad \\text{and} \\quad x_{n+1} = x_n + b \\sin x_n \\quad \\text{for every } n = 0, 1, 2, \\dots\n$$\n\nProve that:\n\n1) For every real number $a$, the sequence $\\{x_n\\}$ corresponding to $(a, b)$ has a finite limit as $n \\to \\infty$. Find this limit.\n\n2) For every $b > 2$, there exists a real number $a$ such that the sequence $\\{x_n\\}$ corresponding to $(a, b)$ does not have a finite limit as $n \\to \\infty$.",
"options": [],
"answer": "See solution",
"solution": "1) For $a = k\\pi$ ($k \\in \\mathbb{Z}$), we have $x_n = k\\pi$ for all $n \\in \\mathbb{N}$, so $\\lim_{n \\to \\infty} x_n = k\\pi$.\n\nFor $a \\neq k\\pi$ ($k \\in \\mathbb{Z}$), consider the function $f(x) = x + \\sin x$ on $\\mathbb{R}$. We have $f'(x) = 1 + \\cos x \\ge 0$ for all $x \\in \\mathbb{R}$.\n\n- If $a \\in (2k\\pi, (2k+1)\\pi)$, the sequence $\\{x_n\\}$ is increasing and bounded above by $(2k+1)\\pi$, so $\\lim_{n \\to \\infty} x_n = (2k+1)\\pi$.\n- If $a \\in ((2k-1)\\pi, 2k\\pi)$, the sequence $\\{x_n\\}$ is decreasing and bounded below by $(2k-1)\\pi$, so $\\lim_{n \\to \\infty} x_n = (2k-1)\\pi$.\n\nThus, for every $a \\in \\mathbb{R}$, the sequence $\\{x_n\\}$ corresponding to $(a, 1)$ is convergent and $\\lim_{n \\to \\infty} x_n = \\left(2\\left[\\frac{a}{2\\pi}\\right] + \\operatorname{sign}\\left(\\left\\{\\frac{a}{2\\pi}\\right\\}\\right)\\right)\\pi$.\n\n2) For $b > 2$, there exists $a_0 \\in (0, \\pi)$ such that $2a_0 = b \\sin a_0$, and the sequence $\\{x_n\\}$ corresponding to $(a = \\pi - a_0, b)$ is periodic with period 2.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21491,
"subject": "Mathematics (Olympiad)",
"question": "\n\n(a) Show that for a polynomial $P(z) = a_0 + a_1 z + \\\\cdots + a_d z^d$, its *reciprocal polynomial* is given by\n$$\nP^*(z) = z^d \\overline{P\\left(\\frac{1}{z}\\right)}.\n$$\n\n(b) Let $q(z)$ be a polynomial of degree $n$ with roots $z_1, z_2, \\dots, z_n$ (not necessarily distinct), all satisfying $|z_i| \\le 1$. Define $q^*(z) = z^n \\overline{q\\left(\\frac{1}{z}\\right)}$ and $Q(z) = z^m q(z) + q^*(z)$ for some $m \\ge 0$. Show that any root $r$ of $Q(z)$ satisfies $|r| = 1$.",
"options": [],
"answer": "See solution",
"solution": "We have\n$$\nr^m (r - z_1) \\cdots (r - z_n) = - (1 - r \\overline{z}_1) \\cdots (1 - r \\overline{z}_n)\n$$\nwhich implies\n$$\n|r|^m |r - z_1| \\cdots |r - z_n| = |1 - r \\overline{z}_1| \\cdots |1 - r \\overline{z}_n|.\n$$\n\n**Case 1:** $|r| > 1$\n\nFor each $i$, $|r - z_i| \\ge |1 - r \\overline{z}_i|$ because\n$$\n|r - z_i|^2 = |r|^2 - r \\overline{z}_i + \\overline{r} z_i + |z_i|^2\n$$\nand\n$$\n|1 - r \\overline{z}_i|^2 = 1 - r \\overline{z}_i - \\overline{r} z_i + (|r||z_i|)^2.\n$$\nSo,\n$$\n|r|^2 + |z_i|^2 \\ge 1 + (|r||z_i|)^2\n$$\nor\n$$\n(|r|^2 - 1)(1 - |z_i|^2) \\ge 0.\n$$\nSince $|r| > 1$ and $|z_i| \\le 1$, this holds. Thus,\n$$\n|r|^m |r - z_1| \\cdots |r - z_n| > |r - z_1| \\cdots |r - z_n| \\ge |1 - r \\overline{z}_1| \\cdots |1 - r \\overline{z}_n|,\n$$\ncontradicting the equality. So $|r| \\le 1$.\n\n**Case 2:** $|r| < 1$\n\nSimilarly, $|r - z_i| \\le |1 - r \\overline{z}_i|$, so\n$$\n|r|^m |r - z_1| \\cdots |r - z_n| < |r - z_1| \\cdots |r - z_n| \\le |1 - r \\overline{z}_1| \\cdots |1 - r \\overline{z}_n|,\n$$\nagain a contradiction. Thus, $|r| = 1$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21492,
"subject": "Mathematics (Olympiad)",
"question": "For all non-negative real numbers $x$, $y$, $z$ with $x \\ge y$, prove the inequality\n\n$$\n\\frac{x^3 - y^3 + z^3 + 1}{6} \\ge (x - y) \\sqrt{xyz}.\n$$",
"options": [],
"answer": "See solution",
"solution": "From the AM-GM inequality, we have $\\sqrt{xyz} \\le \\frac{xy + z}{2}$. Hence, it suffices to prove that\n\n$$\n\\frac{x^3 - y^3 + z^3 + 1}{3} \\ge (x - y)(xy + z),\n$$\n\nwhich is equivalent to\n\n$$\n\\frac{(x - y)^3 + z^3 + 1}{3} \\ge z(x - y).\n$$\n\nThis inequality follows directly from the AM-GM inequality applied to the numbers $(x - y)^3$, $z^3$, and $1$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21493,
"subject": "Mathematics (Olympiad)",
"question": "Prove that for all positive integers $n$, the polynomial\n$$\nP(x) = (x^{2} - 7x + 6)^{2n} + 13\n$$\ncannot be written as a product of $(n + 1)$ non-constant polynomials with integer coefficients.",
"options": [],
"answer": "See solution",
"solution": "Clearly, the polynomial $P(x)$ has degree $4n$ and has no real root. Thus, any factor of $P(x)$ must have even degree. Suppose that $P(x)$ can be expressed as the product of $n + 1$ polynomials with degree greater than $0$:\n$$\nP(x) = P_{1}(x) \\cdot P_{2}(x) \\cdots P_{n+1}(x)\n$$\nThen each $P_i(x)$ for $i = 1, 2, \\dots, n$ has even degree. Since the sum of degrees of $P_i(x)$ is $4n$, there must be at least two polynomials, say $P_1(x)$ and $P_2(x)$, of degree $2$.\n\nSince $P(x)$ has leading coefficient $1$, suppose $P_1(x)$ and $P_2(x)$ have leading coefficient $1$:\n$$\nP_{1}(x) = x^{2} + a x + b,\\quad P_{2}(x) = x^{2} + c x + d \\text{ where } a, b, c, d \\in \\mathbb{Z}.\n$$\nSince $P_1(x)$ and $P_2(x)$ have no real roots, we have $P_1(x) > 0$ and $P_2(x) > 0$ for all integers $x$.\n\nWe have\n$$\n13 = P(1) = P_{1}(1) \\cdot P_{2}(1) \\cdots P_{n+1}(1)\n$$\nand\n$$\n13 = P(6) = P_{1}(6) \\cdot P_{2}(6) \\cdots P_{n+1}(6).\n$$\nFrom this, at least one of $P_1(1)$ and $P_2(1)$ equals $1$. Without loss of generality, assume $P_1(1) = 1$. It follows that $a = -b$, so $P_1(6) = 36 - 5b$. We observe that $36 - 5b > 0$ and cannot be equal to $13$, so $36 - 5b = 1$, thus $b = 7$, $a = -7$. But in this case, the polynomial $P_1(x) = x^2 - 7x + 7$ has a real root, which is a contradiction.\n\nTherefore, the above assumption is wrong, so $P(x)$ cannot be expressed as the product of $(n + 1)$ non-constant polynomials. $\\square$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21494,
"subject": "Mathematics (Olympiad)",
"question": "The Euclidean plane is dissected into several disjoint *bounded* regions (not further subdivided) by $n$ intersecting equilateral triangles of side length 1. Determine the smallest value of $n$ such that the number of parts thus arising can be at least 2023.\n\nThe picture shows four intersecting triangles.\n\n",
"options": [],
"answer": "See solution",
"solution": "We will prove the following claim:\n\n**Claim:** For $n \\in \\mathbb{Z}^{+}$, the maximal number of bounded regions of the plane generated by $n$ triangles (of whatever size and shape) is $1 + 3n(n-1)$.\n\nOnce the claim is established, we can easily see that the inequality $1 + 3n(n-1) \\ge 2023$ is equivalent to $n \\ge 27$, so that we need at least 27 triangles.\n\nBelow, we will establish that 27 equilateral triangles of side length 1 are actually enough.\n\n*Proof.* By induction. The case $n=1$ is trivial.\n\nSuppose that the claim is true for $n$. Let $C = \\{T_1, \\dots, T_n\\}$ be a configuration consisting of triangles $T_1, \\dots, T_n$. Let us see what happens when a new triangle $T$ is added to the configuration. We enumerate the points of intersection of $T$ with the elements of $C$, starting at some point of $T$, in clockwise direction; let the set of these points be $\\mathcal{P} := \\{P_1, \\dots, P_k\\}$. Clearly, each region newly arising through the addition of $T$ will have some part of its border in common with $T$; since the regions are not supposed to be further subdivided, the part of the border of a new region will be the section between two successive elements of $\\mathcal{P}$, where we also count $P_1$ as a successor of $P_k$. Each such line segment can at best generate one new region by either cutting an already bounded region into two parts or by splitting off some bounded part of the unbounded region around $C$. It follows that the number of newly generated regions is at most $k$. Now, two triangles can have at most 6 points of intersection, and $C$ consists of $n$ triangles, so that $k \\le 6n$. Consequently, the addition of $T$ can yield at most $6n$ new regions. By assumption, the number of regions prior to the addition of $T$ was at most $1 + 3n(n-1)$. Thus, after adding $T$, the number of regions is at most $1 + 3n(n-1) + 6n = 1 + 3n(n+1)$.\n\n(To see that two triangles can intersect in at most 6 points, note that a triangle, being convex, can be intersected by a line in at most 2 points, and each triangle consists of three line segments.) $\\square$\n\n*Proof that 27 triangles suffice.* We prove by induction the more general claim that the upper bound $1 + 3n(n-1)$ for the number of bounded regions generated by $n$ triangles can actually be attained. Again, for the case $n=1$, this is trivial.\n\nWe now assume inductively that we have produced a configuration $C = \\{T_1, \\dots, T_n\\}$ of $n$ triangles such that:\n\n1. All elements of $C$ have the same orthocenter.\n2. Each two distinct elements of $C$ intersect in six points.\n3. No point lies on three elements of $C$.\n4. The number of bounded regions is $1 + 3n(n-1)$.\n\nWe will show that $C$ can be extended to a configuration with one extra triangle $T$ such that (1)-(3) still hold for the new configuration and such that the number of bounded regions is increased by $6n$.\n\nTo this end, consider rotations of $T_1$ around its orthocenter. There are only finitely many angles for which the respective rotation will pass through one of the finitely many points of intersection in $C$ (or such that the result will be identical to an element of $C$); thus (because there are infinitely many angles of rotation), there is an angle $\\alpha$ such that the result $T$ of the rotation of $T_1$ around the orthocenter by $\\alpha$ will not pass through any prior point of intersection. Thus (1) and (3) are satisfied for $C \\cup \\{T\\}$. It is easy to see that, rotating an equilateral triangle around its orthocenter will either yield an identical triangle or exactly six points of intersection; the first case is excluded by construction, thus (2) is satisfied as well.\n\nThus, $T$ has $6n$ points of intersection with the elements of $C$, let these be $P_1, \\dots, P_{6n}$. We need to show each polygonal line $P_iP_{i+1}$ (which is either a line segment or comprised of two line segments meeting at a vertex of $T$) generates a new bounded region; let us enumerate these segments as $s_1, \\dots, s_{6n}$. To see this, note that, for each such segment $s_k = P_kP_{k+1}$ (with $s_{6n} = P_{6n}P_1$), one of the following cases holds:\n\n1. $s_k$ runs through a region already bounded by $C$ and $s_1, \\dots, s_{k-1}$, splitting it into two parts.\n2. $s_k$ runs through the unbounded region of $C$, splitting off a bounded part of it.\n3. $s_k$ runs through the unbounded region of $C$ without splitting off a bounded part of it.\n\nIn the cases (1) and (2), a new region is generated, while in case (3), this is not the case. But note that case (3) cannot occur: For if $P_k$ belongs to $T_i$ and $P_{k+1}$ belongs to $T_j$ then, by the case assumption, we have $i \\neq j$ and, by assumption (2) about $C$, $T_i$ and $T_j$ intersect, so that $s_k$ must generate a new bounded region.\n\nIt follows that $6n$ new bounded regions are added, so that we now have $1+3n(n-1) = 1+3(n+1)n$ regions, which finishes the induction. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21495,
"subject": "Mathematics (Olympiad)",
"question": "Given a table with $n$ rows and $12$ columns, where each cell contains a $0$ or a $1$, the table satisfies the following properties:\n\n1. Every two rows are different.\n2. Every row contains exactly $4$ entries equal to $1$.\n3. For every $3$ rows, there is a column that intersects them at three entries equal to $0$.\n\nFind the greatest $n$ for which such a table exists.",
"options": [],
"answer": "See solution",
"solution": "The answer is $\\binom{11}{4} = 330$.\n\n**Construction:**\n\nForm all $\\binom{11}{4} = 330$ unordered quadruples $i, j, k, l$ from $\\{1, 2, \\dots, 11\\}$. For each quadruple, create a row of length $12$ with $1$'s at positions $i, j, k, l$ and $0$'s elsewhere. These $330$ rows form a $330 \\times 12$ table satisfying all conditions. (For condition 3, note that column $12$ contains only zeros.)\n\n**Upper Bound:**\n\nFor any such table $T$, each row $F$ determines a unique quadruple $Q$ of columns with $1$'s (by property 2). Different rows yield different quadruples (by property 1), so there are $n$ admissible quadruples.\n\nFor each admissible quadruple $Q$, consider all partitions of its $8$ complementary columns into two quadruples $Q_1$ and $Q_2$. There are $\\frac{1}{2} \\binom{8}{4}$ such partitions. In each partition, at least one of $Q_1$ or $Q_2$ is non-admissible (otherwise, property 3 is violated). Thus, each admissible $Q$ generates at least $\\frac{1}{2} \\binom{8}{4}$ non-admissible quadruples.\n\nEach non-admissible quadruple $Q'$ can be generated by at most $\\binom{8}{4}$ admissible quadruples. Therefore, the number of distinct non-admissible quadruples is at least $\\frac{n}{2}$.\n\nSince there are $\\binom{12}{4} = 495$ quadruples in total, $n$ admissible and $495-n$ non-admissible, so $495-n \\ge \\frac{n}{2}$, which gives $n \\le 330$.\n\nThus, the greatest $n$ is $330$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21496,
"subject": "Mathematics (Olympiad)",
"question": "Given $n$ points $P_1, P_2, \\dots, P_n$ on a plane, let $M$ be any point on segment $AB$ on the plane. Denote by $|P_iM|$ the distance between $P_i$ and $M$, for $i = 1, 2, 3, \\dots, n$. Prove that\n\n$$\n\\sum_{i=1}^n |P_iM| \\le \\max\\left\\{\\sum_{i=1}^n |P_iA|, \\sum_{i=1}^n |P_iB|\\right\\}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $O$ be the origin. Then $\\overrightarrow{OM} = t\\, \\overrightarrow{OA} + (1-t)\\, \\overrightarrow{OB}$, where $t \\in (0, 1)$.\n\n$$\n\\begin{aligned}\n|P_iM| &= |\\overrightarrow{OM} - \\overrightarrow{OP_i}| \\\\\n&= |t\\, \\overrightarrow{OA} + (1-t)\\, \\overrightarrow{OB} - t\\, \\overrightarrow{OP_i} - (1-t)\\, \\overrightarrow{OP_i}| \\\\\n&= |t(\\overrightarrow{OA} - \\overrightarrow{OP_i}) + (1-t)(\\overrightarrow{OB} - \\overrightarrow{OP_i})| \\\\\n&\\le t|\\overrightarrow{OA} - \\overrightarrow{OP_i}| + (1-t)|\\overrightarrow{OB} - \\overrightarrow{OP_i}| \\\\\n&= t|P_iA| + (1-t)|P_iB|.\n\\end{aligned}\n$$\n\nHence,\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} |P_iM| &\\le t \\sum_{i=1}^{n} |P_iA| + (1-t) \\sum_{i=1}^{n} |P_iB| \\\\\n&\\le \\max\\left\\{\\sum_{i=1}^{n} |P_iA|, \\sum_{i=1}^{n} |P_iB|\\right\\}.\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21497,
"subject": "Mathematics (Olympiad)",
"question": "Let $A = \\{1, 3, 3^2, 3^3, \\dots, 3^{2014}\\}$.\n\nWe obtain a partition of $A$ if $A$ is written as a disjoint union of nonempty subsets.\n\n**a)** Prove that there is no partition of $A$ such that the product of elements in each subset is a square.\n\n**b)** Prove that there exists a partition of $A$ such that the sum of elements in each subset is a square.",
"options": [],
"answer": "See solution",
"solution": "**a)** Assume that such a partition exists. Then the product of all elements of $A$ must be a square as well. But this equals $3^{1+2+3+\\dots+2014} = 3^{2015 \\cdot 1007}$, which is not a square.\n\n**b)** Observe that $3^{2n} + 3^{2n+1} = (3^n \\cdot 2)^2$, hence a possible partition is\n\n$$\nA = \\{1, 3\\} \\cup \\{3^2, 3^3\\} \\cup \\dots \\cup \\{3^{2012}, 3^{2013}\\} \\cup \\{3^{2014}\\}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21498,
"subject": "Mathematics (Olympiad)",
"question": "The sequence $a_1, a_2, \\dots, a_n$ consists of the numbers $1, 2, \\dots, n$ in some order. For which positive integers $n$ is it possible that $0, a_1, a_1 + a_2, \\dots, a_1 + a_2 + \\dots + a_n$ all have different remainders when divided by $n + 1$?",
"options": [],
"answer": "See solution",
"solution": "It is possible if and only if $n$ is odd.\n\nIf $n$ is even, then $a_1 + a_2 + \\dots + a_n = 1 + 2 + \\dots + n = \\frac{n}{2} (n + 1)$, which is congruent to $0 \\pmod{n + 1}$. Therefore, the task is impossible.\n\nNow suppose $n$ is odd. We will show that we can construct $a_1, a_2, \\dots, a_n$ that satisfy the conditions given in the problem. Let $n = 2k + 1$ for some non-negative integer $k$. Consider the sequence: $1, 2k, 3, 2k-2, 5, 2k-3, \\dots, 2, 2k+1$, i.e., for each $1 \\leq i \\leq 2k+1$, $a_i = i$ if $i$ is odd and $a_i = 2k+2-i$ if $i$ is even.\n\nWe first show that each term $1, 2, \\dots, 2k+1$ appears exactly once. Clearly, there are $2k+1$ terms. For each odd number $m$ in $\\{1, 2, \\dots, 2k+1\\}$, $a_m = m$. For each even number $m$ in this set, $a_{2k+2-m} = 2k+2-(2k+2-m)=m$. Hence, every number appears in $a_1, \\dots, a_{2k+1}$.\n\nNow, determine $a_1 + a_2 + \\dots + a_m \\pmod{2k+2}$. Let $b_m = a_1 + a_2 + \\dots + a_m$.\n\nIf $m$ is odd, note that $a_1 \\equiv 1 \\pmod{2k+2}$, $a_2 + a_3 = a_4 + a_5 = \\dots = a_{2k} + a_{2k+1} = 2k+3 \\equiv 1 \\pmod{2k+2}$. Therefore, $\\{b_1, b_3, \\dots, b_{2k+1}\\} = \\{1, 2, 3, \\dots, k+1\\} \\pmod{2k+2}$.\n\nIf $m$ is even, $a_1 + a_2 = a_3 + a_4 = \\dots = a_{2k-1} + a_{2k} = 2k+1 \\equiv -1 \\pmod{2k+2}$. Therefore, $\\{b_2, b_4, \\dots, b_{2k}\\} = \\{-1, -2, \\dots, -k\\} \\pmod{2k+2} \\equiv \\{2k+1, 2k, \\dots, k+2\\} \\pmod{2k+2}$.\n\nTherefore, $b_1, b_2, \\dots, b_{2k+1}$ do indeed have different remainders when divided by $2k+2$. This completes the problem. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21499,
"subject": "Mathematics (Olympiad)",
"question": "Let a square $ABCD$ of side length $8\\ \\text{cm}$ be divided, with lines parallel to its sides, into $64$ small squares of side $1\\ \\text{cm}$. Color $7$ small squares black, and the other $57$ small squares are white. Suppose that there is a positive integer $k$ such that no matter which $7$ squares are black, there is a rectangle of area $k\\ \\text{cm}^2$ with sides parallel to the sides of $ABCD$ and all of its small squares that contains are white. Find the maximum value of $k$.",
"options": [],
"answer": "See solution",
"solution": "We divide $ABCD$ into $8$ rectangles of size $4 \\times 2$. Since we color seven small squares black, by the pigeonhole principle, there will be at least one $4 \\times 2$ rectangle that contains no black square, and its area is $8\\,\\text{cm}^2$.\n\n\n\nIn what follows, we will prove that there is a coloring with $7$ black squares such that there is no rectangle with only white squares and area bigger than $8\\,\\text{cm}^2$. Indeed, we can see such a coloring in the following figure.\n\n\n\n**Note:** For the first part of the solution, we can also consider the $8$ rows or the $8$ columns, and the result follows again by the pigeonhole principle.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21500,
"subject": "Mathematics (Olympiad)",
"question": "In an acute-angled triangle $ABC$, the angle bisector $AL$, the altitude $BH$, and the perpendicular bisector of segment $AB$ intersect at one point. Find the angle $BAC$.",
"options": [],
"answer": "See solution",
"solution": "Let the angle $BAC$ be $2\\alpha$. Since $\\triangle APB$ is isosceles, $\\angle PBA = \\alpha$. In the right triangle $AHB$, $3\\alpha = 90^\\circ$, so $\\angle BAC = 2\\alpha = 60^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21501,
"subject": "Mathematics (Olympiad)",
"question": "In an acute triangle $ABC$, the altitudes $AD$, $BE$, and $CF$ meet at the orthocenter $H$. The circle $\\omega$, centered at $O$, passes through $A$ and $H$ and intersects sides $AB$ and $AC$ again at $Q$ and $P$ (other than $A$), respectively. The circumcircle of triangle $OPQ$ is tangent to segment $BC$ at $R$. Prove that $$\\frac{CR}{BR} = \\frac{ED}{FD}.$$ \n\n*Note:* Let $\\angle CAB = x$, $\\angle ABC = y$, and $\\angle BCA = z$. Without loss of generality, assume $Q$ is between $A$ and $F$, and $P$ is between $C$ and $E$ (since $\\angle FQH = \\angle APH$).",
"options": [],
"answer": "See solution",
"solution": "**First Solution:** (Based on work by Ryan Ko) Let $M$ be the midpoint of segment $AH$. Since $\\angle AEH = \\angle AFH = 90^\\circ$, quadrilateral $AEHF$ is cyclic with $M$ as its circumcenter. Hence triangle $EFM$ is isosceles with vertex angle $\\angle EMF = 2\\angle CAB = 2x$. Likewise, triangle $PQO$ is also isosceles with vertex angle $\\angle POQ = 2x$. Therefore, triangles $EFM$ and $PQO$ are similar.\n\n\n\nSince $AEHF$ and $APHQ$ are cyclic, $\\angle EFH = \\angle EAH = \\angle PQH$ and $\\angle FEH = \\angle FAH = \\angle QPH$. Consequently, triangles $HEF$ and $HPQ$ are similar. Quadrilaterals $EHFM$ and $PHQO$ are also similar. More precisely, if $\\angle QHF = \\theta$, there is a spiral similarity $S$, centered at $H$ with clockwise rotation angle $\\theta$ and ratio $QH/FH$, that sends $FMEH$ to $QOPH$. Let $R_1$ be the point between $B$ and $D$ such that $\\angle R_1HD = \\theta$. Then triangles $QHF$ and $R_1HD$ are similar, so $S(D) = R_1$. It follows that\n\n$$\nS(DFME) = R_1QOP.\n$$\n\nIt is well known that points $D$, $E$, $F$, and $M$ lie on a circle (the *nine-point circle* of triangle $ABC$). (This can be established by noting that $ABDE$ and $ACDF$ are cyclic, implying $\\angle FDB = \\angle CAF = x$, $\\angle EDC = \\angle BAE = x$, and $\\angle EDF = 180^\\circ - 2x = 180^\\circ - \\angle EMF$.) Since $DFME$ is cyclic, $R_1QOP$ must also be cyclic. By the given conditions, $R_1 = R$, implying that\n\n$$\nS(DEF) = RPQ,\n$$\n\nso triangles $DEF$ and $RPQ$ are similar. It follows that\n\n$$\n\\frac{ED}{FD} = \\frac{PR}{QR}.\n$$\n\n\n\nNow, since $ACDF$ and $ABDE$ are cyclic, $\\angle BFD = \\angle AFE = \\angle ACB = z$. Thus $\\angle DFE = 180^\\circ - 2z$. Since triangles $DEF$ and $RPQ$ are similar, $\\angle RQP = 180^\\circ - 2z$. Because $CR$ is tangent to the circumcircle of triangle $PQR$, $\\angle CRP = \\angle RQP = 180^\\circ - 2z$. Thus, in triangle $CPR$, $\\angle CPR = z$, so it is isosceles with $CR = PR$. Likewise, $BR = QR$. Therefore,\n\n$$\n\\frac{ED}{FD} = \\frac{PR}{QR} = \\frac{CR}{BR}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21502,
"subject": "Mathematics (Olympiad)",
"question": "Prove that one can choose 7 pairwise distinct numbers from $1, 2, \\ldots, 10000$, such that none of them is a perfect square and no sum of several numbers out of the 7 chosen ones is a perfect square either.\n\n",
"options": [],
"answer": "See solution",
"solution": "Consider the following 7 numbers: $2^1$, $2^3$, $2^5$, $2^7$, $2^9$, $2^{11}$, $2^{13} = 8192 < 10000$. It is clear that the sum of any subset of them is such that the highest power of 2 dividing the sum is odd, thus, it is not a perfect square.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21503,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$ be positive integers. Prove that if the numbers\n\n$$\n\\frac{a^2}{a+b}, \\frac{b^2}{b+c}, \\frac{c^2}{c+a}\n$$\nare integers and primes, then $a = b = c$.",
"options": [],
"answer": "See solution",
"solution": "We will use the following result:\n\n**Lemma.** If $x$ and $y$ are positive integers such that $\\frac{x^2}{x+y}$ is an integer and prime, then $y \\ge x$.\n\n**Proof of Lemma.** Assume that $\\frac{x^2}{x+y} = p$, where $p$ is a prime.\n\nWe have $py = x(x - p)$, so $p \\mid x(x - p)$, i.e. $p \\mid x$. Let $x = up$, for some positive integer $u$. We get\n\n$$\ny = x(u - 1) \\ge x,\n$$\n\nand we are done, since $u \\ge 2$.\n\nApplying the lemma to our situation, it follows $b \\ge a$, $c \\ge b$, $a \\ge c$, meaning that $a = b = c$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21504,
"subject": "Mathematics (Olympiad)",
"question": "Positive integers $x$ and $y$ satisfy the equation $\\sqrt{x} + \\sqrt{y} = \\sqrt{1183}$. What is the minimum possible value of $x + y$?\n\n(A) 585 (B) 595 (C) 623 (D) 700 (E) 791",
"options": [],
"answer": "See solution",
"solution": "Observe that $1183 = 13^2 \\cdot 7$, so $\\sqrt{1183} = 13\\sqrt{7}$. Because $\\sqrt{x} + \\sqrt{y} = 13\\sqrt{7}$, it follows that $\\sqrt{x}$ and $\\sqrt{y}$ must be of the form $a\\sqrt{7}$ and $b\\sqrt{7}$, respectively, where $a$ and $b$ are positive integers and $a + b = 13$. Then $\\sqrt{x} = \\sqrt{7a^2}$ and $\\sqrt{y} = \\sqrt{7b^2}$, so $x = 7a^2$ and $y = 7b^2$. Substituting gives\n\n$$\nx + y = 7(a^2 + b^2) = 7(a^2 + (13-a)^2) = 14a^2 - 182a + 1183.\n$$\n\nThe minimum value of the quadratic polynomial occurs at $a = \\frac{182}{2 \\cdot 14} = 6.5$. Because $a$ and $b$ must be positive integers, without loss of generality (by symmetry), choose $a = 6$ and $b = 7$. The minimum possible value of $x + y$ is\n\n$$\nx + y = 7 \\cdot (6^2 + 7^2) = 7 \\cdot (36 + 49) = 7 \\cdot 85 = 595.\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21505,
"subject": "Mathematics (Olympiad)",
"question": "\n\n**Variant.** Determine all integers $C$ for which there exists a sequence $(a_1, a_2, \\\\dots)$ of positive integers satisfying\n\n$$\na_{n+1}^2 = C + (n + 2021)a_n\n$$\n\nfor all $n \\ge 1$.",
"options": [],
"answer": "See solution",
"solution": "Clearly for $C = 1$ we have the solution $(a_n)_{n=1}^\\infty = (n + 2019)_{n=1}^\\infty$. Let's prove that this is the only value for $C$ that works.\n\nAssume $(a_n)_{n=1}^\\infty$ is a solution and let $(b_n)_{n=1}^\\infty = (a_n - n)_{n=1}^\\infty$. We claim that for $n > |C| + 2021^2$:\n\n(i) If $b_n < 2019$, then $b_n < b_{n+1} < 2019$.\n\n(ii) If $b_n > 2019$, then $b_n > b_{n+1} > 2019$.\n\nIt is clear that these two claims imply that $b_n = 2019$ for all large $n$ and hence that $C = 1$.\nLet us prove the claims:\n\n(i) First of all, $b_n \\le 2018$ implies that\n\n$$\n\\begin{align*}\na_{n+1}^2 &\\le C + (n + 2021)(n + 2018) \\\\\n&= (n + 2020)^2 - n + C + 2018 \\cdot 2021 - 2020^2 \\\\\n&< (n + 2020)^2\n\\end{align*}\n$$\n\nand hence $a_{n+1} < n + 2020$ so that indeed $b_{n+1} < 2019$.\n\nMoreover, we have\n\n$$\n\\begin{align*}\na_{n+1}^2 &= C + (n + 2021)(n + b_n) \\\\\n&= (n + 1 + b_n)^2 + (2019 - b_n)n + 2021b_n + C - (b_n + 1)^2 \\\\\n&\\ge (n + 1 + b_n)^2 + n + C - 2019^2 \\\\\n&> (n + 1 + b_n)^2\n\\end{align*}\n$$\n\nand hence $a_{n+1} > n + 1 + b_n$ so that indeed $b_{n+1} > b_n$.\n\n(ii) First of all, $b_n \\ge 2020$ implies that\n\n$$\na_{n+1}^2 \\geq C + (n + 2021)(n + 2020) = (n + 2020)^2 + n + C + 2021 > (n + 2020)^2\n$$\n\nand hence $a_{n+1} > n + 2020$ so that indeed $b_{n+1} > 2019$.\n\nMoreover, we have\n\n$$\n\\begin{align*}\na_{n+1}^2 &= C + (n + 2021)(n + b_n) \\\\\n&= (n + 1 + b_n)^2 + (2019 - b_n)n + 2021b_n + C - (b_n + 1)^2 \\\\\n&\\le (n + 1 + b_n)^2 - n + C \\\\\n&< (n + 1 + b_n)^2\n\\end{align*}\n$$\n\nand hence $a_{n+1} < n + 1 + b_n$ so that indeed $b_{n+1} < b_n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21506,
"subject": "Mathematics (Olympiad)",
"question": "Ана замислила еден број. Тој број го помножила со 7, потоа му додала 6, добиениот резултат го поделила со 5 и го добила бројот 53. Откриј кој број го замислила Ана!",
"options": [],
"answer": "See solution",
"solution": "Прв начин: Нека бројот што го замислила Ана се означи со $x$. Тогаш, од условот во задачата се добива равенката $\\frac{7x+6}{5} = 53$, од каде $7x+6 = 265$, па $7x = 259$, односно $x = 37$.\n\n**Втор начин:** Реализирајќи ги условите од задачата од назад кон напред се добива $\\frac{53 \\times 5 - 6}{7} = 37$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21507,
"subject": "Mathematics (Olympiad)",
"question": "Suppose we have six points $P, Q, R, S, T, U$ in the plane, no three of which are collinear. Prove that if for every quadruple of these points, one of the points lies on the circle passing through the other three (i.e., any four points are concyclic), then all six points lie on a common circle.",
"options": [],
"answer": "See solution",
"solution": "Consider two quadruples containing $P$ and $Q$:\n\n$$\nQ_1 = (P, Q, R, S), \\quad Q_2 = (P, Q, T, U).\n$$\n\nIf some point $R$ is in both quadruples, the claim is direct. Otherwise, the remaining four points must fill the four remaining places, so each appears exactly once. Now, since $P$ is a **good** point in at least three quadruples, consider a third quadruple $Q_3$ containing $P$ as a **good** point. The three remaining places must be filled with the other five points $Q, R, S, T, U$. If both $R, S$ or $T, U$ are in $Q_3$, then $Q_1$ or $Q_2$ have three common points with $Q_3$. Otherwise, $Q_3$ contains one from $\\{R, S\\}$, one from $\\{T, U\\}$, and $Q$, so both $Q_1$ and $Q_2$ have three common points with $Q_3$. Thus, in any case, there are two quadruples with exactly three common points and the same **good** point $P$.\n\nAssume these quadruples are $(P, Q, R, S)$ and $(P, Q, R, T)$. Let $\\omega_1$ and $\\omega_2$ be the circumcircles of triangles $QRS$ and $QRT$, respectively. The power of $P$ with respect to both circles is equal:\n\n$$\n\\mathcal{P}_{\\omega_1}(P) = \\mathcal{P}_{\\omega_2}(P) = k.\n$$\n\nIf $\\omega_1 \\neq \\omega_2$, then $P$ lies on the radical axis of these circles, i.e., on $QR$, which is impossible since no three points are collinear. Thus, $\\omega_1 = \\omega_2$, so $Q, R, S, T$ are concyclic. Considering the quadruple $(Q, R, S, T)$, we get $k = 0$.\n\nTherefore, for any quadruple, one point lies on the circle through the other three, so any four points are concyclic. Fixing three points $P, Q, R$, any other point is concyclic with them, so all six points lie on the circumcircle of triangle $PQR$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21508,
"subject": "Mathematics (Olympiad)",
"question": "Given an acute triangle $ABC$ with $BC < CA < AB$. Points $K$ and $L$ lie on segments $AC$ and $AB$ and satisfy $AK = AL = BC$. The perpendicular bisectors of segments $CK$ and $BL$ intersect line $BC$ at points $P$ and $Q$, respectively. Segments $KP$ and $LQ$ intersect at $M$. Prove that $CK + KM = BL + LM$.",
"options": [],
"answer": "See solution",
"solution": "Let $D$ and $E$ be points on rays $ML$ and $MK$, respectively, such that $LD = AB$ and $KE = AC$. As $\\angle DLA = \\angle BLM = \\angle LBC$, $LD = AB$, and $AL = BC$, triangles $DLA$ and $ABC$ are congruent. Analogously, $EAK$ and $ABC$ are congruent. From\n\n$$\n\\angle DAL + \\angle LAK + \\angle KAE = \\angle BCA + \\angle CAB + \\angle ABC = 180^{\\circ},\n$$\n\nit follows that $A$ belongs to the segment $DE$. Therefore, in light of $\\angle LDA = \\angle BAC = \\angle AEK$, triangle $MDE$ is isosceles with $MD = ME$. Note that\n\n$$\n\\begin{aligned}\nBL + LM &= AB + LM - AL = DL + LM - AL = DM - AL, \\\\\nCK + KM &= AC + KM - AK = EK + KM - AK = EM - AK.\n\\end{aligned}\n$$\n\nFrom $DM = EM$ and $AL = AK$, it follows that the right-hand sides of the two equalities above are equal. Hence, the left-hand sides are equal as well, i.e., $CK + KM = BL + LM$.\n\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21509,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\sigma$ be a permutation of $\\{1, 2, \\dots, 2014\\}$. Define $I_\\sigma = \\{|\\sigma_i - i| : i = 1, 2, \\dots, 2014\\}$ and let $|I_\\sigma|$ denote the number of distinct values in $I_\\sigma$.\n\nIt is observed that:\n\n- If $\\sigma = \\{2014, 2013, \\dots, 1008, 1, 1007, 1006, \\dots, 2\\}$, then $|I_\\sigma| = 2013$.\n- If $\\sigma = \\{2013, 2012, \\dots, 1008, 1007, 1, 1006, 1005, \\dots, 2, 2014\\}$, then $|I_\\sigma| = 2012$.\n- If $\\sigma = \\{2k, 2k-1, \\dots, k+1, 1, k, k-1, \\dots, 2, 2k+1, 2k+2, \\dots, 2014\\}$, then $|I_\\sigma| = 2k-1$ for $k = 1007$.\n- If $\\sigma = \\{2k+1, 2k, \\dots, k+2, 1, k+1, k, \\dots, 2, 2k+2, 2k+3, \\dots, 2014\\}$, then $|I_\\sigma| = 2k$ for $k = 1006$.\n- If $\\sigma = \\{1, 2, \\dots, 2014\\}$, then $|I_\\sigma| = 1$.\n\nIn other words, $|I_\\sigma|$ can take the values $1, 2, \\dots, 2013$.\n\nProve that $|I_\\sigma| \\neq 2014$; that is, $|I_\\sigma|$ cannot be $2014$.",
"options": [],
"answer": "See solution",
"solution": "Suppose, for contradiction, that $|I_\\sigma| = 2014$. Then $I_\\sigma = \\{0, 1, 2, \\dots, 2013\\}$.\n\nSince $I_\\sigma = \\{|\\sigma_i - i| : i = 1, 2, \\dots, 2014\\}$, the sum $\\sum_{i=1}^{2014} (\\sigma_i - i) = 0$ because $\\sigma$ is a permutation.\n\nHowever, among the numbers $\\pm 0, \\pm 1, \\pm 2, \\dots, \\pm 2013$, there are $1007$ odd numbers (namely $1, 3, \\dots, 2013$). The sum of any selection of $\\pm$ signs for these numbers cannot be zero, because the sum of an odd number of odd numbers is always odd, and adding or subtracting even numbers does not change the parity. Thus, it is impossible for the sum to be zero.\n\nTherefore, $|I_\\sigma| \\neq 2014$, and the possible values for $|I_\\sigma|$ are $1, 2, \\dots, 2013$.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 21510,
"subject": "Mathematics (Olympiad)",
"question": "Jakob read 210 pages of a book on the first day. On the second day, he read $n$ pages such that the sum of the numbers on these $n$ pages (i.e., pages $211$ to $210 + n$) is $4410$. How many pages did Jakob read on the second day, and how many pages remain if the book has 630 pages in total?",
"options": [],
"answer": "See solution",
"solution": "Let $n$ be the number of pages Jakob read on the second day. The sum of the page numbers from $211$ to $210 + n$ is:\n\n$$\n211 + 212 + \\cdots + (210 + n) = 210n + \\frac{n(n+1)}{2}\n$$\n\nSet this equal to $4410$:\n\n$$\n210n + \\frac{n(n+1)}{2} = 4410\n$$\n\nMultiply both sides by $2$ to clear the fraction:\n\n$$\n420n + n(n+1) = 8820 \\\\\n420n + n^2 + n = 8820 \\\\\nn^2 + 421n - 8820 = 0\n$$\n\nSolve the quadratic equation:\n\n$$\nn = \\frac{-421 \\pm \\sqrt{421^2 + 4 \\times 8820}}{2}\n$$\n\nCalculate the discriminant:\n\n$$\n421^2 + 4 \\times 8820 = 177241 + 35280 = 212521\n$$\n\nSo,\n\n$$\nn = \\frac{-421 + 461}{2} = 20\n$$\n\nThus, Jakob read $20$ pages on the second day.\n\nPages remaining:\n\n$$\n630 - 210 - 20 = 400\n$$\n\n*Answer*: Jakob read $20$ pages on the second day and has $400$ pages left to read.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21511,
"subject": "Mathematics (Olympiad)",
"question": "Sea $a_0 < a_1 < a_2 < \\dots$ una sucesión infinita de números enteros positivos.\n\nDemostrar que existe un único entero $n \\ge 1$ tal que\n\n$$\na_n < \\frac{a_0 + a_1 + a_2 + \\dots + a_n}{n} \\le a_{n+1}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Definamos\n\n$$\nb_n = n a_n - (a_n + a_{n-1} + \\dots + a_1).\n$$\n\nClaramente, $b_1 = 0$, y $b_{n+1} - b_n = n(a_{n+1} - a_n) > 0$, así que la sucesión $b_1, b_2, \\dots$ es una sucesión infinita y estrictamente creciente de enteros. Nótese además que la condición que se desea imponer a $n$ es equivalente a\n\n$$\nb_n < a_0 \\le b_{n+1}.\n$$\n\nSea ahora $C_k = \\{b_k + 1, b_k + 2, \\dots, b_{k+1}\\}$, donde $k$ recorre todos los enteros positivos. Cada uno de estos conjuntos es no vacío por ser $b_{k+1} > b_k$, luego $b_{k+1} \\ge b_k + 1$. Los conjuntos son disjuntos dos a dos porque si $i > j$, el máximo elemento de $C_j$, que es $b_{j+1}$, es menor que el mínimo elemento de $C_i$, que es $b_i + 1 > b_i \\ge b_{j+1}$ por ser la sucesión de los $b_n$ una sucesión creciente de enteros. Como además el mayor elemento de $C_k$ y el menor elemento de $C_{k+1}$ son consecutivos, y los elementos de cada $C_k$ son consecutivos, cada entero mayor o igual que $b_1 + 1 = 1$ está en alguno de los conjuntos. Luego cada entero positivo pertenece a uno y sólo uno de los $C_k$. En concreto, el entero positivo $a_0$ pertenece a uno y sólo uno de los $C_k$, es decir, existe un único entero positivo $n$ tal que $b_n < a_0 \\le b_{n+1}$, como queríamos demostrar.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21512,
"subject": "Mathematics (Olympiad)",
"question": "We say that two real numbers $r$ and $s$ are *close* if $|r - s| = 10^u$ for some integer $u$. Let $y = ax + b$ be a linear function, for which there exist close numbers $x_1$ and $x_2$ so that the corresponding $y_1$ and $y_2$ are also close. Prove that for any close numbers $x'_1$ and $x'_2$, the corresponding $y'_1$ and $y'_2$ are also close.",
"options": [],
"answer": "See solution",
"solution": "From the premises we get $|x_1 - x_2| = 10^u$ and $|y_1 - y_2| = |(ax_1 + b) - (ax_2 + b)| = 10^v$ for some integers $u, v$. So,\n\n$$\n10^v = |(ax_1 + b) - (ax_2 + b)| = |a(x_1 - x_2)| = |a| \\cdot |x_1 - x_2| = |a| \\cdot 10^u,\n$$\n\nwhich gives $|a| = \\frac{10^v}{10^u} = 10^{v-u}$. Let $x'_1, x'_2$ be any close real numbers, $|x'_1 - x'_2| = 10^w$. Then $|y'_1 - y'_2| = |(ax'_1 + b) - (ax'_2 + b)| = |a(x'_1 - x'_2)| = |a| \\cdot |x'_1 - x'_2| = 10^{v-u} \\cdot 10^w = 10^{w+v-u}$. Since $u, v, w$ are integers, $w + v - u$ is also an integer, which shows that $y'_1, y'_2$ are close.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21513,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive real number pairs $(a, b)$ such that the function $f(x) = a x^2 + b$ satisfies\n$$\nf(xy) + f(x + y) \\geq f(x) f(y)\n$$\nfor all real numbers $x, y$.",
"options": [],
"answer": "See solution",
"solution": "The given condition is equivalent to\n$$\n(a x^2 y^2 + b) + (a(x + y)^2 + b) \\geq (a x^2 + b)(a y^2 + b). \\tag{1}\n$$\nLet $y = 0$ in (1):\n$$\nb + (a x^2 + b) \\geq (a x^2 + b) b,\n$$\nwhich simplifies to\n$$\n(1 - b) a x^2 + b(2 - b) \\geq 0.\n$$\nSince $a > 0$ and $a x^2$ can be arbitrarily large, we must have $1 - b \\geq 0$, i.e., $0 < b \\leq 1$.\n\nLet $y = -x$ in (1):\n$$\n(a x^4 + b) + b \\geq (a x^2 + b)^2,\n$$\nor\n$$\n(a - a^2) x^4 - 2 a b x^2 + (2b - b^2) \\geq 0. \\tag{2}\n$$\nLet $g(x)$ denote the left-hand side of (2). If $a - a^2 = 0$, then $a = 1$, but then $g(x) = -2b x^2 + (2b - b^2)$, which can be negative for $b > 0$, a contradiction. Thus, $a - a^2 > 0$, so $0 < a < 1$.\n\nCompleting the square:\n$$\ng(x) = (a - a^2) \\left(x^2 - \\frac{a b}{a - a^2}\\right)^2 - \\frac{(a b)^2}{a - a^2} + (2b - b^2)\n$$\n$$\n= (a - a^2) \\left(x^2 - \\frac{b}{1 - a}\\right)^2 + \\frac{b}{1 - a}(2 - 2a - b) \\geq 0.\n$$\nSo $\\frac{b}{1 - a} > 0$ and $2a + b \\leq 2$.\n\nThus, the necessary conditions are:\n$$\n0 < b \\leq 1, \\quad 0 < a < 1, \\quad 2a + b \\leq 2.\n$$\n\nNow, for any $(a, b)$ satisfying these, for all real $x, y$:\n$$\nh(x, y) = (a - a^2) x^2 y^2 + a(1 - b)(x^2 + y^2) + 2a x y + (2b - b^2) \\geq 0.\n$$\nSince $a(1 - b) \\geq 0$, $a - a^2 > 0$, and $\\frac{b}{1 - a}(2 - 2a - b) \\geq 0$, and using $x^2 + y^2 \\geq -2 x y$:\n$$\nh(x, y) \\geq (a - a^2) x^2 y^2 + 2a b x y + (2b - b^2)\n$$\n$$\n= (a - a^2) \\left(x y + \\frac{b}{1 - a}\\right)^2 + \\frac{b}{1 - a}(2 - 2a - b) \\geq 0.\n$$\n\nTherefore, all pairs $(a, b)$ such that\n$$\n0 < b \\leq 1, \\quad 0 < a < 1, \\quad 2a + b \\leq 2\n$$\nsatisfy the given condition.\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21514,
"subject": "Mathematics (Olympiad)",
"question": "Can the natural numbers $1$ through $12$ be written into the squares of the grid shown below so that all the following conditions are met?\n\n\n\n1. Each square contains exactly one number.\n2. The sums of numbers in both rows consisting of 4 squares, in both columns consisting of 4 squares, and in the 4 middle squares are all equal.\n3. The numbers in any two squares with a common side or vertex differ from each other by at least $2$.",
"options": [],
"answer": "See solution",
"solution": "Yes, it is possible.\n\nA configuration that meets all conditions is shown below:\n\n",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21515,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ and $m$ be positive integers. What is the biggest number of points that can be marked in the vertices of the squares of the $n \\times m$ grid in such a way that no three of the marked points lie in the vertices of any right-angled triangle?",
"options": [],
"answer": "See solution",
"solution": "All vertices of squares lie on $n+1$ horizontal and $m+1$ vertical lines. Suppose that at least $n+m+1$ points are marked in the grid. Because $m > 0$, the number of marked points is greater than $n+1$. Hence, by the pigeonhole principle, at least one horizontal line contains more than one marked point. Hence, at most $n$ marked points are alone on their horizontal lines. Similarly, at most $m$ marked points are alone on their vertical lines. Thus, there exists a marked point that lies neither alone on its horizontal line nor alone on its vertical line. But then there is a right-angled triangle with vertices at marked points. So at most $n+m$ points can be marked according to the conditions of the problem.\n\nBy marking all vertices of squares of the grid that lie at the left and lower edge of the grid except for the lower left corner, we have marked exactly $n+m$ points (see the figure below for a $5 \\times 7$ grid). Any three of the marked points either lie on a common line or are the vertices of an obtuse triangle, so the construction satisfies the conditions of the problem.\n\n\n\nFig. 13",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21516,
"subject": "Mathematics (Olympiad)",
"question": "Triangle $ABC$ is inscribed in a circle. Tangents at points $A$ and $B$ meet at point $T$. The line passing through $T$ and parallel to $AC$ intersects $BC$ at point $D$. Prove that $AD = CD$.",
"options": [],
"answer": "See solution",
"solution": "Since triangle $ATB$ is isosceles, $\\angle TBA$ and $\\angle BCA$ share a common arc, so:\n\n$$\n\\angle TBA = \\angle TAB = \\angle BCA = \\angle BDT.\n$$\n\nThis implies that $BTAD$ is cyclic. Hence $\\angle BTA = \\angle ADC$ and triangles $ABT$ and $ACD$ are similar. Triangle $ATB$ is isosceles, thus so is $\\triangle ACD$. This implies that $AD = CD$.\n\n![](images/Ukrajina_2011_p11_data_01e3ee5bbd.png \"\")",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21517,
"subject": "Mathematics (Olympiad)",
"question": "Determine the smallest possible positive integer $n$ with the following property: For all positive integers $x$, $y$, and $z$ with $x \\mid y^3$, $y \\mid z^3$, and $z \\mid x^3$, we also have $xyz \\mid (x + y + z)^n$.",
"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 $v_p(xyz) \\le v_p((x + y + z)^n)$ 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 $v_p(x + y + z) \\ge v_p(z)$, and from the divisibility constraints we get $v_p(x) \\le 3v_p(y) \\le 9v_p(z)$. It follows that\n\n$$\n\\begin{aligned}\nv_p(xyz) &= v_p(x) + v_p(y) + v_p(z) \\\\\n&\\le 9v_p(z) + 3v_p(z) + v_p(z) = 13v_p(z) \\\\\n&\\le 13v_p(x + y + z) = v_p((x + y + z)^{13}),\n\\end{aligned}\n$$\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 that both inequalities above become equalities), we get\n\n$$\n\\begin{aligned}\n13 &= v_p(p^{13}) = v_p(p^9 \\cdot p^3 \\cdot p^1) = v_p(xyz) \\\\\n &\\le v_p((x + y + z)^n) = v_p((p^9 + p^3 + p^1)^n) = n \\cdot v_p(p(p^8 + p^2 + 1)) = n,\n\\end{aligned}\n$$\n\nwhich yields $n \\ge 13$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21518,
"subject": "Mathematics (Olympiad)",
"question": "Let $x$ be a red positive integer and $y$ a blue positive integer. Each time, we choose the last red number and the last blue number, add them together, and color the resulting number red or blue. We write $m$ red numbers and $n$ blue numbers in total. In the situation where $x = y = 1$, $m = a - 1$, $n = b - 1$, the red numbers are members of $S_{p,q}$ and the blue numbers are members of $T_{p,q}$. Let $f(m, n, x, y)$ be the minimum possible total of the last written red number and the last written blue number. In the original setting, the last red number $s$ and the last blue number $t$ satisfy $sp \\bmod q = (-tp) \\bmod q = 1$, so $s + t = q$. What we need to do is give a lower bound for $f(a, b, 1, 1)$.\n\nWe have the recurrence:\n\n$$\nf(m, n, x, y) = \\min\\big(f(m - 1, n, x + y, y),\\ f(m, n - 1, x, x + y)\\big)\n$$\n\nwhere the first term corresponds to coloring the next number red, and the second term to coloring it blue. Also,\n\n$$\nf(m, 0, x, y) = x + (m + 1)y, \\quad f(0, n, x, y) = (n + 1)x + y.\n$$",
"options": [],
"answer": "See solution",
"solution": "We can induct on $m + n$ and show that\n\n$$\nf(m, n, x, y) = mn \\min(x, y) + (m + 1)x + (n + 1)y.\n$$\n\nSuppose this holds for all smaller $m + n$. Then,\n\n$$\n\\begin{align*}\nf(m, n, x, y) &= \\min \\left( (m-1)ny + m(x+y) + (n+1)y,\\ m(n-1)x + (m+1)x + (n+1)(x+y) \\right) \\\\\n&= mn \\min(x, y) + (m+1)x + (n+1)y.\n\\end{align*}\n$$\n\nThus, $q \\ge f(a-1, b-1, 1, 1) = ab + 1$.\n\nTo show that it is possible for $q = ab + 1$, pick $p = a$. Then $T_{p,q} = \\{1, 2, \\dots, b\\}$ and $S_{p,q} = \\{1, b+1, \\dots, (a-1)b+1\\}$.\n\n**Remark:** The numbers in $S_{p,q}$ are exactly the denominators of terms in the Farey sequence just below $p/q$, and those in $T_{p,q}$ are just above $p/q$. This can also be used to show the claim.",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 21519,
"subject": "Mathematics (Olympiad)",
"question": "The numbers $1, 2, \\ldots, 2020, 2021$ are written on a blackboard. The following operation is executed:\n\nTwo numbers are chosen, both are erased and replaced by the absolute value of their difference.\nThis operation is repeated until there is only one number left on the blackboard.\n\n(a) Show that $2021$ can be the final number on the blackboard.\n\n(b) Show that $2020$ cannot be the final number on the blackboard.",
"options": [],
"answer": "See solution",
"solution": "(a) First, choose the following $1010$ pairs of numbers:\n\n$(1, 2);\\ (3, 4);\\ \\ldots;\\ (2019, 2020)$.\n\nThe absolute value of the difference within each pair is $1$. After applying the operation to each pair, the numbers $2021$ and $1010$ times the number $1$ remain on the blackboard. Now, execute the operation $505$ times with pairs of the form $(1, 1)$. Then, $2021$ and $505$ times the number $0$ remain on the blackboard. As $2021 - 0 = 2021$ and $0 - 0 = 0$, we end up with $2021$ as the final number on the board after an additional $505$ operations, regardless of the pairs chosen at each step.\n\n(b) We prove a more general statement: The final remaining number on the blackboard cannot be even.\n\nAs\n\n$$\na - b \\equiv a + b \\pmod{2},\n$$\n\nthe parity of the sum of all numbers on the board is invariant throughout the game. At the beginning, the sum of the numbers on the blackboard is\n\n$$\n\\frac{2021 \\cdot 2022}{2} = 2021 \\cdot 1011,\n$$\n\nwhich is odd. Therefore, the final number on the board must be odd as well. In particular, $2020$ cannot be the final number on the blackboard.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21520,
"subject": "Mathematics (Olympiad)",
"question": "For each positive integer $n \\geq 3$, we define an $n$-ring to be 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": "The problem splits into three parts (of which the first two are most important): proving that $n$ must be a multiple of three or a perfect cube, giving examples to show that $n$-rings exist in these cases, and calculating that there are 679 possible $n$ in the given range.\n\n**Solution:**\nLet the numbers in the $n$-ring be (in order round the circle) $a_1, a_2, \\dots, a_n$. For every $k$, $a_k a_{k+1} a_{k+2} = n$ (indices modulo $n$).\n\nConsider 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. Let $n = p_1^{b_1} \\cdots p_k^{b_k}$, so\n\n$$\nn^{\\frac{n}{3}} = p_1^{\\frac{n b_1}{3}} \\cdots p_k^{\\frac{n b_k}{3}}.\n$$\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 $n$ 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 $3 \\leq n \\leq 2018$ and $\\lfloor \\sqrt[3]{2018} \\rfloor - 1 = 11$ cubes (excluding $1$). Four cubes are also multiples of $3$ ($3^3, 6^3, 9^3, 12^3$), so the total is $672 + 11 - 4 = 679$.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21521,
"subject": "Mathematics (Olympiad)",
"question": "Determine all sets of six consecutive positive integers such that the product of some two of them, added to the product of some other two of them, is equal to the product of the remaining two numbers.",
"options": [],
"answer": "See solution",
"solution": "Exactly two of the six numbers are multiples of 3, and these two need to be multiplied together; otherwise, two of the three terms of the equality are multiples of 3 but the third one is not.\n\nLet $n$ and $n+3$ denote these multiples of 3. Two of the four remaining numbers give remainder 1 when divided by 3, while the other two give remainder 2, so the two other products are either $\\equiv 1 \\pmod{3}$ and $\\equiv 2 \\pmod{3}$, or they are both $\\equiv 1 \\pmod{2}$ and $\\equiv 2 \\pmod{3}$. In conclusion, the term $n(n+3)$ needs to be on the right-hand side of the equality.\n\nLooking at parity, three of the numbers are odd, and three are even. One of $n$ and $n+3$ is odd, the other even, so exactly two of the other numbers are odd. As $n(n+3)$ is even, the two remaining odd numbers need to appear in different terms.\n\nWe distinguish the following cases:\n\n**I.** The numbers are $n-2, n-1, n, n+1, n+2, n+3$.\n\nThe product of the two numbers on the RHS needs to be larger than $n(n+3)$. The only possibility is $(n-2)(n-1) + n(n+3) = (n+1)(n+2)$, which leads to $n=3$. Indeed,\n\n$$1 \\cdot 2 + 3 \\cdot 6 = 4 \\cdot 5.$$ \n\n**II.** The numbers are $n-1, n, n+1, n+2, n+3, n+4$.\n\nAs $(n+4)(n-1) + n(n+3) = (n+1)(n+2)$ has no solutions, $n+4$ needs to be on the RHS, multiplied with a number having a different parity, so $n-1$ or $n+1$. $(n+2)(n-1) + n(n+3) = (n+1)(n+4)$ leads to $n=3$. Indeed, $2 \\cdot 5 + 3 \\cdot 6 = 4 \\cdot 7$.\n\n$$(n+2)(n+1) + n(n+3) = (n-1)(n+4) \\text{ has no solution.}$$\n\n**III.** The numbers are $n, n+1, n+2, n+3, n+4, n+5$.\n\nWe need to consider the following situations: $(n+1)(n+2) + n(n+3) = (n+4)(n+5)$, which leads to $n=6$; indeed $7 \\cdot 8 + 6 \\cdot 9 = 10 \\cdot 11$; $(n+2)(n+5) + n(n+3) = (n+1)(n+4)$ obviously without solutions, and $(n+1)(n+4) + n(n+3) = (n+2)(n+5)$, which leads to $n=2$ (not a multiple of 3).\n\nIn conclusion, the problem has three solutions:\n\n$$1 \\cdot 2 + 3 \\cdot 6 = 4 \\cdot 5, \\quad 2 \\cdot 5 + 3 \\cdot 6 = 4 \\cdot 7, \\quad \\text{and} \\quad 7 \\cdot 8 + 6 \\cdot 9 = 10 \\cdot 11.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21522,
"subject": "Mathematics (Olympiad)",
"question": "Paul is filling the cells of a rectangular table alternately with crosses and circles (he starts with a cross). When the table is filled in completely, he determines his score as $X + O$ where $X$ is the number of rows containing more crosses than circles and $O$ is the number of columns containing more circles than crosses. In terms of $n$, what is the largest possible score Paul can achieve for a $(2n + 1) \\times (2n + 1)$ table?",
"options": [],
"answer": "See solution",
"solution": "In total, there are $2n(n+1)+1 < (2n+1)(n+1)$ crosses and $2n(n+1)$ circles. Hence, the crosses can dominate in at most $2n$ rows and, similarly, circles can dominate in at most $2n$ columns for the total score $2n+2n=4n$.\n\nSuch a score can be achieved if, for example, Paul draws crosses in the left $n+1$ columns of the first $n$ rows, the right $n+1$ columns of the last $n$ rows, and the middle cell of the middle row. That is precisely $2n(n+1)+1$ crosses, and we easily check that crosses dominate in all rows except for the middle one, while circles dominate in all columns except for the middle one.\n\n\n\nFig. 3",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21523,
"subject": "Mathematics (Olympiad)",
"question": "Даден е остроаголен триаголник $ABC$ таков што аголот во темето $C$ е најголем. Нека $E$ и $G$ се пресечните точки на висината спуштена од $A$ кон $BC$ со опишаната кружница на триаголникот $ABC$ и со $BC$ соодветно, и центарот $O$ на опишаната кружница лежи на нормалата спуштена од $A$ кон $BE$. Точките $M$ и $F$ се подножјата на висините спуштени од $E$ кон $AC$ и $AB$ соодветно. Докажи дека $P_{MFE} < P_{FBEG}$.",
"options": [],
"answer": "See solution",
"solution": "Нека пресекот на $EM$ со $BC$ е $V$ (пресекот секогаш ќе постои бидејќи аголот во $C$ е остар). Од теоремата на Симсон следува дека точките $M$, $G$ и $F$ се колинеарни. Да забележиме дека $\\angle EAC = \\angle EBC$ како агли над ист кружен лак. Уште $\\angle CAE = \\angle BAO$. Четириаголникот $FBEG$ е тетивен. Па добиваме $\\angle GBE = \\angle GFE$. Уште $\\angle GAO = \\angle GBE$ како агли со нормални краци. Добивме $\\angle CAE = \\angle GAO = \\angle BAO$. Па добиваме дека $AO$ е симетрала на аголот и нормала во триаголникот $ABE$. Следува дека $ABE$ е рамнокрак, од каде $GF$ е паралелна со $BE$. Правите $AO$, $BG$ и $EF$ се сечат во една точка ($EF$ и $BG$ се висини во триаголникот $ABE$). Да забележиме дека $AGMV$ е тетивен. Имаме $\\angle MVG = \\angle GAM$. Уште добиваме $\\angle MAV = \\angle VGM = \\angle FGB$. Па $AM$ е висина и симетрала на аголот во триаголникот $EAV$. Следува дека триаголникот $EAV$ е рамнокрак и $M$ е средина на страната $VE$. Уште $\\angle AVE \\cong \\angle AEB$. Јасно е дека $\\angle EGV \\cong \\angle EGB$ и бидејќи и двата се правоаголни, $P_{GEM} = \\frac{1}{2}P_{GEB}$. Од друга страна, $P_{GFE} = P_{GFB}$, од каде се добива бараното неравенство.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21524,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be a triangle and let $\\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2 = BD_1$ and $CE_2 = AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ = D_2P$.",
"options": [],
"answer": "See solution",
"solution": "The key observation is the following lemma.\n\n**Lemma.** *Segment* $D_1Q$ *is a diameter of circle* $\\omega$.\n\n*Proof.* Let $I$ be the center of circle $\\omega$, that is, $I$ is the incenter of triangle $ABC$. Extend segment $D_1I$ through $I$ to intersect circle $\\omega$ again at $Q'$, and extend segment $AQ'$ through $Q'$ to intersect segment $BC$ at $D'$. We show that $D_2 = D'$, which in turn implies that $Q = Q'$, that is, $D_1Q$ is a diameter of $\\omega$.\n\n\n\nLet $\\ell$ be the line tangent to circle $\\omega$ at $Q'$, and let $\\ell$ intersect segments $AB$ and $AC$ at $B_1$ and $C_1$, respectively. Then $\\omega$ is an **excircle** of triangle $AB_1C_1$. Let $\\mathbf{H}_1$ denote the **dilation** with center $A$ and ratio $AD'/AQ'$. Since $\\ell \\perp D_1Q'$ and $BC \\perp D_1Q$, $\\ell \\parallel BC$. Hence, $AB/AB_1 = AC/AC_1 = AD'/AQ'$. Thus, $\\mathbf{H}_1(Q') = D'$, $\\mathbf{H}_1(B_1) = B$, and $\\mathbf{H}_1(C_1) = C$. It also follows that the excircle $\\Omega$ of triangle $ABC$ opposite vertex $A$ is tangent to side $BC$ at $D'$.\n\nIt is well known that\n\n$$\nCD_1 = \\frac{1}{2}(BC + CA - AB). \\qquad (1)\n$$\n\nWe compute $BD'$. Let $X$ and $Y$ denote the points of tangency of circle $\\Omega$ with rays $AB$ and $AC$, respectively. Then by equal tangents, $AX = AY$, $BD' = BX$, and $D'C = YC$. Hence,\n\n$$\n\\begin{align*}\nAX &= AY = \\frac{1}{2}(AX + AY) \\\\\n &= \\frac{1}{2}(AB + BX + YC + CA) \\\\\n &= \\frac{1}{2}(AB + BC + CA).\n\\end{align*}\n$$\n\nIt follows that\n\n$$\nBD' = BX = AX - AB = \\frac{1}{2}(BC + CA - AB). \\qquad (2)\n$$\n\nCombining (1) and (2) yields $BD' = CD_1$. Thus,\n\n$$\nBD_2 = BD_1 - D_2D_1 = D_2C - D_2D_1 = D_1C = BD',\n$$\n\nthat is, $D' = D_2$, as desired. ■\n\n\n\nNow we prove our main result. Let $M_1$ and $M_2$ be the midpoints of segments $BC$ and $CA$, respectively. Then $M_1$ is also the midpoint of segment $D_1D_2$, from which it follows that $IM_1$ is a midline of triangle $D_1QD_2$. Hence,\n\n$$\nQD_2 = 2IM_1 \\qquad (3)\n$$\n\nand $AD_2 \\parallel M_1I$. Similarly, we can prove that $BE_2 \\parallel M_2I$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21525,
"subject": "Mathematics (Olympiad)",
"question": "芙莉蓮與寶箱怪用 $113$ 個寶箱玩一場遊戲。遊戲開始時，所有的寶箱都未上鎖且沒有寶石。每一年，芙莉蓮在其中一個未上鎖的寶箱裡增加一顆寶石，接著寶箱怪依據以下規則行動：\n\n- 如果只剩一個寶箱未上鎖，則將所有寶箱解鎖；\n- 如果多於一個寶箱未上鎖，則寶箱怪從未上鎖的寶箱中擇一上鎖。\n\n試證：存在常數 $C > 0$，使得不論寶箱怪如何行動，芙莉蓮都總是能保證在任意年結束時，任兩個寶箱內的寶石數量相差不超過 $C$。\n\n(備註：寶箱怪不可以吃掉芙莉蓮。)",
"options": [],
"answer": "See solution",
"solution": "一般性地，對於 $n$ 個寶箱，可取 $C = n - 1$。芙莉蓮的策略很簡單：在每一年，從未上鎖的寶箱中，選一個寶石最少的寶箱增加寶石即可。（備註：這並非唯一可行的策略。）\n\n要證明此策略可行，令 $x_1^t \\le x_2^t \\le \\cdots \\le x_n^t$ 為在第 $t$ 年底時各寶箱內的寶石數量，並令 $x_1^0 = x_2^0 = \\cdots = x_n^0 = 0$。我們有以下觀察：\n\n(a) 對於任意 $t$，存在唯一的 $m = m(t)$ 使得 $x_m^{t+1} = x_m^t + 1$；\n\n(b) 對於任意 $j > m$，$x_m^t < x_m^{t+1} \\le x_j^{t+1} = x_j^t$；\n\n(c) 由於 $t$ 年底上鎖的寶箱數量等於 $t$ 除以 $n$ 的餘數 $r$，因此若 $j > r$，必有寶石數不大於 $x_j^t$ 的寶箱尚未上鎖，故 $x_j^t \\ge x_m^t$。\n\n我們的目標是要證明 $x_n^t - x_1^t \\le C = n - 1$ 對於所有 $t$ 皆成立。為此，我們將構造一個滿足 $y_n^t - y_1^t \\le C$ 的遞增序列 $y_n^t$，且\n\n$$\ny_1^t + \\cdots + y_k^t \\le x_1^t + \\cdots + x_k^t \\quad \\text{對於所有 } 1 \\le k \\le n, \\quad (1)\n$$\n\n$$\ny_1^t + \\cdots + y_n^t = x_1^t + \\cdots + x_n^t. \\quad (2)\n$$\n\n若如此，則由 (1) 知 $x_1^t \\ge y_1^t$ 而由 (2) 知 $x_n^t \\le y_n^t$，故 $x_n^t - x_1^t \\le y_n^t - y_1^t \\le C$，即得證。\n\n為此，讓我們來構造 $y_i^t$。令 $y_i^0 = i - \\frac{n+1}{2}$，則易知 $y_i^0$ 遞增且 $\\sum y_i^0 = 0$。我們接著定義\n\n$$\ny_i^{t+1} = \\begin{cases} y_i^t + 1 & \\text{if } t+1 \\equiv i \\pmod{n}, \\\\ y_i^t & \\text{otherwise.} \\end{cases}\n$$\n\n則易知 $\\sum_i y_i^t = t$，且\n\n$$\ny_i^t \\le y_{i+1}^t, \\text{ 且等號僅在 } t \\equiv i \\pmod{n} \\text{ 時可能成立。} \\quad (3)\n$$\n\n現在，(2) 顯然成立，故我們只須證明 (1) 成立即可。對 $t$ 進行數學歸納法。假設原命題對 $t$ 成立，且我們有\n\n$$\ny_1^t + \\cdots + y_k^t \\le x_1^t + \\cdots + x_k^t \\text{ 但 } y_1^{t+1} + \\cdots + y_k^{t+1} > x_1^{t+1} + \\cdots + x_k^{t+1}. \\quad (4)\n$$\n\n注意到從 $t$ 到 $t+1$，不等式兩側都至多加 $1$，因此 (4) 意味著以下三點要同時發生：\n\n(d) 需要存在 $1 \\le j \\le k$ 使得 $y_j^{t+1} = y_j^t + 1$，也就是 $t+1 \\equiv j \\pmod{n}$；\n\n(e) 對於所有 $1 \\le j \\le k$ 都有 $x_j^{t+1} = x_j^t$，也就是 $m(t) > k$；\n\n(f) $y_1^t + \\cdots + y_k^t = x_1^t + \\cdots + x_k^t$。\n\n然而：\n\n- 由歸納假設與 (f) 知 $y_k^t \\ge x_k^t$；\n- 由 (d) 知 $t$ 對 $n$ 的餘數至多為 $k-1$，故由前一點和 (3) 知 $y_{k+1}^t > y_k^t \\ge x_k^t$；\n- 由 (d) 知 $t$ 對 $n$ 的餘數至多為 $k-1$，故由 (c) 知 $x_k^t \\ge x_m^t$；\n- 但由 (e) 的 $m > k$ 與非遞減性知 $x_k^t \\le x_m^t$，結合前項知 $x_k^t = x_{k+1}^t = \\cdots = x_m^t$。\n\n綜合以上，我們有 $y_{k+1}^t > y_k^t \\ge x_k^t = x_{k+1}^t$，故由 (f) 知\n\n$$\ny_1^t + \\cdots + y_{k+1}^t = (y_1^t + \\cdots + y_k^t) + y_{k+1}^t > (x_1^t + \\cdots + x_k^t) + x_{k+1}^t,\n$$\n\n與歸納假設相矛盾。因此 (1) 成立。證畢。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21526,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute triangle. Let $D$ be a point on side $AB$ and $E$ be a point on side $AC$ such that lines $BC$ and $DE$ are parallel. Let $X$ be an interior point of $BCED$. Suppose rays $DX$ and $EX$ meet side $BC$ at points $P$ and $Q$, respectively, such that both $P$ and $Q$ lie between $B$ and $C$. Suppose that the circumcircles of triangles $BQX$ and $CPX$ intersect at a point $Y \\neq X$. Prove that points $A$, $X$, and $Y$ are collinear.",
"options": [],
"answer": "See solution",
"solution": "Let $\\ell$ be the radical axis of circles $BQX$ and $CPX$. Since $X$ and $Y$ are on $\\ell$, it is sufficient to show that $A$ is on $\\ell$. Let line $AX$ intersect segments $BC$ and $DE$ at $Z$ and $Z'$, respectively. Then it is sufficient to show that $Z$ is on $\\ell$. By $BC \\parallel DE$, we obtain\n\n$$\n\\frac{BZ}{ZC} = \\frac{DZ'}{Z'E} = \\frac{PZ}{ZQ},\n$$\n\nthus $BZ \\cdot QZ = CZ \\cdot PZ$, which implies that $Z$ is on $\\ell$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21527,
"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:\n\n$$\n\\log_2(1 + \\sqrt{x}) = (2^x - 1)^2\n$$\n\nDefine $f : [0, \\infty) \\to [0, \\infty)$ by $f(x) = \\log_2(1 + \\sqrt{x})$. This function is one-to-one and onto, and its inverse is $f^{-1}(x) = (2^x - 1)^2$.\n\nSince $f$ is strictly increasing, the equation reduces to $f(x) = f^{-1}(x) = x$, or equivalently:\n\n$$\n2^x = 1 + \\sqrt{x}\n$$\n\nChecking values, $x_1 = 0$ and $x_2 = 1$ are solutions.\n\nBecause $g(x) = 2^x$ is convex and $h(x) = 1 + \\sqrt{x}$ is concave on $[0, \\infty)$, these are the only solutions.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21528,
"subject": "Mathematics (Olympiad)",
"question": "At the robotics workshop, the registered students are grouped into teams made of three boys and a girl. One day, two boys and a girl are missing, so the present students were regrouped into teams made by one girl and four boys.\n\nHow many students are registered at the robotics workshop?",
"options": [],
"answer": "See solution",
"solution": "Let $G$ be the number of girls and $B$ the number of boys registered.\n\nInitially, teams are formed as $(G, B, B, B)$, i.e., each team has 1 girl and 3 boys. Thus, the number of teams is $n = \\min\\left(\\left\\lfloor \\frac{G}{1} \\right\\rfloor, \\left\\lfloor \\frac{B}{3} \\right\\rfloor\\right)$.\n\nWhen two boys and a girl are missing, the remaining students are $G-1$ girls and $B-2$ boys. They are regrouped into teams of 1 girl and 4 boys: $(G, B, B, B, B)$. The number of such teams is $m = \\min\\left(\\left\\lfloor \\frac{G-1}{1} \\right\\rfloor, \\left\\lfloor \\frac{B-2}{4} \\right\\rfloor\\right)$.\n\nLet $n$ be the original number of teams. Then:\n\n$$\nG = n, \\quad B = 3n\n$$\n\nAfter removing 1 girl and 2 boys:\n\n$$\nG - 1 = m, \\quad B - 2 = 4m\n$$\n\nSo:\n\n$$\nG - 1 = m \\implies G = m + 1 \\\\\nB - 2 = 4m \\implies B = 4m + 2\n$$\n\nBut also $B = 3G = 3(m + 1)$.\n\nSet equal:\n\n$$\n3(m + 1) = 4m + 2 \\\\\n3m + 3 = 4m + 2 \\\\\nm = 1\n$$\n\nSo $G = m + 1 = 2$, $B = 3G = 6$.\n\nBut this does not match the solution's numbers. Let's check the solution's logic:\n\nThe solution says that after regrouping, there are 5 teams, each with 4 boys and 1 girl, so 5 girls and 20 boys present. Since 2 boys and 1 girl are missing, the total registered is $20 + 2 = 22$ boys and $5 + 1 = 6$ girls, totaling $28$ students.\n\nThus, the answer is $\\boxed{28}$ students registered at the workshop.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21529,
"subject": "Mathematics (Olympiad)",
"question": "Given a triangle $ABC$, let $D$, $E$, and $F$ denote the midpoints of the sides $BC$, $CA$, and $AB$, respectively. The circle $ABE$, the circle $ACF$, and the line $AD$ meet again at $Q$ and $S$, respectively. Prove that $\\angle AQF = \\angle ASE$ and that $ES = FQ$.\n\n\n\nUpon inversion with pole $A$, the problem reads:\n\nGiven a triangle $AE'F'$, let the symmedian from $A$ meet the medians from $E'$ and $F'$ at $K = Q'$ and $L = S'$, respectively. Prove that the angles $AE'L$ and $AF'K$ are congruent.",
"options": [],
"answer": "See solution",
"solution": "To prove this, denote $E' = X$, $F' = Y$. Let the symmedian from $A$ meet the side $XY$ at $V$, and let the lines $XL$ and $YK$ meet the sides $AY$ and $AX$ at $M$ and $N$, respectively. Since the points $K$ and $L$ lie on the medians, we have $VM \\parallel AX$, $VN \\parallel AY$. Hence $AMVN$ is a parallelogram, and the symmedian $AV$ of triangle $AXY$ supports the median of triangle $AMN$, which implies that the triangles $AMN$ and $AXY$ are similar. Hence the points $M$, $N$, $X$, $Y$ are concyclic, and $\\angle AXM = \\angle AYN$, as required.\n\n**Remark 1.** The points $X$, $Y$, $M$, $N$ are concyclic. Inverting back from $A$ and considering the circles $AFQ$ and $AES$: the former meets $AC$ again at $M'$, and the latter meets $AB$ again at $N'$. Then the points $E$, $F$, $M'$, $N'$ are concyclic.\n\n**Remark 2.** The inversion with pole $A$ also shows that $\\angle AQF$ is the Brocard angle, providing another solution. In this notation, it is equivalent to the fact that the points $Y$, $K$, and $Z$ are collinear, where $Z$ is the Brocard point (so $\\angle ZAX = \\angle ZYA = \\angle ZXY$). This holds because the lines $AV$, $XK$, and $YZ$ are the radical axes of the following circles: (i) passing through $X$ and tangent to $AY$ at $A$; (ii) passing through $Y$ and tangent to $AX$ at $A$; and (iii) passing through $X$ and tangent to $AY$ at $Y$. The point $K$ is the radical center of these three circles.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21530,
"subject": "Mathematics (Olympiad)",
"question": "Let the configuration of cards after round $k \\geq 0$ be represented by a binary sequence $a_1^{(k)}, a_2^{(k)}, \\dots, a_n^{(k)}$, with subscripts taken modulo $n$ (where $k=0$ is the initial deal). The outcome of each round is recursively given by\n\n$$\na_i^{(k+1)} = a_i^{(k)} + a_{i-1}^{(k)} + a_{i-2}^{(k)} \\quad \\forall i = 1, 2, \\dots, n,\n$$\n\nfor $k \\geq 0$, with addition modulo 2.\n\n(a) Show by induction on $r$ that for all $r \\geq 0$ and $k \\geq 0$,\n\n$$\na_i^{(k+2r)} = a_i^{(k)} + a_{i-2r}^{(k)} + a_{i-2^{r+1}}^{(k)} \\quad \\forall i = 1, 2, \\dots, n.\n$$\n\nDeduce that if $n = 2^r$ for $r \\geq 2$, then after $n$ rounds, $a_i^{(n)} = a_i^{(0)} + a_{i-n}^{(0)}$ and $a_{i-2n}^{(0)} = a_i^{(0)}$ for all $i$, so each such $n$ is playable.\n\n(b) For any $n$ divisible by 3, set the initial configuration as\n\n$$\na_i^{(0)} = \\begin{cases} 1 & \\text{if } i \\equiv 0 \\pmod{3} \\\\ 0 & \\text{otherwise.} \\end{cases}\n$$\n\nShow that the next round yields the all-ones sequence, which remains unchanged thereafter, i.e., $a_i^{(k)} = 1$ for all $i$ and $k \\geq 1$. Conclude that every $n$ divisible by 3 is not playable.",
"options": [],
"answer": "See solution",
"solution": "Let us prove part (a) by induction on $r$.\n\n**Base case ($r=0$):**\nThe formula matches the recursive definition:\n$$\na_i^{(k+1)} = a_i^{(k)} + a_{i-1}^{(k)} + a_{i-2}^{(k)}.\n$$\n\n**Inductive step:**\nAssume for some $r \\geq 0$ and all $k \\geq 0$,\n$$\na_i^{(k+2r)} = a_i^{(k)} + a_{i-2r}^{(k)} + a_{i-2^{r+1}}^{(k)}.\n$$\nLet $m = k + 2^r$. Then,\n$$\na_i^{(m+2r)} = a_i^{(m)} + a_{i-2r}^{(m)} + a_{i-2^{r+1}}^{(m)}.\n$$\nSubstituting $m$ and expanding using the inductive hypothesis, we find\n$$\n\\begin{align*}\na_i^{(k+2^{r+1})} &= a_i^{(k+2^r+2^r)} \\\\ &= a_i^{(k+2^r)} + a_{i-2^r}^{(k+2^r)} + a_{i-2^{r+1}}^{(k+2^r)} \\\\ &= (a_i^{(k)} + a_{i-2^r}^{(k)} + a_{i-2^{r+1}}^{(k)}) \\\\ &\\quad + (a_{i-2^r}^{(k)} + a_{i-2^{r+1}}^{(k)} + a_{i-2^{r}-2^{r+1}}^{(k)}) \\\\ &\\quad + (a_{i-2^{r+1}}^{(k)} + a_{i-2^{r}-2^{r+1}}^{(k)} + a_{i-2^{r+2}}^{(k)}) \\\\ &= a_i^{(k)} + a_{i-2^{r+1}}^{(k)} + a_{i-2^{r+2}}^{(k)}.\n\\end{align*}\n$$\nThus, the formula holds for $r+1$.\n\nIf $n = 2^r$ for $r \\geq 2$, after $n$ rounds, $a_i^{(n)} = a_i^{(0)} + a_{i-n}^{(0)}$ and $a_{i-2n}^{(0)} = a_i^{(0)}$ for all $i$, so each such $n$ is playable.\n\n**Part (b):**\nFor $n$ divisible by 3, set\n$$\na_i^{(0)} = \\begin{cases} 1 & i \\equiv 0 \\pmod{3} \\\\ 0 & \\text{otherwise.} \\end{cases}\n$$\nThe next round yields all ones, and this configuration remains unchanged for all subsequent rounds, i.e., $a_i^{(k)} = 1$ for all $i$ and $k \\geq 1$. Thus, every $n$ divisible by 3 is not playable.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21531,
"subject": "Mathematics (Olympiad)",
"question": "In space, there is a convex polyhedron $D$ such that for every vertex of $D$, there are an even number of edges passing through that vertex. We choose a face $F$ of $D$. Then we assign each edge of $D$ a positive integer such that for all faces of $D$ different from $F$, the sum of the numbers assigned to the edges of that face is a positive integer divisible by $2024$. Prove that the sum of the numbers assigned to the edges of $F$ is also a positive integer divisible by $2024$.",
"options": [],
"answer": "See solution",
"solution": "We have the following two observations.\n\n**Lemma 1.** The surface of any convex polyhedron $D$ can be projected to a plane so that the faces of $D$ are in one-to-one correspondence with the faces of a planar graph $G$ and edges of $D$ are in one-to-one correspondence with the edges of $G$.\n\n*Proof.* First, we choose a sufficiently small sphere $H$ within this convex polyhedron and project this polyhedron onto the spherical surface. Then choose a point $P$ on the sphere which is different from the projection images of the vertices of the polyhedron. An inversion in a sphere centered at $P$ turns the sphere $H$ into a plane.\n\n**Lemma 2.** Suppose that all faces of a planar graph are bounded by a cycle of even length, then every cycle is of even length.\n\n*Proof.* Consider the cycle $S$ of the planar graph $G$. Notice that we can subdivide the cycle $S$ into disjoint cycles $S_1, S_2, \\dots, S_n$, where each cycle $S_i$ cannot be divided into smaller cycles. Consider any cycle $S_i$, the region bounded by $S_i$ can be partitioned into a disjoint union of faces of $G$. This implies that each cycle $S_i$ has even length since each face of $G$ has even length. Therefore, the cycle $S$ has even length.\n\nLet $G$ be the planar graph obtained from the polyhedron $D$ by using Lemma 1. We denote its dual graph by $G'$. The vertices of $G'$ are the faces of $G$, and two vertices are adjacent iff the corresponding two faces have a common edge in $G$. This dual graph $G'$ is also a planar graph. The faces of the planar graph $G'$ correspond to the vertices of the graph $G$. Because every vertex of $G$ has even degree, every face of $G'$ is bounded by even cycles. By Lemma 2, $G$ has no odd cycles; it is bipartite. Therefore, the vertices of $G'$ can be colored in red and blue such that no two adjacent vertices have the same color. In other words, the faces of $G$ can be colored in red and blue such that no two faces which share a common edge are of the same color. Combining this with Lemma 1, we see that the faces of the convex polyhedron $D$ can be colored in red and blue such that two faces that share an edge have different colors. Without loss of generality, suppose that $F$ is colored blue.\n\nWe denote by $S_A$ the sum of the numbers assigned to the edges of the face $A$. We have\n\n$$\nS_F = \\sum_{A \\text{ is a red face}} S_A - \\sum_{A \\text{ is a blue face, } A \\neq F} S_A.\n$$\n\nBy the hypothesis, $S_A$ is divisible by $2024$ for all the faces $A \\neq F$. So each term on the right-hand side of the above equality is divisible by $2024$. We deduce that $S_F$ is divisible by $2024$.\n\n$\\boxed{}$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21532,
"subject": "Mathematics (Olympiad)",
"question": "Given a sequence $a_i = a_0 + di$ for $i = 0, 1, 2, 3, 4, 5$, consider the set of six numbers $a_i, a_{i+1}, a_{i+2}, a_{i+3}, a_{i+4}, a_{i+5}$. In how many ways can these numbers be ordered as the side lengths of an equiangular hexagon (with all internal angles $120^\text{\\circ}$), up to congruence?",
"options": [],
"answer": "See solution",
"solution": "If $d = 0$, the sequence is constant and any equiangular hexagon with these side lengths is regular; there is only one such hexagon up to congruence.\n\nIf $d > 0$, the sequence is not constant and the hexagons have six sides of different lengths. For any six positive numbers $a, b, c, a', b', c'$ that satisfy\n\n$$\na + b = a' + b' \\quad \\text{and} \\quad b + c = b' + c',\n$$\nthere exists an equiangular hexagon with these side lengths. It suffices 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. By shifting and scaling, we may assume $a_i = i$ for $i = 0, 1, 2, 3, 4, 5$.\n\nCyclic changes of the order lead to congruent hexagons, so we may fix $b = 5$. Adding the equations above and using $a + b + c + a' + b' + c' = 15$, we find $a + c \\leq 2$, so the only possibilities are $\\{a, c\\} = \\{0, 1\\}$ and $\\{0, 2\\}$.\n\nSwapping $a$ and $c$, as well as $a'$ and $c'$, leads to congruent hexagons. Thus, up to congruence, the two possible orderings are:\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\nTherefore, for each $i \\geq 0$, there are exactly two ways (up to congruence) to order the numbers $a_i, a_{i+1}, a_{i+2}, a_{i+3}, a_{i+4}, a_{i+5}$ as the side lengths of an equiangular hexagon.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21533,
"subject": "Mathematics (Olympiad)",
"question": "Let $f, g : \\mathbb{R} \\to \\mathbb{R}$, where $f$ is continuous. Assume that for all real numbers $a < b < c$, there exists a sequence $(x_n)_{n \\ge 1}$ converging to $b$ such that the limit $\\lim_{n \\to \\infty} g(x_n)$ exists and\n$$\nf(a) < \\lim_{n \\to \\infty} g(x_n) < f(c).\n$$\n\na) Give an example of such functions for which $g$ is discontinuous at every real point.\n\nb) Prove that if $g$ is a monotone function, then $f = g$.",
"options": [],
"answer": "See solution",
"solution": "a) Consider $f(x) = x$ for all $x \\in \\mathbb{R}$, and\n$$\ng(x) = \\begin{cases} x, & x \\in \\mathbb{Q} \\\\ x+1, & x \\in \\mathbb{R} \\setminus \\mathbb{Q} \\end{cases}.\n$$\nThe function $g$ is discontinuous at every real point. For any $a < b < c$ and any sequence $(x_n)_{n \\ge 1}$ of rational numbers converging to $b$, $\\lim_{n \\to \\infty} g(x_n) = \\lim_{n \\to \\infty} x_n = b \\in (a, c) = (f(a), f(c))$.\n\nb) Let $b \\in \\mathbb{R}$ be a continuity point of $g$. We prove $g(b) = f(b)$ by contradiction. If $g(b) < f(b)$, by continuity of $f$ at $b$, there exists $a < b$ such that $f(a) > g(b)$. For any sequence $(x_n)_{n \\ge 1}$ converging to $b$, $\\lim_{n \\to \\infty} g(x_n) = g(b) < f(a)$, contradicting the hypothesis. Similarly, if $g(b) > f(b)$, there exists $c > b$ such that $f(c) < g(b)$, and $\\lim_{n \\to \\infty} g(x_n) = g(b) > f(c)$, again a contradiction. Thus, $g(b) = f(b)$ at every continuity point of $g$.\n\nLet $x$ be any real number. Since $g$ is monotone, it has left and right limits at $x$. The set of discontinuity points of $g$ is at most countable. For each $n \\in \\mathbb{N}^*$, there exist continuity points $u_n \\in (x - 1/n, x)$ and $v_n \\in (x, x + 1/n)$. Then $\\lim_{t \\searrow x} g(t) = \\lim_{n \\to \\infty} g(u_n) = \\lim_{n \\to \\infty} f(u_n) = f(x)$ and $\\lim_{t \\nearrow x} g(t) = \\lim_{n \\to \\infty} g(v_n) = \\lim_{n \\to \\infty} f(v_n) = f(x)$. Thus, $\\lim_{t \\searrow x} g(t) = \\lim_{t \\nearrow x} g(t) = f(x)$. By monotonicity, $g(x) = f(x)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21534,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\Gamma$ be the circumcircle of a triangle $ABC$, and let $D$ and $E$ be two points different from the vertices on the sides $AB$ and $AC$, respectively. Let $A'$ be the second point where $\\Gamma$ intersects the bisector of the angle $\\angle BAC$, and let $P$ and $Q$ be the second points where $\\Gamma$ intersects the lines $A'D$ and $A'E$, respectively. Let $R$ and $S$ be the second points of intersection of the line $AA'$ and the circumcircles of the triangles $APD$ and $AQE$, respectively. Show that the lines $DS$, $ER$ and the tangent line to $\\Gamma$ through $A$ are concurrent.",
"options": [],
"answer": "See solution",
"solution": "Since $\\angle RPD = \\angle RAD = \\angle A'AC = \\angle A'PC = \\angle DPC$, $P$, $R$, $C$ are collinear. Then $\\angle PRD = \\angle PAD = \\angle PAB = \\angle PCB$ implies that $DR \\parallel BC$. Similarly, $SE \\parallel BC$, and consequently, $SE \\parallel DR$ and $\\frac{VD}{DA} = \\frac{SR}{RA}$.\n\nLet $l$ be the tangent line through $A$ to the circumcircle of $ABC$, and let $T$ and $U$ be the points where $l$ intersects the lines $DS$ and $SE$, respectively. Also, let $V$ be the point of intersection of $AB$ and $SE$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21535,
"subject": "Mathematics (Olympiad)",
"question": "Prove that the number $1$ can be represented as the sum of a finite number $n$ of real positive numbers, not necessarily distinct, that only use the digits $0$ and/or $7$ in their decimal representation. What is the smallest possible value of $n$?",
"options": [],
"answer": "See solution",
"solution": "The problem is equivalent to representing\n$$\n\\frac{1}{7} = 0.142857142857\\ldots = 0.\\overline{142857}\n$$\nas the sum of a finite number $n$ of real numbers that only have digits $0$ and $7$ in their decimal representation. To obtain the digit $8$ in the decimal expansion, we need at least $n \\geq 8$. An example with $n = 8$ terms is:\n$$\n\\begin{aligned}\n1 = &\\ 0.\\overline{777777} + 0.\\overline{077777} + 0.\\overline{070777} + 0.\\overline{070777} \\\\\n &+ 0.\\overline{000777} + 0.\\overline{000707} + 0.\\overline{000707} + 0.\\overline{000700}\n\\end{aligned}\n$$\nwhich leads to the following representation of $1$:\n$$\n1 = \\frac{777777 + 77777 + 70777 + 70777 + 777 + 707 + 707 + 700}{10^6 - 1}.\n$$\nThere are other representations for $n = 8$, as well as for $n > 8$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21536,
"subject": "Mathematics (Olympiad)",
"question": "Suppose that a positive integer $n$ is divisible by exactly 36 different prime numbers. For $k = 1, 2, \\dots, 5$, let $c_k$ be the number of integers in the interval $\\left[ \\frac{(k-1)n}{5}, \\frac{kn}{5} \\right]$ that are coprime with $n$. It is known that $c_1, c_2, \\dots, c_5$ are not all equal. Prove that\n$$\n\\sum_{1 \\le i < j \\le 5} (c_i - c_j)^2 \\ge 2^{36}.\n$$\n\n",
"options": [],
"answer": "See solution",
"solution": "Suppose that $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_{36}^{\\alpha_{36}}$, where $p_1, \\dots, p_{36}$ are different prime numbers and $\\alpha_1, \\dots, \\alpha_{36}$ are positive integers. Obviously, if $\\frac{kn}{5}$ ($k = 0, 1, 2, 3, 4, 5$) are integers, then they are not coprime with $n$. Since for any integer $x$, the integers $x$ and $n-x$ are both coprime with $n$ or neither are, it follows that $c_1 = c_5$, $c_2 = c_4$.\n\nFor a positive integer $x$, define the function $\\mu(x)$ as follows: if $x$ is divisible by the square of some prime number, then $\\mu(x) = 0$; if $x$ is the product of $t$ different prime numbers ($t$ can be 0, in which case $x = 1$), then\n$$\n\\mu(x) = (-1)^t.\n$$\nFor a factor $m$ of $n$ and a positive integer $y$, the number of multiples of $m$ in $[1, y]$ is exactly $\\left\\lfloor \\frac{y}{m} \\right\\rfloor$. So by the inclusion-exclusion principle, the number of integers in $[1, y]$ that are coprime with $n$ is\n$$\n\\sum_{m \\mid n} \\mu(m) \\left( \\left\\lfloor \\frac{y}{m} \\right\\rfloor \\right).\n$$\nSubstituting into the problem yields, for $k = 1, \\dots, 5$,\n$$\n\\begin{aligned}\nc_k &= \\sum_{m \\mid n} \\mu(m) \\left( \\left\\lfloor \\frac{kn}{5m} \\right\\rfloor - \\left\\lfloor \\frac{(k-1)n}{5m} \\right\\rfloor \\right) \\\\\n&= \\sum_{d \\mid n} \\mu\\left(\\frac{n}{d}\\right) \\left( \\left\\lfloor \\frac{kd}{5} \\right\\rfloor - \\left\\lfloor \\frac{(k-1)d}{5} \\right\\rfloor \\right).\n\\end{aligned}\n$$\nLet $p_1, \\dots, p_{36}$ be all the prime factors of $n$. Note that the above equation only needs to be summed over $d$ satisfying $\\frac{n}{d} \\mid p_1 p_2 \\cdots p_{36}$. For these $d$, denote $r_k(d) := \\left\\lfloor \\frac{kd}{5} \\right\\rfloor - \\left\\lfloor \\frac{(k-1)d}{5} \\right\\rfloor$ and consider the cases of $d$ modulo 5 with different remainders.\n\nIf $25 \\mid n$, then $d$ that satisfies $\\frac{n}{d}$ divides $p_1 p_2 \\cdots p_{36}$ are all multiples of 5, so all $c_k$ are the same, a contradiction!\n\nIf there exists some prime factor, say $p_1$, that is congruent to 1 modulo 5, then $d$ satisfying $p_1 \\mid \\frac{n}{d}$ and $d$ satisfying $p_1 \\nmid \\frac{n}{d}$ can be paired by quotient $p_1$. The remainder of $d$ modulo 5 of each pair is the same, and $\\mu(\\frac{n}{d})$ is the opposite number of each other, so that all $c_k$ obtained are also exactly the same, a contradiction!\n\nIn the following, we consider the case that $n$ does not have a prime factor that is congruent to 1 modulo 5 and $25 \\nmid n$. Note that\n$$\nc_2 - c_1 = \\sum_{d \\mid n,\\ d \\equiv 3,4 \\pmod{5}} \\mu\\left(\\frac{n}{d}\\right),\n$$\n$$\nc_3 - c_1 = \\sum_{d \\mid n,\\ d \\equiv 2,4 \\pmod{5}} \\mu\\left(\\frac{n}{d}\\right).\n$$\nIf $5 \\nmid n$, the difference of $c_1, c_2, c_3, c_4, c_5$ becomes the opposite of the original after replacing $n$ with $\\frac{n}{5}$, so $\\sum_{1 \\le i < j \\le 5} (c_i - c_j)^2$ remains invariant. In the following, only the cases of $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_{36}^{\\alpha_{36}}$ or $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_{35}^{\\alpha_{35}}$ and $p_i \\equiv 2, 3, 4 \\pmod{5}$ are considered. Let $S_i = \\sum_{d \\mid n,\\ d \\equiv i \\pmod{5}} \\mu(\\frac{n}{d})$, then it is easy to know that\n$$\nS_1 + S_2 + S_3 + S_4 = 0\n$$\nand\n$$\n\\begin{aligned}\n\\sum_{1 \\le i < j \\le 5} (c_i - c_j)^2 &= 4(S_3 + S_4)^2 + 2(S_2 + S_4)^2 + 2(S_2 - S_3)^2 \\\\\n&= (S_3 + S_4 - S_1 - S_2)^2 + \\frac{1}{2}(S_2 + S_4 - S_1 - S_3)^2 \\\\\n&\\quad + 2(S_2 - S_3)^2 \\\\\n&= \\frac{3}{2}(S_1 - S_4)^2 + \\frac{7}{2}(S_2 - S_3)^2 + (S_1 - S_4)(S_2 - S_3).\n\\end{aligned}\n$$\nIf we let $Z = S_1 - S_4 + (S_2 - S_3)i = a + bi$, then\n$$\n\\sum_{1 \\le i < j \\le 5} (c_i - c_j)^2 = \\frac{3}{2}a^2 + \\frac{7}{2}b^2 + ab.\n$$\nLet $n \\equiv 2^t \\pmod{5}$, for different indices $j_1, j_2, \\dots, j_s$, if primes $p_{j_1}, p_{j_2}, \\dots, p_{j_s}$ have $a_2$ numbers congruent to 2 modulo 5, $a_3$ numbers congruent to 3 modulo 5 and $a_4$ numbers congruent to 4 modulo 5, then ...",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21537,
"subject": "Mathematics (Olympiad)",
"question": "Prove that there exists a real constant $c$ such that for any pair $(x, y)$ of real numbers, there exist relatively prime integers $m$ and $n$ satisfying the relation\n\n$$\n\\sqrt{(x-m)^2 + (y-n)^2} < c \\log(x^2 + y^2 + 2).\n$$",
"options": [],
"answer": "See solution",
"solution": "Without loss of generality, consider points $(x, y)$ with $x > y > 0$. For any $c$, let $d = \\frac{c}{2} \\log(x^2 + y^2 + 2)$. Choose $c$ large enough that\n\n$$\nc > \\frac{\\sqrt{2}}{\\log(2)} \\quad \\text{and} \\quad d \\ge \\max\\{9 \\cdot 20 \\cdot 21, 21 + 21 \\log(x)\\}.\n$$\n\nWe claim that $(x, y)$ lies within distance $2d = c \\log(x^2+y^2+2)$ of a lattice point $(m, n)$ with relatively prime coordinates.\n\nIf $y < 1$, then $(x, y)$ is at distance at most $\\sqrt{2} < c \\log(2) < 2d$ from the point $(\\lfloor x \\rfloor, 1)$, which has relatively prime coordinates. Otherwise, consider the points $(a, b) \\in \\mathbb{Z}^2$ with $\\lfloor x \\rfloor \\le a < \\lfloor x + d \\rfloor$ and $\\lfloor y \\rfloor \\le b < \\lfloor y + d \\rfloor$, all of which are within distance $2d$ of $(x, y)$. The number of such pairs with a common factor of $k$ is at most $(d/k + 1)^2$, so the number of pairs with a common factor between $2$ and $d$ is at most\n\n$$\n\\sum_{k=2}^{\\lfloor d \\rfloor} \\left( \\frac{d}{k} + 1 \\right)^2 = d^2 \\sum_{k=2}^{\\lfloor d \\rfloor} \\frac{1}{k^2} + 2d \\sum_{k=2}^{\\lfloor d \\rfloor} \\frac{1}{k} + \\lfloor d \\rfloor.\n$$\n\nWe have the estimates\n\n$$\n\\sum_{k=2}^{\\lfloor d \\rfloor} \\frac{1}{k^2} \\le \\frac{1}{4} + \\sum_{k=3}^{\\lfloor d \\rfloor} \\frac{1}{k(k-1)} = \\frac{1}{4} + \\sum_{k=3}^{\\lfloor d \\rfloor} \\left( \\frac{1}{k-1} - \\frac{1}{k} \\right) \\le \\frac{1}{4} + \\frac{1}{2} - \\frac{1}{\\lfloor d \\rfloor} \\le \\frac{3}{4}\n$$\n\nand\n\n$$\n\\sum_{k=2}^{\\lfloor d \\rfloor} \\frac{1}{k} \\le 4 + \\sum_{k=10}^{\\lfloor d \\rfloor} \\frac{1}{k} \\le 4 + \\frac{d}{10}.\n$$\n\nApplying these estimates, the number of pairs with a common factor between $2$ and $d$ is at most\n\n$$\nd^2 \\sum_{k=2}^{\\lfloor d \\rfloor} \\frac{1}{k^2} + 2d \\sum_{k=2}^{\\lfloor d \\rfloor} \\frac{1}{k} + \\lfloor d \\rfloor \\le \\frac{3}{4}d^2 + \\frac{2d^2}{10} + 8d + d = \\frac{19}{20}d^2 + 9d \\le \\frac{20}{21}d^2,\n$$\n\nwhere the final inequality holds because $d \\ge 9 \\cdot 20 \\cdot 21$.\n\nTherefore, at least $\\frac{d^2}{21}$ of the pairs have no common factor between $2$ and $d$. By the pigeonhole principle, there exists $a$ with $\\lfloor x \\rfloor \\le a < \\lfloor x + d \\rfloor$ such that at least $\\frac{d}{21}$ of the lattice points $(a, b)$ with $\\lfloor y \\rfloor \\le b < \\lfloor y + d \\rfloor$ have no common factor at most $d$. Hence, either some $(a, b)$ is the desired point with relatively prime coordinates, or each such $b$ has a prime factor greater than $d$ in common with $a$. These prime factors must be distinct, since the different values of $b$ differ by at most $d$. Hence $a$ is divisible by their product, which is at least $d^{\\frac{d}{21}}$. But this shows that\n\n$$\ndx > x + d > a \\ge d^{\\frac{d}{21}} \\ge d^{1+\\log(x)} > dx,\n$$\n\nwhere the first inequality holds because $x > y > 1$, the third because $d$ was chosen so that $d \\ge 21 + 21 \\log(x)$, and the last because $d > e$, meaning $d^{\\log(x)} > e^{\\log(x)} = e$. This is a contradiction. Thus, there must have been some point $(a, b)$ with relatively prime coordinates.\n\n**Remark.** It is possible to simplify the proof by using more advanced estimates in the above sums. For instance, it is well-known that\n\n$$\n\\sum_{k=1}^{\\infty} \\frac{1}{k^2} = \\frac{\\pi^2}{6} \\quad \\text{and} \\quad \\sum_{k=1}^{d} \\frac{1}{k} = \\Theta(\\log d).\n$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21538,
"subject": "Mathematics (Olympiad)",
"question": "德克斯特的實驗室裡有 $2024$ 台機器人，每台有德克斯特各自設定好的程式。某天，他調皮搗蛋的姊姊蒂蒂會闖進實驗室，在每個機器人的額頭上寫下一個在 $\\{1, 2, \\ldots, 113\\}$ 內的整數。每台機器人此時偵測到除了它自己以外的所有機器人額頭上的數字，並立即依據其程式，各別且同時猜測自己的數字。\n\n試求最大正整數 $k$，讓德克斯特存在設定程式的方法，使得不論數字如何分布，都至少有 $k$ 台機器人猜對自己的數字。",
"options": [],
"answer": "See solution",
"solution": "$$\nk = \\lfloor 2024/113 \\rfloor = 17\n$$\n\n一般性地，對於 $n$ 台機器人與 $m$ 個數字，$k = \\lfloor n/m \\rfloor$。\n\n估計：將機器人編號 $1$ 到 $n$，數字的集合為 $C = \\{0, 1, \\dots, m-1\\}$，第 $i$ 台機器人戴的數字為 $x_i \\in C$，程式則為\n\n$$\nf_i(x_1, x_2, \\dots, x_{i-1}, x_{i+1}, \\dots, x_n) : C^n \\to C\n$$\n\n假設蒂蒂以隨機的方式讓在第 $i$ 台機器人寫上 $X_i$，其中 $X_i$ 服從 $C$ 上的均勻分布，且獨立於其它 $X_j$。此時由獨立性，易知\n\n$$\nE[1_{f_i=X_i}] = P\\{f_i(X_1, X_2, \\dots, X_{i-1}, X_{i+1}, \\dots, X_n) = X_i\\} = \\frac{1}{m}\n$$\n\n這意味著猜對機器人數量的期望值 $= \\sum E[1_{f_i=X_i}] = n/m$，而這也就代表著蒂蒂必有一種寫數字的方法可以讓猜對機器人的數量至多為 $\\lfloor n/m \\rfloor$。\n\n構造：考慮\n\n$$\nf_i = i - \\sum_{j \\neq i} x_j \\pmod{m}\n$$\n\n注意到第 $i$ 台機器人要猜對，若且唯若\n\n$$\nx_i \\equiv i - \\sum_{j \\neq i} x_j \\pmod{m} \\iff \\sum_j x_j \\equiv i \\pmod{m}\n$$\n\n而必然有至少 $\\lfloor n/m \\rfloor$ 個 $i$ 會讓上述式子成立，故得證。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21539,
"subject": "Mathematics (Olympiad)",
"question": "The set of points in 3-dimensional coordinate space that lie in the plane $x + y + z = 75$ whose coordinates satisfy the inequalities\n$$\nx - yz < y - zx < z - xy\n$$\nforms three disjoint convex regions. Exactly one of those regions has finite area. The area of this finite region can be expressed in the form $a\\sqrt{b}$, where $a$ and $b$ are positive integers and $b$ is not divisible by the square of any prime. Find $a + b$.",
"options": [],
"answer": "See solution",
"solution": "The inequality $x - yz < y - zx$ can be rearranged and factored as $(y-x)(z+1) > 0$. Therefore, the region in question must only contain points $(x, y, z)$ for which $(y-x)(z+1) > 0$. One way to visualize which points in the plane $x+y+z = 75$ satisfy this inequality is to consider the points in terms of the three points where the coordinate axes intersect this plane. These intersections form an equilateral triangle with center $(25, 25, 25)$, as shown in the diagram below.\n\n\n\n\n\nThe planes $y = x$ and $z = -1$ intersect the plane $x+y+z = 75$ at two perpendicular lines that share the common point $(38, 38, -1)$, and the set of points $(x, y, z)$ for which $(y-x)(z+1) > 0$ is bounded by these two lines, as shown shaded in the diagram above.\n\nThe two lines at the boundary of these two regions are parallel and perpendicular to a side of the aforementioned equilateral triangle. The same is true for the set defined by the inequality $y - zx < z - xy$ for analogous reasons, as shown below.\n\n\n\nWhen the solutions of inequalities are graphed simultaneously, there are three regions where all the conditions are satisfied. Two of them are unbounded on the left and the right, while the third is a right triangle $\\mathcal{T}$ with vertices at $(25, 25, 25)$, $(-1, 38, 38)$, and $(-1, -1, 77)$, as shown below. This is the region of points in the plane $x + y + z = 75$ whose coordinates satisfy $x > -1$, $y > x$, and $z > y$.\n\n\n\nThis triangle $\\mathcal{T}$ is the finite region in question. All the sides of $\\mathcal{T}$ are either parallel or perpendicular to a side of the original equilateral triangle, so it must be a $30$-$60$-$90^\\circ$ triangle. The shorter leg has length\n$$\n\\sqrt{(25 - (-1))^2 + (25 - 38)^2 + (25 - 38)^2} = \\sqrt{26^2 + 13^2 + 13^2} = 13\\sqrt{6}.\n$$\nTherefore, the longer leg has length $13\\sqrt{6} \\cdot \\sqrt{3}$, so $\\mathcal{T}$ has area\n$$\n\\frac{1}{2} (13\\sqrt{6})^2 \\sqrt{3} = 507\\sqrt{3}.\n$$\nThe requested sum is $507 + 3 = 510$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21540,
"subject": "Mathematics (Olympiad)",
"question": "How many integers $k$ are there for which $1 \\leq k \\leq n$ and the sum $1 + 2 + \\dots + k$ is divisible by $n$?\n\n(a) $n = 2023^{2024}$\n\n(b) $n = 2024^{2023}$",
"options": [],
"answer": "See solution",
"solution": "Note that $1 + 2 + \\dots + k = \\frac{k(k+1)}{2}$.\n\n**(a)** Since $n$ is odd, $\\frac{k(k+1)}{2}$ is divisible by $n$ if and only if $k(k+1)$ is divisible by $n$.\n\nSince $2023 = 7 \\cdot 17^2$, $n$ can be written as $p_1^{\\alpha_1} p_2^{\\alpha_2}$, where $p_1 = 7$, $p_2 = 17$, and $\\alpha_1, \\alpha_2$ are positive integers. The product $k(k+1)$ is divisible by $n$ if and only if it is divisible by both $p_1^{\\alpha_1}$ and $p_2^{\\alpha_2}$. Since $k$ and $k+1$ are coprime, $k(k+1)$ is divisible by $p_i^{\\alpha_i}$ if and only if $k \\equiv 0 \\pmod{p_i^{\\alpha_i}}$ or $k \\equiv -1 \\pmod{p_i^{\\alpha_i}}$.\n\nBy the Chinese Remainder Theorem, the system\n\n$$\n\\begin{cases}\nk \\equiv d_1 \\pmod{p_1^{\\alpha_1}} \\\\\nk \\equiv d_2 \\pmod{p_2^{\\alpha_2}}\n\\end{cases}\n$$\n\nhas exactly one solution in $1 \\leq k \\leq n$ for each choice of $d_1, d_2 \\in \\{0, -1\\}$. There are $4$ such pairs, so the answer is $4$.\n\n**(b)** Here, $\\frac{k(k+1)}{2}$ is divisible by $n$ if and only if $k(k+1)$ is divisible by $2n$.\n\nSince $2024 = 2^3 \\cdot 11 \\cdot 23$, $n = 2^{\\beta_0} q_1^{\\beta_1} q_2^{\\beta_2}$, where $q_1 = 11$, $q_2 = 23$, and $\\beta_0, \\beta_1, \\beta_2$ are positive integers. $k(k+1)$ is divisible by $2n$ if and only if it is divisible by $2^{\\beta_0+1}$, $q_1^{\\beta_1}$, and $q_2^{\\beta_2}$. As before, $k$ or $k+1$ must be divisible by each prime power, so $k \\equiv 0$ or $-1$ modulo each.\n\nBy the Chinese Remainder Theorem, there are $2^3 = 8$ possible combinations, but not all yield valid $k$ in $1 \\leq k \\leq n$ that also satisfy $k \\equiv d_0 \\pmod{2^{\\beta_0+1}}$. The cases $d_0 = d_1 = d_2 = 0$ and $d_0 = d_1 = d_2 = -1$ do not work, leaving $6$ cases, but each pair $k, k'$ with $d_i$ and $d'_i = -1 - d_i$ are paired, so the number of valid $k$ is $3$.\n\n**Final answers:**\n\n(a) $4$; (b) $3$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21541,
"subject": "Mathematics (Olympiad)",
"question": "Let $D = \\mathbb{R} \\setminus \\{0, 1\\}$. Find all functions $f: D \\to D$ which satisfy, for any $x, y \\in \\mathbb{R}$ with $x, xy \\in D$, the equation\n\n$$\nf(f(xy)) = 1 - \\frac{1}{y f(f(x))}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Plugging in $y = \\frac{a}{x}$ for $a, x \\in D$ gives\n\n$$\nf(f(a)) = 1 - \\frac{x}{a f(f(f(x)))}.\n$$\n\nOn the other hand, $y = 1$ and $x = a$ gives for $a \\in D$\n\n$$\nf(f(a)) = 1 - \\frac{1}{f(f(f(a)))}.\n$$\n\nFrom this, we conclude $\\frac{x}{f(f(f(x)))}$ is constant for all $x \\in D$. Thus, it follows $f(f(f(x))) = Cx$ for some constant $C \\neq 0$. Plugging this into the second equation gives $f(f(x)) = 1 - \\frac{1}{Cx}$. Replace $x$ with $f(x)$ (which is allowed since $f(x) \\in D$ by definition):\n\n$$\nCx = f(f(f(x))) = 1 - \\frac{1}{C f(x)}, \\text{ i.e. } f(x) = \\frac{1}{C(1 - Cx)}.\n$$\n\nThis implies\n\n$$\nf(f(x)) = -\\frac{1 - Cx}{C^2 x}, \\quad f(f(f(x))) = x, \\text{ hence } C = 1.\n$$\n\nSo we get $f(x) = \\frac{1}{1 - x}$ for all $x \\in D$, which clearly solves the given functional equation, so we are done.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21542,
"subject": "Mathematics (Olympiad)",
"question": "Given a foundation in the form of a square $4 \\times 4$, divided into smaller $1 \\times 1$ squares. There is a gap of length $1$ between any two adjacent squares. The foundation is covered with several layers of bricks of size $2 \\times 1$. Every layer consists of $8$ bricks, and each brick fully covers exactly one gap of length $1$. Such a cover is called *strong* if every gap is covered by a brick in at least one of the layers. What is the minimum number of layers needed to make a strong cover?",
"options": [],
"answer": "See solution",
"solution": "4 layers.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21543,
"subject": "Mathematics (Olympiad)",
"question": "An acute-angled scalene triangle $ABC$ is given, with $AC > BC$. Let $O$ be its circumcentre, $H$ its orthocentre, and $F$ the foot of the altitude from $C$. Let $P$ be the point (other than $A$) on the line $AB$ such that $AF = PF$, and $M$ be the midpoint of $AC$. We denote the intersection of $PH$ and $BC$ by $X$, the intersection of $OM$ and $FX$ by $Y$, and the intersection of $OF$ and $AC$ by $Z$. Prove that the points $F$, $M$, $Y$, and $Z$ are concyclic.",
"options": [],
"answer": "See solution",
"solution": "It is enough to show that $OF \\perp FX$. Let $OE \\perp AB$, then it is trivial that $CH = 2OE$.\n\nSince from the hypothesis we have $PF = AF$, then we take $PB = PF - BF$ or $PB = AF - BF$. Also, $\\angle XPB = \\angle HAP$ and $\\angle HAP = \\angle HCX$ since $AFGC$ is inscribable (where $G$ is the foot of the altitude from $A$), so $\\angle XPB = \\angle HCX$ and since $\\angle BXP = \\angle HXC$, the triangles $XHC$ and $XBP$ are similar.\n\nIf $XL$ and $XD$ are respectively the heights of the triangles $XHC$ and $XBP$, we have:\n\n$$\n\\frac{XD}{XL} = \\frac{PB}{CH}\n$$\n\nand from (1) and (2) we get:\n\n$$\n\\frac{XD}{XL} = \\frac{AF - BF}{2OE} = \\frac{FE}{OE} \\Rightarrow \\frac{XD}{FD} = \\frac{FE}{OE}\n$$\n\nTherefore, the triangles $XFD$ and $OEF$ are similar and we get:\n\n$$\n\\angle OFX = \\angle OFC + \\angle LFX = \\angle FOE + \\angle FXD = \\angle XFD + \\angle FXD = 90^\\circ\n$$\n\nso $OF \\perp FX$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21544,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, a_3, a_4, a_5, b_1, b_2, b_3, b_4, b_5$ be positive numbers satisfying the following:\n\n$$\na_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2 = 2013^2,\n$$\n$$\nb_1^2 + b_2^2 + b_3^2 + b_4^2 + b_5^2 = 2012^2,\n$$\n$$\na_1b_1 + a_2b_2 + a_3b_3 + a_4b_4 + a_5b_5 = 2013 \\cdot 2012.\n$$\nFind the ratio $a_1 : b_1$.",
"options": [],
"answer": "See solution",
"solution": "Observe that\n\n$$\n(2012a_1 - 2013b_1)^2 + (2012a_2 - 2013b_2)^2 + (2012a_3 - 2013b_3)^2 + (2012a_4 - 2013b_4)^2 + (2012a_5 - 2013b_5)^2 =\n$$\n\n$$\n\\begin{aligned}\n&= 2012^2 (a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2) + 2013^2 (b_1^2 + b_2^2 + b_3^2 + b_4^2 + b_5^2) \\\\\n&\\quad - 2 \\cdot 2012 \\cdot 2013 (a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4 + a_5b_5) \\\\\n&= 2012^2 \\cdot 2013^2 + 2013^2 \\cdot 2012^2 - 2 \\cdot 2012^2 \\cdot 2013^2 = 0.\n\\end{aligned}\n$$\n\nSince the sum of five squares can only be equal to 0 if each of the squares is zero, we get that $2012a_1 - 2013b_1 = 0$, and so $\\frac{a_1}{b_1} = \\frac{2013}{2012}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21545,
"subject": "Mathematics (Olympiad)",
"question": "Suppose that 12 acrobats labeled 1–12 are divided into two circles A and B, with six persons in each. Each acrobat in B stands on the shoulders of two adjacent acrobats of A. We call it a *tower* if the label of each acrobat of B is equal to the sum of the labels of the acrobats under his feet. How many different towers can they make?\n\n(Remark: Two towers are considered the same if one can be obtained by rotation or reflection of the other. For example, the following towers are the same, where the labels inside the circle refer to the bottom acrobat, and the labels outside the circle refer to the upper acrobat.)\n\n",
"options": [],
"answer": "See solution",
"solution": "Denote the sum of labels of A and B by $x$ and $y$, respectively. Then $y = 2x$. Thus,\n\n$$\n3x = x + y = 1 + 2 + \\cdots + 12 = 78, \\quad x = 26.\n$$\n\nObviously, $1, 2 \\in A$ and $11, 12 \\in B$. Denote $A = \\{1, 2, a, b, c, d\\}$, where $a < b < c < d$. Then $a + b + c + d = 23$, and $a \\geq 3$, $8 \\leq d \\leq 10$ (if $d \\leq 7$, then $a + b + c + d \\leq 4 + 5 + 6 + 7 = 22$, which is a contradiction).\n\n1. If $d = 8$, then $A = \\{1, 2, a, b, c, 8\\}$, $c \\leq 7$, $a + b + c = 15$. Thus, $(a, b, c) = (3, 5, 7)$ or $(4, 5, 6)$, that is, $A = \\{1, 2, 3, 5, 7, 8\\}$ or $A = \\{1, 2, 4, 5, 6, 8\\}$.\n\nIf $A = \\{1, 2, 3, 5, 7, 8\\}$, then $B = \\{4, 6, 9, 10, 11, 12\\}$. Since $B$ contains $11, 4, 6$ and $12$, there is only one tower where, in $A$, 8 and 3, 3 and 1, 1 and 5, 5 and 7 are adjacent.\n\n\n\nIf $A = \\{1, 2, 4, 5, 6, 8\\}$, then $B = \\{3, 7, 9, 10, 11, 12\\}$. Similarly, in $A$, 1 and 2, 5 and 6, 4 and 8 are adjacent, respectively. There are two arrangements, that is, two towers.\n\n\n\n\n\n2. If $d = 9$, then $A = \\{1, 2, a, b, c, 9\\}$, $c \\leq 8$, $a + b + c = 14$, where $(a, b, c) = (3, 5, 6)$ or $(3, 4, 7)$, that is, $A = \\{1, 2, 3, 5, 6, 9\\}$ or $A = \\{1, 2, 3, 4, 7, 9\\}$.\n\nIf $A = \\{1, 2, 3, 5, 6, 9\\}$, then $B = \\{4, 7, 8, 10, 11, 12\\}$. To obtain 4, 10 and 12 in $B$, 1, 3, and 9 in $A$ must be adjacent pairwise, which is impossible!\n\n\n\n\n\nIf $A = \\{1, 2, 3, 4, 7, 9\\}$, then $B = \\{5, 6, 8, 10, 11, 12\\}$. To obtain 6, 8 and 12 in $B$, 2 and 4, 1 and 7, 9 and 3 must be adjacent in $A$, respectively. There are two arrangements, that is, two towers.\n\n3. If $d = 10$, then $A = \\{1, 2, a, b, c, 10\\}$, where $c \\leq 9$, $a + b + c = 13$. Thus, $(a, b, c) = (3, 4, 6)$, that is, $A = \\{1, 2, 3, 4, 6, 10\\}$ and $B = \\{5, 7, 8, 9, 11, 12\\}$.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21546,
"subject": "Mathematics (Olympiad)",
"question": "For a positive integer $n$, let $S(n)$ be the sum of all integers $k$, $1 \\leq k \\leq n$, which are relatively prime to $n$ (i.e., the greatest common divisor of $k$ and $n$ is 1).\n\n1. Find the value of $S(30)$ (no proof required).\n2. Determine all $n$ for which $S(n)$ is a prime.\n\nHere, a prime is an integer greater than or equal to 2, which has no factor other than 1 and itself.",
"options": [],
"answer": "See solution",
"solution": "1. Since $30 = 2 \\times 3 \\times 5$, a positive integer relatively prime to 30 is not a multiple of 2, 3, or 5. The integers between 1 and 30 that are relatively prime to 30 are 1, 7, 11, 13, 17, 19, 23, 29. Therefore,\n\n$$\nS(30) = 1 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 120\n$$\n\n2. First, $S(1) = 1$, and since 1 is not a prime, we consider $n \\geq 2$.\n\nSuppose $k$ with $1 \\leq k \\leq n$ is relatively prime to $n$. Then $k < n$ and $1 \\leq n-k < n$. Let $d = \\gcd(n, n-k)$. Since $d$ divides both $n$ and $k-n = -(n-k)$, and $k$ and $n$ are relatively prime, $d=1$. Thus, $n-k$ is also relatively prime to $n$. Therefore, the integers between 1 and $n$ that are relatively prime to $n$ appear in pairs whose sum is $n$. (For $n=30$, the pairs are $(1,29)$, $(7,23)$, $(11,19)$, $(13,17)$.) Thus, $S(n)$ must be a multiple of $n$.\n\nIf $S(n) = p$ is a prime, then $n$ must divide $p$, so $n = p$ (since $n \\geq 2$ and $p$ is prime). For $n = p$ prime, every $k$ with $1 \\leq k \\leq p-1$ is relatively prime to $p$, so\n$$\nS(p) = 1 + 2 + \\cdots + (p-1) = \\frac{p(p-1)}{2}\n$$\nSetting $\\frac{p(p-1)}{2} = p$ gives $p=3$.\n\nThus, $n=3$ is the only positive integer for which $S(n)$ is a prime.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21547,
"subject": "Mathematics (Olympiad)",
"question": "$n$ is a positive 3-digit number, and its hundreds place and its ones place are not $0$. Make a new number by exchanging the hundreds place and the ones place of $n$, and call it $m$. What is the maximum value of $n - m$?",
"options": [],
"answer": "See solution",
"solution": "Let $n = 100A + 10B + C$. Then,\n\n$$\n n - m = (100A + 10B + C) - (100C + 10B + A) = 99(A - C).\n$$\n\nSince $A$ and $C$ can be any natural number between $1$ and $9$, the maximum value of $n - m$ is $99 \\times (9 - 1) = 792$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21548,
"subject": "Mathematics (Olympiad)",
"question": "What is the smallest number of colors needed to paint a $2011 \\times 2011$ square-tiled board so that every three unit squares on the board forming one of the trimino figures (see below) in any orientation are painted in different colors? Every unit square is painted entirely with one of the colors.\n\n",
"options": [],
"answer": "See solution",
"solution": "We first show that 5 colors are enough. The image below shows how to paint the board: a selected $5 \\times 5$ square is painted properly, and then this pattern is repeated as needed. It is easy to see that such painting satisfies the problem condition.\n\nNow we show that 4 or fewer colors are not enough. Suppose fewer than 5 colors suffice. In every $2 \\times 2$ square, all unit squares must be painted in different colors (see below). Consider the unit squares marked $*$ and $**$. The unit square $*$ can only be painted in color \"2\", and $**$ only in color \"1\". But then the unit square $?$ cannot be painted in any of the four colors.\n\n\n\n$$\n\\begin{array}{|c|c|c|c|c|c|c|}\n\\hline 3 & 4 & 5 & 1 & 2 & 3 & 4 \\\\\n\\hline 5 & 1 & 2 & 3 & 4 & 5 & 1 \\\\\n\\hline 2 & 3 & 4 & 5 & 1 & 2 & 3 \\\\\n\\hline 4 & 5 & 1 & 2 & 3 & 4 & 5 \\\\\n\\hline 1 & 2 & 3 & 4 & 5 & 1 & 2 \\\\\n\\hline 3 & 4 & 5 & 1 & 2 & 3 & 4 \\\\\n\\hline 5 & 1 & 2 & 3 & 4 & 5 & 1 \\\\\n\\hline\n\\end{array}\n$$\n\n*Fig. 39*",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21549,
"subject": "Mathematics (Olympiad)",
"question": "Let $P$ be a path on the vertex set $V = \\{1, 2, \\dots, n\\}$, where $j$ is joined to $j+1$ for $1 \\leq j \\leq n-1$. For each subset $A \\subset V$ and the induced subgraph $G(A)$ of $P$, define $\\mu(A) = |A| + O(G(A))$, where $O(G(A))$ is the number of components of $G(A)$, each with an odd number of vertices; $\\mu(\\emptyset) = 0$.\n\nLet\n\n$$\nT(p, r) = \\{ A \\subset V \\mid |A| = p, \\mu(A) = 2r \\},\n$$\n\nfor $r \\leq p \\leq 2r$. Prove that\n\n$$\n|T(p, r)| = \\binom{n-r}{p-r} \\binom{n-p+1}{2r-p}.\n$$\n\nExample: Let $V = \\{1, 2, \\dots, 9\\}$ and $A = \\{\\{1, 2\\}, \\{4, 5, 6\\}, \\{9\\}\\}$. In this case $\\mu(A) = 6+2=8$.",
"options": [],
"answer": "See solution",
"solution": "If $p = r = 0$, we have $A = \\emptyset$ so that $\\mu(A) = 0$. In this case, $T(0,0) = \\{\\emptyset\\}$ and $|T(0,0)| = 1$, which agrees with the formula.\n\nAssume $p \\geq r \\geq 1$. Add two dummy vertices $0$, $n+1$ to $P$ such that $0$ is joined to $1$ and $n$ is joined to $n+1$. Let\n\n$$\n\\Pi = (l_1, m_1, l_2, m_2, \\dots, l_k, m_k, l_{k+1})\n$$\n\nbe an ordered $(2k+1)$-tuple of positive integers such that\n\n$$\nl_1 + m_1 + l_2 + m_2 + \\cdots + l_k + m_k + l_{k+1} = n + 2. \\tag{*}\n$$\n\nThus, $\\Pi$ is an ordered partition of $n+2$ into $2k+1$ parts. We obtain a subset $A = A(\\Pi)$ of $V$ from this as follows: starting from the left end of the sequence $\\langle 0, 1, 2, \\dots, n+1 \\rangle$, we omit the first $l_1$ numbers; choose the next $m_1$ numbers; omit the next $l_2$ numbers; choose the next $m_2$ numbers, and so on alternately. Finally, we omit the last $l_{k+1}$ numbers. (Observe that $l_1 \\geq 1$ and $l_{k+1} \\geq 1$ implies that the dummy vertices $0$ and $n+1$ are not used at all.) The union of $k$ chosen sets of numbers consisting of $m_1, m_2, \\dots, m_k$ elements respectively is defined as $A$. We see that $A \\subseteq \\{1, 2, 3, \\dots, n\\}$. Conversely, any $A \\subseteq \\{1, 2, 3, \\dots, n\\}$ gives rise to an ordered $2k+1$ tuple of positive integers in a unique way; since $0 \\notin A$ and $n+1 \\notin A$, we have $l_1 \\geq 1$ and $l_{k+1} \\geq 1$. Since the $l$'s are positive, we see that $G(A)$ has $k$ components of vertex sizes $m_1, m_2, \\dots, m_k$. If $u$ of these $k$ numbers, say, $m_{i_1}, m_{i_2}, \\dots, m_{i_u}$ are odd and the remaining $v$ numbers, say, $m_{j_1}, m_{j_2}, \\dots, m_{j_v}$ are even, $0 \\leq u, v \\leq k$, then $u+v = k$, and $O(G(A)) = u$. Thus $\\mu(A) = m_1 + m_2 + \\cdots + m_k + u$. We count $A$ for which $\\mu(A) = 2r$, $1 \\leq r \\leq n$. Let $m_{i_1}, m_{i_2}, \\dots, m_{i_u}$ be respectively equal to $2m'_{i_1} - 1, 2m'_{i_2} - 1, \\dots, 2m'_{i_u} - 1$ and $m_{j_1}, m_{j_2}, \\dots, m_{j_v}$ be equal to $2m'_{j_1}, 2m'_{j_2}, \\dots, 2m'_{j_v}$. Then\n\n$$\n\\mu(A) = 2m'_{i_1} + 2m'_{i_2} + \\cdots + 2m'_{i_u} + 2m'_{j_1} + 2m'_{j_2} + \\cdots + 2m'_{j_v} = 2r.\n$$\n\nThus we get\n\n$$\nm'_{i_1} + m'_{i_2} + \\cdots + m'_{i_u} + m'_{j_1} + m'_{j_2} + \\cdots + m'_{j_v} = r.\n$$\n\nThe number of positive solutions of this is $\\binom{r-1}{k-1}$. Also we have\n\n$$\nl_1 + l_2 + \\cdots + l_{k+1} = n + 2 - (2r - u) = n - 2r + u + 2.\n$$\n\nThe number of positive solutions to this is $\\binom{n-2r+u+1}{k}$. Thus the number of positive solutions of $(*)$ with $\\mu(A) = 2r$ is\n\n$$\n\\sum_{k \\geq 1} \\binom{r-1}{k-1} \\binom{n-2r+u+1}{k}.\n$$\n\nSince $p + u = 2r$, we have\n\n$$\n|T(p, r)| = \\sum_{k \\geq 1} \\binom{r-1}{k-1} \\binom{n-2r+u+1}{k} \\binom{k}{u}.\n$$\n\nas any $u$ of the $k$ $m$'s may be chosen to be odd and the rest even, giving rise to the factor $\\binom{k}{u}$. Using\n\n$$\n\\binom{n}{k} \\binom{k}{m} = \\binom{n}{m} \\binom{n-m}{n-k},\n$$\n\nwe get\n\n$$\n\\begin{align*}\n|T(p, r)| &= \\binom{n-p+1}{u} \\sum_{k \\geq 1} \\binom{r-1}{k-1} \\binom{n-p+1-u}{n-p+1-k} \\\\\n&= \\binom{n-p+1}{u} \\binom{n-p-u+r}{n-p} \\\\\n&= \\binom{n-p+1}{2r-p} \\binom{n-r}{n-p} \\\\\n&= \\binom{n-r}{p-r} \\binom{n-p+1}{2r-p};\n\\end{align*}\n$$\n\nwhere we have used the Vandermonde identity.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21550,
"subject": "Mathematics (Olympiad)",
"question": "A board is called *complete* if for every binary sequence $S$ of length $2018$, there exists a row in the board such that, by filling in the empty cells of that row with 0s and 1s, we can obtain $S$ (the row 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 in 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 $2^{k-1}$ 0s and $2^{k-1}$ 1s. This board satisfies the given condition. We will prove that it is minimal.\n\n\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 $A_s$ is 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 0s. Then the binary sequence of form $s$ concatenated to 1 cannot be formed by any row, which contradicts the complete property of the board. Thus, there exists a subset $A'$ of $A_s$ 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 $A'$ 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 the $2018 - k$ 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 rows in the board, which implies the last $2018 - k$ columns of the board are empty. The remark is proved.\n\nSince the sub-board formed by the last $2018 - k$ columns is totally empty, it can represent any binary sequence of length $2018 - k$. Moreover, the original board is minimal which implies that the sub-board formed by the first $k$ columns is also minimal.\n\nErasing all $2018 - k$ columns, the rest is a sub-board of size $m \\times k$. We number the rows from $1$ to $m$ (from top to bottom) and let $A_i$ ($i = 1, 2, \\dots, m$) be the set of all binary sequences of length $k$ that can be formed by the $i$-th row.\n\nSince the original board is minimal with respect to binary sequences of length $2018$, it follows that the above $m \\times k$ sub-board is also minimal with respect to the binary sequences of length $k$. Set $B = A_1 \\cup A_2 \\cup \\dots \\cup A_m$, it is clear that $|B| = 2^k$ (since the sub-board can generate any binary sequence of length $k$).\n\nFor every $i$ ($i = 1, 2, \\dots, m$), there exists a binary sequence of length $k$ generated by the $i$-th row, otherwise we can omit the $i$-th row and the remaining rows can also generate all binary sequences of length $k$, which contradicts the minimal property of the sub-board. This means for every $i$, there exists a binary sequence $a_i$ such that $a_i \\in A_i \\subset B$ and $a_i \\notin A_j$ for any $j \\neq i$. This implies $|B| \\geq m$.\n\nCombining all above arguments, we have $m \\leq 2^k$, which is our desired conclusion. $\\blacksquare$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21551,
"subject": "Mathematics (Olympiad)",
"question": "Let $B$, $A$, $L$, $T$, $I$, and $C$ be positive numbers. Find all possible values of the expression\n\n$$\n\\frac{BA}{(C+B)(A+L)} + \\frac{LT}{(A+L)(T+I)} + \\frac{IC}{(T+I)(C+B)}\n$$",
"options": [],
"answer": "See solution",
"solution": "The range of the expression is the interval $(0, 1)$. Let $x = \\frac{A}{A+L}$, $y = \\frac{T}{T+I}$, $z = \\frac{C}{C+B}$. Note that $\\frac{L}{A+L} = 1 - x$, etc. The expression becomes\n\n$$\nx(1-z) + y(1-x) + z(1-y) = 1 - xyz - (1-x)(1-y)(1-z),\n$$\n\nwhere $0 < x, y, z < 1$. The expression can be made arbitrarily close to $0$ by letting $x, y, z \\to 0$, and arbitrarily close to $1$ by letting $x, y \\to 0$ and $z \\to 1$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21552,
"subject": "Mathematics (Olympiad)",
"question": "Find all three-digit integers $abc$, in which $a < b < c$, and for which the remainders of $abc$, $bca$, and $cab$ upon division by $27$ are from the set $\\{1, 2, 3, 4, 5\\}$.",
"options": [],
"answer": "See solution",
"solution": "All these numbers $abc$, $bca$, and $cab$ have the same sum of digits, and therefore give the same remainder modulo $9$. Consider a number $n$ which gives remainder $r < 9$ modulo $27$:\n\n$$\nn = 27q + r = 9 \\cdot (3q) + r,\n$$\n\nso this number has the same remainder modulo $9$.\n\nAs our numbers have the same remainder modulo $9$, they have the same remainder modulo $27$. Then:\n\n$$\n27 \\mid \\underline{bca} - \\underline{abc} = 90b + 9c - 99a = 9(a + b + c) + 81b - 108a = 9(a + b + c) + 27(3b - 4a).\n$$\n\nSo, $3 \\mid a + b + c$. From the statement, the only such possibility is $r = 3$. So we need to find such numbers $abc$, for which $a < b < c$ and $a + b + c$ gives remainder $3$ modulo $9$ and $27$. Then, $a + b + c \\in \\{3, 12, 21\\}$.\n\nThe case $a + b + c = 3$ is impossible.\n\nThe case $a + b + c = 12$ gives numbers $129$, $138$, $147$, $156$, $237$, $246$, and $345$, among which modulo $27$ only $138$ and $246$ have remainder $3$.\n\nThe case $a + b + c = 21$ gives numbers $489$, $579$, and $678$, among which modulo $27$ only $489$ and $678$ have remainder $3$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21553,
"subject": "Mathematics (Olympiad)",
"question": "Determine all integers $n \\geq 3$ such that the polynomial\n\n$$\nW(x) = x^n - 3x^{n-1} + 2x^{n-2} + 6\n$$\n\ncan be expressed as a product of two polynomials with positive degrees and integer coefficients.",
"options": [],
"answer": "See solution",
"solution": "We check that for $n = 3$:\n\n$$\nx^3 - 3x^2 + 2x + 6 = (x + 1)(x^2 - 4x + 6).\n$$\n\nSuppose that for $n = 4$ we have\n\n$$\nx^4 - 3x^3 + 2x^2 + 6 = (x^2 + a x + b)(x^2 + c x + d).\n$$\n\nComparing coefficients, we obtain:\n\n$$\na + c = -3, \\quad a c + b + d = 2, \\quad b d = 6.\n$$\n\nThe first equation implies $a$ and $c$ are of different parity, so $b$ and $d$ are of the same parity. This contradicts the third equation. Hence, for $n \\geq 5$, suppose\n\n$$\nW(x) = P(x) Q(x),\n$$\n\nwhere\n\n$$\nP(x) = a_k x^k + a_{k-1} x^{k-1} + \\dots + a_1 x + a_0,\n$$\n$$\nQ(x) = b_{n-k} x^{n-k} + b_{n-k-1} x^{n-k-1} + \\dots + b_1 x + b_0,\n$$\n\nand $a_k = b_{n-k} = \\pm 1$. Without loss of generality, assume $k \\leq \\lfloor \\frac{n}{2} \\rfloor < n-2$ (since $n \\geq 5$). Comparing coefficients, we get:\n\n$$\na_0 b_0 = 6,\n$$\n$$\na_0 b_1 + a_1 b_0 = 0,\n$$\n$$\na_0 b_k + a_1 b_{k-1} + \\dots + a_{k-1} b_1 + a_k b_0 = 0.\n$$\n\nBy induction, $a_0$ divides $a_1, a_2, \\dots, a_k$. For $a_1, a_2, \\dots, a_l$ established, write:\n\n$$\n0 = a_0(a_0 b_{l+1} + a_1 b_l + \\dots + a_l b_1 + a_{l+1} b_0) = a_0^2 b_{l+1} + a_0 a_1 b_l + \\dots + a_0 a_l b_1 + 6 a_{l+1},\n$$\n\nso\n\n$$\n6 a_{l+1} = - (a_0^2 b_{l+1} + a_0 a_1 b_l + \\dots + a_0 a_l b_1).\n$$\n\nAll terms on the right are divisible by $a_0^2$, so $a_0$ divides $a_{l+1}$. But $a_k = \\pm 1$, so $a_0 = \\pm 1$; take $a_0 = 1$, then $b_0 = 6$.\n\nRepeating for $Q$, $b_0 = 6$ divides $b_1, b_2, \\dots, b_{n-3}$ (set $b_l = 0$ if $l > n-k$). This contradicts $b_{n-k} = \\pm 1$, unless $n-k > n-3$. Consider two cases:\n\n**Case $k = 2$:**\n\n$$\na_0 b_{n-k} + a_1 b_{n-k-1} + \\dots + a_{n-k-1} b_1 + a_{n-k} b_0 = 2,\n$$\n\nwhich is a contradiction, since all but the first term are divisible by 6 (even), and the first is $\\pm 1$.\n\n**Case $k = 1$:**\n\nThe problem reduces to finding an integer root of $W$; if $n$ is even, there are no such roots; if $n$ is odd, $W(-1) = 0$.\n\n**Therefore, the answer is:** $n$ is odd.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21554,
"subject": "Mathematics (Olympiad)",
"question": "Solve the equation in integers:\n\n$$\nx^{2010} - 2006 = 4y^{2009} + 4y^{2008} + 2007y.\n$$",
"options": [],
"answer": "See solution",
"solution": "**Lemma.** If $x \\in \\mathbb{Z}$, then every prime divisor of $x^2 + 1$ is of the form $4k + 1$.\n\n**Proof of the lemma.** Let $p \\mid x^2 + 1$. Clearly, $\\gcd(x, p) = 1$. Then\n\n$x^2 + 1 \\equiv 0 \\pmod{p}$, so $x^2 \\equiv -1 \\pmod{p}$. Hence $(x^2)^{\\frac{p-1}{2}} \\equiv (-1)^{\\frac{p-1}{2}} \\pmod{p}$, i.e., $x^{p-1} \\equiv (-1)^{\\frac{p-1}{2}} \\pmod{p}$. By Fermat's theorem, $x^{p-1} \\equiv 1 \\pmod{p}$, so $(-1)^{\\frac{p-1}{2}} = 1$, which implies $\\frac{p-1}{2}$ is even, i.e., $p = 4k + 1$. The lemma is proved.\n\nNow, the given equation is equivalent to\n\n$$\nx^{2010} + 1 = 4y^{2009} + 2007y + 4y^{2008} + 2007\n$$\n\nor\n\n$$\nx^{2010} + 1 = (4y^{2008} + 2007)(y + 1).\n$$\n\nBut $4y^{2008} + 2007 = 4y^{2008} + 2008 - 1$ is of the form $4k - 1$, which implies it must have a prime divisor of the form $4k - 1$. Hence $(x^{1005})^2 + 1$ has a prime divisor of the form $4k - 1$, which contradicts the lemma. So the equation does not have a solution in the set of integers.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21555,
"subject": "Mathematics (Olympiad)",
"question": "Determine which of the following numbers is greater: $2$ or $\\tan 1$?",
"options": [],
"answer": "See solution",
"solution": "Consider the following inequalities:\n$$\n2 > \\sqrt{3} = \\tan \\frac{\\pi}{3} > \\tan 1.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21556,
"subject": "Mathematics (Olympiad)",
"question": "Предположим, что существует такое число $n$ и натуральные числа $a_1, \\dots, a_{11} \\leq 407$, что сумма остатков от деления $n$ на $a_1, \\dots, a_{11}$ и на $4a_1, \\dots, 4a_{11}$ равна $2012$. Докажите, что такого $n$ не существует.",
"options": [],
"answer": "See solution",
"solution": "Заметим, что максимальный возможный остаток от деления на натуральное число $m$ равен $m-1$. Поэтому сумма остатков от деления произвольного числа на числа $a_1, \\dots, a_{11}$ не больше, чем $407 - 11 = 396$, а сумма остатков от деления его на числа $4a_1, \\dots, 4a_{11}$ не больше, чем $4 \\cdot 407 - 11 = 1617$. Если бы все остатки были максимальными возможными, то их сумма равнялась бы $396 + 1617 = 2013$. Поскольку эта сумма для нашего числа $n$ равна $2012$, то все остатки, кроме одного, — максимальные возможные, а один — на единицу меньше максимального возможного.\n\nЗначит, при некотором $k$ один из остатков от деления $n$ на числа $a_k$ и $4a_k$ — максимальный возможный, а другой — на единицу меньше максимального возможного. Тогда одно из чисел $n+1$ и $n+2$ делится на $a_k$, а другое — на $4a_k$, то есть два взаимно простых числа $n+1$ и $n+2$ делятся на $a_k \\ge 2$. Это невозможно.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21557,
"subject": "Mathematics (Olympiad)",
"question": "Consider the sequence $a_1, a_2, \\dots$ defined by\n$$\na_n = 2^n + 3^n + 6^n - 1.\n$$\nfor all positive integers $n$. Determine all positive integers that are relatively prime to every term of the sequence.",
"options": [],
"answer": "See solution",
"solution": "The answer is that $1$ is the only such number. It suffices to show that every prime $p$ divides $a_n$ for some positive integer $n$. Note that both $p=2$ and $p=3$ divide $a_2 = 2^2 + 3^2 + 6^2 - 1 = 48$.\n\nNow we assume that $p \\geq 5$. By \\textbf{Fermat's Little Theorem}, we have $2^{p-1} \\equiv 3^{p-1} \\equiv 6^{p-1} \\equiv 1 \\pmod{p}$. Then\n$$\n3 \\cdot 2^{p-1} + 2 \\cdot 3^{p-1} + 6^{p-1} \\equiv 3 + 2 + 1 \\equiv 6 \\pmod{p},\n$$\nor, $6(2^{p-2} + 3^{p-2} + 6^{p-2} - 1) \\equiv 0 \\pmod{p}$; that is, $6a_{p-2}$ is divisible by $p$. Because $p$ is relatively prime to $6$, $a_{p-2}$ is divisible by $p$, as desired.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21558,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be an odd number. Is it possible to label the points on a circle with black and white colors and assign values to them such that for each diameter, the label at one end is the negative of the label at the other end, and all labels are distinct?",
"options": [],
"answer": "See solution",
"solution": "Since $c = 0 = s$, for each diameter, the label at one end is the negative of the label at the other end.\n\nEach diameter is from a black point to a white point.\n\nIf $n = 3$, we have:\n\n\n\nHence $a + b - c = 0 = a - b + c$. So $b = c$, which is disallowed.\n\nIf $n > 3$, we have:\n\n\n\nHence $b + c + d = 0 = -a - b - c = a + b + c$. So $a = d$, which is disallowed.\n\nSo the required labelling does not exist for odd $n$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21559,
"subject": "Mathematics (Olympiad)",
"question": "Let $G_n = \\{1, 2, \\ldots, n\\}$. Consider sequences of subsets $Q_1, Q_2, \\ldots, Q_{2^n-1}$ of $G_n$ such that:\n\n$$\nQ_1 = \\{1\\}, \\quad |Q_i \\cap Q_{i+1}| = 1 \\text{ for } 1 \\leq i \\leq 2^n-2, \\quad Q_{2^{n-1}} = G_n.\n$$\n\nProve that for all $n \\in \\mathbb{N}^*, n \\geq 4$, there exists a permutation $P_1, P_2, \\ldots, P_{2^n-1}$ of nonempty subsets of $G_n$ such that $|P_i \\cap P_{i+1}| = 1$ for $1 \\leq i \\leq 2^n-2$ and $P_{2^{n-1}} = \\{1, n\\}$.",
"options": [],
"answer": "See solution",
"solution": "First, for $n=3$, the sequence\n\n$$\n\\{1\\}, \\{1,2\\}, \\{2\\}, \\{2,3\\}, \\{1,3\\}, \\{3\\}, \\{1,2,3\\}\n$$\nsatisfies the given conditions.\n\nSuppose the lemma holds for $n$. Let $Q_1, Q_2, \\ldots, Q_{2^{n-1}}$ satisfy the lemma's conditions. Construct the sequence:\n\n$Q_1, Q_{2^{n-1}}, Q_{2^{n-2}}, Q_{2^{n-2}} \\cup \\{n+1\\}, Q_{2^{n-3}}, Q_{2^{n-4}} \\cup \\{n+1\\}, \\ldots, Q_3, Q_2 \\cup \\{n+1\\}, \\{n+1\\}, Q_1 \\cup \\{n+1\\}, Q_2, Q_3 \\cup \\{n+1\\}, Q_4, \\ldots, Q_{2^{n-2}}, Q_{2^{n-1}} \\cup \\{n+1\\}$\n\nThis sequence satisfies the lemma for $n+1$.\n\nFor $n=4$, the sequence\n\n$\\{1,3\\}, \\{1,2,3\\}, \\{2,3\\}, \\{1,2,3,4\\}, \\{1,2\\}, \\{1,2,4\\}, \\{2,4\\}, \\{2,3,4\\}, \\{3,4\\}, \\{1,3,4\\}, \\{1,4\\}$\n\nalso satisfies the conditions.\n\nAssume $P_1, P_2, \\ldots, P_{2^{n-1}}$ satisfies the conditions and $P_{2^{n-1}} = \\{1, n\\}$. Using the lemma, let $Q_1, Q_2, \\ldots, Q_{2^{n-1}}$ satisfy the previous conditions. Then for $n+1$, the sequence\n\n$P_1, P_2, \\ldots, P_{2^{n-1}}, Q_{2^{n-1}} \\cup \\{n+1\\}, Q_{2^{n-2}} \\cup \\{n+1\\}, \\ldots, Q_1 \\cup \\{n+1\\}$\n\nsatisfies the required conditions and $P_{2^{n+1-(n+1)-1}} = Q_1 \\cup \\{n+1\\} = \\{1, n+1\\}$.\n\nTherefore, such a permutation exists for all $n \\geq 4$.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 21560,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f$ from the positive real numbers to the positive real numbers for which $f(x) \\leq f(y)$ whenever $x \\leq y$ and\n\n$$\nf(x^4) + f(x^2) + f(x) + f(1) = x^4 + x^2 + x + 1\n$$\n\nfor all $x > 0$.",
"options": [],
"answer": "See solution",
"solution": "Substituting $x = 1$ gives $f(1) = 1$. Replacing $x$ by $x^2$ gives $f(x^8) + f(x^4) + f(x^2) + 1 = x^8 + x^4 + x^2 + 1$. Now subtract the original equation to get $f(x^8) - f(x) = x^8 - x$. Define $c_x = f(x) - x$. Observe that for all $x$, $c_x = c_{x^8}$, and by induction, this implies $c_x = c_{x^{8k}}$ for all integers $k$.\n\nConsider $x < 1$. If $c_x > 0$ then by choosing a large negative $k$, we can make $c_x + x^{8k} > 1$, but then $f(x^{8k}) = c_x + x^{8k} > 1 = f(1)$, which is a contradiction as $x^{8k} < 1$. If $c_x < 0$, we can choose a large positive $k$ so that $x^{8k} < -c_x$, but then $f(x^{8k}) = c_x + x^{8k} < 0$. So we must have $c_x = 0$, so $f(x) = x$ for $x < 1$.\n\nNow consider $x > 1$. If $c_x < 0$ we can choose a large negative $k$ such that $c_x + x^{8k} < 1$, so $f(x^{8k}) < 1$ which is a contradiction. So $c_x \\geq 0$ for all $x > 1$. If $c_x > 0$, then our original equation, after substituting $f(x) = x + c_x$ and so on, becomes\n\n$$\nx^4 + c_{x^4} + x^2 + c_{x^2} + x + c_x + 1 = x^4 + x^2 + x + 1, \\\\\n\\text{ i.e. } c_{x^4} + c_{x^2} + c_x = 0, \\\\\n\\text{ and hence } c_{x^4} + c_{x^2} < 0,\n$$\n\nbut this is a contradiction as both $c_{x^4} \\geq 0$ and $c_{x^2} \\geq 0$. So $c_x = 0$ for all $x > 1$. Therefore $f(x) = x$ for all $x$, and it is clear that this is a valid solution.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21561,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $f(x + 2\\pi) = f(x)$ for any $x \\in \\mathbb{R}$. Prove that there exist functions $f_i(x)$ ($i = 1, 2, 3, 4$) such that:\n\n1. Each $f_i(x)$ is an even function and $f_i(x + \\pi) = f_i(x)$ for any $x \\in \\mathbb{R}$.\n2. For any $x \\in \\mathbb{R}$,\n $$\n f(x) = f_1(x) + f_2(x) \\cos x + f_3(x) \\sin x + f_4(x) \\sin 2x.\n $$",
"options": [],
"answer": "See solution",
"solution": "**Proof**\n\nLet $g(x) = \\frac{f(x) + f(-x)}{2}$ and $h(x) = \\frac{f(x) - f(-x)}{2}$.\n\nThen $f(x) = g(x) + h(x)$, where $g(x)$ is even and $h(x)$ is odd. Also, $g(x + 2\\pi) = g(x)$ and $h(x + 2\\pi) = h(x)$ for any $x \\in \\mathbb{R}$.\n\nDefine:\n\n$$\nf_1(x) = \\frac{g(x) + g(x + \\pi)}{2}\n$$\n\n$$\nf_2(x) = \\begin{cases}\n \\frac{g(x) - g(x + \\pi)}{2 \\cos x}, & x \\neq k\\pi + \\frac{\\pi}{2} \\\\\n 0, & x = k\\pi + \\frac{\\pi}{2}\n\\end{cases}\n$$\n\n$$\nf_3(x) = \\begin{cases}\n \\frac{h(x) - h(x + \\pi)}{2 \\sin x}, & x \\neq k\\pi \\\\\n 0, & x = k\\pi\n\\end{cases}\n$$\n\n$$\nf_4(x) = \\begin{cases}\n \\frac{h(x) + h(x + \\pi)}{2 \\sin 2x}, & x \\neq \\frac{k\\pi}{2} \\\\\n 0, & x = \\frac{k\\pi}{2}\n\\end{cases}\n$$\n\nwhere $k$ is any integer. It is easy to check that $f_i(x)$ ($i = 1, 2, 3, 4$) satisfy property (1).\n\nNext, we prove that $f_1(x) + f_2(x) \\cos x = g(x)$ for any $x \\in \\mathbb{R}$.\n- When $x \\neq k\\pi + \\frac{\\pi}{2}$, this is clear.\n- When $x = k\\pi + \\frac{\\pi}{2}$,\n $$\n f_1(x) + f_2(x) \\cos x = f_1(x) = \\frac{g(x) + g(x + \\pi)}{2}\n $$\n and\n $$\n \\begin{aligned}\n g(x + \\pi) &= g\\left(k\\pi + \\frac{3\\pi}{2}\\right) = g\\left(k\\pi + \\frac{3\\pi}{2} - 2(k+1)\\pi\\right) \\\\\n &= g\\left(-k\\pi - \\frac{\\pi}{2}\\right) = g\\left(k\\pi + \\frac{\\pi}{2}\\right) = g(x)\n \\end{aligned}\n $$\n so $g(x + \\pi) = g(x)$ and $f_1(x) = g(x)$.\n\nSimilarly, we prove that $f_3(x) \\sin x + f_4(x) \\sin 2x = h(x)$ for any $x \\in \\mathbb{R}$.\n- When $x \\neq \\frac{k\\pi}{2}$, this is clear.\n- When $x = k\\pi$, $h(x) = h(k\\pi) = h(-k\\pi) = -h(k\\pi)$, so $h(x) = 0$ and $f_3(x) \\sin x + f_4(x) \\sin 2x = 0$.\n- When $x = k\\pi + \\frac{\\pi}{2}$,\n $$\n \\begin{aligned}\n h(x + \\pi) &= h\\left(k\\pi + \\frac{3\\pi}{2}\\right) = h\\left(-k\\pi - \\frac{\\pi}{2}\\right) = -h\\left(k\\pi + \\frac{\\pi}{2}\\right) = -h(x)\n \\end{aligned}\n $$\n so\n $$\n f_3(x) \\sin x = \\frac{h(x) - h(x + \\pi)}{2} = h(x)\n $$\n and $f_4(x) \\sin 2x = 0$.\n\nTherefore, $h(x) = f_3(x) \\sin x + f_4(x) \\sin 2x$ in all cases.\n\nIn conclusion, $f_i(x)$ ($i = 1, 2, 3, 4$) satisfy property (2).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21562,
"subject": "Mathematics (Olympiad)",
"question": "Let $S$ be a set of 999 elements and $f: S \\to S$ a mapping such that for every $n \\in S$, there exists a positive integer $\\ell$ with $f^\\ell(n) = n$. For each $n \\in S$, let $\\ell(n)$ be the smallest such positive integer. Prove that there exists $a \\in S$ such that $f(a) = a$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $f^0$ denote the identity map, so $f^0(n) = n$ for all $n \\in S$.\n\n**Lemma 1.** For $n \\in S$ and $i, j \\ge 0$, the following are equivalent:\n\n(a) $f^i(n) = f^j(n)$.\n\n(b) $i \\equiv j \\pmod{\\ell(n)}$.\n\n*Proof:* Let $\\ell = \\ell(n)$. If (b) holds, say $i = j + t\\ell$, then $f^i(n) = f^{j + t\\ell}(n) = f^j(n)$ by induction. Conversely, if (a) holds, write $i = t\\ell + u$, $j = s\\ell + v$ with $0 \\le u, v < \\ell$. Then $f^i(n) = f^u(n)$, $f^j(n) = f^v(n)$, so $f^u(n) = f^v(n)$. Thus $u = v$, so $i \\equiv j \\pmod{\\ell}$.\n\nFor each $n \\in S$, let $C(n) = \\{f^k(n) \\mid k \\ge 0\\}$ (a cycle). By Lemma 1, $|C(n)| = \\ell(n)$. $|C(n)| = 1$ iff $f(n) = n$.\n\n**Lemma 2.** Every $n \\in S$ belongs to exactly one cycle.\n\n*Proof:* If $n \\in C(m)$, then $n = f^i(m)$ for some $i$. Then $C(m) \\subset C(n)$, and similarly $C(n) \\subset C(m)$, so $C(n) = C(m)$.\n\nFrom the problem's condition and Lemma 1, for any $n \\in S$:\n\n$$\nn + f(n) + 1 \\equiv n f(n) \\equiv 0 \\pmod{\\ell(n)}.\n$$\n\nFix $n \\in S$, let $\\ell = \\ell(n)$, $a_i = f^i(n)$. If $\\ell \\ge 2$ and $p$ is a prime divisor of $\\ell$, then for each $i$:\n\n$$\na_i + a_{i+1} \\equiv -1 \\pmod{p}, \\quad a_i a_{i+1} \\equiv 0 \\pmod{p}.\n$$\n\nSo $a_i \\equiv 0$ or $a_{i+1} \\equiv 0 \\pmod{p}$, and $a_i \\equiv 0, a_{i+1} \\equiv -1$ or $a_i \\equiv -1, a_{i+1} \\equiv 0 \\pmod{p}$. Inductively, the cycle alternates between $0$ and $-1$ mod $p$. If $\\ell$ is odd, $a_0 = a_\\ell$ would contradict this, so $\\ell$ must be even.\n\nThus, all cycle sizes are 1 or even. Since $|S| = 999$ is odd, there must be a cycle of size 1, i.e., some $a$ with $f(a) = a$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21563,
"subject": "Mathematics (Olympiad)",
"question": "Isaac attempts all six questions on an Olympiad paper in order. Each question is marked on a scale from 0 to 10. He never scores more in a later question than in any earlier question. How many different possible sequences of six marks can he achieve?\n\nFor example, if there are three questions, and scores can be at most two, then there are ten possibilities:\n\n0.0.0, 1.0.0, 1.1.0, 1.1.1, 2.0.0,\n2.1.0, 2.1.1, 2.2.0, 2.2.1, 2.2.2\n\nand if scores can be at most three, then there are ten more possibilities (for a total of twenty):\n\n3.0.0, 3.1.0, 3.1.1, 3.2.0, 3.2.1,\n3.2.2, 3.3.0, 3.3.1, 3.3.2, 3.3.3\n\nProceeding in this manner, we could gather numerical data into a table (indexed by the number of questions and the maximum permitted score):\n\n| Max score | 1 | 2 | 3 | 4 |\n|-----------|---|---|---|---|\n| 0 | 1 | 1 | 1 | 1 |\n| 1 | 2 | 3 | 4 | 5 |\n| 2 | 3 | 6 | 10| 15|\n| 3 | 4 |10 |20 |35 |\n| 4 | 5 |15 |35 |70 |\n\nThis evidence has no logical status as a proof; however, it makes the following proofs easier to discover.",
"options": [],
"answer": "See solution",
"solution": "The numbers in the table above are binomial coefficients: given numbers $n$ and $k$, the binomial coefficient $\\binom{n}{k}$ counts the ways of choosing $k$ objects from $n$ objects. Its value is $\\frac{n!}{k!(n-k)!}$ for $0 \\leq k \\leq n$. The number in row $i$, column $j$ of the table is $\\binom{i+j}{j}$.\n\nSo we guess that our answer is $\\binom{16}{6}$. To prove this, imagine graphing scores on a bar chart. The following figure shows bar graphs for three sequences of scores: (6, 6, 3, 2, 2, 1), (10, 10, 8, 3, 0, 0), and (6, 5, 5, 4, 2, 0, 1):\n\n\n\nEach bar graph yields a walk from the top left to the bottom right, traveling only right and down.\n\n\n\nEach such walk consists of sixteen steps, of which exactly six are to the right (and the other ten are downwards). Any such walk determines a unique sequence of scores.\n\nThus, the number of different sequences of scores is the same as the number of walks from the top left to the bottom right, which is the number of ways of choosing six steps to be \"rightwards\" out of sixteen. This is $\\binom{16}{6}$.\n\nWe compute:\n\n$$\n\\binom{16}{6} = \\frac{16!}{6!10!} = \\frac{16 \\times 15 \\times 14 \\times 13 \\times 12 \\times 11}{6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1} = 8008.\n$$",
"topic": "Discrete Mathematics",
"subtopic": "Logic"
},
{
"id": 21564,
"subject": "Mathematics (Olympiad)",
"question": "In triangle $ABC$, $\\angle B = 2\\angle C$. Let $P$ and $Q$ be points on the perpendicular bisector of segment $BC$ such that rays $AP$ and $AQ$ trisect $\\angle A$. Prove that $PQ < AB$ if and only if $\\angle B$ is obtuse.",
"options": [],
"answer": "See solution",
"solution": "Let $ABQ_1P$ be a rhombus and let segments $AQ_1$ and $BP$ intersect at $Q_2$. Let $X$ be a point on ray $Q_2Q_1$. Because lines $PQ_2$ and $Q_1Q_2$ are perpendicular, $PX$ increases as $X$ moves away from $Q_2$. Note that $AQ_1$ bisects angle $BAP$, that is, $Q$ is on the ray $Q_2Q_1$. Because $\\angle ABC > \\angle BCA$, $A$ and $B$ are on the same side of line $PQ$. Because $PQ_1 \\parallel AB$ and $PQ \\perp BC$, $\\angle CBA$ is obtuse if and only if the segment $PQ_1$ and $B$ are on different sides of line $PQ$. It follows that $\\angle CBA$ is obtuse if and only if $Q_1$ is farther from $Q_2$ on ray $Q_2Q_1$ than $Q$, that is, $PQ < PQ_1 = AB$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21565,
"subject": "Mathematics (Olympiad)",
"question": "2010 matches are arranged in staircases as shown below:\n\n\n\na) What is the number of the staircase containing the last match?\n\nb) In which level is the head of this last match?",
"options": [],
"answer": "See solution",
"solution": "Let $S(n)$ denote staircase $n$, where $n \\geq 1$. There are 5 matches in $S(1)$. For each $n$, $S(n+1)$ has 4 more matches than $S(n)$ (to obtain $S(n+1)$, add a pair of matches to the left and another pair to the right of $S(n)$). Thus, $S(1), S(2), \\dots, S(n)$ together contain:\n\n$$5 + 9 + \\dots + (5 + 4(n-1)) = 5n + 4 \\cdot \\frac{n(n-1)}{2} = 2n^2 + 3n$$\n\nIf $n = 30$, then $2n^2 + 3n = 1890$; if $n = 31$, then $2n^2 + 3n = 2015$. Since $1890 < 2010 < 2015$, the last match (number 2010) is in $S(31)$. Add 5 more matches to complete $S(31)$. Matches 2014 and 2015 are the last two in $S(31)$, so their heads are in level 0. Therefore, the head of match 2010 is in level 2.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21566,
"subject": "Mathematics (Olympiad)",
"question": "The trapezoid $ABCD$ with parallel sides $BC = a$ and $AD = 2a$ is drawn on the plane. Using only a ruler, construct a triangle with area equal to the area of the trapezoid.",
"options": [],
"answer": "See solution",
"solution": "First, construct the point $O$ where the diagonals intersect, and the point $T$ where sides $AB$ and $CD$ intersect. It is well known that $OT$ contains the midpoints $P$ and $E$ of the parallel sides (see the figure below).\n\nWe have $AE = ED = BC = a$. Hence, $ABCE$ and $BCDE$ are parallelograms, and $M$ and $N$ are the points of intersection of their diagonals, respectively. Draw the line $MN$, which is a midline of the trapezoid and intersects two sides at $K$ and $L$, respectively. Then draw the line $BL$, which intersects $AD$ at point $F$. Thus, $\\Box ABF$ is the desired triangle, since $\\Box ABF = \\Box BCL$.\n\n![](images/Ukrajina_2011_p12_data_7e77f5ec22.png \"Fig. 11\")",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21567,
"subject": "Mathematics (Olympiad)",
"question": "Consider two positive integers $n$ and $d$. Let $S_n(d)$ be the set of ordered tuples $(x_1, \\dots, x_d)$ such that:\n\n1. $x_i \\in \\{1, 2, \\dots, n\\}$ for all $1 \\leq i \\leq d$;\n2. $x_i \\neq x_{i+1}$ for all $1 \\leq i \\leq d-1$;\n3. There do not exist indices $1 \\leq i < j < k < l \\leq d$ such that $x_i = x_k$ and $x_j = x_l$.\n\n(a) Find the number of elements of $S_3(5)$.\n\n(b) Prove that $S_n(d) \\neq \\emptyset$ if and only if $d \\leq 2n-1$.",
"options": [],
"answer": "See solution",
"solution": "(a) Call a tuple satisfying the conditions *nice*. To calculate $|S_3(5)|$, we count the number of tuples $(a, b, c, d, e)$ with $a, b, c, d, e \\in \\{1, 2, 3\\}$ satisfying conditions 2 and 3.\n\n- If the first three terms are distinct, suppose $(a, b, c) = (1, 2, 3)$. For $(1, 2, 3, d, e)$, we must have $d=2$ and $e=1$. There is 1 such tuple for each permutation of $(a, b, c)$, so $3! = 6$ nice tuples.\n- If two of the first three terms are equal, consider $(a, b, c) = (1, 2, 1)$. For $(1, 2, 1, d, e)$, we must have $d=3$ and $e=1$, giving 1 nice tuple. There are $3 \\times 2 = 6$ ways to choose $(a, b, c)$, so 6 nice tuples.\n\nThus, there are $6 + 6 = 12$ nice tuples, so $|S_3(5)| = 12$.\n\n(b) We show that $S_n(d) \\neq \\emptyset$ if and only if $d \\leq 2n - 1$.\n\nFor $d = 2n - 1$, consider the tuple:\n\n$$\n1, 2, 3, \\dots, n-1, n, n-1, \\dots, 3, 2, 1\n$$\n\nThis tuple satisfies condition 3 by symmetry, so $S_n(2n-1) \\neq \\emptyset$. For all $1 \\leq d < 2n-1$, $S_n(d)$ is also nonempty.\n\nTo show $d = 2n - 1$ is maximal, we prove that for $d = 2n$, there is no nice tuple. We proceed by induction.\n\n- For $n = 1$, $d = 2$, the statement is true.\n- Assume for all $1 \\leq k \\leq n$, there is no nice tuple with $d = 2k$. For $k = n+1$, suppose a nice tuple $(x_1, \\dots, x_{2n+2})$ exists. Let $S$ be the number of times $n+1$ appears:\n - If $S = 0$, the tuple uses only $n$ values, contradiction.\n - If $S = 1$, removing $n+1$ yields a nice tuple of length $2n+1$ with values at most $n$, contradiction.\n - If $S = 2$, all values $1, 2, \\dots, n+1$ appear exactly twice. Let $T$ be the number of terms between the two $n+1$'s:\n $$\n \\text{***, } n+1, \\underbrace{\\text{***}}_{T \\text{ numbers}}, n+1, \\text{***}\n $$\n The $T$ numbers must form a nice tuple with fewer than $2n$ terms, contradiction.\n - If $S \\geq 3$, this case is not possible by the above arguments.\n\nThus, $S_n(d) = \\emptyset$ for $d > 2n-1$. This completes the proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21568,
"subject": "Mathematics (Olympiad)",
"question": "The teacher told Peter an odd positive integer $l$, $2013$ and gave him a homework. Peter should put the stars into the cells of the table $2013 \\times 2013$ in such a way that the following condition is fulfilled: if there is a star in some table cell, then there should be no more than $l$ stars (including the given one) in the row or in the column which includes this cell. In addition, the boy can put no more than one star in each table cell. The teacher promised that Peter's mark would be proportional to the number of the stars which the boy would manage to put in the table. What is the greatest number of the stars which Peter will manage to put in the table?\n\n\n\n",
"options": [],
"answer": "See solution",
"solution": "**Answer:**\n\n$$\nN_{\\max} =\n\\begin{cases}\n2l(2013-l), & l \\le 671; \\\\\n\\left(\\frac{2013+l}{2}\\right)^2, & l > 671.\n\\end{cases}\n$$\n\n**Solution.**\n\nWe can swap the rows and columns containing stars arbitrarily; if the arrangement satisfies the condition, it remains valid after permutation, and the total number of stars is unchanged.\n\nConsider a configuration achieving the maximum. By rearranging, let all rows with more than $l$ stars be from $1$ to $m$, and all columns with more than $l$ stars from $1$ to $n$. This divides the table into four zones (see the figures above). Without loss of generality, assume $n \\ge l$.\n\nThere cannot be any star in zone 1, since otherwise both the row and column would have more than $l$ stars, violating the condition.\n\nIf $n \\ge l$, all stars from zone 4 can be moved to zone 3 without increasing the number of stars in any row or column beyond $l$. Thus, the maximum number of stars is $(2013-n)\\min(m, l)$ in zone 2 and $(2013-m)l$ in zone 3 (zones 1 and 4 are empty):\n\n$$\n(2013-n)\\min(m, l) + (2013-m)l \\le \\begin{cases} 2l(2013-l), & l < 1007; \\\\ 2013l, & l \\ge 1007. \\end{cases}\n$$\n\nThe value $2l(2013-l)$ is achieved when $m = n = l$ and all cells in zones 2 and 3 have stars; $2013l$ is achieved when $m = 0$.\n\nIf $n < l$, the maximum is $(2013-n)m$ in zone 2, $(2013-m)n$ in zone 3, and $(2013-m)(l-n)$ in zone 4:\n\n$$\n(2013-n)m + (2013-m)n + (2013-m)(l-n) = (2013-m)(m+l)\n$$\n\nFor $l \\le 671$, this is strictly less than $2l(2013-l)$. For $l > 671$ and $m = \\frac{2013-l}{2} < l$, $2013-m = m+l$, so\n\n$$\n(2013-m)(m+l) = \\left(\\frac{2013+l}{2}\\right)^2.\n$$\n\nThis is achieved when $n = m = \\frac{2013-l}{2}$, all cells in zones 2 and 3 have stars, and zone 4 is filled so that each row and column in it has $l - \\frac{2013-l}{2}$ stars, as shown in the figures.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21569,
"subject": "Mathematics (Olympiad)",
"question": "On a $9 \\times 9$ board, a black cone is placed in the central cell. Four white cones are placed in the middle cells of each side of the border (see figure 14).\n\n- The first player controls the black cone and, on each turn, may move it to an adjacent cell if that cell does not contain a white cone.\n- The second player 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.\n\nWho can win in this game?",
"options": [],
"answer": "See solution",
"solution": "The first player should move toward cell $A$, the middle of a side. After 3 moves, the black cone reaches the cell adjacent to $A$. To the right or left of this cell, there can be at most one white cone. Suppose the white cone is on the left (see figure 15). The first player then moves toward cell $B$.\n\nEven if the second player places a white cone at cell $B$ (see figure 16), they cannot prevent the first player from reaching $B$, since the second player would need to place a cone in the corner, which is not possible. Thus, the first player can always win.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 21570,
"subject": "Mathematics (Olympiad)",
"question": "Шулуун $PA$ нь тойрог $\\gamma$-г $C$ цэгээр ($A$-аас ялгаатай) огтолно. $OC$ шулуун нь $\\gamma$ тойргийг $D$ цэгээр ($C$-ээс ялгаатай) огтолно. $PD$ шулуун нь $\\gamma$ тойргийг $E$ цэгээр ($D$-ээс ялгаатай) тус тус огтолно.\n\nа. Тэгвэл $AE$, $BC$, $PO$ шулуунууд нэг цэгт огтлолцохыг батал. Тэр цэгийг $M$ гэе.\n\nb. $\\gamma$ тойргийн радиусыг $R$ гэвэл $\\triangle AMB$ гурвалжны талбайн авч болох хамгийн их утгыг $R$-ээр илэрхийлж, тэр үед $P$ цэгийн байршлыг тодорхойл.",
"options": [],
"answer": "See solution",
"solution": "(а) $AE$ ба $BP$ шулуунуудын огтлолцолын цэгийг $F$-ээр тэмдэглэе.\n\n\\begin{aligned}\n\\angle ACE &= 90^\\circ + \\angle BCE = 90^\\circ + \\angle FAB = \\angle EFP \\\\\n\\text{гэдгээс} \\quad \\angle EFP + \\angle ECP &= 180^\\circ.\n\\end{aligned}\n\nЭндээс $CEFP$ нь батсан дөрвөн өнцөгт болж $\\angle CFP = \\angle CEP = 90^\\circ$. Эндээс $CF \\parallel AB$ буюу\n\n$$\n\\frac{CA}{CP} = \\frac{FB}{OA}\n$$\n\nболно. $\\triangle ABP$-ийн хувьд\n\n$$\n\\frac{CA}{OA} = \\frac{FB}{OB}\n$$\n\nболж Чевийн урвуу теоремоос $PO$, $AE$ ба $BC$ шулуунууд нэг цэгт огтлолцох нь батлагдана.\n\n(б) $BP = x$ гэе. $\\triangle ABP$-д:\n\n\\begin{aligned}\nPA &= \\sqrt{x^2 + 4R^2}, \\\\\nPC &= \\frac{PB^2}{PA} = \\frac{x^2}{\\sqrt{x^2 + 4R^2}}, \\\\\nAC &= PA - PC = \\frac{4R^2}{\\sqrt{x^2 + 4R^2}}.\n\\end{aligned}\n\nНөгөө талаас (а) хэсгийн баталгаанд $CF \\parallel AB$ гэж баталсан. Эндээс\n\n$$\n\\frac{MC}{PC} = \\frac{CF}{PC}\n$$\n\n\\Rightarrow\n\n$$\n\\frac{BC}{MB} = \\frac{PC}{PA} + 1 = \\frac{PA}{PC + PA}\n$$\n\n\\Rightarrow\n\n$$\nBM = \\frac{PC + PA}{PA \\cdot BC} = \\frac{PC + PA}{PC + PA} = \\frac{R \\times \\sqrt{x^2 + 4R^2}}{x^2 + 2R^2}\n$$\n\nболж\n\n$$\nS_{AMB} = \\frac{1}{2} AB \\cdot BM \\sin \\angle ABM = \\frac{1}{2} \\cdot 2R \\cdot \\frac{R \\times \\sqrt{x^2 + 4R^2}}{x^2 + 2R^2} \\cdot \\frac{AC}{2R} = \\frac{2R^3 x}{x^2 + 2R^2}\n$$\n\nЭндээс\n\n$$\nS_{AMB} \\le \\frac{2R^3 x}{2\\sqrt{2xR}} = \\frac{R^2}{\\sqrt{2}}\n$$\n\nба $S_{AMB} = \\frac{R^2}{2}$.\n\n\\begin{aligned}\n\\Leftrightarrow x^2 &= 2R^2 \\Leftrightarrow x = \\sqrt{2R} \\\\\n\\text{болж } S_{AMB}$-ийн хамгийн бага утга $\\frac{R^2}{\\sqrt{2}}$ ба энэ үед $P$ цэг $B$-ийн баруун (зүүн) талд $\\sqrt{2R}$ зайтай оршино.\n\\end{aligned}",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21571,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a cyclic quadrilateral and $\\omega_1$, $\\omega_2$ the incircles of triangles $ABC$ and $BCD$. Show that the common external tangent line of $\\omega_1$ and $\\omega_2$, the other one than $BC$, is parallel to $AD$.",
"options": [],
"answer": "See solution",
"solution": "Let $I_1$, $I_2$ be the incenters of $ABC$ and $BCD$, and let $t$ be the other external tangent line of $\\omega_1$ and $\\omega_2$. The problem is trivial if lines $AD$ and $BC$ are parallel, so we'll suppose that $AD$ is not parallel to $BC$. Denote by $V$ the intersection point of $AD$ and $BC$.\n\nSince one of the exterior tangents of two circles is the reflection of the other one about the line that passes through the centers, it follows that $t$, $BC$, and $I_1I_2$ are concurrent at a point $U$.\n\nWe have $\\angle BI_1C = 90^\\circ + \\angle BAC = 90^\\circ + \\angle BDC = \\angle BI_2C$, so $BCI_2I_1$ is a cyclic quadrilateral. It follows that\n\n$$\n\\begin{align*}\n\\angle I_1 U B &= \\frac{1}{2} (m(\\widehat{CI_2}) - m(\\widehat{BI_1})) = \\angle I_2 BC - \\angle I_1 CB \\\\\n&= \\frac{1}{2} (\\angle DBC - \\angle ACB) = \\frac{1}{4} (m(\\widehat{DC}) - m(\\widehat{AB})) = \\frac{1}{2} \\angle AVB,\n\\end{align*}\n$$\n\nhence $\\angle (AD, BC) = 2\\angle (I_1I_2, BC)$. This leads to $\\angle (t, BC) = 2\\angle (I_1I_2, BC) = \\angle (AD, BC)$, so $t \\parallel AD$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21572,
"subject": "Mathematics (Olympiad)",
"question": "One number is removed from the set of integers from 1 to $n$. The average of the remaining numbers is $40\\frac{3}{4}$. Which integer was removed?",
"options": [],
"answer": "See solution",
"solution": "Let $Q$ be the sum of all the numbers from 1 to $n$, excluding the one that has been removed. The sum of all the numbers from 1 to $n$ is $\\frac{n(n+1)}{2}$, so:\n\n$$\n\\frac{n(n+1)}{2} - n \\leq Q \\leq \\frac{n(n+1)}{2} - 1.\n$$\n\nThis rearranges to:\n\n$$\n\\frac{n(n-1)}{2} \\leq Q \\leq \\frac{(n+2)(n-1)}{2}.\n$$\n\n$Q$ is the sum of $n-1$ terms whose average is $40\\frac{3}{4}$. Therefore, $Q = 40\\frac{3}{4}(n-1)$, so:\n\n$$\nn \\leq 81\\frac{1}{2} \\leq n+2.\n$$\n\nSince $n$ is an integer, we must have $n = 80$ or $81$. If $n = 80$, then $Q$ is not an integer, which is impossible. If $n = 81$, then $Q = 3260$. The sum of the integers from 1 to 81 is $3321 = Q + 61$, so the missing number must be $61$. It is easy to verify that the average of the numbers from 1 to 81 with 61 excluded is indeed $40\\frac{3}{4}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21573,
"subject": "Mathematics (Olympiad)",
"question": "Let $m$ be the greatest common divisor (g.c.d.) of $E(1), E(2), E(3), \\dots, E(2009)$, where $E(n) = n(n+1)(2n+1)(3n+1)\\cdots(10n+1)$. What is the value of $m$?",
"options": [],
"answer": "See solution",
"solution": "Since $m$ divides $E(1) = 2 \\cdot 3 \\cdots 11$, any prime divisor of $m$ is less than or equal to 11. Let $p$ be a prime number such that $p \\leq 11$ and $p \\nmid m$. Then $m$ divides $E(p) = p(p+1)(2p+1)\\cdots(10p+1)$. The terms $p+1, 2p+1, \\dots, 10p+1$ are all relatively prime to $p$, so $E(p)$ (and thus $m$) is divisible by $p$ but not by $p^2$. Therefore, $m$ is not divisible by the square of any prime number.\n\nSince $m$ divides $E(1) = 2 \\cdot 3 \\cdots 11$, it follows that $m$ divides the product of all primes less than or equal to 11, i.e., $m \\mid 2310$.\n\nTo show that $m = 2310$, it suffices to prove that for all $n \\geq 1$, the number $E(n)$ is divisible by 2310.\n\nFor any $n \\geq 1$:\n- One of $n$ or $n+1$ is divisible by 2, so $2 \\mid E(n)$.\n- One of $n, n+1, 2n+1$ is divisible by 3, so $3 \\mid E(n)$.\n- One of $n, n+1, 2n+1, 3n+1, 4n+1$ is divisible by 5, so $5 \\mid E(n)$.\n- Similarly, $7 \\mid E(n)$ and $11 \\mid E(n)$.\n\nTherefore, $E(n)$ is a multiple of $2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11 = 2310$, so the g.c.d. is $2310$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21574,
"subject": "Mathematics (Olympiad)",
"question": "設 $A$, $B$, $C$ 分別為 $\\triangle A'B'C'$ 的邊 $B'C'$, $C'A'$, $A'B'$ 的中點。設點 $P$ 位於 $\\triangle ABC$ 內部，$AP$, $BP$, $CP$ 分別與 $BC$, $CA$, $AB$ 交於點 $P_a$, $P_b$, $P_c$。\n\n直線 $P_aP_b$, $P_aP_c$ 分別與 $B'C'$ 交於 $R_b$, $R_c$；直線 $P_bP_c$, $P_bP_a$ 分別與 $C'A'$ 交於 $S_c$, $S_a$；直線 $P_cP_a$, $P_cP_b$ 分別與 $A'B'$ 交於 $T_a$, $T_b$。\n\n已知 $S_c$, $S_a$, $T_a$, $T_b$ 皆在圓心為 $O$ 的一圓上。\n\n試證：$OR_b = OR_c$。\n\n",
"options": [],
"answer": "See solution",
"solution": "先證明 $A$ 為 $R_bR_c$ 的中點：因 $BC \\parallel B'C'$，知 $\\frac{AR_c}{BP_a} = \\frac{AP_c}{P_cB}$ 及 $\\frac{AR_b}{CP_a} = \\frac{AP_b}{P_bC}$。由這兩式及西瓦定理知 $\\frac{AR_c}{AR_b} = \\frac{AP_c}{P_cB} \\cdot \\frac{BP_a}{P_aC} \\cdot \\frac{CP_b}{P_bA} = 1$。顯然 $R_b$, $R_c$ 在 $A$ 的異側，所以 $A$ 為 $R_bR_c$ 的中點。\n\n同理可證 $B$ 為 $S_cS_a$ 的中點，$C$ 為 $T_aT_b$ 的中點。因此 $OB \\perp C'A'$，$OC \\perp A'B'$。所以 $O$ 為 $\\triangle A'B'C'$ 的外心（或 $\\triangle ABC$ 的垂心）。得 $OA \\perp B'C'$，以及 $OR_b = OR_c$。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21575,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer and $d$ one of its positive divisors. Integers $1$ to $n$ are written in columns of length $d$ (the first column contains the numbers $1$ to $d$, starting from the top, the second column contains the numbers $d+1$ to $2d$, etc.). Then, one finds the greatest common divisor of each row and finally the least common multiple of the greatest common divisors. Prove that if $d < n$, then the final result is equal to $d$.",
"options": [],
"answer": "See solution",
"solution": "The first two numbers of the $i$-th row are $i$ and $i + d$. Denote the greatest common divisor of this row by $a$; since $a$ divides both $i$ and $i + d$, it must also divide their difference $d$. Thus, all the greatest common divisors divide $d$, meaning their least common multiple is at most $d$. But as the final row is $d, 2d, \\dots, n$ with the greatest common divisor $d$, the least common multiple of the greatest common divisors is exactly $d$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21576,
"subject": "Mathematics (Olympiad)",
"question": "Let\n\n$$\nN = 2^{15} \\cdot 2015.\n$$\n\nHow many divisors of $N^2$ are strictly smaller than $N$ and do not divide $N$?",
"options": [],
"answer": "See solution",
"solution": "A number with prime factorization $p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}$ has\n\n$$\n\\tau(p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}) = (\\alpha_1 + 1)(\\alpha_2 + 1) \\dots (\\alpha_k + 1)\n$$\n\ndivisors. The number $N$ can be factored as $N = 2^{15} \\cdot 5 \\cdot 13 \\cdot 31$, so\n\n$$\n\\tau(N) = (15 + 1) \\cdot (1 + 1) \\cdot (1 + 1) \\cdot (1 + 1) = 2^7 = 128\n$$\n\nand\n\n$$\n\\tau(N^2) = (2 \\cdot 15 + 1) \\cdot (2 \\cdot 1 + 1) \\cdot (2 \\cdot 1 + 1) \\cdot (2 \\cdot 1 + 1) = 31 \\cdot 3^3 = 837.\n$$\n\nIf $d$ divides $N^2$ then $\\frac{N^2}{d}$ also divides $N^2$, so all divisors can be sorted into pairs $(d, \\frac{N^2}{d})$, omitting the number $N$, which would otherwise be paired with itself. In each pair, exactly one divisor is smaller than $N$ and one is greater than $N$. This implies that the number of divisors of $N^2$ which are strictly smaller than $N$ is equal to the number of pairs, which is\n\n$$\n\\frac{\\tau(N^2) - 1}{2}.\n$$\n\nEach divisor of $N$ also divides $N^2$, so we have to subtract the number of divisors of $N$ smaller than $N$. The number of divisors of $N^2$ which are smaller than $N$ and do not divide $N$ is therefore equal to\n\n$$\n\\frac{\\tau(N^2) - 1}{2} - (\\tau(N) - 1) = \\frac{836}{2} - 127 = 291.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21577,
"subject": "Mathematics (Olympiad)",
"question": "As shown below, $\\angle A$ is the largest angle in triangle $ABC$. On the circumcircle of $\\triangle ABC$, the points $D$ and $E$ are the midpoints of $AB$ and $BC$, respectively. Denote by $\\odot O_1$ the circle passing through $A$ and $B$, and tangent to line $AC$; by $\\odot O_2$ the circle passing through $A$ and $E$, and tangent to line $AD$. $\\odot O_1$ intersects $\\odot O_2$ at points $A$ and $P$. Prove that $AP$ is the bisector of $\\angle BAC$.\n\n",
"options": [],
"answer": "See solution",
"solution": "**Proof**\n\nJoin the pairs of points $EP$, $AE$, $BE$, $BP$, and $CD$. For convenience, denote $A = \\angle BAC$, $B = \\angle ABC$, $C = \\angle ACB$, so $A + B + C = 180^{\\circ}$.\n\nLet $X$ be an arbitrary point on the extension of $CA$, and $Y$ a point on the extension of $DA$. It is easy to see that $AD = DC$, $AE = EB$. Since $A$, $B$, $C$, $D$, $E$ are concyclic, we have:\n\n$$\n\\angle BAE = 90^\\circ - \\frac{1}{2} \\angle AEB = 90^\\circ - \\frac{C}{2},\n$$\n\n$$\n\\angle CAD = 90^\\circ - \\frac{1}{2} \\angle ADC = 90^\\circ - \\frac{B}{2}.\n$$\n\nAs line $AC$ and $\\odot O_1$ are tangent at $A$, and line $AD$ and $\\odot O_2$ are tangent at $A$, we get:\n\n$$\n\\angle APB = \\angle BAX = 180^\\circ - A, \\quad \\angle ABP = \\angle CAP,\n$$\n\nand\n\n$$\n\\angle APE = \\angle EAY = 180^\\circ - \\angle DAE = 180^\\circ - (\\angle BAE + \\angle CAD - A)\n$$\n\n$$\n= 180^\\circ - \\left(90^\\circ - \\frac{C}{2}\\right) - \\left(90^\\circ - \\frac{B}{2}\\right) + A = 90^\\circ + \\frac{A}{2}.\n$$\n\nBy computation:\n\n$$\n\\angle BPE = 360^\\circ - \\angle APB - \\angle APE = 90^\\circ + \\frac{A}{2} = \\angle APE.\n$$\n\nIn $\\triangle APE$ and $\\triangle BPE$, applying the Law of Sines and noting $AE = BE$, we obtain:\n\n$$\n\\frac{\\sin \\angle PAE}{\\sin \\angle APE} = \\frac{PE}{AE} = \\frac{PE}{BE} = \\frac{\\sin \\angle PBE}{\\sin \\angle BPE}.\n$$\n\nTherefore $\\sin \\angle PAE = \\sin \\angle PBE$. Since $\\angle APE$ and $\\angle BPE$ are both obtuse, $\\angle PAE$ and $\\angle PBE$ are both acute, so $\\angle PAE = \\angle PBE$. Hence,\n\n$$\n\\angle BAP = \\angle BAE - \\angle PAE = \\angle ABE - \\angle PBE = \\angle ABP = \\angle CAP.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21578,
"subject": "Mathematics (Olympiad)",
"question": "A hotel has ten rooms along each side of a corridor. An olympiad team leader wishes to book seven rooms on the corridor so that no two reserved rooms on the same side of the corridor are adjacent. In how many ways can this be done?",
"options": [],
"answer": "See solution",
"solution": "Number the rooms on each side of the corridor $1$ to $10$. To pick $n$ non-adjacent rooms on one side, select integers $a_1, a_2, \\dots, a_n$ such that $1 \\leq a_1 < a_2 < \\dots < a_n \\leq 11-n$. Then pick rooms $a_1, a_2+1, a_3+2, \\dots, a_n+n-1$. These room numbers are increasing, and since $a_{i+1} > a_i$, $$(a_{i+1} + i) - (a_i + i - 1) \\geq 2$$ so no two rooms are adjacent.\n\nConversely, any arrangement of $n$ non-adjacent rooms on one side can be ordered and transformed into such a sequence. Thus, there are $\\binom{11-n}{n}$ ways to pick $n$ non-adjacent rooms on one side. (It is not possible to pick six or more non-adjacent rooms on one side of the corridor.)\n\nLabel the two sides as A and B. The possible distributions are:\n- 5 rooms from A and 2 from B\n- 4 rooms from A and 3 from B\n- 3 rooms from A and 4 from B\n- 2 rooms from A and 5 from B\n\nSince choices on each side are independent, the total number of ways is:\n$$\n\\begin{aligned}\n& \\binom{6}{5} \\cdot \\binom{9}{2} + \\binom{7}{4} \\cdot \\binom{8}{3} + \\binom{8}{3} \\cdot \\binom{7}{4} + \\binom{9}{2} \\cdot \\binom{6}{5} \\\\\n&= 216 + 1960 + 1960 + 216 \\\\\n&= 4352\n\\end{aligned}\n$$\nways of assigning the rooms.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21579,
"subject": "Mathematics (Olympiad)",
"question": "Place a circle on the Cartesian plane with its center at the origin and $OA$ on the positive horizontal axis. Let the radius of the circle be $1$.\n\n\n\nGiven coordinates:\n- $C (\\cos \\alpha, \\sin \\alpha)$\n- $X (2\\cos \\alpha, 0)$\n- $A (1, 0)$\n\nLet $t = \\cos \\alpha$, and $k = \\dfrac{AX}{XO} = \\dfrac{1 - 2t}{2t}$.\n\nFind the value of $\\alpha$ such that point $B$ with coordinates $(2t + kt, k \\sin \\alpha)$ lies on the circle.",
"options": [],
"answer": "See solution",
"solution": "Since $B$ is on the circle, its coordinates must satisfy:\n\n$$\n\\begin{aligned}\n1 &= (2t + kt)^2 + (k \\sin \\alpha)^2 \\\\\n &= (2t + kt)^2 + k^2(1 - t^2) \\\\\n &= 4t^2 + 4kt^2 + k^2(1 - t^2) \\\\\n\\end{aligned}\n$$\n\nSubstituting $k = \\dfrac{1 - 2t}{2t}$ and simplifying:\n\n$$\n\\begin{aligned}\n1 &= 4t^2 + 4kt^2 + k^2(1 - t^2) \\\\\n &= 4t^2 + 2t(1 - 2t) + \\dfrac{(1 - 2t)^2}{4t^2} \\\\\n4t^2 &= 8t^3 + (1 - 2t)^2 \\\\\n0 &= 8t^3 - 4t + 1\n\\end{aligned}\n$$\n\nThe cubic equation $8t^3 - 4t + 1 = 0$ has three solutions for $t$: $\\dfrac{1}{2}$, $\\dfrac{-1 \\pm \\sqrt{5}}{4}$. Only $t = \\dfrac{\\sqrt{5} - 1}{4}$ fits the geometry.\n\nFrom trigonometry, $\\cos 72^\\circ = \\dfrac{\\sqrt{5} - 1}{4}$, so $\\alpha = 72^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21580,
"subject": "Mathematics (Olympiad)",
"question": "On a plane, $10^6$ points are marked, no three of which are collinear, and all segments between these points are drawn. On each drawn segment, Grisha placed a real number whose absolute value does not exceed $1$. Then, for each six marked points, he calculated the sum of the numbers on all $15$ segments connecting these points. It turned out that the absolute value of each such sum is not less than $C$, and among such sums there exist both positive and negative values. Find the greatest possible $C$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Consider a graph whose vertices are the marked points and edges are the drawn segments.\n\n**Upper bound.** We prove $C \\le \\frac{15}{4}$. The condition states that in our complete graph there exist both $6$-vertex subsets where the sum of edge weights is positive, and $6$-vertex subsets where it's negative. Therefore, there must exist two $6$-vertex subsets differing by exactly one vertex, such that one has positive sum and the other negative. Indeed, start with a positive $6$-vertex subset and transform it into a negative one by changing vertices one at a time—at some step the sign flips, giving the required pair.\n\nNow consider the complete subgraph on a $7$-vertex set $S$—the union of this pair. Examine all seven $6$-vertex subsets obtained by removing one vertex from $S$. Suppose $k$ of them have negative sums (obtained by removing vertices $A_1, \\dots, A_k$, called *A-vertices*), and $7-k$ have positive sums (obtained by removing vertices $B_1, \\dots, B_{7-k}$, called *B-vertices*). We classify edges as:\n- **AA-edges** between two A-vertices\n- **BB-edges** between two B-vertices\n- **AB-edges** between an A-vertex and a B-vertex\n\nWe modify the edge weights on $S$ by replacing:\n- All AA-edge weights with their average $x$\n- All AB-edge weights with their average $y$\n- All BB-edge weights with their average $z$\n\nClearly $|x|, |y|, |z| \\le 1$ since original weights were in $[-1, 1]$.\n\n**Lemma.** The subgraph $S$ with modified weights satisfies the condition with the same constant $C$.\n\n**Proof.** For any edge type, the number of $6$-vertex subsets containing both its endpoints is constant. Similarly, each $6$-vertex subset contains a fixed number of edges of each type. Therefore:\n- The sum $\\Sigma_A$ in any A-$6$-vertex subset is the average of original negative sums, so $\\Sigma_A \\le -C$\n- The sum $\\Sigma_B$ in any B-$6$-vertex subset is the average of original positive sums, so $\\Sigma_B \\ge C$ \\qed\n\nWe analyze cases based on $k$:\n\n*Case $k = 1$:* One A-vertex, six B-vertices. The inequalities are:\n$$15z \\le -C$$\n(A-$6$-vertex subset)\n$$5y + 10z \\ge C$$\n(B-$6$-vertex subset)\nEliminating $z$ gives $15y \\ge 5C \\implies C \\le 3$.\n\n*Case $k = 2$:* Two A-vertices, five B-vertices. Inequalities:\n$$5y + 10z \\le -C$$\n(A-$6$-vertex)\n$$x + 8y + 6z \\ge C$$\n(B-$6$-vertex)\nEliminating $z$ gives $8C \\le 5x + 25y \\le 30 \\implies C \\le \\frac{15}{4}$.\n\n*Case $k = 3$:* Three A-vertices, four B-vertices. Inequalities:\n$$x + 8y + 6z \\le -C$$\n(A-$6$-vertex)\n$$3x + 9y + 3z \\ge C$$\n(B-$6$-vertex)\nEliminating $y$ gives $17C \\le 15x - 30z \\le 45 \\implies C \\le \\frac{45}{17} < \\frac{15}{4}$.\n\nCases $k \\ge 4$ reduce to $k \\le 3$ by symmetry (multiplying all weights by $-1$). Thus $C \\le \\frac{15}{4}$ is proved.\n\n**Construction.** For any number of A-vertices from $2$ to $999,995$ (and remaining as B-vertices), set:\n- All BB-edge weights to $-\\frac{7}{8}$\n- All other edge weights to $1$\n\nThen any $6$-vertex subset with $\\ge 5$ B-vertices has sum $\\le -\\frac{15}{4}$, while others have sum $\\ge \\frac{15}{4}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21581,
"subject": "Mathematics (Olympiad)",
"question": "Find all non-constant functions $f : \\mathbb{Q}^{+} \\to \\mathbb{Q}^{+}$ satisfying the equation\n\n$$\nf(ab + bc + ca) = f(a)f(b) + f(b)f(c) + f(c)f(a)\n$$\n\nfor all $a, b, c \\in \\mathbb{Q}^{+}$.",
"options": [],
"answer": "See solution",
"solution": "Let $c = 1$ in the given condition. Then\n\n$$\nf(ab + a + b) = f(a)f(b) + f(a)f(1) + f(b)f(1), \\quad \\forall a, b \\in \\mathbb{Q}^{+}. \\tag{1}\n$$\n\nSet $b = 3$ in (1):\n\n$$\nf(4a + 3) = f(a)f(3) + f(a)f(1) + f(3)f(1), \\quad \\forall a \\in \\mathbb{Q}^{+}.\n$$\n\nSet $b = 1$ in (1):\n\n$$\nf(2a + 1) = 2f(a)f(1) + f(1)^2, \\quad \\forall a \\in \\mathbb{Q}^{+}.\n$$\n\nThus,\n\n$$\nf(4a + 3) = 2f(2a + 1)f(1) + f(1)^2 = 4f(1)^2f(a) + 2f(1)^3 + f(1)^2, \\quad \\forall a \\in \\mathbb{Q}^{+}.\n$$\n\nComparing,\n\n$$\n[f(3) + f(1)] f(a) + f(3)f(1) = 4f(1)^2f(a) + 2f(1)^3 + f(1)^2, \\quad \\forall a \\in \\mathbb{Q}^{+}.\n$$\n\nIf $f(3) + f(1) \\neq 4f(1)^2$, then $f$ is constant. So $f(3) + f(1) = 4f(1)^2$ and $f(3)f(1) = 2f(1)^3 + f(1)^2$.\n\nThus $f(3), f(1)$ are roots of $t^2 - 2f(1)t + 2f(1)^3 + f(1)^2 = 0$.\n\nSo,\n\n$$\nf(1)^2 - 4f(1)^2 + 2f(1)^3 + f(1)^2 = 0 \\implies f(1)^2(f(1) - 1) = 0.\n$$\n\nThus $f(1) = 1$ and then $f(3) = 3$.\n\nNow, with $c = 1$ in (1):\n\n$$\nf(ab + a + b) = f(a)f(b) + f(a) + f(b), \\quad \\forall a, b \\in \\mathbb{Q}^{+}. \\tag{2}\n$$\n\nSet $b = 1$ and $b = 3$:\n\n$$\nf(4a + 3) = 4f(a) + 3, \\quad f(2a + 1) = 2f(a) + 1.\n$$\n\nSet $a = b = c = \\frac{1}{3}$ in the original condition:\n\n$$\nf\\left(\\frac{1}{3}\\right) = 3f\\left(\\frac{1}{3}\\right)^2 \\implies f\\left(\\frac{1}{3}\\right) = \\frac{1}{3}.\n$$\n\nSet $a = 2$, $b = \\frac{1}{3}$ in (2):\n\n$$\nf(3) = f(2)f\\left(\\frac{1}{3}\\right) + f\\left(\\frac{1}{3}\\right) + f(2) \\implies f(2) = 2.\n$$\n\nSet $b = c = 2$ in the original condition:\n\n$$\nf(4a + 4) = 4f(a) + 4, \\quad \\forall a \\in \\mathbb{Q}^{+}.\n$$\n\nBut also,\n\n$$\nf(4a + 4) = f\\left(4\\left(a + \\frac{1}{4}\\right) + 3\\right) = 4f\\left(a + \\frac{1}{4}\\right) + 3.\n$$\n\nSo $f(a + \\frac{1}{4}) = f(a) + \\frac{1}{4}$, and by induction, $f(x + n) = f(x) + n$ for all $n \\in \\mathbb{Z}^{+}$, $x \\in \\mathbb{Q}^{+}$, so $f(n) = n$ for all $n \\in \\mathbb{Z}^{+}$.\n\nFinally, set $b = n$, $a = \\frac{m}{n+1}$ in (2):\n\n$$\nf(m + n) = f(n)f\\left(\\frac{m}{n+1}\\right) + f\\left(\\frac{m}{n+1}\\right) + f(n) \\implies f\\left(\\frac{m}{n+1}\\right) = \\frac{m}{n+1}.\n$$\n\nThus $f(x) = x$ for all $x \\in \\mathbb{Q}^{+}$. It is easy to check this function satisfies the condition. $\\square$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21582,
"subject": "Mathematics (Olympiad)",
"question": "Are there polynomials $P$ and $Q$ with real coefficients such that $P(P(x)) \\cdot Q(Q(x))$ has exactly 2023 distinct real roots and $P(Q(x)) \\cdot Q(P(x))$ has exactly 2024 distinct real roots?",
"options": [],
"answer": "See solution",
"solution": "There exist such $P$ and $Q$.\n\nLet $M = 2023!$ and $0 < a_1 < a_2 < \\dots < a_{2024} < 1$. We will show that the polynomials $P(x) = (x - a_1) (x - a_2) \\cdots (x - a_{2024}) + M$ and $Q(x) = (x - (M+1)) (x - (M+2)) \\cdots (x - (M + 2023)) + M$ fulfill the stated conditions.\n\nFor $x \\in [0, 1]$, $P$ does not have a real root since $|(x - a_1) \\cdots (x - a_{2024})| < 1$. It is clear that $P(x) > 0$ when $x \\notin [0, 1]$, so $P$ has no real root. In the interval $(-\\infty, M+1)$ the function $Q(x)$ is increasing and can have at most one real root. Obviously, $Q(M) = 0$. For $x \\in [M+1, M+2023]$ we can easily show that $|(x - (M+1)) (x - (M+2)) \\cdots (x - (M + 2023))| < M$, so $Q$ does not have a real root in that interval. As it is obvious that $Q(x) > 0$ for $x > M + 2023$, the only real root of the polynomial $Q$ is equal to $M$.\n\nTherefore,\n\n$$\n|\\{x \\in \\mathbb{R} \\mid P(P(x)) \\cdot Q(Q(x)) = 0\\}| = |\\{x \\in \\mathbb{R} \\mid Q(x) = M\\}|\n$$\n\n$$\n= |\\{M+1, \\dots, M+2023\\}| = 2023\n$$\n\nand\n\n$$\n|\\{x \\in \\mathbb{R} \\mid P(Q(x)) \\cdot Q(P(x)) = 0\\}| = |\\{x \\in \\mathbb{R} \\mid P(x) = M\\}| = |\\{a_1, \\dots, a_{2024}\\}| = 2024.\n$$\n\n**Remark:** The example can be constructed in various ways. The main idea is to choose $P$ with no real roots and $Q$ with a single real root.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21583,
"subject": "Mathematics (Olympiad)",
"question": "Во $\\triangle ABC$, $\\overline{AC} = 5\\ \\text{cm}$ и $\\overline{BC} = 9\\ \\text{cm}$. Од средината на страната $AB$ е издигната нормала која што страната $BC$ ја сече во точка $E$. Точката $E$ е поврзана со темето $A$. Пресметај го периметарот на $\\triangle AEC$.",
"options": [],
"answer": "See solution",
"solution": "Ако $D$ е средина на $AB$, тогаш од $DE \\perp AB$, следува дека\n\n\n\n$\\triangle ABE$ е рамнокрак, со основа $AB$, па $\\overline{AE} = \\overline{EB}$. Тогаш имаме\n\n$$L_{AEC} = \\overline{AC} + (\\overline{CE} + \\overline{EA}) = 5 + 9 = 14\\ \\text{cm.}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21584,
"subject": "Mathematics (Olympiad)",
"question": "Let $P(x) = a_{2016}x^{2016} + a_{2015}x^{2015} + \\dots + a_1x + a_0$ be a degree 2016 polynomial, where its coefficients satisfy\n\n$$\n|a_1 + a_3 + \\dots + a_{2015}| > |a_0 + a_2 + \\dots + a_{2016}|.\n$$\n\nProve that the number of roots of $P(x)$ with absolute value less than 1 is odd (roots are counted with multiplicity).",
"options": [],
"answer": "See solution",
"solution": "The given inequality is equivalent to\n\n$$\n(a_0 + a_2 + \\dots + a_{2016})^2 - (a_1 + a_3 + \\dots + a_{2015})^2 < 0,\n$$\n\nor\n\n$$\n(a_0 - a_1 + a_2 - a_3 + \\dots + a_{2016})(a_0 + a_1 + \\dots + a_{2016}) < 0,\n$$\n\nor equivalently\n\n$$\nP(-1)P(1) < 0.\n$$\n\nSince the coefficients of $P(x)$ are real, if a complex number $z$ is a root of $P$, then so is $\\bar{z}$. Thus, $P(x)$ can be written as\n\n$$\n\\begin{aligned}\nP(x) &= a_{2016} \\prod_{j=1}^{k} (x - z_j)(x - \\bar{z}_j) \\prod_{\\ell=1}^{2016-2k} (x - r_\\ell) \\\\\n&= a_{2016} \\prod_{j=1}^{k} (x^2 + |z_j|^2 - 2 \\Re(z_j)x) \\prod_{\\ell=1}^{2016-2k} (x - r_\\ell)\n\\end{aligned}\n$$\n\nwhere $z_1, z_2, \\dots, z_k$ are complex numbers and $r_1, r_2, \\dots, r_{2016-2k}$ are real numbers, with $0 \\le k \\le 1008$. Now,\n\n$$\nP(1)P(-1) = a_{2016}^2 \\prod_{j=1}^{k} (1 + |z_j|^2 - 2\\Re(z_j)) (1 + |z_j|^2 + 2\\Re(z_j)) \\prod_{\\ell=1}^{2016-2k} (r_{\\ell}^2 - 1).\n$$\n\nSince $1 + |z_j|^2 \\ge 2|z_j| > 2|\\Re(z_j)|$, we have\n\n$$\na_{2016}^2 \\prod_{j=1}^{k} (1 + |z_j|^2 - 2\\Re(z_j)) (1 + |z_j|^2 + 2\\Re(z_j)) > 0.\n$$\n\nIt follows that\n\n$$\n\\prod_{\\ell=1}^{2016-2k} (r_{\\ell}^2 - 1) < 0,\n$$\n\nso there are an odd number of $\\ell$ such that $r_{\\ell}^2 - 1 < 0$. Thus, $P(x)$ has an odd number of roots whose absolute value is less than 1.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21585,
"subject": "Mathematics (Olympiad)",
"question": "Let $F$ be a point on the side $AC$ of a triangle $ABC$. A line through $F$ and parallel to $AB$ intersects side $BC$ at $D$. Similarly, a line through $F$ and parallel to $BC$ intersects side $AB$ at $E$. Assume that the side $AC$ is tangent to the circumcircle of $EDF$. If $AB : BC = k$, then prove that $AF : FC = k^2$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $\\angle A = \\alpha$, $\\angle B = \\beta$, $\\angle C = \\gamma$. Then $\\angle EFA = \\gamma$, because $EF \\parallel BC$. Since $AC$ is tangent to the circumcircle of $EDF$, we have $\\angle EDF = \\angle EFA = \\gamma$.\n\nSimilarly, $\\angle DEF = \\angle DFC = \\alpha$. It follows that $\\triangle EFD \\sim \\triangle ABC$.\n\nOn the other hand, we have $\\triangle AEF \\sim \\triangle FDC \\sim \\triangle ABC$. It follows that $k = \\frac{AB}{BC} = \\frac{EF}{FD} = \\frac{AE}{EF}$. Hence $FD = EF / k$. We also have $\\triangle EFD \\sim \\triangle ABC$. Therefore,\n\n$$\n\\frac{AF}{FC} = \\frac{AE}{FD} = \\frac{AE}{EF / k} = k \\frac{AE}{EF} = k^2.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21586,
"subject": "Mathematics (Olympiad)",
"question": "Wir betrachten ein Spiel mit Murmeln, bei dem Anna und Berta abwechselnd Murmeln nehmen. Es gelten folgende Regeln:\n\n- In jedem Zug darf eine Spielerin eine beliebige gerade oder ungerade Anzahl Murmeln nehmen, jedoch höchstens die Hälfte der verbleibenden Murmeln.\n- Ziel ist es, die letzte Murmel zu nehmen.\n\nBestimme die kleinste Zahl der Form $2^a - 2$, die größer oder gleich $100{,}000$ ist.",
"options": [],
"answer": "See solution",
"solution": "Die Zahlen der Form $2^a - 2$ sind Verlustsituationen, wie im Beweis gezeigt. Um die kleinste solche Zahl zu finden, die mindestens $100{,}000$ ist, bestimmen wir das kleinste $a$ mit $2^a - 2 \\geq 100{,}000$.\n\nSetze $2^a \\geq 100{,}002$. Da $2^{17} = 131{,}072$, ist $a = 17$ die kleinste Lösung.\n\nAlso ist die gesuchte Zahl:\n\n$$2^{17} - 2 = 131{,}072 - 2 = 131{,}070$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21587,
"subject": "Mathematics (Olympiad)",
"question": "Prove that if $n$ is an odd natural number, then $n^3 + 3n^2 - n - 3$ is divisible by $48$.",
"options": [],
"answer": "See solution",
"solution": "We have that\n\n$$\nn^3 + 3n^2 - n - 3 = n^2(n+3) - (n+3) = (n+3)(n^2 - 1) = (n-1)(n+1)(n+3).\n$$\n\nThe numbers $n-1$, $n+1$, and $n+3$ are even, and one of $n-1$ or $n+1$ is divisible by $4$. Therefore, the given number is divisible by $16$.\n\nIf $n$ is divisible by $3$, then $n+3$ is divisible by $3$. If $n$ is not divisible by $3$, then the remainder is $1$ or $-1$, so one of the numbers $n-1$ or $n+1$ is divisible by $3$. Because of this, the given number is divisible by $3 \\cdot 16 = 48$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21588,
"subject": "Mathematics (Olympiad)",
"question": "Sean $a$ y $b$ números enteros positivos tales que $\\frac{5a^4 + a^2}{b^4 + 3b^2 + 4}$ es un número entero. Demostrar que $a$ no es primo.",
"options": [],
"answer": "See solution",
"solution": "Si $b$ es par, entonces $b^2$ y $b^4$ son divisibles por 4, por lo tanto $b^4 + 3b^2 + 4$ es divisible por 4. Si $b$ es impar, entonces $b^2 \\equiv 1 \\pmod{4}$, $b^4 \\equiv 1 \\pmod{4}$ y $3b^2 \\equiv 3 \\pmod{4}$.\n\nAsí que $b^4 + 3b^2 + 4 \\equiv 1 + 3 + 4 \\equiv 0 \\pmod{4}$.\n\nLuego el denominador de la fracción es divisible por 4 para todo entero $b$, por lo que el numerador debe ser divisible por 4.\n\nSi $a$ es impar, entonces $a^2 \\equiv 1 \\pmod{4}$, lo que implica que $5a^4 \\equiv 1 \\pmod{4}$. Por lo tanto, $5a^4 + a^2 \\equiv 2 \\pmod{4}$ y $5a^4 + a^2$ no es divisible por 4. Así que $a$ debe ser par. El único número par que no es compuesto es 2. Si $a = 2$, entonces $5a^4 + a^2 = 84$. Si $b = 1$, vale que $b^4 + 3b^2 + 4 = 8$ y si $b = 2$, $b^4 + 3b^2 + 4 = 32$ y ni $84/8$ ni $84/32$ son enteros. Además, si $b \\geq 3$ entonces $b^4 + 3b^2 + 4 \\geq 112 > 84$, de modo que 84 no puede ser divisible por $b^4 + 3b^2 + 4$ con $b \\geq 3$. Queda demostrado que $a$ no es primo.\n\n**Nota.** Un ejemplo en el que el cociente es un entero: $a = 8$ y $b = 2$. En este caso,\n\n$$\n\\frac{5a^4 + a^2}{b^4 + 3b^2 + 4} = \\frac{20544}{32} = \\frac{2^6 \\cdot 321}{2^5} = 642.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21589,
"subject": "Mathematics (Olympiad)",
"question": "What value of $x$ satisfies\n\n$$\n\\frac{\\log_2 x \\cdot \\log_3 x}{\\log_2 x + \\log_3 x} = 2?\n$$\n\n(A) 25 (B) 32 (C) 36 (D) 42 (E) 48",
"options": [],
"answer": "See solution",
"solution": "**Answer (C):** Observe that\n\n$$\n\\begin{aligned}\n2 &= \\frac{\\log_2 x \\cdot \\log_3 x}{\\log_2 x + \\log_3 x} \\\\\n&= \\frac{1}{\\frac{1}{\\log_3 x} + \\frac{1}{\\log_2 x}} \\\\\n&= \\frac{1}{\\log_4 3 + \\log_4 2} \\\\\n&= \\frac{1}{\\log_4 6} = \\log_6 x.\n\\end{aligned}\n$$\n\nIt follows that $x = 6^2 = 36$.\n\nAlternatively, the given equation is equivalent to\n\n$$\n\\log_2 x \\cdot \\log_3 x = 2 \\log_2 x + 2 \\log_3 x.\n$$\n\nNote that $\\log_3 x = \\frac{\\log_2 x}{\\log_2 3}$, so\n\n$$\n\\log_2 x \\cdot \\frac{\\log_2 x}{\\log_2 3} = 2 \\log_2 x + 2 \\frac{\\log_2 x}{\\log_2 3}.\n$$\n\nMultiplying both sides by $\\frac{\\log_2 3}{\\log_2 x}$ gives\n\n$$\n\\log_2 x = 2 \\log_2 3 + 2 = \\log_2 9 + 2.\n$$\n\nThen\n\n$$\nx = 2^{\\log_2 9+2} = 2^{\\log_2 9} \\cdot 2^2 = 9 \\cdot 4 = 36.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21590,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\mathbb{Z}^+$ be the set of positive integers. Determine all functions $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ such that $a^2 + f(a)f(b)$ is divisible by $f(a) + b$ for all positive integers $a$ and $b$.",
"options": [],
"answer": "See solution",
"solution": "Let $f: \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ such that\n\n$$\nf(a) + b \\mid a^2 + f(a)f(b)\n$$\n\nfor all positive integers $a$ and $b$.\n\nFor any positive integer $n$, set $a = n$ and $b = f(n)$ to get\n\n$$\n2f(n) \\mid n^2 + f(n)f(f(n)).\n$$\n\nThus, $f(n) \\mid n^2$.\n\nFor $n = 1$, $f(1) \\mid 1$, so $f(1) = 1$.\n\nFor $n = p$ (prime), $f(p) \\mid p^2$, so $f(p) \\in \\{1, p, p^2\\}$.\n\n- If $f(p) = 1$, set $a = 1$, $b = p$:\n $$\n 1 + p \\mid 2\n $$\n which is impossible.\n- If $f(p) = p^2$, set $a = p$, $b = 1$:\n $$\n p^2 + 1 \\mid 2p^2\n $$\n which is also impossible.\n\nThus, $f(p) = p$ for all primes $p$.\n\nLet $n$ be any positive integer. Choose a prime $p > n$ and $p > |f(n) - n|$. Set $a = p$, $b = n$:\n\n$$\n\\begin{aligned}\np + n &\\mid p^2 + p f(n) \\\\\n &\\mid p(p + f(n)).\n\\end{aligned}\n$$\n\nSince $\\gcd(p + n, p) = 1$, it follows that\n\n$$\np + n \\mid f(n) - n.\n$$\n\nBut $p + n > |f(n) - n|$, so $f(n) = n$.\n\nIt is easy to check that $f(n) = n$ satisfies the original condition.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21591,
"subject": "Mathematics (Olympiad)",
"question": "The real numbers $a$, $b$, $c$, $d$ satisfy simultaneously the equations\n\n$$\nabc - d = 1, \\quad bcd - a = 2, \\quad cda - b = 3, \\quad dab - c = -6.\n$$\n\nProve that $a + b + c + d \\neq 0$.",
"options": [],
"answer": "See solution",
"solution": "Suppose that $a + b + c + d = 0$. Then\n\n$$\nabc + bcd + cda + dab = 0 \\quad (1)\n$$\n\nIf $abcd = 0$, then one of the numbers, say $d$, must be $0$. In this case $abc = 0$, so at least one of the numbers $a$, $b$, $c$ will be $0$, making one of the given equations impossible. Hence $abcd \\neq 0$ and, from (1),\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} = 0.\n$$\n\nThis implies\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = \\frac{1}{a + b + c}.\n$$\n\nIt follows that $(a + b)(b + c)(c + a) = 0$, which is impossible (for instance, if $a + b = 0$, then adding the second and third given equations would lead to $2 + 3 = 0$, a contradiction). Thus $a + b + c + d \\neq 0$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21592,
"subject": "Mathematics (Olympiad)",
"question": "給定一平面上 100 個半徑為 1 的圓，使得任三個圓心所構成的三角形面積至多為 100。試證：存在一條直線至少與 10 個圓相交。",
"options": [],
"answer": "See solution",
"solution": "我們證明一個更一般的命題：給定一平面上的 $n$ 個圓，若任三個圓心所構成的三角形面積至多為 $n$，則存在一條直線與至少 $\\frac{2n}{\\sqrt{4n+2}}$ 個圓相交。\n\n令 $S$ 為這 $n$ 個圓心所成集合。先證明：存在一條直線 $L$ 使得 $S$ 中的點投影在 $L$ 上會落在一個長度為 $\\sqrt{4n}$ 的區間內。\n\n證明：設 $A, B$ 為 $S$ 中相距最遠的兩個點，令其距離為 $d$。\n\n- 任取 $S$ 中異於 $A, B$ 的一點 $C$，因為 $|\\triangle ABC| \\leq n$，故 $C$ 到直線 $AB$ 的距離至多為 $\\frac{2n}{d}$。\n- 若 $L$ 垂直直線 $AB$ 於 $D$，則 $S$ 中任一點在 $L$ 上的投影點，將落於以 $D$ 為中心，長度為 $\\frac{4n}{d}$ 的區間內。\n- 又因 $S$ 中兩點距離最大為 $d$，故 $S$ 中的點對 $L$ 的投影必落在一個長度為 $d$ 的區間內，故此區間長度至多為 $\\min\\{d, \\frac{4n}{d}\\} \\leq \\sqrt{4n}$。\n\n這 $n$ 個圓投影到 $L$ 上都是長度為 2 的區間，而這些區間都包含在一個長度至多為 $\\sqrt{4n} + 2$ 的區間內。令這個區間為 $I$。令 $C_i$ 為第 $i$ 個圓在 $L$ 上的投影。\n\n所有 $C_i$ 的長度總和為 $2n$，且它們都落於 $I$ 中。這表示 $I$ 中至少有一個點 $x$ 同時屬於至少 $\\frac{2n}{\\sqrt{4n+2}}$ 個 $C_i$（否則依平均值原理將得到矛盾）。\n\n故取垂直於 $AB$ 且過 $x$ 點的直線 $L$ 即可。\n\n將 $n = 100$ 代入，得到至少要 $\\frac{200}{\\sqrt{400+2}} = \\frac{200}{22} > 9$，故至少交 10 個圓。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21593,
"subject": "Mathematics (Olympiad)",
"question": "A ball bounces on a rectangular billiard table with sides of length $a$ and $b$. The ball touches a vertical rail $c$ times and a horizontal rail $c$ times, where $c$ is the least common multiple of $a$ and $b$. If $c$ is even, the ball lands in a pocket on the left side of the table; if $c$ is odd, it lands on the right. Similarly, if $c$ is even, the ball lands in a pocket on the lower side; if $c$ is odd, it lands on the upper side. Thus, the ball lands in the upper left pocket exactly when $c$ is even and $c$ is odd.\n\n\n\nFix an integer $N$ and count the ordered pairs $(a, b)$ with $1 \\leq b \\leq a \\leq N$ for which the ball lands in the upper left pocket.",
"options": [],
"answer": "See solution",
"solution": "Write $a = 2^j a'$ and $b = 2^k b'$, where $a'$ and $b'$ are odd. Then $c = 2^{\\max(j, k)} \\operatorname{lcm}(a', b')$. Since $\\operatorname{lcm}(a', b')$ is odd, $c$ is even and $c$ is odd precisely when $j < k$.\n\nIf $a$ is odd, then the ball lands in the upper left pocket whenever $b$ is even; there are $\\frac{a-1}{2}$ such allowed values of $b$. If $a$ is divisible by 2 but not by 4, then the ball lands in the upper left pocket whenever $b$ is divisible by 4; there are $\\frac{a-2}{4}$ such allowed values of $b$, and so forth.\n\nIn the calculation, we use $C(k, 2) = \\binom{k}{2} = 0 + 1 + 2 + \\dots + (k-2) + (k-1)$.\n\nWhen $a = 1, 3, 5, 7, \\dots$, then there are $0, 1, 2, 3, \\dots$ allowed values of $b$; summing over all such $a$ gives $C\\left(\\left[\\frac{N+1}{2}\\right], 2\\right)$ total ordered pairs. When $a = 2, 6, 10, 14, \\dots$, there are $0, 1, 2, 3, \\dots$ allowed values of $b$; summing over all such $a$ gives $C\\left(\\left[\\frac{N+2}{4}\\right], 2\\right)$ total ordered pairs, and so forth.\n\nFor $N = 2009$, the total is:\n$$\nC(1005, 2) + C(502, 2) + C(251, 2) + C(126, 2) + C(63, 2) + C(31, 2) + C(16, 2) + C(8, 2) + C(4, 2) + C(2, 2) = 672,084\n$$\nallowed ordered pairs. So this is our answer.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21594,
"subject": "Mathematics (Olympiad)",
"question": "Positive integers $a, p$ satisfy $p = 2^a - 1$. Find all $a$ such that $\\frac{1}{2}(p^2+1)$ is a perfect square.",
"options": [],
"answer": "See solution",
"solution": "For $a=1$, $p=1$ and $\\frac{1}{2}(p^2+1)=1$, which is a perfect square.\n\nFor $a=2$, $p=3$ and $\\frac{1}{2}(p^2+1)=5$, which is not a perfect square.\n\nAssume now $a \\geq 3$. Let $\\frac{1}{2}(p^2+1) = p_1^2$, then $p^2 - 2p_1^2 = -1$. Thus:\n\n$$\n2^{2a} - 2^{a+1} + 1 - 2p_1^2 = -1\n$$\n$$\n2^{2a-1} - 2^a = p_1^2 - 1\n$$\nSo $2^a(2^{a-1}-1) = (p_1-1)(p_1+1)$. The left side is even, so the right side is even. Thus $\\gcd(p_1-1, p_1+1) = 2$. Only the following cases are possible:\n\n1) $p_1+1=2l$, $p_1-1=2k$, so $kl = 2^{a-1}-1$.\n\nIf $k \\geq 2$, $p_1 \\geq 2^a+1$ and $l \\geq 2^{a-1}+1$, which contradicts $kl = 2^{a-1}-1$.\n\nIf $k=1$, $p_1 = 2^{a-1}+1$, $l=2^{a-2}+1$, and $kl = 2^{a-2}+1 = 2^{a-1}-1$, so $a=3$.\n\n2) $p_1-1=2k$, $p_1+1=2^{a-1}l$, $kl = 2^{a-1}-1$.\n\nIf $l \\geq 2$, $p_1 \\geq 2^a-1$ and $k \\geq 2^{a-1}-1$, so $2^{a-1}-1 = kl \\geq 2^a-2$, hence $a=1$.\n\nIf $l=1$, $p_1 = 2^{a-1}-1$, $k = 2^{a-2}-1$, and $kl = 2^{a-2}-1 = 2^{a-1}-1$, which is a contradiction.\n\nTherefore, the only solutions are $a=1$ and $a=3$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21595,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a convex quadrilateral such that the line $BD$ bisects the angle $ABC$. Suppose that the circumcircle of triangle $ABC$ intersects the sides $AD$ and $CD$ at the points $P$ and $Q$, respectively. The line through $D$ and parallel to $AC$ intersects the lines $BC$ and $BA$ at the points $R$ and $S$, respectively. Prove that the points $P$, $Q$, $R$, and $S$ lie on a common circle.",
"options": [],
"answer": "See solution",
"solution": "Since $\\angle SDP = \\angle CAP = \\angle RBP$, the quadrilateral $BRDP$ is cyclic (see Figure 1). Similarly, the quadrilateral $BSDQ$ is cyclic. Let $X$ be the second intersection point of the segment $BD$ with the circumcircle of triangle $ABC$. Then\n\n$$\n\\angle AXB = \\angle ACB = \\angle DRB,\n$$\n\nand moreover $\\angle ABX = \\angle DBR$. This means that triangles $ABX$ and $DBR$ are similar. Thus\n\n$$\n\\angle RPB = \\angle RDB = \\angle XAB = \\angle XPB,\n$$\n\nwhich implies that the points $R$, $X$, and $P$ are collinear. Analogously, we show that the points $S$, $X$, and $Q$ are collinear.\n\nThus we obtain $RX \\cdot XP = DX \\cdot XB = SX \\cdot XQ$, which proves that the points $P$, $Q$, $R$, and $S$ lie on a common circle.\n\n\n\nFigure 1",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21596,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute triangle. The line through $A$ perpendicular to $BC$ intersects $BC$ at $D$. Let $E$ be the midpoint of $AD$ and $\\omega$ the circle with center $E$ and radius $AE$. The line $BE$ intersects $\\omega$ at $X$ such that $X$ and $B$ are not on the same side of $AD$, and the line $CE$ intersects $\\omega$ at $Y$ such that $C$ and $Y$ are not on the same side of $AD$. If two intersection points of the circumcircles of triangles $BDX$ and $CDY$ lie on the line $AD$, prove that $AB = AC$.",
"options": [],
"answer": "See solution",
"solution": "Note that the condition of the problem is that $AD$ is the radical axis of the two circles $(BDX)$ and $(CDY)$, which implies that $E$ has the same power to these circles. This gives us\n\n$$\nEB \\cdot EX = EC \\cdot EY.\n$$\n\nHowever, $EX = EY$ because $E$ is the center of $\\omega$, and this means that $BE = CE$. From this, we conclude that $DE$ is the perpendicular bisector of $BC$, which leads to $AB = AC$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21597,
"subject": "Mathematics (Olympiad)",
"question": "求所有函數 $f: \\mathbb{R} \\to \\mathbb{R}$ 滿足\n\n$$\nf(xy + f(y))f(x) = x^2 f(y) + f(xy)\n$$\n\n對於所有實數 $x, y$ 皆成立。",
"options": [],
"answer": "See solution",
"solution": "令 $P(a, b)$ 為 $x = a, y = b$ 代入的結果。\n\n$$\nP(0,0):\nf(f(0))f(0) = f(0) \\Rightarrow f(f(0)) = 1 \\text{ 或 } f(0) = 0\n$$\n\n**Case 1.** $f(f(0)) = 1$\n\n$$\nP(f(0), 0):\nf(f(0))^2 = f(0)^3 + f(0) \\Rightarrow f(0)^3 + f(0) = 1 \\quad \\cdots (1)\n$$\n\n$$\nP(0, f(0)):\nf(1)f(0) = f(0) \\Rightarrow f(1) = 1\n$$\n\n$$\nP(-1, 0):\nf(f(0))f(-1) = 2f(0) \\Rightarrow f(-1) = 2f(0)\n$$\n\n$$\nP(-1, 1):\nf(0)f(-1) = 1 + f(-1) \\Rightarrow 2f(0)^2 = 1 + 2f(0) \\quad \\cdots (2)\n$$\n\n由 (2) 式可得\n\n$$\n2f(0)^2 - 2f(0) - 1 = 0 \\Rightarrow f(0) = \\frac{1 \\pm \\sqrt{3}}{2}\n$$\n\n但代回 (1) 式發現不合，故 $f(0)$ 無解，$f(f(0)) \\neq 1$。\n\n**Case 2.** $f(0) = 0$\n\n$$\n\\begin{aligned}\nP\\left(\\frac{-f(x)}{x}, x\\right): &\\quad f(0)f\\left(\\frac{-f(x)}{x}\\right) = \\frac{f(x)^3}{x^2} + f(-f(x)) \\\\\n&\\Rightarrow 0 = \\frac{f(x)^3}{x^2} + f(-f(x)) \\\\\n&\\Rightarrow f(x)^3 = -x^2 f(-f(x)) \\quad \\cdots (3)\n\\end{aligned}\n$$\n\n**Claim:** $\\forall x \\neq 0, f(x) \\neq 0$，否則 $f(x) = 0, \\forall x \\in \\mathbb{R}$。\n\nProof. 若存在 $c \\neq 0, f(c) = 0$：\n\n$$\nP(c, 1):\n0 = c^2 f(1) \\Rightarrow f(1) = 0\n$$\n\n$$\nP(1, y):\n0 = 2f(y) \\Rightarrow f(y) = 0, \\forall y \\in \\mathbb{R}\n$$\n\n所以 $\\forall x \\neq 0, f(x) \\neq 0$（$f$ 函數零單），否則 $f(x) = 0, \\forall x \\in \\mathbb{R}$（為其中一解）。\n\n與零單矛盾，故不存在此 $d \\Rightarrow \\forall x \\neq 0, f(x) \\neq f(-x)$，所以 $f$ 函數單射。\n\n**Claim:** $f(-1) = 2$ 或 $f(-1) = -1$。\n\nProof.\n\n$$\nP(-1, x):\nf(-x + f(x)) f(-1) = f(x) + f(-x)\n$$\n\n$$\nP(-1, -x):\nf(x + f(-x)) f(-1) = f(x) + f(-x)\n$$\n\n結合以上兩式，得\n\n$$\n\\begin{aligned}\nf(-x + f(x)) &= f(x + f(-x)) \\\\\n\\Rightarrow -x + f(x) &= x + f(-x) \\\\\n\\Rightarrow f(x) - f(-x) &= 2x \\quad \\cdots (4)\n\\end{aligned}\n$$\n\n然後\n\n$$\nP(1, 1):\nf(1 + f(1)) = 2\n$$\n\n$$\nP(-1, -1):\nf(1 + f(-1)) f(-1) = f(-1) + f(1) = 2 - 2f(1)\n$$\n\n$$\nP(-1, 1 + f(1)):\nf(-1 - f(1) + 2) f(-1) = f(1 + f(1)) + f(-1 - f(1))\n$$\n\n(4) 式可得 $f(1) - f(-1) = 2$ 與 $f(1 + f(1)) - f(-1 + f(1)) = 2(1 + f(1))$，與上式結合得\n\n$$\n\\begin{aligned}\nf(1 - (2 + f(-1))) f(-1) &= 2 + (f(1 + f(1)) - 2(1 + f(1))) \\\\\n\\Rightarrow f(-1 + f(-1)) f(-1) &= 2 - 2f(1)\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21598,
"subject": "Mathematics (Olympiad)",
"question": "Find all ordered pairs $(a, b)$ of positive integers for which the numbers $\\frac{a^3 b - 1}{a + 1}$ and $\\frac{b^3 a + 1}{b - 1}$ are both positive integers.",
"options": [],
"answer": "See solution",
"solution": "As $a^3 b - 1 = b(a^3 + 1) - (b + 1)$, and $a + 1 \\mid a^3 + 1$, we have $a + 1 \\mid b + 1$. As $b^3 a + 1 = a(b^3 - 1) + (a + 1)$ and $b - 1 \\mid b^3 - 1$, we have $b - 1 \\mid a + 1$. So $b - 1 \\mid b + 1$, and hence $b - 1 \\mid 2$.\n\n**Case 1.** If $b = 2$, then $a + 1 \\mid b + 1 = 3$ gives $a = 2$. Hence $(a, b) = (2, 2)$ is the only solution in this case.\n\n**Case 2.** If $b = 3$, then $a + 1 \\mid b + 1 = 4$ gives $a = 1$ or $a = 3$. Hence $(a, b) = (1, 3)$ and $(3, 3)$ are the only solutions in this case.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21599,
"subject": "Mathematics (Olympiad)",
"question": "Movement *A* replaces $x^i + x^{i+1}$ with $x^{i-1}$; movement *B* replaces $2x^i$ with $x^{i-1} + x^{i+2}$.\n\nChoose $x$ so that $1 - x - x^2 = 0$, say $x = \\phi^{-1} = \\frac{\\sqrt{5} - 1}{2}$.\n\nShow that the process stabilizes: that is, after finitely many moves, the configuration remains constant, with no two stones in the same glass nor two stones in two consecutive glasses. Also, prove that the final configuration is unique.",
"options": [],
"answer": "See solution",
"solution": "Let $S$ be the sum of the numbers assigned to the stones. Under movement *A*, the change in $S$ is $x^{i-1}(1 - x - x^2)$. Under movement *B*, the change is $x^{i-1}(1 + x^3 - 2x) = x^{i-1}(1 - x - x^2)(1 - x)$.\n\nIf $x$ satisfies $1 - x - x^2 = 0$, then both changes are zero, so $S$ remains constant throughout the process.\n\nThere is a number $x^k$ such that $S < x^k$, so the leftmost stone cannot move indefinitely. By removing this stone and repeating the argument, each stone eventually becomes immobile, so the configuration stabilizes.\n\nTo show uniqueness, note that the representation of $S$ in base $\\phi^{-1}$ with all digits equal to one and no two consecutive ones is unique. Suppose two such representations differ at position $k$; then $\\phi^{-k}$ equals a sum of terms $\\phi^{-m}$ for $m > k + 1$. But $\\sum_{m=k+2}^{\\infty} \\phi^{-m} = \\frac{\\phi^{-k-2}}{1 - \\phi^{-1}} = \\phi^{-k}$, which is impossible. Thus, the final configuration is unique.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21600,
"subject": "Mathematics (Olympiad)",
"question": "A natural number $n \\geq 5$ is called *special* if, no matter how we choose five distinct numbers from $1, 2, 3, \\ldots, n$, we find among them four distinct numbers $a, b, c, d$ such that $a + b = c + d$.\n\n(a) Prove that $n = 6$ is special.\n\n(b) Find all the special numbers.",
"options": [],
"answer": "See solution",
"solution": "a) Since $1 + 6 = 2 + 5 = 3 + 4$, every set of 5 numbers from $1, 2, 3, 4, 5, 6$ contains 4 distinct numbers $a, b, c, d$ such that $a + b = c + d = 7$.\n\nb) No $n \\geq 8$ is special.\n\nIndeed, no 4 numbers out of $1, 2, 3, 5, 8$ provide equal sums: if we do not choose 8, then $5 + a > b + c$ for every $a, b, c \\in \\{1, 2, 3\\}$, and if we choose 8, then $8 + a > b + c$ for every $a, b, c \\in \\{1, 2, 3, 5\\}$.\n\nThe number $n = 7$ is special. Indeed:\n\n* If we do not choose 7, 4, or 1, then the argument from (a) applies to the sums $1 + 6 = 2 + 5 = 3 + 4$, $1 + 7 = 2 + 6 = 3 + 5$, and $2 + 7 = 3 + 6 = 4 + 5$.\n* If we choose 1, 4, or 7 and two of the numbers 2, 3, 5, 6, then we get the equal sums $1 + 4 = 3 + 2$, $1 + 5 = 2 + 4$, $1 + 7 = 2 + 6$, $1 + 7 = 3 + 5$, $3 + 7 = 4 + 6$, or $4 + 7 = 5 + 6$.\n\nSince 5 is, obviously, special, the special numbers are $5$, $6$, and $7$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21601,
"subject": "Mathematics (Olympiad)",
"question": "Given a set $S$ of $n$ variables and a binary operation $\\times$ on $S$ such that $x \\times y \\in \\{x, y\\}$ for all $x, y \\in S$, and $\\times$ is associative, how many equivalence classes are there of full strings (i.e., strings containing each element of $S$ at least once) under the equivalence relation generated by the operation $\\times$? That is, two strings are equivalent if they evaluate to the same element under all such operations $\\times$.",
"options": [],
"answer": "See solution",
"solution": "Given an operation $\\times$, construct a complete directed graph $G(\\times)$ with $n$ vertices, each vertex corresponding to a variable of $S$. Construct an edge marked $L$ going from $u$ to $v$ if $u \\times v = u$, and an edge marked $R$ from $v$ to $u$ if $u \\times v = v$. In this way, each pair of vertices will have two edges between them, either both going in the same direction or both having the same marking.\n\nSuppose we have an edge from $x$ to $y$ marked $a$ and an edge from $y$ to $z$ marked $b$. Through some simple casework on $a, b$ and use of the associative condition on $\\times$, we can verify this implies there is an edge from $x$ to $z$ marked $a$. Call this the *associativity condition* on $G(\\times)$.\n\nNow suppose two vertices $u, v$ are such that there are two edges from $u$ to $v$, one $L$ and one $R$. We claim this means there is no path from $v$ to $u$. Suppose that $v, w_1, w_2, \\dots, w_k, u$ were such a path. Then the associativity condition on $u, v, w_1$ means that $u$ has two edges marked $L$ and $R$ from $u$ to $w_1$. Through the same methods, we can inductively show that the two edges between $u$ to $w_i$ are directed away from $u$ for all $i$. But we need an edge from $w_k$ to $u$, so this is a contradiction.\n\nConsider the strongly connected components of $G(\\times)$, meaning components such that a pair of vertices $u, v$ has a path from $u$ to $v$ and a path from $v$ to $u$ if and only if $u, v$ are in the same component. The above paragraph implies that for any $u, v$ in the same component, the two edges between $u, v$ go in opposite directions and have the same marking. The associativity condition implies the markings between all pairs of vertices in the same component are equal, so we can consider each component to be either $L$ as a whole or $R$ as a whole.\n\nLet $C_1$ be the component such that no other component has an edge going into $C_1$, and call this the top component of $G(\\times)$. Then by the way we have constructed the graph, any full expression evaluates to a variable corresponding to a vertex in $C_1$ under $\\times$. Since all the edges in $C_1$ are either $L$ or $R$, the value of the full expression is either the leftmost variable out of those appearing in $C_1$, or the rightmost variable. In this way, we have completely characterized how a full expression can be evaluated under a simple operation $\\times$: take a nonempty subset $T$ of the variables, then $\\times$ evaluates the expression to either the leftmost occurrence of a variable in $T$, or the rightmost occurrence of a variable in $T$.\n\nFrom here the problem is straightforward to finish. Given a full expression $w$, consider only the leftmost occurrence of each variable in $w$. This gives a permutation $\\sigma_L$ of the variables of $S$. Similarly, construct $\\sigma_R$ from the rightmost occurrence of each variable in $w$. By our above characterization, $w$ is equivalent to $\\sigma_L\\sigma_R$, and there are $n!^2$ ways to make a word from appending two permutations. Finally, suppose we have a different appending $\\sigma'_L\\sigma'_R$ (without loss of generality we may assume that $\\sigma_L \\neq \\sigma'_L$). We have some pair $a, b$ such that $a$ appears to the left of $b$ in $\\sigma_L$ but their order is reversed in $\\sigma'_L$. Then we can construct $\\times$ such that $a, b$ are the only two elements of the top component marked $L$ of $G(\\times)$. This operation gives $\\sigma_L\\sigma_R = a$ and $\\sigma'_L\\sigma'_R = b$, so the two are not equivalent.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21602,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle inscribed in the circle $(O)$. The point $D$ lies on the opposite ray of $BA$. The circle $(K)$ is tangent to $DA$ and $DC$ at $M$ and $N$, respectively, and is also tangent to $(O)$. Prove that the line $MN$ passes through the center of the $A$-excircle of triangle $ABC$.",
"options": [],
"answer": "See solution",
"solution": "Suppose the external bisector at vertex $C$ of triangle $ABC$ cuts $(O)$ at $E$ (distinct from $C$). The line $CE$ meets $MN$ at $J$. The circle $(K)$ is tangent to $(O)$ at $F$. It is easy to see that $E$, $N$, $F$ are collinear.\n\nBecause $E$ is the midpoint of arc $AB$ of $(O)$ containing $C$, $OE \\perp AB$. Also, $MN \\perp AB$, so $MN \\parallel OE$. Since $K$, $O$, $F$ are collinear, $E$, $N$, $F$ are also collinear.\n\nThus, $EAF$ and $ENA$ are similar, which implies:\n\n$$\nEA^2 = EB^2 = EN \\cdot EF.\n$$\n\nAlso,\n\n$$\n\\angle FMN = \\frac{1}{2}\\angle FKN = \\frac{1}{2}\\angle FOE = \\angle FAE = \\angle FCJ\n$$\n\nso $CFMJ$ is a cyclic quadrilateral. Therefore,\n\n$$\n\\begin{aligned}\n\\angle EFJ &= 180^\\circ - \\angle NFJ = 180^\\circ - (\\angle NFM + \\angle MFJ) \\\\\n&= 180^\\circ - (\\angle JMC + \\angle MCJ) = \\angle MJC.\n\\end{aligned}\n$$\n\nSo $\\triangle EFJ \\sim \\triangle EJN$, hence\n\n$$\nEJ^2 = EN \\cdot EF = EA^2 = EB^2.\n$$\n\nBy angle chasing, $AJ$ is the bisector of $\\angle BAC$, so $J$ is the center of the $A$-excircle of triangle $ABC$.\n\nReturning to the original problem:\n\nSuppose $(K)$ is tangent to $ED$ at $G$ and $(L)$ is tangent to $FD$ at $H$. By the lemma, lines $MG$ and $NH$ pass through the center $J$ of the $A$-excircle of $ABC$.\n\nWe will prove that $\\angle MJN = 90^\\circ$. Indeed,\n\n$$\nKEF + \\angle LFE = \\frac{1}{2}\\angle DEB + \\frac{1}{2}\\angle DFC + \\angle DEF + \\angle DFE = 90^\\circ - \\angle BAC + 180^\\circ - \\angle BDC = 90^\\circ,\n$$\n\nso $EK \\perp LF$. Also, $MG \\perp EK$ and $LF \\perp NH$, so $MG$ is perpendicular to $NH$ at $J$. Therefore, the circle with diameter $MN$ always passes through $J$, which is a fixed point.\n\n(b) It is easy to see that triangle $EMG$ is isosceles and $MP \\parallel CE$, so $\\angle EMG = \\angle EGM = \\angle GMP$. Hence, $MJ$ is the external bisector of triangle $AMP$, but $AJ$ is the internal bisector of $\\angle BAC$, so $J$ is the center of the $A$-excircle of triangle $AMP$.\n\nLet $(R)$ be the circle tangent to $CA$, $AB$ at $Y$, $Z$, respectively, and also externally tangent to $(O)$. By the lemma, when $D$ or $E$ moves to infinity, circles $(K)$ and $(L)$ coincide with $(R)$, and the segment connecting the two tangent points passes through $J$.\n\n\n\nSo $J$ is the midpoint of $YZ$. Because $J$ is the center of the $A$-excircle of triangle $AMP$, the circle $(R)$ is tangent to $AP$, $AM$ at $Y$, $Z$, respectively, and also tangent to the circumcircle of triangle $AMP$.\n\nSimilarly, $(R)$ is tangent to the circumcircle of triangle $ANQ$. Therefore, the circumcircles of triangles $AMP$ and $ANQ$ are always tangent to the fixed circle $(R)$.\n\n\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21603,
"subject": "Mathematics (Olympiad)",
"question": "For nonnegative numbers $a, b, c$ that satisfy $a^2b + b^2c + c^2a \\le a + b + c$, prove the following inequality:\n\n$$\nab + bc + ca \\le a + b + c.$$",
"options": [],
"answer": "See solution",
"solution": "We start with the Cauchy-Schwarz inequality for $(a_1, a_2, a_3)$ and $(b_1, b_2, b_3)$:\n\n$$\n(a_1 b_1 + a_2 b_2 + a_3 b_3)^2 \\le (a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2).\n$$\n\nTake $a_1 = a\\sqrt{b}$, $a_2 = b\\sqrt{c}$, $a_3 = c\\sqrt{a}$ and $b_1 = \\sqrt{b}$, $b_2 = \\sqrt{c}$, $b_3 = \\sqrt{a}$. Then we have\n\n$$\n(ab + bc + ca)^2 \\le (a^2b + b^2c + c^2a)(a + b + c) \\le (a + b + c)^2,\n$$\n\nand our inequality is proved.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21604,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$ be pairwise distinct natural numbers.\n\nProve that\n\n$$\n\\frac{a^3 + b^3 + c^3}{3} \\ge abc + a + b + c.\n$$\n\nWhen does equality hold?",
"options": [],
"answer": "See solution",
"solution": "It is well-known and easily verified that\n\n$$\na^3 + b^3 + c^3 - 3abc = \\frac{1}{2}(a + b + c)\\big((a - b)^2 + (b - c)^2 + (c - a)^2\\big).\n$$\n\nAssume without loss of generality that $a > b > c \\ge 0$. Since the numbers are integers, we obtain $a - b \\ge 1$, $b - c \\ge 1$, and $a - c \\ge 2$.\n\nEquation (1) now implies\n\n$$\na^3 + b^3 + c^3 - 3abc \\ge \\frac{1}{2}(a + b + c)(1 + 1 + 4) = 3(a + b + c)\n$$\n\nas desired.\n\nEquality holds for $a = b + 1$, $b = c + 1$, and $a = c + 2$, which are exactly the triples $(t + 2, t + 1, t)$ where $t \\ge 0$ is an integer, and for all their permutations.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21605,
"subject": "Mathematics (Olympiad)",
"question": "Two people play a game. The first writes natural numbers $d_1, d_2, \\ldots, d_n$ with $n \\ge 2$, which are distinct divisors of $2010$ (but not necessarily all of them). The second player selects two of these numbers, $d_i$ and $d_j$. If the fraction $\\frac{d_i}{d_j}$, after simplification, has a numerator not less than $n$, the second player wins; otherwise, the first player wins. Who has a winning strategy in this game?",
"options": [],
"answer": "See solution",
"solution": "The second player has a winning strategy.\n\nLet's prove this. The number $2010$ factors as $2010 = 2 \\cdot 3 \\cdot 5 \\cdot 67$, so it has $2^4 = 16$ divisors. For each prime factor, a divisor may or may not include it.\n\nThe number of divisors divisible by $67$ is $2^3 = 8$ (since for $2,3,5$ there are two choices each, and $67$ must be included). Thus, if $8 < n \\leq 16$, among any $n$ divisors, at least one is divisible by $67$ and at least one is not. The second player can pick $d_i$ divisible by $67$ and $d_j$ not divisible by $67$. Then, after simplification, the numerator is at least $67 > n$.\n\nIf $n \\leq 8$ and all chosen divisors are not divisible by $67$, the possible divisors are $1, 2, 3, 5, 6, 10, 15, 30$ (all products of $2,3,5$). For $n=8$, these are all the divisors, so the second player can pick $d_i = 30$ and $d_j = 1$, yielding $30 > 8 = n$.\n\nFor $n=7$, there is a divisor greater than $6$ (namely $10$, $15$, or $30$), and another relatively prime to it, so the irreducible numerator is at least $7 = n$.\n\nFor $5 \\leq n \\leq 6$, there is a divisor greater than $5$ and another relatively prime to it, so the irreducible numerator is at least $6 \\geq n$.\n\nIf $n \\leq 4$, consider divisibility by $5$. If some divisors are divisible by $5$ and some are not, the second player can pick $d_i$ divisible by $5$ and $d_j$ not, so the irreducible numerator is at least $5 > 4 \\geq n$.\n\nIf all divisors are not divisible by $5$, the possible divisors are $1, 2, 3, 6$. For $n=4$, the second player can pick $d_i = 6$ and $d_j = 1$, so $6 > 4 = n$.\n\nFor $n=3$, there is a divisor greater than $2$ and another relatively prime to it, so the irreducible numerator is at least $3 = n$.\n\nFor $n=2$, the second player picks the larger as $d_i$ and the smaller as $d_j$, so $\\frac{d_i}{d_j} > 1$, and $d_i \\geq 2 = n$.\n\nThus, in all cases, the second player can always win.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21606,
"subject": "Mathematics (Olympiad)",
"question": "Let $n = 8^{2022}$. Which of the following is equal to $\\frac{n}{4}$?\n\n(A) $4^{1010}$ \n(B) $2^{2022}$ \n(C) $8^{2018}$ \n(D) $4^{3031}$ \n(E) $4^{3032}$",
"options": [],
"answer": "See solution",
"solution": "**Answer (E):** Note that $8 = 2^3$, so\n\n$$\nn = 8^{2022} = (2^3)^{2022} = 2^{6066} = (2^2)^{3033} = 4^{3033}.\n$$\n\nThus\n\n$$\n\\frac{n}{4} = \\frac{4^{3033}}{4} = 4^{3032}.\n$$\n\nAll the other choices are less than $4^{3032}$. Choices (A) and (D) have the same base but a lesser exponent. Choice (B) is incorrect because $2^{2022} < 4^{2022} < 4^{3032}$. Choice (C) is incorrect because\n\n$$\n8^{2018} = (4^{\\frac{3}{2}})^{2018} = 4^{3027} < 4^{3032}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21607,
"subject": "Mathematics (Olympiad)",
"question": "Let $P(x) \\in \\mathbb{Z}[x]$ be a polynomial. Determine all polynomials $Q(x) \\in \\mathbb{Z}[x]$ such that for every positive integer $n$, there exists a polynomial $R_n(x) \\in \\mathbb{Z}[x]$ satisfying\n\n$$\nQ(x)^{2n} - 1 = R_n(x) (P(x)^{2n} - 1).\n$$",
"options": [],
"answer": "See solution",
"solution": "By fixing $x \\in \\mathbb{Z}$, we see that $Q(x)^{2n} - 1$ is a multiple of $P(x)^{2n} - 1$ for all positive integers $n$. We now prove the following lemma.\n\n*Lemma.* If $a$ and $b$ are integers greater than $1$ such that $a^n - 1 \\mid b^n - 1$ for all positive integers $n$, then $b = a^k$ for some positive integer $k$.\n\n*Proof.* (Adapted from Andreescu, T., Dospinescu, G, Problems from the Book.)\n\nLet $x_n = \\frac{b^{n-1}}{a^{n-1}}$ and consider the sequence $(x_n^{(i)})_{n>0}$ indexed by $i > 0$ defined by\n\n$$\nx_n^{(i+1)} = b x_n^{(i)} - a^i x_{n+1}^{(i)}, \\quad x_n^{(1)} = x_n.\n$$\n\nBy induction, $x_n^{(i)}$ can be written as\n\n$$\n\\frac{c_i^{(i)} b^n + c_{i-1}^{(i)} a^{(i-1)n} + \\dots + c_1^{(i)} a^n + c_0^{(i)}}{(a^n + i - 1 - a) \\dots (a^n - 1)}.\n$$\n\nFor $i$ such that $a^i > b$, $\\lim_{n \\to \\infty} x_n^{(i)} = 0$. But $x_n^{(1)} \\in \\mathbb{Z}$ for all $n$, so all $x_n^{(i)}$ are integers, which means $x_n^{(i)} = 0$ for large $n$. Let $j$ be minimal such that $x_n^{(j)} = 0$ for all $n \\ge M_j$. Then\n\n$$\nb x_n^{(j-1)} = a^j x_{n+1}^{(j-1)}.\n$$\n\nBy induction,\n\n$$\nx_n^{(j-1)} = \\left(\\frac{b}{a^j}\\right)^{n-M} x_M^{(j-1)}.\n$$\n\nSet $\\frac{b}{a^j} = c > 0$. Then $c^{M-n} x_n^{(j-1)} \\in \\mathbb{Z}$ for all $M > n$, so $c \\in \\mathbb{Z}$ or $x_n^{(j-1)} = 0$. The latter contradicts minimality of $j$. Thus $c \\in \\mathbb{Z}$, so $a$ divides $b$. Write $b = c a$ for $c \\in \\mathbb{Z}_{>0}$, then\n\n$$\na^n - 1 \\mid (a c)^n - 1 \\implies a^n - 1 \\mid c^n - 1.\n$$\n\nSimilarly, $a \\mid c$ or $c = 1$. Repeating, we get $b = a^k$ for some $k$. ■\n\nBack to the problem: since $(Q(x)^2)^n - 1$ is divisible by $(P(x)^2)^n - 1$ for all $(x, n) \\in \\mathbb{Z} \\times \\mathbb{Z}_{>0}$, if $\\deg P > 0$ and $\\deg Q > 0$, there exists $x_0$ such that for all $x > x_0$, $|P(x)|, |Q(x)| > 1$. The lemma gives, for each $x > x_0$, an integer $k_x$ such that\n\n$$\n|Q(x)| = |P(x)|^{k_x}.\n$$\n\nIf $\\frac{\\deg Q}{\\deg P} = k$, then for all $\\varepsilon > 0$,\n\n$$\n\\lim_{n \\to \\infty} \\frac{|P(x)|^{k-\\varepsilon}}{|Q(x)|} = 0, \\quad \\lim_{n \\to \\infty} \\frac{|P(x)|^{k+\\varepsilon}}{|Q(x)|} = +\\infty\n$$\n\nso $k = k_x$ for large $x$, i.e., $|Q|$ is a power of $|P|$. Now consider $\\deg P < 1$ or $\\deg Q < 1$:\n\n- If $\\deg Q < 1$ and $\\deg P > 1$, $\\deg Q^{2n} - 1 < \\deg (P^{2n} - 1)$, so $Q^{2n} = 1$, i.e., $Q(x) \\equiv 1$ or $-1$.\n- If $\\deg P < 1$, write $P(x) = p$. If $|p| = 1$, $Q^{2n} = 1$ so $Q(x) \\equiv 1$ or $-1$. Otherwise, $Q(x)^{2n} - 1$ is a multiple of $p^{2n} - 1$ for all $n$, so $|Q(x)| = p^{k_x}$ for some $k_x$. By Schur's lemma, if $\\deg Q > 0$, the sequence $(Q(n))_{n \\ge 0}$ has infinitely many prime divisors, which is a contradiction, so $\\deg Q = 0$ and $|Q(x)| = p^k$ for some $k$.\n\nThus, the polynomials $Q(x)$ satisfying the problem are:\n\n$$\nQ(x) = \\pm P(x)^k \\text{ for some positive integer } k, \\text{ or } Q(x) \\equiv \\pm 1.\n$$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21608,
"subject": "Mathematics (Olympiad)",
"question": "Let us reformulate the problem in terms of graph theory: A $k$-regular graph $G$ on $2021$ vertices does not contain any triangle. What is the largest value of $k$?\n\nBelow we give an example for $k = 808$:\n\n",
"options": [],
"answer": "See solution",
"solution": "Now we show that $k \\leq 808$.\n\nFirst, suppose that $G$ is bipartite: $G = (A, B)$ where all edges are between $A$ and $B$. Then, since $|A| \\cdot k = |B| \\cdot k$, we get $|A| = |B|$, which is impossible since there are $2021$ vertices. Hence, $G$ is not bipartite and $G$ contains a cycle of odd length. Let $C$ be the shortest odd cycle with $2t+1$ edges. Since $G$ does not contain any triangle, $t \\geq 2$.\n\nLet us show that any vertex $v$ in $G - C$ has at most two neighbours on $C$. On the contrary, suppose that vertices $a, b, c \\in C$ (in clockwise direction) are directly connected to $v$. Let $n_{ab}, n_{bc}, n_{ca}$ be the number of vertices on paths (without endpoints) $(a, b)$, $(b, c)$, $(c, a)$ of $C$, respectively. Since $G$ has no triangles, $n_{ab}, n_{bc}, n_{ca} \\geq 1$. Since $n_{ab} + n_{bc} + n_{ca} = 2t-2$, one of these three numbers, say $n_{ab}$, is even. The vertex $v$ and $(a, b)$ constitute a cycle of length $n_{ab} + 3$. Since $n_{ab} + 3$ is odd and $C$ is a shortest odd cycle, we get $n_{ab} + 3 \\geq 2t+1$ and $n_{ab} \\geq 2t-2$. Hence, we get a contradiction $n_{bc} + n_{ca} = 0$.\n\nLet us consider all edges between $C$ and $G - C$. Since each vertex of $C$ has $k$ neighbours, there are $(k-2)(2t+1)$ edges between $C$ and $G - C$. On the other hand, each vertex of $G - C$ has at most two neighbours in $C$. Therefore,\n\n$$\n2(2021 - 2t - 1) \\geq (k-2)(2t+1) \\implies 4042 \\geq k(2t+1).\n$$\n\nFinally, since $t \\geq 2$, we get $k \\leq 808$. Done.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21609,
"subject": "Mathematics (Olympiad)",
"question": "We are given a tetrahedron with five edges of length $2$ and one edge of length $1$. A point $P$, either in the interior or on the surface (but not outside), has distances $a$, $b$, $c$, and $d$ from the faces of the tetrahedron. For which points $P$ is the value $a + b + c + d$ minimal, and for which is it maximal?",
"options": [],
"answer": "See solution",
"solution": "The tetrahedron has two equilateral faces (sides of length $2$) and two isosceles faces (two sides of length $2$, one of length $1$). Let $F$ be the area of each equilateral face and $G$ the area of each isosceles face, with $F > G$. Let $a$ and $b$ be the distances from $P$ to the equilateral faces, and $c$ and $d$ the distances to the isosceles faces.\n\nIf $V$ is the volume of the tetrahedron, then:\n\n$$\n\\begin{align*}\n3V &= F(a+b) + G(c+d) \\\\\n &= F(a+b+c+d) - (F-G)(c+d) \\\\\n &\\implies a+b+c+d = \\frac{3V + (F-G)(c+d)}{F}.\n\\end{align*}\n$$\n\nSince $F - G > 0$, $a+b+c+d$ is minimal when $c+d = 0$, i.e., $c = d = 0$. This occurs for points $P$ on the common edge of the isosceles faces (the edge of length $1$).\n\nSimilarly,\n\n$$\n\\begin{align*}\n3V &= F(a+b) + G(c+d) \\\\\n &= G(a+b+c+d) + (F-G)(a+b) \\\\\n &\\implies a+b+c+d = \\frac{3V - (F-G)(a+b)}{G}.\n\\end{align*}\n$$\n\nThus, $a+b+c+d$ is maximal when $a+b = 0$, i.e., $a = b = 0$. This occurs for points $P$ on the common edge of the equilateral faces.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21610,
"subject": "Mathematics (Olympiad)",
"question": "Given an acute, scalene triangle $ABC$, let $D$ be a point on side $BC$. Let $E$ and $F$ be points on $AB$ and $AC$ such that $\\angle DEB = \\angle DFC$. Lines $DF$ and $DE$ intersect $AB$ and $AC$ at points $M$ and $N$, respectively. Denote by $(I_1)$ and $(I_2)$ the circumcircles of triangles $DEM$ and $DFN$. The circle $(J_1)$ touches $(I_1)$ internally at $D$ and touches $AB$ at $K$; the circle $(J_2)$ touches $(I_2)$ internally at $D$ and touches $AC$ at $H$. Let $P$ be the intersection of $(I_1)$ and $(I_2)$, and $Q$ be the intersection of $(J_1)$ and $(J_2)$ ($P, Q \\neq D$).\n\n**a)** Prove that $D$, $P$, and $Q$ are collinear.\n\n**b)** The circumcircle of triangle $AEF$ meets the circumcircle of triangle $AHK$ again at $G$ and meets the line $AQ$ again at $L$. Prove that the tangent line from $D$ to the circumcircle of triangle $DQG$ intersects $EF$ at a point on the circumcircle of triangle $DLG$.\n\n",
"options": [],
"answer": "See solution",
"solution": "**a)** Note that $\\angle DEB = \\angle DFC$, so $\\angle DEA = \\angle DFA$, which implies $MNEF$ is a cyclic quadrilateral.\n\nWe have $\\angle DI_2F = 2\\angle DNF = 2\\angle EMF$ and\n\n$$\n\\angle I_2DF = 90^\\circ - \\frac{1}{2} \\angle DI_2F\n$$\n\nso $I_2D \\perp ME$. We also have $J_1K \\perp ME$, so $I_2D \\parallel J_1K$. Similarly, $I_1D \\parallel J_2H$. Hence, $DK$ is the bisector of $\\angle I_2DI_1$. Similarly, $DH$ is the bisector of $\\angle I_2DI_1$. Thus, the three points $D$, $H$, and $K$ are collinear.\n\nSince $MNEF$ is cyclic, $AE \\cdot AM = AF \\cdot AN$, so $A$ belongs to the radical axis of $(I_1)$ and $(I_2)$, which implies $A$, $D$, and $P$ are collinear. Furthermore,\n\n$$\n\\angle AKH = 90^\\circ - \\angle DKJ_1 = 90^\\circ - \\angle DHJ_2 = \\angle DHF = \\angle AHK\n$$\n\nso $AH = AK$. It follows that $A$ has the same power to $(J_1)$ and $(J_2)$, or $A$ lies on the radical axis of $(J_1)$ and $(J_2)$. Hence, $A$, $D$, and $Q$ are collinear.\n\nFrom these, we have four points $A$, $D$, $P$, $Q$ are collinear.\n\n**b)** Since $AK$ is a tangent of $(J_1)$, $\\angle AQK = \\angle AKD = \\angle AHK$, so $AQHK$ is cyclic. We have $\\angle GEF = \\angle GAF = \\angle GKH$, $\\angle GHK = \\angle GAK = \\angle GFE$, so $\\triangle GEF \\sim \\triangle GKH$ (a.a).\n\n\n\nTake the point $S$ on $EF$ such that $\\overline{SE} : \\overline{DF} = \\overline{DK} : \\overline{DH}$, then $\\triangle GES \\sim \\triangle GKD$ (s.a.s). Thus, $\\triangle GEK \\sim \\triangle GSD$ (s.a.s). From here,\n\n$$\n\\angle GDS = \\angle GKE = \\angle GQD\n$$\n\nso $DS$ is the tangent of the circumcircle of triangle $GDQ$. Similarly, $\\triangle LEF \\sim \\triangle QKH$ (a.a), so $\\triangle LES \\sim \\triangle QKD$ (s.a.s). Hence, $\\angle KQD = \\angle ELS$, and\n\n$$\n\\angle KQG = \\angle KHG = \\angle EFG = \\angle ELG\n$$\n\nwhich implies $\\angle SLG = \\angle DQG = \\angle GDS$. It shows that $DLGS$ is a cyclic quadrilateral. Thus, the problem is proved. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21611,
"subject": "Mathematics (Olympiad)",
"question": "A parallelogram has two sides of length $4$ and two sides of length $7$. Also, one of the diagonals has length $7$. (Attention: the picture has not been drawn to scale.) What is the length of the other diagonal?\n\n",
"options": [],
"answer": "See solution",
"solution": "$9$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21612,
"subject": "Mathematics (Olympiad)",
"question": "Show that there exists a positive integer $K$ such that for any prime $p > K$, the number of integers $1 \\leq a \\leq p$ for which $a^{p-1} - 1$ is divisible by $p^2$ is less than or equal to $\\dfrac{p}{2^{2024}}$.",
"options": [],
"answer": "See solution",
"solution": "Define\n\n$$\nS = \\{1 \\leq a \\leq p^2 : p^2 \\mid a^{p-1} - 1\\}, \\quad S_p = S \\cap \\{1, 2, \\dots, p\\}.\n$$\n\nWe observe that $S$ is closed under multiplication modulo $p^2$: if $a, b \\in S$ and $ab \\equiv c \\pmod{p^2}$, then $c \\in S$.\n\n**Lemma 1.** $|S| \\leq p-1$.\n\n*Proof.* (In fact, $|S| = p-1$, but we give a self-contained proof.) If $a, b \\in S$ and $a \\equiv b \\pmod{p}$, then\n\n$$\np^2 \\mid a^{p-1} - b^{p-1} = (a-b)(a^{p-2} + a^{p-3}b + \\dots + b^{p-2}).\n$$\n\nBut\n\n$$\na^{p-2} + a^{p-3}b + \\dots + b^{p-2} \\equiv (p-1)a^{p-2} \\pmod{p}\n$$\n\nis not divisible by $p$, so $p^2 \\mid a-b$, hence $a = b$. Thus, elements of $S$ are all distinct modulo $p$, and $S$ does not contain multiples of $p$, so $|S| \\leq p-1$. $\\square$\n\nNow,\n\n$$\nS_p \\cdot S_p = \\{ab : a, b \\in S_p\\} \\subset S.\n$$\n\nConsider the map $S_p \\times S_p \\to S_p \\cdot S_p$ given by $(a, b) \\mapsto ab$. The preimage of $m \\in S_p \\cdot S_p$ has size at most $\\tau(m)$ (the number of divisors of $m$), since $a \\mid m$ and $b$ is then uniquely determined. Thus,\n\n$$\nM |S_p \\cdot S_p| \\geq |S_p|^2, \\quad M = \\max_{1 \\leq m \\leq p^2} \\tau(m)\n$$\n\nand so\n\n$$\n\\frac{1}{M}|S_p|^2 \\leq |S_p \\cdot S_p| \\leq |S| < p \\implies |S_p| < \\sqrt{pM}.\n$$\n\nMeanwhile, we bound $M$ using the following lemma.\n\n**Lemma 2.** For any $\\epsilon > 0$, there exists a constant $C_\\epsilon > 0$ such that\n\n$$\n\\tau(n) < C_\\epsilon n^\\epsilon\n$$\n\nfor any positive integer $n$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21613,
"subject": "Mathematics (Olympiad)",
"question": "Consider increasing integer sequences with elements from $1, \\dots, 10^9$. Such a sequence is Adriatic if its first element equals $1$ and if every element is at least twice the preceding element. A sequence is Tyrrhenian if its final element equals $10^6$ and if every element is strictly greater than the sum of all preceding elements. Decide whether the number of elements of Adriatic sequences is (i) smaller than, (ii) equal to, or (iii) greater than the number of Tyrrhenian sequences.\n\nEquivalently, a sequence $\\langle a_1, \\dots, a_n \\rangle$ is Adriatic if $a_1 = 1$ and $a_k \\ge 2a_{k-1}$ holds for $k = 2, 3, \\dots, n$. A sequence $\\langle t_1, t_2, \\dots, t_n \\rangle$ is Tyrrhenian if $t_n = 10^6$ and $t_k > t_1 + t_2 + \\dots + t_{k-1}$ holds for $k = 2, 3, \\dots, n$.",
"options": [],
"answer": "See solution",
"solution": "Consider the Adriatic sequence $\\langle a_1, \\dots, a_n \\rangle$ starting with $a_1 = 1$. Construct a new sequence\n\n$$\n\\langle a_2 - 1, a_3 - a_2, \\dots, a_n - a_{n-1}, 10^6 \\rangle\n$$\n\nfrom it. Note that the new sequence is Tyrrhenian, as\n\n$$\n(a_2 - 1) + (a_3 - a_2) + \\dots + (a_n - a_{n-1}) = a_{k-1} - 1 < a_k - a_{k-1}\n$$\nholds for $k = 2, \\dots, n-1$ and as\n\n$$\n(a_2 - 1) + (a_3 - a_2) + \\dots + (a_n - a_{n-1}) = a_n - 1 < 10^6.\n$$\n\nNext, consider the Tyrrhenian sequence $\\langle t_1, t_2, \\dots, t_n \\rangle$ with $t_n = 10^6$. Construct a new sequence\n\n$$\n\\langle 1, 1 + t_1, 1 + t_1 + t_2, \\dots, 1 + t_1 + \\dots + t_{n-1} \\rangle\n$$\n\nfrom it. Note that this new sequence is Adriatic, since\n\n$$\n2(1 + t_1 + t_2 + \\dots + t_{k-1}) \\le 1 + t_1 + \\dots + t_k\n$$\nis equivalent to the Tyrrhenian property $t_1 + \\dots + t_{k-1} < t_k$.\n\nThese two constructions yield two injections and demonstrate that the number of Adriatic sequences equals the number of Tyrrhenian sequences. (In fact, the second injection is the inverse of the first injection, so that we have a clean bijection between the two sets.) $\\Box$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21614,
"subject": "Mathematics (Olympiad)",
"question": "Let $p$ be a prime, and let $a$ and $k$ be positive integers such that $p^a < k < 2p^a$. Prove that there exists a positive integer $n$, with $n < p^{2a}$, such that $$C_n^k \\equiv n \\equiv k \\pmod{p^a}.$$",
"options": [],
"answer": "See solution",
"solution": "We prove a more general statement by induction.\n\nLet $p$ be a prime, and let $a$ and $k$ be positive integers such that $p^a < k < 2p^a$. For any non-negative integer $b$, there exists a positive integer $n < p^{a+b}$ such that $n \\equiv k \\pmod{p^a}$ and $C_n^k \\equiv k \\pmod{p^b}$.\n\nIf $b=0$, $p^b = 1$, take $n = k - p^a$. Assume the statement holds for $b \\ge 0$: there exists $n < p^{a+b}$ with $n \\equiv k \\pmod{p^a}$ and $C_n^k \\equiv k \\pmod{p^b}$.\n\nLet $1 \\le t \\le p-1$. Consider\n\n$$\nC_{n+tp^{a+b}}^{k} = \\prod_{i=0}^{k-1} \\frac{n + tp^{a+b} - i}{k-i}.\n$$\n\nFor integer $m$, let $P(m) = p^{v_p(m)}$, $r(m) = \\frac{m}{P(m)}$, where $v_p(m)$ is the exponent of $p$ in the prime factorization of $m$.\n\nSince $k - i < 2p^a \\le p^{a+1}$, $v_p(k-i) \\le a$ and $n - i \\equiv k - i \\pmod{p^a}$, so\n\n$$\nP(k-i) \\mid n + tp^{a+b} - i.\n$$\n\nThus,\n\n$$\nC_{n+tp^{a+b}}^{k} = \\prod_{i=0}^{k-1} \\frac{\\frac{n-i}{P(k-i)} + tp^{a+b-v_p(k-i)}}{r(k-i)}.\n$$\n\nIf $k - i \\neq p^a$, then $v_p(k - i) \\leq a - 1$ and $a + b - v_p(k - i) \\geq b + 1$. If $k - i = p^a$, then $v_p(k - i) = a$.\n\nTherefore,\n\n$$\nC_{n+tp^{a+b}}^{k} \\equiv \\prod_{i=0}^{k-1} \\frac{n-i}{P(k-i)r(k-i)} + \\left( \\prod_{\\substack{0 \\le i \\le k-1 \\\\ i \\ne k-p^a}} \\frac{n-i}{P(k-i)r(k-i)} \\right) tp^b \\pmod{p^{b+1}}.\n$$\n\nSince $p^a \\mid (n - i) - (k - i)$, $v_p(n-i) = v_p(k-i)$.\n\nThe product $\\prod_{\\substack{0 \\le i \\le k-1 \\\\ i \\ne k-p^a}} \\frac{r(n-i)}{r(k-i)}$ is coprime to $p$, so as $t$ varies, $C_{n+tp^{a+b}}^k$ covers all residues $C_n^k + j p^b$ modulo $p^{b+1}$ for $j = 0, 1, \\dots, p-1$.\n\nSince $C_n^k \\equiv k \\pmod{p^b}$, there exists $j$ such that $C_n^k + j p^b \\equiv k \\pmod{p^{b+1}}$. Thus, for some $t$, $C_{n+tp^{a+b}}^k \\equiv k \\pmod{p^{b+1}}$. Let $N = n + tp^{a+b}$; then $N < p^{a+b+1}$, $N \\equiv n \\equiv k \\pmod{p^a}$, and $C_N^k \\equiv k \\pmod{p^{b+1}}$.\n\nBy induction, the statement is proved. $\\Box$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21615,
"subject": "Mathematics (Olympiad)",
"question": "Integers are written in the cells of a $2010 \\times 2010$ table. Adding $1$ to all the numbers in a row or in a column is called a move. We say that the table is in *equilibrium* if one can obtain, after finitely many moves, a table in which all the numbers are equal.\n\n1. Find the largest positive integer $n$ for which there exists an equilibrium table containing the numbers $2^0, 2^1, \\dots, 2^n$.\n2. For this $n$, find the maximal number that may be contained in such a table.",
"options": [],
"answer": "See solution",
"solution": "1. We claim that for a table of size $m \\times m$ ($m \\geq 2$), the largest such $n$ is $n = 2m - 2$. In particular, for $m = 2010$, $n = 4018$.\n\nLet $a_{ij}$ be the number in cell $(i, j)$ ($1 \\leq i, j \\leq m$). Any move (adding $1$ to a row or column) preserves the value $a - b + c - d$ for any four cells $a, b, c, d$ whose centers form a rectangle with sides parallel to the table. Thus, the table is determined by $2m - 1$ parameters (e.g., the first row and first column). Any such table is equilibrium: by moves on columns, we can equalize the first row; then, by (1), all rows become equal, and finally, by moves on rows, all entries can be made equal. Therefore, for $n = 2m - 2$, we can construct an equilibrium table containing $2m - 1$ numbers: $2^0, 2^1, \\dots, 2^n$.\n\nTo show this is maximal, suppose we try to place $2m$ numbers $2^0, 2^1, \\dots, 2^{2m-1}$ in an equilibrium $m \\times m$ table. A lemma shows that $2m$ integers $b_1, \\dots, b_{2m}$ can be placed in such a table if and only if there exists $p$ ($2 \\leq p \\leq m$) and a permutation so that\n$$\nb_1 + \\dots + b_p = b_{m+1} + \\dots + b_{m+p}.\n$$\nBut for the powers of $2$, dividing both sides by the smallest term yields a contradiction (one side is $1$, the rest are even), so $n = 2m - 2$ is maximal.\n\n2. Suppose the $m \\times m$ equilibrium table contains $2^0, 2^1, \\dots, 2^{2m-1}, k$. By the lemma, $k$ must be involved in the sum above. The maximal value of $k$ is achieved for $p = m$:\n$$\nk = (2^{m-1} + \\dots + 2^{2m-2}) - (2^0 + \\dots + 2^{m-2}) = 2^{2m-1} - 2^m + 1.\n$$\nFor $m = 2010$, this gives $k = 2^{4019} - 2^{2010} + 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21616,
"subject": "Mathematics (Olympiad)",
"question": "Нехай $DF \\parallel BC$. Доведіть, що точки $D$, $E$, $K$ і $N$ лежать на одному колі, якщо $K$ — середина $BC$, $M$ — середина $AN$, а $AD \\cdot AE = AN \\cdot AK$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Оскільки $DF \\parallel BC$, то $\\angle DFA = \\angle CAF$. До того ж, $\\angle DFA = \\angle DFT = \\angle DET$ як вписані, що спираються на одну дугу. Отже, $\\triangle AMT \\sim \\triangle EMA$ за двома кутами, звідки $\\frac{AM}{MT} = \\frac{EM}{AM}$, тобто $AM^2 = EM \\cdot MT$. За властивістю січних, $ME \\cdot MT = MB \\cdot MC$. Тому\n\n$$\nAM^2 = MB \\cdot MC = (AB - AM) \\cdot (AC - AM) = AB \\cdot AC - AM \\cdot (AB + AC) + AM^2.\n$$\n\nЗвідси $AB \\cdot AC = AM \\cdot (AB + AC)$. Оскільки $K$ — середина $BC$, то $AB + AC = 2AK$. За властивістю січних маємо $AB \\cdot AC = AD \\cdot AE$. Отже, $AD \\cdot AE = AM \\cdot 2AK$. Оскільки $M$ — середина $AN$, то останню рівність можна записати у вигляді $AD \\cdot AE = AN \\cdot AK$. Тому точки $D$, $E$, $K$ і $N$ лежать на одному колі.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21617,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle and let the tangent at $B$ to its circumcircle meet the internal bisector of angle $A$ at $P$. The line through $P$ parallel to $AC$ meets $AB$ at $Q$. Assume that $Q$ lies in the interior of segment $AB$ and let the line through $Q$ parallel to $BC$ meet $AC$ at $X$ and $PC$ at $Y$. Prove that $PX$ is tangent to the circumcircle of triangle $XYC$.",
"options": [],
"answer": "See solution",
"solution": "Let $R$ be the point on $AP$ such that $BR$ is parallel to $AC$. Let $D$, $Z$, $L$ be the projections of $P$ on $AB$, $BC$, $AC$ respectively and let $T$, $E$, $M$ be the analogous projections of $R$.\n\n\n\nSince $BP$ is tangent to the circumcircle and $BR \\parallel AC$, we have $\\angle PBZ = \\angle BAC = \\angle TBR$. It follows that the right-angled triangles $RTB$ and $PZB$ are similar and therefore $\\frac{PZ}{RT} = \\frac{PB}{RB}$.\n\nAnalogously, the triangles $REB$ and $PDB$ are also similar and therefore $\\frac{PD}{RE} = \\frac{PB}{RB}$.\n\nSince $P$, $R$ belong on the bisector of $A$, we have that $PD = PL$ and $RT = RM$ so from the results of the previous two paragraphs we get that $\\frac{PL}{RE} = \\frac{PZ}{RM}$.\n\nThe quadrilaterals $ZPLC$ and $ERMC$ are cyclic, therefore $\\angle ZPL = 180^\\circ - \\angle ECL = \\angle ERM$. Together with the previous result we get that the triangles $ZPL$ and $MRE$ are similar. Using this together with properties of cyclic quadrilaterals and the fact that $BC \\parallel XY$ we get that\n\n$$\n\\angle XYC = \\angle ZCP = \\angle ZLP = \\angle MER = \\angle MCR\n$$\n\nSince $BR \\parallel AC \\parallel QP$ and $QX \\parallel BC$ we get\n\n$$\n\\frac{AP}{PR} = \\frac{AQ}{QB} = \\frac{AX}{XC}\n$$\n\nwhich implies that $XP \\parallel CR$. Thus $\\angle PXC = \\angle RCM = \\angle XYC$. So the result follows.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21618,
"subject": "Mathematics (Olympiad)",
"question": "In the isosceles triangle $ABC$, with $AB = AC$, the angle bisector of $\\angle B$ meets the side $AC$ at $B'$. Suppose that $BB' + B'A = BC$.\n\nFind the angles of the triangle $ABC$.",
"options": [],
"answer": "See solution",
"solution": "On side $BC$ take point $M$ such that $BB' = BM$. We have to prove $AB' = MC$. By the angle bisector theorem\n\n$$\n\\frac{AB'}{B'C} = \\frac{AB}{BC} = \\frac{AC}{BC}, \\text{ implying } \\frac{MC}{B'C} = \\frac{AC}{BC}.\n$$\n\nThus triangles $MCB'$ and $ACB$ are similar, so $MC = MB'$. Moreover, $\\angle C = \\angle MCB' = \\angle MB'C$. Then $\\angle BMB' = 2\\angle C$ and, since triangle $BB'M$ is isosceles, we deduce $\\angle BB'M = 2\\angle C$. In triangle $BB'M$ we have $180^\\circ = 2\\angle C + 2\\angle C + \\frac{1}{2}\\angle C$, so $\\angle C = \\angle B = 40^\\circ$ and $\\angle A = 100^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21619,
"subject": "Mathematics (Olympiad)",
"question": "Let $m$ be a positive integer and let the sequence of positive integers $a_1, a_2, \\dots$ be defined by\n\n$$\na_{n+1} = \\begin{cases} a_n^2 + 2^m & \\text{if } a_n < 2^m \\\\ \\frac{1}{2}a_n & \\text{if } a_n \\ge 2^m \\end{cases}\n$$\n\nfor each positive integer $n$, where $a_1$ is a positive integer.\n\nFind all pairs $(m, a_1)$ such that every term of the sequence is a positive integer.",
"options": [],
"answer": "See solution",
"solution": "Let $a_n = 2^{b_n}c_n$ where $c_n$ is an odd positive integer and $b_n$ is a non-negative integer. We analyze two cases:\n\n**Case 1:** $a_n < 2^m$\n\nFrom the recurrence,\n$$\n2^{b_{n+1}}c_{n+1} = 2^{2b_n}c_n^2 + 2^m.\n$$\n- If $m < 2b_n$, then $c_{n+1} = 2^m c_n^2 + 1 > c_n$.\n- If $m > 2b_n$, then $c_{n+1} = c_n^2 + 2^{2b_n} > c_n$.\n- If $m = 2b_n$, then $c_{n+1} = \\frac{1}{2}(c_n^2 + 1) \\ge c_n$, with equality if and only if $c_n = 1$.\n\n**Case 2:** $a_n \\ge 2^m$\n\nThen $a_{n+1} = \\frac{1}{2}a_n$, so $c_{n+1} = c_n$.\n\nThus, $c_1 \\le c_2 \\le \\cdots$, and since the sequence is bounded, $c_n$ is eventually constant. Since case 1(c) occurs infinitely often, $c_n = 1$ for all $n$. Thus, all $a_n$ are powers of $2$.\n\nNow, (1) applies at least once, so the sequence contains a value greater than $2^m$. Since all terms are powers of $2$, there is $n$ such that $a_n = 2^{m-1}$. Then\n$$\na_{n+1} = 2^{2m-2} + 2^m.\n$$\nThis is a power of $2$ if and only if $m = 2m-2$, so $m = 2$. It is routine to check that $m = 2$ and $a_1 = 2^i$ for any $i \\ge 1$ yields a valid sequence.\n\n**Answer:** $m = 2$ and $a_1 = 2^i$ for any positive integer $i$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21620,
"subject": "Mathematics (Olympiad)",
"question": "On a screen formed by $n \\times n$ ($n \\ge 2$) squares, every square displays initially one of the three colors: red, yellow, and blue. Every second, the screen changes the color of each square following the rules below:\n\n- For each square $A$ whose current color is red, if there is a yellow square sharing a side with this square, then $A$ turns yellow in the next second; otherwise, $A$ remains red.\n- For each square $B$ whose current color is yellow, if there is a blue square sharing a side with this square, then $B$ turns blue in the next second; otherwise, $B$ remains yellow.\n- For each square $C$ whose current color is blue, if there is a red square sharing a side with this square, then $C$ turns red in the next second; otherwise, $C$ remains blue.\n\nProve that if the screen does not change to a single color after $2n - 2$ seconds, then it will never change to a single color.",
"options": [],
"answer": "See solution",
"solution": "We first prove the claim: if the screen eventually turns into one color, say blue, then there must be one square which is constantly blue.\n\nSuppose not, namely, suppose that eventually all squares are blue but every square has changed color at some time. We construct an oriented graph $G$ as follows: the vertices are all squares, and we link an oriented arrow $A \\to B$ if $A$ and $B$ are adjacent and there is some time $t_0$ such that square $A$ is not blue at the time $t_0 - 1$ and is constantly blue when $t \\ge t_0$, and square $B$ is blue at the time $t_0 - 1$. In other words, the square $A$ is changed to blue last time because of square $B$.\n\nSince we assume that every square on the screen has changed color, every vertex of the graph has out-degree $\\ge 1$. So $G$ must contain an oriented loop $A_1 \\to A_2 \\to \\dots \\to A_k \\to A_1$. Assume that square $A_i$ turns blue for the last time at the time $t_i$, i.e., $A_i$ is not blue at the time $t_i - 1$, and is constantly blue when $t \\ge t_i$. Without loss of generality, $t_1$ is the largest among $t_1, \\dots, t_k$.\n\nFrom the arrow $A_1 \\to A_2$, the square $A_1$ is yellow at time $t_1 - 1$; from the arrow $A_k \\to A_1$, the square $A_1$ is blue at the time $t_k - 1$. We must have $t_k < t_1$. But a blue square can only turn red first before turning yellow. So there exists $t_k < t' < t_1$, such that $A_1$ is red at the time $t' - 1$. But $t' - 1 \\ge t_k$, $A_k$ is blue at the time $t' - 1$, so $A_1$ will force $A_k$ to turn red at the time $t'$. But this contradicts the definition of $t_k$, namely, $A_k$ is constantly blue from the time $t_k$ onward. The claim is proved.\n\nBack to the original problem, assume that all squares are turned blue eventually, we will show that the squares are all turned blue after $2n - 2$ seconds.\n\nBy the claim, there exists a square $A$ which is constantly blue. Define the distance between two squares by the sum of their horizontal and vertical distances; then the distance between the farthest two squares is $2n - 2$. If $B$ and $A$ have distance 1, then $B$ can never be red. This means: if $B$ is initially blue, then it is always blue; if $B$ is yellow, then ...",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21621,
"subject": "Mathematics (Olympiad)",
"question": "Let $x$ be the least real number greater than $1$ such that $\\sin x = \\sin(x^2)$, where the arguments are in degrees. What is $x$ rounded up to the closest integer?\n\n(A) 10 (B) 13 (C) 14 (D) 19 (E) 20",
"options": [],
"answer": "See solution",
"solution": "**Answer (B):** The given condition means that either $x$ and $x^2$ represent the same angle or they represent supplementary angles. In the first case, $x^2 = x + 360k$ for some integer $k$. By the quadratic formula,\n\n$$\nx = \\frac{1 \\pm \\sqrt{1 + 1440k}}{2}\n$$\n\nThe least value of $x > 1$ occurs when $k = 1$ and the plus sign is used, in which case\n\n$$\nx = \\frac{1 + \\sqrt{1441}}{2}\n$$\n\nIn the second case, $x^2 + x = 180(2k + 1)$ for some integer $k$. By the quadratic formula,\n\n$$\nx = \\frac{-1 \\pm \\sqrt{1 - 720(2k + 1)}}{2}\n$$\n\nThe least value of $x > 1$ occurs when $k = -1$ and the plus sign is used, in which case\n\n$$\nx = \\frac{-1 + \\sqrt{721}}{2}\n$$\n\nBecause $27^2 = 729 > 721$ and $25^2 = 625 < 721$, the value of $x$ rounds up to $\\frac{-1+27}{2} = 13$, which is smaller than the value of $x$ obtained from the first case.\n\n**Note:** The common value of $\\sin x$ and $\\sin (x^2)$ is approximately $0.22$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21622,
"subject": "Mathematics (Olympiad)",
"question": "Find all triplets of integers $ (a, b, c) $ such that each of the fractions\n\n$$\n\\frac{a}{b+c}, \\quad \\frac{b}{c+a}, \\quad \\frac{c}{a+b}\n$$\n\nis an integer.",
"options": [],
"answer": "See solution",
"solution": "The fractions are symmetric: permuting $ (a, b, c) $ or replacing them with $ (-a, -b, -c) $ gives the same set of fractions (possibly reordered).\n\nSuppose all three fractions are integers. If one of them is zero, say $ a = 0 $, then $ \\frac{b}{c} $ and $ \\frac{c}{b} $ must both be integers. Thus, $ b $ and $ c $ are nonzero and $ |b| = |c| $, so $ c = \\pm b $. But $ b = -c $ would make the denominator $ b + c = 0 $, which is not allowed. Therefore, $ c = b $. The admissible triplets in this case are $ (0, c, c) $ and their permutations, for any nonzero integer $ c $.\n\nNow consider $ abc \\ne 0 $. By symmetry, assume at least two of $ a, b, c $ are positive. If all three are positive, the smallest numerator over the sum of the other two is between $ 0 $ and $ 1 $, so cannot be integer.\n\nAssume $ a, b > 0 $ and $ c = -d < 0 $ ($ d > 0 $). The fractions become:\n\n$$\n\\frac{a}{d-b}, \\quad \\frac{b}{d-a}, \\quad \\frac{d}{a+b}\n$$\n\nThe last fraction implies $ d \\ge a + b $. The first requires $ a \\ge d - b $, so $ d \\le a + b $. Thus, $ d = a + b $, so $ c = -a - b $. Therefore, all nonzero integer triplets with $ a + b + c = 0 $ are solutions. For these, each fraction equals $ -1 $.\n\n**Final answer:**\n- All permutations of $ (0, c, c) $ for nonzero integer $ c $.\n- All nonzero integer triplets $ (a, b, c) $ with $ a + b + c = 0 $.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21623,
"subject": "Mathematics (Olympiad)",
"question": "Let $m, n$ be integers greater than $1$, and let $a_1, a_2, \\dots, a_m$ be positive integers not greater than $n^m$. Prove that there exist positive integers $b_1, b_2, \\dots, b_m$ not greater than $n$ such that\n\n$$\n\\gcd(a_1 + b_1, a_2 + b_2, \\dots, a_m + b_m) < n,\n$$\n\nwhere $\\gcd(x_1, x_2, \\dots, x_m)$ denotes the greatest common divisor of $x_1, x_2, \\dots, x_m$.",
"options": [],
"answer": "See solution",
"solution": "Suppose without loss of generality that $a_1$ is the smallest of the $a_i$. If $a_1 \\geq n^m - 1$, then the problem is simple: either all the $a_i$ are equal, or $a_1 = n^m - 1$ and $a_j = n^m$ for some $j$. In the first case, we can take (say) $b_1 = 1$, $b_2 = 2$, and the rest of the $b_i$ can be arbitrary, and we have\n\n$$\n\\gcd(a_1 + b_1, a_2 + b_2, \\dots, a_m + b_m) \\leq \\gcd(a_1 + b_1, a_2 + b_2) = 1.\n$$\n\nIn the second case, we can take $b_1 = 1$, $b_j = 1$, and the rest of the $b_i$ arbitrary, and again\n\n$$\n\\gcd(a_1 + b_1, a_2 + b_2, \\dots, a_m + b_m) \\leq \\gcd(a_1 + b_1, a_j + b_j) = 1.\n$$\n\nSo from now on we can suppose that $a_1 \\leq n^m - 2$.\n\nNow, let us suppose the desired $b_1, \\dots, b_m$ do not exist, and seek a contradiction. Then, for any choice of $b_1, b_2, \\dots, b_m \\in \\{1, 2, \\dots, n\\}$, we have",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21624,
"subject": "Mathematics (Olympiad)",
"question": "Three rectangles are placed on a plane, and any two rectangles have their edges parallel to each other. They divide the plane into several parts. Determine the maximum possible number of such parts. (We consider the area that is contained in no rectangles as one part. For example, if we have only one rectangle, the plane is divided into two parts.)",
"options": [],
"answer": "See solution",
"solution": "We will show that for $n$ rectangles, the answer is $2n^2 - 2n + 2$. We call each divided part a region.\n\nLet $n$ rectangles $R_{1,n}, \\dots, R_{i,n+1-i}, \\dots, R_{n,1}$ divide the plane into $2n^2 - 2n + 2$ regions, where $R_{x,y}$ is a rectangle with corners at $(x, y)$, $(x, -y)$, $(-x, y)$, $(-x, -y)$. Therefore, we only have to show that $n$ rectangles cannot divide the plane into more than $2n^2 - 2n + 2$ regions.\n\n**Induction:**\n\nThe case $n=1$ is trivial. Suppose that the case $n=k$ is true. Let $R_1, \\dots, R_{k+1}$ be $k+1$ rectangles with the property in the problem, and $D_1, \\dots, D_m$ be the regions divided by $R_1, \\dots, R_k$. By the induction hypothesis, we have $m \\le 2k^2 - 2k + 2$.\n\nLet $C_i$ be the border of $R_i$. Suppose that for $i \\ne j$, $C_i$ and $C_j$ share finitely many points. (The proof goes the same way if this is not satisfied.) Let $P_1, \\dots, P_l$ be the intersections of $C_{k+1}$ and $C_1, \\dots, C_n$, ordered along $C_{k+1}$. $l \\le 4k$ since $C_i$ and $C_{k+1}$ cross at 4 or fewer points. Add polygonal lines $P_1P_2, \\dots, P_iP_{i+1}, \\dots, P_lP_1$ one by one to $D_1, \\dots, D_m$. One line increases the number of regions by at most 1, so adding all polygonal lines increases the number of regions by $l$ or less.\n\nSince adding $P_1P_2, \\dots, P_iP_{i+1}, \\dots, P_lP_1$ is equivalent to adding $C_{k+1}$, the number of regions divided by $R_1, \\dots, R_{k+1}$ is\n\n$$\nm + l \\le (2k^2 - 2k + 2) + 4k = 2(k+1)^2 - 2(k+1) + 2.\n$$\n\nSo we have completed the induction. Therefore, for $n=3$, the answer is $2 \\times 3^2 - 2 \\times 3 + 2 = 14$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21625,
"subject": "Mathematics (Olympiad)",
"question": "Call a row of a matrix in $M_n(\\mathbb{C})$ *permutable* if, for any permutation of its entries, the value of the determinant does not change. Prove that any matrix that has two *permutable* rows is singular.",
"options": [],
"answer": "See solution",
"solution": "Consider $A \\in M_n(\\mathbb{C})$ such that row $l$ is *permutable*. Denote by $\\Gamma_{li}$ the algebraic complement of $a_{li}$, $i = 1, \\dots, n$. Suppose $i, j, k, p \\in \\{1, 2, \\dots, n\\}$ are such that $a_{li} \\neq a_{lj}$ and $\\Gamma_{lk} \\neq \\Gamma_{lp}$. Consider matrices $B$ and $C$ obtained from $A$ by permuting elements in row $l$, such that $B$ contains $a_{li}$ in place $(l, k)$ and $a_{lj}$ in place $(l, p)$, and $C$ contains $a_{lj}$ in place $(l, k)$ and $a_{li}$ in place $(l, p)$; the two matrices having in the other positions the initial elements.\n\nDeveloping determinants along row $l$, we get\n$$\n\\det(B) - \\det(C) = (a_{li} - a_{lj})(\\Gamma_{lk} - \\Gamma_{lp}) \\neq 0,\n$$\nin contradiction with the hypothesis.\n\nWe deduce that row $l$ of $A$ is *permutable* if and only if all its elements are equal or all algebraic complements of elements in line $l$ are equal.\n\nConsider now a matrix $A$ that admits two *permutable* rows. Then:\n\n- If both rows are constant, $\\det(A) = 0$.\n- If both rows have constant algebraic complements, then $A^*$ has two constant rows; that is, $\\det(A^*) = 0$, which implies $\\det(A) = 0$.\n- If one row has all elements equal to $a$, and another row has all algebraic complements equal to $b$, then from $AA^* = \\det(A)I_n$ we deduce $ab = 0$, so $a = 0$ or $b = 0$.\n\nThus, in every case, $\\det(A) = 0$.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 21626,
"subject": "Mathematics (Olympiad)",
"question": "Assume that $c \\ge 0$ is such that\n\n$$\nx f(x + y^2) = f(y^2 f(x) + y + c)\n$$\n\nholds for any $x, y \\in \\mathbb{R}^+$. Prove that there are no such non-negative real numbers $c$ with this property.",
"options": [],
"answer": "See solution",
"solution": "Note that letting $y$ tend to $0$, the left-hand side tends to $-\\sqrt{x} f(z) < 0 < 1$. On the other hand, letting $y$ tend to infinity, since $f(x+z) > f(z)$, the expression on the left-hand side tends to $\\infty$. Therefore, the last equation has a positive root $y_0 \\in (0, \\infty)$ and thus:\n\n$$\nf(y_0 f(x+z) + \\sqrt{y_0} + c) = f((x+y_0) f(z) + \\sqrt{x+y_0} + c).\n$$\n\nBut now by (1) with $y = y_0$ we obtain that:\n\n$$\nf((y_0 + z) f(x) + \\sqrt{y_0 + z} + c) = 0\n$$\n\nwhich is a contradiction. Therefore, no $x, z \\in \\mathbb{R}^+$ exist with the property $f(x+z) > f(x)$.\n\n**Step 2.** Now we prove that $f$ is constant on some interval $[x_0, \\infty)$. First, note that if $a > b$ and $f(a) = f(b)$, then, since $f$ is non-increasing, $f(x) = f(a)$ for every $x \\in [b, a]$. Now by $P(1, \\sqrt{y})$ we have:\n\n$$\nf(y+1) = f(y f(1) + \\sqrt{y} + c).\n$$\n\nNext, we make a case distinction with respect to $f(1)$:\n\n- $f(1) \\ge 1$. Then for $y \\ge 4$, we have that $y f(1) + \\sqrt{y} + c \\ge y + 2$. Therefore, by the above remark, we have that $f(y+1) = f(y+2)$ for any $y \\ge 4$. Consequently, $f(y)$ is constant on $[5, \\infty)$.\n\n- $f(1) < 1$. Thus, for $y$ large enough, $y f(1) + \\sqrt{y} < y$ and therefore, again by the remark above, $f(y) = f(y+1)$. Consequently, there is $x_0$ such that on $[x_0, \\infty)$, $f$ is constant.\n\nNow we arrive at a contradiction. Let $x > \\max(x_0, 1)$ and $y > x_0$. Then $x + y^2 > x_0$ and $y^2 f(x) + y + c > x_0$. Consequently:\n\n$$\nf(x + y^2) = f(y^2 f(x) + y + c).\n$$\n\nOn the other hand, by $P(x, y)$, we have:\n\n$$\nx f(x + y^2) = f(y^2 f(x) + y + c).\n$$\n\nFor $x > 1$, by the positivity of $f$, this is impossible. $\\square$\n\n---\n\n**Alternative Solution.**\n\n1. **Finding periods.** Setting $x = 1$ and $u = y^2$, we get:\n\n$$\nf(u + 1) = f(f(1) u + \\sqrt{u} + c).\n$$\n\nThis shows that for any $u > 0$ there are $x_1$ and $x_2$ such that $f(x_1) = f(x_2)$ and:\n\n$$\nx_2 - x_1 = |(f(1) - 1) u + \\sqrt{u} + c - 1| =: T(u).\n$$\n\nFurthermore, $x_1 = f(1) u + \\sqrt{u} + c$ if $f(1) u + \\sqrt{u} + c < u + 1$ and $x_1 = u + 1$, otherwise.\n\n2. **Some structure of the periods.** Whenever $f(x_1) = f(x_2)$, $T = x_2 - x_1$ and $u > x_1$, we can write $u = x_1 + y^2$ for an appropriate $y > 0$. Therefore:\n\n$$\nx_1 f(u) = f(y^2 f(x_1) + y + c), \\quad x_2 f(u + T) = f(y^2 f(x_2) + y + c).\n$$\n\nTaking into account that $f(x_1) = f(x_2)$, the right-hand sides are equal, so:\n\n$$\nx_1 f(u) = x_2 f(u + T)\n$$\nholds for any $u > x_1$. Therefore, for $u > x_1$ and $x_2 = x_1 + T$:\n\n$$\n\\frac{x_1 + T}{x_1} = \\frac{f(u + T)}{f(u)}, \\quad \\left(\\frac{x_1 + T}{x_1}\\right)^2 = \\frac{f(u + 2T)}{f(u)}.\n$$\n\n3. **Combining steps 1 and 2.** Let $t$ be large enough. Since $T(u) = |(f(1) - 1) u + \\sqrt{u} + c - 1|$ tends to infinity as $u$ does, there is $u = U(t)$ such that $T(u) = t$. Let $x_1(t) = U(t) + 1$ if $U(t) + 1 < f(1) U(t) + \\sqrt{U(t)} + c$ and $x_1(t) = U(t) + 1 - t$, otherwise. Then:\n\n$$\n\\left(1 + \\frac{t}{x_1(t)}\\right)^2 = \\frac{f(y + 2t)}{f(y)} = 1 + \\frac{2t}{x_1(2t)},\n$$\n\nwhere $y > \\max(x_1(t) + t, x_1(2t) + 2t)$. Consequently:\n\n$$\n\\frac{2t}{x_1(t)} + \\frac{t^2}{x_1^2(t)} = \\frac{2t}{x_1(2t)}\n$$\n\nor equivalently,\n\n$$\nt x_1(2t) = 2 x_1(t) (x_1(t) - x_1(2t)).\n$$\n\nYet, $x_1(t)$ tends to infinity as $t$ does. It follows that for $t$ large enough, $x_1(t) < x_1(2t)$. This is a contradiction since the left-hand side is positive as is $2 x_1(t)$.\n\nConsequently, there are no non-negative real numbers $c$ with the desired property.\n\n\n\n**Comment.** The original solution and the second alternative solution share the same ideas but are organised differently. This suggests that the structure of the problem can be revealed from different angles and the pieces of the puzzle can be assembled differently to yield a complete solution.\n\nThe first alternative solution shows that the structure of the problem is much more regular and can be revealed by systematically studying the behaviour of the quadratic function $f(x) y^2 + y + c = y^2 + x$ with respect to $y$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21627,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $a$ and $b$ are positive real numbers satisfying $\\frac{1}{a} + \\frac{1}{b} \\le 2\\sqrt{2}$ and $(a-b)^2 = 4(ab)^3$. Then $\\log_a b = \\underline{\\hspace{2cm}}$.",
"options": [],
"answer": "See solution",
"solution": "From $\\frac{1}{a} + \\frac{1}{b} \\le 2\\sqrt{2}$, we have $a + b \\le 2\\sqrt{2}ab$. On the other hand,\n\n$$\n(a + b)^2 = 4ab + (a - b)^2 = 4ab + 4(ab)^3\n$$\n\nand that means\n\n$$\na + b \\ge 2\\sqrt{2}ab.\n$$\n\nTherefore,\n\n$$\na + b = 2\\sqrt{2}ab.\n$$\n\nThe equality holds only when $ab = 1$. Associating it with the previous result, we find\n\n$$\n\\begin{cases} a = \\sqrt{2} - 1, \\\\ b = \\sqrt{2} + 1, \\end{cases} \\quad \\text{and} \\quad \\begin{cases} a = \\sqrt{2} + 1, \\\\ b = \\sqrt{2} - 1. \\end{cases}\n$$\n\nSo the answer is $\\log_a b = -1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21628,
"subject": "Mathematics (Olympiad)",
"question": "Find all $n \\in \\mathbb{N}$ that can be represented in the form\n$$\nn = [a,b] + [b,c] + [c,a]\n$$\nwith $a, b, c \\in \\mathbb{N}$. Here, $[u,v]$ denotes the least common multiple of $u$ and $v$.",
"options": [],
"answer": "See solution",
"solution": "All $n \\in \\mathbb{N}$ are representable except the powers of $2$.\n\nLet $f(a,b,c) = [a,b] + [b,c] + [c,a]$. For any $k \\in \\mathbb{N}$, set $a = k$, $b = c = 1$ to get $f(k,1,1) = 2k + 1$. Thus, all odd $n \\geq 3$ are representable.\n\nIf $n$ is representable, so is $2n$, because $[2u, 2v] = 2[u, v]$ implies $f(2a, 2b, 2c) = 2f(a, b, c)$. Therefore, every $n \\geq 3$ with an odd divisor greater than $1$ is representable.\n\nNow consider powers of $2$, i.e., $n = 2^k$ for $k \\geq 0$. For $k = 0, 1$, $f(a,b,c) \\geq 3$ for all $a, b, c \\in \\mathbb{N}$, so $1$ and $2$ are not representable. Suppose $f(a,b,c) = 2^k$ for $k \\geq 2$ and $k$ minimal. At least two among $a, b, c$ must be even; otherwise, $f(a,b,c)$ is odd. If all are even, dividing by $2$ gives $f(\\frac{a}{2}, \\frac{b}{2}, \\frac{c}{2}) = 2^{k-1}$, contradicting minimality. If two are even (say $a, b$) and $c$ is odd, then $[a,b] = 2[\\frac{a}{2}, \\frac{b}{2}]$, $[a,c] = 2[\\frac{a}{2}, c]$, $[b,c] = 2[\\frac{b}{2}, c]$, so\n$$\nf\\left(\\frac{a}{2}, \\frac{b}{2}, c\\right) = \\frac{1}{2} f(a, b, c) = 2^{k-1},\n$$\nagain contradicting minimality. Thus, powers of $2$ are not representable.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21629,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ and $b$ be positive integers and $c$ be a positive real number satisfying\n$$\n\\frac{a+1}{b+c} = \\frac{b}{a}.\n$$\nProve that $c \\ge 1$ holds.",
"options": [],
"answer": "See solution",
"solution": "The given equation is equivalent to:\n$$\n\\begin{aligned}\na^2 + a &= b^2 + b c \\\\\n4a^2 + 4a + 1 &= 4b^2 + 4b c + 1 \\\\\n(2a + 1)^2 &= 4b^2 + 4b c + 1.\n\\end{aligned}\n$$\n\nAssume, for contradiction, that $c < 1$. Then\n$$\n(2b)^2 = 4b^2 < (2a + 1)^2 = 4b^2 + 4b c + 1 < 4b^2 + 4b + 1 = (2b + 1)^2.\n$$\nThis is impossible, since the square of an integer cannot lie strictly between two consecutive perfect squares. Therefore, $c \\ge 1$ holds. For example, if $a = b$, then $c = 1$, so there is a solution with $c \\ge 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21630,
"subject": "Mathematics (Olympiad)",
"question": "Does there exist a figure on the squared plane that can be split into dominoes in exactly 17 ways?",
"options": [],
"answer": "See solution",
"solution": "Yes, such a figure exists. It is shown in the picture below:\n\n\n\nIf cell A belongs to the horizontal domino, then we have the following forced configuration:\n\n\n\nEach of the $3 \\times 2$ boxes can be split into dominoes in 3 ways, so there are $3 \\times 3 = 9$ ways in this case.\n\nIf cell A belongs to a vertical domino, then we have the following forced configuration:\n\n\n\nEach of the $2 \\times 2$ boxes can be split into dominoes in 2 ways, so there are $2 \\times 2 \\times 2 = 8$ ways in this case.\n\nThus, the total number of ways is $9 + 8 = 17$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21631,
"subject": "Mathematics (Olympiad)",
"question": "Consider the sequence $\\{a_n\\}_{n \\ge 1}$ of positive integers with $a_n a_{n+3} = a_{n+2} a_{n+5}$ for all positive integers $n$. Determine the largest integer that always divides $\\sum_{k=1}^{2550} a_{2k} a_{2k-1}$.",
"options": [],
"answer": "See solution",
"solution": "From $a_n a_{n+3} = a_{n+2} a_{n+5}$ for every positive integer $n$, we have $a_{n+1} a_{n+4} = a_{n+3} a_{n+6}$ and $a_{n+2} a_{n+5} = a_{n+4} a_{n+7}$. Then\n\n$$\na_n a_{n+3} \\cdot a_{n+1} a_{n+4} \\cdot a_{n+2} a_{n+5} = a_{n+2} a_{n+5} \\cdot a_{n+3} a_{n+6} \\cdot a_{n+4} a_{n+7}.\n$$\n\nTherefore, $a_n a_{n+1} = a_{n+6} a_{n+7}$ for every positive integer $n$. Thus,\n\n$$\n\\sum_{k=1}^{2550} a_{2k} a_{2k-1} = \\frac{2550}{3} (a_1 a_2 + a_3 a_4 + a_5 a_6) = 850 (a_1 a_2 + a_3 a_4 + a_5 a_6)\n$$\n\nWe will show that 850 is the largest positive integer that always divides $\\sum_{k=1}^{2550} a_{2k} a_{2k-1}$. Consider the sequence $\\{a_n\\}_{n \\ge 1}$ defined by\n\n$$\na_n = \\begin{cases} 1 & \\text{if } n \\equiv 1, 2, 3 \\pmod{6}, \\\\ 2 & \\text{if } n \\equiv 4, 5, 6 \\pmod{6}. \\end{cases}\n$$\n\nIt can be seen that $a_n a_{n+3} = 2 = a_{n+2} a_{n+5}$ for every positive integer $n$; so, it satisfies the condition in the problem and\n\n$$\n\\sum_{k=1}^{2550} a_{2k} a_{2k-1} = 850(1 \\cdot 1 + 1 \\cdot 2 + 2 \\cdot 2) = 850 \\cdot 7.\n$$\n\nConsider another sequence $\\{a_n\\}_{n \\ge 1}$ defined by $a_n = 1$ for every positive integer $n$. We can see that $a_n a_{n+3} = 1 = a_{n+2} a_{n+5}$ for every positive integer $n$, and\n\n$$\n\\sum_{k=1}^{2550} a_{2k} a_{2k-1} = 850(1 \\cdot 1 + 1 \\cdot 1 + 1 \\cdot 1) = 850 \\cdot 3.\n$$\n\nThus, 850 is the desired largest integer.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21632,
"subject": "Mathematics (Olympiad)",
"question": "Find all integers $n \\ge 2$ for which there exists an integer $m$ and a polynomial $P(x)$ with integer coefficients satisfying the following three conditions:\n\n- $m > 1$ and $\\gcd(m, n) = 1$;\n- the numbers $P(0), P^2(0), \\dots, P^{m-1}(0)$ are not divisible by $n$; and\n- $P^m(0)$ is divisible by $n$.\n\nHere $P^k$ means $P$ applied $k$ times, so $P^1(0) = P(0)$, $P^2(0) = P(P(0))$, etc.",
"options": [],
"answer": "See solution",
"solution": "The answer is that this is possible if and only if there exist primes $p' < p$ such that $p \\mid n$ and $p' \\nmid n$. (Equivalently, the radical $\\operatorname{rad}(n)$ must not be the product of the first several primes.)\n\nFor a polynomial $P$ and an integer $N$, we introduce the notation\n\n$$\n\\mathbf{zord}(P \\bmod N) \\stackrel{\\text{def}}{=} \\min\\{e > 0 \\mid P^e(0) \\equiv 0 \\bmod N\\}\n$$\n\nwhere we set $\\min \\emptyset = 0$ by convention. Note that in general we have\n\n$$\n\\mathbf{zord}(P \\bmod N) = \\operatorname{lcm}_{q \\mid N} (\\mathbf{zord}(P \\bmod q)) \\quad (\\dagger)\n$$\n\nwhere the index runs over all prime powers $q \\mid N$ (by the Chinese remainder theorem). This will be used in both directions.\n\n**Construction:**\n\nFirst, we begin by giving a construction. The idea is to first use the following prime power case.\n\n**Claim** — Let $p^e$ be a prime power, and $1 \\le k < p$. Then\n\n$$\nf(X) = X + 1 - k \\cdot \\frac{X(X-1)(X-2)\\dots(X-(k-2))}{(k-1)!}\n$$\n\nviewed as a polynomial in $(\\mathbb{Z}/p^e)[X]$ satisfies $\\mathbf{zord}(f \\bmod p^e) = k$.\n\n*Proof.* Note $f(0) = 1$, $f(1) = 2$, ..., $f(k-2) = k-1$, $f(k-1) = 0$ as needed. $\\square$\n\n\n\nThis gives us a way to do the construction now. For the prime power $p^e \\mid n$, we choose $1 \\le p' < p$ and require $\\mathbf{zord}(P \\bmod p^e) = p'$ and $\\mathbf{zord}(P \\bmod q) = 1$ for every other prime power $q$ dividing $n$. Then by $(\\dagger)$ we are done.\n\n**Remark.** The claim can be viewed as a special case of Lagrange interpolation adapted to work over $\\mathbb{Z}/p^e$ rather than $\\mathbb{Z}/p$. Thus the construction of the polynomial $f$ above is quite natural.\n\n**Necessity:**\n\nBy $(\\dagger)$ again, it will be sufficient to prove the following claim.\n\n**Claim** — For any prime power $q = p^e$, and any polynomial $f(x) \\in \\mathbb{Z}[x]$, if the quantity $\\mathbf{zord}(f \\bmod q)$ is nonzero then it has all prime factors at most $p$.\n\n*Proof.* This is by induction on $e \\ge 1$. For $e = 1$, the pigeonhole principle immediately implies that $\\mathbf{zord}(P \\bmod p) \\le p$.\n\nNow assume $e \\ge 2$. Let us define\n\n$$\nk \\stackrel{\\text{def}}{=} \\mathbf{zord}(P \\bmod p^{e-1}), \\quad Q \\stackrel{\\text{def}}{=} P^k.\n$$\n\nSince being periodic modulo $p^e$ requires periodic modulo $p^{e-1}$, it follows that\n\n$$\n\\mathbf{zord}(P \\bmod p^e) = k \\cdot \\mathbf{zord}(Q \\bmod p^e).\n$$\n\nHowever, since $Q(0) \\equiv 0 \\bmod{p^{e-1}}$, it follows $\\{Q(0), Q^2(0), \\dots \\}$ are actually all multiples of $p^{e-1}$. There are only $p$ residues modulo $p^e$ which are also multiples of $p^{e-1}$, so $\\mathbf{zord}(Q \\bmod p^e) \\le p$, as needed. $\\square$\n\n**Remark.**\n\nThis is one of these problems where you can make many useful natural observations, and if you make enough of them eventually they cohere into a solution. For example, here are some things I noticed while solving:\n\n- The polynomial $1 - x$ shows that $m = 2$ works for any odd $n$.\n- In general, if $\\zeta$ is a primitive $m$th root of unity modulo $n$, then $\\zeta(x+1) - 1$ has the desired property (assuming $\\gcd(m, n) = 1$). We can extend this using the Chinese remainder theorem to find that if $p \\mid n, m \\mid p-1$, and $\\gcd(m, n) = 1$, then $n$ works. So by this point I already have something about the prime factors of $n$ being sort-of closed downwards.\n- By iterating $P$ we see it is enough to consider $m$ prime.\n- In the case where $n = 2^k$, it is not too difficult to show that no odd prime $m$ works.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21633,
"subject": "Mathematics (Olympiad)",
"question": "For $0 \\le x, y \\le 1$, let\n\n$$\nf(x, y) = x y^2 \\sqrt{1 - x^2} - x^2 y \\sqrt{1 - y^2}\n$$\n\nFind the minimum constant $c$ for which the following condition holds:\n\nFor any integer $n > 1$ and any real numbers $a_1, a_2, \\dots, a_n$ such that $0 \\le a_1 < a_2 < \\dots < a_n \\le 1$,\n\n$$\nf(a_1, a_2) + f(a_2, a_3) + \\dots + f(a_{n-1}, a_n) < c\n$$",
"options": [],
"answer": "See solution",
"solution": "$$\\frac{\\pi}{4}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21634,
"subject": "Mathematics (Olympiad)",
"question": "Given an integer $n > 0$, prove that at least one of the numbers $[2^n\\sqrt{2}]$, $[2^{n+1}\\sqrt{2}]$, ..., $[2^{2n}\\sqrt{2}]$ is even ($[a]$ denotes the integral part of the real number $a$).",
"options": [],
"answer": "See solution",
"solution": "Suppose, by contradiction, that all the numbers are odd. Then there exists a positive integer $a$ such that\n\n$$\n2a - 1 < 2^n \\sqrt{2} < 2a. \\quad (1)\n$$\n\nIt follows that $4a - 2 < 2^{n+1}\\sqrt{2} < 4a$ and, since $[2^{n+1}\\sqrt{2}]$ is odd,\n\n$$\n4a - 1 < 2^{n+1}\\sqrt{2} < 4a.\n$$\n\nContinuing in the same way we arrive at $2^{n+1}a - 1 < 2^{2n}\\sqrt{2} < 2^{n+1}a$, whence $2^{n+1}a - 2^{2n}\\sqrt{2} < 1$, which leads to\n\n$$\n2^{n+1}(a^2 - 2^{2n-1}) < a + 2^{n-1}\\sqrt{2}. \\quad (2)\n$$\n\nOn the other hand, $2a > 2^n\\sqrt{2}$ gives $a^2 > 2^{2n-1}$, implying $a^2 - 2^{2n-1} \\ge 1$. Thus (2) leads to\n\n$$\na > 2^{n+1} - 2^{n-1}\\sqrt{2} = 2^{n-1}(4 - \\sqrt{2}) > 2^{n-1}(\\sqrt{2} + 1) > 2^{n-1}\\sqrt{2} + 0.5,\n$$\n\nwhich contradicts (1).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21635,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\geq 3$ be an integer, and consider a circle with $n+1$ equally spaced points marked on it. Consider all labellings of these points with the numbers $0, 1, \\dots, n$ such that each label is used exactly once; two such labellings are considered to be the same if one can be obtained from the other by a rotation of the circle. A labeling is called *beautiful* if, for any four labels $a < b < c < d$ with $a + d = b + c$, 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 $x + y \\leq n$ and $\\gcd(x, y) = 1$. Prove that\n\n$$M = N + 1.$$",
"options": [],
"answer": "See solution",
"solution": "We will prove that there are $N + 1$ beautiful labellings for all $n \\geq 2$.\n\nLet $0 < x < 1$ be a real number. Define the beautiful labelling $C_n(x)$ as follows. For $0 \\leq k \\leq n$, 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 $0 \\leq a < b < c < d \\leq n$ 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 $q_1, q_2 \\leq n-1$, 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 $A \\neq C_n(x)$, 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 $\\frac{p_2}{q_2} - \\frac{p_1}{q_1} \\leq \\frac{1}{n-1}$, 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 $1 \\leq i \\leq q$ 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": 21636,
"subject": "Mathematics (Olympiad)",
"question": "Let $s_1, s_2, \\ldots$ and $t_1, t_2, \\ldots$ be two nonconstant sequences of rational numbers. Suppose that for all $i, j$, the product $(s_i - s_j)(t_i - t_j)$ is an integer. Prove that there exists a rational number $r$ such that $s_i r$ and $t_i / r$ are integers for all $i$.",
"options": [],
"answer": "See solution",
"solution": "For $p$ a prime, define the $p$-adic norm $\\| \\cdot \\|_p$ on rational numbers as follows: for $r \\neq 0$, $\\|r\\|_p$ is the unique integer $n$ for which we can write $r = p^n a/b$ with $a, b$ integers not divisible by $p$. (By convention, $\\|0\\|_p = +\\infty$.) We will repeatedly use the well-known (or easy to prove) fact that for any rational numbers $r_1, r_2$, we have $\\|r_1 \\pm r_2\\|_p \\ge \\min(\\|r_1\\|_p, \\|r_2\\|_p)$, with equality whenever $\\|r_1\\|_p \\neq \\|r_2\\|_p$. The condition of the problem implies that\n\n$$\n\\|s_i - s_j\\|_p \\geq -\\|t_i - t_j\\|_p\n$$\n\nfor all $i, j$ and all prime $p$.\n\nWe claim in fact that\n\n$$\n\\|s_i - s_j\\|_p \\geq -\\|t_k - t_l\\|_p\n$$\n\nfor all $i, j, k, l$ and all prime $p$. Suppose otherwise; then there exist $i, j, k, l, p$ for which $\\|s_i - s_j\\|_p < -\\|t_k - t_l\\|_p$. Since $\\|s_i - s_j\\|_p = \\|(s_i - s_k) - (s_j - s_k)\\|_p \\ge \\min(\\|s_i - s_k\\|_p, \\|s_j - s_k\\|_p)$, at least one of $\\|s_i - s_k\\|_p$ and $\\|s_j - s_k\\|_p$, say the former, is strictly less than $-\\|t_k - t_l\\|_p$. By the previous inequality, it follows that $\\|t_i - t_k\\|_p > \\|t_k - t_l\\|_p$, and thus $\\|t_i - t_l\\|_p = \\|(t_i - t_k) + (t_k - t_l)\\|_p = \\|t_k - t_l\\|_p$. Then by the previous inequality again, $\\|s_i - s_l\\|_p \\ge -\\|t_k - t_l\\|_p$ and $\\|s_k - s_l\\|_p \\ge -\\|t_k - t_l\\|_p$, whence $\\|s_i - s_k\\|_p = \\|(s_i - s_l) - (s_k - s_l)\\|_p \\ge -\\|t_k - t_l\\|_p$, contradicting the assumption that $\\|s_i - s_k\\|_p < -\\|t_k - t_l\\|_p$. This proves the claim.\n\nNow for each prime $p$, define the integer $f(p) = \\min_{i,j} \\|s_i - s_j\\|_p$. Choose $i_0, j_0, k_0, l_0$ such that $s_{i_0} \\neq s_{j_0}$ and $t_{k_0} \\neq t_{l_0}$; then $f(p)$ exists since it is bounded below by $-\\|t_{k_0} - t_{l_0}\\|_p$ (by the claim) and above by $\\|s_{i_0} - s_{j_0}\\|_p$. Moreover, if $p$ does not divide the numerator or denominator of either $s_{i_0} - s_{j_0}$ or $t_{k_0} - t_{l_0}$, then $\\|s_{i_0} - s_{j_0}\\|_p = \\|t_{k_0} - t_{l_0}\\|_p = 0$ and thus $f(p) = 0$. It follows that $f(p) = 0$ for all but finitely many primes.\n\nWe can now define $r = \\prod_p p^{-f(p)}$, where the product is over all primes. For any $i, j$, we have $\\|s_i - s_j\\|_p \\ge f(p)$ for all $p$ by construction, and so $(s_i - s_j)r$ is an integer. On the other hand, for any $k, l$ and any prime $p$, $\\|t_k - t_l\\|_p \\ge -\\|s_i - s_j\\|_p$ for all $i, j$ by the claim, and so $\\|t_k - t_l\\|_p \\ge -f(p)$. It follows that $(t_k - t_l)/r$ is an integer for all $k, l$, whence $r$ is the desired rational number.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21637,
"subject": "Mathematics (Olympiad)",
"question": "Let $A$ be a set with $n \\ge 2$ elements and let $f_1, \\dots, f_k : A \\to \\mathbb{R}$ be functions. Prove that there exist functions $g_1, \\dots, g_{n-1} : A \\to \\mathbb{R}$ such that\n\n$$\n\\max_{1 \\le i \\le k} |f_i(x) - f_i(y)| = \\max_{1 \\le i \\le n-1} |g_i(x) - g_i(y)|\n$$\n\nfor any $x, y \\in A$.",
"options": [],
"answer": "See solution",
"solution": "Let $A = \\{a_1, a_2, \\dots, a_n\\}$ and set\n\n$$\nd(x, y) = \\max_{1 \\le i \\le k} |f_i(x) - f_i(y)|, \\quad g_i(x) = d(x, a_i) - d(a_i, a_i), \\quad 1 \\le i \\le n-1\n$$\n\nand $d'(x, y) = \\max_{1 \\le i \\le n-1} |g_i(x) - g_i(y)|$. Since\n\n$$\n|g_i(x) - g_i(y)| = |d(x, a_i) - d(y, a_i)| \\le d(x, y)\n$$\n\nthen $d'(x, y) \\le d(x, y)$. On the other hand,\n\n$$\nd(a_j, a_i) = g_i(a_j) - g_i(a_i) \\le d'(a_j, a_i), \\quad 1 \\le i \\le n-1,\\ 1 \\le j \\le n\n$$\n\n$$\nd(a_n, a_n) = 0 = d'(a_n, a_n) \\text{ and hence } d'(x, y) = d(x, y).\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21638,
"subject": "Mathematics (Olympiad)",
"question": "Let $I$ be the incenter of a triangle $ABC$. Let $F$ and $G$ be the feet of the perpendiculars drawn from $A$ to the lines $BI$ and $CI$, respectively. Rays $AF$ and $AG$ intersect the circumcircles of the triangles $CFI$ and $BGI$ a second time at points $K$ and $L$, respectively. Prove that line $AI$ bisects the segment $KL$.",
"options": [],
"answer": "See solution",
"solution": "Since $\\angle IFK = 90^\\circ$, $IK$ is the diameter of the circumcircle of $CFI$, so $\\angle ICK = 90^\\circ$. Similarly, $IL$ is the diameter of the circumcircle of $BGI$ and $\\angle IBL = 90^\\circ$. Therefore, the lines $CK$ and $GL$ are parallel, and $BL$ and $FK$ are also parallel.\n\nLet the lines $CK$ and $BL$ intersect at $D$ (see the figure below). From the above, $DKAL$ is a parallelogram. Note that $D$ is the excenter with respect to vertex $A$ of triangle $ABC$, since the lines $BL$ and $CK$ are perpendicular to the corresponding internal angle bisectors. The excenter lies on the internal angle bisector $AI$, hence $AI$ bisects the diagonal $KL$.\n\n\n\nFigure 17",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21639,
"subject": "Mathematics (Olympiad)",
"question": "We have two integers consisting of two digits, and both numbers do not start with a 0. If you add these numbers, you get the number $S$. If you interchange the two digits of both numbers and add the new numbers, you get $4S$.\n\nDetermine all possible pairs of two-digit numbers satisfying these constraints. Make sure to clearly indicate in your answer which numbers form a pair.",
"options": [],
"answer": "See solution",
"solution": "$\\{14, 19\\}$, $\\{15, 18\\}$, and $\\{16, 17\\}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21640,
"subject": "Mathematics (Olympiad)",
"question": "a) Let $(I)$ touch $BC$ at $D$. Prove that\n$$\n\\frac{DB}{DC} = \\sqrt{\\frac{\\cot B}{\\cot C}}.\n$$\n\nb) Suppose that $(I)$ intersects $BC$ at $M$ and $N$. Let $H$ be the orthocenter of triangle $ABC$ and $P, Q$ be the intersections of $(I)$ and the circumcircle of triangle $HBC$. Let $(K)$ be the circle passing through $P, Q$ and touching $(O)$ at $T$ ($T$ is on the same side as $A$ with respect to $PQ$). Prove that the interior angle bisector of $\\angle MTN$ passes through a fixed point.",
"options": [],
"answer": "See solution",
"solution": "In the case $AB = AC$, it is obvious that $DB = DC$ and the bisector of $\\angle MTN$ passes through the midpoint of the minor arc $BC$ of $(O)$ (which is the fixed point revealed in part b), so we only need to consider the case where $AB \\neq AC$.\n\na) Let $R, S$ be the second intersections of $(I)$ with $AB, AC$ respectively. Since $E, F$ lie on a circle with diameter $BC$, and quadrilateral $EFRS$ is cyclic, we conclude $BC \\parallel RS$ by Reim's theorem.\n\n\n\nBy the assumption, $(I)$ touches $BC$ at $D$ then\n\n$$\n\\frac{BD^2}{CD^2} = \\frac{BF \\cdot BR}{CE \\cdot CS} = \\frac{BF \\cdot AB}{CE \\cdot AC} = \\frac{BF \\cdot BE}{CE \\cdot CF} = \\frac{\\cot B}{\\cot C}\n$$\n\nwhich means\n$$\n\\frac{BD}{CD} = \\sqrt{\\frac{\\cot B}{\\cot C}}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21641,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs of prime numbers $p$ and $q$ such that $p^2 - p - 1 = 2q + 3$.",
"options": [],
"answer": "See solution",
"solution": "If $p = q$, then $p^2 - p - 1 = 2p + 3$, which simplifies to $p^2 - 3p - 4 = 0$. The solutions are $p_1 = -1$ and $p_2 = 4$, neither of which are prime numbers.\n\nAssume $p \\neq q$. Then $p$ divides $2q + 3$, so there exists a natural number $k$ such that $2q + 3 = kp$, i.e., $q = \\frac{kp - 3}{2}$. Substituting into the original equation gives:\n\n$$\n2p^2 - 2p - 2 = (kp - 3)k, \\quad \\text{i.e.} \\quad 2p^2 - (k^2 + 2)p + (3k - 2) = 0.\n$$\n\nThis is a quadratic in $p$, so the discriminant $D = (k^2 + 2)^2 - 8(3k - 2)$ must be a perfect square.\n\nFor $k \\geq 6$, $2k^2 - 12k + 10 > 0$ implies $D > k^4$, so the only possibility is $D = (k^2 + 1)^2$, i.e.,\n\n$$\n(k^2 + 2)^2 - 8(3k - 2) = (k^2 + 1)^2,\n$$\n\nwhich simplifies to $2k^2 - 12k + 11 = 0$. This has no integer solutions, so $k < 6$.\n\nTesting $k = 1, 2, 3, 4, 5$, only $k = 5$ makes $D$ a perfect square. For $k = 5$, $p = 13$ and $q = 31$.\n\nThus, the only solution is $p = 13$, $q = 31$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21642,
"subject": "Mathematics (Olympiad)",
"question": "A positive integer is written in each $1 \\times 1$ cell of a rectangular board. The following two steps are allowed:\n\n1. The numbers in the cells of an arbitrarily chosen row are multiplied by $2$.\n2. The numbers in the cells of an arbitrarily chosen column are decreased by $1$.\n\nIs it possible, after a finite repetition of steps 1 and 2, to get a board containing only zeroes?",
"options": [],
"answer": "See solution",
"solution": "We will show that it is possible.\n\nIf there are $1$'s in the first column, perform step 1 on those rows that have $1$ in their first cell. Then perform step 2 on the first column. Repeat this procedure until all the numbers in the first column are $1$'s. Then perform step 2 on the first column, so now all the numbers in the first column are zeroes.\n\nIf there aren't any $1$'s in the first column, perform step 2 on it until at least one $1$ appears. Then proceed as above.\n\nIn the same manner, obtain zeroes in all other columns.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21643,
"subject": "Mathematics (Olympiad)",
"question": "Given $n \\geq 3$ pairwise distinct real numbers, prove that there are either three numbers whose sum is positive or two numbers whose sum is negative.",
"options": [],
"answer": "See solution",
"solution": "Suppose there are no positive numbers in the set. Then, at most one number is zero, and all others are negative, so there are at least two negative numbers. Thus, any two of them have a negative sum.\n\nOtherwise, there is at least one positive number. Isolate this positive number. Among the remaining $n-1$ numbers, pick any two. If their sum is negative, we have found two numbers with a negative sum. If their sum is non-negative, add the isolated positive number. If the total sum is positive, we have found three numbers with a positive sum. If the total sum is non-positive, then the sum of the two numbers must be negative, so again, we have two numbers with a negative sum.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21644,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(x+y)f(x-y) = x^2 f(x) + y^2 f(y) - 2f(xy)\n$$\nfor all real numbers $x$ and $y$.",
"options": [],
"answer": "See solution",
"solution": "Let $x = y = 0$. Then $(f(0))^2 = -2f(0)$, so $f(0) = -2$ or $f(0) = 0$.\n\n**Case 1:** $f(0) = -2$.\n\nLet $x = y = 1$. Then $-2f(2) = 2f(1) - 2f(1)$, so $f(2) = 0$.\n\nLet $y = 0$. Then $(f(x))^2 = x^2 f(x) + 4$, so $f(x) = \\frac{x^2 \\pm \\sqrt{x^4 + 16}}{2}$. For $x = 2$, $f(2) = 0 = \\frac{4 \\pm \\sqrt{32}}{2}$, which is a contradiction.\n\n**Case 2:** $f(0) = 0$.\n\nLet $y = 0$. Then $(f(x))^2 = x^2 f(x)$, so $f(x) = 0$ or $f(x) = x^2$.\n\nSuppose for some $t \\neq 0$, $f(t) = 0$. For $x + y = t$ with $x, y \\neq 0$, we have $0 = x^2 f(x) + y^2 f(y) - 2f(xy)$.\n\nIf $f(y) = 0$ and $f(x) = x^2$, then $0 \\neq x^4 = 2f(xy)$, so $x^4 = 2x^2 y^2$, which is a contradiction unless $y = \\pm \\frac{x}{\\sqrt{2}}$.\n\nIf $f(x) = x^2$ and $f(y) = y^2$, then $0 < x^4 + y^4 = 2f(xy)$, so $x^4 + y^4 = 2x^2 y^2$, which only holds if $x^2 = y^2$.\n\nThis implies that for every real $x$ different from $\\frac{t}{2}$ and $\\frac{\\sqrt{2} t}{\\pm \\sqrt{2} - 1}$, $f(x) = 0$. But then for $t_1 = 2t$, $f(t_1) = 0$, and also $f\\left(\\frac{t}{2}\\right) = 0$ and $f\\left(\\frac{t}{1 \\pm \\sqrt{2}}\\right) = 0$.\n\nTherefore, the only solutions are $f(x) = 0$ for all $x$, and $f(x) = x^2$ for all $x$. (It is easy to check that these functions satisfy the equation.)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21645,
"subject": "Mathematics (Olympiad)",
"question": "Prove that the sequence $a_2, a_4, a_6, \\dots$ is geometric if the sequence $(a_n)$ of real numbers satisfies $a_{n+1} a_{n-1} = -a_n^2$ for all $n \\ge 2$.",
"options": [],
"answer": "See solution",
"solution": "For each $n \\ge 2$, we can rearrange the given recursive formula to $\\frac{a_{n+1}}{a_n} = -\\frac{a_n}{a_{n-1}}$. Using this, we calculate\n\n$$\n\\frac{a_{2n+2}}{a_{2n}} = \\frac{a_{2n+2}}{a_{2n+1}} \\cdot \\frac{a_{2n+1}}{a_{2n}} = \\left(-\\frac{a_{2n+1}}{a_{2n}}\\right) \\cdot \\left(-\\frac{a_{2n}}{a_{2n-1}}\\right) = \\frac{a_{2n+1}}{a_{2n-1}}.\n$$\n\nUsing this equation twice in a row, we get\n\n$$\n\\frac{a_{2n+2}}{a_{2n}} = \\frac{a_{2n+1}}{a_{2n-1}} = \\frac{a_{2n}}{a_{2n-2}},\n$$\n\nwhich means that the sequence $a_2, a_4, a_6, \\dots$ is geometric.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21646,
"subject": "Mathematics (Olympiad)",
"question": "Do there exist positive integers $a_1, a_2, \\dots, a_{2017}$ such that the product\n$$\n(a_1^{2017} + a_2)(a_2^{2017} + a_3)\\dots(a_{2016}^{2017} + a_{2017})(a_{2017}^{2017} + a_1)\n$$\nis a power of a prime with exponent\n\na) $2017 \\cdot 2018$,\nb) $2017 \\cdot 2023$.",
"options": [],
"answer": "See solution",
"solution": "Assume that there are positive integers $a_1, a_2, \\dots, a_{2017}$ and a prime $p$ as required. Then, for each $i$, there is a positive integer $k_i$ so that\n$$\na_i^{2017} + a_{i+1} = p^{k_i}.\n$$\nHere $a_{2018} = a_1$. The sum of all $p^{k_i}$ equals the sum of all $a_i^{2017} + a_i$, which is even, so $p = 2$.\n\nWe claim that $a_i$ is odd for each $i$. Write $a_i = 2^{\\alpha_i} (2b_i + 1)$ for some integer $\\alpha_i$ and $b_i$. Since $a_i^{2017} + a_{i+1} = 2^{k_i}$, we get $2017\\alpha_i = \\alpha_{i+1}$ for each $i$, so $\\alpha_i = 0$.\n\nWe claim that $(a, m) = (1, 1)$ is the only solution of $a^{2017} + 1 = 2^m$. By contradiction, suppose there is a solution with $m > 1$. Then $a > 1$ and $a^{(2 \\cdot 2017, 2^{m-1})} \\equiv 1 \\pmod{2^m}$ since $a^{2 \\cdot 2017} \\equiv 1 \\pmod{2^m}$ and $a^{2^{m-1}} \\equiv 1 \\pmod{2^m}$. Thus $a^2 \\equiv 1 \\pmod{2^m}$ and so $a^2 - 1 \\ge 2^m = a^{2017} + 1$, which is a contradiction.\n\nThe claim implies that $a_i > 1$ for each $i$. Indeed, if there is $i$ so that $a_i = 1$, then $a_{i-1} = 1$ by the claim, and so $a_i = k_i = 1$ for each $i$. In this case, the sum of all $k_i$ is $2018$.\n\nThus $a_i^{2017} + a_{i+1} \\ge 2^{2017} + a_{i+1}$ and so $k_i \\ge 2018$. Let $k = \\min(k_i)$. For each $1 \\le i \\le 2017$, it follows from above that\n$$\na_i^{20172017} \\equiv -a_i \\pmod{2^k}.\n$$\nSince $a_i$ is odd, we get $a_i^{(2(2017^{2017}-1), 2^{k-1})} \\equiv 1 \\pmod{2^k}$, which implies $a_i^{2^6} \\equiv 1 \\pmod{2^k}$. Thus $a_i^{2^6} \\ge 2^{2017}$ and so $k_i \\ge 30 \\cdot 2017$, which means the sum of all $k_i$ is more than $2017 \\cdot 2023$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21647,
"subject": "Mathematics (Olympiad)",
"question": "Let $M$ be the midpoint of the side $AC$ of triangle $ABC$ and $N \\in (AM)$. The parallel from $N$ to $AB$ intersects $BM$ at $P$, the parallel from $M$ to $BC$ intersects $BN$ at $Q$, and the parallel from $N$ to $AQ$ intersects $BC$ at $S$. Prove that the straight lines $PS$ and $AC$ are parallel.",
"options": [],
"answer": "See solution",
"solution": "Denote $MQ \\cap AB = \\{E\\}$ and $D = NP \\cap ME$. Then $EA = EB$ and $ND = DP$. Since $ANPB$ is a trapezoid, points $A$, $Q$, $P$ are collinear.\n\nFrom $\\triangle ADP \\equiv \\triangle SDN$ (A.S.A.) it follows that $[AP] \\equiv [SN]$ and, since $AP \\parallel SN$, quadrilateral $ANSP$ is a parallelogram. Hence, $PS \\parallel AC$.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21648,
"subject": "Mathematics (Olympiad)",
"question": "Let $x_1, x_2, x_3, y_1, y_2, y_3, z_1, z_2, z_3$ be real numbers. Find the maximum value of\n\n$$\nA = \\frac{(x_1^3 + x_2^3 + x_3^3 + 1)(y_1^3 + y_2^3 + y_3^3 + 1)(z_1^3 + z_2^3 + z_3^3 + 1)}{(x_1 + y_1 + z_1)(x_2 + y_2 + z_2)(x_3 + y_3 + z_3)}\n$$\n\nand determine when equality holds.",
"options": [],
"answer": "See solution",
"solution": "We use the following inequality:\n\n$$\n\\begin{align*}\n& (p_1^3 + \\cdots + p_n^3)(q_1^3 + \\cdots + q_n^3)(r_1^3 + \\cdots + r_n^3)(p_1 q_1 r_1 + \\cdots + p_n q_n r_n) \\\\\n\\ge & \\left((p_1^{3/2} q_1^{3/2} + \\cdots + p_n^{3/2} q_n^{3/2})(p_1^{1/2} q_1^{1/2} r_1^2 + \\cdots + p_n^{1/2} q_n^{1/2} r_n^2)\\right)^2 \\\\\n\\ge & \\left((p_1 q_1 r_1 + \\cdots + p_n q_n r_n)^2\\right)^2 \\\\\n= & (p_1 q_1 r_1 + \\cdots + p_n q_n r_n)^4\n\\end{align*}\n$$\n\nLet $t = 6^{-1/3}$. Using this inequality and the AM-GM inequality,\n\n$$\n\\begin{align*}\n& (x_1^3 + x_2^3 + x_3^3 + 1)(y_1^3 + y_2^3 + y_3^3 + 1)(z_1^3 + z_2^3 + z_3^3 + 1) \\\\\n&= (x_1^3 + x_2^3 + x_3^3 + t^3 + t^3 + t^3 + t^3 + t^3 + t^3) \\\\\n& \\quad \\times (t^3 + t^3 + t^3 + y_1^3 + y_2^3 + y_3^3 + t^3 + t^3 + t^3) \\\\\n& \\quad \\times (t^3 + t^3 + t^3 + t^3 + t^3 + t^3 + z_1^3 + z_2^3 + z_3^3) \\\\\n&\\ge (t^2 x_1 + t^2 x_2 + t^2 x_3 + t^2 y_1 + t^2 y_2 + t^2 y_3 + t^2 z_1 + t^2 z_2 + t^2 z_3)^3 \\\\\n&= \\frac{1}{36} \\left( (x_1 + y_1 + z_1) + (x_2 + y_2 + z_2) + (x_3 + y_3 + z_3) \\right)^3 \\\\\n&\\ge \\frac{27}{36} (x_1 + y_1 + z_1)(x_2 + y_2 + z_2)(x_3 + y_3 + z_3).\n\\end{align*}\n$$\n\nEquality holds if and only if $x_1 = x_2 = x_3 = y_1 = y_2 = y_3 = z_1 = z_2 = z_3 = t$.\n\nTherefore, the maximum value of $A$ is $\\frac{3}{4}$, and equality holds when all variables equal $t$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21649,
"subject": "Mathematics (Olympiad)",
"question": "Consider the sequence $a_n$ defined by $a_1 = \\frac{1}{2}$ and $2n \\cdot a_{n+1} = (n+1)a_n$ for any positive integer $n$.\n\na) Determine the general formula for $a_n$.\n\nb) Let $b_n = a_1 + a_2 + \\dots + a_n$. Prove that $b_n$, $b_{n+1}$, and $b_{n+2}$ cannot be consecutive terms of an arithmetic progression for any positive integer $n$.",
"options": [],
"answer": "See solution",
"solution": "a) We have $\\frac{a_{n+1}}{n+1} = \\frac{1}{2} \\cdot \\frac{a_n}{n}$ for any positive integer $n$. This means the sequence $\\left(\\frac{a_n}{n}\\right)$ is a geometric progression with ratio $\\frac{1}{2}$, so $\\frac{a_n}{n} = \\left(\\frac{1}{2}\\right)^{n-1} \\cdot a_1 = \\left(\\frac{1}{2}\\right)^n$ for any positive integer $n$. Thus, $a_n = \\frac{n}{2^n}$ for any positive integer $n$.\n\nb) We have:\n\n$$\nb_n = a_1 + a_2 + \\dots + a_n = \\sum_{k=1}^n \\frac{k}{2^k}\n$$\n\nFor $n \\ge 3$, it can be shown that $2^n > n+2$, so $b_n = 1 - \\frac{n+2}{2^n}$ for $n \\ge 3$.\n\nAssume, for contradiction, that $b_n$, $b_{n+1}$, and $b_{n+2}$ are consecutive terms of an arithmetic progression for some $n \\ge 3$. Then:\n\n$$\n\\frac{n+2}{2^n} + \\frac{n+4}{2^{n+2}} = \\frac{n+3}{2^n}\n$$\n\nwhich simplifies to $5n+12 = 4n+12$, a contradiction. For $n = 1$ and $n = 2$, the numbers $\\frac{1}{2}$, $0$, $\\frac{3}{8}$, and $0$, $\\frac{3}{8}$, $\\frac{5}{8}$ are not consecutive terms of an arithmetic progression.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21650,
"subject": "Mathematics (Olympiad)",
"question": "Equilateral triangle $ABC$ is inscribed in circle $w$. Points $F$ and $E$ are chosen on sides $AB$ and $AC$, respectively, so that $\\angle ABE + \\angle ACF = 60^\\circ$. The circumscribed circle of $\\triangle AFE$ intersects circle $w$ at point $D$. Rays $DE$ and $DF$ intersect line $BC$ at points $X$ and $Y$, respectively. Prove that the center of the inscribed circle of $\\triangle DXY$ does not depend on the choice of points $F$ and $E$.\n\n\n\n*Fig. 27*",
"options": [],
"answer": "See solution",
"solution": "Let $P$ be the intersection point of $CF$ and $BE$. Then, $\\angle CBE = 60^\\circ - \\angle ABE = \\angle FCA$. Analogously, $\\angle FCB = \\angle ABE$ (see Fig. 27). Then, $\\angle FKE = \\angle CKB = 180^\\circ - \\angle CBE - \\angle FCB = 120^\\circ$.\n\nThus, quadrilateral $AFPE$ is inscribed in circle $\\gamma$, which point $D$ lies on. Draw the bisector of $\\angle CAB$ until its intersection with $\\gamma$ at some point $O$. Obviously, $O$ is the midpoint of the smaller arc $EF$ of circle $\\gamma$. Let us prove that $O$ is the center of equilateral $\\triangle ABC$. $\\triangle AFC = \\triangle BEC$ by a side and two adjacent angles. Hence, $FC = BE$. Naturally, $\\angle OAFC = \\angle BEO$ and $OF = OE$. Thus, $\\triangle OFC = \\triangle BEO$ by two pairs of sides and the included angle, which yields $\\angle PBO = \\angle PCO$, i.e., quadrilateral $BPOC$ is inscribed, hence $\\angle BOC = 120^\\circ$. Therefore, $O$ lies on the bisector of $\\angle CAB$ in the same half-plane as $P$ with respect to line $BC$, and $\\angle BOC = 120^\\circ$. Obviously, then $O$ is the center of $\\triangle ABC$.\n\nNow, we prove that $O$ is the incenter of $\\triangle DXY$. Clearly, then this center does not depend on the points $E$ and $F$. Obviously, $\\angle YDO = \\angle XDO = 30^\\circ$, i.e., $DO$ is a bisector of $\\angle XDY$. It suffices to show that $YO$ is a bisector of $\\angle DYX$. $\\angle YDO = \\angle OBY = 30^\\circ$, hence, quadrilateral $DOYB$ is inscribed, and $DO = OB$, as radii. Therefore,\n\n$$\n\\angle OYX = \\angle ODB = \\angle DBO = \\angle DYO,\n$$\n\ni.e., $YO$ is a bisector of $\\angle DYX$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21651,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a natural number. We define sequences $a_i$ and $b_i$ as follows. Let $a_0 = 1$ and $b_0 = n$. For $i > 0$, define\n\n$$\n(a_i, b_i) = \\begin{cases} (2a_{i-1} + 1,\\ b_{i-1} - a_{i-1} - 1) & \\text{if } a_{i-1} < b_{i-1}, \\\\ (a_{i-1} - b_{i-1} - 1,\\ 2b_{i-1} + 1) & \\text{if } a_{i-1} > b_{i-1}, \\\\ (a_{i-1},\\ b_{i-1}) & \\text{if } a_{i-1} = b_{i-1}. \\end{cases}\n$$\n\nGiven that $a_k = b_k$ for some natural number $k$, prove that $n + 3$ is a power of 2.",
"options": [],
"answer": "See solution",
"solution": "We first note that $a_i + b_i = n + 1$ for all natural numbers $i$. Define three sequences $\\{r_i\\}$, $\\{s_i\\}$, $\\{t_i\\}$ as follows: $r_0 = 1$, $s_0 = t_0 = 0$. For $i > 0$,\n\n$$\n(r_i, s_i, t_i) = \\begin{cases} (2r_{i-1},\\ 2s_{i-1},\\ 2t_{i-1} + 1) & \\text{if } a_{i-1} < b_{i-1}, \\\\ (2r_{i-1} - 1,\\ 2s_{i-1} - 1,\\ 2t_{i-1} - 1) & \\text{if } a_{i-1} > b_{i-1}, \\\\ (r_{i-1},\\ s_{i-1},\\ t_{i-1}) & \\text{if } a_{i-1} = b_{i-1}. \\end{cases}\n$$\n\nIt is easy to verify that $a_i = r_i a_0 + s_i b_0 + t_i$ and $b_i = (1 - r_i)a_0 + (1 - s_i)b_0 - t_i$ for all $i \\ge 0$.\n\nLet $k$ be the smallest non-negative integer such that $a_k = b_k$. For any integer $l$ with $0 \\le l \\le k$, we claim that $r_l - s_l = t_l + 1 - 2s_l = 2^l$. We prove this by induction on $l$. The claim is straightforward for $l = 0$. Suppose it holds for some $l < k$. Since $l + 1 \\le k$, $(r_{l+1}, s_{l+1}, t_{l+1})$ equals either $(2r_l, 2s_l, 2t_l + 1)$ or $(2r_l - 1, 2s_l - 1, 2t_l - 1)$. In either case, $r_{l+1} - s_{l+1} = 2(r_l - s_l)$ and $t_{l+1} + 1 - 2s_{l+1} = 2(t_l + 1 - 2s_l)$. Hence the claim.\n\nIf $a_k = b_k$, then $r_k a_0 + s_k b_0 + t_k = (1 - r_k)a_0 + (1 - s_k)b_0 - t_k$. This implies $(2r_k - 1) + 2t_k = (1 - 2s_k)n$. Therefore, $(n+3)(1 - 2s_k) = 2(r_k - s_k) + 2t_k - 4s_k + 2 = 2^{k+2}$. Hence $n+3$ is a power of 2. Since $1 - 2s_k$ is also a power of 2, it follows that $s_k = 0$ and thus $n + 3 = 2^{k+2}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21652,
"subject": "Mathematics (Olympiad)",
"question": "Find the least real number $k$ with the following property: if the real numbers $x$, $y$, and $z$ are not all positive, then\n\n$$\nk(x^2 - x + 1)(y^2 - y + 1)(z^2 - z + 1) \\geq (xyz)^2 - xyz + 1.\n$$",
"options": [],
"answer": "See solution",
"solution": "The answer is $k = \\frac{16}{9}$.\n\nWe start with a lemma.\n\n**Lemma 1.** If real numbers $s$ and $t$ are not both positive, then\n\n$$\n\\frac{4}{3}(s^2 - s + 1)(t^2 - t + 1) \\geq (st)^2 - st + 1. \\quad (*)\n$$\n\n*Proof:* Without loss of generality, assume $s \\geq t$.\n\nFirst, suppose $s \\geq 0 \\geq t$. Setting $u = -t$, (*) becomes\n\n$$\n\\frac{4}{3}(s^2 - s + 1)(u^2 + u + 1) \\geq (su)^2 + su + 1,\n$$\n\nor\n\n$$\n4(s^2 - s + 1)(u^2 + u + 1) \\geq 3s^2u^2 + 3su + 3.\n$$\n\nExpanding the left-hand side gives\n\n$$\n4s^2u^2 + 4s^2u - 4su^2 - 4su + 4s^2 + 4u^2 - 4s + 4u + 4 \\geq 3s^2u^2 + 3su + 3.\n$$\n\nOr,\n\n$$\ns^2u^2 + 4u^2 + 4s^2 + 1 + 4s^2u + 4u \\geq 4su^2 + 4s + 7su.\n$$\n\nThis is evident as $s^2u^2 + 4u^2 \\geq 4su^2$, $4s^2 + 1 \\geq 4s$, and $4s^2u + 4u \\geq 8su \\geq 7su$.\n\nSecond, suppose $0 \\geq s \\geq t$. Let $v = -s$. By the previous argument,\n\n$$\n\\frac{4}{3}(v^2 - v + 1)(t^2 - t + 1) \\geq (vt)^2 - vt + 1.\n$$\n\nIt is clear that $t^2 - t + 1 > 0$, $s^2 - s + 1 \\geq v^2 - v + 1$, and $(vt)^2 - vt + 1 \\geq (st)^2 - st + 1$. Combining these gives (*), completing the proof of the lemma.\n\nNow, for $x, y, z$ not all positive real numbers,\n\n$$\n\\frac{16}{9}(x^2 - x + 1)(y^2 - y + 1)(z^2 - z + 1) \\geq (xyz)^2 - xyz + 1. \\quad (**)\n$$\n\nConsider three cases:\n\n(a) If $y \\geq 0$, set $(s, t) = (y, z)$ and then $(s, t) = (x, yz)$ in the lemma to get the result.\n\n(b) If $0 \\geq y$, set $(s, t) = (x, y)$ and then $(s, t) = (xy, z)$ in the lemma to get the result.\n\nFinally, the minimum value of $k$ is $\\frac{16}{9}$, as equality holds in (**) when $(x, y, z) = (\\frac{1}{2}, \\frac{1}{2}, 0)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21653,
"subject": "Mathematics (Olympiad)",
"question": "In the figure below, $WXYZ$ is a rectangle with $WX = 4$ and $WZ = 8$. Point $M$ lies on $\\overline{XY}$, point $A$ lies on $\\overline{YZ}$, and $\\angle WMA$ is a right angle. The areas of triangles $\\triangle WXM$ and $\\triangle WAZ$ are equal. What is the area of $\\triangle WMA$?\n\n\n\n(A) 13 (B) 14 (C) 15 (D) 16 (E) 17",
"options": [],
"answer": "See solution",
"solution": "Label the diagram as shown, where $MX = a$ and $ZA = b$.\n\n\n\nThe Pythagorean Theorem on $\\triangle WMA$ gives $WM^2 + MA^2 = WA^2$, which implies that\n\n$$\n4^2 + a^2 + (8-a)^2 + (4-b)^2 = 8^2 + b^2.\n$$\n\nExpanding and simplifying yields $a^2 - 8a - 4b + 16 = 0$. Because the areas of triangles $\\triangle WXM$ and $\\triangle WAZ$ are equal, $\\frac{1}{2} \\cdot 4a = \\frac{1}{2} \\cdot 8b$, so $a = 2b$. Substituting into the previous equation and factoring gives $4(b-1)(b-4) = 0$. Therefore $b = 1$ or $b = 4$. But $b = 4$ would require $A = Y = M$, and $\\angle WMA$ would not exist, so it must be that $b = 1$ and $a = 2$. The area of $\\triangle WMA$ can be found by subtracting the three other triangle areas from the area of the rectangle:\n\n$$\n8 \\cdot 4 - \\frac{1}{2} \\cdot 4 \\cdot 2 - \\frac{1}{2} \\cdot 6 \\cdot 3 - \\frac{1}{2} \\cdot 8 \\cdot 1 = 32 - 4 - 9 - 4 = 15.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21654,
"subject": "Mathematics (Olympiad)",
"question": "At 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 $k \\ge 2$, the move $M_k$ can be replaced by moves involving only $M_0$ and $M_1$ in such a way that the total increase in the blue number is the same, but the red number increases by less than $k^2+2$.\n\n*Case 1:* $k = 2p + 1$ is odd. Then $M_k$ increases the blue number by $2p + 2$ and the red number by $(2p+1)^2 + 2$. Instead, $p+1$ moves of type $M_1$ increase the blue number also by $2p+2$, but the red number increases by $3(p+1)$. Since $(2p+1)^2 + 2 > 3(p+1)$ for all $p \\ge 1$, we can replace $M_k$ with $p+1$ moves of $M_1$.\n\n*Case 2:* $k = 2p$ is even. Then $M_k$ increases the blue number by $2p+1$, and the red number by $(2p)^2 + 2$. Alternatively, $p$ moves of $M_1$ and one move of $M_0$ increase the blue number by $2p+1$ and the red number by $3p+2$. Since $(2p)^2 + 2 > 3p+2$ for all $p \\ge 1$, we conclude that $M_k$ can be replaced by $p$ moves of $M_1$ and one $M_0$.\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": 21655,
"subject": "Mathematics (Olympiad)",
"question": "Prove that in an arithmetic progression consisting of 40 distinct positive integers, at least one of the numbers cannot be written as $2^k + 3^l$, where $k, l$ are nonnegative integers.\n\n",
"options": [],
"answer": "See solution",
"solution": "Suppose, for contradiction, that there exist 40 distinct positive integers in arithmetic progression such that each term can be written as $2^k + 3^l$. Denote this sequence by $a, a+d, a+2d, \\ldots, a+39d$, where $a, d$ are positive integers.\n\nLet\n$$\nm = \\lfloor \\log_2 (a + 39d) \\rfloor, \\quad n = \\lfloor \\log_3 (a + 39d) \\rfloor.\n$$\n\nWe first show that at most one of $a+26d, a+27d, \\ldots, a+39d$ cannot be written as $2^m + 3^l$ or $2^k + 3^n$ ($k, l$ nonnegative integers).\n\nSuppose $a+hd$ cannot be written as $2^m + 3^l$ or $2^k + 3^n$ for some $26 \\le h \\le 39$. Then, by assumption, $a+hd = 2^b + 3^c$ for some nonnegative integers $b, c$. By the definition of $m$ and $n$, $b \\le m$, $c \\le n$. Since $a+hd$ cannot be written as $2^m + 3^l$ or $2^k + 3^n$, we have $b \\le m-1$, $c \\le n-1$.\n\nIf $b \\le m-2$, then\n$$\n\\begin{aligned}\na + hd &\\le 2^{m-2} + 3^{n-1} = \\frac{1}{4} \\times 2^m + \\frac{1}{3} \\times 3^n \\\\\n&\\le \\frac{7}{12} \\times (a + 39d) < a + 26d,\n\\end{aligned}\n$$\na contradiction.\n\nIf $c \\le n-2$, then\n$$\n\\begin{aligned}\na + hd &\\le 2^{m-1} + 3^{n-2} = \\frac{1}{2} \\times 2^m + \\frac{1}{9} \\times 3^n \\\\\n&\\le \\frac{11}{18} \\times (a + 39d) < a + 26d,\n\\end{aligned}\n$$\nalso a contradiction.\n\nIt follows that $b = m - 1$, $c = n - 1$, which implies that at most one of $a + 26d, a + 27d, \\ldots, a + 39d$ cannot be written as $2^m + 3^l$ or $2^k + 3^n$.\n\nIn these 14 numbers, at least 13 can be written as $2^m + 3^l$ or $2^k + 3^n$. By the pigeonhole principle, at least 7 numbers can be written in the same form. Consider two cases:\n\n*Case 1:* There are 7 numbers in the form $2^m + 3^l$, denoted by\n$$\n2^m + 3^{l_1}, 2^m + 3^{l_2}, \\ldots, 2^m + 3^{l_7},\n$$\nwhere $l_1 < l_2 < \\ldots < l_7$. Thus, $3^{l_1}, 3^{l_2}, \\ldots, 3^{l_7}$ are 7 terms of an arithmetic progression with 14 terms and common difference $d$. However,\n$$\n13d \\ge 3^{l_7} - 3^{l_1} \\ge (3^5 - \\frac{1}{3}) \\times 3^{l_2} > 13(3^{l_2} - 3^{l_1}) \\ge 13d,\n$$\na contradiction.\n\n*Case 2:* There are 7 numbers in the form $2^k + 3^n$, denoted by\n$$\n2^{k_1} + 3^n, 2^{k_2} + 3^n, \\ldots, 2^{k_7} + 3^n,\n$$\nwhere $k_1 < k_2 < \\ldots < k_7$. Thus $2^{k_1}, 2^{k_2}, \\ldots, 2^{k_7}$ are 7 terms of an arithmetic progression with 14 terms and common difference $d$. However,\n$$\n13d \\ge 2^{k_7} - 2^{k_1} \\ge (2^5 - \\frac{1}{2}) \\times 2^{k_2} > 13(2^{k_2} - 2^{k_1}) \\ge 13d,\n$$\na contradiction.\n\nTherefore, our initial assumption is false, which completes the proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21656,
"subject": "Mathematics (Olympiad)",
"question": "In an acute triangle $ABC$ with $AC > AB$, let $D$ be the projection of $A$ on $BC$, let $E$ and $F$ be the projections of $D$ on $AB$ and $AC$, and let $G$ and $H$ be the second intersections of the line $AD$ with $EF$ and the circumcircle of triangle $ABC$. Prove\n\n$$\nAG \\cdot AH = AD^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "\n\nFrom similar right triangles, we get\n\n$$\n\\frac{BE}{AE} = \\frac{\\frac{BD}{AD} ED}{\\frac{AD}{BD} ED} = \\left(\\frac{BD}{AD}\\right)^2\n$$\n\nand analogously,\n\n$$\n\\frac{CF}{AF} = \\left(\\frac{CD}{AD}\\right)^2.\n$$\n\nNow, because $AC > AB$, the lines $BC$ and $EF$ intersect at a point $X$ on the extension of segment $BC$ beyond $B$. By Menelaus' theorem,\n\n$$\nBX \\frac{CF}{AF} = CX \\frac{BE}{AE}, \\quad CX \\frac{DG}{AG} = DX \\frac{CF}{AF}, \\quad DX \\frac{BE}{AE} = BX \\frac{DG}{AG}.\n$$\n\nAdding these relations and rearranging terms, we arrive at\n\n$$\nBC \\frac{DG}{AG} = BD \\frac{CF}{AF} + CD \\frac{BE}{AE} = BC \\frac{BD \\cdot CD}{AD^2} = BC \\frac{AD \\cdot HD}{AD^2} = BC \\frac{HD}{AD},\n$$\n\nwhence\n\n$$\n\\frac{DG}{AG} = \\frac{HD}{AD}.\n$$\n\nHence,\n\n$$\n\\frac{AD}{AG} = \\frac{AG + DG}{AG} = \\frac{AD + HD}{AD} = \\frac{AH}{AD},\n$$\n\nwhich is equivalent to the problem's assertion.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21657,
"subject": "Mathematics (Olympiad)",
"question": "Let $K$ and $N > K$ be fixed positive integers. Let $n$ be a positive integer and let $a_1, a_2, \\dots, a_n$ be distinct integers. Suppose that whenever $m_1, m_2, \\dots, m_n$ are integers, not all equal to $0$, such that $|m_i| \\le K$ for each $i$, then the sum\n\n$$\n\\sum_{i=1}^{n} m_i a_i\n$$\n\nis not divisible by $N$. What is the largest possible value of $n$?",
"options": [],
"answer": "See solution",
"solution": "The answer is $n = \\lceil \\log_{K+1} N \\rceil$.\n\nNote first that for $n \\le \\lceil \\log_{K+1} N \\rceil$, taking $a_i = (K+1)^{i-1}$ works. Indeed, let $r$ be maximal such that $m_r \\ne 0$. Then, on the one hand, we have\n\n$$\n\\left| \\sum_{i=1}^{n} m_i a_i \\right| \\le \\sum_{i=1}^{n} K (K+1)^{i-1} = (K+1)^n - 1 < N.\n$$\n\nOn the other hand, we have\n\n$$\n\\left| \\sum_{i=1}^{n} m_i a_i \\right| \\ge |m_r a_r| - \\left| \\sum_{i=1}^{r-1} m_i a_i \\right| \\ge (K+1)^{r-1} - \\sum_{i=1}^{r-1} K (K+1)^{i-1} = 1 > 0.\n$$\n\nSo the sum is indeed not divisible by $N$.\n\nAssume now that $n \\ge \\lceil \\log_{K+1} N \\rceil$ and look at all $n$-tuples of the form $(t_1, \\dots, t_n)$ where each $t_i$ is a non-negative integer with $t_i \\le K$. There are $(K+1)^n > N$ such tuples, so there are two of them, say $(t_1, \\dots, t_n)$ and $(t'_1, \\dots, t'_n)$ such that\n\n$$\n\\sum_{i=1}^{n} t_i a_i \\equiv \\sum_{i=1}^{n} t'_i a_i \\pmod{N}.\n$$\n\nNow taking $m_i = t_i - t'_i$ for each $i$ satisfies the requirements on the $m_i$'s, but $N$ divides the sum\n\n$$\n\\sum_{i=1}^{n} m_i a_i,\n$$\n\na contradiction.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21658,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c, d, e, f, g$ be 7 distinct positive integers less than or equal to 7. Determine all the prime numbers which can be represented in the form\n\n$$\na \\times b \\times c \\times d + e \\times f \\times g.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $A = \\{a, b, c, d\\}$ and $B = \\{e, f, g\\}$. Suppose $X = abcd + efg$ is a prime.\n\nThe numbers 2, 4, 6 must belong to the same set $A$ or $B$, otherwise both $abcd$ and $efg$ are even, making $X$ even. Since $X \\ne 2$, $X$ cannot be prime if even.\n\nSimilarly, 3 and 6 must be in the same set, otherwise $X$ is divisible by 3, and since $X \\ne 3$, $X$ cannot be prime.\n\nThus, $A = \\{2, 3, 4, 6\\}$ and $B = \\{1, 5, 7\\}$, so the only prime number of the form $abcd + efg$ is:\n\n$$\n2 \\times 3 \\times 4 \\times 6 + 1 \\times 5 \\times 7 = 179\n$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21659,
"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 $\\{1, 2, 3, \\ldots, 2015\\}$ 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": 21660,
"subject": "Mathematics (Olympiad)",
"question": "Let $n_0, n_1, \\dots, n_8$ be integers, and let $r_0, r_1, \\dots, r_8$ be their remainders after division by $5$.\n\nShow that among these nine integers, there exist five whose sum is divisible by $5$.",
"options": [],
"answer": "See solution",
"solution": "*Case 4a.* Suppose $r_1$ occurs at least $3$ times, say $r_1 = r_3 = r_4$.\n\n$$\n\\sum_{i=0}^{4} n_i \\equiv (r_1 - s) + 3r_1 + (r_1 + s) = 5r_1 \\equiv 0 \\pmod{5},\n$$\nso the sum is divisible by $5$.\n\n*Case 4b.* Suppose $r_1$ occurs once or twice. Then $r_0$ and $r_2$ must each occur $3$ or $4$ times, say $r_0 = r_5 = r_6$, $r_2 = r_7 = r_8$. Then\n\n$$\n\\begin{aligned}\n n_0 + n_1 + n_2 + n_5 + n_7 &\\equiv (r_1 - s) + r_1 + (r_1 + s) + (r_1 - s) + (r_1 + s) \\\\\n &= 5r_1 \\equiv 0 \\pmod{5},\n\\end{aligned}\n$$\nso the sum is divisible by $5$.\n\nAs these cases exhaust all possibilities, the proof is complete.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21661,
"subject": "Mathematics (Olympiad)",
"question": "A passenger wanted to know the speed of the bus in which he was traveling, so he looked through the window and saw on a road sign (denoting the distance from the town from where he started his traveling) a two-digit number. After one hour of driving, he saw on another road sign a three-digit number written with the same two digits as one hour before but in opposite order and with a zero between them. The passenger evaluated the speed of the bus and fell asleep. Two hours later he woke up and saw on a road sign a three-digit number with the first and last digit the same as the number two hours ago but a different middle digit. If we know that the bus had constant speed, determine the speed and the numbers which the passenger saw on the road signs.",
"options": [],
"answer": "See solution",
"solution": "Let the first road sign show the number $10x + y$, where $0 < x \\leq 9$, $0 \\leq y \\leq 9$. One hour later, the passenger saw the number $100y + x$, with $y \\neq 0$. The distance traveled in one hour is:\n\n$$\n(100y + x) - (10x + y) = 100y + x - 10x - y = 99y - 9x = 9(11y - x) \\text{ km}\n$$\n\nSince the bus had constant speed, in the next two hours it traveled $2 \\cdot 9(11y - x) = 18(11y - x)$ km. The passenger then saw the number $100y + 10z + x$, where $0 < z \\leq 9$. Thus:\n\n$$\n18(11y - x) = (100y + 10z + x) - (100y + x) = 10z\n$$\n\nSo:\n$$\n18(11y - x) = 10z \\implies 9(11y - x) = 5z\n$$\n\nSince $z$ is a digit, $z = 9$ is the only value making $5z$ divisible by $9$. Thus, $9(11y - x) = 45 \\implies 11y - x = 5$.\n\nWith $0 < x, y \\leq 9$, we get $x = 6$, $y = 1$.\n\nTherefore, the speed is $9(11y - x) = 9(11 \\cdot 1 - 6) = 9 \\cdot 5 = 45$ km/h.\n\nThe numbers seen on the road signs are:\n- First: $10x + y = 10 \\cdot 6 + 1 = 61$\n- Second: $100y + x = 100 \\cdot 1 + 6 = 106$\n- Third: $100y + 10z + x = 100 \\cdot 1 + 10 \\cdot 9 + 6 = 100 + 90 + 6 = 196$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21662,
"subject": "Mathematics (Olympiad)",
"question": "A regular polygon has 180 sides. What is the measure of each interior angle?",
"options": [],
"answer": "See solution",
"solution": "It is easier to work with exterior angles, since the sum of the exterior angles of a polygon is $360\\degree$. There are 180 exterior angles, which are all equal since the polygon is regular. Thus, each exterior angle is equal to $\\frac{360\\degree}{180} = 2\\degree$, and each interior angle is equal to $180\\degree - 2\\degree = 178\\degree$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21663,
"subject": "Mathematics (Olympiad)",
"question": "For $T \\ge 0$ and $x \\in \\mathbb{Z}$, let $F_{T,x}$ be the number of feasible paths between $0$ and $T$ such that $W_T \\ge x$. These are paths that start at $(0, 0)$ and end at $(T, y)$ for some $y \\ge x$.\n\nNote that $F_{T,x} = F_{T,-1}$ for all $x \\le -1$ and $F_{T,x} = 0$ for $T < x$ since $|W_T| \\le T$. Also, $y \\equiv T \\pmod{2}$, so $F_{T,x} = F_{T,x-1}$ if $x \\equiv T \\pmod{2}$.\n\nThe relationship to $T_{n,k}$ and $T_n$ is:\n\n$$\nF_{2n,2r} = \\sum_{k=r}^{n} T_{n,k} \\quad \\text{and} \\quad F_{2n,0} = T_n.\n$$\n\nShow that\n\n$$\nF_{20,0} \\equiv 4 \\pmod{8}.\n$$\n\nA recurrence for $F_{t,x}$ is:\n\n$$\nF_{0,x} = \\begin{cases} 1 & \\text{if } x \\le 0 \\\\ 0 & \\text{if } x > 0 \\end{cases}\n$$\n\nand for $t > 0$, $x \\ge -1$, $x \\equiv t \\pmod{2}$:\n\n$$\nF_{t,x} = F_{t-1,x-1} + F_{t-1,x+1}.\n$$\n\nFor even $t$ and $x \\ge 0$:\n\n$$\nF_{2t,2x} = F_{2t-2,2x-2} + 2F_{2t-2,2x} + F_{2t-2,2x+2}.\n$$\n\nFor $x = 0$:\n\n$$\nF_{2t,0} = 3F_{2t-2,0} + F_{2t-2,2}.\n$$\n\nCalculate $F_{20,0} \\pmod{8}$.",
"options": [],
"answer": "See solution",
"solution": "We construct a table of values for $F_{2t,2x}$:\n\n| $2t$ | 0 | 2 | 4 | 6 | 8 | 10 |\n|------|---|---|---|---|---|----|\n| 0 | 1 |\n| 2 | 3 | 1 |\n| 4 | 2 | 5 | 1 |\n| 6 | 3 | 5 | 7 | 1 |\n| 8 | 6 | 4 | 4 | 1 | 1 |\n| 10 | 6 | 2 | 5 | 7 | 3 | 1 |\n| 12 | 4 | 7 | 3 | 6 | 6 |\n| 14 | 3 | 5 | 3 | 5 |\n| 16 | 6 | 0 | 0 |\n| 18 | 2 | 6 |\n| 20 | 4 |\n\nThe number $4$ in the bottom row shows $F_{20,0} \\equiv 4 \\pmod{8}$.\n\nA second approach uses the reflection principle for random walks. For $t \\ge 0$, $H < 0$, $x > H$, $t \\equiv x \\pmod{2}$:\n\n- The number of paths from $W_0 = 0$ to $W_t = x$ touching $H$ at least once equals the number from $W_0 = 0$ to $W_t = 2H - x$.\n\nThis is shown by reflecting the path after the first time it hits $H$.\n\n\n\nCounting such paths, the number with $u$ up steps and $d$ down steps ($u + d = t$, $u - d = 2H - x$) is:\n\n$$\n\\binom{t}{u} = \\binom{t}{H + \\frac{t-x}{2}}.\n$$\n\nSo, the number of paths from $W_0 = 0$ to $W_t = x$ never touching $H$ is:\n\n$$\n\\binom{t}{\\frac{t-x}{2}} - \\binom{t}{H + \\frac{t-x}{2}}.\n$$\n\nFor $H = -2$, $x = t - 2k$:\n\n$$\n\\binom{t}{k} - \\binom{t}{k-2} = \\binom{t+1}{k} - \\binom{t+1}{k-1}.\n$$\n\nSumming telescopically, for $F_{t,t-2k}$:\n\n$$\nF_{t,t-2k} = \\binom{t+1}{k}.\n$$\n\nFor $t = 20$, $k = 10$:\n\n$$\nF_{20,0} = \\binom{21}{10} = \\frac{21!}{10!11!}.\n$$\n\nThe $2$-adic valuation is $v_2(F_{20,0}) = 2$, so $F_{20,0} \\equiv 4 \\pmod{8}$ by Legendre's formula:\n\n$$\nv_2 \\left( \\frac{21!}{10!11!} \\right) = 18 - 8 - 8 = 2.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21664,
"subject": "Mathematics (Olympiad)",
"question": "На стороне $AB$ отложен отрезок $BE = BC$. Даны равнобедренные треугольники $EBK$ и $KBC$, которые равны по двум сторонам и углу между ними. Докажите, что $AD \\parallel EK$ и $\\angle KAD = \\angle KCD$.",
"options": [],
"answer": "See solution",
"solution": "$EK = KC$, а $\\angle AEK = 180^\\circ - \\angle BEK = 180^\\circ - \\angle BKC = \\angle CKD$. Кроме того, $KD = BD - BK = BA - BE = EA$. Следовательно, треугольники $AEK$ и $DKC$ равны, откуда $\\angle KCD = \\angle EKA$. Далее, поскольку оба треугольника $BEK$ и $BAD$ — равнобедренные, $\\angle BEK = 90^\\circ - \\angle EBD/2 = \\angle BAD$. Поэтому $AD \\parallel EK$, откуда $\\angle KAD = \\angle EKA = \\angle KCD$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21665,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for all $x, y \\in \\mathbb{R}$,\n$$\nf(y + 2f(x)) = f(x + yf(x)).\n$$",
"options": [],
"answer": "See solution",
"solution": "If $f(0) = 0$, then inserting $x = 0$ into the equation gives $f(y) = 0$ for all $y$, so $f(x) = 0$ is a solution.\n\nIf $f(0) \\neq 0$, further analysis shows $f(x) = x - 1$ is the only other solution. Thus, the solutions are $f(x) = 0$ and $f(x) = x - 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21666,
"subject": "Mathematics (Olympiad)",
"question": "One side of an equilateral triangle of height $24$ lies on line $l$. A circle of radius $12$ is tangent to $l$ and is externally tangent to the triangle. The area of the region exterior to the triangle and the circle and bounded by the triangle, the circle, and line $l$ can be written as $a\\sqrt{b} - c\\pi$, where $a$, $b$, and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is $a + b + c$?\n\n\n\n(A) 72 (B) 73 (C) 74 (D) 75 (E) 76",
"options": [],
"answer": "See solution",
"solution": "Because $\\angle BAC = 60^\\circ$ and $\\angle DEA = \\angle DGF = 90^\\circ$, both $\\triangle AEF$ and $\\triangle DGF$ are $30$-$60$-$90^\\circ$ right triangles. Therefore $DF = 24$, $EF = 24 - 12 = 12$, $AE = 4\\sqrt{3}$, $AF = 8\\sqrt{3}$, $FG = 12\\sqrt{3}$, and $AG = 12\\sqrt{3} - 8\\sqrt{3} = 4\\sqrt{3}$. The area of kite $GAED$ is twice the area of $\\triangle GAD$, so it is $48\\sqrt{3}$. The area of the $60^\\circ$-sector $EDG$ of the circle is $\\frac{1}{6} \\cdot \\pi \\cdot 12^2 = 24\\pi$. Thus the required area, shaded in the figure, is $48\\sqrt{3} - 24\\pi$, and the requested sum is $48 + 3 + 24 = 75$.\n\nAlternatively, the required area is $\\frac{1}{6}$ of the difference between the area of a circle of radius $12$ and a circumscribed regular hexagon. The hexagon is the union of $6$ equilateral triangles of side length $s = 12 \\cdot \\frac{2}{\\sqrt{3}}$. The area of the hexagon is $6 \\cdot \\frac{\\sqrt{3}}{4}s^2 = 288\\sqrt{3}$. The area of the circle is $144\\pi$. Then $\\frac{1}{6}$ of the difference is $48\\sqrt{3} - 24\\pi$, and the requested sum is $48 + 3 + 24 = 75$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21667,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be positive real numbers. Find the largest integer $n$ such that\n\n$$\n\\frac{1}{ax + b + c} + \\frac{1}{a + bx + c} + \\frac{1}{a + b + cx} \\ge \\frac{n}{a + b + c},\n$$\nfor all $x \\in [0, 1]$.",
"options": [],
"answer": "See solution",
"solution": "Let $x = 1$. Then\n$$\n\\frac{1}{a + b + c} + \\frac{1}{a + b + c} + \\frac{1}{a + b + c} = \\frac{3}{a + b + c} \\ge \\frac{n}{a + b + c},\n$$\nso $n \\le 3$.\n\nWe claim $n = 3$ works. It suffices to show\n$$\nE(x) = \\frac{1}{ax + b + c} + \\frac{1}{a + bx + c} + \\frac{1}{a + b + cx} \\ge \\frac{3}{a + b + c}\n$$\nfor all $x \\in [0, 1]$.\n\nBy the AM--GM inequality, for the numbers $ax + b + c$, $a + bx + c$, and $a + b + cx$,\n$$\nE(x) \\ge \\frac{9}{(ax + b + c) + (a + bx + c) + (a + b + cx)} = \\frac{9}{3a + 3b + 3c + (a + b + c)(x - 1)}.\n$$\nBut for $x \\in [0, 1]$, $ax + b + c \\le a + b + c$, so $E(x) \\ge \\frac{3}{a + b + c}$.\n\nTherefore, the largest integer $n$ is $\\boxed{3}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21668,
"subject": "Mathematics (Olympiad)",
"question": "令 $a, b, c, d$ 為任意實數且滿足 $a + b + c + d = 0$。試證：\n\n$$\n1296(a^7 + b^7 + c^7 + d^7)^2 \\leq 637(a^2 + b^2 + c^2 + d^2)^7.\n$$",
"options": [],
"answer": "See solution",
"solution": "由題設的對稱性，不妨設 $a$ 最大，$d$ 最小。依題設知 $a \\ge 0$，$d \\le 0$ 且 $d = -(a+b+c) \\le 0$，即 $a+b+c \\ge 0$。\n\n令 $S_k = a^k + b^k + c^k + d^k$，其中 $k$ 為正整數，則 $S_7 = a^7 + b^7 + c^7 - (a+b+c)^7$。\n\n因為當 $a = -b$ 或 $b = -c$ 或 $c = -a$ 時，$S_7 = 0$。故可假設\n\n$$\nS_7 = (a+b)(b+c)(c+a)[x(a^4 + b^4 + c^4) + y(a^3b + ab^3 + b^3c + bc^3 + c^3a + ca^3) + z(a^2b^2 + b^2c^2 + c^2a^2) + wabc(a+b+c)].\n$$\n\n（因為 $S_7$ 是關於 $a, b, c$ 對稱的齊次多項式。）\n\n在上式中分別取 $a = b = 1, c = 0$；$a = b = c = 1$；$a = b = 1, c = 2$；$a = b = 1, c = 3$ 得到\n\n$$\n\\begin{cases}\n2x + 2y + z + 63 = 0, \\\\\nx + 2y + z + w + 91 = 0, \\\\\n54x + 66y + 27z + 24w + 2709 = 0, \\\\\n332x + 248y + 76z + 60w + 9492 = 0.\n\\end{cases}\n$$\n\n解得 $x = -7, y = -14, z = -21, w = -35$。故可得\n\n$$\n\\begin{align*}\nS_7 &= -7(a+b)(b+c)(c+a)[(a^4 + b^4 + c^4) \\\\\n&\\quad + 2(a^3b + ab^3 + b^3c + bc^3 + c^3a + ca^3) \\\\\n&\\quad + 3(a^2b^2 + b^2c^2 + c^2a^2) + 5abc(a+b+c)] \\\\\n&= -7(a+b)(b+c)(c+a)[(a^2 + b^2 + c^2)^2 \\\\\n&\\quad + 2(ab + bc + ca)(a^2 + b^2 + c^2) + (ab + bc + ca)^2 + abc(a+b+c)] \\\\\n&= -7(a+b)(b+c)(c+a)[(a^2 + b^2 + c^2 + ab + bc + ca)^2 + abc(a+b+c)].\n\\end{align*}\n$$\n\n$$\n\\begin{align*}\nS_2 &= a^2 + b^2 + c^2 + (a+b+c)^2 \\\\\n&= 2(a^2 + b^2 + c^2 + ab + bc + ca) \\\\\n&= (a+b)^2 + (b+c)^2 + (c+a)^2.\n\\end{align*}\n$$\n\n由算幾不等式，得\n\n$$\n27(a + b)^2(b + c)^2(c + a)^2 \\leq [(a + b)^2 + (b + c)^2 + (c + a)^2]^3.\n$$\n\n又由\n\n$$\nabc(a + b + c) \\leq \\frac{1}{3}(ab + bc + ca)^2, \\quad ab + bc + ca \\leq a^2 + b^2 + c^2,\n$$\n\n得\n\n$$\n\\begin{align*}\n& 48[(a^2 + b^2 + c^2 + ab + bc + ca)^2 + abc(a + b + c)] \\\\\n& \\leq 48(a^2 + b^2 + c^2 + ab + bc + ca)^2 + 16(ab + bc + ca)^2 \\\\\n& \\leq 52(a^2 + b^2 + c^2 + ab + bc + ca)^2 \\\\\n&= 13[(a + b)^2 + (b + c)^2 + (c + a)^2]^2.\n\\end{align*}\n$$\n\n$$\n1296S_7^2\n$$\n\n$$\n\\begin{align*}\n&= 49 \\cdot 27(a+b)^2(b+c)^2(c+a)^2 \\times \\\\\n&\\quad \\{48[(a^2+b^2+c^2+ab+bc+ca)^2+abc(a+b+c)]\\}^2 \\\\\n&\\leq 49[(a+b)^2+(b+c)^2+(c+a)^2]^3 \\cdot \\{13[(a+b)^2+(b+c)^2+(c+a)^2]^2\\}^2 \\\\\n&= 637[(a+b)^2+(b+c)^2+(c+a)^2]^7.\n\\end{align*}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21669,
"subject": "Mathematics (Olympiad)",
"question": "Let $D$ and $E$ be points on sides $AB$ and $AC$ of triangle $ABC$, respectively, such that $BD = CE$. Let $M$ and $N$ be midpoints of $BC$ and $DE$, respectively. Prove that $MN$ is parallel to the bisector of $\\angle BAC$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $P$ and $Q$ be midpoints of the sides $BE$ and $CD$, respectively. Then $PN$ and $MQ$ are midlines of the triangles $BDE$ and $BDC$. Thus\n\n$$\nPN = QM = \\frac{BD}{2}\n$$\n\nand similarly\n\n$$\nNQ = PM = \\frac{EC}{2}.\n$$\n\nFrom the hypothesis, $BD = EC$. So $PNQM$ is a rhombus. Thus $MN$ is a bisector of $\\angle PNQ$, and $PN \\parallel BA$, $NQ \\parallel AC$ implies $AL \\parallel MN$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21670,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute triangle with a circumcircle $\\omega$. Let $\\ell$ be the tangent line to $\\omega$ at $A$. Let $X$ and $Y$ be the projections of $B$ onto lines $\\ell$ and $AC$, respectively. Let $H$ be the orthocenter of $BXY$. Let $CH$ intersect $\\ell$ at $D$. Prove that $BA$ bisects angle $CBD$.",
"options": [],
"answer": "See solution",
"solution": "\n\nNote that $XH \\perp BY \\perp AC$ and $YH \\perp BX \\perp AD$. Therefore $XH \\parallel AC$ and $YH \\parallel AD$. It follows that\n\n$$\n\\frac{AD}{AX} = \\frac{CD}{CH} = \\frac{CA}{CY} \\implies \\frac{AD}{CA} = \\frac{AX}{CY} = \\frac{AB \\sin \\angle XAB}{CB \\sin \\angle YCB}.\n$$\n\nSince $\\ell$ is tangent to $\\omega$, we have $\\angle XAB = \\angle YCB$. Thus the sines in the equality above cancel out and we obtain\n\n$$\n\\frac{AD}{CA} = \\frac{AB}{CB}.\n$$\n\nThis, along with $\\angle DAB = \\angle ACB$, proves that $\\triangle DAB \\sim \\triangle ACB$ by SAS. Therefore $\\angle CBA = \\angle ABD$. This shows that $BA$ bisects angle $CBD$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21671,
"subject": "Mathematics (Olympiad)",
"question": "How many 4-digit palindromes are divisible by 99? \n(A *palindrome* is a number, not starting with 0, that is the same when its digits are written in reverse order.)",
"options": [],
"answer": "See solution",
"solution": "A 4-digit palindrome has the form $abba$ where $a$ and $b$ are digits. Since the sum of the first and third digits, $a + b$, is equal to the sum of the second and fourth digits, all 4-digit palindromes are multiples of 11.\n\nSo we seek numbers of the form $abba$ which are divisible by 9. The sum of the digits is $2a + 2b = 2(a + b)$, so we seek $a, b$ such that $a + b = 9$ or $18$. These pairs are $\\{1,8\\}, \\{2,7\\}, \\{3,6\\}, \\{4,5\\}, \\{5,4\\}, \\{6,3\\}, \\{7,2\\}, \\{8,1\\}, \\{9,0\\}, \\{9,9\\}$, giving the palindromes $1881$, $2772$, $3663$, $4554$, $5445$, $6336$, $7227$, $8118$, $9009$, $9999$, a total of **10**.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21672,
"subject": "Mathematics (Olympiad)",
"question": "В шахматном турнире участвуют 8 игроков, каждый играет с каждым один раз. Какое наибольшее возможное число ничейных партий может быть в таком турнире?",
"options": [],
"answer": "See solution",
"solution": "Будем оценивать число $S$ — сумму количеств ничьих всех 8 шахматистов. Эта сумма ровно в 2 раза больше числа ничьих в турнире (каждую ничью посчитали 2 раза — у обоих игроков). Докажем, что $S \\le 41$ — тогда число ничейных партий в турнире не превосходит 20, так как оно целое. Это число и будет ответом, так как пример с 20 ничьими существует (см. рис. 8; шахматисты обозначены буквами A, B, C, D, E, F, G и H).\n\nЗаметим, что если два человека никому не проиграли, то они сыграли друг с другом вничью и потому, по условию, имеют разное число очков; то же самое верно, если они ни у кого не выиграли. Отсюда сразу следует, что есть не больше одного человека с 7 ничьими и не больше 2 человек с 6 ничьими (такой человек либо никому не проиграл, либо ни у кого не выиграл). Кроме того, людей с 5 ничьими, у которых обе результативные партии выиграны или обе проиграны — тоже не больше, чем по одному. Заметим, что все остальные игроки с 5 ничьими имеют по $3\\frac{1}{2}$ очка.\n\nПусть есть человек X с 7 ничьими; тогда других игроков с 5 ничьими быть не может, ибо у них будет столько же очков, сколько у X, причем они с X сыграют вничью. Значит, людей с 5 ничьими в этом случае не больше 2, и $S \\le 7+2\\cdot6+2\\cdot5+3\\cdot4 = 41$.\n\nПусть теперь человека с 7 ничьими нет. Оценим в этом случае количество игроков с 5 ничьими и $3\\frac{1}{2}$ очками. Каждый из них должен был сыграть с каждым не вничью; однако у них всего по 2 результативных партии — значит, их не больше 3, а всего игроков с 5 ничьими не больше $3+2=5$. Значит, и в этом случае $S \\le 2\\cdot6+5\\cdot5+4 = 41$, что и требовалось.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21673,
"subject": "Mathematics (Olympiad)",
"question": "Integers $a$, $b$, $c$ are such that $a + b + c$ is divisible by $6$, and $a^2 + b^2 + c^2$ is divisible by $36$. Does it imply that $a^3 + b^3 + c^3$ is divisible by:\n\na) $8$\n\nb) $27$?",
"options": [],
"answer": "See solution",
"solution": "a) Since $a + b + c$ is divisible by $6$, it is even, so there must be either $0$ or $2$ odd numbers among $a$, $b$, $c$. If there were $2$ odd numbers, $a^2 + b^2 + c^2$ would give a remainder of $0 + 1 + 1 = 2$ when divided by $4$, which is impossible since $a^2 + b^2 + c^2$ is divisible by $36$ (and thus by $4$). Therefore, all of $a$, $b$, $c$ are even. Hence, $a^3$, $b^3$, $c^3$ are all divisible by $8$, so their sum is divisible by $8$.\n\nb) Consider $a = 8$, $b = c = 2$. Then $8 + 2 + 2 = 12$ is divisible by $6$, and $8^2 + 2^2 + 2^2 = 64 + 4 + 4 = 72$ is divisible by $36$. But $8^3 + 2^3 + 2^3 = 512 + 8 + 8 = 528$, which is not divisible by $9$ (and thus not by $27$). Therefore, it does not necessarily follow that $a^3 + b^3 + c^3$ is divisible by $27$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21674,
"subject": "Mathematics (Olympiad)",
"question": "Given $n$ pairwise distinct positive numbers. Vasia wrote on the board all possible numbers of the form $a + \\frac{b}{c}$, where $a, b, c$ are distinct numbers from the given set. Can one always find two numbers on the board such that the larger one is not more than twice the other one, if \n\na) $n=4$? \n\nb) $n=3$?",
"options": [],
"answer": "See solution",
"solution": "**Answer:** a) yes, b) no.\n\n**Solution.**\n\na) We will show that for four given numbers one can always find two numbers on the board such that one is not more than twice the other one. Suppose by contradiction, it is not true. Let the given numbers be $a < b < c < d$. Then $d + \\frac{b}{a} > c + \\frac{b}{a} \\Rightarrow d + \\frac{b}{a} > 2(c + \\frac{b}{a}) \\Rightarrow d + \\frac{b}{a} > 1$.\n\nOn the other hand, $d + \\frac{b}{c} > d + \\frac{a}{c} \\Rightarrow d + \\frac{b}{c} > 2(d + \\frac{a}{c}) \\Rightarrow d < \\frac{b}{c} < 1$. Thus, we get a contradiction.\n\n\n\nb) We will give an example of three numbers for which the condition doesn't hold: $a = 10^{-1}$, $b = 10^{-3}$ and $c = 10^{-9}$. Then, the following numbers are written on the board:\n\n$10^{-1} + 10^{6}$, $10^{-1} + 10^{-6}$, $10^{-3} + 10^{8}$, $10^{-3} + 10^{-8}$, $10^{-9} + 10^{2}$, $10^{-9} + 10^{-2}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21675,
"subject": "Mathematics (Olympiad)",
"question": "Each of the positive rational numbers $a$ and $b$ has a minimal period of length 30 in decimal notation. The number $a-b$ has a minimal period of length 15 in decimal notation. Find the least positive integer $k$ for which it may happen that the length of the minimal period of the number $a+kb$ in decimal notation is also 15.",
"options": [],
"answer": "See solution",
"solution": "Assume the decimal representations of $a$, $b$, $a-b$, and $a+kb$ are purely periodic. Then $a = \\frac{m}{10^{30}-1}$ and $b = \\frac{n}{10^{30}-1}$, while $a-b$ and $a+kb$ can be written as fractions with denominator $10^{15}-1$. Thus, $\\gcd(k+1, 10^{15}+1) > 1$, so $k+1 \\geq 7$. For $k=6$, an example is $a = \\frac{8}{7(10^{15}-1)}$ and $b = \\frac{1}{7(10^{15}-1)}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21676,
"subject": "Mathematics (Olympiad)",
"question": "Consider a field $\\mathbb{L}$ with $q$ elements.\n\n(a) If $q \\equiv 3 \\pmod{4}$ and $n \\ge 2$ is a positive integer divisible by $q-1$, prove that $x^n = (x^2+1)^n$ for any $x \\in \\mathbb{L}^*$. \n\n(b) If there is an integer $n \\ge 2$ such that $x^n = (x^2+1)^n$ for any $x \\in \\mathbb{L}^*$, prove that $q \\equiv 3 \\pmod{4}$ and $q-1$ divides $n$.",
"options": [],
"answer": "See solution",
"solution": "a) The multiplicative group $\\mathbb{L}^*$ has order $q-1$, so for any $x \\in \\mathbb{L}^*$, $x^{q-1} = 1$. Since $n$ is a multiple of $q-1$, $x^n = 1$ for any $x \\in \\mathbb{L}^*$.\n\nBecause $4$ does not divide $q-1$, there are no elements of order $4$ in $\\mathbb{L}^*$, so $x^2 \\ne -1$ for any $x \\in \\mathbb{L}^*$. Thus $x^2+1 \\ne 0$, so $(x^2+1)^n = 1$ for all $x \\in \\mathbb{L}^*$. Therefore, $x^n = (x^2+1)^n$ for all $x \\in \\mathbb{L}^*$.\n\nb) If $\\operatorname{char}(\\mathbb{L}) = 2$, then $1 = 1^n = (1^2 + 1)^n = 0$, a contradiction. So $q$ is odd.\n\nIf $q \\equiv 1 \\pmod{4}$, let $a$ be a generator of $\\mathbb{L}^*$ and $b = a^{\\frac{q-1}{4}}$. Then $b$ has order $4$, so $b^2 \\ne 1$ and $(b^2)^2 = 1$. This implies $b^2 = -1$ and $b^n = (b^2+1)^n = 0$, a contradiction. Thus $q \\equiv 3 \\pmod{4}$.\n\nLet $A = \\{x + x^{-1} \\mid x \\in \\mathbb{L}^*\\}$ and $H = \\{x \\in \\mathbb{L}^* \\mid x^n = 1\\}$. $H$ is a subgroup of $\\mathbb{L}^*$. For any $y \\in A$, there is $x \\in \\mathbb{L}^*$ with $y = x + x^{-1} = x^{-1}(x^2 + 1)$, so $y^n = x^{-n}(x^2 + 1)^n = 1$. Thus $A \\subseteq H$.\n\nDefine $f : \\mathbb{L}^* \\to A$ by $f(x) = x + x^{-1}$. $f$ is onto, so $q-1 = |\\mathbb{L}^*| = \\sum_{y \\in A} |f^{-1}(y)|$.\n\nIf $u, v \\in \\mathbb{L}^*$ and $f(u) = f(v)$, then $u-v = v^{-1} - u^{-1}$, so $(u-v)(uv-1) = 0$, which means $v \\in \\{u, u^{-1}\\}$. For $u = u^{-1}$, $u^2 = 1$, so $u \\in \\{1, -1\\}$. Thus $|f^{-1}(2 \\cdot 1)| = 1$, $|f^{-1}(2 \\cdot (-1))| = 1$, and $|f^{-1}(y)| = 2$ for all $y \\in A \\setminus \\{2, -2\\}$.\n\nTherefore,\n$$\nq-1 = 1 + 1 + 2(|A| - 2) = 2|A| - 2,\n$$\nso $|A| = \\frac{q+1}{2}$. Since $A \\subseteq H$, $|H| \\ge \\frac{q+1}{2} > \\frac{1}{2}|\\mathbb{L}^*|$. As $H$ is a subgroup, $H = \\mathbb{L}^*$. Thus $x^n = 1$ for all $x \\in \\mathbb{L}^*$.\n\nTherefore, for a generator $a$ with $\\operatorname{ord}(a) = q-1$, $a^n = 1$, so $q-1$ divides $n$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21677,
"subject": "Mathematics (Olympiad)",
"question": "In a convex quadrilateral $ABCD$, diagonal $BD$ bisects neither $\\angle ABC$ nor $\\angle CDA$. Point $P$ lies inside quadrilateral $ABCD$ such that\n$$\n\\angle PBC = \\angle DBA \\quad \\text{and} \\quad \\angle PDC = \\angle BDA.\n$$\nProve that quadrilateral $ABCD$ is cyclic if and only if $AP = CP$.\n\n*Note:* The given conditions on the construction of point $P$ are symmetric with respect to $B$ and $D$, so we may assume without loss of generality that $P$ lies inside triangle $ACB$ (including the boundary).\n\n\n\nNext, note that $\\angle PBC = \\angle DBA$ if and only if $\\angle PBA = \\angle DBC$, and $\\angle PDC = \\angle BDA$ if and only if $\\angle PDA = \\angle DBA$. Hence, the given conditions are symmetric with respect to vertices $A$ and $C$. Without loss of generality, we may assume that $\\angle DBA < \\angle DBC$. Then $P$ lies inside triangle $BDC$; that is, we assume in all our solutions that we have the configuration shown on the left-hand side of the above figure. Set $\\alpha = \\angle PBC = \\angle DBA$ and $\\beta = \\angle PDC = \\angle BDA$.",
"options": [],
"answer": "See solution",
"solution": "**First Solution:**\n\n\n\nWe first prove the “only if” part by assuming that $ABCD$ is cyclic. Let lines $DP$ and $BP$ meet segment $AC$ at $R$ and $Q$, respectively. Then $\\angle RCD = \\angle ACD = \\angle ABD = \\alpha$ and $\\angle RCB = \\angle ACB = \\angle ADB = \\beta$. Hence, triangles $ABD$, $QBC$, and $RCD$ are similar to each other. Because triangles $QBC$ and $RCD$ are similar, $\\angle CQB = \\angle DRC$, implying that $\\angle QRP = \\angle RQP$. It follows that triangle $PQR$ is isosceles with $PQ = PR$. We conclude that\n$$\nPQ = PR \\quad \\text{and} \\quad \\angle PQA = \\angle PRC. \\qquad (*)\n$$\nBecause triangles $ABD$ and $QBC$ are similar, we have $\\frac{AB}{BD} = \\frac{QB}{BC}$. Note that $\\angle ABQ = \\angle DBC$. It follows that triangle $ABQ$ is similar to $DBC$. (That is, we obtain two pairs of triangles, namely $\\{ABD, QBC\\}$ and $\\{ABC, DBC\\}$, with \\textbf{spiral similarity}.) Hence\n$$\n\\frac{AQ}{DC} = \\frac{BQ}{BC}.\n$$\nBecause triangles $DRC$ and $CQB$ are similar, we have\n$$\n\\frac{RC}{DC} = \\frac{QB}{CB}.\n$$\nCombining the last two equations gives $AQ = RC$. Putting $AQ = RC$ together with the equations in $(*)$ shows that triangle $APQ$ is congruent to triangle $CPR$, implying that $AP = CP$.\n\nNow we prove the “if” part by assuming that $AP = CP$. Without loss of generality, we assume that $P$ lies inside triangle $ABC$. Then $D$ must lie on the ray $BD$ such that $\\angle DBA = \\angle PBC$. Let $\\omega$ denote the circumcircle of triangle $ABC$, and let $D'$ be the (unique) point on $\\widehat{AC}$ of $\\omega$ such that $\\angle D'BA = \\angle PBC$. We will first prove that $ABCD'$ and $P$ satisfy the conditions of the problem, that is, $\\angle AD'B = \\angle PD'C$.\n\n\n\nExtend segment $BP$ through $P$ to meet segment $AC$ at $Q$. Set $\\alpha = \\angle D'BA$ and $\\beta = \\angle BD'A$. Then we have $\\angle QCB = \\angle ACB = \\beta$ and $\\angle QBC = \\angle ACD' = \\alpha$. Thus, triangles $AD'B$ and $QCB$ are similar, and so triangles $ABQ$ and $D'BC$ are similar, implying that\n$$\n\\frac{AQ}{BQ} = \\frac{D'C}{BC}.\n$$\nConstruct the point $R'$ on segment $AC$ such that $CR' = AQ$. Substitute $CR' = AQ$ into the above equation:\n$$\n\\frac{CR'}{BQ} = \\frac{D'C}{BC} \\quad \\text{or} \\quad \\frac{D'C}{CR'} = \\frac{CB}{BQ}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21678,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs of natural numbers $(m, n)$ for which $2^n - 13^m$ is a cube of a natural number.",
"options": [],
"answer": "See solution",
"solution": "Let $2^n - 13^m = k^3$. Then $2^n \\equiv k^3 \\pmod{13}$. If $n$ is not divisible by $3$, then $2^n$ can be written as $4l^3$ or $2l^3$. Then either $4$ or $2$ is a cube of some integer modulo $13$. But by simple checking, only remainders $0$, $\\pm1$, $\\pm5$ can be cubes of integers modulo $13$. Therefore, $n$ is divisible by $3$, so let $n = 3t$. The condition can be rewritten as\n\n$$\n13^m = (2^t - k)(4^t + k \\cdot 2^t + k^2).\n$$\n\nThen both factors are powers of $13$, and the left one is smaller than the right one, so $2^t - k$ divides $4^t + k \\cdot 2^t + k^2$. Since $2^t \\equiv k \\pmod{(2^t - k)}$, we have $4^t + k \\cdot 2^t + k^2 \\equiv 3 \\cdot 4^t \\pmod{(2^t - k)}$. Therefore, $3 \\cdot 4^t$ is divisible by $2^t - k$, which is a power of $13$, so $2^t - k = 13^0 = 1$ and $13^m = 3 \\cdot 4^t - 3 \\cdot 2^t + 1$. Suppose that $t \\geq 4$. Then $13^m \\equiv 1 \\pmod{16}$, so $m$ is divisible by $4$. Therefore, $m = 4s$ and\n\n$$\n(13^{2s} + 1)(13^{2s} - 1) = 3 \\cdot 2^t \\cdot (2^t - 1).\n$$\n\nBut $13^{2s} + 1$ is not divisible by $4$ and $3$, so $13^{2s} + 1$ divides $2 \\cdot (2^t - 1)$. If these numbers are not equal, then\n\n$$\n13^{2s} + 1 \\leq 2^t - 1 < \\sqrt{3 \\cdot 2^t \\cdot (2^t - 1)} < 13^{2s}.\n$$\n\nOtherwise, $13^{2s} + 1 = 2 \\cdot (2^t - 1)$ and $13^{2s} - 1 = 3 \\cdot 2^{t-1}$. Subtracting the second equation from the first gives $2^{t-1} - 2 = 2$, so $t = 3$, which contradicts the assumption that $t > 3$. By checking for $t = 1$, $t = 2$, and $t = 3$, we verify that only $t = 3$ satisfies the condition. Therefore, $n = 9$ and $m = 2$ is the unique solution.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21679,
"subject": "Mathematics (Olympiad)",
"question": "For every positive integer $x$, denote by $\\kappa(x)$ the number of composite numbers not greater than $x$. Find all positive integers $n$ such that\n\n$$\n(\\kappa(n))! \\cdot \\operatorname{lcm}(1,2,\\dots,n) > (n-1)!\n$$",
"options": [],
"answer": "See solution",
"solution": "The inequality holds for $n = 2, 3, 4, 5, 7, 9$ and does not hold for $n = 1, 6, 8, 10, 11, 12$. Assume in the rest that $n \\ge 13$. By definition of $\\kappa(n)$, there exist exactly $n-1-\\kappa(n)$ prime numbers not greater than $n$; these are the primes dividing $\\operatorname{lcm}(1,2,\\dots,n)$. We have $n-1-\\kappa(n) \\ge 6$ as $n \\ge 13$. Let $q_1, \\dots, q_{n-1-\\kappa(n)}$ be all prime powers in the canonical representation of $\\operatorname{lcm}(1,2,\\dots,n)$. W.l.o.g., $q_1 > q_2 > \\dots > q_{n-1-\\kappa(n)}$. As at least 5 numbers among $q_1, \\dots, q_6$ are odd, in the case of odd $n$ we have\n\n$$\nq_1 q_2 q_3 q_4 q_5 q_6 \\le n(n-1)(n-2)(n-4)(n-6)(n-8)\n$$\n\nand in the case of even $n$ similarly\n\n$$\nq_1 q_2 q_3 q_4 q_5 q_6 \\le n(n-1)(n-3)(n-5)(n-7)(n-9).\n$$\n\nBut since $(n-3)(n-5)(n-7)(n-9) < (n-2)(n-4)(n-6)(n-8)$ and $n(n-8) < (n-3)(n-5)$, we anyway obtain\n\n$$\nq_1 q_2 q_3 q_4 q_5 q_6 < (n-1)(n-2)(n-3)(n-4)(n-5)(n-6) = \\frac{(n-1)!}{(n-7)!}\n$$\n\nAs $q_6 < n-6$, the inequality $q_i < n-i$ holds for every $i > 6$, whence\n\n$$\nq_7 \\cdots q_{n-1-\\kappa(n)} \\le (n-7) \\cdots (\\kappa(n)+1) = \\frac{(n-7)!}{(\\kappa(n))!}\n$$\n\nConsequently,\n\n$$\n\\operatorname{lcm}(1,2,\\dots,n) = q_1 q_2 \\cdots q_{n-1-\\kappa(n)} < \\frac{(n-1)!}{(\\kappa(n))!},\n$$\n\ncontradicting the original inequality. Hence the inequality holds for $n = 2, 3, 4, 5, 7, 9$ only.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21680,
"subject": "Mathematics (Olympiad)",
"question": "Sea $A$ un conjunto de 11 enteros positivos distintos. Demuestra que siempre es posible seleccionar 6 de ellos cuya suma es divisible por 6.",
"options": [],
"answer": "See solution",
"solution": "Primero, observamos que en cualquier conjunto de 3 o más enteros positivos distintos, siempre hay dos cuya suma es par, ya que el conjunto debe contener al menos dos números con el mismo residuo módulo 2 (ambos pares o ambos impares), y su suma será par.\n\nLuego, en cualquier conjunto de 5 o más enteros positivos distintos, siempre hay tres cuya suma es divisible por 3. Si hay tres números con residuos diferentes módulo 3, su suma es divisible por 3. Si no, al menos tres tienen el mismo residuo, y su suma también es divisible por 3.\n\nAplicando esto al conjunto $A$ de 11 números:\n- Podemos formar 5 parejas de números con suma par: $a+b$, $c+d$, $e+f$, $g+h$, $i+j$ (quedando uno sin emparejar).\n- De estas 5 sumas, por el argumento anterior, podemos seleccionar tres cuya suma es divisible por 3.\n- Estas tres sumas corresponden a seis números distintos, y la suma de estos seis números es divisible tanto por 2 como por 3, es decir, por 6.\n\nPor lo tanto, siempre es posible seleccionar 6 números de $A$ cuya suma es divisible por 6.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21681,
"subject": "Mathematics (Olympiad)",
"question": "Given $f(500) = f(100 \\times 5) = \\frac{f(100)}{5}$ and $f(500) = 3$, find $f(100)$.",
"options": [],
"answer": "See solution",
"solution": "We have $3 = f(500) = \\frac{f(100)}{5}$, so $f(100) = 3 \\times 5 = 15$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21682,
"subject": "Mathematics (Olympiad)",
"question": "Genji the ninja is to jump along the real axis, starting at the point $0$. In doing so, he alternates between the following two types of jumps, his first jump being type 1.\n\n*Type 1:* Genji jumps from the current position $x$ to a point $y$ in the set\n\n$$\n\\{x, x+1, x+2, x+3, x+4, x+5, x+6\\}.\n$$\n\n*Type 2:* Genji jumps from the current position $y$ to $(\\sqrt{2}-3)y$.\n\nLet $a$ and $b$ be integers. Prove that there exists a finite sequence of jumps that allows Genji to land on the point $a + b\\sqrt{2}$.",
"options": [],
"answer": "See solution",
"solution": "We first prove the following lemma.\n\n**Lemma.** Let $P(x) \\in \\mathbb{Z}[x]$. Then, there exists a polynomial $Q(x) \\in \\mathbb{Z}[x]$ with the property that all of its coefficients are in $\\{0, 1, \\dots, 6\\}$ and that $x^2 + 6x + 7$ divides the difference $P(x) - Q(x)$.\n\n**Proof.** (due to Wijit Yangjit) First, by the division algorithm, we can reduce the problem to the case where $\\deg P \\le 1$. Write $P(x) = ax + b$. Note that\n\n$$\nax + b \\equiv -\\left\\lfloor \\frac{b}{7} \\right\\rfloor x^2 + \\left( a - 6 \\left\\lfloor \\frac{b}{7} \\right\\rfloor \\right) x + \\left( b - 7 \\left\\lfloor \\frac{b}{7} \\right\\rfloor \\right) \\pmod{x^2 + 6x + 7}.\n$$\n\nSince $b - 7 \\left\\lfloor \\frac{b}{7} \\right\\rfloor \\in \\{0, 1, \\dots, 6\\}$, if we can solve the problem for $P_1(x) = a_1x + b_1$ where $a_1 = -\\left\\lfloor \\frac{b}{7} \\right\\rfloor$ and $b_1 = a - 6 \\left\\lfloor \\frac{b}{7} \\right\\rfloor$, we will finish.\n\nDefine $a_0 = a$ and $b_0 = b$. For all $n \\ge 0$,\n\n$$\n(a_{n+1}, b_{n+1}) = \\left( -\\left\\lfloor \\frac{b_n}{7} \\right\\rfloor, a_n - 6 \\left\\lfloor \\frac{b_n}{7} \\right\\rfloor \\right).\n$$\n\nObserve that for all $n \\ge 0$, we have\n\n$$\n0 \\le a_n + 6a_{n+1} + 7a_{n+2} \\le 6,\n$$\n\nfrom which we derive the monovariant relation:\n\n$$\n\\max(|a_{n+1}|, |a_{n+2}|) \\le \\max(|a_n|, |a_{n+1}|).\n$$\n\nThe integral sequence $\\{\\max(|a_n|, |a_{n+1}|)\\}_{n=0}^{\\infty}$ is non-increasing and bounded from below. Therefore, it eventually stabilizes. After certain casework, we deduce that it must stabilize at $0$, and thus for a sufficiently large $K$, $(a_K, b_K) = (0, 0)$. This finishes the proof of the lemma.\n\nNow, to solve the problem, let $\\alpha = \\sqrt{2} - 3$. We need to prove that every element in\n\n$$\nS = \\{a + b\\sqrt{2} \\mid a, b \\in \\mathbb{Z}\\} = \\{a + b\\alpha \\mid a, b \\in \\mathbb{Z}\\}\n$$\n\ncan be written in base $\\alpha$ using the digits $0, 1, 2, 3, 4, 5, 6$. That is, for $a, b \\in \\mathbb{Z}$, we need to prove that\n\n$$\na + b\\sqrt{2} = c_0 + c_1\\alpha + c_2\\alpha^2 + \\dots + c_k\\alpha^k\n$$\n\nfor some non-negative integer $k$ and $c_0, c_1, \\dots, c_k \\in \\{0, 1, 2, 3, 4, 5, 6\\}$. One may change the variables to\n\n$$\na + b\\alpha = c_0 + c_1\\alpha + c_2\\alpha^2 + \\dots + c_k\\alpha^k,\n$$\n\nand this is equivalent to proving that $\\alpha$ is a root of the equation\n\n$$\na + bx = c_0 + c_1x + c_2x^2 + \\dots + c_kx^k.\n$$\n\nSince the minimal polynomial of $\\alpha$ is $x^2 + 6x + 7$, it now suffices to show that, for any $a, b \\in \\mathbb{Z}$,\n\n$$\na + bx \\equiv c_0 + c_1x + c_2x^2 + \\dots + c_kx^k \\pmod{x^2 + 6x + 7}\n$$\n\nfor some positive integer $k$ and $c_0, c_1, \\dots, c_k \\in \\{0, 1, 2, 3, 4, 5, 6\\}$. This is done by the lemma.\n\n**Remark 1.** Most students who solve this problem actually do so by defining some norm on $S = \\{a + b\\sqrt{2} \\mid a, b \\in \\mathbb{Z}\\}$. Such approaches typically require a bashing of around 100 small cases.\n\n**Remark 2.** This problem is a special case of the results given in the paper W. J. Gilbert, Radix Representations of Quadratic Fields, Journal of Mathematical Analysis and Applications, 83 (1981), 264-274.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21683,
"subject": "Mathematics (Olympiad)",
"question": "Given roots $\\alpha$, $\\beta$, $\\gamma$ of a cubic equation, the sum of the squares of the roots is $s_2 = \\alpha^2 + \\beta^2 + \\gamma^2$. Express $s_2$ in terms of the elementary symmetric sums $\\alpha + \\beta + \\gamma$ and $\\alpha\\beta + \\beta\\gamma + \\gamma\\alpha$, and determine the condition for $s_2 > 0$ if the roots are distinct.",
"options": [],
"answer": "See solution",
"solution": "$$\n\\begin{align*}\ns_2 &= \\alpha^2 + \\beta^2 + \\gamma^2 \\\\\n &= (\\alpha + \\beta + \\gamma)^2 - 2(\\alpha\\beta + \\beta\\gamma + \\gamma\\alpha) \\\\\n &= 0^2 - 2p \\\\\n &= -2p\n\\end{align*}\n$$\nSince the roots are distinct, $s_2 > 0$ if and only if $p < 0$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21684,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an isosceles triangle ($AB = AC$). Let $D$ and $E$ be two points on the side $BC$ such that $D \\in BE$, $E \\in DC$ and $2\\angle DAE = \\angle BAC$. Prove that we can construct a triangle $XYZ$ such that $XY = BD$, $YZ = DE$ and $ZX = EC$. Find $\\angle BAC + \\angle YXZ$.",
"options": [],
"answer": "See solution",
"solution": "Let $\\omega$ be the circle of center $A$ and radius $AB$. Let $A'$ be a point on the circle $\\omega$, lying on the minor arc $\\widehat{BC}$, such that $\\angle BAD = \\angle A'AD$. Since $2\\angle DAE = \\angle A$, it is easy to see that $\\angle CAE = \\angle A'AE$.\n\n\n\nWe deduce that the triangles $BAD$ and $A'AD$ are congruent by SAS postulate, and thus $BD = A'D$. Similarly, $EC = A'E$, and thus, the triangle $A'DE$ has its sides of lengths $BD$, $DE$ and $EC$ respectively, and we can choose $X = A'$, $Y = D$ and $Z = E$.\n\nMoreover,\n\n$$\n\\begin{align*}\n\\angle BAC + \\angle YXZ &= \\angle BAC + \\angle DA'E \\\\\n&= \\angle BAC + \\angle DA'A + \\angle AA'E \\\\\n&= \\angle BAC + \\angle DBA + \\angle ACE = 180^{\\circ}.\n\\end{align*}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21685,
"subject": "Mathematics (Olympiad)",
"question": "For all triplets $a, b, c$ of pairwise distinct real numbers, prove the inequality\n\n$$\n\\left| \\frac{a}{b-c} \\right| + \\left| \\frac{b}{c-a} \\right| + \\left| \\frac{c}{a-b} \\right| \\ge 2\n$$\n\nand determine all cases of equality.\n\nProve that if we also impose $a, b, c$ positive, then all equality cases disappear, but the value $2$ remains the best constant possible.",
"options": [],
"answer": "See solution",
"solution": "Denote $x = \\frac{a}{b-c}$, $y = \\frac{b}{c-a}$, $z = \\frac{c}{a-b}$. It is easily seen that\n\n$$\n\\prod (x-1) = \\frac{\\prod (a-b+c)}{\\prod (b-c)} = \\frac{\\prod (a+b-c)}{\\prod (b-c)} = \\prod (x+1),\n$$\n\nwhence $xy + yz + zx = -1$. This relation can also be obtained by writing the homogeneous system with $a, b, c$ as unknowns, of determinant\n\n$$\n\\Delta = \\begin{vmatrix} 1 & -x & x \\\\ y & 1 & -y \\\\ -z & z & 1 \\end{vmatrix} = (1 + yz) + x(y - yz) + x(yz + z),\n$$\n\nwhich computes to $\\Delta = 1 + xy + yz + zx$. Since the system originates with $a, b, c$ pairwise distinct, it must also have some other solution than the trivial one $a = b = c = 0$, thus $\\Delta = 0$, yielding the relation $xy + yz + zx = -1$.\n\nBut since $(xy)(yz)(zx) = (xyz)^2 \\ge 0$, we must have at least one of the factors being non-negative, say $xy \\ge 0$. Then\n\n$$\n|x| + |y| + |z| \\ge |x+y| + |z| \\ge 2\\sqrt{|zx + yz|} = 2\\sqrt{1+xy} \\ge 2.\n$$\n\nEquality occurs for $xy = 0$, say $x = 0$, and also $|y| = 1$, with $z = -y$, when $a = 0$ and $b = -c$. Thus all equality cases are $\\{a, b, c\\} = \\{0, t, -t\\}$, for $t \\ne 0$. However, since the value of the expression for $a, b, c$ is the same as that for $ta, tb, tc$ with $t \\ne 0$, the only essential solution is $\\{0, 1, -1\\}$.\n\nIf we also impose $a, b, c > 0$, then the same inequality with respect to $2$ holds, but there is no case of equality. The fact we can approach $2$ as close as wanted is argued by taking an arbitrary $0 < \\varepsilon < 1/2$, and $a = \\varepsilon^2$, $b = 1 + \\varepsilon$, $c = 1 - \\varepsilon$, when\n\n$$\n\\left| \\frac{a}{b-c} \\right| + \\left| \\frac{b}{c-a} \\right| + \\left| \\frac{c}{a-b} \\right| = \\varepsilon \\left( \\frac{1}{2} + \\frac{2+\\varepsilon}{1-\\varepsilon-\\varepsilon^2} - \\frac{2-\\varepsilon}{1+\\varepsilon-\\varepsilon^2} \\right) + 2,\n$$\n\nwith $\\lim_{\\varepsilon \\to 0} \\varepsilon \\left( \\frac{1}{2} + \\frac{2+\\varepsilon}{1-\\varepsilon-\\varepsilon^2} - \\frac{2-\\varepsilon}{1+\\varepsilon-\\varepsilon^2} \\right) = 0$.\n\n*Remarks.* Notice for $a + b + c = 0$ the inequality becomes\n\n$$\n\\left| \\frac{a+b}{a-b} \\right| + \\left| \\frac{b+c}{b-c} \\right| + \\left| \\frac{c+a}{c-a} \\right| \\ge 2,\n$$\n\na relatively easier inequality. Thus, in this case, the two inequalities are equivalent. This can be pushed further. For a new variable $\\sigma$, the inequality\n\n$$\n\\left| \\frac{a - \\sigma}{b - c} \\right| + \\left| \\frac{b - \\sigma}{c - a} \\right| + \\left| \\frac{c - \\sigma}{a - b} \\right| \\ge 2\n$$\n\ncan be proved in the same manner, with all equality cases given by $\\{a, b, c\\} = \\{\\sigma, \\tau, 2\\sigma - \\tau\\}$, for $\\tau \\neq \\sigma$ (this allows equality cases even when $a, b, c$ are positive, by taking $0 < \\sigma, 0 < \\tau < 2\\sigma$). Taking $\\sigma = a + b + c$ yields the alternative inequality in all its generality.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21686,
"subject": "Mathematics (Olympiad)",
"question": "Let $n = 2012^{2011}$. How many divisors of $n$ give a remainder of $1$ when divided by $3$?",
"options": [],
"answer": "See solution",
"solution": "First, note that $2012 \\equiv 2 \\pmod{3}$, so $n = 2012^{2011} \\equiv 2^{2011} \\pmod{3}$. Since $2^1 \\equiv 2$, $2^2 \\equiv 1$, $2^3 \\equiv 2$, $2^4 \\equiv 1$, the powers of $2$ modulo $3$ cycle every $2$ terms. $2011$ is odd, so $2^{2011} \\equiv 2 \\pmod{3}$, thus $n \\equiv 2 \\pmod{3}$.\n\nIf $d$ is a divisor of $n$, then $\\frac{n}{d}$ is also a divisor. The product $d \\cdot \\frac{n}{d} = n \\equiv 2 \\pmod{3}$. Therefore, the remainders of $d$ and $\\frac{n}{d}$ modulo $3$ multiply to $2$. The possible pairs are $(1,2)$ and $(2,1)$, so half of the divisors give remainder $1$ and half give remainder $2$ modulo $3$.\n\nThe prime factorization of $2012$ is $2^2 \\cdot 503$, so $n = 2^{4022} \\cdot 503^{2011}$. The number of divisors of $n$ is $(4022+1) \\times (2011+1) = 4023 \\times 2012$.\n\nThus, the number of divisors of $n$ that give remainder $1$ when divided by $3$ is:\n\n$$\n\\frac{4023 \\times 2012}{2} = 4023 \\times 1006\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21687,
"subject": "Mathematics (Olympiad)",
"question": "Given are 51 natural numbers written in a row. Their sum is 100. An integer is *representable* if it can be expressed as the sum of several consecutive numbers in the given row. Prove that for each $k \\in \\{1, 2, \\dots, 100\\}$, one of the numbers $k$ and $100 - k$ is representable.",
"options": [],
"answer": "See solution",
"solution": "Let the given row be $a_1, \\dots, a_{51}$. Take a circle $\\gamma$ of length 100. Mark 51 blue points on it so that they determine 51 consecutive arcs of lengths $a_1, \\dots, a_{51}$. One may imagine each number $a_i$ written next to an arc of $\\gamma$ with length $a_i$. So, starting from a certain blue point $B$, the numbers are arranged around $\\gamma$ as in the given row.\n\nThe claim that $k$ or $100 - k$ is representable is equivalent to saying that there is an arc $\\alpha$ of length $k$ with blue endpoints. Indeed, consider $\\alpha$ and its complementary arc $\\alpha'$, of length $100 - k$. Point $B$ cannot be interior to both $\\alpha$ and $\\alpha'$, hence $k$ or $100 - k$ equals the sum of several consecutive numbers in the initial row.\n\nWe show that for each $k \\leq 50$ there is an arc of length $k$ with blue endpoints (once this is proven, the case $k > 50$ follows trivially). Mark 49 more red points on $\\gamma$ so that the total of marked 100 points, blue and red, yields a division into arcs of length 1. Now $k = 50$ is almost immediate. To each blue point assign its diametrically opposite point (the number 100 of division points is even). The 51 assigned points are distinct. Since there are $49 < 51$ red points, some assigned point is blue, which is enough.\n\nThe essential case $k < 50$ needs an appropriate modification. For each blue point $X$, write down the endpoints of the arc with length $2k$ and midpoint $X$ (this arc is \"proper\", it does not overlap itself). One obtains a list of $2 \\cdot 51 = 102$ points. It is clear that no point occurs in it more than twice.\n\nTherefore, the list contains at least 51 distinct points. There are only $49 < 51$ red points. Hence some blue point is listed, completing the argument.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21688,
"subject": "Mathematics (Olympiad)",
"question": "In a school with $n$ students, a team of students must be chosen to clean the school. A team of students is called *good* if the school principal believes that they can manage to clean the school. It is known that at least one team is good, and that if a team is good, then it is also good after adding any other student. Show that the average number of students in good teams is at least $\\frac{n}{2}$.",
"options": [],
"answer": "See solution",
"solution": "Let $S$ be the set of students and let $G$ be the set of good teams. For any student $s$, consider the trivial injection from the set of good teams without $s$ to the set of good teams containing $s$ by taking a team $T$ and mapping it to $T \\cup \\{s\\}$. Since this is an injection, we conclude that there are at least as many good teams with $s$ as without $s$. This means that\n\n$$\n\\sum_{T \\in G} 1_{(s \\in T)} \\geq \\frac{|G|}{2}.\n$$\n\nThe average number of students in good teams is then\n\n$$\n\\frac{1}{|G|} \\sum_{T \\in G} |T| = \\frac{1}{|G|} \\sum_{T \\in G} \\sum_{s \\in S} 1_{(s \\in T)} = \\frac{1}{|G|} \\sum_{s \\in S} \\sum_{T \\in G} 1_{(s \\in T)} \\geq \\frac{1}{|G|} \\sum_{s \\in S} \\frac{|G|}{2} = \\frac{n}{2}.\n$$\n\nNote: The last part is just linearity of expectation in disguise.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21689,
"subject": "Mathematics (Olympiad)",
"question": "Let a tortoise and a hare move along the boundary of a square of side length $pq$ units, where $p$ and $q$ are odd positive integers. In each time-step, the tortoise moves $p$ units and the hare moves $q$ units. For $0 \\leq t < pq$, let $T$ and $H$ denote their positions at time $t$.\n\nDetermine the proportion of times $T$ and $H$ are on the same side of the square, in the following cases:\n\n(a) $\\frac{1}{4}$, when $p-q$ is odd;\n\n(b) $\\frac{1}{4} + \\frac{1}{4pq}$ when $p-q$ is even but not divisible by $4$;\n\n(c) $\\frac{1}{4} + \\frac{3}{4pq}$ when $4 \\mid p-q$.",
"options": [],
"answer": "See solution",
"solution": "(b) We combine aspects of the two previous cases. Here, the two vertices are opposite. Therefore, for any time $0 \\leq t < pq$, $T$ and $H$ are on the same side or on opposite sides of the square at time $t$ if and only if they are on the same side or on opposite sides at time $t + pq$. Furthermore, on this event, they are on the same side at precisely one of the times $\\{t, t + pq\\}$.\n\nSo we calculate the proportion of times the runners are on the same side or opposite sides before timestep $pq$. Again from (1), using $p \\equiv -q \\pmod{4}$,\n\n$$\na - b \\equiv m'p - mq \\equiv (m' + m)p \\pmod{4}.\n$$\n\nSince $p$ is odd, $2 \\mid m' + m$ precisely if $2 \\mid a - b$. We enumerate\n\n$$\n\\begin{aligned}\nL(p, q) &:= \\left| \\{ (a, b) \\in [0, q-1] \\times [0, p-1] : 2 \\mid a-b \\} \\right| \\\\\n&= \\frac{p+1}{2} \\times \\frac{q+1}{2} + \\frac{p-1}{2} \\times \\frac{q-1}{2}.\n\\end{aligned}\n$$\n\nSo $\\frac{L(p,q)}{2pq} = \\frac{1}{4} + \\frac{1}{4pq}$ as required.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21690,
"subject": "Mathematics (Olympiad)",
"question": "Points $P$ and $Q$ lie inside parallelogram $ABCD$ and are such that triangles $ABP$ and $BCQ$ are equilateral. Prove that the line through $P$ perpendicular to $DP$ and the line through $Q$ perpendicular to $DQ$ meet on the altitude from $B$ in triangle $ABC$.",
"options": [],
"answer": "See solution",
"solution": "Let $\\angle ABC = m$ and let $O$ be the circumcenter of triangle $DPQ$. Since $P$ and $Q$ are in the interior of $ABCD$, it follows that $m = \\angle ABC > 60^\\circ$ and $\\angle DAB = 180^\\circ - m > 60^\\circ$, which together imply that $60^\\circ < m < 120^\\circ$.\n\nNow note that $\\angle DAP = \\angle DAB - 60^\\circ = 120^\\circ - m$, $\\angle DCQ = \\angle DCB - 60^\\circ = 120^\\circ - m$, and that $\\angle PBQ = 60^\\circ - \\angle ABQ = 60^\\circ - (\\angle ABC - 60^\\circ) = 120^\\circ - m$. This, combined with the facts that $AD = BQ = CQ$ and $AP = BP = CD$, implies that triangles $DAP$, $QBP$, and $QCD$ are congruent. Therefore $DP = PQ = DQ$ and triangle $DPQ$ is equilateral.\n\nThis implies that $\\angle ODA = \\angle PDA + 30^\\circ = \\angle DQC + 30^\\circ = \\angle OQC$. Combining this fact with $OQ = OD$ and $CQ = AD$ implies that triangles $ODA$ and $OQC$ are congruent. Therefore $OA = OC$ and, if $M$ is the midpoint of segment $AC$, it follows that $OM$ is perpendicular to $AC$. Since $ABCD$ is a parallelogram, $M$ is also the midpoint of $DB$.\n\nIf $K$ denotes the intersection of the line through $P$ perpendicular to $DP$ and the line through $Q$ perpendicular to $DQ$, then $K$ is diametrically opposite $D$ on the circumcircle of $DPQ$ and $O$ is the midpoint of segment $DK$. This implies that $OM$ is a midline of triangle $DBK$ and hence that $BK$ is parallel to $OM$, which is perpendicular to $AC$. Therefore $K$ lies on the altitude from $B$ in triangle $ABC$, as desired. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21691,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCDE$ be a convex pentagon with all sides of length $1$, such that some pair of diagonals intersect perpendicularly. Find the maximum possible area of $ABCDE$.\n\n",
"options": [],
"answer": "See solution",
"solution": "We consider two cases: $AC \\perp AD$ and $AC \\perp BD$.\n\n**Case 1:** $AC \\perp AD$\n\nIn $\\triangle ACD$, $\\angle CAD = 90^\\circ$ and $AC, AD \\leq CD = 1$. In $\\triangle ABC$, $AB = BC \\geq CA$, so $\\angle CAB \\geq 60^\\circ$. Similarly, in $\\triangle AED$, $\\angle DAE \\geq 60^\\circ$. Thus, $\\angle BAE = \\angle CAB + \\angle CAD + \\angle DAE \\geq 60^\\circ + 90^\\circ + 60^\\circ = 210^\\circ > 180^\\circ$, which contradicts convexity. So this case is impossible.\n\n**Case 2:** $AC \\perp BD$\n\nLet $P$ be the intersection of $AC$ and $BD$. Since $\\angle BPC = \\angle BPA = 90^\\circ$, $\\triangle APB \\cong \\triangle CPB$ and $AP = CP$. Similarly, $\\triangle APD \\cong \\triangle CPD$, so $AD = CD = 1$. Thus, $AD = DE = EA = 1$, so $\\triangle ADE$ is equilateral with side $1$. Also, $AB = BC = CD = DA = 1$, so $ABCD$ is a rhombus with side $1$. The area of $ADE$ is $\\frac{\\sqrt{3}}{4}$, and the maximal area of $ABCD$ is $1$ (when it is a square). The pentagon is convex in this configuration.\n\nTherefore, the maximum area is:\n\n$$1 + \\frac{\\sqrt{3}}{4}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21692,
"subject": "Mathematics (Olympiad)",
"question": "已知多項式函數 $f, g$ 為實數映至實數。試求出所有的多項式函數對 $(f(x), g(x))$ 使得：對任意實數 $x$，\n\n$$\nf(f(f(f(x)))) = g(g(g(g(x))))\n$$\n\n成立。",
"options": [],
"answer": "See solution",
"solution": "所有滿足條件的多項式函數對為：\n\n1. $f(x) = g(x)$。\n2. $f(x) = S(x+a) - a$, $g(x) = -S(x+a) - a$，其中 $a$ 為任意實數，$S$ 為任意奇多項式函數（即 $S(x) = \\sum a_i x^{2k+1}$）。\n\n證明如下：\n\n設 $F(x) = f(f(x))$，$G(x) = g(g(x))$。則 $F(F(x)) = G(G(x))$。\n\n考察 $F(x)$ 和 $G(x)$ 的首項係數，設 $F(x) = a_0 x^m + \\cdots$，$G(x) = b_0 x^m + \\cdots$。展開 $F(F(x))$，最高次項係數為 $a_0^{m+1}$，同理 $G(G(x))$ 為 $b_0^{m+1}$。由 $F(F(x)) = G(G(x))$，得 $a_0^{m+1} = b_0^{m+1}$，所以 $a_0 = b_0$ 或 $a_0 = -b_0$（後者僅當 $m$ 為奇數時成立）。\n\n用數學歸納法可證 $F(x) = G(x)$ 或 $F(x) = -G(x) + r$（$r$ 為常數）。\n\n若 $F(x) = G(x)$，則 $f(f(x)) = g(g(x))$，進而 $f(x) = g(x)$。\n\n若 $F(x) = -G(x) + r$，則 $G(x)$ 必為以 $\\left(\\frac{r}{2}, \\frac{r}{2}\\right)$ 為中心的奇函數，即 $G(x) = S(x+a) - a$，$F(x) = -S(x+a) - a$，其中 $S$ 為奇多項式，$a = \\frac{r}{2}$。\n\n但回代原式，只有 $f(x) = g(x)$ 和 $f(x) = S(x+a) - a$, $g(x) = -S(x+a) - a$ 這兩種情形成立。\n\n因此，所有解如上。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21693,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be positive real numbers such that $abc = 1$. Prove that the following inequality holds:\n\n$$\n\\frac{1}{2}(\\sqrt{a} + \\sqrt{b} + \\sqrt{c}) + \\frac{1}{1+a} + \\frac{1}{1+b} + \\frac{1}{1+c} \\ge 3\n$$\n\nWhen does equality hold?",
"options": [],
"answer": "See solution",
"solution": "Since $(1 - \\sqrt{bc})^2 \\ge 0$, it follows that $1 + bc \\ge 2\\sqrt{bc}$, i.e., $\\frac{1}{2\\sqrt{bc}} \\ge \\frac{1}{1+bc}$. We get that\n\n$$\n\\frac{\\sqrt{a}}{2} + \\frac{1}{1+a} = \\frac{1}{2\\sqrt{bc}} + \\frac{1}{1+a} \\ge \\frac{1}{1+bc} + \\frac{1}{1+a} = \\frac{1}{1+\\frac{1}{a}} + \\frac{1}{1+a} = 1.\n$$\n\nIn the same way, we prove that\n\n$$\n\\frac{\\sqrt{b}}{2} + \\frac{1}{1+b} \\ge 1\n$$\n\nand\n\n$$\n\\frac{\\sqrt{c}}{2} + \\frac{1}{1+c} \\ge 1.\n$$\n\nBy adding these three inequalities, we get the required result. Equality holds if and only if $a = 1$, $b = 1$, and $c = 1$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21694,
"subject": "Mathematics (Olympiad)",
"question": "The distance between the parabola $y = ax^2 + bx + c$ and the axis of abscissae is equal to\n$$\nd = \\frac{4ac - b^2}{4a} \\quad (*)\n$$\nTherefore, the problem is equivalent to the following: find the greatest and the smallest value of the expression $(*)$ if positive integers $a, b, c$ satisfy the inequalities $1 \\leq a \\leq 50$, $1 \\leq b \\leq 50$, $1 \\leq c \\leq 50$ and $b^2 - 4ac < 0$.",
"options": [],
"answer": "See solution",
"solution": "We find the greatest value. We have\n$$\nd = \\frac{4ac - b^2}{4a} = c - \\frac{b^2}{4a} \\leq 50 - \\frac{b^2}{4a} \\leq 50 - \\frac{1}{4a} \\leq 50 - \\frac{1}{200}.\n$$\nThe first inequality follows from $c \\leq 50$, the second from $a \\leq 50$ and $b^2 \\geq 1$. All inequalities become equalities for $c = 50$, $b = 1$, $a = 50$. For these values, $b^2 - 4ac = 1^2 - 4 \\cdot 50 \\cdot 50 < 0$ is valid. Therefore, the greatest value is $50 - 1/200$.\n\nNow, for the smallest value: in $(*)$ the numerator and denominator are positive integers. The numerator cannot be 1 or 2, since $b^2 = 4ac - 1$ or $b^2 = 4ac - 2$ would make $b^2$ congruent to 2 or 3 mod 4, which is impossible.\n\nSo the smallest numerator is 3, i.e., $b^2 = 4ac - 3$. Then $b$ is odd, $b = 2k + 1$, so $(2k+1)^2 = 4ac - 3$, i.e., $k^2 + k + 1 = ac$. Thus $a$ and $c$ are odd. Try $a = 49$; $c = \\frac{b^2+3}{4a} \\leq \\frac{2503}{196} < 13$. For $c = 1, 3, 5, 7, 9, 11$, only $c = 7$ gives $196 \\cdot 7 - 3 = 1369 = 37^2$. So for $a = 49$, $c = 7$, $b = 37$, $d = \\frac{3}{196}$.\n\nIf the denominator is at least 4, $d \\geq \\frac{4}{200} = \\frac{1}{50} > \\frac{3}{196}$. Therefore, the smallest value is $\\frac{3}{196}$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21695,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be an isosceles triangle with $AB = AC = 3$ and $BC = 4$. Squares $PQBA$, $RSCB$, and $TUAC$ are constructed externally on sides $AB$, $BC$, and $CA$, respectively. Let $P, Q, R, S, T, U$ be the vertices of these squares not shared with $\\triangle ABC$. Find the area of the hexagon $PQRSTU$.",
"options": [],
"answer": "See solution",
"solution": "First, the areas of triangles $AUP$, $BQR$, and $CST$ are each equal to the area of $\\triangle ABC$. For example, since $\\angle QBR + \\angle ABC = 180^\\circ$, $QB = AB$, and $BR = BC$, rotating $\\triangle BQR$ $90^\\circ$ clockwise about $B$ maps it to a triangle with the same base and height as $\\triangle ABC$, so their areas are equal. The same holds for $AUP$ and $CST$.\n\nTherefore, the area of hexagon $PQRSTU$ equals the sum of the areas of the three squares $PQBA$, $RSCB$, $TUAC$ plus four times the area of $\\triangle ABC$.\n\nThe areas of the squares are $3^2 + 3^2 + 4^2 = 9 + 9 + 16 = 34$.\n\nTo find the area of $\\triangle ABC$: since $AB = AC = 3$ and $BC = 4$, $\\triangle ABC$ is isosceles. Let $M$ be the midpoint of $BC$. Then $AM \\perp BC$, and by the Pythagorean theorem,\n\n$$\nAM = \\sqrt{3^2 - 2^2} = \\sqrt{9 - 4} = \\sqrt{5}\n$$\n\nSo the area is\n\n$$\n\\frac{1}{2} \\cdot BC \\cdot AM = \\frac{1}{2} \\cdot 4 \\cdot \\sqrt{5} = 2\\sqrt{5}\n$$\n\nThus, the area of hexagon $PQRSTU$ is\n\n$$\n34 + 4 \\times 2\\sqrt{5} = 34 + 8\\sqrt{5}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21696,
"subject": "Mathematics (Olympiad)",
"question": "a. How many distinct flags can be made if there are 5 colours available, and each flag consists of a central region, top-left, top-right, bottom-left, and bottom-right regions, with each region assigned a different colour?\n\nb. How many distinct flags can be made if each of the five regions (central, top-left, top-right, bottom-right, bottom-left) must have a different colour, chosen from 5 available colours?\n\nc. How many flags are there with certain symmetry properties? Specifically:\n- (i) Flags that look the same when flipped top to bottom or side to side, with two different outer colours.\n- (ii) Flags that look the same when flipped top to bottom (type $H$), or side to side (type $V$), or both.\n\nd. How many flags are there with diagonal symmetry? Specifically:\n- (i) Flags that look the same when flipped about the forward diagonal, the backward diagonal, or both.\n- (ii) Flags that look the same when flipped about the forward diagonal, the backward diagonal, or both, considering all possible cases.",
"options": [],
"answer": "See solution",
"solution": "a. There are 5 colours to choose from for the central region, then 4 for the top-left (and bottom-right) region, then 3 for the top-right region, then 2 for the bottom-left region. So the number of required flags is $5 \\times 4 \\times 3 \\times 2 = 120$.\n\nb. There are 5 colours to choose from for the central region, then 4 for the top-left region, then 3 for the top-right region, then 2 for the bottom-right region, then 1 for the bottom-left region. So the number of required flags is $5 \\times 4 \\times 3 \\times 2 \\times 1 = 120$.\n\nc. **Alternative i**\n\nA flag looks the same when flipped top to bottom if and only if the bottom-left and top-left colours are the same and the bottom-right and top-right colours are the same. The central region can be any colour. Hence the number of these flags with the two outer colours different is $5 \\times 4 \\times 5 = 100$.\n\nSimilarly, the number of flags that look the same when flipped side to side and have two different outer colours is $5 \\times 4 \\times 5 = 100$.\n\nA flag looks the same when flipped top to bottom and when flipped side to side if and only if the four outer regions have the same colour. The central region can be any colour. Hence the number of these flags is $5 \\times 5 = 25$.\n\nSo the number of required flags is $100 + 100 + 25 = 225$.\n\n**Alternative ii**\n\nA flag is the same when flipped top to bottom if and only if the bottom-left and top-left colours are the same and the bottom-right and top-right colours are the same. The central region can be any colour. Hence the number of these flags is $5 \\times 5 \\times 5 = 125$. Call these flags type $H$.\n\nSimilarly, the number of flags that are the same when flipped side to side is 125. Call these flags type $V$.\n\nIf we add the number of type $H$ flags and the number of type $V$ flags, then we will have counted twice the flags that are both types.\n\nA flag of both types must have all non-central colours the same. The central region can be any colour. Hence the number of flags of both types is $5 \\times 5 = 25$.\n\nSo the number of required flags is $125 + 125 - 25 = 225$.\n\nd. **Alternative i**\n\nA flag looks the same when flipped about its forward diagonal if and only if the bottom-right and top-left colours are the same.\n\nA flag looks the same when flipped about its backward diagonal if and only if the bottom-left and top-right colours are the same.\n\nIn either case, the centre region can have any colour.\n\nSo the number of flags that look the same when flipped about a given diagonal but not the other is $5 \\times 5 \\times 5 \\times 4 = 500$ and the number of flags that have both diagonal symmetries is $5 \\times 5 \\times 5 = 125$.\n\nSo the number of required flags is $500 + 500 + 125 = 1125$.\n\n**Alternative ii**\n\nFor a flag to be the same when flipped about its forward diagonal, the bottom-right and top-left colours must be the same. The other three regions can be any colour. Hence the number of flags in this case is $5 \\times 5 \\times 5 \\times 5 = 625$.\n\nSimilarly, the number of flags that are the same when flipped about its backward diagonal is 625.\n\nIf we add the number of flags with one of these diagonal symmetries to the number of flags with the other diagonal symmetry, then we will count twice the flags with both diagonal symmetries.\n\nA flag with both diagonal symmetries has the same bottom-left and top-right colours and the same top-left and bottom-right colours. The central region can be any colour. Hence the number of these flags is $5 \\times 5 \\times 5 = 125$.\n\nSo the number of required flags is $625 + 625 - 125 = 1125$.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21697,
"subject": "Mathematics (Olympiad)",
"question": "Consider the alternating sum:\n\n$$\n1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\dots - \\frac{1}{1334} + \\frac{1}{1335}\n$$\n\nExpress this sum as a single reduced fraction $\\frac{p}{q}$, where $p$ and $q$ are positive integers with $\\gcd(p, q) = 1$. Prove that $p$ is divisible by $2003$.",
"options": [],
"answer": "See solution",
"solution": "$$\n\\begin{align*}\n& 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\dots - \\frac{1}{1334} + \\frac{1}{1335} \\\\\n&= \\left(1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\dots + \\frac{1}{1334} + \\frac{1}{1335}\\right) - 2\\left(\\frac{1}{2} + \\frac{1}{4} + \\dots + \\frac{1}{1334}\\right) \\\\\n&= \\left(1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\dots + \\frac{1}{1334} + \\frac{1}{1335}\\right) - \\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{667}\\right) \\\\\n&= \\frac{1}{668} + \\frac{1}{669} + \\dots + \\frac{1}{1335} \\\\\n&= \\left(\\frac{1}{668} + \\frac{1}{1335}\\right) + \\left(\\frac{1}{669} + \\frac{1}{1334}\\right) + \\dots + \\left(\\frac{1}{1001} + \\frac{1}{1002}\\right) \\\\\n&= 2003 \\cdot \\left(\\frac{1}{668 \\cdot 1335} + \\frac{1}{669 \\cdot 1334} + \\dots + \\frac{1}{1001 \\cdot 1002}\\right).\n\\end{align*}\n$$\n\nSince $2003$ is a prime, this factor cannot be cancelled out by the denominator. Thus, $p$ must be divisible by $2003$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21698,
"subject": "Mathematics (Olympiad)",
"question": "Given an odd number $n \\geq 3$, let\n\n$$\nS = \\{k : 1 \\leq k \\leq n, \\gcd(k, n) = 1\\}\n$$\n\nand let\n\n$$\nT = \\{k : k \\in S, \\gcd(k+1, n) = 1\\}.\n$$\n\nFor each $k \\in S$, let $r_k$ be the remainder left by $\\frac{k^{|S|} - 1}{n}$ upon division by $n$. Show that\n\n$$\n\\prod_{k \\in T} (r_k - r_{n-k}) \\equiv |S|^{|T|} \\pmod{n}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Since $n$ is odd,\n\n$$\n|S| = \\varphi(n) = n \\prod_{p \\mid n} \\left(1 - \\frac{1}{p}\\right)\n$$\n\nis even. Given an element $k$ of $S$, write\n\n$k^{|S|} \\equiv 1 + n r_k \\pmod{n^2}$ and $(n - k)^{|S|} \\equiv 1 + n r_{n - k} \\pmod{n^2}$, and notice that\n\n$$\n(n - k)^{|S|} \\equiv k^{|S|} \\cdot n \\cdot k^{|S| - 1} \\pmod{n^2},\n$$\n\nto get\n\n$$\nr_k - r_{n - k} \\equiv |S| \\cdot k^{|S| - 1} \\pmod{n}.\n$$\n\nHence\n\n$$\n\\prod_{k \\in T} (r_k - r_{n - k}) \\equiv |S|^{|T|} \\left(\\prod_{k \\in T} k\\right)^{|S| - 1} \\pmod{n}.\n$$\n\nFinally, notice that the product in the right-hand member is congruent to $1$ modulo $n$. To see this, let $k'$ denote the modulo $n$ multiplicative inverse of an element $k$ of $S$, and notice that\n\n$$\nk' + 1 \\equiv k'(k + 1) \\pmod{n} \\quad \\text{and} \\quad (k - 1)(k' + 1) \\equiv k - k' \\pmod{n}.\n$$\n\nThe first congruence shows that if $k$ belongs to $T$, then so does $k'$, and the second shows that $k = k'$ if and only if $k = 1$. Consequently, if $k \\neq 1$, the factors $k$ and $k'$ in the product can be paired off and the conclusion follows.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21699,
"subject": "Mathematics (Olympiad)",
"question": "Bestimme alle Polynome $P(x)$ mit reellen Koeffizienten, die folgende Bedingungen erfüllen:\n\n(a) $P(2017) = 2016$\n\n(b) $(P(x) + 1)^2 = P(x^2 + 1)$ für alle reellen Zahlen $x$.",
"options": [],
"answer": "See solution",
"solution": "Sei $Q(x) := P(x) + 1$. Dann gilt $Q(2017) = 2017$ und $(Q(x))^2 = Q(x^2 + 1) - 1$ für alle $x \\in \\mathbb{R}$, also $Q(x^2 + 1) = Q(x)^2 + 1$ für alle $x \\in \\mathbb{R}$.\n\nDefiniere die Folge $(x_n)_{n \\ge 0}$ rekursiv durch $x_0 = 2017$ und $x_{n+1} = x_n^2 + 1$ für alle $n \\ge 0$. Mit vollständiger Induktion zeigt man, dass $Q(x_n) = x_n$ für alle $n \\ge 0$, denn $Q(x_{n+1}) = Q(x_n)^2 + 1 = x_n^2 + 1 = x_{n+1}$.\n\nDa $x_0 < x_1 < x_2 < \\dots$, stimmt das Polynom $Q(x)$ an unendlich vielen Stellen mit $x$ überein. Daher ist $Q(x) = x$ und somit $P(x) = x - 1$. Da $x - 1$ beide Bedingungen erfüllt, ist dies die einzige Lösung.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21700,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. Prove that every integer between $1$ and $n!$ can be written as the sum of at most $n$ distinct positive divisors of $n!$.",
"options": [],
"answer": "See solution",
"solution": "We use induction on $n$; the base case $n=1$ is obvious.\n\nSuppose now that the conclusion holds for some positive integer $n$ and consider $m \\leq (n+1)!$. Then $m = (n+1)q + r$, where $q, r \\in \\mathbb{N}$ and $0 \\leq r \\leq n$. It is easily noticed that $q \\leq n!$, therefore $q$ can be written as a sum of at most $n$ divisors of $n!$—denote them $d_1, \\dots, d_k$, with $k \\leq n$. Thus,\n\n$$\nm = (n+1)d_1 + (n+1)d_2 + \\dots + (n+1)d_k + r\n$$\n\nand each $(n+1)d_i$ is a divisor of $(n+1)!$.\n\nIf $r = 0$, the induction step is finished. Otherwise, if $r > 0$, $r$ divides $(n+1)!$ and $r < n+1 \\leq (n+1)d_i$, therefore $m$ can be written as the sum of $k+1$ distinct divisors of $(n+1)!$, with $k+1 \\leq n+1$, and we are done.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21701,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle with circumcircle $\\Omega$ and incenter $I$. Let the line through $I$ and perpendicular to $CI$ intersect segment $BC$ and the arc $BC$ (not containing $A$) of $\\Omega$ at points $U$ and $V$, respectively. Let the line through $U$ and parallel to $AI$ intersect $AV$ at $X$, and let the line through $V$ and parallel to $AI$ intersect $AB$ at $Y$. Let $W$ and $Z$ be the midpoints of $AX$ and $BC$, respectively. Prove that if the points $I$, $X$, and $Y$ are collinear, then the points $I$, $W$, and $Z$ are also collinear.",
"options": [],
"answer": "See solution",
"solution": "We begin with some general observations. Let $\\alpha = \\angle A/2$, $\\beta = \\angle B/2$, $\\gamma = \\angle C/2$. Clearly, $\\alpha + \\beta + \\gamma = 90^\\circ$. Since $\\angle UIC = 90^\\circ$, we have $\\angle IUC = \\alpha + \\beta$. Thus,\n\n$$\n\\angle BIV = \\angle IUC - \\angle IBC = \\alpha = \\angle BAI = \\angle BYV,\n$$\n\nso $B$, $Y$, $I$, $V$ are concyclic, as shown in the figure below.\n\n\n\nNow, assume $I$, $X$, $Y$ are collinear. We will prove that $\\angle YIA = 90^\\circ$.\n\nLet the line $XU$ intersect $AB$ at $N$. Since $AI$, $UX$, $VY$ are parallel, we have\n\n$$\n\\frac{NX}{AI} = \\frac{YN}{YA} = \\frac{VU}{VI} = \\frac{XU}{AI},\n$$\n\nso $NX = XU$. Also, $\\angle BIU = \\alpha = \\angle BNU$. Thus, $BUIN$ is a cyclic quadrilateral, and since $BI$ is the angle bisector of $\\angle UBN$, we have $NI = UI$. Therefore, in the isosceles triangle $NIU$, $X$ is the midpoint of base $NU$. Thus, $\\angle IXN = 90^\\circ$, i.e., $\\angle YIA = 90^\\circ$.\n\nLet $S$ be the midpoint of $VC$, and let $AX$ and $SI$ intersect at $T$. Let $x = \\angle BAV = \\angle BCV$. Since $\\angle CIA = 90^\\circ + \\beta$ and $SI = SC$, we get\n\n$$\n\\begin{align*}\n\\angle TIA &= 180^\\circ - \\angle AIS = 90^\\circ - \\beta - \\angle CIS \\\\\n&= 90^\\circ - \\beta - \\gamma - x = \\alpha - x \\\\\n&= \\angle TAI,\n\\end{align*}\n$$\n\nso $TI = TA$. Also, since $\\angle XIA = 90^\\circ$, $T$ is the midpoint of $AX$, i.e., $T = W$.\n\nTo complete the proof, it suffices to show that the intersection of $IS$ and $BC$ is the midpoint of $BC$. Since $S$ is the midpoint of $VC$, it suffices to show that $BV$ and $IS$ are parallel.\n\nSince $BYIV$ is cyclic, $\\angle VBI = \\angle VYI = \\angle YIA = 90^\\circ$. This shows that $BV$ is the external angle bisector of $\\triangle ABC$, so $\\angle VAC = \\angle VCA$. Thus, $2\\alpha - x = 2\\gamma + x$, so $\\alpha = \\gamma + x$. Therefore, $\\angle SCI = \\alpha$, and $\\angle VSI = 2\\alpha$.\n\nOn the other hand, $\\angle BVC = 180^\\circ - \\angle BAC = 180^\\circ - 2\\alpha$, so $BV$ and $IS$ are parallel. The proof is complete.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21702,
"subject": "Mathematics (Olympiad)",
"question": "A regular hexagon of side length $1$ is inscribed in a circle. Each minor arc of the circle determined by a side of the hexagon is reflected over that side. What is the area of the region bounded by these 6 reflected arcs?\n\n(A) $\\frac{5\\sqrt{3}}{2} - \\pi$\n(B) $3\\sqrt{3} - \\pi$\n(C) $4\\sqrt{3} - \\frac{3\\pi}{2}$\n(D) $\\pi - \\frac{\\sqrt{3}}{2}$\n(E) $\\frac{\\pi + \\sqrt{3}}{2}$",
"options": [],
"answer": "See solution",
"solution": "Note that the reflected arcs do not overlap except at their endpoints. The area of the region can be found by subtracting from the area of the hexagon the difference between the areas of the circle and the hexagon. This is equivalent to twice the area of the hexagon minus the area of the circle. Therefore, the requested area is\n\n$$\n2 \\cdot 6 \\cdot \\frac{1^2 \\cdot \\sqrt{3}}{4} - \\pi \\cdot 1^2 = 3\\sqrt{3} - \\pi.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21703,
"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 $2017$. Is it true that $2017$ necessarily divides all the coefficients of $f(x)$?",
"options": [],
"answer": "See solution",
"solution": "Yes.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21704,
"subject": "Mathematics (Olympiad)",
"question": "Let $B, C$ be two fixed points on the fixed circle $(O)$ ($BC$ is not the diameter of $(O)$). Point $A$ moves on $(O)$ such that $AB > BC$ and $M$ is the midpoint of $AC$. The circle of diameter $BM$ intersects $(O)$ at $R$. Suppose that $RM$ intersects $(O)$ at the second point $Q$ and cuts $BC$ at $P$. The circle of diameter $BP$ intersects $AB$ and $BO$ at the second points $K$ and $S$, respectively.\n\n1. Prove that $SR$ passes through the midpoint of $KP$.\n\n2. Denote $N$ as the midpoint of $BC$. The radical axis of the two circles of diameter $AN$ and $BM$ intersects $SR$ at $E$. Prove that $ME$ always passes through a certain fixed point as $A$ moves on $(O)$.",
"options": [],
"answer": "See solution",
"solution": "1) It is easy to check that $BQ$ is the diameter of circle $(O)$. Denote $I$ as the intersection of $SR$ and $PK$. We have\n\n$$\n\\angle SPI = \\angle SBK = \\angle QCA\n$$\n\nand\n\n$$\n\\angle PSI = \\angle PBR = \\angle CQR.\n$$\n\nThis implies that the triangles $PSI$ and $CQM$ are similar. Similarly, the triangles $KSI$ and $AQM$ are also similar. Since $M$ is the midpoint of $AC$, $I$ is the midpoint of $PK$.\n\n2) Redefine point $E$ as the projection of $C$ onto $AB$. Denote $H$ as the intersection of the altitudes $AD$ and $BL$ of triangle $ABC$. Note that the quadrilateral $LMND$ is cyclic, so $CM \\cdot CL = CN \\cdot CD$ and $HA \\cdot HD = HB \\cdot HL$, which implies that $CH$ is the radical axis of the two circles of diameter $AN$ and $BM$. We have $E$ belongs to $CH$, so it also belongs to the radical axis of the two circles of diameter $AN$ and $BM$. Since $\\angle BEH = \\angle BRP = 90^\\circ$, the quadrilateral $BHER$ is cyclic, thus\n\n$$\n\\angle HRE = \\angle EBH = \\angle OBC = \\angle PBS = \\angle PRS\n$$\n\nwhich implies that $S, R, E$ are collinear. Thus, $E$ is the intersection of $SR$ and the radical axis of the two circles of diameter $AN$ and $BM$. Denote $X$ as the intersection of $EM$ and $BQ$. Let $T$ be the midpoint of $BC$, then the quadrilateral $ETML$ is cyclic. Thus\n\n$$\n\\angle MEC = 90^\\circ - \\angle MET = 90^\\circ - \\angle ALT = 90^\\circ - \\angle BAC = \\angle BAL = \\angle BCQ.\n$$\n\n\n\nThis means that the quadrilateral $BCXE$ is cyclic, and then $\\angle BXC = BEC = 90^\\circ$. Therefore, $X$ is the projection of $C$ onto $BQ$, which is a fixed point, and the line $EM$ passes through $X$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21705,
"subject": "Mathematics (Olympiad)",
"question": "Sequence $a_n$ is defined as follows:\n\n$$\na_1 = 1, \\quad a_2 = 2, \\quad a_{n+2} = (n+1)(a_n + a_{n+1})\n$$\n\nfor each natural $n$. How many zeros does $a_{2011}$ end with?",
"options": [],
"answer": "See solution",
"solution": "**Answer:** 501.\n\n**Solution.** We first prove that $a_n = n!$ by induction:\n\n$$\na_{n+2} = (n+1)(n! + (n+1)!) = n!(n+1)(n+2) = (n+2)!\n$$\n\nNow, the number of zeros at the end of $a_{2011}$ is given by the well-known formula:\n\n$$\n\\left\\lfloor \\frac{2011}{5} \\right\\rfloor + \\left\\lfloor \\frac{2011}{5^2} \\right\\rfloor + \\left\\lfloor \\frac{2011}{5^3} \\right\\rfloor + \\dots = 402 + 80 + 16 + 3 = 501.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21706,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive even integer, and let $c_1, c_2, \\dots, c_{n-1}$ be real numbers satisfying\n\n$$\n\\sum_{i=1}^{n-1} |c_i - 1| < 1.\n$$\n\nProve that\n\n$$\n2x^n - c_{n-1}x^{n-1} + c_{n-2}x^{n-2} - \\dots - c_1x + 2\n$$\n\nhas no real roots.",
"options": [],
"answer": "See solution",
"solution": "Denote $P(x) = 2x^n - c_{n-1}x^{n-1} + c_{n-2}x^{n-2} - \\dots - c_1x + 2$. Since $c_i \\in (0, 2)$ for all $1 \\le i \\le n-1$, it follows $P(x) \\ge 2 > 0$ for all $x \\le 0$. It remains to prove $P(x) > 0$ also for all $x > 0$. Let $c_i = 1 + \\epsilon_i$, so $\\sum_{i=1}^{n-1} |\\epsilon_i| < 1$. Then for $x > 0$ we have\n\n$$\nP(x) = (x^n - x^{n-1} + x^{n-2} - \\dots - x + 1) + x^n + 1 + \\sum_{i=1}^{n-1} (-1)^i \\epsilon_i x^i.\n$$\n\nWe have $x^n - x^{n-1} + x^{n-2} - \\dots - x + 1 = \\frac{x^{n+1}+1}{x+1} > \\frac{1}{2} > 0$ since $2x^{n+1} - x + 1 > 0$ (for $x \\ge 1$ we have $x^{n+1} - x \\ge 0$ while for $x \\le 1$ we have $1 - x \\ge 0$).\n\nFor $0 < x \\le 1$ we have (by the triangle inequality)\n\n$$\n\\left| \\sum_{i=1}^{n-1} (-1)^i \\epsilon_i x^i \\right| \\le \\sum_{i=1}^{n-1} |(-1)^i \\epsilon_i x^i| \\le \\sum_{i=1}^{n-1} |\\epsilon_i| < 1,\n$$\n\nand so\n\n$$\nx^n + 1 + \\sum_{i=1}^{n-1} (-1)^i \\epsilon_i x^i \\ge x^n + 1 - \\left| \\sum_{i=1}^{n-1} (-1)^i \\epsilon_i x^i \\right| > x^n > 0.\n$$\n\nFor $1 \\le x$ we have essentially the same argument, but with the role of $x$ above replaced by $1/x$:\n\n$$\n\\left| \\sum_{i=1}^{n-1} (-1)^i \\epsilon_i (1/x)^{n-i} \\right| \\le \\sum_{i=1}^{n-1} |(-1)^i \\epsilon_i (1/x)^{n-i}| \\le \\sum_{i=1}^{n-1} |\\epsilon_i| < 1,\n$$\n\nand so\n\n$$\nx^n+1+\\sum_{i=1}^{n-1}(-1)^i \\epsilon_i x^i = x^n \\left( 1 + (1/x)^n + \\sum_{i=1}^{n-1} (-1)^i \\epsilon_i (1/x)^{n-i} \\right) \\ge x^n \\left( 1 + (1/x)^n - \\left| \\sum_{i=1}^{n-1} (-1)^i \\epsilon_i (1/x)^{n-i} \\right| \\right) > 1 > 0.\n$$\n\nThus $P(x) > \\frac{1}{2} > 0$, with room to spare, so we may either relax the condition $\\sum_{i=1}^{n-1} |c_i - 1| < 1$ or lower the free term of $P(x)$ to $\\frac{3}{2}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21707,
"subject": "Mathematics (Olympiad)",
"question": "$a_0, a_1, \\dots$ дараалал нь $a_0 = 1$ ба $k \\ge 1$ үед $a_{2k} = (-1)^k a_k$, $k \\ge 0$ үед $a_{2k+1} = -a_k$ гэж тодорхойлогдсон бол $n \\ge 0$ бүрийн хувьд $a_0 + a_1 + \\dots + a_n \\ge 0$ гэдгийг батал.",
"options": [],
"answer": "See solution",
"solution": "Короллар 1. $S_n = a_0 + a_1 + \\dots + a_n$ гэж тэмдэглэе.\n\n$$\n\\left\\{\n\\begin{array}{l}\na_{2k+1} = -a_k \\\\\na_{2k} = (-1)^k a_k\n\\end{array}\n\\right. \\implies a_{2k} + a_{2k+1} =\n\\begin{cases}\n0 & ;\\ k \\text{ - тэгш} \\\\\n2a_k & ;\\ k \\text{ - сондгой}\n\\end{cases}\n$$\n\nМатематик индукцийн аргаар баталъя.\n\n$$\n\\begin{align*}\nS_0 &= a_0 = 1 \\\\\nS_1 &= a_0 + a_1 = 1 + (-1) = 0 \\\\\nS_2 &= a_0 + a_1 + a_2 = 1 + (-1) + 1 = 1 \\\\\nS_3 &= a_0 + a_1 + a_2 + a_3 = 1 + (-1) + 1 + 1 = 2\n\\end{align*}\n$$\n\nСуурь тохиолдлууд үнэн. Одоо $n < 4k,\\ k \\in \\mathbb{N}$ үед $S_n \\ge 0$ гэж үзээд $S_{4k}$, $S_{4k+1}$, $S_{4k+2}$, $S_{4k+3}$-ийг баталъя.\n\n$$\n\\begin{align*}\nS_{4k+3} &= (a_0 + a_1) + (a_2 + a_3) + \\dots + (a_{4k+2} + a_{4k+3}) \\\\\n&= 0 + (-2a_1) + 0 + (-2a_3) + \\dots + 0 + (-2a_{2k+1}) \\\\\n&= 2a_0 + 2a_1 + \\dots + 2a_k = 2S_k\n\\end{align*}\n$$\n\n$$\nS_{4k} = S_{4(k-1)+3} + a_{4k} = 2S_{k-1} + a_{4k}\n$$\n\n$$\nS_{4k+1} = 2S_{k-1} + a_{4k} + a_{4k+1} = 2S_{k-1}\n$$\n\n$$\nS_{4k+2} = 2S_{k-1} + a_{4k+2}\n$$\n\nИндукцээр $S_{k-1} = 0$ бол $a_{4k} = a_k = 1$, $a_{4k+2} = a_k = 1$ тул $S_{4k}, S_{4k+2} \\ge 0$.\n\n---\n\nБодолт 2. $n \\ge 0$ үед\n\n$$\nf_n(z) = \\sum_{k=0}^{n} a_k z^k, \\quad S_n = f_n(1)\n$$\n\nТэгш ба сондгой зэргийн коэффициентүүдийг өгсөн нөхцлөөр хувиргавал:\n\n$$\n\\begin{align*}\nf_n(z) &= [a_0 + a_2 z^2 + a_4 z^4 + \\dots] + z[a_1 + a_3 z^2 + a_5 z^4 + \\dots] \\\\\n&= f_{[n/2]}(-z^2) - z f_{[(n-1)/2]}(z^2)\n\\end{align*}\n$$\n\nҮүнийг хоёр дахин хэрэглэхэд:\n\n$$\n\\begin{align*}\nf_{4n}(z) &= f_n(-z^4) - z f_{n-1}(-z^4) + (z^2 + z^3) f_{n-1}(z^4) \\\\\nf_{4n+1}(z) &= (1 - z) f_n(-z^4) + (z^2 + z^3) f_{n-1}(z^4) \\\\\nf_{4n+2}(z) &= (1 - z) f_n(-z^4) + z^2 f_n(z^4) + z^3 f_{n-1}(z^4) \\\\\nf_{4n+3}(z) &= (1 - z) f_n(-z^4) + (z^2 + z^3) f_n(z^4)\n\\end{align*}\n$$\n\n$z = 1$ үед:\n\n$$\n\\begin{array}{l}\nS_{4n} = f_n(-1) - f_{n-1}(-1) + 2S_{n-1} = (-1)^n a_n + 2S_{n-1} \\\\\nS_{4n+1} = S_{n-1} \\\\\nS_{4n+2} = S_n + S_{n-1} \\\\\nS_{4n+3} = 2S_n\n\\end{array}\n$$\n\n$a_k = \\pm 1$ ба $S_{n-1} \\equiv n \\pmod{2}$ тул $n \\ge 0$ үед $S_n \\ge 0$ байна.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21708,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\{x_n\\}$ and $\\{y_n\\}$ be two positive sequences defined by $x_1 = 1$, $y_1 = \\sqrt{3}$, and\n$$\n\\begin{cases}\nx_{n+1} y_{n+1} - x_n = 0 \\\\\nx_{n+1}^2 + y_n = 2\n\\end{cases}\n$$\nfor all positive integers $n$. Prove that these sequences are convergent and find their limits.",
"options": [],
"answer": "See solution",
"solution": "We observe that $x_1 = 1 = 2 \\sin \\frac{\\pi}{6}$ and $y_1 = \\sqrt{3} = 2 \\cos \\frac{\\pi}{6}$.\n\nWe will prove by induction that for all positive integers $n$,\n$$\nx_n = 2 \\sin \\frac{\\pi}{3 \\cdot 2^n}, \\quad y_n = 2 \\cos \\frac{\\pi}{3 \\cdot 2^n}. \\qquad (1)\n$$\nFor $n=1$, the statement holds. Assume (1) is true for $n$. Applying the recurrence relations:\n$$\n\\begin{aligned}\nx_{n+1} &= \\sqrt{2 - y_n} = \\sqrt{2 - 2 \\cos \\frac{\\pi}{3 \\cdot 2^n}} \\\\\n&= \\sqrt{4 \\sin^2 \\frac{\\pi}{3 \\cdot 2^{n+1}}} = 2 \\sin \\frac{\\pi}{3 \\cdot 2^{n+1}}\n\\end{aligned}\n$$\nand\n$$\ny_{n+1} = \\frac{x_n}{x_{n+1}} = \\frac{2 \\sin \\frac{\\pi}{3 \\cdot 2^n}}{2 \\sin \\frac{\\pi}{3 \\cdot 2^{n+1}}} = 2 \\cos \\frac{\\pi}{3 \\cdot 2^{n+1}}.\n$$\nThus, (1) holds for $n+1$. By induction, (1) is true for all $n$.\n\nTherefore,\n$$\n\\lim_{n \\to \\infty} x_n = 2 \\sin 0 = 0\n$$\nand\n$$\n\\lim_{n \\to \\infty} y_n = 2 \\cos 0 = 2.\n$$\nHence, $\\{x_n\\}$ and $\\{y_n\\}$ are convergent and $\\lim x_n = 0$, $\\lim y_n = 2$. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21709,
"subject": "Mathematics (Olympiad)",
"question": "Let $O$ be an interior point of an acute-angled $\\triangle ABC$. Denote by $A_1$, $B_1$, and $C_1$ its projections onto the sides $BC$, $AC$, and $AB$, respectively. Let $P$ be the intersection point of the lines through $A$ and $B$ orthogonal to $B_1C_1$ and $A_1C_1$, respectively. If $H$ is the projection of $P$ onto $AB$, prove that the points $A_1$, $B_1$, $C_1$, and $H$ are concyclic.",
"options": [],
"answer": "See solution",
"solution": "Since $\\angle PBC_1 = \\angle OC_1A_1$ and $\\angle OC_1A_1 = \\angle OBA_1$ (why?), then $\\angle PBA = \\angle OBC$. Analogously, $\\angle PAB = \\angle OAC$. Denote by $M$ and $N$ the projections of $P$ onto $BC$ and $AC$. Since\n\n$$\nBC_1 \\cdot BH = BO \\cdot BP \\cos \\angle OBA \\cos \\angle PBA = BA_1 \\cdot BM,\n$$\n\nthen the points $H$, $C_1$, $A_1$, and $M$ lie on a circle with center at the midpoint $Q$ of $OP$ (since the bisectors of $HC_1$ and $MA_1$ pass through $Q$). Similarly, the points $H$, $C_1$, $N$, and $B_1$ lie on a circle with center $Q$. Hence, the points $A_1$, $B_1$, $C_1$, $H$, $M$, and $N$ are concyclic.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21710,
"subject": "Mathematics (Olympiad)",
"question": "Prove by induction that for every positive integer $n$, there exists a unique $n$-digit integer with all digits odd that is divisible by $5^n$.",
"options": [],
"answer": "See solution",
"solution": "We proceed by induction.\n\n*Base case*: For $n=1$, $A(1) = 5$ is the unique 1-digit odd integer divisible by $5$.\n\n*Inductive step*: Assume the assertion holds for $n = m$, i.e., there is a unique $A(m) = \\overline{a_1a_2\\cdots a_m}$ with all digits odd and $5^m \\mid A(m)$.\n\nFor $n = m+1$, consider $A(d, m) = \\overline{d a_1 a_2 \\cdots a_m}$ for $d = 1, 3, 5, 7, 9$. The differences between these numbers are multiples of $2 \\times 10^m$, $4 \\times 10^m$, etc., none of which is divisible by $5^{m+1}$. Thus, their remainders modulo $5^{m+1}$ are distinct. Since $A(m)$ is divisible by $5^m$, so are all $A(d, m)$. Their remainders modulo $5$ are also distinct, covering $0, 1, 2, 3, 4$. Therefore, exactly one $A(d, m)$ is divisible by $5^{m+1}$.\n\n*Uniqueness*: Suppose $\\overline{b_1b_2\\cdots b_{m+1}}$ is a multiple of $5^{m+1}$ with all digits odd. Then $\\overline{b_1b_2\\cdots b_{m+1}} = b_1 \\times 10^m + \\overline{b_2\\cdots b_{m+1}}$. Since $b_1 \\times 10^m$ is a multiple of $5^m$, $\\overline{b_2\\cdots b_{m+1}}$ must also be the unique such multiple of $5^m$ with all digits odd, i.e., $A(m)$. Thus, the $n$-digit number is unique.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21711,
"subject": "Mathematics (Olympiad)",
"question": "For each positive integer $n$ and each integer $i$ ($0 \\leq i \\leq n$), let $C_n^i \\equiv c(n,i) \\pmod{2}$, where $c(n,i) \\in \\{0, 1\\}$, and define\n$$\nf(n,q) = \\sum_{i=0}^{n} c(n,i)q^i.\n$$\nLet $m, n$ and $q$ be positive integers with $q+1$ not a power of 2. Suppose that $f(m,q) \\mid f(n,q)$. Prove that\n\n$$\nf(m,r) \\mid f(n,r)\n$$\nfor every positive integer $r$.",
"options": [],
"answer": "See solution",
"solution": "For each positive integer $n$, write $n$ in binary as $n = 2^{a_1} + 2^{a_2} + \\dots + 2^{a_k}$, where $0 \\le a_1 < a_2 < \\dots < a_k$. Define $T(n) = \\{2^{a_1}, \\dots, 2^{a_k}\\}$, and $T(0)$ as the empty set.\n\nBy Lucas' theorem, $C_n^i$ is odd if and only if $T(i) \\subseteq T(n)$, so\n$$\nf(n,q) = \\sum_{A \\subseteq T(n)} q^{\\sigma(A)} = \\prod_{a \\in T(n)} (1+q^a),\n$$\nwhere $\\sigma(A)$ is the sum of elements of $A$.\n\nGiven $m, n, q$ as in the problem, if\n$$\nf(m,q) = \\prod_{a \\in T(m)} (1+q^a) \\mid \\prod_{a \\in T(n)} (1+q^a) = f(n,q),\n$$\nthen $T(m) \\subseteq T(n)$, so $f(m,r) \\mid f(n,r)$ for every $r$.\n\nFor any integers $i, j$ with $0 \\le i < j$,\n$$\nq^{2j} - 1 = (q^{2j-1} + 1) \\cdots (q^2 + 1)(q^2 - 1),\n$$\nso\n$$\n\\gcd(q^{2j} + 1, q^{2i} + 1) = \\gcd(q^{2i} + 1, 2) \\mid 2.\n$$\nLet $s(k)$ be the largest odd divisor of $k$. Then $s(q^{2i} + 1)$ and $s(q^{2j} + 1)$ are coprime. Clearly $q > 1$. If $i > 0$, $q^{2i} + 1 \\equiv 1 \\pmod{2}$, and $q^{2j} + 1 > 2$, so $s(q^{2i} + 1) > 1$. If $i = 0$, since $q + 1$ is not a power of 2, $s(q + 1) > 1$. For any $a \\in T(m)$, $s(q^a + 1)$ divides $\\prod_{b \\in T(n)} s(q^b + 1)$. Since $s(1 + q^a) > 1$, $a \\in T(n)$, so $T(m) \\subseteq T(n)$, completing the proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21712,
"subject": "Mathematics (Olympiad)",
"question": "Let $c_1 = b_1 = 0$ and $c_i = b_i - b_{i-1} \\ge 0$ for $2 \\le i \\le n$. For $j > i$, we have $|b_i - b_j| = b_j - b_i = c_{i+1} + c_{i+2} + \\cdots + c_j$. Prove that for all $c_i \\ge 0$ and all real $a_i$,\n\n$$\n\\sum_{i 0$ for some $i$, since the inequality is invariant under scaling of the $c_i$ by any positive real, we may assume $c_1 + \\cdots + c_n = 1$.\n\nHold all the $a_i$ constant, so the right-hand side is constant. To maximize the left-hand side, note that it is linear in the $c_i$, so it is maximized at a vertex of the polytope defined by $c_i \\ge 0$ and $c_1 + \\cdots + c_n = 1$. Thus, $c_k = 1$ for some $k$ and $c_i = 0$ for all $i \\ne k$.\n\nSubstituting into the inequality, it suffices to show\n\n$$\n\\sum_{i 0$。因此，若 $S$ 不滿足 (1)，則存在*無窮多*正有理數 $r < 1$，使得對所有有限子集 $F$，$\\sum_{x \\in F} 1/x \\ne r$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21719,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle with circumcenter $O$ and orthocenter $H$. Let $Y$ and $Z$ be points constructed as follows: $Z$ is the reflection of $A$ over $BC$, and $Y$ is the reflection of $A$ over $CA$. Consider the quadrilateral $BCYZ$ and the isogonal conjugate $Q$ of $O$ in $BCYZ$.\n\nProve that $Q = H$, and that the reflection $P$ of $H$ in $YZ$ lies on both the circumcircle $(ABC)$ and the circle $(AYZ)$. Furthermore, show that the circumcenter $O_1$ of $AYZ$ lies on the perpendicular bisector of $BC$, and that $O$ is the center of spiral similarity sending $YZ$ to $QR$. Finally, prove that $OD \\perp YZ$, where $D$ is the foot of the $A$-altitude, and that $O, K, YZ \\cap QR$ are collinear, where $K$ is the intersection of $OO_1$ with $YZ$.",
"options": [],
"answer": "See solution",
"solution": "Observe that $\\angle BOZ + \\angle COY = 180^\\circ$, hence the isogonal conjugate $Q$ of point $O$ in quadrilateral $BCYZ$ exists. Since $H$ and $O$ are isogonal conjugates in $ABC$, $Q = H$ follows.\n\nIn particular, $O$ is the center of the circle passing through the reflections of $H$ in $BC$, $CY$, $YZ$, $ZB$, which coincides with the circumcircle of $ABC$. Thus, the reflection $P$ of $H$ in $YZ$ lies on $(ABC)$.\n\nHowever, $\\angle YHZ = \\angle A$ as $\\angle YOZ = 180^\\circ - 2\\angle A$ and $H, O$ are isogonal conjugates in $BCYZ$, hence $P$ lies on $(AYZ)$ as well.\n\nNow $BZ = OB / \\sin C$, hence $BZ \\cdot BA = 2R^2 = CY \\cdot CA$, where $R$ is the circumradius of $ABC$. Thus, $B$ and $C$ have equal powers in $(AYZ)$, or the circumcenter $O_1$ of triangle $AYZ$ lies on the perpendicular bisector of $BC$, which passes through $O$.\n\nFurther, $O$ is the center of spiral similarity sending $YZ$ to $QR$, so it suffices to show that it sends $O_1$ to $T$, where $T$ is the intersection of the tangents from $Y$ and $Z$ to $(AYZ)$.\n\nLet $OO_1 \\cap YZ = K$, and the line through $O$ parallel to $BC$ meet $QR$ at $S$. Now the same spiral similarity maps $K$ to $S$, so we just need to show that $OS / OK = OT / OO_1$. Both are equal to $\\tan \\angle (YZ, BC)$, so we are done.\n\nTo conclude, it suffices to show $OD \\perp YZ$, where $D$ is the foot of the $A$-altitude, which will allow $O, K, YZ \\cap QR$ to be collinear. This follows as $HP \\perp YZ$ since $H, O$ are isogonal conjugates in $BCYZ$, hence\n\n$$\n\\angle PAY = \\angle PZY = \\angle HZY = \\angle OZB = \\angle C.\n$$\n\n□\n\nWe prove Lemma 2 of Solution A using complex numbers, and then finish the same way as in Solution A.\n\nWe place the diagram on the complex plane. Let the circumcircle of $\\triangle ABC$ be the unit circle, so $O$ is the origin, and let $A = a$, $B = b$, $C = 1/b$ without loss of generality. Use $z - a = -b(z\\bar{x} - 1)$ and $z = -b^2\\bar{z}$ to get that $Z = z = \\frac{b(a + b)}{b - a}$. By symmetry, $Y = y = \\frac{c(a + c)}{c - a} = \\frac{(ab + 1)}{b(1 - ab)}$.\n\nNow, redefine $K$ to be the intersection of the perpendicular from $O$ to $BC$ with $YZ$. We need to show that $HK \\perp YZ$.\n\nWe calculate\n\n$$\n y - z = \\frac{(1 - b^2)(ab(a + b) + a - b)}{b(1 - ab)(a - b)}, \\quad \\text{so} \\quad \\frac{y - z}{\\bar{y} - \\bar{z}} = \\frac{ab(a + b) + a - b}{a + b + ab(b - a)}\n$$\n\nTherefore, $k = z - \\frac{ab(a + b) + a - b}{a + b + ab(b - a)}(\\bar{k} - \\bar{z})$. Also $k = \\bar{k}$ since it lies on the real axis. Therefore,\n\n$$\n\\frac{2a + 2ab^2}{a + b + ab(b - a)} \\cdot k = z \\left(1 - \\frac{ab(a + b) + a - b}{b^2(a + b + ab(b - a))}\\right) \\Rightarrow k = \\frac{b(a + b)}{b - a} \\cdot \\frac{(b^2 + 1)(1 + ab)(b - a)}{2ab^2(1 + b^2)} = \\frac{(a + b)(ab + 1)}{2ab}\n$$\n\nFinally, the orthocenter $H = h = a + b + \\frac{1}{b}$. Thus, $h - k = \\frac{ab(a + b) + (a - b)}{2ab}$. Finally,\n\n$$\n\\frac{y - z}{h - k} = \\frac{(1 - b^2)2a}{(1 - ab)(a - b)}\n$$\n\nwhich is easily checked to be equal to the negative of its conjugate.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21720,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\mathbb{R}_+$ be the set of all positive integers. Find all functions $g: \\mathbb{R}_+ \\to \\mathbb{R}_+$ such that for all $x, y \\in \\mathbb{R}_+$,\n$$\nxg(x + g(y)) = g(g(xy) + 1).\n$$",
"options": [],
"answer": "See solution",
"solution": "**Answer:** $g(x) = \\frac{c}{x}$ for an arbitrary fixed $c \\in \\mathbb{R}_+$.\n\nLet $xy = z$, so $y$ and $z$ can take all positive real values independently. The equation becomes\n$$\n\\frac{z}{y}g\\left(\\frac{z}{y} + g(y)\\right) = g(g(z) + 1). \\qquad (1)\n$$\n\nFix $z$ and vary $y$ over $\\mathbb{R}_+$; then $g(g(z) + 1)$ is fixed while $\\frac{z}{y}$ attains all positive real values, so $g\\left(\\frac{z}{y} + g(y)\\right)$ attains all positive real values. Thus, $g$ is surjective.\n\nSuppose $\\frac{g(y_2) - g(y_1)}{y_2 - y_1} > 0$ for some $y_2 \\neq y_1$. Substitute $z = y_2 y_1 \\cdot \\frac{g(y_2) - g(y_1)}{y_2 - y_1}$ into (1) with $y = y_2$ and $y = y_1$. In both cases, $g\\left(\\frac{z}{y} + g(y)\\right) = g\\left(\\frac{y_2 g(y_2) - y_1 g(y_1)}{y_2 - y_1}\\right)$, so $y_1 = y_2$—a contradiction. Thus, $\\frac{g(y_2) - g(y_1)}{y_2 - y_1} \\le 0$ for all $y_2 \\neq y_1$, so $g$ is non-strictly decreasing.\n\nSince $g$ is surjective and non-strictly decreasing, it is continuous and $\\lim_{y \\to +\\infty} g(y) = 0$. Now consider the original equation:\n$$\nxg(x + g(y)) = g(g(xy) + 1),\n$$\nas $y \\to +\\infty$. Then $g(x + g(y)) \\to g(x)$ and $g(xy) + 1 \\to 1$. Hence $xg(x) = g(1)$, so $g(x) = \\frac{c}{x}$, where $c = g(1)$. This function satisfies the original condition for any $c \\in \\mathbb{R}_+$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21721,
"subject": "Mathematics (Olympiad)",
"question": "Let $q$ be a positive rational number. Two ants are initially at the same point $X$ in the plane. In the $n$-th minute ($n=1,2,3,\\ldots$) each of them chooses whether to walk due north, east, south, or west and then walks the distance of $q^n$ meters. After a whole number of minutes, they are at the same point in the plane (not necessarily $X$), but have not taken exactly the same route within that time. Determine all possible values of $q$.",
"options": [],
"answer": "See solution",
"solution": "$q = 1$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21722,
"subject": "Mathematics (Olympiad)",
"question": "Given a $4 \\times 4$ grid with entries $a_{i,j}$, define $A_i$ as the sum of the $i$-th row and $B_j$ as the sum of the $j$-th column. What is the maximum possible value of $m = \\max\\{A_1, A_2, A_3, A_4, B_1, B_2, B_3, B_4\\} - \\min\\{A_1, A_2, A_3, A_4, B_1, B_2, B_3, B_4\\}$, given that the grid contains the numbers $1$ to $16$ exactly once?\n\n",
"options": [],
"answer": "See solution",
"solution": "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. Then:\n\n$$\n\\begin{aligned}\nm &\\leq A_4 - A_1,\\quad m \\leq B_4 - B_1 \\\\\nm &\\leq \\frac{(A_4 - A_1) + (B_4 - B_1)}{2} \\\\\n&= \\frac{1}{2}(a_{4,4} - a_{1,1}) + \\frac{1}{2}(a_{4,4} + a_{4,2} + a_{4,3} + a_{2,4} + a_{3,4} - a_{1,1} - a_{1,2} - a_{1,3} - a_{2,1} - a_{3,1}) \\\\\n&\\leq \\frac{1}{2}(16 - 1) + \\frac{1}{2}(16 + 15 + 14 + 13 + 12 - 1 - 2 - 3 - 4 - 5) \\\\\n&= 35\n\\end{aligned}\n$$\n\nThere exists a way of writing if $m = 35$ (see table below). So the maximum $m$ is $35$.\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21723,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle in which $BC < AC$. Let $M$ be the midpoint of $AB$; $AP$ be the altitude from $A$ onto $BC$; and $BQ$ be the altitude from $B$ onto $AC$. Suppose $QP$ produced meets $AB$ (extended) at $T$. If $H$ is the orthocentre of $ABC$, prove that $TH$ is perpendicular to $CM$.",
"options": [],
"answer": "See solution",
"solution": "Complete the parallelogram $ADBC$. Join $CD$, $CH$, and $HD$. Let $S$ and $L$ be the midpoints of $CH$ and $HD$, respectively. Observe that $CPHQ$ is a cyclic quadrilateral and $CH$ is a diameter of the circumscribing circle. Thus, $S$ is the centre of a circle $\\Gamma_1$ passing through $H$, $P$, $Q$, and $C$. Similarly, $L$ is the centre of a circle $\\Gamma_2$ passing through $A$, $D$, $B$, and $H$. Hence, the radical axis of these circles, which passes through $H$, is perpendicular to $SL$. However, $SL$ is parallel to $CD$ (which passes through $M$), and hence the radical axis of $\\Gamma_1$ and $\\Gamma_2$ is perpendicular to $CM$. But $BPQA$ is also cyclic so that $TP \\cdot TQ = TB \\cdot TA$. This shows that $T$ has the same power with respect to $\\Gamma_1$ and $\\Gamma_2$. Thus, $T$ is on the radical axis of these circles. It follows that $TH$ is perpendicular to $CM$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21724,
"subject": "Mathematics (Olympiad)",
"question": "Prove that a sequence $\\{a_n\\}_{n \\ge 1}$ with the property\n$$\na_n - \\frac{1}{m} \\le a_m \\le \\frac{m}{n}\n$$\nfor all $m, n \\in \\mathbb{N}^*$ with $m \\ge n$, is convergent.",
"options": [],
"answer": "See solution",
"solution": "For $m = n \\in \\mathbb{N}^*$, we get $a_n \\le 1$, so the sequence is upper bounded.\n\nLet $L = \\sup\\{a_n \\mid n \\in \\mathbb{N}^*\\}$. Then $L \\le 1$. For any $\\varepsilon > 0$, choose $n_0 \\in \\mathbb{N}^*$ such that $L - \\varepsilon < a_{n_0} \\le L$.\n\nLet $m_0 = \\max\\left\\{n_0, 1 + \\left\\lfloor \\frac{1}{a_{n_0} - L + \\varepsilon} \\right\\rfloor\\right\\}$. For $m \\ge m_0$,\n$$\na_m \\ge a_{n_0} - \\frac{1}{m} \\ge a_{n_0} - \\frac{1}{m_0} > L - \\varepsilon,\n$$\nso $|a_m - L| < \\varepsilon$ for all $m \\ge m_0$, proving that the sequence converges to $L$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21725,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be a right-angled triangle with $\\angle BAC = 90^\\circ$. Let the height from $A$ cut its side $BC$ at $D$. Let $I$, $I_B$, $I_C$ be the incenters of triangles $ABC$, $ABD$, $ACD$ respectively. Let $E_B$, $E_C$ be the excenters of $ABC$ with respect to vertices $B$ and $C$ respectively. If $K$ is the point of intersection of the circumcircles of $E_C I B_I$ and $E_B I C_I$, show that $KI$ passes through the midpoint $M$ of side $BC$.",
"options": [],
"answer": "See solution",
"solution": "Since $\\angle E_C B I = 90^\\circ = I C E_B$, we conclude that $E_C B C E_B$ is cyclic. Moreover, we have that\n\n$$\n\\angle B A I_B = \\frac{1}{2} \\angle BAD = \\frac{1}{2} \\hat{C},\n$$\n\nso $A I_B \\perp C I$. Similarly, $A I_C \\perp B I$. Therefore $I$ is the orthocenter of triangle $A I_B I_C$. It follows that\n\n$$\n\\angle I I_B I_C = 90^\\circ - \\angle A I_C B = \\angle I A I_C = 45^\\circ - \\angle I_C A C = 45^\\circ - \\frac{1}{2} \\hat{B} = \\frac{1}{2} \\hat{C}.\n$$\n\nTherefore $I_B I_C C B$ is cyclic. Since $A E_B C I$ is also cyclic (on a circle of diameter $I E_B$), then\n\n\n\n$$\n\\angle E_C E_B B = \\angle A C I = \\frac{1}{2} \\hat{C} = \\angle I I_B I_C,\n$$\n\ntherefore $I_B I_C \\parallel E_B E_C$.\n\nFrom the inscribed quadrilaterals we get that\n\n$$\n\\angle K I_C I = \\angle K E_B I \\quad \\text{and} \\quad \\angle K E_C I = \\angle K I_B I,\n$$\n\nwhich implies that the triangles $K E_C I_C$ and $K I_B E_B$ are similar. So\n\n$$\n\\frac{d(K, E_C I_C)}{d(K, E_B I_B)} = \\frac{E_C I_C}{E_B I_B}.\n$$\n\nBut $I_B I_C \\parallel E_B E_C$ and $I_B I_C C B$ is cyclic, therefore\n\n$$\n\\frac{E_C I_C}{E_B I_B} = \\frac{I I_C}{I I_B} = \\frac{I B}{I C}.\n$$\n\nWe deduce that\n\n$$\n\\frac{d(K, I C)}{d(K, I B)} = \\frac{I B}{I C},\n$$\n\ni.e., the distances of $K$ to the sides $I C$ and $I B$ are inversely proportional to the lengths of these sides. So by a well-known property of the median, $K$ lies on the median of the triangle $I B C$. (The last property of the median can be proved either by the law of sines, or by taking the distances from the median $M$ to the sides and proving by Thales' theorem that $M$, $I$, $K$ are collinear.)",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21726,
"subject": "Mathematics (Olympiad)",
"question": "Decide if there is a permutation $a_1, a_2, \\ldots, a_{6666}$ of the numbers $1, 2, \\ldots, 6666$ with the property that the sum $k + a_k$ is a perfect square for all $k = 1, 2, \\ldots, 6666$.",
"options": [],
"answer": "See solution",
"solution": "Such a permutation exists. We construct one in several steps, by reducing the problem for $n = 6666$ to the analogous problem for smaller $n$.\n\nStart with the perfect square $82^2 = 6724$ and note that $82^2 - 6666 = 58$. So let $a_k = 82^2 - k$ for $k = 58, 59, \\ldots, 6666$; then $k + a_k = 82^2$ is a perfect square for these values of $k$.\n\nHence, it suffices to find a permutation $a_1, a_2, \\ldots, a_{57}$ of $1, 2, \\ldots, 57$ such that $k + a_k$ is a perfect square for $k = 1, 2, \\ldots, 57$, i.e., to solve the same problem for $n_1 = 57$. Now $10^2 - 57 = 43$ lets us define $a_k = 10^2 - k$ for $k = 43, 44, \\ldots, 57$. This reduces the question to $n_2 = 42$. Here we use $8^2 - 42 = 22$ to define $a_k = 8^2 - k$ for $k = 22, 23, \\ldots, 42$. Now we face $n_3 = 21$ where $6^2 - 21 = 15$ resolves the issue for $k = 15, 16, \\ldots, 21$. One can make more steps for the remaining $n_4 = 14$, but let us write a feasible permutation directly:\n\n$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 \\rightarrow 3, 2, 1, 5, 4, 10, 9, 8, 7, 6, 14, 13, 12, 11$\n\nThe sum of every two corresponding numbers is a perfect square.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 21727,
"subject": "Mathematics (Olympiad)",
"question": "It is given the sequence $\\{a_n\\}$: $a_1 = 1$,\n\n$$\na_{n+1} = 2a_n + n \\cdot (1 + 2^n), \\quad n = 1, 2, 3, \\dots\n$$\n\nFind the general term $a_n$.",
"options": [],
"answer": "See solution",
"solution": "Divide the recursion formula by $2^{n+1}$ throughout:\n\n$$\n\\frac{a_{n+1}}{2^{n+1}} = \\frac{a_n}{2^n} + \\frac{n}{2^{n+1}} + \\frac{n}{2}\n$$\n\nThat is,\n\n$$\n\\frac{a_{n+1}}{2^{n+1}} - \\frac{a_n}{2^n} = \\frac{n}{2^{n+1}} + \\frac{n}{2}\n$$\n\nSum both sides from $i = 1$ to $n$:\n\n$$\n\\sum_{i=1}^{n} \\left( \\frac{a_{i+1}}{2^{i+1}} - \\frac{a_i}{2^i} \\right) = \\sum_{i=1}^{n} \\frac{i}{2^{i+1}} + \\sum_{i=1}^{n} \\frac{i}{2}\n$$\n\nThis telescopes to:\n\n$$\n\\frac{a_{n+1}}{2^{n+1}} - \\frac{a_1}{2^1} = \\frac{n(n+1)}{4} + \\sum_{i=1}^{n} \\frac{i}{2^{i+1}}\n$$\n\nSo,\n\n$$\na_{n+1} = 2^{n+1} \\left[ \\frac{n(n+1)}{4} + \\frac{1}{2^n} + \\frac{1}{2} \\sum_{i=1}^{n} \\frac{i}{2^i} \\right]\n$$\n\nLet $S_n = \\sum_{i=1}^{n} \\frac{i}{2^i}$. Then $2S_n = \\sum_{i=1}^{n} \\frac{i}{2^{i-1}}$, and\n\n$$\n\\begin{align*}\nS_n &= 2S_n - S_n = \\sum_{i=1}^{n} \\frac{i}{2^{i-1}} - \\sum_{i=1}^{n} \\frac{i}{2^i} \\\\\n&= \\sum_{i=1}^{n} \\frac{i}{2^{i-1}} - \\sum_{i=2}^{n+1} \\frac{i-1}{2^{i-1}} \\\\\n&= \\frac{1}{2^{0}} - \\frac{n}{2^n} + \\sum_{i=2}^{n} \\left( \\frac{i}{2^{i-1}} - \\frac{i-1}{2^{i-1}} \\right) \\\\\n&= 1 - \\frac{n}{2^n} + \\sum_{i=2}^{n} \\frac{1}{2^{i-1}}\n\\end{align*}\n$$\n\n$$\n\\begin{aligned}\nS_n &= 1 - \\frac{n}{2^n} + \\frac{1}{2} \\left[ 1 - \\left( \\frac{1}{2} \\right)^{n-1} \\right] \\\\\n&= 1 - \\frac{n}{2^n} + 1 - \\frac{1}{2^{n-1}} \\\\\n&= 2 - \\frac{n+2}{2^n}\n\\end{aligned}\n$$\n\nThus,\n\n$$\n\\begin{aligned}\na_{n+1} &= 2^{n+1} \\left[ \\frac{n(n+1)}{4} + \\frac{1}{2^n} + \\frac{1}{2} \\left( 2 - \\frac{n+2}{2^n} \\right) \\right] \\\\\n&= 2^{n+1} \\left[ \\frac{3}{2} + \\frac{n(n+1)}{4} - \\frac{n+2}{2^{n+1}} \\right], \\quad n \\ge 1\n\\end{aligned}\n$$\n\nConsequently,\n\n$$\na_n = 2^{n-2}(n^2 - n + 6) - n - 1, \\quad n \\ge 2\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21728,
"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 $x_1, \\ldots, x_n$ 為整數點的座標，將所有滿足\n\n$$\n(2n + 1) \\mid (x_1 + 2x_2 + \\cdots + nx_n)\n$$\n\n的點標記，即可達成條件。對於每個點，$x_1 + 2x_2 + \\cdots + nx_n$ 可以唯一地表示為 $(2n + 1)l \\pm k$，其中 $l$ 為整數，$k = 0, 1, \\ldots, n$。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": 21729,
"subject": "Mathematics (Olympiad)",
"question": "Determine positive integers $x, y, z$ which satisfy the system\n\n$$\n\\begin{aligned}\nx + y + z &= xy + yz + zx \\\\\nxyz &= 1\n\\end{aligned}\n$$\n\nand have the least possible sum.",
"options": [],
"answer": "See solution",
"solution": "We write the system as\n\n$$\nxy + yz + zx = x + y + z \\quad (1)\n$$\n$$\nxyz = 1 \\quad (2)\n$$\n\nSubtracting the two equations by parts, we find\n\n$$\n\\begin{aligned}\nxyz - (xy + yz + zx) &= 1 - (x + y + z) \\\\\nxyz - xy - yz - zx + x + y + z - 1 &= 0 \\\\\nxy(z - 1) - x(z - 1) - y(z - 1) + (z - 1) &= 0 \\\\\n(z - 1)(xy - x - y + 1) &= 0 \\\\\n(z - 1)(x - 1)(y - 1) &= 0 \\\\\nx = 1 \\text{ or } y = 1 \\text{ or } z = 1\n\\end{aligned}\n$$\n\nFor $x=1$, from (1) and (2) we have $yz=1$, which gives\n\n$$\n(y, z) = \\left( a, \\frac{1}{a} \\right), \\quad a > 0\n$$\n\nSo the solutions are\n\n$$\n(x, y, z) = \\left( 1, a, \\frac{1}{a} \\right), \\quad a > 0\n$$\n\nSimilarly, for $y=1$ or $z=1$ we find\n\n$$\n(x, y, z) = \\left( a, 1, \\frac{1}{a} \\right) \\text{ or } (x, y, z) = \\left( a, \\frac{1}{a}, 1 \\right), \\quad a > 0\n$$\n\nSince for each $a > 0$,\n\n$$\nx + y + z = 1 + a + \\frac{1}{a} \\ge 3\n$$\n\nEquality holds for $a=1$, so $(x, y, z) = (1, 1, 1)$ is the solution with the least possible sum $x + y + z$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21730,
"subject": "Mathematics (Olympiad)",
"question": "Define a sequence $(a_n)$ as follows:\n\n$$\n\\begin{cases}\n a_1 = 1 \\\\\n a_{n+1} = 3 - \\frac{a_n + 2}{2^{a_n}} & \\text{for } n \\ge 1\n\\end{cases}\n$$\n\nProve that this sequence has a finite limit and find this limit.",
"options": [],
"answer": "See solution",
"solution": "Firstly, by induction, one can prove that\n\n$$\na_n > 1, \\quad \\forall n > 1\n$$\n\nbecause the function $u(x) = 2^{x+1} - x - 2$ is increasing on $(1, +\\infty)$. It implies that\n\n$$\na_{n+1} = 3 - \\frac{a_n + 2}{2^{a_n}} < 3, \\quad \\forall n \\ge 1.\n$$\n\nNext, we will prove that $(a_n)$ is an increasing sequence. Considering the function $f(x) = 3 - \\frac{2+x}{2^x}$ where $1 < x < 3$. We have\n\n$$\nf'(x) = \\frac{\\ln 4 + x \\ln 2 - 1}{2^x} > 0,\n$$\n\nso $f(x)$ is increasing on $(1, 3)$. Moreover, $a_2 = \\frac{3}{2} > a_1$ so $(a_n)$ is an increasing sequence. Note that $(a_n)$ is bounded above by $3$ so $(a_n)$ has a finite limit. Let $L \\in (1, 3)$ be that limit. Letting $n$ tend to infinity, we have $L = 3 - \\frac{L+2}{2^L}$. We will prove the equation\n\n$$\nx = 3 - \\frac{x+2}{2^x} \\quad (1)\n$$\n\nhas a unique solution on $(1, 3)$. Indeed, consider the function $g(x) = 3 - \\frac{x+2}{2^x} - x$ where $x \\in (1, 3)$, we have\n\n$$\ng'(x) = \\frac{\\ln 4 + x \\ln 2 - 1}{2^x} - 1 = \\frac{\\ln 4 + x \\ln 2 - 1 - 2^x}{2^x},\n$$\n\nSetting $h(x) = \\ln 4 + x \\ln 2 - 1 - 2^x$, for $x \\in (1, 3)$ then\n\n$$\nh'(x) = \\ln 2(1 - 2^x) < 0, \\quad \\forall x \\in (1, 3),\n$$\n\nso $h(x)$ is a decreasing function on $(1, 3)$. Thus, $h(x) < h(1) = \\ln 8 - 3 < 0$, or $\\ln 4 + x \\ln 2 - 1 - 2^x < 0$ for all $x \\in (1, 3)$. Thus, the function $g(x)$ is decreasing on $(1, 3)$ and the equation $g(x) = 0$ has no more than one solution.\n\nFurthermore, $g(2) = 0$ so $x = 2$ is the unique solution of (1), which means $L = 2$ is the limit of the given sequence. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21731,
"subject": "Mathematics (Olympiad)",
"question": "We order the positive odd integers as follows:\n\n| | column 1 | column 2 | column 3 | column 4 | column 5 | column 6 | ... |\n|--------|----------|----------|----------|----------|----------|----------|-----|\n| row 1 | 1 | 3 | 11 | 13 | 29 | 31 | ... |\n| row 2 | 5 | 9 | 15 | 27 | 33 | ... | |\n| row 3 | 7 | 17 | 25 | 35 | ... | | |\n| row 4 | 19 | 23 | 37 | ... | | | |\n| row 5 | 21 | 39 | ... | | | | |\n| row 6 | 41 | ... | | | | | |\n| ... | ... | | | | | | |\n\nFor each odd number we can determine in which row and column it is placed. For example, the number $35$ is placed in row $3$ and column $4$. What number is placed in row $22$ and column $24$?\n\nA) $2021$ \nB) $2023$ \nC) $2025$ \nD) $2027$ \nE) $2029$",
"options": [],
"answer": "See solution",
"solution": "D) $2027$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21732,
"subject": "Mathematics (Olympiad)",
"question": "A boardgame board consists of 10 squares in a row that are numbered 1 to 10. On some square there is a button. In one move, it is allowed to move the button to a square whose number is either smaller by 2 or 2 times bigger. Does there exist an initial location for the button that allows the player to visit all squares of the board? It is allowed to visit one square several times.",
"options": [],
"answer": "See solution",
"solution": "No move allows the button to be placed on square number 9. Therefore, the button should start from there to have any hope. If on some later move the button is placed on an even-numbered square, then it will also stay on an even-numbered square on every move that follows. Therefore, all the odd-numbered squares must be visited right in the beginning, i.e., the button must be moved to 7, 5, 3, 1. On the next move, there is no other option but to step to square number 2. But now it is impossible to reach square number 5, since it is odd-numbered, and therefore it is also impossible to reach 10. Therefore, it is not possible to visit all the squares.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21733,
"subject": "Mathematics (Olympiad)",
"question": "Let\n$$\n2013n = a^3 - b^3.\n$$\nRewrite as\n$$\n2013n = (a-b)^3 + 3ab(a-b).\n$$\nShow that the smallest possible value of $n$ is $39$.",
"options": [],
"answer": "See solution",
"solution": "Note that if $k=11$, $b=10$, $a=43$, then $m=13$, i.e. $n=39$. Show that $n=39$ is the smallest possible value of $n$, i.e. $m=13$ is the smallest possible value of $m$. Indeed, from (2) it follows that $k(3k^2+3kb+b^2) \\ge 11$, and it is easy to show that $k \\ge 11$. But for $k \\ge 22$ we have $m > 13$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21734,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle and $D$ a point on its side $BC$. Points $E, F$ lie on the lines $AB, AC$ beyond vertices $B, C$, respectively, such that $BE = BD$ and $CF = CD$. Let $P$ be a point such that $D$ is the incenter of triangle $PEF$. Prove that $P$ lies inside the circumcircle $\\Omega$ of triangle $ABC$ or on it.",
"options": [],
"answer": "See solution",
"solution": "Let $\\omega$ be the circumcircle of triangle $AEF$ and let $I_A$ be the $A$-excenter of triangle $ABC$. First, we prove that $I_A$ is the midpoint of the arc $EF$ of $\\omega$ that does not contain point $A$ (see the left figure).\n\nTo that end, note that since $I_A$ lies on the external angle bisector of $\\angle B$ and $BE = BD$, triangles $BEI_A$ and $BDI_A$ are congruent (SAS). Similarly, triangles $CFI_A$ and $CDI_A$ are congruent, so in particular $I_A E = I_A F$. Moreover, $\\angle BEI_A + \\angle CFI_A = \\angle BDI_A + \\angle CDI_A = 180^\\circ$, hence the points $A, E, I_A, F$ lie on a single circle in this order.\n\n\n\nNext, we prove that $P$ is the second intersection of $I_A D$ and $\\omega$ (see the middle figure). Let $I$ be the incenter of triangle $ABC$. Then $ED \\parallel BI$ and $DF \\parallel IC$. Setting $\\angle EAF = \\alpha$, we get $\\angle EDF = \\angle BIC = 90^\\circ + \\frac{1}{2}\\alpha$, thus $\\angle EPF = \\alpha = \\angle EAF$, so $P$ lies on $\\omega$. Since $I_A$ is the midpoint of arc, it lies on the angle bisector $PD$, so $P$ lies both on $I_A D$ and on $\\omega$ as claimed.\n\nFinally, we show that $P$ lies on that arc of $\\omega$ which lies inside $\\Omega$ (see the right figure). Let $M \\neq A$ be the second intersection of $\\omega$ and $\\Omega$ (if they are tangent, we set $M = A$). Then $M$ is the center of the spiral similarity that maps $BE$ to $CF$ (alternatively, we angle-chase that triangles $MBE$ and $MCF$ are similar by AA). Thus $MB/MC = BE/CF = BD/DC$, so $MD$ is the angle bisector of $BMC$, and so it passes through the midpoint $S$ of the arc $BC$ of $\\Omega$ that does not contain $A$.\n\nNow forget about points $B, C, E, F$ and focus on circles $\\Omega, \\omega$ and on the points $A, M, I_A, S, D, P$. Circles $\\Omega$ and $\\omega$ share points $A$ and $M$. Being the $A$-excenter of $ABC$, point $I_A$ belongs to that arc $AM$ of $\\omega$ which lies outside of $\\Omega$ (e.g. since $AI_A > AS$). Point $S$ lies on the segment $AI_A$ and point $D$ lies on the segment $SM$, so point $D$ lies inside the angle $AI_A M$. Thus, point $P = I_A D \\cap \\omega$ belongs to the other arc $AM$ of $\\omega$ than $I_A$, namely to the one which lies inside $\\Omega$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21735,
"subject": "Mathematics (Olympiad)",
"question": "Solve the equation\n\n$$\n\\log_{3x+4}(2x+1)^2 + \\log_{2x+1}(6x^2 + 11x + 4) = 4.\n$$",
"options": [],
"answer": "See solution",
"solution": "By definition, $2x+1 > 0$, $3x+4 > 0$, $6x^2+11x+4 > 0$, and $2x+1,\\ 3x+4 \\neq 1$. Therefore, $x \\in \\left(-\\frac{1}{2}, \\infty\\right)$. Since $6x^2+11x+4 = (2x+1)(3x+4)$, the equation becomes:\n\n$$\n2 \\log_{3x+4}(2x+1) + \\log_{2x+1}((2x+1)(3x+4)) = 4\n$$\n\nExpanding:\n\n$$\n2 \\log_{3x+4}(2x+1) + \\log_{2x+1}(2x+1) + \\log_{2x+1}(3x+4) = 4\n$$\n\nCombine terms:\n\n$$\n2 \\log_{3x+4}(2x+1) + \\log_{2x+1}(3x+4) = 3\n$$\n\nLet $u = \\log_{3x+4}(2x+1)$. Then $\\log_{2x+1}(3x+4) = \\frac{1}{u}$, so the equation is:\n\n$$\n2u + \\frac{1}{u} = 3\n$$\n\nMultiply both sides by $u$:\n\n$$\n2u^2 - 3u + 1 = 0\n$$\n\nThe roots are $u_1 = 1$ and $u_2 = \\frac{1}{2}$.\n\nFor $u_1 = 1$:\n\n$$\n3x+4 = 2x+1 \\implies x = -3\n$$\n\nBut $x = -3 \\notin \\left(-\\frac{1}{2}, \\infty\\right)$.\n\nFor $u_2 = \\frac{1}{2}$:\n\n$$\n3x+4 = (2x+1)^2\n$$\n\n$$\n4x^2 + 4x - 3 = 0\n$$\n\nThe roots are $x_1 = -1$ and $x_2 = \\frac{3}{4}$.\n\nBut $x_1 = -1 \\notin \\left(-\\frac{1}{2}, \\infty\\right)$. Therefore, the solution is $x = \\frac{3}{4}$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21736,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, a_3, a_4, a_5$ be five real numbers with zero sum, such that $|a_i - a_j| \\leq 1$ for all $i, j \\in \\{1, 2, 3, 4, 5\\}$. Prove that\n$$a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2 \\leq \\frac{6}{5}.$$",
"options": [],
"answer": "See solution",
"solution": "Since $0 = (a_1 + a_2 + a_3 + a_4 + a_5)^2 = a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2 + 2 \\sum_{i 2$，所以 $x_1$ 與 $y_1$ 中至少有一個大於 $1$。而 $n > 1$，因此 $A > 1$。由 (1) 推得\n\n$$\nA(x_1 + y_1) = p^{k-m}.\n$$\n\n因為 $x_1 + y_1 > 1$ 且 $A > 1$，所以它們都能被 $p$ 整除，且存在某個正整數 $\\beta$，使得 $x_1 + y_1 = p^\\beta$。所以\n\n$$\n\\begin{aligned}\nA &= x_1^{n-1} - x_1^{n-2}(p^\\beta - x_1) + x_1^{n-3}(p^\\beta - x_1)^2 - \\cdots \\\\\n&\\quad - x_1(p^\\beta - x_1)^{n-2} + (p^\\beta - x_1)^{n-1} \\\\\n&= n x_1^{n-1} + B \\beta \\quad (B \\text{ 是某個整數}).\n\\end{aligned}\n$$\n\n因為 $p|A$，且 $x^{pq} + y^{pq} = p^k$，即 $(x^p)^q + (y^p)^q = p^k$。\n\n如果 $q > 1$，同理可證 $p|q$。如果 $q = 1$，則 $p = q$。如此重複上述步驟，可得 $n = p^l$，對某個正整數 $l$。\n\n*回到本題*：$n$ 的可能值只有 $2$。\n\n設 $3^n = x^s + y^s$，其中 $(x, y) = 1$。不妨設 $x > y$。由於 $s > 1$，$n$ 為正整數，顯然 $3 \\nmid x$ 且 $s \\nmid y$。討論如下：\n\n1. 若 $s$ 是偶數，則\n $$\n x^s \\equiv 1 \\pmod{3}, \\quad y^s \\equiv 1 \\pmod{3},\n $$\n 於是 $x^s + y^s \\equiv 2 \\pmod{3}$，$x^s + y^s$ 不是 $3$ 的整數乘幂，矛盾！\n\n2. 若 $s$ 是奇數且 $s > 1$，則\n $$\n 3^n = (x + y)(x^{s-1} - x^{s-2} y + \\cdots + y^{s-1}),\n $$\n 於是 $x + y = 3^m, m \\ge 1$。\n\n 以下證明 $n \\ge 2m$。\n\n 由引理知 $3|s$。取 $x_1 = x^{1/3}, y_1 = y^{1/3}$，代入後可得 $s = 3$。於是 $x^3 + y^3 = 3^n, x + y = 3^m$。\n\n 只要證明 $x^3 + y^3 \\ge (x + y)^2$，即證明 $x^2 - x y + y^2 \\ge x + y$。\n\n 由於 $x \\ge y + 1$，則 $x^2 - x = x(x - 1) \\ge x y$。因此 $(x^2 - x y - x) + (y^2 - y) \\ge 0$。\n\n 於是 $n \\ge 2m$ 得證。\n\n 由 $(x + y)^3 - (x^3 + y^3) = 3 x y (x + y)$，推出 $3^{2m-1} - 3^{n-m-1} = x y$。\n\n 而 $2m-1 \\ge 1$ 且 $n-m-1 \\ge n-2m \\ge 0$。\n\n 如果 (2) 中至少有一個不等號是嚴格不等號，則 $3|(3^{2m-1} - 3^{n-m-1})$。\n 由 $3 \\nmid x y$ 推出矛盾。可知\n $$\n n - m - 1 = n - 2m = 0.\n $$\n 則 $m = 1, n = 2$ 且 $3^2 = 2^3 + 1^3$。\n\n故 $n = 2$ 是唯一滿足條件的值。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21738,
"subject": "Mathematics (Olympiad)",
"question": "A positive integer $n$ is given. There are $n$ computers in a network, numbered $1, 2, \\ldots, n$. The computers are connected with one-way communication lines so that information can be sent from any computer to all others, either directly or through other computers. Initially, each computer only knows its own number. When a procedure is initiated on a computer, it outputs all the numbers it knows and communicates them to all computers it has direct lines to. The operator can initiate this procedure on any selected computers, in any sequence (possibly multiple times on the same computer).\n\nProve that the minimum total number of procedure initiations required so that all computers have output the numbers of all computers at least once does not depend on the network of lines, but only on $n$, and find this number of initiations.",
"options": [],
"answer": "See solution",
"solution": "We will show that $2n - 1$ initiations are necessary regardless of the network of lines.\n\nConsider any sequence of initiations after which all computers have output the numbers of all computers. Let $A$ be the computer with the latest time of the first procedure initiation. Since the procedure has been initiated at least once in all other computers by that time, the order number of this initiation is at least $n$. Therefore, after the first $n - 1$ initiations, no computer has output the number of computer $A$. To ensure that each computer outputs the number of computer $A$ at least once, $n$ more initiations are needed. Thus, at least $2n - 1$ initiations are necessary.\n\nNow we show that $2n - 1$ initiations are sufficient, regardless of the network. Choose any computer $A$ and construct a sequence of the remaining computers $B_1, B_2, \\dots, B_{n-1}$, such that each $B_i$ can receive information directly from some preceding computer in the sequence or from $A$. Such a sequence exists, since otherwise, some computers would be unreachable, contradicting the problem's condition. Similarly, construct a sequence $C_1, C_2, \\dots, C_{n-1}$, such that each $C_i$ can send information directly to some preceding computer in the sequence or to $A$; by symmetry, this sequence also exists.\n\nA suitable sequence of procedure initiations of length $2n - 1$ is:\n\n$$C_{n-1}, C_{n-2}, \\dots, C_1, A, B_1, B_2, \\dots, B_{n-1}.$$ \n\nLet $C_0 = A$. By construction, each $C_i$ ($i > 0$) sends its number to at least one $C_{i'}$ ($0 \\leq i' < i$). Continuing recursively, the number of $C_i$ reaches $A$ in at most $n-1$ steps. Thus, after the first $n-1$ initiations, all numbers reach $A$. In the subsequent $n$ initiations, $A$ and the $B_i$ output all numbers, and by construction, each $B_i$ can receive all numbers. Therefore, after $2n-1$ initiations, all computers have output the numbers of all computers at least once.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21739,
"subject": "Mathematics (Olympiad)",
"question": "If $n$ is a positive integer and $n^2$ equals the 4-digit number $aabb$, find $n$.",
"options": [],
"answer": "See solution",
"solution": "Method 1\n\nSince $a - a + b - b = 0$, we know that $aabb$ is divisible by $11$. Therefore, $n$ is divisible by $11$.\nSince $n^2$ has 4 digits, $n = 33, 44, 55, 66, 77, 88$, or $99$.\nOf these, only $88^2 = 7744$ has the form $aabb$. So $n = 88$.\n\nMethod 2\n\nA square number ends in $0, 1, 4, 5, 6,$ or $9$ and has remainder $0$ or $1$ when divided by $4$.\nWhen divided by $4$, $aa11$, $aa55$, and $aa99$ all have remainder $3$, and $aa66$ has remainder $2$.\nSo $n^2$ must be of the form $aa00$ or $aa44$. If $aa00$ is a square number, then $aa$ must be a square number, which is impossible. So $n^2$ must be of the form $aa44$.\n\nSince $aa44 = 11 \\times a04$ and $11$ is prime, $11$ divides $a04 = 100a + 4 = 99a + (a + 4)$.\nTherefore $11$ divides $a + 4$. Hence $a = 7$, $n^2 = 7744$, and $n = \\mathbf{88}$.\n\nMethod 3\n\nSince $n^2 = aabb = a0b \\times 11$ and $11$ is prime, $11$ also divides $a0b$ and $a0b/11$ is a square number.\nFrom the divisibility test for $11$, we have $a + b = 11$.\n\nSince square numbers end in $0, 1, 4, 5, 6,$ or $9$, the only possibilities for $a0b$ are $209, 506, 605, 704$, and these give $19, 46, 55, 64$ when divided by $11$. The only square among these is $64$, so $n^2 = 8^2 \\times 11^2 = 7744$ and $n = 8 \\times 11 = 88$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21740,
"subject": "Mathematics (Olympiad)",
"question": "Дурын аргаар 3 өнгийн аль нэгээр нь квадрат матрицын нүднүүдийг будахад, дор хаяж 3 нүд нь ижил өнгөөр будагдсан мөр эсвэл багана ямагт олддог байх $n$-ийн хамгийн бага утгыг ол.",
"options": [],
"answer": "See solution",
"solution": "Хариу: $n = 7$.\n\n$$\n n = 7 \\text{ үед } 7^2 = 49 = 3 \\cdot 16 + 1 \\Rightarrow \\text{дор хаяж } 17 \\text{ квадрат ижил} \n\\text{өнгөөр будагдана. } 17 = 7 \\cdot 2 + 3 \\Rightarrow 7 \\text{ мөрөөс нэг мөр нь}\n$$\n\nДор хаяж 3 ижил өнгийн нүд агуулна. $n = 6$ үед эсрэг жишээ:\n\n$$\nA = \\begin{pmatrix} 1 & 2 & 3 & 1 & 2 & 3 \\\\ 2 & 3 & 1 & 2 & 3 & 1 \\\\ 3 & 1 & 2 & 3 & 1 & 2 \\end{pmatrix}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21741,
"subject": "Mathematics (Olympiad)",
"question": "設 $O$, $H$ 分別為銳角三角形 $ABC$ 的外心及垂心。分別在邊 $AB$, $AC$ 上取兩點 $D$, $E$，滿足 $A$, $D$, $O$, $E$ 共圓。令 $P$ 為三角形 $ABC$ 的外接圓上一點，過 $P$ 分別作平行於 $OD$, $OE$ 的直線並分別交 $AB$, $AC$ 於 $X$, $Y$。設 $\\,\\overline{HP}$ 的中垂線與 $XY$ 不重合並交於 $Q$，且 $A$, $Q$ 位於 $DE$ 異側。\n\n證明：$\\angle EQD = \\angle BAC$。\n\n",
"options": [],
"answer": "See solution",
"solution": "令 $P_B$, $P_C$ 分別為 $P$ 關於 $CA$, $AB$ 的對稱點，$O_B$, $O_C$ 分別為 $O$ 關於 $CA$, $AB$ 的對稱點，則\n\n$$\n\\triangle PXP_C \\sim \\triangle PYP_B \\sim \\triangle ODO_C \\sim \\triangle OEO_B.\n$$\n\n由施坦納定理知 $H$, $P_B$, $P_C$ 共線。取 $Q'$ 滿足 $\\triangle PQ'H \\sim \\triangle PXP_C$，那麼 $Q'$ 位於 $\\overline{HP}$ 的中垂線上，且由旋似可得 $Q'XY \\sim HP_C P_B$，故 $X$, $Q'$, $Y$ 共線。所以由同一法知 $Q = Q'$。\n\n取 $K$ 滿足 $\\triangle OKH \\sim \\triangle ODO_C$，那麼由 $\\triangle OKH \\sim \\triangle PQH$ 及旋似可得\n\n$$\n\\triangle DKO \\sim \\triangle O_C HO \\quad \\text{及} \\quad \\triangle HKQ \\sim \\triangle HOP,\n$$\n\n所以有\n\n$$\n\\frac{KQ}{OP} = \\frac{HK}{HO} = \\frac{OK}{OH} = \\frac{KD}{HO_C} \\quad (1)\n$$\n\n作 $H$ 關於 $AB$ 的對稱點 $H_C$，則 $HO_C = H_C O = OP$，再由 (1) 知 $KQ = KD$。\n\n同理，由 $\\triangle OKH \\sim \\triangle OEO_B$ 有 $\\overline{KQ} = \\overline{KE}$，所以 $K$ 為 $\\triangle QDE$ 的外心。\n\n注意到 $O_B O_C$ 為 $\\overline{AH}$ 的中垂線，所以由 $\\triangle KDE \\sim \\triangle HO_C O_B$，\n\n$$\n\\angle EQD = 90^\\circ - \\angle KDE = 90^\\circ - \\angle HO_C O_B = 90^\\circ - \\angle HBA = \\angle BAC,\n$$\n\n得證。\n\n註：難度約為 G5。本題的解法不長，但利用相當多的旋似變換，是相對不容易觀察的，因此難度較高。另外，若固定 $D$, $E$，使 $P$ 在三角形 $ABC$ 的外接圓上動時，$XY$ 的包絡線是以 $K$ 為中心的三尖瓣線，且 $Q$ 為 $XY$ 與該三尖瓣線的內切圓的其中一個交點。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21742,
"subject": "Mathematics (Olympiad)",
"question": "What is the size of the largest set of numbers we can choose among $1, 2, \\ldots, 2n$ in such a way that any two of them have a common divisor greater than $1$?\n\n\n\nFig. 30",
"options": [],
"answer": "See solution",
"solution": "$n$.\n\nIf we choose all even numbers, they satisfy the required condition and there are exactly $n$ of them.\n\nNow, consider the following pairs of integers: $(1, 2), (3, 4), \\ldots, (2n-1, 2n)$. In each pair, the numbers are relatively prime, so we cannot choose both of them for our set. Therefore, it is impossible to choose more than $n$ integers.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21743,
"subject": "Mathematics (Olympiad)",
"question": "Let $p \\neq 2, 5$. Prove that if $p$ divides $m = 5^{4n} - 5^{3n} + 5^{2n} - 5^n + 1$ for some integer $n$, then $p \\equiv 1 \\pmod{4}$.",
"options": [],
"answer": "See solution",
"solution": "Clearly, $p \\neq 2, 5$. Let $m = 5^{4n} - 5^{3n} + 5^{2n} - 5^n + 1$.\n\n$$\n(2 \\cdot 5^{2n} - 5^n + 2)^2 - 5 \\cdot 5^{2n} = 4m \\equiv 0 \\pmod{p}.\n$$\n\nThis gives $5 \\equiv (5^{-n}(2 \\cdot 5^{2n} - 5^n + 2))^2 \\pmod{p}$. Using the Legendre symbol, we have $\\left(\\frac{5}{p}\\right) = 1$.\n\nOn the other hand,\n$$\n(5^{2n} - 5^n + 1)^2 + 5^n(5^n - 1)^2 = m \\equiv 0 \\pmod{p}.\n$$\n\nAs above, this implies $\\left(\\frac{-5^n}{p}\\right) = 1$. It follows that\n$$\n\\left(\\frac{-1}{p}\\right) = \\left(\\frac{-5^n}{p}\\right) \\left(\\frac{5^n}{p}\\right) = 1 \\cdot 1^n = 1.\n$$\n\nTherefore, $p \\equiv 1 \\pmod{4}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21744,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a cyclic quadrilateral. The diagonals $AC$ and $BD$ meet at $P$, and $DA$ and $CB$ produced meet at $Q$. The midpoint of $AB$ is $E$. Prove that if $PQ$ is perpendicular to $AC$, then $PE$ is perpendicular to $BC$.",
"options": [],
"answer": "See solution",
"solution": "Let circle $APQ$ meet $BC$ at $X$, so $\\angle AXQ = \\pi/2$.\n\n\n\nNow, $ABCD$ is cyclic, so $\\angle PAQ = \\angle PBQ$. Also, since $APXQ$ is cyclic, $\\angle PAQ = \\angle PXB$.\n\nHence, $\\angle PXB = \\angle PBQ$, so $PBX$ is isosceles, and so $P$ lies on the perpendicular bisector of $BX$.\n\nNow, since $AX$ and $BC$ are perpendicular, and $E$ is the midpoint of $AB$, $E$ also lies on the perpendicular bisector of $BX$.\n\nTherefore, $PE$ is the perpendicular bisector of $BX$, so $PE$ and $BC$ are perpendicular, as required.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21745,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle and let $O$ be its circumcentre. The internal and external bisectors of the angle $BAC$ meet the line $BC$ at points $D$ and $E$, respectively. Let $M$ and $L$ denote the midpoints of the segments $BC$ and $DE$, respectively. The circles $ABC$ and $ALO$ meet again at point $N$. Show that the angles $BAN$ and $CAM$ are equal.",
"options": [],
"answer": "See solution",
"solution": "We must show that $AN$ is the symmedian in triangle $ABC$.\n\nLet $R$ denote the radius of the circle $ABC$. Notice that the cross-ratio $(EDBC) = -1$, so $LA^2 = LD^2 = LB \\cdot LC = LO^2 - R^2 = LO^2 - AO^2$, which means $L$ and $O$ are antipodal on the circle $ALO$. It follows that $N$ is the reflection of $A$ in the diameter $LO$ and lies on the Apollonius circle $ADE$.\n\nLet $AN$ and $BC$ meet at $K$. We shall prove that $AK$ is the symmedian through $A$ in triangle $ABC$. Notice that $K$ has equal powers relative to the circles $ABC$ and $ADE$, $KB \\cdot KC = KA \\cdot LK = KD \\cdot KE$, so\n\n$$\nKB = \\frac{BD \\cdot BE}{BE + CD} \\quad \\text{and} \\quad KC = \\frac{CD \\cdot CE}{BD + CE}.\n$$\n\nRecall that\n\n$$\nBD = \\frac{AB \\cdot BC}{AB + AC}, \\quad CD = \\frac{AC \\cdot BC}{AB + AC},\n$$\n\n$$\nBE = \\frac{AB \\cdot BC}{|AB - AC|} \\quad \\text{and} \\quad CE = \\frac{AC \\cdot BC}{|AB - AC|},\n$$\n\nfrom the bisector theorem, so\n\n$$\nKB = \\frac{AB^2 \\cdot BC}{AB^2 + AC^2} \\quad \\text{and} \\quad KC = \\frac{AC^2 \\cdot BC}{AB^2 + AC^2},\n$$\n\nso $\\frac{KB}{KC} = \\frac{AB^2}{AC^2}$, and by Steiner's theorem, $AK$ is the symmedian from $A$ in triangle $ABC$.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21746,
"subject": "Mathematics (Olympiad)",
"question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a continuous function such that $f(x) + \\sin(f(x)) \\geq x$ for all $x \\in \\mathbb{R}$. Prove that\n$$\n\\int_{0}^{\\pi} f(x) \\, dx \\geq \\frac{\\pi^2}{2} - 2.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $g : \\mathbb{R} \\to \\mathbb{R}$ be defined by $g(x) = x + \\sin(x)$ for any $x \\in \\mathbb{R}$. Clearly, $g$ is differentiable on $\\mathbb{R}$, and $g'(x) > 0$ for all $x \\in ((2k-1)\\pi, (2k+1)\\pi)$, $k \\in \\mathbb{Z}$. Thus, $g$ is increasing on each interval $[(2k-1)\\pi, (2k+1)\\pi]$, so $g$ is increasing on $\\mathbb{R}$.\n\nAs $\\lim_{x \\to -\\infty} g(x) = -\\infty$ and $\\lim_{x \\to \\infty} g(x) = \\infty$, $g$ is one-to-one, with $g^{-1} : \\mathbb{R} \\to \\mathbb{R}$ continuous and increasing on $\\mathbb{R}$. The given inequality can thus be written as\n$$\ng(f(x)) \\geq x, \\text{ for any } x \\in \\mathbb{R} \\iff f(x) \\geq g^{-1}(x), \\text{ for any } x \\in \\mathbb{R}.\n$$\nTherefore,\n$$\n\\int_{0}^{\\pi} f(x) \\, dx \\geq \\int_{0}^{\\pi} g^{-1}(x) \\, dx.\n$$\nSince $g(0) = 0$ and $g(\\pi) = \\pi$, by Young's inequality, we have\n$$\n\\begin{align*}\n\\int_{0}^{\\pi} g^{-1}(x) \\, dx &= \\pi \\cdot g(\\pi) - 0 \\cdot g(0) - \\int_{0}^{\\pi} g(x) \\, dx \\\\\n&= \\pi^2 - \\left( \\frac{\\pi^2}{2} - \\cos(\\pi) \\right) + (0 - \\cos(0)) \\\\\n&= \\frac{\\pi^2}{2} - 2\n\\end{align*}\n$$\nwhich gives the result.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21747,
"subject": "Mathematics (Olympiad)",
"question": "Suppose the image of the integers under the polynomial\n\n$$\nP(x) = a_n x^n + a_{n-1} x^{n-1} + \\dots + a_1 x + a_0\n$$\n\ncontains an infinite geometric progression with common ratio $a \\in \\mathbb{Z} \\setminus \\{0\\}$. Find all polynomials $P(x) \\in \\mathbb{Z}[x]$ with this property.",
"options": [],
"answer": "See solution",
"solution": "Let $P(x)$ be a polynomial such that $P(\\mathbb{Z})$ contains an infinite geometric progression with integer ratio $a \\neq 0$.\n\nFor each $b \\in \\mathbb{Z}$,\n\n$$\nP(ax + b) = a_n a^n x^n + (n a_n a^{n-1} b + a_{n-1} a^{n-1}) x^{n-1} + \\dots\n$$\n\nand\n\n$$\na^n P(x) = a_n a^n x^n + a_{n-1} a^n x^{n-1} + \\dots + a_0 a^n.\n$$\n\nWe can find $b_1, b_2 \\in \\mathbb{Z}$ and $N \\in \\mathbb{N}$ such that for every $x \\ge N$,\n\n$$\nP(ax + b_2) < a^n P(x) < P(ax + b_1),\n$$\n\nand for every $x \\le -N$,\n\n$$\nP(ax + b_2) < a^n P(x) < P(ax + b_1), \\quad \\text{or} \\quad P(ax + b_1) < a^n P(x) < P(ax + b_2).\n$$\n\nFor each $P(x)$ in the geometric progression, $a P(x), a^2 P(x), \\dots, a^n P(x), \\dots$ are all in $P(\\mathbb{Z})$, so there exists $y \\in \\mathbb{Z}$ such that $a^n P(x) = P(y)$. For large $|x|$, using the above inequalities, $ax + b_1 < y < ax + b_2$, so $y - ax$ is a constant between $b_1$ and $b_2$. Thus, there is a constant $c$ such that $a^n P(x) = P(ax + c)$ for infinitely many $x$, so $P(ax + c)$ and $a^n P(x)$ are equal as polynomials.\n\nNow, we seek all $P(x) \\in \\mathbb{Z}[x]$ such that $a^n P(x) = P(ax + c)$. Let $Q(x) = a x + c$. If $\\alpha$ is a root of $P(x)$, then $Q(\\alpha)$, $Q^2(\\alpha)$, ... are also roots. Since $P(x)$ has finitely many roots, there exist $m_1 > m_2$ with $Q^{m_1}(\\alpha) = Q^{m_2}(\\alpha)$, so $Q^{m_1 - m_2}(\\alpha) = \\alpha$. Since $Q$ is injective, the only fixed point is $\\beta = \\frac{c}{1 - a}$, so $\\alpha = \\beta$ is the only root. Thus, $P(x) = r(x - \\beta)^n = \\frac{r}{q^n}(q x - p)^n$, where $p, q, r \\in \\mathbb{Z}$, $(p, q) = 1$.\n\nSince $P(x) \\in \\mathbb{Z}[x]$, we require $\\frac{r}{q^n} p^n \\in \\mathbb{Z}$, so $q^n \\mid r$. Let $r = q^n s$ for $s \\in \\mathbb{Z}$, so $P(x) = s(q x - p)^n$ with $(p, q) = 1$.\n\nEach such polynomial satisfies the property: for $P(x) = q x - p$, the sequence $\\{-p (q + 1)^m\\}_{m \\ge 0}$ is a geometric progression with ratio $q + 1$, and all terms are congruent to $-p$ mod $q$.\n\nThus, all polynomials of the form $P(x) = s(q x - p)^n$ with $s, p, q \\in \\mathbb{Z}$, $(p, q) = 1$, are the solutions.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21748,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer and $a_1, a_2, \\dots, a_n$ be real numbers such that\n\n$$a_m + a_{m+1} + \\dots + a_n \\ge m + (m+1) + \\dots + n$$\n\nfor every $m = 1, 2, \\dots, n$.\n\nProve that\n\n$$a_1^2 + \\dots + a_n^2 \\ge \\frac{n(n+1)(2n+1)}{6}.$$",
"options": [],
"answer": "See solution",
"solution": "Let $b_k = a_k - k$ for $k = 1, 2, \\dots, n$. Then $b_m + b_{m+1} + \\dots + b_n \\ge 0$ for $m = 1, 2, \\dots, n$.\n\nIt follows that\n\n$$\n\\sum_{i=1}^{n} a_i^2 = \\sum_{i=1}^{n} b_i^2 + 2 \\sum_{i=1}^{n} i b_i + \\sum_{i=1}^{n} i^2 \\ge \\sum_{i=1}^{n} i^2 = \\frac{n(n+1)(2n+1)}{6},\n$$\n\nbecause\n\n$$\n\\sum_{i=1}^{n} i b_i = (b_1 + \\dots + b_n) + (b_2 + \\dots + b_n) + \\dots + b_n \\ge 0.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21749,
"subject": "Mathematics (Olympiad)",
"question": "Let $P$ be a point on the side $BC$ of an acute triangle $ABC$. Let $H$ be the orthocenter of $\\triangle ABC$ and $D$ be the foot of the perpendicular from $H$ to $AP$. Let $\\Gamma_1, \\Gamma_2$ be the circumcircles of $\\triangle ABD$ and $\\triangle ACD$, respectively. Let $l$ be the line parallel to $BC$ and passing through $D$. The line $l$ intersects $\\Gamma_1$ and $\\Gamma_2$ at $X$ and $Y$, different from $D$, respectively. The line $l$ intersects $AB$ and $AC$ at $E$ and $F$, respectively. Two lines $XB$ and $YC$ meet at $Z$. Show that $BP = CP$ if and only if $ZE = ZF$.",
"options": [],
"answer": "See solution",
"solution": "Observation 1. $A$, $B$, $C$, $Z$ are cyclic.\n\n$$\n\\angle BAD = \\angle BXD = \\angle ZBC \\text{ and } \\angle DAC = \\angle DYC = \\angle BCZ.\n$$\n\nHence $\\angle BAC = \\angle BAD + \\angle BCZ = 180^\\circ - \\angle BZC$, that is, $\\angle BAC + \\angle BZC = 180^\\circ$.\n\nObservation 2. Both $A$, $E$, $Z$, $Y$ and $A$, $X$, $Z$, $F$ are cyclic.\n\nBy Observation 1, we have\n\n$$\n\\angle AEY = \\angle ABC = \\angle AZC = \\angle AZY.\n$$\n\nSimilarly, $\\angle AFX = \\angle AZX$.\n\nLemma 1. $ZE = ZF$ if and only if $\\frac{BD}{CD} = \\frac{AB}{AC}$.\n\n*Proof.* We can easily see that $\\triangle AEZ$ is similar to $\\triangle ADC$, and symmetrically, $\\triangle AFZ$ is similar to $\\triangle ADB$. Then we have\n\n$$\nZE = \\frac{AE \\cdot CD}{AD} \\quad \\text{and} \\quad ZF = \\frac{AF \\cdot BD}{AD}.\n$$\n\nTherefore, we have $ZE = ZF \\Leftrightarrow \\frac{BD}{CD} = \\frac{AE}{AF} = \\frac{AB}{AC}$.\n\nNow let $Q$, $R$, $S$ be the feet of perpendiculars from $A$, $B$, $C$ to $BC$, $CA$, $AB$, respectively. Then $A$, $S$, $D$, $H$, $F$ are cyclic. Since $AD \\cdot AP = AH \\cdot AQ = AS \\cdot AB$, $S$, $D$, $P$, $B$ are cyclic.\n\nSimilarly, $R$, $D$, $P$, $C$ are cyclic.\n\nSince $AS : AP = AD : AB$, $\\triangle ASP$ is similar to $\\triangle ADB$.\n\nNow we have\n\n$$\n(1) \\qquad PS = BD \\frac{AP}{AB}, \\quad RP = CD \\frac{AP}{AC}.\n$$\n\nLemma 2. Let $P$ be a point on $BC$. Then\n\n$$\nBP = PC \\Leftrightarrow PR = PS.\n$$\n\n*Proof.* Suppose $P$ is a midpoint of $BC$. Then $P$, $Q$, $R$, $S$ are on the nine-point circle, so we have\n\n$$\n\\angle PSR = \\angle RQC = \\angle A = \\angle SQB = \\angle PRS.\n$$\n\nThat is, $PR = PS$. Hence $P$ is the intersection of the perpendicular bisector of $SR$ and $BC$, so it is unique. Thus $PR = PS$ implies that $P$ is a midpoint of $BC$. $\\square$\n\nNow we are ready to prove that $BP = PC \\Leftrightarrow ZE = ZF$.\n\nSuppose $BP = PC$, then by Lemma 2, $PR = PS$. Then by (1), we have $\\frac{BD}{CD} = \\frac{AB}{AC}$. Then by Lemma 1, $ZE = ZF$.\n\nConversely, assume that $ZE = ZF$, then by Lemma 1, $\\frac{BD}{CD} = \\frac{AB}{AC}$. Then by (1), we have $PS = PR$. By Lemma 2, $BP = PC$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21750,
"subject": "Mathematics (Olympiad)",
"question": "Let $a \\geq b$ and $c \\geq d$ be real numbers. Prove that the equation\n$$\n(x + a)(x + d) + (x + b)(x + c) = 0\n$$\nhas real roots.",
"options": [],
"answer": "See solution",
"solution": "Let $f(x) = (x + a)(x + d) + (x + b)(x + c)$. The leading coefficient of $f(x)$ is $2$, so $f(x)$ is positive for large $x$.\n\nWe have\n$$\n\\begin{aligned}\nf(-a) + f(-b) &= (b-a)(c-a) + (a-b)(d-b) = (a-b)(-c + a + d - b), \\\\\nf(-c) + f(-d) &= (a-c)(d-c) + (b-d)(c-d) = (c-d)(-a + c + b - d).\n\\end{aligned}\n$$\n\nSince $a \\geq b$ and $c \\geq d$, if $f(-a) + f(-b) \\leq 0$, then either $f(-a) \\leq 0$ or $f(-b) \\leq 0$, so $f(x)$ crosses the $x$-axis. Otherwise, $-c + a + d - b > 0$, which implies $-a + c + b - d < 0$. Then\n$$\nf(-c) + f(-d) = (c - d)(-a + c + b - d) \\leq 0.\n$$\nSo either $f(-c) \\leq 0$ or $f(-d) \\leq 0$, and again $f(x)$ crosses the $x$-axis somewhere.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21751,
"subject": "Mathematics (Olympiad)",
"question": "給定一個平行四邊形，其四個角的角度都是 $360^\\circ$ 的有理數倍。是否可能將座標平面上除了原點的每一點著黑白兩色，使得當某三點和原點構成一個該四邊形的相似形時，該三點必定不是同色？",
"options": [],
"answer": "See solution",
"solution": "可以。\n\n首先，若該平行四邊形不是菱形，可令 $t > 1$ 為長邊和短邊的比值。對平面上的任意點，若該點與原點距離在 $[t^{2k}, t^{2k+1})$ 內（$k$ 為任意整數），則著黑色；否則著白色。如此平行四邊形最靠近原點的兩個點一定是不同色。\n\n以下考慮平行四邊形是菱形。設其小於或等於 $90^\\circ$ 的內角角度為 $2a$，即 $a \\le 45^\\circ$。則其邊與對角線的夾角為 $a$ 或 $90^\\circ - a$。現在著色如下：以圓心及 $x$ 軸方向量仰角，將區間 $[0^\\circ, 90^\\circ)$ 分割為 $[0, a)$、$[a, 2a)$、...、$[ka, 90^\\circ)$，將幅角落在第一個區間的點著黑色，第二個著白色，這樣黑白交錯。在 $90^\\circ$ 以後的角的顏色，則遵從以下原則：若兩個點的幅角差 $90^\\circ$，則這兩點同色。這麼一來，任何之間幅角相差 $a$ 的連續三個點（依逆時序為 $A, B, C$）必不全同色，特別是：$A, B$ 之仰角在 $[0, 90^\\circ)$ 時，$A, B$ 不同色；若 $A$ 之仰角在 $[0, 90^\\circ)$ 但 $B, C$ 之仰角在 $[90^\\circ, 180^\\circ)$ 時，$B, C$ 不同色。（不會有 $A, B, C$ 在三個不同象限的狀況，因為 $a \\le 45^\\circ$。）而其他狀況都是對稱的。又相差 $90^\\circ - a$ 與相差 $a$ 對於著色是等價的，因此得證。\n\n註：這個解答不需要“有理數倍”的假設，然而有其他解法是需要的。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21752,
"subject": "Mathematics (Olympiad)",
"question": "$$\n\\min_{1 \\leq k \\leq n} \\{|z_{k+1} - z_k|^2\\} = 1.\n$$\n\nUnder this assumption, find the minimum value of $\\sum_{k=1}^n |z_k|^2$ for complex numbers $z_1, z_2, \\dots, z_n$.\n",
"options": [],
"answer": "See solution",
"solution": "When $n$ is even:\n\n$$\n\\sum_{k=1}^{n} |z_k|^2 = \\frac{1}{2} \\sum_{k=1}^{n} (|z_k|^2 + |z_{k+1}|^2) \\geq \\frac{1}{4} \\sum_{k=1}^{n} |z_{k+1} - z_k|^2 \\geq \\frac{n}{4} \\min_{1 \\leq k \\leq n} \\{|z_{k+1} - z_k|^2\\} = \\frac{n}{4}.\n$$\n\nEquality holds when $(z_1, z_2, \\dots, z_n) = (\\frac{1}{2}, -\\frac{1}{2}, \\dots, \\frac{1}{2}, -\\frac{1}{2})$, so the minimum value is $\\frac{n}{4}$.\n\nWhen $n$ is odd:\n\nLet $\\theta_k = \\arg \\frac{z_{k+1}}{z_k} \\in [0, 2\\pi)$ for $k = 1, 2, \\dots, n$.\n\nIf $\\theta_k \\leq \\frac{\\pi}{2}$ or $\\theta_k \\geq \\frac{3\\pi}{2}$, then\n$$\n|z_k|^2 + |z_{k+1}|^2 \\geq |z_k - z_{k+1}|^2 \\geq 1.\n$$\n\nIf $\\theta_k \\in (\\frac{\\pi}{2}, \\frac{3\\pi}{2})$, then\n$$\n|z_k|^2 + |z_{k+1}|^2 \\geq \\frac{1}{2 \\sin^2 \\frac{\\theta_k}{2}}.\n$$\n\nIf for all $k$, $\\theta_k \\in (\\frac{\\pi}{2}, \\frac{3\\pi}{2})$, then by Jensen's Inequality and the convexity of $f(x) = \\frac{1}{\\sin^2 x}$,\n$$\n\\sum_{k=1}^{n} |z_k|^2 \\geq \\frac{n}{4} \\cdot \\frac{1}{\\sin^2 \\frac{m\\pi}{n}} \\geq \\frac{n}{4} \\cdot \\frac{1}{\\cos^2 \\frac{\\pi}{2n}}.\n$$\n\nIf there exists $j$ such that $\\theta_j \\notin (\\frac{\\pi}{2}, \\frac{3\\pi}{2})$, then\n$$\n\\sum_{k=1}^{n} |z_k|^2 \\geq \\frac{n+1}{4} \\geq \\frac{n}{4} \\cdot \\frac{1}{\\cos^2 \\frac{\\pi}{2n}}.\n$$\n\nTherefore, for odd $n \\geq 3$, the minimum value is $\\lambda_0(n) = \\frac{n}{4} \\cdot \\frac{1}{\\cos^2 \\frac{\\pi}{2n}}$.\n\nThe minimum is attained when $z_k = \\frac{1}{2\\cos \\frac{\\pi}{2n}} \\cdot e^{\\frac{i(n-1)k\\pi}{n}}$ for $k = 1, \\dots, n$.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21753,
"subject": "Mathematics (Olympiad)",
"question": "Let $p_0(x) = 0$, $p_1(x) = x - 2013$, and for $n \\ge 2$, $p_n(x) = (x - 2013)p_{n-1}(x) + (2014 - x)p_{n-2}(x)$. For which real numbers $x$ is $p_n(x) = 0$ for all positive integers $n$?",
"options": [],
"answer": "See solution",
"solution": "We have, for $n \\ge 2$,\n\n$$\np_n(x) - p_{n-1}(x) = (x - 2014)(p_{n-1}(x) - p_{n-2}(x)).\n$$\n\nHence, by induction,\n\n$$\np_n(x) - p_{n-1}(x) = (x - 2014)^{n-1} (p_1(x) - p_0(x)) = (x - 2014)^{n-1} (x - 2013),\n$$\n\ni.e.\n\n$$\np_n(x) = p_{n-1}(x) + (x - 2014)^{n-1}(x - 2013), \\text{ for } n \\ge 2,\n$$\n\nwhich also holds for $n = 1$.\n\nHence, by induction, for $n \\ge 1$,\n\n$$\np_n(x) = \\left((x - 2014)^{n-1} + (x - 2014)^{n-2} + \\dots + 1\\right)(x - 2013).\n$$\n\nWe note $p_n(2015) = 2n \\ne 0$ and, for $x \\ne 2015$,\n\n$$\np_n(x) = \\frac{(x - 2014)^n - 1}{x - 2015} (x - 2013).\n$$\n\nThus, for $x \\ne 2015$, $p_n(x) = 0$ if and only if either $x = 2013$ or $(x - 2014)^n = 1$.\n\nNow $(x - 2014)^n = 1$ only if $x = 2015$ (which gives $p_n(2015) \\ne 0$) or $n$ is even and $x = 2013$.\n\nTherefore, $x = 2013$ is the only real solution for all $n$.\n\nSubstituting back, $p_0(2013) = p_1(2013) = 0$ and, by induction, $p_n(2013) = 0$ for all $n \\ge 0$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21754,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ be a positive integer. For all positive integers $n$, define\n\n$$\na_n = 1 + a + a^2 + \\dots + a^{n-1}.\n$$\n\nLet $s$ and $t$ be two different positive integers such that: if $p$ is a prime divisor of $s-t$, then $p$ also divides $a-1$. Prove that\n\n$$\n\\frac{a_s - a_t}{s - t}\n$$\n\nis an integer.",
"options": [],
"answer": "See solution",
"solution": "Without loss of generality, assume $s > t$. Then\n\n$$\na_s - a_t = a^t + a^{t+1} + \\dots + a^{s-1} = a^t a_{s-t}.\n$$\n\nTo show $s-t \\mid a_s - a_t$, it suffices to show $s-t \\mid a_{s-t}$. Let $s-t = m$ and $m = p^k r$, where $p$ is a prime and $r$ is a positive integer. Set $b = a^r$. Then,\n\n$$\n\\begin{aligned}\n1 + a + \\dots + a^{m-1} &= \\frac{a^m - 1}{a-1} = \\frac{a^r - 1}{a-1} \\cdot \\frac{a^m - 1}{a^r - 1} \\\\\n&= \\frac{a^r - 1}{a-1} \\cdot \\frac{a^{p^k r} - 1}{a^r - 1} = \\frac{a^r - 1}{a-1} \\cdot \\frac{b^{p^k} - 1}{b-1} \\\\\n&= \\frac{a^r - 1}{a-1} \\prod_{i=1}^{k} \\frac{b^{p^i} - 1}{b^{p^{i-1}} - 1} \\\\\n&= \\frac{a^r - 1}{a-1} \\prod_{i=1}^{k} \\left( 1 + b^{p^{i-1}} + b^{2p^{i-1}} + \\dots + b^{(p-1)p^{i-1}} \\right).\n\\end{aligned}\n$$\n\nEach term in the product is divisible by $p$ because $b = a^r \\equiv 1 \\pmod{p}$, so\n\n$$\n1 + b^{p^{i-1}} + b^{2p^{i-1}} + \\dots + b^{(p-1)p^{i-1}} \\equiv p \\equiv 0 \\pmod{p}.\n$$\n\nThus, $a_{s-t}$ is divisible by $p^k$, and so $\\frac{a_s - a_t}{s-t}$ is an integer. $\\square$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21755,
"subject": "Mathematics (Olympiad)",
"question": "Given the equality\n\n$$\n6 = \\frac{a+b}{a-b} + \\frac{a-b}{a+b} = \\frac{2a^2 + 2b^2}{a^2 - b^2}\n$$\n\nfind the value of\n\n$$\n\\frac{a^3 + b^3}{a^3 - b^3} + \\frac{a^3 - b^3}{a^3 + b^3}.\n$$",
"options": [],
"answer": "See solution",
"solution": "The given equality\n\n$$\n6 = \\frac{a+b}{a-b} + \\frac{a-b}{a+b} = \\frac{2a^2 + 2b^2}{a^2 - b^2}\n$$\n\nimplies $6a^2 - 6b^2 = 2a^2 + 2b^2$, or $4a^2 = 8b^2$. Thus, $a = \\pm b\\sqrt{2}$.\n\nNow,\n\n$$\n\\frac{a^3 + b^3}{a^3 - b^3} + \\frac{a^3 - b^3}{a^3 + b^3} = \\frac{2a^6 + 2b^6}{a^6 - b^6}\n$$\n\nSubstituting $a = b\\sqrt{2}$:\n\n$$\n\\frac{2(b\\sqrt{2})^6 + 2b^6}{(b\\sqrt{2})^6 - b^6} = \\frac{2 \\cdot 8b^6 + 2b^6}{8b^6 - b^6} = \\frac{16b^6 + 2b^6}{7b^6} = \\frac{18}{7}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21756,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\mathbb{R}^+$ be the set of all positive real numbers. Find all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$ satisfying\n\n$$\n(1 + y f(x))(1 - y f(x + y)) = 1\n$$\n\nfor any $x, y \\in \\mathbb{R}^+$.",
"options": [],
"answer": "See solution",
"solution": "We start by manipulating the given equation:\n\n$$\n(1 + y f(x))(1 - y f(x + y)) = 1\n$$\n\nExpanding and simplifying:\n\n$$\n1 + y f(x) - y f(x + y) - y^2 f(x) f(x + y) = 1\n$$\n$$\ny f(x) - y f(x + y) - y^2 f(x) f(x + y) = 0\n$$\n$$\ny (f(x) - f(x + y)) = y^2 f(x) f(x + y)\n$$\n$$\nf(x) - f(x + y) = y f(x) f(x + y)\n$$\n\nRearranging:\n\n$$\nf(x) - f(x + y) = y f(x) f(x + y)\n$$\n$$\nf(x) = f(x + y) + y f(x) f(x + y)\n$$\n$$\nf(x) = f(x + y) (1 + y f(x))\n$$\n$$\nf(x + y) = \\frac{f(x)}{1 + y f(x)}\n$$\n\nTaking reciprocals:\n\n$$\n\\frac{1}{f(x + y)} = \\frac{1 + y f(x)}{f(x)} = \\frac{1}{f(x)} + y\n$$\n\nSo,\n\n$$\n\\frac{1}{f(x + y)} = \\frac{1}{f(x)} + y\n$$\n\nLet $g(x) = \\frac{1}{f(x)}$. Then $g(x + y) = g(x) + y$ for all $x, y > 0$, so $g(x)$ is linear:\n\n$$\ng(x) = x + c\n$$\nfor some constant $c \\in \\mathbb{R}$.\n\nThus,\n\n$$\nf(x) = \\frac{1}{x + c}\n$$\n\nSince $f(x) > 0$ for all $x > 0$, we require $x + c > 0$ for all $x > 0$, so $c \\ge 0$.\n\nTherefore, all solutions are:\n\n$$\nf(x) = \\frac{1}{x + c}, \\quad c \\ge 0\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21757,
"subject": "Mathematics (Olympiad)",
"question": "(a) How many sunny numbers are there such that twice the number is again a sunny number?\n\n(b) Prove that every sunny number greater than 2000 is divisible by a three-digit number with a 9 in the middle.",
"options": [],
"answer": "See solution",
"solution": "(a) First, we look at the last two digits of a sunny number. There are nine possibilities: 01, 12, 23, 34, 45, 56, 67, 78, and 89. Doubling these gives, respectively, the last two digits: 02, 24, 46, 68, 90, 12, 34, 56, and 78. Thus, twice a number can only be sunny if the original sunny number ends in 56, 67, 78, or 89. In all four cases, doubling causes a carryover to the hundreds.\n\nNow, consider the first two digits of a sunny number: 10, 21, 32, 43, 54, 65, 76, 87, and 98. If the first digit is 5 or higher, twice the number has more than four digits, so it cannot be sunny. The possibilities 10, 21, 32, and 43 remain. After doubling and adding the carried-over 1 to the hundreds, we get 21, 43, 65, and 87, respectively. Thus, twice a sunny number is sunny if the first two digits are 10, 21, 32, or 43, and the last two digits are 56, 67, 78, or 89. In total, there are $4 \\times 4 = 16$ such numbers. $\\square$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21758,
"subject": "Mathematics (Olympiad)",
"question": "Given the following moves on pairs $(x, y)$:\n\n- $(c') = (a) \\circ (c) \\circ (a)$: If $x \\ge 4$, sends $(x, y)$ to $(x - 4, y + 4)$.\n- $(d') = (a) \\circ (d) \\circ (a)$: If $x \\ge 5$, sends $(x, y)$ to $(x - 3, y + 5)$.\n- $(e)$: Sends $(x, y)$ to $(x - 1, y - 1)$ if $x$ and $y$ are both positive and one of them is at least 4:\n $$\n (e) = \\begin{cases} (d') \\circ (c) & \\text{if } x \\ge 1,\\ y \\ge 4 \\\\ (d) \\circ (c') & \\text{if } x \\ge 4,\\ y \\ge 1 \\end{cases}\n $$\n- $(f) = (c) \\circ (e)^4$: If $x \\ge 4$ and $y \\ge 8$, sends $(x, y)$ to $(x, y - 8)$.\n- $(f') = (c') \\circ (e)^4$: If $x \\ge 8$ and $y \\ge 4$, sends $(x, y)$ to $(x - 8, y)$.\n\nProve that for any $n \\ge 0$, the pair $(2024, 8n)$ can be reached from $(1, 1)$ after a finite number of moves.",
"options": [],
"answer": "See solution",
"solution": "To reach $(2024, 8n)$ from $(1, 1)$:\n\n1. Apply move $(b)$ repeatedly to reach $(3^k, 3^k)$ for some $k$ with $3^k \\ge a$.\n2. Use move $(e)$ repeatedly to decrease both coordinates, reaching any $(a, a)$ with $a \\ge 3$.\n3. For $n \\ge 1$, reach $(8n, 8n)$, then:\n - If $8n \\ge 2024$ (i.e., $n \\ge 253$), apply $(f')$ repeatedly to reach $(2024, 8n)$.\n - If $n < 253$, start from $(2024, 2024)$ and apply $(f)$ repeatedly to reach $(2024, 8n)$.\n\nThus, for any $n \\ge 0$, $(2024, 8n)$ is reachable from $(1, 1)$ in finitely many moves.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21759,
"subject": "Mathematics (Olympiad)",
"question": "設 $f(x)$ 為實係數多項式，滿足：\n\n$$\nx + 1 \\leq f(x) \\leq 3x^2 - 5x + 4\n$$\n\n對所有實數 $x$ 均成立。試求 $f(11)$ 的所有可能值。",
"options": [],
"answer": "See solution",
"solution": "$f(11) \\in [12, 312]$\n\n易知 $f(x)$ 至多為二次多項式。注意到 $y = x + 1$ 為二次函數 $y = 3x^2 - 5x + 4$ 在點 $(1, 2)$ 處的切線。由於 $3x^2 - 5x + 4 = 3(x - 1)^2 + x + 1$，所以存在實數 $a \\in [0, 3]$ 使得\n\n$$\nf(x) = a(x - 1)^2 + x + 1\n$$\n\n故 $f(11)$ 的所有可能值為 $[11 + 1, 3(11 - 1)^2 + 11 + 1] = [12, 312]$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21760,
"subject": "Mathematics (Olympiad)",
"question": "Prove that the following inequality is true for any triangle:\n\n$$\n2R + (3\\sqrt{3} - 4)r \\ge p,\n$$\n\nwhere $R$ and $r$ are the radii of the circumcircle and incircle of the triangle, respectively, and $p$ is the semiperimeter of the triangle.",
"options": [],
"answer": "See solution",
"solution": "Let's use the *u*, *v*, *w* method. For a triangle with $x = p - a$, $y = p - b$, $z = p - c$:\n\n$$\nR = \\frac{abc}{4S} = \\frac{(x+y)(y+z)(z+x)}{4\\sqrt{xyz}(x+y+z)}, \\quad r = \\frac{S}{p} = \\sqrt{\\frac{xyz}{x+y+z}}.\n$$\n\nWe need to prove the equivalent inequality:\n\n$$\n(x+y)(y+z)(z+x) + 2(3\\sqrt{3}-4)xyz \\geq 2(x+y+z)\\sqrt{xyz}(x+y+z).\n$$\n\nLet:\n$$\n\\begin{cases}\nx + y + z = 3u, \\\\\nxy + yz + zx = 3v^2, \\\\\nxyz = w^3.\n\\end{cases}\n$$\n\nThen the inequality becomes:\n$$\n9uv^2 + (6\\sqrt{3}-9)w^3 \\ge 6\\sqrt{3}\\sqrt{uw^3}u.\n$$\n\nOr, equivalently:\n$$\n(3uv^2 + (2\\sqrt{3}-3)w^3)^2 \\ge 12u^3w^3.\n$$\n\nLet $f(w^3) = (2\\sqrt{3}-3)^2w^6 + w^3(6(2\\sqrt{3}-3)uv^2 - 12u^3) + 9u^2v^4$. If $u, v^2$ are fixed, $f$ is convex, so its minimum occurs when two of $\\{x, y, z\\}$ coincide. Without loss of generality, let $y = z$.\n\nThe inequality reduces to:\n\n$$\n(x + y)((2x + y) - xy) + 2(3\\sqrt{3} - 4)xy \\geq 2\\sqrt{x(x + 2y)^3}.\n$$\n\nLet $t = \\frac{x}{y}$, then:\n\n$$\n(t+2)(t+1) + 2(3\\sqrt{3}-4)t \\ge 2\\sqrt{t(t+2)^3}.\n$$\n\nThis is equivalent to:\n\n$$\nt^2 + 3(3\\sqrt{3}-2)t + 1 \\ge \\sqrt{t(t+2)^3},\n$$\n\nwhich further reduces to:\n\n$$\nt^3(6\\sqrt{3}-10) + t^2(21-12\\sqrt{3}) + t(6\\sqrt{3}-12) + 1 \\ge 0,\n$$\n\nor\n\n$$\n(t-1)^2((6\\sqrt{3}-10)t+1) \\ge 0.\n$$\n\nThus, the inequality holds, with equality when $x = y = z$ (i.e., $a = b = c$).",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21761,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, b_1, a_2, b_2, \\ldots, a_{2016}, b_{2016}$ be sequences of real numbers such that for all $n \\geq 1$,\n\n- $a_{n+1} + b_{n+1} = \\frac{a_n + b_n}{2}$,\n- $a_{n+1} b_{n+1} = \\sqrt{a_n b_n}$,\n\nand $b_{2016} = 1$. What is the value of $a_1$?",
"options": [],
"answer": "See solution",
"solution": "Let $s_n = a_n + b_n$ and $p_n = a_n b_n$ for all $n \\ge 1$. The relations become $s_{n+1} = \\frac{s_n}{2}$ and $p_{n+1} = \\sqrt{p_n}$. Inductively, $s_n = \\frac{s_1}{2^{n-1}}$ and $p_n = \\sqrt[2^{n-1}]{p_1}$.\n\nSince $a_n$ and $b_n$ are real roots of $x^2 - s_n x + p_n = 0$, we have $s_n^2 - 4p_n \\ge 0$ for any $n$. This implies\n\n$$\n\\frac{s_1^2}{2^{2n-2}} - 4 \\sqrt[2^{n-1}]{p_1} \\ge 0.\n$$\n\nIf $p_1 > 0$, then as $n \\to \\infty$, the left side approaches $0 - 4 = -4 < 0$, a contradiction. Thus, $p_1 = 0$, so $p_n = 0$ for all $n$, and $a_n = 0$ or $b_n = 0$.\n\nGiven $b_{2016} = 1$, we need $a_{2016} = 0$. Working backwards, $s_{2015} = 2s_{2016} = 2$ and $p_{2015} = 0$, so $\\{a_{2015}, b_{2015}\\} = \\{0, 2\\}$. By induction, $\\{a_n, b_n\\} = \\{0, 2^{2016-n}\\}$ for $1 \\le n \\le 2016$. In particular, $a_1 = 2^{2015}$ since $a_1 > 0$.\n\nThus, $a_1 = 2^{2015}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21762,
"subject": "Mathematics (Olympiad)",
"question": "Consider the numbers $1, 2, \\dots, n$. Find, in terms of $n$, the largest $t$ such that these numbers can be arranged in a row so that all consecutive terms differ by at least $t$.",
"options": [],
"answer": "See solution",
"solution": "First, consider the case when $n$ is even, $n = 2m$. Then the sequence\n\n$$\nm,\\ 2m,\\ m-1,\\ 2m-1,\\ \\dots,\\ 1,\\ m+1\n$$\n\nhas every consecutive term differing by $m$ or $m+1$. So we can take $t = m = \\frac{n}{2}$. We can't have $t > m$ because the sequence would have to contain $m$, which is distance at most $m$ from all the other terms.\n\nNow consider the case where $n$ is odd; $n = 2m + 1$. We can take the sequence\n\n$$\n2m+1,\\ m+1,\\ 1,\\ m+2,\\ 2,\\ m+3,\\ \\dots,\\ 2m,\\ m\n$$\n\nwhich has every term differing by $m$ or $m+1$. So we can take $t = m = \\frac{n-1}{2}$. We can't have $t > m$ because $m+1$ differs by at most $m$ from each other term in the sequence.\n\nThe best value of $t$ is therefore\n\n$$\nt = \\begin{cases} \\frac{n}{2} & \\text{if } n \\text{ is even} \\\\ \\frac{n-1}{2} & \\text{if } n \\text{ is odd.} \\end{cases}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21763,
"subject": "Mathematics (Olympiad)",
"question": "An ant starts from the midpoint of a certain edge and crawls on the surface of $P$ along a closed path $L$ composed of the edges of $P$. It passes through each point on $L$ exactly once and finally returns to the starting point.\n\nIt is known that $L$ divides the surface of $P$ into two regions and that for any $k$, the number of faces of $k$-sided polygon in both regions is equal.\n\nProve that during the above process of crawling, the number of times the ant turns to the left at vertex $P$ is the same as the number of times it turns to the right.\n\n",
"options": [],
"answer": "See solution",
"solution": "We consider the graph $G$ consisting of $L$ and all the vertices and edges on its left side. It is a planar graph.\n\nAccording to Euler's formula for planar graphs:\n\n$$\nV - E + F = 1,\n$$\n\nwhere $V$, $E$, and $F$ denote the number of vertices, edges, and faces (excluding the infinite face), respectively.\n\nIn graph $G$, note:\n\n1. Except for some vertices on $L$ whose degree is 2, all other vertices in this graph have degree 3.\n2. The number of vertices with degree 2, say $a$, is exactly the number of left turns along $L$.\n3. The number of right turns along $L$ is exactly the number of vertices with degree 3 on $L$ in $G$, denoted as $b$.\n\nNow consider graph $G'$ consisting of $L$ and all the vertices and edges on the other side. It is also a planar graph. We have:\n\n$$\nV' - E' + F' = 1,\n$$\n\nwhere $V'$, $E'$, and $F'$ denote the number of vertices, edges, and faces, respectively.\n\nSuppose that on convex polyhedron $P$, the $k$-sided polygon has $N_k$ faces, $k = 3, 4, \\dots$. By the condition, the number of faces of $k$-sided polygon on each side of $L$ is $\\frac{N_k}{2}$. Let the number of vertices on $L$ be $\\ell$. Then:\n\n$$\nF' - F = \\frac{N_3}{2} + \\frac{N_4}{2} + \\dots\n$$\n\n$$\n2E' - \\ell = 2E - \\ell = \\frac{N_3}{2} \\times 3 + \\frac{N_4}{2} \\times 4 + \\dots\n$$\n\n1. Except for some vertices on $L$ whose degree is 2, all other vertices in $G'$ have degree 3.\n2. The number of vertices with degree 2, say $a'$, is exactly the number of right turns, namely $b$, along $L$.\n3. The number of left turns along $L$ is exactly the number of vertices with degree 3 on $L$ in $G'$, denoted as $b'$.\n\nUsing the relationship between the number of edges and degrees in the graph:\n\n$$\n2E = 2a + 3(V - a) = 3V - a\n$$\n\n$$\n2E' = 2a' + 3(V' - a') = 3V' - a' = 3V - b\n$$\n\nCombining the above, we get $E = E'$. Therefore, $a = b$ and the proposition holds. $\\Box$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21764,
"subject": "Mathematics (Olympiad)",
"question": "Prove that there exist two positive real numbers $C$ and $\\alpha > \\frac{1}{2}$ such that, for any positive integer $n$, there is a subset $A$ of $\\{1, 2, \\dots, n\\}$ with $|A| \\ge C n^\\alpha$, such that the difference between any pair of different numbers in $A$ is not a perfect square.",
"options": [],
"answer": "See solution",
"solution": "For $n \\ge 25$, write $5^{2t} \\le n < 5^{2t+2}$ for $t \\in \\mathbb{N}^*$. Take the set\n\n$$\nA = \\{ (\\alpha_{2t}, \\dots, \\alpha_1)_5 \\mid \\alpha_{2i} \\in \\{0, 1, 2, 3, 4\\},\\ \\alpha_{2i-1} \\in \\{1, 3\\}\\ \\text{for}\\ i = 1, \\dots, t \\},\n$$\n\nwhere $(\\alpha_{2t}, \\dots, \\alpha_1)_5$ denotes the number $m = 5^{2t-1}\\alpha_{2t} + \\dots + 5\\alpha_2 + \\alpha_1$ in base 5. Clearly, $A \\subset \\{1, 2, \\dots, n\\}$.\n\nFor each pair $u_1, u_2 \\in A$, write $u_1 = (a_{2t}, \\dots, a_1)$ and $u_2 = (b_{2t}, \\dots, b_1)$. Assume $u_1 > u_2$ and consider $u_1 - u_2$. Let $s$ be the minimal index such that $a_s \\ne b_s$. That is, $a_1 = b_1, \\dots, a_{s-1} = b_{s-1}$, $a_s \\ne b_s$. Then $u_1 - u_2 = (a_{2t} - b_{2t})5^{2t-1} + \\dots + (a_s - b_s)5^{s-1}$.\n\nIf $2 \\mid s$, then $5^{s-1} \\mid (u_1 - u_2)$ and $a_s - b_s \\ne 0$ with $-4 \\le a_s - b_s \\le 4$. Thus, $u_1 - u_2$ cannot be a perfect square.\n\nIf $2 \\nmid s$, then $\\frac{u_1 - u_2}{5^{s-1}} = (a_{2t} - b_{2t})5^{2t-s} + \\dots + (a_s - b_s)$ is an integer. If $u_1 - u_2$ is a perfect square, then so is $\\frac{u_1 - u_2}{5^{s-1}}$. But $2 \\mid s-1$ and $a_s \\ne b_s$ imply $\\{a_s, b_s\\} = \\{1, 3\\}$ and $\\frac{u_1 - u_2}{5^{s-1}} \\equiv 2, 3 \\pmod{5}$, which is never a perfect square. Contradiction.\n\nThus, for any distinct $u_1, u_2 \\in A$, $|u_1 - u_2|$ is not a perfect square. Also, $|A| = 10^t$. Since $\\alpha = \\log_{25} 10 > 1/2$, we have\n\n$$\nn^{\\alpha} < 5^{(2t+2)\\log_{25} 10} = 10^{t+1} = 10|A|.\n$$\n\nSo it suffices to take $C = \\frac{1}{24}$ and $\\alpha = \\log_{25} 10 \\in (0, 1)$. For $n \\le 24$, take $A = \\{1\\}$, so $|A| \\ge \\frac{1}{24} n \\ge \\frac{1}{24} n^\\alpha$.\n\nTo sum up, for all $n \\in \\mathbb{N}^*$, one can find a desired subset $A$ with $|A| \\ge C n^\\alpha$.\n\n**Remark.** One may give a similar proof by considering base 16 expressions, e.g.,\n\n$$\nA = \\{ (\\alpha_t, \\dots, \\alpha_1)_{16} \\mid \\alpha_i \\in \\{2, 5, 7, 13, 15\\}\\ \\text{for}\\ i = 1, \\dots, t \\}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21765,
"subject": "Mathematics (Olympiad)",
"question": "Sea $ABC$ un triángulo de incentro $I$, $M$ un punto sobre el lado $AB$ y $N$ un punto sobre el lado $AC$, y llamamos $x = \\frac{BM}{MA}$, $y = \\frac{CN}{NA}$. Sean $a$, $b$ y $c$ las longitudes de los lados de $ABC$. Demostrar que la recta $MN$ pasa por $I$ si y solamente si $bx + cy = a$.",
"options": [],
"answer": "See solution",
"solution": "Con las notaciones del problema tenemos\n$$\nAM = \\frac{c}{1+x}, \\quad AN = \\frac{b}{1+y}\n$$\nLa condición de que $M$, $I$, $N$ sean colineales es equivalente a la igualdad de áreas siguiente:\n$$\n[AMN] = [AMI] + [AIN]\n$$\nEsta igualdad se escribe sucesivamente como\n$$\n\\frac{1}{2} \\cdot AM \\cdot AI \\cdot \\sin\\left(\\frac{A}{2}\\right) + \\frac{1}{2} \\cdot AN \\cdot AI \\cdot \\sin\\left(\\frac{A}{2}\\right) = \\frac{1}{2} \\cdot AM \\cdot AN \\cdot \\sin A\n$$\nque puede ponerse como\n$$\nAI \\cdot \\sin\\left(\\frac{A}{2}\\right) \\cdot \\left(\\frac{c}{1+x} + \\frac{b}{1+y}\\right) = \\frac{bc}{(1+x)(1+y)} \\cdot \\sin A\n$$\nLlamando $S$ al área del triángulo,\n$$\nbc \\cdot \\sin A = 2S, \\quad AI \\cdot \\sin\\left(\\frac{A}{2}\\right) = r = \\frac{S}{p}\n$$\nllegamos a que la igualdad inicial de áreas es equivalente a\n$$\n\\frac{S}{p} \\left( \\frac{c}{1+x} + \\frac{b}{1+y} \\right) = \\frac{2S}{(1+x)(1+y)}\n$$\npor lo que\n$$\n\\frac{c}{1+x} + \\frac{b}{1+y} = \\frac{2p}{(1+x)(1+y)} \\Leftrightarrow c(1+y) + b(1+x) = a + b + c\n$$\nes decir,\n$$\nbx + cy = a\n$$\ncomo queríamos demostrar.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21766,
"subject": "Mathematics (Olympiad)",
"question": "If $a_1, a_2, \\dots, a_n$ are $n$ nonzero complex numbers (not necessarily distinct), and $k, l$ are distinct positive integers such that $a_1^k, a_2^k, \\dots, a_n^k$ and $a_1^l, a_2^l, \\dots, a_n^l$ are two identical collections of numbers, prove that each $a_j$ ($1 \\leq j \\leq n$) is a root of unity.",
"options": [],
"answer": "See solution",
"solution": "The hypothesis implies there is a bijection $f : \\{1, 2, \\dots, n\\} \\to \\{1, 2, \\dots, n\\}$ such that $f(j) = m$ means $a_j^k = a_m^l$. Consider the sequence\n\n$$\n1, f(1), f^{(2)}(1), \\dots\n$$\n\nSince $f$ is a bijection on a finite set, there exist positive integers $r, s$ with $0 \\leq r < s \\leq n-1$ such that $f^{(r)}(1) = f^{(s)}(1)$. Thus, $f^{(t)}(1) = 1$ for $t = s - r$. We have\n\n$$\na_1^k = a_{f(1)}^l, \\quad a_{f(1)}^k = a_{f^{(2)}(1)}^l.\n$$\n\nSo $a_1^{k^2} = a_{f^{(2)}(1)}^{l^2}$. By induction,\n\n$$\na_1^{k^t} = a_{f^{(t)}(1)}^{l^t} = a_1^{l^t}.\n$$\n\nSince $l \\neq k$, it follows that $a_1$ is a root of unity. The same argument applies to any $a_j$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21767,
"subject": "Mathematics (Olympiad)",
"question": "Given triangle $ABC$ with $M$ as the midpoint of $AB$. Given $\\angle ABC = 30^\\circ$ and $\\angle BCM = 105^\\circ$. Prove that $CM \\cdot AC = BM \\cdot BC$.",
"options": [],
"answer": "See solution",
"solution": "Let $AA_1$ be the altitude from $A$ in $\\triangle ABC$. Since the triangle is obtuse, $AA_1$ lies on the extension of $BC$. The triangle $\\triangle AA_1B$ is right-angled with angle $30^\\circ$, so $AA_1 = AM = A_1M = BM = x$. Thus, $\\angle CMB = 45^\\circ$, $\\angle A_1MA = 60^\\circ$, $\\angle A_1MC = 75^\\circ$, $\\angle MCA_1 = 75^\\circ$. Therefore, $\\triangle MCA_1$ is isosceles, so $CA_1 = x$, and $\\triangle AA_1C$ is isosceles and right-angled, so $\\angle ACA_1 = 45^\\circ$ and $\\angle ACM = 30^\\circ$.\n\nWe calculate the area of $\\triangle ACM$ in two ways:\n\n$$S_{ACM} = \\frac{AC \\cdot CM}{4}$$\nbecause $\\angle ACM = 30^\\circ$, so the height to $AC$ is $\\frac{MC}{2}$.\n\nAlso, since $M$ is the midpoint of $AB$, $S_{ACM} = \\frac{S_{ABC}}{2} = \\frac{BC \\cdot AA_1}{4} = \\frac{BC \\cdot BM}{4}$.\n\nEquating the two expressions gives the desired result:\n\n$$CM \\cdot AC = BM \\cdot BC.$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21768,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ and $b$ be real numbers such that $a + b = 1$. Prove the following inequality:\n\n$$\n\\sqrt{1 + 5a^2} + 5\\sqrt{2 + b^2} \\geq 9\n$$",
"options": [],
"answer": "See solution",
"solution": "By the Cauchy-Schwarz inequality, we have\n\n$$\n3\\sqrt{1 + 5a^2} = \\sqrt{2^2 + (\\sqrt{5})^2} \\cdot \\sqrt{1 + 5a^2} \\geq 2 + 5a\n$$\n\nand\n\n$$\n3\\sqrt{2 + b^2} = \\sqrt{(2\\sqrt{2})^2 + 1} \\cdot \\sqrt{2 + b^2} \\geq 4 + b.\n$$\n\nHence,\n\n$$\n\\sqrt{1 + 5a^2} + 5\\sqrt{2 + b^2} \\geq \\frac{2 + 5a}{3} + \\frac{5(4 + b)}{3} = 9.\n$$\n\nEquality holds for $a = b = \\frac{1}{2}$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21769,
"subject": "Mathematics (Olympiad)",
"question": "Let $H$ be the orthocenter of an acute triangle $ABC$. The circumcircle of $\\triangle BCH$ intersects $AB$ and $AC$ again at points $A_1$ and $A_2$ respectively. Define points $B_1$, $B_2$, $C_1$, and $C_2$ analogously. Prove that the circumcenter of the triangle formed by lines $A_1A_2$, $B_1B_2$, and $C_1C_2$ is on the Euler line with respect to $\\triangle ABC$.",
"options": [],
"answer": "See solution",
"solution": "Let $D$, $E$, and $F$ be the feet of the altitudes from $A$, $B$, and $C$, respectively, with respect to $\\triangle ABC$. Let $B_1B_2$ intersect $C_1C_2$ at $K$, $C_1C_2$ intersect $A_1A_2$ at $L$, and $A_1A_2$ intersect $B_1B_2$ at $M$. Let $O$ be the circumcenter of $\\triangle ABC$.\n\n\n\nWe have\n\n$$\n\\begin{align*}\n\\angle(KL, DE) &= \\angle(KL, BC) + \\angle(BC, DE) \\\\\n&= \\angle(C_1C_2, C_2B) + \\angle(BD, DE) \\\\\n&= \\angle(C_1A, AB) + \\angle(BA, AE) \\\\\n&= \\angle(CA, AB) + \\angle(AB, CA) \\\\\n&= 0\n\\end{align*}\n$$\n\nThus, $KL \\parallel DE$, so $LM \\parallel EF$ and $MK \\parallel FD$ analogously.\n\nTherefore, there is a homothety sending $\\triangle DEF$ to $\\triangle KLM$. Call its center $X$.\n\nConsider\n\n$$\n\\begin{align*}\n\\angle(FD, DH) &= \\angle(FB, BH) \\\\\n&= \\angle(FB, BE) \\\\\n&= \\angle(FC, CE) \\\\\n&= \\angle(HC, CE) \\\\\n&= \\angle(HD, DE)\n\\end{align*}\n$$\n\nThus, $HD$ is an angle bisector of $\\angle FDE$. Similarly, we can show that $HE$ and $HF$ are angle bisectors of $\\angle DEF$ and $\\angle EFD$, respectively. Hence, $H$ is either the incenter or an excenter of $\\triangle DEF$. Since $\\triangle ABC$ is acute, we observe that $H$ is inside $\\triangle DEF$, and thus $H$ must be the incenter of $\\triangle DEF$. We have\n\n$$\n\\begin{align*}\n\\angle(HC_2, C_2C) &= \\angle(HC_2, C_2B) \\\\\n&= \\angle(HA, AB) \\\\\n&= \\angle(DA, AF) \\\\\n&= \\angle(DC, CF) \\\\\n&= \\angle(C_2C, CH)\n\\end{align*}\n$$\n\nThus, $\\overrightarrow{C_2D} = \\overrightarrow{DC}$, so $\\overrightarrow{B_1D} = \\overrightarrow{DB}$, $\\overrightarrow{A_2E} = \\overrightarrow{EA}$, $\\overrightarrow{C_1E} = \\overrightarrow{EC}$, $\\overrightarrow{B_2F} = \\overrightarrow{FB}$, and $\\overrightarrow{A_1F} = \\overrightarrow{FA}$ analogously (where $\\overrightarrow{XY}$ denotes the vector $XY$).\n\nWe have\n\n$$\n\\begin{align*}\n\\angle(KC_2, C_2B_1) &= \\angle(ED, DB) \\\\\n&= \\angle(EA, AB) \\\\\n&= \\angle(CA, AF) \\\\\n&= \\angle(CD, DF) \\\\\n&= \\angle(C_2B_1, B_1K)\n\\end{align*}\n$$\n\nThus, the internal angle bisector of $\\angle MKL$ is the perpendicular bisector of $B_1C_2$. Since $B_1C_2$ is the image of $BC$ reflected about $HD$ and the perpendicular bisector of $BC$ passes through $O$, the perpendicular bisector of $B_1C_2$ passes through the reflection of $O$ about $H$, which we denote $O'$. Similarly, the internal angle bisectors of $\\angle KLM$ and $\\angle LMK$ also pass through $O'$. Therefore, $O'$ is the incenter of $\\triangle KLM$.\n\nThe incenters of $\\triangle DEF$ and $\\triangle KLM$ both are on $\\ell$, the Euler line of $\\triangle ABC$, so $X$ lies on $\\ell$. Since the circumcenter of $\\triangle DEF$ (the nine-point center of $\\triangle ABC$) lies on $\\ell$, and $X$ lies on $\\ell$, the circumcenter of $\\triangle KLM$ also lies on $\\ell$, as required.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21770,
"subject": "Mathematics (Olympiad)",
"question": "Let $x$, $y$, $z$ be real numbers such that\n\n$$\nx^2 + 4y^2 + 9z^2 + 20 = 4x + 12y + 24z.\n$$\n\nProve that $x \\in [-1, 5]$, $y \\in [0, 3]$, and $z \\in \\left[\\frac{1}{3}, \\frac{7}{3}\\right]$.",
"options": [],
"answer": "See solution",
"solution": "We rewrite the given equality as\n\n$$\n(x-2)^2 + (2y-3)^2 + (3z-4)^2 = 9.\n$$\n\nThis implies $(x-2)^2 \\leq 9$, $(2y-3)^2 \\leq 9$, and $(3z-4)^2 \\leq 9$, therefore $|x-2| \\leq 3$, $|2y-3| \\leq 3$, $|3z-4| \\leq 3$. These inequalities are equivalent to the conclusion.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21771,
"subject": "Mathematics (Olympiad)",
"question": "Find all integers $a, b$ for which there exist integers $x, y$ such that the following equation holds:\n\n$$\n8x^4 + 8y^4 = a^4 + 6a^2b^2 + b^4.\n$$",
"options": [],
"answer": "See solution",
"solution": "If $a$ and $b$ have the same parity, define $x, y$ as:\n\n$$\nx = \\frac{a+b}{2}, \\quad y = \\frac{a-b}{2}.\n$$\n\nThen $x$ and $y$ are integers, and substituting these values verifies that the equation holds.\n\nIf $a$ and $b$ do not have the same parity, the right-hand side of the equation is odd. For example, if $a$ is even and $b$ is odd, then $a^4 + 6a^2b^2$ is even and $b^4$ is odd, so the sum is odd. Therefore, the equality cannot hold.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21772,
"subject": "Mathematics (Olympiad)",
"question": "Eva, Igor, Marko, and Maruša each wrote a positive integer on a piece of paper. If we erase the last digit of Eva's number, we get Igor's number. If we erase the last digit of Igor's number, we get Marko's number. If we erase the last digit of Marko's number, we get Maruša's number. The sum of all four written numbers was 3838. Which numbers did Eva, Igor, Marko, and Maruša write?",
"options": [],
"answer": "See solution",
"solution": "Eva's number must be at least four digits so that we can erase the last digit three times in a row and still get a positive integer. Eva's number also cannot be more than four digits since the sum of all four numbers is a four-digit number. Denote Eva's number by $abcd$ where $a, b, c$, and $d$ are the digits. Then Igor's number is $abc$, Marko's number is $ab$, and Maruša's number is $a$. The sum of these four numbers is:\n\n$$\n1000a + 100(a + b) + 10(a + b + c) + (a + b + c + d) = 3838.\n$$\n\nSince $a, b, c$, and $d$ are digits, $a + b + c + d \\leq 36$. However, since the ones digit of 3838 is 8, we must have $a + b + c + d \\leq 28$. Thus, in the calculation, at most 2 carries over from ones to tens. Since $a + b + c + 2 \\leq 29$, at most 2 carries over from tens to hundreds. Since $a + b + 2 \\leq 20$ and the hundreds digit of 3838 is 3, at most 1 carries over from hundreds to thousands. It follows that $a$ equals 2 or 3. The case $a = 2$ is not possible since we would have $10 \\leq a + b = 2 + b \\leq 11$ and at least 7 would have to carry over from tens to hundreds. Thus $a = 3$.\n\nSince at most 2 carries over from tens to hundreds, $b$ must equal 3, 4, or 5. The case $b = 5$ is not possible since we would have $8 \\leq 8 + c = a + b + c \\leq 9$ and at least 4 would have to carry over from ones to tens. The case $b = 3$ is also impossible. In this case, exactly 2 would have to carry over from ones to tens, which would mean $a + b + c + 2 = 8 + c < 20$, and thus at most 1 would carry over from tens to hundreds, which yields a contradiction. Therefore $b = 4$.\n\nSince at most 2 carries over from ones to tens, $c$ equals 4, 5, or 6. The case $c = 6$ is not possible since we would have $13 + c = a + b + c + d \\leq 9$, a contradiction. If we had $c = 4$, exactly 2 would carry over from ones to tens. In this case, we would have $a + b + c + d = 11 + d$ and hence $d = 9$. But then the ones digit of the sum would equal 0, a contradiction. Thus $c = 5$ and hence $d = 6$.\n\nEva, Igor, Marko, and Maruša wrote the numbers 3456, 345, 34, and 3.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21773,
"subject": "Mathematics (Olympiad)",
"question": "Real nonzero numbers $a, b, c, d$ satisfy the conditions $a^3 + b^3 + c^3 + d^3 = 0$ and $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} \\neq 0$. Prove that $a + b + c + d \\neq 0$.",
"options": [],
"answer": "See solution",
"solution": "We will use a proof by contradiction. Suppose $a + b + c + d = 0$. Thus, $a + b = -(c + d)$. Therefore, we can obtain the following equalities:\n\n$$\n(a + b)^3 = -(c + d)^3 \\Leftrightarrow a^3 + b^3 + 3ab(a + b) = -c^3 - d^3 - 3cd(c + d) \\\\\n\\Leftrightarrow 3ab(a + b) + 3cd(c + d) = 0 \\\\\n\\Leftrightarrow -3ab(c + d) - 3cd(a + b) = 0 \\\\\n\\Leftrightarrow ab(c + d) + cd(a + b) = 0 \\\\\n\\Leftrightarrow abc + abd + acd + bcd = 0.\n$$\n\nThat contradicts the conditions since:\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} \\neq 0 \\Leftrightarrow \\frac{abc + abd + acd + bcd}{abcd} \\neq 0 \\Leftrightarrow abc + abd + acd + bcd \\neq 0.\n$$\n\nTherefore, the statement is proved.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21774,
"subject": "Mathematics (Olympiad)",
"question": "How many sextuples $(a, b, c, d, e, f)$ of positive integers are there such that $a > b > c > d > e > f$ and $a + f = b + e = c + d = 22$?",
"options": [],
"answer": "See solution",
"solution": "For each integer $n$ from $1$ to $10$, the pair $(m, n)$ with $m > n$ and $m + n = 22$ is possible. Thus, there are $10$ such pairs $(m, n)$. Finding sextuples $(a, b, c, d, e, f)$ satisfying the problem's conditions is equivalent to selecting $3$ distinct pairs from these $10$ pairs. If we order the chosen pairs as $(a, f)$, $(b, e)$, $(c, d)$ so that $f < e < d$, then $(a, b, c, d, e, f)$ satisfies $a > b > c > d > e > f$. Conversely, any such sextuple arises from a triplet of pairs from the $10$ pairs. Therefore, the answer is $\\binom{10}{3} = 120$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21775,
"subject": "Mathematics (Olympiad)",
"question": "A person wants to plant two different kinds of tree on a tabular grid of size $m \\times n$ (exactly one tree is planted on each square). A planting way is called *impressive* if the following two conditions are satisfied:\n\n1. The number of trees of each kind is equal.\n2. The difference between the number of the two kinds of tree in each column and each row is at least $\\frac{m}{2}$ and $\\frac{n}{2}$, respectively.\n\na) Find an impressive planting way when $m = n = 2016$.",
"options": [],
"answer": "See solution",
"solution": "For convenience, consider this problem on an $m \\times n$ table and write $+1$ or $-1$ in each square to represent the two kinds of trees.\n\na) For a $4 \\times 4$ table, the arrangement below satisfies the conditions:\n\n![]()\n\nIt is clear that we can merge such $4 \\times 4$ tables to get a $2016 \\times 2016$ table satisfying part (a).\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21776,
"subject": "Mathematics (Olympiad)",
"question": "Solve the system of equations:\n\n$$\n\\begin{cases}\nx + \\frac{1}{y-x} = 1 \\\\\ny + \\frac{1}{x-y} = 3\n\\end{cases}\n$$\n\nFind all real solutions $(x, y)$.",
"options": [],
"answer": "See solution",
"solution": "We add the equations to obtain $x + y = 3$. Thus, $y = 3 - x$. Substitute into the first equation:\n$$x + \\frac{1}{3 - 2x} = 1$$\nRearrange:\n$$x - 1 + \\frac{1}{3 - 2x} = 0$$\nMultiply both sides by $3 - 2x$:\n$$ (x - 1)(3 - 2x) + 1 = 0 $$\nExpand:\n$$ 3x - 2x^2 - 3 + 2x + 1 = 0 $$\n$$ -2x^2 + 5x - 2 = 0 $$\n$$ 2x^2 - 5x + 2 = 0 $$\nFactor:\n$$ 2(x - 2)(x - \\frac{1}{2}) = 0 $$\nSo $x = 2$ or $x = \\frac{1}{2}$. Then $y = 1$ or $y = \\frac{5}{2}$, respectively. Since $x - y \\neq 0$ in both cases, the solutions are $(2, 1)$ and $\\left(\\frac{1}{2}, \\frac{5}{2}\\right)$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21777,
"subject": "Mathematics (Olympiad)",
"question": "Four boys—Andrej, Bojan, Vasko, and Goce—are collecting post stamps. Andrej has as many post stamps as Bojan and Vasko have together. Goce has five times fewer post stamps than Andrej, and Bojan has four times more post stamps than Vasko. If they together have 5016 post stamps, how many post stamps does each of the four boys have?",
"options": [],
"answer": "See solution",
"solution": "Let $A$, $B$, $V$, and $G$ be the numbers of post stamps that Andrej, Bojan, Vasko, and Goce have, respectively.\n\nFrom the problem:\n- $A = B + V$\n- $A = 5G$\n- $B = 4V$\n- $A + B + V + G = 5016$\n\nSubstitute $B = 4V$ into $A = B + V$:\n$$A = 4V + V = 5V$$\n\nAlso, $A = 5G$ so $5V = 5G$ which gives $V = G$.\n\nNow, sum all:\n$$A + B + V + G = 5V + 4V + V + V = 11V = 5016$$\n$$V = \\frac{5016}{11} = 456$$\n\nSo:\n- $V = 456$\n- $G = 456$\n- $B = 4 \\times 456 = 1824$\n- $A = 5 \\times 456 = 2280$\n\n**Final answer:**\n- Andrej has 2280 post stamps\n- Bojan has 1824 post stamps\n- Vasko has 456 post stamps\n- Goce has 456 post stamps",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21778,
"subject": "Mathematics (Olympiad)",
"question": "Given four numbers, the sums of each group of three numbers are $20$, $22$, $24$, and $27$. Find the sum of all four numbers.",
"options": [],
"answer": "See solution",
"solution": "Since each of the given sums involves exactly three of the numbers, the sum of the four sums is three times the sum of the four numbers. Thus, the required sum is $$\\frac{20 + 22 + 24 + 27}{3} = \\frac{93}{3} = 31.$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21779,
"subject": "Mathematics (Olympiad)",
"question": "There is a heap of 50 fishes. Two players take turns; on each turn, a player may eat 1, 3, or 4 fishes from the heap. The player who eats the last fish wins. Who will win if both play optimally: the first or the second player?",
"options": [],
"answer": "See solution",
"solution": "It is easy to see that if either 1 or 4 fishes remain in the heap, then the player who must move wins, but if 2 fishes remain, then this player loses. So we will solve the problem moving backward. We write all numbers from 1 to 50 and mark them with \"+\" or \"-\". If $k$ fishes remain in the heap before the move of a player and he can win, then we write $+k$, otherwise we write $-k$.\n\nWe have the following table:\n\n$$\n\\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\n\\hline\n+1 & -2 & +3 & +4 & -5 & +6 & +7 & -8 & +9 & -10 & +11 & +12 & -13 & +14 & +15 & -16 & \\ldots \\\\\n\\hline\n1 & - & 1 & 4 & - & 4 & 7 & - & 1 & - & 1 & 4 & - & 4 & 7 & - & \\ldots \\\\\n\\hline\n\\end{array}\n$$\n\nThe second row of the table contains the possible winning moves (to win, the player can eat so many fishes that after his move the number of fishes remaining in the heap would be marked with \"+\"). We see that the signs in the first row of the table are repeated with period 8 (it is easy to prove by induction).\n\nSince 50 leaves remainder 2 when divided by 8, it follows that 50 and 2 are marked with the same sign, i.e., \"-\". Thus, the number 50 is a losing number for the beginning player. To win, Tom can eat so many fishes as it is written in the second row of the table.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21780,
"subject": "Mathematics (Olympiad)",
"question": "Find all rational numbers $r$ such that $r^{\\frac{1}{r-1}}$ is rational.",
"options": [],
"answer": "See solution",
"solution": "Let $r$ be any number of the form $1 + \\frac{1}{k}$, where $k$ is an integer different from $0$ and $-1$.\n\nIf $r > 1$, let $r = 1 + \\frac{m}{n}$ where $m, n \\in \\mathbb{Z}^+$ and $(m, n) = 1$. Since\n\n$$\nr^{\\frac{1}{r-1}} = \\left(\\frac{n+m}{n}\\right)^{\\frac{n}{m}}\n$$\n\nand $(n+m, n) = (m, n) = 1$, we must have $n+m = a^m$ and $n = b^m$ for some $a, b \\in \\mathbb{Z}^+$. This gives $a^m - b^m = m$. Clearly, $a \\ge b + 1$. Therefore,\n\n$$\nm = a^m - b^m = (a-b)(a^{m-1} + a^{m-2}b + \\dots + a b^{m-2} + b^{m-1}) \\ge (1)(1+1+\\dots+1) = m.\n$$\n\nEquality must hold. This means $m = 1$ (since otherwise $a = b = 1$, contradiction) and $a - b = 1$. Conversely, when $m = 1$, $r = 1 + \\frac{1}{n}$, so that $r^{\\frac{1}{r-1}} = \\left(\\frac{n+1}{n}\\right)^n \\in \\mathbb{Q}$.\n\nIf $r < 1$, let $r = 1 - \\frac{m}{n}$ where $m, n \\in \\mathbb{Z}^+$ and $(m, n) = 1$. Since\n\n$$\nr^{\\frac{1}{r-1}} = \\left(\\frac{n-m}{n}\\right)^{-\\frac{n}{m}} = \\left(\\frac{n}{n-m}\\right)^{\\frac{n}{m}}\n$$\n\nand $(n, n-m) = (n, -m) = 1$, we must have $n = a^m$ and $n-m = b^m$ for some $a, b \\in \\mathbb{Z}^+$. This gives $a^m - b^m = m$. Again, the only solution is $m = 1$ and $a-b = 1$.\n\nWhen $m = 1$, $r = 1 - \\frac{1}{n}$. Since $r > 0$, we need $n > 1$. In that case,\n\n$$\nr^{\\frac{1}{r-1}} = \\left(\\frac{n}{n-1}\\right)^n \\in \\mathbb{Q}.\n$$\n\nTherefore, $r = 1 + \\frac{1}{n}$ for $n \\ge 1$ or $r = 1 - \\frac{1}{n}$ for $n > 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21781,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\geq 2$ teams participate in an underwater polo tournament, where each pair of teams plays exactly once against each other. A team receives 2 points for a win, 1 point for a draw, and 0 points for a loss. All teams end up with distinct numbers of points, and the final standings are ordered by total points. Later, organizers discover a technical error: every match recorded as a draw actually had a winner, and every match recorded as a win actually ended in a draw. After correcting the results, all teams still have distinct numbers of points, and the standings are reordered by total points. For which values of $n$ could the correct order turn out to be the reverse of the initial order?\n\n\n\n*Fig. 4*",
"options": [],
"answer": "See solution",
"solution": "**Answer:** For odd $n$.\n\n**Solution.** Let $n = 2k + 1$ be odd. We show the described situation is possible. Consider $2k + 1$ teams $T_1, T_2, \\dots, T_{2k+1}$. Define the result of the match between $T_i$ and $T_j$ for $i < j$ as a win for $T_i$ if $j \\leq i + k$, and as a draw otherwise. Then, for $1 \\leq i \\leq k + 1$, team $T_i$ has $k$ wins and $k + 1 - i$ draws; for $k + 2 \\leq i \\leq 2k + 1$, team $T_i$ has $2k + 1 - i$ wins and $i - k - 1$ draws. In both cases, $T_i$ has $3k + 1 - i$ points, so the order is $T_1, T_2, \\dots, T_{2k+1}$. After correction, define the result of the match $T_i$ and $T_j$ for $i < j$ as a draw if $j \\leq i + k$, and as a loss for $T_i$ otherwise. After correction, $T_i$ has $k - 1 + i$ points, so the order is $T_{2k+1}, T_{2k}, \\dots, T_1$. See Fig. 4 for $n = 7$.\n\nSuppose $n = 2k$ is even. Each match adds 2 to the total points, so the total is $2k(2k-1)$. Assume team $T$ was first, and after correction became last. Before or after correction, $T$ had at least $k$ draws. If $T$ had at least $k$ draws before correction, it would have at most $3k-2$ points, so the total wouldn't exceed\n\n$$\n(3k-2) + (3k-3) + \\dots + (k-1) = 2k(3k-2) - \\frac{2k(2k-1)}{2} = 4k^2 - 3k < 2k(2k-1),\n$$\n\ncontradiction. If $T$ has at least $k$ draws after correction, it has at least $k$ points, so the total isn't less than\n\n$$\nk + (k + 1) + \\dots + (3k - 1) = 2k^2 + \\frac{2k(2k - 1)}{2} = 4k^2 - k > 2k(2k - 1),\n$$\n\nagain contradiction, finishing the proof.\n\n**Alternative solution for the odd case.** The number of points for a team can't change by more than $n-1$. If the first team had $x$ points, and the last had $y$, and now they have $x, y$ correspondingly, then\n\n$$\ny_1 \\geq x_1 + (n-1), \\quad x \\geq y + (n-1), \\quad x_1 \\geq x - (n-1), \\quad y_1 \\leq y + (n-1),\n$$\n\nso $y_1 = x = x_1 + (n-1) = y + (n-1)$, so the numbers of points are $n$ consecutive integers. Then\n\n$$\ny + (y + 1) + \\dots + (y + n - 1) = n(n - 1) \\Rightarrow ny = \\frac{n(n - 1)}{2}, \\quad y = \\frac{n - 1}{2},\n$$\n\nwhich is not an integer for even $n$. This contradiction finishes the proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21782,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\alpha_1 = 11$, $\\alpha_2 = 1111$, $\\alpha_3 = \\underbrace{111111}_{6\\text{-digits}}$, $\\dots$, $\\alpha_n = \\underbrace{111\\dots111}_{2n\\text{-digits}}$ for integer $n > 8$.\n\nLet $q_i = \\frac{\\alpha_i}{11}$ for $i = 1, 2, \\dots, n$.\n\nProve that the sum of nine successive quotients\n\n$$\nS_i = q_i + q_{i+1} + q_{i+2} + \\dots + q_{i+8}\n$$\n\nis a multiple of nine, for every $i = 1, 2, 3, \\dots, (n-8)$.",
"options": [],
"answer": "See solution",
"solution": "We shall prove that every quotient $q_i = \\frac{\\alpha_i}{11}$, $i = 1, 2, \\dots, n$ is of the form $\\underbrace{101\\dots101}_{(2i-1)\\text{-digits}}$.\n\nBy considering the decimal representation of $\\alpha_i$ we get:\n\n$$\n\\begin{align*}\n\\alpha_i &= 10^{2i-1} + 10^{2i-2} + 10^{2i-3} + \\dots + 10^2 + 10 + 1 \\\\\n&= 10^{2i-2}(10+1) + 10^{2i-4}(10+1) + \\dots + 10^2(10+1) + (10+1) \\\\\n&= 11(10^{2i-2} + 10^{2i-4} + \\dots + 10^2 + 1).\n\\end{align*}\n$$\n\nHence $q_i = 10^{2i-2} + 10^{2i-4} + \\dots + 10^2 + 1 = \\underbrace{101\\dots101}_{(2i-1)\\text{-digits}}$, which consists of $i-1$ zeros and $i$ ones in alternating succession starting with 1.\n\nThe sum of the digits of $q_i$ is $i$, of $q_{i+1}$ is $i+1$, and so on. Thus, for $S_i = q_i + q_{i+1} + \\dots + q_{i+8}$, the sum of the digits is\n\n$$\n(i) + (i+1) + (i+2) + \\dots + (i+8) = 9i + \\frac{8 \\cdot 9}{2} = 9(i+4).\n$$\n\nTherefore, $S_i$ is a multiple of nine for every $i = 1, 2, 3, \\dots, (n-8)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21783,
"subject": "Mathematics (Olympiad)",
"question": "An organization has 30 employees, 20 of whom have a brand A computer while the other 10 have a brand B computer. For security, the computers can only be connected to each other and only by cables. The cables can only connect a brand A computer to a brand B computer. Employees can communicate with each other if their computers are directly connected by a cable or by relaying messages through a series of connected computers. Initially, no computer is connected to any other. A technician arbitrarily selects one computer of each brand and installs a cable between them, provided there is not already a cable between that pair. The technician stops once every employee can communicate with every other. What is the maximum possible number of cables used?\n\n(A) 190 \n(B) 191 \n(C) 192 \n(D) 195 \n(E) 196",
"options": [],
"answer": "See solution",
"solution": "**Answer (B):**\n\nLet $P$ be one of the computers of brand A. If every brand B computer is connected to every brand A computer other than $P$, then there will be $10 \\times 19 = 190$ cables, but the employee using computer $P$ is completely isolated and cannot communicate with any other. Therefore the requested answer is greater than 190.\n\nTo see that 191 cables is the requested maximum, suppose that the technician has installed 191 cables. First note that every brand A computer must have at least one cable attached, because otherwise there would be at most $19 \\times 10 = 190$ cables. Furthermore, if each of the brand A computers is attached to at most 9 cables, this accounts for at most $20 \\times 9 = 180$ cables. Therefore at least one brand A computer has 10 cables attached—one leading to each of the brand B machines. Call this brand A computer $R$. A symmetric argument shows that at least one of the brand B computers, say $S$, is attached by 20 cables to every computer of brand A. It now follows that there is a path of length 2, going through $R$, joining any pair of brand B computers; there is a path of length 2, going through $S$, joining any pair of brand A computers; and there is a path of length at most 3, going through $R$ and $S$, joining any brand B computer to any brand A computer. Therefore 191 cables guarantee that every employee can communicate with every other, regardless of which pairs of computers are directly connected.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21784,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive real solutions of the system of equations:\n\n$$\nx + \\frac{1}{x} - w = 2, \\quad y + \\frac{1}{y} - w = 2, \\quad z + \\frac{1}{z} + w = 2, \\quad y + \\frac{1}{z} + w = 2.\n$$",
"options": [],
"answer": "See solution",
"solution": "Subtracting the second equation from the first and multiplying by $xy$ gives $x^2y + y - xy^2 - x = 0$, which implies either $x = y$ or $x = \\frac{1}{y}$.\n\nIf $x = y$, then subtracting the fourth equation from the third and multiplying by $xz$ gives either $x = z$ or $x = -\\frac{1}{z}$. If $x = z$, then subtracting the third equation from the first gives $-2w = 0$, so $w = 0$, which is not positive. If $x = -\\frac{1}{z}$, then $x$ and $z$ cannot both be positive.\n\nIf $x = \\frac{1}{y}$, then subtracting the fourth equation from the third gives $z - \\frac{1}{z} = 0$, so $z = 1$. Adding the first equation to the third gives $x + \\frac{2}{x} = 3$, or $x^2 - 3x + 2 = 0$. This has solutions $x = 1$ (which leads to $x = y = z$ already considered) and $x = 2$, which gives $y = \\frac{1}{2}$ and $w = \\frac{1}{2}$.\n\nThus, the positive real solution is:\n\n$$x = 2, \\quad y = w = \\frac{1}{2}, \\quad z = 1.$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21785,
"subject": "Mathematics (Olympiad)",
"question": "給定 $6 \\times 6$ 的方格，並記第一列六個方格座標為 $(1, 1), (1, 2), \\ldots, (1, 6)$，其餘類推。對於任意的 $k = 0, 1, \\ldots, 5$，滿足 $i - j \\equiv k \\pmod{6}$ 的六個格子 $(i, j)$ 稱為在同一條對角線上（故共有六條對角線）。試問：能否將 $1, 2, \\ldots, 36$ 填入 $6 \\times 6$ 的方格中，同時滿足：\n\n1. 每一列的和都相等。\n2. 每一行的和都相等。\n3. 每一條對角線的和都相等。\n\nThere's a $6 \\times 6$ chess board, which we label the squares in the first column by $(1, 1), (1, 2), \\ldots, (1, 6)$, and label the other squares similarly. For every $k = 0, 1, \\ldots, 5$, all squares $(i, j)$ satisfying $i-j \\equiv k \\pmod{6}$ form a diagonal; therefore, there are six diagonals. Decide whether we can write $1, 2, \\ldots, 36$ on the chess board so that all the following conditions hold:\n\n1. The sums for each column are the same.\n2. The sums of each row are the same.\n3. The sums of each diagonal are the same.",
"options": [],
"answer": "See solution",
"solution": "不可能。\n\n歸謬證法，假設可以填成功，則這個和必為\n\n$$\nS = \\frac{1}{6}(1 + 2 + \\cdots + 36) = 111.\n$$\n\n將 $6 \\times 6$ 的方格分成四類：\n\n1. $A$: 座標為 $(\\text{奇}, \\text{奇})$ 的格子。\n2. $B$: 座標為 $(\\text{奇}, \\text{偶})$ 的格子。\n3. $C$: 座標為 $(\\text{偶}, \\text{奇})$ 的格子。\n4. $D$: 座標為 $(\\text{偶}, \\text{偶})$ 的格子。\n\n則依題目假設，$A + B = 3S$，$B + D = 3S$，$A + D = 3S$，全部加起來得\n\n$$\n2(A + B + D) = 9S,\n$$\n\n但左邊為偶數而右邊為奇數，矛盾！",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21786,
"subject": "Mathematics (Olympiad)",
"question": "Let $k$ be a positive integer. Two players $A$ and $B$ play a game on an infinite grid of regular hexagons. Initially all the grid cells are empty. Then the players alternately take turns with $A$ moving first. On his move, $A$ may choose two adjacent hexagons in the grid which are empty and place a counter in both of them. On his move, $B$ may choose any counter on the board and remove it. If at any time there are $k$ consecutive grid cells in a line all of which contain a counter, $A$ wins. Find the minimum value of $k$ for which $A$ cannot win in a finite number of moves, or prove that no such minimum value exists.",
"options": [],
"answer": "See solution",
"solution": "The answer is $k = 6$.\n\nFirst, we show that $A$ cannot win for $k \\ge 6$. Color the grid in three colors so that no two adjacent spaces have the same color, and arbitrarily pick one color $C$. $B$ will play by always removing a counter from a space colored $C$ that $A$ just played. If there is no such counter, $B$ plays arbitrarily. Because $A$ cannot cover two spaces colored $C$ simultaneously, it is possible for $B$ to play in this fashion. Now note that any line of six consecutive squares contains two spaces colored $C$. For $A$ to win he must cover both, but $B$'s strategy ensures at most one space colored $C$ will have a counter at any time.\n\nNow we show that $A$ can obtain 5 counters in a row. Take a set of cells in the grid forming the shape shown below. We will have $A$ play counters only in this set of grid cells until this is no longer possible. Since $B$ only removes one counter for every two $A$ places, the number of counters in this set will increase each turn, so at some point it will be impossible for $A$ to play in this set anymore. At that point any two adjacent grid spaces in the set have at least one counter between them.\n\n\n\nConsider only the top row of cells in the set, and take the lengths of each consecutive run of cells. If there are two adjacent runs that have a combined length of at least 4, then $A$ gets 5 counters in a row by filling the space in between. Otherwise, a bit of case analysis shows that there exists a run of 1 counter which is neither the first nor last run. This single counter has an empty space on either side of it on the first row. As a result, the four spaces of the second row touching these two empty spaces all must have counters. Then $A$ can play in the 5th cell on either side of these 4 to get 5 counters in a row. So in all cases $A$ can win with $k \\le 5$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21787,
"subject": "Mathematics (Olympiad)",
"question": "Let $p$, $q$, and $r$ be distinct prime numbers such that\n$$\nrp^3 = -(p^2 + p) + 2rq^2 + (q^2 + q).\n$$\nFind the value of $pqr$.",
"options": [],
"answer": "See solution",
"solution": "By the given condition, $rp^3 = -(p^2 + p) + 2rq^2 + (q^2 + q)$. Note that $p^2 + p$, $q^2 + q$, and $2rq^2$ are even for all natural $p$ and $q$, so $rp^3$ is also even. Since $r$ and $p$ are prime, either $p = 2$ or $r = 2$.\n\nIf $p = 2$, the equation becomes $8r + 4 + 2 = 2rq^2 + q^2 + q$. But for $q > 2$, the right side exceeds the left, so this case is impossible.\n\nThus, $r = 2$, and the equation becomes $2p^3 + p^2 + p = 4q^2 + q^2 + q$, or\n$$\np(2p^2 + p + 1) = q(5q + 1). \\quad (1)\n$$\nSince $p$ and $q$ are distinct primes, they are coprime. From (1), $2p^2 + p + 1$ is divisible by $q$, and $5q + 1$ is divisible by $p$. Let $2p^2 + p + 1 = mq$ and $5q + 1 = mp$ for some $m \\in \\mathbb{N}$. Substitute $q = (mp - 1)/5$ into the first equation:\n$$\n10p^2 + (5 - m^2)p + (m + 5) = 0. \\quad (2)\n$$\nTreating (2) as a quadratic in $p$, its discriminant must be a perfect square:\n$$\nD = (m^2 - 5)^2 - 4 \\cdot 10 \\cdot (m + 5) = m^4 - 10m^2 - 40m - 175 = n^2 \\quad (3)\n$$\nFor $m = 1, 2, 3, 4$, $D < 0$, so $m \\ge 5$. Also, $(m^2 - 5)^2 > n^2 > (m^2 - 11)^2$ for $m \\ge 7$, and $n$ must have opposite parity to $m$, so $n = m^2 - 9$ or $n = m^2 - 7$.\n\nIf $n = m^2 - 9$, then $m^2 - 5m - 32 = 0$ has no integer solutions. If $n = m^2 - 7$, then $m^2 - 10m - 56 = 0$ gives $m = 14$.\n\nThus, $m = 14$, $n = 189$. From (2), $p = (191 \\pm 189)/20$, so $p = 19$ (prime). Then $q = (14 \\cdot 19 - 1)/5 = 53$ (prime).\n\nTherefore, the required product is $pqr = 19 \\cdot 53 \\cdot 2 = 2014$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21788,
"subject": "Mathematics (Olympiad)",
"question": "Let $m$ and $n$ be positive integers of different parity. Prove that\n\n$$\n\\frac{3m^2 + 5mn}{3n^2 + mn}\n$$\n\nis not a positive integer.",
"options": [],
"answer": "See solution",
"solution": "Let $d$ be the greatest common divisor of $m$ and $n$, i.e., $m = d m'$ and $n = d n'$, where $m'$ and $n'$ are relatively prime positive integers of different parity.\n\nWe need to prove that\n\n$$\n\\frac{3d^2 m'^2 + 5d^2 m' n'}{3d^2 n'^2 + d m' n'} = \\frac{m'(3m' + 5n')}{n'(m' + 3n')}\n$$\n\nis not a positive integer.\n\nNote that $3m' + 5n'$ and $m' + 3n'$ are both odd.\n\nIf $m'$ is odd and $n'$ is even, the even $n'(m' + 3n')$ cannot divide the odd $m'(3m' + 5n')$.\n\nOtherwise, let the odd $k$ be the greatest common divisor of $3m' + 5n'$ and $m' + 3n'$. Then:\n\n$$\n\\begin{aligned}\nk &\\mid 3(3m' + 5n') - 5(m' + 3n') \\implies k \\mid 4m', \\\\\nk &\\mid (3m' + 5n') - 3(m' + 3n') \\implies k \\mid -4n'.\n\\end{aligned}\n$$\n\nThus, $k$ divides both $m'$ and $n'$, so $k = 1$ and both factors in\n\n$$\n\\frac{m'}{n'} \\cdot \\frac{3m' + 5n'}{m' + 3n'}\n$$\n\nare irreducible fractions.\n\nSince $m' + 3n' > m'$, the proof is finished.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21789,
"subject": "Mathematics (Olympiad)",
"question": "The teacher drew a $3 \\times 3$ table in Juku's exercise book and wrote a number in every position of the table. Then he gave Juku the following task:\n\n1. Turn the next page and draw a similar table. Write in the first row the numbers obtained by subtracting the numbers in the third row of the corresponding column from the numbers in the second row of the corresponding column in the previous table. Similarly, the numbers in the second and third row are obtained as differences of the third and the first, and the first and the second row.\n\n2. Turn the next page and draw a new table. Write in the first column the numbers obtained by subtracting the numbers in the third column from the numbers in the second column in the corresponding row in the previous table. Similarly, the numbers in the second and third column are obtained as differences of the third and the first, and the first and the second column.\n\nRepeat in turns steps 1 and 2 until you reach a table where all the numbers are zeroes. Juku has reached the end of the third page and has not yet reached the table with all zeroes in it. Prove that his task never ends.",
"options": [],
"answer": "See solution",
"solution": "First, note that after step 1 we get a table where the column sums are $0$, and after step 2 we get a table where the row sums are $0$.\n\nSuppose that after some steps we reach the table with all zeroes. By symmetry, we can consider the case where we get this table after step 2. Then the table on the previous step was\n\n$$\n\\begin{array}{ccc}\na & a & a \\\\\nb & b & b \\\\\nc & c & c\n\\end{array}\n$$\n\nwhere at least one of the numbers $a$, $b$, $c$ is not zero. This table was obtained after step 1, hence $a + b + c = 0$.\n\nThe table on the previous step was\n\n$$\n\\begin{array}{ccc}\nd & e & f \\\\\nd-c & e-c & f-c \\\\\nd+b & g+b & f+b\n\\end{array}\n$$\n\nThis was also written by Juku, because he computed at least 2 tables. Since this table was obtained after step 2, we must have $d + e + f = d - c + e - c + f - c = d + b + g + b + f + b = 0$.\n\nFrom the first equality it follows that $c = 0$, from the equality of the first and third expression $b = 0$, and since $a + b + c = 0$, we have $a = 0$, which contradicts the assumption that at least one of the numbers $a, b, c$ is not zero.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21790,
"subject": "Mathematics (Olympiad)",
"question": "For a convex pentagon $ABCDE$, $AB = DE = EA$, $AB \\neq EA$, and $B$, $C$, $D$, $E$ are concyclic. Prove that $A$, $B$, $C$, $D$ are concyclic if and only if $AC = AD$.",
"options": [],
"answer": "See solution",
"solution": "First, if $A$, $B$, $C$, $D$ are concyclic, by $AB = DE$ and $BC = EA$ we have $\\angle BAC = \\angle EDA$, $\\angle ACB = \\angle DAE$, so $\\angle ABC = \\angle DEA$, which means that $AC = AD$.\n\n\n\nSecond, if $AC = AD$, let $O$ be the center of the circle (with $B$, $C$, $D$, $E$ on the circle). Then $O$ is on the perpendicular bisector ($AH$) of $CD$. Let $F$ be the symmetric point of $B$ with respect to the line $AH$. Then $F$ is on the circle $O$. $AB \\neq EA$, so $DE \\neq DF$, and thus $E$, $F$ are not the same points. Now one can see that $\\triangle AFD \\cong \\triangle ABC$, with $AB = DE$, $BC = EA$, and one can get $\\triangle AED \\cong \\triangle CBA$, and so $\\triangle AED \\cong \\triangle DFA$, which indicates $\\angle AED = \\angle DFA$, and therefore $A$, $E$, $F$, $D$ are concyclic. This means that $A$ is on the circle $O$, and $B$, $C$, $D$ are concyclic.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21791,
"subject": "Mathematics (Olympiad)",
"question": "Let $q$ be an odd prime and $p$ a prime divisor of $2^q - 1$. Show that there exist $p$-sequences $(x_n)$ and $(y_n)$ such that $x_0 < y_0$ but $x_n > y_n$ for all sufficiently large $n$.",
"options": [],
"answer": "See solution",
"solution": "Let $T = q$. The numbers $2^q - 1$ for distinct odd primes $q$ are pairwise coprime, so there are infinitely many such primes $p$. For $p \\nmid x$, $d_p(x) + d_p(p-x) = p$, so $S_p^+ + S_p^- = pq$ is odd, implying $S_p^+ \\neq S_p^-$. Assume $(x_n)$ and $(y_n)$ are $p$-sequences with $S_p(x_0) > S_p(y_0)$ but $x_0 < y_0$. Then:\n\n$$\nx_{M_{q+r}} - y_{M_{q+r}} = (x_r - y_r) + M(S_p(x_0) - S_p(y_0)) \\geq (x_r - y_r) + M\n$$\n\nfor all $M \\ge 0$ and $r = 0, 1, \\dots, q-1$. Thus, $x_n > y_n$ for all $n \\ge q + q \\cdot \\max\\{y_r - x_r : r = 0, 1, \\dots, q-1\\}$. Since $x_0 < y_0$, there exists $n_0$ with $x_{n_0} < y_{n_0}$. Then $a_n = x_{n-n_0}$ and $b_n = y_{n-n_0}$ have the desired property. To construct such sequences, since $S_p^+ \\neq S_p^-$, if $S_p^+ > S_p^-$, set $x_0 = 1$, $y_0 = p-1$; otherwise, set $x_0 = p-1$, $y_0 = p+1$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21792,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a convex quadrilateral with the shortest side $AB$ strictly less than the longest side $CD$. Show that there exists a point $E$ on the segment $CD$ such that for any point $P$ (different from $E$) on the segment $CD$, the length of $O_1O_2$ is constant, where $O_1$ and $O_2$ are the circumcenters of the triangles $APD$ and $BPE$, respectively.",
"options": [],
"answer": "See solution",
"solution": "Claim: The point $E$ is the intersection point of the line parallel to $AD$ through $B$ and the line $CD$.\n\nLet $E$ be the intersection point of the line parallel to $AD$ through $B$ and the line $CD$. Firstly, we will show that $E$ is on the segment $CD$. Since $AB \\leq AD$ and $BC \\leq CD$ (because $AB$ and $CD$ are the shortest and longest sides, respectively), then $\\angle ABD \\geq \\angle ADB$ and $\\angle CBD \\geq \\angle BDC$. Therefore $\\angle ABC = \\angle ABD + \\angle DBC \\geq \\angle ADB + \\angle BDC = \\angle ADC$. Similarly, we will have that $\\angle BAD \\geq \\angle BCD$. Thus, $\\angle ABC + \\angle BAD \\geq \\angle BCD + \\angle CDA$, and then $\\angle ABC + \\angle BAD \\geq 180^\\circ$. Moreover, one can show that $\\angle ABC + \\angle BAD > 180^\\circ$. (If $\\angle ABC + \\angle BAD = 180^\\circ = \\angle BCD + \\angle CDA$, we would have that $\\angle ABC = \\angle ADC$ and $\\angle BAD = \\angle BCD$, and then $ABCD$ would be a parallelogram; contradicting the fact that $AB < CD$.) Therefore, we can move the point $C$ along the line $CD$ closer to $D$ to the point $E'$, ...",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21793,
"subject": "Mathematics (Olympiad)",
"question": "А олонлог дараах нөхцөлүүдийг хангах олонлог болно. Хэрэв $x \\in A$ ($x \\ne 0,\\ x \\ne 1$) бол $\\frac{x+1}{x} \\in A$ ба $\\frac{2x-1}{x-1} \\in A$. Мөн $2 \\in A$ бол $A$ олонлогт $1$-ээс их бүх рациональ тоонууд орно гэдгийг батал.",
"options": [],
"answer": "See solution",
"solution": "$$\\exists\\ \\frac{p}{q} > 1,\\ \\frac{p}{q} \\notin A$$ гэж үзье. Тэгвэл $p > q$ ба $\\frac{p}{q} \\neq 2 \\Leftrightarrow p - 2q \\neq 0$ байна. $x \\mapsto \\frac{x+1}{x}$ үйлдлийг (1), $x \\mapsto \\frac{2x-1}{x-1}$ үйлдлийг (2) гэж тэмдэглэе.\n\n(1), (2) үйлдлээр $\\frac{p}{q}$-г гаргаж болох $x$-ийг олъё:\n\n$$\n\\frac{x+1}{x} = \\frac{p}{q} \\Leftrightarrow x = \\frac{p}{p-q}\n$$\n\n$$\n\\frac{2x-1}{x-1} = \\frac{p}{q} \\Leftrightarrow x = \\frac{p-q}{p-2q}\n$$\n\nӨөрөөр хэлбэл, $\\frac{p}{p-q} \\xrightarrow{(1)} \\frac{p}{q}$ ба $\\frac{p-q}{p-2q} \\xrightarrow{(2)} \\frac{p}{q}$.\n\n$A$-д ордоггүй $1$-ээс их рационал тоонуудаас (үл хураагдах энгийн бутархай) хүртвэр нь хамгийн бага байхаар $\\frac{p}{q}$-г сонгож болно.\n\n$\\frac{p}{q} \\notin A \\Rightarrow \\frac{q}{p-q} \\notin A$ ба $\\frac{p-q}{p-2q} \\notin A$ байна. Мөн:\n\n- $1 < \\frac{p}{q} < 2$ бол $\\frac{q}{p-q} > 1$, $0 < q < p$;\n- $\\frac{p}{q} > 2$ бол $\\frac{p-q}{p-2q} > 1$, $0 < p-q < p$.\n\nИймд $A$-д ордоггүй, хүртвэр нь $p$-ээс бага бөгөөд $1$-ээс их тоо олдож зөрчил үүснэ. Тэгэхээр $1$-ээс их дурын рационал тоо $A$ олонлогт орно.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21794,
"subject": "Mathematics (Olympiad)",
"question": "For real numbers $a, b, c$ satisfying $0 \\leq a \\leq b \\leq c$ and $a + b + c = 1$, prove that the inequality\n\n$$\nab\\sqrt{b-a} + bc\\sqrt{c-b} + ca\\sqrt{c-a} < \\frac{1}{4}$$\n\nholds.",
"options": [],
"answer": "See solution",
"solution": "For any $x \\leq y$, we have $\\sqrt{y-x} \\leq \\frac{y-x+1}{2}$. Thus, it suffices to prove that\n\n$$ab(b-a+1) + bc(c-b+1) + ca(c-a+1) = ab(2b+c) + bc(2c+a) + ca(2c+b) < \\frac{1}{2}.$$ \n\nSince $(a+b+c)^3 = a^3 + b^3 + c^3 + 3(ab^2 + bc^2 + ac^2) + 3(a^2b + b^2c + a^2c) + 6abc$, this is equivalent to proving the following:\n\n$$ab^2 + ac^2 + bc^2 < a^3 + b^3 + c^3 + 3(a^2b + a^2c + b^2c).$$\n\nHere $ab^2 \\leq b^3$ and $bc^2 + ac^2 \\leq 2bc^2 \\leq c^3 + b^2c$ implies that $ab^2 + ac^2 + bc^2 \\leq b^3 + c^3 + b^2c$. Since it is impossible to have $a^3 = b^2c = c^3 = 0$ and $a + b + c = 1$, the inequality never holds.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21795,
"subject": "Mathematics (Olympiad)",
"question": "Let $r, s \\in [1, \\infty)$ be numbers such that for every positive integers $a, b$ with $a$ dividing $b$, it holds that $[ar]$ divides $[bs]$.\n\n(a) Prove that $\\frac{s}{r}$ is a positive integer.\n\n(b) Show that $r$ and $s$ are positive integers.\n\n*Remark.* By $[x]$ we denote the floor of the real number $x$.",
"options": [],
"answer": "See solution",
"solution": "a) Suppose $\\frac{s}{r} \\notin \\mathbb{N}$. Then there exists $k \\in \\mathbb{N}$ such that $k < \\frac{s}{r} < k+1$, i.e., $kr < s < (k+1)r$. Choosing $b = a \\in \\mathbb{N}^*$ arbitrarily, we obtain $[ar] \\mid [as]$, and thus $[ar] \\mid [as] - k[ar]$. (1)\n\nFrom $s > kr$, there exists $u > 0$ such that $us > ukr + 2$, and for every $a > u$ we get $as > akr + 2 \\implies [as] \\geq [akr] + 2 > akr + 1 > k[ar]$, so $[as] > k[ar]$.\n\nFrom (1), $[as] - k[ar] \\geq [ar] \\iff [as] \\geq (k+1)[ar]$, so\n\n$$\nas > (k+1)(ar - 1) \\iff k+1 > a((k+1)r - s),\n$$\n\nfor every $a > u$.\n\nThus, $a < \\frac{k+1}{(k+1)r - s}$ for every $a > u$, which is a contradiction, so the assumption is false.\n\nb) Let us show that $s$ is a positive integer.\n\nWe will show that for every $a \\in \\mathbb{N}$ such that $ar \\geq 2$, we get $as \\in \\mathbb{N}$.\n\nIf $as \\notin \\mathbb{N}$, then there exists $n \\in \\mathbb{N}^*$ such that $\\frac{1}{n+1} \\leq \\{as\\} < \\frac{1}{n}$, so $1 \\leq (n+1)\\{as\\} < \\frac{n+1}{n} \\leq 2$, thus $[(n+1)\\{as\\}] = 1$.\n\nWe obtain\n\n$$\n[(n+1)as] = (n+1)[as] + [(n+1)\\{as\\}] = (n+1)[as] + 1.\n$$\n\nSince $[ar] \\mid [as]$ and $[ar] \\mid [(n+1)as]$, we obtain $[ar] \\mid 1 \\implies [ar] = 1$, which is a contradiction.\n\nThus $as \\in \\mathbb{N}$ for every $a \\in \\mathbb{N}$ with $ar \\geq 2$, from which we get $(a+1)s \\in \\mathbb{N}$, so $(a+1)s - as = s \\in \\mathbb{N}$.\n\nLet us prove that $r$ is a positive integer.\n\nLet $p$ be an arbitrary prime number with $p[r] > s$ and $m = [p\\{r\\}]$. Since $p\\{r\\} < p$, we get $m < p$.\n\nIf $m \\neq 0$, then $(m, p) = 1$. Since\n\n$$\n[pr] \\mid ps \\implies [p([r] + \\{r\\})] \\mid ps \\implies p[r] + m \\mid ps.\n$$\n\nSince $(p[r] + m, p) = 1 \\implies p[r] + m \\mid s$, we obtain a contradiction as $p[r] > s$.\n\nThus, $m = 0 \\implies p\\{r\\} < 1 \\implies \\{r\\} < \\frac{1}{p}$ for every prime number $p$ with $p > \\frac{s}{[r]} \\implies \\{r\\} = 0$, and thus $r \\in \\mathbb{N}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21796,
"subject": "Mathematics (Olympiad)",
"question": "a) Prove, by induction on the number $n$ of dominoes, that any chessboard polygon that can be tiled by $n$ dominoes can be tastefully tiled.\n\nb) Show that any given chessboard polygon admits at most one tasteful domino tiling.",
"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 $n = 1$. 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\nb) 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. (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 right. Hence 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 $R$. Hence 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 towards 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 contradiction completes the proof.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21797,
"subject": "Mathematics (Olympiad)",
"question": "給定一圓及圓上的四個點 $B, C, X, Y$，設 $A$ 為線段 $BC$ 的中點，$Z$ 為線段 $XY$ 的中點。過 $B, C$ 分別作垂直 $BC$ 的直線 $L_1, L_2$。設過 $X$ 且垂直 $AX$ 的直線分別交 $L_1, L_2$ 於 $X_1, X_2$ 兩點，過 $Y$ 且垂直 $AY$ 的直線分別交 $L_1, L_2$ 於 $Y_1, Y_2$ 兩點。令 $X_1Y_2$ 與 $X_2Y_1$ 相交於 $P$ 點。證明：$\\angle AZP = 90^\\circ$。",
"options": [],
"answer": "See solution",
"solution": "作 $A$ 對 $X_2Y_1$ 垂足 $D$ 點。因為 $\\angle AYY_1 = \\angle ABY_1 = \\angle AXX_2 = \\angle ACX_2 = 90^\\circ$，所以 $DX_2XAC$ 五點共圓，$DY_1YAB$ 五點亦共圓。考慮此兩圓與一開始的給定圓等三圓的根心，得 $AD, BY, CX$ 三線共點。設此共點為 $S$，且有\n\n$$\nSA \\cdot SD = SB \\cdot SY = SC \\cdot SX.\n$$\n\n考慮以 $S$ 為中心，$SA \\cdot SD$ 為反演幂的變換。此變換將 $AD, BY, CX$ 互換，且 $BAC$ 共線且 $A$ 為 $BC$ 中點，故 $DXSY$ 四點共圓且為調和四邊形。有\n\n$$\n\\angle DZX = \\angle DYX + \\angle ZDY = \\angle DSX + \\angle XDS = 180^\\circ - \\angle DXS \\\\\n\\angle SZX = \\angle SYX + \\angle ZSY = \\angle SDX + \\angle XSD = 180^\\circ - \\angle DXS.\n$$\n\n（其中用到 $DXSY$ 是調和的，所以 $DS, DZ$ 對 $\\angle XDY$ 等角共軛，$SD, SZ$ 對 $\\angle XSY$ 等角共軛。）所以 $\\angle DZX = \\angle SZX$。又因 $XYBC$ 四點共圓，所以 $\\triangle SXY$ 與 $\\triangle SBC$ 相似，從而 $\\angle SAB = \\angle SZX = \\angle DZX$。令 $XY$ 與 $BC$ 交於 $T$ 點，則 $DZAT$ 四點共圓。\n\n令一開始給定圓的圓心為 $O$ 點。由於 $OZ \\perp ZT, OA \\perp AT$，所以 $DZOAT$ 五點共圓。令過 $O$ 且平行於 $BC$ 的直線為 $L_3$，過 $Z$ 且與 $AZ$ 垂直的直線為 $L_4$。則 $A$ 對圓 $DZOAT$ 的對徑點為 $X_2Y_1, L_3, L_4$ 的共同\n\n\n\n交點，也就是 $X_2Y_1, L_3, L_4$ 共點。同理 $X_1Y_2, L_3, L_4$ 共點。所以這點即為 $P$，也就有 $\\angle AZP = 90^\\circ$。證畢。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21798,
"subject": "Mathematics (Olympiad)",
"question": "Let $x_0, x_1, \\dots, x_n$ be a non-decreasing sequence of positive integers with $x_0 = 1$. Suppose the subsequence $x_1, x_2, \\dots, x_{2017}$ contains exactly $m$ distinct positive integers. Show that\n\n$$\n\\sum_{i=2}^{n} x_i (x_i - x_{i-2}) \\ge m^2 - 2.\n$$",
"options": [],
"answer": "See solution",
"solution": "Since the sequence is non-decreasing, we have $x_i - x_{i-2} - 1 \\ge -1$ for $i \\ge 2$. If $x_i - x_{i-2} - 1 = -1$, then $x_i = x_{i-1}$. Thus, for $i = 2, 3, \\dots, n$:\n\n$$\n(x_i - x_{i-2} - 1)(x_i - x_{i-1}) \\ge 0\n$$\n\nThis is equivalent to:\n\n$$\nx_i^2 - x_i x_{i-2} + x_{i-1} x_{i-2} - x_i x_{i-1} \\geq x_i - x_{i-1}\n$$\n\nSumming for $i = 2, 3, \\dots, n$:\n\n$$\n\\sum_{i=2}^{n} x_i(x_i - x_{i-2}) \\geq x_n(1 + x_{n-1}) - 2x_1\n$$\n\nSince $\\{x_1, x_2, \\dots, x_n\\}$ includes $m$ distinct elements:\n\n$$\nx_n \\geq x_1 + m - 1\n$$\n$$\nx_{n-1} \\geq x_1 + m - 2\n$$\n\nTherefore:\n\n$$\n\\sum_{i=2}^{n} x_i(x_i - x_{i-2}) \\geq (x_1 + m - 1)^2 - 2x_1\n$$\n\nAnd:\n\n$$\n(x_1 + m - 1)^2 - 2x_1 - (m^2 - 2) = (x_1 - 1)(x_1 + 2m - 3) \\geq 0\n$$\n\nThus, the required inequality holds. For equality, we must have $x_1 = 1$, $x_n = m$, $x_{n-1} = m - 1$, and for all $i = 2, 3, \\dots, n$:\n\n$$\nx_i - x_{i-2} = 1 \\quad \\text{or} \\quad x_i = x_{i-1}\n$$\n\nSince $x_n \\neq x_{n-1}$, $x_{n-2} = m-1$. For $2 \\le j \\le n$, if $x_j - x_{j-1} > 1$, then $x_j - x_{j-2} \\ge x_j - x_{j-1} > 1$, $x_j \\ne x_{j-1}$. The difference between consecutive terms can be at most 1. Furthermore, if $x_j - x_{j-1} = x_{j-1} - x_{j-2} = 1$, then $x_j - x_{j-2} = 2 > 1$, $x_j \\ne x_{j-1}$. Thus, +1 increases cannot be in neighbouring consecutive pairs. All conditions are satisfied if +1 increases are not in neighbouring consecutive pairs. We need to count all possible places for +1 increases. The corresponding indices are between 2 and $n-2$, so there are $n-3$ possibilities. The total number of +1 increases is $m-2$. This is equivalent to placing $m-2$ identical balls into $n-3$ boxes such that no neighbouring boxes both contain a ball. The answer is $\\binom{n-m}{m-2}$ for $n \\ge 2m-2 \\ge 2$, and 0 otherwise. In our case, the answer is $\\binom{1992}{23}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21799,
"subject": "Mathematics (Olympiad)",
"question": "Long John Silverman has captured a treasure map from Adam McBones. Adam has buried the treasure at the point $ (x, y) $ with integer coordinates (not necessarily positive). He has indicated on the map the values of $ x^2 + y $ and $ x + y^2 $, and these numbers are distinct. Prove that Long John has to dig only in one place to find the treasure.",
"options": [],
"answer": "See solution",
"solution": "We are to show that the equations $ x^2 + y = a $ and $ x + y^2 = b $ can have at most one solution in integers if $ a $ and $ b $ are distinct.\n\nSuppose not, so that there are two different solutions $ (x_1, y_1) $ and $ (x_2, y_2) $. Then:\n\n$$\nx_1^2 + y_1 = x_2^2 + y_2 \\implies x_1^2 - x_2^2 = y_2 - y_1 \\quad (1)\n$$\n$$\ny_1^2 - y_2^2 = x_2 - x_1 \\quad (2)\n$$\nMultiplying these and factorizing, we get:\n\n$$\n(x_1 - x_2)(x_1 + x_2)(y_1 - y_2)(y_1 + y_2) = (x_1 - x_2)(y_1 - y_2)\n$$\n\nIf $ x_1 = x_2 $, then from (1) and (2), $ y_1 = y_2 $, so the solutions are not different. Thus, $ x_1 \\neq x_2 $ and $ y_1 \\neq y_2 $, so we can cancel $ (x_1 - x_2) $ and $ (y_1 - y_2) $:\n\n$$\n(x_1 + x_2)(y_1 + y_2) = 1\n$$\n\nSo $ x_1 + x_2 = y_1 + y_2 = \\pm 1 $. Plugging into (1):\n\n$$\nx_1 - x_2 = \\pm (y_2 - y_1) \\quad (3)\n$$\n$$\nx_1 + x_2 = y_1 + y_2 \\quad (4)\n$$\nAdding (3) and (4) shows that either $ x_1 = y_2 $ and $ x_2 = y_1 $, or $ x_1 = y_1 $ and $ x_2 = y_2 $. In either case, this contradicts the condition that $ x^2 + y $ and $ x + y^2 $ are distinct for the two solutions. Therefore, the equations can have at most one integer solution.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21800,
"subject": "Mathematics (Olympiad)",
"question": "Prove that for any positive real numbers $a$, $b$, and $c$,\n\n$$\n\\frac{1 + ab}{c} + \\frac{1 + bc}{a} + \\frac{1 + ca}{b} > \\sqrt{a^2 + 2} + \\sqrt{b^2 + 2} + \\sqrt{c^2 + 2}.\n$$",
"options": [],
"answer": "See solution",
"solution": "By manipulating the left-hand side, we get\n\n$$\n\\frac{1 + ab}{c} + \\frac{1 + bc}{a} + \\frac{1 + ca}{b} = \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) + abc \\left( \\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{c^2} \\right). \\quad (4)\n$$\n\nFor any real numbers $x$, $y$, $z$, it holds that $x^2 + y^2 + z^2 \\ge xy + yz + zx$, because we can get it by adding together the AM-GM inequalities $\\frac{1}{2}x^2 + \\frac{1}{2}y^2 \\ge xy$, $\\frac{1}{2}y^2 + \\frac{1}{2}z^2 \\ge yz$, and $\\frac{1}{2}z^2 + \\frac{1}{2}x^2 \\ge zx$. Thus,\n\n$$\n\\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{c^2} \\ge \\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca},\n$$\n\nfrom which\n\n$$\nabc \\left( \\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{c^2} \\right) \\ge abc \\left( \\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca} \\right) = a + b + c.\n$$\n\nTogether with equality (4), this gives\n\n$$\n\\frac{1+ab}{c} + \\frac{1+bc}{a} + \\frac{1+ca}{b} \\ge \\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right) + (a+b+c). \\quad (5)\n$$\n\nAs $\\left(x + \\frac{1}{x}\\right)^2 = x^2 + 2 + \\frac{1}{x^2} > x^2 + 2$ for any $x > 0$, we have\n\n$$\n(a + \\frac{1}{a}) + (b + \\frac{1}{b}) + (c + \\frac{1}{c}) > \\sqrt{a^2 + 2} + \\sqrt{b^2 + 2} + \\sqrt{c^2 + 2}.\n$$\n\nUsing this together with (5), we get the necessary inequality.\n\n*Remark.* The intermediate result (5) can be proven more directly using the rearrangement inequality. Because of symmetry, we can assume the ordering $a \\ge b \\ge c$, which implies $\\frac{1}{c} \\ge \\frac{1}{b} \\ge \\frac{1}{a}$, as well as $1 + ab \\ge 1 + ca \\ge 1 + bc$. Thus, the rearrangement inequality gives\n\n$$\n\\frac{1+ab}{c} + \\frac{1+bc}{a} + \\frac{1+ca}{b} \\ge \\frac{1+ab}{a} + \\frac{1+bc}{b} + \\frac{1+ca}{c} = \\frac{1}{a} + b + \\frac{1}{b} + c + \\frac{1}{c} + a.\n$$\n\n\n\nFigure 17",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21801,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be an integer greater than or equal to $2$, and let $K$ be a closed convex set of area greater than or equal to $n$, contained in the open square $(0, n) \\times (0, n)$. Prove that $K$ contains some point of the integral lattice $\\mathbb{Z} \\times \\mathbb{Z}$.",
"options": [],
"answer": "See solution",
"solution": "Transform $K$ by a suitable two-step Steiner-Edler symmetrization. First, perform a horizontal translation of each slice $K \\cap (\\mathbb{R} \\times y)$ to place it symmetrically about the vertical line $x = \\frac{1}{2}$. It is readily checked that the image $K'$ of $K$ under this transformation is a closed convex subset of the open square $(1/2 - n/2, 1/2 + n/2) \\times (0, n)$, symmetrical about the vertical line $x = 1/2$, and $\\text{area } K' = \\text{area } K \\ge n$. (Also, the transformation does not increase perimeters, a fact which will not be used in the sequel.) In addition, if $K$ contains no lattice point, so does $K'$. Indeed, if $K'$ contained one such, say $i \\times j$, then the line $y = j$ would intersect $K'$ in a closed line segment of length at least $1$, so it would also intersect $K$ in a closed line segment of the same length; the latter, in turn, would contain at least one lattice point—a contradiction.\n\nNext, apply the pattern vertically to further symmetrize $K'$ about the horizontal line $y = 1/2$. The resulting image $K''$ is a closed convex subset of the open square $(1/2 - n/2, 1/2 + n/2) \\times (1/2 - n/2, 1/2 + n/2)$, symmetrical about both lines $x = 1/2$ and $y = 1/2$—whence centrally symmetric about the point $1/2 \\times 1/2$—and of course $\\text{area } K'' = \\text{area } K' = \\text{area } K \\ge n$. Then if $K$ contains no lattice point, so does $K''$.\n\nNow let $a = \\sup\\{x : x \\times 1/2 \\in K''\\}$ and $b = \\sup\\{y : 1/2 \\times y \\in K''\\}$ and notice that $a - 1/2 < n/2$ and $b - 1/2 < n/2$—both inequalities are strict, for $K''$ is closed. Let further $K_0 = K'' \\cap \\{x \\times y : x \\ge 1/2 \\text{ and } y \\ge 1/2\\}$ be the upper right quarter of $K''$. Clearly, $K_0$ is a closed convex set, and $\\text{area } K_0 = (\\text{area } K'')/4 \\ge n/4$. Suppose, if possible, that $K$ contains no lattice point. Then so does $K_0$, so by convexity it must lie below some line through the lattice point $1 \\times 1$ with a non-positive slope; that is,\n\n$$\nK_0 \\subset H = \\{x \\times y : \\alpha(x - 1) + \\beta(y - 1) < 0\\},\n$$\n\nwhere $\\alpha$ and $\\beta$ are non-negative real numbers such that $\\alpha + \\beta > 0$. Notice that $K_0 \\subset H \\cap ([1/2, a] \\times [1/2, b])$, so $\\max\\{a, b\\} > 3/2$, for otherwise\n\n$$\n1/2 \\le n/4 \\le \\text{area } K_0 < \\text{area}(H \\cap ([1/2, 3/2] \\times [1/2, 3/2])) = 1/2,\n$$\n\nwhich is a contradiction—the strict inequality above is due to the fact that $K_0$ is closed. Without loss of generality, assume that $a > 3/2$, so $\\beta \\ne 0$ and\n\n$$\n\\begin{aligned}\n\\text{area } K_0 &< \\text{area}(H \\cap ([1/2, a] \\times [1/2, \\infty))) \\\\\n&= \\frac{1}{2} (a - \\frac{1}{2}) \\left(1 - \\frac{\\alpha}{\\beta} (a - \\frac{3}{2})\\right) \\le \\frac{1}{2} (a - \\frac{1}{2});\n\\end{aligned}\n$$\n\nas before, the inequality is strict because $K_0$ is closed. Finally, recall that $a - 1/2 < n/2$ and $\\text{area } K_0 \\ge n/4$ to derive a contradiction.\n\n**Remarks.**\n\n1. The argument applies mutatis mutandis to show that an open convex set of area greater than $n$, contained in the open square $(0, n) \\times (0, n)$, contains some lattice point. By a similar but more careful argument, it can be shown that an open convex set in the plane whose area is greater than half its perimeter contains some lattice point.\n\n2. The condition on the area of $K$ is essential for the conclusion to hold. Given any positive $\\varepsilon$, there exist closed convex sets $K \\subset (0, n) \\times (0, n)$ such that $n - \\varepsilon < \\text{area } K < n$, which contain no lattice point—for a suitable positive $\\delta$, the set $K = [\\delta, n - \\delta] \\times [\\delta, 1 - \\delta]$ is one such. Similarly, $K = (0, n) \\times (0, 1) \\subset (0, n) \\times (0, n)$ is an open convex set of area $n$, which contains no lattice point.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21802,
"subject": "Mathematics (Olympiad)",
"question": "What is the greatest possible number of distinct positive integers that can be written on the blackboard such that the difference between any two of them (the smaller number subtracted from the larger) is a prime number?",
"options": [],
"answer": "See solution",
"solution": "An example of 4 numbers that satisfy the condition is: 1, 3, 6, 8. Their pairwise differences are 2, 3, 5, and 7, all of which are prime numbers.\n\nSuppose there are $n \\geq 5$ such numbers, ordered as $a_1 < a_2 < \\dots < a_n$. Since all pairwise differences must be prime, $a_3 - a_2 \\geq 2$ and $a_4 - a_3 \\geq 2$. Thus, $a_3 - a_1 > 2$ and $a_5 - a_3 > 2$ are both primes greater than 2, so they are odd. Then $a_5 - a_1 = (a_5 - a_3) + (a_3 - a_1)$ is the sum of two odd numbers, which is even and greater than 2, so it cannot be prime. This is a contradiction. Therefore, the greatest possible number is 4.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21803,
"subject": "Mathematics (Olympiad)",
"question": "Let $f$ be a function from the set of integers to the set of positive integers. Suppose that, for any two integers $m$ and $n$, the difference $f(m) - f(n)$ is divisible by $f(m - n)$. Prove that, for any two integers with $f(m) < f(n)$, the number $f(n)$ is divisible by $f(m)$.\n",
"options": [],
"answer": "See solution",
"solution": "According to the problem, we have $f(k) \\mid f(x+k) - f(x)$ for all $x \\in \\mathbb{Z}$ and any integer $k \\ne 0$. Consequently, $f(k) \\mid f(x+tk) - f(x)$ for all $x \\in \\mathbb{Z}$ and $t \\in \\mathbb{Z}$. So $f(k) \\mid f(m) - f(n)$ for any $m, n$ such that $k \\mid m - n$.\n\n$$\nf(n) \\mid f(m + n) - f(m) \\qquad \\textcircled{1}\n$$\n\n$$\nf(m - (-n)) \\mid f(m) - f(-n) \\qquad \\textcircled{2}\n$$\n\nWe now show that for any integer $n$, $f(n) = f(-n)$. Otherwise, suppose there is an integer $n \\neq 0$ such that $f(n) > f(-n) > 0$. Then $0 < f(n) - f(-n) < f(n)$, which contradicts $f(n) \\mid f(n) - f(-n)$.\n\nTherefore, by ②, we have\n\n$$\nf(m + n) \\mid f(m) - f(n) \\qquad \\textcircled{3}\n$$\n\nIf $0 < f(m) < f(n)$, then by ③, $0 < f(m+n) < f(n)$. Combining ① and ③, we have $f(m+n) = f(m)$. Then by ③, we get the result $f(m) \\mid f(n)$. For $f(m) = f(n)$, the result is obvious. $\\square$\n",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21804,
"subject": "Mathematics (Olympiad)",
"question": "A class consists of 7 boys and 13 girls. During the first three months of the school year, each boy has communicated with each girl at least once. Prove that there exist two boys and two girls such that both boys communicated with both girls for the first time in the same month.",
"options": [],
"answer": "See solution",
"solution": "Call the first communication between a boy and a girl their \\textit{acquaintance}. During the 3 months, there are $7 \\cdot 13 = 91$ acquaintances in total. Thus, there exists a month when there were at least 31 acquaintances. Let the boys be denoted by $p_1$ through $p_7$ and let $T_i$, $i = 1, \\dots, 7$, be the set of girls to whom $p_i$ became acquainted in this month. We have to show that there exist distinct $i$ and $j$ such that $T_i \\cap T_j$ contains at least 2 girls. Without loss of generality, assume the inequalities $|T_1| \\ge |T_2| \\ge \\dots \\ge |T_7|$.\n\nConsider two cases.\n\n1. The case $|T_1| + |T_2| + |T_3| + |T_4| \\ge 20$. Suppose that all intersections $T_1 \\cap T_2, T_1 \\cap T_3, \\dots, T_3 \\cap T_4$ contain at most one girl. Let $k \\le 6$ be the number of non-empty intersections. Then the first four boys became acquainted with\n\n$$\n|T_1 \\cup T_2 \\cup T_3 \\cup T_4| \\ge |T_1| + |T_2| + |T_3| + |T_4| - k \\ge 20 - 6 = 14\n$$\n\ngirls in the month under consideration. (For proving the first inequality, note that girls that belong to one or two subsets count once in the right-hand side, girls belonging to three subsets do not count, and girls in all four subsets count $-2$ times.)\n\nThis is a contradiction since there are only 13 girls.\n\n2. The case $|T_1| + |T_2| + |T_3| + |T_4| \\le 19$. As $|T_5| + |T_6| + |T_7| \\ge 12$, we have $|T_5| \\ge 4$. Now $|T_4| \\le 4$ implies $|T_4| = |T_5| = 4$ and hence also $|T_6| = |T_7| = 4$. Now $|T_1| + |T_2| + |T_3| = 15$ in order to get 31 in total.\n\nSuppose that all intersections $T_i \\cap T_j$, $i, j = 1, \\dots, 7$, contain at most one girl. If boys $p_1, p_2$ and $p_3$ altogether became acquainted with all girls in this month, then at least one intersection $T_i \\cap T_4$, $i = 1, 2, 3$, contains at least two girls. Otherwise,\n\n$$\n|T_1| + |T_2 \\setminus T_1| + |T_3 \\setminus (T_1 \\cup T_2)| = 15 - 1 - 2 = 12\n$$\n\n(Figure 33 shows all possibilities), since only then there is a girl, say $t_{13}$, with whom none of $p_1, p_2, p_3$ became acquainted. Clearly, all boys $p_4, p_5, p_6, p_7$ must have become acquainted with her, and each of them also became acquainted with one girl from sets $T_1$, $T_2 \\setminus T_1$, and $T_3 \\setminus (T_1 \\cup T_2)$. As the last set contains at most three elements, two of the four boys became acquainted with the same girl from $T_3 \\setminus (T_1 \\cup T_2)$.\n\n![](images/estonian-2012-2013_p28_data_9078c4dc2e.png \"Figure 33\")",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21805,
"subject": "Mathematics (Olympiad)",
"question": "Find all perfect squares $\\overline{aabcd}$ with $a \\neq 0$ and $d \\neq 0$ such that the number $\\overline{dcbaa}$ is also a perfect square. ($a, b, c, d$ are decimal digits.)",
"options": [],
"answer": "See solution",
"solution": "The solutions are $44521$ and $44944$.\n\nSuppose $\\overline{aabcd}$ and $\\overline{dcbaa}$ are both squares. Then $a$ and $d$ must belong to $\\{1, 4, 5, 6, 9\\}$.\n\nThe numbers $\\overline{dcb55}$ are multiples of $5$, but not of $25$, hence they cannot be squares. The numbers $\\overline{dcb66}$ are even but not multiples of $4$, hence they cannot be squares. The numbers $\\overline{dcb11}$ and $\\overline{dcb99}$ leave the remainder $3$ when divided by $4$ – again, they cannot be squares.\n\nWe are left with $a = 4$. As $209^2 < 44000$ and $213^2 > 45000$, we consider the cases $\\overline{aabcd} \\in \\{210^2, 211^2, 212^2\\}$. Since $210^2 = 44100$ implies $d = 0$, we reject this case. Finally, given that $211^2 = 44521$, $44521$ reversed is $12544 = 112^2$, and $212^2 = 44944$, $44944$ reversed is $44944 = 212^2$, we find the two solutions: $44521$ and $44944$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21806,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\Gamma$ be the circumcircle of an acute-angled triangle $ABC$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively, such that $AD = AE$. The perpendicular bisectors of $BD$ and $CE$ intersect the minor arcs $AB$ and $AC$ of $\\Gamma$ at points $F$ and $G$, respectively. Prove that the lines $DE$ and $FG$ are parallel (or are the same line).",
"options": [],
"answer": "See solution",
"solution": "Let the line through $F$ and $D$ intersect $\\Gamma$ for a second time at $X$, and let the line through $G$ and $E$ intersect $\\Gamma$ for a second time at $Y$.\n\n\n\nSince $FAXB$ is cyclic, we have $\\angle DXA = \\angle FXA = \\angle FBA = \\angle FBD$. Since also $\\angle FDB = \\angle XDA$, it follows that $\\triangle FBD \\sim \\triangle AXD$ (AA). Since $F$ lies on the perpendicular bisector of $BD$, we have $\\triangle FBD$ is isosceles with $FB = FD$. Thus $\\triangle AXD$ is isosceles with $AX = AD$.\n\nA similar argument shows that $AE = AY$. But since $AD = AE$, it follows that $AX = AD = AE = AY$, and so $XEDY$ is cyclic with centre $A$. Using this and the fact that $FGXY$ is also cyclic, we deduce\n\n$$\n\\angle YED = \\angle YXD = \\angle YXF = \\angle YGF,\n$$\n\nand so $DE$ is parallel to $FG$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21807,
"subject": "Mathematics (Olympiad)",
"question": "Let $I$, $O$, and $H$ be the incenter, circumcenter, and orthocenter of an acute triangle $ABC$, respectively. The incircle of $\\triangle ABC$ touches side $BC$ at $D$. Suppose that $\\angle B < \\angle C$, and $AO$ is parallel to $HD$. Show that the four points $E$, $F$, $I$, $O$ lie on a common circle, where $E$ is the intersection point of lines $OD$ and $AH$, and $F$ is the midpoint of segment $CI$.",
"options": [],
"answer": "See solution",
"solution": "Let $E' \\neq A$ be the intersection point of line $AH$ and the circumscribed circle of $\\triangle ABC$, and $D'$ be the intersection point of line $OE'$ and side $BC$. We have\n\n$$\n\\angle OEA' = \\angle D'E'A = \\angle D'HE'\n$$\n\nsince points $H$ and $E'$ are symmetric about side $BC$. It also holds that\n\n$$\n\\angle OEA' = \\angle OAE'\n$$\n\nsince $AO$ and $E'O$ are radii of the circumscribed circle of $\\triangle ABC$.\n\nHence $\\angle OAE' = \\angle D'HE'$, and thus $AO \\parallel HD'$. By the given $AO \\parallel HD$, we have $D = D'$ and $E = E'$, since both $D$ and $D'$ lie on $BC$.\n\nLet $P \\neq D$ be the intersection of $ID$ and the incircle of $\\triangle ABC$, and $Q$ be the intersection of $AP$ and $BC$. By homothety, the incircle touches $BC$ at $Q$ because $A$ is the center of similarity for the incircle and the incircle of $\\triangle ABC$. Thus, the midpoint $M$ of $BC$ is also the midpoint of $QD$. Now let $P'$ be the intersection of $OQ$ and $PD$. Since $QM = MD$ and $OM \\parallel P'D$, we have\n\n$$\n2OM = P'D.\n$$\n\nUsing properties of the Euler line,\n\n$$\n2OM = AH.\n$$\n\nHence,\n\n$$\nP'D = AH.\n$$\n\nSince $P'D \\parallel AH$, quadrilateral $AP'DH$ is a parallelogram, so $AP' \\parallel HD$. Therefore, $O$ and $Q$ lie on $AP'$, so $P = P'$. Hence $2OM = P'D = PD = 2ID$ and $OM = ID$. Therefore,\n\n$$\nOI \\parallel BC.\n$$\n\nBy the relationship between the circumcenter and orthocenter, $\\angle BAO = \\angle CAE$ and $\\angle CAE = 90^\\circ - \\angle C$. Thus,\n\n$$\n\\angle OAH = \\angle A - (\\angle CAE + \\angle BAO) = \\angle A - 2(90^\\circ - \\angle C).\n$$\n\nBy properties of inscribed angles, $\\angle AEC = \\angle B$. Also, $\\angle DEA = \\angle OAH$ because $\\overline{AO} = \\overline{OE}$. Thus,\n\n$$\n\\angle DEC = \\angle DEA + \\angle AEC = \\angle A - 2(90^\\circ - \\angle C) + \\angle B = \\angle C.\n$$\n\nOn the other hand, $F$ is the center of the circle passing through $I$, $D$, $C$, so $FD = FC$. Hence,\n\n$$\n\\angle DFC = 180^\\circ - \\angle C.\n$$\n\nTherefore, $E$, $C$, $F$, $D$ are concyclic, so $\\angle EDC = \\angle EFC$.\n\nFrom $OI \\parallel BC$, $\\angle EDC = \\angle EOI$, i.e., $\\angle EFC = \\angle EOI$. Hence, the four points $E$, $F$, $I$, $O$ lie on a circle. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21808,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be positive real numbers. Prove that\n\n$$\n\\frac{a^2}{(b+c)^2} + \\frac{b^2}{(a+c)^2} + \\frac{c^2}{(a+b)^2} \\geq \\frac{3}{4}.\n$$",
"options": [],
"answer": "See solution",
"solution": "By the rearrangement inequality or Chebyshev's inequality, we have\n\n$$\n\\frac{a^2}{(b+c)^2} + \\frac{b^2}{(a+c)^2} + \\frac{c^2}{(a+b)^2} \\geq \\frac{1}{3} \\left( \\frac{a}{b+c} + \\frac{b}{a+c} + \\frac{c}{a+b} \\right)^2.\n$$\n\nBy Nesbitt's inequality,\n\n$$\n\\frac{a}{b+c} + \\frac{b}{a+c} + \\frac{c}{a+b} \\geq \\frac{3}{2},\n$$\n\nwhich proves the claim.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21809,
"subject": "Mathematics (Olympiad)",
"question": "(a) Prove that if $a - \\frac{1}{b} + b\\left(b + \\frac{3}{a}\\right)$ is an integer for some positive integers $a$ and $b$, then it is a perfect square.\n\n(b) Find two integers $a$ and $b$ such that $a - \\frac{1}{b} + b\\left(b + \\frac{3}{a}\\right)$ is a positive integer but not a perfect square.",
"options": [],
"answer": "See solution",
"solution": "Let $I = a - \\frac{1}{b} + b\\left(b + \\frac{3}{a}\\right)$. For $I = a + b^2 + \\frac{3b}{a} - \\frac{1}{b}$ to be an integer, $N = \\frac{3b}{a} - \\frac{1}{b}$ must also be an integer. Since $N = \\frac{3b^2 - a}{ab}$, $ab$ must divide $3b^2 - a$. This implies $b$ divides $3b^2 - a$, so $b$ divides $a$.\n\nLet $a = k b$. Then $N = \\frac{3b - k}{k b}$, so $k b$ divides $3b - k$. Thus, $b$ divides $k$. Let $k = l b$. Then $N = \\frac{3 - l}{l b}$. We see $l$ must divide $3 - l$, so $l$ divides $3$. Since $a$ and $b$ are positive integers, $l$ is also positive, so $l = 1$ or $l = 3$.\n\nIf $l = 1$, then $N = \\frac{2}{b}$ and $b = 1$ or $b = 2$. In the first case, $a = 1$; in the second, $a = 4$. If $l = 3$, then $N = 0$ and $a = 3b^2$.\n\nNow, compute $I$:\n- If $a = b = 1$, $I = 4$.\n- If $a = 4$, $b = 2$, $I = 9$.\n- If $a = 3b^2$, $I = 4b^2 = (2b)^2$.\nThus, $I$ is always a perfect square in these cases.\n\nFor part (b), consider $a = 4$, $b = -2$ or $a = -4$, $b = -2$. In these cases, $l$ is a positive integer but $I$ is not a perfect square. More generally, $l$ divides $3$, so $l = 1, -1, 3, -3$. For $l = -1$, $a = -4$, $b = -2$; for $l = 1$, $a = 4$, $b = -2$. In both cases, $I$ is a positive integer but not a perfect square.\n\n*Remark*: The only cases where $a - \\frac{1}{b} + b\\left(b + \\frac{3}{a}\\right)$ is a positive integer but not a perfect square are $a = 4$, $b = -2$ and $a = -4$, $b = -2$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21810,
"subject": "Mathematics (Olympiad)",
"question": "In a country there are $n$ airports and $n$ air companies operating return flights. Each company operates an odd number of flights forming a closed route. Prove that a traveller can complete a closed route consisting of an odd number of flights operated by pairwise distinct companies.",
"options": [],
"answer": "See solution",
"solution": "In graph-theoretic terms, consider a collection of $n$ odd cycles (not necessarily distinct), all on the same vertex set of size $n$. Prove that at most one edge can be chosen from each of these cycles to form a collection that contains the edges of an odd cycle.\n\nCall a set of edges *rainbow* if it is formed by choosing at most one edge from each cycle. We have to prove that there exists a rainbow cycle of odd length.\n\nBegin by choosing a maximal rainbow forest $F$, that is, an acyclic rainbow set of edges.\n\nSince $F$ is acyclic, its size is less than $n$, so there is a cycle $C$ in the collection no edge of which lies in $F$. The forest $F$ contains every vertex of $C$, for otherwise an edge of $C$ incident with a vertex outside $F$ could be added to $F$ to form a larger rainbow forest, contradicting maximality. Moreover, no edge of $C$ joins different components of $F$, for one such could again be added to $F$ to contradict maximality.\n\nConsequently, the vertices of $C$ all lie in some component of $F$, a tree $T$. As such, $T$ is bipartite, that is, its vertices split into two disjoint parts, and all edges are between the two. Since $C$ is an odd cycle, it has an edge whose endpoints both lie in the same part. This edge then completes an odd rainbow cycle.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21811,
"subject": "Mathematics (Olympiad)",
"question": "Let the set $M = \\{1, 2, 3, \\ldots, 50\\}$. Find all positive integers $n$ such that, in any subset of $M$ with 35 elements, there exist at least two distinct elements $a$ and $b$ such that $a + b = n$ or $a - b = n$.",
"options": [],
"answer": "See solution",
"solution": "Consider $A = \\{1, 2, 3, \\ldots, 35\\}$. For any $a, b \\in A$:\n\n$$\na - b \\leq 34, \\quad a + b \\leq 34 + 35 = 69.\n$$\n\nWe show that $1 \\leq n \\leq 69$ works. Let $A = \\{a_1, a_2, \\ldots, a_{35}\\}$, with $a_1 < a_2 < \\ldots < a_{35}$.\n\n**Case (i): $1 \\leq n \\leq 19$**\n\nSince $1 \\leq a_1 < a_2 < \\ldots < a_{35} \\leq 50$ and $2 \\leq a_1 + n < a_2 + n < \\ldots < a_{35} + n \\leq 69$, by the pigeonhole principle, there exist $i \\neq j$ such that $a_i + n = a_j$, i.e., $a_j - a_i = n$.\n\n**Case (ii): $51 \\leq n \\leq 69$**\n\nSimilarly, $1 \\leq n - a_{35} < n - a_{34} < \\ldots < n - a_1 \\leq 68$. By the pigeonhole principle, there exist $i \\neq j$ such that $n - a_i = a_j$, i.e., $a_i + a_j = n$.\n\n**Case (iii): $20 \\leq n \\leq 24$**\n\nSince $50 - 2n \\leq 10$, at least 25 elements of $A$ are in $[1, 2n]$. There are at most 24 pairs $\\{i, n+i\\}$ in $\\{1, n+1\\}, \\ldots, \\{n, 2n\\}$, so by the pigeonhole principle, $a_j - a_i = n$ for some $i \\neq j$.\n\n**Case (iv): $25 \\leq n \\leq 34$**\n\nThe pairs $\\{1, n+1\\}, \\ldots, \\{n, 2n\\}$ have at most 34 elements, so among 35 elements, there exist $i \\neq j$ such that $a_j - a_i = n$.\n\n**Case (v): $n = 35$**\n\nThere are 33 pairs $\\{1, 34\\}, \\{2, 33\\}, \\ldots, \\{17, 18\\}$ and the remaining elements $\\{35\\}, \\{36\\}, \\ldots, \\{50\\}$. Thus, there exist $i \\neq j$ such that $a_i + a_j = 35$.\n\n**Case (vi): $36 \\leq n \\leq 50$**\n\nIf $n = 2k + 1$, consider pairs $\\{1, 2k\\}, \\{2, 2k-1\\}, \\ldots, \\{k, k+1\\}$ and remaining elements $\\{2k+1\\}, \\ldots, \\{50\\}$. For $18 \\leq k \\leq 20$, $50 - 2k \\leq 14$; for $21 \\leq k \\leq 24$, $50 - 2k \\leq 8$. Thus, there exist $i \\neq j$ such that $a_i + a_j = n$.\n\nIf $n = 2k$, consider pairs $\\{1, 2k-1\\}, \\{2, 2k-2\\}, \\ldots, \\{k-1, k+1\\}$ and remaining elements $\\{k\\}, \\{2k\\}, \\{2k+1\\}, \\ldots, \\{50\\}$. For $18 \\leq k \\leq 19$, $50 - 2k + 3 \\leq 18$; for $20 \\leq k \\leq 23$, $50 - 2k + 2 \\leq 22$; for $24 \\leq k \\leq 25$, $50 - 2k + 2 \\leq 24$. Thus, there exist $i \\neq j$ such that $a_i + a_j = n$.\n\nTherefore, all $n$ with $1 \\leq n \\leq 69$ satisfy the condition. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21812,
"subject": "Mathematics (Olympiad)",
"question": "Для любых точек $A_i = (x_i, y_i)$ ($1 \\leq i \\leq n$) на плоскости, никакие три из которых не лежат на одной прямой, найдётся такой многочлен $P(x, y)$ степени не больше $[n/2]$, что $P(x_n, y_n) = 1$ и $P(x_i, y_i) = 0$ при $i = 1, \\dots, n-1$.\n\nДокажите, что минимально возможная степень $k$ такого многочлена равна $[n/2]$.",
"options": [],
"answer": "See solution",
"solution": "Пусть $d = [n/2]$. Существуют такие $d$ прямых, что точка $A_n$ не лежит ни на одной из них, а каждая из точек $A_1, \\dots, A_{n-1}$ лежит хотя бы на одной (при нечётном $n$ это прямые $A_1A_2, A_3A_4, \\dots, A_{n-2}A_{n-1}$, а при чётном $n$ — прямые $A_1A_2, A_3A_4, \\dots, A_{n-3}A_{n-2}, A_{n-2}A_{n-1}$). Пусть $k_i x + \\ell_i y + m_i = 0$ — уравнение $i$-й прямой ($i = 1, \\dots, d$). Тогда многочлен\n\n$$\nQ(x, y) = \\frac{(k_1x + \\ell_1y + m_1) \\dots (k_d x + \\ell_d y + m_d)}{(k_1x_n + \\ell_1y_n + m_1) \\dots (k_d x_n + \\ell_d y_n + m_d)}\n$$\n\nявляется искомым.\n\nПокажем, что $k = [n/2]$ подходит. Для каждого $i = 1, \\dots, n$ найдём многочлен $P_i(x, y)$, обращающийся в ноль во всех точках $A_1, \\dots, A_n$, кроме $A_i$, причём $P_i(x_i, y_i) = 1$. Тогда многочлен $P(x, y) = c_1P_1(x, y) + \\dots + c_nP_n(x, y)$ принимает требуемые значения во всех точках $A_1, \\dots, A_n$.\n\nПокажем, что при $k < [n/2]$ утверждение неверно. Рассмотрим точки $A_i = (i, i^2)$ ($i = 1, \\dots, n$), лежащие на параболе $y = x^2$, и положим $c_1 = \\dots = c_{n-1} = 0$, $c_n = 1$. Парабола пересекается с прямой не более чем в двух точках, так что условие выполнено. Пусть существует многочлен $P(x, y)$ степени не выше $k$, для которого $P(x_i, y_i) = c_i$. Положим $Q(x) = P(x, x^2)$; тогда степень $Q(x)$ не превосходит $2k$. По предположению, $Q(1) = Q(2) = \\dots = Q(n-1) = 0$ и $Q(n) = 1$. Значит, ненулевой многочлен $Q(x)$ имеет $n-1$ корень, то есть его степень не меньше $n-1$, а значит $2k \\geq n-1$, то есть $k \\geq [n/2]$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21813,
"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$$\n\\begin{bmatrix} \\cdot & ? & \\cdot & \\cdot & \\cdot \\\\ \\cdot & \\cdot & \\cdot & 48 & \\cdot \\\\ 12 & \\cdot & \\cdot & \\cdot & \\cdot \\\\ \\cdot & \\cdot & 16 & \\cdot & \\cdot \\\\ \\cdot & \\cdot & \\cdot & \\cdot & 0 \\end{bmatrix}\n$$\n\n(A) 19 \n(B) 24 \n(C) 29 \n(D) 34 \n(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$$\na_{43} = 16 = \\frac{2a_{44} + a_{41}}{3} = \\frac{\\frac{4}{3}d + 32 + 6 + 2d}{3},\n$$\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$$\n\\begin{bmatrix} 12 & \\underline{\\mathbf{29}} & 46 & 63 & 80 \\\\ 12 & 24 & 36 & \\mathbf{48} & 60 \\\\ \\mathbf{12} & 19 & 26 & 33 & 40 \\\\ 12 & 14 & \\mathbf{16} & 18 & 20 \\\\ 12 & 9 & 6 & 3 & \\mathbf{0} \\end{bmatrix}.\n$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 21814,
"subject": "Mathematics (Olympiad)",
"question": "حدد كل الأعداد الصحيحة الموجبة $k$ التي لأجلها توجد دالة $f: \\mathbb{N} \\to \\mathbb{N}$ تحقق الشرط: لكل عدد صحيح موجب $n$، لدينا $f(n) < f(n + 1)$ و $f(f(n)) = k n$.",
"options": [],
"answer": "See solution",
"solution": "1. إذا لم يكن $f(n+1)$ من مضاعفات $k$: هذا يعني أن $n+1$ تم اختياره كـ $m$ في الخوارزمية، وتم تعيين $f(n+1)$ ليكون أكبر من $f(n)$.\n\n2. إذا كان $f(n+1)$ من مضاعفات $k$: لنفرض أن $x$ هو أكبر عدد صحيح أقل من أو يساوي $n$ بحيث أن $f(x)$ من مضاعفات $k$. بما أن جميع مضاعفات $k$ تُرسَل إلى مضاعفات $k$ تحت $f$، إذًا $n - k < x \\leq n$. لنفرض $f(x) = k a$ و $f(n+1) = k b$. إذًا $f(a) = x$ و $f(b) = n + 1$. بما أن $x < n + 1$ و $a, b < n$، فإن فرضية الاستقراء تعطي $a < b$. لاحظ أنه لكل $m \\in \\{n - x + 1, n - x + 2, \\dots, n\\}$، تم تعيين $f(m) = f(m - 1) + 1$. إذًا الخوارزمية تعطي $f(n) = f(x) + (n - x) = k a + (n - x)$. وبما أن $n - x < k$، إذًا $k a + (n - x) < k(a + 1) \\leq k b$. إذًا $f(n) < f(n + 1)$ كما هو مطلوب.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21815,
"subject": "Mathematics (Olympiad)",
"question": "令 $a_i > 0$，$i = 1, 2, \\dots, n$，且 $\\sum_{i=1}^{n} a_i = 1$。\n\n試證：對任意正整數 $k$，\n\n$$\n(a_1^k + \\frac{1}{a_1^k})(a_2^k + \\frac{1}{a_2^k}) \\cdots (a_n^k + \\frac{1}{a_n^k}) \\ge (n^k + \\frac{1}{n^k})^n.\n$$\n\nLet $a_i > 0$, $i = 1, 2, \\dots, n$, and $\\sum_{i=1}^{n} a_i = 1$.\n\nProve that for any positive integer $k$,\n\n$$\n(a_1^k + \\frac{1}{a_1^k})(a_2^k + \\frac{1}{a_2^k}) \\cdots (a_n^k + \\frac{1}{a_n^k}) \\ge (n^k + \\frac{1}{n^k})^n.\n$$",
"options": [],
"answer": "See solution",
"solution": "*證明：*\n\n$$\n\\begin{aligned}\na_1^k + \\frac{1}{a_1^k} &= a_1^k + \\frac{1}{n^{2k} a_1^k} + \\cdots + \\frac{1}{n^{2k} a_n^k} \\\\\n&\\ge (n^{2k} + 1) \\left( \\frac{a_1^k}{(n^{2k} a_1^{2k})^{n^{2k}}} \\right)^{\\frac{1}{n^{2k}+1}} \\\\\n&= (n^k + \\frac{1}{n^k}) \\left( \\frac{1}{n a_1} \\right)^{\\frac{k(n^{2k}-1)}{n^{2k}+1}}.\n\\end{aligned}\n$$\n\n$$\n\\begin{aligned}\na_2^k + \\frac{1}{a_2^k} &\\ge (n^k + \\frac{1}{n^k}) \\left(\\frac{1}{n a_2}\\right)^{\\frac{k(n^{2k}-1)}{n^{2k}+1}}, \\\\\n\\vdots & \\\\\na_n^k + \\frac{1}{a_n^k} &\\ge (n^k + \\frac{1}{n^k}) \\left(\\frac{1}{n a_n}\\right)^{\\frac{k(n^{2k}-1)}{n^{2k}+1}}\n\\end{aligned}\n$$\n\n同理可得，將 $n$ 個不等式相乘，得\n\n$$\n\\begin{aligned}\n& (a_1^k + \\frac{1}{a_1^k})(a_2^k + \\frac{1}{a_2^k}) \\cdots (a_n^k + \\frac{1}{a_n^k}) \\\\\n&\\ge (n^k + \\frac{1}{n^k})^n \\left(\\frac{1}{n^n a_1 \\cdots a_n}\\right)^{\\frac{k(n^{2k}-1)}{n^{2k}+1}}\n\\end{aligned}\n$$\n\n因 $a_i > 0$，$i = 1, \\dots, n$，且 $\\sum_{i=1}^n a_i = 1$，則\n\n$$\na_1 \\cdots a_n \\le \\left( \\frac{1}{n} \\sum_{i=1}^n a_i \\right)^n = \\frac{1}{n^n},\n$$\n\n則\n\n$$\n\\frac{1}{n^n a_1 \\cdots a_n} \\ge 1.\n$$\n\n又因\n\n$$\n\\frac{k(n^{2k} - 1)}{n^{2k} + 1} > 0, \\text{ 所以}\n$$\n\n$$\n\\left(\\frac{1}{n^n a_1 \\cdots a_n}\\right)^{\\frac{k(n^{2k}-1)}{n^{2k}+1}} \\ge 1\n$$\n\n故\n$$\n(a_1^k + \\frac{1}{a_1^k})(a_2^k + \\frac{1}{a_2^k}) \\cdots (a_n^k + \\frac{1}{a_n^k}) \\ge (n^k + \\frac{1}{n^k})^n\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21816,
"subject": "Mathematics (Olympiad)",
"question": "Let $f: \\mathbb{Z}_{>0} \\to \\mathbb{Z}$ be a function with the following properties:\n\n1. $f(1) = 0$,\n2. $f(p) = 1$ for all prime numbers $p$,\n3. $f(xy) = y f(x) + x f(y)$ for all $x, y$ in $\\mathbb{Z}_{>0}$.\n\nDetermine the smallest integer $n \\ge 2015$ that satisfies $f(n) = n$.",
"options": [],
"answer": "See solution",
"solution": "We claim that\n\n$$\nf(q_1 \\cdots q_s) = q_1 \\cdots q_s \\left( \\frac{1}{q_1} + \\cdots + \\frac{1}{q_s} \\right)\n$$\n\nholds for (not necessarily distinct) prime numbers $q_1, \\dots, q_s$.\n\nWe prove the claim by induction on $s$. For $s=0$, the claim reduces to $f(1) = 0$, which is true by assumption.\n\nIf the formula holds for some $s$, then\n\n$$\n\\begin{aligned}\nf(q_1 \\cdots q_s q_{s+1}) &= f((q_1 \\cdots q_s)q_{s+1}) \\\\\n&= q_{s+1} f(q_1 \\cdots q_s) + q_1 \\cdots q_s f(q_{s+1}) \\\\\n&= q_1 \\cdots q_{s+1} \\left( \\frac{1}{q_1} + \\cdots + \\frac{1}{q_s} \\right) + q_1 \\cdots q_s \\\\\n&= q_1 \\cdots q_{s+1} \\left( \\frac{1}{q_1} + \\cdots + \\frac{1}{q_s} + \\frac{1}{q_{s+1}} \\right).\n\\end{aligned}\n$$\n\nIt is easily verified that the function given by this formula fulfills the given functional equation.\n\nLet $p_1, \\dots, p_r$ be distinct primes and $\\alpha_1, \\dots, \\alpha_r$ be positive integers. Then collecting equal primes leads to\n\n$$\nf(p_1^{\\alpha_1} \\cdots p_r^{\\alpha_r}) = p_1^{\\alpha_1} \\cdots p_r^{\\alpha_r} \\sum_{j=1}^{r} \\frac{\\alpha_j}{p_j}.\n$$\n\nNow, we determine all $n \\ge 2015$ with $f(n) = n$. Write $n = p_1^{\\alpha_1} \\cdots p_r^{\\alpha_r}$. Then\n\n$$\n\\frac{\\alpha_1}{p_1} + \\cdots + \\frac{\\alpha_r}{p_r} = 1.\n$$\n\nLet\n\n$$\n\\frac{\\alpha_1}{p_1} + \\cdots + \\frac{\\alpha_{r-1}}{p_{r-1}} = \\frac{a}{p_1 \\cdots p_{r-1}}\n$$\n\nfor some non-negative integer $a$. Then\n\n$$\n\\frac{a}{p_1 \\cdots p_{r-1}} + \\frac{\\alpha_r}{p_r} = 1 \\iff a p_r + \\alpha_r p_1 \\cdots p_{r-1} = p_1 \\cdots p_r.\n$$\n\nAs $p_r$ is coprime to $p_1 \\cdots p_{r-1}$, we conclude that $p_r \\mid \\alpha_r$. Since $\\alpha_r \\le p_r$, we must have $r=1$ and $\\alpha_r = p_r$.\n\nThus, $f(n) = n$ holds if and only if $n = p^p$ for some prime number $p$.\n\nWe have\n\n$$\n2^2 = 4 < 3^3 = 27 < 2015 < 5^5 = 3125,\n$$\n\nso the smallest such $n$ is $3125$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21817,
"subject": "Mathematics (Olympiad)",
"question": "(a) Prove that if the ratio of the side lengths of the floor of the room is rational, then the beam eventually returns to the point of entrance.\n\n(b) Prove that if the ratio of the side lengths of the floor of the room is irrational, then the beam never returns to the point of entrance.",
"options": [],
"answer": "See solution",
"solution": "Let $a$ and $b$ be the side lengths of the floor. We use a planar coordinate system whose axes are aligned with the sides of the floor.\n\n\n\nAfter reflection from a wall, one coordinate of the beam reverses direction while retaining its speed. As the beam enters the room along a corner bisector, both coordinates change with the same speed. Without loss of generality, let the speed be $1$, so the coordinates change with periods $2a$ and $2b$, respectively.\n\n(a) If $\\frac{a}{b}$ is rational, then $\\frac{a}{b} = \\frac{m}{n}$ for some integers $m$ and $n$. Let $t = 2a n = 2b m$. At time $t$, both coordinates have completed a whole number of periods, so the beam returns to the point of entrance.\n\n(b) Suppose the beam returns to the point of entrance at time $t$. Then $\\frac{t}{2a}$ and $\\frac{t}{2b}$ are integers; let $n = \\frac{t}{2a}$ and $m = \\frac{t}{2b}$, so $a = \\frac{t}{2n}$ and $b = \\frac{t}{2m}$. Thus, $\\frac{a}{b} = \\frac{m}{n}$, which is rational. Therefore, for the beam to return to the point of entrance, $\\frac{a}{b}$ must be rational.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21818,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCDE$ be a convex pentagon. For each vertex, consider the triangle formed by that vertex and its two adjacent vertices: $ABC$, $BCD$, $CDE$, $DEA$, and $EAB$. Let $S$ denote the area of pentagon $ABCDE$, and let $S'$ denote the sum of the areas of these five triangles:\n\n$$\nS' = S(ABC) + S(BCD) + S(CDE) + S(DEA) + S(EAB)\n$$\n\nProve that:\n\n$$\nS \\leq S' \\leq 2S.\n$$\n\n% \n% ",
"options": [],
"answer": "See solution",
"solution": "The right-hand inequality $S' \\leq 2S$ is clear, since each region in the pentagon $ABCDE$ is covered at most twice by the corner triangles.\n\nTo prove $S \\leq S'$, consider a convex pentagon $ABCDE$. There exist two adjacent angles, say $\\angle D$ and $\\angle E$, with sum greater than $180^\\circ$. Fix $B$, $C$, $D$, and $E$, and let $A$ vary in the convex region determined by the pentagon. Define\n\n$$\nf(A) = S(ABC) + S(BCD) + S(CDE) + S(DEA) - S(BCDE)\n$$\n\nWe must show $f(A) \\geq 0$ for all $A$ in the region. Since $f$ is linear in $A$ and the region is convex, the minimum occurs at a corner ($B$, $E$, or at infinity):\n\n$$\nf(\\infty) = +\\infty > 0\n$$\n$$\nf(B) = 0 + S(BCD) + S(CDE) + S(DEB) - S(BCDE) \\geq 0\n$$\n$$\nf(E) = S(EBC) + S(BCD) + S(CDE) + 0 - S(BCDE) \\geq 0\n$$\n\nThus, $f(A) \\geq 0$ everywhere in the region, so $S' \\geq S$. $\\square$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21819,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a right triangle with hypotenuse $BC$ and altitude $AD$. Let the midpoints of $AD$ and $AC$ be $E$ and $F$, respectively. Let point $M$ be the circumcenter of $\\triangle BEF$. Prove that $AC \\parallel BM$.",
"options": [],
"answer": "See solution",
"solution": "Since $\\triangle ADB \\sim \\triangle CAB$, we have $\\frac{AD}{AB} = \\frac{CA}{CB}$. As $AE = \\frac{1}{2}AD$ and $CF = \\frac{1}{2}CA$, it follows that $\\frac{AF}{AB} = \\frac{CF}{CB}$. From this similarity, we obtain:\n\n$$\n\\angle BAE = \\angle BAD = \\angle BCA = \\angle BCF\n$$\n\nTherefore, by angle and side ratios, $\\triangle AEB \\sim \\triangle CFB$, so $\\angle ABE = \\angle CBF$.\n\nSince $EF$ is the midline of $\\triangle CAD$, we also have $EF \\parallel BC$, which implies $\\angle BFE = \\angle CBF$. Next, we get the following equalities of angles: $\\angle BFE = \\frac{1}{2}\\angle BME = 90^\\circ - \\angle EBM$, so $\\angle ABE = \\angle CBF = 90^\\circ - \\angle EBM$. Therefore, $\\angle ABM = \\angle ABE + \\angle EBM = 90^\\circ$, so $AC \\parallel BM$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21820,
"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",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21821,
"subject": "Mathematics (Olympiad)",
"question": "Suppose that $X$ is a machine with inputs from the set $\\{0, 1, \\dots, k-1\\}$. Show that each $k$-computable sequence is $k$-left computable. (The converse can be proved similarly.)\n\nConstruct a machine $Y$ with the same input set as $X$ such that the output of $Y$ after entering digits of a number from left to right equals the output of $X$ after entering digits of that same number from right to left.",
"options": [],
"answer": "See solution",
"solution": "* Let $\\{s_1, s_2, \\dots, s_m\\}$ be the state set of $X$, with $s_1$ as its starting state and $x_i$ the output of state $s_i$.\n* **Inputs:** $\\{0, 1, \\dots, k-1\\}$.\n* **States:** All $m$-tuples $(A_1, A_2, \\dots, A_m)$ of partitions of $\\{s_1, s_2, \\dots, s_m\\}$.\n* **Outputs:** The output of state $(A_1, A_2, \\dots, A_m)$ is $A_i$, where $s_1 \\in A_i$.\n* **Starting State:** $(\\{s_1\\}, \\{s_2\\}, \\dots, \\{s_m\\})$.\n* **Transition Function:** When $r$ is entered, the state changes from $(A_1, A_2, \\dots, A_m)$ to $(B_1, B_2, \\dots, B_m)$, where $B_i$ is the set of all states of $X$ that change to a state in $A_i$ after entering $r$ in $X$.\n\nThis construction ensures that $Y$ simulates $X$ on reversed input, so every $k$-computable sequence is $k$-left computable.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21822,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\{a_n\\}$ be a sequence such that $a_0 = 0 = a_m$ for some $m > 0$, and $P(x)$ is a polynomial with integer coefficients such that $a_{n+1} = P(a_n)$ for all $n$. Prove that $a_1 = 0$.",
"options": [],
"answer": "See solution",
"solution": "Since $a_0 = 0 = a_m$, we can prove by induction that $a_k = a_{m+k}$ for all $k$. If there are two consecutive terms of $\\{a_n\\}$ which are the same, then all subsequent terms are the same. By the periodicity, all terms are the same, and they are equal to $a_0 = 0$. This gives $a_1 = 0$ and we are done.\n\nNow, assume $d_n = a_{n+1} - a_n \\ne 0$ for all $n$. Using the fact $x - y \\mid P(x) - P(y)$ for all $x, y \\in \\mathbb{Z}$, we know that $d_n$ divides $P(a_{n+1}) - P(a_n) = a_{n+2} - a_{n+1} = d_{n+1}$ for all $n$. This gives\n\n$$\n|a_1 - a_0| = |d_0| \\le |d_1| \\le |d_2| \\le \\dots \\le |d_m| = |a_{m+1} - a_m| = |a_1 - a_0|.\n$$\n\nTherefore, all equalities hold, and we have $|d_n| = |d_0|$ for all $n$, which means $d_n = \\pm d_0$ for all $n$. Now, note that\n\n$$\nd_0 + d_1 + \\cdots + d_{m-1} = a_m - a_0 = 0.\n$$\n\nSince $d_n \\neq 0$, there are both positive and negative terms among $d_0, d_1, \\dots, d_{m-1}$. WLOG assume $d_k$ and $d_{k+1}$ have different signs for some $k < m$. Then\n\n$$\na_{k+2} = a_k + d_k + d_{k+1} = a_k \\pm d_0 \\mp d_0 = a_k.\n$$\n\nThis implies $a_{n+2} = a_n$ for all $n \\ge k$, and so $a_2 = a_{m+2} = a_m = 0$ as desired.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21823,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle with incentre $I$ such that $AB < AC < BC$. The second intersections of $AI$, $BI$ and $CI$ with the circumcircle of triangle $ABC$ are $M_A$, $M_B$ and $M_C$, respectively. Lines $AI$ and $BC$ intersect at $D$, and lines $BM_C$ and $CM_B$ intersect at $X$. Suppose the circumcircles of triangles $XM_B M_C$ and $XBC$ intersect again at $S \\neq X$. Lines $BX$ and $CX$ intersect the circumcircle of triangle $SXM_A$ again at $P \\neq X$ and $Q \\neq X$, respectively. Prove that the circumcentre of triangle $SID$ lies on $PQ$.",
"options": [],
"answer": "See solution",
"solution": "Let $O$ be the circumcentre of triangle $ABC$. First, we note from standard properties of the Miquel point $S$:\n\n$$\n\\bullet \\ \\triangle SM_C M_B \\sim \\triangle SBC \\sim \\triangle SPQ; \\quad (*).\n$$\n\n\n\n* $I$ and $S$ are inverses with respect to circle $ABC$;\n* $\\angle OSX = 90^{\\circ}$.\n\nFrom the above, we have $\\triangle OM_A I \\sim \\triangle OSM_A$ and\n\n$$\n\\angle M_A PB = \\angle M_A SX = 90^{\\circ} + \\angle M_A SO = 90^{\\circ} + \\angle OM_A I = \\angle M_A BA = \\angle CDA.\n$$\n\nObserve that $\\angle PM_C M_A = \\angle BM_C M_A = \\angle DAC$ and $\\angle M_C M_A B = \\angle ICD$. Combining these with $\\angle M_A PB = \\angle CDA$ shows $M_C PM_A B \\sim ADCI$. Therefore, we have $\\frac{M_C B}{BP} = \\frac{AI}{ID}$. Similarly, $\\frac{M_B C}{CQ} = \\frac{AI}{ID}$. Thus we get\n\n$$\n\\frac{M_C B}{BP} = \\frac{M_B C}{CQ} = \\frac{AI}{ID}. \\qquad (**)\n$$\n\nNow, observe that $\\angle ICB = \\angle AM_B M_C$ and $\\angle CBI = \\angle M_B M_C A$ which gives that $\\triangle IBC \\sim \\triangle AM_C M_B$. This, combined with $(**)$, is enough to show $\\triangle DPQ \\sim \\triangle IBC$ by linearity, thus $\\frac{DP}{DQ} = \\frac{IB}{IC}$.\n\nCombining $\\triangle IBM_C \\sim \\triangle ICM_B$ with $(***)$ shows $IBM_C P \\sim ICM_B Q$, thus $\\frac{IP}{IQ} = \\frac{IB}{IC}$. Finally, we have that\n\n$$\n\\frac{SP}{SQ} = \\frac{SB}{SC} = \\frac{BM_C}{CM_B} = \\frac{IB}{IC}\n$$\n\nfrom $(*)$ and $\\triangle IBM_C \\sim \\triangle ICM_B$. Putting this together with above ratios, we have\n\n$$\n\\frac{IB}{IC} = \\frac{DP}{DQ} = \\frac{IP}{IQ} = \\frac{SP}{SQ}\n$$\n\nwhich shows that circle *SID* is an Apollonius circle with respect to *P* and *Q*, giving the desired conclusion.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21824,
"subject": "Mathematics (Olympiad)",
"question": "Let $f$ be a continuous function on $[0, 1]$. Define $F(x) = \\int_{0}^{x} f(t)\\,dt$ for $x \\in [0, 1]$, and define $\\tilde{f}(x) = \\frac{F(x)}{x}$ for $x \\in (0, 1]$, with $\\tilde{f}(0) = f(0)$.\n\n**a)** Prove that $\\tilde{f}$ is continuous on $[0, 1]$.\n\n**b)** Let $I = \\int_{0}^{1} f(x)\\,dx$. Prove that\n$$\n\\int_{0}^{1} f^2(x)\\,dx - \\left(\\int_{0}^{1} f(x)\\,dx\\right)^2 = \\int_{0}^{1} (f(x) - \\tilde{f}(x))^2\\,dx.\n$$",
"options": [],
"answer": "See solution",
"solution": "a) Since $f$ is continuous on $[0, 1]$, the function $F(x) = \\int_{0}^{x} f(t)\\,dt$ is differentiable on $[0, 1]$ with $F'(x) = f(x)$. Thus, $\\tilde{f}(x) = \\frac{F(x)}{x}$ is differentiable on $(0, 1]$ as a quotient of differentiable functions.\n\nTo show $\\tilde{f}$ is continuous at $0$, note that $\\lim_{x \\to 0} f(x) = f(0)$. By l'Hôpital's rule:\n$$\n\\lim_{x \\to 0} \\tilde{f}(x) = \\lim_{x \\to 0} \\frac{F(x)}{x} = \\lim_{x \\to 0} \\frac{F'(x)}{1} = \\lim_{x \\to 0} f(x) = f(0) = \\tilde{f}(0),\n$$\nso $\\tilde{f}$ is continuous at $0$.\n\nb) Let $I = \\int_{0}^{1} f(x)\\,dx$, so $\\tilde{f}(1) = I$. Define $G(x) = x (\\tilde{f}(x) - I)^2$ for $x \\in [0, 1]$. $G$ is continuous at $0$ and differentiable on $(0, 1]$, with $G(0) = G(1) = 0$.\n\nCompute $G'(x)$ for $x \\in (0, 1]$:\n$$\n\\begin{aligned}\nG'(x) &= (\\tilde{f}(x) - I)^2 + 2x (\\tilde{f}(x) - I) \\tilde{f}'(x) \\\\\n&= (\\tilde{f}(x) - I) \\left( \\tilde{f}(x) - I + 2x \\left( -\\frac{1}{x^2} \\int_0^x f(t)\\,dt + \\frac{1}{x} f(x) \\right) \\right) \\\\\n&= (\\tilde{f}(x) - I)(2f(x) - \\tilde{f}(x) - I) \\\\\n&= 2f(x)(\\tilde{f}(x) - I) + I^2 - \\tilde{f}^2(x) \\\\\n&= I^2 - 2I f(x) + f^2(x) - (f(x) - \\tilde{f}(x))^2.\n\\end{aligned}\n$$\n\nSince $\\lim_{x \\to 0} G'(x) = (I - f(0))^2$, $G'$ is continuous on $[0, 1]$.\n\nTherefore,\n$$\n\\begin{aligned}\n&\\int_{0}^{1} f^2(x)\\,dx - \\left(\\int_{0}^{1} f(x)\\,dx\\right)^2 - \\int_{0}^{1} (f(x) - \\tilde{f}(x))^2\\,dx \\\\\n&= \\int_{0}^{1} G'(x)\\,dx = G(1) - G(0) = 0,\n\\end{aligned}\n$$\nwhich proves the required relation.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21825,
"subject": "Mathematics (Olympiad)",
"question": "a) Find all positive integers $a$ for which\n\n$$\n\\frac{1}{4} < \\frac{1}{a+1} + \\frac{1}{a+2} + \\frac{1}{a+3} < \\frac{1}{3}.\n$$\n\nb) Prove that for any integer $p \\ge 2$ there exist $p$ consecutive positive integers $a_1, a_2, \\dots, a_p$ such that\n\n$$\n\\frac{1}{p+1} < \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_p} < \\frac{1}{p}.\n$$",
"options": [],
"answer": "See solution",
"solution": "a) Set $S = \\frac{1}{a+1} + \\frac{1}{a+2} + \\frac{1}{a+3}$. Notice that\n$$\n\\frac{3}{a+3} < S < \\frac{3}{a+1}\n$$\nso from $\\frac{1}{4} < S < \\frac{1}{3}$ we get $\\frac{1}{4} < \\frac{3}{a+1}$ and $\\frac{3}{a+3} < \\frac{1}{3}$. Thus, $6 < a < 11$, so $a \\in \\{7, 8, 9, 10\\}$. Checking, $a = 7$ fails, and $a = 8, 9, 10$ are solutions.\n\nb) Select $a_1 = p^2 + 1, a_2 = p^2 + 2, \\dots, a_p = p^2 + p$. Then\n$$\nS = \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_p} = \\frac{1}{p^2+1} + \\frac{1}{p^2+2} + \\dots + \\frac{1}{p^2+p}\n$$\nSince $\\frac{1}{p^2+1} > \\frac{1}{p^2+2} > \\dots > \\frac{1}{p^2+p}$,\n$$\n\\frac{p}{p^2+p} < S < \\frac{p}{p^2+1} < \\frac{p}{p^2} = \\frac{1}{p}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21826,
"subject": "Mathematics (Olympiad)",
"question": "在三角形 $ABC$ 中，$AB = AC \\neq BC$，點 $I$ 為其內心。直線 $BI$ 交 $AC$ 於點 $D$。已知過 $D$ 且與 $AC$ 垂直的直線交 $AI$ 於點 $E$。證明：$I$ 對直線 $AC$ 的反射點落在三角形 $BDE$ 的外接圓上。",
"options": [],
"answer": "See solution",
"solution": "令 $\\Gamma$ 為以 $E$ 為圓心、過 $B, C$ 兩點的圓。因為 $DE \\perp AC$，$C$ 對 $D$ 的反射點 $F$ 會落在 $\\Gamma$ 上。由 $\\angle DCI = \\angle ICB = \\angle CBI$，知直線 $DC$ 為三角形 $IBC$ 外接圓的切線。設 $J$ 為 $I$ 對 $D$ 的反射點。使用有向線段，知\n\n$$\nDC \\cdot DF = -DC^2 = -DI \\cdot DB = DJ \\cdot DB,\n$$\n\n得 $J$ 亦落在 $\\Gamma$ 上。\n\n\n\n令 $I'$ 為 $I$ 對 $AC$ 的反射點。因為 $IJ$ 與 $CF$ 互相平分，$CJFI$ 為平行四邊形。由 $\\angle FI'C = \\angle CIF = \\angle FJC$，得 $I'$ 落在 $\\Gamma$ 上。由此知 $EI' = EB$。\n\n注意到 $AC$ 是 $\\angle BDI'$ 的內角平分線。因為 $DE \\perp AC$，所以 $DE$ 是 $\\angle BDI'$ 的外角平分線。加上 $EI' = EB$，知 $E$ 落在三角形 $BDI'$ 的外接圓上。得證。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21827,
"subject": "Mathematics (Olympiad)",
"question": "Let $AM$ be a median in an acute triangle $ABC$. Its extension intersects the circumcircle $w$ of $ABC$ at $P$. Let $AH_1$ be an altitude of $\\triangle ABC$, and $H$ its orthocenter. The rays $MH$ and $PH_1$ intersect $w$ at $K$ and $T$, respectively. Prove that the circumcircle of $\\triangle AKTH_1$ is tangent to $BC$.",
"options": [],
"answer": "See solution",
"solution": "It suffices to show that $\\angle TKH_1 = \\angle TH_1B$. Let us extend $KH_1$ and let it intersect $w$ at $S$. Then\n\n$$\n\\angle TKS = \\angle TAB + \\angle BAS, \\quad \\angle TH_1B = \\angle TAB + \\angle PAC.\n$$\n\nSo it is sufficient to show that $\\angle PAC = \\angle BAS$.\n\nDenote by $A_1$ the point such that $AA_1$ is the diameter of $w$. Then $\\angle A_1CA = \\angle ABA_1 = 90^\\circ$, so $BH \\parallel A_1C$ and $CH \\parallel A_1B$. Thus, $BHCA_1$ is a parallelogram, so $HA_1$ passes through the point $M$.\n\nThen $K$ lies on $HA_1$, hence $\\angle A_1KA = 90^\\circ$. Thus, the quadrilateral $AKH_1M$ is inscribed. So $\\angle KH_1A = \\angle KMA$. Suppose $AH_1$ intersects $w$ again at $F$. Then\n\n$$\n\\angle KH_1A = \\angle KCA + \\angle FAS, \\quad \\angle KMA = \\angle KCA + \\angle PAA_1.\n$$\n\nHence, $\\angle FAS = \\angle PAA_1$. Also, $\\angle ABC = \\angle AA_1C$, so $\\angle BAF = 90^\\circ - \\angle ABC = \\angle A_1AC$. Then\n\n$$\n\\angle BAS = \\angle BAF + \\angle FAS = \\angle PAA_1 + \\angle A_1AC = \\angle PAC.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21828,
"subject": "Mathematics (Olympiad)",
"question": "On a switchboard there are $nm$ lamps arranged in an $n \\times m$ array. In the beginning all lamps are off. At each step one can switch three consecutive lamps in one row or in one column, changing the state of each lamp to the opposite. For which pairs of positive integers $(n, m)$ is it possible to achieve the situation where all the lamps are switched on?",
"options": [],
"answer": "See solution",
"solution": "If $n$ (or $m$) is a multiple of 3, then we can divide all lamps in each column (or row) into groups of 3 and switch the lamps on by the groups.\n\nIf neither $n$ nor $m$ is a multiple of 3, then color all lamps by diagonals with three colors (see below). Then each switching changes the state of exactly one lamp of each color, therefore each switching changes the parity of the number of lamps switched on for each color. Since in the beginning all lamps are off, the parity of the lamps switched on for each color always stays the same. Let $n = 3a + b$ and $m = 3c + d$, where $a$ and $c$ are nonnegative integers and $b$ and $d$ are either 1 or 2. In regions of sizes $3a \\times 3c$, $b \\times 3c$ and $3a \\times d$ the numbers of lamps of each color are equal and hence their parities are equal, but in the remaining $b \\times d$ region there is either 1 lamp, 2 lamps of different colors, or 4 lamps (2 of one color and 2 of the other two colors), hence the parities of the numbers of lamps of each color are not equal. Consequently, it is not possible to switch on all lamps.\n\n\n\n**Answer:** All pairs $(n, m)$, where either $n$ or $m$ is a multiple of 3.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21829,
"subject": "Mathematics (Olympiad)",
"question": "We are given a tetrahedron with five edges of length $2$ and one edge of length $1$. A point $P$, either in the interior or on the surface (but not outside), has distances $a$, $b$, $c$, and $d$ from the four faces of the tetrahedron. For which points $P$ is the value of $a + b + c + d$ minimal, and for which is it maximal?",
"options": [],
"answer": "See solution",
"solution": "The tetrahedron has two equilateral faces (sides of length $2$) and two isosceles faces (two sides of length $2$, one of length $1$). Let $F$ be the area of each equilateral face and $G$ the area of each isosceles face, so $F > G$. Let $a$ and $b$ be the distances from $P$ to the equilateral faces, and $c$ and $d$ the distances to the isosceles faces.\n\nIf $V$ is the volume of the tetrahedron, then:\n\n$$\n3V = F(a + b) + G(c + d) = F(a + b + c + d) - (F - G)(c + d) \\\\\n\\implies a + b + c + d = \\frac{3V + (F - G)(c + d)}{F}.\n$$\n\nSince $F - G > 0$, $a + b + c + d$ is minimal when $c + d = 0$, i.e., $c = d = 0$. This occurs for points $P$ on the common edge of the isosceles faces (the edge of length $1$).\n\nSimilarly,\n\n$$\n3V = F(a + b) + G(c + d) = G(a + b + c + d) + (F - G)(a + b) \\\\\n\\implies a + b + c + d = \\frac{3V - (F - G)(a + b)}{G}.\n$$\n\nAgain, since $F - G > 0$, $a + b + c + d$ is maximal when $a + b = 0$, i.e., $a = b = 0$. This occurs for points $P$ on the common edge of the equilateral faces.\n\n*QED*",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21830,
"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, depending on the starting number. What are all possible values of $n$?\n\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 possible reduction is by $2^k$ 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 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 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": 21831,
"subject": "Mathematics (Olympiad)",
"question": "Suppose there are initially 201 blue amoebas and 112 red amoebas. Is it possible, by a sequence of allowed moves, to reach:\n\n(a) 100 blue amoebas and 314 red amoebas?\n\n(b) 99 blue amoebas and 314 red amoebas?",
"options": [],
"answer": "See solution",
"solution": "Let $b$ be the number of blue amoebas and $r$ the number of red amoebas. The quantity $2b + r$ is invariant: whenever one blue amoeba appears/disappears, two red amoebas disappear/appear. Initially, $2 \\cdot 201 + 112 = 514$.\n\n(a) $2 \\cdot 100 + 314 = 514$, so it is possible. For example, if 50 pairs of blue amoebas turn into 200 red amoebas and a pair of one amoeba of each color turns into 3 red amoebas, we have $201 - 2 \\cdot 50 - 1 = 100$ blue amoebas and $112 + 200 - 1 + 3 = 314$ red amoebas.\n\n(b) $2 \\cdot 99 + 314 = 512$, which does not equal 514, so it is not possible.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21832,
"subject": "Mathematics (Olympiad)",
"question": "Find the largest possible value of the expression\n\n$$\n\\frac{a+b-c}{a^3+b^3+abc} + \\frac{b+c-a}{b^3+c^3+abc} + \\frac{c+a-b}{c^3+a^3+abc},\n$$\n\nwhere $a$, $b$, $c$ are positive real numbers such that $a + b + c \\ge \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}$.",
"options": [],
"answer": "See solution",
"solution": "Denote by $E(a, b, c)$ the given expression. First, we find the maximum value of $E(a, b, c)$ under the assumption that $a + b - c$, $b + c - a$, $c + a - b$ are non-negative.\n\nSince $a^2 + b^2 - ab \\ge ab$, it follows that $a^3 + b^3 = (a + b)(a^2 + b^2 - ab) \\ge ab(a + b)$, so $a^3 + b^3 + abc \\ge ab(a + b + c)$. Analogously, $b^3 + c^3 + abc \\ge bc(a + b + c)$ and $c^3 + a^3 + abc \\ge ca(a + b + c)$. Therefore:\n\n$$\nE(a, b, c) = \\sum \\frac{a+b-c}{a^3+b^3+abc} \\le \\sum \\frac{a+b-c}{ab(a+b+c)} = \\frac{1}{a+b+c} \\cdot \\sum \\frac{a+b-c}{ab}\n$$\n\nSince $\\sum \\frac{a+b-c}{ab} = \\frac{2(ab+bc+ca)-(a^2+b^2+c^2)}{abc} \\le \\frac{2(ab+bc+ca)-(ab+bc+ca)}{abc}$ (since $a^2 + b^2 + c^2 \\ge ab + bc + ca$), we infer that\n\n$$\nE(a, b, c) \\le \\frac{1}{a+b+c} \\cdot \\frac{ab+bc+ca}{abc} = \\frac{1}{a+b+c} \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) \\le 1.\n$$\n\nNow consider the case when at least one of $a + b - c$, $b + c - a$, $c + a - b$ is negative. Without loss of generality, suppose $a + b - c < 0$. Then $c > a + b > |a - b|$, so both $b + c - a$ and $c + a - b$ are positive. It follows that:\n\n$$\n\\begin{align*}\n\\sum \\frac{a+b-c}{a^3+b^3+abc} &\\le \\frac{b+c-a}{b^3+c^3+abc} + \\frac{c+a-b}{c^3+a^3+abc} \\\\\n&\\le \\frac{b+c-a}{bc(a+b+c)} + \\frac{c+a-b}{ca(a+b+c)} \\\\\n&= \\frac{1}{a+b+c} \\cdot \\frac{ac+bc-(a-b)^2}{abc} < \\frac{1}{a+b+c} \\cdot \\frac{ab+bc+ca}{abc} \\\\\n&= \\frac{1}{a+b+c} \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) \\le 1.\n\\end{align*}\n$$\n\nIn summary, the maximum value of $E(a, b, c)$ is $1$, and it is attained if and only if $a = b = c = 1$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21833,
"subject": "Mathematics (Olympiad)",
"question": "There are five piles made of 42, 70, 105, 462, and 2009 stones respectively. You are allowed to do the following:\n\n- Instead of two piles consisting of $a$ and $b$ stones, you can make:\n - two piles of $a$ and $(a+b)$ stones;\n - two piles of $a$ and $|a-b|$ stones.\n- You can also choose $k$ different piles, in which the total number of stones equals $nk$ where $n$ is some natural number, and create $k$ piles of $n$ stones in each one instead.\n\nCan we make five piles each of which will consist of\n\na) 2008 stones;\n\nb) 2009 stones\n\nin a finite number of steps?",
"options": [],
"answer": "See solution",
"solution": "a) The greatest common divisor (GCD) of 42, 70, 105, 462, and 2009 is 7. For every allowed operation, the number of stones in each pile remains divisible by 7. Since 2008 is not divisible by 7, the answer is **No**.\n\nb) In this case, the answer is **Yes**. One way to achieve this is:\n\n$$(42, 70, 105, 462, 2009) \\rightarrow (42, 35, 105, 462, 2009) \\rightarrow (7, 35, 105, 462, 2009)$$\n\nNow, with a pile of 7 stones, we can increase another pile by 7 stones 1061 times:\n\n$$(7, 35 + 7 \\times 1061, 105, 462, 2009)$$\n\nAfter that, we divide the five resulting piles into equal parts, where the total number of stones is 10045, and obtain five piles of 2009 stones each.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21834,
"subject": "Mathematics (Olympiad)",
"question": "Let $SH$ be an altitude in a tetrahedron $SABC$ and $H$ be inside the base $ABC$. A point $O$ on $SH$ is chosen so that $\\angle AOS + \\alpha = \\angle BOS + \\beta = \\angle COS + \\gamma = 180^\\circ$, where $\\alpha, \\beta, \\gamma$ are the dihedral angles corresponding to the edges $BC, AC, AB$ respectively. Let $A_1, B_1, C_1$ be the points of intersection of the following lines and planes: $A_1 = AO \\cap SBC$, $B_1 = BO \\cap SAC$, $C_1 = CO \\cap SBA$. If the planes $ABC$ and $A_1B_1C_1$ are parallel to each other, prove that $SA = SB = SC$.",
"options": [],
"answer": "See solution",
"solution": "Let $A_2$ be a projection of the point $H$ onto the edge $BC$. Then $\\angle AOS = 180^\\circ - \\angle HA_2S = 180^\\circ - \\alpha$, and so $\\angle AOH = \\alpha$. This implies that $\\triangle AOH \\sim \\triangle SA_2H \\Rightarrow \\frac{AH}{SH} = \\frac{OH}{HA_2} \\Rightarrow AH \\cdot HA_2 = OH \\cdot SH$. Analogously, $BH \\cdot HB_2 = CH \\cdot HC_2 = OH \\cdot SH$. \n\nWe will next prove that $H$ is the orthocenter of $\\triangle ABC$. Since $\\frac{AH}{HC_2} = \\frac{CH}{HA_2}$ we have that $\\sin \\angle HAC_2 = \\sin \\angle HCA_2$. In view of the fact that $\\angle HAC_2 + \\angle HCA_2 < 180^\\circ$ this implies that $\\angle HAB = \\angle HCB$, from which $\\angle AHC_2 = \\angle CHA_2$. Similarly we obtain that $\\angle AHB_2 = \\angle BHA_2$ and $\\angle BHC_2 = \\angle CHB_2$. From this it easily follows that $\\angle HAB = \\angle HCB = 90^\\circ - \\angle B$, which implies that $H$ belongs to the altitude of $\\triangle ABC$ starting at $A$. Analogously, $H$ belongs to the other two altitudes of $\\triangle ABC$. So, we see that the projection $H$ of the point $S$ onto the base of the tetrahedron is the orthocenter of the base, and the line segments $AA_2, BB_2$ and $CC_2$ are the altitudes of the base.\n\nConsider the $\\triangle ASA_2$. Since $\\triangle AOH \\sim \\triangle SA_2H$, we have that $AA_1 \\perp SA_2$.\n\n\n\nSo the lines $SA_2$ and $AA_2$ are both perpendicular to the line $BC$, which means that the whole plane $ASA_2$ is perpendicular to $BC$. This implies that $AO \\perp BC$. Similarly, in the plane $SBC$ we have two non-parallel lines $BC$ and $SA_2$ that are perpendicular to $AO$, and so $AO \\perp SBC$. Analogously, $BO \\perp SAC, CO \\perp SBA$.\n\nNow, since $ABC \\parallel A_1B_1C_1$, we have that $AB \\parallel A_1B_1$, and so the triangles $AOB$ and $A_1OB_1$ are similar. It follows that $\\frac{OA_1}{AO} = \\frac{OB_1}{BO} = t$. Also we have that $OA_1 = SO \\sin \\angle OSA_1$ and $OH = AO \\sin \\angle OAH = AO \\sin \\angle OSA_1$. This implies that $OA_1 \\cdot OA = SO \\cdot OH$. Analogously we can obtain that $SO \\cdot OH = OA \\cdot OA_1 = OB \\cdot OB_1 = OC \\cdot OC_1 \\Rightarrow SO \\cdot OH = t \\cdot OA^2 = t \\cdot OB^2 = t \\cdot OC^2$.\n\nIt follows that $OA = OB = OC \\Rightarrow AH = BH = CH$ and finally that $SA = SB = SC$.\n\n",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21835,
"subject": "Mathematics (Olympiad)",
"question": "The sequence of positive integers $a_1, a_2, \\dots, a_{2025}$ is defined as follows:\n\n- $a_1 = 2^{2024} + 1$\n- For each $n = 1, 2, \\dots, 2024$, define $a_{n+1}$ to be the largest prime divisor of $a_n^2 - 1$.\n\nDetermine the value of $a_{2024} + a_{2025}$.",
"options": [],
"answer": "See solution",
"solution": "We show that $\\{a_{2024}, a_{2025}\\} = \\{2, 3\\}$ and thus the answer is $5$.\n\nFirst, observe that if $a_k = 2$ for some $k$, then from that point on the sequence values alternate between $2$ and $3$. We will show that $a_{2024}$ is equal to either $2$ or $3$.\n\nIf for some $n \\le 2024$, $a_n$ is even and thus equal to $2$, then we are done. Thus, we may suppose $a_n$ is odd (and greater than $3$) for $n \\le 2024$.\n\nAs $a_n$ is odd, $a_n^2 - 1 = (a_n - 1)(a_n + 1) = 2 \\times 2 \\times \\frac{a_n - 1}{2} \\times \\frac{a_n + 1}{2}$, and $a_{n+1}$ is the maximum of $2$, the largest prime divisor of $\\frac{a_n - 1}{2}$, and the largest prime divisor of $\\frac{a_n + 1}{2}$. In particular,\n\n$$\na_{n+1} \\le \\frac{a_n + 1}{2}.\n$$\n\nBy repeated use of the above inequality, we obtain $a_2 \\le 2^{2023} + 1$, $a_3 \\le 2^{2022} + 1$, and so on until we reach $a_{2024} \\le 2^1 + 1 = 3$.\n\nHence, the value $2$ or $3$ has been reached at some point in the sequence no later than $a_{2024}$, forcing the sum $a_{2024} + a_{2025}$ to equal $5$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21836,
"subject": "Mathematics (Olympiad)",
"question": "A chooses 13 different 3-digit numbers. Then B selects several of these 13 and tries to obtain, by using each selected number once and the operations $+, -, \\times$, an expression with value strictly between 3 and 4. B wins if he succeeds in doing so; otherwise A wins. Who has a winning strategy?",
"options": [],
"answer": "See solution",
"solution": "Player B has a winning strategy. One way to see this is to divide the 3-digit numbers into 8 groups with the following property: For every two numbers $a, b$ from the same group, $a > b$, one has $\\frac{a}{b} < \\frac{4}{3}$.\n\n![alt](path \"title\")\n\n$$\n\\begin{align*}\nG_1 &= \\{100, \\dots, 133\\}, \\\\\nG_2 &= \\{134, \\dots, 178\\}, \\\\\nG_3 &= \\{179, \\dots, 238\\}, \\\\\nG_4 &= \\{239, \\dots, 318\\}, \\\\\nG_5 &= \\{319, \\dots, 425\\}, \\\\\nG_6 &= \\{426, \\dots, 567\\}, \\\\\nG_7 &= \\{568, \\dots, 757\\}, \\\\\nG_8 &= \\{758, \\dots, 999\\}.\n\\end{align*}\n$$\n\nFor justification, note that the ratio of the last and first number in a group is less than $\\frac{4}{3}$, for instance, $\\frac{567}{426} < \\frac{4}{3}$.\n\nSince there are 13 different numbers chosen by A and $13 > 8$, some two of them are in the same group $G_i$, $1 < i < 8$. Let them be $a$ and $b$ with $a > b$; then $1 < \\frac{a}{b} < \\frac{4}{3}$. Among the remaining 11 numbers, B can find another two with the same property, say $c$ and $d$ with $1 < \\frac{c}{d} < \\frac{4}{3}$, because $11 > 8$. Finally, B can select one more analogous pair $e, f$, satisfying $1 < \\frac{e}{f} < \\frac{4}{3}$, since there are still $9 > 8$ numbers left. Now adding up gives $3 = 1+1+1 < \\frac{a}{b} + \\frac{c}{d} + \\frac{e}{f} < 3 \\cdot \\frac{4}{3} = 4$, and B's task is complete.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21837,
"subject": "Mathematics (Olympiad)",
"question": "Represent the fraction $\\frac{1}{2022}$ as a difference of two regular fractions with smaller denominators.",
"options": [],
"answer": "See solution",
"solution": "One possible representation is:\n\n$$\n\\frac{1}{2022} = \\frac{1}{2 \\cdot 3 \\cdot 337} = \\frac{3-2}{2 \\cdot 3 \\cdot 337} = \\frac{3}{2 \\cdot 3 \\cdot 337} - \\frac{2}{2 \\cdot 3 \\cdot 337} = \\frac{1}{2 \\cdot 337} - \\frac{1}{3 \\cdot 337} = \\frac{1}{674} - \\frac{1}{1011}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21838,
"subject": "Mathematics (Olympiad)",
"question": "Find all prime numbers $p$, $q$, and $r$ such that\n\n$$\n15p + 7pq + qr = pqr.\n$$",
"options": [],
"answer": "See solution",
"solution": "The identities $qr = pqr - 15p - 7pq = p(qr - 15 - 7q)$ imply that $qr$ is divisible by $p$. Since $p$, $q$, and $r$ are prime numbers, there are only two possibilities: $p = q$ or $p = r$.\n\nIf $p = q$, we get $15 + 7q + r = qr$, which can be rewritten as\n\n$$\n22 = qr - 7q - r + 7 = (q - 1)(r - 7).\n$$\n\nExactly one of the numbers $q-1$ and $r-7$ is odd, so one of the primes $q$ and $r$ is even. The only possible case is $q = 2$, and from this we get $r = 29$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21839,
"subject": "Mathematics (Olympiad)",
"question": "Find all real numbers $x$, $y$, and $z$ which are solutions of the system of equations:\n\n$$\nx + y - 2z = 0, \\quad xy - z^2 = 0, \\quad y^2 + 5z + 6 = 0.\n$$",
"options": [],
"answer": "See solution",
"solution": "From the first equation, we get $x = -2z - y$. Inserting this into the second equation, we get $(-2z - y)y - z^2 = 0$, or equivalently, $-(z + y)^2 = 0$. This implies that $z = -y$.\n\nSubstituting $z = -y$ into the third equation gives:\n$$\ny^2 + 5(-y) + 6 = 0 \\implies y^2 - 5y + 6 = 0\n$$\nwhich factors as $(y - 2)(y - 3) = 0$. Thus, $y = 2$ or $y = 3$.\n\nFor $y = 2$, $z = -2$, and $x = -2(-2) - 2 = 4 - 2 = 2$.\nFor $y = 3$, $z = -3$, and $x = -2(-3) - 3 = 6 - 3 = 3$.\n\nTherefore, the solutions are:\n- $x = 2$, $y = 2$, $z = -2$\n- $x = 3$, $y = 3$, $z = -3$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21840,
"subject": "Mathematics (Olympiad)",
"question": "Suppose we have a blackboard with pairs of numbers, each pair being either $(a_j, a_j)$ for some $j$ or $(-a_k, -a_\\ell)$ for $1 \\leq k < \\ell \\leq 8$. For each $i = 1, \\dots, n$, we can choose to underline $a_i$ with probability $\\phi$ or $-a_i$ with probability $1-\\phi$, independently. A strategy $s$ is a choice of underlining for each $i$, and its weight is $w(s) = \\prod_{i=1}^n s_i$. For each pair $p$ and strategy $s$, define a cost coefficient $c(p, s)$: if $s$ scores a point on $p$, then $c(p, s) = w(s)$; otherwise, $c(p, s) = 0$. Let $c(p) = \\sum_s c(p, s)$. Prove that there always exists a strategy that scores at least 43 points, but it is not always possible to score 44 points.",
"options": [],
"answer": "See solution",
"solution": "First, the sum of weights of all $2^n$ strategies is $\\sum_s w(s) = \\prod_{i=1}^n [\\phi + (1-\\phi)] = 1$.\n\nFor a fixed pair $p = (x, y)$, we consider two cases:\n\n(a) If $x = y = a_j$, every strategy that erases $a_j$ scores a point on this pair. Thus, $c(p) = \\phi \\prod_{i \\neq j} [\\phi + (1-\\phi)] = \\phi$.\n\n(b) If $x \\neq y$, we have:\n\n$$\nc(p) = \\begin{cases} \\phi^2 + \\phi(1-\\phi) + (1-\\phi)\\phi = 3\\phi - 1, & (x, y) = (a_k, a_\\ell); \\\\ \\phi(1-\\phi) + (1-\\phi)\\phi + (1-\\phi)^2 = \\phi, & (x, y) = (-a_k, -a_\\ell); \\\\ \\phi^2 + \\phi(1-\\phi) + (1-\\phi)^2 = 2 - 2\\phi, & (x, y) = (\\pm a_k, \\mp a_\\ell). \\end{cases}\n$$\n\nSince $\\phi \\approx 0.618$ and $\\frac{1}{2} < \\phi < \\frac{2}{3}$, we have $c(p) \\geq \\phi$ in all cases.\n\nLet $C = \\sum_{p,s} c(p,s) = \\sum_p c(p)$. There are 68 pairs, so $C \\geq 68\\phi > 42$.\n\nIf every strategy scores at most 42 points, then $C \\leq 42 \\sum_s w(s) = 42$, contradicting $C > 42$. Therefore, some strategy scores at least 43 points.\n\nTo show 44 is not always possible, consider a blackboard with 40 pairs $(m, m)$ (five copies for each $m = 1, \\dots, 8$) and 28 pairs $(-m, -n)$ for $1 \\leq m < n \\leq 8$. If $k$ numbers are underlined, the maximum points is $5k + 28 - \\binom{k}{2}$. This quadratic is maximized at $k = 5$ or $6$, giving at most 43 points. Thus, 44 points cannot always be achieved.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21841,
"subject": "Mathematics (Olympiad)",
"question": "Given that $0 < x, y < 1$, determine, with proof, the maximum value of\n$$\n\\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)}.\n$$",
"options": [],
"answer": "See solution",
"solution": "When $x = y = \\frac{1}{3}$, the value of the expression is $\\frac{1}{8}$.\n\nWe will prove that\n$$\n\\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)} \\leq \\frac{1}{8}\n$$\nfor any $0 < x, y < 1$ as follows.\n\nIf $x + y \\geq 1$, then\n$$\n\\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)} \\leq 0 < \\frac{1}{8}.\n$$\n\nIf $x + y < 1$, let $1 - x - y = z > 0$. It follows from the AM-GM inequality that\n$$\n\\begin{aligned}\n\\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)} &= \\frac{xyz}{(x+y)(y+z)(z+x)} \\\\\n&\\leq \\frac{xyz}{2\\sqrt{xy} \\cdot 2\\sqrt{yz} \\cdot 2\\sqrt{zx}} = \\frac{1}{8}.\n\\end{aligned}\n$$\n\nIn conclusion, the maximum value of the expression is $\\frac{1}{8}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21842,
"subject": "Mathematics (Olympiad)",
"question": "Let $x_1, x_2, \\ldots, x_n$ be complex numbers. Show that\n$$\n\\sum_{i,j=1}^{n} |x_i + x_j|^2 - \\sum_{i,j=1}^{n} |x_i - x_j|^2 \\geq 0,\n$$\nand determine when equality holds.",
"options": [],
"answer": "See solution",
"solution": "Note that $|a+b|^2 - |a-b|^2 = 4 \\operatorname{Re}(\\bar{a}b)$ for any complex numbers $a, b$. Hence,\n\n$$\n\\begin{align*}\n\\sum_{i,j=1}^{n} |x_i + x_j|^2 - \\sum_{i,j=1}^{n} |x_i - x_j|^2 &= 4 \\sum_{i,j=1}^{n} \\operatorname{Re} (\\bar{x}_i x_j) = 4 \\operatorname{Re} \\sum_{i,j=1}^{n} \\bar{x}_i x_j \\\\\n&= 4 \\operatorname{Re} \\sum_{i=1}^{n} \\bar{x}_i \\sum_{j=1}^{n} x_j = 4 \\left| \\sum_{i=1}^{n} x_i \\right|^2 \\geq 0,\n\\end{align*}\n$$\n\nand equality holds iff $\\sum_{i=1}^{n} x_i = 0$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21843,
"subject": "Mathematics (Olympiad)",
"question": "Two dissections of a square into three rectangles are considered essentially different if one cannot be switched to the other by simple rearrangement of the pieces.\n\nHow many essentially different dissections of the $2010 \\times 2010$ square into three rectangles with integer side lengths exist such that the area of one rectangle is equal to the arithmetic mean of the areas of the other two?",
"options": [],
"answer": "See solution",
"solution": "There are two distinct possibilities for dissections to consider:\n\n1. **Three strips:** All rectangles have one side of length $2010$. The areas are $2010$ times their widths. The middle strip's width must be the arithmetic mean of the other two. Since $2010 \\div 3 = 670$, the middle strip's width is $670$, and the smallest strip's width can be any integer from $1$ to $670$. Thus, there are $670$ possible dissections in this case.\n\n2. **One strip, two rectangles from a perpendicular cut:**\n - Let the areas be $A$, $B$, and $C$ with $B = \\frac{A+C}{2}$ and $A \\leq C$.\n - If the strip has area $B$, its dimensions are $670 \\times 2010$. The other two together have dimensions $1340 \\times 2010$, and the common edge is $1340$. The shorter edge of the rectangle with area $A$ can be any integer from $1$ to $1005$ ($2010 \\div 2$), giving $1005$ possible dissections.\n - If the strip has area $A$ or $C$, there must exist an integer $b < 2010$ such that $B = (2010 - a) \\cdot b = 670 \\cdot 2010$. Since $67^2$ divides $670 \\cdot 2010$ and $67^2 > 2010$, both $a$ and $b$ must be divisible by $67$. Let $a = 67c$ and $b = 67d$. Substituting, $(30 - c) d = 300$ with $d < 30$ and $d > 10$. The divisors of $300$ between $10$ and $30$ are $12, 15, 20, 25$, yielding $c = 5, 10, 15, 18$, and thus $a = 335, 670, 1005, 1206$. The value $a = 670$ was already counted, so only the other three are new.\n\nIn total, the number of essentially different dissections is:\n\n$$670 + 1005 + 3 = 1678$$\n\n*QED*",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21844,
"subject": "Mathematics (Olympiad)",
"question": "Let $P$ be a point inside an equilateral triangle. The lengths of the perpendiculars from $P$ to the three sides are $1$, $2$, and $3$. What is the length of one edge of the triangle?",
"options": [],
"answer": "See solution",
"solution": "Let $A$, $B$, and $C$ be the vertices of the triangle, and let $a$ be the length of one edge. The area of the triangle is\n\n$$\n\\triangle ABC = \\frac{\\sqrt{3}}{4} a^2.\n$$\n\nAlternatively, the area can be expressed as the sum of the areas of the three smaller triangles formed by $P$ and the sides:\n\n$$\n\\triangle ABC = \\triangle ABP + \\triangle BCP + \\triangle CAP = \\frac{1}{2} (1 + 2 + 3) a = 3a.\n$$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21845,
"subject": "Mathematics (Olympiad)",
"question": "令 $n$ 為一正整數。求最小的正整數 $k$，滿足存在一種將 $2n \\times 2n$ 白色棋盤上的 $k$ 格塗黑的方法，使得只有唯一一種用 $1 \\times 2$ 和 $2 \\times 1$ 骨牌覆蓋棋盤的方法，且：\n\n1. 所有骨牌都貼齊棋盤格線且不超出棋盤；\n2. 任兩個骨牌都不重疊；\n3. 任一骨牌最多只蓋到 $1$ 格黑色格子。",
"options": [],
"answer": "See solution",
"solution": "1. 構造 $k=2n$ 的情形：將棋盤的格子以 $(i, j)$ 標記，並將所有 $\\{(i, j) : j = i \\text{ 或 } j = i + 1,\\ i \\le n\\}$ 塗黑，如下圖：\n\n\n\n證明只有一種骨牌排列方式：考慮棋盤上由兩條對角線切割出的四個區域。\n\n- 對於所有 $i \\le n$，覆蓋 $(i, i)$ 的骨牌必須同時覆蓋 $(i, i+1)$，即直放，這會迫使左區域的骨牌都直放（可用歸納法證明）。\n- 同理，覆蓋 $(i, i+1)$ 的骨牌必須橫放，並迫使上方區域的骨牌都橫放。\n- 左側骨牌直放將迫使下側骨牌都橫放。\n- 最後，右側骨牌都被迫直放。換言之，只有一種放法。\n\n2. 證明 $k=2n$ 為最小值。假設 $k<2n$，棋盤上有 $k$ 格被塗黑，且存在一種覆蓋方法 $P$，則要證明存在另一種覆蓋方法 $P'$。設 $D=\\{(i,i):1\\le i\\le 2n\\}$ 為主對角線。\n\n(a) 建構一個圖，點為棋盤格子，邊分紅、藍兩色：\n- 兩格被同一骨牌覆蓋，連紅線；\n- 兩格對主對角線對稱且被同一骨牌覆蓋，連藍線。\n\n注意：\n- 兩點可能同時有紅、藍線。\n- 每點紅、藍線度數皆為 1。\n- 整張圖可拆為若干互不相交的 cycle，每個 cycle 由紅、藍線交替組成，允許長度為 2。\n\n(b) 考慮 $d\\in D$。$d$ 不可能對同一點同時連紅藍線，必屬於長度至少為 4 的 cycle，記為 $C(d)$。\n\n(c) 設 $C(d)$ 點為 $c_0, c_1, \\ldots, c_m$，$c_0=d$。令 $m$ 為最小正整數使 $c_m\\in D$，$c_m\\neq d$。由構造，連接 $c_0,\\ldots,c_m$ 的 path 對 $D$ 鏡射後也在原圖中，故 $C(d)$ 上恰有兩點屬於 $D$。\n\n總結，$D$ 中 $2n$ 點分屬 $n$ 個 cycle $C_1,\\ldots,C_n$，每個 cycle 長度至少 4。\n\n(d) 由鴿籠原理，必有一個 $C_i$ 僅至多一格被塗黑（否則黑格數 $\\ge 2n > k$，矛盾）。可將 $C_i$ 上紅線對應骨牌移除，依藍線放入骨牌，得新覆蓋 $P'$。因 $|C_i|\\ge 4$，$P'$ 與 $P$ 不同，且 $C_i$ 僅至多一格黑格，所有骨牌仍最多蓋到一格黑格。\n\n故若 $k<2n$，覆蓋方法不唯一。證畢。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21846,
"subject": "Mathematics (Olympiad)",
"question": "Let $AX$ meet $\\Gamma$ again at $P$, and let $AY$ meet $\\Gamma$ again at $Q$. If $PQ \\parallel l$, the figure is symmetric with respect to $AB$, and so $(AXY)$ and $(AX'Y')$ are tangent at $A$. In the following, we only consider the configuration as shown.\n\nLet $PQ$ meet $l$ at $C$. Prove that the circles $(AXY)$, $(AX'Y')$, and $\\Gamma$ are coaxial.",
"options": [],
"answer": "See solution",
"solution": "Firstly, since\n\n$$\n\\angle AQP = \\angle ABP = 90^\\circ - \\angle PAB = \\angle YXP,\n$$\n\nthe points $Q, P, X, Y$ are concyclic. Similarly, $P, Q, X', Y'$ are concyclic.\n\nNow, let $PQ$ meet $l$ at $C$. Then we have\n\n$$\nCX \\times CY = CP \\times CQ = CX' \\times CY'.\n$$\n\nHence, $C$ has the same power with respect to $(AXY)$, $(AX'Y')$, and $\\Gamma$. As all three circles pass through $A$, the line $AC$ is the common radical axis of these circles. Thus, the circles are coaxial as desired.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21847,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $p, m, n$ such that $p^m = n^3 + 8$ and $p$ is a prime number.",
"options": [],
"answer": "See solution",
"solution": "By moving $n^3$, we get a sum of cubes on the right-hand side:\n\n$$\np^m = n^3 + 8 = (n+2)(n^2 - 2n + 4).\n$$\n\nSince $p$ is prime, each of the factors on the right-hand side must be a power of $p$:\n\n$$\nn + 2 = p^{\\alpha}, \\quad n^2 - 2n + 4 = p^{\\beta},\n$$\n\nwhere $\\alpha$ and $\\beta$ are positive integers.\n\nNote that $n^2 - 2n + 4 \\ge n + 2$, since that is equivalent to $n^2 - 3n + 2 \\ge 0$, i.e., $(n-1)(n-2) \\ge 0$, which holds for $n \\ge 2$ or $n \\le 1$. Therefore, $\\beta \\ge \\alpha$.\n\nWe can conclude that $p^\\alpha$ divides both $n+2$ and $n^2 - 2n + 4$, so it also divides\n\n$$\nn \\cdot (n+2) - (n^2 - 2n + 4) = 4n - 4,\n$$\n\nand then it also divides $4 \\cdot (n+2) - (4n - 4) = 12$, so $p=2$ or $p=3$.\n\nIf $p=2$, then $n+2 = 2^\\alpha$ is at most 4, and since $n > 0$, it follows that $n=2$, which gives the solution $(2, 4, 2)$.\n\nIf $p=3$, then $n+2 = 3^\\alpha$ is 3, so $n=1$, which gives the second solution $(3, 2, 1)$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21848,
"subject": "Mathematics (Olympiad)",
"question": "Let $P_1, P_2, \\dots, P_{1000}$ be the vertices of a convex polygon, and let $Q$ be a point strictly inside the polygon. For each $i = 1, 2, \\dots, 1000$, let $X_i$ be the second intersection point (other than $Q$) of the ray $\\overrightarrow{P_iQ}$ with the polygon. Prove that there exists at least one edge of the polygon that does not contain any of the points $X_i$.\n\n",
"options": [],
"answer": "See solution",
"solution": "The diagonal $\\ell = P_1P_{501}$ splits the polygon into two halves. Since $Q$ does not lie on $\\ell$ (otherwise $X_1 = P_{501}$ which is not permitted), it lies strictly inside one of these halves. Without loss of generality, assume that $Q$ lies inside $P_1P_2\\dots P_{501}$.\n\nOrient the original polygon so that $\\ell$ is horizontal and so that $P_2, \\dots, P_{500}$ and $Q$ lie below $\\ell$. Note that 500 edges of the polygon lie on each side of $\\ell$.\n\n\n\nEach of the rays $\\overrightarrow{P_{501}Q}, \\overrightarrow{P_{502}Q}, \\dots, \\overrightarrow{P_{1000}Q}, \\overrightarrow{P_1Q}$ intersects $\\ell$. Therefore, they intersect the polygon for a second time below $\\ell$. So the 501 points $X_{501}, X_{502}, \\dots, X_{1000}, X_1$ all lie below $\\ell$. Hence, at most 499 of the $X_i$ lie above $\\ell$. But 500 edges of the polygon lie above $\\ell$. So at least one of these edges does not contain an $X_i$, as required. $\\square$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21849,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c$ be non-negative real numbers. Denote\n\n$$\nS = a + 2b + 3c, \\quad T = a + b^2 + c^3.\n$$\n\n1. Find the minimum of $T - S$.\n\n2. If $S = 4$, find the maximum of $T$.",
"options": [],
"answer": "See solution",
"solution": "1. Note that $b, c \\ge 0$. By the inequality of arithmetic and geometric means, we get\n\n$$\nb^2 + 1 \\ge 2b, \\quad c^3 + 1 + 1 \\ge 3\\sqrt[3]{c^3 \\cdot 1 \\cdot 1} = 3c.\n$$\n\nThus,\n\n$$\nT - S = (b^2 - 2b) + (c^3 - 3c) \\ge -1 - 2 = -3.\n$$\n\nWhen $a$ is any non-negative real number and $b = c = 1$, $T - S$ takes the minimum $-3$.\n\n2. Since $a, b, c \\ge 0$ and $S = a + 2b + 3c = 4$, we know that $0 \\le b \\le 2$, $0 \\le c \\le \\frac{4}{3}$. Therefore,\n\n$$\n2b - b^2 = b(2 - b) \\ge 0, \\quad 3c - c^3 = c(3 - c^2) \\ge 0.\n$$\n\nThus,\n\n$$\nT = a + b^2 + c^3 \\le a + 2b + 3c = S = 4.\n$$\n\nWhen $a = 4, b = c = 0$ (or $a = 0, b = 2, c = 0$), $T$ takes the maximum $4$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21850,
"subject": "Mathematics (Olympiad)",
"question": "Доведіть, що:\n\n$$\n\\frac{a^2}{b} \\ge 2a - b, \\quad \\frac{b^3}{c^2} \\ge 3b - 2c, \\quad \\frac{c^4}{a^3} \\ge 4c - 3a.\n$$\n\n*Підказка:* Кожна з цих нерівностей випливає з невід'ємності відповідного квадрату або добутку квадрату на додатний вираз.",
"options": [],
"answer": "See solution",
"solution": "Доведемо кожну нерівність окремо:\n\n$$\n\\frac{a^2}{b} \\ge 2a - b \\Leftrightarrow a^2 \\ge 2ab - b^2 \\Leftrightarrow (a-b)^2 \\ge 0.\n$$\n\n$$\n\\frac{b^3}{c^2} \\ge 3b - 2c \\Leftrightarrow b^3 \\ge 3bc^2 - 2c^3 \\Leftrightarrow (b-c)^2 (b+2c) \\ge 0.\n$$\n\n$$\n\\frac{c^4}{a^3} \\ge 4c - 3a \\Leftrightarrow c^4 \\ge 4a^3c - 3a^4 \\Leftrightarrow (c-a)^2 (c^2 + 2ca + 3a^2) \\ge 0.\n$$\n\nОскільки всі ці вирази невід'ємні, доводжувана нерівність є очевидним наслідком цих трьох нерівностей.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21851,
"subject": "Mathematics (Olympiad)",
"question": "Розглянемо вектори $\\overrightarrow{MA} = \\vec{a}$, $\\overrightarrow{MB} = \\vec{b}$, $\\overrightarrow{MC} = \\vec{c}$, $\\overrightarrow{MS} = \\vec{s}$, $\\overrightarrow{SA} = \\vec{a} - \\vec{s}$, $\\overrightarrow{SB} = \\vec{b} - \\vec{s}$ і $\\overrightarrow{SC} = \\vec{c} - \\vec{s}$. За умовою задачі, $|\\vec{a}| > 1$, $|\\vec{b}| > 1$ і $|\\vec{c}| > 1$. Нам потрібно довести нерівність\n\n$$\n|\\vec{a} - \\vec{s}| + |\\vec{b} - \\vec{s}| + |\\vec{c} - \\vec{s}| > 3.\n$$\n",
"options": [],
"answer": "See solution",
"solution": "Розглянемо тепер випадок $|\\vec{s}| < 1$. Тоді\n\n$$\n\\begin{aligned}\n& |\\vec{a} - \\vec{s}| + |\\vec{b} - \\vec{s}| + |\\vec{c} - \\vec{s}| \\ge \\frac{(\\vec{a} - \\vec{s}) \\cdot \\vec{a}}{|\\vec{a}|} + \\frac{(\\vec{b} - \\vec{s}) \\cdot \\vec{b}}{|\\vec{b}|} + \\frac{(\\vec{c} - \\vec{s}) \\cdot \\vec{c}}{|\\vec{c}|} = \\\\\n&= |\\vec{a}| + |\\vec{b}| + |\\vec{c}| - \\vec{s} \\left( \\frac{\\vec{a}}{|\\vec{a}|} + \\frac{\\vec{b}}{|\\vec{b}|} + \\frac{\\vec{c}}{|\\vec{c}|} \\right) = \\\\\n&= 3 + \\left( |\\vec{a}| - 1 \\right) \\left( 1 - \\frac{\\vec{s} \\cdot \\vec{a}}{|\\vec{a}|} \\right) + \\left( |\\vec{b}| - 1 \\right) \\left( 1 - \\frac{\\vec{s} \\cdot \\vec{b}}{|\\vec{b}|} \\right) + \\left( |\\vec{c}| - 1 \\right) \\left( 1 - \\frac{\\vec{s} \\cdot \\vec{c}}{|\\vec{c}|} \\right) > 3.\n\\end{aligned}\n$$\n\nДля останньої оцінки ми врахували, що $1 > |\\vec{s}| \\ge \\frac{\\vec{s} \\cdot \\vec{a}}{|\\vec{a}|}$, $1 > |\\vec{s}| \\ge \\frac{\\vec{s} \\cdot \\vec{b}}{|\\vec{b}|}$, $1 > |\\vec{s}| \\ge \\frac{\\vec{s} \\cdot \\vec{c}}{|\\vec{c}|}$.\n\n*Розв'язання 2.* Якщо застосувати ортогональне проектування точки $S$ на площину $ABC$, то стає очевидним, що твердження задачі достатньо довести для точки $S$ цієї площини, для якої сума відстаней $SA + SB + SC$ є мінімальною. Добре відомо, що коли всі кути трикутника $ABC$ менші за $120^\\\text{o}$, то потрібною точкою є точка Торічеллі (така точка $T$ всередині трикутника $ABC$, що $\\angle ATB = \\angle ATC = \\angle BTC = 120^\\\text{o}$), а якщо, наприклад, $\\angle BAC \\ge 120^\\\text{o}$, то — вершина $A$. У першому випадку позначимо $x = TA$, $y = TB$, $z = TC$ (без обмеження загальності вважаємо, що $x \\le y \\le z$), і з використанням формули для довжини медіани та теореми косинусів матимемо:\n\n$$\n\\begin{aligned}\nMA^2 &= \\frac{1}{9}(2AC^2 + 2AB^2 - BC^2) = \\\\\n&= \\frac{1}{9}(2x^2 + 2z^2 + 2xz + 2x^2 + 2y^2 + 2xy - y^2 - z^2 - yz) = \\\\\n&= \\frac{1}{9}(4x^2 + y^2 - z^2 + 2xz + 2xy - yz) = \\\\\n&= \\frac{1}{9}((x + y + z)^2 + 3x^2 - 2z^2 - 3yz) \\le \\frac{1}{9}(x + y + z)^2.\n\\end{aligned}\n$$\n\nЗвідси дістанемо потрібну нерівність $x + y + z > 3$. У другому випадку, з урахуванням нерівності $\\cos \\angle BAC \\le -\\frac{1}{2}$, одержуємо:\n\n$$\nMA^2 = \\frac{1}{9}(2AC^2 + 2AB^2 - BC^2) \\le \\frac{1}{9}(AC^2 + AB^2 + AB \\cdot AC) \\le \\frac{1}{9}(AB + AC)^2, \\\\\n\\text{тобто } AB + AC > 3.\n$$\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21852,
"subject": "Mathematics (Olympiad)",
"question": "Let $p$ be an odd prime. Show that for every integer $c$, there exists an integer $a$ such that\n\n$$\na^{\\frac{p+1}{2}} + (a+c)^{\\frac{p+1}{2}} \\equiv c \\pmod{p}.\n$$",
"options": [],
"answer": "See solution",
"solution": "If $p \\mid c$, then we can choose $a = 0$.\n\nNext, consider the case where $p$ does not divide $c$.\n\nLet $b$ be an integer coprime to $p$. Define $f(b) = b^{\\frac{p+1}{2}} \\pmod{p}$. By Fermat's little theorem, $b^{p-1} \\equiv 1 \\pmod{p}$, so $b^{p+1} \\equiv b^2 \\pmod{p}$. Thus, $(f(b))^2 \\equiv b^2 \\pmod{p}$, and the equation $x^2 \\equiv b^2 \\pmod{p}$ has at most two solutions modulo $p$: $b$ and $-b$. Therefore, $f(b) \\equiv \\pm b \\pmod{p}$.\n\nAssume, for contradiction, that there is no required integer $a$.\n\n1. If $f(c) \\equiv c$, then $f(0) + f(c) \\equiv c$, so $f(c) \\equiv -c$.\n2. If $f(2c) \\equiv 2c$, then $f(c) + f(2c) \\equiv c$, so $f(2c) \\equiv -2c$.\n\nSince the sets $\\{1, 2, \\dots, p-1\\}$ and $\\{c, 2c, \\dots, (p-1)c\\}$ are equal modulo $p$, by induction, for every integer $b$ coprime to $p$, $f(b) \\equiv -b \\pmod{p}$. Thus, $f(p-c) \\equiv -p + c \\equiv c \\pmod{p}$. Now $f(p-c) + f(p) \\equiv c \\pmod{p}$, which is a contradiction. Therefore, for every integer $c$, there exists an integer $a$ such that $a^{\\frac{p+1}{2}} + (a+c)^{\\frac{p+1}{2}} \\equiv c \\pmod{p}$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21853,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be an acute triangle and $P$ be a point inside the triangle such that $\\triangle APB = \\triangle BPC = \\triangle CPA$. Denote with $S$ the area and with $\\alpha, \\beta, \\gamma$ the angles of $\\triangle ABC$. Prove that\n\n$$\n\\frac{1}{\\sin \\alpha} + \\frac{1}{\\sin \\beta} + \\frac{1}{\\sin \\gamma} \\ge \\frac{PA^2 + PB^2 + PC^2}{2S} + \\frac{4}{\\sqrt{3}}\n$$\n\nWhen does the equality occur?",
"options": [],
"answer": "See solution",
"solution": "The inequality can be rewritten as\n\n$$\n2S \\left( \\frac{1}{\\sin \\alpha} + \\frac{1}{\\sin \\beta} + \\frac{1}{\\sin \\gamma} - \\frac{4}{\\sqrt{3}} \\right) \\ge PA^2 + PB^2 + PC^2.\n$$\n\nNote that $AB \\cdot AC \\cdot \\sin \\alpha = AB \\cdot BC \\cdot \\sin \\beta = AC \\cdot BC \\cdot \\sin \\gamma = 2S$ and\n\n$$\n2S = 2(S_{PAB} + S_{PBC} + S_{PCA}) = (PA \\cdot PB + PB \\cdot PC + PC \\cdot PA) \\frac{\\sqrt{3}}{2}.\n$$\n\nThe inequality can be further rewritten as\n\n$$\nAB \\cdot AC + AC \\cdot BC + BC \\cdot AB - 2(PA \\cdot PB + PB \\cdot PC + PC \\cdot PA) \\ge PA^2 + PB^2 + PC^2\n$$\n\nthat is equivalent to\n\n$$\nAB \\cdot AC + AC \\cdot BC + BC \\cdot AB \\ge (PA + PB + PC)^2\n$$\n\nConsider the points $X$ and $Y$ such that $\\triangle XAB$ and $\\triangle YAC$ are equilateral ($X$ and $C$ lie on different halfplanes with respect to $AB$, similarly $Y$ and $B$ with respect to $AC$).\n\n\n\n$\\triangle AXB + \\triangle APB = 180^\\circ = \\triangle AYC + \\triangle APC \\Rightarrow XAPB$ and $YAPC$ are cyclic quadrilaterals. Also note that $\\triangle BPA + \\triangle APY = 120^\\circ + \\triangle ACY = 120^\\circ + 60^\\circ = 180^\\circ$ hence $B, P$ and $Y$ are collinear. Similarly, points $C, P$ and $X$ are collinear.\n\nFrom Ptolemy's Theorem in cyclic quadrilateral $XAPB$ we have that $PA \\cdot XB + PB \\cdot XA = PX \\cdot AB$, but since $\\triangle XAB$ is equilateral, then $XA = XB = AB$ and we have that $PA + PB = PX$. From here, $CX = PX + PC = PA + PB + PC$. Similarly, $BY = PA + PB + PC$.\n\nNow, we apply Ptolemy's Inequality in quadrilateral $XBCY$ and get that $XB \\cdot YC + XY \\cdot BC \\ge CX \\cdot BY$. From Triangle Inequality we have that $AX + AY \\ge XY$ so $XB \\cdot YC + (AX + AY)BC \\ge CX \\cdot BY$. Rewriting the inequality based on the above relations we have that $AB \\cdot AC + AB \\cdot BC + AC \\cdot BC \\ge (PA + PB + PC)^2$.\n\nThe equality occurs if and only if both equality cases of Ptolemy's Inequality and Triangle's Inequality occur. The equality case of Triangle's Inequality occurs when $X, A$ and $Y$ are collinear $\\iff 180^\\circ = \\triangle XAB + \\triangle BAC + \\triangle CAY = 60^\\circ + \\triangle BAC + 60^\\circ \\Rightarrow \\triangle BAC = 60^\\circ$. The Ptolemy's Inequality equality case occurs if and only if $XBCY$ is cyclic $\\iff 180^\\circ = \\triangle BXY + \\triangle BCY = 60^\\circ + \\triangle BCA + \\triangle ACY = 60^\\circ + \\triangle BCA + 60^\\circ \\Rightarrow \\triangle BCA = 60^\\circ$. So the equality case happens if and only if $\\triangle ABC$ is equilateral. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21854,
"subject": "Mathematics (Olympiad)",
"question": "Let $w = \\frac{\\sqrt{3} + i}{2}$ and $z = \\frac{-1 + i\\sqrt{3}}{2}$, where $i = \\sqrt{-1}$. Find the number of ordered pairs $(r, s)$ of positive integers not exceeding $100$ that satisfy the equation\n\n$$\ni \\cdot w^r = z^s.$$",
"options": [],
"answer": "See solution",
"solution": "Note that $w = \\cos \\frac{\\pi}{6} + i \\sin \\frac{\\pi}{6}$ is a primitive twelfth root of unity, so $w^3 = i$ and $w^r = w^{r + 12m}$ for all integers $m$ and $r$. Furthermore, $z = \\cos \\frac{2\\pi}{3} + i \\sin \\frac{2\\pi}{3}$ is a primitive cube root of unity, so $w^4 = z$ and\n\n$$\ni \\cdot w^r = w^{r + 3} = z^s = w^{4s}.$$\n\nHence, the given equation requires that $r + 3 \\equiv 4s \\pmod{12}$. In particular, $r + 3 \\equiv 0, 4, \\text{ or } 8 \\pmod{12}$.\n\n- If $r + 3 \\equiv 0 \\pmod{12}$, then $r = 9 + 12m$ and $s = 3n$, where $0 \\leq m \\leq 7$ and $1 \\leq n \\leq 33$, accounting for $8 \\times 33 = 264$ ordered pairs.\n- If $r + 3 \\equiv 4 \\pmod{12}$, then $r = 1 + 12m$ and $s = 1 + 3n$, where $0 \\leq m \\leq 8$ and $0 \\leq n \\leq 33$, accounting for $9 \\times 34 = 306$ ordered pairs.\n- If $r + 3 \\equiv 8 \\pmod{12}$, then $r = 5 + 12m$ and $s = 2 + 3n$, where $0 \\leq m \\leq 7$ and $0 \\leq n \\leq 32$, accounting for $8 \\times 33 = 264$ ordered pairs.\n\nThe requested number of ordered pairs $(r, s)$ is therefore $264 + 306 + 264 = 834$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21855,
"subject": "Mathematics (Olympiad)",
"question": "For each $m \\in \\mathbb{N}$, let $f(m)$ be the product of the primes that appear with odd exponent in the prime factorization of $m$.\n\nGiven two positive integers $a$ and $b$, the product $ab$ is a perfect square if and only if $f(a) = f(b)$.\n\nFor each $k \\in \\mathbb{N}$, let $S$ be the set of all $m$ in $\\{1, 2, \\dots, 4n\\}$ such that $f(m) = k$. Note that if $k$ is not squarefree, then $S$ is empty. Also, as $f(m) \\leq m$ for all $m$, each element of $\\{1, 2, \\dots, 4n\\}$ belongs to exactly one of the sets $S_1, S_2, \\dots, S_{4n}$.\n\nWhat is the maximum number of pairs $(a, b)$ with $a, b \\in \\{1, 2, \\dots, 4n\\}$, $a \\neq b$, such that $ab$ is a perfect square and each element is used in at most one pair?",
"options": [],
"answer": "See solution",
"solution": "Let each set $S_j$ have $a_j$ elements. To maximize the number of pairs $(a, b)$ with $ab$ a perfect square, we must pair elements within the same $S_j$ (since $f(a) = f(b)$ is required).\n\nThus, the maximum number of pairs is:\n$$\n\\left\\lfloor \\frac{a_1}{2} \\right\\rfloor + \\left\\lfloor \\frac{a_2}{2} \\right\\rfloor + \\dots + \\left\\lfloor \\frac{a_{4n}}{2} \\right\\rfloor.\n$$\n\nWe claim that $\\left\\lfloor \\frac{a_j}{2} \\right\\rfloor$ equals the number of multiples of $4$ in $S_j$. If $j$ is squarefree, the elements of $S_j$ are $j \\cdot s^2$ for $s$ in a suitable range. Since $j$ has at most one factor of $2$, $j \\cdot s^2$ is a multiple of $4$ if and only if $s$ is even. Thus, the number of multiples of $4$ in $S_j$ is $\\left\\lfloor \\frac{a_j}{2} \\right\\rfloor$.\n\nSumming over all $j$, the total number of such pairs is the number of multiples of $4$ in $\\{1, 2, \\dots, 4n\\}$, which is $n$.\n\n**Answer:** $n$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21856,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\{a_n\\}_{n \\ge 1}$ be a sequence defined by $a_1 = 45$ and $a_n = a_{n-1}^2 + 15a_{n-1}$ for $n > 1$. Prove that the sequence contains no perfect squares.",
"options": [],
"answer": "See solution",
"solution": "Suppose, for contradiction, that there exists a positive integer $n$ such that $a_n$ is a perfect square.\n\n$$\na_n = a_{n-1}^2 + 15a_{n-1} = a_{n-1}(a_{n-1} + 15)\n$$\n\nThus, $a_{n-1}$ divides $a_n$ for all $n$. In particular, $a_1 = 45$ divides $a_n$ for all $n$.\n\nLet $k > 1$ be the smallest integer such that $a_k$ is a perfect square, and write $a_{k-1} = 45x$ for some positive integer $x$. Then:\n\n$$\na_k = 45x(45x + 15) = 15^2 \\cdot 3x(3x + 1)\n$$\n\nIf $a_k$ is a perfect square, then $3x(3x + 1)$ must be a perfect square. Since $3x$ and $3x + 1$ are coprime, both must be perfect squares, say $3x = m^2$ and $3x + 1 = n^2$ for integers $m, n$. But $n^2 - m^2 = 1$, which only has the solution $m = 0$, $n = 1$, contradicting $a_n > 0$ for all $n$. Therefore, the sequence contains no perfect squares. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21857,
"subject": "Mathematics (Olympiad)",
"question": "Determine the largest value the expression\n$$\n\\sum_{1 \\le i < j \\le 4} (x_i + x_j) \\sqrt{x_i x_j}\n$$\nmay achieve, as $x_1, x_2, x_3, x_4$ run through the non-negative real numbers that add up to $1$. Determine also the $x_i$ at which the maximum is achieved.",
"options": [],
"answer": "See solution",
"solution": "The required maximum is $\\frac{3}{4}$ and is achieved if and only if the $x_i$ are all equal to $\\frac{1}{4}$. To prove this, use the binomial expansion of $(\\sqrt{x_i} - \\sqrt{x_j})^4$ to write\n\n$$\n4(x_i + x_j)\\sqrt{x_i x_j} = x_i^2 + 6x_i x_j + x_j^2 - (\\sqrt{x_i} - \\sqrt{x_j})^4,\n$$\n\nthen sum over $1 \\le i < j \\le 4$ and refer to the constraint $x_1 + x_2 + x_3 + x_4 = 1$, to get\n\n$$\n\\begin{aligned}\n4 \\sum_{1 \\le i < j \\le 4} (x_i + x_j) \\sqrt{x_i x_j} &= 3(x_1 + x_2 + x_3 + x_4)^2 - \\sum_{1 \\le i < j \\le 4} (\\sqrt{x_i} - \\sqrt{x_j})^4 \\\\\n&\\le 3(x_1 + x_2 + x_3 + x_4)^2 = 3;\n\\end{aligned}\n$$\n\nClearly, equality holds if and only if the $x_i$ are all equal, and the constraint forces them all equal to $\\frac{1}{4}$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21858,
"subject": "Mathematics (Olympiad)",
"question": "A segment $S$ of length $50$ is covered by several segments of length $1$, all of them contained in $S$. If any of these unit segments is removed, $S$ is not completely covered any more. Find the maximum number of unit segments with this property. Assume that the segments include their endpoints.",
"options": [],
"answer": "See solution",
"solution": "Label the unit segments $S_1, S_2, S_3, \\ldots$ in the order they appear on $S$ from left to right. Suppose that $S_k$ and $S_{k+2}$ have a common point for some $k$. Then their union is a longer segment that contains $S_{k+1}$. So the latter can be removed and $S$ will still be completely covered, contrary to the hypothesis. It follows that $S_k$ and $S_{k+2}$ have no points in common (not even one). In particular, the odd-indexed segments $S_1, S_3, S_5, \\ldots$ are disjoint, with segments of positive length separating $S_{2i-1}$ and $S_{2i+1}$ for each $i$. There can be at most $49$ such unit segments on a segment of length $50$. This implies that there are at most $98$ segments in our covering system. Indeed, if there were at least $99$ of them, then at least $50$ would be odd-indexed, which is impossible.\n\nConsider the interval $I = [0, 50]$ on the numerical line. Mark on it the terms of the arithmetic progression with first term $\\frac{1}{2}$, last term $\\frac{99}{2}$, and length $98$; its common difference is $d = \\frac{49}{97} > \\frac{1}{2}$.\n\nPlace $98$ unit segments $S_1, S_2, \\ldots, S_{98}$ on $I$ so that their midpoints coincide with the terms of the progression. It is clear that they\n\n\n\ncover $I$ completely; also $S_1$ and $S_{98}$ are the only segments containing $0$ and $50$ respectively. Consider $S_k$ and $S_{k+2}$ where $1 \\leq k \\leq 96$. Their midpoints are at distance $2d = \\frac{98}{97} > 1$, hence they are separated by a gap of length $\\frac{1}{97}$. The gap is covered only by segment $S_{k+1}$. It follows that no segment $S_k$ with $2 \\leq k \\leq 97$ can be removed without spoiling the covering. By the same reason, neither is it possible to remove $S_1$ or $S_{98}$. So $S_1, S_2, \\ldots, S_{98}$ is a covering system with the desired properties. It has a maximum number of segments, equal to $98$, which is the answer to our question.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21859,
"subject": "Mathematics (Olympiad)",
"question": "Malle drew a rhombus $ABCD$ and chose points $E$ and $F$ on sides $AB$ and $BC$, respectively, such that the triangle $DEF$ is equilateral. Malle was very surprised when she discovered that there is another possibility to choose points $E$ and $F$ on sides $AB$ and $BC$, respectively, such that $DEF$ is equilateral. What can be the measures of the angles of this rhombus?\n\nWhat can be the measures of the angles of this rhombus?",
"options": [],
"answer": "See solution",
"solution": "The angles of the rhombus can be $60^\\circ$ and $120^\\circ$.\n\n---\n\nThere is clearly only one way to choose points $E$ and $F$ on sides $AB$ and $BC$, respectively, such that $E$ and $F$ would be symmetrical with respect to the diagonal $BD$ and $\\angle EDF = 60^\\circ$. Thus it is possible in Malle's rhombus to choose $E$ and $F$ asymmetrically with respect to diagonal $BD$ such that triangle $DEF$ is equilateral; let us consider one of the possible setups. W.l.o.g., we can assume that $|EB| < |BF|$ (otherwise we can switch the roles of $A$ and $C$ and the roles of $E$ and $F$). Let $\\alpha = \\angle BAD = \\angle BCD$ and $\\beta = \\angle AED$. Also let $E'$ be the point symmetrical to $E$ with respect to diagonal $BD$ (see fig. 17); then $|DE'| = |DE| = |DF|$ gives that triangle $E'DF$ is isosceles and $\\angle BFD = \\angle FE'D = \\angle AED = \\beta$. Therefore\n\n$$\n\\angle CDF = 180^\\circ - \\angle DCF - \\angle DFC = 180^\\circ - \\alpha - (180^\\circ - \\beta) = \\beta - \\alpha.\n$$\n\nNow\n\n$$\n\\begin{aligned}\n180^\\circ - \\alpha &= \\angle ADC = \\angle ADE + \\angle EDF + \\angle FDC \\\\\n&= (180^\\circ - \\alpha - \\beta) + 60^\\circ + (\\beta - \\alpha) = 240^\\circ - 2\\alpha.\n\\end{aligned}\n$$\n\nFrom here $\\alpha = 60^\\circ$, that is the angles of the rhombus are $60^\\circ$ and $120^\\circ$.\n\n\n\n---\n\nLet the first choices of Malle be $E_1$ and $F_1$ and second ones $E_2$ and $F_2$ (see fig. 18). As $\\angle E_1DF_1 = 60^\\circ = \\angle E_2DF_2$ and $|DE_1| = |DF_1|$ and $|DE_2| = |DF_2|$, the rotation of the plane by $60^\\circ$ around point $D$ that takes $E_1$ to $F_1$ must also take $E_2$ to $F_2$. The same rotation must then take the line $E_1E_2$ to $F_1F_2$ or line $AB$ to line $BC$. Therefore $\\angle ABC = 120^\\circ$, thus the angles of the rhombus are $60^\\circ$ and $120^\\circ$.\n\n_Remark_: In a rhombus with angles $60^\\circ$ and $120^\\circ$ there are actually infinitely many possibilities to draw an equilateral triangle with one vertex in the vertex of the rhombus and the other two on the sides.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21860,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge 2$ be a fixed integer, and $x_1, x_2, \\dots, x_n$ be any real numbers. Find the maximum value of\n$$\n2 \\sum_{1 \\le i < j \\le n} [x_i x_j] - (n-1) \\sum_{i=1}^n [x_i^2],\n$$\nwhere $[x]$ represents the largest integer less than or equal to $x$.",
"options": [],
"answer": "See solution",
"solution": "Notice that\n$$\n2 \\sum_{1 \\le i < j \\le n} [x_i x_j] - (n-1) \\sum_{i=1}^{n} [x_i^2] = \\sum_{1 \\le i < j \\le n} (2[x_i x_j] - [x_i^2] - [x_j^2]).\n$$\nSince\n$$\n2[x_i x_j] \\le 2x_i x_j \\le x_i^2 + x_j^2 < [x_i^2] + [x_j^2] + 2,\n$$\nwhere the leftmost and rightmost quantities are both integers, it follows that\n$$\n2[x_i x_j] \\le [x_i^2] + [x_j^2] + 1,\n$$\nand equality is possible only when $[x_i^2]$ and $[x_j^2]$ have different parities (since $2[x_i x_j]$ is even).\n\nAs a result, when $[x_i^2]$ and $[x_j^2]$ have different parities,\n$$\n2[x_i x_j] - [x_i^2] - [x_j^2] \\le 1;\n$$\nwhen $[x_i^2]$ and $[x_j^2]$ have the same parity,\n$$\n2[x_i x_j] - [x_i^2] - [x_j^2] \\le 0.\n$$\nSuppose, among $[x_1^2], [x_2^2], \\dots, [x_n^2]$, there are $k$ odd numbers and $n-k$ even numbers. Then\n$$\n\\sum_{1 \\le i < j \\le n} (2[x_i x_j] - [x_i^2] - [x_j^2]) \\le k(n-k) \\le \\left\\lfloor \\frac{n^2}{4} \\right\\rfloor.\n$$\nOn the other hand, let $m = \\left\\lfloor \\frac{n}{2} \\right\\rfloor$ and\n$$\nx_1 = x_2 = \\cdots = x_m = 1.4, \\quad x_{m+1} = x_{m+2} = \\cdots = x_n = 1.5.\n$$\nFor $1 \\le i < j \\le m$ or $m+1 \\le i < j \\le n$,\n$$\n2[x_i x_j] - [x_i^2] - [x_j^2] = 0;\n$$\nwhile for $1 \\le i \\le m$ and $m+1 \\le j \\le n$, $2[x_i x_j] - [x_i^2] - [x_j^2] = 1$. Together, they give\n$$\n\\sum_{1 \\le i < j \\le n} (2[x_i x_j] - [x_i^2] - [x_j^2]) = m(n-m) = \\left\\lfloor \\frac{n^2}{4} \\right\\rfloor.\n$$\nTherefore, the desired maximum value is $\\left\\lfloor \\frac{n^2}{4} \\right\\rfloor$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21861,
"subject": "Mathematics (Olympiad)",
"question": "For each integer $i$, define $k_i \\in \\{0, 1, \\dots, b-1\\}$ and $\\ell_i \\in \\{0, 1, \\dots, a-1\\}$ such that\n\n$$\nn - ic = k_i a \\pmod{b}, \\quad n - ic = \\ell_i b \\pmod{a}.\n$$\n\nLet $a, b, c$ be pairwise coprime. Let $f(n)$ be defined by\n\n$$\nf(n) = \\sum_{i=0}^{\\lfloor \\frac{n}{c} \\rfloor} g(n - ic)\n$$\nwhere $g(x)$ is a function such that\n$$\ng(x) = \\frac{x - k_i a + \\ell_i b}{ab} + 1\n$$\nfor $x = n - ic$.\n\nShow that for all $n$, there exist constants $\\alpha, \\beta, \\gamma$ (depending on $a, b, c$) such that\n$$\n|f(n) - (\\alpha n^2 + \\beta n + \\gamma)| < \\frac{a+b+c}{12}\n$$\nfor all $n$.",
"options": [],
"answer": "See solution",
"solution": "*Case 1:* $a = b = 1$ and $c > 1$. An investigation on the convexity and the axis of symmetry yields\n\n$$\nA + B = c - 1, \\quad A - B \\geq -\\frac{(c-2)^2}{4},\n$$\n\nand hence $B \\leq \\frac{c^2}{8}$. In addition, the left-hand sides of (1) and (2) are constantly 0, thereby making no contribution to the error. It follows that\n\n$$\n|f(n) - (\\alpha n^2 + \\beta n + \\gamma)| \\leq \\frac{c^2}{8 \\cdot 2abc} = \\frac{c}{16} < \\frac{a+b+c}{12}\n$$\n\nwhich is (5).\n\n*Case 2:* $a = 1$, $b > 1$, and $c \\geq a + b$. Again, by considering the convexity and the axis of symmetry, we find\n\n$$\nA + B = b(c - 1), \\quad A - B \\geq -\\frac{(c - 1 - b)^2}{4},\n$$\n\nand $B \\leq \\frac{(b+c-1)^2}{8}$. The left-hand side of (1) is constantly 0, contributing zero error. Therefore,\n\n$$\n\\begin{aligned}\n|f(n) - (\\alpha n^2 + \\beta n + \\gamma)| &\\le \\frac{b}{8} + \\frac{(b+c-1)^2}{8 \\cdot 2abc} \\\\\n&= \\frac{b}{8} + \\frac{b^2 + 2bc + c^2 - 2b - 2c + 1}{16bc} \\\\\n&= \\frac{b}{8} + \\frac{3bc + c^2 - 2b - 2c + 1 - b(c-b)}{16bc} \\\\\n&\\le \\frac{b}{8} + \\frac{3bc + c^2 - 2b - 2c + 1 - (c-1)}{16bc} \\\\\n&< \\frac{b}{8} + \\frac{2bc + c^2 - 3c}{16bc} \\\\\n&= \\frac{b}{8} + \\frac{3}{16} + \\frac{c-3}{16b} \\\\\n&\\le \\frac{b}{8} + \\frac{3}{16} + \\frac{c-3}{32} \\\\\n&= \\frac{a+b+c}{12} - \\frac{5a-4b-1}{96} \\\\\n&< \\frac{a+b+c}{12},\n\\end{aligned}\n$$\n\nand (5) is verified.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21862,
"subject": "Mathematics (Olympiad)",
"question": "We start with $n$ ones. In each step, one is allowed to replace two numbers $a, b$ with the number $\\frac{a+b}{4}$. Show that the final number (left over after $n-1$ steps) is no less than $\\frac{1}{n}$.\n\nNow, consider starting with arbitrary $n$ numbers $x_1, \\dots, x_n$, but change $4$ to $3$ (i.e., $a, b \\rightarrow \\frac{a+b}{3}$). What is the set of possible final numbers? It turns out to be the set of numbers of the form $\\sum_{i=1}^{n} \\frac{x_i}{3^i}$, where $\\sum_{i=1}^{n} \\frac{1}{2^i} = 1$.\n\nIf we start with $1, 2, \\dots, n$, can we obtain $1$ as the final number? For $n = 1, 2, 5$ we can, but not for $n = 3, 4$. For which $n$ is this possible?",
"options": [],
"answer": "See solution",
"solution": "Key ideas:\n\n1. If both $a$ and $2a$ are present, we can erase $2a$ by the operation $a, 2a \\rightarrow a$.\n2. If $a, b$ and $\\frac{a+b}{6}$ are present, we can erase both $a$ and $b$ ($a, b, \\frac{a+b}{6} \\rightarrow \\frac{a+b}{3}$, $\\frac{a+b}{6} \\rightarrow \\frac{a+b}{6}$).\n\nUsing these, we can eliminate numbers $6k-i, \\dots, 6k+i$ provided $1, 2, \\dots, 6k-i-1$ are present and $i < 3k$.\n\nThis reduces the problem to an inductive step from $n$ to $n+12$. The involvement of moduli $3$ and $4$ makes smaller steps unlikely to work. The same idea applies for other divisors, but the inductive basis becomes more complicated.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21863,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a cyclic quadrilateral with circumcenter $\\omega$, and let $E$ be the intersection of the diagonals $AC$ and $BD$. A line passing through $E$ intersects lines $AB$ and $BC$ at $P$ and $Q$, respectively. Let $R$ ($R \\ne D$) be the intersection point of $\\omega$ and a circle that passes through $D$ and $E$ and is tangent to the line $PQ$ at $E$. Prove that $B$, $P$, $Q$, $R$ are cyclic.",
"options": [],
"answer": "See solution",
"solution": "Let $\\omega_1$ be the circle that passes through $D$ and $E$ and is tangent to the line $PQ$ at $E$. Since $PQ$ is tangent to the circle $\\omega_1$ at $E$, we have $\\angle EDR = \\angle QER$. So it follows from $\\angle EDR = \\angle BAR$ that $\\angle PAR = \\angle PER$, i.e., $P$, $A$, $E$, $R$ are cyclic. Thus, since $\\angle RBC = \\angle RAE$, we get $\\angle RPQ = \\angle RBQ$, and so $P$, $B$, $Q$, $R$ are cyclic.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21864,
"subject": "Mathematics (Olympiad)",
"question": "Дробката $\\frac{59}{143}$ да се претстави како збир на две прави нескратливи дробки.",
"options": [],
"answer": "See solution",
"solution": "Бројот $143$ можеме да го запишеме како $143 = 11 \\cdot 13$. Според тоа, дробката $\\frac{59}{143}$ може да се претстави во облик:\n\n$$\n\\frac{59}{143} = \\frac{59}{11 \\cdot 13} = \\frac{x}{11} + \\frac{y}{13}.\n$$\n\nАко десната страна на последното равенство ја сведеме на најмал заеднички именител, добиваме:\n\n$$\n\\frac{13x + 11y}{143} = \\frac{59}{143}.\n$$\n\nДве дробки со еднаков именител се еднакви ако и само ако им се еднакви броителите. Значи, треба да важи:\n\n$$\n13x + 11y = 59.\n$$\n\nСо директно пребарување се добива дека единствено решение на последната равенка е $x = 2$, $y = 3$. Значи,\n\n$$\n\\frac{59}{143} = \\frac{2}{11} + \\frac{3}{13}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21865,
"subject": "Mathematics (Olympiad)",
"question": "Given a triangle $ABC$, let $D$ be the point where the incircle of triangle $ABC$ touches the side $BC$. A circle through the vertices $B$ and $C$ is tangent at point $E$ to the incircle of triangle $ABC$. Show that the line $DE$ passes through the excentre of triangle $ABC$ corresponding to vertex $A$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $I$ be the incentre of triangle $ABC$, and let $I_A$ be the excentre corresponding to vertex $A$. Notice that the vertices $B$ and $C$ both lie on the circle of diameter $II_A$. The line $DI_A$ meets this circle again at point $K$, and the lines $BC$ and $IK$ meet at point $L$ (unless $AB = AC$, in which case the conclusion is obvious).\n\nNotice that the line $BC$ is the radical axis of the circles $BEC$ and $BIC$, so $LB \\cdot LC = LI \\cdot LK$. On the other hand, $LI \\cdot LK = LD^2$, since $K$ is the perpendicular foot dropped from the right-angled vertex $D$ of triangle $DIL$. Consequently, point $L$ is the radical centre of the following three circles: the incircle of triangle $ABC$, the circle $BEC$, and the circle $BIC$. Since the common tangent at $E$ of the first two circles is their radical axis, it must pass through $L$. It follows that $E$ is the reflection of $D$ across the line $IL$, so the lines $DE$ and $IL$ are perpendicular, and we are done.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21866,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$, $d$ be positive real numbers. Prove that\n\n$$\n\\sum_{cyclic} \\frac{a - \\sqrt[3]{bcd}}{a + 3(b + c + d)} \\geq 0\n$$",
"options": [],
"answer": "See solution",
"solution": "By the AM-GM inequality, $\\sqrt[3]{bcd} \\leq \\frac{b + c + d}{3}$, so\n\n$$\n\\frac{a - \\sqrt[3]{bcd}}{a + 3(b + c + d)} \\geq \\frac{a - \\frac{b + c + d}{3}}{a + 3(b + c + d)} = \\frac{3a - (b + c + d)}{3(a + 3(b + c + d))}\n$$\n\nSumming cyclically and simplifying, we get\n\n$$\n\\sum_{cyc} \\frac{a - \\sqrt[3]{bcd}}{a + 3(b + c + d)} \\geq \\frac{1}{3} \\sum_{cyc} \\frac{3a - (b + c + d)}{a + 3(b + c + d)}\n$$\n\nLet $S = a + b + c + d$. Then,\n\n$$\n\\sum_{cyc} \\frac{a}{a + 3(b + c + d)} = \\sum_{cyc} \\frac{a}{3S - 2a} = \\sum_{cyc} \\frac{a^2}{3aS - 2a^2}\n$$\n\nBy Bergström's inequality,\n\n$$\n\\sum_{cyc} \\frac{a^2}{x} \\geq \\frac{\\left(\\sum_{cyc} a\\right)^2}{\\sum_{cyc} x}\n$$\n\nSo,\n\n$$\n\\sum_{cyc} \\frac{a^2}{3aS - 2a^2} \\geq \\frac{(a + b + c + d)^2}{3S^2 - 2\\sum_{cyc} a^2}\n$$\n\nWe need to show\n\n$$\n\\frac{(a + b + c + d)^2}{3S^2 - 2\\sum_{cyc} a^2} \\geq \\frac{2}{5}\n$$\n\nSince $\\sum_{cyc} a^2 \\geq \\frac{1}{4}(a + b + c + d)^2$, it follows that\n\n$$\n5(a + b + c + d)^2 \\geq 6S^2 - 4\\sum_{cyc} a^2\n$$\n\nwhich reduces to $3(a + b + c + d)^2 \\geq 0$, always true. Thus, the original inequality holds.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21867,
"subject": "Mathematics (Olympiad)",
"question": "Find all injective functions $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ such that\n\n$$\nf(f(n)) \\leq \\frac{n + f(n)}{2}, \\quad \\forall n \\in \\mathbb{N}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $f$ be such a function. Then\n\n$$\nf(f(n)) \\leq \\max\\{n, f(n)\\}, \\quad \\forall n \\in \\mathbb{N}. \\tag{*}\n$$\n\nDefine $f^k(a)$ as $f$ applied $k$ times to $a$. Suppose there exists $a$ such that $a > f(a)$. Then $f^2(a) < a$, and by induction, $f^k(a) < a$ for all $k > 0$:\n\n$$\nf^k(a) < a, \\quad \\forall k > 0. \\tag{1}\n$$\n\nSince the set $\\{1, 2, \\ldots, a-1\\}$ is finite, there exist $i < j$ such that $f^i(a) = f^j(a)$. Because $f$ is injective, this implies $f^{j-i}(a) = a$, contradicting (1). Therefore,\n\n$$\na \\leq f(a), \\quad \\forall a > 0. \\tag{2}\n$$\n\nAlso, $f(a) \\leq f(f(a))$ for all $a \\in \\mathbb{N}$.\n\nFrom (*) and (2), $f(f(a)) \\leq \\max\\{a, f(a)\\} = f(a)$, so $f(f(a)) = f(a)$.\n\nThus, $f(a) = a$ for all $a \\in \\mathbb{N}$. Therefore, the only solution is $f(n) = n$ for all $n \\in \\mathbb{N}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21868,
"subject": "Mathematics (Olympiad)",
"question": "There are $n$ rays $OA_1, OA_2, \\ldots, OA_n$ starting at point $O$ and enumerated counterclockwise on a plane so that $\\angle A_1OA_n < 180^\\circ$. Determine the smallest $n$ such that there exists a pair of $60^\\circ$ angles, a pair of $45^\\circ$ angles, and a pair of $30^\\circ$ angles among the angles $\\angle A_iOA_j$, $1 \\leq i < j \\leq n$.\n\n\n\nFig. 1",
"options": [],
"answer": "See solution",
"solution": "Firstly, we will show that $n = 5$ satisfies the conditions (see Fig. 1).\n\nSuppose there exists a configuration that satisfies the conditions with 4 rays (see Fig. 2). Then these rays form 6 different angles. Thus, if the conditions were satisfied, all the pairs of angles would be among these 6 angles. But if the greatest angle $\\angle A_1OA_4$ equals $60^\\circ$, then another $60^\\circ$ angle will not exist. If $\\angle A_1OA_4 < 60^\\circ$, then there are no $60^\\circ$ angles. If $\\angle A_1OA_4 > 60^\\circ$, then there won't be enough angles to satisfy all the conditions.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21869,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute triangle with circumcircle $\\Gamma$. Let $A_1$, $B_1$, and $C_1$ be respectively the midpoints of the arcs $BAC$, $CBA$, and $ACB$ of $\\Gamma$. Show that the inradius of triangle $A_1B_1C_1$ is not less than the inradius of triangle $ABC$.",
"options": [],
"answer": "See solution",
"solution": "Rotate the triangle around the centre of $\\Gamma$ by $180^\\circ$. Then $A_1$, $B_1$, $C_1$ respectively go to $A_2$, $B_2$, $C_2$, which are the midpoints of the minor arcs $BC$, $CA$, and $AB$. The triangles $A_1B_1C_1$ and $A_2B_2C_2$ are congruent and thus have the same inradii. Note that $A_2$ is the intersection of the angle bisector of $\\angle BAC$ with $\\Gamma$, and similarly for the other points.\n\n\n\nObserve that $\\angle C_2A_2A = \\angle C_2CA = \\angle C/2$ and $\\angle B_2A_2A = \\angle B_2BA = \\angle B/2$. Hence $\\angle A_2 = (\\angle B/2) + (\\angle C/2) = (\\pi - \\angle A)/2$. Similarly, we get other angles:\n\n$$\n\\angle B_2 = (\\pi - \\angle B)/2, \\quad \\angle C_2 = (\\pi - \\angle C)/2.\n$$\n\nFor any triangle $ABC$, its inradius is given by\n\n$$\nr = 4R \\sin(A/2) \\sin(B/2) \\sin(C/2),\n$$\n\nwhere $R$ is its circumradius. If $r$ and $r_1$ are the inradii of $\\triangle ABC$ and $\\triangle A_2B_2C_2$, then\n\n$$\n\\frac{r}{r_1} = \\frac{\\sin(A/2) \\sin(B/2) \\sin(C/2)}{\\sin((\\pi - A)/4) \\sin((\\pi - B)/4) \\sin((\\pi - C)/4)}.\n$$\n\nUsing the identity:\n\n$$\n1 + 4 \\prod \\sin \\left( \\frac{\\pi - A}{4} \\right) = \\sum \\sin \\frac{A}{2},\n$$\n\nwe obtain\n\n$$\n\\frac{r}{r_1} = \\frac{4 \\prod \\sin(A/2)}{(\\sum \\sin(A/2)) - 1}.\n$$\n\nWe need to show $r \\le r_1$, or equivalently,\n\n$$\n1 + 4 \\prod \\sin(A/2) \\le \\sum \\sin(A/2).\n$$\n\nBut\n\n$$\n\\sum \\sin(A/2) \\ge 3 \\left( \\prod \\sin(A/2) \\right)^{1/3}.\n$$\n\nSo it suffices to prove\n\n$$\n1 + 4 \\prod \\sin(A/2) \\le 3 \\left( \\prod \\sin(A/2) \\right)^{1/3}.\n$$\n\nLet $x = \\left( \\prod \\sin(A/2) \\right)^{1/3}$, then the inequality becomes\n\n$$\n4x^3 - 3x + 1 \\ge 0.\n$$\n\nBut $4x^3 - 3x + 1 = (2x - 1)^2(x + 1)$, which is non-negative for all positive $x$. This completes the proof.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21870,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\geq 3$ be an integer such that $4n + 1$ is a prime number. Prove that $4n + 1$ divides $n^{2n} - 1$.",
"options": [],
"answer": "See solution",
"solution": "Since $p = 4n + 1$ is a prime number, each nonzero remainder modulo $p$ has a unique multiplicative inverse. Note that $-4n \\equiv 1 \\pmod{p}$, so $n \\equiv (-4)^{-1} \\pmod{p}$. Thus, $n \\equiv -(2^{-1})^2 \\pmod{p}$. Consequently,\n\n$$\nn^{2n} - 1 \\equiv (-(2^{-1})^2)^{2n} - 1 \\equiv (2^{-1})^{4n} - 1 \\equiv 0 \\pmod{p},$$\n\nby Fermat's Little Theorem.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21871,
"subject": "Mathematics (Olympiad)",
"question": "Prove the following inequality for positive $a, b, c$:\n\n$$\n\\frac{a^3 + b^3 + c^3}{abc} + 6 \\ge 9 \\cdot \\frac{a^2 + b^2 + c^2}{ab + bc + ca}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Subtract $9$ from both sides and use the well-known identity\n\n$$\na^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca).\n$$\n\nThen apply the following transformations:\n\n$$\n\\frac{a^3 + b^3 + c^3}{abc} - 3 \\ge 9 \\cdot \\frac{a^2 + b^2 + c^2}{ab + bc + ca} - 9 \\Leftrightarrow \\\\\n\\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc}{abc} - 3 \\ge 9 \\cdot \\frac{a^2+b^2+c^2}{ab+bc+ca} - 9 \\Leftrightarrow\n$$\n\n$$\n\\begin{aligned}\n\\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc-3abc}{abc} &\\ge \\frac{9(a^2+b^2+c^2-ab-bc-ca)}{ab+bc+ca} \\\\\n&\\iff \\\\\n\\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}{abc} &\\ge \\frac{9(a^2+b^2+c^2-ab-bc-ca)}{ab+bc+ca} \\\\\n&\\iff \\\\\n\\frac{(a^2+b^2+c^2-ab-bc-ca)((a+b+c)(ab+bc+ca)-9abc)}{abc(ab+bc+ca)} &\\ge 0.\n\\end{aligned}\n$$\n\nThe last inequality holds, since\n\n$$\na^2 + b^2 + c^2 - ab - bc - ca \\ge 0 \\quad \\text{and} \\quad (a+b+c)(ab+bc+ca) - 9abc \\ge 0.\n$$\n\nBy the inequality with the sum of three squares:\n\n$$\n\\begin{aligned}\n(a-b)^2 + (b-c)^2 + (c-a)^2 \\ge 0 &\\iff a^2 + b^2 + c^2 \\ge ab + bc + ca \\\\\n(a+b+c)^2 \\ge 3(ab+bc+ca) &\\Rightarrow (a+b+c)^6 \\ge 27(ab+bc+ca)^3 \\ge 27 \\cdot 27(abc)^2 \\\\\n(a+b+c)^3 &\\ge 27abc.\n\\end{aligned}\n$$\n\nMoreover,\n\n$$\n\\left(\\frac{1}{3}(a+b+c)\\right)^3 \\ge abc \\implies (ab+bc+ca)^3 \\ge 27a^2b^2c^2.\n$$\n\nHence,\n\n$$\n(a+b+c)^3(ab+bc+ca)^3 \\ge (9abc)^3.\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21872,
"subject": "Mathematics (Olympiad)",
"question": "Initially, there are $m$ balls in one bag and $n$ balls in the other, where $m, n > 0$.\n\nTwo different operations are allowed:\n\n1. Remove an equal number of balls from each bag.\n2. Double the number of balls in one bag.\n\nIs it always possible to empty both bags after a finite sequence of operations?\n\nNow, operation 2 is replaced by:\n\n2'. Triple the number of balls in one bag.\n\nIs it now always possible to empty both bags after a finite sequence of operations?",
"options": [],
"answer": "See solution",
"solution": "We claim that in the first case, it is always possible.\n\nIf $m = n$, we can simply remove $m$ balls from each bag and empty both.\n\nOtherwise, assume $m < n$. Repeatedly double the number of balls in the bag with $m$ balls until it has at least $\\frac{n}{2}$ balls. Now, let the bags have $m'$ and $n$ balls, where $\\frac{n}{2} \\leq m' < n$.\n\nRemove $2m' - n$ balls from each bag. The bags now have $n - m'$ and $2(n - m')$ balls. Double the number of balls in the bag with fewer balls, and finally remove all balls from both bags.\n\nWith the modified operation, it is not always possible. Operations 1 and 2' both change the total number of balls by an even number. Thus, if the total number of balls is odd (e.g., 2 balls in one bag and 1 in the other), it is impossible to empty both bags.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21873,
"subject": "Mathematics (Olympiad)",
"question": "Petryk programmed a calculator so that if there is a number $x$ on the screen and the button «=>» is pressed, the number $$\\frac{x-1}{x+1}$$ appears on the screen. Petryk pressed the button «=>» 2014 times, and then 2016 appeared on the screen. What number was on the screen at the beginning? The screen can show not only integer numbers.",
"options": [],
"answer": "See solution",
"solution": "Let us examine how the number changes after pressing the «=>» button. Let the initial number be $x$.\n\n- After the first press: $$\\frac{x-1}{x+1}$$\n- After the second press: $$\\frac{\\frac{x-1}{x+1}-1}{\\frac{x-1}{x+1}+1} = \\frac{x-1-x-1}{x-1+x+1} = -\\frac{1}{x}$$\n- After the third press: $$\\frac{-\\frac{1}{x}-1}{-\\frac{1}{x}+1} = \\frac{-1-x}{-1+x}$$\n- After the fourth press: $$\\frac{\\frac{-1-x}{-1+x}-1}{\\frac{-1-x}{-1+x}+1} = x$$\n\nThus, after every fourth press, the original number $x$ reappears. Therefore, after 2012 presses, the screen shows $x$. After 2 more presses, the screen shows $-\\frac{1}{x} = 2016$, so $x = -\\frac{1}{2016}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21874,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying\n$$\n(a-b)f(a+b) + (b-c)f(b+c) + (c-a)f(c+a) = 0\n$$\nfor all $a, b, c \\in \\mathbb{R}$.",
"options": [],
"answer": "See solution",
"solution": "The solutions are all functions of the form $f(x) = Ax + B$ for real numbers $A$ and $B$.\n\nFor $x \\in \\mathbb{R}$, let $a = \\frac{x-1}{2}$, $b = \\frac{x+1}{2}$, and $c = \\frac{1-x}{2}$. Then:\n$$\n\\begin{cases}\na+b = x \\\\\nb+c = 1 \\\\\nc+a = 0\n\\end{cases}\n$$\nand\n$$\n\\begin{cases}\na-b = -1 \\\\\nb-c = x \\\\\nc-a = 1-x\n\\end{cases}\n$$\nTherefore,\n$$\n- f(x) + x f(1) + (1-x) f(0) = 0.\n$$\nLetting $A = f(1) - f(0)$ and $B = f(0)$, we get $f(x) = Ax + B$.\n\nConversely, for any $A$ and $B$, the function $f(x) = Ax + B$ satisfies\n$$\n(a-b)f(a+b) + (b-c)f(b+c) + (c-a)f(c+a) = \\sum_{\\text{cyc}} A(a^2 - b^2) + B(a-b) = 0.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21875,
"subject": "Mathematics (Olympiad)",
"question": "Solve in the set of integers the equation:\n\n$$\ny = 2x^2 + 5xy + 3y^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "The equation can be rewritten as\n\n$$\ny = 2x^2 + 2xy + 3xy + 3y^2 = (x + y)(2x + 3y). \\quad (1)\n$$\n\nLet $x + y = z \\in \\mathbb{Z}$. Then equation (1) becomes\n\n$$\ny = z(2z + y) \\implies y = 2z^2 + yz \\implies (z - 1)y = -2z^2. \\quad (2)\n$$\n\nFor $z = 1$, equation (2) gives $0 \\cdot y = -2$, which is impossible. For $z \\neq 1$,\n\n$$\ny = -\\frac{2z^2}{z - 1} = -\\frac{2(z^2 - 1) + 2}{z - 1} = -2(z + 1) - \\frac{2}{z - 1}. \\quad (3)\n$$\n\nFor $y \\in \\mathbb{Z}$, $z - 1$ must divide $2$, so\n\n$$\nz - 1 \\in \\{-1, 1, -2, 2\\} \\implies z \\in \\{0, 2, -1, 3\\}.\n$$\n\nFor $z = 0$, $(x, y) = (0, 0)$. \nFor $z = 2$, $(x, y) = (10, -8)$. \nFor $z = -1$, $(x, y) = (-2, 1)$. \nFor $z = 3$, $(x, y) = (12, -9)$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21876,
"subject": "Mathematics (Olympiad)",
"question": "Join $AX$, $BY$, and $CZ$. These lines are the bisectors of the angles of $\\triangle ABC$ and intersect at its incentre $I$. Label the vertices of the hexagon as shown below and join $KI$ and $ZA$.\n\n\n\nShow that the three diagonals $KF$, $DG$, and $EH$ of the hexagon are concurrent at $I$.",
"options": [],
"answer": "See solution",
"solution": "We have $\\angle KZI = \\angle XZC = \\angle XAC = \\angle KAI$, hence $AZKI$ is a cyclic quadrilateral. This implies $\\angle ZIK = \\angle ZAK = \\angle ZAB = \\angle ZCB$, from which we infer that $KI$ and $BC$ are parallel. Similarly, it follows that $IF$ and $BC$ are parallel. This shows that $K$, $I$, and $F$ are collinear, i.e., the diagonal $KF$ passes through $I$.\n\nThe same argument shows that the other two diagonals, $DG$ and $EH$, also pass through $I$, thus the three diagonals of the hexagon are concurrent.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21877,
"subject": "Mathematics (Olympiad)",
"question": "Find all triples $(p, x, y)$ consisting of a prime number $p$ and two positive integers $x$ and $y$ such that $x^{p-1} + y$ and $x + y^{p-1}$ are both powers of $p$.",
"options": [],
"answer": "See solution",
"solution": "All solutions are $$(p, x, y) \\in \\{(3, 2, 5), (3, 5, 2)\\} \\cup \\{(2, n, 2^k - n) \\mid 0 < n < 2^k\\}.$$ \n\n(1) When $p=2$, clearly all pairs $(x, y)$ whose sum is a power of $2$ satisfy the condition, so we only need to consider $p > 2$.\n\n(2) Suppose $x^{p-1} + y = p^a$ and $x + y^{p-1} = p^b$. Without loss of generality, assume $x \\le y$, so $a \\le b$. Thus,\n\n$$\np^b = y^{p-1} + x = (p^a - x^{p-1})^{p-1} + x\n$$\n\nTake both sides modulo $p^a$, noting $p-1$ is even:\n\n$$\n0 = x^{(p-1)^2} + x \\pmod{p^a} \\qquad (1)\n$$\n\nIf $p \\nmid x$, then $x^{(p-1)^2-1} + 1$ is not divisible by $p$, contradicting (1) (since $x \\le x^{p-1} < p^a$).\n\nSo $p$ does not divide $x$, i.e.,\n\n$$\np^a \\mid x^{(p-1)^2-1} + 1 = x^{p(p-2)} + 1.\n$$\n\n(3) By Fermat's little theorem, $x^{(p-1)^2} \\equiv 1 \\pmod{p}$, so\n\n$$\nx + 1 = x + x^{(p-1)^2} = x(1 + x^{(p-1)^2-1}) \\equiv 0 \\pmod{p},\n$$\n\nso $p \\nmid x + 1$. Let $p^r$ be the highest power dividing $x + 1$.\n\n(4) Consider the highest power of $p$ dividing $x^{p(p-2)} + 1$. Expand $x^{p(p-2)} = (x + 1 - 1)^{p(p-2)}$ by binomial theorem as a sum of $(x+1)^k$ terms. For $k \\ge 3$, each term is divisible by $p^{3r}$. The $k=2$ term is\n\n$$\n-\\frac{p(p-2)(p^2-2p-1)}{2}(x+1)^2,\n$$\n\nwhich is divisible by $p^{2r+1}$. The $k=1$ term is\n\n$$\np(p-2)(x+1),\n$$\n\nwhich is divisible by $p^{r+1}$ but not $p^{r+2}$ (since $r$ is maximal). The last term is $-1$. Thus, the highest power dividing $x^{p(p-2)} + 1$ is $p^{r+1}$.\n\n(5) But we assumed $p^a \\mid x^{p(p-2)} + 1$, so $a \\le r+1$. Also,\n\n$$\np^r \\le x+1 \\le x^{p-1} + y = p^a \\quad (2)\n$$\n\nSo $a = r$ or $a = r+1$.\n\n(6) If $a = r$, then all equalities in (2) must hold, so $x = y = 1$, which is impossible for $p > 2$. Thus $a = r+1$. Since $p^r \\le x+1$, we have\n\n$$\nx = \\frac{x^2 + x}{x+1} \\le \\frac{x^{p-1} + y}{x+1} = \\frac{p^a}{x+1} \\le \\frac{p^a}{p^r} = p.\n$$\n\nSince $p \\mid x+1$, $x = p-1$; thus $r=1, a=2$.\n\n(7) If $p \\ge 5$,\n\n$$\np^a = x^{p-1} + y > (p-1)^4 = (p^2 - 2p + 1)^2 > (3p)^2 > p^2 = p^a\n$$\n\nContradiction! So $p=3$, $x=p-1=2$, and $y = p^a - x^{p-1} = 5$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21878,
"subject": "Mathematics (Olympiad)",
"question": "The opposite sides of a convex hexagon of unit area are pairwise parallel. The lines of support of three alternate sides meet pairwise to form a triangle. Similarly, the lines of support of the other three alternate sides meet pairwise to form another triangle. Show that the area of at least one of these two triangles is greater than or equal to $\\frac{3}{2}$.",
"options": [],
"answer": "See solution",
"solution": "Unless otherwise stated, throughout the proof indices take on values from $0$ to $5$ and are reduced modulo $6$. Label the vertices of the hexagon in circular order, $A_0, A_1, \\dots, A_5$, and let the lines of support of the alternate sides $A_iA_{i+1}$ and $A_{i+2}A_{i+3}$ meet at $B_i$. To show that the area of at least one of the triangles $B_0B_2B_4$, $B_1B_3B_5$ is greater than or equal to $\\frac{3}{2}$, it is sufficient to prove that the total area of the six triangles $A_{i+1}B_iA_{i+2}$ is at least $1$:\n\n$$\n\\sum_{i=0}^{5} \\text{area } A_{i+1}B_{i}A_{i+2} \\geq 1.\n$$\n\nTo begin with, reflect each $B_i$ through the midpoint of the segment $A_{i+1}A_{i+2}$ to get the points $B'_i$. We shall prove that the six triangles $A_{i+1}B'_iA_{i+2}$ cover the hexagon. To this end, reflect $A_{2i+1}$ through the midpoint of the segment $A_{2i}A_{2i+2}$ to get the points $A'_{2i+1}$, $i=0,1,2$. The hexagon splits into three parallelograms, $A_{2i}A_{2i+1}A_{2i+2}A'_{2i+1}$, $i=0,1,2$, and a (possibly degenerate) triangle, $A'_1A'_3A'_5$. Notice first that each parallelogram $A_{2i}A_{2i+1}A_{2i+2}A'_{2i+1}$ is covered by the pair of triangles $(A_{2i}B'_{2i+5}A_{2i+1}, A_{2i+1}B'_{2i}A_{2i+2})$, $i=0,1,2$. The proof is completed by showing that at least one of these pairs contains a triangle that covers the triangle $A'_1A'_3A'_5$. To this end, it is sufficient to prove that $A_{2i}B'_{2i+5} \\geq A_{2i}A'_{2i+5}$ and $A_{2j+2}B'_{2j} \\geq A_{2j+2}A'_{2j+3}$ for some indices $i, j \\in \\{0,1,2\\}$. To establish the first inequality, notice that\n\n$$\nA_{2i}B'_{2i+5} = A_{2i+1}B_{2i+5}, \\quad A_{2i}A'_{2i+5} = A_{2i+4}A_{2i+5}, \\quad i=0,1,2, \\\\\n\\frac{A_1B_5}{A_4A_5} = \\frac{A_0B_5}{A_5B_3} \\quad \\text{and} \\quad \\frac{A_3B_1}{A_0A_1} = \\frac{A_2A_3}{A_0B_5},\n$$\n\nto get\n\n$$\n\\prod_{i=0}^{2} \\frac{A_{2i}B'_{2i+5}}{A_{2i}A'_{2i+5}} = 1.\n$$\n\nSimilarly,\n\n$$\n\\prod_{j=0}^{2} \\frac{A_{2j+2}B'_{2j}}{A_{2j+2}A'_{2j+3}} = 1,\n$$\n\nwhence the conclusion.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21879,
"subject": "Mathematics (Olympiad)",
"question": "Let $S = \\{1, 2, \\dots, 2017\\}$.\n\nFind the maximal $n$ with the property that there exist $n$ distinct subsets of $S$ such that for no two subsets their union equals $S$.",
"options": [],
"answer": "See solution",
"solution": "Answer: $n = 2^{2016}$.\n\n**Proof:**\n\nThere are $2^{2016}$ subsets of $S$ which do not contain $2017$. The union of any two such subsets does not contain $2017$ and is thus a proper subset of $S$. Thus $n \\geq 2^{2016}$.\n\nTo show the other direction, we group the subsets of $S$ into $2^{2016}$ pairs so that every subset forms a pair with its complement. If $n > 2^{2016}$, then the $n$ subsets would contain such a pair. Their union would be $S$, a contradiction.\n\nThus $n = 2^{2016}$.\n\n$\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21880,
"subject": "Mathematics (Olympiad)",
"question": "Determine all integers $n \\ge 3$ such that there exist pairwise distinct real numbers $a_1, a_2, \\dots, a_n$ for which the $\\frac{1}{2}n(n-1)$ sums $a_i + a_j$ (for $1 \\le i < j \\le n$), when ordered increasingly, form an arithmetic sequence (i.e., the difference between every two consecutive sums is the same).",
"options": [],
"answer": "See solution",
"solution": "The required integers are $n=3$ and $n=4$.\n\nFor $n=3$, let $(a_1, a_2, a_3) = (1, 2, 3)$; the pairwise sums are $3, 4, 5$, which form an arithmetic sequence.\n\nFor $n=4$, let $(a_1, a_2, a_3, a_4) = (1, 3, 4, 5)$; the pairwise sums are $4, 5, 6, 7, 8, 9$, which also form an arithmetic sequence.\n\nIf $n \\ge 5$, suppose such $a_1, \\dots, a_n$ exist. Assume $a_1 < a_2 < \\dots < a_n$ and let $d$ be the common difference of the sequence of sums.\n\nThe smallest sum is $a_1 + a_2$, the next is $a_1 + a_3$, so $a_3 - a_2 = d$. The largest sum is $a_{n-1} + a_n$, the next largest is $a_{n-2} + a_n$, so $a_{n-1} - a_{n-2} = d$. Thus, $a_2 + a_{n-1} = a_3 + a_{n-2}$.\n\nIf $n \\ge 6$, these sums correspond to distinct pairs, so they must be at least $d$ apart, which is a contradiction. Thus, $n=5$ is the only remaining case.\n\nFor $n=5$, $a_3 - a_2 = d$ and $a_4 - a_3 = d$, so $2a_3 = a_2 + a_4$. Consider the sum $s = \\sum_{1 \\le i < j \\le 5} (a_i + a_j)$. Each $a_k$ appears in $4$ sums, so $s = 4(a_1 + a_2 + a_3 + a_4 + a_5)$. Alternatively, $s = 5(a_1 + a_2 + a_4 + a_5)$. Equating and simplifying gives $a_1 + a_2 + a_4 + a_5 = 4a_3 = 2(a_2 + a_4)$, so $a_1 + a_5 = a_2 + a_4$, a contradiction.\n\nAlternatively, analyzing the order of sums for $n=5$ leads to $a_5 - a_4 = d$ and also $a_5 - a_4 = 2d$, which is impossible. Thus, no such $a_1, \\dots, a_5$ exist.\n\nTherefore, the only possible values are $n=3$ and $n=4$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21881,
"subject": "Mathematics (Olympiad)",
"question": "Let $0 \\leq c_1 \\leq \\cdots \\leq c_t$ and $0 \\leq d_1 \\leq \\cdots \\leq d_t$. Prove that\n\n$$\n(c_1 + d_1)(c_2 + d_2) \\cdots (c_t + d_t) \\leq (c_1 + d_t)(c_2 + d_{t-1}) \\cdots (c_t + d_1).\n$$\n\nLet $L_{i,j} = \\sum_{q=1}^{j} a_{i,q}$, $R_{i,j} = \\sum_{q=j}^{n} a_{i,q}$, $U_{i,j} = \\sum_{p=1}^{i} a_{p,j}$, and $D_{i,j} = \\sum_{p=i}^{m} a_{p,j}$. Define\n\n$$\nX_{i,j} = L_{i,j} + U_{i,j} - a_{i,j}, \\quad Y_{i,j} = D_{i,j} + R_{i,j} - a_{i,j}.\n$$\n\nShow that\n\n$$\n\\prod_{i=1}^{m} \\prod_{j=1}^{n} X_{i,j} \\geq \\prod_{i=1}^{m} \\prod_{j=1}^{n} Y_{i,j}.\n$$",
"options": [],
"answer": "See solution",
"solution": "$$\n\\begin{align*}\n\\prod_{i=1}^{m} \\prod_{j=1}^{n} X_{i,j} &= \\prod_{i=1}^{m} \\prod_{j=1}^{n} (L_{i,j} + (U_{i,j} - a_{i,j})) \\\\\n&\\geq \\prod_{i=1}^{m} \\prod_{j=1}^{n} (L_{i,n+1-j} + (U_{i,j} - a_{i,j})) \\\\\n&\\quad \\text{(since $\\{L_{i,j}\\}_{j=1}^{n}$ increases while $\\{U_{i,j} - a_{i,j}\\}_{j=1}^{n}$ decreases)} \\\\\n&\\geq \\prod_{i=1}^{m} \\prod_{j=1}^{n} (R_{i,j} + (U_{i,j} - a_{i,j})) \\quad \\text{(since $L_{i,n+1-j} \\geq R_{i,j}$)} \\\\\n&= \\prod_{j=1}^{n} \\prod_{i=1}^{m} ((R_{i,j} - a_{i,j}) + U_{i,j}) \\\\\n&\\geq \\prod_{j=1}^{n} \\prod_{i=1}^{m} ((R_{i,j} - a_{i,j}) + U_{m+1-i,j}) \\\\\n&\\quad \\text{(since $\\{R_{i,j} - a_{i,j}\\}_{i=1}^{m}$ decreases while $\\{U_{i,j}\\}_{i=1}^{m}$ increases)} \\\\\n&\\geq \\prod_{j=1}^{n} \\prod_{i=1}^{m} ((R_{i,j} - a_{i,j}) + D_{i,j}) \\quad (\\text{since $U_{m+1-i,j} \\geq D_{i,j}$}) \\\\\n&= \\prod_{i=1}^{m} \\prod_{j=1}^{n} Y_{i,j}.\n\\end{align*}\n$$\n\n$\\boxed{}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21882,
"subject": "Mathematics (Olympiad)",
"question": "Solve in real numbers the following equation:\n\n$$\n3^{\\log_5(5x-10)} - 2 = 5^{-1+\\log_3 x}.\n$$",
"options": [],
"answer": "See solution",
"solution": "We notice that the equation has solutions $x_1 = 3$ and $x_2 = 27$. We show that the equation does not have any other solutions.\n\n**Continuation A.** Using the properties of logarithms, the equation becomes $15 \\cdot 3^{\\log_5(x-2)} = 10 + 5^{\\log_3 x}$, or $15(x-2)^{\\log_5 3} = 10 + x^{\\log_3 5}$, with $x > 2$. Since $\\log_3 5 > 1$ and $0 < \\log_5 3 < 1$, the function $2 < x \\mapsto 15(x-2)^{\\log_5 3}$ is strictly concave, and the function $0 < x \\mapsto 10 + x^{\\log_3 5}$ is strictly convex. Hence, the function $f : (2, \\infty) \\to \\mathbb{R}$, $f(x) = 15(x-2)^{\\log_5 3} - 10 - x^{\\log_3 5}$ is strictly concave, and the equation has at most two solutions.\n\n**Continuation B.** From the existence condition of logarithms, we have $x > 2$. We observe that the function $f : (2, \\infty) \\to (0, \\infty)$, $f(x) = 3^{\\log_5(5x-10)}$ has inverse which is $f^{-1}(x) = \\frac{5^{\\log_3 x} + 10}{5}$, and the equation from the statement becomes $f(x) = f^{-1}(x)$.\n\nSince $f$ is strictly increasing, the equation from the statement is equivalent to $f(x) = x$, meaning $3^{\\log_5(5x-10)} = x$, or $\\log_5(5x-10) = \\log_3 x$. If $\\log_5(5x-10) = \\log_3 x = t$, then $x = 3^t = \\frac{1}{5}(5^t + 10)$, which is equivalent to solving the equation $5 \\cdot 3^t = 5^t + 10$, or $5 = \\left(\\frac{5}{3}\\right)^t + 10\\left(\\frac{1}{3}\\right)^t$, with $t > \\log_3 2$. Since the function $g : (\\log_3 2, \\infty) \\to \\mathbb{R}$, $g(t) = \\left(\\frac{5}{3}\\right)^t + 10\\left(\\frac{1}{3}\\right)^t$ is strictly convex, being the sum of strictly convex functions, the equation $g(t) = 5$ admits at most two solutions.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21883,
"subject": "Mathematics (Olympiad)",
"question": "By a strip of breadth $b$ we mean a closed part of the plane consisting of all points that lie between two parallel lines at distance $b$ from each other. Let $S$ be a finite set of $n$ ($n \\ge 4$) points in the plane, such that any three points from $S$ can be covered by a strip of breadth $1$. Prove that $S$ can be covered by a strip of breadth $2$.",
"options": [],
"answer": "See solution",
"solution": "Firstly, we shall 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. $\\square$\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\nChoose now 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 $C$ on $AB$ 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 $S$ will be covered by a strip of breadth $2$ with borders parallel to $AB$, at distance $1$ on both sides of $AB$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21884,
"subject": "Mathematics (Olympiad)",
"question": "Let $X = \\sum_{i=1}^{n} ia_{i}$ and $Y = \\sum_{i=1}^{n} \\frac{a_{i}}{i}$. Find the maximum possible value of $XY^{2}$.",
"options": [],
"answer": "See solution",
"solution": "First, for each $i \\in \\{1, 2, \\dots, n\\}$, we have $i + \\frac{n}{i} \\le n + 1$, since $$(n+1) - \\left(i + \\frac{n}{i}\\right) = \\frac{1}{i}(i-1)(n-i) \\ge 0.$$ Thus,\n\n$$\nX + nY = \\sum_{i=1}^{n} \\left(i + \\frac{n}{i}\\right) a_i \\le \\sum_{i=1}^{n} (n+1)a_i = n+1.\n$$\n\nApplying the inequality of arithmetic and geometric means to $X$, $\\frac{nY}{2}$, $\\frac{nY}{2}$:\n\n$$\nX \\cdot \\frac{nY}{2} \\cdot \\frac{nY}{2} \\le \\left(\\frac{X+nY}{3}\\right)^3 \\le \\left(\\frac{n+1}{3}\\right)^3 = \\frac{(n+1)^3}{27}.\n$$\n\nTherefore,\n\n$$\nXY^2 \\le \\frac{4(n+1)^3}{27n^2}.\n$$\n\nEquality is achieved by choosing\n\n$$\na_1 = \\frac{2n-1}{3(n-1)}, \\quad a_2 = \\dots = a_{n-1} = 0, \\quad a_n = \\frac{n-2}{3(n-1)}.\n$$\n\nWith this choice, $XY^2 = \\frac{4(n+1)^3}{27n^2}$, which is the maximum value.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21885,
"subject": "Mathematics (Olympiad)",
"question": "Given three pairwise distinct positive integers $a$, $b$, and $c$ whose product is $320$, determine the smallest possible prime value of $a + b + c$.",
"options": [],
"answer": "See solution",
"solution": "Since $a$, $b$, and $c$ are distinct positive integers with $abc = 320$, and we seek the smallest prime value of $a + b + c$.\n\nThe product $320 = 2^6 \\times 5$ has only two odd divisors: $1$ and $5$. Consider cases where one of the numbers is odd ($c = 1$ or $c = 5$), and the other two are even.\n\n**Case 1:** $c = 1$\n- $a = 2$, $b = 160$; $a + b + c = 163$ (prime)\n- $a = 4$, $b = 80$; $a + b + c = 85$ (not prime)\n- $a = 8$, $b = 40$; $a + b + c = 49$ (not prime)\n- $a = 16$, $b = 20$; $a + b + c = 37$ (prime, smaller than 163)\n- $a = 32$, $b = 10$; $a + b + c = 43$ (greater than 37)\n\n**Case 2:** $c = 5$\n- $a = 2$, $b = 32$; $a + b + c = 39$ (not prime)\n- $a = 4$, $b = 16$; $a + b + c = 25$ (not prime)\n\nThus, the smallest possible prime sum is $37$, achieved with $a = 16$, $b = 20$, $c = 1$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21886,
"subject": "Mathematics (Olympiad)",
"question": "Let $r$ be a positive integer and let $N_r$ be the smallest positive integer such that the numbers\n$$\n\\frac{N_r}{n+r} \\binom{2n}{n}, \\quad n = 0, 1, 2, \\dots,\n$$\nare all integers. Show that\n$$\nN_r = \\frac{r}{2} \\binom{2r}{r}.\n$$",
"options": [],
"answer": "See solution",
"solution": "We first show that\n$$\nN_r \\leq \\frac{r}{2} \\binom{2r}{r}\n$$\nby proving that\n$$\nK(n, r) = \\frac{r}{2(n+r)} \\binom{2n}{n} \\binom{2r}{r}\n$$\nis an integer for all $n \\geq 0$ and $r \\geq 1$. Notice that\n$$\nK(0, r) = \\binom{2r-1}{r} \\quad \\text{and} \\quad K(n, 1) = \\frac{1}{n+1} \\binom{2n}{n},\n$$\nthe latter being a Catalan number, so $K(0, r)$ and $K(n, 1)$ are integers for all $n$ and $r$.\n\nNext, the recurrence relation (which will be proved at the end of the solution)\n$$\nK(n, r + 1) - K(n + 1, r) = 2K(n, 1)K(0, r)\n$$\nshows by induction on $r$ that $K(n, r)$ is indeed an integer for all $n$ and $r$.\n\nSuppose now that, for some $r$, $N_r < M_r = \\frac{r}{2} \\binom{2r}{r}$. The integer $N_r$ is the least common multiple, over all $n \\geq 0$, of the denominators of the numbers $\\frac{1}{n+r} \\binom{2n}{n}$ when written in lowest terms. By the argument above, $M_r$ is a multiple of all these denominators. Hence $N_r$ divides $M_r$. By the definition of $N_r$, any prime $p$ that divides $M_r/N_r$ also divides $K(n, r)$ for each $n \\geq 0$. Since $K(n, s+1) = K(n+1, s) + 2K(n, 1)K(0, s)$, induction on $m$ shows that $p$ divides $K(n, m)$ for all $n \\geq 0$ and $m \\geq r$.\n\nNow choose $k$ such that $p^k \\geq r$. Since $p$ divides $\\binom{p^k}{j}$, $j = 1, 2, \\dots, p^k - 1$, the identity $\\binom{2n}{n} = \\sum_{j=1}^n \\binom{n}{j}^2$ yields $\\binom{2p^k}{p^k} \\equiv 2 \\pmod{p}$. Therefore $p$ does not divide $\\frac{1}{2}\\binom{2p^k}{p^k} = K(0, p^k)$. This contradicts the preceding paragraph, so $N_r = \\frac{r}{2}\\binom{2r}{r}$ for all $r$.\n\nTo prove the recurrence relation, notice that\n$$\n\\binom{2m+2}{m+1} = 2\\binom{2m+1}{m} = 2\\frac{2m+1}{m+1}\\binom{2m}{m},\n$$\nto get\n$$\n\\begin{align*}\nK(n, r + 1) - K(n + 1, r) &= \\\\\n&\\frac{r+1}{2(n+r+1)} \\binom{2n}{n} \\binom{2r+2}{r+1} - \\frac{r}{2(n+r+1)} \\binom{2n+2}{n+1} \\binom{2r}{r} = \\\\\n&\\frac{1}{n+r+1} \\binom{2n}{n} \\binom{2r}{r} \\left(2r+1 - r\\frac{2n+1}{n+1}\\right) = \\\\\n&\\frac{1}{n+1} \\binom{2n}{n} \\binom{2r}{r} = 2K(n, 1)K(0, r).\n\\end{align*}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21887,
"subject": "Mathematics (Olympiad)",
"question": "Given $\\triangle ABC$ with $CA > BC > AB$. Let $O$ and $H$ be the circumcentre and orthocentre of $\\triangle ABC$ respectively. Denote $D$ and $E$ to be the midpoints of arcs $AB$ and $AC$ of the circumcircle of $\\triangle ABC$ not containing the opposite vertices. Let $D'$ be the reflection of $D$ about side $AB$ and $E'$ be the reflection of $E$ about side $AC$. Prove that $O, H, D', E'$ are concyclic if and only if $A, D', E'$ are collinear.",
"options": [],
"answer": "See solution",
"solution": "From $CA > BC > AB$, we have $\\angle B > 60^\\circ > \\angle C$. In particular, $\\angle C < 60^\\circ$ yields $AD < AO$, so that $D'$ lies between $D$ and $O$. Similarly, $\\angle B > 60^\\circ$ implies that $O$ lies between $E$ and $E'$. Also, from $CA > BC$, point $H$ lies on the same side as $B$ to line $DO$. Similarly, $BC > AB$ implies that $H$ lies on the",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21888,
"subject": "Mathematics (Olympiad)",
"question": "Consider positive integers $n$ and $m$ such that $m \\ge n \\ge 2$. Find the number of one-to-one functions $f : \\{1, 2, \\dots, n\\} \\to \\{1, 2, \\dots, m\\}$, having the property that there is a unique number $i \\in \\{1, 2, \\dots, n-1\\}$ such that $f(i) > f(i+1)$.",
"options": [],
"answer": "See solution",
"solution": "Equivalently, the problem asks for the number of functions that are increasing on each of the sets $\\{1, 2, \\dots, i\\}$ and $\\{i+1, i+2, \\dots, n\\}$, but are not increasing on the whole set $\\{1, 2, \\dots, n\\}$.\n\nThe image of a one-to-one function $f : \\{1, 2, \\dots, n\\} \\to \\{1, 2, \\dots, m\\}$ is a set with $n$ elements. For a function $f : \\{1, 2, \\dots, n\\} \\to \\{1, 2, \\dots, m\\}$ having the given property, denote by $A$ its image and let $g$ be the unique increasing function from $A$ to $\\{1, 2, \\dots, n\\}$. Evidently, $g$ is also onto.\n\nAs a consequence, $h = g \\circ f : \\{1, 2, \\dots, n\\} \\to \\{1, 2, \\dots, n\\}$ is bijective and has the given property. Indeed, let $i \\in \\{1, 2, \\dots, n-1\\}$ such that $f(1) < \\dots < f(i)$, $f(i) > f(i+1)$, and $f(i+1) < \\dots < f(n)$. As $g$ is increasing, we get $g(f(1)) < \\dots < g(f(i))$, $g(f(i)) > g(f(i+1))$, and $g(f(i+1)) < \\dots < g(f(n))$, that is, $h(1) < \\dots < h(i)$, $h(i) > h(i+1)$, and $h(i+1) < \\dots < h(n)$.\n\nWe shall prove that for $h$ there is a unique subset $M$ of $\\{0, 1, 2, \\dots, n\\}$, different from the sets $\\emptyset$, $\\{1\\}$, $\\{1, 2\\}$, ..., $\\{1, 2, \\dots, n\\}$. For each $i \\in \\{0, 1, 2, \\dots, n\\}$, choose $M$ a subset of $\\{1, 2, \\dots, n\\}$ having $i$ elements. The function $h$ is uniquely determined by the $n$-tuple $(h(1), h(2), \\dots, h(n))$, obtained by ordering increasingly first the elements in $M$, and then the elements in $\\{1, 2, \\dots, n\\} \\setminus M$.\n\nFor $h$ to fail in being increasing on $\\{1, 2, \\dots, n\\}$, the set $M$ with $\\text{card } M = i$ must be different from $\\{1, 2, \\dots, i\\}$. Because we have $2^n$ subsets of $\\{1, 2, \\dots, n\\}$, and $n+1$ of them: $\\emptyset$, $\\{1\\}$, $\\{1, 2\\}$, ..., $\\{1, 2, \\dots, n\\}$ are not convenient, there are $2^n - n - 1$ bijective functions $h = g \\circ f : \\{1, 2, \\dots, n\\} \\to \\{1, 2, \\dots, n\\}$ with the given property.\n\nAs a subset $A \\subset \\{1, 2, \\dots, m\\}$ having $n$ elements can be chosen in $\\binom{m}{n}$ ways, we conclude that there are $\\binom{m}{n}(2^n - n - 1)$ one-to-one functions $f = g^{-1} \\circ h : \\{1, 2, \\dots, n\\} \\to \\{1, 2, \\dots, m\\}$ with the given property.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21889,
"subject": "Mathematics (Olympiad)",
"question": "Mrs Habana will be $45 + x$ years old and her son will be $17 + x$ years old, where $x$ is the number of years from 2016. In how many years from 2016 will Mrs Habana be twice as old as her son, and what will the year be?",
"options": [],
"answer": "See solution",
"solution": "We set up the equation: \n$$45 + x = 2(17 + x)$$\nExpanding gives:\n$$45 + x = 34 + 2x$$\nSubtract $x$ from both sides:\n$$45 = 34 + x$$\nSo:\n$$x = 45 - 34 = 11$$\nThe year will then be:\n$$2016 + 11 = 2027$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21890,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an isosceles triangle with $\\angle BAC = 100^\\circ$. Extend $AB$ through $B$ to $D$ and $E$ such that $BC = AD = BE$. Show that $BC \\cdot DE = BD \\cdot CE$.",
"options": [],
"answer": "See solution",
"solution": "Let $G$ and $F$ be points on the segments $EC$ and $BC$, respectively, so that $BDGF$ is a parallelogram with $DG \\parallel BF$ and $BD \\parallel FG$ (see figure).\n\n\n\nSince $\\triangle ABC$ is isosceles, $\\angle ABC = \\angle ACB = 40^\\circ$. Since $BE = BC$, $\\triangle BEC$ is isosceles, with $\\angle BEC = \\angle BCE = 20^\\circ$. Since $DG \\parallel BF$, $\\angle DGE = \\angle BCE = \\angle BEC$, so $\\triangle DEG$ is isosceles with $DE = DG$.\n\nNote that $DE = BE - BD = AD - BD = AB$, and by parallelogram $BDGF$, $BF = DG = DE = AB = AC$, and $FG = BD = BE - DE = BC - BF = FC$. Hence, $\\triangle FGC$ is isosceles with $\\angle FGC = \\angle FCG = 20^\\circ$, and is similar to $\\triangle BEC$. Since $BC \\cdot DE = BD \\cdot CE$ is equivalent to $\\frac{BC}{CE} = \\frac{BD}{DE}$ and $\\frac{BC}{CE} = \\frac{FC}{CG} = \\frac{BD}{CG}$ from similarity of $\\triangle BEC$ and $\\triangle FGC$, it suffices to show that $CG = DE$.\n\nConsider triangles $ABC$ and $GDA$: $\\angle ABC = \\angle GDA = 40^\\circ$, $AB = GD$, and $BC = DA$. Thus, $\\triangle ABC \\cong \\triangle GDA$. This gives $GA = AC$, so $\\triangle GAC$ is isosceles with $\\angle ACG = \\angle ACB + \\angle BCG = 60^\\circ$, i.e., $\\triangle GAC$ is an equilateral triangle.\n\nTherefore, $CG = AC = AB = DE$.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21891,
"subject": "Mathematics (Olympiad)",
"question": "Let $I$ be the incenter of triangle $ABC$ and let $\\Gamma$ be its circumcircle. Let the line $AI$ intersect $\\Gamma$ again at $D$. Let $E$ be a point on the arc $\\widehat{BDC}$ and $F$ a point on the side $BC$ such that\n$$\n\\angle BAF = \\angle CAE < \\frac{1}{2} \\angle BAC.\n$$\nFinally, let $G$ be the midpoint of the segment $IF$. Prove that the lines $DG$ and $EI$ intersect on $\\Gamma$.",
"options": [],
"answer": "See solution",
"solution": "",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21892,
"subject": "Mathematics (Olympiad)",
"question": "Броевите $m$ и $n$ се взаемно прости. Дробката $\\frac{3n - m}{5n + 2m}$ може да се скрати со некој природен број. Определи го бројот со кој може да се скрати.",
"options": [],
"answer": "See solution",
"solution": "Нека претпоставиме дека $k$, каде $k > 1$, е бројот со кој може да се скрати дробката. Тогаш постојат природни броеви $p$ и $s$ такви што $(p, s) = 1$ и $3n - m = kp$, $5n + 2m = ks$. Ако го решиме системот\n\n$$\n\\begin{cases}\n3n - m = kp \\\\\n5n + 2m = ks\n\\end{cases}\n$$\n\nпо $n$ и $m$, ќе добиеме:\n\n$$\nn = \\frac{k(2p + s)}{11}, \\quad m = \\frac{k(3s - 5p)}{11}.\n$$\n\nБроевите $m$ и $n$ се взаемно прости, па според тоа $k = 11$. Навистина, ако претпоставиме дека $k \\neq 11$, тогаш за било кој делител $d$ на $k$ поголем од 1 и различен од 11 имаме $(m, n) \\geq d > 1$, што е спротивно на претпоставката $(m, n) = 1$.\n\nЗначи, $k = 11$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21893,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\mathbb{Z}^+$ be the set of positive integers. Find all strictly increasing functions $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ with $f(1) = 1$ that satisfy the equation\n\n$$\n3 \\cdot (f(1) + f(2) + \\dots + f(n)) = f(n+1) + f(n+2) + \\dots + f(2n)\n$$\n\nfor all $n \\in \\mathbb{Z}^+$.",
"options": [],
"answer": "See solution",
"solution": "The strictly increasing function $f(n) = 2n-1$ for all $n \\in \\mathbb{Z}^+$ satisfies $f(1) = 1$ and solves the functional equation, since $1+3+\\dots+(2n-1) = n^2$ and $(2n+1)+(2n+3)+\\dots+(4n-1) = (2n)^2-n^2 = 3n^2$ for all $n \\in \\mathbb{Z}^+$.\n\nWe claim that no other function is suitable. Let $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ be a function that meets all requirements of the problem.\n\nLet $k \\in \\mathbb{Z}^+$. The given functional equation for $k$ and $k+1$ implies\n\n$$\n3 \\cdot \\sum_{l=1}^{k} f(l) = \\sum_{l=k+1}^{2k} f(l), \\\\\n3 \\cdot \\sum_{l=1}^{k+1} f(l) = \\sum_{l=k+2}^{2k+2} f(l).\n$$\n\nThe difference of the two equations yields $3f(k+1) = -f(k+1) + f(2k+1) + f(2k+2)$. In other words,\n\n$$\n4f(k+1) = f(2k+1) + f(2k+2) \\quad (*)\n$$\n\nholds for all $k \\in \\mathbb{Z}^+$.\n\nEquation $(*)$ implies that $f(2k+1)$ and $f(2k+2)$ have the same parity for every $k$. Since $f$ is strictly increasing, $f(2k+2) \\ge f(2k+1)+2$. Shifting indices, $4f(k+2) = f(2k+3) + f(2k+4)$. Note $f(2k+3) \\ge f(2k+2) + 1 \\ge f(2k+1) + 3$, and $f(2k+4) \\ge f(2k+3) + 2 \\ge f(2k+2) + 3$, so\n\n$$\n\\begin{aligned}\n4f(k+2) &= f(2k+3) + f(2k+4) \\\\\n&\\ge (f(2k+1) + 3) + (f(2k+2) + 3) \\\\\n&= 4f(k+1) + 6.\n\\end{aligned}\n$$\n\nThus,\n\n$$\nf(k+2) \\ge f(k+1) + 2 \\text{ for all } k \\in \\mathbb{Z}^+.\n$$\n\nNow, we show $f(n) = 2n - 1$ for all $n$. Use strong induction to show $f(2k-1) = 4k-3$ and $f(2k) = 4k-1$ for all $k$. This implies $f(n) = 2n-1$ for all $n$.\n\nBase case: $f(1) = 1$ by definition; for $n=1$, the condition gives $f(2) = 3f(1) = 3$. So $f(2k-1) = 4k-3$ and $f(2k) = 4k-1$ for $k=1$.\n\nInduction step: Assume $f(2l-1) = 4l-3$ and $f(2l) = 4l-1$ for $l = 1, \\dots, k$. We want $f(2k+1) = 4k+1$ and $f(2k+2) = 4k+3$.\n\nSince $k+1 \\le 2k$, the hypothesis gives $f(k+1) = 2k+1$. Equation $(*)$ gives $f(2k+1)+f(2k+2) = 8k+4$. By hypothesis, $f(2k) = 4k-1$, and from above, $f(2k+1) \\ge 4k+1$, $f(2k+2) \\ge 4k+3$. Since their sum is $8k+4$, we must have $f(2k+1) = 4k+1$ and $f(2k+2) = 4k+3$.\n\nTherefore, the only strictly increasing function is $f(n) = 2n-1$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21894,
"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": "Let $v_2(n)$ be the greatest integer such that $2^{v_2(n)}$ 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\nWe define 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": 21895,
"subject": "Mathematics (Olympiad)",
"question": "Suppose that the sequence $\\{a_n\\}$ is defined by $a_1 = a_2 = 1$, and for $n \\geq 3$, $a_n = 7a_{n-1} - a_{n-2}$. Prove that $a_n + a_{n+1} + 2$ is a perfect square for any positive integer $n$.\n\n",
"options": [],
"answer": "See solution",
"solution": "It is well known that the solution of the sequence can be obtained by solving two geometric sequences. We get $a_n = C_1\\lambda_1^n + C_2\\lambda_2^n$, where\n\n$$\n\\lambda_1 = \\frac{7 + \\sqrt{45}}{2} = \\left( \\frac{3 + \\sqrt{5}}{2} \\right)^2, \\quad \\lambda_2 = \\frac{7 - \\sqrt{45}}{2} = \\left( \\frac{3 - \\sqrt{5}}{2} \\right)^2\n$$\n\n$\\lambda_1$ and $\\lambda_2$ are solutions of the equation $\\lambda^2 - 7\\lambda + 1 = 0$, $a_1 = 1 = C_1\\lambda_1 + C_2\\lambda_2$, $a_2 = 1 = C_1\\lambda_1^2 + C_2\\lambda_2^2$.\n\nTherefore,\n\n$$\n\\begin{aligned}\na_n + a_{n+1} + 2 &= \\lambda_1^{n-1}C_1(\\lambda_1 + \\lambda_1^2) + \\lambda_2^{n-1}C_2(\\lambda_2 + \\lambda_2^2) + 2 \\\\\n&= \\lambda_1^{n-1} + \\lambda_2^{n-1} + 2 \\\\\n&= \\left( \\left( \\frac{3+\\sqrt{5}}{2} \\right)^{n-1} \\right)^2 + \\left( \\left( \\frac{3-\\sqrt{5}}{2} \\right)^{n-1} \\right)^2 + 2 \\\\\n&= \\left[ \\left( \\frac{3+\\sqrt{5}}{2} \\right)^{n-1} + \\left( \\frac{3-\\sqrt{5}}{2} \\right)^{n-1} \\right]^2 = x_n^2\n\\end{aligned}\n$$\n\nwhere $x_n = \\left(\\frac{3+\\sqrt{5}}{2}\\right)^{n-1} + \\left(\\frac{3-\\sqrt{5}}{2}\\right)^{n-1}$ is the solution of the sequence $\\{x_n\\}$ of positive integers: $x_1 = 2$, $x_2 = 3$, $x_n = 3x_{n-1} - x_{n-2}$ for $n \\geq 3$. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21896,
"subject": "Mathematics (Olympiad)",
"question": "On a southern island, there are $n$ fortresses lying on one line ($n > 0$). Each fortress is guarded by two elephants, both watching along the line of fortresses but in opposite directions. For a fortress $A$ to be able to conquer a fortress $B$, the elephant from $A$ who is watching towards $B$ must go along the line of fortresses to $B$, fight with all elephants who stand on his way and are watching opposite to him (including that of fortress $B$ but not the other elephant from $A$), and win them all. Given that all elephants have different constant weights and a heavier elephant always wins a lighter elephant, prove that there exists exactly one fortress that cannot be conquered by any other fortress.",
"options": [],
"answer": "See solution",
"solution": "The elephants from the outermost fortresses who watch in the direction where there are no more fortresses never take part in any fight, so we can discount them. Until only one fortress remains on the island, repeat the following: remove the heaviest elephant together with all fortresses and elephants on his watching direction. Thanks to the initial assumption, at least one fortress is removed on each step. It is easy to see that the fortress guarded by the heaviest elephant can conquer all fortresses that are subject to removal on the current step, while no fortress subject to removal can conquer any of the remaining fortresses since that would require winning the heaviest elephant. Hence the fortress that remains after all other fortresses are removed can be conquered by none of the other fortresses, while all other fortresses can be conquered by at least one of the other fortresses.\n\nA fortress being able to conquer another fortress implies the former fortress also being able to conquer all fortresses between the two fortresses. Let $F_1, \\dots, F_n$ be the fortresses along the line. There definitely exists a fortress which cannot be conquered by any fortress with a smaller number (for instance, $F_1$ is such). Let $F_i$ be the one with the largest number among those. We show that this fortress is the one we are looking for.\n\nSuppose that $F_j$ can conquer $F_i$. Then $j > i$ and $F_j$ can conquer also $F_{i+1}, \\dots, F_{j-1}$. As $j > i$, there must exist $l < j$ such that $F_l$ can conquer $F_j$. If $l < i$, then $F_l$ can conquer $F_i$ which contradicts the choice of $i$. If $i \\le l$, then because $l < j$, $F_j$ can conquer $F_l$, while $F_l$ can conquer $F_j$ as well. This is impossible, since in both cases, the two elephants from $F_j$ and $F_l$ looking in the opposite direction must fight against each other.\n\nIt remains to show that there are no more unconquerable fortresses. Suppose some $F_j$, $j \\ne i$, cannot be conquered. If $j > i$, then this contradicts the choice of $i$. Hence assume that $j < i$. Let among the elephants of $F_{j+1}, \\dots, F_i$ watching towards $F_j$, that of $F_k$ be the heaviest. As even $F_k$ cannot conquer $F_j$, there must exist $l$ such that $j \\le l < k$ and the elephant from $F_l$ watching towards $F_k$ is heavier than the elephant from $F_k$ watching towards $F_l$. But this means that $F_l$ can conquer $F_i$ which is impossible.\n\nWe proceed by induction on $n$. The claim obviously holds if $n = 1$. Consider a situation with $n + 1$ fortresses and assume that the claim holds for $n$ fortresses. Let $A$ be the fortress on one end of the line and let $B$ be the next fortress. Let $x$ be the weight of the elephant from $A$ watching towards $B$, let $y$ be the weight of the elephant from $B$ watching towards $A$ and let $z$ be the weight of the other elephant from $B$. Consider three cases.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21897,
"subject": "Mathematics (Olympiad)",
"question": "Тоон $a > 1$, $b > 1$ бүхэл тоонуудын хувьд $1 < a^b < 1000$ байх хэдэн $(a, b)$ хос байна вэ?",
"options": [],
"answer": "See solution",
"solution": "$$\na = 2: \\quad 1 < 2^b < 1000 \\implies b = 2, 3, 4, 5, 6, 7, 8, 9 \\quad (8~\\text{боломж})\n$$\n\n$$\na = 3: \\quad 1 < 3^b < 1000 \\implies b = 2, 3, 4, 5, 6 \\quad (5~\\text{боломж})\n$$\n\n$$\na = 4: \\quad 1 < 4^b < 1000 \\implies b = 2, 3, 4 \\quad (3~\\text{боломж})\n$$\n\n$$\na = 5: \\quad 1 < 5^b < 1000 \\implies b = 2, 3 \\quad (2~\\text{боломж})\n$$\n\n$$\na = 6: \\quad 1 < 6^b < 1000 \\implies b = 2, 3 \\quad (2~\\text{боломж})\n$$\n\n$$\na = 7, 8, 9: \\quad b = 2, 3 \\quad (2~\\text{боломж~бүрд~тус~бүр})\n$$\n\n$$\na = 10, 11, \\ldots, 33: \\quad b = 2 \\quad (1~\\text{боломж~бүрд})\n$$\n\n$$\n\\text{Нийт боломж: } 8 + 5 + 3 + 5 \\times 2 + 24 \\times 1 = 50\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21898,
"subject": "Mathematics (Olympiad)",
"question": "Let the length of the longer side of each of the 7 congruent rectangles be $a$, and the shorter side be $b$. Given that $|AB| = |DC| = 4b = 3a$, it follows that $b = \\frac{3}{4}a$. What is the ratio $\\frac{|BC|}{|AB|}$?",
"options": [],
"answer": "See solution",
"solution": "The ratio is:\n\n$$\n\\frac{|BC|}{|AB|} = \\frac{a + b}{4b} = \\frac{a + \\frac{3}{4}a}{3a} = \\frac{7}{4} \\div 3 = \\frac{7}{12}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21899,
"subject": "Mathematics (Olympiad)",
"question": "a) Prove that there are infinitely many positive integers $t$ such that both $2012t + 1$ and $2013t + 1$ are perfect squares.\n\nb) Suppose that $m, n$ are positive integers such that both $mn + 1$ and $mn + n + 1$ are perfect squares. Prove that $8(2m + 1)$ divides $n$.",
"options": [],
"answer": "See solution",
"solution": "a) Let $d = \\gcd(2012t + 1, 2013t + 1)$. We can easily prove that $d = 1$. Therefore, $2012t + 1$ and $2013t + 1$ are both perfect squares if and only if $(2012t + 1)(2013t + 1) = y^2$ for some integer $y$. We can rewrite this equation as\n$$\n(2 \\cdot 2012 \\cdot 2013 t + 4025)^2 - 1 = 4 \\cdot 2012 \\cdot 2013 \\cdot y^2.\n$$\nLet $x = 2 \\cdot 2012 \\cdot 2013 t + 4025$, so we have\n$$\nx^2 - 4 \\cdot 2012 \\cdot 2013 \\cdot y^2 = 1.\n$$\nSince $4 \\cdot 2012 \\cdot 2013$ is not a perfect square, this Pell equation has infinitely many solutions. The smallest solution is $(x, y) = (4025, 1)$, so the solutions are given by:\n$$\n\\begin{cases}\nx_0 = 1,\\ x_1 = 4025,\\ x_{n+2} = 8050 x_{n+1} - x_n & (n \\ge 0) \\\\\ny_0 = 1,\\ y_1 = 1,\\ y_{n+2} = 8050 y_{n+1} - y_n &\n\\end{cases}\n$$\nBy induction, $x_{2i+1}$ has remainder $4025$ when divided by $2 \\cdot 2012 \\cdot 2013$ for every $i$. Thus, $t = \\dfrac{x_{2i+1} - 4025}{2 \\cdot 2012 \\cdot 2013}$ gives infinitely many positive integers $t$ such that $2012t + 1$ and $2013t + 1$ are both perfect squares.\n\nb) Let $d = \\gcd(mn + 1, mn + n + 1)$. Then $d \\mid (mn + n + 1 - mn - 1) = n$, and $d \\mid (mn + 1 - mn) = 1$, so $d = 1$. Thus, $mn + 1$ and $(m + 1)n + 1$ are coprime. Both are perfect squares if and only if $(mn + 1)[(m + 1)n + 1] = y^2$ for some integer $y$. Rewriting,\n$$\n[2m(m+1)n + (2m+1)]^2 - 1 = 4m(m+1)y^2.\n$$\nLet $x = 2m(m+1)n + (2m+1)$, so\n$$\nx^2 - 4m(m+1)y^2 = 1. \\quad (1)\n$$\nSince $4m(m+1)$ is not a perfect square for any positive integer $m$, Pell's equation (1) has infinitely many solutions. The smallest solution is $(x, y) = (2m + 1, 1)$, so the recurrence is\n$$\n\\begin{cases}\nx_0 = 1,\\ x_1 = 2m + 1,\\ x_{i+2} = 2(2m + 1)x_{i+1} - x_i & (i \\ge 0) \\\\\ny_0 = 0,\\ y_1 = 1,\\ y_{i+2} = 2(2m + 1)y_{i+1} - y_i &\n\\end{cases}\n$$\nBy induction, $x_{2i}$ has remainder $1$ and $x_{2i+1}$ has remainder $2m + 1$ modulo $2m(m+1)$ for all $i$.\n\nDefine $r_i = x_{2i+1}$. Note that\n$$\n\\begin{aligned}\nr_{i+2} &= x_{2i+5} = 2(2m+1)x_{2i+4} - x_{2i+3} \\\\\n&= 2(2m+1)[2(2m+1)x_{2i+3} - x_{2i+2}] - x_{2i+3} \\\\\n&= [4(2m+1)^2 - 1] x_{2i+3} - 2(2m+1)x_{2i+2} \\\\\n&= [4(2m+1)^2 - 2] r_{i+1} - r_i.\n\\end{aligned}\n$$\nLet $r_i = 2m(m+1)s_i + (2m+1)$, then $(s_i)$ for $i > 0$ are distinct positive integers. Substituting, we get\n$$\n2m(m+1)s_{i+2} = 2m(m+1)[4(2m+1)^2 - 2]s_{i+1} - 2m(m+1)s_i + 4(2m+1)[(2m+1)^2 - 1]\n$$\nor\n$$\ns_{i+2} = [4(2m+1)^2 - 2]s_{i+1} - s_i + 8(2m+1).\n$$\nWe have $r_0 = x_1 = 2m+1$ so $s_0 = 0$, and\n$$\nr_1 = 2(2m+1)[2(2m+1)^2 - 1] - (2m+1) = 16m(m+1)(2m+1) + 2m+1\n$$\nso $s_1 = 8(2m+1)$. Thus,\n$$\n\\begin{cases}\ns_0 = 0,\\ s_1 = 8(2m+1), \\\\\ns_{i+2} = [4(2m+1)^2 - 2]s_{i+1} - s_i + 8(2m+1), & i \\ge 0\n\\end{cases}\n$$\nAll terms of $(s_i)$ are divisible by $8(2m+1)$. Since $x = 2m(m+1)n + (2m+1)$, $n = s_i$ for $i = 1, 2, 3, \\dots$ (since $s_0 = 0$ is not positive). Hence, all such $n$ are divisible by $8(2m+1)$. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21900,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, \\dots, a_{2023}$ be positive real numbers with\n$$\na_1 + a_2^2 + a_3^3 + \\dots + a_{2023}^{2023} = 2023.\n$$\nShow that\n$$\na_1^{2023} + a_2^{2022} + \\dots + a_{2022}^2 + a_{2023} > 1 + \\frac{1}{2023}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let us prove that conversely, the condition\n$$\na_1^{2023} + a_2^{2022} + \\dots + a_{2023} \\le 1 + \\frac{1}{2023}\n$$\nimplies that\n$$\nS := a_1 + a_2^2 + \\dots + a_{2023}^{2023} < 2023.\n$$\nThis is trivial if all $a_i$ are less than 1. So suppose that there is an $i$ with $a_i \\ge 1$, clearly it is unique and $a_i < 1 + \\frac{1}{2023}$. Then we have\n$$\n\\begin{aligned}\na_i^i &< \\left(1 + \\frac{1}{2023}\\right)^{2023} = 1 + \\sum_{k=1}^{2023} \\frac{1}{k!} \\cdot \\frac{2023}{2023} \\cdot \\frac{2022}{2023} \\dots \\cdot \\frac{2023-k+1}{2023} \\\\\n&< 1 + \\sum_{k=1}^{2023} \\frac{1}{k!} \\le 1 + \\sum_{k=0}^{2022} \\frac{1}{2^k} < 3,\n\\end{aligned}\n$$\n$$\n\\sum_{\\substack{k=1, \\\\ k \\ne i}}^{1011} a_k^k \\le 1011 \\quad \\text{and} \\quad \\sum_{\\substack{k=1012, \\\\ k \\ne i}}^{2023} a_k^k \\le \\sum_{\\substack{k=1012, \\\\ k \\ne i}}^{2023} a_k^{2024-k} < \\frac{1}{2023}.\n$$\nHence we have\n$$\nS = a_i^i + \\sum_{\\substack{k=1, \\\\ k \\ne i}}^{1011} a_k^k + \\sum_{\\substack{k=1012, \\\\ k \\ne i}}^{2023} a_k^k < 3 + 1011 + \\frac{1}{2023} < 2023.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21901,
"subject": "Mathematics (Olympiad)",
"question": "Let $(a_n)$ be a sequence of real numbers satisfying the inequality\n$$\na_{n+2}^2 + a_n a_{n+2} < a_{n+1}^2 + a_n + a_{n+2}\n$$\nfor all $n \\in \\mathbb{N}$. Prove that $a_{2023} \\leq 1$.",
"options": [],
"answer": "See solution",
"solution": "The given relation can be rewritten as:\n$$\na_n \\geq a_{n+2}(a_n - 1) + a_{n+1}^2 \\tag{1}\n$$\nor as\n$$\na_{n+2} \\geq a_n(a_{n+2} - 1) + a_{n+1}^2 \\tag{2}\n$$\nSuppose there exists a positive integer $k$ such that $a_{k+1} > 1$ and $a_{k+2} > 1$. From (2) with $n = k$ we get\n$$\na_{k+2} \\geq a_k(a_{k+2} - 1) + a_{k+1}^2 > a_{k+1}^2 > a_{k+1},\n$$\nwhile (1) with $n = k + 1$ gives\n$$\na_{k+1} \\geq a_{k+3}(a_{k+1} - 1) + a_{k+2}^2 > a_{k+2}^2 > a_{k+2},\n$$\nwhich is a contradiction.\n\nSuppose $a_{2023} > 1$. Then from the previous discussion, $a_{2022} \\leq 1$ and $a_{2024} \\leq 1$. Plugging $n = 2022$ into the given inequality, we get\n$$\n1 + a_{2022} a_{2024} < a_{2023}^2 + a_{2022} a_{2024} \\leq a_{2022} + a_{2024},\n$$\nso\n$$\n(1 - a_{2022})(1 - a_{2024}) < 0,\n$$\nwhich is impossible since $1 - a_{2022} \\geq 0$ and $1 - a_{2024} \\geq 0$. Therefore, $a_{2023} \\leq 1$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21902,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle, and let $r$ denote its inradius. Let $R_A$ denote the radius of the circle internally tangent at $A$ to the circumcircle of $ABC$ and tangent to the line $BC$; the radii $R_B$ and $R_C$ are defined similarly. Show that\n$$\n\\frac{1}{R_A} + \\frac{1}{R_B} + \\frac{1}{R_C} \\le \\frac{2}{r}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $a$, $h_A$, and $\\Delta$ denote the length of side $BC$, the length of the altitude from $A$ in triangle $ABC$, and the area of triangle $ABC$, respectively. Consider the circle internally tangent at $A$ to the circumcircle of $ABC$ and tangent at $T$ to the line $BC$. Then $2R_A \\ge AT \\ge h_A = \\frac{2\\Delta}{a}$, and equality holds throughout if and only if triangle $ABC$ is isosceles with apex at $A$. Similar inequalities hold for $R_B$ and $R_C$, and the conclusion follows at once; equality holds if and only if triangle $ABC$ is equilateral.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21903,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = 3$. Prove that\n\n$$\n\\sqrt{\\frac{b}{a^2 + 3}} + \\sqrt{\\frac{c}{b^2 + 3}} + \\sqrt{\\frac{a}{c^2 + 3}} \\leq \\frac{3}{2} \\sqrt[4]{abc}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $\\vec{u} = \\left( \\frac{1}{\\sqrt{a^2 + 3}}, \\frac{1}{\\sqrt{b^2 + 3}}, \\frac{1}{\\sqrt{c^2 + 3}} \\right)$ and $\\vec{v} = (\\sqrt{b}, \\sqrt{c}, \\sqrt{a})$. By the Cauchy–Schwarz inequality,\n\n$$\n\\left( \\sqrt{\\frac{b}{a^2 + 3}} + \\sqrt{\\frac{c}{b^2 + 3}} + \\sqrt{\\frac{a}{c^2 + 3}} \\right)^2 \\leq \\left( \\frac{1}{a^2 + 3} + \\frac{1}{b^2 + 3} + \\frac{1}{c^2 + 3} \\right)(a + b + c) = 3 \\left( \\frac{1}{a^2 + 3} + \\frac{1}{b^2 + 3} + \\frac{1}{c^2 + 3} \\right)\n$$\n\nSince $a^2 + 3 \\geq 4\\sqrt{a}$ (by AM-GM), similarly for $b$ and $c$,\n\n$$\n\\frac{1}{a^2 + 3} + \\frac{1}{b^2 + 3} + \\frac{1}{c^2 + 3} \\leq \\frac{1}{4} \\left( \\frac{1}{\\sqrt{a}} + \\frac{1}{\\sqrt{b}} + \\frac{1}{\\sqrt{c}} \\right)\n$$\n\nBy AM-GM,\n\n$$\n\\frac{1}{\\sqrt{a}} + \\frac{1}{\\sqrt{b}} + \\frac{1}{\\sqrt{c}} \\leq \\frac{a + b + c}{\\sqrt[4]{abc}}\n$$\n\nThus,\n\n$$\n\\left( \\sqrt{\\frac{b}{a^2 + 3}} + \\sqrt{\\frac{c}{b^2 + 3}} + \\sqrt{\\frac{a}{c^2 + 3}} \\right)^2 \\leq 3 \\cdot \\frac{a + b + c}{4 \\sqrt{abc}} = \\frac{9}{4 \\sqrt{abc}}\n$$\n\nTaking square roots,\n\n$$\n\\sqrt{\\frac{b}{a^2 + 3}} + \\sqrt{\\frac{c}{b^2 + 3}} + \\sqrt{\\frac{a}{c^2 + 3}} \\leq \\frac{3}{2} \\sqrt[4]{abc}\n$$\n\nEquality holds when $a = b = c = 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21904,
"subject": "Mathematics (Olympiad)",
"question": "The angles of a triangle add up to $180^\\circ$. Given that the angles are $\\hat{C}$, $5\\hat{C}$, and $120^\\circ$, find the value of $\\hat{C}$.",
"options": [],
"answer": "See solution",
"solution": "We have:\n\n$$\n\\hat{C} + 5\\hat{C} + 120^\\circ = 180^\\circ\n$$\n\nSo,\n\n$$\n6\\hat{C} = 60^\\circ \\implies \\hat{C} = 10^\\circ\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21905,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $k$ such that there exists a finite odd positive integer $n$ satisfying $n \\mid k^n + 1$.",
"options": [],
"answer": "See solution",
"solution": "We will prove that $k$ must have the form $2^m - 1$ where $m$ is a positive integer. Consider two cases:\n\nIf $k+1$ has an odd prime divisor $p$, consider $n = p^m$. Then:\n\n$$\nv_p(k^n + 1) = v_p(k+1) + v_p(n) > m\n$$\n\nso $n \\mid k^n + 1$. Therefore, there are infinitely many odd values of $n$ such that $n \\mid k^n + 1$.\n\nIf $k+1$ is a power of $2$, assume $n > 1$ and $n \\mid k^n + 1$. Let $p$ be the smallest prime divisor of $n$ and $h = \\operatorname{ord}_p k$. It is well-known that $2n$ and $p-1$ are divisible by $h$. Since $p$ is the smallest prime divisor, $\\gcd(n, p-1) = 1$ and $\\gcd(2n, p-1) = d \\mid 2$. So:\n\n- If $d=1$, then $p \\mid k-1$, but also $p \\mid k^n+1$ so $p \\mid 2$. But $n$ is odd, a contradiction.\n- If $d=2$, then $p \\mid k+1$, but $k+1$ is a power of $2$, so $p=2$. Again, a contradiction.\n\nTherefore, $k$ must have the form $2^m - 1$ where $m$ is a positive integer. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21906,
"subject": "Mathematics (Olympiad)",
"question": "In the figure below, $WXYZ$ is a rectangle with $WX = 4$ and $WZ = 8$. Point $M$ lies on $\\overline{XY}$, point $A$ lies on $\\overline{YZ}$, and $\\angle WMA$ is a right angle. The areas of triangles $\\triangle WXM$ and $\\triangle WAZ$ are equal. What is the area of $\\triangle WMA$?\n\n\n\n(A) 13 (B) 14 (C) 15 (D) 16 (E) 17",
"options": [],
"answer": "See solution",
"solution": "**Answer (C):** Label the diagram as shown, where $MX = a$ and $ZA = b$.\n\n\n\nThe Pythagorean Theorem on $\\triangle WMA$ gives $WM^2 + MA^2 = WA^2$, which implies that\n\n$$\n4^2 + a^2 + (8-a)^2 + (4-b)^2 = 8^2 + b^2.\n$$\n\nExpanding and simplifying yields $a^2 - 8a - 4b + 16 = 0$. Because the areas of triangles $\\triangle WXM$ and $\\triangle WAZ$ are equal, $\\frac{1}{2} \\cdot 4a = \\frac{1}{2} \\cdot 8b$, so $a = 2b$. Substituting into the previous equation and factoring gives $4(b-1)(b-4) = 0$. Therefore $b = 1$ or $b = 4$. But $b = 4$ would require $A = Y = M$, and $\\angle WMA$ would not exist, so it must be that $b = 1$ and $a = 2$. The area of $\\triangle WMA$ can be found by subtracting the three other triangle areas from the area of the rectangle:\n\n$$\n8 \\cdot 4 - \\frac{1}{2} \\cdot 4 \\cdot 2 - \\frac{1}{2} \\cdot 6 \\cdot 3 - \\frac{1}{2} \\cdot 8 \\cdot 1 = 32 - 4 - 9 - 4 = 15.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21907,
"subject": "Mathematics (Olympiad)",
"question": "Determine, with proof, all pairs $(a, b)$ of integers such that for any positive integer $n$, one has $n \\mid (a^n + b^{n-1})$.",
"options": [],
"answer": "See solution",
"solution": "The solution pairs consist of $(0, 0)$ and $(-1, -1)$.\n\nIf one of $a$ and $b$ is $0$, it is obvious that the other is also $0$.\n\nNow assume that $ab \\neq 0$. Select a large prime $p$ such that $p > |a + b^2|$. By Fermat's little theorem,\n\n$$\na^p + b^{p+1} \\equiv a + b^2 \\pmod{p}.\n$$\n\nAs $p \\mid (a^p + b^{p+1})$ and $p > |a + b^2|$, we have $a + b^2 = 0$.\n\nNext, select another prime $q$ such that $q > |b + 1|$ and $(q, b) = 1$.\n\nLet $n = 2q$. Then,\n\n$$\na^n + b^{n+1} = (-b^2)^{2q} + b^{2q+1} = b^{4q} + b^{2q+1} = b^{2q+1}(b^{2q-1} + 1).\n$$\n\nSince $n \\mid (a^n + b^{n+1})$ and $(q, b) = 1$, it follows that $q \\mid (b^{2q-1} + 1)$.\n\nBut\n\n$$\nb^{2q-1} + 1 \\equiv (b^{q-1})^2 \\cdot b + 1 \\equiv b + 1 \\pmod{q},\n$$\n\nand $q > |b + 1|$, so $b + 1 = 0$, i.e., $b = -1$, and thus $a = -b^2 = -1$.\n\nIn conclusion, the only solution pairs are $(0, 0)$ and $(-1, -1)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21908,
"subject": "Mathematics (Olympiad)",
"question": "As shown in the figure, $\\triangle ABC$ is an acute triangle with $AB < AC$. $AD$ is the altitude from $A$ to $BC$, and $P$ is a point on $AD$. Through $P$, draw $PE \\perp AC$ with $E$ as the foot, and $PF \\perp AB$ with $F$ as the foot. Let $O_1$ and $O_2$ be the circumcenters of $\\triangle BDF$ and $\\triangle CDE$, respectively. Prove that $O_1$, $O_2$, $E$, and $F$ are concyclic if and only if $P$ is the orthocenter of $\\triangle ABC$.\n\n",
"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 $CP$. Then $O_1O_2 \\parallel BC$, and $\\angle PO_2O_1 = \\angle PCB$.\n\nAs\n\n$$\nAF \\cdot AB = AP \\cdot AD = AE \\cdot AC,\n$$\n\nwe conclude that $B$, $C$, $E$, and $F$ are concyclic.\n\n**Sufficiency proof.** Assume $P$ is the orthocenter of $\\triangle ABC$. As $PE \\perp AC$ and $PF \\perp AB$, we know that points $B$, $O_1$, $P$, 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 proof.** 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.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21909,
"subject": "Mathematics (Olympiad)",
"question": "For a fixed $N > 1$, consider the integer $M = N^6 + N^3 + 1$. Show that $N^{4028} - N^{1166}$ is divisible by $M$.",
"options": [],
"answer": "See solution",
"solution": "Observe that $N^6 \\equiv -(N^3 + 1) \\pmod{M}$. This implies that\n\n$$\nN^9 \\equiv N^6 \\cdot N^3 \\equiv -(N^3 + 1)N^3 \\equiv -(N^6 + N^3) \\equiv 1 \\pmod{M}\n$$\n\nTherefore,\n\n$$\nN^{4028} - N^{1166} \\equiv N^5 - N^5 \\equiv 0 \\pmod{M}.\n$$\n\nThus, $N^{4028} - N^{1166}$ is divisible by $M$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21910,
"subject": "Mathematics (Olympiad)",
"question": "Given a positive integer $n$, divide a large equilateral triangle of side length $3n$ into $9n^2$ small equilateral triangles of side length $1$. Then color each small equilateral triangle with one of three colors: red, yellow, or blue, so that each color appears exactly $3n^2$ times. A trapezoid composed of three small equilateral triangles is called a *standard trapezoid*. For example, in the case $n = 2$ shown below, the two shaded parts are standard trapezoids. A standard trapezoid is called a *multicolored trapezoid* if the three small equilateral triangles in it are of different colors. Determine the maximum possible number of multicolored trapezoids. (These multicolored trapezoids do not need to be pairwise non-overlapping.)\n\n",
"options": [],
"answer": "See solution",
"solution": "The maximum number of multicolored trapezoids is $18n^2 - 9n$.\n\nUse the numbers $1$, $2$, and $3$ to represent the three colors. First, consider a small equilateral triangle of side length $2$ formed by $4$ small equilateral triangles. Let the middle small equilateral triangle be labeled $a$, and the other three be labeled $b$, $c$, and $d$. Then the three standard trapezoids within this triangle have the labels $(a, b, c)$, $(a, b, d)$, and $(a, c, d)$.\n\n\n\n1. If $a = b$, then at most $1$ of these three standard trapezoids can be multicolored. The same applies if $a = c$ or $a = d$.\n2. If $a$ is different from $b$, $c$, and $d$, then at least two of $b$, $c$, and $d$ must be equal, so at most $2$ of these standard trapezoids can be multicolored. Equality holds if and only if $a$ is different from $b$, $c$, and $d$, and $b$, $c$, and $d$ are not all equal.\n\nIt is easy to see that the number of small equilateral triangles of side length $2$ is\n\n$$\n(1 + 2 + \\cdots + (3n - 1)) + (1 + 2 + \\cdots + (3n - 3)) = 9n^2 - 9n + 3.\n$$\n\nThese equilateral triangles each contain distinct standard trapezoids, because the way to complete a standard trapezoid to a side length $2$ equilateral triangle is unique.\n\nApart from these standard trapezoids, there are some standard trapezoids that require an additional small equilateral triangle outside the large triangle to form a side length $2$ equilateral triangle.\n\nThe base of length $1$ of these trapezoids must be on the boundary of the large triangle. It is easy to see that the number of such standard trapezoids is\n\n$$3(3n - 2) = 9n - 6.$$\n\nBecause the number of multicolored trapezoids within a side length $2$ equilateral triangle does not exceed $2$, the number of multicolored trapezoids does not exceed\n\n$$\n2(9n^2 - 9n + 3) + (9n - 6) = 18n^2 - 9n.\n$$\n\nEquality holds if and only if the following two conditions are met simultaneously:\n\n(A) For any side length $2$ equilateral triangle formed by $4$ small equilateral triangles, the middle small equilateral triangle is different from the other three, and the other three are not all the same.\n\n(B) Each standard trapezoid with a base of length $1$ on the boundary of the large triangle must be multicolored.\n\nWe now give a construction that meets these two conditions simultaneously. Center the large triangle at its center and draw concentric equilateral triangles with side lengths $3$, $6$, ..., $3(n - 1)$ in a counterclockwise manner, dividing all small triangles into $n$ \"layers.\" As shown in the figure below:\n\n\n\nIn each layer of small equilateral triangles, remove the three corner small equilateral triangles (call these special triangles), so that the remaining small equilateral triangles form a loop (each adjacent pair shares a side). Starting from the innermost, the $k$th loop has $18k - 12$ small equilateral triangles, so we can start from a certain triangle and label them $1, 2, 3, 1, 2, 3, \\ldots$ in a clockwise manner until the loop is complete. This way, all standard trapezoids in the loop are multicolored.\n\nNow condition (B) is met, and for condition (A), consider the side length $2$ equilateral triangles. These triangles can be classified into three types: the first type consists of a standard trapezoid on a loop plus a special triangle; the second type consists of a standard trapezoid on a loop plus a small equilateral triangle from another loop; and the third type consists of an equilateral triangle with a special triangle at the center. Therefore, for the first and third types of side length $2$ equilateral triangles, the three corner small equilateral triangles are not all the same color.\n\nTherefore, we only need to simultaneously satisfy:\n\n(a) Any two adjacent small equilateral triangles from different loops have different labels;\n\n(b) Each special triangle has a different label from its neighbors, and except for the three corner special triangles, the three neighbors of each special triangle are not all the same color.\n\nFirst, we set the numbers in the innermost loop, and ensure that for any adjacent pair of small equilateral triangles from different loops, the number of the outer loop triangle is $1$ more than the number of the inner loop triangle (mod $3$). This can be achieved because moving $2$ steps clockwise on the inner loop reaches the next contact position, while the outer loop moves $2$ or $8$ steps (skipping a special triangle), ensuring that the outer loop triangle is always $1$ more than the inner loop triangle. This satisfies condition (a). For condition (b), we set the number of each special triangle to be $1$ more than its neighbor from the inner loop (mod $3$). Thus, if the number of the special triangle is $x$, the numbers of its three neighbors should be $x-1$, $x+1$, $x-1 \\pmod{3}$, satisfying (b).\n\nFinally, we count the occurrences of each color. Each color clearly appears equally in each loop. For the three small equilateral triangles in the loop that contact the outer special triangle, their labels are different since $\\frac{18k-12}{3} = 6k-4$ is not divisible by $3$. Thus, the three symmetrically placed special triangles have different labels, ensuring that each color appears equally.\n\nIn summary, the maximum number of multicolored trapezoids is $18n^2 - 9n$. $\\Box$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21911,
"subject": "Mathematics (Olympiad)",
"question": "Determine all positive integers $n \\ge 2$ for which the following statement is true:\n\nGiven any $n$ distinct points on the plane such that the distance between each pair of points is distinct, there exists a pair of points $A$, $B$ for which the difference between the number of points lying on either side of the perpendicular bisector of segment $AB$ is not greater than $1$.",
"options": [],
"answer": "See solution",
"solution": "The statement trivially holds for $n = 2$ and $n = 3$. We will show that it does not hold for any $n \\ge 4$.\n\nFirst, suppose that $n$ is even. Set up a coordinate system and place a point at $(0,0)$. Place $n/2 - 1$ points on the negative $x$-axis and $n/2 - 1$ points on the positive $x$-axis at $(-1/2^i, 0)$ and $(2^i, 0)$ for $i = 1, 2, \\dots, n-1$. Place the last point on the positive $y$-axis, so that the perpendicular bisector of the segment joining that point and any of the previously placed points divides the plane into two parts, one of which contains only the last point. (It is clear that we can do this, and that this construction serves as a counterexample to the given statement.)\n\nIf $n$ is odd, we simply add another point on the negative $y$-axis, so that the perpendicular bisector of the segment joining that point and any of the previously placed points on the $x$-axis divides the plane into two parts, one of which contains only the last point. Also, the distance from the $x$-axis of the point on the positive $y$-axis and the point on the negative $y$-axis should be distinct.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21912,
"subject": "Mathematics (Olympiad)",
"question": "Find possible positive solutions $(x, y, z)$ for the given system of equations:\n\n$$\n\\begin{cases}\nx + y^2 = 2z^3, \\\\\ny + z^2 = 2x^3, \\\\\nz + x^2 = 2y^3.\n\\end{cases}\n$$",
"options": [],
"answer": "See solution",
"solution": "Without loss of generality, let $z \\ge \\max\\{x, y\\}$.\n\n**Case 1.** $z > 1$. Then, due to the first equation, $x + y^2 \\le z + z^2 < 2z^3$ — contradiction.\n\n**Case 2.** $z = 1$. From the first equation, $2 = x + y^2$.\n\nSince $\\max\\{x, y\\} \\le 1$, it is possible only if $x = y = 1$. It is easy to check that $(1, 1, 1)$ is a solution.\n\n**Case 3.** $z < 1$. Then for the smallest variable, say $x$, $x < 1$ holds, then by the second equation:\n\n$x + y^2 \\le z + z^2 < 2z^3$ — contradiction.\n\nA similar contradiction holds if $y$ is the smallest variable.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21913,
"subject": "Mathematics (Olympiad)",
"question": "Given positive integers $k$ and $n$ ($n \\ge 2$), find the minimum constant $c$ satisfying this assertion: if $G$ is a simple $kn$-regular graph (the degree of each vertex is $kn$) with $m$ vertices, then each vertex can be coloured one of $n$ colours, such that the number of \"mono edges\" is at most $cm$. Here, a mono edge is an edge incident to two vertices of the same colour.\n\n",
"options": [],
"answer": "See solution",
"solution": "$$c = \\frac{k(kn - n + 2)}{2(kn + 1)}.$$ \n\nFirst, we prove $c \\ge \\frac{k(kn - n + 2)}{2(kn + 1)}$. Consider a complete graph $K_{kn+1}$ with any $n$-colouring of the vertices. Suppose that the numbers of vertices coloured by $1, 2, \\ldots, n$ are $a_1, a_2, \\ldots, a_n$, respectively. Then\n\n$$\n\\begin{aligned}\na_1^2 + a_2^2 + \\dots + a_n^2 &\\ge \\frac{(a_1 + a_2 + \\dots + a_n)^2}{n} \\\\\n&= \\frac{(kn + 1)^2}{n} \\\\\n&= k^2 n + 2k + \\frac{1}{n}.\n\\end{aligned}\n$$\n\nAs $a_1, a_2, \\ldots, a_n$ are integers,\n\n$$\na_1^2 + a_2^2 + \\dots + a_n^2 \\ge k^2 n + 2k + 1.\n$$\n\nTherefore, the number of mono edges is:\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} \\binom{a_i}{2} &= \\frac{1}{2} \\left( \\sum_{i=1}^{n} a_i^2 - \\sum_{i=1}^{n} a_i \\right) \\\\\n&\\ge \\frac{1}{2} (k^2 n + 2k + 1 - (kn + 1)) \\\\\n&= \\frac{1}{2} k(kn - n + 2),\n\\end{aligned}\n$$\n\nwhich implies $c \\ge \\frac{k(kn - n + 2)}{2(kn + 1)}$.\n\nNext, we give three proofs that $c = \\frac{k(kn - n + 2)}{2(kn + 1)}$ satisfies the problem assertion.\n\n**Proof 1** For any $kn$-regular graph $G$, it is well known that $G$ has a $(kn+1)$-colouring such that adjacent vertices are of different colours. Divide the $kn+1$ colours randomly into $n$ groups such that one group contains $k+1$ colours while each of the others contains $k$ colours, and for all these divisions, they have equal probabilities. Now change all colours of a group into one colour. For any edge of $G$, initially, it is incident to vertices of different colours; after changing the colours, it is incident to vertices of the same colour, namely, it becomes a mono edge with a probability\n\n$$\n\\frac{\\binom{k+1}{2} + (n-1)\\binom{k}{2}}{\\binom{kn+1}{2}} = \\frac{k(k+1) + k(k-1)(n-1)}{(kn+1)(kn)} = \\frac{kn - n + 2}{n(kn+1)}.\n$$\n\nTherefore, after changing the colours, the expected number of mono edges is\n\n$$\n\\frac{kn - n + 2}{n(kn + 1)} \\cdot \\frac{knm}{2} = m \\cdot \\frac{k(kn - n + 2)}{2(kn + 1)} = cm.\n$$\n\nThere must exist an $n$-colouring of $G$ such that the number of mono edges is less than or equal to $cm$. This verifies the problem assertion.\n\n**Proof 2** Suppose that among all $n$-colourings of $G$, the minimum number of mono edges is $T(G)$. Let $V$ consist of vertices of $G$. For a permutation $\\pi = (v_1, v_2, \\ldots, v_m)$ of $V$, colour $v_1, v_2, \\ldots, v_m$ in order (by one of $n$ colours) as follows. For a vertex $v$, suppose that in $\\pi$, $A_{\\pi}(v)$ of $kn$ neighbouring vertices of $v$ appear before $v$. We colour $v$ with the colour that appears least frequently among those $A_{\\pi}(v)$ neighbours (it appears at most $\\lfloor A_{\\pi}(v)/n \\rfloor$ times). According to this greedy algorithm, it is guaranteed that\n\n$$\nT(G) \\leq F(\\pi) := \\sum_{v \\in V} \\left\\lfloor \\frac{A_{\\pi}(v)}{n} \\right\\rfloor.\n$$\n\nThe above inequality holds for any permutation $\\pi$. It suffices to prove, for all $m!$ permutations, the average of $F(\\pi)$ is $\\frac{k(kn - n + 2)}{2(kn + 1)} \\cdot m$, so that some permutation $\\pi$ gives a desired $n$-colouring. For a vertex $v$, let its neighbours be $\\{u_1, u_2, \\ldots, u_{kn}\\}$. Consider $B = \\{v, u_1, u_2, \\ldots, u_{kn}\\}$: among the $m!$ relative orders of elements of $B$ given by all $m!$ permutations, $v$ appears at position $1, 2, \\ldots, kn + 1$ with equal probability $1/(kn + 1)$. Hence, as $\\pi$ traverses all $m!$ permutations, the average of $\\left\\lfloor A_{\\pi}(v)/n \\right\\rfloor$ is\n\n$$\n\\begin{aligned}\n\\frac{1}{kn+1} \\sum_{a=0}^{kn} \\left\\lfloor \\frac{a}{n} \\right\\rfloor &= \\frac{1}{kn+1} (0 \\times n + 1 \\times n + \\dots + (k-1) \\times n + k) \\\\\n&= \\frac{k(kn - n + 2)}{2(kn + 1)}.\n\\end{aligned}\n$$\n\nAs a result, the average of $F(\\pi)$ is $\\frac{k(kn - n + 2)}{2(kn + 1)} \\cdot m = cm$.\n\n**Proof 3** For a $kn$-regular graph $G$, let $A = |V(G)|$ be the number of vertices of $G$. Since there are only finitely many $n$-colourings of $G$, there must be one with the minimum number of mono edges, say $C$. We have the following claims on $C$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21914,
"subject": "Mathematics (Olympiad)",
"question": "Determine all positive integers $k$ and $n$ satisfying the equation\n\n$$k^2 - 2016 = 3^n$$",
"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 $l^2 - 224 = 3^{n-2}$. Modulo $3$, 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": 21915,
"subject": "Mathematics (Olympiad)",
"question": "Given a positive integer $k \\ge 2$, set $a_1 = 1$ and, for every integer $n \\ge 2$, let $a_n$ be the smallest solution of the equation\n\n$$\nx = 1 + \\sum_{i=1}^{n-1} \\lfloor \\sqrt[k]{x/a_i} \\rfloor\n$$\n\nthat exceeds $a_{n-1}$. Prove that all primes are among the terms of the sequence $a_1, a_2, \\dots$.",
"options": [],
"answer": "See solution",
"solution": "We prove that the $a_n$ are precisely the $k$th-power-free positive integers, that is, those divisible by the $k$th power of no prime. The conclusion then follows.\n\nLet $B$ denote the set of all $k$th-power-free positive integers. We first show that, given a positive integer $c$,\n\n$$\n\\sum_{b \\in B,\\ b \\le c} \\lfloor \\sqrt[k]{c/b} \\rfloor = c.\n$$\n\nTo this end, notice that every positive integer has a unique representation as a product of an element in $B$ and a $k$th power. Consequently, the set of all positive integers less than or equal to $c$ splits into\n\n$$\nC_b = \\{x : x \\in \\mathbb{Z}_{>0},\\ x \\le c,\\ \\text{and}\\ x/b\\ \\text{is a}\\ k\\text{th power} \\},\\ \\quad b \\in B,\\ b \\le c.\n$$\n\nClearly, $|C_b| = \\lfloor \\sqrt[k]{c/b} \\rfloor$, whence the desired equality.\n\nFinally, enumerate $B$ according to the natural order: $1 = b_1 < b_2 < \\dots < b_n < \\dots$. We prove by induction on $n$ that $a_n = b_n$. Clearly, $a_1 = b_1 = 1$, so let $n \\ge 2$ and assume $a_m = b_m$ for all indices $m < n$. Since $b_n > b_{n-1} = a_{n-1}$ and\n\n$$\nb_n = \\sum_{i=1}^{n} \\lfloor \\sqrt[k]{b_n/b_i} \\rfloor = \\sum_{i=1}^{n-1} \\lfloor \\sqrt[k]{b_n/b_i} \\rfloor + 1 = \\sum_{i=1}^{n-1} \\lfloor \\sqrt[k]{b_n/a_i} \\rfloor + 1,\n$$\n\nthe definition of $a_n$ forces $a_n \\le b_n$. Were $a_n < b_n$, a contradiction would follow:\n\n$$\na_n = \\sum_{i=1}^{n-1} \\lfloor \\sqrt[k]{a_n/b_i} \\rfloor = \\sum_{i=1}^{n-1} \\lfloor \\sqrt[k]{a_n/a_i} \\rfloor = a_n - 1.\n$$\n\nConsequently, $a_n = b_n$. This completes the proof.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21916,
"subject": "Mathematics (Olympiad)",
"question": "Consider integers $m \\ge 2$ and $n \\ge 1$. Show that there is a polynomial $P(x)$ of degree equal to $n$ with integer coefficients such that $P(0), P(1), \\dots, P(n)$ are all perfect powers of $m$.",
"options": [],
"answer": "See solution",
"solution": "Let $a_0, a_1, \\dots, a_n$ be integers to be chosen later, and consider the polynomial $P(x) = \\frac{1}{n!} Q(x)$ where\n\n$$\nQ(x) = \\sum_{k=0}^{n} (-1)^{n-k} \\binom{n}{k} a_k \\prod_{\\substack{0 \\le i \\le n \\\\ i \\ne k}} (x-i).\n$$\n\nObserve that for $l \\in \\{0, 1, \\dots, n\\}$ we have\n\n$$\n\\begin{aligned}\nP(l) &= \\frac{1}{n!} (-1)^n \\binom{n}{l} a_l \\prod_{\\substack{0 \\le i \\le n \\\\ i \\ne l}} (l-i) \\\\\n&= \\frac{1}{n!} (-1)^{n-l} \\binom{n}{l} a_l l! (-1)^{n-l} (n-l)! \\\\\n&= a_l\n\\end{aligned}\n$$\n\nSo $P(x)$ is the unique polynomial of degree at most $n$ such that $P(l) = a_l$. (Any two polynomials of degree at most $n$ agreeing on $n+1$ distinct values are equal.) Note in particular that\n\n$$\n\\sum_{k=0}^{n} (-1)^{n-k} \\binom{n}{k} \\prod_{\\substack{0 \\le i \\le n \\\\ i \\ne k}} (x-i) = n! \\quad (*)\n$$\n\nIf $p$ is a prime dividing $n!$, let $r_p$ be maximal such that $p^{r_p}$ divides $n!$. If $p$ divides $m$, then there is an integer $d_p$ such that $m^{d_p} \\equiv 0 \\pmod{p^{r_p}}$, for example $d_p = r_p$ will do. If $p$ does not divide $m$, then there is an integer $d_p$ such that $m^{d_p} \\equiv 1 \\pmod{p^{r_p}}$, for example, by Euler's theorem, $d_p = \\varphi(p^{r_p})$ will do. Let $d = d_1 d_2 \\dots d_p$ and observe that for every positive integer $t$ we have $m^{td} \\equiv 0 \\pmod{p^{r_p}}$ if $p \\mid m$ and $m^{td} \\equiv 1 \\pmod{p^{r_p}}$ if $p \\nmid m$.\n\nNow let $t_0, \\dots, t_n$ be positive integers to be chosen later and define $a_k = m^{t_k d}$. We will show that the polynomial $P(x)$ has integer coefficients. We will also show that there is an appropriate choice of $t_0, \\dots, t_n$ such that $P(x)$ has degree exactly equal to $n$.\n\nTo show that $P(x)$ has integer coefficients, it is enough to show that for every $p$ dividing $n!$, all coefficients of $Q(x)$ are multiples of $p^{r_p}$. This is immediate if $p$ divides $m$ as all $a_k$'s are multiples of $p^{r_p}$. If $p$ does not divide $m$ then we have $a_k \\equiv 1 \\pmod{p^{r_p}}$ for every $0 \\le k \\le n$ and so by $(*)$\n\n$$\nQ(x) = \\sum_{k=0}^{n} (-1)^{n-k} \\binom{n}{k} \\prod_{\\substack{0 \\le i \\le n \\\\ i \\ne k}} (x-i) \\equiv n! \\pmod{p^{r_p}}.\n$$\n\nThis shows that all coefficients of $Q(x)$ are indeed multiples of $p^{r_p}$. It remains to show that there is a choice of $t_0, \\dots, t_n$ guaranteeing that the degree of $P(x)$ is exactly equal to $n$. One such choice is $t_0 = 2$ and $t_1 = \\dots = t_n = 1$. This works because if $P(x)$ had degree less than $n$, then looking at the values $P(1), \\dots, P(n)$ we would get that $P(x)$ is constant. But this is impossible as $P(0) \\ne P(1)$.\n\nNote: Looking at the coefficient of $x^n$ in the definition of $P(x)$, it is not difficult to see that if we fix any $t_1, \\dots, t_n$ and pick $t_0$ large enough we will get that this coefficient is non-zero. In particular, we can additionally guarantee that $P(0), P(1), \\dots, P(n)$ are distinct perfect powers of $m$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21917,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer and let $a_1, a_2, \\dots, a_k$ ($k \\ge 2$) be distinct integers in the set $\\{1, \\dots, n\\}$ such that $n$ divides $a_i(a_{i+1} - 1)$ for $i = 1, \\dots, k-1$. Prove that $n$ does not divide $a_k(a_1 - 1)$.",
"options": [],
"answer": "See solution",
"solution": "We first prove by induction that for any integer $2 \\leq i \\leq k$, $n \\mid a_1(a_i - 1)$.\n\nIf $i = 2$, we have $n \\mid a_1(a_2 - 1)$ by assumption. Suppose that $n \\mid a_1(a_i - 1)$ for some $2 \\leq i \\leq k-1$; then\n\n$$\nn \\mid a_1(a_i - 1)(a_{i+1} - 1).\n$$\n\nBy assumption, $n \\mid a_i(a_{i+1} - 1)$, so $n \\mid a_1 a_i(a_{i+1} - 1)$, and therefore\n\n$$\nn \\mid \\big(a_1 a_i(a_{i+1} - 1) - a_1(a_i - 1)(a_{i+1} - 1)\\big),\n$$\n\ni.e., $n \\mid a_1(a_{i+1} - 1)$. By induction, $n \\mid a_1(a_i - 1)$ holds for every integer $2 \\leq i \\leq k$. In particular, $n \\mid a_1(a_k - 1)$.\n\nSince $a_1(a_k - 1) - a_k(a_1 - 1) = a_k - a_1$ is not divisible by $n$ (because $a_1$ and $a_k$ are distinct elements in $\\{1, \\dots, n\\}$), we conclude that $a_k(a_1 - 1)$ is not divisible by $n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21918,
"subject": "Mathematics (Olympiad)",
"question": "$$\nx^2 + y^2 + z^2 \\geq x^3 + y^3 + z^3 + 6xyz.\n$$\n\nIt 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. Based on this, prove that if $x + y + z = 1$, then\n\n$$\nx^2 + y^2 + z^2 \\geq x^3 + y^3 + z^3 + 6xyz.\n$$",
"options": [],
"answer": "See solution",
"solution": "To prove the inequality, observe that when $x + y + z = 1$, we have:\n\n$$\nxy + yz + zx \\geq 9xyz\n$$\n\nApplying 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 the constraint and the previous inequality:\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\nFrom which\n\n$$\nx^2 + y^2 + z^2 \\geq x^3 + y^3 + z^3 + 6xyz\n$$\nfollows. Equality holds when $x = y = z = \\frac{1}{3}$, i.e., when $P$ is the centroid of the triangle.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21919,
"subject": "Mathematics (Olympiad)",
"question": "a) Let $z_1, z_2, z_3, z_4$ be distinct complex numbers with zero sum and equal absolute values. Prove that the points with affixes $z_1, z_2, z_3, z_4$ are the vertices of a rectangle.\n\nb) Let $x, y, z, t$ be real numbers such that $\\sin x + \\sin y + \\sin z + \\sin t = 0$ and $\\cos x + \\cos y + \\cos z + \\cos t = 0$. Prove that, for every integer $n$,\n\n$$\n\\sin((2n + 1)x) + \\sin((2n + 1)y) + \\sin((2n + 1)z) + \\sin((2n + 1)t) = 0.\n$$",
"options": [],
"answer": "See solution",
"solution": "a) The equality $z_1 + z_2 + z_3 + z_4 = 0$ implies $\\bar{z}_1 + \\bar{z}_2 + \\bar{z}_3 + \\bar{z}_4 = 0$ and, furthermore, $\\frac{1}{z_1} + \\frac{1}{z_2} + \\frac{1}{z_3} + \\frac{1}{z_4} = 0$ for $|z_1| = |z_2| = |z_3| = |z_4| \\neq 0$.\n\nSuppose $z_1 + z_2 = -z_3 - z_4 \\neq 0$. The previous relation gives $z_1 z_2 = z_3 z_4$, so $\\{z_1, z_2\\} = \\{-z_3, -z_4\\}$. On the other hand, if $z_1 + z_2 = 0$, then $z_3 + z_4 = 0$. In both cases, the numbers $z_1, z_2, z_3, z_4$ form two pairs of equal sum, hence the conclusion.\n\nb) Let $z_1 = \\cos x + i \\sin x$, $z_2 = \\cos y + i \\sin y$, $z_3 = \\cos z + i \\sin z$, and $z_4 = \\cos t + i \\sin t$ so that $z_1 + z_2 + z_3 + z_4 = 0$ and $|z_1| = |z_2| = |z_3| = |z_4| = 1$.\n\nAs before, the numbers $z_1, z_2, z_3, z_4$ form two pairs of opposite numbers, so the same goes for $z_1^{2n+1}, z_2^{2n+1}, z_3^{2n+1}, z_4^{2n+1}$. Therefore,\n$$\nz_1^{2n+1} + z_2^{2n+1} + z_3^{2n+1} + z_4^{2n+1} = 0,\n$$\nwhich implies the claim.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21920,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be a triangle such that $AB \\neq AC$. We denote its orthocentre by $H$, its circumcentre by $O$, and the midpoint of $BC$ by $D$. The extensions of $HD$ and $AO$ meet at $P$. Prove that triangles $AHP$ and $ABC$ have the same centroid.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $AO$ intersect the circumcircle at $X$. Since $AX$ is a diameter, $\\angle ABX$ is a right angle, hence $BX \\parallel CH$ (both are perpendicular to $AB$). For the same reason, $CX \\parallel BH$, so $BHCX$ is a parallelogram. This means that $XH$ passes through $D$, the midpoint of $BC$. Hence $X$ coincides with $P$ (it lies on both $AO$ and $HD$), and $D$ is also the midpoint of $HX = HP$.\n\nBut this implies that $AD$ is a median in both triangles $ABC$ and $AHP$. Since the centroid always divides the median in a $2:1$ ratio, we conclude that the centroids of $ABC$ and $AHP$ coincide as well.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21921,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c \\ge 0$ be non-negative numbers satisfying $a^3 + b^3 + c^3 = abc + 2$. Prove that\n$$\na^4 + b^4 + c^4 \\ge a + b + c.\n$$",
"options": [],
"answer": "See solution",
"solution": "It suffices to prove\n$$\nF(a, b, c) = 2(a^4 + b^4 + c^4) - (a + b + c)(a^3 + b^3 + c^3 - abc) \\ge 0\n$$\nfor all $a, b, c \\ge 0$. We may assume $b \\ge a$ and $c \\ge a$, and then we have\n$$\nF(a, b, c) = (b - c)^2 \\big[ b(b - a) + bc + c(c - a) \\big] + a^2(c - a)(b - a) \\ge 0.\n$$\nEquality holds for $(a, b, c) = (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 1, 1)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21922,
"subject": "Mathematics (Olympiad)",
"question": "Tetra has a $M \\times N$ square grid and a tetromino as shown in the figure, where $M$ and $N$ are positive integers. What is the lowest number of squares Tetra has to color, so that it is impossible for Tetra to place the tetromino on the grid without covering a colored square? The piece can be rotated and flipped freely, but must cover four squares completely.\n\n",
"options": [],
"answer": "See solution",
"solution": "For $M, N \\le 2$, the answer is $0$ since the tetromino does not fit on the grid.\n\nFor larger grids, the minimum number of colored squares required is $\\left\\lfloor \\frac{M}{2} \\right\\rfloor \\left\\lfloor \\frac{N}{2} \\right\\rfloor$.\n\nA coloring with $\\left\\lfloor \\frac{M}{2} \\right\\rfloor \\left\\lfloor \\frac{N}{2} \\right\\rfloor$ colored squares makes placing the piece impossible:\n\n\n\nFor even $M, N$, the grid can be subdivided into disjoint $4 \\times 2$ and $6 \\times 2$ rectangles (assuming $M > 2$). A $4 \\times 2$ rectangle requires at least $2$ colored squares to block the tetromino, and a $6 \\times 2$ requires at least $3$, as can be shown by brute force. Thus, on average, every fourth square must be colored, so the total number of colored squares must be at least $\\frac{MN}{4}$, as desired.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21923,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle with $\\angle AC > \\angle AB$ and circumcenter $O$. The tangents to the circumcircle at $A$ and $B$ intersect at $T$. The perpendicular bisector of the side $BC$ intersects $AC$ at $S$.\n\n(a) Prove that the points $A$, $B$, $O$, $S$ and $T$ lie on a common circle.\n\n(b) Prove that the line $ST$ is parallel to the side $BC$.\n",
"options": [],
"answer": "See solution",
"solution": "Since $\\angle AT$ and $\\angle BT$ are perpendicular to $\\angle AO$ and $\\angle BO$, the points $A$, $B$, $T$ and $O$ lie on a circle $k_1$ by Thales' theorem. By the central angle theorem we have $\\angle AOB = 2\\gamma$. Since $BCS$ is an isosceles triangle, we find $\\angle BCS = \\angle CBS = \\gamma$. Now $\\angle ASB = 2\\gamma$, because an exterior angle of a triangle equals the sum of the other two interior angles. Thus\n\n$$\\angle ASB = 2\\gamma = \\angle AOB.$$\n\nBy the inscribed angle theorem, the points $A$, $B$, $S$ and $O$ lie on a circle $k_2$. Since the circles $k_1$ and $k_2$ have the three points $A$, $B$ and $O$ in common, we have $k_1 = k_2$, and the points $A$, $B$, $O$, $S$ and $T$ lie on a circle.\n\nFinally, we have $\\angle TSB = \\angle TOB = \\gamma = \\angle SBC$, from which it follows that $ST$ is parallel to $BC$.\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21924,
"subject": "Mathematics (Olympiad)",
"question": "令 $n > 3$ 為一正整數。有 $n$ 個寶寶坐成一圈，且有 $n$ 片餅乾被分配給這些寶寶（有些寶寶可能沒有餅乾）。在每一回合，我們讓其中一個有至少 2 片餅乾的寶寶給他左右兩邊的寶寶各 1 片餅乾。試決定能在有限回合內讓 $n$ 個寶寶同時有 1 片餅乾的所有起始餅乾配置。（註：假設寶寶不會把餅乾吃掉。）",
"options": [],
"answer": "See solution",
"solution": "答案為所有滿足\n\n$$\n\\sum_{i=1}^{n} i c_{i} \\equiv \\frac{n(n+1)}{2} \\pmod{n}\n$$\n\n的起始配置，其中 $c_i$ 為第 $i$ 個寶寶一開始有的餅乾個數。\n\n**解法一** 首先證明必要性。定義 $c_{n+i} = c_i$。注意到\n\n$$\n(i-1)(c_{i-1} + 1) + i(c_i - 2) + (i+1)(c_{i+1} + 1) = (i-1)c_{i-1} + ic_i + (i+1)c_{i+1},\n$$\n\n因此 $\\sum ic_i$ 在模 $n$ 下是一個不變量。故若寶寶們最終可以每人一片餅乾，必然有\n\n$$\n\\sum_{i=1}^{n} i c_i \\equiv \\sum_{i=1}^{n} i \\times 1 = \\frac{n(n+1)}{2} \\pmod{n}.\n$$\n\n接著證明充分性。對於所有 $M \\ge 1$，若 $c_n = M$ 且 $c_i \\le 1$ 對於所有 $i \\ne n$ 皆成立，則我們稱寶寶們 $M$-uniform。以下分引理說明：\n\n**引理一：** 對於每個起始配置，都存在 $M \\ge 1$ 使得我們可以讓寶寶們 $M$-uniform。\n\n*Proof.* 考慮以下策略：在任一回合，選編號 1 到 $n-1$ 的寶寶中，有 2 片以上餅乾的任何一個人執行操作。易知這樣的操作終會在有限回合內無法執行，而那時 1 到 $n-1$ 的寶寶至多都只能有 1 片餅乾，故為 $M$-uniform。□\n\n**引理二：** 對於每個起始配置，都存在 $N \\le 2$ 使得我們可以讓寶寶們 $N$-uniform。\n\n*Proof.* 先用引理一操作到寶寶是 $M$-uniform。若此時 $M \\le 2$ 則證畢，故只需考慮 $M \\ge 3$ 的情況。我們將證明，此時我們總是可以從 $M$-uniform 操作成 $(M-1)$-uniform 或 $(M-2)$-uniform，從而證明引理。\n\n例如考慮 $\\sum i^2 c_i$，且注意到這個量在此操作下遞增（因 $(i-1)^2 + (i+1)^2 > 2i^2$）。\n\n注意此時必然有至少 3 個寶寶是沒有餅乾的，考慮其中在 $M$ 左右兩側最靠近的兩位，其間寶寶的餅乾數量必形如\n\n$$\n0, \\overbrace{1, 1, \\dots, 1}^{x}, M, \\overbrace{1, 1, \\dots, 1}^{y}, 0.\n$$\n\n我們將用數學歸納法證明，對於任何 $(x, y) \\in \\mathbb{Z}_{\\ge 0} \\times \\mathbb{Z}_{\\ge 0}$，我們都可以讓 $M$ 傳遞一片餅乾到至少一個 0 的位置，從而降成 $(M-1)$ 或 $(M-2)$-uniform。以下不失一般性假設 $x \\ge y$。\n\n- $(0,0)$：此時只要對 $M$ 操作，讓 $0, M, 0$ 變成 $1, M-2, 1$ 便得證。\n- $(x,0)$：先對 $M$ 操作，之後從 $M$ 左邊的嬰兒開始依次向左操作一次，直到左邊的 $0$ 變成 $1$，此時的分佈為\n\n$$\n1, 0, \\overbrace{1, 1, \\dots, 1}^{x-1}, M-1, 1\n$$\n\n故為 $(M-1)$-uniform，得證。\n- $(x, y)$：假設命題對所有 $y < N$ 成立。則當 $y = N$ 時，先對 $M$ 操作，之後從 $M$ 左邊的嬰兒開始依次向左操作一次，直到左邊的 $0$ 變成 $1$，再從 $M$ 右邊的嬰兒開始依次向右操作一次，直到右邊的 $0$ 變成 $1$，此時的分佈為\n\n$$\n1, 0, \\overbrace{1, 1, \\dots, 1}^{x-1}, M, \\overbrace{1, 1, \\dots, 1}^{y-1}, 0, 1.\n$$\n\n但由歸納假設，對於中間從 0 到 0 的子列，我們必可讓 $M$ 傳遞一片餅乾到至少一個 0 的位置，故得證。\n\n\n\n**引理三：** 對於滿足 (1) 的起始配置，我們都可以讓寶寶們 1-uniform，從而滿足題意。\n\n*Proof.* 由引理二知可以讓寶寶們 1-uniform 或 2-uniform。但對於 2-uniform，必然存在唯一的一個 $c_j = 0$，從而\n\n$$\n\\sum_{i=1}^{n} i c_{i} \\equiv \\sum_{i=1}^{n-1} i c_{i} + n \\equiv \\sum_{i=1}^{n} i - j \\equiv \\frac{n(n+1)}{2} - j \\pmod{n}\n$$\n\n此與 (1) 不合。故，我們必會將寶寶操作成 1-uniform。證明完畢。\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21925,
"subject": "Mathematics (Olympiad)",
"question": "Call two triangles in $T$ adjacent if they share an edge. Consider a triangulation $T$ of a 100-gon, and let $P(T)$ be the set of diagonals drawn in $T$.\n\nLet $G$ be a graph with 98 vertices, each corresponding to a triangle in $T$, and edges connecting vertices whose triangles are adjacent (i.e., share an edge). For each vertex $i$ in $G$, let $d_i$ be its degree ($1 \\leq d_i \\leq 3$). Let $r, s, t$ be the number of vertices of degree 1, 2, and 3, respectively, so $r + s + t = 98$.\n\nDefine $f(T)$ as the number of pairs of adjacent triangles sharing a vertex:\n\n$$\nf(T) = \\sum_{i=1}^{98} \\binom{d_i}{2}\n$$\n\nFind the minimum and maximum possible values of $f(T)$ over all triangulations $T$ of the 100-gon.",
"options": [],
"answer": "See solution",
"solution": "We have $r + s + t = 98$ and $r + 2s + 3t = 2 \\times 97 = 194$. Also, $f(T) = s + 3t$.\n\nExpress $s$ and $t$ in terms of $r$:\n\nFrom $r + s + t = 98$ and $r + 2s + 3t = 194$:\nSubtract the first from the second:\n$$\n(r + 2s + 3t) - (r + s + t) = 194 - 98 \\\\\n(s + 2t) = 96\n$$\nSo $s = 96 - 2t$.\n\nBut $f(T) = s + 3t = (96 - 2t) + 3t = 96 + t$.\n\nThe minimum $t$ occurs when as many vertices as possible have degree 1. Since each triangle with degree 1 corresponds to a triangle sharing two edges with the 100-gon, and there are 100 sides, at most 50 such triangles can exist, so $r \\leq 50$. By induction, $r \\geq 2$.\n\nSince $r + s + t = 98$ and $s + 2t = 96$, we can solve for $t$:\nLet $r = x$.\nThen $s + t = 98 - x$ and $s + 2t = 96$.\nSubtract:\n$(s + 2t) - (s + t) = 96 - (98 - x)$\n$t = x - 2$\nSo $t = r - 2$.\n\nTherefore,\n$$\nf(T) = 96 + t = 96 + (r - 2) = r + 94\n$$\nWith $2 \\leq r \\leq 50$,\n$$\n96 \\leq f(T) \\leq 144\n$$\n\nFor the minimum, draw all diagonals from one vertex (\"fan\" triangulation), so $r = 2$, $f(T) = 96$.\nFor the maximum, cut off 50 triangles sharing two edges with the 100-gon, then triangulate the remaining 50-gon arbitrarily, so $r = 50$, $f(T) = 144$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21926,
"subject": "Mathematics (Olympiad)",
"question": "As seen in the figure below, $AB$ is a chord of circle $\\omega$, $P$ is a point on arc $AB$, and $E, F$ are two points on $AB$ such that $AE = EF = FB$. Connect $PE$ and $PF$ and extend them to intersect $\\omega$ at $C$ and $D$, respectively. Prove that\n\n$$\nEF \\cdot CD = AC \\cdot BD.\n$$\n\n\n\n",
"options": [],
"answer": "See solution",
"solution": "As shown in the figure below, connect $AD$, $BC$, $CF$, and $DE$. Since $AE = EF = FB$, we have\n\n$$\n\\frac{BC \\cdot \\sin \\angle BCE}{AC \\cdot \\sin \\angle ACE} = \\frac{\\text{distance between } B \\text{ and } CP}{\\text{distance between } A \\text{ and } CP} = \\frac{BE}{AE} = 2. \\qquad \\textcircled{1}\n$$\n\nSimilarly,\n\n$$\n\\frac{AD \\cdot \\sin \\angle ADF}{BD \\cdot \\sin \\angle BDF} = \\frac{\\text{distance between } A \\text{ and } PD}{\\text{distance between } B \\text{ and } PD} = \\frac{AF}{BF} = 2. \\qquad \\textcircled{2}\n$$\n\nOn the other hand, since\n\n$$\n\\angle BCE = \\angle BCP = \\angle BDP = \\angle BDF,\n$$\n\n$$\n\\angle ACE = \\angle ACP = \\angle ADP = \\angle ADF,\n$$\n\nMultiplying (1) by (2), we have $\\frac{BC \\cdot AD}{AC \\cdot BD} = 4$, or\n\n$$\nBC \\cdot AD = 4AC \\cdot BD. \\qquad \\textcircled{3}\n$$\n\nBy Ptolemy's Theorem,\n\n$$\nAD \\cdot BC = AC \\cdot BD + AB \\cdot CD. \\qquad \\textcircled{4}\n$$\n\nCombining (3) and (4), we get $AB \\cdot CD = 3AC \\cdot BD$, and thus\n\n$$\nEF \\cdot CD = AC \\cdot BD.\n$$\n\nThe proof is complete.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21927,
"subject": "Mathematics (Olympiad)",
"question": "We consider points with integer coordinates in the rectangle with corners at $(0,0)$, $(n,0)$, $(n,2)$, and $(0,2)$. It is possible to move from a point $(a,b)$ in the rectangle to either $(a+1,b)$, $(a+1,b+1)$, or $(a,b-1)$ if the second point is also in the given rectangle.\n\nHow many possible paths are there from $(0,0)$ to $(n,2)$ under these rules?",
"options": [],
"answer": "See solution",
"solution": "Let $a_k$, $b_k$, and $c_k$ be the number of possible paths leading from $(0,0)$ to $(k,0)$, $(k,1)$, and $(k,2)$ respectively. It is obvious that $a_0 = 1$, $b_0 = 0$, and $c_0 = 0$. Furthermore, for $k \\ge 1$ we have the recursive equations\n\n$$\n\\begin{aligned}\nc_k &= b_{k-1} + c_{k-1}, \\\\\nb_k &= a_{k-1} + b_{k-1} + c_k, \\\\\na_k &= a_{k-1} + b_k.\n\\end{aligned}\n$$\n\nFrom the first equation, we obtain $b_m = c_{m+1} - c_m$ for $m \\ge 0$. Substituting in the second equation therefore yields\n\n$$\na_m = c_{m+2} - c_{m+1} - c_{m+1} + c_m - c_{m+1} = c_{m+2} - 3c_{m+1} + c_m\n$$\n\nfor $m \\ge 0$, and substitution in the third equation finally yields\n\n$$\nc_{m+2} - 3c_{m+1} + c_m - (c_{m+1} - 3c_m + c_{m-1}) - (c_{m+1} - c_m) = c_{m+2} - 5c_{m+1} + 5c_m - c_{m-1} = 0\n$$\nfor $m \\ge 1$. The characteristic equation of the recursion is $q^3 - 5q^2 + 5q - 1 = 0$, and since $q^3 - 5q^2 + 5q - 1 = (q-1)(q^2 - 4q + 1)$, the roots of the characteristic equation are $q_1 = 1$, $q_2 = 2 + \\sqrt{3}$, and $q_3 = 2 - \\sqrt{3}$.\n\nIt follows that the required values are given by expressions of the form $c_n = A + B (2+\\sqrt{3})^n + C (2-\\sqrt{3})^n$. Since we know $c_0 = c_1 = 0$ and $c_2 = 1$, we obtain the system of equations\n\n$$\n\\begin{aligned}\nA + B + C &= 0 \\\\\nA + (2 + \\sqrt{3})B + (2 - \\sqrt{3})C &= 0 \\\\\nA + (7 + 4\\sqrt{3})B + (7 - 4\\sqrt{3})C &= 1.\n\\end{aligned}\n$$\n\nSolving this system of equations yields $A = -\\frac{1}{2}$, $B = \\frac{1}{4} - \\frac{\\sqrt{3}}{12}$, and $C = \\frac{1}{4} + \\frac{\\sqrt{3}}{12}$, and the number of possible paths is therefore given by the expression\n\n$$\nc_n = -\\frac{1}{2} + \\frac{(3 - \\sqrt{3})(2 + \\sqrt{3})^n}{12} + \\frac{(3 + \\sqrt{3})(2 - \\sqrt{3})^n}{12}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21928,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a cyclic quadrilateral such that $DB = DC$. Let $M$ and $N$ be the midpoints of $AB$ and $AC$, respectively. Let $J$, $E$, and $F$ be the touchpoints of the incircle $(I)$ of $\\triangle ABC$ with $BC$, $CA$, and $AB$, respectively. The line $MN$ intersects $JE$ and $JF$ at $K$ and $H$, respectively. The lines $IJ$ and $DG$ intersect the circumcircle of $BIC$ at $G$ and $T$, respectively.\n\n**a)** Prove that $JA$ passes through the midpoint of $HK$ and is perpendicular to $IT$.\n\n**b)** Let $R$ and $S$ be the perpendicular projections of $D$ onto $AB$ and $AC$, respectively. Take points $P$ and $Q$ on $IF$ and $IE$, respectively, such that $KP$ and $HQ$ are both perpendicular to $MN$. Prove that the three lines $MP$, $NQ$, and $RS$ are concurrent.",
"options": [],
"answer": "See solution",
"solution": "a) Let $K'$ be the intersection of the lines $BI$ and $JE$. We have\n\n$$\n\\begin{aligned}\n\\angle BIA &= 90^\\circ + \\frac{\\angle BCA}{2} = 180^\\circ - \\left(90^\\circ - \\frac{\\angle BCA}{2}\\right) \\\\\n&= 180^\\circ - \\angle JEC = \\angle AEK'.\n\\end{aligned}\n$$\n\n\n\nHence $AIK'E$ is cyclic, which implies that $\\angle AK'I = \\angle AEI = 90^\\circ$. Combined with $M$ being the midpoint of $AB$, this deduces that triangle $MBK'$ is isosceles. We have\n\n$$\n\\angle MK'B = \\angle MBK' = \\angle K'BC,\n$$\n\nso $MK' \\parallel BC$, and thus $M$, $N$, $K'$ are collinear. Therefore, $K' \\equiv K$, and $BK \\perp AK$, which implies $KA \\parallel JH$. Similarly, $HA \\parallel JK$, so $AKJH$ is a parallelogram, and $AJ$ bisects $MN$.\n\nSuppose that the lines $IT$ and $BC$ intersect at $L$. Let $Z$ be the midpoint of arc $BC$ that does not contain $A$; then $Z$ is the center of the circle $(BIC)$, where $\\angle DBZ = \\angle DCZ = 90^\\circ$, so $DB$ and $DC$ are tangent to $(BIC)$, and $BTCG$ is a harmonic quadrilateral. Then\n\n$$\n(BC, LJ) = I(BC, TG) = -1\n$$\n\nand thus $L$ lies on the line $EF$, so $IL \\perp AJ$ and $IT \\perp AJ$.\n\n\n\nb) Let $U$ be the midpoint of $BC$. Since $RS$ is the Simson line of $D$ with respect to triangle $ABC$, $R$, $S$, and $U$ are collinear and $RS \\parallel AI$.\n\nWe show that $RS$, $MP$, and $NQ$ are concurrent at the incenter of triangle $ZMN$. It is well-known that $AI$, $JE$, and $MU$ concur at $V$, and $B$, $A$, $K$, $V$ lie on the circle with diameter $AB$. We have\n\n$$\n\\angle IVK = \\angle ABI = \\frac{1}{2}\\angle ABC = \\frac{1}{2}\\angle AMN = \\frac{1}{2}\\angle IPK. \\quad (1)\n$$\n\nOn the other hand,\n\n$$\n\\begin{aligned}\n\\angle IKP &= 90^\\circ - \\angle MKP = 90^\\circ - \\angle IBC = 90^\\circ - \\angle ABK \\\\\n&= \\angle BAK = \\angle PIK\n\\end{aligned}\n$$\n\nso $PI = PK$. Combining with (1), we conclude that $P$ is the circumcenter of triangle $IVK$, so $PK = PV$. We have\n\n$$\n\\angle PVM = \\angle PVI + \\angle AVM = \\angle IAC + 90^\\circ - \\angle BAI = 90^\\circ.\n$$\n\nThus, $PV = PK$ and $PK$, $PV$ are perpendicular to $MN$, $MU$, respectively. Hence, $P$ belongs to the angle bisector of $\\angle UMN$. It follows that $MP$ is the angle bisector of $\\angle UMN$. Similarly, $NQ$ is the angle bisector of $\\angle UNM$. Since $RS$ passes through $U$ and $RS \\parallel AI$, then $UR$ is the angle bisector of $\\angle NUM$.\n\nSo $RS$, $MP$, and $NQ$ concur at the incenter of triangle $ZMN$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21929,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{Q} \\to \\mathbb{Q}$ such that for all $x, y \\in \\mathbb{Q}$,\n$$\nf(xy) = f(x)f(y) + f(x + y) - 1.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let us analyze the given functional equation:\n$$\nf(xy) = f(x)f(y) + f(x + y) - 1.\n$$\n\nFirst, substitute $x = 0$:\n$$\nf(0) = f(0)f(y) + f(y) - 1.\n$$\nLet $f(0) = a$. Then:\n$$\na = a f(y) + f(y) - 1 = (a + 1)f(y) - 1.\n$$\nSo for all $y$, $(a + 1)f(y) = a + 1$. If $a \\neq -1$, then $f(y) = 1$ for all $y$, but this does not satisfy the original equation. Thus, $a = -1$, so $f(0) = -1$.\n\nNow, substitute $x = y = 1$:\n$$\nf(1) = f^2(1) + f(2) - 1.\n$$\nLet $f(1) = b$, $f(2) = c$:\n$$\nb = b^2 + c - 1 \\implies b^2 - b + c - 1 = 0.\n$$\n\nSubstitute $x = y = 2$:\n$$\nf(4) = f^2(2) + f(4) - 1 \\implies f^2(2) + f(4) - 1 = f(4).\n$$\nSo $f^2(2) = 1 \\implies f(2) = 1$ or $f(2) = -1$.\n\nSuppose $f(2) = -1$.\nThen $b^2 - b - 2 = 0 \\implies (b - 2)(b + 1) = 0 \\implies b = 2$ or $b = -1$.\n\nIf $b = 2$, substitute $x = x$, $y = 1$:\n$$\nf(x) = f(x)f(1) + f(x + 1) - 1 \\implies f(x + 1) = 1 - f(x).\n$$\nThen $f(x + 2) = 1 - f(x + 1) = f(x)$, so $f$ is periodic of period $2$.\nBut $f(1) = 2$, $f(3) = f(1) = 2$, $f(2) = 1 - f(1) = -1$, which matches our assumption. However, substituting $x = y = 2$ gives $f(4) = f^2(2) + f(4) - 1 \\implies 1 + f(4) - 1 = f(4)$, so $1 = 0$, a contradiction.\n\nIf $b = -1$, substitute $x = x$, $y = 1$:\n$$\nf(x) = f(x)f(1) + f(x + 1) - 1 \\implies f(x + 1) = 2f(x) + 1.\n$$\nThis is a linear recurrence. Solving, $f(n) = 2f(n - 1) + 1$ for $n \\geq 1$ with $f(1) = -1$.\nBut this leads to contradictions with other values, as shown in the original analysis.\n\nThus, $f(2) = 1$.\nNow, $b^2 - b + 1 - 1 = 0 \\implies b^2 - b = 0 \\implies b = 0$ or $b = 1$.\n\nIf $b = 1$, substitute $x = x$, $y = 1$:\n$$\nf(x) = f(x)f(1) + f(x + 1) - 1 \\implies f(x + 1) = 1.\n$$\nSo $f(0) = 1$, contradicting $f(0) = -1$.\n\nThus, $b = 0$, so $f(1) = 0$.\nNow, for $x \\in \\mathbb{Q}$:\n$$\nf(x) = f(x)f(1) + f(x + 1) - 1 \\implies f(x + 1) = f(x) + 1.\n$$\nSo $f(n) = n - 1$ for $n \\in \\mathbb{N}$, and by similar arguments for rationals, $f(x) = x - 1$ for all $x \\in \\mathbb{Q}$.\n\nAlternatively, the constant function $f(x) = -1$ also works.\n\n**Conclusion:**\nThe solutions are $f(x) = x - 1$ and $f(x) = -1$ for all $x \\in \\mathbb{Q}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21930,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer and let $S_i = (a_i, a_{i+1}, \\dots, a_n, a_1, a_2, \\dots, a_{i-1})$ for all $i = 1, \\dots, n$ (indices modulo $n$).\n\nFor each $n$-tuple $r = (b_1, b_2, \\dots, b_n)$, define\n$$\n\\omega(r) = b_1 \\cdot 2^{n-1} + b_2 \\cdot 2^{n-2} + \\dots + b_n.\n$$\n\nSuppose the numbers $\\omega(S_1), \\omega(S_2), \\dots, \\omega(S_n)$ attain exactly $k$ distinct values.\n\n**a)** Prove that $k$ divides $n$ and $\\frac{2^n - 1}{2^k - 1}$ divides $\\omega(S_i)$ for all $i = 1, \\dots, n$.\n\n**b)** Let $M = \\max_{i=1,\\dots,n} \\omega(S_i)$ and $m = \\min_{i=1,\\dots,n} \\omega(S_i)$. Prove that\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 $S_d = (a_d, a_{d+1}, \\dots, a_{d+n-1})$. By the binary representation, $\\omega(S_i) = \\omega(S_j)$ if and only if $S_i = S_j$.\n\nLet $t$ be the smallest positive integer such that there exist $i$ and $j$ with $\\omega(S_i) = \\omega(S_j)$ and $j - i = t$. This implies $(a_d)_{d=1}^{\\infty}$ 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\nBut $(a_d)_{d=1}^{\\infty}$ is $n$-periodic, so $k = t$ divides $n$.\n\nMoreover, $S_i$ has the form $(a_i, \\dots, a_{i+k-1})$ repeated $n/k$ times. Thus,\n$$\n\\omega(S_i) = \\omega(a_i, \\dots, a_{i+k-1}) \\cdot (2^{n-k} + 2^{n-2k} + \\dots + 2^0) = \\omega(a_i, \\dots, a_{i+k-1}) \\cdot \\frac{2^n - 1}{2^k - 1}.\n$$\nSo $\\omega(S_i)$ is divisible by $\\frac{2^n - 1}{2^k - 1}$ for all $i$.\n\nb) Consider the $k$ tuples $S_1^* = (a_1, \\dots, a_k), \\dots, S_k^* = (a_k, \\dots, a_{2k})$.\nClearly, $\\omega(S_i) = \\omega(S_i^*) \\cdot \\frac{2^n - 1}{2^k - 1}$.\n\nIt suffices to prove\n$$\n\\max \\omega(S_i^*) - \\min \\omega(S_i^*) \\geq 2^{k-1} - 1. \\quad (\\heartsuit)\n$$\n\nLet $T_i^* = (1 - a_i, \\dots, 1 - a_{i+k-1})$. Then\n$$\n\\max \\omega(T_i^*) - \\min \\omega(T_i^*) = \\max \\omega(S_i) - \\min \\omega(S_i)\n$$\n\nConsider two cases:\n\n* If there exist $i, j < k$ such that $a_i = a_{i+1} = 1$ and $a_j = a_{j+1} = 0$, then\n$$\n\\omega(S_i^*) - \\omega(S_j^*) > 2^k + 2^{k-1} - 2^{k-1} > 2^k - 1,\n$$\nwhich proves $(\\heartsuit)$.\n\n* Otherwise, for every $i$ with $a_i = 1$, $a_{i+1} = 0$. If $a_i = 0$ for all $i$ or only one $a_i = 1$, the case is trivial. Otherwise, for $i < j$ with $a_i = a_j = 0$, for each $i$ with $a_i = 1$, let $i'$ be the smallest index $> i$ with $a_{i'} = 1$. Since $(a_1, \\dots, a_k)$ is not periodic, there is a pair $(j < j')$ with $j' - j - 1 \\geq i' - i$. Then\n$$\n\\begin{aligned}\n& \\omega(S_i) - \\omega(S_{j+1}) \\\\\n&= 2^k + \\omega(a_{i'+1}, \\dots, a_{k+i-1}) - \\omega(a_{j'}, \\dots, a_{k+j-1}) \\\\\n&\\geq 2^k + 2^{k+i-i'} - (2^{k+j-j'+1} - 1) \\\\\n&\\geq 2^k + 2^{k+j-j'+1} - (2^{k+j-j'+1} - 1) > 2^k - 1,\n\\end{aligned}\n$$\nwhich completes the proof.",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 21931,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ be non-negative numbers satisfying the following conditions simultaneously:\n\n$$\n\\sum_{i=1}^{n} (a_i + b_i) = 1\n$$\n\n$$\n\\sum_{i=1}^{n} i(a_i - b_i) = 0\n$$\n\n$$\n\\sum_{i=1}^{n} i^2 (a_i + b_i) = 10\n$$\n\nProve that $\\max\\{a_k, b_k\\} \\le \\frac{10}{10 + k^2}$ for all $1 \\le k \\le n$.",
"options": [],
"answer": "See solution",
"solution": "For any $1 \\le k \\le n$, it follows from the given conditions and Cauchy's Inequality that\n\n$$\n\\begin{align*}\n(ka_k)^2 &\\le \\left(\\sum_{i=1}^n i a_i\\right)^2 = \\left(\\sum_{i=1}^n i b_i\\right)^2 \\\\\n&\\le \\left(\\sum_{i=1}^n i^2 b_i\\right) \\left(\\sum_{i=1}^n b_i\\right) \\\\\n&= \\left(10 - \\sum_{i=1}^n i^2 a_i\\right) \\left(1 - \\sum_{i=1}^n a_i\\right) \\\\\n&\\le (10 - k^2 a_k) (1 - a_k) \\\\\n&= 10 - (10 + k^2)a_k + k^2 a_k^2.\n\\end{align*}\n$$\n\nIt follows that $a_k \\le \\frac{10}{10 + k^2}$. Similarly, $b_k \\le \\frac{10}{10 + k^2}$, and hence the result follows. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21932,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $\\mathcal{P}$ is a polygon of area $1$, and $\\varepsilon > 0$ is a constant, such that for any translate $Q = \\mathcal{P} + v$, where $v$ has length exactly $\\frac{1}{100}$, the intersection of $\\mathcal{P}$ and $Q$ has area at least $1 - \\varepsilon$. Prove a lower bound on $\\varepsilon$.",
"options": [],
"answer": "See solution",
"solution": "We prove the result for any measurable set $\\mathcal{P}$ (not just polygons). For a vector $v$ in the plane, let $\\mathcal{P} + v$ denote the translate of $\\mathcal{P}$ by $v$.\n\n**Lemma**\n\nFix a sequence of $n$ vectors $v_1, v_2, \\dots, v_n$, each of length $\\frac{1}{100}$. A grasshopper starts at a random point $x$ of $\\mathcal{P}$, and makes $n$ jumps to $x + v_1 + \\dots + v_n$. Then it remains in $\\mathcal{P}$ with probability at least $1 - n\\varepsilon$.\n\n*Proof.* For the grasshopper to leave $\\mathcal{P}$ at step $i$, its position before step $i$ must be inside the difference set $\\mathcal{P} \\setminus (\\mathcal{P} - v_i)$. This set has area at most $\\varepsilon$, so the probability of leaving at step $i$ is at most $\\varepsilon$. Summing over $n$ steps, the probability of ever leaving $\\mathcal{P}$ is at most $n\\varepsilon$. $\\square$\n\n**Corollary**\n\nFix a vector $w$ of length at most $8$. A grasshopper starts at a random point $x$ of $\\mathcal{P}$, and jumps to $x + w$. Then it remains in $\\mathcal{P}$ with probability at least $1 - 800\\varepsilon$.\n\n*Proof.* Apply the lemma with $800$ jumps. Any vector $w$ of length at most $8$ can be written as $w = v_1 + v_2 + \\dots + v_{800}$, where each $v_i$ has length exactly $\\frac{1}{100}$. $\\square$\n\nNow, select a random starting point $x \\in \\mathcal{P}$ and a random vector $w$ of length at most $8$ (sampled uniformly from the closed disk of radius $8$). Let $q$ be the probability of staying inside $\\mathcal{P}$.\n\n- If we pick $w$ first, by the corollary, $q \\ge 1 - 800\\varepsilon$ (for any $w$).\n- If we pick $x$ first, the landing points $x + w$ are uniformly distributed over a disk of radius $8$ (area $64\\pi$). The probability of landing in $\\mathcal{P}$ is at most $\\frac{1}{64\\pi}$.\n\nThus,\n\n$$\n1 - 800\\varepsilon \\le q \\le \\frac{1}{64\\pi} \\implies \\varepsilon > \\frac{1 - \\frac{1}{64\\pi}}{800} > 0.001\n$$\n\nas desired.\n\n**Remark.** The choice of $800$ jumps is for concreteness; any $n$ with $\\pi(n/100)^2 > 1$ works. $n = 98$ gives a better bound.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21933,
"subject": "Mathematics (Olympiad)",
"question": "Let $v(n)$ denote the highest power of $2$ that divides $n$.\n\nSuppose $a_0, a_1, \\dots, a_{3030}$ are positive integers satisfying the recurrence\n$$\n2a_{n+2} = a_{n+1} + 4a_n\n$$\nfor $n = 0, 1, \\dots, 3028$. Prove that at least one of $a_0, a_1, \\dots, a_{3030}$ is divisible by $2^{2020}$.",
"options": [],
"answer": "See solution",
"solution": "From the recurrence, we have\n$$\na_{n+2} = \\frac{1}{2}a_{n+1} + 2a_n\n$$\nfor $n = 0, 1, \\dots, 3028$. Thus,\n$$\nv(a_{n+2}) = v\\left(\\frac{1}{2}a_{n+1} + 2a_n\\right).\n$$\n\nConsider two cases:\n\n*Case 1:* $v(\\frac{1}{2}a_{n+1}) = v(2a_n)$, so $v(a_{n+1}) - 1 = v(a_n) + 1$, which implies\n$$v(a_{n+1}) = v(a_n) + 2.$$\n\n*Case 2:* $v(\\frac{1}{2}a_{n+1}) \\neq v(2a_n)$, so\n$$v(a_{n+2}) = \\min(v(\\frac{1}{2}a_{n+1}), v(2a_n)) \\leq v(a_{n+1}) - 1.$$\n\nFor $n = 0, 1, \\dots, 3028$, the sequence $v(a_0), v(a_1), \\dots, v(a_{3030})$ is strictly increasing by $2$ until some index $k$, then strictly decreasing by at least $1$:\n$$\n\\underbrace{v(a_0), v(a_1), \\dots, v(a_k)}_{\\text{increasing, by 2}} \\underbrace{v(a_{k+1}), v(a_{k+2}), \\dots, v(a_{3030})}_{\\text{decreasing, by at least 1}}\n$$\n\nSuppose for contradiction that $0 \\le v(a_0), v(a_1), \\dots, v(a_{3030}) < 2020$. Then the increasing part can have at most $\\lfloor \\frac{2019}{2} \\rfloor + 1 = 1010$ terms, and the decreasing part at most $2019 + 1 = 2020$ terms, totaling $1010 + 2020 = 3030$ terms. But there are $3031$ terms, a contradiction.\n\nTherefore, at least one $v(a_i) \\ge 2020$, so at least one $a_i$ is divisible by $2^{2020}$.\n\n$\\square$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21934,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c, d, e$ be the lengths of the sides of a pentagon, $f$ be the length of one of its diagonals (see the figure below), and $S$ be its area. Prove that:\n\n$$\n10S \\le 3B + C\n$$\n\nwhere $B$ is the sum of all five pairwise products of neighboring sides, and $C$ is the sum of all five pairwise products of non-neighboring sides.\n\n",
"options": [],
"answer": "See solution",
"solution": "We use the following lemma: Let $a, b, c, d$ be the lengths of the sides of a quadrilateral, and $S$ its area. Then $2S \\le ab + cd$ and $2S \\le ac + bd$.\n\nApplying this to the pentagon, and considering the diagonal $f$, we have $2S \\le ab + cf + de < ab + c(d + e) + de$, i.e., $2S \\le (ab + cd + de) + ce$. All terms in parentheses are products of neighboring sides, while $ce$ is a product of non-neighboring sides.\n\nSumming similar inequalities obtained by cyclically permuting the side labels, we get:\n\n$$\n10S \\le 3B + C. \\quad (1)\n$$\n\nFurther, $2S \\le ac + bf + de < ac + b(d+e) + de = (ac + bd + be) + de$. Similarly, we obtain:\n\n$$\n10S \\le 3C + B. \\quad (2)\n$$\n\nSumming (1) and (2) gives the required inequality.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21935,
"subject": "Mathematics (Olympiad)",
"question": "$$\n(a+b+c) \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) + \\frac{x^2 + y^2 + z^2}{xy + yz + zx} \\cdot \\lambda \\geq 9 + \\lambda\n$$\nтэнцэтгэл биш биелэх $\\lambda$-ийн хамгийн их утгыг ол.",
"options": [],
"answer": "See solution",
"solution": "$$\n4\\sqrt{2} \\ge \\frac{x^2 + y^2 + z^2}{xy + yz + zx}\n$$\n\nyлдсэн тэнцэтгэл бишүүдийг хэрэглэвэл:\n\n$$\n4\\sqrt{2}(xy + yz + zx) \\geq x^2 + y^2 + z^2\n$$\n\n$$\n\\left\\{ \\begin{array}{l}\nx^2 + y^2 \\geq \\frac{(x+y)^2}{4} \\\\\nxy \\leq \\frac{x^2}{4}\n\\end{array} \\right.\n$$\n\n$$\n4\\sqrt{2} \\left( \\frac{(x+y)^2}{4} + z(x+y) \\right) \\geq z^2 + \\frac{(x+y)^2}{2} \\cdot \\frac{x+y}{z} = t\n$$\n\n$$\n(2\\sqrt{2}-1)t^2 + 3\\sqrt{2}t - 2 \\geq 0 \\Rightarrow x+y \\geq \\frac{\\sqrt{30+4\\sqrt{2}-4\\sqrt{2}}}{2\\sqrt{2}-1} \\cdot z.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21936,
"subject": "Mathematics (Olympiad)",
"question": "Let $I$ be the incenter of a triangle $ABC$. Let $L$, $M$, and $N$ be the midpoints of $AI$, $AC$, and $CI$, respectively. Assume there is a point $D$ in the interior of segment $AM$ such that $BD = BC$. The incircle of $\\triangle ABD$ touches $AD$ and $BD$ at $E$ and $F$, respectively. Let $J$ be the circumcenter of $\\triangle AIC$. Let $\\omega$ denote the circumcircle of $\\triangle DMJ$. The line $MN$ intersects $\\omega$ at point $P \\ne M$, and the line $JL$ intersects $\\omega$ at point $Q \\ne J$.\n\nProve that the three lines $EF$, $PQ$, and $LN$ pass through a common point.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $K$ be the midpoint of $ID$, then $LN$ passes through $K$. Next, we prove that both $EF$ and $PQ$ pass through $K$.\n\nLet $X$ and $Y$ be the reflections of $D$ across $E$ and $F$, respectively. Let $Z$ be the projection of $I$ onto $AC$. Then,\n\n$$\nCX = 2DE + CD = AD + BD - AB + CD = AC + BC - AB = 2CZ.\n$$\n\nThus, $CZ = ZX$. Combining this with $IZ \\perp CX$, we know $IC = IX$. Therefore,\n\n$$\n2\\angle IXC = 2\\angle ICX = \\angle ACB = \\angle BDC = 2\\angle FEC = 2\\angle YXC.\n$$\n\nThus, $I$, $Y$, and $X$ are collinear, which implies that $EF$ passes through $K$.\n\nSince $MN \\parallel AI$, we have $\\angle PQD = \\angle PMD = \\angle IAD$.\n\nFrom $\\angle DMJ = 90^\\circ$, we know that $DJ$ is the diameter of $\\omega$, so $DQ \\perp QJ$. Combined with $AL \\perp LJ$, we conclude that $AI \\parallel DQ$. Let $T$ be the reflection of $D$ across $Q$. Then $ADTI$ forms an isosceles trapezoid. Since $KQ \\parallel IT$, we have\n\n$$\n\\angle DQK = \\angle DTI = 180^\\circ - \\angle IAD = 180^\\circ - \\angle PQD,\n$$\n\nwhich implies that $PQ$ passes through $K$. Thus, the proof is complete.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21937,
"subject": "Mathematics (Olympiad)",
"question": "For which integers $n \\geq 2$ does $n$ divide $\\binom{2n-3}{n-1}$?",
"options": [],
"answer": "See solution",
"solution": "We claim that all $n \\geq 2$ that are not a power of $2$ satisfy the property.\n\nFirst, consider $n = 2^k$ for $k \\geq 1$. In this case, the numerator of $\\binom{2n-3}{n-1}$ is the product $2^k, 2^k+1, \\dots, 2^{k+1}-3$, and the denominator is $1, 2, \\dots, 2^k-2$. For each $1 \\leq i \\leq k-1$, the number of factors of $2^i$ in the numerator and denominator are equal, and for $i \\geq k$ both are zero. Thus,\n\n$$\ne_2(1 \\cdot 2 \\cdot 3 \\cdots (2^k - 2)) = e_2((2^k + 1)(2^k + 2) \\cdots (2^{k+1} - 2)).$$\n\nNow, the numerator also includes $2^k$ (with $k$ factors of $2$), and the denominator does not. The difference in the number of factors of $2$ is $k-1$, so $\\binom{2n-3}{n-1}$ contains exactly $k-1$ factors of $2$, which is not enough for divisibility by $n = 2^k$. Thus, $n = 2^k$ does not work.\n\nNow, suppose $n$ is not a power of $2$. Let $2^k$ be the largest power of $2$ less than $n$, and let $\\ell$ be maximal such that $2^\\ell \\mid n$. For $i = 1$, the number of factors of $2$ in numerator and denominator are equal. For $2 \\leq i \\leq \\ell$, the numerator has one more factor of $2^i$ than the denominator. For $\\ell+1 \\leq i \\leq k$, the numerator has at least as many as the denominator. For $i = k+1$, the numerator has one factor, the denominator none. For $i > k+1$, both have none. Summing up, the difference is $\\ell$, which equals $e_2(n)$. Thus, $n$ divides $\\binom{2n-3}{n-1}$.\n\nTherefore, all $n \\geq 2$ that are not a power of $2$ satisfy the property. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21938,
"subject": "Mathematics (Olympiad)",
"question": "We number the members with $1, 2, \\dots, 6$. Consider the following decision method:\n\nIf person number 1 to the third person have the same vote, their vote is considered as the final result; otherwise, the result is based on the majority of the votes between the fourth to the sixth persons.\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 play the same role, and persons 4–6 also play 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 is $2b + a$, so $2b + a > 2a + b$, which implies $b > a$.\n\nThis contradicts the earlier inequality $a > b$. Therefore, the method cannot be represented as a weighted voting system.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21939,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $f(x) = x^2 + a$. Define $f^1(x) = f(x)$, $f^n(x) = f(f^{n-1}(x))$ for $n = 2, 3, \\dots$, and let\n$$\nM = \\{ a \\in \\mathbb{R} \\mid |f^n(0)| \\le 2 \\text{ for any } n \\in \\mathbb{N} \\}.\n$$\nProve that $M = [-2, \\frac{1}{4}]$.",
"options": [],
"answer": "See solution",
"solution": "(a) When $a < -2$, $|f^1(0)| = |a| > 2$, so $a \\notin M$.\n\n(b) When $-2 \\le a < 0$, $|f^1(0)| = |a| \\le 2$. Assume $|f^{k-1}(0)| \\le |a| \\le 2$ for $k \\ge 2$. Since $a^2 \\le -2a$ for $-2 \\le a < 0$, we get\n$$\n-2 \\le -|a| = a \\le f^k(0) = (f^{k-1}(0))^2 + a \\le a^2 + a \\le |a| \\le 2.\n$$\nBy induction, $|f^n(0)| \\le a \\le 2$ for all $n \\ge 1$.\n\n(c) When $0 \\le a \\le \\frac{1}{4}$, $|f^1(0)| = |a| \\le \\frac{1}{2}$. Assume $|f^{k-1}(0)| \\le \\frac{1}{2}$ for $k \\ge 2$. Then\n$$\n|f^k(0)| \\le |f^{k-1}(0)|^2 + a \\le \\left(\\frac{1}{2}\\right)^2 + \\frac{1}{4} = \\frac{1}{2}.\n$$\nBy induction, $|f^n(0)| \\le \\frac{1}{2}$ for all $n \\ge 1$.\n\nFrom (b) and (c), $[-2, \\frac{1}{4}] \\subseteq M$.\n\n(d) When $a > \\frac{1}{4}$, let $a_n = f^n(0)$. Then\n$$\na_{n+1} = f(a_n) = a_n^2 + a.\n$$\nSince $a_n > a > \\frac{1}{4}$ for all $n \\ge 1$,\n$$\na_{n+1} - a_n = a_n^2 - a_n + a = \\left(a_n - \\frac{1}{2}\\right)^2 + a - \\frac{1}{4} \\ge a - \\frac{1}{4}.\n$$\nSo,\n$$\n\\begin{aligned}\na_{n+1} - a &= (a_{n+1} - a_n) + \\dots + (a_2 - a_1) \\\\ &\\ge n \\left(a - \\frac{1}{4}\\right).\n\\end{aligned}\n$$\nTherefore, when $n > \\frac{2-a}{a-\\frac{1}{4}}$,\n$$\na_{n+1} \\ge n\\left(a - \\frac{1}{4}\\right) + a > 2.\n$$\nThus, $a \\notin M$.\n\nFrom (a)—(d), $M = [-2, \\frac{1}{4}]$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21940,
"subject": "Mathematics (Olympiad)",
"question": "Agustin and Lucas take turns marking one cell at a time in a $101 \\times 101$ grid. Agustin starts the game. A cell cannot be marked if there are already two marked cells in its row or in its column. The player who cannot move loses. Decide which player has a winning strategy.",
"options": [],
"answer": "See solution",
"solution": "We describe a winning strategy for the second player, Lucas. This strategy applies to any $n \\times n$ grid with odd $n \\geq 3$.\n\nCall a row or column *empty*, *incomplete*, or *full* at a certain moment of the game if it contains respectively 0, 1, or 2 marked cells.\n\n**Stage 1:**\n\nAgustin's first move is in an empty row $R$ and an empty column $C$, and after it there are still empty rows left. Call such a move *standard*. Lucas's answer to a standard move is to choose an empty row $R' \\neq R$ and mark the intersection of $R'$ and column $C$. This makes $C$ a full column, so it cannot be used later. Lucas gives standard answers as long as Agustin makes standard moves (at least Agustin's first move is standard). After each standard answer, each column is either empty or full, so Agustin's next move is in an empty column. Since there is a standard answer to any standard move, Lucas cannot lose if all of Agustin's moves are standard.\n\nSuppose Agustin makes a non-standard move $M$. As above, $M$ is in an empty column $C$. There are two possibilities:\n\n1. $M$ is in an incomplete row $R$ (with empty rows remaining). Lucas gives a standard answer: take an empty row $R' \\neq R$ and mark the intersection of $R'$ and column $C$.\n2. $M$ is in the last empty row. Lucas takes any row $R' \\neq R$ and marks the intersection of $R'$ and column $C$, making $R'$ full.\n\nAfter stage 1, the table has:\n\n(i) The number $E$ of empty rows is even.\n\n(ii) The number $I$ of incomplete rows is even.\n\n(iii) There are no incomplete columns.\n\n**Stage 2:**\n\nLucas ensures that after each of his moves, properties (i)-(iii) are preserved.\n\n- If Agustin marks a cell in an empty row $R$ and column $C$, both were empty before. Lucas marks the intersection of an empty row $R' \\neq R$ and column $C$, making $C$ full and $R'$ incomplete. The combined move decreases $E$ by 2 and increases $I$ by 2, preserving (i)-(iii).\n- If Agustin marks a cell in an incomplete row $R$ and empty column $C$, $M$ makes $R$ full and $C$ incomplete. Lucas marks the intersection of an incomplete row $R' \\neq R$ and column $C$, making both $R'$ and $C$ full. The combined move decreases $I$ by 2 and does not change $E$.\n\nThus, Lucas can always respond, and Agustin will eventually be unable to move. Therefore, Lucas has a winning strategy.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21941,
"subject": "Mathematics (Olympiad)",
"question": "Determine all possible pairs of integers $a, b$ so that exactly one of them is even and so that there exist non-integer $x, y$ such that both $x + y$ and $a x + b y$ are integers.",
"options": [],
"answer": "See solution",
"solution": "The condition holds for numbers such that $|a - b| \\neq 1$.\n\nSince $a x + b y = a(x + y) + (b - a) y$, the value of $(b - a) y$ must be integer. If $|a - b| = 1$, then $y$ must be integer—a contradiction.\n\nFor $|a - b| > 1$, we can let $y = \\frac{1}{b - a}$ and $x = -y$. Then $x + y = 0$, $(b - a) y$ is integer, hence $a x + b y$ is also integer.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21942,
"subject": "Mathematics (Olympiad)",
"question": "How many 4-digit numbers are there whose digit product is $60$?",
"options": [],
"answer": "See solution",
"solution": "Since $60 = 2^2 \\times 3 \\times 5$, only the digits $1, 2, 3, 4, 5,$ and $6$ 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": 21943,
"subject": "Mathematics (Olympiad)",
"question": "Each of the $n^2$ cells of an $n \\times n$ grid is colored either black or white. Let $a_i$ denote the number of white cells in the $i$-th row, and let $b_i$ denote the number of black cells in the $i$-th column. Determine the maximum value of $\\sum_{i=1}^{n} a_i b_i$ over all coloring schemes of the grid.",
"options": [],
"answer": "See solution",
"solution": "The answer is $\\frac{n^3-n}{3}$.\n\nWe refer to cells by their coordinates, i.e., $(x, y)$ is the cell in row $x$ and column $y$.\n\nThe maximum $\\sum_{i=1}^{n} a_i b_i = \\frac{n^3-n}{3}$ can be attained by coloring the cell $(x, y)$ white if and only if $x \\ge y$. With this coloring, $a_i = i$ and $b_i = i - 1$ for each $i$. Thus,\n\n$$\n\\sum_{i=1}^{n} a_i b_i = \\sum_{i=1}^{n} i(i-1) = \\frac{n^3-n}{3}.\n$$\n\nTo show that no coloring can achieve more than $\\frac{n^3-n}{3}$, consider the set $S$ of all ordered triples $(x, y, z)$ with $1 \\le x, y, z \\le n$. Define the subsets:\n\n$$\nA := \\{(x, y, z) \\in S : (x, y) \\text{ is black and } (y, z) \\text{ is white}\\}\n$$\n$$\nB := \\{(x, y, z) \\in S : (y, z) \\text{ is black and } (z, x) \\text{ is white}\\}\n$$\n$$\nC := \\{(x, y, z) \\in S : (z, x) \\text{ is black and } (x, y) \\text{ is white}\\}\n$$\n\nThese subsets are pairwise disjoint.\n\nFor each $1 \\le i \\le n$, the number of elements $(x, y, z) \\in A$ with $y = i$ is $b_i \\cdot a_i$ (there are $b_i$ choices for $x$ and $a_i$ choices for $z$). Thus, $|A| = \\sum_{i=1}^{n} a_i b_i$, and similarly for $B$ and $C$.\n\nFor each $1 \\le i \\le n$, the triple $(i, i, i)$ is not in any of $A$, $B$, or $C$, since $(i, i)$ cannot be both white and black. Thus,\n\n$$\n|A| + |B| + |C| + n \\le |S|\n$$\n\nand hence\n\n$$\n3 \\sum_{i=1}^{n} a_i b_i \\le n^3 - n,\n$$\n\nwhich rewrites to\n\n$$\n\\sum_{i=1}^{n} a_i b_i \\le \\frac{n^3 - n}{3},\n$$\n\nas desired.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21944,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_n$ be the position of the grasshopper after the $n$th jump:\n\n$$\na_1 = 1, \\quad a_n = 1 + k + \\dots + k^{n-1}, \\quad n \\ge 2.\n$$\n\nFind all numbers $k$ such that $2015$ does not divide $a_n$ for any $n = 1, \\dots, 2015$.",
"options": [],
"answer": "See solution",
"solution": "Suppose $\\gcd(k, 2015) = d > 1$. Then every $a_n$ divided by $d$ leaves remainder $1$, and since $2015$ is divisible by $d$, we have $2015 \\nmid a_n$ for all $n$. Therefore, all positive integers $k$ not relatively prime to $2015$ satisfy the condition.\n\nIf $\\gcd(k, 2015) = 1$, consider the remainders of $a_1, \\dots, a_{2015}$ modulo $2015$. If none is divisible by $2015$, then by the pigeonhole principle, at least two $a_l$ and $a_m$ ($m > l$) have the same remainder. Their difference is divisible by $2015$:\n\n$$\na_m - a_l = k^l + \\dots + k^{m-1} = k^l (1 + \\dots + k^{m-l-1}) = k^l \\cdot a_{m-l}.\n$$\n\nSince $2015 \\mid k^l \\cdot a_{m-l}$ and $\\gcd(k, 2015) = 1$, it follows that $2015 \\mid a_{m-l}$, contradicting the assumption. Thus, only $k$ not relatively prime to $2015$ are suitable.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21945,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle. $D$ is the foot of the internal bisector of angle $A$. The perpendicular from $D$ to the tangent $AT$ ($T$ belongs to $BC$) to the circumscribed circle of $ABC$ intersects the altitude $AH$ at the point $I$ ($H$ belongs to $BC$). If $P$ is the midpoint of $AB$ and $O$ is the circumcenter of $\\triangle ABC$, $TI$ intersects $AB$ at $M$ and $PT$ intersects $AD$ at $F$. Prove that $MF$ is perpendicular to $AO$.",
"options": [],
"answer": "See solution",
"solution": "Let $Q$ be the midpoint of $AC$ and $N$ the intersection of $AD$ and $PQ$. Then $N$ is the midpoint of $AD$. As $DE$ is perpendicular to $AT$, with $E$ the intersection point of $DI$ and $AT$, and as $OA$ is perpendicular to $AT$, we get that $DE$ is parallel to $OA$, so the angles $OAN$ and $ADE$ are equal. Indeed, $\\angle OAQ = \\angle BAH$ because $\\angle OAQ = \\frac{180^\\circ - \\angle COA}{2} = 90^\\circ - \\frac{\\angle COA}{2} = 90^\\circ - B = \\angle BAH$. Moreover, $\\angle BAD = \\angle DAC$ and hence $\\angle OAN = \\angle HAD = \\angle ADE$. As a consequence, triangles $ADE$ and $DAH$ are congruent.\n\n\n\nIn particular, angle $DAT$ equals angle $HAD$, that is, $ATD$ is isosceles and point $I$ is the orthocenter of $\\triangle ABC$. So, $TI$ is perpendicular to $AD$, and the intersection point of $TI$ and $AD$ is the midpoint of $AD$, say $N$. The four points $M, N, I, T$ are collinear. We will apply the Ceva theorem in triangle $APT$ with the cevians $PN$, $AD$, and $TM$. We get\n\n$$\n\\frac{FP}{FT} \\cdot 1 \\cdot \\frac{MA}{PM} \\Leftrightarrow \\frac{PF}{TF} = \\frac{MP}{MA}\n$$\n\n(Observe that *NP* cuts *AT* at its midpoint.) So, *MF* is parallel to *AT*, and from this, *MF* is perpendicular to *AO*, as claimed.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21946,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a scalene 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 $AL = AC$. The circumcircle of triangle $DKL$ intersects segment $AB$ at a second point $T$ (other than $D$). Prove that $\\angle ACT = \\angle BCT$.",
"options": [],
"answer": "See solution",
"solution": "Let $T'$ be the point on $AB$ such that $\\angle ACT' = \\angle BCT'$. Let $\\omega$ be the circle centered at $T'$ tangent to segments $AC$ and $BC$, and denote the points of tangency to $AC$ and $BC$ by $M$ and $N$ respectively. Note that $\\omega$ exists because $CT'$ is an angle bisector.\n\n\n\n**Lemma 1.** If $RDT'S$ is cyclic and $R$ and $S$ are on the boundary of $\\omega$, then $BR$, $AS$, and $CD$ concur.\n\n*Proof.* Let $P$ be the intersection of the lines tangent to $\\omega$ that intersect $R$ and $S$. Since $\\angle T'SP = \\angle T'PP = 90^\\circ$, $PRST'$ is cyclic. Since we assumed $RDT'S$ is cyclic, then $PRDT'S$ is cyclic. Thus $\\angle T'DP = 90^\\circ$. Thus $P$ is on $CD$. Now, let $M$ and $N$ meet at $C$. If $MN$ and $RS$ meet at $Q$, $PC$ is the polar of $Q$, so $MS$ and $RN$ lie on $PQ$. Let $MN$ and $RN$ meet at $G$.\n\nBy the law of sines, we have\n$$\n\\begin{aligned}\n\\left( \\frac{\\sin \\angle CAS}{\\sin \\angle SAB} \\right) \\cdot \\left( \\frac{\\sin \\angle RBA}{\\sin \\angle RBC} \\right) &= \\frac{\\frac{MS \\sin \\angle AMG}{AS}}{\\frac{SD \\sin \\angle ADS}{AS}} \\cdot \\frac{\\frac{RD \\sin \\angle RDB}{RB}}{\\frac{RN \\sin \\angle BNG}{RB}} \\\\\n&= \\frac{MS}{SD} \\cdot \\frac{RD}{RN} \\cdot \\frac{\\sin \\angle AMG}{\\sin \\angle BNG} \\cdot \\frac{\\sin \\angle RDB}{\\sin \\angle SDA} \\\\\n&= \\frac{\\sin \\angle MDS}{\\sin \\angle SMD} \\cdot \\frac{\\sin \\angle RND}{\\sin \\angle RDN} \\cdot \\frac{\\sin \\angle CMG}{\\sin \\angle CNG} \\cdot \\frac{\\sin \\angle RDA}{\\sin \\angle SDB} \\\\\n&= \\frac{\\sin \\angle RND}{\\sin \\angle SMD} \\cdot \\frac{\\sin \\angle CMG}{\\sin \\angle CNG} \\cdot \\frac{\\sin \\angle MDS}{\\sin \\angle RDN} \\cdot \\frac{\\sin \\angle RDA}{\\sin \\angle SDB}.\n\\end{aligned}\n$$\n\nDenote this expression by $S$. For convenience, let $x = \\frac{\\sin \\angle MDS}{\\sin \\angle RDN} \\cdot \\frac{\\sin \\angle RDA}{\\sin \\angle SDB}$. Then, we have\n$$\n\\begin{aligned}\nS &= \\frac{\\sin \\angle RND}{\\sin \\angle SMD} \\cdot \\frac{\\sin \\angle CMG}{\\sin \\angle CNG} x = \\frac{\\frac{GD \\cdot \\sin \\angle CDN}{GN}}{\\frac{GD \\sin \\angle CDM}{GM}} \\cdot \\frac{\\frac{CG \\cdot \\sin \\angle MCD}{MG}}{\\frac{CG \\sin \\angle NCD}{NG}} x = \\frac{\\sin \\angle CDN}{\\sin \\angle NCD} \\cdot \\frac{\\sin \\angle MCD}{\\sin \\angle CDM} x \\\\\n&= \\frac{CN}{DN} \\cdot \\frac{DM}{CM} x = \\frac{DM}{DN} x = \\frac{\\frac{CD \\sin \\angle ACD}{\\sin \\angle DMC}}{\\frac{CD \\sin \\angle BCD}{\\sin \\angle DNC}} x = \\frac{\\sin \\angle ACD}{\\sin \\angle BCD} \\cdot \\frac{\\sin \\angle DNC}{\\sin \\angle DMC} x.\n\\end{aligned}\n$$\n\nSince $\\angle T'NC = \\angle T'MC = \\angle T'DC = 90^\\circ$, $CMDN$ is cyclic, so $\\angle DNC = 180^\\circ - \\angle DMC$, and\n$$\n\\angle MDC = \\angle MNC = \\angle NMC = \\angle NDC.\n$$\n\nFurther, notice that\n$$\n\\angle RDA = \\angle RST' = \\angle SRT' = \\angle SDB\n$$\nbecause $T'R = T'S$. Hence, we conclude that\n$$\n\\angle MDS = \\angle MDC + 90^\\circ - \\angle SDB = \\angle NDC + 90^\\circ - \\angle RDA = \\angle RDN.\n$$\n\nTherefore $x = \\frac{\\sin \\angle MDS}{\\sin \\angle RDN} \\cdot \\frac{\\sin \\angle RDA}{\\sin \\angle SDB} = 1$, so we have\n$$\n\\left( \\frac{\\sin \\angle CAS}{\\sin \\angle SAB} \\right) \\cdot \\left( \\frac{\\sin \\angle RBA}{\\sin \\angle RBC} \\right) \\cdot \\left( \\frac{\\sin \\angle BCD}{\\sin \\angle ACD} \\right) = 1,\n$$\nwhich implies that lines $AS$, $RB$, and $CD$ concur by trig Ceva. This concludes the proof of the lemma. $\\square$\n\nDenote the intersection point of $BX$ and $\\omega$ farther from $B$ by $R$, and let the circumcircle of $RDT'$ meet $\\omega$ again at $S$. By Lemma 1, $AS$ and $BR$ meet on $CD$, so $A$, $S$, and $K$ are collinear. Let $\\omega_A$ and $\\omega_B$ be the circles centered at $A$, $B$ with radii $AC$ and $BC$ respectively. Let $AK$ meet $\\omega_B$ again at $S'$, and let $BL$ meet $\\omega_A$ again at $R'$.\n\nA homothety centered at $B$ takes $\\omega$ to $\\omega_A$, so\n$$\n\\frac{BR}{BR'} = \\frac{BN}{BC} = \\frac{BT'}{BA} = \\frac{a}{a+b}.\n$$\n\nThus $BL \\cdot BR = \\frac{a}{a+b} \\cdot BL \\cdot BR' = \\frac{a}{a+b} \\cdot BC^2 = \\frac{a^3}{a+b}$ and $BT' \\cdot BD = c\\left(\\frac{a}{a+b}\\right) \\cdot \\left(c \\cdot \\frac{a^2}{a^2+b^2}\\right) = \\frac{a^3}{a+b}$. Thus $BLRDT'$ is cyclic, and similarly $SKDT'$ is cyclic. Since $LRDT'$, $KDT'S$, and $RDT'S$ are cyclic, then $LRDT'SK$ is cyclic, so $RDT'S$ is cyclic. Therefore $T = T'$, as desired.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21947,
"subject": "Mathematics (Olympiad)",
"question": "Is it possible to define a notion of a \"good team\" among students such that a team is good if and only if it contains an odd number of exceptional students? (For example, choose three exceptional students, or any odd number between 3 and 21.)",
"options": [],
"answer": "See solution",
"solution": "Yes.\n\nTo construct such an example, select three exceptional students. Define a team as good if and only if it contains an odd number of these exceptional students. Alternatively, you may choose any odd number of exceptional students between 3 and 21.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21948,
"subject": "Mathematics (Olympiad)",
"question": "Let $MN$ be a line parallel to the side $BC$ of a triangle $ABC$, with $M$ on the side $AB$ and $N$ on the side $AC$. The lines $BN$ and $CM$ meet at point $P$. The circumcircles of triangles $BMP$ and $CNP$ meet at two distinct points $P$ and $Q$. Prove that $\\angle BAQ = \\angle CAP$.",
"options": [],
"answer": "See solution",
"solution": "We have:\n\n$$\n\\frac{AN}{AM} \\sin \\angle QAN = \\frac{S_{AQN}}{S_{AQM}} = \\frac{AN}{AM} \\cdot \\frac{QN}{QM} \\cdot \\frac{AN}{AM} \\sin \\angle QNC \\\\\n\\frac{\\sin \\angle QAN}{\\sin \\angle QAM} = \\frac{QN}{QM} \\cdot \\frac{\\sin \\angle QNC}{\\sin \\angle QMB} = \\frac{QM}{QB} \\cdot \\frac{QM}{QM} = \\frac{QN}{QB} = \\frac{AN}{AM} \\quad (1)\n$$\n\nIt is because of the fact that $\\triangle QCN \\sim \\triangle QMB$ and since $\\frac{QN}{QB} = \\frac{AN}{AM}$.\n\nFrom (1), it follows that $AQ$ is the symmedian of $\\triangle AMN$, which implies the result.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21949,
"subject": "Mathematics (Olympiad)",
"question": "Find all polynomials $P$ with integer coefficients such that the number $P(a) - P(b)$ is divisible by $a + b$ for all integers $a, b$, provided that $a + b \\neq 0$.",
"options": [],
"answer": "See solution",
"solution": "All polynomials whose every odd-degree term has zero coefficient.\n\n*Solution:* Let $P(x) = P_0(x) + P_1(x)$, where $P_0$ and $P_1$ are polynomials whose nonzero terms have even and odd degrees, respectively.\n\nWe can write $P_0(x) = Q(x^2)$ for some polynomial $Q$. For any integers $a, b$, $P_0(a) - P_0(b) = Q(a^2) - Q(b^2)$ is divisible by $a^2 - b^2$, and thus by $a + b$. Therefore, if every odd-degree term of $P$ has zero coefficient, the condition is satisfied.\n\nConversely, suppose $P$ satisfies the condition. Then $P_1 = P - P_0$ must also satisfy it. Note that $P_1(-x) = -P_1(x)$, so $P_1$ is odd. Substituting $b$ by $-b$ gives $a - b \\mid P_1(a) + P_1(b)$ for any distinct integers $a, b$. Since also $a - b \\mid P_1(a) - P_1(b)$, we have $a - b \\mid 2P_1(a)$. For any $a$, there exists $b$ such that $|a - b| > 2|P_1(a)|$, so $P_1(a) = 0$ for all integers $a$. Thus, $P = P_0$, i.e., all odd-degree coefficients are zero.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21950,
"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": "Substitute $x=1$ and $x=-1$ into the equation:\n\n$$\n\\begin{aligned}\na_0 + a_1 + a_2 + \\dots + a_8 + a_9 &= 2^3 \\cdot 3^3 \\cdot 4^3 \\\\\na_0 - a_1 + a_2 - a_3 + a_4 - a_5 + a_6 - a_7 + a_8 - a_9 &= 0.\n\\end{aligned}\n$$\n\nAdding these equations gives:\n\n$$\na_0 + a_2 + a_4 + a_6 + a_8 = \\frac{1}{2}(2^3 \\cdot 3^3 \\cdot 4^3 + 0) = 6912.\n$$\n\nSubstituting $x=0$ yields:\n\n$$\na_0 = 1^3 \\cdot 2^3 \\cdot 3^3 = 216.\n$$\n\nTherefore,\n\n$$\na_2 + a_4 + a_6 + a_8 = 6912 - 216 = 6696.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21951,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ and $b$ be complex numbers with $m = |a+b|$ and $n = |a-b|$. Let $c = x + yi$, where $x, y \\in \\mathbb{R}$. Show that\n\n$$\n(\\max\\{|ac+b|, |a+bc|\\})^2 \\ge \\frac{m^2 n^2}{m^2 + n^2}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Or equivalently,\n\n$$\n\\max\\{|ac+b|, |a+bc|\\} \\ge \\frac{mn}{\\sqrt{m^2 + n^2}}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21952,
"subject": "Mathematics (Olympiad)",
"question": "There are different positive integers written on the board. Their arithmetic mean is a decimal number, with the decimal part exactly $0.2016$. What is the least possible value of the mean?\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $s$ be the sum, $n$ the number, and $p$ the integer part of the mean of the numbers on the board. Then:\n\n$$\n\\frac{s}{n} = p + \\frac{2{,}016}{10{,}000} = p + \\frac{126}{625}\n$$\n\nwhich gives\n\n$$\n625(s - pn) = 126n.\n$$\n\nNumbers $126$ and $625$ are coprime, thus $625 \\mid n$. Therefore $n \\geq 625$.\n\nThe numbers on the board are different, that is:\n\n$$\np = \\frac{s}{n} - \\frac{126}{625} \\geq \\frac{1+2+\\cdots+n}{n} - \\frac{126}{625} = \\frac{n(n+1)}{2n} - \\frac{126}{625} = \\frac{n+1}{2} - \\frac{126}{625}\n$$\n\nFor $n = 625$:\n\n$$\np \\geq \\frac{625+1}{2} - \\frac{126}{625} > 312\n$$\n\nThe integer $p$ is thus at least $313$ and the value of the mean at least $313.2016$.\n\nThis value can be attained by the numbers $1, 2, \\ldots, 624$ and $751$. We get:\n\n$$\n\\frac{1+2+\\cdots+624+751}{625} = \\frac{312 \\times 625 + 751}{625} = 313 + \\frac{126}{625} = 313.2016\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21953,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCDEF$ be a regular octahedron with lower vertex $E$, upper vertex $F$, middle slice plane $ABCD$, center $M$ and circumsphere $k$. Furthermore, let $X$ be an arbitrary point within side $ABF$. The line $EX$ intersects $k$ in $E$ and $Z$, and the plane $ABCD$ in $Y$.\n\n$$\n\\text{Show that } \\angle EMZ = \\angle EYF.\n$$",
"options": [],
"answer": "See solution",
"solution": "\n\nWe intersect the entire figure with the plane through the points $X$, $E$, and $F$. Since $M$ is on line $EF$, it is part of that plane. Also, points $Y$ and $Z$ are part of that plane because they are on line $EX$. The intersection of the circumsphere $k$ and the plane results in a circle $k'$, which also has $M$ as its center. The intersection of $ABCD$ with the plane is the perpendicular bisector of line $EF$, and $Y$ is on that bisector.\n\n\n\nWe denote $\\angle ZEF = \\alpha$.\n\nSince $ZM$ and $EM$ are both radii of $k$ (and $k'$), the triangle $ZME$ is isosceles, therefore\n$\\angle EZM = \\angle ZEM = \\alpha$.\n\nSince $Y$ is on the bisector of $EF$, also triangle $EYF$ is isosceles, therefore $\\angle YFE = \\angle YEF = \\alpha$.\n\nTherefore, the triangles $ZME$ and $EYF$ are similar, and their corresponding angles $\\angle EMZ$ and $\\angle EYF$ are identical. $\\square$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21954,
"subject": "Mathematics (Olympiad)",
"question": "For any $a \\in \\mathbb{Z}$ and $X = (x_1, x_2) \\in T$, define $aX = (x'_1, x'_2) \\in T$ such that $x'_1 \\equiv a x_1 \\pmod{41}$ and $x'_2 \\equiv a x_2 \\pmod{41}$. Let $G$ be a graph with vertex set $T$ where each vertex $X$ is connected to $2X$.\n\nGiven that $2^{20} \\equiv 1 \\pmod{41}$ and $2^{10} \\equiv -1 \\pmod{41}$, the graph $G$ can be decomposed into edge-disjoint cycles of length 20. Each such cycle is of the form\n\n$$\n(X, 2X, \\dots, 2^9X, -X, -2X, \\dots, -2^9X)\n$$\n\nSuppose we color 10 vertices in each cycle so that for every $i = 0, 1, \\dots, 9$, exactly one of $2^iX$ and $-2^iX$ is colored. Let $N$ be the total number of pairs $(U, 2U)$ in all cycles such that both $U$ and $2U$ are colored. Determine all possible values of $N$.",
"options": [],
"answer": "See solution",
"solution": "If we color $X, 2X, \\dots, 2^9X$ in a cycle, there are 9 pairs $(U, 2U)$ where both are colored. If we color $X, 2^2X, 2^4X, 2^6X, 2^8X, 2^9X, -2X, -2^3X, -2^5X, -2^7X$, then only one such pair exists. Changing the coloring from $U$ to $-U$ changes the number of such pairs by 0, 2, or $-2$, so all odd values from 1 to 9 are possible for a single cycle. Since there are $\\frac{41^2 - 1}{20} = 84$ cycles, $N$ can take all even values from $84$ to $84 \\times 9 = 756$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21955,
"subject": "Mathematics (Olympiad)",
"question": "Given an acute triangle $ABC$ with circumcircle $(O)$ and A-excircle $(J)$, which touches $BC$, $CA$, and $AB$ at $D$, $E$, and $F$ respectively.\n\n**a)** Let $L$ be the midpoint of $BC$. The circle with diameter $LJ$ cuts $DE$ and $DF$ at $K$ and $H$. Prove that the circumcircles of triangles $BDK$ and $CDH$ meet again at a point that belongs to $(J)$.\n\n**b)** Let $G$ be the intersection of $EF$ and $BC$. $GJ$ cuts $AB$ and $AC$ at $M$ and $N$ respectively. Let $P$ and $Q$ be the points on $JB$ and $JC$ such that $\\angle PAB = \\angle QAC = 90^\\circ$. The line $PM$ intersects $QN$ at $T$. Let $S$ be the midpoint of the major arc $BC$ of $(O)$ and let $(I)$ be the incircle of triangle $ABC$. Prove that $SI$ meets $AT$ on $(O)$.\n\n",
"options": [],
"answer": "See solution",
"solution": "**a)** Let $D'$ and $F'$ be the tangency points of the incircle $(I)$ in triangle $ABC$ with $BC$ and $BA$. Construct diameter $DW$ of circle $(J)$. We have\n\n$$\n\\frac{AI}{AJ} = \\frac{IF'}{JF} = \\frac{ID'}{JW}\n$$\n\nso $A$, $D'$, and $W$ are collinear.\n\nLet $D_1$ be the intersection of $ID'$ and $AD$. Since $JD = JW$, so $ID' = ID_1$. But $BD' = CD$, then $LD = LD'$, this implies that $IL \\parallel DD_1$. Let $X$ be the midpoint of $D'W$, then the quadrilateral $DLXJ$ is a rectangle. It follows that $\\angle XHD = 90^\\circ$, or $XH \\parallel JB$. From there it turns out that $XH \\perp BI$.\n\nConstruct the rectangle $CDJU$. Then $JU \\parallel CD$ and $JU = CD$, so $JU \\parallel BD'$ and $JU = BD'$, hence $BD'UJ$ is a parallelogram. From there, we get $D'U \\parallel XH$. Let $XH$ meet $DJ$, $UW$, and $UC$ at $Y$, $V$, and $Z$ respectively. It is clear that $V$ is the midpoint of $UW$, but we also have $UZ \\parallel YW$, it follows that $V$ is the midpoint of $YZ$.\n\nConsider the cyclic quadrilateral $CEUJ$ with $CE = CD = UJ$, so $UE \\parallel CJ$ or $UE \\perp DE$. Note that $EW \\perp ED$, so $W$, $U$, and $E$ are collinear. Consider two triangles $BIC$ and $YVW$, by angle chasing,\n\n$$\n\\angle YWV = \\angle CED = \\angle ICB\n$$\n\nand\n\n$$\n\\angle WYV = \\angle XYJ = \\angle BJD = \\angle IBC\n$$\n\nso $\\triangle BIC \\sim \\triangle YVW$ (a-a). Note that $V$ is the midpoint of $YZ$ and $L$ is the midpoint of $BC$, so $\\triangle BIL \\sim \\triangle YZW$ (c-g-c). Hence, $\\angle YWZ = \\angle BLI = \\angle BDA = 90^\\circ - \\angle WDU'$. It follows that $\\angle DU'W = 90^\\circ$ where $U'$ is the intersection of $WZ$ and $AD$. On the other hand, we have $\\angle DU'W = \\angle DU'Z = 90^\\circ$, so $U'$ lies on $(CDH)$ and $(J)$. Similarly, circle $(BDK)$ also passes through $U'$.\n\n**b)** Let $S'$ and $R$ be the second intersection of the lines $AI$, $SI$ with $(O)$. We have $S'I^2 = S'L \\cdot S'S$, so $\\angle S'IL = \\angle S'SI = \\angle SAR$. We also get $IL \\parallel AD$, so $\\angle S'IL = \\angle S'AD$, then $\\angle S'AR = \\angle S'AD$ or\n\n$$\n\\angle BAR = \\angle CAD. \\quad (1)\n$$\n\nDraw $DH' \\perp JG$ ($H' \\in JG$). Note that $GH' \\cdot GJ = GD^2 = GE \\cdot GF$, so $EFJH'$ is a cyclic quadrilateral. It also obtains that $AEJF$ is cyclic, hence $A$, $E$, $F$, $J$, and $H'$ are concyclic. It implies $\\angle AH'J = 90^\\circ$ or $A$, $D$, and $H'$ are collinear.\n\n\n\nOn the other hand, it is easy to see that $\\angle PAJ = \\angle QAJ = 90^\\circ + \\frac{1}{2} \\angle BAC = \\angle BIC$ and $\\angle IBC = \\angle IJC$, $\\angle ICB = \\angle IJB$, thus $\\triangle IBC \\sim \\triangle APJ \\sim \\triangle AJQ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21956,
"subject": "Mathematics (Olympiad)",
"question": "There are $n$ values of $x$ in the interval $0 < x < 2\\pi$ where $f(x) = \\sin(7\\pi \\cdot \\sin(5x)) = 0$. For $t$ of these $n$ values of $x$, the graph of $y = f(x)$ is tangent to the $x$-axis. Find $n + t$.",
"options": [],
"answer": "See solution",
"solution": "The function $\\sin(7\\pi \\cdot \\sin(5x))$ is zero if and only if $\\sin(5x)$ equals one of $0, \\pm\\frac{1}{7}, \\pm\\frac{2}{7}, \\pm\\frac{3}{7}, \\pm\\frac{4}{7}, \\pm\\frac{5}{7}, \\pm\\frac{6}{7}$, or $\\pm 1$.\n\nAs $x$ increases from $0$ to $\\frac{\\pi}{10}$, the value of $\\sin(5x)$ starts at $0$ and passes successively through $\\frac{1}{7}, \\frac{2}{7}, \\frac{3}{7}, \\frac{4}{7}, \\frac{5}{7}, \\frac{6}{7}, 1$, and then as $x$ increases from $\\frac{\\pi}{10}$ to $\\frac{\\pi}{5}$, the value of $\\sin(5x)$ passes successively through $\\frac{6}{7}, \\frac{5}{7}, \\frac{4}{7}, \\frac{3}{7}, \\frac{2}{7}, \\frac{1}{7}$, before ending at $0$ when $x = \\frac{\\pi}{5}$.\n\nThe same occurs as $x$ traverses the interval $\\left(\\frac{\\pi}{5}, \\frac{2\\pi}{5}\\right]$, except that the corresponding values of $\\sin(5x)$ are negated. Thus there are $28$ values of $x$ in the interval $(0, \\frac{2\\pi}{5}]$ for which $y = 0$. Because $f$ is periodic with a period of $\\frac{2\\pi}{5}$, it follows that there are $5 \\cdot 28 = 140$ values of $x$ in $(0, 2\\pi]$ for which $y = 0$, so there are $n = 139$ values once $x = 2\\pi$ has been excluded.\n\nIt remains to determine for which of these values the curve is tangent to the $x$-axis. Because the graph of $\\sin x$ is never tangent to the $x$-axis, it is impossible for the graph of $y = \\sin(7\\pi \\cdot \\sin(5x))$ to be tangent to the $x$-axis on an interval where $7\\pi \\cdot \\sin(5x)$ is monotonically increasing or monotonically decreasing. Therefore the points of tangency are exactly at the local extrema of $7\\pi \\cdot \\sin(5x)$, and in the interval $(0, \\frac{2\\pi}{5}]$, these occur at $x = \\frac{\\pi}{10}$ and $x = \\frac{3\\pi}{10}$.\n\n\n\nIndeed, the graph is tangent to the $x$-axis from above at $x = \\frac{\\pi}{10}$ and from below at $x = \\frac{3\\pi}{10}$. Therefore there are a total of $t = 5 \\cdot 2 = 10$ points of tangency in the interval $(0, 2\\pi)$. The requested sum is $139 + 10 = 149$.",
"topic": "Discrete Mathematics",
"subtopic": "Other"
},
{
"id": 21957,
"subject": "Mathematics (Olympiad)",
"question": "Од точка $M$ кон кружница $k$ се повлечени две тангенти со допирни точки $G$ и $H$. Ако $O$ е центарот на $k$ и $K$ е ортоцентарот на триаголникот $MGH$, докажи дека $\\angle GMH = \\angle OGK$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Да забележиме дека $K$ мора да лежи на $OM$. Од $\\overline{HK} \\perp \\overline{GM}$ и $\\overline{OG} \\perp \\overline{GM}$, следува $\\overline{HK} \\parallel \\overline{OG}$. Аналогно, $\\overline{OH} \\parallel \\overline{GK}$. Од $\\overline{OG} = \\overline{OH}$, следува дека $OHKG$ е ромб. Да забележиме дека $O$, $H$, $M$ и $G$ лежат на кружница со дијаметар $OM$. Оттука $\\angle OGH = \\angle OMH$. Сега тврдењето на задачата следува од $\\angle OGK = 2\\angle OGH$ и $\\angle GMH = 2\\angle OMH$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21958,
"subject": "Mathematics (Olympiad)",
"question": "117 spies are divided into three missions. Each password allows two spies to communicate directly. What is the minimum number of passwords needed to ensure that at least one mission forms a connected network (i.e., all its spies can communicate, possibly via intermediaries), regardless of how the spies are assigned to missions?",
"options": [],
"answer": "See solution",
"solution": "Let the sizes of the three missions be $a$, $b$, and $c$ with $a + b + c = 117$. To guarantee that at least one mission is connected, we need to issue enough passwords so that, no matter how the spies are assigned, at least one mission's communication graph is connected.\n\nFor a group of $a$ spies, the maximum number of edges in a disconnected graph is $\\frac{(a-1)(a-2)}{2}$. Thus, to ensure connectivity, we need at least $\\frac{(a-1)(a-2)}{2} + 1$ edges (passwords) for that group. Similarly for $b$ and $c$.\n\nTherefore, the minimum total number of passwords required is:\n\n$$\nn = \\frac{a^2 - 3a + 4}{2} + \\frac{b^2 - 3b + 4}{2} + \\frac{c^2 - 3c + 4}{2} - 2\n$$\n\nTo minimize $n$ under $a + b + c = 117$, we set $a = b = c = 39$ (since the sum of squares is minimized when the numbers are equal):\n\n$$\n2n = a^2 + b^2 + c^2 - 3(a + b + c) + 8 = 3 \\times 39^2 - 3 \\times 117 + 8 = 4563 - 351 + 8 = 4220\n$$\n\nSo $n = 2110$.\n\n**Answer:** The minimum number of passwords needed is $2110$, achieved when the three missions are of equal size ($39$ spies each).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21959,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an isosceles right-angled triangle of area $1$. Find the length of the shortest segment that divides the triangle into two parts of equal area.",
"options": [],
"answer": "See solution",
"solution": "There are two cases as shown in the figures below:\n\n\n\nFig. 1\n\n\n\nFig. 2\n\nIn Fig. 1,\n\n$$\nXY^2 = AX^2 + AY^2 = (AX - AY)^2 + 2AX \\cdot AY = (AX - AY)^2 + 2.\n$$\n\nThus, $XY$ is minimum when $AX = AY$, so $XY^2 = 2$.\n\nIn Fig. 2, let $B$ be the origin and $BA$ be the $x$-axis, $X$ be $(x, 0)$ and $Y$ be $(y, y)$. Then $2[BXY] = xy = 1$ and\n\n$$\nXY^2 = (x - y)^2 + y^2 = x^2 + 2y^2 - 2xy = x^2 + 2y^2 - 2 = (x - \\sqrt{2}y)^2 + 2\\sqrt{2} - 2.\n$$\n\nSo $XY$ is minimum when $x = \\sqrt{2}y$ and the minimum value is $2\\sqrt{2} - 2$.\n\nSince $2\\sqrt{2} - 2 < 2$, the length of the shortest segment is $\\sqrt{2\\sqrt{2} - 2}$.\n\nAlternatively, we have $2[BXY] = BX \\cdot BY \\sin 45^\\circ = BX \\cdot BY \\cos 45^\\circ = 1$. Therefore $BX \\cdot BY = \\sqrt{2}$. Using the cosine rule,\n\n$$\nXY^2 = BX^2 + BY^2 - 2BX \\cdot BY \\cos 45^\\circ = BX^2 + BY^2 - 2 = (BX - BY)^2 + 2\\sqrt{2} - 2.\n$$\n\nTherefore, $XY$ is minimum when $BX = BY$ and $XY^2 = 2\\sqrt{2} - 2$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21960,
"subject": "Mathematics (Olympiad)",
"question": "Given a $2n$-gon, assign numbers $1, 2, \\dots, 2n$ to its vertices so that for each pair of neighboring vertices, the difference of the numbers assigned is at least $n-1$. Is it possible to do this? If so, provide a construction; if not, explain why.",
"options": [],
"answer": "See solution",
"solution": "Pick a vertex $P$ and label the vertices consecutively as $1, 2, \\dots, 2n$ starting with $P$ as $1$ and going clockwise. For $1 \\leq k \\leq n$, reassign the number $k$ to the vertex labeled $2k-1$, and the number $n+k$ to the vertex labeled $2k$. For $1 \\leq j < 2n$, the difference between the numbers reassigned to vertices originally labeled $j$ and $j+1$ is $n$ if $j$ is odd, and $n-1$ if $j$ is even. The difference between the numbers at vertices $1$ and $2n$ is $2n-1$. Thus, the requirement is satisfied. However, if we try to assign $n$ to a vertex, it is impossible to assign numbers to both its neighbors so that the difference is $n$ or greater, except in the construction above.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21961,
"subject": "Mathematics (Olympiad)",
"question": "Let $p$ be a prime of the form $13k + 1$ with $p > 8191$. Show that $$\\frac{2^{p-1}-1}{8191p}$$ is an integer.",
"options": [],
"answer": "See solution",
"solution": "By Dirichlet's theorem, there are infinitely many primes $p$ of the form $13k + 1$. For such $p > 8191$:\n\n- By Fermat's little theorem, $p$ divides $2^{p-1} - 1$.\n- Note that $2^{p-1} - 1 = 2^{13k} - 1 = (2^{13} - 1)(2^{13(k-1)} + 2^{13(k-2)} + \\dots + 1)$, so $2^{13} - 1 = 8191$ divides $2^{p-1} - 1$.\n- Since $p$ and $8191$ are relatively prime, both divide $2^{p-1} - 1$ and thus $\\frac{2^{p-1}-1}{8191p}$ is an integer.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21962,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\geq 3$ be an integer. Consider a grid consisting of $n \\times n$ squares, where each square can be either white or black. In each step, we change the colours of the five squares that make up the pattern\n\n\n\nin any rotation. Initially, all squares are white. For which values of $n$ can all squares be made black after a finite number of steps?",
"options": [],
"answer": "See solution",
"solution": "Let us call the individual squares of the grid *fields*. We prove that recoloring all $n^2$ white fields in finitely many steps is possible if and only if $n > 3$ and $n$ is divisible by two or three.\n\nFirst, for $n=3$, the desired recoloring is impossible. Consider the three fields A, B, C as shown below:\n\n\n\nIn each step, field A is recolored and exactly one of fields B or C is recolored. If all 9 fields were black, the number of recolorings of B and C would be odd, so the number of recolorings of A would be even, meaning A would remain white—a contradiction. Thus, for $n \\geq 4$, we proceed as follows:\n\n1. **If $n$ is even:**\n Use the procedure in Figure 2 repeatedly, where in four steps we change the colors of exactly 4 fields in a $4 \\times 4$ square. Divide the $n \\times n$ grid into $(\\frac{n}{2})^2$ squares of size $2 \\times 2$ and apply the procedure to each. For border squares, rotate Figure 2 as needed so the $4 \\times 4$ square fits inside the grid.\n\n\n\n2. **If $n$ is divisible by three:**\n Use the procedure in Figure 3, which first uses the construction from step 1 twice. This changes the colors of 9 fields forming a $3 \\times 3$ square inside a $4 \\times 5$ rectangle (see the purple border). Apply this to each $3 \\times 3$ square, rotating as needed for boundary squares.\n\n\n\n3. **If $n$ is not divisible by two or three:**\n Number the fields in each line as 0, 1, 2, 0, ... as in Figure 4.\n\n\n\nLet $a_i$ be the number of black fields with number $i$, $i \\in \\{0, 1, 2\\}$. The parity of each $a_i$ changes in each step, since we change the color in three adjacent columns, each with an odd number of cells. Initially, $a_0 = a_1 = a_2 = 0$, so after any number of steps, $a_0 \\equiv a_1 \\equiv a_2 \\pmod{2}$.\n\nSince $3 \\nmid n$, the number of squares labeled 0 is $n$ more than those labeled 2. If all boxes were black, $a_0 - a_2 = n$, and since $2 \\nmid n$, $a_0$ and $a_2$ would have different parity, contradicting the previous conclusion.\n\nTherefore, all squares can be made black after a finite number of steps if and only if $n > 3$ and $n$ is divisible by 2 or 3.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21963,
"subject": "Mathematics (Olympiad)",
"question": "Confirm that\n\n$$\n\\binom{n}{\\lfloor n/2 \\rfloor} < 2^{n-2}, \\quad \\forall n \\ge 9.\n$$\n\nProve this inequality by induction, considering separately the cases when $n$ is even and when $n$ is odd.",
"options": [],
"answer": "See solution",
"solution": "We use induction on $n$, considering even and odd cases separately.\n\n**Case 1:** $n = 2m$, $m \\ge 5$.\n\nWe have\n$$\n\\binom{2m}{m} = \\frac{(2m)!}{(m!)^2}.\n$$\nFor $m = 5$ ($n = 10$):\n$$\n\\binom{10}{5} = 252 < 2^8 = 256 = 2^{10-2}.\n$$\nAssume for $m \\ge 5$,\n$$\n\\frac{(2m)!}{(m!)^2} < 2^{2m-2}.\n$$\nThen for $m+1$:\n$$\n\\frac{(2(m+1))!}{((m+1)!)^2} = \\frac{2(m+1)(2m+1)(2m)!}{(m+1)^2(m!)^2} = \\frac{2m+1}{m+1} \\cdot 2 \\cdot \\frac{(2m)!}{(m!)^2}.\n$$\nBy the induction hypothesis,\n$$\n< \\frac{2m+1}{m+1} \\cdot 2 \\cdot 2^{2m-2} = \\frac{2m+1}{m+1} 2^{2m-1} < 2^{2m}.\n$$\nSo the inequality holds for even $n$.\n\n**Case 2:** $n = 2m+1$, $m \\ge 4$.\n\nWe have\n$$\n\\binom{2m+1}{m} = \\frac{(2m+1)!}{m!(m+1)!}.\n$$\nFor $m = 4$ ($n = 9$):\n$$\n\\binom{9}{4} = 126 < 2^7 = 128 = 2^{9-2}.\n$$\nAssume for $m \\ge 4$,\n$$\n\\frac{(2m+1)!}{m!(m+1)!} < 2^{2m-1}.\n$$\nThen for $m+1$:\n$$\n\\frac{(2m+3)!}{(m+2)!(m+1)!} = \\frac{(2m+3)(2m+2)}{(m+2)(m+1)} \\cdot \\frac{(2m+1)!}{m!(m+1)!}.\n$$\nBy the induction hypothesis,\n$$\n< \\frac{2m+3}{m+2} 2^{2m} < 2^{2m+1}.\n$$\nThus, the inequality holds for odd $n$ as well.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 21964,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\mathbb{N}_+$ be the set of all positive integers. Find all functions $f : \\mathbb{N}_+ \\to \\mathbb{N}_+$ satisfying that for any $x, y \\in \\mathbb{N}_+$, $f(f(x) + y)$ divides $x + f(y)$.",
"options": [],
"answer": "See solution",
"solution": "All maps that satisfy the question are one of the following three categories:\n\n1. For any $x \\in \\mathbb{N}_+$, $f(x) = x$.\n\n2. For any $x > 1$, $f(x) = 1$ and $f(1)$ can be any positive integer.\n\n3. For any $x > 1$, $f(x) = \\begin{cases} 1, & x \\text{ is odd} \\\\ 2, & x \\text{ is even} \\end{cases}$ and $f(1)$ can be any odd positive integer.\n\nIt is easy to verify that all the three types of functions satisfy the conditions. In the following, we will prove that only the above three types of functions satisfy the conditions. The proof is done in three steps.\n\n**Step 1:** We will prove that either $f(x) = x$ or $f$ is not an injection.\n\nFor example, if $f(1) > 1$, by taking $x = 1$ in the problem, we know that $f(f(1) + y)$ divides $1 + f(y)$. In particular,\n\n$$\nf(f(1) + y_0) \\leq 1 + f(y_0).\n$$\n\nBy induction, it is easy to see that for fixed $y_0 \\in \\{1, \\dots, f(1)\\}$ and any $t \\in \\mathbb{Z}_+$, there is\n\n$$\nf(t \\cdot f(1) + y) \\leq t + f(y).\n$$\n\nTherefore, for any $y$ large enough, $f(y) < y$. So when $y$ is large enough, $f : \\{1, \\dots, y\\} \\to \\{1, \\dots, y-1\\}$ cannot be an injection.\n\nIn the following, we will discuss the case $f(1) = 1$. By taking $x = 1$ in the problem, we know that $f(1+y)$ divides $1+f(y)$. In particular, $f(1+y) \\leq 1+f(y)$. By induction, it is easy to know that for any $x \\in \\mathbb{Z}_+$, there is $f(x) = x$ or $f(y) < y$ for $y$ large enough. Similarly, we obtain that $f$ cannot be an injection.\n\n**Step 2:** We will show that if $f$ is not an injection, then for $x$ large enough, either $f(x) = 1$ or $f(x) = \\begin{cases} 1, & x \\text{ is odd} \\\\ 2, & x \\text{ is even} \\end{cases}$.\n\nSuppose $f$ is not an injection and we denote $A$ as the smallest positive integer such that there exists a positive integer $x_0 \\in \\mathbb{N}_+$ satisfying\n\n$$\nf(x_0 + A) = f(x_0).\n$$\n\nBy substituting $x = x_0$ and $x = x_0 + A$ respectively into the problem, we get\n\n$$\nf(f(x_0) + y) \\mid x_0 + f(y)\n$$\n\nand\n\n$$\nf(f(x_0 + A) + y) \\mid x_0 + A + f(y).\n$$\n\nSubtracting the above two, we get\n\n$$\nf(f(x_0) + y) \\mid A.\n$$\n\nThis shows that $f(z)$ can only take values in the factors of $A$ when $z$ is greater than $f(x_0)$. Suppose that $A$ has a total of $D$ factors. According to the Pigeonhole principle, we investigate $D+1$ consecutive positive integers $z, z+1, \\dots, z+D$ that are greater than $f(x_0)$, of which there must be two numbers that have equal images under $f$ and the difference between these two numbers is less than or equal to $D$. This leads to $A \\leq D$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21965,
"subject": "Mathematics (Olympiad)",
"question": "Let $f(x)$ be a monic polynomial of even degree with integer coefficients. It is known that there exist infinitely many integers $x$ for which $f(x)$ is a perfect square. Prove that there exists a polynomial $g(x)$ with integer coefficients such that $f(x) = g^2(x)$.",
"options": [],
"answer": "See solution",
"solution": "Let $n = 2k$ and $f(x) = x^{2k} + a_{2k-1}x^{2k-1} + \\dots + a_1x + a_0$, where $a_i$ are integers. First, we prove that $f(x)$ can be written in the form\n\n$$\nf(x) = (x^k + b_{k-1}x^{k-1} + \\dots + b_1x + b_0)^2 + r(x),\n$$\n\nwhere $b_0, b_1, \\dots, b_{k-1}$ are rational numbers and $r(x)$ is a polynomial with rational coefficients and degree at most $k-1$. Indeed, the coefficient of $x^{k+t}$, $t = k-1, k-2, \\dots, 1, 0$, of the polynomial $(x^k + b_{k-1}x^{k-1} + \\dots + b_1x + b_0)^2$ is of the form $c_{k+t} = 2b_t + \\sum_{i=1}^{k-1-t} b_{t+i}b_{k-i}$. Inductively, we find the values of $b_{k-1}, b_{k-2}, \\dots, b_1, b_0$ such that $c_{k+t} = a_{k+t}$ for $t = k-1, k-2, \\dots, 1, 0$. After that, we compute the coefficients of $r(x)$.\n\nIf $f(x) = y^2$ has infinitely many integer solutions for which $x < 0$, then $f_1(x) = y^2$ for $f_1(x) = f(-x)$ has infinitely many solutions for which $x > 0$. Therefore, we may assume that $f(x) = y^2$ has infinitely many solutions for which $x > 0$. The equality $(x^k + b_{k-1}x^{k-1} + \\dots + b_1x + b_0)^2 + r(x) = y^2$ can be written in the form $h^2(x) + M^2r(x) = (My_x)^2$, where $M$ is the least common multiple of the denominators of $b_i$, $0 \\le i \\le k-1$, and $h(x)$ is a polynomial with integer coefficients and leading coefficient $M$. Suppose that $r(x)$ is not identically $0$ (if $b_i = 0$ for all $i$, $0 \\le i \\le k-1$, we set $M=1$).\n\n**Case 1.** Let the leading coefficient of $r(x)$ be positive. For large enough values of $x$, we have $(My_x)^2 > h^2(x)$, implying that $(My_x)^2 \\ge (h(x)+1)^2$. Therefore, $h^2(x) + M^2r(x) \\ge (h(x)+1)^2$, i.e., $2h(x) \\le M^2r(x)-1$. The latter inequality is not true for large enough values of $x$ since $h(x)$ is of degree $k$ whereas $r(x)$ is of degree at most $k-1$.\n\n**Case 2.** Let the leading coefficient of $r(x)$ be negative. For large enough values of $x$, we have $(My_x)^2 < h^2(x)$, implying $(My_x)^2 \\le (h(x)-1)^2$. Therefore, $h^2(x) + M^2r(x) \\le (h(x)-1)^2$, i.e., $2h(x) \\le -M^2r(x)+1$, which is not true for large enough values of $x$.\n\nTherefore, $r(x) \\equiv 0$, i.e., $f(x) = (x^k + b_{k-1}x^{k-1} + \\dots + b_1x + b_0)^2$. Since $f(x)$ has integer coefficients, the same is true for $x^k + b_{k-1}x^{k-1} + \\dots + b_1x + b_0$ (Gauss Lemma).",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 21966,
"subject": "Mathematics (Olympiad)",
"question": "Deslizamos un cuadrado de $10\\ \\text{cm}$ de lado por el plano $OXY$ de forma que los vértices de uno de sus lados estén siempre en contacto con los ejes de coordenadas, uno con el eje $OX$ y otro con el eje $OY$. Determina el lugar geométrico que en ese movimiento describen:\n\n1. El punto medio del lado de contacto con los ejes.\n2. El centro del cuadrado.\n3. Los vértices del lado de contacto y del opuesto en el primer cuadrante.",
"options": [],
"answer": "See solution",
"solution": "Sean $PQRS$ el cuadrado de lado $10\\ \\text{cm}$, $PQ$ el lado de apoyo, $M(m_1, m_2)$ el punto medio de dicho lado y $C(c_1, c_2)$ el centro del cuadrado tal y como muestra la figura donde, además, señalamos los puntos $A, B, D$ y $E$.\n\n\n\n**a) Caso del punto medio $M$.**\n\n$$\nOM = PM = \\frac{1}{2}PQ = 5,\n$$\n\nluego $m_1^2 + m_2^2 = 25$.\n\n**b) Caso del centro del cuadrado $C$.**\n\nLos triángulos $AQM$, $AOM$, $BMO$ y $DMC$ son claramente congruentes:\n\n$$\nAM = OB = DC, \\quad AQ = OA = MD = BM, \\text{ y } OM = MQ = MC = 5.\n$$\n\nAsí, resulta que las coordenadas del centro del cuadrado, en su deslizamiento, son iguales:\n\n$$\nc_1 = OE = OB + BE = m_1 + MD = m_1 + m_2,\n$$\n$$\nc_2 = EC = ED + DC = OA + AM = m_2 + m_1.\n$$\n\nLuego, el centro del cuadrado se mueve, en este primer cuadrante, sobre un segmento de la bisectriz. Las posiciones extremas se dan cuando el lado $PQ$ se apoya sobre alguno de los ejes, $C(5, 5)$, y cuando forma una escuadra, esto es, un triángulo rectángulo isósceles, con ellos, $C(5\\sqrt{2}, 5\\sqrt{2})$. Trabajando análogamente en los demás cuadrantes podemos afirmar que el centro del cuadrado recorre el segmento de sus bisectrices que viene dado por la expresión:\n\n$$\nC(c_1, c_2) = (\\pm 5\\lambda, \\pm 5\\lambda) \\text{ con } \\lambda \\in [1, \\sqrt{2}].\n$$\n\n**c) Caso de los vértices del cuadrado en el lado de contacto: $P$ y $Q$.**\n\nLos vértices $P$ y $Q$ se mueven sobre segmentos de los ejes coordenados, esto es, de las líneas $x = 0$ y $y = 0$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 21967,
"subject": "Mathematics (Olympiad)",
"question": "Given integer $n \\geqslant 4$. Find the maximum of\n$$\n\\frac{\\sum_{i=1}^{n} a_i (a_i + b_i)}{\\sum_{i=1}^{n} b_i (a_i + b_i)}\n$$\nfor non-negative real numbers $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ satisfying\n$$\na_1 + a_2 + \\dots + a_n = b_1 + b_2 + \\dots + b_n > 0.\n$$",
"options": [],
"answer": "See solution",
"solution": "The maximum is $n - 1$.\n\nBy homogeneity, we can assume without loss of generality that $\\sum_{i=1}^{n} a_i = \\sum_{i=1}^{n} b_i = 1$.\n\nFirst, if $a_1 = 1$, $a_2 = a_3 = \\dots = a_n = 0$ and $b_1 = 0$, $b_2 = b_3 = \\dots = b_n = \\frac{1}{n-1}$, then\n$$\n\\sum_{i=1}^{n} a_i (a_i + b_i) = 1,\n$$\n$$\n\\sum_{i=1}^{n} b_i (a_i + b_i) = \\frac{1}{n-1},\n$$\nso\n$$\n\\frac{\\sum_{i=1}^{n} a_i (a_i + b_i)}{\\sum_{i=1}^{n} b_i (a_i + b_i)} = n - 1.\n$$\n\nNow, we prove that for any real numbers $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ satisfying $\\sum_{i=1}^{n} a_i = \\sum_{i=1}^{n} b_i = 1$, we have\n$$\n\\frac{\\sum_{i=1}^{n} a_i (a_i + b_i)}{\\sum_{i=1}^{n} b_i (a_i + b_i)} \\leqslant n - 1.\n$$\n\nSince the denominator is positive, it suffices to show\n$$\n\\sum_{i=1}^{n} a_i (a_i + b_i) \\le (n-1) \\sum_{i=1}^{n} b_i (a_i + b_i),\n$$\ni.e.,\n$$\n(n-1) \\sum_{i=1}^{n} b_i^2 + (n-2) \\sum_{i=1}^{n} a_i b_i \\ge \\sum_{i=1}^{n} a_i^2.\n$$\n\nBy symmetry, assume $b_1$ is the smallest among $b_1, b_2, \\dots, b_n$. Then\n$$\n\\begin{align*}\n& (n-1) \\sum_{i=1}^{n} b_i^2 + (n-2) \\sum_{i=1}^{n} a_i b_i \\\\\n&\\ge (n-1) b_1^2 + (n-1) \\sum_{i=2}^{n} b_i^2 + (n-2) \\sum_{i=1}^{n} a_i b_1 \\\\\n&\\ge (n-1) b_1^2 + \\left( \\sum_{i=2}^{n} b_i \\right)^2 + (n-2) b_1 \\\\\n&= (n-1) b_1^2 + (1-b_1)^2 + (n-2) b_1 \\\\\n&= n b_1^2 + (n-4) b_1 + 1 \\\\\n&\\ge 1 = \\sum_{i=1}^{n} a_i \\ge \\sum_{i=1}^{n} a_i^2.\n\\end{align*}\n$$\n\n$\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21968,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$ be positive real numbers such that $abc = 1$. Prove that\n\n$$\n\\frac{1}{a^5(b+2c)^2} + \\frac{1}{b^5(c+2a)^2} + \\frac{1}{c^5(a+2b)^2} \\geq \\frac{1}{3}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $x = \\frac{1}{a}$, $y = \\frac{1}{b}$, $z = \\frac{1}{c}$. The condition is now $xyz = 1$ and the inequality becomes: for $xyz = 1$,\n\n$$\n\\frac{x^3}{(2y+z)^2} + \\frac{y^3}{(2z+x)^2} + \\frac{z^3}{(2x+y)^2} \\geq \\frac{1}{3}. \\qquad (1)\n$$\n\nBy Hölder's inequality:\n\n$$\n(u_1^3 + v_1^3 + w_1^3)(u_2^3 + v_2^3 + w_2^3)(u_3^3 + v_3^3 + w_3^3) \\geq (u_1v_1w_1 + u_2v_2w_2 + u_3v_3w_3)^3\n$$\n\ntake\n\n$$\n\\begin{aligned}\n u_1 &= u_2 = \\sqrt[3]{2y+z}, & u_3 &= \\frac{x}{\\sqrt[3]{(2y+z)^2}}, \\\\\n v_1 &= v_2 = \\sqrt[3]{2z+x}, & v_3 &= \\frac{y}{\\sqrt[3]{(2z+x)^2}}, \\\\\n w_1 &= w_2 = \\sqrt[3]{2x+y}, & w_3 &= \\frac{z}{\\sqrt[3]{(2x+y)^2}},\n\\end{aligned}\n$$\n\nyielding\n\n$$\n[3(x + y + z)]^2 \\left( \\frac{x^3}{(2y+z)^2} + \\frac{y^3}{(2z+x)^2} + \\frac{z^3}{(2x+y)^2} \\right) \\geq (x + y + z)^3.\n$$\n\nRewriting,\n\n$$\n\\frac{x^3}{(2y+z)^2} + \\frac{y^3}{(2z+x)^2} + \\frac{z^3}{(2x+y)^2} \\geq \\frac{x+y+z}{9},\n$$\n\nand since $x + y + z \\geq 3\\sqrt[3]{xyz} = 3$, by the AM-GM inequality, $\\frac{x+y+z}{9} \\geq \\frac{1}{3}$. Thus, (1) holds.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21969,
"subject": "Mathematics (Olympiad)",
"question": "It is given that the bookshelf can fit 9 of the same thick books, but the 10th one will not fit anymore. Similarly, it can hold 15 of the same thin books, but the 16th will not fit anymore. Is it possible for that shelf to hold simultaneously:\n\n\n\nFig. 3\n\na) 7 thick and 5 thin books?\n\nb) 6 thick and 6 thin books?",
"options": [],
"answer": "See solution",
"solution": "a) No.\n\nb) Let us denote the length of the shelf by $S$, the width of the thick book by $x$, and the width of the thin book by $y$. Then, we have the conditions:\n\n$$\n9x \\leq S < 10x \\quad \\text{and} \\quad 15y \\leq S < 16y\n$$\nwhich is equivalent to\n$$\n\\frac{1}{10}S < x \\leq \\frac{1}{9}S \\quad \\text{and} \\quad \\frac{1}{16}S < y \\leq \\frac{1}{15}S.\n$$\nWe can re-write these as:\n$$\n\\frac{72}{720}S < x \\leq \\frac{80}{720}S \\quad \\text{and} \\quad \\frac{45}{720}S < y \\leq \\frac{48}{720}S.\n$$\nSelect $x = \\frac{73}{720}S$ and $y = \\frac{46}{720}S$. Then,\n$$\n6x + 6y = 6 \\cdot \\frac{73}{720}S + 6 \\cdot \\frac{46}{720}S = \\frac{438}{720}S + \\frac{276}{720}S = \\frac{714}{720}S < S.\n$$\nThus, it is possible.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21970,
"subject": "Mathematics (Olympiad)",
"question": "At noon on a certain day, Minneapolis is $N$ degrees warmer than St. Louis. At 4:00 the temperature in Minneapolis has fallen by 5 degrees while the temperature in St. Louis has risen by 3 degrees, at which time the temperatures in the two cities differ by 2 degrees. What is the product of all possible values of $N$?\n\n(A) 10 \n(B) 30 \n(C) 60 \n(D) 100 \n(E) 120",
"options": [],
"answer": "See solution",
"solution": "At 4:00, the amount by which the temperature in Minneapolis exceeds the temperature in St. Louis has decreased by $5 + 3 = 8$. Therefore, $|N - 8| = 2$. The two solutions to this equation are $N = 6$ and $N = 10$. The requested product is $6 \\cdot 10 = 60$.\n\nAlternatively, let $M$ and $S$ be the temperatures in the two cities at noon. It is given that $M - S = N$. At 4:00, the temperatures are $M - 5$ and $S + 3$, respectively. The difference in temperature at 4:00 is therefore\n\n$$\n|(M - 5) - (S + 3)| = |(M - S) - 8| = |N - 8|.\n$$\n\nThe equation $|N - 8| = 2$ has two solutions: $N = 6$ and $N = 10$. The requested product is $6 \\cdot 10 = 60$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21971,
"subject": "Mathematics (Olympiad)",
"question": "Let $x$, $y$, and $z$ be positive real numbers with $x + y + z = 3$. 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\nis 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 $x \\geq y, z$. Consequently, we have $x \\geq \\frac{x + y + z}{3} = 1$. We now show that $a := y(y + z - x) = y(3 - 2x)$ satisfies $a \\leq 1$.\n\n*Case a*: For $\\frac{3}{2} \\leq x < 3$, clearly $a \\leq 0 < 1$.\n\n*Case b*: For $1 \\leq x < \\frac{3}{2}$, the factor $3 - 2x$ is positive. Therefore, $a \\leq x(3 - 2x)$. Hence, it suffices to prove $x(3 - 2x) \\leq 1$, which is equivalent to $2x^2 - 3x + 1 \\geq 0$, i.e., $(2x - 1)(x - 1) \\geq 0$.\n\nThis completes the proof.\n\n$\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21972,
"subject": "Mathematics (Olympiad)",
"question": "Given a positive integer $n$ and the function $f: \\mathbb{N} \\to \\mathbb{N}$ defined by\n$$\nf(x) = \\begin{cases} x/2 & \\text{if } x \\text{ is even} \\\\ (x-1)/2 + 2^{n-1} & \\text{if } x \\text{ is odd} \\end{cases}.\n$$\n\nDetermine the set $A = \\{x \\in \\mathbb{N} \\mid (\\underbrace{f \\circ f \\circ \\dots \\circ f}_{n \\text{ times}})(x) = x\\}$.",
"options": [],
"answer": "See solution",
"solution": "For $x \\in \\{0, 1, \\dots, 2^n - 1\\}$, we have $f(x) \\in \\{0, 1, \\dots, 2^n - 1\\}$, and for $x \\ge 2^n$, $f(x) < x$. Thus, $f(x) \\le \\max(x, 2^n - 1)$. If $x \\in A$, then\n$$\nx = f^{[n]}(x) \\le \\max(f^{[n-1]}(x), 2^n - 1) \\le \\dots \\le \\max(f(x), 2^n - 1).\n$$\nIf $x \\ge 2^n$, then $x \\le f(x)$, a contradiction. Therefore, $A \\subseteq \\{0, 1, \\dots, 2^n - 1\\}$.\n\nWe now prove $A = \\{0, 1, \\dots, 2^n - 1\\}$.\n\n*Proof 1.* Since $f(2^n - 1) = 2^n - 1$, we have $2^n - 1 \\in A$. Also, $2f(x) \\in \\{x, x + 2^n - 1\\}$, so $2f(x) \\equiv x \\pmod{2^n - 1}$. Inductively, $x \\equiv 2^n f^{[n]}(x) \\equiv f(x) \\pmod{2^n - 1}$. Because $f(x) < 2^n - 1$ for $x < 2^n - 1$, we obtain $f^{[n]}(x) = x$ for $x \\in \\{0, 1, \\dots, 2^n - 2\\}$.\n\n*Proof 2.* Represent any $x \\in \\{0, 1, \\dots, 2^n - 1\\}$ in base 2 with $n$ digits (adding leading zeros if necessary). To compute $f(x)$, move the last digit of $x$ to the first position. After $n$ such moves, the digits return to their original positions, so $f^{[n]}(x) = x$.\n\nTherefore, $A = \\{0, 1, \\dots, 2^n - 1\\}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21973,
"subject": "Mathematics (Olympiad)",
"question": "A triangle $\\triangle ABC$ is given together with a segment $PQ$ of length $t$ on the segment $BC$, so that $P$ is between $B$ and $Q$ and $Q$ is between $P$ and $C$. We draw parallel lines from the point $P$ to $AB$ and $AC$ which intersect $AC$ and $AB$ in $P_1$ and $P_2$, respectively. We draw parallel lines from the point $Q$ to $AB$ and $AC$ which intersect $AC$ and $AB$ in $Q_1$ and $Q_2$, respectively. Prove that the sum of the areas of $PQQ_1P_1$ and $PQQ_2P_2$ doesn't depend on the position of $PQ$ on $BC$.",
"options": [],
"answer": "See solution",
"solution": "\n\nLet $D$ be the intersection of $PP_1$ and $QQ_2$. Note that $P_1 D Q_2 P_2 = 2 \\cdot \\triangle ADP$ and $P_1 Q_1 D Q_1 = 2 \\cdot \\triangle ADQ$. So now we have:\n\n$$\n\\begin{aligned}\nP_1 P Q Q_2 P_2 + P_1 P_1 Q_1 Q &= P_1 D Q_2 P_2 + P_1 Q_1 D Q_1 + 2 \\cdot \\triangle PQD \\\\\n&= 2 \\cdot \\triangle ADP + 2 \\cdot \\triangle ADQ + 2 \\cdot \\triangle PQD \\\\\n&= 2 \\cdot \\triangle APQ = \\overline{PQ} \\cdot h_a\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21974,
"subject": "Mathematics (Olympiad)",
"question": "Consider the following operation. Given a positive integer $n$, if $n$ is a multiple of $3$, then you replace $n$ by $\\frac{n}{3}$. If $n$ is not a multiple of $3$, then you replace $n$ by $n + 10$. Then continue this process. For example, beginning with $n = 4$, this procedure gives:\n\n$$\n4 \\rightarrow 14 \\rightarrow 24 \\rightarrow 8 \\rightarrow 18 \\rightarrow 6 \\rightarrow 2 \\rightarrow 12 \\rightarrow \\dots\n$$\n\nSuppose you start with $n = 100$. What value results if you perform this operation exactly $100$ times?\n\n(A) 10 (B) 20 (C) 30 (D) 40 (E) 50",
"options": [],
"answer": "See solution",
"solution": "The first several iterations give:\n\n$$\n100 \\rightarrow 110 \\rightarrow 120 \\rightarrow 40 \\rightarrow 50 \\rightarrow 60 \\rightarrow 20 \\rightarrow 30 \\rightarrow 10 \\rightarrow 20 \\rightarrow 30 \\rightarrow 10 \\rightarrow \\dots\n$$\n\nThe values then cycle through $20 \\rightarrow 30 \\rightarrow 10$. Note that $20$ occurs after 6, 9, 12, 15, ... operations; $30$ occurs after 7, 10, 13, 16, ... operations; and $10$ occurs after 8, 11, 14, 17, ... operations. Because $100$ has remainder $1$ when divided by $3$, the value after $100$ operations is $30$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21975,
"subject": "Mathematics (Olympiad)",
"question": "試求滿足下列條件的最大正整數 $K$：\n\n任給有限多個長度皆為 1 的閉區間 $A_1, A_2, \\dots, A_N$（$N$ 為任意正整數）。若其聯集為 $[0, 2021]$，則我們必定可以在 $A_1, \\dots, A_N$ 中找到 $K$ 個兩兩交集皆為空集合的區間。",
"options": [],
"answer": "See solution",
"solution": "最大的 $K$ 為 $1011$。\n\n首先證明 $K \\geq 1011$。令 $\\epsilon \\in (0, \\frac{1}{1009})$，並考慮集合 $\\mathcal{A} = \\{0, 2 + \\epsilon, 4 + 2\\epsilon, \\dots, 2018 + 1009\\epsilon\\}$。顯然 $\\mathcal{A}$ 的每個點都必須存在一個 $A_i$ 包含之，且這些 $A_i$ 兩兩互斥（因為 $A_i$ 的長度皆為 $1$），故這邊共選到 $1010$ 個 $A_i$。又基於 $\\cup A_i = [0, 2021]$，我們必須有一個 $A_j$ 是 $[2020, 2021]$，而這與前述的 $1010$ 個 $A_i$ 都不相同（基於 $2018 + 1009\\epsilon + 1 < 2020$），故我們得到 $1011$ 個兩兩互斥的 $A_i$。\n\n接著我們證明 $K \\leq 1011$。考慮 $A_i = [i - 1, i]$，$i = 1, 2, \\dots, 2021$。易知我們最多僅能找到 $1011$ 個兩兩互斥的空集合 $[0, 1], [2, 3], \\dots, [2020, 2021]$，故 $K \\leq 1011$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21976,
"subject": "Mathematics (Olympiad)",
"question": "Let $R$ be the radius of a given circle. Let $A$, $B$, $C$ be points with complex coordinates $a$, $b$, $c$ respectively. Define $M$ as the midpoint of $AC$, $N$ as the midpoint of $BM$, $T$ as the midpoint of $BK$, and $K$ as the midpoint of $CX$. Let $X$ be a point with complex coordinate $x$. Find the locus of the point $T$ as $X$ varies on the circle $|x - a| = R$.",
"options": [],
"answer": "See solution",
"solution": "Let $m = \\frac{a + c}{2}$, $n = \\frac{b + m}{2}$, $k = \\frac{c + x}{2}$, $t = \\frac{k + b}{2}$. We can find $x = 4t - 2b - c$, $n = \\frac{2b + c + a}{2}$. The complex equation of the given circle is $|x - a| = R$, i.e. $|4t - 2b - c - a| = R$, or $|t - \\frac{2b + c + a}{4}| = \\frac{R}{4}$, or $|t - n| = \\frac{R}{4}$. Thus, the locus of points $T$ is the circle of radius $R/4$ centered at $N$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21977,
"subject": "Mathematics (Olympiad)",
"question": "Let $P_1, P_2, \\dots, P_{2556}$ be distinct interior points of a regular hexagon $ABCDEF$ with side length $1$. Assume that no three points of the set\n$$\nS = \\{A, B, C, D, E, F, P_1, P_2, \\dots, P_{2556}\\}\n$$\nare collinear. Show that there is a triangle with area less than $\\frac{1}{1700}$ all of whose vertices belong to $S$.",
"options": [],
"answer": "See solution",
"solution": "Draw line segments from $P_1$ to the points $A, B, C, D, E, F$ to create six smaller triangles. Since no three points in $S$ are collinear, $P_2$ must be inside one of these six triangles. Draw line segments joining $P_2$ with the vertices of this triangle to divide it into three triangles, increasing the total number of triangles by $2$. Repeat this process for $P_3, P_4, \\dots, P_{2556}$, each time increasing the number of triangles by $2$. Thus, after all points are added, the total number of triangles is $6 + (2555 \\times 2) = 5116$. If the area of each triangle were $\\ge \\frac{1}{1700}$, then the total area would be $\\ge 5116 \\times \\frac{1}{1700} > 3 > \\frac{3\\sqrt{3}}{2}$, which is the area of the regular hexagon $ABCDEF$. This is a contradiction, so there must exist a triangle with area less than $\\frac{1}{1700}$ whose vertices are in $S$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21978,
"subject": "Mathematics (Olympiad)",
"question": "Let $F$ be the set of all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$ satisfying the condition $f(3x) \\geq f(f(2x)) + x$ for every real positive number $x$.\n\nFind the greatest real number $\\alpha$ such that for all $f \\in F$, we have\n\n$$\nf(x) \\geq \\alpha x\n$$\nfor every real positive number $x$.\n\n$\\mathbb{R}^+$ denotes the set of all real positive numbers.",
"options": [],
"answer": "See solution",
"solution": "It is clear that the function $f(x) = x/2$, $x \\in \\mathbb{R}^+$, is a function belonging to $F$. Thus $\\alpha \\leq 1/2$.\n\nLet $f$ be an arbitrary function in $F$. It is easy to see that\n\n$$\nf(x) \\geq \\frac{x}{3} \\quad \\forall x \\in \\mathbb{R}^+.\n$$\n\nConsider the sequence of numbers $\\{\\alpha_n\\}$ defined by:\n\n$$\n\\alpha_1 = \\frac{1}{3}, \\quad \\alpha_{n+1} = \\frac{2\\alpha_n^2 + 1}{3} \\quad \\forall n = 1, 2, 3, \\dots\n$$\n\nBy induction on $n$, we shall prove that for all $n \\in \\mathbb{N}^+$,\n\n$$\nf(x) \\geq \\alpha_n x \\quad \\forall x \\in \\mathbb{R}^+.\n$$\n\nIndeed, the base case shows that this holds for $n = 1$.\n\nSuppose it holds for $n = k$. Then:\n\n$$\nf(x) \\geq \\alpha_k f(2x/3) + \\frac{x}{3} \\geq \\alpha_k \\cdot \\alpha_k \\left(\\frac{2x}{3}\\right) + \\frac{x}{3} = \\frac{2\\alpha_k^2 + 1}{3} x = \\alpha_{k+1} x \\quad \\forall x \\in \\mathbb{R}^+\n$$\n\nSo, by induction, the inequality holds for all $n$.\n\nNow, we shall prove that $\\lim \\alpha_n = 1/2$.\n\nIndeed, by induction, the sequence $\\{\\alpha_n\\}$ is bounded above by $1/2$. Therefore,\n\n$$\n\\alpha_{n+1} - \\alpha_n = \\frac{1}{3}(\\alpha_n - 1)(2\\alpha_n - 1) > 0,\n$$\n\nshowing that $\\{\\alpha_n\\}$ is increasing and convergent.\n\nPassing to the limit, with $\\alpha_n < 1/2$, we find $\\lim \\alpha_n = 1/2$, and thus $f(x) \\geq x/2$ for all $x \\in \\mathbb{R}^+$.\n\n*Consequently, the answer to the problem is $\\alpha = 1/2$.*",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21979,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute-angled triangle. The feet of the altitudes from $A$, $B$, and $C$ are $D$, $E$, and $F$ respectively. Prove that $DE + EF \\leq BC$ and determine the triangles for which equality holds.\n\nThe altitude through $A$ is the line through $A$ which is perpendicular to $BC$. The foot of this altitude is the point $D$ where it meets $BC$. The other altitudes are similarly defined.\n\nLet $A$, $B$, and $C$ be the angles of triangle $ABC$ and let $a$, $b$, and $c$ be the lengths of $BC$, $CA$, and $AB$ respectively.\n\n",
"options": [],
"answer": "See solution",
"solution": "First, we compute $DE$ and $DF$. From right-angled triangles $BEC$ and $ADC$, we have $CE = a \\cos C$. By the sine rule in triangle $CED$,\n\n$$\nDE = \\sin C \\frac{CE}{\\sin EDC} = \\sin C \\frac{a \\cos C}{\\sin A}\n$$\n\nBy the sine rule in triangle $ABC$, $\\frac{a}{\\sin A} = \\frac{c}{\\sin C}$, so $DE = c \\cos C$. Similarly, $DF = b \\cos B$.\n\nSo we have $DE + DF = c \\cos C + b \\cos B$, while $BC = BD + DC = c \\cos B + b \\cos C$. Therefore,\n\n$$\n\\begin{aligned}\nDE + DF - BC &= c \\cos C + b \\cos B - c \\cos B - b \\cos C \\\\\n&= (c - b)(\\cos C - \\cos B).\n\\end{aligned}\n$$\n\nIf $c > b$, then $C > B$, so $\\cos C < \\cos B$ and therefore $DE + DF < BC$. If $c = b$, then $DE + DF = BC$. If $c < b$, then $C < B$, so $\\cos C > \\cos B$ and therefore $DE + DF < BC$.\n\nWe have shown that $DE + DF \\leq BC$ with equality if and only if $b = c$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21980,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a convex quadrilateral inscribed in a circle $O$. Suppose that the line $AB$ meets the line $CD$ at $M$ and the line $AD$ meets the line $BC$ at $N$. Let $P$, $Q$, $S$, and $T$ be the intersection points of the bisectors of the angles $\\widehat{MAN}$ and $\\widehat{MBN}$, $\\widehat{MBN}$ and $\\widehat{MCN}$, $\\widehat{MCN}$ and $\\widehat{MDN}$, $\\widehat{MDN}$ and $\\widehat{MAN}$, respectively. Suppose that $P$, $Q$, $S$, and $T$ are pairwise different.\n\n1. Prove that the four points $P$, $Q$, $S$, and $T$ are cyclic. Let $I$ be the center of the circle passing through $P$, $Q$, $S$, and $T$.\n\n2. Let $E$ be the intersection of $AC$ and $BD$. Prove that the three points $E$, $O$, and $I$ are collinear.",
"options": [],
"answer": "See solution",
"solution": "We consider the case that $B$ lies between $A$ and $M$, and $C$ lies between $B$ and $N$. In this case, the points $P$ and $T$ are in the same half-plane of the line $SQ$. The proofs for other cases are similar.\n\n\n\n1. We have\n\n$$\n\\widehat{QPT} = \\widehat{BPA} = \\widehat{MBP} - \\widehat{MAP} = \\frac{1}{2}(\\widehat{MBN} - \\widehat{MAN}) = \\frac{1}{2}\\widehat{BNA}.\n$$\n\nSimilarly,\n\n$$\n\\widehat{QST} = \\widehat{CSD} = \\widehat{MCS} - \\widehat{MDS} = \\frac{1}{2}(\\widehat{MCN} - \\widehat{MDN}) = \\frac{1}{2}\\widehat{CND}.\n$$\n\nSince $\\widehat{BNA} = \\widehat{CND}$, we have $\\widehat{QPT} = \\widehat{QST}$, which implies that $PQST$ is cyclic.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21981,
"subject": "Mathematics (Olympiad)",
"question": "Solve the equation\n\n$$\n(x + 1)\\sqrt{x^2 + 2x + 2} + x\\sqrt{x^2 + 1} = 0.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $a = \\sqrt{x^2 + 2x + 2} > 0$ and $b = \\sqrt{x^2 + 1} > 0$. Therefore,\n\n$$\nx = \\frac{(x^2 + 2x + 2) - (x^2 + 1) - 1}{2} = \\frac{a^2 - b^2 - 1}{2}\n$$\n$$\nx + 1 = \\frac{a^2 - b^2 + 1}{2}.\n$$\n\nThe equation is equivalent to:\n\n$$\n\\frac{a^2 - b^2 + 1}{2} \\cdot a + \\frac{a^2 - b^2 - 1}{2} \\cdot b = 0\n$$\n$$\n(a^2 - b^2)a + a + (a^2 - b^2)b - b = 0\n$$\n$$\n(a - b)((a + b)^2 + 1) = 0.\n$$\n\nHence, $a = b$ and $x = -\\frac{1}{2}$. $\\square$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 21982,
"subject": "Mathematics (Olympiad)",
"question": "Let $A = \\{2, 2^2, \\dots, 2^n, \\dots\\}$.\n\n1. For any $a \\in A$, $b \\in \\mathbb{N}^*$, if $b < 2a - 1$, then $b(b+1)$ will not be a multiple of $2a$.\n2. For any $a \\in \\bar{A} = \\mathbb{N}^* - A$ with $a \\neq 1$, there exists $b \\in \\mathbb{N}^*$ with $b < 2a - 1$ such that $b(b+1)$ is a multiple of $2a$.",
"options": [],
"answer": "See solution",
"solution": "1. For any $a \\in A$, $a = 2^k$ for some $k \\in \\mathbb{N}^*$. Then $2a = 2^{k+1}$. Let $b$ be any positive integer strictly less than $2a - 1$. Then $b+1 \\leq 2a - 1$.\n\nBetween $b$ and $b+1$, one is odd (contains no factor of 2), and the other is even (contains at most the $k$th power of 2). Therefore, $b(b+1)$ is not a multiple of $2a$.\n\n2. For $a \\in \\bar{A}$ and $a \\neq 1$, suppose $a = 2^k m$ where $k$ is a non-negative integer and $m$ is an odd number greater than 1. Then $2a = 2^{k+1} m$. We present three proofs:\n\n**Proof 1.** Let $b = mx$, $b+1 = 2^{k+1} y$. Eliminating $b$, we have $2^{k+1}y - mx = 1$. Since $(2^{k+1}, m) = 1$, this equation has integer solutions:\n\n$$\n\\begin{cases}\nx = x_0 + 2^{k+1}t, \\\\\ny = y_0 + mt\n\\end{cases} \\quad (t \\in \\mathbb{Z},\\ (x_0, y_0) \\text{ is a particular solution})\n$$\n\nLet $(x^*, y^*)$ be the minimal solution with $x^* < 2^{k+1}$. Then $b = mx^* < 2a - 1$ and $b(b+1)$ is a multiple of $2a$.\n\n**Proof 2.** Since $(2^{k+1}, m) = 1$, by the Chinese Remainder Theorem, the system\n\n$$\n\\begin{cases}\nx \\equiv 0 \\pmod{2^{k+1}} \\\\\nx \\equiv m-1 \\pmod{m}\n\\end{cases}\n$$\n\nhas a solution $x = b$ with $b \\in (0, 2^{k+1}m)$. Clearly, $b < 2a-1$ and $b(b+1)$ is a multiple of $2a$.\n\n**Proof 3.** Since $(2^{k+1}, m) = 1$, there exists $r \\in \\mathbb{N}^*$, $r \\leq m-1$, such that $2^r \\equiv 1 \\pmod{m}$. Take $t \\in \\mathbb{N}^*$ with $tr > k+1$. Then $2^{tr} \\equiv 1 \\pmod{m}$. There exists\n\n$$\nb = (2^{tr} - 1) - q \\cdot 2^{k+1}m > 0 \\quad (q \\in \\mathbb{N})\n$$\n\nwith $0 < b < 2a - 1$. Then $m \\mid b$, $2^{k+1} \\mid b+1$, so $b(b+1)$ is a multiple of $2a$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21983,
"subject": "Mathematics (Olympiad)",
"question": "Call a polynomial $f(x_1, \\dots, x_n)$ with integer coefficients *nice* if $f(0, 0, \\dots, 0) = 0$ and $f(x_{\\pi_1}, \\dots, x_{\\pi_n}) = f(x_1, \\dots, x_n)$ for any permutation $\\pi$ of $1, \\dots, n$ (i.e., $f$ is symmetric and its constant term is zero).\n\nDenote by $\\mathcal{I}$ the set of polynomials of the form\n\n$$\np_1q_1 + p_2q_2 + \\dots + p_mq_m,\n$$\n\nwhere $m$ is an integer, $q_1, \\dots, q_m$ are polynomials with integer coefficients, and $p_1, \\dots, p_m$ are nice polynomials.\n\nFind the least $N$ for which any monomial of degree at least $N$ belongs to $\\mathcal{I}$.",
"options": [],
"answer": "See solution",
"solution": "The answer is $N = n(n-1)/2 + 1$.\n\n**Lower bound:**\nConsider the polynomial\n$$\nF(x_1, \\dots, x_n) = x_2 x_3^2 x_4^3 \\dots x_n^{n-1}\n$$\nThis does not belong to $\\mathcal{I}$.\n\nSuppose $F = \\sum p_i q_i$ as in (2). By considering only monomials of degree $n(n-1)/2$, we may assume all $p_i$ and $q_i$ are homogeneous, $\\deg p_i > 0$, and $\\deg p_i + \\deg q_i = n(n-1)/2$ for all $i$.\n\nConsider the alternating sum\n$$\n\\sum_{\\pi} \\mathrm{sign}(\\pi) F(x_{\\pi_1}, \\dots, x_{\\pi_n}) = \\sum_{i=1}^{m} p_i \\sum_{\\pi} \\mathrm{sign}(\\pi) q_i(x_{\\pi_1}, \\dots, x_{\\pi_n}) := S_1\n$$\nwhere the sum is over all permutations $\\pi$ of $1, \\dots, n$, and $\\mathrm{sign}(\\pi)$ is the sign of $\\pi$.\n\nSince $\\deg q_i = n(n-1)/2 - \\deg p_i < n(n-1)/2$, in any monomial $Q$ of $q_i$, at least two variables have equal exponents. Thus,\n$$\n\\sum_{\\pi} \\mathrm{sign}(\\pi) Q(x_{\\pi_1}, \\dots, x_{\\pi_n}) = 0\n$$\nbecause terms corresponding to permutations differing by a transposition cancel. This holds for all $i$ and all monomials of $q_i$, so $S_1 = 0$. But the left side is a nonzero polynomial, a contradiction.\n\n**Upper bound:**\nWe prove by induction on $n$ that any monomial $h = x_1^{c_1} \\dots x_n^{c_n}$ of degree $n(n-1)/2 + 1$ belongs to $\\mathcal{I}$, and all $p_i, q_i$ in (2) can be chosen homogeneous with degrees summing to $n(n-1)/2 + 1$. (Any monomial of degree at least $n(n-1)/2 + 1$ is divisible by one of degree exactly $n(n-1)/2 + 1$.)\n\nThe base case $n=1$ is clear. Assume $n > 1$ and the claim holds for smaller $n$.\n\nProceed by induction on $S = |\\{i : c_i = 0\\}|$. If $S=0$, $h$ is divisible by the nice polynomial $x_1 \\cdots x_n$, so $h \\in \\mathcal{I}$. If $S > 0$, let $T = n-S$. Assume $c_{T+1} = \\dots = c_n = 0$ and $h = x_1 \\cdots x_T g(x_1, \\dots, x_{n-1})$ with $\\deg g = n(n-1)/2 - T + 1 \\ge (n-1)(n-2)/2 + 1$. By the outer induction hypothesis, $g$ can be written as $p_1 q_1 + \\dots + p_m q_m$ with $p_i$ nice in $n-1$ variables. There exist nice homogeneous $P_i(x_1, \\dots, x_n)$ such that $P_i(x_1, \\dots, x_{n-1}, 0) = p_i(x_1, \\dots, x_{n-1})$. Then $\\Delta_i := p_i(x_1, \\dots, x_{n-1}) - P_i(x_1, \\dots, x_{n-1}, x_n)$ is divisible by $x_n$, say $\\Delta_i = x_n g_i$.\n\nThus,\n$$\nh = x_1 \\cdots x_T \\sum p_i q_i = x_1 \\cdots x_T \\sum (P_i + x_n g_i) q_i = (x_1 \\cdots x_T x_n) \\sum g_i q_i + \\sum P_i q_i \\in \\mathcal{I}.\n$$\nThe first term is in $\\mathcal{I}$ by the inner induction. This completes the proof.\n\nThe argument also works for real coefficients. For integer coefficients, a similar parity argument shows $N \\ge n(n-1)/2 + 1$.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 21984,
"subject": "Mathematics (Olympiad)",
"question": "The cube is cut into 2010 smaller cubes. 2008 of them are unit cubes (cubes of side 1), and the edges of the other 2 cubes are assigned integer numbers (different from 1). Find the volume of the big cube.",
"options": [],
"answer": "See solution",
"solution": "**Answer:** $13^3 = 2197$.\n\nLet the edge of the big cube be $a$, and the edges of the two non-unit cubes be $b$ and $c$.\n\nWe have:\n\n$$\nb^3 + c^3 + 2008 = a^3.\n$$\n\nSince $12^3 = 1728 < 2008$, $a > 12$, so $a \\geq 13$.\n\nTo find an upper bound for $a$, note that $a \\geq b + c$.\n\n$$\nb^3 + 3b^2c + 3bc^2 + c^3 \\leq a^3 = b^3 + c^3 + 2008,\n$$\n\nwhich implies $b^2c + bc^2 \\leq \\frac{2008}{3}$, so $b^2c + bc^2 \\leq 669$.\n\nThen $2c^3 \\leq b^2c + bc^2 \\leq 669$, so $c^3 \\leq 334$.\n\nMoreover,\n\n$$\n2b^2 + 8 \\leq b^2c + bc^2 \\leq 669 \\Rightarrow 2b^2 \\leq 661 \\Rightarrow b^2 \\leq 330 \\Rightarrow b \\leq 18 \\Rightarrow b^3 \\leq 5832 \\Rightarrow b^3 + c^3 \\leq 6166 \\Rightarrow a^3 = b^3 + c^3 + 2008 \\leq 6166 + 2008 = 8174 \\Rightarrow a \\leq 20.\n$$\n\nTesting integer values $b \\geq c \\geq 2$ for $a$ from 13 to 20, the unique solution is $5^3 + 4^3 + 2008 = 13^3$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 21985,
"subject": "Mathematics (Olympiad)",
"question": "Given 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 sequence is:\n\n$a_2, a_4, \\dots, a_{2010}, a_{2009}, a_{2007}, \\dots, a_3, a_1.$\n\nFind which term is at the 1000-th and 2000-th positions in this sequence.",
"options": [],
"answer": "See solution",
"solution": "It is straightforward that the 1000-th term is $a_{2000}$, and the 2000-th term is $-a_{21}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21986,
"subject": "Mathematics (Olympiad)",
"question": "Solve $\\frac{2 \\cos 2x}{6 - 3 \\cos 3x} = \\frac{\\cos 2x + 1}{\\cos 3x + 2}$ for $-\\pi \\le x \\le \\pi$.",
"options": [],
"answer": "See solution",
"solution": "**Answer:** $x = 0$.\n\n**Solution.** The equation can be written as:\n$$\n\\frac{2 \\cos 2x}{\\cos 2x + 1} = \\frac{6 - 3 \\cos 3x}{\\cos 3x + 2}.\n$$\nConsider the following two functions: $f(x) = \\frac{2x}{x+1}$ and $g(x) = \\frac{6-3x}{x+2}$, $x \\in (-1, 1]$.\n\nIf $x \\in (-1, 1)$, then\n$$\nf(x) = \\frac{2x}{x+1} < 1 \\Leftrightarrow 2x < x+1 \\Leftrightarrow x < 1, \\\\\ng(x) = \\frac{6-3x}{x+2} > 1 \\Leftrightarrow 6-3x > x+2 \\Leftrightarrow x < 1.\n$$\nThus, the equality holds only if $\\cos 2x = \\cos 3x = 1$. A sketch of the graphs of the functions can lead to the same conclusion.\n\nIn order for the last equation to be true, the following has to hold:\n\n$2x = 2\\pi k$, and $3x = 2\\pi l$, $k, l \\in \\mathbb{Z}$. From the first equation, $x \\in \\{-\\pi, 0, \\pi\\}$. Substituting into the second equation, it is easy to check that the only answer is $x = 0$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21987,
"subject": "Mathematics (Olympiad)",
"question": "Determine all nonnegative integers $n$ having two distinct positive divisors with the same distance from $\\frac{n}{2}$.",
"options": [],
"answer": "See solution",
"solution": "Let $n$ be a nonnegative integer. Consider its positive divisors. The distance from a divisor $d$ to $\\frac{n}{2}$ is $|d - \\frac{n}{2}|$. For two distinct divisors $d_1$ and $d_2$ to have the same distance from $\\frac{n}{2}$, we require $|d_1 - \\frac{n}{2}| = |d_2 - \\frac{n}{2}|$ with $d_1 \\neq d_2$.\n\nThe only way for two distinct positive divisors to be equidistant from $\\frac{n}{2}$ is if they are symmetric about $\\frac{n}{2}$, i.e., $d_1 + d_2 = n$. For $d_1$ and $d_2$ to both be divisors of $n$, and $d_1 \\neq d_2$, this occurs when $n$ is a multiple of $6$ (since $1$ and $n-1$ are not both divisors except for $n=2$, and for $n=6$, $2$ and $4$ are both divisors and $2 + 4 = 6$).\n\nTherefore, all positive multiples of $6$ have two distinct positive divisors with the same distance from $\\frac{n}{2}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21988,
"subject": "Mathematics (Olympiad)",
"question": "A rhombus has one diagonal of length $7.2\\ \\text{cm}$ and an area of $34.56\\ \\text{cm}^2$. Calculate the perimeter of the rhombus and the radius of its inscribed circle.",
"options": [],
"answer": "See solution",
"solution": "Given $d_1 = 7.2\\ \\text{cm}$ and $A = 34.56\\ \\text{cm}^2$. The area formula is $A = \\frac{d_1 \\cdot d_2}{2}$, so:\n\n$$d_2 = \\frac{2A}{d_1} = \\frac{2 \\times 34.56}{7.2} = 9.6\\ \\text{cm}$$\n\nThe side length:\n\n$$a = \\sqrt{\\left(\\frac{d_1}{2}\\right)^2 + \\left(\\frac{d_2}{2}\\right)^2} = \\sqrt{3.6^2 + 4.8^2} = 6\\ \\text{cm}$$\n\nPerimeter:\n\n$$L = 4a = 4 \\times 6 = 24\\ \\text{cm}$$\n\nFor the radius of the inscribed circle, let $x$ be the length of the orthogonal projection of one side onto its neighbor:\n\n$$x = \\frac{2a^2 - d_1^2}{2a} = \\frac{2 \\times 6^2 - 7.2^2}{2 \\times 6} = 1.68\\ \\text{cm}$$\n\nAltitude:\n\n$$h = \\sqrt{a^2 - x^2} = \\sqrt{6^2 - 1.68^2} = 5.76\\ \\text{cm}$$\n\nRadius:\n\n$$r = \\frac{1}{2}h = \\frac{1}{2} \\times 5.76 = 2.88\\ \\text{cm}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21989,
"subject": "Mathematics (Olympiad)",
"question": "Let $x$, $y$, and $z$ be positive real numbers such that $xy + yz + zx = 3xyz$. Prove that\n\n$$\nx^2 y + y^2 z + z^2 x \\geq 2(x + y + z) - 3\n$$",
"options": [],
"answer": "See solution",
"solution": "The given condition can be rearranged to $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = 3$. Using this, we obtain:\n\n$$\n\\begin{aligned}\nx^2 y + y^2 z + z^2 x - 2(x + y + z) - 3 &= x^2 y - 2x + \\frac{1}{y} + y^2 z - 2y + \\frac{1}{z} + z^2 x - 2z + \\frac{1}{x} \\\\\n&= y \\left(x - \\frac{1}{y}\\right)^2 + z \\left(y - \\frac{1}{z}\\right)^2 + x \\left(z - \\frac{1}{x}\\right)^2 \\geq 0\n\\end{aligned}\n$$\n\nEquality holds if and only if $x = y = z = 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21990,
"subject": "Mathematics (Olympiad)",
"question": "Let positive real number sequences $\\{a_n\\}$, $\\{b_n\\}$ satisfy: for any integer $n \\ge 101$,\n$$\na_n = \\sqrt{\\frac{1}{100} \\sum_{j=1}^{100} b_{n-j}^2}, \\quad b_n = \\sqrt{\\frac{1}{100} \\sum_{j=1}^{100} a_{n-j}^2}.\n$$\nProve that there exists a positive integer $m$ such that $|a_m - b_m| < 0.001$.",
"options": [],
"answer": "See solution",
"solution": "Let $c_n = a_n^2 - b_n^2$ for $n = 1, 2, \\dots$. Then for $n \\ge 101$,\n$$\nc_n = a_n^2 - b_n^2 = \\frac{1}{100} \\left( \\sum_{j=1}^{100} b_{n-j}^2 - \\sum_{j=1}^{100} a_{n-j}^2 \\right) = -\\frac{1}{100} \\sum_{j=1}^{100} c_{n-j}.\n$$\nIf there exists a $c_n = 0$, then the conclusion holds. Otherwise, assume all $c_n \\ne 0$.\n\n*Lemma.* There exists a constant $\\lambda \\in (0, 1)$ such that: if $|c_n|, |c_{n+1}|, \\dots, |c_{n+99}| \\le M$ for some $M > 0$, then there exists $k > n + 99$ with $|c_k|, |c_{k+1}|, \\dots, |c_{k+99}| \\le \\lambda M$.\n\n*Proof of the lemma.*\n\n(1) If not all of $c_n, \\dots, c_{n+99}$ have the same sign, then\n$$\n|c_{n+100}| \\le \\frac{99}{100} M = \\lambda_1 M,\n$$\nwhere $\\lambda_1 = \\frac{99}{100}$. Repeatedly applying the recurrence and triangle inequality,\n$$\n\\begin{align*}\n|c_{n+101}| &\\le \\frac{1}{100}(99 + \\lambda_1)M = \\lambda_2 M, \\\\\n|c_{n+102}| &\\le \\frac{1}{100}(99 + \\lambda_1 + \\lambda_2)M = \\lambda_3 M, \\\\\n&\\vdots \\\\\n|c_{n+199}| &\\le \\frac{1}{100}(1 + \\lambda_1 + \\dots + \\lambda_{99})M = \\lambda_{100} M.\n\\end{align*}\n$$\nSet $\\lambda = \\max\\{\\lambda_1, \\dots, \\lambda_{100}\\} \\in (0, 1)$ and $k = n + 100$.\n\n(2) If $c_n, \\dots, c_{n+99}$ have the same sign, then $c_{n+100}$ must differ in sign from them, and $|c_{n+100}| \\le M$. Then, by (1), $|c_{n+101}|, \\dots, |c_{n+200}| \\le \\lambda M$ for the same $\\lambda$.\n\nThus, the lemma holds.\n\nReturning to the problem: Let $M = \\max\\{|c_1|, \\dots, |c_{100}|\\}$. Repeatedly applying the lemma, for any $t$, there exists $m$ with $|c_m| \\le \\lambda^t M$. Since $0 < \\lambda < 1$, $\\lambda^t M \\to 0$ as $t \\to \\infty$. Thus, there exists $m$ with $|c_m| < (0.001)^2$.\n\nBut $|c_m| = |a_m^2 - b_m^2| = |a_m - b_m|\\,|a_m + b_m| > |a_m - b_m|^2$, so $|a_m - b_m| < 0.001$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21991,
"subject": "Mathematics (Olympiad)",
"question": "Let $x$, $y$, $z$ be positive real numbers such that $xyz = 1$. Prove that\n$$\n\\frac{x^2 + y^2 + z^2}{x^2 + 2} + \\frac{y^2 + z^2 + x^2}{y^2 + 2} + \\frac{z^2 + x^2 + y^2}{z^2 + 2} \\ge 3.\n$$\n\nWhen does equality hold?",
"options": [],
"answer": "See solution",
"solution": "From the AM-GM inequality and the condition $xyz = 1$, we have:\n$$\n\\frac{x^2 + y^2 + z^2}{x^2 + 2} + \\frac{y^2 + z^2 + x^2}{y^2 + 2} + \\frac{z^2 + x^2 + y^2}{z^2 + 2} \\ge 3.\n$$\nEquality holds when $x = y = z = 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21992,
"subject": "Mathematics (Olympiad)",
"question": "設實數 $u_1, \\ldots, u_n$（$n \\ge 3$）滿足：\n\n$$\n\\sum_{i=1}^{n} u_{i}^{2018} = 1, \\quad \\sum_{i=1}^{n} u_{i}^{2019} = 0.\n$$\n\n證明：必存在 $1 \\le k_1 < k_2 \\le n$ 滿足：\n\n$$\n\\sum_{i=1}^{n} u_{i}^{2020} \\le |u_{k_1} u_{k_2}|.\n$$",
"options": [],
"answer": "See solution",
"solution": "將原不等式記為 $(*)$。不失一般性，假設 $u_i \\le u_{i+1}$，$i = 1, \\ldots, n-1$。令\n\n$$\nP = \\{i : u_i > 0\\}, \\quad N = \\{i : u_i \\le 0\\}.\n$$\n\n根據條件，顯然 $u_n \\in P$，$u_1 \\in N$，$u_1 < 0$，且\n\n$$\n\\max_{i \\in P} u_i = u_n, \\quad \\max_{i \\in N} |u_i| = |u_1|. \\quad (1)\n$$\n\n以下證明當 $(u_{k_1}, u_{k_2}) = (u_1, u_n)$ 時，不等式 $(*)$ 恆成立。\n\n由 $\\sum_{i=1}^{n} u_{i}^{2019} = 0$，可得\n\n$$\n\\sum_{i \\in P} u_{i}^{2019} = \\sum_{i \\in N} |u_i|^{2019}. \\quad (2)\n$$\n\n因此，\n\n$$\n\\begin{aligned}\n0 < \\sum_{i \\in P} u_i^{2020} &\\le u_n \\sum_{i \\in P} u_i^{2019} = u_n \\sum_{i \\in N} |u_i|^{2019} \\\\\n&\\implies u_n \\ge \\frac{\\sum_{i \\in P} u_i^{2020}}{\\sum_{i \\in N} |u_i|^{2019}}, \\\\\n0 < \\sum_{i \\in N} u_i^{2020} &\\le |u_1| \\sum_{i \\in N} |u_i|^{2019} = |u_1| \\sum_{i \\in P} u_i^{2019} \\\\\n&\\implies |u_1| \\ge \\frac{\\sum_{i \\in N} u_i^{2020}}{\\sum_{i \\in P} u_i^{2019}}.\n\\end{aligned}\n$$\n\n進一步利用 (1) 得\n\n$$\n\\begin{align}\nu_n|u_1| &\\ge \\frac{\\sum_{i \\in P} u_i^{2020}}{\\sum_{i \\in N} |u_i|^{2019} \\sum_{i \\in P} u_i^{2019}} \\sum_{i \\in N} u_i^{2020} \\\\\n&= \\frac{\\sum_{i \\in P} u_i^{2020}}{\\left(\\sum_{i \\in P} u_i^{2019}\\right)^2} \\sum_{i \\in N} u_i^{2020} \\ge \\frac{\\sum_{i \\in N} u_i^{2020}}{\\sum_{i \\in P} u_i^{2018}}. \\tag{3}\n\\end{align}\n$$\n\n這裡用到柯西不等式：$(\\sum_{i \\in P} u_i^{2019})^2 \\le \\sum_{i \\in P} u_i^{2018} \\sum_{i \\in P} u_i^{2020}$。\n\n由 (3) 推得\n\n$$\nu_n|u_1| \\sum_{i \\in P} u_i^{2018} \\geq \\sum_{i \\in N} u_i^{2020}. \\quad (4)\n$$\n\n同理，\n\n$$\nu_n|u_1| \\sum_{i \\in N} u_i^{2018} \\geq \\sum_{i \\in P} u_i^{2020}. \\quad (5)\n$$\n\n由 (4)、(5) 及 $\\sum_{i=1}^{n} u_i^{2018} = 1$，得不等式 $(*)$，其中 $(u_{k_1}, u_{k_2}) = (u_1, u_n)$。\n\n證畢！\n\n*註：* 若僅考慮特殊情況（如 $|u_{k_1}| = |u_{k_2}| = \\max_{1 \\le i \\le n} |u_i|$）來證明 $(*)$，則僅能部分得分。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21993,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $n^2 = T_m = \\frac{m(m+1)}{2}$. Find all perfect squares that are also triangular numbers, i.e., all $n$ and $m$ such that $n^2 = \\frac{m(m+1)}{2}$.",
"options": [],
"answer": "See solution",
"solution": "Multiplying both sides by $8$ gives:\n\n$$\n8n^2 = 4m^2 + 4m = (2m + 1)^2 - 1 \\implies (2m + 1)^2 - 8n^2 = 1.\n$$\n\nLet $A = 2m + 1$ and $B = 2n$, so the equation becomes the Pell equation:\n\n$$\nA^2 - 2B^2 = 1.\n$$\n\nThe fundamental solution is $A = 3$, $B = 2$. The general solution is:\n\n$$\na_k + \\sqrt{2}b_k = (3 + 2\\sqrt{2})^k,\n$$\n\nwhere\n\n$$\na_k = \\frac{(3 + 2\\sqrt{2})^k + (3 - 2\\sqrt{2})^k}{2}, \\quad b_k = \\frac{(3 + 2\\sqrt{2})^k - (3 - 2\\sqrt{2})^k}{2\\sqrt{2}}.\n$$\n\nThe corresponding squares are:\n\n$$\nx^2 = \\frac{1}{32}\\left((3 + 2\\sqrt{2})^k - (3 - 2\\sqrt{2})^k\\right)^2, \\quad k \\in \\mathbb{N}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21994,
"subject": "Mathematics (Olympiad)",
"question": "We operate on piles of cards placed at $n+1$ positions $A_1, A_2, \\dots, A_n$ ($n \\ge 3$) and $O$. In one operation, we can do either of the following:\n\n1. If there are at least three cards at $A_i$, we may take three cards from $A_i$ and place one at each of $A_{i-1}$, $A_{i+1}$, and $O$ (assume that $A_0 = A_n$, $A_{n+1} = A_1$).\n2. If there are at least $n$ cards at $O$, we may take $n$ cards from $O$ and place one at each of $A_1, A_2, \\dots, A_n$.\n\nProve that if the total number of cards is at least $n^2 + 3n + 1$, we can take some operations such that there are at least $n+1$ cards at each position.",
"options": [],
"answer": "See solution",
"solution": "We only need to consider the case where the total number of cards is exactly $n^2 + 3n + 1$. We use the following strategy:\n\n- If there are at least three cards at some $A_i$, use operation (1) at this position. Such operations can be performed only finitely many times. Eventually, there will be no more than two cards at each $A_i$ and at least $n^2 + n + 1$ at $O$.\n\n- Next, perform operation (2) $n+1$ times. Then, there are at least $n+1$ cards at each $A_i$.\n\nNow, we show that we can increase the number of cards at $O$ to at least $n+1$, while keeping at least $n+1$ cards at each $A_i$.\n\nArrange $A_1, A_2, \\dots, A_n$ evenly and in increasing order on a circle centered at $O$. Define a team $G = \\{A_i, A_{i+1}, \\dots, A_{i+l-1}\\}$ ($1 \\leq i \\leq n$, $1 \\leq l \\leq n$), where for $j > n$ we set $A_j = A_{j-n}$. A team is *good* if, after performing operation (1) once at each point in $G$, there are at least $n+1$ cards at every point in $G$.\n\nLet $a_1, a_2, \\dots, a_n$ be the number of cards at $A_1, A_2, \\dots, A_n$, with $a_i \\geq n+1$ for all $i$. For a team $G$:\n- If $G = \\{A_i\\}$, it is good iff $a_i \\geq n+4$.\n- If $G = \\{A_i, A_{i+1}\\}$, it is good iff $a_i, a_{i+1} \\geq n+3$.\n- If $G = \\{A_i, \\dots, A_{i+l-1}\\}$ with $3 \\leq l \\leq n-1$, it is good iff $a_i, a_{i+l-1} \\geq n+3$ and $a_j \\geq n+2$ for $i+1 \\leq j \\leq i+l-2$.\n- If $G = \\{A_1, \\dots, A_n\\}$, it is good iff $a_j \\geq n+2$ for all $j$.\n\nWe claim that if $a_1 + \\dots + a_n \\geq n^2 + 2n + 1$, there must exist at least one good team.\n\nSuppose not. Then each $a_i \\in \\{n+1, n+2, n+3\\}$, otherwise there is a good team of one point at any $A_i$ with $a_i \\geq n+4$. Let $x, y, z$ be the counts of $n+1, n+2, n+3$ among $a_1, \\dots, a_n$. We show $x \\geq z$. Since $n^2 + 2n + 1 > n(n+2)$, $z \\geq 1$. If $z = 1$, then $x \\geq 1$, otherwise all $a_i \\geq n+2$ and $G = \\{A_1, \\dots, A_n\\}$ is a good team. If $z \\geq 2$, the $z$ points with $n+3$ cards divide the circle into $z$ arcs (no two of these $z$ points are adjacent). Since there is no good team, there is at least one point on each arc with $n+1$ cards. So $x \\geq z$.\n\nThe total number of cards is\n\n$$\nx(n+1) + y(n+2) + z(n+3) \\leq (x+y+z)(n+2) = n(n+2) < n^2 + 2n + 1,\n$$\n\nwhich is a contradiction. Thus, when the number of cards at $O$ is less than $n+1$, there exists a good team. We use operation (1) on each point of a good team; the number of cards at $O$ increases while each $A_i$ still has at least $n+1$ cards. Repeat until $O$ has at least $n+1$ cards.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21995,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $\\sum_i p_i = 1$ and we want to have two random variables $X, Y$ such that $\\mathbb{P}(X = i) = p_i$ and $\\mathbb{P}(Y = i) = q_i$ for $1 \\leq i \\leq n$. What is the maximum possible value of $\\mathbb{P}(X = Y)$ over all possible joint distributions of $X$ and $Y$?",
"options": [],
"answer": "See solution",
"solution": "We use induction on $k$. For $k=1$ the lemma is obvious. If $b_k > a$, we can set $a_1 = a_2 = \\cdots = a_{k-1} = 0$ and $a_k = a$, so suppose that $b_k \\leq a$. Let $a_k := b_k$. Now, we must have $a_1 + a_2 + \\cdots + a_{k-1} = a - b_k$. But $a - b_k \\leq b_1 + b_2 + \\cdots + b_k - b_k = b_1 + b_2 + \\cdots + b_{k-1}$ and the assertion follows by the induction hypothesis. $\\square$\n\nWe have\n\n$$\n\\sum_{i=2}^{n} (q_i - \\min(p_i, q_i)) = \\sum_{i=1}^{n} (q_i - \\min(p_i, q_i)) = \\sum_{i=1}^{n} (p_i - \\min(p_i, q_i)) \\\\\n\\geq p_1 - \\min(p_1, q_1) = p_1 - q_1.\n$$\n\nHence, by the lemma we can find nonnegative real numbers $\\beta_2, \\beta_3, \\dots, \\beta_n$ such that equation 1 holds. Let $A_{1i} = \\beta_i$ for $2 \\leq i \\leq n$. The remaining entries of $A$ should satisfy\n\n* $A_{i2} + A_{i3} + \\cdots + A_{in} = p_i$ for $2 \\leq i \\leq n$,\n* $A_{2i} + A_{3i} + \\cdots + A_{ni} = q_i - \\beta_i$ for $1 \\leq i \\leq n$.\n\nWe have $q_i - \\beta_i \\geq \\min(p_i, q_i)$ by equation 1, so $\\min(p_i, q_i - \\beta_i) = \\min(p_i, q_i)$. Hence, the rest of the matrix can be filled using the induction hypothesis. $\\square$\n\n**Second Solution.** We can change the order of $p_1, p_2, \\dots, p_n$ and $q_1, q_2, \\dots, q_n$ arbitrarily by changing corresponding rows and columns of the matrix. So we can suppose $p_1 \\leq q_1, p_2 \\leq q_2, \\dots, p_k \\leq q_k$ and $p_{k+1} \\geq q_{k+1}, \\dots, p_n \\geq q_n$. Let $A_{ii} = \\min(p_i, q_i)$ for $1 \\leq i \\leq n$. Also let $A_{ij} = 0$ if $i \\neq j$ and either $i \\leq k$ or $j > k$. The remaining entries $\\{A_{ij} : i > k, j \\leq k\\}$ should satisfy:\n\n* $A_{i1} + A_{i2} + \\cdots + A_{ik} = p_i - q_i$ for $i > k$,\n* $A_{(k+1)i} + \\cdots + A_{ni} = q_i - p_i$ for $i \\leq k$.\n\ni.e., we should find a matrix with nonnegative entries for the given row sums and column sums. We can construct such a matrix using induction on the number of rows as follows.\n\nBy the lemma above, we can find $A_{n1}, A_{n2}, \\dots, A_{nk}$ such that their sum is $p_n - q_n$ and $A_{ni} \\leq q_i - p_i$ for each $i \\leq k$. Then, we delete the $n$th row and replace $q_i - p_i$ by $q_i - p_i - A_{ni}$ for each $i \\leq k$. We can proceed by using induction and the desired matrix will be constructed. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21996,
"subject": "Mathematics (Olympiad)",
"question": "Three positive integers have sum $1810$. In how many zeros can their product end? Find all possibilities.",
"options": [],
"answer": "See solution",
"solution": "Let $a, b, c \\in \\mathbb{N}$ with $a + b + c = 1810$, and let $abc$ end in $N$ zeros. The following triples $(a, b, c)$ show that $N$ can be $0, 1, \\dots, 7$:\n\n$(1808, 1, 1)$, $(1799, 10, 1)$, $(1709, 100, 1)$, $(1765, 40, 5)$,\n$(1200, 605, 5)$, $(1700, 100, 10)$, $(1250, 520, 40)$, $(1250, 400, 160)$.\n\nWe prove that $N \\leq 7$ always holds. Clearly, $N$ is at most the total number of 5's in the prime factorizations of $a, b, c$. Let $5^k, 5^l, 5^m$ be the highest powers of 5 dividing $a, b, c$ respectively. Assume $k \\geq l \\geq m$ by symmetry. We have $k \\leq 4$ as $5^5 > 1810$. If $m = 0$ (i.e., $5 \\nmid c$), then $l = 0$ because $5 \\mid 1810$, so $N \\leq k \\leq 4$. If $m \\geq 1$, then $m = 1$ since $1810$ is exactly divisible by $5$. Hence $N \\leq 4 + 4 + 1 = 9$. However, $N = 8$ and $N = 9$ cannot occur.\n\nIndeed, $N \\geq 8$ only if $k = 4$ and $m = 1$, so $k + l + m \\geq N \\geq 8$ yields $l \\geq 3$. Thus $(k, l, m) = (4, 3, 1)$ or $(4, 4, 1)$. In the first case, $a$ is divisible by $5^4 = 625$; in the second, so are $a$ and $b$. Some of $a, b, c$ can be a multiple of $625$ only if it is even. Since $a + b + c = 1810$ is even, having an odd summand means having exactly two odd ones. Since $N$ is at most the total number of 2's in the prime factorizations of $a, b, c$, $N \\geq 8$ implies that the third number must be divisible by $5 \\cdot 2^8 = 1280$. This cannot hold as $625 + 1280 > 1810$.\n\nSo $a \\geq 2 \\cdot 625 = 1250$ for $(k, l, m) = (4, 3, 1)$ and $a, b \\geq 1250$ for $(4, 4, 1)$. Thus, the latter is rejected together with $N = 9$ ($1250 + 1250 > 1810$). There remains $N = 8$ with $(k, l, m) = (4, 3, 1)$ and $a = 1250$ (an even multiple of $625$ and $4 \\cdot 625 > 1810$). Here $b + c = 560$ and $5^3 \\mid b$; moreover, $b$ is even. Thus $b = 2 \\cdot 125 = 250$ or $b = 4 \\cdot 125 = 500$, but neither gives a product $abc$ divisible by $10^8$. The desired $N \\leq 7$ is established.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21997,
"subject": "Mathematics (Olympiad)",
"question": "For a triangle $ABC$, let $B_1$ and $C_1$ be the excenters corresponding to $B$ and $C$, respectively. The line $B_1C_1$ intersects the circumcircle of $ABC$ at $D \\ne A$. Let $E$ be the intersection of the perpendiculars from $B_1$ to $CA$ and from $C_1$ to $AB$. Let $\\omega$ be the circumcircle of triangle $ADE$. The tangent to $\\omega$ at $D$ intersects the line $AE$ at $F$. The perpendicular from $D$ to $AE$ intersects $AE$ at $G$, and this line intersects $\\omega$ at $H \\ne D$. The circumcircle of triangle $HGF$ intersects $\\omega$ at $I \\ne H$. Let $J$ be the foot of the perpendicular from $D$ to $AH$. Show that $AI$ passes through the midpoint of $DJ$.",
"options": [],
"answer": "See solution",
"solution": "We first show that $\\angle ADE = 90^\\circ$. $\\angle EB_1C_1 = \\frac{1}{2}\\angle A = \\angle EC_1B_1$, hence $EB_1 = EC_1$. Similarly, $EB_1 = EA_1$, so $E$ is the circumcenter of $\\triangle A_1B_1C_1$. On the other hand, $A$, $B$, $C$ are the perpendicular feet of $\\triangle A_1B_1C_1$ because, for example, $C_1$ lies on the bisector of $\\angle ACB$ and $\\angle B_1CC_1 = \\angle B_1CA + \\angle ACC_1 = \\left(\\frac{\\pi}{2} - \\frac{1}{2}\\angle C\\right) + \\left(\\frac{1}{2}\\angle C\\right) = \\frac{\\pi}{2}$. Hence, the circumcircle of $\\triangle ABC$ is the nine-point circle of $\\triangle A_1B_1C_1$, which means that $D$ is the midpoint of $B_1C_1$. Therefore, $\\angle ADE = 90^\\circ$.\n\nBecause $AE$ is a diameter of $\\omega$ and $H$ is symmetric to $D$ with respect to $AE$, $HF$ is also tangent to $\\omega$. Let $O_1$ be the center of $\\omega$. Let $O_2$ be the circumcenter of $\\triangle HGF$; then $O_2$ is the midpoint of $HF$. Since\n\n$$\n\\angle HO_1F = \\angle HO_1D = \\angle HAD = \\angle JAD,\n$$\n\nthe triangles $\\triangle HO_1F$ and $\\triangle JAD$ are similar. For these triangles we have\n\n$$\n\\angle HO_1O_2 = \\frac{1}{2}\\angle HO_1I = \\angle HAI = \\angle JAK,\n$$\n\nwhere $K = DJ \\cap AI$. Since $HO_2 = O_2F$ in $\\triangle HO_1F$, we have $JK = KD$ in $\\triangle JAD$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 21998,
"subject": "Mathematics (Olympiad)",
"question": "A 3-digit number has the following properties:\n\n- If its tens and units digits are interchanged, its value increases by 36.\n- If its hundreds and units digits are interchanged, its value decreases by 198.\n\nWhen its hundreds and tens digits are interchanged, its value decreases. By how much?",
"options": [],
"answer": "See solution",
"solution": "Suppose the 3-digit number has the form $abc$ (where $a$, $b$, $c$ are its digits).\n\nFrom the given properties:\n\n$$\n(10c + b) - (10b + c) = 36 \\quad (1)\n$$\n\n$$\n(100a + c) - (100c + a) = -198 \\quad (2)\n$$\n\nFrom (1):\n\n$9(c - b) = 36 \\implies c - b = 4$\n\nFrom (2):\n\n$99(a - c) = 198 \\implies a - c = 2$\n\nSo $a - b = (a - c) + (c - b) = 2 + 4 = 6$.\n\nWhen hundreds and tens digits are interchanged, the value changes by:\n\n$$(100a + 10b + c) - (100b + 10a + c) = 90(a - b) = 90 \\times 6 = \\mathbf{540}$$\n\nTherefore, the value decreases by $540$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 21999,
"subject": "Mathematics (Olympiad)",
"question": "For every integer $n \\geq 4$, consider $m$ subsets $A_1, A_2, \\dots, A_m$ of $\\{1, 2, \\dots, n\\}$ such that:\n\n- $A_1$ has 1 element,\n- $A_2$ has 2 elements,\n- $A_m$ has $m$ elements,\n- none of these subsets is contained in another.\n\nFind the maximum possible value of $m$.",
"options": [],
"answer": "See solution",
"solution": "The answer is $m = n - 2$.\n\nWe will first show how to construct $n-2$ subsets of $\\{1, 2, \\dots, n\\}$ satisfying the required conditions. Such subsets will be called *nice*.\n\nFor $n=4$, we can take $A_1 = \\{1\\}$ and $A_2 = \\{2, 3\\}$.\n\nFor $n=5$, we can take $A_1 = \\{1\\}$, $A_2 = \\{2, 3\\}$, and $A_3 = \\{2, 4, 5\\}$.\n\nWe now show that if there exist $n-2$ nice subsets for $n$, then there exist $n$ nice subsets for $n+2$. Let $A_1, A_2, \\dots, A_{n-2}$ be nice subsets for $n$. Consider:\n\n$$\n\\begin{aligned}\n&\\bullet\\ B_1 = \\{n+2\\}, \\\\\n&\\bullet\\ B_{i+1} = A_i \\cup \\{n+1\\}, \\text{ for every } 1 \\leq i \\leq n-2, \\\\\n&\\bullet\\ B_n = \\{1, 2, \\dots, n\\}.\n\\end{aligned}\n$$\n\nIt is easy to check that $B_1, B_2, \\dots, B_n$ are nice subsets for $n+2$.\n\nNow, it remains to be seen that it is not possible to construct $n-1$ nice subsets for $n$.\n\nAssume, by contradiction, that $A_1, \\dots, A_{n-1}$ are nice subsets for $n$. If $A_1 = \\{x\\}$, then $A_{n-1} = \\{1, \\dots, n\\} \\setminus \\{x\\}$. Since $A_2$ has two elements that are different from $x$, it follows that they are elements of $A_{n-1}$ and so, $A_2 \\subset A_{n-1}$, which is a contradiction.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22000,
"subject": "Mathematics (Olympiad)",
"question": "Let $T$ be the set of ordered pairs $(i, j)$ of teachers and students who are acquainted with each other. With this notation, $|\\{j : (i, j) \\in T\\}| = a_i \\ge 0$ and $|\\{i : (i, j) \\in T\\}| = b_j \\ge 1$.\n\nAssume that $(i_0, j_0) \\in T$ satisfies $\\dfrac{a_{i_0}}{b_{j_0}} \\ge \\dfrac{a_i}{b_j}$ for all $(i, j) \\in T$. Show that $t \\ge \\dfrac{2006}{14}$, where $t = \\dfrac{a_{i_0}}{b_{j_0}}$.",
"options": [],
"answer": "See solution",
"solution": "Adding the inequalities $\\dfrac{a_{i_0}}{b_{j_0}} \\cdot \\dfrac{1}{a_i} \\ge \\dfrac{1}{b_j}$ for all $(i, j) \\in T$, we obtain:\n\n$$\n\\frac{a_{i_0}}{b_{j_0}} \\sum_{(i,j) \\in T} \\frac{1}{a_i} \\ge \\sum_{(i,j) \\in T} \\frac{1}{b_j}.\n$$\n\nNow,\n\n$$\n\\sum_{(i,j) \\in T} \\frac{1}{a_i} = \\sum_{i=1}^{14} \\sum_{\\{j:(i,j) \\in T\\}} \\frac{1}{a_i} = \\sum_{\\{i:a_i \\ne 0\\}} \\frac{a_i}{a_i} = \\sum_{\\{i:a_i \\ne 0\\}} 1 \\le 14,\n$$\n\nand similarly,\n\n$$\n\\sum_{(i,j) \\in T} \\frac{1}{b_j} = \\sum_{j=1}^{2006} \\sum_{\\{i:(i,j) \\in T\\}} \\frac{1}{b_j} = \\sum_{j=1}^{2006} \\frac{b_j}{b_j} = \\sum_{j=1}^{2006} 1 = 2006.\n$$\n\nTherefore,\n\n$$\n\\frac{a_{i_0}}{b_{j_0}} \\ge \\frac{2006}{14},\n$$\n\nso $t \\ge \\dfrac{2006}{14}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22001,
"subject": "Mathematics (Olympiad)",
"question": "設有兩個圓 $O_1$ 和 $O_2$，其半徑分別為 $R_1$ 和 $R_2$。這兩圓交於兩點 $A$ 和 $D$。過 $D$ 作一直線 $L$，$L$ 分別再交圓 $O_1$、$O_2$ 於 $B$、$C$ 兩點。現在允許兩圓圓心的距離和直線 $L$ 皆可變動。當 $\\triangle ABC$ 的面積達到最大時，求 $AD$ 的長度。\n\nLet $O_1$ and $O_2$ be circles with radii $R_1$ and $R_2$, respectively. The two circles intersect at points $A$ and $D$. Let $L$ be a straight line passing through $D$, intersecting $O_1$ and $O_2$ again at $B$ and $C$, respectively. Suppose the distance between the centers of $O_1$ and $O_2$, and the line $L$, can be varied. When the area of $\\triangle ABC$ is maximized, find the length of $AD$.",
"options": [],
"answer": "See solution",
"solution": "觀察當 $AD$ 為某一定值時，$\\angle B$、$\\angle C$（或其補角，下同）為一定角，因此 $\\angle A$ 為一定角。\n\n$$\n\\begin{aligned}\n|\\triangle ABC| &= \\frac{1}{2} AB \\cdot AC \\sin \\angle A \\\\\n&= 2R_1 R_2 \\sin \\angle ADB \\sin \\angle ADC \\sin \\angle A \\\\\n&= 2R_1 R_2 \\sin^2 \\angle ADB \\sin \\angle A.\n\\end{aligned}\n$$\n\n當 $\\angle ADB = 90^\\circ$ 時，面積達到最大值。若允許 $AD$ 變動，則當 $\\angle A = 90^\\circ$ 時，面積最大。而當兩圓正交，即 $\\angle O_1AO_2 = 90^\\circ$ 時，$\\angle A = 90^\\circ$。令此時 $AD = l$，則\n\n$$\n\\begin{aligned}\nl^2 &= \\sqrt{4R_1^2 - l^2} \\sqrt{4R_2^2 - l^2} \\\\\n\\Rightarrow \\quad l^4 &= (4R_1^2 - l^2)(4R_2^2 - l^2) = 16R_1^2 R_2^2 - 4(R_1^2 + R_2^2)l^2 + l^4,\n\\end{aligned}\n$$\n\n因此\n\n$$\nl = \\frac{2}{\\sqrt{\\frac{1}{R_1^2} + \\frac{1}{R_2^2}}}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22002,
"subject": "Mathematics (Olympiad)",
"question": "Im gleichschenkeligen Dreieck $ABC$ mit $\\overline{AC} = \\overline{BC}$ ist $D$ der Fußpunkt der Höhe durch $C$ und $M$ der Mittelpunkt der Strecke $CD$. Die Gerade $BM$ schneidet $AC$ in $E$. Beweise, dass $AC$ dreimal so lang wie $CE$ ist.",
"options": [],
"answer": "See solution",
"solution": "Wir zeichnen eine Parallele zu $BE$ durch den Punkt $D$ und schneiden sie mit $AC$. Den Schnittpunkt nennen wir $G$. Weil $M$ der Mittelpunkt der Strecke $DC$ ist, ist $E$ der Mittelpunkt von $GC$ (Strahlensatz).\n\nWeil das Dreieck $ABC$ gleichschenklig ist und $AB$ die Basis, ist $D$ der Mittelpunkt von $AB$. Daher ist $G$ der Mittelpunkt von $AE$ (Strahlensatz).\n\nDie Strecken $AG$, $GE$ und $EC$ sind also gleich lang. Daher ist $AC$ dreimal so lang wie $EC$. $\boxed{}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22003,
"subject": "Mathematics (Olympiad)",
"question": "Each of 2009 distinct points in the plane is coloured blue or red, so that on every blue-centered unit circle lie exactly two red points. Determine the largest possible number of blue points.",
"options": [],
"answer": "See solution",
"solution": "Suppose there are $r$ red points among $n$ points. Any pair of red points can lie on at most two blue-centered unit circles, so the number $b$ of blue points is at most $2\\binom{r}{2} = r(r-1)$. Since $b + r = n$, we have $r + r(r-1) = r^2 \\geq n$, i.e., $r \\geq \\lceil \\sqrt{n} \\rceil$, so $b \\leq n - \\lceil \\sqrt{n} \\rceil$.\n\nA model: take $r = \\lceil \\sqrt{n} \\rceil$ red points at coordinates $R_i(r_i, 0)$, with $0 < r_i < 2$ for $i = 1, \\ldots, r$. Take $n - r$ blue points among the $r(r-1)$ given by all pairs $(i, j)$, $1 \\leq i < j \\leq r$, with coordinates $B_{i,j}(x_{i,j}, b_{i,j})$ and $B'_{i,j}(x_{i,j}, -b_{i,j})$, where\n\n$$\nx_{i,j} = \\frac{r_i + r_j}{2} \\quad \\text{and} \\quad b_{i,j} = \\sqrt{1 - \\left(\\frac{r_i - r_j}{2}\\right)^2}.\n$$\n\nAll blue points $B_{i,j}$, $B'_{i,j}$ are distinct, and on unit circles centered at these points lie only $R_i$ and $R_j$.\n\nThe choices ensure $n - r \\leq r(r-1)$, since $n \\leq \\lceil \\sqrt{n} \\rceil^2$, and $(r_i - r_j)^2/4 < 1$. For $n = 2009$, the answer is $2009 - \\lceil \\sqrt{2009} \\rceil = 2009 - 45 = 1964$.\n\n**Remarks:**\n- If red points are allowed inside the circles, the answer is $b \\leq n - 2$. A model: $n - 2$ blue points on a circle of center $O$ and radius $\\varrho < 1$, with 2 red points inside the disk of radius $1 - \\varrho$.\n- For the original problem, a configuration of $r$ red points accommodates the maximum $b$ if the distance between any two is less than 2, and the circumradius of any three is different from 1. Another model: vertices of a regular $r$-gon of circumradius less than 1.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22004,
"subject": "Mathematics (Olympiad)",
"question": "Розглянемо точку $B(100, 100010101)$, яка лежить на графіку функції $g$, і точку $A(\\sqrt[4]{100010101}, 100010101)$, яка лежить на графіку функції $f$. Відстань між цими точками дорівнює...",
"options": [],
"answer": "See solution",
"solution": "Між цими точками відстань\n\n$$\n\\sqrt[4]{100010101} - 100 = \\frac{100010101 - 100^4}{(\\sqrt[4]{100010101} + 100)(\\sqrt{100010101} + 100^2)} < \\frac{10101}{200 \\cdot 20000} < \\frac{1}{100}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22005,
"subject": "Mathematics (Olympiad)",
"question": "The checkered plane is painted black and white, after a chessboard fashion. A polygon $\\Pi$ of area $S$ and perimeter $P$ consists of some of these unit squares (i.e., its sides go along the borders of the squares). Prove that polygon $\\Pi$ contains not more than $\\frac{S}{2} + \\frac{P}{8}$, and not less than $\\frac{S}{2} - \\frac{P}{8}$ squares of a same color.\n\n",
"options": [],
"answer": "See solution",
"solution": "Consider all $b$ black squares inside $\\Pi$; they have area $b$, and the number of their sides is $4b$. At least $4b - P$ of these sides do not belong to the boundary of $\\Pi$ and therefore lie on the border of some white squares inside $\\Pi$. This means there are at least $\\frac{4b-P}{4}$ white squares inside $\\Pi$, and $S \\ge b + \\frac{4b-P}{4} = 2b - \\frac{P}{4}$, which is equivalent to $b \\le \\frac{S}{2} + \\frac{P}{8}$ as desired. The same proof applies to the white squares, and since the total number of squares is clearly equal to $S$, the other inequality $b \\ge \\frac{S}{2} - \\frac{P}{8}$ follows. A same inequality holds for the white squares.\n\nNotice that the estimate is sharp for those polygons, and only those, whose boundary is monochromatic.\n\n\n\nA polygon with $S = 14, P = 20$ and $8 < \\frac{S}{2} + \\frac{P}{8}$ black squares.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22006,
"subject": "Mathematics (Olympiad)",
"question": "Given a positive integer $k$, define the function $f$ on the set of all positive integers by\n\n$$\nf(n) = \\begin{cases} 1, & \\text{if } n \\leq k+1, \\\\ f(f(n-1)) + f(n - f(n-1)), & \\text{if } n > k+1. \\end{cases}\n$$\n\nShow that the preimage of every positive integer under $f$ is a finite non-empty set of consecutive positive integers.",
"options": [],
"answer": "See solution",
"solution": "It is sufficient to show that the difference $\\Delta(n) = f(n) - f(n-1)$ is $0$ or $1$ for all integers $n \\geq 2$, and that $f$ is unbounded.\n\nClearly, $\\Delta(n) = 0$ for $2 \\leq n \\leq k+1$, which provides the basis for an inductive proof. If $n > k+1$, apply the recurrence. By the induction hypothesis, $\\Delta(n-1)$ is $0$ or $1$ and $1 \\leq f(n-1) \\leq n-1$. Notice that $\\Delta(n) = \\Delta(n-f(n-1))$ if $\\Delta(n-1) = 0$, and $\\Delta(n) = \\Delta(f(n-1))$ if $\\Delta(n-1) = 1$, so $\\Delta(n)$ is indeed $0$ or $1$.\n\nTo show that $f$ is unbounded, suppose, for contradiction, that $f$ achieves its maximum value $N$ for the first time at $m$. Then recursively, $N = f(m+n) = f(N) + f(m-N+n)$ for all positive integers $n$, so $f(m-N+n) = N-f(N)$ for all $n$. Setting $n \\geq N$ yields a contradiction: $N = N-f(N)$. The conclusion follows.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22007,
"subject": "Mathematics (Olympiad)",
"question": "Given integers $m > 1$ and $n > 1$, prove that there exist positive integers $N_1, \\dots, N_m$ such that\n$$\n\\sqrt{m} = \\sum_{k=1}^{m} (\\sqrt{N_k} - \\sqrt{N_k - 1})^{1/n}.\n$$",
"options": [],
"answer": "See solution",
"solution": "For $k = 1, \\dots, m$:\n$$\n(\\sqrt{k} \\pm \\sqrt{k-1})^n = a_k \\pm b_k \\sqrt{k(k-1)},\n$$\nif $n$ is even, or\n$$\n(\\sqrt{k} \\pm \\sqrt{k-1})^n = c_k \\sqrt{k} \\pm d_k \\sqrt{k-1},\n$$\nif $n$ is odd, where $a_k, b_k, c_k, d_k$ are positive integers. In both cases,\n$$\n(\\sqrt{k} \\pm \\sqrt{k-1})^n = \\sqrt{N_k} \\pm \\sqrt{M_k},\n$$\nfor positive integers $N_k, M_k$. Thus,\n$$\nN_k - M_k = (\\sqrt{N_k} + \\sqrt{M_k})(\\sqrt{N_k} - \\sqrt{M_k}) = (\\sqrt{k} + \\sqrt{k-1})^n (\\sqrt{k} - \\sqrt{k-1})^n = 1,\n$$\nfor $k = 1, \\dots, m$. Consequently,\n$$\n\\sqrt{k} - \\sqrt{k-1} = (\\sqrt{N_k} - \\sqrt{N_{k-1}})^{1/n},\n$$\nfor $k = 1, \\dots, m$, and summing over $k$ yields the result.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22008,
"subject": "Mathematics (Olympiad)",
"question": "What is the smallest possible value the expression $ab + a + b$ can take, if real numbers $a, b$ satisfy the condition $a^2 + b^2 = 25$.",
"options": [],
"answer": "See solution",
"solution": "Since $(a + b + 1)^2 \\ge 0$,\n\n$$\na^2 + b^2 + 1 + 2ab + 2a + 2b \\ge 0 \\implies ab + a + b \\ge -\\frac{1}{2}(a^2 + b^2 + 1) = -13.\n$$\n\nBy choosing $a = -4$ and $b = 3$, we obtain:\n\n$$\nab + a + b = (-4) \\times 3 + (-4) + 3 = -12 - 4 + 3 = -13,\n$$\n\ni.e., the smallest possible value is $-13$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 22009,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle and $P$ be a point in its interior. Let $AP$ meet the circumcircle of $ABC$ again at $A'$. The points $B'$ and $C'$ are similarly defined. Let $O_A$ be the circumcentre of $BCP$. The circumcentres $O_B$ and $O_C$ are similarly defined. Let $O'_A$ be the circumcentre of $B'C'P$. The circumcentres $O'_B$ and $O'_C$ are similarly defined. Prove that the lines $O_AO'_A$, $O_BO'_B$ and $O_CO'_C$ are concurrent.\n\n",
"options": [],
"answer": "See solution",
"solution": "Consider a cyclic quadrilateral $ABCD$ with circumcentre $O$. Let $P$ be the intersection of the diagonals, $O_1$ and $O_2$ be centres of circles $DAP$ and $BCP$ respectively, $H_1$ and $H_2$ defined similarly for orthocentres. Since $\\angle O_1PD = \\angle CPO_2$ by similarity, and $\\angle CPO_2 = \\angle H_2PB$ by the isogonal conjugacy of $O_2$ and $H_2$, we find that $O_1$, $P$ and $H_2$ are collinear, so $O_1P$ is perpendicular to $BC$. Similarly, $O_2P$ is perpendicular to $AD$. Since $O$ and $O_2$ both lie on the perpendicular bisector of $BC$, we have that $O_2O$ is parallel to $PO_1$, and similarly $O_1O$ is parallel to $PO_2$. Hence, $O_1PO_2O$ is a parallelogram, and so $O_1O_2$ bisects $PO$.\n\nWe can apply this to the problem, considering cyclic quadrilaterals $BC'B'C$, $CA'C'A$ and $AB'A'B$ in turn. It follows that each of $O_AO'_A$, $O_BO'_B$ and $O_CO'_C$ bisect $OP$, and thus they concur at the midpoint.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22010,
"subject": "Mathematics (Olympiad)",
"question": "Let $M$ and $N$ be the midpoints of diagonals $AC$ and $BD$ of a quadrilateral, respectively. Let $K$ be the midpoint of side $AD$. Suppose $AB = CD$. Let $P$ be an arbitrary point on segment $MN$, and let $Q$ and $R$ be the feet of the perpendiculars from $P$ to sides $AB$ and $CD$, respectively.\n\n\n\nProve that $PQ + PR$ is constant as $P$ varies along $MN$.",
"options": [],
"answer": "See solution",
"solution": "Let $QP \\cap KN = Q'$ and $RP \\cap MK = R'$. Since $QQ' + RR'$ is constant, it suffices to show that $PR' + PQ'$ is constant. In fact, $PR' + QQ' = \\frac{4 \\cdot S_{MNK}}{AB}$, and $S_{MNK}$ is constant. Thus, $PR' + QQ'$ is constant, and the result follows.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22011,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs of positive rational numbers such that the sum of the numbers in a pair is an integer and the sum of their multiplicative inverses is also an integer.",
"options": [],
"answer": "See solution",
"solution": "Let the numbers in the pair be represented as reduced fractions $\\frac{a}{b}$ and $\\frac{c}{d}$. For\n\n$$\n\\frac{a}{b} + \\frac{c}{d} = \\frac{ad + bc}{bd}\n$$\nto be an integer, we must have\n\n$$\nad + bc = k \\cdot bd\n$$\nwith $k$ being some integer. By writing the equality as $bc = (kb - a) d$ and noting that $c$ and $d$ are relatively prime, $b$ must be divisible by $d$. Similarly, writing $ad = (kd - c) b$ and noting $a$ and $b$ are relatively prime, $d$ must be divisible by $b$. Therefore, $b = d$.\n\nFor $\\frac{b}{a} + \\frac{d}{c}$ to also be an integer, we analogously must have $a = c$. Therefore $\\frac{a}{b} = \\frac{c}{d}$.\n\nConsequently, $\\frac{a}{b} + \\frac{c}{d} = \\frac{2a}{b}$ and $\\frac{b}{a} + \\frac{d}{c} = \\frac{2b}{a}$ are both integers. If $a = b$, then $\\frac{a}{b} = 1$. If $a < b$, then $\\frac{2a}{b} < 2$, implying $\\frac{2a}{b} = 1$ and $\\frac{a}{b} = \\frac{1}{2}$ as the only possibility. If $a > b$, then similarly $\\frac{a}{b} = 2$.\n\nThus, the pairs are $(1, 1)$, $(2, 2)$, and $\\left(\\frac{1}{2}, \\frac{1}{2}\\right)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22012,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be a triangle with medians $AA_1$, $BB_1$, and $CC_1$ intersecting at point $T$. Let $\\overline{AD_1} = \\overline{AT}$. On the extension of $CC_1$, choose a point $C_2$ such that $\\overline{C_1C_2} = \\frac{CC_1}{3}$, and on the extension of $BB_1$, choose a point $B_2$ such that $\\overline{B_1B_2} = \\frac{BB_1}{3}$. Prove that the quadrilateral $TB_2AC_2$ is a rectangle.",
"options": [],
"answer": "See solution",
"solution": "Since $AA_1$ is a median in $\\triangle ABC$ and $\\overline{BA_1} = \\overline{AT}$, we get $\\overline{AT} = \\frac{BC}{2}$. $A_1$ is the circumcenter of the circumcircle of $\\triangle BC_1T$. By Thales' theorem, $\\angle BT_1C = 90^\\circ$. We have $\\angle B_2T_1C_2 = 90^\\circ$ (as vertically opposite angles). Since $T$ is the centroid of $\\triangle ABC_1$, $\\overline{CT_1T} = \\frac{CC_1}{3} = \\overline{C_1C_2}$. From $\\overline{BC_1} = \\overline{C_1A}$, the quadrilateral $BT_1AC_2$ is a parallelogram. Then $BT \\parallel AC_2$, so $\\angle T_1C_2A = \\angle C_1TB = 180^\\circ - \\angle B_2T_1C_2 = 90^\\circ$ (as angles on the transversal).\n\nSimilarly, we can show that the quadrilateral $CB_2A$ is a parallelogram, i.e., $\\angle TB_2A = \\angle C_2T'B_2 = 90^\\circ$ (as angles on the transversal). Thus, $\\angle C_2AB_2 = 360^\\circ - 270^\\circ = 90^\\circ$, so the quadrilateral is a rectangle.\n\n*Second proof for the statement that $\\angle TB_2A = 90^\\circ$.*\n\nConsider $\\triangle BB_2C_1$. Since $C_1T$ is both a median and an altitude in the triangle, $\\triangle BB_2C_1$ is isosceles, so $\\overline{BC_1} = \\overline{CB_2}$ (by SAS: $C_1T$ is common, $\\angle BT'C_1 = \\angle B_2T'C_1 = 90^\\circ$, and $\\overline{BT'} = \\overline{TB_2}$). Thus, $C_1$ is the circumcenter of the circumcircle of $\\triangle BB_2A$, so by Thales' theorem $\\angle BB_2A = 90^\\circ$. Therefore, $\\angle C_2AB_2 = 360^\\circ - 270^\\circ = 90^\\circ$, so the quadrilateral is a rectangle.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22013,
"subject": "Mathematics (Olympiad)",
"question": "Let $O'$ be the circumcenter of $\\triangle ABC$ and let $M$, $N$, and $P$ be the respective midpoints of $AB$, $AC$, and $AD$. \n\nNotice that $O'$, $M$, and $N$ are fixed points and, by Thales' theorem, $P$ is a variable point on the segment $MN$.\n\nLet $O_1$ be the intersection of the perpendicular bisectors of $AB$ and $AD$, $O_2$ the intersection of the perpendicular bisectors of $AD$ and $AC$, and $O'$ the intersection of the perpendicular bisectors of $AB$ and $AC$.\n\n\n\nFind the locus of the circumcenter $O$ of $AO_1O'O_2$ as $D$ varies along $BC$.",
"options": [],
"answer": "See solution",
"solution": "Since $\\angle ANO_2 = \\angle APO_2 = 90^\\circ$, $APNO_2$ is cyclic; analogously, $AMO_1P$ is cyclic as well. So $\\angle O_2AN = \\angle O_2PN = \\angle O_1PM = \\angle O_1AM$. Hence $\\angle O_1AO_2 = \\angle BAC$.\n\nQuadrilateral $AMO'N$ is cyclic as well, so $\\angle MO'N = 180^\\circ - \\angle MAN = 180^\\circ - \\angle BAC$. Thus $\\angle O_1O'O_2 + \\angle O_1AO_2 = 180^\\circ$, which implies that $AO_1O'O_2$ is cyclic. Its circumcenter is $O$, and so $O$ lies on the perpendicular bisector of $AO'$, which is fixed. So the locus is a segment contained in such perpendicular bisector.\n\nIt remains to find the vertices of this segment. When $D$ tends to $B$, the perpendicular bisectors of $AD$ and $AB$ tend to coincide, so $O_2$ and $O'$ tend to coincide and $O$ tends to be the circumcenter of $AO'B$; analogously, when $D$ tends to $C$, $O$ tends to be the circumcenter of $AO'C$. So the locus is the open segment with vertices at the circumcenters of $AO'B$ and $AO'C$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22014,
"subject": "Mathematics (Olympiad)",
"question": "Determine the numbers $\\overline{abc}$, with $a < b < c$, knowing that the remainders of divisions of the numbers $\\overline{abc}$, $\\overline{bca}$, $\\overline{cab}$ by $27$ belong to the set $\\{1, 2, 3, 4, 5\\}$.",
"options": [],
"answer": "See solution",
"solution": "The numbers $\\overline{abc}$, $\\overline{bca}$, $\\overline{cab}$ have the same sum of digits, so they will have the same remainder $r$ when divided by $9$.\n\nSince the remainders modulo $27$ are small, they are preserved modulo $9$. Indeed, if $n = 27k + r$, then $n = 9 \\cdot 3k + r$, so $r$ will be a common remainder.\n\n$27$ must divide the difference $\\overline{bca} - \\overline{abc} = 90b + 9c - 99a = 9(a + b + c) + 81b - 108a = 9(a + b + c) + 27(3b - 4a)$, therefore $3 \\mid a + b + c$, which implies $r = 3$.\n\nWe only have to find the numbers $\\overline{abc}$, with $a < b < c$, and the remainder $3$ modulo $27$ (and modulo $9$).\n\nThe sum of digits has the form $9k + 3$, leaving the cases $a + b + c \\in \\{3, 12, 21\\}$.\n\nSince $a < b < c$, the case $a + b + c = 3$ is not valid.\n\nThe numbers $abc$, with $a < b < c$ and $a + b + c = 12$ are $129$, $138$, $147$, $156$, $237$, $246$, and $345$. Only $138$ and $246$ have the remainder $3$ modulo $27$.\n\nThe numbers $abc$, with $a < b < c$ and $a + b + c = 21$ are $489$, $579$, and $678$. Convenient are $489$ and $678$.\n\nIn the end, we have $4$ solutions: $138$, $246$, $489$, and $678$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22015,
"subject": "Mathematics (Olympiad)",
"question": "Let $A$ be a point outside a given circle $\\Gamma$ with diameter $BC$. Find the locus of the orthocenter $H$ of triangle $ABC$ when $BC$ changes.",
"options": [],
"answer": "See solution",
"solution": "Let $O$ be the center of $\\Gamma$ and let $AA_1$, $BB_1$, and $CC_1$ be the altitudes of $\\triangle ABC$ with orthocenter $H$. Let $\\Gamma_1$ be a circle with diameter $AO$. Since $\\angle A_1$ is right, $\\Gamma_1$ passes through the points $A$, $O$, $A_1$. Likewise, points $B$, $C$, and $B_1$ lie on circle $\\Gamma$.\n\nThe power of $H$ with respect to $\\Gamma$ is $HB \\cdot HB_1$ and the power of $H$ with respect to $\\Gamma_1$ is $HA \\cdot HA_1$. Since triangles $BHA_1$ and $AHB_1$ are similar, we have\n\n$$\n\\frac{HB}{HA} = \\frac{HA_1}{HB_1} \\Leftrightarrow HB \\cdot HB_1 = HA \\cdot HA_1\n$$\n\nThis means that $H$ lies on the radical axis $\\ell$ of $\\Gamma$ and $\\Gamma_1$.\n\nConversely, each point of $\\ell$ is the orthocenter of a triangle of the given type.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22016,
"subject": "Mathematics (Olympiad)",
"question": "Let $A = \\{a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n\\}$ be a set with $2n$ elements. For $i = 1, 2, \\dots, m$, let $B_i \\subseteq A$. If $\\bigcup_{j=1}^m B_j = A$, the ordered $m$-tuple $(B_1, B_2, \\dots, B_m)$ is called an *ordered m-covering* of $A$. If, for every $i = 1, 2, \\dots, m$ and every $j = 1, 2, \\dots, n$, $B_i$ does not contain both $a_j$ and $b_j$, then $(B_1, B_2, \\dots, B_m)$ is called a *no-matching ordered m-covering* of $A$. Let $a(n, m)$ be the number of ordered $m$-coverings of $A$, and $b(n, m)$ be the number of no-matching ordered $m$-coverings of $A$.\n\n1. Find $a(n, m)$ and $b(n, m)$.\n2. Suppose for some integers $m \\ge 2$ and $n \\ge 1$, $\\frac{a(n, m)}{b(n, m)} \\le 2021$. Find the largest possible value of $m$.",
"options": [],
"answer": "See solution",
"solution": "1. To find $a(n, m)$: For each of the $2n$ elements in $A$, we must assign it to at least one of the $m$ sets $B_1, \\dots, B_m$. For each element, there are $2^m - 1$ choices (all non-empty subsets of $\\{B_1, \\dots, B_m\\}$). Thus,\n\n$$\na(n, m) = (2^m - 1)^{2n}.$$\n\nTo find $b(n, m)$: For each $j = 1, \\dots, n$, we must assign $a_j$ and $b_j$ to the $m$ sets so that no $B_i$ contains both $a_j$ and $b_j$. For each $j$, there are $3^m - 2^{m+1} + 1$ valid assignments. Thus,\n\n$$\nb(n, m) = (3^m - 2^{m+1} + 1)^n.$$\n\n2. Given $\\frac{a(n, m)}{b(n, m)} \\le 2021$, substitute the formulas:\n\n$$\\frac{(2^m - 1)^{2n}}{(3^m - 2^{m+1} + 1)^n} \\le 2021.$$\n\nThis simplifies to\n\n$$\\frac{(4^m - 2^{m+1} + 1)^n}{(3^m - 2^{m+1} + 1)^n} \\le 2021.$$\n\nSince the ratio increases with $n$, set $n = 1$ for the largest $m$:\n\n$$\\frac{4^m - 2^{m+1} + 1}{3^m - 2^{m+1} + 1} \\le 2021.$$\n\nThis expression increases with $m$ for $m \\ge 4$. Solving $\\frac{4^m}{3^m} > 2021$ gives $m > \\log_{4/3} 2021$, so the largest possible $m$ is $\\left\\lfloor \\log_{4/3} 2021 \\right\\rfloor = 26$.\n\n**Conclusion:** The largest possible value of $m$ is $26$, attained when $n = 1$.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 22017,
"subject": "Mathematics (Olympiad)",
"question": "Fiona wants to arrange 6 red buttons, 3 green buttons, and 3 yellow buttons in a row on her dress, such that no two red buttons are adjacent. In how many ways can she arrange the buttons?",
"options": [],
"answer": "See solution",
"solution": "There are five spaces to fill between the six red buttons. Since there are 6 non-red buttons, there are at most two non-red buttons between any two otherwise adjacent red buttons.\n\nSuppose there is exactly one non-red button between any two otherwise adjacent red buttons. Then there must be a red button at the top or at the bottom. The two cases are symmetrical. With a red button at the top, we must choose 3 out of 6 places for the three green buttons and then the three yellow buttons must occupy the remaining three places. The number of such arrangements is $\\binom{6}{3} = 20$. So the number of arrangements with exactly one non-red button between any two otherwise adjacent red buttons is $20 + 20 = 40$.\n\nNow suppose there are two non-red buttons between two otherwise adjacent red buttons. Then there are only 5 possible locations for the non-red pair and the remaining four non-red buttons must be placed separately in the remaining four places. The non-red pair must be green and yellow. They can be arranged in 2 ways and the number of arrangements for the two remaining green buttons is $\\binom{4}{2} = 6$. Then the remaining two yellow buttons must occupy the remaining two places. So the number of arrangements with two non-red buttons between two otherwise adjacent red buttons is $5 \\times 2 \\times 6 = 60$.\n\nThus the total number of ways Fiona can arrange the buttons on her dress is $40 + 60 = 100$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22018,
"subject": "Mathematics (Olympiad)",
"question": "Find all values of the parameter $a$ for which the equation\n\n$$\nx^2 - 3x[x] + 2x = a\n$$\n\nhas two positive roots.",
"options": [],
"answer": "See solution",
"solution": "Let's analyze the function $y = x^2 - 3x[x] + 2x$ for $x > 0$.\n\n$$\n\\begin{align*}\nx \\in [0, 1) &\\Rightarrow [x] = 0 \\Rightarrow y = x^2 + 2x \\\\\nx \\in [1, 2) &\\Rightarrow [x] = 1 \\Rightarrow y = x^2 - x \\\\\nx \\in [2, 3) &\\Rightarrow [x] = 2 \\Rightarrow y = x^2 - 4x \\\\\nx \\in [3, 4) &\\Rightarrow [x] = 3 \\Rightarrow y = x^2 - 7x\n\\end{align*}\n$$\n\nFor $x \\ge 4$, the function on each interval $[n, n+1)$ with $n \\ge 4$ is downward. Thus, exactly two positive solutions can exist only when $x \\le 4$.\n\nBy analyzing the graph, we see there are exactly two positive solutions when $0 < a < 2$, and also when $-\\frac{49}{4} < a < -12$.\n\n\n\n**Answer:** $\\left(-\\frac{49}{4}, -12\\right) \\cup (0, 2)$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22019,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_j = x_{j+1} - x_j$ and $b_j = y_{j+1} - y_j$ for $j = 1, 2, \\dots, n$, where the indices are taken modulo $n$. It is given that $a_j^2 + b_j^2 = c$ for some constant $c$.\n\nIs it possible to construct an $n$-sided polygon with integer coordinates such that every side has length $\\sqrt{c}$? For which $n$ is this possible?\n\n",
"options": [],
"answer": "See solution",
"solution": "* If $c \\equiv 0 \\pmod{4}$, then both $a_j, b_j$ are even. Since $x_0, y_0$ are even, we easily deduce that all $x_j, y_j$ are even. This contradicts our assumption.\n\n* If $c$ is odd, then $a_j + b_j \\equiv 1 \\pmod{2}$. Thus, we have\n\n$$\n1 \\equiv \\sum_{j=1}^{n} (a_j + b_j) = \\sum_{j=1}^{n} (x_{j+1} - x_j + y_{j+1} - y_j) = 0 \\pmod{2},\n$$\n\nwhich is a contradiction.\n\n* If $c \\equiv 2 \\pmod{4}$, then $a_j, b_j$ are odd. Thus, we have\n\n$$\n1 \\equiv \\sum_{j=1}^{n} a_j = \\sum_{j=1}^{n} (x_{j+1} - x_j) = 0 \\pmod{2},\n$$\n\nwhich is a contradiction.\n\nTherefore, it is impossible that $n$ is odd.\n\nFor even $n \\ge 4$, we provide the following construction. The following quadrilaterals have the properties that each side has length 5, and two of the sides are vertical:\n\n* (type I) a square with vertices $(0,0), (5,0), (5,5), (0,5)$\n* (type II) a rhombus with vertices $(0,0), (4,3), (4,8), (0,5)$\n* (type III) a rhombus with vertices $(0,0), (4,-3), (4,2), (0,5)$\n\nWe can place $\\frac{n}{2}-1$ copies of these quadrilaterals in the order types I, II, III, II, III, II, III, ... by translating them so that the left edge of the next quadrilateral overlaps with the right edge of the previous one as shown. This gives an $n$-sided polygon whose every side has length 5.",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 22020,
"subject": "Mathematics (Olympiad)",
"question": "For non-empty sets $S, T$ of numbers, we define\n$$\nS + T = \\{s + t \\mid s \\in S, t \\in T\\}, \\quad 2S = \\{2s \\mid s \\in S\\}.\n$$\nLet $n$ be a positive integer, and $A, B$ be non-empty subsets of $\\{1, 2, \\dots, n\\}$. Prove that there exists a subset $D$ of $A + B$ such that\n$$\nD + D \\subseteq 2(A + B), \\quad \\text{and} \\quad |D| \\ge \\frac{|A| \\cdot |B|}{2n},\n$$\nwhere $|X|$ denotes the number of elements of a finite set $X$.",
"options": [],
"answer": "See solution",
"solution": "Let $S_y = \\{(a, b) \\mid a - b = y,\\ a \\in A,\\ b \\in B\\}$. Since $\\sum_{y=1-n}^{n-1} |S_y| = |A| \\cdot |B|$, there exists an integer $y_0$ such that $1 - n \\le y_0 \\le n - 1$ and $|S_{y_0}| \\ge \\frac{|A| \\cdot |B|}{2n - 1} > \\frac{|A| \\cdot |B|}{2n}$.\n\nLet $D = \\{2b + y_0 \\mid (a, b) \\in S_{y_0}\\}$, then\n$$\n|D| = |S_{y_0}| > \\frac{|A| \\cdot |B|}{2n}.\n$$\nFrom the definition of $S_{y_0}$, for each $d \\in D$, there exists $(a, b) \\in S_{y_0}$ such that $d = 2b + y_0 = a + b \\in A + B$. So $D \\subseteq A + B$. For any $d_1, d_2 \\in D$, let $d_1 = 2b_1 + y_0 = 2a_1 - y_0$, $d_2 = 2b_2 + y_0$ ($b_1, b_2 \\in B$, $a_1 \\in A$), then\n$$\nd_1 + d_2 = 2a_1 - y_0 + 2b_2 + y_0 = 2(a_1 + b_2) \\subseteq 2(A + B).\n$$\nTherefore, $D$ satisfies the condition. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22021,
"subject": "Mathematics (Olympiad)",
"question": "Let $p > 3$ be a prime, and let $x, y, z$ be integers such that $0 < x, y, z < p$ and $p \\mid x^3 - y^3$, $p \\mid x^3 - z^3$, and $p \\mid y^3 - z^3$. Prove that $p \\mid x^2 + y^2 + z^2$.",
"options": [],
"answer": "See solution",
"solution": "Note that $p > 3$ and $p \\mid x^3 - y^3 = (x-y)(x^2 + xy + y^2)$. Since $x, y < p$, $p \\nmid x-y$. Therefore $p \\mid x^2 + xy + y^2$. Similarly, $p \\mid x^2 + xz + z^2$ and $p \\mid y^2 + yz + z^2$.\n\nThus, $p \\mid (x^2 + xy + y^2) - (y^2 + yz + z^2) = (x-z)(x+y+z)$, so $p \\mid x+y+z$.\n\nSince $x, y, z < p$ and $p$ is a prime, $x+y+z = p$ or $2p$.\n\nSince the parity of $x+y+z$ and $x^2+y^2+z^2$ are the same, it suffices to prove that $p \\mid x^2+y^2+z^2$.\n\nNow $p \\mid x^2 + xy + y^2 = x(x+y+z) + y^2 - xz$, so $p \\mid y^2 - xz$. Therefore, $p \\mid (x^2 + xy + y^2) + (y^2 - xz) = x^2 + y^2 + z^2$, and we are done.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22022,
"subject": "Mathematics (Olympiad)",
"question": "Let $p(x) = a_nx^n + a_{n-1}x^{n-1} + \\dots + a_0$, where $a_n \\neq 0$. Find all polynomials $p(x)$ such that the polynomial $p(x)^3 - p(p(x))$ has only non-negative values for all $x \\in \\mathbb{R}$.",
"options": [],
"answer": "See solution",
"solution": "The leading term of $p(p(x))$ is $a_n(a_nx^n)^n = a_n^{n+1}x^{n^2}$, and the leading term of $p(x)^3$ is $(a_nx^n)^3 = a_n^3x^{3n}$. Both are of odd degree. Since $p(x)^3 - p(p(x))$ must be non-negative for all $x$, it must be of even degree, so the leading terms must cancel: $a_n^{n+1}x^{n^2} = a_n^3x^{3n}$, i.e., $n^2 = 3n$ and $a_n^{n+1} = a_n^3$. Thus, $n=3$ and $a_n=1$ (since $a_n \\neq 0$), so $p(x)$ is degree 3 with leading coefficient 1. Let $p(x) = x^3 + ax + b$. Substituting into the inequality and simplifying gives $ax^3 + a^2x + ab + b \\le 0$ for all $x \\in \\mathbb{R}$. This is only possible if $a=0$, since otherwise the polynomial of odd degree takes positive values. Thus, $b \\le 0$. The polynomials that satisfy the conditions are exactly those of the form $p(x) = x^3 + b$, where $b \\le 0$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 22023,
"subject": "Mathematics (Olympiad)",
"question": "Determine all triples $ (a, b, c) $ of positive real numbers satisfying the following system of equations:\n\n$$\n\\begin{aligned}\na\\sqrt{b} - c &= a, \\\\\nb\\sqrt{c} - a &= b, \\\\\nc\\sqrt{a} - b &= c.\n\\end{aligned}\n$$",
"options": [],
"answer": "See solution",
"solution": "Without loss of generality, suppose $a = \\max\\{a, b, c\\}$. From the first equation:\n\n$$\na\\sqrt{b} - c = a \\implies a(\\sqrt{b} - 1) = c\n$$\n\nSince $a \\ge c$, $\\sqrt{b} - 1 \\ge 0$, so $b \\ge 1$. Now, from the second equation:\n\n$$\nb\\sqrt{c} - a = b \\implies b(\\sqrt{c} - 1) = a\n$$\n\nSince $a \\ge b$, $\\sqrt{c} - 1 \\ge 1$, so $c \\ge 4$. Similarly, from the third equation:\n\n$$\nc\\sqrt{a} - b = c \\implies c(\\sqrt{a} - 1) = b\n$$\n\nCombining these, we get:\n\n$$\n4 \\le c \\le c(\\sqrt{c} - 1) \\le c(\\sqrt{a} - 1) = b \\le 4\n$$\n\nThis yields $a = b = c = 4$.\n\n**Conclusion.** The unique solution to the system is $(a, b, c) = (4, 4, 4)$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22024,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge 100$ be an integer. Ivan writes the numbers $n, n+1, \\dots, 2n$ each on different cards. He then shuffles these $n+1$ cards and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.",
"options": [],
"answer": "See solution",
"solution": "If we can find three cards numbered $a, b, c$ such that $a + b$, $a + c$, and $b + c$ are perfect squares, then we will be done because two of $a, b, c$ will be in the same pile.\n\nSolving the system\n\n$$\n\\begin{aligned}\na + b &= (2k - 1)^2 \\\\\na + c &= (2k)^2 \\\\\nb + c &= (2k + 1)^2\n\\end{aligned}\n$$\n\nyields\n\n$$\n\\begin{aligned}\na &= 2k^2 - 4k \\\\\nb &= 2k^2 + 1 \\\\\nc &= 2k^2 + 4k\n\\end{aligned}\n$$\n\nUsing these values of $a, b, c$, it suffices to show that for each integer $n \\ge 100$ there exists a positive integer $k$ such that $n \\le 2k^2 - 4k$ and $2k^2 + 4k \\le 2n$. That is,\n\n$$\nk^2 + 2k \\le n \\le 2k^2 - 4k.\n$$\n\nTo this end, let $k$ be the greatest integer such that $k^2 + 2k \\le n$. Note that $k \\ge 9$ because $n \\ge 100$. Since $k$ is the greatest integer satisfying $k^2 + 2k \\le n$, it follows that\n\n$$\n(k + 1)^2 + 2(k + 1) \\ge n + 1 \\Leftrightarrow n \\le k^2 + 4k + 4.\n$$\n\nTo complete the proof it suffices to show that $k^2 + 4k + 4 \\le 2k^2 - 4k$. This is equivalent to $(k - 4)^2 \\ge 20$. But this is true since $k \\ge 9$. $\\square$\n\n*Comment 1*\n\nThe above proof shows that the conclusion of the problem is true for all $n \\ge 99$.\n\n*Comment 2*\n\nThe conclusion of the problem is false for $n = 98$. Indeed, a counterexample occurs if the first pile contains the even numbers from 98 to 126, the odd numbers from 129 to 161, and the even numbers from 162 to 196, while the second pile contains the rest of the numbers.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22025,
"subject": "Mathematics (Olympiad)",
"question": "The range of $y = \\sin^2 x + \\sqrt{3} \\sin x \\cos x$ when $\\frac{\\pi}{4} \\le x \\le \\frac{\\pi}{2}$ is ______.",
"options": [],
"answer": "See solution",
"solution": "$$\ny = \\sin^2 x + \\sqrt{3} \\sin x \\cos x \n= \\frac{1 - \\cos 2x}{2} + \\frac{\\sqrt{3}}{2} \\sin 2x \n= \\sin\\left(2x - \\frac{\\pi}{6}\\right) + \\frac{1}{2}.\n$$\n\nWhen $\\frac{\\pi}{4} \\le x \\le \\frac{\\pi}{2}$, we have $\\frac{\\pi}{3} \\le 2x - \\frac{\\pi}{6} \\le \\frac{5\\pi}{6}$, so the range of $\\sin\\left(2x - \\frac{\\pi}{6}\\right)$ is $\\left[\\frac{1}{2}, 1\\right]$. Therefore, the range of $y = \\sin^2 x + \\sqrt{3}\\sin x \\cos x$ is $\\left[1, \\frac{3}{2}\\right]$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22026,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, \\dots, a_n$ and $b_1, \\dots, b_n$ be $2n$ real numbers. Prove that there exists an integer $k$ with $1 \\leq k \\leq n$ such that\n\n$$\n\\sum_{i=1}^{n} |a_i - a_k| \\leq \\sum_{i=1}^{n} |b_i - a_k|.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $A_{\\min}$ and $A_{\\max}$ denote respectively the minimum and maximum value among $a_1, a_2, \\dots, a_n$. Then the triangle inequality implies for every $i$ with $1 \\leq i \\leq n$ that\n\n$$\n\\begin{aligned}\n|A_{\\max} - a_i| + |a_i - A_{\\min}| &= (A_{\\max} - a_i) + (a_i - A_{\\min}) = A_{\\max} - A_{\\min} \\\\\n&= |A_{\\max} - A_{\\min}| \\leq |A_{\\max} - b_i| + |b_i - A_{\\min}|.\n\\end{aligned}\n$$\n\nSumming these inequalities up over all $i = 1, 2, \\dots, n$ yields\n\n$$\n\\sum_{i=1}^{n} |A_{\\max} - a_i| + \\sum_{i=1}^{n} |a_i - A_{\\min}| \\leq \\sum_{i=1}^{n} |A_{\\max} - b_i| + \\sum_{i=1}^{n} |b_i - A_{\\min}|.\n$$\n\nThis implies that the integer $k$ with $a_k = A_{\\max}$ or the integer $k$ with $a_k = A_{\\min}$ must satisfy the desired inequality. $\\square$\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22027,
"subject": "Mathematics (Olympiad)",
"question": "The integer $x$ is at least $3$ and $n = x^6 - 1$. Let $p$ be a prime and $k$ be a positive integer such that $p^k$ is a factor of $n$. Show that $p^{3k} < 8n$.",
"options": [],
"answer": "See solution",
"solution": "In what follows, we will use the notation $p^i \\mid N$ to mean \"$p^i$ divides $N$ but $p^{i+1}$ does not divide $N$\". We assume that $p^k \\mid n$, since choosing smaller values of $k$ only makes the problem easier.\n\nClearly $p$ does not divide $x$, or else $n$ will be $-1 \\pmod p$ and therefore not divisible by $p^k$.\n\nWe can factorise\n\n$$\nn = x^6 - 1 = (x^2 - 1)(x^2 + x + 1)(x^2 - x + 1).\n$$\n\nNote that\n\n$$\nx^2 - x + 1 < x^2 - 1 < x^2 + x + 1 < 2x^2 - 1. \\quad (1)\n$$\n\nSuppose (for a contradiction) that\n\n$$\np^{3k} \\geq 8n = 8(x^6 - 1) > 8(x^2 - 1)^3.\n$$\n\nwhich means that $p^k > 2(x^2 - 1)$ and so $p^k \\ge 2x^2 - 1$.\n\nThen, by Equation (1), $p$ must divide at least two of the numbers $x^2-x+1$, $x^2-1$ and $x^2+x+1$.\n\nWe consider three cases, one for each pair.\n\n* Suppose $p$ divides $x^2 - x + 1$ and $x^2 + x + 1$. Then $p$ also divides their difference $2x$. Since $p \\nmid x$, this means $p = 2$. But then $x$ is odd, so both $-x-1$ and $x^2+x+1$ are odd, which contradicts their being divisible by the even prime $p$.\n\n* Suppose $p$ divides $x^2 + x + 1$ and $x^2 - 1 = (x + 1)(x - 1)$. As above, $p$ cannot be $2$, since it divides the odd number $x^2 + x + 1$. Thus $p$ cannot divide both $x - 1$ and $x + 1$. We split into subcases:\n\n - $p \\mid x+1$: In this case $p$ divides $(x^2 + x + 1) - (x+1) = x^2$, so $p \\mid x$, which is a contradiction.\n - $p \\mid x - 1$: In this case $p$ divides $(x^2 + x + 1) + (x - 1) = x(x + 2)$, so $p \\mid x + 2$. Since we also have $p \\mid x - 1$, we must have $p = 3$. But then $9 \\mid (x - 1)(x + 2) = x^2 + x - 2$, so $x^2 + x + 1 \\equiv 3 \\pmod 9$. If $3^j \\mid x - 1$, then $3^{j+1} \\mid (x - 1)(x^2 + x + 1) = x^3 - 1$, so\n\n $$\n 3^{j+1} \\mid (x^3 - 1)(x^3 + 1) = (x^6 - 1).\n $$\n\n and so $k = j + 1$. Then $3^k = 3 \\cdot 3^j \\le 3(x - 1) \\le 2x^2 - 1$ (the latter inequality holds for all $x \\ge 3$), which is a contradiction.\n\n* Suppose, finally, that $p$ divides $x^2 - x + 1$ and $x^2 - 1$. As before, we can split into two cases:\n\n - $p \\mid x - 1$: In this case $p$ divides $(x^2 - x + 1) + (x - 1) = x^2$, so $p \\mid x$, which is a contradiction.\n - $p \\nmid x - 1$: In this case $p$ divides $(x^2 - x + 1) - (x + 1) = x(x - 2)$, so $p \\nmid x - 2$. Since we also have $p \\mid x + 1$, we must have $p = 3$. But then $9 \\mid (x - 1)(x - 2) = x^2 - x - 2$, so $x^2 - x + 1 \\equiv 3 \\pmod 9$.\n\n If $3^j \\mid x - 1$, then $3^{j+1} \\mid (x + 1)(x^2 - x + 1)$, so $3^{j+1} \\mid x^3 + 1$, so $3^{j-1} \\mid (x^3 + 1)(x^3 - 1)$, so $k = j + 1$. Then\n\n $$\n 3^k = 3 \\cdot 3^j \\le 3(x + 1) \\le 2x^2 - 1\n $$\n\n (the latter inequality holds for all $x \\ge 3$), which is another contradiction.\n\nIn all cases we have reached a contradiction, so we are done.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22028,
"subject": "Mathematics (Olympiad)",
"question": "Find the greatest positive integer $n$ such that the system of equations\n\n$$\n(x+1)^2 + y_1^2 = (x+2)^2 + y_2^2 = \\dots = (x+k)^2 + y_k^2 = \\dots = (x+n)^2 + y_n^2\n$$\n\nhas an integral solution $(x, y_1, y_2, \\dots, y_n)$.",
"options": [],
"answer": "See solution",
"solution": "*Lemma:* For arbitrary integers $a, b$, we have:\n\n$$\na^2 + b^2 \\equiv \\begin{cases}\n2 \\pmod{8} & \\text{if } a \\equiv \\pm 1 \\pmod{4} \\\\\n1 \\pmod{8} & \\text{if } a \\equiv 0 \\pmod{4} \\\\\n5 \\pmod{8} & \\text{if } a \\equiv 2 \\pmod{4}\n\\end{cases}\n$$\n\nReturning to the problem:\n\nFor $n=3$: The system has an integral solution $(x = -2, y_1 = 0, y_2 = 1, y_3 = 0)$.\n\nFor $n=4$: Suppose the system has an integral solution $(x, y_1, y_2, y_3, y_4)$. Since $x+1, x+2, x+3, x+4$ form a complete residue system modulo $4$, by the lemma, there must exist an integer $m$ such that\n\n$$\nm \\in \\{2, 1, 5\\} \\cap \\{1, 0, 4\\} \\cap \\{5, 4, 0\\} = \\emptyset.\n$$\n\nThis contradiction shows there is no integral solution for $n=4$. Thus, the greatest such $n$ is $n=3$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22029,
"subject": "Mathematics (Olympiad)",
"question": "A big rectangle is divided into small rectangles that are twice as high as they are wide. The rectangle is 10 of these small rectangles wide, as in the figure on the right. In this figure, you can see some squares of different sizes.\n\nHow many small rectangles high is the figure if we can find exactly 345 squares in it?\n\n",
"options": [],
"answer": "See solution",
"solution": "15",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22030,
"subject": "Mathematics (Olympiad)",
"question": "設三角形 $ABC$ 的外心為 $O$，並令三角形 $BOC$ 的外接圓為 $\\omega$。直線 $AO$ 與圓 $\\omega$ 的第二個交點為 $G$。設 $BC$ 邊的中點為 $M$，且 $BC$ 的中垂線交 $\\omega$ 於 $O, N$ 兩點。\n\n證明：線段 $AN$ 的中點，位於三角形 $OMG$ 的外接圓與以 $AO$ 為直徑的圓的根軸上。\n\nLet $O$ be the circumcenter of triangle $ABC$, and $\\omega$ be the circumcircle of triangle $BOC$. Line $AO$ intersects with circle $\\omega$ again at the point $G$. Let $M$ be the midpoint of side $BC$, and the perpendicular bisector of $BC$ meets circle $\\omega$ at the points $O$ and $N$.\n\nProve that the midpoint of the segment $AN$ lies on the radical axis of the circumcircle of triangle $OMG$, and the circle whose diameter is $AO$.\n\n",
"options": [],
"answer": "See solution",
"solution": "令 $H$ 為 $A$ 到 $BC$ 的垂足。易見 $AN$、$AO$ 直徑圓與圓 $BOC$ 共點於一點 $S$。設 $AM$ 交圓 $OMG$ 於另一點 $V$，$GV$ 交圓 $BOC$ 於另一點 $U$，且交 $AN$ 於點 $T$。因為 $\\angle NUG = \\angle MOG = \\angle MVU$，也就是 $UN \\parallel AV$，所以\n\n$$\n\\angle TGA = \\angle AMO = \\angle HAM = \\angle SAO = \\angle TAG\n$$\n\n（因為 $AN$ 是共軛中線，故 $AM, AN$ 等角共軛，$AH, AO$ 等角共軛）。又 $\\angle AGN = 90^\\circ$，故 $T$ 為 $AN$ 中點，綜合上述，得 $AVNU$ 為一平行四邊形，且 $T$ 為 $VU$ 的中點。於是 $TS \\cdot TA = TS \\cdot TN = TU \\cdot TG = TV \\cdot TG$，故 $T$ 在兩圓根軸上。證畢。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22031,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle with $AB < AC$. A point $P$ on the circumcircle of $ABC$ (on the same side of $BC$ as $A$) is chosen such that $BP = CP$. Let $BP$ and the angle bisector of $\\angle BAC$ intersect at $Q$, and let the line through $Q$ and parallel to $BC$ intersect $AC$ at $R$. Prove that $BR = CR$.",
"options": [],
"answer": "See solution",
"solution": "Let $AQ$ extended meet the circumcircle of triangle $ABC$ in $T$, and let the line through $Q$ and $R$ intersect $PC$ in $D$.\n\n\n\nPut $\\angle BAQ = \\angle CAQ = \\alpha$, so that also $\\angle BCT = \\alpha$ and $\\angle TBC = \\alpha$. Put $\\angle BCA = \\theta$, so that also $\\angle BTA = \\theta$. Put $\\angle ACP = \\phi$, so that also $\\angle ABP = \\phi$.\n\nThen $\\angle PBC = \\angle PCB$ (since $BP = CP$) $= \\theta + \\phi$. From $\\angle PBT + \\angle TCP = 180^{\\circ}$ (since $BPCT$ is a cyclic quadrilateral), we therefore have $2(\\alpha + \\theta + \\phi) = 180^{\\circ}$, giving $\\alpha + \\theta + \\phi = 90^{\\circ}$.\n\nSince triangle $QPD$ is isosceles ($QD \\parallel BC$, so that $\\angle PQD = \\angle PBC = \\angle PCB = \\angle PDQ$), we have $PQ = PD$, forcing $BQ = CD$. Furthermore, chords $BT$ and $CT$ both subtend the same angle $\\alpha$ at $A$, hence are equal in length. It follows that $\\triangle TBQ \\equiv \\triangle TCD$, hence $\\angle DTC = \\angle QTB = \\theta$.\n\nBut since $\\angle DRC = \\angle BCR = \\theta$ (recall that $QD \\parallel BC$), we see that $RDCT$ is a cyclic quadrilateral. This implies that $\\angle TRD = 90^{\\circ}$ (because $\\angle TCD = \\alpha + \\theta + \\phi = 90^{\\circ}$). So $\\angle TRQ = 90^{\\circ}$, and we also have that $RQBT$ is a cyclic quadrilateral (recall that $\\angle QBT = 90^{\\circ}$).\n\nIt follows that $\\angle QRB = \\angle QTB = \\theta$, and we thus have $\\angle RBC = \\theta$, since $QD \\parallel BC$. Finally, $BR = RC$ follows from the fact that $\\angle RBC = \\angle RCB = \\theta$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22032,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_n > 1$ for all $n$, and $2(a_{n+1} - 1) = a_n^2 - 1 = (a_n - 1)(a_n + 1)$. Define $s_n = \\sum_{k=1}^{n} \\frac{1}{1+a_k}$ for $n \\ge 1$.\n\nProve by induction that\n$$\ns_n = \\frac{1}{2} - \\frac{2}{a_n^2 - 1}, \\quad n = 1, 2, \\dots\n$$\nwhere $a_1 = 3$.",
"options": [],
"answer": "See solution",
"solution": "Note that $a_n > 1$ for all $n$, and $2(a_{n+1} - 1) = a_n^2 - 1 = (a_n - 1)(a_n + 1)$, hence\n$$\n\\frac{1}{a_{n+1}-1} = \\frac{2}{a_n^2-1} > 0 \\quad \\text{for } n=1,2,\\dots\n$$\n\nBecause $a_1 = 3$ we easily get the start of induction:\n$$\ns_1 = \\frac{1}{1+a_1} = \\frac{1}{4} = \\frac{1}{2} - \\frac{2}{8} = \\frac{1}{2} - \\frac{2}{a_1^2 - 1}\n$$\n\nAssuming the inductive hypothesis,\n$$\n\\begin{aligned}\ns_{n+1} &= s_n + \\frac{1}{1+a_{n+1}} = \\frac{1}{2} - \\frac{2}{a_n^2-1} + \\frac{1}{1+a_{n+1}} \\\\\n&= \\frac{1}{2} - \\frac{1}{a_{n+1}-1} + \\frac{1}{1+a_{n+1}} = \\frac{1}{2} - \\frac{2}{a_{n+1}^2-1} < \\frac{1}{2}.\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22033,
"subject": "Mathematics (Olympiad)",
"question": "For a positive integer $k > 1$, let $f(k)$ be the number of ways of factoring $k$ into a product of positive integers greater than 1. (The order of factors is not counted; for example, $f(12) = 4$, as $12$ can be factored in these four 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 $f(n) \\leq \\frac{n}{p}$.",
"options": [],
"answer": "See solution",
"solution": "Let $P(n)$ 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 a prime $p \\mid n$, we have $f(n) \\leq \\sum_{d \\mid \\frac{n}{p}} f(d)$.\n\n*Proof of Lemma 1.* For convenience, call a valid factoring simply a factoring.\n\nFor any factoring of $n$, write $n = n_1 n_2 \\cdots n_k$. Since $p \\mid n$, there exists $i \\in \\{1, \\dots, k\\}$ such that $p \\mid n_i$ (if there is more than one 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 $n = n'_1 n'_2 \\cdots n'_l$ (where $p$ divides $n_1$ and $n'_1$):\n- If $n_1 = n'_1$, then $d = n_2 \\cdots n_k$ and $d = n'_2 \\cdots n'_l$ are two different factorizations of $d$ (where $d$ is a divisor of $\\frac{n}{p}$).\n- If $n_1 \\neq n'_1$, then $d = \\frac{n}{n_1} \\neq \\frac{n}{n'_1} = d'$, so these two factorizations map to factorizations of $d$ and $d'$, respectively (with $d, d'$ divisors of $\\frac{n}{p}$).\n\nThus, $f(n) \\leq \\sum_{d \\mid \\frac{n}{p}} f(d)$. 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 $g(n) \\leq n$.\n\n*Proof of Lemma 2.* Induct on the number of distinct prime divisors of $n$.\n\n- If $n = 1$, then $g(1) = 1$.\n- If $n = p^a$ for a prime $p$, then\n $$\ng(n) = 1 + 1 + p + \\cdots + p^{a-1} = 1 + \\frac{p^a - 1}{p-1} \\leq 1 + p^a - 1 = n.\n $$\n\nAssume the result holds when $n$ has $k$ distinct prime divisors. 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\n$$\ng(n) = g(m) + \\sum_{d \\mid m} \\sum_{i=1}^{a_{k+1}} \\frac{d p_{k+1}^i}{p_{k+1}} = g(m) + \\sigma(m) \\frac{p_{k+1}^{a_{k+1}} - 1}{p_{k+1} - 1},\n$$\n\nwhere $\\sigma(m)$ is the sum of positive divisors of $m$. By induction, $g(m) \\leq m$. Since\n\n$$\n\\begin{aligned}\n\\sigma(m) \\frac{p_{k+1}^{a_{k+1}} - 1}{p_{k+1} - 1} &= \\left( \\prod_{i=1}^{k} \\frac{p_i^{a_i+1} - 1}{p_i - 1} \\right) \\frac{p_{k+1}^{a_{k+1}} - 1}{p_{k+1} - 1} \\\\\n&\\leq \\left( \\prod_{i=1}^{k} \\frac{p_i^{a_i+1} - 1}{p_{i+1} - 1} \\right) (p_{k+1}^{a_{k+1}} - 1)\n\\end{aligned}\n$$\n\n[The proof continues, but the main structure and lemmas are established to show $f(n) \\leq \\frac{n}{p}$.]",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22034,
"subject": "Mathematics (Olympiad)",
"question": "Determine all positive integers $a, b, c, d, e, f$ satisfying the following condition: for any two of them, $x$ and $y$, two of the remaining four numbers, $z$ and $t$, exist such that $\\frac{x}{y} = \\frac{z}{t}$.",
"options": [],
"answer": "See solution",
"solution": "For any positive integer $a$, the 6-tuple $(a, a, a, a, a, a)$ satisfies the required condition. We are looking for solutions $(a, b, c, d, e, f)$ for which not all the components are equal. Suppose $a \\leq b \\leq c \\leq d \\leq e \\leq f$. Obviously, we have $a < f$.\n\nIf $x = a$ and $y = f$, for all $z, t \\in \\{b, c, d, e\\}$ we have $\\frac{a}{f} \\leq \\frac{z}{t}$, where the equality holds for $z = a$ and $t = f$. Since $z \\in \\{b, c, d, e\\}$, it follows that $z \\geq b$, hence $a \\geq b$. Therefore, $a = b$. Similarly, we infer that $e = f$.\n\nNow choose $x = c$, $y = d$. Then $\\frac{c}{d} = \\frac{z}{t}$, where $z, t \\in \\{a, f\\}$. But $c \\leq d$, hence $c = d$, or $\\frac{c}{d} = \\frac{a}{f}$.\n\nIf $c = d$, the 6-tuple $(a, a, c, c, f, f)$ is obviously a solution. Indeed, if $x \\neq y$, we choose $z = x$ and $t = y$, and if $x = y$, we choose $z = t \\neq x$.\n\nIf $c \\neq d$, we have $\\frac{c}{d} = \\frac{a}{f}$, thus $d = \\frac{c f}{a} \\geq f$. It follows that $a = c$ and $d = f$. Similarly, it is easy to prove that the 6-tuple $(a, a, a, f, f, f)$ is a solution.\n\nThus, the 6-tuples which satisfy the required condition are $(a, a, b, b, c, c)$, with $a \\leq b \\leq c$, $(a, a, a, b, b, b)$, with $a \\leq b$, and all their permutations.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22035,
"subject": "Mathematics (Olympiad)",
"question": "Define\n\n$$\nS(n, r) := \\frac{n+1-2r}{n+1-r} \\binom{n}{r},\n$$\n\nfor all pairs $r, n$ with $-1 \\le r$ and $2r \\le n+1$. In particular, $S(n, r) = 0$ if $r = -1$ or if $2r = n+1$. Note that $S(1, 0) = 1$.\n\nProve that if $n \\ge 2$ and $0 \\le r \\le n/2$, then\n\n$$\nS(n, r) = S(n-1, r-1) + S(n-1, r).\n$$\n\nFurther, show that\n\n$$\n\\sum_{r=0}^{\\lfloor n/2 \\rfloor} S(n, r) < 2^{n-2},\n$$\n\nfor $n \\ge 9$.",
"options": [],
"answer": "See solution",
"solution": "We prove the recurrence by direct computation:\n\nFor $n \\ge 2$ and $0 \\le r \\le n/2$,\n\n$$\n\\begin{align*}\nS(n-1, r-1) + S(n-1, r)\n&= \\frac{n+2-2r}{n+1-r} \\frac{(n-1)!}{(r-1)!(n-r)!} + \\frac{n-2r}{n-r} \\frac{(n-1)!}{r!(n-1-r)!} \\\\\n&= \\frac{[r(n+2-2r) + (n-2r)(n+1-r)](n-1)!}{(n+1-r)r!(n-r)!} \\\\\n&= \\frac{[(n+1-2r)n](n-1)!}{(n+1-r)r!(n-r)!} \\\\\n&= S(n, r)\n\\end{align*}\n$$\n\nFor the sum, the base case $n=9$ gives $S(9, 0) = 1$, $S(9, 1) = 8$, $S(9, 2) = 27$, $S(9, 3) = 48$, $S(9, 4) = 42$, summing to $126 < 2^{9-2} = 128$.\n\nFor the inductive step:\n\n$$\n\\sum_{r=0}^{\\lfloor n/2 \\rfloor} S(n, r) = \\sum_{r=0}^{\\lfloor n/2 \\rfloor} [S(n-1, r) + S(n-1, r-1)] \\le 2 \\sum_{r=0}^{\\lfloor (n-1)/2 \\rfloor} S(n-1, r) < 2 \\times 2^{\\lfloor (n-1)/2 \\rfloor} = 2^{n-2}.\n$$\n\nEach $S(n-1, r-1)$ with $r < n/2$ contributes twice to the sum, and when $r = n/2$, $S(n-1, n/2-1)$ contributes only once.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22036,
"subject": "Mathematics (Olympiad)",
"question": "Consider two positive integers $a$ and $b$ such that $a + 2b = 2020$.\n\n(a) Determine the largest possible value of the greatest common divisor of $a$ and $b$.\n\n(b) Determine the smallest possible value of the least common multiple of $a$ and $b$.",
"options": [],
"answer": "See solution",
"solution": "We observe that $2020 = 2^2 \\cdot 5 \\cdot 101$. Whenever $a + 2b = 2020$, $d = \\gcd(a, b)$ must divide $2020$. To find the largest possible value of $d$, we look at the large divisors of $2020$.\n\n$d = 2020$ is not possible, since $a + 2b \\geq 2020 + 2 \\cdot 2020 > 2020$. The next largest divisor is $1010$. Since $3 \\cdot 1010 > 2020$, $d = 1010$ is also not possible. Next is $505$, and here $a = 1010$ and $b = 505$ work, since $\\gcd(1010, 505) = 505$, and $1010 + 2 \\cdot 505 = 2020$. Thus, the largest possible value for $\\gcd(a, b)$ is $505$.\n\nFor part (b), consider two cases for pairs $(a, b)$ such that $a + 2b = 2020$:\n\n1. $a \\mid b$: Let $b = k a$ for some integer $k \\geq 1$. Then $a + 2ka = 2020$, so $a = \\frac{2020}{1 + 2k}$. The least common multiple is $\\operatorname{lcm}(a, b) = k a = \\frac{2020 k}{1 + 2k}$. To minimize $\\operatorname{lcm}(a, b)$, $k$ should be minimized. Since $1 + 2k \\mid 2020$, the smallest possible $k$ is $2$ (since $3 \\nmid 2020$). Thus, $a + 4a = 5a = 2020$, so $a = 404$ and $b = 808$, giving $\\operatorname{lcm}(404, 808) = 808$.\n\n2. $b \\mid a$: Let $a = k b$ for some integer $k \\geq 1$. Then $k b + 2b = 2020$, so $b = \\frac{2020}{k + 2}$. The least common multiple is $\\operatorname{lcm}(a, b) = k b = \\frac{2020 k}{k + 2}$. Again, $k$ should be minimized. Since $k + 2 \\mid 2020$, the smallest possible $k$ is $2$ (since $3 \\nmid 2020$). Thus, $2b + 2b = 4b = 2020$, so $b = 505$ and $a = 1010$, giving $\\operatorname{lcm}(1010, 505) = 1010$.\n\nWe may assume that either $a \\mid b$ or $b \\mid a$ gives the minimal least common multiple. For any $a, b$ with $a + 2b = 2020$ and $\\gcd(a, b) = d$, write $a = \\hat{a} d$ and $b = \\hat{b} d$ with $\\gcd(\\hat{a}, \\hat{b}) = 1$. Then $\\operatorname{lcm}(a, b) = d \\hat{a} \\hat{b}$. The minimum occurs when either $\\hat{a} = 1$ or $\\hat{b} = 1$, i.e., $a \\mid b$ or $b \\mid a$.\n\nTherefore, the smallest possible least common multiple of $a$ and $b$ with $a + 2b = 2020$ is $\\operatorname{lcm}(404, 808) = 808$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22037,
"subject": "Mathematics (Olympiad)",
"question": "Suppose Ben, Aisha, and Cheng compete weekly, with points awarded for first, second, and third place. Ben's total score after 5 weeks is 34, Aisha's is 12, and Cheng's is 9. The total points awarded each week is 11. \n\n(a) What are the possible point distributions for first, second, and third place each week? \n\n(b) Given that Aisha won in the first week, what were the weekly scores for each competitor? \n\n(c) What is the minimum number of additional weeks required for all three to have equal total scores, and what must the weekly results be?",
"options": [],
"answer": "See solution",
"solution": "**(a) & (b)**\n\nThe only possible weekly point distribution is 8 for first, 2 for second, and 1 for third place. Since Aisha won in the first week, her scores were, in order: $8, 1, 1, 1, 1$ (total $12$). Ben's scores were $2, 8, 8, 8, 8$ (total $34$), and Cheng's were $1, 2, 2, 2, 2$ (total $9$).\n\n**(c)**\n\nTo have equal total scores, the combined scores must be divisible by 3. Since $11$ is not divisible by 3, the total number of weeks must be a multiple of 3. After 12 weeks, each competitor can have $44$ points. One possible allocation is:\n\nAisha: $8, 1, 1, 1, 1, 8, 8, 8, 2, 2, 2, 2$\n\nBen: $2, 8, 8, 8, 8, 2, 2, 2, 1, 1, 1, 1$\n\nCheng: $1, 2, 2, 2, 2, 1, 1, 1, 8, 8, 8, 8$\n\nThus, the minimum number of additional weeks required is $7$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22038,
"subject": "Mathematics (Olympiad)",
"question": "Find positive integers $a$ and $b$ such that\n$$\n\\frac{2^a}{23^b} = \\frac{2^{24} + 3 \\times 2^{21}}{2024}.\n$$\nWhat is $a + b$?",
"options": [],
"answer": "See solution",
"solution": "We have\n\n$$\n\\frac{2^a}{23^b} = \\frac{2^{24} + 3 \\times 2^{21}}{2024} = \\frac{2^{21}(2^3 + 3)}{2^3 \\times 11 \\times 23} = \\frac{2^{21} \\times 11}{2^3 \\times 11 \\times 23} = \\frac{2^{18}}{23}.\n$$\n\nHence $23 \\times 2^a = 23^b \\times 2^{18}$. Since 2 and 23 are primes and $a$ and $b$ are positive integers, $b = 1$ and $a = 18$. Therefore $a + b = 19$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22039,
"subject": "Mathematics (Olympiad)",
"question": "En un torneo participan $p^2$ jugadores. En cada partida juegan $p$ jugadores. El torneo se divide en rondas, y las rondas se dividen en partidas. Cada jugador juega una, o ninguna, partida en cada ronda. Al final del torneo, cada jugador se ha enfrentado exactamente una vez con cada uno de los otros jugadores.\n\n¿Es posible diseñar un torneo así? En caso afirmativo, determina el mínimo número de rondas que puede tener el torneo.",
"options": [],
"answer": "See solution",
"solution": "El número de partidas que disputa cada jugador es\n\n$$\n\\frac{\\text{número de jugadores a los que se enfrenta}}{\\text{número de jugadores a los que se enfrenta en cada partida}} = \\frac{p^2 - 1}{p - 1}.\n$$\n\nEs decir, cada jugador juega $p+1$ partidas. Por tanto, el número de rondas es al menos $p+1$, y es exactamente $p+1$ cuando todos los jugadores juegan en todas las rondas. En ese caso, en cada ronda se disputan $\\frac{p^2}{p} = p$ partidas.\n\nVamos a probar que es posible organizar un torneo con $p+1$ rondas. Representamos a cada jugador como un número de dos cifras escrito en base $p$, es decir, cada jugador se escribe como $C_iC_d$, donde $C_i$ (cifra izquierda) y $C_d$ (cifra derecha) toman valores enteros entre $0$ y $p-1$.\n\nLa planificación es la siguiente:\n\n- **Ronda 0:** Agrupamos los jugadores en los que $C_i$ coincide.\n- **Ronda 1:** Agrupamos los jugadores en los que $C_i + C_d$ tiene el mismo resto al dividir por $p$.\n- ...\n- **Ronda $k$:** Agrupamos los jugadores en los que $C_i + k \\cdot C_d$ tiene el mismo resto al dividir por $p$.\n- ...\n- **Ronda $p-1$:** Agrupamos los jugadores en los que $C_i + (p-1) \\cdot C_d$ tiene el mismo resto al dividir por $p$.\n- **Ronda $p$:** Agrupamos los jugadores en los que $C_d$ coincide.\n\nEn los argumentos que siguen, diremos que $D \\equiv E$ si ambos números tienen el mismo resto al dividir por $p$.\n\nLa planificación propuesta agrupa a los jugadores en conjuntos de $p$ elementos. En las rondas 0 y $p$ eso es claro. Consideremos una ronda $k$ con $0 < k < p$. Para cada $C_i$ fijado, al mover $C_d$, el resto de $C_i + k \\cdot C_d$ es siempre distinto (si no lo fuera, tendríamos dos jugadores $C_iC_d$ y $C_iC'_d$ para los que $C_i + k \\cdot C_d \\equiv C_i + k \\cdot C'_d$, y restando llegaríamos a que $k(C'_d - C_d)$ es múltiplo de $p$, sin que ni $k$ ni $C'_d - C_d$ lo sean). Por tanto, obtenemos una vez, y sólo una, todos los posibles restos. Al variar $C_i$ obtenemos $p$ veces cada uno de los restos.\n\nSupongamos que dos jugadores, $C_iC_d$ y $C'_iC'_d$, se enfrentan dos veces.\n\nSi lo hacen en rondas $0 \\le j < k \\le p-1$, tenemos que $C_i + k \\cdot C_d \\equiv C'_i + k \\cdot C'_d$ y $C_i + j \\cdot C_d \\equiv C'_i + j \\cdot C'_d$. Restando, $(k-j) \\cdot C_d \\equiv (k-j) \\cdot C'_d$. Como $k-j$ es primo con $p$, $C_d \\equiv C'_d$, luego $C_d = C'_d$. Llevando esta igualdad a nuestra hipótesis y restando, obtenemos fácilmente $C_i \\equiv C'_i$, luego $C_i = C'_i$.\n\nSi los jugadores se enfrentan en rondas $k$ y $p$ con $0 \\le k < p$, obtenemos directamente que $C_d = C'_d$, y repetimos el argumento anterior.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22040,
"subject": "Mathematics (Olympiad)",
"question": "死靈法師和聖騎士在一個 $666 \\times 666$ 的方陣上大戰。方陣上原本空無一物。每一回合，死靈法師先在方陣上指定一格，並在以該格為中心的 $3 \\times 3$ 範圍內的所有格子（至多九格）各增加一隻骷髏，然後聖騎士指定方陣上任意四格，並從這四格中各消滅一隻骷髏（格子中的骷髏數是非負整數）。\n\n如果在某回合，一個格子內有 $10^6$ 或更多骷髏，則稱這個格子在該回合處於完蛋狀態。試求最大的正整數 $K$，使得死靈法師有策略能保證，不論聖騎士如何行動，都能在某個有限回合內，讓方陣上同時存在至少 $K$ 個完蛋的格子。",
"options": [],
"answer": "See solution",
"solution": "最大的 $K = 5 \\times 222^2$；一般性的，對於 $3N \\times 3N$ 的方陣，$K = 5N^2$。\n\n\n\n首先證明 $K \\le 5N^2$。如上圖將方陣著色，並注意到每回合死靈法師只能在至多四個白色格子上增加骷髏，因此聖騎士永遠可以讓白色區域的骷髏數為 0，從而死靈法師有可能保證製造出完蛋的格子，至多只能是灰色格子數，也就是 $5N^2$。\n\n接著證明 $K \\ge 5N^2$。事實上，我們可以考慮以下的加強版遊戲：當輪到聖騎士時，他改為選擇死靈法師剛剛增加骷髏的 $3 \\times 3$ 區域中的五個格子，並將全方陣上除了這五個格子以外的所有格子都減少一隻骷髏。顯然，如果在加強版遊戲中，死靈法師可以保證 $K \\ge 5N^2$，則在原版遊戲中也可以。注意到，對於一個 $3 \\times 3$ 區域，聖騎士選擇其中五個格子的方式共有 $M = \\binom{9}{5}$ 種。同時注意到，雙方行動加總，等於讓聖騎士選擇的那五格骷髏數加 $1$，$3 \\times 3$ 區域中剩餘四格的骷髏數不變，其餘所有格子的骷髏數減 $1$。\n\n考慮死靈法師對著同一個 $3 \\times 3$ 區域連續增加 $M\\ell$ 次。由鴿籠原理，$3 \\times 3$ 中選五格的全部 $M$ 種選項中，必有一種被聖騎士選了至少 $\\ell$ 次。由前段討論，這表示這個 $3 \\times 3$ 區域中，有至少五格的骷髏數量 $\\ge \\ell$。\n\n現在，將方陣切割為不相交的 $N^2$ 個 $3 \\times 3$ 區域，並編號 $0, 1, \\dots, N^2 - 1$。接著，從最大號碼的區域開始，我們依次執行以下動作：對於編號 $b$ 的區域，讓死靈法師連續選該區域 $C = 10^6 M (M+1)^b$ 次。由前述討論，這 $C$ 個回合會保證該區域有五格的骷髏數量增加至少 $10^6 (M+1)^b$，而之後的每個回合會讓其數量減 $1$。然而，在執行完編號 $b$ 的 $C$ 回合後，我們僅會對編號 $b-1$ 到 $0$ 的區域進行動作，而這些動作的總回合數為 $10^6 M \\{(M+1)^{b-1} + (M+1)^{b-2} + \\dots + 1\\} = 10^6 \\{(M+1)^b - 1\\}$，因此編號 $b$ 的區域至少有五格的骷髏數量不小於 $10^6 (M+1)^b - 10^6 \\{(M+1)^b - 1\\} = 10^6$。以上推論對於所有 $b$ 都成立，因此有 $5N^2$ 格完蛋。證畢。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22041,
"subject": "Mathematics (Olympiad)",
"question": "Let the incircle of a scalene triangle $ABC$ touch the sides $BC$, $CA$, $AB$ at the points $A'$, $B'$, $C'$, respectively. Let $D$ be the midpoint of the arc $B'C'$ that does not contain $A'$ (so $D$ lies on the opposite side of $A'$). If the common tangent of the incircles of triangles $A'CB'$ and $A'C'B$ (opposite to the common tangent $BC$ with respect to the line joining their centers) intersects the lines $A'D$ and $A'C'$ at the points $E$ and $F$ respectively, show that $C'EF$ is an isosceles triangle.",
"options": [],
"answer": "See solution",
"solution": "Since $D$ is the midpoint of the arc $B'C'$, and $AC$ touches the incircle at $B'$, we see that\n\n$$\n\\therefore \\angle AB'D = \\angle BA'D = \\angle DA'C' = \\angle DB'C'\n$$\n\nThus, $B'D$ bisects $\\angle AB'C'$, and similarly, $DC'$ bisects $\\angle B'C'A$. Hence $D$ is the incenter of $\\triangle AB'C'$. By the same arguments, we can conclude that the incenters of $A'CB'$ and $A'BC'$, denoted by $O_1$ and $O_2$, are the midpoints of the arcs $A'B'$ and $A'C'$, respectively.\n\nSince $O_1$, $O_2$, $D$ are the midpoints of the arcs $A'B'$, $A'C'$, $B'C'$, then $C'O_1$, $A'D$, and $B'O_2$ intersect at the incenter of $\\triangle A'B'C'$.\n\n$$\n\\therefore \\angle A'DO_1 + \\angle DO_1C' + \\angle C_1O_1O_2 = \\frac{1}{2}(\\angle A'C'B' + \\angle B'A'C' + \\angle A'B'C') = \\frac{\\pi}{2}\n$$\n\nThus, $A'D \\perp O_1O_2$. Since $A'$ lies on $BC$, the common tangent of the circles $O_1$ and $O_2$, the reflection of $A'$ with respect to the line $O_1O_2$ is $E$. By the reflection, $\\angle EO_2O_1 = \\angle A'O_2O_1 = \\angle O_1O_2B'$. Since $E$ and $B'$ lie on the same side with respect to $O_1O_2$, $E$ must lie on $B'O_2$. Therefore, $E$ is the incenter of $\\triangle A'B'C'$.\n\nNow, it is left to show that the common tangent is parallel to $B'C'$; this will imply that $\\angle FEC' = \\angle B'C'E = \\angle FC'E$. This follows from the reflection of the tangent $BC$ with respect to $O_1O_2$ that,\n\n$$\n\\angle FEA' = \\angle EA'B = \\angle A'B'D = \\angle A'B'C' + \\angle C'B'D = \\angle A'DC' + \\angle BC'D = \\angle EGC'\n$$\n\nwhere $G$ is the intersection of $A'D$ and $B'C'$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22042,
"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": "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": 22043,
"subject": "Mathematics (Olympiad)",
"question": "Let $m$ be a positive integer. Prove that there exists a positive integer $N$ such that for every integer base $b$ with $2 \\leq b \\leq 1389$, the sum of the digits of $N$ in base $b$ is at least $m+1$.",
"options": [],
"answer": "See solution",
"solution": "We claim that $N = (1389!)^{m+1} - 1$ has the desired properties.\n\nLet $2 \\leq b \\leq 1389$. Since $1389!$ is divisible by every $b$ in this range, $(1389!)^{m+1}$ is divisible by $b^{m+1}$. Thus, $(1389!)^{m+1}$ ends with at least $m+1$ zeros in base $b$, so $(1389!)^{m+1} - 1$ ends with $m+1$ digits equal to $b-1$ in base $b$. Therefore, the sum of the digits of $N$ in base $b$ is at least $(m+1)(b-1) \\geq m+1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22044,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs of positive odd integers $(m, n)$ such that:\n\n$$\n\\begin{cases}\n n \\mid 3m + 1 \\\\\n m \\mid n^2 + 3\n\\end{cases}\n$$",
"options": [],
"answer": "See solution",
"solution": "Let us analyze the given conditions:\n\n1. $n \\mid 3m + 1$\n2. $m \\mid n^2 + 3$\n\nFirst, note that $3$ and $n$ must be coprime, i.e., $(3, n) = 1$.\n\nLet $n \\leq 9$. Since $(3, n) = 1$, possible values for $n$ are $1, 5, 7$.\n\n- If $n = 1$, then $m \\mid 4$. Since $m$ is odd, $m = 1$. The pair $(m, n) = (1, 1)$ satisfies both conditions.\n- If $n = 5$, then $5 \\mid 3m + 1$ and $m \\mid 28$. Since $m$ is odd, $m = 1$ or $7$, but neither satisfies $5 \\mid 3m + 1$.\n- If $n = 7$, then $7 \\mid 3m + 1$ and $m \\mid 52$. Since $m$ is odd, $m = 1$ or $13$, but neither satisfies $7 \\mid 3m + 1$.\n\nNow consider $n > 9$. From the conditions, there exist positive integers $p, q$ such that $np = 3m + 1$ and $mq = n^2 + 3$.\n\nWe claim $m \\geq n + 1$. Suppose $m \\leq n$. Then $4n > 3m + 1$, so $p$ can only be $1, 2,$ or $3$.\n- If $p = 3$, then $3 \\mid 1$, which is impossible.\n- If $p = 1$, then $np \\equiv 1 \\pmod{2}$, but $3m + 1$ is even, so contradiction.\n- Thus, $p = 2$.\n\nNow, $m \\mid n^2 + 3 \\implies m \\mid 4n^2 + 12 = (2n)^2 + 12 = (3m + 1)^2 + 12 \\implies m \\mid 13$.\nSo $m = 1$ or $13$. But $n = \\frac{3m + 1}{2}$, so $n$ is even in both cases, contradiction.\n\nThus, $m \\geq n + 1$.\n\nNow, $n(n + 1) > n^2 + 3 = mq \\geq (n + 1)q \\implies q < n$.\nFrom (1), $n \\mid 3m + 1 \\implies n \\mid 3mq + q = 3(n^2 + 3) + q \\implies n \\mid q + 9$.\nThus, $n \\leq q + 9$.\nSince $q < n$ and $9 < n$, we get $n \\leq q + 9 < 2n$, so $q + 9 = n$.\n\nTherefore, $n^2 + 3 = m q = m(n - 9)$, so $m(n - 9) - n^2 - 3 = 0 \\implies (m - n - 9)(n - 9) = 84 = 3 \\cdot 4 \\cdot 7$.\n\nSince $m - n - 9$ is odd and $n - 9$ is even, $4 \\mid n - 9$. Also, $(n - 9, 3) = 1$.\n\nThus, possible cases:\n- $m - n - 9 = 3$, $n - 9 = 28$ ($4 \\cdot 7$): $(m, n) = (49, 37)$\n- $m - n - 9 = 21$ ($3 \\cdot 7$), $n - 9 = 4$: $(m, n) = (43, 13)$\n\nThese pairs satisfy the original conditions.\n\n**Final answer:**\n\nThe solutions are $(m, n) = (1, 1)$, $(49, 37)$, and $(43, 13)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22045,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge 2$ be an integer. A Welsh darts board is a disc divided into $2n$ equal sectors, half of them being red and the other half being white. Two Welsh darts boards are matched if they have the same radius and they are superimposed so that each sector of the first board comes exactly over a sector of the second board.\n\nSuppose that two given Welsh darts boards can be matched so that more than half of the pairs of superimposed sectors have different colors. Prove that these Welsh darts boards can be matched so that at least $2\\left\\lfloor\\frac{n}{2}\\right\\rfloor + 2$ pairs of superimposed sectors have the same color.",
"options": [],
"answer": "See solution",
"solution": "For any two Welsh darts boards that can be matched, if two of their superimposed sectors have the same color, we call it a *concordance*, and if two of their superimposed sectors have different colors, we call it a *non-concordance*.\n\nOn each of the two Welsh darts boards, write $+1$ on every red sector and $-1$ on every white sector. At any matching, the product of the numbers in overlapping sectors is $+1$ for a concordance and $-1$ for a non-concordance.\n\nLet $t$ be the number of concordances. The sum $S$ of the products of the numbers on overlapping sectors is the difference between the number of concordances and non-concordances:\n\n$$\nS = t \\cdot 1 + (2n - t) \\cdot (-1) = 2(t - n).\n$$\n\nAt any matching, let $k$ be the number of concordances between white sectors and $j$ the number between red sectors. Then $n-k$ white sectors of the first board overlap $n-k$ red sectors of the second, and $n-j$ red sectors overlap $n-j$ white sectors. Thus, on the second board, there are $k + (n-j)$ white sectors and $j + (n-k)$ red sectors, so $k + (n-j) = j + (n-k) = n$, hence $k = j$. Therefore, $t = k + j$ is even.\n\nLet $u_1, u_2, \\dots, u_{2n} \\in \\{-1, +1\\}$ be the numbers on the first board's sectors and $v_1, v_2, \\dots, v_{2n} \\in \\{-1, +1\\}$ on the second. By rotating the second board, we get all possible matchings, and the sums:\n\n$$\nS_1 = u_1v_1 + u_2v_2 + \\dots + u_{2n}v_{2n},\\\\\nS_2 = u_1v_2 + u_2v_3 + \\dots + u_{2n}v_1,\\\\\n\\vdots\\\\\nS_{2n} = u_1v_{2n} + u_2v_1 + \\dots + u_{2n}v_{2n-1}.\n$$\n\nMoreover,\n\n$$\nS_1 + S_2 + \\dots + S_{2n} = (u_1 + u_2 + \\dots + u_{2n})(v_1 + v_2 + \\dots + v_{2n}) = 0.\n$$\n\nSince initially there were more than $n$ non-concordances, $S_1 < 0$. Therefore, there exists $j$ such that $S_j > 0$. If $t$ is the number of concordances for $S_j$, then $2(t-n) > 0$, so $2(t-n) \\ge 2$. Thus, $t \\ge n+1$, and since $t$ is even, $t \\ge 2\\left\\lfloor\\frac{n}{2}\\right\\rfloor + 2$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22046,
"subject": "Mathematics (Olympiad)",
"question": "In the diagram below, a number is written in each circle. The numbers do not have to be integers or positive. Next to each line segment, the sum of the two numbers in the circles at its ends is written. There are two quadruples of numbers that can be written in the circles such that the numbers next to the line segments are exactly $0$, $1$, $2$, $3$, $4$, and $5$. For both of these quadruples, we multiply the four numbers in the circles together. Which two results can we get from this multiplication?\n\n",
"options": [],
"answer": "See solution",
"solution": "$-6$ and $-\\frac{21}{16}$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22047,
"subject": "Mathematics (Olympiad)",
"question": "Given an acute, scalene triangle $ABC$ with circumcircle $(O)$. The line passing through $O$ and the midpoint $I$ of $BC$ intersects $AB$ and $AC$ at $E$ and $F$, respectively. Let $D$ and $G$ be the reflections of $A$ over $O$ and the circumcenter of triangle $AEF$. Let $K$ be the reflection of $O$ over the circumcenter of triangle $OBC$.\n\na) Prove that $D$, $G$, and $K$ are collinear.\n\nb) Take $M$ on $KB$ and $N$ on $KC$ such that $IM \\perp AC$ and $IN \\perp AB$. The perpendicular bisector of $IK$ intersects $MN$ at $H$. Suppose that $IH$ meets $AB$ and $AC$ at $P$ and $Q$, respectively. Prove that the circumcircle of triangle $APQ$ intersects $(O)$ again at a point on $AI$.",
"options": [],
"answer": "See solution",
"solution": "a) Let $T$ be the projection of $A$ on $GD$. It is well known that $T$ is the second intersection of $(AEF)$ and $(O)$. We observe that $\\triangle TEB \\sim \\triangle TFC$, then\n$$\n\\frac{TB}{TC} = \\frac{BE}{CF}.\n$$\n\nOn the other hand,\n$$\n\\frac{BE}{CF} = \\frac{BE}{IB} \\cdot \\frac{IC}{CF} = \\frac{\\cos ACB}{\\cos ABC} = \\frac{BD}{CD'}\n$$\nwhich implies that $TBDC$ is a harmonic quadrilateral, or $TD$ passes through $K$, which is the intersection of the tangents at $B$ and $C$ of $(O)$.\n\nb) Let $X$ and $Y$ be the intersections of $KB$, $KC$ with $IN$, $IM$, respectively. The tangents at $B$ and $C$ of $(O)$ meet the tangent at $A$ at $X'$ and $Y'$. Because $IX \\parallel OX'$ and $IY \\parallel OY'$, then\n$$\n\\triangle KX'Y' \\sim \\triangle KXY\n$$\nwith $O$ and $I$ corresponding. Note that $O$ is the incenter of $KX'Y'$, so $I$ is the incenter of triangle $KXY$.\n\nLet $(KXY)$ meet $IY$ and $IX$ at $R$ and $S$, respectively. Clearly, $R$ and $S$ are the circumcenters of triangles $KIX$ and $KIY$. Hence, $RS$ is the perpendicular bisector of $IK$, so $H$ lies on $RS$. Applying Pascal's theorem for\n$$\n\\begin{pmatrix} K & X & R \\\\ S & Y & K \\end{pmatrix}\n$$\nwe obtain that the tangent at $K$ of $(KXY)$, $RS$, and $MN$ are concurrent, which means $KH$ is the tangent of $(KXY)$. By angle chasing, we have\n$$\n\\begin{aligned}\n\\angle HIK &= \\angle HKI = \\angle HKB + \\angle BKI \\\\\n&= \\angle KYX + 90^\\circ - \\angle BAC = 270^\\circ - 2\\angle ABC - \\angle BAC \\\\\n&= 90^\\circ + \\angle ACB - \\angle ABC\n\\end{aligned}\n$$\nAlso, $\\angle(AO, BC) = \\angle OAC + \\angle ACB = 90^\\circ - \\angle ABC + \\angle ACB = \\angle HIK$. Note that $IK \\perp BC$, thus $IH \\perp AO$. Hence, $PBQC$ is cyclic or $I$ has the same power to $(ABC)$ and $(APQ)$, which means $AI$ passes through the second intersection of $(APQ)$ and $(O)$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22048,
"subject": "Mathematics (Olympiad)",
"question": "Solve the equation $3^x - 5^y = z^2$ in positive integers.",
"options": [],
"answer": "See solution",
"solution": "Looking modulo 4, $x$ is even. Let $x = 2a$.\n\n$$9^a - 5^y = z^2$$\n\n$$(3^a - z)(3^a + z) = 5^y$$\n\nSo let $3^a - z = 5^b$, $3^a + z = 5^c$. Then $3^a = \\frac{5^b + 5^c}{2}$, which is divisible by 5 unless $b=0$ (since $b \\leq c$). So $2 \\cdot 3^a = 5^c + 1$.\n\nWe find that $c$ is divisible by 3 as follows: From $2 \\cdot 3^a = 5^c + 1$, we conclude that $c$ is odd (modulo 3), so $c = 2m+1$ and we get:\n\n$$2 \\cdot 3^a = (5+1)(5^{2m} - 5^{2m-1} + \\dots + 5^2 - 5 + 1)$$\n\nFrom this, it is obvious that (for $a > 1$, the case $a = 1$ is easy) $3$ divides $5^{2m} - 5^{2m-1} + \\dots + 5^2 - 5 + 1$, and since each of these $2m+1$ summands is congruent to 1 modulo 3, we get $2m+1 \\equiv 0 \\pmod{3}$, which is equivalent to $c = 3d$.\n\nNow, the right-hand side of $2 \\cdot 3^a = 5^c + 1$ is divisible by $5^3 + 1 = 126 = 2 \\cdot 3 \\cdot 3 \\cdot 7$, but the left-hand side isn't divisible by 7, so the conclusion follows.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22049,
"subject": "Mathematics (Olympiad)",
"question": "Denote by $a_{ij}$ the entry that lies in the $i$-th column and the $j$-th row. The given conditions yield that each of $1$, $2$, $3$, and $4$ appear exactly once in each column.\n\nHow many $4 \\times 4$ matrices with entries from $\\{1,2,3,4\\}$ are there such that each number appears exactly once in each column?",
"options": [],
"answer": "See solution",
"solution": "We may assume without loss of generality that $a_{1j} = j$ for $j = 1, 2, 3, 4$. We would multiply the answer by $4!$ afterwards. We may also assume that $a_{22} = a_{33} = 1$. We would multiply the answer by $3 \\times 2$ afterwards.\n\nIf $a_{21}, a_{31} \\neq 4$, then $a_{23} = a_{32} = 4$. We can choose which of $a_{24}$ and $a_{34}$ to be $2$, and the rest entries would be determined uniquely. So we have two ways.\n\nOtherwise, if $a_{21} = 4$, there is only one way to fill the other squares. The case $a_{31} = 4$ is similar. So we have two ways.\n\nTherefore, the answer is $$4! \\times 3 \\times 2 \\times (2+2) = 576.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22050,
"subject": "Mathematics (Olympiad)",
"question": "Given a positive integer $k$, a pair of positive integers $(a, b)$ is said to be $k$-nice if $a \\mid b^k + 1$ and $b \\mid a^k + 1$. For which positive integers $k$ are there infinitely many $k$-nice pairs?",
"options": [],
"answer": "See solution",
"solution": "First, consider the case $k = 1$. A pair $(a, b)$ is 1-nice if $a \\mid b + 1$ and $b \\mid a + 1$. Assume without loss of generality that $a \\leq b$ and set $d = b - a$. Since $b \\mid a + 1$, we have $b \\leq a + 1$, i.e., $d \\leq 1$. Also, $a$ divides $(b - a) + 1 = d + 1$, so $a \\leq 2$, and thus $b \\leq 3$. Therefore, only finitely many pairs are 1-nice.\n\nNow, consider $k \\geq 2$. The pair $(1, 1)$ is $k$-nice. Given a $k$-nice pair $(a, b)$ with $1 \\leq a \\leq b$, let $c = \\frac{b^k + 1}{a}$. By construction, $c$ is a positive integer dividing $b^k + 1$. We must have $\\gcd(a, b) = 1$, since otherwise any common divisor $p > 1$ would divide $b^k + 1$ and $a^k + 1$, leading to $p \\mid 1$, a contradiction.\n\nNow observe:\n\n$$\nc^k + 1 = \\frac{(b^k + 1)^k + a^k}{a^k}\n$$\n\nSince $(b^k + 1)^k + a^k \\equiv 1 + a^k \\equiv 0 \\pmod{b}$ and $\\gcd(a, b) = 1$, it follows that $b \\mid c^k + 1$, so $(b, c)$ is $k$-nice. Moreover, $c \\geq \\frac{b^2 + 1}{b} > b$, so $b + c > a + b$.\n\nTherefore, there is no $k$-nice pair $(a, b)$ with maximal $a + b$, and since $(1, 1)$ is $k$-nice, there are infinitely many $k$-nice pairs for $k \\geq 2$.\n\n*Conclusion*: For all $k \\geq 2$, there exist infinitely many $k$-nice pairs.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22051,
"subject": "Mathematics (Olympiad)",
"question": "平面上給定一個半徑為 $1$ 的圓 $\\omega$。對於一組三角形所成的集合 $T$，如果滿足下列兩條件：\n\n1. $T$ 中的每一個三角形都內接於 $\\omega$；\n2. $T$ 中任兩個三角形都沒有共同內部點；\n\n就說 $T$ 是好棒棒。\n\n試決定所有的正實數 $t$，使得對於每一個正整數 $n$，都可以找到一組由 $n$ 個三角形所組成的好棒棒集合 $T$，其中的每一個三角形的周長都大於 $t$。",
"options": [],
"answer": "See solution",
"solution": "$t \\in (0, 4]$。\n\n首先，我們說明如何構造一組 $n$ 個三角形的好棒棒集合，且每個三角形的周長都大於 $4$。這將證明所有 $t \\le 4$ 都滿足條件。\n\n以歸納法構造一個內接於 $\\omega$ 的 $(n+2)$ 邊形 $BA_1A_2\\dots A_nC$，其中 $BC$ 是直徑，並且 $BA_1A_2, BA_2A_3, \\dots, BA_{n-1}A_n, BA_nC$ 構成 $n$ 個三角形的好棒棒集合。當 $n=1$ 時，取任意內接於 $\\omega$ 且 $BC$ 為直徑的三角形 $BA_1C$，其周長大於 $2 \\cdot BC = 4$。歸納步驟：假設已構造出 $BA_1A_2\\dots A_nC$，由於 $A_nB + A_nC + BC > 4$，可以在小弧 $CA_n$ 上選擇一點 $A_{n+1}$，使得 $A_nB + A_nA_{n+1} + BA_{n+1}$ 仍大於 $4$。因此，新的三角形 $BA_nA_{n+1}$ 和 $BA_{n+1}C$ 的周長都大於 $4$，完成歸納。\n\n接下來證明不存在 $t > 4$ 滿足題意。假設存在一組 $n$ 個三角形的好棒棒集合 $T$，每個三角形的周長都大於 $t$，我們將對 $n$ 作上界。\n\n取 $\\varepsilon > 0$，使 $t = 4 + 2\\varepsilon$。\n\n**斷言：** 存在正數 $\\sigma = \\sigma(\\varepsilon)$，使得任何內接於 $\\omega$、周長 $2s \\ge 4 + 2\\varepsilon$ 的三角形 $\\Delta$，其面積 $S(\\Delta) \\ge \\sigma$。\n\n*證明：* 設 $a, b, c$ 為 $\\Delta$ 的邊長。由於 $\\Delta$ 內接於 $\\omega$，每邊長至多為 $2$，因此 $s - a \\ge (2 + \\varepsilon) - 2 = \\varepsilon$，同理 $s - b \\ge \\varepsilon$，$s - c \\ge \\varepsilon$。由海龍公式，\n$$\nS(\\Delta) = \\sqrt{s(s-a)(s-b)(s-c)} \\ge \\sqrt{(2+\\varepsilon)^3\\varepsilon}\n$$\n因此可取 $\\sigma(\\varepsilon) = \\sqrt{(2+\\varepsilon)^3\\varepsilon}$。$\\square$\n\n因此，所有三角形的總面積 $S \\ge n \\cdot \\sigma(\\varepsilon)$，但 $S$ 不超過圓 $\\omega$ 的面積，即 $n\\sigma(\\varepsilon) \\le \\pi$，所以 $n$ 有上界。\n\n_註：也可用 $S = \\frac{abc}{4R}$ 代替海龍公式證明斷言。_",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22052,
"subject": "Mathematics (Olympiad)",
"question": "已知 $x, y$ 為滿足 $x + y = 1$ 的正實數。試證：\n\n$$\n\\frac{x}{x^2 + y^3} + \\frac{y}{x^3 + y^2} \\le 2 \\left( \\frac{x}{x+y^2} + \\frac{y}{x^2+y} \\right).\n$$\n\nLet $x, y$ be positive real numbers such that $x + y = 1$. Prove that\n\n$$\n\\frac{x}{x^2 + y^3} + \\frac{y}{x^3 + y^2} \\le 2 \\left( \\frac{x}{x+y^2} + \\frac{y}{x^2+y} \\right).\n$$",
"options": [],
"answer": "See solution",
"solution": "解：令 $t = xy$，則\n\n$$\nx^2 + y^2 = 1 - 2t,\n$$\n\n$$\nx^3 + y^3 = 1 - 3t,\n$$\n\n$$\nx^4 + y^4 = 1 - 4t + t^2,\n$$\n\n$$\nx^5 + y^5 = 1 - 5t + 5t^2.\n$$\n\n因為 $x^2 + y = x + y^2$，所以原不等式等價於\n\n$$\n\\begin{aligned}\n& \\frac{x}{x^2 + y^3} + \\frac{y}{x^3 + y^2} \\le \\frac{2}{x+y^2} \\\\\n\\Leftrightarrow & \\frac{x^4 + y^4 + xy}{(x^2 + y^3)(x^3 + y^2)} \\le \\frac{4}{x+y+x^2+y^2} \\\\\n\\Leftrightarrow & (1+x^2+y^2)(x^4+y^4+xy) \\le 4(x^2+y^3)(x^3+y^2) \\\\\n\\Leftrightarrow & (2-2t)(1-3t+2t^2) \\le 4(1-5t+6t^2+t^3) \\\\\n\\Leftrightarrow & (4t-1)(t^2+2t-1) \\ge 0 \\\\\n\\Leftrightarrow & \\left(t-\\frac{1}{4}\\right)\\left(t-(\\sqrt{2}-1)\\right)\\left(t+\\sqrt{2}+1\\right) \\ge 0.\n\\end{aligned}\n$$\n\n由\n\n$$\n0 < t \\le \\left(\\frac{x+y}{2}\\right)^2 = \\frac{1}{4}\n$$\n\n知，上式中之最後的不等式成立。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22053,
"subject": "Mathematics (Olympiad)",
"question": "Suppose the product $N$ of all the positive factors of a positive integer $n$ equals $24^{240}$. Given that the prime factors of $24^{240}$ are $2$ and $3$, $n$ must be of the form $2^a 3^b$ for some non-negative integers $a$ and $b$. Find $n$.",
"options": [],
"answer": "See solution",
"solution": "Let $n = 2^a 3^b$. The number of positive factors of $n$ is $(a+1)(b+1)$. The product $N$ of all positive factors of $n$ is $n^{(a+1)(b+1)/2}$. Setting this equal to $24^{240} = 2^{720}3^{240}$, we have:\n\n$$\n(2^a 3^b)^{\\frac{(a+1)(b+1)}{2}} = 2^{720}3^{240}\n$$\n\nEquating exponents:\n\n$$\n\\frac{a(a+1)(b+1)}{2} = 720 \\\\\n\\frac{b(b+1)(a+1)}{2} = 240\n$$\n\nDividing the first by the second gives $a = 3b$. Substitute into the second equation:\n\n$$(3b+1)b(b+1) = 480$$\n\nSolving, $b = 5$ and $a = 15$. Thus, $n = 2^{15}3^5$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22054,
"subject": "Mathematics (Olympiad)",
"question": "Suppose the altitudes of a triangle are all integers whose sum does not exceed 20, and its inradius is also an integer. Find all possible values for the area of the triangle.",
"options": [],
"answer": "See solution",
"solution": "We begin with the standard expression:\n\n$$\n\\frac{1}{r} = \\frac{1}{h_a} + \\frac{1}{h_b} + \\frac{1}{h_c},\n$$\n\nwhere $r$ is the inradius and $h_a, h_b, h_c$ are the altitudes of the triangle.\n\nThe arithmetic mean-harmonic mean inequality gives\n\n$$\n\\frac{1}{r} \\geq \\frac{9}{h_a + h_b + h_c} \\geq \\frac{9}{20}.\n$$\n\nThus $r \\leq \\frac{20}{9}$, so $r = 1$ or $2$ (since $r$ is an integer).\n\n**Case 1:** $r = 1$\n\nAssume $h_a \\leq h_b \\leq h_c$. Then $3h_a \\leq h_a + h_b + h_c \\leq 20$, so $h_a \\leq 6$. We can rule out $h_a = 1$ and $h_a = 2$ (since $1/h_a = 1/h_b + 1/h_c$ is impossible for integers $h_b, h_c \\geq h_a$).\n\nIf $h_a = 3$:\n\n$$\n\\frac{1}{h_b} + \\frac{1}{h_c} = 1 - \\frac{1}{3} = \\frac{2}{3}\n$$\n\nThis leads to $(2h_b - 3)(2h_c - 3) = 9$, giving $(h_b, h_c) = (3, 3)$. So $(h_a, h_b, h_c) = (3, 3, 3)$.\n\nFor $h_a = 4, 5, 6$, no integer solutions exist.\n\n**Case 2:** $r = 2$\n\nHere $3 \\leq h_a \\leq 6$. For $h_a = 5$, we get $(h_a, h_b, h_c) = (5, 5, 10)$. For $h_a = 6$, we get $(6, 6, 6)$.\n\nIn the cases $(3, 3, 3)$ and $(6, 6, 6)$, the triangle is equilateral. The area is $\\Delta = \\frac{h^2}{\\sqrt{3}}$, so the areas are $3\\sqrt{3}$ and $12\\sqrt{3}$, respectively.\n\nFor $(5, 5, 10)$, $a = b = 2c$. Using Heron's formula, $\\Delta = \\frac{100}{\\sqrt{15}}$.\n\n**Thus, the possible values for the area $\\Delta$ are $3\\sqrt{3}$, $12\\sqrt{3}$, and $\\frac{100}{\\sqrt{15}}$.**",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22055,
"subject": "Mathematics (Olympiad)",
"question": "Given a checked square. One draws a big diagonal and paints black all the cells such that their centers belong to this diagonal (see figure below).\n\nAfterward, the cells on the upper side are cut into two pieces and the lower side is cut into three pieces. It turns out that the areas of these figures are $70$, $80$, $90$, and $100$. What is the possible area of the last figure?\n\n\n\nFig. 01.",
"options": [],
"answer": "See solution",
"solution": "Let the areas of the pieces be $a = 70$, $b = 80$, $c = 90$, $d = 100$, and let the unknown area be $x$.\n\nSince the areas of the lower and upper parts of the square are equal, the sum of some three numbers among them equals the sum of the other two plus $x$. Thus, one of the following equalities holds:\n\n$$\na + c + x = b + d \\implies 160 + x = 180 \\implies x = 20;\n$$\n$$\na + b + x = c + d \\implies 150 + x = 190 \\implies x = 40;\n$$\n$$\na + x = b + c + d \\implies 70 + x = 270 \\implies x = 200;\n$$\n$$\nb + x = a + c + d \\implies 80 + x = 260 \\implies x = 180;\n$$\n$$\nc + x = a + b + d \\implies 90 + x = 250 \\implies x = 160;\n$$\n$$\nd + x = a + b + c \\implies 100 + x = 240 \\implies x = 140.\n$$\nBut the total area of all pieces must equal the area of the square without its diagonal. Denote the side of the square by $n$. Then the sum must be $n^2 - n = n(n-1)$.\n\nNow, consider the cases above:\n\n1. $a + b + c + d + x = 360$. Impossible, since $18 \\times 19 = 342 < 360 < 19 \\times 20 = 380$.\n2. $a + b + c + d + x = 380$. Possible if $n = 20$.\n3. $a + b + c + d + x = 540$. $22 \\times 23 = 506 < 540 < 23 \\times 24 = 552$.\n4. $a + b + c + d + x = 520$. $22 \\times 23 = 506 < 520 < 23 \\times 24 = 552$.\n5. $a + b + c + d + x = 500$. $21 \\times 22 = 462 < 500 < 22 \\times 23 = 506$.\n6. $a + b + c + d + x = 480$. $21 \\times 22 = 462 < 480 < 22 \\times 23 = 506$.\n\nTherefore, the possible area of the last figure is $\\boxed{40}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22056,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$ be positive real numbers with $a + b + c = 1$.\n\n*Prove that*\n\n$$\n\\frac{a}{2a+1} + \\frac{b}{3b+1} + \\frac{c}{6c+1} \\leq \\frac{1}{2}.\n$$\n\n*When does equality hold?*",
"options": [],
"answer": "See solution",
"solution": "We use the following inequalities:\n\n$$\n\\frac{a}{2a+1} \\leq \\frac{2a+1}{8}, \\quad \\frac{b}{3b+1} \\leq \\frac{3b+1}{12}, \\quad \\text{and} \\quad \\frac{c}{6c+1} \\leq \\frac{6c+1}{24}.\n$$\n\nThese follow from the arithmetic-geometric mean inequality applied to the pairs $(1, 2a)$, $(1, 3b)$, and $(1, 6c)$, so equality holds for $a = \\frac{1}{2}$, $b = \\frac{1}{3}$, $c = \\frac{1}{6}$.\n\nUsing these inequalities and $a + b + c = 1$, we get\n\n$$\n\\frac{a}{2a+1} + \\frac{b}{3b+1} + \\frac{c}{6c+1} \\leq \\frac{2a+1}{8} + \\frac{3b+1}{12} + \\frac{6c+1}{24} = \\frac{1}{2}.\n$$\n\nTherefore, the inequality holds, and equality occurs exactly when $a = \\frac{1}{2}$, $b = \\frac{1}{3}$, $c = \\frac{1}{6}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22057,
"subject": "Mathematics (Olympiad)",
"question": "For a non-empty finite set $B$ of real numbers and a real number $x$, define\n\n$$\nd_B(x) = \\min_{b \\in B} |x - b|.\n$$\n\n1. Given a positive integer $m$, find the smallest real number $\\lambda$ such that for any positive integer $n$ and any real numbers $x_1, x_2, \\dots, x_n \\in [0, 1]$, there exists an $m$-element set $B$ of real numbers satisfying\n\n$$\nd_B(x_1) + d_B(x_2) + \\dots + d_B(x_n) \\le \\lambda n.\n$$\n\n2. Given a positive integer $m$ and a positive real number $\\varepsilon$, prove that there exist a positive integer $n$ and non-negative real numbers $x_1, x_2, \\dots, x_n$ such that for any $m$-element set $B$ of real numbers, we have\n\n$$\nd_B(x_1) + d_B(x_2) + \\dots + d_B(x_n) > (1 - \\varepsilon)(x_1 + x_2 + \\dots + x_n).\n$$",
"options": [],
"answer": "See solution",
"solution": "(1) Consider two $m$-element sets\n\n$$\nB_1 = \\left\\{ \\frac{0}{2m-1}, \\frac{2}{2m-1}, \\dots, \\frac{2m-2}{2m-1} \\right\\}, \\quad B_2 = \\left\\{ \\frac{1}{2m-1}, \\frac{3}{2m-1}, \\dots, \\frac{2m-1}{2m-1} \\right\\}.\n$$\n\nFor any real number $x \\in [0, 1]$, there exists $k \\in \\{0, 1, \\dots, m-1\\}$ such that $\\frac{2k}{2m-1} \\le x \\le \\frac{2k+1}{2m-1}$, or there exists $k \\in \\{1, 2, \\dots, m-1\\}$ such that $\\frac{2k-1}{2m-1} \\le x \\le \\frac{2k}{2m-1}$, in both cases we have $d_{B_1}(x) + d_{B_2}(x) = \\frac{1}{2m-1}$. Thus,\n\n$$\n\\sum_{i=1}^{n} d_{B_1}(x_i) + \\sum_{i=1}^{n} d_{B_2}(x_i) = \\sum_{i=1}^{n} (d_{B_1}(x_i) + d_{B_2}(x_i)) = \\frac{n}{2m-1}.\n$$\n\nTherefore, we can take $B = B_1$ or $B = B_2$ such that $\\sum_{i=1}^{n} d_B(x_i) \\le \\frac{n}{4m-2}$. That is, $\\lambda = \\frac{1}{4m-2}$ satisfies the condition.\n\nOn the other hand, let us take $n = 2m$ and an arithmetic sequence of $n$ numbers $X = \\left\\{ \\frac{0}{n-1}, \\frac{1}{n-1}, \\dots, \\frac{n-1}{n-1} \\right\\}$. For any $m$-element set $B = \\{b_1, b_2, \\dots, b_m\\}$ of real numbers, we group the $n$ numbers in $X$ according to their proximity to each $b_k$. Let $a_k$ be the number of elements in $X$ that are closest to $b_k$. The sum of $|x - b_k|$ for these $a_k$ numbers is no less than the difference between the largest and smallest of these $a_k$ numbers, which is at least $\\frac{1}{n-1} \\times (a_k - 1)$.\n\nWe have $a_1 + a_2 + \\dots + a_m = n$ and the sum of $d_B(x)$ for the numbers in $X$ is\n\n$$\n\\sum_{i=1}^{n} d_B(x_i) \\ge \\sum_{k=1}^{m} \\frac{1}{n-1} \\times (a_k - 1) = \\frac{1}{n-1} \\times (n-m) = \\frac{n}{4m-2}.\n$$\n\nHence, it can be seen that $\\lambda < \\frac{1}{4m-2}$ does not satisfy the condition. Therefore, the smallest real number $\\lambda$ that satisfies the condition is $\\frac{1}{4m-2}$.\n\n(2) Let us take larger positive integers $L$ and $t$ such that $\\frac{L-m}{L} \\times \\frac{t-2}{t} > 1 - \\varepsilon$. Set\n\n$$\nn = 1 + t + \\cdots + t^{L-1} = \\frac{t^L - 1}{t - 1},\n$$\n\nand choose $n$ real numbers $x_1, x_2, \\dots, x_n$ as follows:\n\ntake $t^{L-1}$ times $t$, $t^{L-2}$ times $t^2$, \\dots, $t^1$ times $t^{L-1}$, $1$ time $t^L$. (The sum is $L \\times t^L$).\n\nFor any $m$-element set $B = \\{b_1, b_2, \\dots, b_m\\}$ of real numbers, if there exists $k = 1, 2, \\dots, L$ such that the interval $[2t^{k-1}, (2t-2)t^{k-1}]$ does not contain any number from $B$, then $d_B(t^k) > (t-2)t^{k-1}$, which means that the sum of $d_B(x)$ for $t^{L-k}$ numbers $t^k$ is at least $(t-2)t^{k-1} \\times t^{L-k} = (t-2)t^{L-1}$.\n\nConsider the $L$ disjoint intervals $[2t^{k-1}, (2t-2)t^{k-1}]$ ($k = 1, 2, \\dots, L$), where at least $L-m$ of these intervals do not contain any number from $B$. Hence, the sum of $d_B(x_i)$ for the $n$ real numbers is\n\n$$\n\\sum_{i=1}^{n} d_{B}(x_{i}) \\geq (L-m) \\times (t-2)t^{L-1} = \\frac{L-m}{L} \\times \\frac{t-2}{t} \\times \\sum_{i=1}^{n} x_{i} > (1-\\varepsilon) \\cdot \\sum_{i=1}^{n} x_{i}.\n$$\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22058,
"subject": "Mathematics (Olympiad)",
"question": "Let $P(z) = a_n z^n + a_{n-1} z^{n-1} + \\dots + a_m z^m$ be a polynomial with complex coefficients such that $a_m \\neq 0$, $a_n \\neq 0$, and $n > m$. Prove that\n\n$$\n\\max_{|z|=1} \\{|P(z)|\\} \\geq \\sqrt{\\sum_{k=m}^{n} |a_k|^2 + 2|a_m a_n|}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Note that we may assume $m = 0$, since $P(z) = z^m Q(z)$ for some polynomial $Q(z)$ where $|P(z)| = |Q(z)|$ on the unit circle. Thus, we may write\n\n$$\nP(z) = \\sum_{k=0}^{n} a_k z^k, \\quad a_n \\neq 0,\\ a_0 \\neq 0,\n$$\n\nand we have to prove that\n\n$$\n\\max_{|z|=1} \\{|P(z)|\\} \\geq \\sqrt{\\sum_{k=0}^{n} |a_k|^2 + 2|a_0 a_n|}.\n$$\n\nObserve that $|P(z)|^2 = \\sum_{k=0}^{n} \\sum_{j=0}^{n} a_k \\bar{a}_j z^{k-j}$. Let $\\omega$ be a primitive $n$-th root of unity. Then\n\n$$\n\\sum_{l=0}^{n-1} |P(\\omega^l z)|^2 = \\sum_{l=0}^{n-1} \\left( \\sum_{k=0}^{n} \\sum_{j=0}^{n} a_k \\bar{a}_j \\omega^{l(k-j)} z^{k-j} \\right) = \\sum_{k=0}^{n} \\sum_{j=0}^{n} a_k \\bar{a}_j z^{k-j} \\sum_{l=0}^{n-1} \\omega^{l(k-j)}.\n$$\n\nThe last sum is $n$ if $k-j \\equiv 0 \\pmod{n}$ and zero otherwise. Hence\n\n$$\n\\frac{1}{n} \\sum_{l=0}^{n-1} |P(\\omega^l z)|^2 = \\sum_{k=0}^{n} |a_k|^2 + a_n \\bar{a}_0 z^n + \\bar{a}_n a_0 z^{-n}.\n$$\n\nChoosing $z_0$ such that $z_0^n = \\frac{a_0 |a_n|}{|a_0| a_n}$, we obtain\n\n$$\n\\frac{1}{n} \\sum_{l=0}^{n-1} |P(\\omega^l z_0)|^2 = \\sum_{k=0}^{n} |a_k|^2 + 2|a_0 a_n|.\n$$\n\nHence, we can find $l$ such that\n\n$$\n|P(\\omega^l z_0)|^2 \\geq \\sum_{k=0}^{n} |a_k|^2 + 2|a_0 a_n|.\n$$\n\nThe result follows.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 22059,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be non-negative real numbers. Prove that\n\n$$\n3(a^2 + b^2 + c^2) \\geq (a+b+c)(\\sqrt{ab} + \\sqrt{bc} + \\sqrt{ca}) + (a-b)^2 + (b-c)^2 + (c-a)^2 \\geq (a+b+c)^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "It is well-known that $\\sqrt{ab} + \\sqrt{bc} + \\sqrt{ca} - a - b - c \\leq 0$, so we have\n\n$$\n\\begin{aligned}\n& (a+b+c)(\\sqrt{ab} + \\sqrt{bc} + \\sqrt{ca}) + (a-b)^2 + (b-c)^2 + (c-a)^2 \\\\\n= (a+b+c)(\\sqrt{ab} + \\sqrt{bc} + \\sqrt{ca}) + 3(a^2 + b^2 + c^2) - (a+b+c)^2 \\\\\n= (a+b+c)(\\sqrt{ab} + \\sqrt{bc} + \\sqrt{ca} - a - b - c) + 3(a^2 + b^2 + c^2) \\\\\n\\leq 3(a^2 + b^2 + c^2).\n\\end{aligned}\n$$\n\nTo prove the other inequality, let $a = x^2$, $b = y^2$, $c = z^2$ where $x, y, z > 0$. The inequality can be rewritten as\n\n$$\n(x^2 + y^2 + z^2)(xy + yz + zx) + \\sum x^4 \\geq 4(x^2y^2 + y^2z^2 + z^2x^2)\n$$\n\nwhich is equivalent to\n\n$$\n\\sum x^4 + xyz \\sum x + \\sum xy(x^2 + y^2) \\geq 4 \\sum x^2 y^2. \\quad (1)\n$$\n\nBy Schur's inequality, we have\n\n$$\n\\sum x^2(x-y)(x-z) \\geq 0\n$$\n\nhence\n\n$$\n\\sum x^4 + xyz \\sum x \\geq \\sum xy(x^2 + y^2). \\quad (2)\n$$\n\nNote that by Cauchy's inequality,\n\n$$\n\\sum xy(x^2 + y^2) \\geq \\sum 2x^2y^2, \\quad (3)\n$$\n\nthen the result follows from (1), (2), and (3).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22060,
"subject": "Mathematics (Olympiad)",
"question": "Suppose you have coins of denominations $a$, $b$, $c$, and $d$ (all positive integers), and you can use each coin at most once in any sum or difference. Is it possible to choose four such coins so that every integer value from 1 to 20 finbars can be obtained as the sum or difference of some subset of these coins, with no two combinations yielding the same value?",
"options": [],
"answer": "See solution",
"solution": "To cover all values from 1 to 20 finbars, no two of these values can be the same. The largest value is $2a$ so this must be 20 finbars. Hence $a = 10$.\n\nThe only way to achieve 19 finbars is by addition, so $b = 9$.\n\nThese two coins between them cover the values 20, 19, 18, 10, 9, and 1. The next value to consider is 17 finbars, which also has to be made by addition. We cannot use an 8-finbar coin, because $9 - 8 = 1$ and we will have duplicated the 1 finbar value already obtained with $10 - 9$. So the only way to get 17 finbars is with a 7-finbar coin added to the 10-finbar coin. Hence $c = 7$.\n\nAt this point we have covered the following values: 20, 19, 18, 17, 16, 14, 10, 9, 7, 3, 2, 1.\n\nThere is only one coin left to choose, and it will have to be used to cover all the remaining values. The next value to consider is 15 finbars. Again, it has to be obtained by addition. It cannot be made by $10 + 5$, because $7 - 5 = 2$ and we already have a 2 finbar combination with $9 - 7$. Similarly, it cannot be made with a 6-finbar coin (added to 9) or an 8-finbar coin (added to 7) because 1 finbar will be duplicated in both these cases. That exhausts our possible choices for making 15 finbars. Therefore, four coins are not enough.\n\nAlternatively, if $d$ is odd, then 11 of the 20 values above will be even, which is too many. So $d$ is even and must be used to give 15. The only possible even numbers to give 15 are 8 and 6 and again they fail the duplicate value test.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 22061,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ such that\n$$\nf(2m + f(m) + f(m)f(n)) = nf(m) + m\n$$\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 obtain that $f$ is injective. On the other hand, substitute $n = 0$ to get\n$$\nf(2m + (a+1) f(m)) = m, \\quad \\forall m \\in \\mathbb{Z}. \\tag{2}\n$$\nThis implies that $f$ is also surjective. Therefore, there exists an integer $b$ such that $f(b) = -1$. Substitute $m = n = b$ to get $f(2b) = 0$. Next, replace $m = n = 0$ in the original equation to get $f(a^2 + a) = 0$. Combining this with injectivity, we have\n$$\nb = \\frac{a^2 + a}{2}.\n$$\nNow, replace $n = b$ in the original equation:\n$$\nf(2m) = \\frac{a^2 + a}{2} f(m) + m, \\quad \\forall m \\in \\mathbb{Z}. \\tag{3}\n$$\nNext, replace $m = 0$ to get $f(a f(n) + a) = a n$, $\\forall n \\in \\mathbb{Z}$.\nSubstitute $n = b$ to get $\\frac{a(a^2 + a)}{2} = f(0) = a$, so $a \\in \\{0, 1, -2\\}$.\nLet $m = a n$ in (2) 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:\n1. If $a = 1$, then $f(n) = 1 - 2n$, $\\forall n \\in \\mathbb{Z}$. This function does not satisfy the original condition.\n2. If $a = 0$, from (3), $f(2m) = m$, $\\forall m$. Comparing with (2), $f(m) = 0$, $\\forall m$, which contradicts $f$ being non-constant.\n3. If $a = -2$, we have\n$$\nf(-2n) + 4n = 2 f(n) + 2, \\quad \\forall n.\n$$\nBy (3), $f(-2n) = f(-n) - n$. Thus,\n$$\nf(-n) + 3n = 2 f(n) + 2, \\quad \\forall n.\n$$\nReplace $n$ by $-n$:\n$$\nf(n) - 3n = 2 f(-n) + 2, \\quad \\forall n.\n$$\nCombining, we find $f(n) = n - 2$, $\\forall n$. Direct checking shows this function satisfies the given condition.\n\nIn conclusion, the unique function is\n$$\nf(n) = n - 2, \\quad \\forall n \\in \\mathbb{Z}.\n$$\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22062,
"subject": "Mathematics (Olympiad)",
"question": "Two circles, $\\omega_1$ and $\\omega_2$, of equal radius intersect at different points $X_1$ and $X_2$. Consider a circle $\\omega$ externally tangent to $\\omega_1$ at a point $T_1$, and internally tangent to $\\omega_2$ at a point $T_2$. Prove that lines $X_1T_1$ and $X_2T_2$ intersect at a point lying on $\\omega$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let the line $X_kT_k$ and $\\omega$ meet again at $X'_k$, for $k=1,2$. Notice that the tangent $t_k$ to $\\omega_k$ at $X_k$ and the tangent $t'_k$ to $\\omega$ at $X'_k$ are parallel. Since the $\\omega_k$ have equal radii, the $t_k$ are parallel, so the $t'_k$ are parallel, and consequently the points $X'_1$ and $X'_2$ coincide (they are not antipodal, since they both lie on the same side of the line $T_1T_2$). The conclusion follows.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22063,
"subject": "Mathematics (Olympiad)",
"question": "Find all prime numbers $a$, $b$, $c$ and positive integers $k$ satisfying the equation\n\n$$\na^2 + b^2 + 16c^2 = 9k^2 + 1\n$$",
"options": [],
"answer": "See solution",
"solution": "The relation $9k^2 + 1 \\equiv 1 \\pmod{3}$ implies\n$$a^2 + b^2 + 16c^2 \\equiv 1 \\pmod{3} \\iff a^2 + b^2 + c^2 \\equiv 1 \\pmod{3}.$$ \n\nSince $a^2 \\equiv 0,1 \\pmod{3}$, $b^2 \\equiv 0,1 \\pmod{3}$, $c^2 \\equiv 0,1 \\pmod{3}$, we have:\n\n\n\nFrom the previous table it follows that two of the three prime numbers $a$, $b$, $c$ are equal to $3$.\n\n**Case 1.** $a = b = 3$. We have\n$$a^2 + b^2 + 16c^2 = 9k^2 + 1 \\iff 9k^2 - 16c^2 = 17 \\iff (3k - 4c)(3k + 4c) = 17$$\n\nIf $\\begin{cases} 3k - 4c = 1 \\\\ 3k + 4c = 17 \\end{cases}$, then $\\begin{cases} c = 2 \\\\ k = 3 \\end{cases}$ and $(a, b, c, k) = (3, 3, 2, 3)$.\n\nIf $\\begin{cases} 3k - 4c = -1 \\\\ 3k + 4c = -17 \\end{cases}$, then $\\begin{cases} c = 2 \\\\ k = -3 \\end{cases}$ and $(a, b, c, k) = (3, 3, 2, -3)$.\n\n**Case 2.** $c = 3$. If $(3, b_0, c, k)$ is a solution of the given equation, then $(b_0, 3, c, k)$ is a solution, too.\n\nLet $a = 3$. We have\n$$a^2 + b^2 + 16c^2 = 9k^2 + 1 \\iff 9k^2 - b^2 = 152 \\iff (3k - b)(3k + b) = 152.$$ \n\nBoth factors shall have the same parity and we obtain only 4 cases:\n\nIf $\\begin{cases} 3k - b = 2 \\\\ 3k + b = 76 \\end{cases}$, then $\\begin{cases} b = 37 \\\\ k = 13 \\end{cases}$ and $(a, b, c, k) = (3, 37, 3, 13)$.\n\nIf $\\begin{cases} 3k - b = 4 \\\\ 3k + b = 38 \\end{cases}$, then $\\begin{cases} b = 17 \\\\ k = 7 \\end{cases}$ and $(a, b, c, k) = (3, 17, 3, 7)$.\n\nIf $\\begin{cases} 3k - b = -76 \\\\ 3k + b = -2 \\end{cases}$, then $\\begin{cases} b = 37 \\\\ k = -13 \\end{cases}$ and $(a, b, c, k) = (3, 37, 3, -13)$.\n\nIf $\\begin{cases} 3k - b = -38 \\\\ 3k + b = -4 \\end{cases}$, then $\\begin{cases} b = 17 \\\\ k = -7 \\end{cases}$ and $(a, b, c, k) = (3, 17, 3, -7)$.\n\nIn addition, $(a, b, c, k) \\in \\{(37, 3, 3, 13), (17, 3, 3, 7), (37, 3, 3, -13), (17, 3, 3, -7)\\}$.\n\nSo, the given equation has 10 solutions:\n$$S = \\{(37, 3, 3, 13), (17, 3, 3, 7), (37, 3, 3, -13), (17, 3, 3, -7), (3, 37, 3, 13), (3, 17, 3, 7), (3, 37, 3, -13), (3, 17, 3, -7), (3, 3, 2, 3), (3, 3, 2, -3)\\}.$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22064,
"subject": "Mathematics (Olympiad)",
"question": "Show that the set $\\{1, 2, \\ldots, 2n\\}$ can be divided into two subsets $A$ and $B$ such that for each $i = 1, \\ldots, n$, the sum $a_i + b_i$ is a prime number, where $a_i \\in A$, $b_i \\in B$. Furthermore, prove that for the product $(a_1 + b_1)\\cdots(a_n + b_n) = p_1^{\\alpha_1} p_2^{\\alpha_2}\\cdots p_s^{\\alpha_s}$, the following inequality holds:\n\n$$\n\\prod_{j=1}^s (\\alpha_j + 1) \\le 2^{n-1} (\\alpha_j + 1)\n$$\nfor any $j$, and that\n$$\n\\prod_{i=1}^{s} (\\alpha_i + 1) \\le \\left( \\frac{n + s}{s} \\right)^{s}.\n$$",
"options": [],
"answer": "See solution",
"solution": "It is sufficient to show that $$\\left(\\frac{n+s}{s}\\right)^s \\le 2^n.$$ Since $\\alpha_i > 1$ implies $\\alpha_i \\ge 2$ and $\\sum_{i=1}^s \\alpha_i = n$, we have $n \\ge 2s$, so $s \\le \\frac{n}{2}$.\n\nConsider the function $f(x) = \\left(\\frac{n+x}{x}\\right)^x$ for $x \\in \\mathbb{R}$. Taking the derivative:\n\n$$\n\\begin{align*}\nf'(x) &= (1 + \\frac{n}{x})^x \\ln(1 + \\frac{n}{x}) - \\frac{n}{x}(1 + \\frac{n}{x})^{x-1} \\\\\n&= (1 + \\frac{n}{x})^{x-1} \\left[ (1 + \\frac{n}{x}) \\ln(1 + \\frac{n}{x}) - \\frac{n}{x} \\right].\n\\end{align*}\n$$\n\nFor $x \\le \\frac{n}{2}$, $\\frac{n}{x} \\ge 2$, so $\\ln(1 + \\frac{n}{x}) \\ge \\ln 3 > 1$. Thus, $(1 + \\frac{n}{x}) \\ln(1 + \\frac{n}{x}) - \\frac{n}{x} > 1 > 0$, so $f$ is increasing on $(0, n/2]$. The maximum is at $x = \\frac{n}{2}$, so $f(\\frac{n}{2}) = 3^{n/2} < 2^n$. Thus, the inequality holds and the proof is complete.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22065,
"subject": "Mathematics (Olympiad)",
"question": "Two positive integers $m$ and $n$ are called *anagrams* if each decimal digit $a$ appears as many times in the decimal representation of $m$ as in that of $n$. Is it possible to find four different positive integers such that each of them is an anagram of the sum of the other three?",
"options": [],
"answer": "See solution",
"solution": "Let $p$ be a prime number such that its index modulo $10$ is equal to $p-1$ (i.e., the numbers $0, 1, 10, \\ldots, 10^{p-2}$ form a complete residue system modulo $p$). Let $N(p)$ be the number $\\frac{10^{p-1}-1}{p}$ with added leading zeroes to make it a $(p-1)$-digit number. Then the numbers $iN(p)$ for $1 \\leq i \\leq p-1$, each with added leading zeroes to make them $(p-1)$-digit numbers, are anagrams of $N(p)$.\n\nIndeed, for $1 \\leq i \\leq p-1$, we can find $k$ such that $10^k \\equiv i \\pmod{p}$. Now $iN(p)$ is a period of the repeating decimal $\\frac{i}{p}$, and hence of $\\frac{10^k}{p}$. However, the period of the latter is a cyclic permutation of the period of $\\frac{1}{p}$, which equals $N(p)$.\n\nThe index of $17$ modulo $10$ is $16$. If leading zeroes were allowed, the numbers $N(17)$, $2N(17)$, $3N(17)$, and $4N(17)$ would provide a suitable example. To bypass the leading zeroes problem, we can prepend to each of these numbers a number $s$ with no leading zeroes which is an anagram of $3s$. Since the index of $7$ modulo $10$ is $6$ and $7 < 10$, a suitable example is $N(7) = 142857$. Thus, a possible example is given by the numbers $10^{16}N(7) + iN(17)$ for $1 \\leq i \\leq 4$.\n\n$\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22066,
"subject": "Mathematics (Olympiad)",
"question": "Знайдіть усі прості числа $p$, для яких існує натуральне число $n$ таке, що $p^5 - p = 2n!$.",
"options": [],
"answer": "See solution",
"solution": "Число $p = 2$ не задовольняє умову, оскільки $p^5 - p = 30 = 2 \\cdot 15$ та $3! < 15 < 4!$.\n\nЧисло $p = 3$ задовольняє умову, оскільки $p^5 - p = 240 = 2 \\cdot 5!$.\n\nНехай $p \\ge 5$ та $p^5 - p = 2n!$ для деякого натурального $n$, тоді $n! = (p-1)p(p+1) \\frac{p^2+1}{2}$.\n\nОскільки $p$ просте та $n!$ ділиться на $p$, то $n \\ge p$. Отже,\n\n$$\n1 \\cdot 2 \\cdots (p-2) \\cdot (p-1) \\cdot p \\cdots n = (p-1)p(p+1) \\frac{p^2+1}{2}.\n$$\n\nЗвідси $(p+1)(p^2 + 1)$ ділиться на $p-2$. Але $(p+1)(p^2 + 1) = p^3 + p^2 + p + 1 = (p-2)(p^2 + 3p + 7) + 15$, тому 15 ділиться на $p-2$.\n\nОскільки $p-2 \\ge 3$, то $p-2 = 3$, $5$, або $15$, тобто $p = 5$, $7$, або $17$. Перевірка показує, що ці числа не задовольняють умову.\n\nВідповідь: $p = 3$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22067,
"subject": "Mathematics (Olympiad)",
"question": "On the sides $BC$, $CA$, and $AB$ of triangle $ABC$, respectively, points $D$, $E$, and $F$ are chosen. Prove that\n$$\n\\frac{1}{2} (BC + CA + AB) < AD + BE + CF < \\frac{3}{2} (BC + CA + AB).\n$$",
"options": [],
"answer": "See solution",
"solution": "Without loss of generality, let $BC \\ge CA \\ge AB$.\n\nLet us first prove the second inequality. Consider the circle with center $A$ that passes through the vertex $C$ and consider the extension of the side $CB$ past the vertex $B$ up to this circle\n\n(Fig. 5). As a chord lies inside the circle, we have $AD \\le AC$. We also have $BE \\le \\max(BC, BA) = BC$ and $CF \\le \\max(CA, CB) = CB$. Since by the triangle inequality $\\frac{1}{2}BC < \\frac{1}{2}CA + \\frac{1}{2}AB$, we obtain $AD + BE + CF \\le 2BC + CA < \\frac{3}{2}BC + \\frac{3}{2}CA + \\frac{1}{2}AB < \\frac{3}{2}(BC + CA + AB)$.\n\nLet us now prove the first inequality. If the triangle is not acute, then $BE \\ge BA$ and $CF \\ge CA$\n\n(Fig. 6). Since by the triangle inequality we have $\\frac{1}{2}BC < \\frac{1}{2}CA + \\frac{1}{2}AB$, we obtain $AD + BE + CF \\ge AD + BA + CA > BA + CA > \\frac{1}{2}(BC + CA + AB)$. For an acute triangle, we have $AD \\ge AK$, $BE \\ge BL$, and $CF \\ge CM$, where $AK$, $BL$, and $CM$ are the altitudes of triangle $ABC$. Let $H$ be the intersection point of the altitudes of triangle $ABC$\n\n(Fig. 7). We obtain $AD + BE + CF \\ge AK + BL + CM > AH + BH + CH = \\frac{1}{2}(BH + CH) + \\frac{1}{2}(CH + AH) + \\frac{1}{2}(AH + BH) > \\frac{1}{2}(BC + CA + AB)$, where the last inequality follows from the triangle inequality.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22068,
"subject": "Mathematics (Olympiad)",
"question": "Let $k$ be the incircle of a $\\triangle ABC$. A line, parallel to $BC$, touches $k$ and intersects the sides $AB$ and $AC$ at points $A_1$ and $A_2$. Define the points $B_1, B_2$ and $C_1, C_2$ in a similar way. Prove that\n\n$$\n9(\\overline{AA_1} \\cdot \\overline{AA_2} + \\overline{BB_1} \\cdot \\overline{BB_2} + \\overline{CC_1} \\cdot \\overline{CC_2}) \\geq \\overline{AB}^2 + \\overline{BC}^2 + \\overline{CA}^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $D, E$, and $F$ be the points where the incircle touches the sides $BC$, $CA$, and $AB$, respectively, and let $AE = AF = x$, $BF = BD = y$, and $CD = CE = z$.\n\nSince $\\triangle AA_1A_2 \\sim \\triangle ABC$, we have $\\frac{AA_1}{AB} = \\frac{AA_2}{AC} = \\frac{x}{x+y+z}$, so\n$$\nAA_1 = \\frac{x(x+y)}{x+y+z}, \\quad AA_2 = \\frac{x(x+z)}{x+y+z}.\n$$\nAnalogously,\n$$\nBB_1 = \\frac{y(y+z)}{x+y+z}, \\quad BB_2 = \\frac{y(y+x)}{x+y+z}, \\\\\nCC_1 = \\frac{z(z+x)}{x+y+z}, \\quad CC_2 = \\frac{z(z+y)}{x+y+z}.\n$$\n\nThe given inequality becomes\n$$\n9 \\sum x^2 (x+y)(x+z) \\geq (x+y+z)^2 \\sum (x+y)^2,\n$$\ni.e.\n$$\n9 \\sum x^4 + 2(\\sum x^2)(\\sum xy) \\geq 2(\\sum x^2)^2 + 4(\\sum xy)^2.\n$$\nThe last inequality follows from the well-known inequalities\n$$\n3(x^4 + y^4 + z^4) \\geq (x^2 + y^2 + z^2)^2 \\quad \\text{and} \\quad x^2 + y^2 + z^2 \\geq xy + yz + zx.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22069,
"subject": "Mathematics (Olympiad)",
"question": "For all triplets $a, b, c$ of pairwise distinct real numbers, prove the inequality\n\n$$\n\\left| \\frac{a+b}{a-b} \\right| + \\left| \\frac{b+c}{b-c} \\right| + \\left| \\frac{c+a}{c-a} \\right| \\ge 2\n$$\n\nand determine all cases of equality.\n\nProve that if we also impose $a, b, c \\ge 0$, then\n\n$$\n\\left| \\frac{a+b}{a-b} \\right| + \\left| \\frac{b+c}{b-c} \\right| + \\left| \\frac{c+a}{c-a} \\right| > 3,\n$$\n\nwith the value 3 being the best constant possible.",
"options": [],
"answer": "See solution",
"solution": "Denote $x = \\frac{a+b}{a-b}$, $y = \\frac{b+c}{b-c}$, $z = \\frac{c+a}{c-a}$. It is easily seen that\n\n$$\n\\prod (x-1) = \\frac{8abc}{\\prod (a-b)} = \\prod (x+1),\n$$\n\nwhence $xy + yz + zx = -1$.\n\nBut since $(xy)(yz)(zx) = (xyz)^2 \\ge 0$, we must have at least one of the factors being non-negative, say $xy \\ge 0$. Then\n\n$$\n|x| + |y| + |z| \\ge |x+y| + |z| \\ge 2\\sqrt{|zx + yz|} = 2\\sqrt{1 + xy} \\ge 2.\n$$\n\nEquality occurs for $xy = 0$, say $x = 0$, and also $|y| = 1$, with $z = -y$, when $b = -a$ and $c = 0$. Thus, equality cases are $\\{a, b, c\\} = \\{t, -t, 0\\}$ for $t \\ne 0$. However, since the value of the expression for $a, b, c$ is the same as that for $ta, tb, tc$ with $t \\ne 0$, the only essential solution is $\\{1, -1, 0\\}$.\n\nIf we also impose $a, b, c \\ge 0$, then, since the expression is symmetric, we may assume $0 \\le a < b < c$. Consequently,\n\n$$\n\\left| \\frac{a+b}{a-b} \\right| + \\left| \\frac{b+c}{b-c} \\right| + \\left| \\frac{c+a}{c-a} \\right| = 3 + \\frac{2a}{b-a} + \\frac{2b}{c-b} + \\frac{2a}{c-a} > 3.\n$$\n\nThe fact we can approach 3 as closely as desired is argued by taking $a = 0$, $c = (2n + 1)b$ for some $n \\in \\mathbb{N}^*$, when\n\n$$\n\\frac{2a}{b-a} + \\frac{2b}{c-b} + \\frac{2a}{c-a} = \\frac{1}{n}\n$$\n\ncan be made as small as desired for large enough $n$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22070,
"subject": "Mathematics (Olympiad)",
"question": "For every non-negative integer $n$, let $s_n$ be the sum of the digits in the decimal expansion of $2^n$. Is the sequence $(s_n)_{n \\in \\mathbb{N}}$ eventually increasing?",
"options": [],
"answer": "See solution",
"solution": "The answer is negative. To prove this, notice that the sequence is periodic modulo $9$, with period $6$; the first block of values is $1, 2, 4, 8, 7, 5$.\n\nSuppose, for contradiction, that the sequence is eventually increasing from some index $n_0$ onward. Fix $m$ such that $6m \\geq n_0$ and write:\n\n$$\n\\begin{cases}\ns_{6m+1} \\geq s_{6m} + 1, \\\\\ns_{6m+2} \\geq s_{6m+1} + 2, \\\\\ns_{6m+3} \\geq s_{6m+2} + 4, \\\\\ns_{6m+4} \\geq s_{6m+3} + 8, \\\\\ns_{6m+5} \\geq s_{6m+4} + 7, \\\\\ns_{6m+6} \\geq s_{6m+5} + 5,\n\\end{cases}\n$$\n\nThus, $s_{6m+6} \\geq s_{6m} + 27$, so\n\n$$\ns_{6m+6n} \\geq s_{6m} + 27n, \\quad n \\in \\mathbb{N}.\n$$\n\nHowever, the number of nonzero digits in $2^{6m+6n}$ is at most $\\lceil (6m + 6n) \\log_{10} 2 \\rceil < 2m + 2n$, so $s_{6m+6n} \\leq 18m + 18n$, which contradicts the previous inequality for large $n$. Therefore, the sequence is not eventually increasing.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22071,
"subject": "Mathematics (Olympiad)",
"question": "Given an integer $n \\ge 2$, let non-negative real numbers $a_1, a_2, \\dots, a_n$ satisfy\n$$\na_1 \\ge a_2 \\ge \\dots \\ge a_n, \\quad a_1 + a_2 + \\dots + a_n = n.\n$$\nFind the minimum of\n$$\na_1 + a_1 a_2 + a_1 a_2 a_3 + \\dots + a_1 a_2 \\dots a_n.\n$$",
"options": [],
"answer": "See solution",
"solution": "For integer $m \\ge 1$, define\n$$\nf_m(x_1, x_2, \\dots, x_m) = x_1 + x_1 x_2 + \\dots + x_1 x_2 \\dots x_m.\n$$\nWe prove the following lemma by induction on $m$.\n\n**Lemma**: If the average of non-negative real numbers $x_1 \\ge x_2 \\ge \\dots \\ge x_m$ is $A$ and $A \\le 1$, then\n$$\nf_m(x_1, x_2, \\dots, x_m) \\ge f_m(A, A, \\dots, A).\n$$\n**Proof of the lemma:** The conclusion clearly holds for $m = 1$. Assume it holds for $m$, and consider $m + 1$.\n\nLet the average of non-negative real numbers $x_1 \\ge x_2 \\ge \\dots \\ge x_{m+1}$ be $A \\le 1$.\n\nSince $x_1 \\ge A$, the average of $x_2, x_3, \\dots, x_{m+1}$, denoted $B$, satisfies $B \\le A \\le 1$. By the induction hypothesis,\n$$\n\\begin{aligned}\nf_{m+1}(x_1, x_2, \\dots, x_{m+1}) &= x_1(1 + f_m(x_2, x_3, \\dots, x_{m+1})) \\\\\n&\\ge x_1(1 + f_m(B, B, \\dots, B)) \\\\\n&= ((m+1)A - mB)(1 + B + B^2 + \\dots + B^m).\n\\end{aligned}\n$$\nWe now show that\n$$\n((m+1)A - mB)(1 + B + B^2 + \\dots + B^m) \\ge f_{m+1}(A, A, \\dots, A).\n$$\nIndeed,\n$$\n\\begin{align*}\n&((m+1)A - mB)(1 + B + B^2 + \\dots + B^m) - f_{m+1}(A, A, \\dots, A) \\\\\n&= ((m+1)A - mB)(1 + B + B^2 + \\dots + B^m) \\\\\n&\\quad - A(1 + A + A^2 + \\dots + A^m) \\\\\n&= m(A - B)(1 + B + B^2 + \\dots + B^m) \\\\\n&\\quad + A(B + B^2 + \\dots + B^m - A - A^2 - \\dots - A^m)\n\\end{align*}\n$$\n$$\n\\begin{align*}\n&= (A - B)(m(1 + B + B^2 + \\cdots + B^m) \\\\\n&\\quad - A(1 + (A + B) + (A^2 + AB + B^2) \\\\\n&\\quad + \\cdots + (A^{m-1} + A^{m-2}B + \\cdots + B^{m-1})) \\\\\n&\\ge (A - B)(m(1 + B + B^2 + \\cdots + B^m) - 1 - (1 + B) - (1 + B + B^2) \\\\\n&\\quad - \\cdots - (1 + B + \\cdots + B^{m-1})) \\quad (\\text{Here } A \\le 1 \\text{ is used}) \\\\\n&= (A - B)(B + 2B^2 + 3B^3 + \\cdots + mB^m) \\\\\n&\\ge 0.\n\\end{align*}\n$$\nThe lemma is proven.\n\nSince the average of non-negative real numbers $a_1 \\ge a_2 \\ge \\dots \\ge a_n$ in the original question is $1$, by the lemma,\n$$\nf_n(a_1, a_2, \\dots, a_n) \\ge f_n(1, 1, \\dots, 1) = n.\n$$\nTherefore, the desired minimum is $n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22072,
"subject": "Mathematics (Olympiad)",
"question": "There are $2k+1$ people with pairwise distinct heights. Let their heights in increasing order be $a_1 < a_2 < \\dots < a_{2k+1}$. These people are placed arbitrarily in a table of $M \\times N$ ($M$ rows and $N$ columns). In each row, select the person with the middle height. From these $M$ middle-height people, select the one with the middle height and call this person the *middling*.\n\n(a) What is the minimum possible height (i.e., position in the ordered list) for the middling if $2k+1 = 2015$?\n\n(b) What is the minimum possible height for the middling for arbitrary $k$?",
"options": [],
"answer": "See solution",
"solution": "**Answer:**\n\n(a) $528$\n\n(b) The minimum number is $(m+1)(n+1)$, where $2k+1 = (2m+1)(2n+1)$, and the factors are the closest possible to each other.\n\n**Solution:**\n\nSince $2k+1 = M \\cdot N$ is odd, both $M$ and $N$ must be odd. Let $M = 2m+1$, $N = 2n+1$. Let $b_1, b_2, \\dots, b_M$ be the middle-height people in each row. By rearranging the rows, we can ensure $b_1 < b_2 < \\dots < b_M$, so $b_{m+1}$ is the middling. We count how many people $b_{m+1}$ is taller than:\n\nIn each of the first $m+1$ rows, the middle person is taller than $n$ people in their row. Since $b_{m+1}$ is taller than $b_i$ for $i = 1, \\dots, m$, he is also taller than $m$ more people. Thus, $b_{m+1}$ is taller than $(m+1)n + m = mn + m + n$ people, so his position is at least $mn + m + n + 1 = (m+1)(n+1)$.\n\nTo minimize $(m+1)(n+1)$, given $2k+1 = (2m+1)(2n+1)$, choose $m$ and $n$ so that $2m+1$ and $2n+1$ are the closest possible odd divisors of $2k+1$.\n\nIf $2k+1$ is a perfect square, the closest divisors are equal, so the minimum is $i^2$ for $2k+1 = (2i+1)^2$. If not, list all odd divisors $1 = d_1 < d_2 < \\dots < d_{2l} = 2k+1$; the closest pair $d_l = 2j+1$, $d_{l+1} = 2t+1$ gives the minimum $(j+1)(t+1)$.\n\nFor (a), $2k+1 = 2015 = 5 \\cdot 13 \\cdot 31$, so the odd divisors are:\n\n$$\n1 < 5 < 13 < 31 < 65 < 155 < 403 < 2015\n$$\n\nThe closest are $31 = 2 \\times 15 + 1$ and $65 = 2 \\times 32 + 1$, so the minimum position is:\n\n$$\n(15+1) \\times (32+1) = 528\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22073,
"subject": "Mathematics (Olympiad)",
"question": "Let $S$ be the sum of all invertible elements of a finite ring.\n\nProve that $S^2 = S$ or $S^2 = 0$.",
"options": [],
"answer": "See solution",
"solution": "If $1 + 1 \\ne 0$, then $x \\ne -x$ for every invertible $x$, so $S = 0$ (we can group the invertible elements into pairs $(x, -x)$).\n\nIf $1 + 1 = 0$, notice that $xS = S$ for every invertible $x$. Adding all these relations, $S^2 = kS$, where $k$ is the number of invertible elements. Then $S^2 = S$ for odd $k$ and $S^2 = 0$ for even $k$ (since $1 + 1 = 0$).\n\n**Remarks:**\n- $S = 0$ for every finite ring of characteristic different from $2$, and also for the finite fields $\\mathbb{F}_{2^n}$, $n > 1$ (in a field, $xS = S$ and $S \\neq 0$ implies $x = 1$).\n- $S = 1$ for the Boolean rings $\\mathbb{Z}_2^n$ with $n \\ge 1$ (including the field $\\mathbb{F}_2$).\n- $S^2 = 0$, but $S \\neq 0$, for the ring of upper triangular $2 \\times 2$ matrices over $\\mathbb{F}_2$.\n- $S^2 = S$, but $S \\neq 0$ and $S \\neq 1$, for the rings $\\mathbb{F}_{2^n} \\times \\mathbb{F}_2$ with $n > 1$.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 22074,
"subject": "Mathematics (Olympiad)",
"question": "Show that $2010$ cannot be written as the difference of two squares.",
"options": [],
"answer": "See solution",
"solution": "Assume there are integers $x$ and $y$ such that\n\n$$\n2010 = x^2 - y^2 = (x - y)(x + y).\n$$\n\nThe factors $x - y$ and $x + y$ have the same parity.\n\n- If both are even, their product is divisible by $4$, but $2010$ is not.\n- If both are odd, their product is odd, but $2010$ is even.\n\nTherefore, no such integers exist.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22075,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute triangle inscribed in the circle $O$, and let $I$ be the circumcenter of triangle $OBC$. Point $G$ belongs to the arc $BC$ (not containing $O$) of $(I)$. The circle $(ABG)$ intersects $AC$ at $E$, and the circle $(ACG)$ intersects $AB$ at $F$ (with $E, F$ different from $A$).\n\n1. Let $K$ be the intersection of $BE$ and $CF$. Prove that $AK$, $BC$, and $OG$ are concurrent.\n\n2. Let $D$ be a fixed point on the arc $BC$ that contains $O$ of $(I)$, and let $GB$ meet $CD$ at $M$, $GC$ meet $BD$ at $N$. Suppose that $MN$ intersects $(O)$ at $P, Q$. Prove that as $G$ moves on $(I)$, the circumcircle of triangle $GPQ$ always passes through two certain fixed points.",
"options": [],
"answer": "See solution",
"solution": "1) We have\n\n$$\n\\begin{aligned}\n\\angle EGF &= \\angle BGE + \\angle CGF - \\angle EGF \\\\\n&= 360^{\\circ} - 2\\angle BAC - (180^{\\circ} - 2\\angle BAC) = 180^{\\circ}\n\\end{aligned}\n$$\n\nThus, the three points $E$, $G$, $F$ are collinear. Since $\\angle ABK + \\angle ACK = \\angle AGE + \\angle AGF = 180^{\\circ}$, $K$ belongs to the circle $(O)$.\n\n\n\nIt is easy to check that $G$ is the Miquel point, so triangles $GBA$ and $GKC$ are similar, which implies $\\angle BGA = \\angle KGC$. We also have $GO$ is the angle bisector of $\\angle BGC$, so $GO$ is the angle bisector of $\\angle AGK$. Combined with $OA = OK$, we conclude that $AOKG$ is a cyclic quadrilateral. Considering the radical axis of circles $(O)$, $(AOKG)$, $(BOC)$, we see that $AK$, $OG$, and $BC$ are concurrent.\n\n2) In this part, $(O)$ and $(I)$ are two fixed circles passing through $B$, $C$, and $D$ is fixed on $(I)$ while $G$ moves on $(I)$. By applying Pascal's theorem for the tuple\n\n$$\n\\begin{pmatrix}\nB & C & D \\\\\nC & B & G\n\\end{pmatrix}\n$$\n\none can check that the line $MN$ passes through the intersection $J$ of the tangent lines at $B$, $C$ of the fixed circle $(I)$. Suppose $JD$ meets $(I)$ at the second point $X$, then $X$ is fixed and the quadrilateral $BCDX$ is harmonic, so $G(BC, DX) = -1$. Let $T$ be the intersection of $MN$ and $BC$, then $G(BC, DT) = -1$, which means $GX$ passes through $T$. Thus,\n\n$$\n\\overline{TX} \\cdot \\overline{TG} = \\overline{TB} \\cdot \\overline{TC} = \\overline{TP} \\cdot \\overline{TQ}.\n$$\n\n\n\nThis implies that $(GPQ)$ passes through the fixed point $X$. Suppose $Y$ is the intersection of $(GPQ)$ and $DX$ (different from $X$), then\n\n$$\n\\overline{JX} \\cdot \\overline{JY} = \\overline{JP} \\cdot \\overline{JQ} = \\mathcal{P}_{J/(O)}\n$$\n\nwhich is constant, so $Y$ is fixed. Therefore, the circles $(GPQ)$ pass through two fixed points $X$, $Y$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22076,
"subject": "Mathematics (Olympiad)",
"question": "找出所有實係數多項式 $P$，滿足存在無窮多對互質正整數對 $(m, n)$，使得\n\n$$\nP\\left(\\frac{m}{n}\\right) = \\frac{1}{n}.\n$$",
"options": [],
"answer": "See solution",
"solution": "容易證明 $P$ 必須有有理係數。\n\n若 $P$ 的次數 $d \\ge 2$，設 $P(x) = \\frac{p}{q}x^d + \\frac{1}{M}Q(x)$，其中 $p, q \\in \\mathbb{Z}$，$Q(x)$ 為次數小於 $d$ 的整係數多項式。假設 $P\\left(\\frac{m}{n}\\right) = \\frac{1}{n}$，$m, n$ 互質。\n\n則 $Mn^{d-1}P\\left(\\frac{m}{n}\\right) \\in \\mathbb{N}$，因此 $qn \\mid pMm^d$，所以 $n \\mid pM$。但 $n$ 可任意大，故 $pM = 0$，矛盾。因此必有 $d = 1$。\n\n設 $P(x) = \\frac{p x}{q} + \\frac{r}{s}$，$\\gcd(p, q) = \\gcd(r, s) = 1$。條件等價於存在無窮多對互質正整數 $(m, n)$，使得\n\n$$\npsm + rqn = qs.\n$$\n\n令 $A = ps,\\ B = qr,\\ C = qs$。則 $\\gcd(A, B) \\mid C$。又必有 $AB \\le 0$，否則 $|C| = |Am + Bn| \\ge |m + n|$，與存在無窮多解矛盾。\n\n反之，若 $\\gcd(A, B) \\mid C$ 且 $AB \\le 0$，可取整數 $a, b$ 使 $Aa + Bb = \\gcd(A, B)$。令 $m = |B|t + Ca,\\ n = |A|t + Cb$，則 $Am + Bn = C$，且 $\\gcd(m, n) \\mid \\gcd(C, am - bn) = \\gcd(C, t + s)$，其中 $s$ 為與 $t$ 無關的整數。故可取足夠大的 $t$ 使 $m, n$ 為正且互質。\n\n因此，所有 $P(x)$ 的解為 $P(x) = \\frac{p x}{q} + \\frac{r}{s}$，其中 $\\gcd(ps, qr) \\mid qs$ 且 $\\frac{pr}{qs} \\le 0$。由 $\\gcd(p, q) = \\gcd(r, s) = 1$，可知 $\\gcd(ps, qr) = \\gcd(p, r)\\gcd(q, s)$，故第一條件等價於 $\\gcd(p, r) = 1$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22077,
"subject": "Mathematics (Olympiad)",
"question": "Find all integer pairs $(n, m)$ such that\n$$\n n^2 + n = m^2 + 2m - 9.\n$$",
"options": [],
"answer": "See solution",
"solution": "The integer solutions $(n, m)$ are:\n$$\n(-10, -11),\\quad (-10, 9),\\quad (-3, -5),\\quad (-3, -3),\\quad (2, -5),\\quad (2, 3),\\quad (9, -11),\\quad (9, 9).\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22078,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\in \\mathbb{N}$, $n > 2$, and suppose $a_1, a_2, \\dots, a_{2n}$ is a permutation of the numbers $1, 2, \\dots, 2n$ such that $a_1 < a_3 < \\dots < a_{2n-1}$ and $a_2 > a_4 > \\dots > a_{2n}$. Prove that\n\n$$\n(a_1 - a_2)^2 + (a_3 - a_4)^2 + \\dots + (a_{2n-1} - a_{2n})^2 > n^3.\n$$",
"options": [],
"answer": "See solution",
"solution": "Denote $S = (a_1 - a_2)^2 + (a_3 - a_4)^2 + \\dots + (a_{2n-1} - a_{2n})^2$. We have\n\n$$\n\\begin{aligned}\nS &= \\sum_{i=1}^{2n} i^2 - 2(a_1a_2 + a_3a_4 + \\dots + a_{2n-1}a_{2n}) \\\\\n &= \\frac{n(2n+1)(4n+1)}{3} - 2(a_1a_2 + a_3a_4 + \\dots + a_{2n-1}a_{2n}).\n\\end{aligned}\n$$\n\nNext, observe that for each $j = 1, 2, \\dots, n$, one of the numbers $a_{2j-1}, a_{2j}$ is greater than $n$, and the other is at most $n$. Indeed, suppose $a_{2j-1}, a_{2j} \\le n$. Then $a_1 < a_3 < \\dots < a_{2j-1} \\le n$ and $a_{2n} < a_{2n-2} < \\dots < a_{2j} \\le n$, yielding $j + (n - j + 1) = n + 1$ distinct positive integers not exceeding $n$, a contradiction. The case $a_{2j-1}, a_{2j} > n$ is handled similarly. It follows that $a_1a_2 + a_3a_4 + \\dots + a_{2n-1}a_{2n}$ has form $1 \\cdot b_1 + 2 \\cdot b_2 + \\dots + n \\cdot b_n$, where $b_1, b_2, \\dots, b_n$ is some permutation of $n+1, n+2, \\dots, 2n$. But it is known that such an expression will be maximal if and only if $b_1 < b_2 < \\dots < b_n$. Therefore,\n\n$$\na_1a_2 + a_3a_4 + \\dots + a_{2n-1}a_{2n} \\le 1(n+1) + 2(n+2) + \\dots + n \\cdot 2n = n \\cdot \\frac{n(n+1)}{2} + \\frac{n(n+1)(2n+1)}{6}. \\quad (2)\n$$\n\nFrom (1) and (2), we find\n\n$$\nS \\ge \\frac{n(2n+1)(4n+1)}{3} - n^2(n+1) - \\frac{n(n+1)(2n+1)}{3} = n^3. \\quad (3)\n$$\n\nBy the above arguments, for equality to hold, there would have to exist indices $i, j, k$ (since $n \\ge 3$) such that $\\{a_{2i-1}, a_{2i}\\} = \\{1, n+1\\}$, $\\{a_{2j-1}, a_{2j}\\} = \\{2, n+2\\}$ and $\\{a_{2k-1}, a_{2k}\\} = \\{3, n+3\\}$. It is easy to check that this is impossible, given the assumptions on the permutation $a_1, a_2, \\dots, a_{2n}$. Therefore, equality cannot hold in (3) and $S > n^3$.\n\n*Comment.* The assumption $n > 2$ ensures that the inequality is strict.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 22079,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle and let $X$, $Y$, $Z$ be interior points on the sides $BC$, $CA$, and $AB$, respectively. Show that the magnified image of the triangle $XYZ$ under a homothety of factor $4$ from its centroid covers at least one of the vertices $A$, $B$, or $C$.",
"options": [],
"answer": "See solution",
"solution": "Since the problem is of an affine nature, we may (and will) assume that the triangle $XYZ$ is equilateral. The triangle $ABC$ has at least one vertex angle, say at $A$, greater than or equal to $60^{\\circ}$, so $A$ is covered by the closed circumdisc $OYZ$, where $O$ is the center of the triangle $XYZ$. Since the latter is covered by the $4$-fold blow-up of the triangle $XYZ$ from $O$, the conclusion follows.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22080,
"subject": "Mathematics (Olympiad)",
"question": "Using the inequality\n$$ \\frac{1}{2^2} + \\frac{1}{3^2} + \\dots + \\frac{1}{n^2} < 1 $$\nfor all $n > 1$, show by induction on $N$ that the board may get at most $N^2$ numbers, with equality achieved only for $N=1$.",
"options": [],
"answer": "See solution",
"solution": "$N=1$.\n\nRecall that\n$$ \\frac{1}{2^2} + \\frac{1}{3^2} + \\dots + \\frac{1}{n^2} < 1 $$\nfor all $n > 1$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22081,
"subject": "Mathematics (Olympiad)",
"question": "In triangle $ABC$, points $D$ and $E$ lie on sides $BC$ and $AC$ respectively such that $AD \\perp BC$ and $DE \\perp AC$. The circumcircle of triangle $ABD$ meets segment $BE$ at point $F$ (other than $B$). Ray $AF$ meets segment $DE$ at point $P$. Prove that\n$$\n\\frac{DP}{PE} = \\frac{CD}{DB}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Since $AED$ and $ADC$ are both right triangles, $\\widehat{ADP} = 90^\\circ - \\widehat{DAC} = \\widehat{ECB}$. Also, since $AFDB$ is a cyclic quadrilateral, $\\widehat{DAP} = \\widehat{EBC}$.\n\nTherefore, $ADP$ and $BCE$ are similar and $\\frac{BC}{EC} = \\frac{AD}{DP}$. Also, $ADC$ and $DEC$ are similar since both are right-angled triangles that share an acute angle. Therefore,\n$$\n\\frac{AD}{DC} = \\frac{DE}{EC}.\n$$\nThese two equations imply that\n$$\nBC \\cdot DP = AD \\cdot EC = DC \\cdot DE,\n$$\nso we have\n$$\n\\frac{BC}{DC} = \\frac{DE}{DP}.\n$$\nSince $BC = BD + DC$ and $DE = DP + PE$, subtracting one from both sides gives us $\\frac{BD}{DC} = \\frac{PE}{DP}$, as desired.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22082,
"subject": "Mathematics (Olympiad)",
"question": "Let $d$ be a positive integer. Show that for every integer $S$, there exists an integer $n > 0$ and a sequence $\\epsilon_1, \\epsilon_2, \\dots, \\epsilon_n$, where for any $k$, $\\epsilon_k = 1$ or $\\epsilon_k = -1$, such that\n\n$$\nS = \\epsilon_1(1+d)^2 + \\epsilon_2(1+2d)^2 + \\epsilon_3(1+3d)^2 + \\dots + \\epsilon_n(1+nd)^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $U_k = (1 + kd)^2$. We calculate $U_{k+3} - U_{k+2} - U_{k+1} + U_k$, which turns out to be $4d^2$, a constant. Changing signs, we obtain the sum $-4d^2$.\n\nThus, if we have found an expression for a certain number $S_0$ as a sum of the desired type, we can obtain an expression of the desired type for $S_0 + (4d^2)q$, for any integer $q$.\n\nIt remains to show that for any $S$, there exists an integer $S'$ such that $S' \\equiv S \\pmod{4d^2}$ and $S'$ can be expressed in the desired form. Consider the sum\n\n$$\n(1+d)^2 + (1+2d)^2 + \\dots + (1+Nd)^2,\n$$\n\nwhere $N$ is large. We can choose $N$ so that the sum is odd or even.\n\nBy changing the sign in front of $(1 + kd)^2$ to a minus sign, we decrease the sum by $2(1 + kd)^2$. In particular, if $k \\equiv 0 \\pmod{2d}$, we decrease the sum by $2$ (modulo $4d^2$). So,\n\nif $N$ is large enough, there are many $k < N$ such that $k$ is a multiple of $2d$. By switching the sign in front of $r$ of these, we change (downward) the congruence class modulo $4d^2$ by $2r$. By choosing $N$ so that the original sum is odd, and choosing suitable $r < 2d^2$, we can obtain numbers congruent to all odd numbers modulo $4d^2$. By choosing $N$ so that the original sum is even, we can obtain numbers congruent to all even numbers modulo $4d^2$. This completes the proof. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22083,
"subject": "Mathematics (Olympiad)",
"question": "In an acute triangle $ABC$ with $AB < AC$, the altitudes $BE$ and $CF$ intersect at $H$. The tangent to the circumcircle of $ABC$ at $A$ intersects the circumcircle of $AEF$ at $T \\neq A$. The circumcircles of $TBE$ and $TCF$ intersect at $K \\neq T$. Prove that $\\angle KHB = \\angle ABC$.",
"options": [],
"answer": "See solution",
"solution": "From $\\angle HEA = 90^\\circ = 180^\\circ - 90^\\circ = 180^\\circ - \\angle HFA$, we deduce that $AEHF$ is cyclic. By definition, $T$ lies on this circle as well, so\n\n$$\n\\angle THE = 180^\\circ - \\angle TAE = 180^\\circ - \\angle TAC = \\angle ABC,\n$$\n\nwhere the final equality is due to the tangent-chord theorem.\n\nLet $K'$ be the intersection of the lines $TH$ and $BC$. It's sufficient to show that $K' = K$ or that $K'$ lies on the circumcircles of $TBE$ and $TCF$, as we have $\\angle K'HB = \\angle THE = \\angle ABC$. To show this we notice that\n\n$$\n\\angle ETK' = \\angle ETH = \\angle EAH = \\angle CAH = 90^\\circ - \\angle ACB = \\angle EBC = \\angle EBK',\n$$\n\n\n\n\n\nwhich shows that the points $T, B, K'$ and $E$ are concyclic. Analogously we show that $T, C, K'$ and $F$ are concyclic. Thus $K' = K$, as desired.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22084,
"subject": "Mathematics (Olympiad)",
"question": "Let $S = a + b$ and $P = ab$. Find all real numbers $a$ and $b$ such that\n\n$$\n\\begin{cases}\nS(P + S + 1) = 2 \\\\\nS(S^2 - 3P) = 1\n\\end{cases}\n$$",
"options": [],
"answer": "See solution",
"solution": "We have $S = a + b$ and $P = ab$.\n\n$(a + 1)(b + 1) = ab + a + b + 1 = P + S + 1$.\n\n$a^3 + b^3 = (a + b)(a^2 - ab + b^2) = S(S^2 - 3P)$.\n\nSo,\n\n$$\n\\begin{cases}\nS(P + S + 1) = 2 \\\\\nS(S^2 - 3P) = 1\n\\end{cases}\n$$\n\nFrom the second equation:\n\n$$\nS(S^2 - 3P) = 1 \\implies S^3 - 3SP = 1\n$$\n\nFrom the first equation:\n\n$$\nS(P + S + 1) = 2 \\implies SP + S^2 + S = 2\n$$\n\nNow, solve for $SP$ from the first equation:\n\n$$\nSP = 2 - S^2 - S\n$$\n\nSubstitute $SP$ into the second equation:\n\n$$\nS^3 - 3SP = 1 \\\\\nS^3 - 3(2 - S^2 - S) = 1 \\\\\nS^3 - 6 + 3S^2 + 3S = 1 \\\\\nS^3 + 3S^2 + 3S = 7\n$$\n\nNotice that $(S + 1)^3 = S^3 + 3S^2 + 3S + 1$, so:\n\n$$\n(S + 1)^3 = 8 \\implies S + 1 = 2 \\implies S = 1\n$$\n\nNow, substitute $S = 1$ into $SP = 2 - S^2 - S$:\n\n$$\nSP = 2 - 1 - 1 = 0 \\implies P = 0\n$$\n\nSo $a + b = 1$ and $ab = 0$, which means $\\{a, b\\} = \\{0, 1\\}$.\n\n*Remark:* It can easily be shown that $P = 0$, so $\\{a, b\\} = \\{0, 1\\}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22085,
"subject": "Mathematics (Olympiad)",
"question": "The vertices of a cube are numbered $1, 2, \\ldots, 8$. Someone chose three faces of the cube and told Pete the numbers written on their vertices:\n\n- $\\{1, 4, 6, 8\\}$\n- $\\{1, 2, 6, 7\\}$\n- $\\{1, 2, 5, 8\\}$\n\nIs it possible to determine which number is on the vertex opposite to the one numbered $5$?",
"options": [],
"answer": "See solution",
"solution": "**Answer:** Yes, its number is $6$.\n\n**Solution.** Three edges go from each vertex of a cube, and every edge is part of two faces. Consider vertex $1$. It lies on all three given faces, so these faces meet at vertex $1$. The three edges from $1$ are $1$-$2$, $1$-$6$, and $1$-$8$. By analyzing the cube's structure, we see that $6$ is opposite to $5$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 22086,
"subject": "Mathematics (Olympiad)",
"question": "In January, Petro bought between one and three toy cars each day. On February 1, he tried to arrange all his cars into a rectangle. When he arranged them into rows of 7 cars, one car remained. When he arranged them into rows of 10, there were 2 cars left over. Can Petro arrange them into rows of 4 cars?",
"options": [],
"answer": "See solution",
"solution": "Let the total number of cars be $N$. We are told:\n\n- $N \\equiv 1 \\pmod{7}$\n- $N \\equiv 2 \\pmod{10}$\n\nWe seek $N$ such that $29 \\leq N \\leq 92$ (since Petro could have bought between 29 and 92 cars in January).\n\nSet $N = 7k + 1 = 10n + 2$ for integers $k, n$.\n\nSolving $7k + 1 = 10n + 2$ gives $7k = 10n + 1$.\n\nWe look for $N$ in the range $29 \\leq N \\leq 92$ that satisfies both conditions. The possible values are $N = 91$ (since $91 \\equiv 1 \\pmod{7}$ and $91 \\equiv 1 \\pmod{10}$, but we need $N \\equiv 2 \\pmod{10}$). Try $N = 92$:\n\n- $92 \\div 7 = 13$ remainder $1$ ($92 \\equiv 1 \\pmod{7}$)\n- $92 \\div 10 = 9$ remainder $2$ ($92 \\equiv 2 \\pmod{10}$)\n\nSo $N = 92$ is the only possible value in the range.\n\nSince $92 = 4 \\times 23$, Petro can arrange his cars into rows of 4.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22087,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer and let\n\n$$\nS_n = \\{(x, y) \\in \\mathbb{Z}^2 \\mid 1 \\le x, y \\le 5n\\}.\n$$\n\nA subset $A \\subseteq S_n$ is said to be _sumfree_ if\n\n$$\n(x_1, y_1), (x_2, y_2) \\in A \\implies (x_1 + x_2, y_1 + y_2) \\notin A.\n$$\n\nShow that there exists a sumfree subset with at least $15n^2$ elements.",
"options": [],
"answer": "See solution",
"solution": "Let $A := \\{(x, y) \\in S_n \\mid 4n+1 \\le x+y \\le 8n+1\\}$. Then it is clear that $A$ is sumfree.\n\nNow we show that $|A| \\ge 15n^2$.\n\nFor $k \\in \\mathbb{Z}_{\\ge 1}$, let $T_k := \\{(x, y) \\in S_n \\mid x + y = k\\}$. Then\n\n$$\n|T_k| = \\begin{cases} k-1 & k \\le 5n \\\\ 10n-k+1 & k \\ge 5n+1. \\end{cases}\n$$\n\nSince $A$ is the disjoint union of $T_k$ for $4n+1 \\le k \\le 8n+1$, we have\n\n$$\n|A| = 4n + (4n + 1) + \\dots + (5n - 1) + 5n + (5n - 1) + \\dots + 2n = 15n^2 + 3n.\n$$\n\nClearly $|A| \\ge 15n^2$.\n\n![alt](path \"title\")",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22088,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be a triangle with perimeter $100$ and incenter $I$. The parallel to $AB$ through $I$ divides the median through $A$ in ratio $7:3$, counted from $A$. Find the length of side $AB$.",
"options": [],
"answer": "See solution",
"solution": "Let $AM$ be the median through $A$, $CJ$ the bisector through $C$, and let the parallel to $AB$ through $I$ intersect $AM$ at $P$. Set $AP:PM = \\lambda$; in our problem, $\\lambda = \\frac{7}{3}$.\n\nIf $N$ is the midpoint of $CL$ then $MN \\parallel AB$ as $M$ is the midpoint of $BC$. Hence $MN \\parallel IP$. Now\n\nThales' theorem yields $\\frac{IL}{IN} = \\frac{AP}{PM} = \\lambda$. Indeed, let $AM$ and $CL$ meet at $Q$. Then\n\n$$\n\\frac{IL}{AP} = \\frac{QI}{QP} = \\frac{QN}{QM} = \\frac{IN}{PM}, \\quad \\text{implying} \\quad \\frac{IL}{IN} = \\frac{AP}{PM} = \\lambda.\n$$\n\nBecause $N$ is the midpoint of $CL$, the equality\n\n$$\n\\frac{IL}{IN} = \\lambda \\text{ gives } NL = CN = (\\lambda + 1)IN,\n$$\n\n$$\nCI = (\\lambda + 2)IN. \\text{ Hence, } \\frac{CI}{LI} = \\frac{\\lambda + 2}{\\lambda}.\n$$\n\nOn the other hand $\\frac{CI}{LI} = \\frac{AC}{AL}$ by the bisector theorem in triangle $ACL$. Under standard notation $BC = a$, $CA = b$, $AB = c$ we have $AL = \\frac{bc}{a+b}$, so\n\n$$\n\\frac{CI}{LI} = \\frac{a+b}{c}. \\text{ (The last equality is generally known as a fact). It follows that } \\frac{\\lambda+2}{\\lambda} = \\frac{a+b}{c}, \\text{ implying}\n$$\n\n$$\nc = \\frac{\\lambda}{\\lambda + 2}(a + b), \\quad c = \\frac{\\lambda}{2\\lambda + 2}(a + b + c).\n$$\n\nFor $\\lambda = \\frac{7}{3}$ and $a+b+c = 100$ the outcome is $c = 35$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22089,
"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 $120\\degree$ 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$$\n6t - \\frac{t}{120} = \\frac{360 \\times 719}{11}k = (360 \\times 65 + \\frac{360 \\times 4}{11})k\n$$\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": 22090,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1 > a_2 > \\dots > a_n$ be pairwise distinct positive integers. Prove that it is always possible to choose two of these numbers so that, for any other number from the set, its triple is different from their sum; that is, there exist $a_i, a_j$ such that $3a_k \\neq a_i + a_j$ for all $k = 1, \\dots, n$.",
"options": [],
"answer": "See solution",
"solution": "Denote the numbers by $a_1, a_2, \\dots, a_n$. Without loss of generality, assume $a_1 > a_2 > \\dots > a_n$. We may also assume not all $a_j$ are divisible by $3$. If all $b_j$ are divisible by $3$, let $3^k$ be the highest power dividing all $b_j$, and $3^{k+1}$ does not divide some $b_l$. Then $a_j = b_j / 3^k$ are still distinct positive integers, not all divisible by $3$.\n\nSuppose for contradiction that for every pair $a_i, a_j$, there exists $k$ such that $3a_k = a_i + a_j$. Consider $i = 1$: for each $a_i + a_1$, there is $k_{i1}$ with $3a_{k_{i1}}$ divisible by $a_i + a_1$. If $a_i + a_1$ is not divisible by $3$, then $a_i + a_1$ divides $a_{k_{i1}}$, but $a_i + a_1 > a_{k_{i1}}$, a contradiction. Thus, $3$ divides $a_i + a_1$ for all $2 \\leq i \\leq n$. Since not all $a_j$ are divisible by $3$, $a_2, \\dots, a_n$ all have the same nonzero remainder modulo $3$.\n\nNow, for $i = 3, \\dots, n$, $a_i + a_2$ is not divisible by $3$, so $a_i + a_2$ divides $a_{k_{i2}}$, which is only possible if $k_{i2} = 1$, so $a_3 + a_2, \\dots, a_n + a_2$ all divide $a_1$.\n\nConsider $a_1 + a_2 = 3a_m$ for some $m$. If $l \\geq 3$, $(a_1 + a_2)l > 3a_1 > 3a_m$, so $l = 1$ or $2$. If $l = 2$, $3a_m = 2(a_1 + a_2) > 4a_2$, so $m = 1$ and $a_1 = 2a_2$. But then $a_2 + a_3$ divides $a_1$, and $2(a_2 + a_3) = a_1 + 2a_3 > a_1$, so $a_2 + a_3 = a_1 = 2a_2$ or $a_2 = a_3$, contradicting distinctness. Thus, $l = 1$.\n\nSo $a_1 + a_2 = 3a_m$ for some $m$. Since $3a_m = a_1 + a_2 \\geq a_2 + a_3 + a_3 > 3a_3$, $m < 3$. If $m = 1$, $a_1 = 2a_2$, not possible since $a_1 > a_2$. Thus $m = 2$ and $a_1 = 2a_2$. As above, $a_2 + a_3$ divides $a_1$, but $a_2 + a_3 = a_1 = 2a_2$ or $a_2 = a_3$, again a contradiction.\n\nTherefore, it is always possible to choose two numbers so that no triple of any number equals their sum.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22091,
"subject": "Mathematics (Olympiad)",
"question": "找出所有有下列性質的正整數 $n$：存在一個 $n$ 的所有正因數（共 $k$ 個）的排列 $(d_1, d_2, \\dots, d_k)$，使得對於每個 $i = 1, 2, \\dots, k$，整數 $\\sum_{j=1}^{i} d_j = d_1 + \\dots + d_i$ 都是一個平方數。",
"options": [],
"answer": "See solution",
"solution": "答案是 $n = 1, 3$。\n\n對於 $i = 1, 2, \\dots, k$，令 $d_1 + \\dots + d_i = s_i^2$，並定義 $s_0 = 0$。顯然有 $0 = s_0 < s_1 < s_2 < \\dots < s_k$，因此 $s_i \\ge i$ 且\n\n$$\nd_i = s_i^2 - s_{i-1}^2 = (s_i + s_{i-1})(s_i - s_{i-1}) \\ge s_i + s_{i-1} \\ge 2i - 1. \\quad (1)\n$$\n\n數字 1 是其中一個因數，因此必有 $d_1 = 1$。\n\n考慮 $d_2$ 且 $s_2 \\ge 2$。由定義，$d_2 = s_2^2 - 1 = (s_2 - 1)(s_2 + 1)$，所以 $s_2 - 1$ 和 $s_2 + 1$ 也是 $n$ 的因數。特別地，存在某個 $j$ 使得 $d_j = s_2 + 1$。\n\n注意到\n\n$$\ns_2 + s_1 = s_2 + 1 = d_j \\ge s_j + s_{j-1}; \\qquad (2)\n$$\n\n由於序列 $s_0 < s_1 < \\dots < s_k$ 遞增，$j$ 不可能大於 2。因此，$s_2 - 1$ 和 $s_2 + 1$ 必定在 $d_1$ 和 $d_2$ 之中。於是 $s_2 - 1 = d_1 = 1$ 且 $s_2 + 1 = d_2$，因此 $s_2 = 2$ 且 $d_2 = 3$。重複上述過程，可證 $d_i = 2i - 1$ 且 $s_i = i$ 對所有 $i = 1, 2, \\dots, k$ 成立。\n\n假設對所有 $j \\le i$ 已有 $d_j = 2j - 1$ 且 $s_j = j$。考慮 $d_{i+1} = s_{i+1}^2 - s_i^2 = s_{i+1}^2 - i^2 = (s_{i+1} - i)(s_{i+1} + i)$，其中 $s_{i+1} + i$ 是 $n$ 的因數，所以存在某個 $j$ 使得 $d_j = s_{i+1} + i$。\n\n同理，由 (1) 有\n\n$$\ns_{i+1} + s_i = s_{i+1} + i = d_j \\ge s_j + s_{j-1} \\qquad (3)\n$$\n\n由於序列 $s_0 < s_1 < \\dots < s_k$ 遞增，(3) 強制 $j \\le i + 1$。另一方面，$d_j = s_{i+1} + i > 2i > d_i > d_{i-1} > \\dots > d_1$，所以 $j \\le i + 1$ 只可能 $j = i + 1$。\n\n因此，\n\n$$\n\\begin{aligned}\ns_{i+1} + i &= d_{i+1} = s_{i+1}^2 - s_i^2 = s_{i+1}^2 - i^2; \\\\\ns_{i+1}^2 - s_{i+1} &= i(i+1).\n\\end{aligned}\n$$\n\n解此方程得 $s_{i+1} = i + 1$ 且 $d_{i+1} = 2i + 1$。\n\n因此，$n$ 的因數必為 $1, 3, \\dots, n-2, n$，且 $d_k = 2k-1 = n$，為奇數。又 $d_{k-1} = n-2$ 也是 $n$ 的因數，故 $n \\le 4$。唯一可能的 $n$ 為 1 或 3，且可直接驗證兩者皆滿足條件。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22092,
"subject": "Mathematics (Olympiad)",
"question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a continuous function such that for all $x \\ge c$ and all $y \\in \\mathbb{R}$, we have\n$$\nf(f(x) + y + 2023f(y)) = x + 2024f(y).\n$$\nProve that $f(x) = x$ for all $x \\in \\mathbb{R}$.",
"options": [],
"answer": "See solution",
"solution": "We analyze the behavior of $f$ on $[c, +\\infty)$, where $f$ is truly monotone. There are two cases.\n\n**Case 1:** $\\lim_{x \\to +\\infty} f(x) = -\\infty$. We will prove $\\lim_{x \\to -\\infty} f(x) = +\\infty$. Suppose there exist real numbers $L, \\ell$ such that $f(x) \\le L$ for all $x \\le \\ell$. Fix $y_0$. Since $\\lim_{x \\to +\\infty} f(x) = -\\infty$, there exists $x_0 \\ge c$ such that $x_0 > L - 2024f(y_0)$ and $f(x_0) + y_0 + 2023 \\le \\ell$. Then, from (2),\n$$\nL < x_0 + 2024f(y_0) = f(f(x_0) + y_0 + 2023f(y_0)) \\le L.\n$$\nThis is a contradiction, so $\\lim_{x \\to -\\infty} f(x) = +\\infty$. Since $f$ is continuous, $f$ is an all-reflection on $\\mathbb{R}$.\n\nNow, fix $x_1 \\ge c$. There exists $y_1$ such that $f(y_1) = -\\frac{f(x_1)}{2023}$. From (2),\n$$\nx_1 - \\frac{2024}{2023} f(x_1) = x_1 + 2024 f(y_1) = f(f(x_1) + y_1 + 2023f(y_1)) = f(y_1) = -\\frac{f(x_1)}{2023},\n$$\nso $f(x_1) = x_1$. Thus, $f(x) = x$ for all $x \\ge c$, which contradicts $\\lim_{x \\to +\\infty} f(x) = -\\infty$.\n\n**Case 2:** $\\lim_{x \\to +\\infty} f(x) = +\\infty$. Then $f$ is strictly increasing on $[c, +\\infty)$. Suppose $f(a) = f(b)$ for some $a, b$. Since $\\lim_{x \\to +\\infty} f(x) = +\\infty$, there exists $d \\ge c$ such that $f(d)+a+2023f(a) \\ge c$ and $f(d)+b+2023f(b) \\ge c$. From (2),\n$$\nf(f(d) + a + 2023f(a)) = d + 2024f(a) = d + 2024f(b) = f(f(d) + b + 2023f(b)).\n$$\nBecause $f$ is strictly increasing, $f(d) + a + 2023f(a) = f(d) + b + 2023f(b)$, so $a = b$. Thus, $f$ is one-to-one on $\\mathbb{R}$. Since $f$ is continuous and strictly increasing on $[c, +\\infty)$, $f$ is strictly increasing on $\\mathbb{R}$.\n\nNext, we prove $\\lim_{x \\to -\\infty} f(x) = -\\infty$. Suppose there exist $V, \\nu$ such that $f(x) \\ge V$ for all $x \\le \\nu$. Substitute $y = -f(x)$ into (2):\n$$\nf(2023f(-f(x))) = x + 2024f(-f(x)), \\quad \\forall x \\ge c.\n$$\nSince $\\lim_{x \\to +\\infty} f(x) = +\\infty$, there exists $x_2 \\ge c$ such that $x_2 > f(2023f(\\nu)) - 2024V$ and $f(x_2) \\ge -\\nu$. We have $-f(x_2) \\le \\nu$ so $f(-f(x_2)) \\ge V$, thus\n$$\nx_2 + 2024f(-f(x_2)) \\ge x_2 + 2024V.\n$$\nAlso, $-f(x_2) \\le \\nu$ so $f(-f(x_2)) \\le f(\\nu)$, so\n$$\nf(2023f(-f(x_2))) \\le f(2023f(\\nu)).\n$$\nCombined with the previous equation,\n$$\n\\begin{aligned}\nf(2023f(\\nu)) &\\ge f(2023f(-f(x_2))) = x_2 + 2024f(-f(x_2)) \\\\\n&\\ge x_2 + 2024V > f(2023f(\\nu)).\n\\end{aligned}\n$$\nThis is a contradiction, so $\\lim_{x \\to -\\infty} f(x) = -\\infty$. Now $f$ is continuous, strictly increasing, and $\\lim_{x \\to \\pm\\infty} f(x) = \\pm\\infty$, so $f$ is an all-reflection. As in case 1, we have $f(x) = x$ for all $x \\ge c$. In (2), fix $x$ and choose $x \\ge c$ so that $f(x) + y + 2023f(y) \\ge c$. Then\n$$\n\\begin{aligned}\nx + y + 2023f(y) &= f(x) + y + 2023f(y) \\\\\n&= f(f(x) + y + 2023f(y)) = x + 2024f(y).\n\\end{aligned}\n$$\nIt follows that $f(y) = y$. Thus, $f(x) = x$ for all $x \\in \\mathbb{R}$. It is easy to verify that $f(x) = x$ satisfies the condition.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22093,
"subject": "Mathematics (Olympiad)",
"question": "The points $M \\in (AB)$, $N \\in (BC)$, and $P \\in (CD)$ are chosen on three sides of the rhombus $ABCD$. Prove that the centroid of the triangle $MNP$ belongs to the line $AC$ if and only if $AM + DP = BN$.",
"options": [],
"answer": "See solution",
"solution": "Let $a$ be the side length of the rhombus. Then there exist $u, v, t \\in (0, 1)$ such that $AM = au$, $BN = av$, and $DP = at$.\n\nDenote by $G$ the centroid of $MNP$, and suppose the diagonals of the rhombus intersect at $O$. Then $G \\in AC$ if and only if there exists $k \\in \\mathbb{R}$ such that $\\overline{OG} = k\\overline{OA}$.\n\nOn the other hand,\n\n$$\n\\begin{align*}\n3\\overline{OG} &= \\overline{OM} + \\overline{ON} + \\overline{OP} \\\\\n&= u\\overline{OB} + (1-u)\\overline{OA} + v\\overline{OC} + (1-v)\\overline{OB} + t\\overline{OC} + (1-t)\\overline{OD} \\\\\n&= (1-u-v-t)\\overline{OA} + (u+1-v-1+t)\\overline{OB} \\\\\n&= (1-u-v-t)\\overline{OA} + (u-v+t)\\overline{OB}.\n\\end{align*}\n$$\n\nWe conclude that $G \\in AC$ if and only if $u - v + t = 0$, which is equivalent to $AM + DP = BN$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22094,
"subject": "Mathematics (Olympiad)",
"question": "桌面上有 $n$ 张牌排成一圈。每张牌都有一面是黑的，另一面是白的。一次操作定义为：选择一张黑面朝上的牌，将它和与它相邻的两张牌同时翻面。假设一开始时，只有一张黑面朝上的牌。\n\n(a) 若 $n = 2015$，是否能通过有限次操作让所有牌都白面朝上？\n\n(b) 若 $n = 2016$，是否能通过有限次操作让所有牌都白面朝上？\n\n证明你的结论。\n\nLet $n$ cards are placed in a circle. Each card has a white side and a black side. On each move, you pick one card with black side up, flip it over, and also flip over the two neighbouring cards. Suppose initially, there is only one black-side-up card.\n\n(a) If $n = 2015$, can you make all cards white-side-up through a finite number of moves?\n\n(b) If $n = 2016$, can you make all cards white-side-up through a finite number of moves?\n\nProve your claim for both cases.",
"options": [],
"answer": "See solution",
"solution": "解：\n\n**(a)** 对于 $n = 2015$，可以通过有限次操作使所有牌都白面朝上。\n\n将牌从 1 到 2015 编号。假设最初只有第 2 张牌是黑面朝上。考虑如下操作（B=黑，W=白）：\n\n- 1B, 2W, 3B\n- 1B, 2B, 3W, 4B\n- 1B, 2B, 3B, 4W, 5B\n- ...\n- 1B, 2B, ..., 2013B, 2014W, 2015B\n- 1W, 2B, 3B, ..., 2014B, 2015W\n\n此时有 2013 张连续的黑面朝上牌。将它们每三张分为一组，分别翻面（共需 $2013/3 = 671$ 次），最终所有牌都变为白面朝上。\n\n**(b)** 对于 $n = 2016$，无法通过有限次操作使所有牌都白面朝上。\n\n将牌依次标记为 $\\alpha, \\beta, \\gamma, \\alpha, \\beta, \\gamma, \\dots$，每种标记数量相等。每次操作总是翻转一个 $\\alpha$、一个 $\\beta$ 和一个 $\\gamma$。假设初始黑面朝上的牌标记为 $\\alpha$。则需要奇数次操作才能让所有 $\\alpha$ 牌白面朝上，但需要偶数次操作才能让所有 $\\beta$（或 $\\gamma$）牌白面朝上，因此不可能全部变为白面朝上。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22095,
"subject": "Mathematics (Olympiad)",
"question": "A country has 2015 cities, some of which are connected by two-way airlines. For any $n > 3$, there are no $n$ distinct cities $A_1, A_2, \\ldots, A_n$ that form a cyclic route $A_1 \\rightarrow A_2 \\rightarrow \\cdots \\rightarrow A_n \\rightarrow A_1$ (for three cities such routes can exist). What is the maximal possible number of pairs of cities that are connected by a direct flight?",
"options": [],
"answer": "See solution",
"solution": "**Answer:** 3021.\n\nRewrite the problem in the language of graphs.\n\n\n\nFig. 10\n\n\n\nFig. 11\n\nA graph $G$ consists of vertices $\\{v_1, v_2, \\ldots, v_{2n+1}\\}$ and edges $v_i v_j$ connecting some of them. Call a sequence of edges $v_{i_1}v_{i_2}, v_{i_2}v_{i_3}, \\ldots, v_{i_{m-1}}v_{i_m}, v_i v_i$ such that $v_{i_k} \\neq v_{i_l}$, $1 \\leq k < l \\leq m$, a cycle of length $m$. If $G$ has no cycles of length more than 3, how many edges can it have at most?\n\nFirst, we shall show that a graph on $3n$ vertices satisfying the condition exists. See Fig. 11.\n\nLet us prove that this is the maximum possible number.\n\nObviously, the graph is connected; otherwise, we can draw an edge between any two connected components, which won't contradict the condition and will only increase the number of edges.\n\nDenote by $s(v_i, v_j)$ the number of edges in the shortest path between the corresponding vertices. By $L(G)$ we denote the largest possible value for all $1 \\leq i < j \\leq N$.\n\nConsider the graph with the maximal possible number of edges.\n\nAmong all of them, choose one with the minimum value of $L(G)$.\n\nAssume $L(G) \\geq 3$ and suppose one of the longest paths consists of consecutive vertices $A_1, A_2, \\ldots, A_{l+1}$, $l \\geq 3$.\n\nIf $A_1$ has degree 1 (terminal vertex, Fig. 12), that is, $A_1A_2$ is the only edge adjacent to $A_1$, we replace $A_1A_2$ with $A_1A_3$. This will not increase the length of the longest path. Do the same with all terminal vertices in all longest paths. Assume that again $L(G) \\geq 3$ and the longest path goes along vertices $A_1, A_2, \\ldots, A_{l+1}$, $l \\geq 3$.\n\nSuppose that along with $A_1A_2$ the edge $A_1B$ also exists. Here are all possible cases:\n\n- $A_2B$ is also an edge (Fig. 13). Then we modify the graph as follows. Instead of $A_1A_2$ and $A_2B$ we draw edges $A_1A_3$ and $A_3B$. The maximal length of a path will not increase.\n\n- $A_2B$ is not an edge. The shortest path from $B$ to $A_{l+1}$ cannot pass through the vertices $B, A_1, A_2, \\ldots, A_{l+1}$, because its length exceeds $L(G)$. Therefore, there is another path from $B$ to $A_{l+1}$. But then $B$ is connected at least to one of the vertices $A_3, A_4, \\ldots, A_{l+1}$. Suppose the first vertex on the shortest way between $B$ and $A_{l+1}$ is $A_k$, $k \\geq 3$. Then we have a cycle $B, A_1, A_2, \\ldots, A_k, B$ with more than three edges. Hence, this case is impossible.\n\nSo we can assume $L(G) = 2$. Suppose there is no vertex of degree $2n$. Choose a vertex $A$ with the maximal degree $m < 2n$, then we can choose another vertex $B$ not connected to $A$ (Fig. 14). Then the shortest path between $B$ and $A$ is 2, because $L(G) = 2$. Call this path $BCA$. If there is some vertex $D$ for which edges $DA$ and $DB$ exist, the cycle $BCDAB$ has length 4. Then choose an arbitrary vertex $E$ connected to $A$ but not to $B$. In this case, we must have the edge $CE$, otherwise the shortest path between $B$ and $E$ will be longer than 2. But this means the degree of $C$ is greater than that of $A$, because it is still connected to $B$, unlike $A$. Or, for the next vertex $G$, for which edges $GA$ and $GC$ exist, we have a cycle $GCEAG$. Contradiction.\n\n\n\nFig. 12\n\n\n\nFig. 13\n\n\n\nFig. 14\n\nTherefore, there is a vertex connected to all others. Denote it by $A_{2n+1}$. If the degree of another vertex $B \\neq A_{2n+1}$ is more than 2, there are edges $BC$ and $BD$, where $A_{2n+1} \\notin \\{C, D\\}$. But then we have a prohibited cycle $CBDA_{2n+1}C$, since $A_{2n+1}$ is connected to all other vertices.\n\nTherefore, besides $A_{2n+1}$, all other vertices can have degree no more than 2. But this is precisely the example from Fig. 11, where we have the maximal number of edges.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22096,
"subject": "Mathematics (Olympiad)",
"question": "A triangle $ABC$ is given. The circle $w$ with center at point $Q$ touches the side $BC$ and is internally tangent to the circumscribed circle of $\\Delta ABC$ at point $A$. Let $M$ be the midpoint of side $BC$, and $N$ be the midpoint of the arc $BAC$ of the circumscribed circle of $\\Delta ABC$. On the side $BC$, the point $S$ is selected such that $\\angle BAM = \\angle SAC$. Prove that the points $N$, $Q$, and $S$ are collinear.",
"options": [],
"answer": "See solution",
"solution": "Denote by $\\Gamma$ the circumscribed circle of $\\triangle ABC$. Let $W$ be the midpoint of the smaller arc $BC$ of the circle $\\Gamma$, and $O$ be the center of $\\Gamma$. Obviously, points $N$, $M$, $W$, $O$ lie on the perpendicular bisector of segment $BC$. By the Lemma of Archimedes, the points of contact of $w$ to $BC$, to $\\Gamma$, and the point $W$ are collinear. As $w$ is tangent to the circle $\\Gamma$ at point $A$, the circle $w$ touches $BC$ at point $L$, which is the intersection of $BC$ and $AW$. Note that\n\n\n\n$L$ is the base of the bisector of angle $\\angle BAC$. Draw the rays $AM$ and $AS$ until they intersect the circle $\\Gamma$ at points $X$ and $P$ respectively. By the statement of the problem, $\\angle BAX = \\angle PAC$, hence $P$ and $X$ are symmetric with respect to $WN$. Also, $\\angle XAW = \\angle WAP$. $WN$ is the diameter of $\\Gamma$, therefore $\\angle NML = \\angle NAL = 90^\\circ$, and thus the quadrilateral $ALMN$ is cyclic. Then $\\angle WNP = \\angle WAP = \\angle XAW = \\angle MNL$, and hence $N$, $L$, $P$ are collinear. Note that $LM$, $NA$, $WP$ are the altitudes of $\\triangle NLW$, and their intersection point is the orthocenter $K$. Let $T$ be the intersection point of the ray $AN$ and $w$. There exists a homothetic transformation mapping $\\Gamma$ to $w$. Under this transformation, point $N$ maps to $T$, point $W$ to $L$, i.e., $TL$ is the diameter of $w$, because $WN$ is the diameter of $\\Gamma$. Hence, $Q$ is the midpoint of segment $TL$. Apply Menelaus' theorem to $\\triangle TLK$ and points $N$, $Q$, $S$. To prove $N$, $Q$, $S$ are collinear, it is sufficient to show that\n\n$$\n\\frac{TN}{NK} \\cdot \\frac{KS}{SL} \\cdot \\frac{LQ}{QT} = 1, \\text{ i.e. (as } TQ = QL) \\frac{TN}{NK} = \\frac{SL}{KS}.\n$$\n\nLet us prove this equality. It is clear that the lines $AL$ and $AK$ are the internal and external bisectors of angle $\\angle MAS$, hence $\\frac{ML}{LS} = \\frac{MK}{SK}$. Then $\\frac{ML}{MK} = \\frac{SL}{SK}$. Obviously, $TL \\parallel NW$, hence by Thales' theorem $\\frac{TN}{NK} = \\frac{LM}{MK}$, therefore $\\frac{TN}{NK} = \\frac{SL}{SK}$, Q.E.D.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22097,
"subject": "Mathematics (Olympiad)",
"question": "$n$ students arrived to the summer camp. Every child (boy or girl) in the camp knows exactly one boy and one girl. Is it possible if\n\na) $n = 2000$;\n\nb) $n = 2018$.",
"options": [],
"answer": "See solution",
"solution": "**Answer:** a) yes, b) no.\n\n**Solution.**\n\na) Suppose 1000 boys and 1000 girls arrived at the camp. We can split them into groups of 4 students each: 2 boys and 2 girls. They know each other in the following way: $B_1 \\leftrightarrow B_2 \\leftrightarrow G_2 \\leftrightarrow G_1 \\leftrightarrow B_1$. One can check that all conditions are satisfied.\n\nb) Suppose by contradiction that it is possible. Then every boy can be paired with a girl if they know each other. Thus, there must be exactly 1009 boys and 1009 girls. Similarly, all the boys can be divided into pairs if they know each other, so the number of boys must be even. That leads to a contradiction.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22098,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle. The point $X$ lies on the extension of $AC$ beyond $A$, such that $AX = AB$. Similarly, the point $Y$ lies on the extension of $BC$ beyond $B$ such that $BY = AB$.\n\nProve that the circumcircles of $ACY$ and $BCX$ intersect a second time in a point different from $C$ that lies on the bisector of the angle $\\angle BCA$.",
"options": [],
"answer": "See solution",
"solution": "Let the angles at $A$, $B$, and $C$ be $\\alpha$, $\\beta$, and $\\gamma$, respectively.\n\nIt is sufficient to show that the center $I_c$ of the excircle touching the line $AB$ lies on both circles. To do this, consider the respective inscribed angles.\n\nSince triangle $AYB$ is isosceles, we have:\n\n$$\n\\angle CYA = \\angle BYA = 90^\\circ - \\frac{1}{2} \\angle YBA = 90^\\circ - 90^\\circ + \\frac{\\beta}{2} = \\frac{\\beta}{2}.\n$$\n\nAlso,\n\n$$\n\\angle CI_cA = 180^\\circ - \\frac{180^\\circ - \\alpha}{2} - \\alpha - \\frac{\\gamma}{2} = 90^\\circ - \\frac{\\alpha}{2} - \\frac{\\gamma}{2} = \\frac{\\beta}{2}.\n$$\n\nSo $I_c$ lies on the circumcircle of $ACY$ by the inverse of the inscribed angle theorem. Similarly, $I_c$ also lies on the circumcircle of $BCX$. Thus, $I_c$ is the second point of intersection, which lies on the angle bisector through $C$ as required.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22099,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. We want to make up a collection of cards with the following properties:\n\n- Each card has a number of the form $m!$ written on it, where $m$ is a positive integer.\n- For any positive integer $t \\leq n!$, we can select some card(s) from this collection such that the sum of the number(s) on the selected card(s) is $t$.\n\nDetermine the smallest possible number of cards needed in this collection.",
"options": [],
"answer": "See solution",
"solution": "We need at least $\\frac{n(n-1)}{2} + 1$ cards.\n\nFor example, we can have $i$ cards with number $i!$ for each $i = 1, 2, \\dots, n-1$ and another card with number $n!$. For $t = n!$, we can simply choose the card with number $n!$. Suppose that $1 \\leq t < n!$. Let $r_0 = t$, and for $i = 1, 2, \\dots, n-1$, define integers $q_i$ and $r_i$ inductively, using the division algorithm:\n\n$$\nr_{i-1} = q_i (n-i)! + r_i, \\quad 0 \\leq r_i < (n-i)!.\n$$\n\nThen $t = \\sum_{i=1}^{n-1} q_i (n-i)!$. Since $r_{i-1} < (n-i+1)!$, we have $q_i \\leq n-i$. Thus, we can choose $q_i$ cards with number $(n-i)!$ for each $i = 1, 2, \\dots, n-1$ so that the sum of the numbers is $t$. So the required properties are satisfied.\n\nNext, consider the smallest set of cards we can make with numbers adding up to $n! - 1$. Clearly, this set cannot contain any card with number greater than $(n-1)!$. For each $i = 1, 2, \\dots, n-1$, let $c_i$ be the number of cards with number $i!$ in this set. Then $c_i \\leq i$ for all $i$, for if $c_i \\geq i+1$ then we can replace $i+1$ cards with number $i!$ in this set with just one card with number $(i + 1)!$, contradicting the minimality of the set. So now we have that\n\n$$\nn! - 1 = \\sum_{i=1}^{n-1} c_i i! \\leq \\sum_{i=1}^{n-1} i(i!) = \\sum_{i=1}^{n-1} ((i + 1)! - i!) = n! - 1.\n$$\n\nThis implies that all inequalities involved must be equality; that is, $c_i = i$ for all $i$. Thus, this set has $1 + 2 + \\dots + (n - 1) = \\frac{n(n-1)}{2}$ cards.\n\nSuppose now that we have a collection of cards with the required properties. Then some of these cards have numbers adding up to $n! - 1$. So by what we have just shown, this collection must contain at least $\\frac{n(n-1)}{2}$ cards. However, if we have exactly $\\frac{n(n-1)}{2}$ cards, then we must select all these cards for the sum of the numbers to be $n! - 1$, but this means that we cannot select cards for the sum of the numbers to be $n!$, a contradiction. Therefore we need at least $\\frac{n(n-1)}{2} + 1$ cards. $\\square$\n\n**Remark.** There are exactly $n$ collections of $\\frac{n(n-1)}{2} + 1$ cards with such properties: such a collection must consist of $i$ cards with number $i!$ for each $i = 1, 2, \\dots, n-1$ and one additional card with number $m!$ where $m$ can be any integer from $1$ to $n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22100,
"subject": "Mathematics (Olympiad)",
"question": "Find the smallest positive integer $n$ such that $n^2$ is a perfect cube, and $n^3$ is a perfect square.",
"options": [],
"answer": "See solution",
"solution": "Let $n$ be a positive integer. We require $n^2$ to be a perfect cube and $n^3$ to be a perfect square.\n\nLet $n = p_1^{a_1} p_2^{a_2} \\cdots p_k^{a_k}$ be the prime factorization of $n$.\n\n- $n^2$ is a perfect cube $\\implies$ for each $i$, $2a_i$ is divisible by $3$.\n- $n^3$ is a perfect square $\\implies$ for each $i$, $3a_i$ is divisible by $2$.\n\nSo $a_i$ must satisfy both $2a_i \\equiv 0 \\pmod{3}$ and $3a_i \\equiv 0 \\pmod{2}$.\n\nThe smallest positive $a_i$ satisfying both is $a_i = 6$ (since $2 \\times 6 = 12$ is divisible by $3$, and $3 \\times 6 = 18$ is divisible by $2$).\n\nThus, the smallest $n$ is $n = p^6$ for any prime $p$. The smallest such $n$ is $2^6 = 64$.\n\nBut we can also use two primes: $n = 2^6 \\cdot 3^6 = 64 \\cdot 729 = 46656$, but $64$ is smaller.\n\nTherefore, the smallest positive integer $n$ is $64$.\n\nCheck:\n- $64^2 = 4096$, which is $16^3$ (a perfect cube).\n- $64^3 = 262144$, which is $512^2$ (a perfect square).\n\nSo, $n = 64$ is the answer.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22101,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. Is $$(2n-3)(2n-1)(2n+1)(2n+3) + 16$$ a square of a positive integer?\n\nIs the number $2005 \\times 2007 \\times 2009 \\times 2011 + 16$ a square of a positive integer?",
"options": [],
"answer": "See solution",
"solution": "Since $(2n-3)(2n+3) = 4n^2 - 9$ and $(2n-1)(2n+1) = 4n^2 - 1$, we have:\n\n$$(2n-3)(2n-1)(2n+1)(2n+3) + 16 = (4n^2 - 9)(4n^2 - 1) + 16$$\n\nExpanding:\n\n$$(4n^2 - 9)(4n^2 - 1) = 16n^4 - 4n^2 - 36n^2 + 9 = 16n^4 - 40n^2 + 9$$\n\nSo,\n\n$$16n^4 - 40n^2 + 9 + 16 = 16n^4 - 40n^2 + 25 = (4n^2 - 5)^2$$\n\nTherefore, $(2n-3)(2n-1)(2n+1)(2n+3) + 16$ is a square of a positive integer.\n\nFor $n = 1004$:\n\n$$(2 \\times 1004 - 3)(2 \\times 1004 - 1)(2 \\times 1004 + 1)(2 \\times 1004 + 3) + 16 = 2005 \\times 2007 \\times 2009 \\times 2011 + 16$$\n\nThis equals $$(4 \\times 1004^2 - 5)^2 = 1008016^2$$\n\nSo $2005 \\times 2007 \\times 2009 \\times 2011 + 16$ is a square of a positive integer.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22102,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f: \\mathbb{R}^2 \\rightarrow \\mathbb{R}$ for which\n\n$$\nf(xf(x,z), yf(y,z)) = f(x,z)f(y,z)\n$$\nfor every real numbers $x$, $y$, and $z$, and $f(0,0) \\neq 0$.",
"options": [],
"answer": "See solution",
"solution": "Let $f(0,0) \\neq 0$. Setting $x = y = z = 0$ gives $f(0,0) = (f(0,0))^2$, so $f(0,0) = 1$.\n\nSetting $x = y = 0$ gives $f(0,0) = (f(0,z))^2$, so $f(0,z) = \\pm 1$ for all $z$.\n\nSetting $x = 0$ gives $\\pm 1 = \\pm f(y,z)$, so $f(y,z) = \\pm 1$ for all $y, z$.\n\nSuppose $f(t,t) = -1$ for some $t$. Then for $x = y = z = t$, $f(-t,-t) = (-1)^2 = 1$. Thus, for every $t$, either $f(t,t) = 1$ or $f(-t,-t) = 1$.\n\nLet $t$ be arbitrary and $s$ such that $f(s,s) = 1$ and $|s| = t$. If $f(t,t) = -1$, then $f(-t,-t) = 1$.\n\nFor $x = z = t$, $y = -t$, $f(s,-sf(-s,s)) = f(-s,s)$. If $f(-s,s) = -1$, then $f(s,s) = -1$, a contradiction. So $f(-s,s) = 1$ and $f(s,-s) = f(-s,s) = 1$.\n\nIf $f(-s,-s) = -1$, for $x = s$, $y = z = -s$, $f(s,s) = f(s,-s)f(-s,-s) = -1$, which is impossible. Thus, for every $t$, $f(t, \\pm t) = 1$.\n\nIf $f(u,v) = -1$ for some $u, v$, then for $x = u$, $y = z = v$, $f(-u,v) = -1$. For $x = z = v$, $y = u$, $f(v,-u) = -1$. For $x = z = -u$, $y = v$, $f(-u,-v) = -1$. For $x = -u$, $y = z = -v$, $f(u,-v) = -1$. For $x = v$, $y = z = -u$, $f(-v,-u) = -1$. For $x = z = -v$, $y = -u$, $f(-v,u) = -1$. For $x = z = v$, $y = -u$, $f(v,u) = -1$.\n\nThus, for all $u, v$ with $f(u,v) = -1$:\n$$\nf(u, \\pm v) = f(\\pm u, v) = f(v, \\pm u) = f(\\pm v, u) = -1,\n$$\nand if $f(u,v) = 1$:\n$$\nf(u, \\pm v) = f(\\pm u, v) = f(v, \\pm u) = f(\\pm v, u) = 1.\n$$\n\nTherefore, $f(xf(x,z), yf(y,z)) = f(x, y)$ for all $x, y, z$, so\n$$\nf(x,y) = f(x,z)f(y,z).\n$$\n\nIf $f(a,b) = -1$, then for any $z$, $x = a$, $y = b$, $-1 = f(a,z)f(b,z)$, so either $f(a,z) = 1$, $f(b,z) = -1$ or $f(a,z) = -1$, $f(b,z) = 1$.\n\nIf $f(a,c) = -1$, $f(a,d) = 1$, then for $x = c$, $y = d$, $z = a$:\n$$\nf(c,d) = f(c,a)f(d,a) = -1.\n$$\nIf $f(a,c) = f(a,d) = 1$, then $f(c,d) = 1$; if $f(a,c) = f(a,d) = -1$, then $f(c,d) = 1$. Thus, if $f(a,c)$ and $f(a,d)$ have the same sign, $f(c,d) = 1$; if opposite, $f(c,d) = -1$.\n\nLet $M \\subseteq \\mathbb{R}^+ \\cup \\{0\\}$ be $M = \\{x \\mid f(a,x) = 1\\}$ and $N = \\mathbb{R}^+ \\cup \\{0\\} \\setminus M$. Then $f(x,y) = 1$ iff $|x|$ and $|y|$ are in the same set, and $f(x,y) = -1$ iff $|x|$ and $|y|$ are in different sets. If $N$ is empty, $f(x,y) = 1$ for all $x, y$.\n\nThus, for any subset $M \\subseteq \\mathbb{R}^+ \\cup \\{0\\}$, define $f(x,y) = 1$ iff $|x|, |y| \\in M$ or $|x|, |y| \\in N$; $f(x,y) = -1$ otherwise. These functions satisfy the given equation and are unique.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22103,
"subject": "Mathematics (Olympiad)",
"question": "All my maths tests are out of the same total, and the average of my marks in the four tests so far is 75%. What percentage do I need in a fifth test so that my overall average will be 80%?\n\n(A) 100\n\n(B) 90\n\n(C) 85\n\n(D) 80\n\n(E) 75",
"options": [],
"answer": "See solution",
"solution": "Let the five marks be $a$, $b$, $c$, $d$, and $e$.\n\n$$\n\\frac{a + b + c + d + e}{5} = 80 \\implies a + b + c + d + e = 400\n$$\n\nBut $\\frac{a + b + c + d}{4} = 75 \\implies a + b + c + d = 300$, so $e = 400 - 300 = 100$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22104,
"subject": "Mathematics (Olympiad)",
"question": "Let $k$ be a non-negative integer and let $A$ be a nonempty finite set. We define a *word from the alphabet $A$ of length $k$* as a sequence of elements of $A$ of length $k$. (Notice that by definition the empty sequence is a word of length $0$.)\n\nLet $B$ be the set of all words from the nine-element alphabet\n$$\n\\{b, @, l, t, i, c, w, a, y\\},\n$$\nwhose lengths are at most $2023$.\n\nMoreover, let $M$ be the set of all words from the four-element alphabet\n$$\n\\{m, a, t, h\\},\n$$\nwhose lengths are at most $2025$ and which contain the element $t$ exactly twice.\n\n(For example, $tata \\in M$, because it has length $4$ and contains the letter $t$ twice, but $mahatma \\notin M$, because it contains the letter $t$ only once.)\n\nProve that the difference $|B| - |M|$ is divisible by $3^{2023}$.",
"options": [],
"answer": "See solution",
"solution": "For $k \\in \\mathbb{N}_0$, let $b_k$ be the number of words from the alphabet $\\{b, @, l, t, i, c, w, a, y\\}$ of maximum length $k$, so $b_{2023} = |B|$. Let $m_k$ be the number of words of maximum length $k+2$ from the alphabet $\\{m, a, t, h\\}$ that contain the letter $t$ exactly twice, so $m_{2023} = |M|$.\n\nFor example, $b_0 = 1$ (the empty word), $b_1 = b_0 + 9 = 10$ (adding $9$ one-letter words), and $b_2 = b_1 + 9^2 = 91$ (adding $9^2$ two-letter words). More generally,\n$$\nb_k = b_{k-1} + 9^k\n$$\nfor all $k \\ge 1$, since there are exactly $9^k$ words of length $k$ from the given alphabet. Alternatively,\n$$\nb_k = 1 + 9b_{k-1}\n$$\nfor all $k \\ge 1$, since every word of maximum length $k$ is either the empty word or the concatenation of a word of maximum length $k-1$ and one of the nine letters.\n\nFor $m_k$, $m_0 = 1$ (the word $tt$), $m_1 = m_0 + 9 = 10$ (adding $9$ words of the form $tx$, $txt$, $xtt$ with $x \\in \\{h, a, m\\}$), and $m_2 = m_1 + 54 = 64$ (adding $6 \\cdot 9 = 54$ words of the form $ttxy$, $txty$, $txyt$, $xtty$, $xtyt$, $xytt$ with $x, y \\in \\{h, a, m\\}$).\n\n**Lemma.** The recursion\n$$\nm_{k+2} = 9m_{k+1} - 27m_k + 27m_{k-1} + 1\n$$\nholds for $k \\ge 1$.\n\n*Proof of the lemma.*\nLet $p_k$ be the number of words of maximum length $k+2$ from $\\{m, a, t, h\\}$ with exactly one $t$, and $q_k$ the number with no $t$ (i.e., from $\\{h, a, m\\}$).\n\nA word of maximum length $k+2$ with exactly two $t$'s is either:\n- $t$ concatenated with a word of maximum length $k+1$ with exactly one $t$ ($p_{k-1}$ cases), or\n- $x \\in \\{h, a, m\\}$ concatenated with a word of maximum length $k+1$ with exactly two $t$'s ($3m_{k-1}$ cases).\nSo,\n$$\nm_k = p_{k-1} + 3m_{k-1}.\n$$\nSimilarly,\n$$\np_k = q_{k-1} + 3p_{k-1},\n$$\n$$\nq_k = 1 + 3q_{k-1}.\n$$\nFrom above, $p_{k-1} = m_k - 3m_{k-1}$, so $p_k = m_{k+1} - 3m_k$. Plugging into the previous equation and shifting indices gives the desired recursion.\n\nNow, we show $b_k \\equiv m_k \\pmod{3^k}$ for all $k \\ge 0$, which gives the result for $k = 2023$.\n\nWe proceed by strong induction on $k$.\n- For $k=0$: $b_0 - m_0 = 1 - 1 = 0$ is divisible by $3^0 = 1$.\n- For $k=1$: $b_1 - m_1 = 10 - 10 = 0$ is divisible by $3^1 = 3$.\n- For $k=2$: $b_2 - m_2 = 91 - 64 = 27$ is divisible by $3^2 = 9$.\n\nSuppose for $k \\ge 3$ that $b_l \\equiv m_l \\pmod{3^l}$ for all $l \\in [0, k-1]$. Then,\n$$\nm_k = 9m_{k-1} - 27m_{k-2} + 27m_{k-3} + 1.\n$$\nBy similar recursions for $b_k$, the difference $b_k - m_k$ is divisible by $3^k$ by induction. Thus, $|B| - |M|$ is divisible by $3^{2023}$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22105,
"subject": "Mathematics (Olympiad)",
"question": "Prove that, among any ten consecutive positive integers, there are five numbers such that no two of them have a common factor larger than $1$.",
"options": [],
"answer": "See solution",
"solution": "Among any ten consecutive positive integers, five of them are odd and the remaining five are even.\n\n- The five odd integers contain either one or two multiples of $3$. We remove one multiple of $3$ and select the remaining four integers.\n- The five even integers contain at most two multiples of $3$, at most one multiple of $5$, and at most one multiple of $7$. Therefore, there is at least one of them that is not a multiple of $3$, $5$, or $7$. We select this integer.\n\nWe now prove that, among the five integers selected, no two of them have a common factor larger than $1$. Since the largest possible difference between two of the numbers is $9$, the largest possible common factor that two of the numbers can have is $9$. So it suffices to show that we have not selected a pair of numbers with a common factor of $2$, $3$, $5$, or $7$.\n\n- By construction, we have chosen exactly one number that is divisible by $2$.\n- By construction, we have chosen at most one number that is divisible by $3$.\n- At most one of the odd integers selected is divisible by $5$ and, by construction, the even integer selected is not divisible by $5$.\n- At most one of the odd integers selected is divisible by $7$ and, by construction, the even integer selected is not divisible by $7$.\n\nTherefore, we have selected five numbers such that no two of them have a common factor larger than $1$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22106,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\geq 3$. For a sequence $a_1, a_2, \\dots, a_n$ of integers, the following operations are allowed:\n\nFor any $1 \\leq k \\leq n-2$:\n\n1. Increase $a_k$ and $a_{k+2}$ by $1$ each, and decrease $a_{k+1}$ by $1$.\n2. Decrease $a_k$ and $a_{k+2}$ by $1$ each, and increase $a_{k+1}$ by $1$.\n\nStarting from the sequence $1, 2, \\dots, 2022$, is it possible to arrive at a constant sequence after performing a finite number of operations?",
"options": [],
"answer": "See solution",
"solution": "No, it is not possible.\n\nFor a sequence $a_1, a_2, \\dots, a_n$, define the invariant:\n\n$$\nS = \\sum_{k=1}^{n} 3^k a_k \\pmod{7}.\n$$\n\nFor $c = \\pm 1$, the operation $(a_k, a_{k+1}, a_{k+2}) \\to (a_k + c, a_{k+1} - c, a_{k+2} + c)$ leaves $S$ unchanged modulo $7$.\n\nLet $N = 2022$ and consider $S = S(1, 2, \\dots, N)$. Calculations show $S \\equiv 2 \\pmod{7}$.\n\nFor a constant sequence $b, \\dots, b$, $S(b, \\dots, b) \\equiv 0 \\pmod{7}$.\n\nSince the invariant differs, it is impossible to reach a constant sequence.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22107,
"subject": "Mathematics (Olympiad)",
"question": "Determine whether there exists a positive integer $k$ such that $p = 6k + 1$ is prime and\n\n$$\n\\binom{3k}{k} \\equiv 1 \\pmod{p}.\n$$",
"options": [],
"answer": "See solution",
"solution": "No such $k$ exists. Suppose $k$ and $p$ are as described. Consider\n\n$$\nA = \\sum_{i=0}^{p-1} (i^3 - 1)^{3k}.\n$$\n\nSince $p-1 = 6k$ is divisible by 3, there are three cube roots of 1 modulo $p$. Thus, three terms in the sum are $0$ modulo $p$, and the others are $[(p-1)/2]$th powers of nonzero residues, so are congruent to either $1$ or $-1$ modulo $p$. Therefore, $A$ is congruent to one of $p-3, p-5, p-7, \\dots, -(p-3)$ modulo $p$, and cannot be congruent to $1$ or $-1$ modulo $p$.\n\nOn the other hand, by the binomial theorem and changing the order of summation:\n\n$$\nA = \\sum_{i=0}^{p-1} \\left( \\sum_{j=0}^{3k} \\binom{3k}{j} (-1)^j i^{3(3k-j)} \\right) = \\sum_{j=0}^{3k} \\left( (-1)^j \\binom{3k}{j} \\sum_{i=0}^{p-1} i^{3(3k-j)} \\right),\n$$\n\nwhere $0^0 = 1$ for $i = 0, j = 3k$.\n\nWe claim $\\sum_{i=0}^{p-1} i^d \\equiv 0 \\pmod{p}$ if $p-1 \\nmid d$. For this, let $g$ be a primitive root modulo $p$; then $g^d \\not\\equiv 1 \\pmod{p}$, and\n\n$$\n(g^d - 1) \\sum_{i=0}^{p-1} i^d \\equiv \\sum_{i=0}^{p-1} (g i)^d - \\sum_{i=0}^{p-1} i^d \\equiv 0 \\pmod{p},\n$$\n\nso the claim holds. Also, $\\sum_{i=0}^{p-1} i^d \\equiv 0 \\pmod{p}$ when $d = 0$.\n\nSince $3 \\cdot (3k) < 2(p-1)$, the only $j$ such that $3 \\cdot (3k-j)$ is a multiple of $p-1$ is $j = k$, where $3 \\cdot (3k-j) = 6k = p-1$. Thus, $\\sum_i i^{3(3k-j)} \\equiv 0 \\pmod{p}$ unless $j = k$. For $j = k$,\n\n$$\n\\sum_{i=0}^{p-1} i^{6k} \\equiv \\sum_{i=0}^{p-1} i^{p-1} \\equiv \\sum_{i=1}^{p-1} 1 \\equiv -1 \\pmod{p}.\n$$\n\nTherefore,\n\n$$\nA \\equiv (-1)^k \\binom{3k}{k} (-1) \\pmod{p}.\n$$\n\nSince $A$ is not congruent to $1$ or $-1$ modulo $p$, $\\binom{3k}{k} \\not\\equiv 1 \\pmod{p}$, a contradiction.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22108,
"subject": "Mathematics (Olympiad)",
"question": "Given integer $n \\ge 2$. Find the least positive integer $m$, such that there are $n^2$ distinct positive real numbers $x_{i,j}$ ($1 \\le i, j \\le n$) satisfying the following conditions:\n\n1. For every $i, j$,\n\n$$\nx_{i,j} = \\max\\{x_{i,1}, x_{i,2}, \\dots, x_{i,j}\\} \\text{ or}\n$$\n$$\nx_{i,j} = \\max\\{x_{1,j}, x_{2,j}, \\dots, x_{i,j}\\}\n$$\n\n2. For every $i$, there are at most $m$ indices $k$, with\n$$\nx_{i,k} = \\max\\{x_{i,1}, x_{i,2}, \\dots, x_{i,k}\\}\n$$\n\n3. For every $j$, there are at most $m$ indices $k$, with\n$$\nx_{k,j} = \\max\\{x_{1,j}, x_{2,j}, \\dots, x_{k,j}\\}\n$$\n\nFind the least $m$ for which such an arrangement is possible.",
"options": [],
"answer": "See solution",
"solution": "$m = \\left\\lfloor \\frac{n+3}{2} \\right\\rfloor$.\n\nPut these $n^2$ numbers $x_{i,j}$ in an $n \\times n$ table: call $x_{i,j}$ a \"row pivot\" if $x_{i,j} = \\max\\{x_{i,1}, x_{i,2}, \\dots, x_{i,j}\\}$; call $x_{i,j}$ a \"column pivot\" if $x_{i,j} = \\max\\{x_{1,j}, x_{2,j}, \\dots, x_{i,j}\\}$.\n\nFirst, prove $m \\ge \\left\\lfloor \\frac{n+3}{2} \\right\\rfloor$. Consider all $(n-1)^2$ numbers $x_{i,j}$ ($2 \\le i, j \\le n$): every column (from $2$ to $n$) contains at most $m-1$ column pivots (by condition (3), and notice that $x_{1,j}$ ($2 \\le j \\le n$) is a pivot); every row (from $2$ to $n$) contains at most $m-1$ row pivots (by condition (2), and notice that $x_{i,1}$ ($2 \\le i \\le n$) is a pivot). Due to condition (1), every number is a pivot (row or column or both), and hence\n\n$$\n(n-1)(m-1) + (n-1)(m-1) \\ge (n-1)^2.\n$$\n\nWe claim the equality above cannot be attained. Let $x_{i_0,j_0} = \\max x_{i,j}$ ($2 \\le i, j \\le n$). If $x_{i_0,j_0}$ is both row and column pivots, then $x_{i_0,j_0}$ is counted twice on the left-hand side, and the equality cannot hold. Suppose $x_{i_0,j_0}$ is not a row pivot. Then $x_{i_0,1} > x_{i_0,j_0}$ and there are no other row pivots in this row, so the equality cannot hold. Similarly, if $x_{i_0,j_0}$ is not a column pivot, then there are no other column pivots in this column, and the equality cannot hold. Therefore, $2(m-1) > n-1$, so $m \\ge \\left\\lfloor \\frac{n+3}{2} \\right\\rfloor$.\n\nNext, construct tables for which $m = \\left\\lfloor \\frac{n+3}{2} \\right\\rfloor$. When $n = 2t$, divide the $n \\times n$ table into four $t \\times t$ tables as follows:\n\n\n\nTake $\\varepsilon_1, \\varepsilon_2 > 0$ sufficiently small, such that $\\varepsilon_1 < \\frac{1}{2n}\\varepsilon_2 < \\frac{1}{4n^2}$. For $k=1, 4$, let the number in row $i$ and column $j$ of $\\Omega_k$ be $k - i\\varepsilon_1 + j\\varepsilon_2$. Note that in $\\Omega_1$ and $\\Omega_4$, the numbers decrease from the top right to the bottom left. For $k = 2, 3$, let the number in row $i$ and column $j$ of $\\Omega_k$ be $k + i\\varepsilon_1 - j\\varepsilon_2$. Note that in $\\Omega_2$ and $\\Omega_3$ the numbers decrease from the bottom left to the top right.\n\nIt is easy to check that all numbers in $\\Omega_1$ and $\\Omega_4$ are row pivots, and all numbers in $\\Omega_2$ and $\\Omega_3$ are column pivots. In addition, except for the first and the $t+1$ rows, there are no other column pivots in $\\Omega_1$ and $\\Omega_4$; except for the first and the $t+1$ columns, there are no other row pivots in $\\Omega_2$ and $\\Omega_3$. Hence, conditions (2) and (3) are fulfilled, and $m = t+1$.\n\nWhen $n = 2t - 1$, take the construction for $n = 2t$ as above and remove the rightmost column and the bottom row.\n\n$\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22109,
"subject": "Mathematics (Olympiad)",
"question": "Let $n > 0$ be an integer. We are given a balance and $n$ weights of weight $2^0, 2^1, \\dots, 2^{n-1}$. We are to place each of the $n$ weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed.\n\nDetermine the number of ways in which this can be done.",
"options": [],
"answer": "See solution",
"solution": "The answer is $$(2n - 1)!! = 1 \\cdot 3 \\cdot 5 \\cdots (2n - 1).$$\n\nCall a sequence of moves valid if the right pan is never heavier than the left pan when making these moves. It suffices to give a $(2n + 1)$-to-1 mapping between valid sequences for weights $2^0, \\dots, 2^{n+1}$ and weights $2^0, \\dots, 2^n$.\n\nFor a valid sequence of moves of weights $2^0, 2^1, \\dots, 2^{n+1}$, if we remove the move of putting weight $2^0$ in this sequence and divide the remaining weights by 2, we obtain a valid sequence of moves of weights $2^0, \\dots, 2^n$. On the other hand, for a valid sequence $S$ of weights $2^0, \\dots, 2^n$, doubling each weight gives a valid sequence $S'$ of weights $2^1, \\dots, 2^{n+1}$. Note that the difference in weight between the left and right pans is always at least 2 after the first move in $S'$. Therefore, modifying $S'$ by adding weight $2^0$ to the left pan on the first move or to either pan on any move after the first yields $2n + 1$ valid sequences of weights $2^0, \\dots, 2^{n+1}$. These two constructions give the desired mapping.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22110,
"subject": "Mathematics (Olympiad)",
"question": "Three circles touch externally. $A_1A_2$, $B_1B_2$, $C_1C_2$ are their diameters, all oriented in the same direction. Prove that the lines $A_1B_2$, $B_1C_2$, and $C_1A_2$ are concurrent.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let the circles with diameters $B_1B_2$ and $C_1C_2$ touch at point $A$. Similarly, define points $B$ and $C$ (see figure). Since the touching point is the center of homothety mapping one circle to another, $A = B_1C_2 \\cap B_2C_1$, and similarly for the other points. Let $D = A_2C_1 \\cap A_1B_2$. The points $A$, $B$, $C$, and $D$ are cyclic because\n\n$$\n\\angle CDB = \\angle B_2DC_1 = \\pi - \\angle DB_2C_1 - \\angle DC_1B_2 = \\pi - \\angle CAE - \\angle BAE = \\pi - \\angle CAB,\n$$\n\nwhere $E$ is a point on the common external tangent to the circles with diameters $B_1B_2$ and $C_1C_2$. Thus, $A_2C_1$ and $A_1B_2$ intersect on the circumcircle of quadrilateral $ABCD$. By analogy, the intersection point of $A_2B_1$ and $B_2C_1$ also lies on this circle. We show that the points $A$, $B$, $C$, and $F = A_1B_2 \\cap D_1C_2$ are cyclic:\n\n$$\n\\begin{align*}\n\\angle CFA &= \\angle FA_1C_2 + \\angle FC_2A_1 \\\\\n&= \\angle CA_1B + \\angle FC_2B \\\\\n&= \\angle CA_1B + \\angle BC_1A \\\\\n&= \\angle CBE_1 + \\angle E_1BA \\\\\n&= \\angle CBA.\n\\end{align*}\n$$\n\nExcept for point $C$, the line $A_1B_2$ intersects the circle at point $D$, hence $A_1B_2$, $B_1C_2$, and $C_1A_2$ are concurrent.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22111,
"subject": "Mathematics (Olympiad)",
"question": "The squares of an $8\\times8$ board are coloured alternatingly black and white. A rectangle consisting of some of the squares of the board is called *important* if its sides are parallel to the sides of the board and all its corner squares are coloured black. The side lengths can be anything from 1 to 8 squares. On each of the 64 squares of the board, we write the number of important rectangles in which it is contained. The sum of the numbers on black squares is $B$, and the sum of the numbers on white squares is $W$. Determine the difference $B-W$.",
"options": [],
"answer": "See solution",
"solution": "In each important rectangle, the number of black squares is one more than the number of white squares. Hence, each important rectangle contributes $+1$ to the difference $B-W$. The value of $B-W$ is thus the same as the number of important rectangles on the board.\n\nLet us number the rows on the board $1, 2, \\ldots, 8$ from the top downwards and the columns $1, 2, \\ldots, 8$ from the left to the right. So $(1, 1)$ is the upper left square and $(8, 8)$ denotes the lower right square. Assume $(1, 1)$ is a black square. Then all $(i, j)$ with both $i$ and $j$ odd, as well as all those with both $i$ and $j$ even, are black squares. All other squares are white. By focusing only on the four odd-numbered rows and the four odd-numbered columns, we find that they determine $(4 + \\binom{4}{2})^2 = 100$ important rectangles. Similarly, the four even-numbered rows and the four even-numbered columns determine another $100$ important rectangles, giving a total of $200$ important rectangles on the board. It follows that $B-W = 200$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22112,
"subject": "Mathematics (Olympiad)",
"question": "Let $A_0B_0C_0$ be a triangle with area equal to $\\sqrt{2}$. We consider the excenters $A_1, B_1$, and $C_1$, then we consider the excenters, say $A_2, B_2$, and $C_2$, of the triangle $A_1B_1C_1$. By continuing this procedure, examine if it is possible to arrive at a triangle $A_nB_nC_n$ with all coordinates rational.",
"options": [],
"answer": "See solution",
"solution": "The answer is no. Suppose that it is possible. We assert that the previous triangle $A_{n-1}B_{n-1}C_{n-1}$ has rational coordinates. In fact, the points $A_{n-1}, B_{n-1}, C_{n-1}$ are the feet of the altitudes of the triangle $A_nB_nC_n$. Therefore, it is enough to show that, if a line segment has its ends with rational coordinates, then the foot of the perpendicular from a point with rational coordinates also has rational coordinates. This is true because the coordinates $(x, y)$ of the foot of the perpendicular are the solutions of the system $y = ax + b$, $y = -\\frac{1}{a}x + c$ with $a, b, c$ rational. Therefore, at every step, the coordinates must be rational, so the coordinates of the triangle $A_0B_0C_0$ must be rational. Then, from the area formula using coordinates of the vertices, we find that the area of the triangle is a rational number. This contradicts the supposition that the area of the triangle is equal to $\\sqrt{2}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22113,
"subject": "Mathematics (Olympiad)",
"question": "令 $S$ 是 $[0, 1]$ 區間中所有有理數的集合。給定一個無限長的實數數列\n\n$$\n\\{x_1, x_2, \\dots, x_k, \\dots\\},\n$$\n\n如果存在一個由 $S$ 映至實數的函數 $H(x)$ 滿足：\n\n1. $H$ 在 $[0, \\frac{1}{2}]$ 遞增。即：對任意有理數 $0 \\le a \\le b \\le \\frac{1}{2}$，有 $H(a) \\le H(b)$。\n2. 對任意兩個整數 $0 < p, 0 \\le q \\le p$，都有\n\n$$\nH\\left(\\frac{q}{p}\\right) = \\frac{\\sum_{k=1}^{q} x_{p+1-k} - \\sum_{k=1}^{q} x_k}{p}\n$$\n\n則我們稱這個實數數列是“超乎想像”的。試求所有超乎想像的數列。",
"options": [],
"answer": "See solution",
"solution": "所有超乎想像的數列是：$x_k = a(k \\log k - (k-1) \\log(k-1)) + c$，其中 $a$ 是非負實數，$c$ 是任意實數。\n\n首先，若 $\\{x_1, x_2, \\cdots, x_k, \\cdots\\}$ 是超乎想像的，則對任意實數 $c$ 及正實數 $r$，$\\{\\frac{x_1}{r} + c, \\frac{x_2}{r} + c, \\cdots, \\frac{x_k}{r} + c, \\cdots\\}$ 也是超乎想像的。\n\n因此不妨假設 $x_1 = 0$。令 $f(p) = \\sum_{i=1}^{p} x_i$，則易知 $H\\left(\\frac{q}{p}\\right) = \\frac{f(p) - f(q) - f(p-q)}{p}$。\n\n由於 $H\\left(\\frac{q}{p}\\right) = H\\left(\\frac{kp}{kp}\\right)$，因此\n\n$$\n\\frac{f(p) - f(q) - f(p-q)}{p} = \\frac{f(kp) - f(kq) - f(k(p-q))}{kp}\n$$\n\n移項後整理可得\n\n$$\n(f(kp) - k f(p)) = (f(kq) - k f(q)) + (f(k(p-q)) - k f(p-q))\n$$\n\n對所有正整數 $k, p$ 和非負整數 $0 \\le q \\le p$ 都成立。現在固定正整數 $k$，定義 $g_k(p) = f(kp) - kf(p)$，則上式即是\n\n$$\ng_k(p) = g_k(q) + g_k(p - q)\n$$\n\n此式是標準的柯西方程。令 $q=1$ 並利用數學歸納法即可得 $g_k(p) = p g_k(1) = p(f(k) - kf(1)) = p f(k)$（因為 $f(1) = x_1 = 0$），因此\n\n$$\nf(kp) = kf(p) + p f(k)\n$$\n\n更進一步，若固定此式中的 $p$ 並利用數學歸納法，便可得到對任意正整數 $p, k$，有 $f(p^k) = p^{k-1} k f(p)$。\n\n對於兩個正整數 $q < p$，由於 $\\frac{q}{p+q+1} < \\frac{q+1}{p+q+1} \\le \\frac{1}{2}$ 以及 $H(x)$ 的遞增性，我們有\n\n$$\n\\frac{f(p+q+1) - f(q) - f(p+1)}{p+q+1} \\le \\frac{f(p+q+1) - f(q+1) - f(p)}{p+q+1}\n$$\n\n整理後得到\n\n$$\nf(p+1) - f(p) \\ge f(q+1) - f(q)\n$$\n\n對所有 $p > q$ 都成立。注意到此即表示 $f(p)$ 是個凸函數。\n\n由此，我們可以得到一個顯然的不等式：對於正整數 $p > q > r$，有\n\n$$\n\\frac{f(p) - f(q)}{p - q} \\ge \\frac{f(p) - f(r)}{p - r} \\ge \\frac{f(q) - f(r)}{q - r}\n$$\n\n而又因\n\n$$\n\\frac{f(p) - f(r)}{p - r} = \\frac{p - q}{p - r} \\frac{f(p) - f(q)}{p - q} + \\frac{q - r}{p - r} \\frac{f(q) - f(r)}{q - r}\n$$\n\n故其值界在兩者之間。\n\n現在假設 $f(2) = a$ 是已知的，則 $a = f(2) = \\frac{f(2)-f(1)-f(1)}{2} = H\\left(\\frac{1}{2}\\right) \\ge H(0) = 0$。根據 $f(p^k) = p^{k-1} k f(p)$，我們可以推得 $f(2^k) = 2^{k-1} k a$，即當 $p = 2^k$ 時，$f(p) = \\frac{a}{2} p \\log 2p$。當 $p \\neq 2^k$ 時，我們希望能夠用不等式估計出 $f(p)$ 的值。對任意正整數 $m$，存在正整數 $n$ 使得 $2^{n+1} > p^m > 2^n$，因此\n\n$$\n\\frac{f(2^{n+1}) - f(2^n)}{2^{n+1} - 2^n} \\ge \\frac{f(p^m) - f(2^n)}{p^m - 2^n}\n$$\n\n整理後並代入 $f(2^k) = 2^{k-1} k a$ 我們得到\n\n$$\na(2^{n-1} n + (\\frac{n}{2} + 1)(p^m - 2^n)) \\geq f(p^m) = p^{m-1} m f(p)\n$$\n\n進一步有\n\n$$\nf(p) \\le \\frac{a p n + 2 - \\frac{2^n}{p^m}}{2m} \\le \\frac{a p n + 2}{2m}\n$$\n\n由 $2^{n+1} > p^m > 2^n$ 可推得 $\\frac{n+1}{m} > \\log_2 p > \\frac{n}{m}$，所以 $\\frac{n}{m}$ 是一個 $\\log_2 p$ 的有理數估計。所以\n\n$$\nf(p) \\le \\frac{a p n + 2}{2m} \\le \\frac{a p}{2}(\\log_2 p + \\frac{2}{m})\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22114,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f: \\mathbb{N} \\to \\mathbb{N}$ such that for every positive integer $m$, the following holds: If $d_1, d_2, \\dots, d_n$ are all the divisors of $m$, then\n\n$$\nf(d_1) \\cdot f(d_2) \\cdots f(d_n) = m.\n$$",
"options": [],
"answer": "See solution",
"solution": "We will show that the only solution is the function $f$ defined by\n\n$$\nf(m) = \\begin{cases} p, & \\text{if } m \\text{ is a non-trivial power of a prime } p,\\ \\text{i.e. } m = p^k,\\ k \\ge 1, \\\\ 1, & \\text{otherwise.} \\end{cases}\n$$\n\nNumber $1$ has the unique divisor $1$, so plugging $m = 1$ into the given equality gives $f(1) = 1$.\n\nLet $m = p$ be a prime. Then\n\n$$\nf(1) \\cdot f(p) = p \\implies f(p) = p.\n$$\n\nFor $m = p^2$ we get\n\n$$\nf(1) \\cdot f(p) \\cdot f(p^2) = p^2 \\implies f(p^2) = p\n$$\n\nand in general, for $k > 1$ and $m = p^k$,\n\n$$\nf(1) \\cdot f(p) \\cdot f(p^2) \\cdots f(p^k) = p^k.\n$$\n\nBy induction, $f(p^k) = p$ for all positive integers $k$.\n\nNow, let $m$ be a positive integer with at least two distinct prime factors, with factorization $m = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}$, where $k \\ge 2$ and $\\alpha_i \\ge 1$ for every $i \\in \\{1, 2, \\dots, k\\}$. The divisors of $m$ include the powers\n\n$$\np_1, p_1^2, \\dots, p_1^{\\alpha_1},\\ p_2, p_2^2, \\dots, p_2^{\\alpha_2},\\ \\dots,\\ p_k, p_k^2, \\dots, p_k^{\\alpha_k},\n$$\n\nof its prime factors, and the product of the corresponding function values\n\n$$\n\\begin{aligned}\n& f(p_1)f(p_1^2)\\dots f(p_1^{\\alpha_1})f(p_2)f(p_2^2)\\dots f(p_2^{\\alpha_2})\\dots f(p_k)f(p_k^2)\\dots f(p_k^{\\alpha_k}) \\\\\n&= \\underbrace{p_1p_1\\dots p_1}_{\\alpha_1} \\underbrace{p_2p_2\\dots p_2}_{\\alpha_2} \\dots \\underbrace{p_kp_k\\dots p_k}_{\\alpha_k} = p_1^{\\alpha_1}p_2^{\\alpha_2}\\dots p_k^{\\alpha_k} = m\n\\end{aligned}\n$$\n\nyields $m$. This means that the product of all the other function values (none of which is a non-trivial power of a prime), including $f(m)$, is equal to $1$, so all the other function values, including $f(m)$, are equal to $1$.\n\nThis specifies $f$ uniquely and also shows that it has the desired properties.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22115,
"subject": "Mathematics (Olympiad)",
"question": "We assign coordinates $(i, j)$ to a unit square of a chessboard if it lies in the $i$th column from the left and the $j$th row from the bottom.\n\nA configuration of rooks is called *k-open* if there is a $k \\times k$ sub-square that does not contain any rooks. Conversely, it is called *k-closed* if each $k \\times k$ sub-square contains at least one rook.\n\nFind the greatest positive integer $k$ such that every peaceful configuration of rooks is $k$-open.",
"options": [],
"answer": "See solution",
"solution": "The answer is $\\lfloor \\sqrt{n-1} \\rfloor$.\n\nOur proof splits into two parts. We prove that $k \\geq \\lfloor \\sqrt{n-1} \\rfloor$ and that $k \\leq \\lfloor \\sqrt{n-1} \\rfloor$.\n\n**Part 1.** Proof that $k \\geq \\lfloor \\sqrt{n-1} \\rfloor$.\n\nLet $m = \\lfloor \\sqrt{n-1} \\rfloor$. Note that $m^2 < n$. We claim that any peaceful configuration is $m$-open. Assume, for contradiction, that there is an $m$-closed peaceful configuration.\n\nConsider the bottom-left $m^2 \\times m^2$ sub-board. It can be partitioned into $m^2$ squares of side length $m$. Each contains a rook. The $m^2$ rooks must be peaceful for the $m^2 \\times m^2$ sub-board. The remaining $n - m^2$ rooks must be in the top-right $(n - m^2) \\times (n - m^2)$ sub-board, also in peaceful configuration.\n\nSimilarly, the bottom-right $m^2 \\times m^2$ and top-left $(n - m^2) \\times (n - m^2)$ sub-boards each contain peaceful configurations. Thus, the left column of both the bottom-left and top-left sub-boards each contain a rook, so there are two rooks in the left column of the $n \\times n$ chessboard—a contradiction. Therefore, $k \\geq \\lfloor \\sqrt{n-1} \\rfloor$. $\\square$\n\n**Part 2.** Proof that $k \\leq \\lfloor \\sqrt{n-1} \\rfloor$.\n\nDivide the board into $m^2$ squares of size $m \\times m$ (called *boxes*), by $m$ columns of width $m$ and $m$ rows of height $m$. A box has coordinates $(i, j)$ according to the wider columns and rows. For example, in the diagram below, the shaded box has coordinates (3, 2). The rook in it has coordinates (2, 3) within the box and (8, 6) with respect to the whole chessboard.\n\nPlace a rook in each box $S$ at position $(j, i)$ within $S$ (where $S$ has coordinates $(i, j)$). This is illustrated below:\n\n\n\n**Lemma 1.** The configuration is peaceful.\n\n*Proof.* Suppose two rooks $R_1$ and $R_2$ are in the same column. Their boxes $S_1$ and $S_2$ are in the same column, and their positions within the boxes are the same, so $S_1 = S_2$ and $R_1 = R_2$, a contradiction. Similarly, no two rooks are in the same row. $\\square$\n\n**Lemma 2.** Every $m \\times m$ square contains a rook.\n\n*Proof.* Imagine a movable $m \\times m$ square frame. When the frame is around the bottom-left box, it contains a rook. When moved up or right by one unit (remaining within the $n \\times n$ board), it still contains a rook.\n\nSuppose the old frame bounds $F$ containing rook $R$. Consider $F'$ one unit to the right. Let $S$ be the box containing $R$ with coordinates $(i, j)$; $R$ is at $(j, i)$ within $S$.\n\n- **Case 1.** $R$ is not in the left column of $F$: $R$ is also in $F'$.\n- **Case 2.** $R$ is in the leftmost column of $F$ but not the top row: $i < n$. Let $S'$ be the box to the right of $S$, with coordinates $(i+1, j)$. The rook $R'$ in $S'$ is at $(j, i+1)$ within $S'$, $m$ units right and 1 unit above $R$ in the board, so $F'$ contains $R'$.\n- **Case 3.** $R$ is in the top-left square of $F$:\n\n\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22116,
"subject": "Mathematics (Olympiad)",
"question": "We color the numbers $1, 2, 3, \\ldots, 20$ with two colors: white and blank, in such a way that both colors are used. Find the number of ways we can perform this coloring if the product of the white numbers and the product of the blank numbers have greatest common divisor equal to $1$.",
"options": [],
"answer": "See solution",
"solution": "Number $1$ can be colored in two ways: white or blank. Number $2$ also can be colored white or blank. Then all even numbers $2, 4, 6, 8, 10, 12, 14, 16, 18, 20$ have to be colored with the color of $2$.\n\nAlso, all numbers having common divisor greater than $1$ with the above numbers must be colored with the color of $2$. The remaining numbers greater than $10$, that is, $11, 13, 17, 19$, can be colored in two ways each. Therefore, the coloring can be made with $2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 = 2^6 = 64$ different ways. However, we must delete the two cases where we color all numbers white or all blank. So we have finally $62$ different ways.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 22117,
"subject": "Mathematics (Olympiad)",
"question": "Let $T$ be a set of $k m$ integers. Prove that it is possible to partition $T$ into $m$ subsets of $k$ elements each, so that in every subset $A$, the sum $\\sum_{x \\in A} x$ is not divisible by $k$, provided that $k$ does not divide the sum of all elements of $T$.\n\n*Comment.* We cannot dispense with the condition that $n$ does not divide the sum of all elements. Indeed, for each $n > 1$ and the set consisting of $1$, $-1$, and $n-2$ elements divisible by $n$, the required permutation does not exist.",
"options": [],
"answer": "See solution",
"solution": "If no $k$ elements of $T$ have the same residue modulo $k$, there are three elements $a, b, c \\in T$ leaving pairwise distinct remainders upon division by $k$. Let $t$ be the sum of elements of $T$. It suffices to find $A \\subset T$ such that $|A| = k$ and $\\sum_{x \\in A} \\not\\equiv 0, t \\pmod{k}$: then neither the sum of elements of $A$ nor the sum of elements of $B = T \\setminus A$ is divisible by $k$.\n\nConsider $U' \\subset T \\setminus \\{a, b, c\\}$ with $|U'| = k-1$. The sums of elements of three sets $U' \\cup \\{a\\}$, $U' \\cup \\{b\\}$, $U' \\cup \\{c\\}$ leave three different remainders upon division by $k$, and at least one of them is not congruent either to $0$ or to $t$.\n\nNow let $m > 2$. If $T$ contains $k$ elements leaving the same remainder upon division by $k$, we form one subset $A$ of these elements and apply the inductive hypothesis to the remaining $k(m-1)$ elements. Otherwise, we choose any $U \\subset R$, $|U| = k-1$. Since all the remaining elements cannot be congruent modulo $k$, there is $a \\in T \\setminus U$ such that $a \\not\\equiv -\\sum_{x \\in U} x \\pmod{k}$. Now we can take $A = U \\cup \\{a\\}$ and apply the inductive hypothesis to $T \\setminus A$.\n\nNow we are ready to prove the statement of the problem for all odd $n$ and $n = 2^k$. The proof is by induction.\n\nIf $n$ is prime, the statement follows immediately from Lemma 1, since in this case $(n, s) = 1$. Turning to the general case, we can find a prime $p$ and an integer $t$ such that $p^t \\mid n$ and $p^t \\nmid s$. By Lemma 2, we can partition $S$ into $p$ sets of $\\frac{n}{p} = k$ elements so that in every set either the sum of numbers is not divisible by $k$ or all numbers have the same residue modulo $k$.\n\nFor sets in the first category, by the inductive hypothesis, there is a permutation $(a_i)$ such that $k = \\sum_{i=1}^k i a_i$.\n\nIf $n$ (and therefore $k$) is odd, then for each permutation $(b_i)$ of a set in the second category we have\n\n$$\n\\sum_{i=1}^{k} i b_i \\equiv b_1 \\frac{k(k+1)}{2} \\equiv 0 \\pmod{k}.\n$$\n\nBy combining such permutations for all sets of the partition, we get a permutation $(c_i)$ of $S$ such that $k = \\sum_{i=1}^n i c_i$. Since this sum is divisible by $k$, and $k$ is divisible by $(n, s)$, we are done by Lemma 1.\n\nIf $n = 2^s$, we have $p = 2$ and $k = 2^{s-1}$. Then for each of the subsets there is a permutation $(a_1, \\dots, a_k)$ such that $\\sum_{i=1}^k i a_i$ is divisible by $2^{s-2} = \\frac{k}{2}$: if the subset belongs to the first category, the expression is divisible even by $k$, and if it belongs to the second one,\n\n$$\n\\sum_{i=1}^{k} i a_i \\equiv a_1 \\frac{k(k+1)}{2} \\equiv 0 \\pmod{\\frac{k}{2}}.\n$$\n\nNow the numbers of each permutation should be multiplied by all the odd or all the even numbers not exceeding $n$ in increasing order so that the resulting sums are divisible by $k$:\n\n$$\n\\sum_{i=1}^{k} (2i - 1)a_i \\equiv \\sum_{i=1}^{k} 2i a_i \\equiv 2 \\sum_{i=1}^{k} i a_i \\equiv 0 \\pmod{k}.\n$$\n\nCombining these two sums, we again get a permutation $(c_i)$ of $S$ such that $k = \\sum_{i=1}^n i c_i$, and finish the case by applying Lemma 1.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22118,
"subject": "Mathematics (Olympiad)",
"question": "The equation\n\n$$\nn! + A \\cdot n = n^k\n$$\n\nhas $(n, k) = (0, 0)$ as a solution for every non-negative integer $A$. Determine all non-negative integer solutions of this equation for $A = 7$ and $A = 2012$.",
"options": [],
"answer": "See solution",
"solution": "First, note that for any $A$, $n = 0$ implies $k = 0$ and vice versa. Also, for any $A > 0$, there are no solutions for $n = 1$ or $k = 1$, since $n! + A n > n^k$ in these cases. Thus, we consider $n, k \\ge 2$.\n\nDividing both sides by $n$ gives:\n\n$$\n(n - 1)! + A = n^{k-1}.\n$$\n\n**Case $A = 7$:**\n\n- For $n = 2$: $1 + 7 = 2^{k-1} \\implies 8 = 2^{k-1} \\implies k = 4$. So $(n, k) = (2, 4)$.\n- For $n = 3$: $2 + 7 = 3^{k-1} \\implies 9 = 3^{k-1} \\implies k = 3$. So $(n, k) = (3, 3)$.\n- For $n > 3$: For $n \\in \\{5, 7, 11, 13\\}$, Wilson's theorem gives $(n-1)! + 7 \\equiv 6 \\pmod{n}$, so this cannot be a power of $n$. For $n = 9$, $8! + 7 \\equiv 1 \\pmod{3}$, which is not $0$.\n- For $n \\ge 15$, $(n-1)! + 7 \\equiv n^{k-1} \\pmod{7}$, so $7 \\mid n$. But then $7^2 \\mid (n-1)!$, so $(n-1)! + 7 \\equiv 7 \\pmod{7}$, which is not $0$.\n- For $k = 2$, $(n-1)! + 7 = n$ cannot hold since $(n-1)! + 7 > n$ for $n \\ge 2$.\n\nThus, the only non-trivial solutions for $A = 7$ are $(2, 4)$ and $(3, 3)$.\n\n**Case $A = 2012$:**\n\n- For $n = 2$: $1 + 2012 = 2^{k-1} \\implies 2013 = 2^{k-1}$, which has no integer solution.\n- For $n > 2$: $(n-1)!$ is even, so $n$ must be even. For $n = 4$: $3! + 2012 = 6 + 2012 = 2018 = 4^{k-1}$, but $4^{k-1}$ cannot be $2018$ for integer $k$.\n- For $n \\ge 6$ and $k > 3$, $n$ even implies $n^{k-1} \\equiv 4 \\pmod{8}$, which is a contradiction.\n- For $k = 2$, $(n-1)! + 2012 = n$ cannot hold since $(n-1)! + 2012 > n$.\n- For $k = 3$, $(n-1)! + 2012 = n^2$ contradicts $(n-1)! + 2012 > n^2$ for $n \\ge 6$.\n\nTherefore, there are no non-trivial solutions for $A = 2012$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22119,
"subject": "Mathematics (Olympiad)",
"question": "Let $f(x) = (x-a)(x-b)(x-c)$ where $a, b, c \\ge 0$ and $abc = -f(0) = 64$. What is the largest possible value of $f(-1)$?",
"options": [],
"answer": "See solution",
"solution": "The largest possible value of $f(-1)$ is $-125$.\n\n$$\n\\begin{aligned}\nf(-1) &= -(1+a)(1+b)(1+c) \\\\\n&= -1 - (a+b+c) - (ab+bc+ca) - abc \\\\\n&\\le -1 - 3\\sqrt[3]{abc} - 3\\sqrt[3]{a^2b^2c^2} - abc \\\\\n&= -125.\n\\end{aligned}\n$$\n\nEquality holds when $a = b = c = 4$, i.e. $f(x) = (x-4)^3$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22120,
"subject": "Mathematics (Olympiad)",
"question": "Find the smallest positive integer $n$ for which there exist two distinct pairs of positive integers $(x, y)$ such that $(x^2 - 1)(y^2 - 1) = n$ and $x \\leq y$.",
"options": [],
"answer": "See solution",
"solution": "Notice that for $n = 360$, the pairs $x = 2, y = 11$ and $x = 4, y = 5$ satisfy the condition.\n\nWe will show that for smaller numbers $n$, there do not exist two distinct suitable pairs $(x, y)$. If $x = 1$, then $(x^2 - 1)(y^2 - 1)$ is not positive. Thus, we can assume that $1 < x \\leq y$. Now, as $x$ or $y$ increases, $(x^2 - 1)(y^2 - 1)$ also increases. Let's examine the cases.\n\n- If $x = 2$, then $y = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, \\dots$ give us respectively $(x^2 - 1)(y^2 - 1) = 9, 24, 45, 72, 105, 144, 189, 240, 297, 360, \\dots$.\n- If $x = 3$, then $y = 3, 4, 5, 6, 7, \\dots$ give similarly $(x^2 - 1)(y^2 - 1) = 64, 120, 192, 280, 384, \\dots$.\n- If $x = 4$, then $y = 4, 5, \\dots$ give $(x^2 - 1)(y^2 - 1) = 225, 360, \\dots$.\n\nNo positive number smaller than 360 appeared repeatedly. If we continue the inspection for $x = 5, 6, \\dots$, even the first case $y = x$ would give $(x^2 - 1)(y^2 - 1) > 360$, since $x = 4$ gives 360 or a larger number for the same $y$. In conclusion, 360 is the smallest number with the required property.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22121,
"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 find the value of $\\tan \\angle ABF_1 \\cdot \\tan \\angle ABF_2$.",
"options": [],
"answer": "See solution",
"solution": "By symmetry, suppose the equation of $\\Gamma$ is $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$ ($a > b > 0$), and $A(a, 0)$, $B(0, b)$, $F_1(-c, 0)$, $F_2(c, 0)$, where $c = \\sqrt{a^2 - b^2}$.\n\nBy the given conditions, we know that\n\n$$\n\\begin{aligned}\n\\overrightarrow{AF_1} \\cdot \\overrightarrow{AF_2} + \\overrightarrow{BF_1} \\cdot \\overrightarrow{BF_2} &= (-c-a)(c-a) + (-c^2+b^2) \\\\\n&= a^2 + b^2 - 2c^2 = 0.\n\\end{aligned}\n$$\n\nThus, $a^2 + b^2 - 2c^2 = -a^2 + 3b^2 = 0$, and hence $a = \\sqrt{3}b$, $c = \\sqrt{2}b$.\n\nLet $O$ be the origin of the coordinates, and then\n\n$$\n\\begin{aligned}\n\\tan \\angle ABO &= \\frac{a}{b} = \\sqrt{3}, \\\\\n\\tan \\angle OBF_1 &= \\tan \\angle OBF_2 = \\frac{c}{b} = \\sqrt{2}.\n\\end{aligned}\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\n& \\tan \\angle ABF_1 \\cdot \\tan \\angle ABF_2 \\\\\n&= \\tan(\\angle ABO + \\angle OBF_1) \\cdot \\tan(\\angle ABO - \\angle OBF_1) \\\\\n&= \\frac{\\sqrt{3} + \\sqrt{2}}{1 - \\sqrt{3} \\cdot \\sqrt{2}} \\cdot \\frac{\\sqrt{3} - \\sqrt{2}}{1 + \\sqrt{3} \\cdot \\sqrt{2}} = -\\frac{1}{5}.\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22122,
"subject": "Mathematics (Olympiad)",
"question": "Solve the equation $xyz + yzt + xzt + xyt = xyzt + 3$ in the set of natural numbers.",
"options": [],
"answer": "See solution",
"solution": "After dividing the equation by $xyzt$, we get\n$$\n\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} + \\frac{1}{t} = 1 + \\frac{3}{xyzt}.\n$$\nBecause of symmetry, without loss of generality, we can assume that\n$$\nx \\le y \\le z \\le t.\n$$\nFrom this, $\\frac{1}{x} \\ge \\frac{1}{y} \\ge \\frac{1}{z} \\ge \\frac{1}{t}$. Thus,\n$$\n\\frac{4}{x} \\ge 1 + \\frac{3}{xyzt} > 1 \\implies x < 4.\n$$\n\n**Case 1:** $x=3$\n\nThe equation becomes $3yz + yzt + 3zt + 3yt = 3yzt + 3$, or $3(yz + zt + yt) = 2yzt + 3$. Dividing by $yzt$:\n$$\n3\\left(\\frac{1}{y} + \\frac{1}{z} + \\frac{1}{t}\\right) = 2 + \\frac{3}{yzt} > 2 \\implies y \\le 4.\n$$\nPossible $y$: $3, 4$.\n\n- For $y=4$:\n $$\n 12(z + t) = 5zt + 3 \\implies 12\\left(\\frac{1}{z} + \\frac{1}{t}\\right) = 5 + \\frac{3}{zt} > 5 \\implies z \\le 4.\n $$\n From $x \\le y \\le z \\le t$, $z=4$. The equation becomes $12(4 + t) = 20t + 3$, or $8t = 45$, which has no natural solution.\n\n- For $y=3$:\n $$\n 3(z + t) = zt + 1 \\implies 3\\left(\\frac{1}{z} + \\frac{1}{t}\\right) = 1 + \\frac{1}{zt} > 1 \\implies z < 6.\n $$\n Possible $z$: $3, 4, 5$.\n - $z=3$: $3(3 + t) = 3t + 1$ (impossible).\n - $z=4$: $3(4 + t) = 4t + 1 \\implies t=11$.\n - $z=5$: $3(5 + t) = 5t + 1 \\implies t=7$.\n\nSo, solutions: $(3,3,4,11)$, $(3,3,5,7)$.\n\n**Case 2:** $x=2$\n\nEquation: $2(yz + zt + yt) = yzt + 3$.\n\nDividing by $yzt$:\n$$\n2\\left(\\frac{1}{y} + \\frac{1}{z} + \\frac{1}{t}\\right) = 1 + \\frac{3}{yzt} > 1 \\implies y < 6.\n$$\nPossible $y$: $3, 5$.\n\n- $y=5$: $10(z + t) = 3zt + 3 \\implies z \\le 6$. Only $z=5$ possible, but $t$ is not a natural number.\n- $y=3$: $6(z + t) = zt + 3$. Possible $z$: $3, 5, 7, 9, 11$.\n - $z=3$: $t=-5$ (not natural).\n - $z=5$: $t=-27$ (not natural).\n - $z=7$: $t=39$.\n - $z=9$: $t=17$.\n\nSo, solutions: $(2,3,7,39)$, $(2,3,9,17)$.\n\n**Case 3:** $x=1$\n\nEquation: $yz + zt + yt = 3$. From $x \\le y \\le z \\le t$, $y=1$, $z=1$, $t=1$.\n\nSo, solution: $(1,1,1,1)$.\n\n**Final solutions:** All permutations of $(3,3,4,11)$, $(3,3,5,7)$, $(2,3,7,39)$, $(2,3,9,17)$, $(1,1,1,1)$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22123,
"subject": "Mathematics (Olympiad)",
"question": "Consider an $m \\times n$ table, where $m, n \\ge 2$ (rows are numbered $1, 2, \\ldots, m$ and columns $1, 2, \\ldots, n$), filled with positive integers. Let $b_i$ be the least common multiple (LCM) of all numbers in the $i$th row ($1 \\le i \\le m$), and let $B$ be the greatest common divisor (GCD) of the numbers $b_1, b_2, \\ldots, b_m$. Also, let $c_j$ be the GCD of all numbers in the $j$th column ($1 \\le j \\le n$), and let $C$ be the LCM of the numbers $c_1, c_2, \\ldots, c_n$. Is it true that $B$ is divisible by $C$, or is $C$ divisible by $B$?",
"options": [],
"answer": "See solution",
"solution": "**Answer:** $B$ is divisible by $C$.\n\n**Solution:**\n\nConsider any prime number $p$, and examine its power in each number in the table. Replace all the numbers in the table with the power of the chosen prime number. Let the $m \\times n$ table be filled with $\\alpha_{i,j}$, where $i = 1, \\ldots, m$ and $j = 1, \\ldots, n$. Now, $\\beta_i$ is the greatest number from the $i$th row, and $B$ is the smallest of the $\\beta_i$. Similarly, $\\gamma_j$ is the smallest number of the $j$th column, and $\\Gamma$ is the greatest of the $\\gamma_j$. Thus, both $B$ and $\\Gamma$ are among the table entries. If they belong to the same row or column, then $B \\geq \\Gamma$ by construction. If they are in different rows and columns, consider $\\Delta$, the number at the intersection of the column of $\\Gamma$ and the row of $B$. By construction, $\\Gamma \\leq \\Delta \\leq B$. Thus, the power of $p$ in $B$ is at least as large as its power in $C$. Since $p$ is arbitrary, $C \\mid B$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22124,
"subject": "Mathematics (Olympiad)",
"question": "Let\n\n$$\nA = \\left\\{ 1 + \\frac{1}{k} : k = 1, 2, 3, \\dots \\right\\} = \\left\\{ \\frac{k+1}{k} : k = 1, 2, 3, \\dots \\right\\}.\n$$\n\nFor every integer $x \\ge 2$,\n\n$$\nx = \\frac{2}{1} \\times \\frac{3}{2} \\times \\dots \\times \\frac{x}{x-1} = \\prod_{k=1}^{x-1} \\frac{k+1}{k}.\n$$\n\nDefine $f(n)$ as the minimal number of elements of $A$ whose product is $n$ (for $n \\ge 2$). Prove that there exist infinitely many pairs of integers $x, y \\ge 2$ such that\n\n$$\nf(xy) < f(x) + f(y).\n$$",
"options": [],
"answer": "See solution",
"solution": "For every integer $x \\ge 2$,\n$$\nx = \\frac{2}{1} \\times \\frac{3}{2} \\times \\dots \\times \\frac{x}{x-1} = \\prod_{k=1}^{x-1} \\frac{k+1}{k}.\n$$\nThus, every integer $x \\ge 2$ can be written as a product of one or more elements of $A$.\n\nThe largest element of $A$ is $2$, so for integers $n \\ge 0$,\n$$\n3 \\times 2^n > 2^{n+1} \\implies f(3 \\times 2^n) \\ge n + 2.\n$$\nSince $3 \\times 2^n = \\frac{3}{2} \\times 2^{n+1}$, the product of $n+2$ elements of $A$, $f(3 \\times 2^n) = n+2$.\n\nSimilarly,\n$$\n33 \\times 2^n > 2^{n+5} \\implies f(33 \\times 2^n) \\ge n + 6\n$$\nand since $33 \\times 2^n = \\frac{33}{32} \\times 2^{n+5}$, the product of $n+6$ elements of $A$, $f(33 \\times 2^n) = n+6$.\n\nAlso, $f(11) \\ge 5$ since 11 cannot be written as the product of four or fewer elements of $A$: $2^4 > 11$, $2^3 \\times \\frac{3}{2} > 11$, and any other product of at most four elements of $A$ does not exceed $2^3 \\times \\frac{4}{3} = 10^{2/3} < 11$. Since $11 = \\frac{11}{10} \\times \\frac{5}{4} \\times 2 \\times 2 \\times 2$, we have $f(11) = 5$.\n\nPutting these results together,\n$$\nf(33 \\times 2^n) = n + 6 < n + 2 + 5 = f(3 \\times 2^n) + f(11)\n$$\nfor all integers $n \\ge 0$. Hence, there are infinitely many pairs $x = 3 \\times 2^n$, $y = 11$, with $x \\ge 2$, $y \\ge 2$ and $f(xy) < f(x) + f(y)$.\n\n$\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22125,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for all $x, y \\in \\mathbb{R}$,\n$$\nf(f(x) + f(y)) = f(f(x)) + \\frac{1}{8} f(x) f(4y) + f(f(y)).\n$$",
"options": [],
"answer": "See solution",
"solution": "Let us analyze the given functional equation:\n\nSuppose $f(x) = 0$ for all $x$. Then the equation holds trivially.\n\nAssume there exists $t$ such that $f(t) \\neq 0$.\n\nBy substituting and manipulating the equation, we deduce:\n- $f(4x) = 8a f(x)$ for some $a$.\n- $f(x) + f(-x) = a x^2$.\n- $a = 2$.\n- $f(4x) = 16 f(x)$.\n- $f(x + f(y)) = f(x) + 2x f(y) + f^2(y)$.\n- $f(f(x)) = f^2(x)$.\n\nFrom these, we find $f(x) = x^2$ is a solution. Thus, all solutions are:\n$$\nf(x) = 0 \\quad \\text{and} \\quad f(x) = x^2.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22126,
"subject": "Mathematics (Olympiad)",
"question": "Could it happen that in a tournament with 73 teams, each team won either $m$ or $n$ games (where $m \\neq n$)?",
"options": [],
"answer": "See solution",
"solution": "*Answer.* It could not happen.\n\nSuppose $m \\neq n$. In a tournament with 73 teams, there are a total of $\\frac{73 \\cdot 72}{2} = 36 \\cdot 73$ games. Let $x$ teams have $n$ wins each, and the remaining $73 - x$ teams have $m$ wins each. Then:\n\n$$\nx \\cdot n + (73 - x) \\cdot m = 36 \\cdot 73\n$$\n\nwhich gives:\n\n$$\nx \\cdot (n - m) = (36 - m) \\cdot 73\n$$\n\nSince 73 is prime, either $x$ or $n - m$ must be divisible by 73. The first is impossible because $x < 73$. The second is impossible because $n < 73$, $m < 73$, so $0 < |n - m| < 73$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22127,
"subject": "Mathematics (Olympiad)",
"question": "A domino is a rectangle formed by two unit squares that share a common side. A number of 18 dominoes fit together to tile a $6 \\times 6$ square.\n\nShow that some line crossing the interior of the square crosses the interior of no domino. Is it possible that such a line be unique?",
"options": [],
"answer": "See solution",
"solution": "Let the square be $[0, 6] \\times [0, 6]$.\n\nWe first show that either some grid-vertical $x = i$, $i = 1, 2, 3, 4, 5$, or some grid-horizontal $y = j$, $j = 1, 2, 3, 4, 5$, crosses no tile.\n\nLet $m_i$ and $n_j$ be the number of tiles crossed by the grid-vertical $x = i$ and the grid-horizontal $y = j$, respectively. Clearly, a grid-vertical crosses only horizontal tiles, and a grid-horizontal crosses only vertical tiles.\n\nSince every horizontal tile is crossed by a single grid-vertical, the number of horizontal tiles is $m_1 + m_2 + m_3 + m_4 + m_5$. Similarly, the number of vertical tiles is $n_1 + n_2 + n_3 + n_4 + n_5$. Hence the number of tiles is $$m_1 + m_2 + m_3 + m_4 + m_5 + n_1 + n_2 + n_3 + n_4 + n_5 = 18.$$ Consequently, either some $m_i \\le 1$, $i = 1, 2, 3, 4, 5$, or some $n_j \\le 1$, $j = 1, 2, 3, 4, 5$.\n\nNow, for each positive integer $i \\le 5$, the rectangle $[0, i] \\times [0, 6]$ consists of a certain number of tiles and $m_i$ unit cells, the left halves of the horizontal tiles the grid-vertical $x = i$ crosses. Since the area of each $[0, i] \\times [0, 6]$ and the area of each tile are both even, so is each $m_i$. Similarly, each $n_j$ is even.\n\nFinally, by the conclusion of the preceding paragraph, either some $m_i = 0$, $i = 1, 2, 3, 4, 5$, in which case the corresponding grid-vertical $x = i$ crosses no tile; or some $n_j = 0$, $j = 1, 2, 3, 4, 5$, in which case the corresponding grid-horizontal $y = j$ crosses no tile.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22128,
"subject": "Mathematics (Olympiad)",
"question": "Given a real number $a \\ge 3$ and a polynomial $P$ with real coefficients, prove that\n\n$$\n\\max \\{ |a^k - P(k)| : k = 0, 1, 2, \\dots, \\deg P + 1 \\} \\ge 1,\n$$\n\nwhere $\\deg P$ is the degree of $P$.",
"options": [],
"answer": "See solution",
"solution": "Proceed by induction on $\\deg P$.\n\nIf $\\deg P = 0$, then $P = c$, a constant. If the assertion were to fail, then both $|1-c| < 1$ and $|a-c| < 1$, so $|a-1| \\leq |a-c| + |c-1| < 2$, contradicting $a \\geq 3$.\n\nIf $\\deg P > 0$, let\n\n$$\nM = \\max \\{ |a^k - P(k)| : k = 0, 1, 2, \\dots, \\deg P + 1 \\},\n$$\n\nand consider the polynomial\n\n$$\nQ(x) = \\frac{P(x+1) - P(x)}{x-1}\n$$\n\nof degree $\\deg Q < \\deg P$. Then, for $k = 0, 1, 2, \\dots, \\deg P$,\n\n$$\n\\begin{align*}\n|a^k - Q(k)| &= \\frac{|a^{k+1} - a^k - P(k+1) + P(k)|}{a-1} \\\\\n&\\leq \\frac{|a^{k+1} - P(k+1)|}{a-1} + \\frac{|a^k - P(k)|}{a-1} \\\\\n&\\leq \\frac{2M}{a-1} \\leq M.\n\\end{align*}\n$$\n\nSince $\\deg Q < \\deg P$, the induction assumption applies and we are done.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 22129,
"subject": "Mathematics (Olympiad)",
"question": "Positive integers from $1$ to $n$ are written on the blackboard. The first player chooses a number and erases it. Then the second player chooses two consecutive numbers and erases them. After that, the first player chooses three consecutive numbers and erases them. Finally, the second player chooses four consecutive numbers and erases them. What is the smallest value of $n$ for which the second player can ensure that he completes both his moves?",
"options": [],
"answer": "See solution",
"solution": "The answer is $n = 14$.\n\nFor $n = 13$, the first player can prevent the second player from erasing four consecutive numbers in the final move. For example, the first player can erase $4$ in the first move, and then erase $8$, $9$, and $10$ in the third move, so no interval of length $4$ remains.\n\nFor $n = 14$, the second player can always complete both his moves. Suppose the first player erases number $k$ in his first move (by symmetry, assume $k \\leq 7$). If $k \\geq 5$, the second player erases $k+1$ and $k+2$, leaving two intervals of length at least $4$: $1$ to $k-1$ and $k+3$ to $14$. The first player can only destroy one of these intervals in his next move, so the second player can erase four consecutive numbers in the final move. If $k \\leq 4$, the second player erases $9$ and $10$, leaving intervals $(k+1)$ to $8$ and $11$ to $14$, both of length at least $4$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22130,
"subject": "Mathematics (Olympiad)",
"question": "Let $x', y', z'$ be an arbitrary permutation of $x, y, z$. Consider the triples $(\\sqrt{x}, \\sqrt{y}, \\sqrt{z})$ and $\\left(\\frac{1}{\\sqrt{x+8}}, \\frac{1}{\\sqrt{y+8}}, \\frac{1}{\\sqrt{z+8}}\\right)$. These triples are inversely monotonic. Prove that\n\n$$\n\\sqrt{\\frac{x}{x+8}} + \\sqrt{\\frac{y}{y+8}} + \\sqrt{\\frac{z}{z+8}} \\leq \\sqrt{\\frac{x'}{y'+8}} + \\sqrt{\\frac{y'}{z'+8}} + \\sqrt{\\frac{z'}{x'+8}}\n$$\n\nfor all permutations $(x', y', z')$ of $(x, y, z)$.",
"options": [],
"answer": "See solution",
"solution": "By the rearrangement inequality, since $(\\sqrt{x}, \\sqrt{y}, \\sqrt{z})$ and $\\left(\\frac{1}{\\sqrt{x+8}}, \\frac{1}{\\sqrt{y+8}}, \\frac{1}{\\sqrt{z+8}}\\right)$ are inversely monotonic, the sum\n\n$$\n\\sqrt{\\frac{x}{x+8}} + \\sqrt{\\frac{y}{y+8}} + \\sqrt{\\frac{z}{z+8}}\n$$\n\nis maximized when the variables are paired in the opposite order. Thus, for any permutation $(x', y', z')$ of $(x, y, z)$,\n\n$$\n\\sqrt{\\frac{x}{x+8}} + \\sqrt{\\frac{y}{y+8}} + \\sqrt{\\frac{z}{z+8}} \\leq \\sqrt{\\frac{x'}{y'+8}} + \\sqrt{\\frac{y'}{z'+8}} + \\sqrt{\\frac{z'}{x'+8}}.\n$$\n\nEquality holds when $x = y = z$ (for example, $x = y = z = 1$).",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 22131,
"subject": "Mathematics (Olympiad)",
"question": "We observe how the height of people changes as we go around a circle in the clockwise direction. For every pair of neighbours $A$ and $B$ (where $B$ is after $A$ in the clockwise direction), we put the symbol $\\clubsuit$ between them if $B$ is taller than $A$ and the symbol $\\spadesuit$ if $B$ is shorter than $A$. In this way, we obtain a sequence of $N$ symbols $\\clubsuit$ or $\\spadesuit$.\n\nA person is *average* if and only if the symbol before that person is equal to the symbol after that person. Hence, the number of people who are not average is equal to the number of alternations of the symbols $\\clubsuit$ and $\\spadesuit$. The number of changes from the symbol $\\clubsuit$ to $\\spadesuit$ is equal to the number of changes from $\\spadesuit$ to $\\clubsuit$. So the number of people who are not average is even.\n\nFrom this, we conclude that the number of average people has the same parity as $N$. Since the tallest person in the circle is not average, the number of average people is certainly smaller than $N$.\n\nLet $a_1 < a_2 < \\dots < a_N$ denote the heights of people. Exhibit an example showing that for all $k = 1, \\dots, \\lfloor N/2 \\rfloor$, there exists a configuration of $N$ people in a circle such that exactly $N - 2k$ are average. More precisely, consider a configuration where the people are ordered in a circle so that their heights are respectively\n\n$$\n\\underbrace{a_1, a_N, a_2, a_{N-1}, \\dots, a_k, a_{N-k+1}}_{\\text{A}}, \\underbrace{a_{N-k}, \\dots, a_{k+1}}_{\\text{G}}\n$$",
"options": [],
"answer": "See solution",
"solution": "The first $2k$ people are not average and the remaining $N - 2k$ are average.\n\nThis shows that the number of average people can be any number strictly less than $N$ and having the same parity as $N$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22132,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute-angled triangle inscribed in a circle with center $O$ and radius $R$, and let $H$ be the orthocenter of triangle $ABC$. Let $A_1$ be a point on the side $BC$ such that $HA_1 + A_1O = R$. Similarly, define points $B_1$ on $CA$ and $C_1$ on $AB$. If\n\n$$\n\\overrightarrow{AA_1} + \\overrightarrow{BB_1} + \\overrightarrow{CC_1} = \\overrightarrow{0},\n$$\n\nprove that triangle $ABC$ is equilateral.",
"options": [],
"answer": "See solution",
"solution": "Let $H_A$ be the reflection of $H$ across line $BC$. Then $H_A$ lies on the circumcircle of triangle $ABC$. Since $A_1H = A_1H_A$, we get $A_1O + A_1H_A = R = OH_A$. Hence, $A_1$ lies on segment $OH_A \\cap BC$, and is uniquely defined by the given condition.\n\nWe now prove that $\\angle BHA_1 = \\angle B$ and $\\angle CHA_1 = \\angle C$. Suppose without loss of generality that $\\angle B \\leq \\angle C$. Let $A_O$ be the point diametrically opposite to $A$ on the circle. Then:\n\n$$\n\\angle BHA_1 = \\angle BHH_A - \\angle A_1HH_A = \\angle C - \\angle OH_A A = \\angle C - \\angle A_O A H_A = \\angle C - (\\angle A - \\angle BAA_O - \\angle H_A A C) = \\angle C - (\\angle A - (90^\\circ - \\angle C) - \\angle BCA_O) = \\angle C - (\\angle A - (90^\\circ - \\angle C) - (90^\\circ - \\angle C)) = \\angle C - (\\angle A - 180^\\circ + 2\\angle C) = \\angle B.\n$$\n\nSimilarly, we find that $\\angle CHA_1 = \\angle C$.\n\nTherefore, using area ratios:\n\n$$\n\\frac{BA_1}{CA_1} = \\frac{A_{BHA_1}}{A_{CHA_1}} = \\frac{BH \\cdot HA_1 \\cdot \\sin(\\angle BHA_1)}{CH \\cdot HA_1 \\cdot \\sin(\\angle CHA_1)} = \\frac{2R \\cos B \\sin B}{2R \\cos C \\sin C} = \\frac{\\sin(2B)}{\\sin(2C)}.\n$$\n\nApplying the same process for $B_1$ and $C_1$, we obtain:\n\n$$\n\\frac{CB_1}{AB_1} = \\frac{\\sin(2C)}{\\sin(2A)}, \\quad \\frac{AC_1}{BC_1} = \\frac{\\sin(2A)}{\\sin(2B)}.\n$$\n\nMultiplying the three ratios gives:\n\n$$\n\\frac{BA_1}{CA_1} \\cdot \\frac{CB_1}{AB_1} \\cdot \\frac{AC_1}{BC_1} = \\frac{\\sin(2B)}{\\sin(2C)} \\cdot \\frac{\\sin(2C)}{\\sin(2A)} \\cdot \\frac{\\sin(2A)}{\\sin(2B)} = 1.\n$$\n\nThus, the cevians $AA_1, BB_1$, and $CC_1$ are concurrent. Given that $\\overrightarrow{AA_1} + \\overrightarrow{BB_1} + \\overrightarrow{CC_1} = \\overrightarrow{0}$, the concurrency point must be the centroid, implying these are medians. Therefore, $A_1, B_1, C_1$ are midpoints of $BC, CA, AB$ respectively, and hence:\n\n$$\n\\frac{\\sin(2A)}{\\sin(2B)} = 1 \\implies \\sin(2A) = \\sin(2B),\n$$\n\nand, similarly, $\\sin(2A) = \\sin(2C)$.\n\nSince triangle $ABC$ is acute and the sine function is injective in $(0^\\circ, 180^\\circ)$, so $2A = 2B = 2C \\implies A = B = C$. Therefore, triangle $ABC$ is equilateral.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22133,
"subject": "Mathematics (Olympiad)",
"question": "Enumerate the cells of a $3 \\times 3$ board as shown below. Note that the upper left corner of a red $2 \\times 2$ square can only be one of the cells 1, 2, 4, or 5.\n\n\\begin{tabular}{|c|c|c|}\\hline\n1 & 2 & 3 \\\\\n\\hline\n4 & 5 & 6 \\\\\n\\hline\n7 & 8 & 9 \\\\\n\\hline\\end{tabular}\n\nHow many colorings of the $3 \\times 3$ board are there such that there is no red $2 \\times 2$ square whose upper left corner is at one of the cells 1, 2, 4, or 5?",
"options": [],
"answer": "See solution",
"solution": "Let $A_i$ ($i = 1, 2, 4, 5$) be the set of colorings where there is at least one red $2 \\times 2$ square with its upper left corner at cell $i$. By the inclusion-exclusion principle, the number of such colorings is:\n\n$$\n|A| = |A_1| + |A_2| + |A_4| + |A_5| - \\sum |A_i \\cap A_j| + \\sum |A_i \\cap A_j \\cap A_k| - |A_1 \\cap A_2 \\cap A_4 \\cap A_5|\n$$\n\nCalculating each term:\n- $|A_1| = |A_2| = |A_4| = |A_5| = 2^5$\n- $|A_i \\cap A_j|$ (for adjacent $i, j$): $2^3$ (4 pairs),\n- $|A_i \\cap A_j|$ (for diagonal $i, j$): $2^2$ (2 pairs),\n- $|A_i \\cap A_j \\cap A_k|$: $2$ (4 triples),\n- $|A_1 \\cap A_2 \\cap A_4 \\cap A_5| = 1$\n\nSo,\n$$\n|A| = 4 \\cdot 2^5 - (4 \\cdot 2^3 + 2 \\cdot 2^2) + 4 \\cdot 2 - 1 = 128 - (32 + 8) + 8 - 1 = 128 - 40 + 8 - 1 = 95\n$$\n\nThe total number of colorings is $2^9 = 512$, so the desired number is $512 - 95 = 417$.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 22134,
"subject": "Mathematics (Olympiad)",
"question": "Let $S = \\{1, 2, \\dots, 2n\\}$, where $n$ is a positive integer.\n\n(i) How many subsets of $S$ are *good*, where a subset is called *good* if the number of even elements is greater than the number of odd elements?\n\n(ii) What is the sum of all elements in all good subsets of $S$?",
"options": [],
"answer": "See solution",
"solution": "For part (i):\n\nA subset $T$ of $S$ is called *bad* if the number of odd elements in $T$ is greater than the number of even elements. A subset is neither good nor bad if it has exactly $k$ odd and $k$ even elements for some $k = 0, 1, \\dots, n$. Since there are $n$ odd and $n$ even elements in $S$, the number of subsets which are neither good nor bad is\n\n$$\n\\sum_{k=0}^{n} \\binom{n}{k}^2 = \\binom{2n}{n}\n$$\n\nby Vandermonde's identity.\n\nBy symmetry, the number of good subsets equals the number of bad subsets. Since there are $2^{2n}$ subsets of $S$ in total, the number of good subsets is\n\n$$\n\\frac{1}{2}\\left[2^{2n} - \\binom{2n}{n}\\right] = 2^{2n-1} - \\frac{1}{2}\\binom{2n}{n}.\n$$\n\nFor part (ii):\n\nLet $N_1$ be the number of good subsets containing a particular even number. We have\n\n$$\nN_1 = \\sum_{i=0}^{n-1} \\sum_{j=0}^{i} \\binom{n-1}{i} \\binom{n}{j} = \\sum_{i+k \\ge n} \\binom{n-1}{i} \\binom{n}{k} = \\sum_{m=n}^{2n-1} \\binom{2n-1}{m} = 2^{2n-2}.\n$$\n\nSimilarly, let $N_2$ be the number of good subsets containing a particular odd number:\n\n$$\n\\begin{align*}\nN_2 &= \\sum_{i=2}^{n} \\sum_{j=0}^{i-2} \\binom{n}{i} \\binom{n-1}{j} \\\\\n&= \\sum_{i+k \\ge n+1} \\binom{n}{i} \\binom{n-1}{k} \\\\\n&= \\sum_{m=n+1}^{2n-1} \\binom{2n-1}{m} \\\\\n&= 2^{2n-2} - \\binom{2n-1}{n}.\n\\end{align*}\n$$\n\nThe sum of all elements in all good subsets of $S$ is\n\n$$\n\\begin{align*}\n&N_1(2+4+\\cdots+2n) + N_2(1+3+\\cdots+(2n-1)) \\\\\n&= 2^{2n-2} \\cdot n(n+1) + \\left[2^{2n-2} - \\binom{2n-1}{n}\\right] \\cdot n^2 \\\\\n&= 2^{2n-2}(2n^2+n) - n^2 \\binom{2n-1}{n}.\n\\end{align*}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22135,
"subject": "Mathematics (Olympiad)",
"question": "The side of a triangle is subdivided by the bisector of its opposite angle into two segments of lengths $1$ and $3$. Determine all possible values of the area of that triangle.",
"options": [],
"answer": "See solution",
"solution": "Call the triangle $ABC$ and let $AP$ be the angle bisector, with $P$ on $BC$, $BP = 3$ and $CP = 1$. By the Angle Bisector Theorem, we get $\\frac{AB}{AC} = 3$. Fixing the points $B$ and $C$, the locus of all points $A$ satisfying this is an Apollonius circle, whose centre lies on the line $BC$. This circle passes through $P$ itself and a point on $BC$ that lies $2$ units beyond $C$. Consequently, that circle has radius $\\frac{3}{2}$.\n\n\n\nIt is clear that the maximal height of the triangle, as measured from the base line $BC$ of length $4$, is $\\frac{3}{2}$, but that there is no minimal height. The area of the triangle may therefore take any positive value that is at most $3$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22136,
"subject": "Mathematics (Olympiad)",
"question": "$\\{1, 2, \\ldots, 10\\}$ тоонуудаас хэдэн тоог сонгон авч тэдгээрийг \"багц\" гэе. Хэрэв багцуудын олонлог нь\n\n1. Аль ч 2 багц дор хаяж нэг ерөнхий элементтэй,\n2. Аль ч багц өөр ямар ч багцыг агуулаагүй,\n\nгэсэн 2 чанарыг хангаж байвал \"цогц\" гэе. Тэгвэл \"цогц\" хамгийн олондоо хэдэн \"багц\"-аас тогтож болох вэ?",
"options": [],
"answer": "See solution",
"solution": "а) Дан 1-р нөхцөлийг хангах \"багц\"-уудын олонлогийг тогтооё.\n\n$A_l \\subset A$ олонлог энэ багцуудын олонлогт ордог бол $A_r = A_r \\setminus A$ олонлог орж болохгүй. Нийт $A$ олонлогт $2^{10}$ ширхэг дэд олонлог байх учир 1-р нөхцөлийг хангах $B = \\{A_1, \\ldots, A_k\\}$ олонлогийн чадал $|B| = k \\leq 2^{n-1}$ болно. Харин $A' \\subset A$, $|A'| = 6, 10$ байх бүх $A'$ олонлог, $A' \\subset A$, $|A'| = 5$ байх ба аль нь ч алиныхаа гүйцээлт болохгүй олонлогуудаас тогтох $B$ олонлогийн чадал $2^{n-1}$ ба бодлогын нөхцөлийг хангана.\n\nб) Дан 2-р нөхцөлийг хангах $C$ олонлог нь\n\n$C = \\{A'' \\subset A, |A''| = i\\}$ олонлогууд болох бөгөөд тэдгээрийн дотроос хамгийн их чадалтай нь $|C| = \\binom{10}{5}$ болно.\n\nв) Хэрэв нөхцөлүүдийг давхцуулан тавьваас дээрх хоёр нөхцөлийг зэрэг хангадаг хамгийн олон \"багц\"-аас бүрдэх олонлог болох юм. Энэ олонлог нь $A''' \\subset A$, $|A'''| = 6$ байх багцуудын олонлог юм. Энэ олонлогийн чадал $\\binom{10}{6} = 210$ бөгөөд $A = \\{A_r : |A_r| = 6\\}$ болно.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22137,
"subject": "Mathematics (Olympiad)",
"question": "For $n$ a positive integer, let $f(n)$ be the quotient obtained when the sum of all positive divisors of $n$ is divided by $n$. For example, $f(14) = (1 + 2 + 7 + 14) \\div 14 = \\frac{12}{7}$. What is $f(768) - f(384)$?\n\n(A) $\\frac{1}{768}$ (B) $\\frac{1}{192}$ (C) $1$ (D) $\\frac{4}{3}$ (E) $\\frac{8}{3}$",
"options": [],
"answer": "See solution",
"solution": "Suppose all the positive divisors of $n$ are $d_1 < d_2 < \\dots < d_k$. Then\n\n$$\nf(n) = \\frac{d_1 + d_2 + \\dots + d_k}{n} = \\frac{d_1}{n} + \\frac{d_2}{n} + \\dots + \\frac{d_k}{n} = \\frac{1}{d_k} + \\frac{1}{d_{k-1}} + \\dots + \\frac{1}{d_1}.\n$$\n\nThus $f(n)$ is the sum of the reciprocals of the positive divisors of $n$. Because $768 = 384 \\cdot 2$, all the divisors of $384$ also divide $768$, so it is necessary to consider only the divisors of $768$ that do not divide $384$, which are $2^8$ and $2^8 \\cdot 3$. Therefore\n\n$$\nf(768) - f(384) = \\frac{1}{2^8} + \\frac{1}{2^8 \\cdot 3} = \\frac{4}{2^8 \\cdot 3} = \\frac{1}{192}.\n$$\n\nAlternatively,\n\nLet $\\sigma(n)$ be the sum of the divisors of $n$. Then $f(n) = \\frac{\\sigma(n)}{n}$, and because $\\sigma$ is multiplicative (meaning that $\\sigma(ab) = \\sigma(a)\\sigma(b)$ whenever $\\gcd(a, b) = 1$), $f$ is multiplicative as well. That is,\n\n$$\nf(768) - f(384) = f(2^8 \\cdot 3) - f(2^7 \\cdot 3) = f(3)(f(2^8) - f(2^7)).\n$$\n\nNote that $f(3) = \\frac{4}{3}$ and\n\n$$\nf(2^k) = \\frac{\\sigma(2^k)}{2^k} = \\frac{2^{k+1} - 1}{2^k} = 2 - \\frac{1}{2^k}.\n$$\n\nThus\n\n$$\nf(3)(f(2^8) - f(2^7)) = \\frac{4}{3} \\cdot \\frac{1}{2^8} = \\frac{1}{2^6 \\cdot 3} = \\frac{1}{192}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22138,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer and let $x$ and $y$ be positive divisors of $2n^2 - 1$. Prove that $x + y$ is not divisible by $2n + 1$.",
"options": [],
"answer": "See solution",
"solution": "Suppose, if possible, that $x + y$ is divisible by $2n + 1$. Note that $2n + 1$ and $2n^2 - 1$ are relatively prime, as $(2n - 1)(2n + 1) - 2(2n^2 - 1) = 1$. Hence $2n + 1$ is coprime to $\\gcd(x, y)$, so $\\frac{x}{\\gcd(x, y)}$ and $\\frac{y}{\\gcd(x, y)}$ are coprime divisors of $2n^2 - 1$ whose sum is divisible by $2n + 1$.\n\nTo reach a contradiction, we may and will therefore assume $x$ coprime to $y$. Then $2n^2-1$ is divisible by $xy$, say, $2n^2-1 = kxy$; write $x+y = \\ell(2n+1)$. Express $n$ from this latter and plug it into the former to get $kxy = \\frac{1}{2} \\left(\\frac{x+y}{\\ell} - 1\\right)^2 - 1$; alternatively, but equivalently,\n$$2k\\ell^2xy = (x + y - \\ell)^2 - 2\\ell^2.$$ \n\nFix $k$ and $\\ell$ to regard the above equality as an equation in positive integers $x$ and $y$. By assumption, it has at least one solution. Consider one such with a minimal $x + y$. Without loss of generality, assume $x \\ge y$.\n\nRewrite the equation as a quadratic in $x$:\n$$x^2 + 2(y-\\ell-k\\ell^2y)x + (y-\\ell)^2 - 2\\ell^2 = 0,$$\nand consider the other root $x'$. Then $x + x' = 2(k\\ell^2y - y + \\ell)$ and $xx' = (y-\\ell)^2 - 2\\ell^2$. The former shows that $x'$ is also integer. The latter implies $x' < 0$: Clearly, $x' \\ne 0$, as $(y-\\ell)^2 = 2\\ell^2$ cannot hold in integers; and if $x' > 0$, then $x'y \\le xx' = (y-\\ell)^2 - 2\\ell^2 = y^2 - \\ell(2y+\\ell) < y^2$, so $x' < y$, whence $x' + y < 2y \\le x + y$, contradicting the minimality of $x + y$.\n\nThus, $(y-\\ell)^2 < 2\\ell^2$ and $x$ divides $2\\ell^2 - (y-\\ell)^2$, so $x \\le 2\\ell^2$. On the other hand, as $x'$ is negative, $x > x + x' = 2((k\\ell^2 - 1)y + \\ell) \\ge 2(\\ell^2 + \\ell - 1) \\ge 2\\ell^2$. This is a contradiction, so the sum $x + y$ is not divisible by $2n + 1$, as required.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22139,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ which for any real numbers $x$ and $y$ satisfy\n$$\n(f(x+y))^2 = x f(x) + 2 f(xy) + (f(y))^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "$f(x) = 0$ and $f(x) = x$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22140,
"subject": "Mathematics (Olympiad)",
"question": "Prove that all real numbers $x \\neq -1$, $y \\neq -1$ with $xy = 1$ satisfy the following inequality:\n$$\n\\left(\\frac{2+x}{1+x}\\right)^2 + \\left(\\frac{2+y}{1+y}\\right)^2 \\ge \\frac{9}{2}\n$$",
"options": [],
"answer": "See solution",
"solution": "Since $xy = 1$, we may assume that $x \\neq 0$ and $y \\neq 0$. By substituting $y = \\frac{1}{x}$, we have\n$$\n\\left(\\frac{2+x}{1+x}\\right)^2 + \\left(\\frac{2+y}{1+y}\\right)^2 = \\left(\\frac{2+x}{1+x}\\right)^2 + \\left(\\frac{2x+1}{x+1}\\right)^2 = \\frac{5x^2 + 8x + 5}{x^2 + 2x + 1}\n$$\nIt remains to show that\n$$\n\\frac{5x^2 + 8x + 5}{x^2 + 2x + 1} \\ge \\frac{9}{2}\n$$\nThis inequality is equivalent to\n$$\n(x-1)^2 \\ge 0\n$$\nand hence the statement is proved.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22141,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle such that $AB \\neq AC$, $\\angle B \\neq 90^\\circ$. Let $I$ be the incenter of $ABC$, and let $D$, $E$, $F$ be the perpendicular feet from $I$ to $BC$, $CA$, and $AB$, respectively. Let $S$ be the intersection of $AB$ and $DI$, $T$ be the intersection of $DE$ and the line passing through $F$ and perpendicular to $DF$, and $R$ be the intersection of $ST$ and $EF$. Let $P_{ABC}$ be the intersection point of the circle with diameter $IR$ and the incircle of $ABC$, that is on the other side to $A$ with respect to the line $IR$.\n\nLet $XYZ$ be an isosceles triangle with $XZ = YZ > XY$ and $W$ be a point on the side $YZ$ with $WY < XY$. For $K = P_{YXW}$, $L = P_{ZXW}$, show that $2KL \\leq XY$.",
"options": [],
"answer": "See solution",
"solution": "**Lemma 1** Let $Q$ be the intersection of $DI$ and the incircle of $\\triangle ABC$. Then the three points $A$, $Q$, $P_{ABC}$ are collinear.\n\n**Proof:** By applying Pascal's theorem for the hexagon $QQDEFF$, we get that the point $R$ is the intersection of the line $EF$ and the tangent to the incircle at $Q$. Since $Q$ is on the circle $O$, inversion with respect to the incircle $I$ maps the circle $O$ to the line $QP_{ABC}$. Moreover, the inversion maps the point $A$ to the midpoint $A'$ of $EF$, and $A'$ lies on the circle $O$ since $\\angle IA'Q = 90^\\circ$. Thus, the three points $A$, $Q$, $P_{ABC}$ are collinear. $\\square$\n\n**Lemma 2** Let $U$ be the intersection of $BC$ and the line passing through $A$, $Q$, $P_{ABC}$. Then $BD = UC$.\n\n**Proof:** Let $I_a$ be the center of the $A$-excircle, $r_a$ its radius, and $J$, $K$ the perpendicular feet from $I_a$ to $AB$, $BC$, respectively. Then:\n\n$$\nIQ : I_aK = r : r_a = IF : I_aJ = AI : AI_a.\n$$\n\nSince $IQ \\parallel I_aK$, the three points $A$, $Q$, $K$ are collinear and thus $U = K$. It is clear that $BD = UC = \\frac{c + a - b}{2}$. $\\square$\n\n**Lemma 3** Let $M$ be the midpoint of $BC$. Then $MP_{ABC} = \\frac{|b - c|}{2}$.\n\n**Proof:** By Lemmas 1 and 2, $MD = MU = \\frac{b - c}{2}$ (or $\\frac{c - b}{2}$), and $MP_{ABC} = \\frac{b - c}{2}$ (or $\\frac{c - b}{2}$). $\\square$\n\nLet $N$ be the midpoint of $XW$. By Lemma 3 (since $XY > YW$, $XZ > ZW$), it is easy to see that\n\n$$\nNP_{YWX} = \\frac{XY - YW}{2}, \\quad NP_{ZXW} = \\frac{XZ - ZW}{2}\n$$\n\nSince $YW + ZW = YZ = XZ$, we have\n\n$$\nP_{YWX}P_{ZXW} \\leq NP_{YWX} + NP_{ZXW} = \\frac{XY}{2}\n$$\n\n$\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22142,
"subject": "Mathematics (Olympiad)",
"question": "One of the cells of a $20 \\times 20$ torus contains a buried treasure. To find the treasure, we select several rectangles of size $1 \\times 4$ or $4 \\times 1$ on this torus and ask a sapper to investigate them with a mine detector. For each rectangle, the sapper will tell us whether the treasure is in that rectangle. What is the minimal number of rectangles we should select in order to determine the exact cell containing the treasure?",
"options": [],
"answer": "See solution",
"solution": "The minimal number of rectangles required is **160**.\n\nLet each cell be identified by coordinates $(i, j)$, where $1 \\leq i, j \\leq 20$. Consider the following selection of 160 rectangles:\n\n$$\n(a-1, b),\\ (a-2, b),\\ (a-3, b),\\ (a-4, b) \\pmod{20}, \\quad \\text{where } 5 \\mid (a+b),\n$$\n$$\n(a, b-1),\\ (a, b-2),\\ (a, b-3),\\ (a, b-4) \\pmod{20}, \\quad \\text{where } 5 \\mid (a+b-1).\n$$\n\nIf the sapper reports that the treasure is in only one rectangle, the cell is uniquely determined, since each rectangle contains a unique cell not covered by the other rectangles. If the treasure is in two rectangles, it must be at their intersection cell.\n\nIf only 159 rectangles are selected, the torus is covered except possibly one cell. At least $399 \\cdot 2 - 159 \\cdot 4 = 162$ cells are covered by only one rectangle, so two such cells must share a rectangle. If the treasure is in that rectangle, we cannot distinguish which cell contains it.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22143,
"subject": "Mathematics (Olympiad)",
"question": "En el triángulo isósceles $ABC$, sean $D$ y $E$ puntos en los lados $AB$ y $AC$, respectivamente, tales que las rectas $BE$ y $CD$ se cortan en $F$. Además, los triángulos $AEB$ y $ADC$ son iguales y tienen $AD = AE = 10$ y $AB = AC = 30$.\n\nCalcular $\\frac{\\text{área}(ADFE)}{\\text{área}(ABC)}$.",
"options": [],
"answer": "See solution",
"solution": "Como $AD = AE$, el punto $F$ está a igual distancia de los lados $AC$ y $AB$; llamemos $h$ a esa distancia.\n\nEntonces $\\frac{\\text{área}(ADF)}{\\text{área}(AEF)} = \\frac{10h}{2}$, por lo tanto $\\frac{\\text{área}(ADFE)}{\\text{área}(ADF)} = 2 \\frac{\\text{área}(ADF)}{\\text{área}(ADF)} = 10h$.\n\nPor otra parte,\n\n$$\n\\text{área}(ACF) = \\text{área}(ABF) = \\frac{30h}{2}\n$$\n\n$$\n\\text{área}(ACD) = \\text{área}(ACF) + \\text{área}(ADF) = 20h\n$$\n\n$$\n\\text{área}(DBF) = \\text{área}(ABF) - \\text{área}(ADF) = 10h\n$$\n\nLuego $\\frac{\\text{área}(ADF)}{\\text{área}(ACF)} = \\frac{\\frac{10h}{2}}{\\frac{30h}{2}} = \\frac{1}{3} = \\frac{DF}{FC} = \\frac{\\text{área}(DBF)}{\\text{área}(BCF)}$, y resulta que\n\n$$\n\\text{área}(BCF) = 3 \\text{área}(DBF) = 30h\n$$\n\nFinalmente,\n\n$$\n\\frac{\\text{área}(ADFE)}{\\text{área}(ABC)} = \\frac{10h}{\\text{área}(BCF) + 2\\text{área}(ACF)} = \\frac{10h}{30h + 30h} = \\frac{1}{6}\n$$\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22144,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a trapezium with the bottom edge $BC$ ($BC \\parallel AD$ and $BC > AD$) inscribed in the circle $(O)$ (where $O$ is the center of $(O)$). Let $P$ be a point moving on the line $BC$ outside the segment $BC$ such that $PA$ does not touch the circle $(O)$. The circle with diameter $PD$ intersects $(O)$ at $E$ ($E \\neq D$). Let $M$ be the point of intersection of $BC$ and $DE$, and $N$ ($N \\neq A$) be the second point of intersection of $PA$ and $(O)$. Prove that the line $MN$ passes through a fixed point.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $A'$ be the reflection of point $A$ through $O$. We prove that $N$, $M$, and $A'$ are collinear, so the line $MN$ passes through the fixed point $A'$. First, $DE$ is the radical axis of the circle $(O)$ and the circle $\\gamma_1$ with diameter $PD$.\n\nSince $\\angle PNA' = 90^\\circ$, the line $NA'$ is the radical axis of $(O)$ and the circle $\\gamma_2$ with diameter $PA'$. The line $DA'$ meets $BC$ at point $F$; since $\\angle PFA' = 90^\\circ$ and $\\angle ADA' = 90^\\circ$, it follows that $\\angle PFA' = 90^\\circ$. Therefore, $BC$ is the radical axis of $\\gamma_1$ and $\\gamma_2$, so the radical axes $DE$, $BC$, and $NA'$ are concurrent at the radical center $M$. Thus, the points $M$, $N$, and $A'$ are collinear.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 22145,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCDEF$ be a convex hexagon with $\\angle A = \\angle C = \\angle E$ and $\\angle B = \\angle D = \\angle F$, such that there is a point $P$ in its interior that is equidistant from the sides $AB$, $CD$, and $EF$. If $G_1 \\neq G_2$ are the centroids of triangles $ACE$ and $BDF$, respectively, prove that $\\angle G_1 P G_2 = 60^\\circ$.",
"options": [],
"answer": "See solution",
"solution": "Let $\\angle A = \\angle C = \\angle E = \\alpha$ and $\\angle B = \\angle D = \\angle F = \\beta$. The sum of the angles of hexagon $ABCDEF$ is $3\\alpha + 3\\beta = 720^\\circ$, so $\\alpha + \\beta = 240^\\circ$. Since $\\angle B + \\angle C = \\alpha + \\beta = 240^\\circ$, the lines $AB$ and $CD$ intersect at a point outside the hexagon. Let $X = AB \\cap CD$, $Y = CD \\cap EF$, and $Z = EF \\cap AB$. The point $P$, equidistant from the sides $AB$, $CD$, and $EF$, is the center of the equilateral triangle $XYZ$.\n\nIn the complex plane, place the origin at $P$. From $\\triangle BCX \\sim \\triangle DEY \\sim \\triangle FAZ$, there exists $k > 0$ such that\n\n$$\n\\frac{c-x}{b-x} = \\frac{e-y}{d-y} = \\frac{a-z}{f-z} = k \\cdot \\varepsilon,\n$$\n\nwhere $\\varepsilon = \\cos \\frac{\\pi}{3} \\pm i \\sin \\frac{\\pi}{3}$.\n\nThus,\n\n$$\nc-x = (b-x)k \\cdot \\varepsilon, \\quad e-y = (d-y)k \\cdot \\varepsilon, \\quad a-z = (f-z)k \\cdot \\varepsilon.\n$$\n\nSince $p = \\frac{x+y+z}{3} = 0$, by addition,\n\n$$\nc+e+a = (b+d+f)k \\cdot \\varepsilon \\implies g_1 = g_2 \\cdot k \\cdot \\varepsilon.\n$$\n\nSince $G_1 \\neq G_2$, $g_1$ and $g_2$ are nonzero, so $G_1$, $G_2$, and $P$ are distinct. Therefore,\n\n$$\n\\frac{g_1 - p}{g_2 - p} = k \\cdot \\varepsilon,\n$$\n\nwhich implies $\\angle G_1 P G_2 = 60^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22146,
"subject": "Mathematics (Olympiad)",
"question": "If $v$ is a positive integer and the number $v(v+3)$ is a perfect square, prove that $v$ is not a multiple of $3$.",
"options": [],
"answer": "See solution",
"solution": "Let $v(v+3) = w^2$, where $w \\in \\mathbb{Z}$. Assume $v = 3\\kappa$, with $\\kappa \\in \\mathbb{N}^*$. Then:\n\n$$\nv(v+3) = 3\\kappa(3\\kappa+3) = 9\\kappa(\\kappa+1) = w^2\n$$\n\nSo $3 \\cdot 3\\kappa(\\kappa+1) = w^2$, which implies $3 \\mid w^2$. Since $3$ is prime, $3 \\mid w$.\n\nTherefore,\n$$\n\\kappa(\\kappa+1) = \\left(\\frac{w}{3}\\right)^2, \\quad \\frac{w}{3} \\in \\mathbb{Z}\n$$\n\nBut $\\kappa^2 < \\kappa(\\kappa+1) = \\left(\\frac{w}{3}\\right)^2 < (\\kappa+1)^2$, so $\\kappa < \\left|\\frac{w}{3}\\right| < \\kappa+1$, which is impossible for integer $\\frac{w}{3}$.\n\nHence, $v$ is not a multiple of $3$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22147,
"subject": "Mathematics (Olympiad)",
"question": "Find the number of 2013-digit numbers $d_1d_2\\dots d_{2013}$ with odd digits $d_1, d_2, \\dots, d_{2013}$ such that\n\n$$\nd_1 \\cdot d_2 + d_2 \\cdot d_3 + \\dots + d_{1809} \\cdot d_{1810} \\equiv 1 \\pmod{4},\n$$\n\nand\n\n$$\nd_{1810} \\cdot d_{1811} + d_{1811} \\cdot d_{1812} + \\dots + d_{2012} \\cdot d_{2013} \\equiv 1 \\pmod{4}.\n$$",
"options": [],
"answer": "See solution",
"solution": "We use the following observation: Any odd numbers $x_1, \\dots, x_k$ satisfy\n\n$$\nx_1x_2 + x_2x_3 + \\dots + x_{k-1}x_k + x_kx_1 \\equiv k \\pmod{4}. \\quad (*)\n$$\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 $+1$ or $-1$ as odd numbers are congruent to $\\pm 1 \\pmod{4}$. Replacing an $x_i = -1$ by $x_i = 1$ does not change the mod 4 remainder of $S = \\sum_{j=1}^k x_jx_{j+1}$. 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 $d_1d_2\\dots d_{2013}$ satisfy the stated conditions. By $(*)$ we have\n\n$$\n\\sum_{j=1}^{1810} d_j d_{j+1} \\equiv 1810 \\pmod{4}\n$$\n\n(here $d_{1810+1} = d_1$), hence\n\n$$\n\\sum_{j=1}^{1809} d_j d_{j+1} \\equiv 1 \\pmod{4}\n$$\n\nif and only if $1810 - d_{1810}d_1 \\equiv 1 \\pmod{4}$, i.e. $d_1 d_{1810} \\equiv 1 \\pmod{4}$.\n\nSimilarly,\n\n$$\n\\sum_{j=1810}^{2012} d_j d_{j+1} \\equiv 1 \\pmod{4}\n$$\n\nif and only if $(2013-1809) - d_{1810} d_{2013} \\equiv 1 \\pmod{4}$, i.e.\n\n$$\nd_{1810} d_{2013} \\equiv -1 \\pmod{4}.\n$$\n\nWe 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 $3, 7$.\n\nLet $d_{1810} \\in \\{1,5,9\\}$. 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 $d_1, d_{1810}, d_{2013}$.\n\nBecause there are 5 choices for each of the remaining 2010 digits $d_i$ (they can be arbitrary), 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 $d_1, d_{1810}, d_{2013}$.\n\nLike in the previous case, we obtain $12 \\cdot 5^{2010}$ admissible numbers with $d_{1810} \\in \\{3,7\\}$.\n\nIn summary, there are\n\n$$\n18 \\cdot 5^{2010} + 12 \\cdot 5^{2010} = 6 \\cdot 5^{2011}\n$$\nadmissible numbers in all.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22148,
"subject": "Mathematics (Olympiad)",
"question": "A country has $n$ cities, labeled $1, 2, 3, \\ldots, n$. It wants to build exactly $n-1$ roads between certain pairs of cities so that every city is reachable from every other city via some sequence of roads. However, it is not permitted to put roads between pairs of cities that have labels differing by exactly $1$, and it is also not permitted to put a road between cities $1$ and $n$. Let $T_n$ be the total number of possible ways to build these roads.\n\n(a) For all odd $n$, prove that $T_n$ is divisible by $n$.\n\n(b) For all even $n$, prove that $T_n$ is divisible by $n/2$.",
"options": [],
"answer": "See solution",
"solution": "*Solution 1* (by Jon Schneider). Let $A$ be the set of good trees. Consider the operation $S$ which takes a tree, and replaces each edge connecting vertices $x$ and $y$ with an edge connecting vertices $x+1$ and $y+1$ (labels taken modulo $n$). Note that $S$ is invertible and sends good trees to good trees, so it partitions the set $A$ into several *orbits* (where two trees are in the same orbit if you can reach one from the other by repeatedly applying $S$). We will show that all orbits either have size $n$ or $n/2$, thus proving the required statement.\n\nGiven any tree $T$, note that $S^n T = T$ (because $x+n \\equiv x \\pmod{n}$). Now, if the orbit that $T$ belongs to has size $r$, then $r$ is the smallest positive value satisfying $S^r T = T$, and it follows that $r$ must divide $n$. We now have two cases.\n\n*Case 1*: Assume that $r$ does not divide $n/2$. Consider any edge $(x, y)$ of $T$. If $S^r T = T$, then the edge $(x+r, y+r)$ is also in $T$, so the $n/r$ edges\n\n$$\n(x, y), (x+r, y+r), (x+2r, y+2r), \\ldots, (x-r, y-r)\n$$\nmust belong to $T$. All these edges are distinct; if $(x, y)$ and $(x+ar, y+ar)$ are the same for $1 \\leq a < \\frac{n}{r}$, then $x \\equiv x + 2ar$ so $n | 2ar$. If $n$ is odd, $n | ar$, which cannot happen for $1 \\leq a < \\frac{n}{r}$. Similarly for even $n$ with $r$ not dividing $n/2$. Thus, we can partition the edges of $T$ into groups of $\\frac{n}{r}$ edges. Since $T$ has $n-1$ edges and $\\gcd(n-1, n) = 1$, $\\frac{n}{r}$ cannot divide $n-1$ unless $r = n$. Therefore, the orbit has size $n$.\n\n*Case 2*: Assume $r$ divides $\\frac{n}{2}$ (so $n$ is even). If $y = x + \\frac{n}{2}$, then we have a group of $\\frac{n}{2r}$ distinct edges. We can partition the edges of $T$ into groups of either $\\frac{n}{r}$ or $\\frac{n}{2r}$ edges, so $n-1$ must be divisible by $\\frac{n}{2r}$. Again, since $\\gcd(n-1, n) = 1$, this can only happen if $r = \\frac{n}{2}$, so the orbit has size $n/2$.\n\n*Solution 2* (by the poser). There are many spanning trees which avoid edges between consecutive labels. Separate them into groups as follows. Given such a labeled spanning tree $T$, consider the tree $T'$ obtained by taking $T$, and replacing each label $i$ with $i+1$, and replacing the label $n$ with $1$. This cycles the labels on the vertices. Two trees are of the same type if they can be reached from each other by performing this cycling operation some number of times. The cycling operation respects the property that no consecutively-labeled vertices span an edge.\n\nThis partitions the spanning trees counted by $T_n$ into some collection of types. For each fixed type, the number of spanning trees of that type is divisible by $n$ when $n$ is odd, and divisible by $n/2$ when $n$ is even. For this, we use the following lemma:\n\n*Lemma 4.* Let $T$ be a tree with vertices labeled $1, \\ldots, n$, and let $\\pi : [n] \\to [n]$ be an automorphism of $T$, i.e., where vertices $i$ and $j$ are adjacent in $T$ if and only if $\\pi(i)$ and $\\pi(j)$ are adjacent in $T$. Then $\\pi$ must have a fixed point $\\pi(i) = i$ or a fixed edge $\\{\\pi(i), \\pi(j)\\} = \\{i, j\\}$.\n\n*Proof.* Suppose for contradiction that $\\pi$ does not have a fixed point. For every vertex $i$, $\\pi$ maps it to some vertex $\\pi(i)$ elsewhere in the tree. Since trees are acyclic and connected, there is a unique path from $i$ to $\\pi(i)$, departing from $i$ along an edge. Draw an arrow on that edge indicating the direction of departure for every vertex.\n\nThere are $n$ vertices and $n-1$ edges, so by the pigeonhole principle, some edge $\\{i, j\\}$ receives two arrows. These arrows must be in different directions, indicating that $\\pi(i)$ is on the $j$-side and $\\pi(j)$ is on the $i$-side. Since $\\pi$ is an automorphism, $\\pi(i)$ and $\\pi(j)$ are adjacent in $T$, so $\\pi(i) = j$ and $\\pi(j) = i$, producing the fixed edge. $\\square$\n\nApplying lemma 4: For a labeled tree $T$ counted by $T_n$, the number of other trees of the same type is the minimum number of times $k$ that one must apply the cycling operation to $T$ before obtaining $T$ again. This corresponds to an automorphism $\\pi$ which is a cyclic shift $i \\mapsto i + k$ (with $n + 1$ corresponding to $1$, etc). Fixed points do not arise unless $n \\mid k$, and fixed edges (transpositions) never arise for odd $n$, immediately proving that $n \\mid T_n$ when $n$ is odd.\n\nFor even $n$, the only way a cyclic shift can produce a transposition is when $k$ is an odd multiple of $n/2$, since $i + k + k = i$. Therefore, the period of cycling is either $n/2$ or $n$, both divisible by $n/2$, so we are done.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22149,
"subject": "Mathematics (Olympiad)",
"question": "Find all $3 \\times 3$ matrices whose rows and columns each contain the same three numbers $\\{a, b, c\\}$ (possibly in different orders), and such that the sum and product of the entries in each row and each column are the same. Describe all possible forms of such matrices.",
"options": [],
"answer": "See solution",
"solution": "The key observation is that two numbers are determined by their product and sum.\n\n**Lemma.** If $a$, $b$, $u$, $v$ are real numbers that satisfy $a + b = u + v$ and $ab = uv$, then $\\{a, b\\} = \\{u, v\\}$.\n\n**Proof.** Because\n$$\n\\begin{align*}\n(x - a)(x - b) &= x^2 - (a + b)x + ab \\\\\n&= x^2 - (u + v)x + uv = (x - u)(x - v),\n\\end{align*}\n$$\nthis quadratic has roots $a$, $b$ as well as roots $u$, $v$.\n\nConsidering the first row and third column of the matrix\n$$\n\\begin{bmatrix}\na & b & c \\\\\n\\star & \\star & u \\\\\n\\star & \\star & v\n\\end{bmatrix}\n$$\nwhich have the same sum and the same product, the lemma implies that $\\{u, v\\} = \\{a, b\\}$. After swapping rows two and three if necessary, we even get $u = a$ and $v = b$:\n$$\n\\begin{bmatrix}\na & b & c \\\\\n\\star & \\star & a \\\\\n\\star & \\star & b\n\\end{bmatrix}\n$$\nFor the same reason, the missing entries in the first column are $b, c$ and in the second column $a, c$ are missing. We can also work with the third column and the second or third row and obtain that the missing entries in the second row are $b, c$ and in the third row are $a, c$.\n\nThis already proves the statement of the problem. From the information collected, it is not hard to describe the possible shapes of the matrix.\n\nAs there is no $a$ missing in the first column, it now is clear that the bottom left corner must be occupied by $c$. In a similar way, we can now fill in the remaining numbers. As we may earlier have swapped the second and the third row, we now see that the matrix must have one of these shapes\n$$\n\\begin{bmatrix} a & b & c \\\\ b & c & a \\\\ c & a & b \\end{bmatrix} \\quad \\text{or} \\quad \\begin{bmatrix} a & b & c \\\\ c & a & b \\\\ b & c & a \\end{bmatrix}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22150,
"subject": "Mathematics (Olympiad)",
"question": "Let $f : \\mathbb{Z}_+ \\to \\mathbb{Z}_+$ be a function such that for all $x, y \\in \\mathbb{Z}_+$, $f(f(x) + y)$ divides $x + f(y)$. Determine all such functions $f$.",
"options": [],
"answer": "See solution",
"solution": "*Step 1:* We will prove that either $f(x) = x$ or $f$ is not an injection.\n\nFor example, $f(1) > 1$, by taking $x = 1$ in the problem, we know that $f(f(1) + y)$ divides $1 + f(y)$. In particular,\n\n$$\nf(f(1) + y_0) \\leq 1 + f(y_0).\n$$\n\nBy induction, it is easy to see that for fixed $y_0 \\in \\{1, \\dots, f(1)\\}$ and any $t \\in \\mathbb{Z}_+$, there is\n\n$$\nf(t \\cdot f(1) + y) \\leq t + f(y).\n$$\n\nTherefore, for any $y$ large enough, $f(y) < y$. So when $y$ is large enough, $f : \\{1, \\dots, y\\} \\to \\{1, \\dots, y-1\\}$ cannot be an injection.\n\nIn the following, we will discuss the case $f(1) = 1$. By taking $x = 1$ in the problem, we know that $f(1+y)$ divides $1+f(y)$. In particular, $f(1+y) \\leq 1+f(y)$. By induction, it is easy to know that for any $x \\in \\mathbb{Z}_+$, there is $f(x) = x$ or $f(y) < y$ for $y$ large enough. Similarly, we obtain that $f$ cannot be an injection.\n\n*Step 2:* We will show that if $f$ is not an injection, then for $x$ large enough, either $f(x) = 1$ or $f(x) = \\begin{cases} 1, & x \\text{ is odd}, \\\\ 2, & x \\text{ is even.} \\end{cases}$\n\nSuppose $f$ is not an injection and we denote $A$ as the smallest positive integer such that there exists a positive integer $x_0 \\in \\mathbb{N}_+$ satisfying\n\n$$\nf(x_0 + A) = f(x_0).\n$$\n\nBy substituting $x = x_0$ and $x = x_0 + A$ respectively into the problem, we get\n\n$$\nf(f(x_0) + y) \\mid x_0 + f(y)\n$$\n\nand\n\n$$\nf(f(x_0 + A) + y) \\mid x_0 + A + f(y).\n$$\n\nSubtracting the above two, we get\n\n$$\nf(f(x_0) + y) \\mid A.\n$$\n\nThis shows that $f(z)$ can only take values in the factors of $A$ when $z$ is greater than $f(x_0)$. Suppose that $A$ has a total of $D$ factors. According to the Pigeonhole principle, we investigate $D+1$ consecutive positive integers $z, z+1, \\dots, z+D$ that are greater than $f(x_0)$, of which there must be two numbers that have equal images under $f$ and the difference between these two numbers is less than or equal to $D$. This leads to $A \\leq D$.\n\nSuch positive integer $A$ can only be $A = 1$ or $A = 2$. If $A = 1$, according to the previous result we know that for $x$ large enough, $f(x) = 1$. If $A = 2$, we see that for $x$ large enough, $f(x) = 1$ or $f(x) = 2$. But $A = 2$ is the smallest $A$ that makes the previous condition valid, from which we know that\n\n\n\nHowever, (b) is impossible because taking two even numbers $x$ and $y$ large enough and substituting them into the problem gives\n\n$$\n2 \\mid x + 1,\n$$\n\na contradiction. This completes the proof of *Step 2*.\n\n*Step 3:* We will prove that if $f$ is not an injection, then for $x > 1$, $f(x) = 1$ or $f(x) = \\begin{cases} 1, & x \\text{ is odd}, \\\\ 2, & x \\text{ is even}. \\end{cases}$\n\nLet $A$ be as set in *Step 2*. Taking $x_0$ large enough so that $f(x_0) = 1$ and substituting $x = x_0$ and $x = x_0 + A$ into the problem, we get\n\n$$\nf(1+y) \\mid x_0 + f(y) \\quad \\text{and} \\quad f(1+y) \\mid x_0 + A + f(y).\n$$\n\nSubtracting the above two, we get\n\n$$\nf(1+y) \\mid A.\n$$\n\nWhen $A = 1$, this equation immediately leads to $f(y+1) = 1$, and thus we obtain the function of (2) mentioned before.\n\nWhen $A = 2$, this equation gives $f(y+1) = 1$ or $f(y+1) = 2$. However, the minimality of $A = 2$ shows that $f(x)$ must be $1, 2, 1, 2, \\dots$ alternately in the case of $x > 1$. Combining the results of *Step 2* yields the function of the above mentioned (3). $\\Box$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22151,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a cyclic quadrilateral with $AB < CD$, whose diagonals intersect at point $F$, and $AD$ and $BC$ intersect at point $E$. Let $K$ and $L$ be the projections of $F$ onto the sides $AD$ and $BC$, respectively. Let $M$, $S$, and $T$ be the midpoints of $EF$, $CF$, and $DF$. Prove that the second intersection point of the circumcircles of triangles $MKT$ and $MLS$ lies on the side $CD$.",
"options": [],
"answer": "See solution",
"solution": "Let $N$ be the midpoint of $CD$. We will prove that the circumcircles of triangles $MKT$ and $MLS$ pass through $N$.\n\nWe will first 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\n\n\nWe will prove that\n\n$$\n\\angle SLQ = \\angle QNS \\qquad (1)\n$$\n\nSince $FLC$ is right-angled and $LS$ is its median, we have $SL = SC$ and\n\n$$\n\\angle SLC = \\angle SCL = \\angle ABC \\qquad (2)\n$$\n\nIn addition, since $N$ and $S$ are the midpoints of $DC$ and $FC$, we have $SN \\parallel FD$.\n\nAlso, $Q$ and $S$ are the midpoints of $EC$ and $CD$, so $QN \\parallel ED$.\n\nIt follows that the angles $\\angle EDB$ and $\\angle QNS$ have parallel sides, and since $AB < CD$, they are acute. As a result,\n\n$$\n\\angle EDB = \\angle QNS \\qquad (3)\n$$\n\nFrom the cyclic quadrilateral $ABCD$, we get\n\n$$\n\\angle EDB = \\angle ACB \\qquad (4)\n$$\n\nNow, from (2), (3), and (4), we immediately obtain (1), so $\\angle SLQ = \\angle QNS$ and the quadrilateral $LNSQ$ is cyclic. Since this circle also passes through $M$, the points $M$, $L$, $Q$, $S$, $N$ are concyclic, which means 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": 22152,
"subject": "Mathematics (Olympiad)",
"question": "Two incongruent triangles $ABC$ and $XYZ$ are called a pair of *pals* if they satisfy the following conditions:\n\n(a) The two triangles have the same area.\n\n(b) Let $M$ and $W$ be the respective midpoints of sides $BC$ and $YZ$. The two sets of lengths $\\{AB, AM, AC\\}$ and $\\{XY, XW, XZ\\}$ are identical 3-element sets of pairwise relatively prime integers.\n\nDetermine if there are infinitely many pairs of triangles that are pals of each other.",
"options": [],
"answer": "See solution",
"solution": "The answer is *yes*.\n\nWe start with the following observations.\n\n**Lemma 2.** The following statement and its converse are both true: If $q, r, s$ are three distinct positive real numbers such that $q, r, 2s$ are side lengths of a triangle, then there is a unique triangle $PQR$ with $PQ = q$, $PR = r$, and $PS = s$, where $S$ is the midpoint of side $QR$.\n\n*Proof.* If $q, r, 2s$ are side lengths of a triangle, we construct the unique triangle $PRQ_1$ with $PR = r$, $RQ_1 = q$, and $PQ_1 = 2s$. Let $S$ be the midpoint of side $PQ_1$. Extend segment $RS$ through $S$ to $Q$ so that $PQQ_1R$ is a parallelogram. It is clear that $PQR$ is a triangle with $PQ = q$, $PR = r$, and $PS = s$, where $S$ is the midpoint of side $QR$. To prove the converse statement, we only need to note that the above procedure can be reversed. $\\Box$\n\n**Lemma 3.** Let $PQR$ be a triangle with $PQ = q$, $PR = r$, and $PS = s$, where $S$ is the midpoint of side $QR$. Then the area of triangle $PQR$ is\n\n$$\n\\frac{1}{4}\\sqrt{2q^2r^2 + 8r^2s^2 + 8s^2q^2 - q^4 - r^4 - 16s^4}.\n$$\n\n*Proof.* Because $PQQ_1R$ is a parallelogram, triangles $PQR$ and $PRQ_1$ have the same area. Because $(PR, RQ_1, Q_1P) = (r, q, 2s)$, the desired result follows directly from Heron's formula. $\\Box$\n\nIn view of Lemma 2, if triangles $ABC$ and $XYZ$ are a pair of pals, we may assume without loss of generality that $(AB, AC, AM) = (n, s, t)$ and $(XY, XZ, XW) = (n, t, s)$. By Lemma 3, we have\n\n$$\n2n^2s^2 + 8s^2t^2 + 8t^2n^2 - n^4 - s^4 - 16t^4 = 2n^2t^2 + 8t^2s^2 + 8s^2n^2 - n^4 - t^4 - 16s^4\n$$\n\nwhich simplifies to\n\n$$\n6n^2t^2 - 6n^2s^2 = 15t^4 - 15s^4.\n$$\n\nBecause $ABC$ and $XYZ$ are incongruent, we deduce that $2n^2 = 5(t^2 + s^2)$. We set $n = 5(k^2 + (k+1)^2) = 10k^2 + 10k + 1$ for some positive integer $k$. By applying the identity $(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 - (ad - bc)^2$ repeatedly, we have\n\n$$\n\\begin{aligned}\nt^2 + s^2 &= 10n^2 = (1^2 + 3^2)(k^2 + (k+1)^2)(k^2 + (k+1)^2) \\\\\n&= ((4k+3)^2 + (2k-1)^2)(k^2 + (k+1)^2) \\\\\n&= (6k^2 + 4k - 1)^2 + (2k^2 + 8k + 3)^2.\n\\end{aligned}\n$$\n\nWe set $(n, s, t) = (10k^2 + 10k + 1, 6k^2 + 4k - 1, 2k^2 + 8k + 3)$ for positive integer $k$. For large $k$, it is easy to see that each of $(n, 2s, t)$ and $(n, s, 2t)$ is a set of side lengths of a triangle. Furthermore, by the Euclidean algorithm, it is not difficult to check these numbers are pairwise relatively prime to each other, hence $(n, 2s, t)$ and $(n, s, 2t)$ are the side lengths of a pair of pals, completing the solution.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22153,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute triangle such that $AB \\neq AC$, with circumcircle $\\Gamma$ and circumcenter $O$. Let $M$ be the midpoint of $BC$ and $D$ be a point on $\\Gamma$ such that $AD \\perp BC$. Let $T$ be a point such that $BDCT$ is a parallelogram and $Q$ a point on the same side of $BC$ as $A$, such that\n\n$$\n\\angle BQM = \\angle BCA \\quad \\text{and} \\quad \\angle CQM = \\angle CBA.\n$$\n\nLet the line $AO$ intersect $\\Gamma$ at $E$ ($E \\neq A$), and let the circumcircle of $\\triangle ETQ$ intersect $\\Gamma$ at point $X \\neq E$. Prove that the points $A$, $M$, and $X$ are collinear.",
"options": [],
"answer": "See solution",
"solution": "Let $X'$ be the symmetric point to $Q$ with respect to the line $BC$. Since $\\angle CBA = \\angle CQM = \\angle CX'M$ and $\\angle BCA = \\angle BQM = \\angle BX'M$, we have\n\n$$\n\\angle BX'C = \\angle BX'M + \\angle CX'M = \\angle CBA + \\angle BCA = 180^{\\circ} - \\angle BAC\n$$\n\nso $X' \\in \\Gamma$. Since $\\angle AX'B = \\angle ACB = \\angle MX'B$, it follows that $A$, $M$, $X'$ are collinear. Note that\n\n$$\n\\angle DCB = \\angle DAB = 90^{\\circ} - \\angle ABC = \\angle OAC = \\angle EAC\n$$\n\nso $DBCE$ is an isosceles trapezoid.\n\nSince $BDCT$ is a parallelogram, $MT = MD$, with $M$, $D$, $T$ collinear, $BD = CT$, and since $BDEC$ is an isosceles trapezoid, $BD = CE$ and $ME = MD$. Also,\n\n$$\n\\angle BTC = \\angle BDC = \\angle BED, \\quad CE = BD = CT\n$$\n\n\n\nand $ME = MT$, so $E$ and $T$ are symmetric with respect to the line $BC$. Since $Q$ and $X'$ are symmetric with respect to $BC$ as well, $QX'ET$ is an isosceles trapezoid, so $Q$, $X'$, $E$, $T$ are concyclic. Since $X' \\in \\Gamma$, this means $X \\equiv X'$, and therefore $A$, $M$, $X$ are collinear.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22154,
"subject": "Mathematics (Olympiad)",
"question": "令 $n$ 與 $k$ 為滿足 $k \\le 2n^2$ 的正整數。小李與小晴用一張 $2n \\times 2n$ 的方格紙玩一個遊戲。首先，小李在紙上的每一個方格中寫上一個至多為 1 的非負數，使得整張方格紙上的數字總和為 $k$。接著，小晴照著方格紙上的格線，將方格紙切成數片，使得每一片都是由若干個完整的方格所組成，且每一片上的數字總和至多為 1。小晴所切出的每一片的形狀沒有限制。\n\n令遊戲最後得到的片數為 $M$。小李的目標是極大化 $M$，而小晴的目標是極小化 $M$。\n\n試求當小李與小晴都以最佳策略進行下，遊戲結束時的 $M$ 值。",
"options": [],
"answer": "See solution",
"solution": "答案是 $2k-1$。\n\n讓我們依序考慮兩人的策略：\n\n- 對於小李，他可以在 $2k-1$ 個格子裡寫 $\\frac{1}{2}+\\epsilon$，一個格子裡寫 $\\frac{1}{2}-(2k-1)\\epsilon$，其餘填 0。基於每一片至多只能有一個寫 $\\frac{1}{2}+\\epsilon$ 的格子，小李可以用此策略保證 $M \\ge 2k-1$。\n\n- 對於小晴，他先選擇一個一筆劃通過所有方格的路徑，假設此路徑上的數字依序為 $a_1, a_2, \\dots, a_{4n^2}$。小晴的策略是，先從 $a_1$ 開始，找到最大的正整數 $M$ 使得 $\\sum_{i=1}^M a_i \\le 1$，並將 $a_1$ 到 $a_M$ 切成一片。接著，從 $a_{M+1}$ 開始，重複相同的動作，直到結束。\n\n讓我們證明以上策略最多只會切出 $2k-1$ 片。假設最後切出了 $t \\ge 2k$ 片，令每一片的數字和為 $b_1, b_2, \\dots, b_t$。基於策略，我們知道 $b_i + b_{i+1} > 1$（否則它們會被分在同一片）。但如此一來，\n\n$$\nk = \\sum_{i=1}^{t} b_i \\ge \\sum_{i=1}^{k} (b_{2i-1} + b_{2i}) > k,\n$$\n\n矛盾！故小晴可以保證 $M \\le 2k-1$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22155,
"subject": "Mathematics (Olympiad)",
"question": "As shown in the figure, in the right-angle triangle $ABC$, $D$ is the midpoint of the hypotenuse $AB$, $MB \\perp AB$, $MD$ intersects $AC$ at $N$, and $MC$ is extended to intersect $AB$ at $E$. Prove: $\\angle DBN = \\angle BCE$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Extend $ME$ to intersect the circumscribed circle of $\\triangle ABC$ at $F$, and $MD$ to intersect $AF$ at $K$. Draw $CG \\parallel MK$, which meet $AF$, $AB$ at $G$ and $P$ respectively. On the other hand, draw $DH \\perp CF$ with $H$ as the foot of the perpendicular, then $H$ is the midpoint of $CF$. Connect $HB$, $HP$, then $D$, $H$, $B$, $M$ are concyclic, that is,\n\n$$\n\\angle HBD = \\angle HMD = \\angle HCP.\n$$\n\nHence, $H$, $B$, $C$, $P$ are also concyclic, which means\n\n$$\n\\angle PHC = \\angle ABC = \\angle AFC\n$$\n\nand $PH \\parallel AF$. Therefore, $PH$ is the midline of $\\triangle CFG$ and $P$ is the midpoint of $CG$, then $AP$ is the median to $CG$ in $\\triangle ACG$.\n\nMoreover, from $NK \\parallel CG$, we get that $D$ is the midpoint of $NK$. In other words, $AB$ and $NK$ bisect each other. Therefore, $\\angle DBN = \\angle DAK$, and\n\n$$\n\\angle DAK = \\angle BAF = \\angle BCF = \\angle BCE,\n$$\n\nthen we prove $\\angle DBN = \\angle BCE$ as desired.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22156,
"subject": "Mathematics (Olympiad)",
"question": "A set of points in the plane is called *good* if the distance between any two points in it is at most $1$. Let $f(n, d)$ be the largest positive integer such that in any good set of $3n$ points, there is a circle of diameter $d$ which contains at least $f(n, d)$ points. Prove that there exists a positive real $\\epsilon$ such that for all $d \\in (1 - \\epsilon, 1)$, the value of $f(n, d)$ does not depend on $d$, and find that value as a function of $n$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Since $f(n, d)$ is an increasing function of $d$ and cannot be larger than $3n$, it becomes constant as $d$ approaches $1$. Fix some $d < 1$, possibly very close to $1$. Place $3n$ points in an equilateral triangle of side length $1$, with $n$ points near each vertex, each within distance less than $(1 - d)/2$ from its vertex. This set is good. A disk of diameter $d$ cannot cover points near two different vertices, so the maximum number of points such a disk can cover is $n$. Thus, $f(n, d) \\leq n$ for all $d$ close to $1$.\n\nWe now show $f(n, d) = n$ for $d$ close to $1$. Any good set of $3n$ points lies in a disk of radius $1$, but not necessarily in a smaller disk. However, any good set can be placed inside a disk of radius $1/\\sqrt{3}$ (the circumcircle of an equilateral triangle of side $1$).\n\n**Lemma.** Any good set can be placed inside a disk of radius $1/\\sqrt{3}$.\n\n*Proof.* Let $X$ be a good set. Consider its convex hull and the minimal enclosing circle. Tighten the circle until it touches $1$ or $2$ points; if $2$ points are on a diameter, the diameter is at most $1$. Otherwise, continue until the circle touches $3$ points, say $A, B, C$. If $\\triangle ABC$ is acute, the circle is its circumcircle, which has radius at most $1/\\sqrt{3}$. If not, further tightening is possible until either a new point is touched or a diameter is formed.\n\nTherefore, for all $d$ sufficiently close to $1$, $f(n, d) = n$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22157,
"subject": "Mathematics (Olympiad)",
"question": "Let $EF$ be the crease such that $E$ is on $AB$ and $F$ is on $BC$. Join $H$ to $C$, $F$, and $E$.\n\n\n\nThe crease $EF$ lies along the perpendicular bisector of $HB$ and the triangles $\\triangle EHF$ and $\\triangle EBF$ are congruent. If we let $x = |AE|$ and $y = |CF|$, then $|EH| = |EB| = 4 - x$ and $|HF| = |BF| = 4 - y$.\n\nFind the length $|EF|$ in terms of the given information.",
"options": [],
"answer": "See solution",
"solution": "In triangle $EHA$ we have $|AH| = 2$ and $\\angle HAE = 120^\\circ$, thus the cosine rule reads\n\n$$\n(4 - x)^2 = 2^2 + x^2 - 4x \\cos(120^\\circ) = x^2 + 2x + 4\n$$\n\nand so $10x = 12$, i.e. $x = \\frac{6}{5}$, thus $|EB| = 4 - x = \\frac{14}{5}$.\n\nIn triangle $HDC$ we have $|HD| = 2$, $|DC| = 4$ and $\\angle HDC = 60^\\circ$, and the cosine rule gives\n\n$$\n|HC|^2 = 4 + 16 - 16 \\cos(60^\\circ) = 12.\n$$\n\nIn particular, $|HC|^2 + |HD|^2 = |CD|^2$ and so $\\angle FCH = \\angle CHD = 90^\\circ$. Therefore $|HF|^2 = |HC|^2 + |CF|^2$, i.e. $(4 - y)^2 = 12 + y^2$ and so $4 = 8y$, that is $y = \\frac{1}{2}$. This implies $|BF| = 4 - y = \\frac{7}{2}$.\n\nThe cosine rule for triangle $BEF$ now gives\n\n$$\n\\begin{aligned}\n|EF|^2 &= |BF|^2 + |EB|^2 - 2|BF| \\cdot |EB| \\cos(60^\\circ) \\\\\n&= \\left(\\frac{7}{2}\\right)^2 + \\left(\\frac{14}{5}\\right)^2 - 2 \\cdot \\frac{7}{2} \\cdot \\frac{14}{5} \\cdot \\frac{1}{2} = \\frac{21 \\cdot 49}{100}, \\quad \\text{and so} \\\\\n|EF| &= \\frac{7}{10}\\sqrt{21}\n\\end{aligned}\n$$\n\nas required.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22158,
"subject": "Mathematics (Olympiad)",
"question": "Find the smallest positive integer $n$ such that $10 + n$ and $10n$ are both perfect squares.",
"options": [],
"answer": "See solution",
"solution": "Let $n$ be a positive integer such that $n + 10$ and $10n$ are both perfect squares.\n\nThen, there exists a positive integer $k$ such that $10n = k^2$. Since $k$ must be divisible by both $2$ and $5$, we can write $k = 10\\ell$ for some positive integer $\\ell$, so $n = 10\\ell^2$.\n\nFor $\\ell = 1$, $n + 10 = 10 + 10 = 20$ (not a perfect square). For $\\ell = 2$, $n + 10 = 40 + 10 = 50$ (not a perfect square). For $\\ell = 3$, $n = 10 \\times 9 = 90$, and $n + 10 = 100 = 10^2$ (a perfect square), and $10n = 900 = 30^2$ (a perfect square).\n\nTherefore, the smallest such $n$ is $\\boxed{90}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22159,
"subject": "Mathematics (Olympiad)",
"question": "Let $p > 3$ be a prime. Find the number of ordered sextuples $(a, b, c, d, e, f)$ of positive integers, whose sum is $3p$, and all the fractions\n\n$$\n\\frac{a+b}{c+d}, \\quad \\frac{b+c}{d+e}, \\quad \\frac{c+d}{e+f}, \\quad \\frac{d+e}{f+a}, \\quad \\frac{e+f}{a+b}\n$$\n\nare integers.",
"options": [],
"answer": "See solution",
"solution": "Taking the product of the 1st, 3rd, and 5th fractions reveals that their value must be 1, that is,\n\n$$\na + b = c + d = e + f = p. \\tag{1}\n$$\n\nThe form of the second and fourth fractions implies\n\n$$\nf + a \\mid d + e \\quad \\text{and} \\quad d + e \\mid b + c. \\tag{2}\n$$\n\nThat is, first $f + a$ is at most the arithmetic mean of its multiples,\n\n$$\nf + a \\le \\frac{1}{3}((f+a) + (d+e) + (b+c)) = p. \\tag{3}\n$$\n\nAnd\n\n$$\nf + a \\mid (f + a) + (d + e) + (b + c) = 3p.\n$$\n\nThus $f + a$ divides $3p$ and is in the interval $[2, p]$. Consequently, either $f + a = p$ or $f + a = 3$. We deal separately with these cases.\n\n**Case (i):** $f + a = p$.\n\nBecause of (3), we have $f + a = d + e = b + c = p$, which together with (1) gives $p - 1$ solutions of the form\n\n$$\n(a, b, c, d, e, f) = (a, p-a, a, p-a, a, p-a), \\quad \\text{where } a \\in \\{1, 2, \\dots, p-1\\}.\n$$\n\n**Case (ii):** $f + a = 3$. Then $\\{a, f\\} = \\{1, 2\\}$.\n\nFirst, let $a = 1$ and $f = 2$. According to (1), $b = p - 1$ and $e = p - 2$, and (2) has the form\n\n$$\n3 \\mid d + (p-2) \\quad \\text{and} \\quad d + (p-2) \\mid (p-1) + c. \\tag{4}\n$$\n\nIn analyzing (4), distinguish between $d = 1$ and $d \\ge 2$.\n\nIf $d = 1$, then $c = p - 1$ and (4) reads\n\n$$\n3 \\mid p-1 \\quad \\text{and} \\quad p-1 \\mid 2(p-1).\n$$\n\nThe right relation always holds, the left one holds only for $p = 3q + 1$ (for some positive integer $q$). For such primes, we get the solution\n\n$$\n(a, b, c, d, e, f) = (1, p-1, p-1, 1, p-2, 2).\n$$\n\nIf $d \\ge 2$, the right relation in (4) is satisfied if and only if $d + (p-2) = (p-1) + c$, or $d = c + 1$. Since $c + d = p$ and $d = c + 1$, we get $c = \\frac{1}{2}(p - 1)$ and $d = \\frac{1}{2}(p + 1)$. Since $d + (p - 2) = \\frac{3}{2}(p - 1)$, the left relation in (4) is fulfilled and\n\n$$\n(a, b, c, d, e, f) = (1, p-1, \\frac{1}{2}(p-1), \\frac{1}{2}(p+1), p-2, 2)\n$$\n\nis a solution.\n\nFinally, $a = 2$ and $f = 1$. In this case, $b = p - 2$ and $e = p - 1$, and (2) reads\n\n$$\n3 \\mid d + (p-1) \\quad \\text{and} \\quad d + (p-1) \\mid (p-2) + c. \\tag{5}\n$$\n\nBecause\n\n$$\nd + (p - 1) \\ge p \\quad \\text{and} \\quad (p - 2) + c < 2p,\n$$\n\nthe right relation in (5) holds if and only if $d + (p-1) = (p-2) + c$, that is, if $c = d+1$. Together with $c+d=p$, we get $c = \\frac{1}{2}(p+1)$ and $d = \\frac{1}{2}(p-1)$, so the right relation in (5) holds as well, and the last solution is\n\n$$\n(a, b, c, d, e, f) = (2, p-2, \\frac{1}{2}(p+1), \\frac{1}{2}(p-1), p-1, 1).\n$$\n\n**Conclusion:** All the solutions found are mutually different and their number depends on $p$ modulo $3$ ($p > 3$):\n- If $p = 3q + 1$, then there are $p + 2$ sextuples.\n- If $p = 3q + 2$, there are $p + 1$ sextuples.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22160,
"subject": "Mathematics (Olympiad)",
"question": "Let $h(0) = a$. Set $x = 0$ in the initial equation\n$$\nf(x^2 + y h(x)) = x h(x) + f(xy) \\quad \\text{for all } x, y \\in \\mathbb{R}.\n$$\nFind all functions $f: \\mathbb{R} \\to \\mathbb{R}$ and $h: \\mathbb{R} \\to \\mathbb{R}$ satisfying this equation.",
"options": [],
"answer": "See solution",
"solution": "Either\n$$\nf(x) = c, \\quad h(x) = \\begin{cases} 0, & x \\neq 0, \\\\ a, & x = 0, \\end{cases}\n$$\nwhere $a$ and $c$ are arbitrary constants,\nor\n$$\nf(x) = x + b, \\quad h(x) = x,\n$$\nwhere $b$ is an arbitrary constant.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22161,
"subject": "Mathematics (Olympiad)",
"question": "Prove that there exists an infinite set of points\n\n$\\ldots, P_{-3}, P_{-2}, P_{-1}, P_0, P_1, P_2, P_3, \\ldots$\n\nin the plane with the following property: For any three distinct integers $a, b,$ and $c$, points $P_a, P_b,$ and $P_c$ are collinear if and only if $a + b + c = 2014$.",
"options": [],
"answer": "See solution",
"solution": "We claim that defining $P_n$ to be the point with coordinates $(n, n^3 - 2014n^2)$ will satisfy the conditions of the problem. Recall that points $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ are collinear if and only if\n\n$$\n\\begin{vmatrix} x_1 & y_1 & 1 \\\\ x_2 & y_2 & 1 \\\\ x_3 & y_3 & 1 \\end{vmatrix} = 0.\n$$\n\nTherefore we examine the determinant\n\n$$\n\\begin{vmatrix} a & a^3 - 2014a^2 & 1 \\\\ b & b^3 - 2014b^2 & 1 \\\\ c & c^3 - 2014c^2 & 1 \\end{vmatrix} = \\begin{vmatrix} a & a^3 & 1 \\\\ b & b^3 & 1 \\\\ c & c^3 & 1 \\end{vmatrix} - 2014 \\begin{vmatrix} a & a^2 & 1 \\\\ b & b^2 & 1 \\\\ c & c^2 & 1 \\end{vmatrix}.\n$$\n\nThe first determinant on the right is a homogeneous polynomial of degree four divisible by $(a-b)(b-c)(c-a)$. The remaining factor has degree one, is symmetric, and yields an $ab^3$ term when the product is expanded, hence must be $(a+b+c)$. The second determinant is a homogeneous polynomial of degree three divisible by $(a-b)(b-c)(c-a)$, and comparing coefficients of the $ab^2$ term we see that this is the desired polynomial. Thus\n\n$$\n\\begin{vmatrix} a & a^3 - 2014a^2 & 1 \\\\ b & b^3 - 2014b^2 & 1 \\\\ c & c^3 - 2014c^2 & 1 \\end{vmatrix} = (a-b)(b-c)(c-a)(a+b+c-2014).\n$$\n\nIt follows that for distinct $a, b$ and $c$ this expression will equal zero if and only if $a + b + c = 2014$, as desired.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22162,
"subject": "Mathematics (Olympiad)",
"question": "Після 100 розрізань опуклого многокутника отримано 101 опуклий многокутник. Доведіть, що серед них є принаймні 101 фігуру, які є трикутниками або чотирикутниками.",
"options": [],
"answer": "See solution",
"solution": "Очевидно, що отримані під час розрізання шматки є опуклими многокутниками. Одним розрізанням загальна кількість вершин збільшується щонайбільше на 4. Тому після 100 розрізань многокутники матимуть не більше за 404 вершини.\n\nЗ іншого боку, після 100 розрізань утворився 101 многокутник. Нехай серед них $n$ трикутників та $m$ чотирикутників. Тоді утворені многокутники мають не менше за $3n + 4m + 5(101 - m - n) = 505 - 2n - m$ вершин. Отже,\n\n$$\n505 - 2n - m \\le 404\n$$\n\nзвідки\n\n$$\n2n + m \\ge 101 > 100.\n$$\n\nТобто, серед отриманих многокутників принаймні 101 є трикутниками або чотирикутниками.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22163,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle with circumcenter $U$ such that $\\angle CBA = 60^\\circ$ and $\\angle CBU = 45^\\circ$. Let $D$ be the point of intersection of the lines $BU$ and $AC$.\n\nProve that $AD = DU$.\n\n",
"options": [],
"answer": "See solution",
"solution": "In the isosceles triangle $AUB$, we have\n\n$$\n\\angle BAU = \\angle UBA = 60^\\circ - 45^\\circ = 15^\\circ,\n$$\n\nand therefore\n\n$$\n\\angle AUB = 180^\\circ - \\angle BAU - \\angle UBA = 150^\\circ.\n$$\n\nThe inscribed angle theorem implies\n\n$$\n\\angle BCA = \\frac{1}{2} \\angle BUA = 75^\\circ,\n$$\n\nand therefore\n\n$$\n\\angle BAC = 180^\\circ - 60^\\circ - 75^\\circ = 45^\\circ.\n$$\n\nWe can finally compute the two angles of interest:\n\n$$\n\\angle UAD = \\angle BAD - \\angle BAU = \\angle BAC - \\angle BAU = 45^\\circ - 15^\\circ = 30^\\circ\n$$\n\n$$\n\\angle DUA = 180^\\circ - \\angle AUB = 180^\\circ - 150^\\circ = 30^\\circ\n$$\n\nTherefore, the triangle $AUD$ is isosceles with apex $D$ and we have $AD = DU$ as desired.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22164,
"subject": "Mathematics (Olympiad)",
"question": "Euclid has a tool called _cyclos_ which allows him to do the following:\n\n- Given three non-collinear marked points, draw the circle passing through them.\n- Given two marked points, draw the circle with them as endpoints of a diameter.\n- Mark any intersection points of two drawn circles or mark a new point on a drawn circle.\n\nShow that given two marked points, Euclid can draw a circle centered at one of them and passing through the other, using only the cyclos.",
"options": [],
"answer": "See solution",
"solution": "We begin by proving a series of lemmas.\n\n**Lemma 1.** Given a non-right angled triangle $ABC$, we can draw the nine-point circle and mark the orthocentre $H$ using only a cyclos.\n\n**Proof.** Draw circles $(BC)$, $(CA)$, $(AB)$ and mark their intersections to get the three feet of altitudes $D$, $E$, $F$ opposite $A$, $B$, $C$. Now draw the circle $(DEF)$ to get the nine-point circle. Draw $(BDF)$, $(CDE)$, $(AEF)$ and they meet at $H$, which we can also mark.\n\n**Lemma 2.** Given points $A$, $B$, we can mark the midpoint $M$ of $AB$ using only a cyclos.\n\n**Proof.** Draw the circle $(AB)$ and choose a point $X$ on it. Draw circles $(XA)$, $(XB)$ and mark their intersection $Y$. Now mark a point $Z$ on the circle $(XA)$ apart from the marked points. Clearly, $Z$ does not lie on $AB$ nor on $(AB)$, hence we can draw $(AZB)$. Mark five points $Z_1, \\ldots, Z_5$ on this circle, each different from all previous points and verify if either $A$ lies on $(Z_iB)$ or $B$ lies on $(Z_iA)$ for each $1 \\leq i \\leq 5$ before marking the new point. By pigeonhole principle, for some three indices $i$, $j$, $k$, the three triangles $AZ_iB$ are non-right angled, hence we can draw their nine-point circles by Lemma 1. All of them pass through $M$, and their centres are not collinear, else homothety at the centre of $(ABZ)$ implies the orthocentres of the three triangles are collinear; but they all lie on the reflection of $(AZB)$ in $AB$, a contradiction! Thus, these three nine-point circles meet at only $M$, and we mark this point.\n\n**Lemma 3.** Given points $A$, $B$, $C$, $D$ on the plane in general position, we can mark the intersection point $E$ of lines $AB$ and $CD$ using only a cyclos.\n\n**Proof.** Draw $(AB)$ and mark five points on it, all different from previously marked points. For each marked point $X$, draw $(CX)$ and $(DX)$ and check whether they have an intersection apart from $X$ (i.e., if they are tangent, or if $X$ lies on $CD$). We can find three points $X_1$, $X_2$, $X_3$ among them not lying on $CD$. Denote by $Y_i$ the second intersection of $(AX_i)$, $(BX_i)$ and by $Z_i$ the second intersection of $(CX_i)$, $(DX_i)$ and mark them, for each $1 \\leq i \\leq 3$. Draw the circles $(X_iY_iZ_i)$ and note that they all pass through $E$ and have diameters $EX_i$ for all $i$; so they are not coaxial as $X_1$, $X_2$, $X_3$ are not collinear; all lying on $(AB)$. Thus we mark $E$ as the unique point common to them all. (Note: if $C$, $D$ lie on $(AB)$, we can pick a point $T$ on it other than these four, then a point $X$ on $(AT)$, and continue the same argument again, avoiding all edge cases.)\n\n**Lemma 4.** Given a circle $\\Gamma$, we can mark the centre of $\\Gamma$ using only a cyclos.\n\n**Proof.** Mark points $A, B, C \\in \\Gamma$ and mark the midpoints of $BC, CA, AB$ to get $A_1, B_1, C_1$ according to Lemma 2. Draw the circles $(AB_1C_1)$, $(BA_1C_1)$, $(CB_1A_1)$ and mark the intersection to get the centre of $\\Gamma$.\n\n**Lemma 5.** Given a circle $\\Gamma$ and point $A$ on $\\Gamma$, we can mark a point $K$ such that line $AK$ is tangent to $\\Gamma$ using only a cyclos.\n\n**Proof.** Mark points $B_1, B_2, B_3$ and $C$ on $\\Gamma$. By Lemma 4, mark the point $O$, the centre of $\\Gamma$. Draw $(B_iO)$ and $(AOC)$ and mark the intersection denoted $F_i$; there exists an index $j$ for which $F_j \\neq O$; mark the point $K$ which is the intersection of $B_jF_j$ and $OM$ where $M$ is the midpoint of $AC$ (which we mark by Lemma 2) by Lemma 3. Clearly, $K$ lies on $A$ tangent to $\\Gamma$. Note that we can do this again to get multiple such points $K$ by choosing different $C$ each time.\n\n**Lemma 6.** Given a circle $\\Gamma$ and a point $A$ on $\\Gamma$, and a point $B$ not on $\\Gamma$, we can mark the point $C$ which is the second intersection of line $AB$ and $\\Gamma$ using only a cyclos.\n\n**Proof.** Mark the foot of perpendicular $M$ from $O$ onto line $AB$ as done in Lemma 1. Mark the intersection of line $OM$ and $AK$ by Lemma 3, where $K$ is a point on the $A$-tangent to $\\Gamma$ as constructed in Lemma 5. Draw $(OAK)$ and mark the second intersection with $\\Gamma$ to obtain $C$.\n\n**Lemma 7.** Given points $A, B, C$ not all on a line, we can draw the reflection of $A$ in $BC$ using only a cyclos.\n\n**Proof.** Draw $(ABC)$ and mark the orthocentre $H$ of $ABC$ by Lemma 1. Mark the intersection $A'$ of line $AH$ with $(BHC)$ using Lemma 6, which is the $A$-reflection in line $BC$.\n\n**Lemma 8.** Given points $A, B$, we can mark the point $C$ which is the reflection of $A$ in $B$ using only a cyclos.\n\n**Proof.** Draw $(AB)$, and by Lemma 6, mark two points $K_1, K_2$ such that $BK_i$ is tangent to $(AB)$ for $i \\in \\{1, 2\\}$. By Lemma 7, mark the reflection of $A$ in line $K_1K_2$ as desired.\n\nThus, a cyclos can do everything a compass can: to draw a circle with given centre $A$ and given radius $B$, we use Lemma 8 to mark the reflection $C$ of $B$ in $A$ and use the cyclos to draw $(BC)$ which has centre $A$ and passes through $B$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22165,
"subject": "Mathematics (Olympiad)",
"question": "Consider a sequence $\\{a_n\\}_{n=1}^{\\infty}$ of positive integers satisfying, for each $n \\ge 3$,\n$$\na_n = a_1 a_2 + a_2 a_3 + \\dots + a_{n-2} a_{n-1} - 1.\n$$\n\na) Prove that some prime number is a divisor of infinitely many terms of this sequence.\n\nb) Prove that there are infinitely many such prime numbers.",
"options": [],
"answer": "See solution",
"solution": "Since all terms $a_i$ are positive integers, we have\n$$\na_5 = a_1 a_2 + a_2 a_3 + a_3 a_4 - 1 \\ge 1 + 1 + 1 - 1 = 2, \\text{ and therefore } a_5 \\ne 1.\n$$\nThe number $a_5$ is thus divisible by at least one prime number.\n\nFor every $n \\ge 4$, the following holds:\n$$\n\\begin{align*}\na_n &= (a_1 a_2 + a_2 a_3 + \\dots + a_{n-3} a_{n-2}) + a_{n-2} a_{n-1} - 1 \\\\\n &= (a_{n-1} + 1) + a_{n-2} a_{n-1} - 1 \\\\\n &= a_{n-1}(a_{n-2} + 1).\n\\end{align*}\n$$\nTherefore, $a_{n-1} \\mid a_n$. Let $p$ be any prime divisor of $a_5$. Then the relation $a_{n-1} \\mid a_n$ implies $p \\mid a_6$, from where $p \\mid a_7$, and so on. By mathematical induction, $p \\mid a_n$ for every $n \\ge 5$. Thus, the prime $p$ divides infinitely many terms of the given sequence. This completes the proof of part a).\n\nLet $\\mathcal{P}$ denote the set of all primes that divide infinitely many terms of the sequence. Suppose that the set $\\mathcal{P}$ is finite, i.e., $\\mathcal{P} = \\{p_1, \\dots, p_k\\}$ for some $k$. For every $i \\in \\{1, 2, \\dots, k\\}$, there exists a term $a_{n_i}$ divisible by $p_i$ with $n_i \\ge 5$. Due to the relation $a_{n-1} \\mid a_n$ (proved earlier for each $n \\ge 4$), we have $p_i \\mid a_n$ for all $n \\ge n_i$. If we denote $N = \\max(n_1, \\dots, n_k)$, then $a_N$ is divisible by all primes $p_1, \\dots, p_k$. Therefore, $a_N + 1 > 1$ is not divisible by any primes from $\\mathcal{P}$, so there must be a prime $q \\notin \\mathcal{P}$ such that $q \\mid a_N + 1$. This prime $q$ is then also a divisor of $a_{N+2} = a_{N+1}(a_N + 1)$, so $q \\mid a_n$ holds for each $n \\ge N+2$, and therefore $q \\in \\mathcal{P}$. Thus, we get a contradiction, which proves the statement in part b).\n\n**Remark.** From solving part a), we know that every prime number that divides one member of the sequence starting with the third one divides all the following members. So it is enough to prove that there are infinitely many primes that divide at least one member of a given sequence. This observation is a consequence of a stronger claim, namely that *for every $n \\ge 1$, the number $a_{2n+3}$ is divisible by at least $n$ different primes*. Proof of this statement will not be given here.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22166,
"subject": "Mathematics (Olympiad)",
"question": "A $5\\times5$ 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 $2\\times2$ 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": "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.* Indeed, 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 $2\\times2$ 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\n$$\n\\begin{pmatrix} \\bullet & x & y & z & \\bullet \\\\ \\bullet & z & t & x & \\bullet \\\\ \\bullet & x & y & z & \\bullet \\\\ \\bullet & z & t & x & \\bullet \\\\ \\bullet & x & y & z & \\bullet \\end{pmatrix}\n$$\n\nCompleting all the array, it easily follows that each column contains exactly two of the numbers and our claim is proven. \\hfill (1)\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 $4\\times4$ array, that can be divided into four $2\\times2$ 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 $C_1$. \\hfill (2)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22167,
"subject": "Mathematics (Olympiad)",
"question": "Prove that there exists a positive number $C$ such that the following statement holds: for any infinite arithmetic progression $a_1, a_2, a_3, \\dots$ of positive integers, if the greatest common divisor of $a_1$ and $a_2$ is square-free, then there exists some positive integer $m \\le C \\cdot a_2^2$ such that $a_m$ is square-free.",
"options": [],
"answer": "See solution",
"solution": "*Proof.* We prove that $C = 8$ satisfies the requirement.\n\n(1) First, consider the case where $a_1$ and $a_2$ are coprime. Let $d = a_2 - a_1$ be the common difference. For any prime $p$, if $p \\mid d$, then $p \\nmid a_1$ and hence $p$ does not divide any $a_n$. If $p \\nmid d$, then any consecutive $p^2$ terms of the sequence $a_n$ form a complete set of residues modulo $p^2$, among which exactly one term is divisible by $p^2$.\n\nLet $N = 4a_2$. We prove the existence of $1 \\le n \\le N$ such that $a_n$ has no square factors. If $1 \\le m \\le N$ and $a_m$ has a square factor, then there exists a prime $p$ such that $p^2 \\mid a_m$. Thus, $p \\nmid d$ and $p \\le \\sqrt{a_N}$. Moreover, the number of terms in $a_1, a_2, \\dots, a_N$ that are divisible by $p^2$ is at most $\\lceil \\frac{N}{p^2} \\rceil$. Therefore, the number $M$ of terms in $a_1, a_2, \\dots, a_N$ that have square factors satisfies:\n\n$$\nM \\le \\sum_{p \\le \\sqrt{a_N}} \\left\\lceil \\frac{N}{p^2} \\right\\rceil < \\sum_{p \\le \\sqrt{a_N}} \\left( \\frac{N}{p^2} + 1 \\right) \\le N \\sum_{p \\le \\sqrt{a_N}} \\frac{1}{p^2} + \\sqrt{a_N} < \\frac{N}{2} + \\sqrt{a_N}.\n$$\n\nThe last inequality is due to\n\n$$\n\\sum_{p} \\frac{1}{p^2} < \\frac{1}{4} + \\sum_{k=2}^{\\infty} \\frac{1}{(2k-1)^2} < \\frac{1}{4} + \\sum_{k=2}^{\\infty} \\frac{1}{(2k-2)2k} = \\frac{1}{2}.\n$$\n\nWhen $N = 4a_2$, we have $\\sqrt{a_N} < \\sqrt{Na_2} = 2a_2 = \\frac{N}{2}$. Thus, $M < N$. Therefore, there exist numbers in $a_1, a_2, \\dots, a_N$ that have no square factors.\n\n(2) Assume $\\gcd(a_1, a_2) = \\gcd(a_1, d) = q = q_1 \\cdots q_l > 1$, where $q_1, \\cdots, q_l$ are pairwise distinct prime factors. Note that every $a_i$ is divisible by $q$. For each prime factor $q_i$, there exists an index $t_i$ such that $v_{q_i}(a_{t_i}) = 1$. In fact, we can take $t_i \\in 1, 2$ since $a_1$ and $a_2$ cannot both be divisible by $q_i^2$. By Chinese Remainder Theorem, there exists $k \\in 1, 2, \\cdots, q$ such that $k \\equiv t_i \\pmod{q_i}$ for $1 \\le i \\le l$. Thus, for $1 \\le i \\le l$, we have $a_k \\equiv a_{t_i} \\pmod{q_id}$, which implies $v_{q_i}(a_k) = 1$.\n\nConsider the subsequence $a_k, a_{k+q}, a_{k+2q}, \\dots$, which has common difference $qd$ and is divisible by $q^2$. By construction, $q_i^2 \\nmid a_k$ for $1 \\le i \\le l$, i.e., each term in this subsequence has no square factors that are equal to $q_1, q_2, \\dots, q_l$. Let $b_i = \\frac{a_{k+(i-1)q}}{q}$, then $b_1 < b_2 < \\dots$ is an arithmetic sequence with common difference $d$, and $b_1$ is coprime to $d$. By the conclusion of (1), there exists $i \\le 4b_2$ such that $b_i$ has no square factor, and since $q$ is coprime to $b_i$, $a_{k+(i-1)q} = qb_i$ also has no square factor. Finally,\n\n$$\nk + (i - 1)q \\le iq \\le 4b_2q = 4a_{k+q} < 4(k + q)a_2 \\le 8da_2 < 8a_2^2.\n$$\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22168,
"subject": "Mathematics (Olympiad)",
"question": "Each point of the plane is colored either red or blue. Show that there exists a triangle with side lengths $1$, $2$, and $\\sqrt{3}$ whose three vertices are all the same color.",
"options": [],
"answer": "See solution",
"solution": "Assume, for contradiction, that in every triangle with side lengths $1$, $2$, and $\\sqrt{3}$, there is at least one red vertex and one blue vertex. Consider an equilateral triangle $ABC$ with side length $2$.\n\n\n\nAmong $A$, $B$, $C$, at least two must share the same color; suppose $B$ and $C$ are red.\n\nLet $D$ and $E$ be the midpoints of $AB$ and $AC$, respectively, and let $D'$, $E'$ be their reflections over the line $BC$. The triangles $BDC$, $BEC$, $BD'C$, and $BE'C$ all have side lengths $1$, $2$, and $\\sqrt{3}$, so $D$, $E$, $D'$, and $E'$ must be blue. But then triangle $DD'E$ has side lengths $1$, $2$, and $\\sqrt{3}$, and all its vertices are blue—a contradiction.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22169,
"subject": "Mathematics (Olympiad)",
"question": "There are $n \\ge 3$ players, numbered $1, 2, \\ldots, n$, participating in a tennis tournament. The rules are as follows:\n\n- At the start, all players stand in a line in order from $1$ to $n$.\n- Players $1$ and $2$ play the first game.\n- The loser moves to the end of the queue and will be the last to play in the next game; the winner becomes the first to play in the next game.\n- This process continues. After $N$ matches, the tournament ends.\n\nLet $a_1, a_2, \\ldots, a_n$ be the number of matches won by players $1$ through $n$, respectively, so that $a_1 + a_2 + \\cdots + a_n = N$.\n\n**Question:**\nHow many matches did each player lose?",
"options": [],
"answer": "See solution",
"solution": "Represent the tournament as a table with $n$ rows (one per player) and $N$ columns (one per match). In each column, write $1$ for a win, $0$ for a loss, and a dash for not participating. Each column has exactly one $1$ and one $0$; the rest are dashes.\n\nAfter a player loses (a $0$), they move to the end of the queue and will not play for the next $n-2$ matches. Thus, after each $0$, there are $n-2$ dashes in that row before the next entry.\n\nLet $b_j$ be the number of losses for player $j$ ($j = 1, \\ldots, n$). For player $1$, after removing their $a_1$ wins from the $N$ matches, $N - a_1$ matches remain, which are either losses or non-participations. Each loss is followed by $n-2$ non-participations, so the number of losses is:\n\n$$\nb_1 = \\left\\lceil \\frac{N - a_1}{n-1} \\right\\rceil\n$$\n\nSimilarly, for player $2$:\n$$\nb_2 = \\left\\lceil \\frac{N - a_2}{n-1} \\right\\rceil\n$$\n\nFor player $j$ ($j \\ge 3$), they do not participate in the first $j-2$ matches, so:\n$$\nb_j = \\left\\lceil \\frac{N - a_j - (j-2)}{n-1} \\right\\rceil\n$$\nwhere $\\lceil x \\rceil$ denotes the smallest integer not less than $x$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22170,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, \\dots, a_n$ be positive real numbers whose product is $1$. Show that the sum\n\n$$\n\\frac{a_1}{1+a_1} + \\frac{a_2}{(1+a_1)(1+a_2)} + \\frac{a_3}{(1+a_1)(1+a_2)(1+a_3)} + \\dots + \\frac{a_n}{(1+a_1)(1+a_2)\\dots(1+a_n)}\n$$\n\nis greater than or equal to $\\frac{2^n-1}{2^n}$.",
"options": [],
"answer": "See solution",
"solution": "Note that for every positive integer $m$,\n\n$$\n\\begin{aligned}\n\\frac{a_m}{(1+a_1)(1+a_2)\\cdots(1+a_m)} &= \\frac{1+a_m}{(1+a_1)(1+a_2)\\cdots(1+a_m)} - \\frac{1}{(1+a_1)(1+a_2)\\cdots(1+a_m)} \\\\\n&= \\frac{1}{(1+a_1)\\cdots(1+a_{m-1})} - \\frac{1}{(1+a_1)\\cdots(1+a_m)}.\n\\end{aligned}\n$$\n\nTherefore, if we let $b_j = (1+a_1)(1+a_2)\\cdots(1+a_j)$, with $b_0 = 1$, then by telescoping sums,\n\n$$\n\\sum_{j=1}^{n} \\frac{a_j}{(1+a_1)\\cdots(1+a_j)} = \\sum_{j=1}^{n} \\left( \\frac{1}{b_{j-1}} - \\frac{1}{b_j} \\right) = 1 - \\frac{1}{b_n}.\n$$\n\nNote that $b_n = (1+a_1)(1+a_2)\\cdots(1+a_n) \\geq (2\\sqrt{a_1})(2\\sqrt{a_2})\\cdots(2\\sqrt{a_n}) = 2^n$, with equality if and only if all $a_i = 1$. Therefore,\n\n$$\n1 - \\frac{1}{b_n} \\geq 1 - \\frac{1}{2^n} = \\frac{2^n - 1}{2^n}.\n$$\n\nTo check that this minimum can be obtained, substitute all $a_i = 1$ to yield\n\n$$\n\\frac{1}{2} + \\frac{1}{2^2} + \\frac{1}{2^3} + \\dots + \\frac{1}{2^n} = \\frac{2^{n-1} + 2^{n-2} + \\dots + 1}{2^n} = \\frac{2^n - 1}{2^n},\n$$\n\nas desired.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22171,
"subject": "Mathematics (Olympiad)",
"question": "For which positive integers $n \\ge 3$ is it possible to mark $n$ points of a plane in such a way that, starting from one marked point and moving on each step to the marked point which is the second closest to the current point, one can walk through all the marked points and return to the initial one? For each point, the second closest marked point must be uniquely determined.",
"options": [],
"answer": "See solution",
"solution": "To construct such a configuration for any $n \\ge 4$, choose $\\varepsilon < \\frac{2\\pi}{n^3}$. Place the points $A_1, A_2, \\dots, A_{n-1}$ on a circle so that the angle between the radii to $A_i$ and $A_{i+1}$ for $i = 1, \\dots, n-2$ is $\\alpha_i = \\frac{2\\pi}{n-2} - (n-2-i)\\varepsilon$. The angle between the radii to $A_{n-1}$ and $A_1$ is $\\alpha_{n-1} = \\frac{(n-2)(n-3)}{2}\\varepsilon$. Place $A_n$ outside the circle, on the extension of the radius through $A_{n-1}$, at the same distance from $A_{n-1}$ as between $A_1$ and $A_2$. It can be verified that $A_1, \\dots, A_n$ satisfy the problem's condition.\n\nFor $n = 3$, suppose such a construction exists with a cyclic walk $A_1 \\to A_2 \\to A_3 \\to A_1$. Then $d(A_1, A_2) > d(A_1, A_3)$, $d(A_2, A_3) > d(A_2, A_1)$, and $d(A_3, A_1) > d(A_3, A_2)$, where $d(X, Y)$ is the distance between $X$ and $Y$. These three inequalities cannot all hold at once.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22172,
"subject": "Mathematics (Olympiad)",
"question": "We are given 60 arbitrary points in a unit disc. Prove that there exists a point on the boundary of the disc such that the sum of its distances from all 60 given points does not exceed 80.",
"options": [],
"answer": "See solution",
"solution": "Let us inscribe an equilateral triangle $PQR$ into the boundary of the unit circle. If we prove that any point $X$ of our unit disc satisfies\n\n$$\n|PX| + |QX| + |RX| \\le 4, \\quad (1)\n$$\n\nthen summing (1) for the given points $X = X_k$, $1 \\leq k \\leq 60$, we get\n\n$$\n\\sum_{k=1}^{60} |PX_k| + \\sum_{k=1}^{60} |QX_k| + \\sum_{k=1}^{60} |RX_k| \\leq 4 \\cdot 60 = 240.\n$$\n\nConsequently, one of the sums on the left-hand side does not exceed $240 \\div 3 = 80$, hence some of the points $P$, $Q$, or $R$ always has the required property.\n\nBy symmetry, it suffices to prove (1) if $X$ lies in the sector $PSQ$, where $S$ denotes the center of the disc. We are going to show that, in this case,\n\n$$\n|PX| + |QX| \\leq 2, \\quad (2)\n$$\n\nwhich together with $|RX| \\leq 2$ will lead to (1).\n\nLet $S'$ be the midpoint of the arc $PQ$ (opposite to the arc $PRQ$; see Fig. 1). Clearly, $PS'QS$ is a rhombus, so it is sufficient to prove (2) only for points $X$ in the region $PS'Q$ of the disc (bounded by segment $PQ$ and arc $PS'Q$).\n\nLet $\\alpha = |\\angle XPQ|$, $\\beta = |\\angle XQP|$ with $\\alpha + \\beta \\leq 60^\\circ$. Using the law of sines in triangle $PQX$:\n\n$$\n\\begin{aligned}\n|PX| + |QX| &= \\frac{|PQ|(\\sin \\alpha + \\sin \\beta)}{\\sin(\\alpha + \\beta)} = \\frac{\\sqrt{3} \\cdot 2 \\sin \\frac{\\alpha+\\beta}{2} \\cos \\frac{\\alpha-\\beta}{2}}{2 \\sin \\frac{\\alpha+\\beta}{2} \\cos \\frac{\\alpha+\\beta}{2}} \\\\\n&= \\frac{\\sqrt{3} \\cos \\frac{\\alpha-\\beta}{2}}{\\cos \\frac{\\alpha+\\beta}{2}} \\leq \\frac{\\sqrt{3} \\cdot 1}{\\sqrt{3}} = 2, \\quad \\text{as } \\frac{\\alpha+\\beta}{2} \\leq 30^\\circ.\n\\end{aligned}\n$$\n\nThus, (2) is proven.\n\n\n\n*Fig. 1*\n\n\n\n*Fig. 2*\n\n_Remarks._ The second part of the solution can also be shown as follows: For any point $X$ in the considered region, take point $Q'$ on the ray $PX$ outside segment $PX$ such that $|XQ'| = |XQ|$. The angle $QXQ'$ is at most $60^\\circ$, i.e., $|\\angle XQQ'| = |\\angle XQ'Q| \\geq 60^\\circ$, so $Q'$ lies in the unit disc, which is a reflection of the given disc across line $PQ$ (see Fig. 2). Thus,\n\n$$\n|PX| + |XQ| = |PX| + |XQ'| = |PQ'| \\leq 2,\n$$\n\nthe diameter of the unit disc.\n\nAlternatively, the region of the disc bounded by arcs $PSQ$ and $PS'Q$ is contained in the region bounded by the ellipse with foci $P$ and $Q$, which is the set of all points $X$ with $|PX| + |XQ| \\leq |PS'| + |QS'| = |PS| + |QS| = 2$.\n\nNotice that $|PX| + |XQ| = \\sqrt{3} < 2$ if $X$ lies on side $PQ$ and triangle $PQX$ does not exist.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22173,
"subject": "Mathematics (Olympiad)",
"question": "Given is a table with $n$ rows and 12 columns. Each cell in it contains a 0 or a 1. The table has the following properties:\n\n(a) Every two rows are different.\n\n(b) Every row contains exactly 4 entries equal to 1.\n\n(c) For every 3 rows there is a column that intersects them at three entries equal to 0.\n\nFind the greatest $n$ for which such a table exists.\n\n",
"options": [],
"answer": "See solution",
"solution": "The answer is $\\binom{11}{4} = 330$; here and later on $\\binom{n}{k}$ denotes a binomial coefficient.\n\nHere is an example with $n=330$. Form the $\\binom{11}{4} = 330$ (unordered) quadruples $i, j, k, l$ with elements from $\\{1, 2, \\ldots, 11\\}$. To every such quadruple assign a row of length 12 in which there are 1's exactly at positions $i, j, k, l$; the remaining entries are 0's. The obtained 330 rows can be arranged to form a $330 \\times 12$ table, which satisfies conditions (a)–(c) by construction. (For condition (c) note that column 12 contains only zeros.)\n\nLet us show that $n \\leq 330$ for each table $T$ with the given properties. For each row $F$ consider the 4 columns that intersect it at 1's. We say that they form an *admissible quadruple* $Q$. It is uniquely determined by $F$ in view of condition (b). In addition, different rows generate different admissible quadruples by condition (a). Hence there is a bijection between the rows and the admissible quadruples; in particular there are exactly $n$ admissible quadruples.\n\nLet $Q$ be an admissible quadruple. Consider all partitions of its complementary 8 columns into 2 quadruples $Q_1$ and $Q_2$. Since 4 columns out of 8 can be chosen in $\\binom{8}{4}$ ways, there are $\\frac{1}{2}\\binom{8}{4}$ such partitions. Clearly the quadruples $Q_1$ and $Q_2$ are different for different partitions. Observe also that in each partition at least one of $Q_1$ and $Q_2$ is non-admissible. Indeed, if $Q_1$, $Q_2$ are admissible then no column intersects their respective rows at 3 zeros, which contradicts condition (c). Hence the specified $\\frac{1}{2}\\binom{8}{4}$ partitions generate at least $\\frac{1}{2}\\binom{8}{4}$ non-admissible quadruples associated to the initial admissible quadruple $Q$. Because there are $n$ admissible quadruples, this argument yields a list of $l \\geq \\frac{1}{2}\\binom{8}{4} n$ non-admissible quadruples. We are about to see that the repetitions in it are not too numerous.\n\nEach non-admissible quadruple $Q'$ on the list occurs in it as many times as there are admissible quadruples $Q$ that generate $Q'$ in the way explained above. Every such $Q$ occupies 4 columns among the 8 complementary columns of $Q'$, which gives at most $\\binom{8}{4}$ possibilities for $Q$.\n\nConsequently every non-admissible quadruple on the list occurs at most $\\binom{8}{4}$ times in it.\n\nGiven the length $l \\geq \\frac{1}{2}\\binom{8}{4} n$ of the list and the maximum number $\\binom{8}{4}$ of repetitions of an item, we find at least $\\frac{n}{2}$ different non-admissible quadruples in $T$. The total number of quadruples of columns is $\\binom{12}{4} = 495$. Exactly $n$ of them are admissible and $495-n$ are non-admissible. Hence $495-n \\geq \\frac{n}{2}$, which yields the desired $n \\leq 330$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22174,
"subject": "Mathematics (Olympiad)",
"question": "Let $A$ be a set of real numbers satisfying the following:\n\n(a) $\\sqrt{n^2+1} \\in A$ for all positive integers $n$;\n\n(b) If $x \\in A$ and $y \\in A$, then $x - y \\in A$.\n\nProve that every integer can be written as a product of two different elements in $A$.",
"options": [],
"answer": "See solution",
"solution": "We first observe some natural consequences of property (b):\n\n1. $0 \\in A$, since if $x \\in A$ is arbitrary, then $0 = x - x \\in A$.\n2. If $x \\in A$, then $-x = 0 - x \\in A$.\n3. If $x, y \\in A$, then $x + y = x - (-y) \\in A$.\n4. If $x \\in A$, then $k x \\in A$ for any integer $k$. This follows by induction on $k$ (for positive $k$), using (3). For negative $k$, apply (2). For $k = 0$, use (1).\n\nSo now we know that $k\\sqrt{m^2+1} + l\\sqrt{n^2+1} \\in A$ for all integers $k$ and $l$, and all positive integers $m$ and $n$.\n\nLet us first try to write the integer $1$ as a product of two different elements in $A$:\n\nWe can find (by inspection) positive integers $k, l, m, n$ such that $1 = k^2(m^2+1) - l^2(n^2+1)$, for example, $k=3$, $m=1$, $l=1$, $n=4$. (There are many other possibilities as well.) Hence,\n\n$$\n\\begin{aligned}\n1 &= 3^2(1^2 + 1) - 1^2(4^2 + 1) \\\\\n &= (3\\sqrt{1^2 + 1} - 1\\sqrt{4^2 + 1})(3\\sqrt{1^2 + 1} + 1\\sqrt{4^2 + 1}),\n\\end{aligned}\n$$\n\na product of two different elements of $A$.\n\nIf $p$ is an arbitrary integer, then we have\n\n$$\n\\begin{aligned}\np = p \\cdot 1 &= p \\cdot (3\\sqrt{1^2+1} - 1\\sqrt{4^2+1})(3\\sqrt{1^2+1} + 1\\sqrt{4^2+1}) \\\\\n&= (3p\\sqrt{1^2+1} - p\\sqrt{4^2+1})(3\\sqrt{1^2+1} + 1\\sqrt{4^2+1}),\n\\end{aligned}\n$$\n\nagain a product of two elements of $A$. These two elements of $A$ are indeed different, since if $3p\\sqrt{1^2+1} - p\\sqrt{4^2+1} = 3\\sqrt{1^2+1} + 1\\sqrt{4^2+1}$, then we would have $p = 35 + 6\\sqrt{34}$, which is not an integer.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22175,
"subject": "Mathematics (Olympiad)",
"question": "Suppose a $2m \\times (2n+1)$ board is colored like a chessboard. Define a *cycle* as a sequence of cells $c_1, \\dots, c_k$ such that each $c_i$ and $c_{i+1}$ (with $c_{k+1} = c_1$) share a side. There are two possible domino covers of a cycle: one joining each black square to the next, and one joining each black square to the previous. These covers are *inverse* to each other.\n\n*Lemma.* Given a covering $T$ of a $2m \\times (2n+1)$ board, each square in the $(n+1)^{\\text{st}}$ column is contained in a single symmetrical cycle with respect to it and covered by the pieces of $T$ in one of the two ways described above.\n\n% IMAGE: \n\nProve that the number of domino tilings of a $2m \\times (2n+1)$ board, denoted $N^*(2m, 2n+1)$, satisfies:\n\n$$\nN^*(2m, 2n+1) = 2^m N(2m, n)N(2m, n-1)\n$$\n\nwhere $N(2m, n)$ is the number of domino tilings of a $2m \\times n$ board.",
"options": [],
"answer": "See solution",
"solution": "To recover $T$ from its normal form $T^*$, we must know both $T^*$ and the subset of special cells of $T$. There are $m$ black cells in the middle column, giving $2^m$ possible sets of special cells. Thus, each normal form $T^*$ comes from $2^m$ possible coverings. In summary, we obtain:\n\n$$\nN^*(2m, 2n+1) = 2^m N(2m, n)N(2m, n-1)\n$$\n\nas required.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 22176,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $n$ with the following property: the $k$ positive divisors of $n$ have a permutation $(d_1, d_2, \\dots, d_k)$ such that for every $i = 1, 2, \\dots, k$ the number $d_1 + d_2 + \\dots + d_i$ is a perfect square.",
"options": [],
"answer": "See solution",
"solution": "For $i = 1, 2, \\dots, k$, let $d_1 + \\dots + d_i = s_i^2$ and define $s_0 = 0$ as well. Obviously $0 = s_0 < s_1 < \\dots < s_k$, so\n\n$$\ns_i \\le i \\quad \\text{and} \\quad d_i = s_i^2 - s_{i-1}^2 = (s_i + s_{i-1})(s_i - s_{i-1}) \\ge s_i + s_{i-1} \\ge 2i - 1.\n$$\n\nThe number 1 is one of the divisors $d_1, d_2, \\dots, d_k$ but, due to $d_i \\le 2i - 1$, the only possibility is $d_1 = 1$. Now consider $d_2$ and $s_2 \\ge 2$. By definition, $d_2 = s_2^2 - 1 = (s_2 - 1)(s_2 + 1)$, so the numbers $s_2 - 1$ and $s_2 + 1$ are divisors of $n$. In particular, there is some index $j$ such that $d_j = s_2 + 1$.\n\nNotice that\n\n$$\ns_2 + s_1 = s_2 + 1 = d_j \\le s_j + s_{j-1};\n$$\n\nsince the sequence $s_0 < s_1 < \\dots < s_k$ increases, the index $j$ cannot be greater than 2. Hence, the divisors $s_2 - 1$ and $s_2 + 1$ are listed among $d_1$ and $d_2$. That means $s_2 - 1 = d_1 = 1$ and $s_2 + 1 = d_2$; therefore $s_2 = 2$ and $d_2 = 3$.\n\nWe can repeat the above process in general.\n\n*Claim.* $d_i = 2i - 1$ and $s_i = i$ for $i = 1, 2, \\dots, k$.\n\n*Solution.* Apply induction on $i$. The claim has been proved for $i = 1, 2$. Suppose that we have already proved $d_1 = 1, d_2 = 3, \\dots, d_i = 2i - 1$, and consider the next divisor $d_{i+1}$:\n\n$$\nd_{i+1} = s_{i+1}^2 - s_i^2 = s_{i+1}^2 - i^2 = (s_{i+1} - i)(s_{i+1} + i).\n$$\n\nThe number $s_{i+1} + i$ is a divisor of $n$, so there is some index $j$ such that $d_j = s_{i+1} + i$. Similarly to above arguments, we have\n\n$$\ns_{i+1} + s_i = s_{i+1} + i = d_j \\le s_j + s_{j-1};\n$$\n\nsince the sequence $s_0 < s_1 < \\dots < s_k$ increases, it is easy to check that $j \\le i + 1$. On the other hand, $d_j = s_{i+1} + i > 2i > d_i > \\dots > d_1$, so $j \\le i$ is not possible. The only possibility is $j = i + 1$.\n\nHence,\n\n$$\n\\begin{aligned}\ns_{i+1} + i &= d_{i+1} = s_{i+1}^2 - s_i^2 = s_{i+1}^2 - i^2; \\\\\ns_{i+1}^2 - s_i &= i(i+1).\n\\end{aligned}\n$$\n\nBy solving this equation we get $s_{i+1} = i + 1$ and $d_{i+1} = 2i + 1$, that finishes the proof. $\\square$\n\nNow we know that the positive divisors of the number $n$ are $1, 3, 5, \\dots, n-2, n$. The greatest divisor is $d_k = 2k - 1 = n$ itself, so $n$ must be odd. The second greatest divisor is $d_{k-1} = n - 2$; then $n - 2 \\mid n$ therefore $n$ must be 1 or 3. Obviously, these values satisfy the given conditions. $\\square$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 22177,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a cyclic quadrilateral with perpendicular diagonals and circumcenter $O$. Let $g$ be the line obtained by reflecting the diagonal $AC$ about the angle bisector of $\\angle BAD$. Prove that the point $O$ lies on the line $g$.",
"options": [],
"answer": "See solution",
"solution": "Denote by $X$ the point of intersection of the diagonals $AC$ and $BD$, i.e., $AX$ is an altitude in triangle $ABD$. By the inscribed angle theorem, we have $\\angle ABX = \\frac{1}{2}\\angle DOA$. Hence\n$$\n\\angle XAB = 90^\\circ - \\angle ABX = \\frac{1}{2} (180^\\circ - \\angle DOA) = \\angle OAD.\n$$\nIn the last step, the angle sum in the equilateral triangle $DAU$ has been used. Since the lines $AB$ and $AD$ are symmetric with respect to the angle bisector $w_\\alpha$, the same is true for $AX$ and $AU$. Hence, the assertion follows.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22178,
"subject": "Mathematics (Olympiad)",
"question": "Let the cubes in picture 1 be numbered as in picture 2 below (from top to bottom by layers). In one move, it is allowed for an empty cube to exchange place with a neighbouring cube (two cubes are neighbouring if they share a common side). Is it possible, after finitely many moves, to get an arrangement of the numbers as in picture 3?\n\n\n\nPicture 1\n\n\n\nPicture 2\n\n\n\nPicture 3",
"options": [],
"answer": "See solution",
"solution": "The cubes numbered $1, 7, 11, 15, 19,$ and $22$ cannot be moved, because in order to move them the empty cube must be in the corner, and then when returning the empty cube, this cube moves back to its place. Therefore, we consider the following problem:\n\nWe focus on the movable cubes and their arrangements. Suppose after finitely many moves we can achieve the sought-after arrangement. Then the same can be done if cubes are added to make a big cube with side of length $5$ little cubes. A new beginning and final configuration are achieved with the additional numbering of the newly-added cubes.\n\nWe consider a permutation for every move we've made so far. We number the cubes with consecutive integers, and we skip the empty cube in the moves when it's not at the end. In the beginning, the permutation is identical. In every step, the parity of the permutation does not change because we always move a cube a few steps backward or forward, or the permutation remains the same.\n\nOn the other hand, the beginning and final permutation are of different parity, therefore we cannot achieve the final configuration.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 22179,
"subject": "Mathematics (Olympiad)",
"question": "Solve, in the set of natural numbers, the equation\n\n$$\nx^2 + y^2 + xy(x - y) = 17.\n$$",
"options": [],
"answer": "See solution",
"solution": "The equation can be rewritten as $x^2 + y^2 + xy(x - y) = 17$.\n\nLet $d = x - y$ and $p = xy$. Then:\n\n$$\nx^2 + y^2 = (x - y)^2 + 2xy = d^2 + 2p\n$$\n\nSo the equation becomes:\n\n$$\nd^2 + 2p + p d = 17\n$$\n\nor\n\n$$\np(d + 2) = 17 - d^2\n$$\n\nThus,\n\n$$\np = \\frac{17 - d^2}{d + 2}\n$$\n\nFor $p$ to be a natural number, $d + 2$ must divide $17 - d^2$.\n\nChecking possible integer values:\n\n- If $d + 2 = 1$, $d = -1$, $p = 16$. No natural numbers $x, y$ with $xy = 16$ and $x - y = -1$.\n- If $d + 2 = -1$, $d = -3$, $p = -8 < 0$ (not natural).\n- If $d + 2 = 13$, $d = 11$, $p = -8 < 0$ (not natural).\n- If $d + 2 = -13$, $d = -15$, $p = 16$. $x - y = -15$, $xy = 16$.\n\nSolving $x - y = -15$, $xy = 16$:\n\nLet $x = y - 15$, so $(y - 15)y = 16$.\n\n$y^2 - 15y - 16 = 0$\n\nSolving the quadratic:\n\n$$\ny = \\frac{15 \\pm \\sqrt{225 + 64}}{2} = \\frac{15 \\pm 17}{2}\n$$\n\nSo $y = 16$ or $y = -1$. Only $y = 16$ is natural, so $x = 1$.\n\n**Final answer:**\n\nThe unique solution in natural numbers is $x = 1$, $y = 16$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22180,
"subject": "Mathematics (Olympiad)",
"question": "Let the excircle of triangle $ABC$ opposite to vertex $A$ be tangent to side $BC$ at $A_1$. Define the points $B_1$ and $C_1$ analogously, using the excircles opposite $B$ and $C$, respectively. Suppose that the circumcenter of triangle $A_1B_1C_1$ lies on the circumcircle of triangle $ABC$. Prove that triangle $ABC$ is right-angled.\n\nThe excircle of triangle $ABC$ opposite vertex $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$. The excircles opposite $B$ and $C$ are similarly defined.",
"options": [],
"answer": "See solution",
"solution": "Let $\\omega$ be the circumcircle of $ABC$, and let $O_1$ be the circumcenter of $A_1B_1C_1$. Because $A_1, B_1$, and $C_1$ are on the boundary of $ABC$ and $O_1$ is outside of $ABC$, $A_1B_1C_1$ is obtuse. Without loss of generality, assume that $\\angle B_1A_1C_1$ is obtuse so that $O_1$ and $A$ lie on the same side of line $B_1C_1$.\n\n**Lemma 1.** The second intersection $A_0$ of $\\omega$ and the circumcircle of triangle $AB_1C_1$ is the midpoint of arc $\\widehat{BAC}$.\n\n*Proof.* By the definition of $A_0$, we have $\\angle A_0BC_1 = \\angle A_0BA = \\angle A_0CA = \\angle A_0CB_1$ and $\\angle A_0C_1A = \\angle A_0B_1A$, hence $AC_1B$ and $AB_1C$ are similar. But $BC_1 = CB_1$, so these two triangles are congruent, hence $A_0B = A_0C$. Because $AA_0B_1C_1$ is cyclic, we have $\\angle C_1A_0B_1 = \\angle C_1AB_1 = \\angle BAC$, so $A_0$ lies on $\\widehat{BAC}$ with $BA_0 = CA_0$, implying that $A_0$ is the midpoint of $\\widehat{BAC}$. $\\square$\n\nBy Lemma 1, a spiral similarity centered at $A_0$ sends $B_1C_1$ to $CB$, so $A_0$ is the intersection of $\\omega$ and the perpendicular bisector of $B_1C_1$ which is on the same side of $BC$ as $A$. Recalling that $A_0$ is the circumcenter of $A_1B_1C_1$ and using this result for the analogous points $B_0$ and $C_0$, we obtain that $A_0C_1B_0A_1$ and $A_0A_1C_0B_1$ are kites with symmetry axes $A_0B_0$ and $A_0C_0$. Recalling that $C_1B_1AA_0$ is cyclic, we have $\\angle CAB = \\angle C_1A_0B_1 = 2\\angle B_0A_0C_0 = \\widehat{B_0C_0}$. By Lemma 1, $B_0$ and $C_0$ are the midpoints of $\\widehat{ABC}$ and $\\widehat{BCA}$, hence\n\n$$\n\\begin{aligned}\n\\angle CAB &= \\widehat{B_0C_0} = 360^\\circ - \\widehat{ACC_0} - \\widehat{B_0A} = 360^\\circ - \\frac{\\widehat{BCA} + \\widehat{ABC}}{2} \\\\\n&= 360^\\circ - \\frac{360^\\circ - 2\\angle BCA + 360^\\circ - 2\\angle ABC}{2} = \\angle BCA + \\angle ABC,\n\\end{aligned}\n$$\n\nimplying that $\\angle CAB = 90^\\circ$, so $ABC$ has right angle at vertex $A$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 22181,
"subject": "Mathematics (Olympiad)",
"question": "Let $N$ denote the number of ordered triples of positive integers $(a, b, c)$ such that $a, b, c \\leq 3^6$ and $a^3 + b^3 + c^3$ is a multiple of $3^7$. Find the remainder when $N$ is divided by 1000.",
"options": [],
"answer": "See solution",
"solution": "By the Lifting the Exponent Lemma, if integers $a$ and $b$ satisfy $a \\equiv b \\pmod{3}$ and $a$ is not a multiple of 3, then the power of 3 dividing $a^3 - b^3$ is exactly 1 greater than the power of 3 dividing $a - b$.\n\nFor a positive integer $n \\geq 2$, let $S_n$ be the set of $2 \\cdot 3^{n-2}$ positive integers $a$ such that $1 \\leq a \\leq 3^{n-1}$ and $\\gcd(a, 3) = 1$. From the above, the following $3^{n-2}$ cubes of members of $S_n$ of the form $3k + 1$ give different remainders when divided by $3^n$:\n\n$$\n1^3, 4^3, 7^3, \\dots, (3^{n-1} - 2)^3.\n$$\n\nBecause there are $3^{n-2}$ such cubes and all these cubes are congruent to 1 modulo 9, it follows that every remainder modulo $3^n$ of the form $9k + 1$ occurs exactly once on the above list. Similarly, every remainder modulo $3^n$ of the form $9k - 1$ occurs exactly once in the list\n\n$$\n2^3, 5^3, 8^3, \\dots, (3^{n-1} - 1)^3.\n$$\n\nFurthermore, for any integer $a$, $(a + 3^{n-1})^3 \\equiv a^3 \\pmod{3^n}$ by expansion.\n\nBecause any cube is congruent to $-1, 0, \\text{ or } 1 \\pmod{9}$, it follows that if $3^7 \\mid a^3 + b^3 + c^3$, at least one of $a, b$, or $c$ is a multiple of 3.\n\n* If exactly one of $a, b$, or $c$ is a multiple of 3, then one of $a, b$, or $c$ is a multiple of 3, one is of the form $3k + 1$, and one is of the form $3k - 1$. There are 6 ways to choose which of $a, b$, or $c$ gives which remainder when divided by 3. There are $3^5$ ways to choose $a \\equiv 0 \\pmod{3}$ and also $3^5$ ways to choose $b \\equiv 1 \\pmod{3}$. For each choice, $a^3 + b^3 \\equiv 1 \\pmod{9}$ and thus there is a unique choice of $c$ such that $3^7 \\mid a^3 + b^3 + c^3$. Hence there are $6 \\cdot 3^5 \\cdot 3^5 = 2 \\cdot 3^{11}$ ways to choose $a, b$, and $c$ in this case.\n\n* If $a, b$, and $c$ are all divisible by 3 and exactly one of $a, b$, or $c$ is a multiple of $3^2$, then write $a = 3a_1$, $b = 3b_1$, and $c = 3c_1$, with $3^4 \\mid a_1^3 + b_1^3 + c_1^3$ and $1 \\leq a_1, b_1, c_1 \\leq 3^5$. There are again 6 ways to choose which of $a_1, b_1$, or $c_1$ gives which remainder when divided by 3. There are $3^4$ ways to choose $a_1 \\equiv 0 \\pmod{3}$ and also $3^4$ ways to choose $b_1 \\equiv 1 \\pmod{3}$. For each choice, $a_1^3 + b_1^3 \\equiv 1 \\pmod{9}$ and there is therefore a unique choice of $d$ with $1 \\leq d \\leq 3^3$ such that $3^4 \\mid a_1^3 + b_1^3 + d^3$. Because $(d + m \\cdot 3^3)^3 \\equiv d^3 \\pmod{3^4}$ for all integers $m$, it follows that there are $\\frac{3^5}{3^3} = 3^2$ choices for $c_1$ between 1 and $3^5$ such that $3^4 \\mid a_1^3 + b_1^3 + c_1^3$. Hence there are $6 \\cdot 3^4 \\cdot 3^4 \\cdot 3^2 = 2 \\cdot 3^{11}$ ways to choose $a, b$, and $c$ in this case.\n\n* If all three of $a, b$, and $c$ are divisible by $3^2$, then write $a = 9a_2$, $b = 9b_2$, and $c = 9c_2$ with $3 \\mid a_2^3 + b_2^3 + c_2^3$ and $1 \\leq a_2, b_2, c_2 \\leq 3^4$. For any choice of $a_2$ and $b_2$, there is a unique remainder $r$ modulo 3 such that $3 \\mid a_2^3 + b_2^3 + r^3$ and thus there are $3^3$ choices for $c_2$ such that $3 \\mid a_2^3 + b_2^3 + c_2^3$. Hence there are $3^4 \\cdot 3^4 \\cdot 3^3 = 3^{11}$ ways to choose $a, b$, and $c$ in this case.\n\nThus there are a total of $2 \\cdot 3^{11} + 2 \\cdot 3^{11} + 3^{11} = 5 \\cdot 3^{11}$ such triples. Substituting in\n\n$$\n3^{10} = (10 - 1)^5 = -\\binom{5}{2}10^2 + \\binom{5}{1}10 - 1 \\equiv 49 \\pmod{1000}\n$$\n\ngives $5 \\cdot 3^{11} \\equiv 5 \\cdot 3 \\cdot 49 \\equiv 735 \\pmod{1000}$. Hence the requested remainder is 735.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22182,
"subject": "Mathematics (Olympiad)",
"question": "Given a triangle $ABC$, the internal bisector of $\\hat{A}$ intersects $BC$ at $D$ and the external bisector of $\\hat{A}$ intersects $BC$ at $E$. Let $C_1$ be the circumcircle of $ADE$. $AC$ intersects $C_1$ at $F$. Let $C_2$ be the circumcircle of $ABF$, and the tangent to $C_2$ at $A$ intersects $C_1$ at $G$. Prove that $AF = AG$.",
"options": [],
"answer": "See solution",
"solution": "Let $\\widehat{ABC} = 2\\beta$, $\\widehat{BAC} = 2\\alpha$, $\\widehat{ACB} = 2\\gamma$. Thus, the law of sines on triangle $AEF$ gives us\n\n$$\n\\frac{AE}{AF} = \\frac{\\sin(\\alpha + 2\\beta)}{\\sin(\\alpha - \\beta + \\gamma)}.\n$$\n\n\n\nOn the other hand, using the law of sines on $ABD$, $ADF$ and the fact that $\\widehat{BAD} = 180^\\circ - \\widehat{DAF}$, we get that\n\n$$\n\\frac{AB}{AF} = \\frac{BD \\sin(\\widehat{BDA})}{DF \\sin(\\widehat{FDA})} = \\frac{BD \\sin(\\alpha + 2\\beta)}{DF \\sin(\\alpha - \\beta + \\gamma)}.\n$$\n\nCombining these two identities, we obtain that $\\frac{AE}{AB} = \\frac{DF}{DB}$. Then since we also have $\\widehat{FDB} = \\widehat{FAE} = \\widehat{EAB}$, triangle $EAB$ is similar to triangle $FDB$. Thus we find that $\\widehat{FBE} = 2\\beta$. Using this and the fact that $A, D, F, E, G$ are on the same circle, we are able to compute that $\\widehat{GEA} = \\alpha - \\beta + \\gamma$. But this was also computed to be equal to $\\widehat{AEF}$. Then $\\widehat{GFA} = \\widehat{GEA} = \\widehat{AEF} = \\widehat{AGF}$, so $AF = AG$ as desired.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22183,
"subject": "Mathematics (Olympiad)",
"question": "A $3 \\times 3$ square is divided into $9$ squares. A positive integer is inscribed in every square, such that the sums of the numbers in each row, column, and diagonal are equal. Prove that this sum cannot be $2008$.",
"options": [],
"answer": "See solution",
"solution": "Let $m$ be the common sum in each row, column, and diagonal. Consider the arrangement:\n\n| $x$ | $y$ | $m - x - y$ |\n|-----------|-----------|---------------|\n| $a$ | $b$ | $m - a - b$ |\n| $m - x - a$ | $m - b - y$ | $x + a + b + y - m$ |\n\nBecause the sum of the elements in the diagonals must be $m$, we obtain:\n\n$$\n\\begin{cases}\nx + b + x + a + b + y - m = m \\\\\nm - x - a + b + m - x - y = m\n\\end{cases}\n$$\n\nwhich is equivalent to\n\n$$\n\\begin{cases}\n2x + a + y + 2b = 2m \\\\\n2x + a + y - b = m\n\\end{cases}\n$$\n\nSubtracting the second equation from the first gives $2b + b = 2m - m$, or $3b = m$. Hence, $3$ divides $m$. But $3$ does not divide $2008$, so $m \\neq 2008$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22184,
"subject": "Mathematics (Olympiad)",
"question": "Let $S_i$ be the cardinality of the set of numbers colored $C_i$. What is the maximum possible number of colors $k$ such that the numbers $1$ through $170$ can be colored with $k$ colors, and for each color $C_i$, the set of numbers colored $C_i$ forms a set whose sum is a power of $2$, and the sets are ordered so that each set divides the next one?",
"options": [],
"answer": "See solution",
"solution": "We begin by bounding the number of colors for which $S_i = 1$. Let there be $\\ell$ such colors. If $a_j$ are the numbers with these colors and $a_1 < a_2 < \\dots < a_\\ell$, then $a_1 \\mid a_2 \\mid \\dots \\mid a_\\ell$. Therefore, $a_j \\ge 2a_{j-1}$, so $a_\\ell \\ge 2^{\\ell-1}$. Since $a_\\ell \\le 170$, we get $2^{\\ell-1} \\le 170$, so $\\ell \\le 8$.\n\nNow we bound $k$. We have $170 = \\sum_{i=1}^k S_i = \\sum_{S_i=1} S_i + \\sum_{S_i>1} S_i$. If we bound the cardinality of the sets that do not have only one number by $2$, we get $170 \\ge \\ell + 2(k-\\ell) = 2k - \\ell \\ge 2k - 8$. From here, $k \\le 89$. This is the maximum possible value. It remains to show an example with $k = 89$ colors.\n\nThe proof guides the construction. For $8$ colors, color only one number, and those numbers are the powers of $2$. For the rest, color exactly $2$ numbers.\n\nLet $D_i$ be the set of numbers colored $C_i$. We construct:\n\n$$\nD_1 = \\{1\\},\\quad D_2 = \\{2\\},\\quad D_3 = \\{4\\},\n$$\n$$\nD_4 = \\{3, 5\\},\\quad D_5 = \\{8\\},\n$$\n$$\nD_6 = \\{6, 10\\},\\quad D_7 = \\{7, 9\\},\\quad D_8 = \\{16\\},\n$$\n$$\nD_9 = \\{11, 21\\},\\quad D_{10} = \\{12, 20\\},\\quad \\dots,\\quad D_{13} = \\{15, 17\\},\\quad D_{14} = \\{32\\},\n$$\n$$\nD_{15} = \\{22, 42\\},\\quad D_{16} = \\{23, 41\\},\\quad \\dots,\\quad D_{24} = \\{31, 33\\},\\quad D_{25} = \\{64\\},\n$$\n$$\nD_{26} = \\{43, 85\\},\\quad D_{27} = \\{44, 84\\},\\quad \\dots,\\quad D_{46} = \\{63, 65\\},\\quad D_{47} = \\{128\\},\n$$\n$$\nD_{48} = \\{86, 170\\},\\quad D_{49} = \\{87, 169\\},\\quad \\dots,\\quad D_{89} = \\{127, 129\\}.\n$$\n\nThe sum of every set is a power of $2$, and they are ordered so each divides the next. The example has the desired properties.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22185,
"subject": "Mathematics (Olympiad)",
"question": "Prove the inequality\n\n$$\n\\frac{1}{k} + \\frac{1}{k+1} + \\dots + \\frac{1}{n} \\geq \\frac{2(n-k+1)}{n+k}\n$$\n\nfor all pairs $n, k$ of positive integers such that $k \\leq n$.",
"options": [],
"answer": "See solution",
"solution": "Applying the AM-HM inequality to $k, k+1, \\dots, n$ gives\n\n$$\n\\frac{n-k+1}{\\frac{1}{k} + \\frac{1}{k+1} + \\dots + \\frac{1}{n}} \\leq \\frac{k + (k+1) + \\dots + n}{n-k+1}\n$$\n\nSince $k, k+1, \\dots, n$ form an arithmetic sequence, the right-hand side equals the average of the first and last terms, i.e., $\\frac{k+n}{2}$. Taking reciprocals and multiplying both sides by $n-k+1$ yields the desired result.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22186,
"subject": "Mathematics (Olympiad)",
"question": "試求最小正整數 $n$ 滿足下列條件：\n\n存在一組相異的正整數 $s_1, s_2, \\dots, s_n$ 使得\n\n$$\n(1 - \\frac{1}{s_1})(1 - \\frac{1}{s_2})\\cdots(1 - \\frac{1}{s_n}) = \\frac{17}{670}.\n$$",
"options": [],
"answer": "See solution",
"solution": "答：$n = 39$。\n\n不妨假設 $s_1 < s_2 < \\cdots < s_n$。顯然，$s_1 > 1$，否則 $1 - \\frac{1}{s_1} = 0$。因此，$2 \\leq s_1 \\leq s_2 - 1 \\leq \\cdots \\leq s_n - (n-1)$。故 $s_i \\geq i + 1$，對所有 $i = 1, \\dots, n$。\n\n題設可得：\n\n$$\n\\begin{aligned}\n\\frac{17}{670} &= (1 - \\frac{1}{s_1})(1 - \\frac{1}{s_2})\\cdots(1 - \\frac{1}{s_n}) \\\\\n&\\geq (1 - \\frac{1}{2})(1 - \\frac{1}{3})\\cdots(1 - \\frac{1}{n+1}) \\\\\n&= \\frac{1}{n+1}\n\\end{aligned}\n$$\n\n所以 $n + 1 \\geq \\frac{670}{17} > 39$，即 $n \\geq 39$。\n\n底下證明 $n = 39$ 滿足所求。\n\n考慮 39 個相異正整數：$2, 3, \\dots, 33, 35, 36, 40, 67$。可得：\n\n$$\n\\frac{1}{2} \\cdot \\frac{2}{3} \\cdot \\frac{32}{33} \\cdot \\frac{34}{35} \\cdot \\frac{39}{40} \\cdot \\frac{66}{67} = \\frac{17}{670}.\n$$\n\n故得證。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22187,
"subject": "Mathematics (Olympiad)",
"question": "Let $M$ be the least common multiple of all the integers $10$ through $30$, inclusive. Let $N$ be the least common multiple of $M$, $32$, $33$, $34$, $35$, $36$, $37$, $38$, $39$, and $40$. What is the value of $\\frac{N}{M}$?\n\n(A) 1 \n(B) 2 \n(C) 37 \n(D) 74 \n(E) 2886",
"options": [],
"answer": "See solution",
"solution": "The value of $M$ is the product of the greatest prime powers found in the prime factorizations of $10, 11, \\ldots, 30$. (Specifically, $M = 2^4 \\cdot 3^3 \\cdot 5^2 \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 17 \\cdot 19 \\cdot 23 \\cdot 29$, although it is not necessary to compute this.)\n\nBecause $32 = 2^5$, $33 = 3 \\cdot 11$, $34 = 2 \\cdot 17$, $35 = 5 \\cdot 7$, $36 = 2^2 \\cdot 3^2$, $37$ is prime, $38 = 2 \\cdot 19$, $39 = 3 \\cdot 13$, and $40 = 2^3 \\cdot 5$, the new greatest prime powers introduced in the least common multiple of $M$, $32, 33, 34, 35, 36, 37, 38, 39, 40$ are $2^5$ and $37$.\n\nSo,\n$$\n\\frac{N}{M} = \\frac{\\text{lcm}(M, 32, \\ldots, 40)}{\\text{lcm}(10, \\ldots, 30)} = \\frac{2 \\cdot 37 \\cdot M}{M} = 74.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22188,
"subject": "Mathematics (Olympiad)",
"question": "Sean $x$ e $y$ números reales entre 0 y 1. Probar que\n\n$$\nx^3 + x y^2 + 2 x y \\leq 2 x^2 y + x^2 + x + y\n$$",
"options": [],
"answer": "See solution",
"solution": "La desigualdad equivale a\n\n$$\nP = 2 x^2 y + x^2 + x + y - x^3 - x y^2 - 2 x y \\geq 0.\n$$\n\nEscribimos $P$ como un polinomio en la variable $x$:\n\n$$\nP = -x^3 + (2y + 1)x^2 + (1 - 2y - y^2)x + y.\n$$\n\nDividimos este polinomio entre $x - 1$, mediante el algoritmo de Ruffini:\n\n$$\n\\begin{array}{c|cccc}\n1 & -1 & 2y+1 & 1-2y-y^2 & y \\\\\n\\cline{2-5}\n & -1 & -1 & 2y & 1-y^2 \\\\\n\\cline{2-5}\n & -1 & 2y & 1-y^2 & 1+y-y^2\n\\end{array}\n$$\n\nEs decir,\n\n$$\nP = (-x^2 + 2 x y + 1 - y^2)(x - 1) + 1 + y - y^2.\n$$\n\nTambién,\n\n$$\n\\begin{align*}\nP &= (-x^2 + 2 x y - y^2)(x - 1) + x - 1 + 1 + y - y^2 \\\\\n &= (x^2 - 2 x y + y^2)(1 - x) + x + y - y^2 \\\\\n &= (x - y)^2 (1 - x) + x + y (1 - y).\n\\end{align*}\n$$\n\nPuesto que las cinco cantidades $(x - y)^2$, $1 - x$, $x$, $y$, $1 - y$ son no negativas, $P \\geq 0$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22189,
"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 $90^\\circ$. 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\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\nTo maximize, $h_1$ is largest when $P$ is at the midpoint of the arc between $A$ and $B$. For $h_2 + h_3$, consider the rectangle $QXRY$ with sides parallel to $ABCD$ 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 at the midpoint 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 at the midpoint of the arc between $A$ and $B$.\n\nTo minimize, $h_1$ is smallest when $P$ approaches $A$ 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": 22190,
"subject": "Mathematics (Olympiad)",
"question": "Beto has a set of $N$ cards, each with a distinct integer. He repeatedly selects two cards, writes the difference of their numbers in his notebook, and removes the two cards from the table. He continues until fewer than two cards remain. What is the largest integer $N$ such that, no matter how Beto chooses the pairs, the product of all the differences he writes down is always divisible by $7^{100}$?",
"options": [],
"answer": "See solution",
"solution": "The answer is $N = 128$.\n\n**Proof:**\n\n*First, we show that with $128$ cards, Beto can always achieve his goal.*\n\nAs long as there are more than $49$ cards on the table, by the pigeonhole principle, there must be two cards with the same remainder modulo $49$. Their difference is divisible by $49 = 7^2$. Beto can repeat this process $40$ times (after $39$ such operations, $128 - 2 \\times 39 = 50$ cards remain). Thus, he writes $40$ numbers divisible by $7^2$, contributing at least $80$ factors of $7$.\n\nWith the remaining $48$ cards, as long as there are more than $7$ cards, there must be two with the same remainder modulo $7$. Their difference is divisible by $7$. This can be done $20$ more times, leaving $8$ cards. So, Beto writes $20$ more multiples of $7$, contributing $20$ more factors of $7$.\n\nIn total, the product is divisible by $7^{80} \\cdot 7^{20} = 7^{100}$.\n\n*Now, we show that with $127$ cards, Beto may not be able to achieve his goal.*\n\nConsider a set of $127$ cards constructed so that for each residue modulo $7$, there is an odd number of cards. Then, at most $\\frac{127-7}{2} = 60$ pairs can have differences divisible by $7$. Since none of these differences is divisible by $7^3 = 343$, each contributes at most two factors of $7$. To reach $7^{100}$, at least $40$ differences must be divisible by $7^2$, but in this construction, at most $39$ such pairs exist. Thus, the highest power of $7$ dividing the product is $7^{99}$, which is insufficient.\n\nTherefore, the largest such $N$ is $128$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22191,
"subject": "Mathematics (Olympiad)",
"question": "On a table there are $k \\ge 2$ piles containing $n_1, n_2, \\dots, n_k$ pens. A _move_ consists in choosing two piles having $a$ and $b$ pens, with $a \\ge b$, and transferring $b$ pens from the first pile to the second.\n\nFind the necessary and sufficient condition for $n_1, n_2, \\dots, n_k$ so that there exists a sequence of moves which puts all the pens in the same pile.",
"options": [],
"answer": "See solution",
"solution": "The necessary and sufficient condition is\n\n$$\n\\frac{n_1 + n_2 + \\dots + n_k}{d} = 2^m, \\quad m \\in \\mathbb{N}^*,\n$$\n\nwhere $d$ is the greatest common divisor of $n_1, n_2, \\dots, n_k$.\n\nIndeed, if $a, b$ are positive integers, then $(a-b, 2b) = (a, b)$ or $(a-b, 2b) = 2(a, b)$, so after each move the GCD of the numbers of pens in the piles remains the same or is multiplied by 2. Finally, the remaining pile has $n_1 + \\dots + n_k = 2^m d$, $m \\in \\mathbb{N}^*$ pens.\n\nConversely, if $n_1 + n_2 + \\dots + n_k = 2^m d$, $m \\in \\mathbb{N}^*$, then we can use induction on $m$ to prove that there exists a sequence of moves such that all the pens can be transferred to the same pile.\n\nThe case $m=1$ means that there are two piles with $n_1 = n_2$ pens, so after one move we have only one pile.\n\nSuppose now the property is true for all $m \\le p$ and every $d$, and consider the case $n_1 + n_2 + \\dots + n_k = 2^{p+1}d$. The set\n\n$$\nA = \\{i \\mid 1 \\le i \\le k,\\ n_i/d \\text{ is odd}\\}\n$$\n\nhas an even number of elements, so we can group the piles with $n_i$, $i \\in A$ two by two, and, making a move in each group, we get piles with $n'_1, \\dots, n'_l$ pens, where $n'_1 + \\dots + n'_l = 2^q (n'_1, \\dots, n'_l)$, $q \\le p$. By the induction hypothesis, from this point on there exists a sequence of moves which puts all the pens into the same pile, so the statement is true also for $m = p + 1$.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 22192,
"subject": "Mathematics (Olympiad)",
"question": "Let $m$ be a positive integer. Determine the smallest positive integer $n$ for which there exist real numbers $x_1, x_2, \\dots, x_n \\in (-1, 1)$ such that\n$$\n|x_1| + |x_2| + \\dots + |x_n| = m + |x_1 + x_2 + \\dots + x_n|.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let us consider $x_1, x_2, \\dots, x_n$ a solution of the equation above. We may suppose, without loss of generality, that $x_1 + x_2 + \\dots + x_n \\ge 0$ (otherwise we change the signs of all the numbers) and that $x_1 \\le \\dots \\le x_p \\le 0 < x_{p+1} \\le \\dots \\le x_n$.\n\nThen the equation becomes\n$$\n-2(x_1 + x_2 + \\dots + x_p) = m.\n$$\nFrom $x_i \\in (-1, 1)$ we obtain that $m < 2p$, i.e. $p > \\frac{m}{2}$. Moreover, $n-p > x_{p+1} + \\dots + x_n \\ge -x_1 - \\dots - x_p = \\frac{m}{2}$.\n\nIf $m$ is even, then $p \\ge \\frac{m}{2} + 1$, $n-p \\ge \\frac{m}{2} + 1$, and therefore $n \\ge m + 2$. A convenient choice for $n = m + 2$ is\n$$\nx_1 = \\dots = x_{\\frac{m}{2} + 1} = -\\frac{m}{m + 2}, \\quad x_{\\frac{m}{2} + 2} = \\dots = x_n = \\frac{m}{m + 2}.\n$$\nIf $m$ is odd, from $p \\ge \\frac{m + 1}{2}$, $n - p \\ge \\frac{m + 1}{2}$, we obtain $n \\ge m + 1$. A convenient choice for $n = m + 1$ is\n$$\nx_1 = \\dots = x_{\\frac{m + 1}{2}} = -\\frac{m}{m + 1}, \\quad x_{\\frac{m + 1}{2} + 1} = \\dots = x_n = \\frac{m}{m + 1}.\n$$\n\nIn conclusion, the smallest value of $n$ for which the equation has solutions is\n$$\nn = m + 1 \\text{ if $m$ is odd, and } n = m + 2 \\text{ if $m$ is even.}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22193,
"subject": "Mathematics (Olympiad)",
"question": "Given a sequence of real numbers $x_n$:\n\n$$\nx_1 = 1 \\quad \\text{and} \\quad x_n = \\frac{2n}{(n-1)^2} \\sum_{i=1}^{n-1} x_i \\quad \\text{for all } n \\ge 2.\n$$\n\nFor each positive integer $n$, let $y_n = x_{n+1} - x_n$.\n\nShow that the sequence $(y_n)$ has a finite limit as $n \\to +\\infty$.",
"options": [],
"answer": "See solution",
"solution": "For all $n \\ge 1$, we have\n\n$$\nx_{n+1} = \\frac{2(n+1)}{n^2} \\sum_{i=1}^{n} x_i = \\frac{2(n+1)}{n^2} \\left( \\frac{(n-1)^2}{2n} + 1 \\right) x_n = \\frac{(n+1)(n^2+1)}{n^3} x_n.\n$$\n\nConsequently,\n$$\n\\frac{x_{n+1}}{n+1} = \\left(1 + \\frac{1}{n^2}\\right) \\frac{x_n}{n} \\quad \\forall n \\ge 1.\n$$\n\nHence, for all $n \\ge 2$,\n$$\ny_n = x_{n+1} - x_n = \\left( \\frac{(n+1)(n^2+1)}{n^3} - 1 \\right) x_n = \\frac{n^2+n+1}{n^2} \\cdot \\frac{x_n}{n} = \\left(1 + \\frac{n+1}{n^2}\\right) \\prod_{k=1}^{n-1} \\left(1 + \\frac{1}{k^2}\\right). \\quad (1)\n$$\n\nNoting that $y_1 = x_2 - x_1 = 3$, we have $y_n > 0$ for all $n \\ge 1$, $y_1 < y_2$, and for all $n \\ge 3$,\n$$\n\\frac{y_n}{y_{n-1}} = \\frac{n^2 + n + 1}{n^2} \\cdot \\frac{(n-1)^2}{(n-1)^2 + n} \\cdot \\left(1 + \\frac{1}{(n-1)^2}\\right) = 1 + \\frac{2}{n^4 - n^3 + n^2} > 1.\n$$\n\nConsequently, $(y_n)$ is an increasing sequence. (2)\n\nSince for all $n \\ge 2$ we have $n + 1 < n^2$ and\n$$\n\\prod_{k=1}^{n-1} \\left(1 + \\frac{1}{k^2}\\right) \\le \\left(1 + \\frac{\\sum_{k=1}^{n-1} \\frac{1}{k^2}}{n-1}\\right)^{n-1},\n$$\nit follows from (1) that\n$$\ny_n < 2 \\left( 1 + \\frac{\\sum_{k=1}^{n-1} \\frac{1}{k^2}}{n-1} \\right)^{n-1} \\quad \\forall n \\ge 2. \\quad (3)\n$$\n\nBut\n$$\n\\sum_{k=1}^{n-1} \\frac{1}{k^2} < 1 + \\sum_{k=2}^{n-1} \\frac{1}{k(k-1)} = 1 + \\sum_{k=2}^{n-1} \\left(\\frac{1}{k-1} - \\frac{1}{k}\\right) = 2 - \\frac{1}{n-1} < 2 \\quad \\forall n \\ge 3.\n$$\n\nHence, (3) implies\n$$\ny_n < 2 \\left(1 + \\frac{2}{n-1}\\right)^{n-1} < 2e^2 \\quad \\forall n \\ge 2.\n$$\n\nHence $(y_n)$ is bounded from above. Together with (2), this implies that $(y_n)$ is convergent. $\\blacksquare$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22194,
"subject": "Mathematics (Olympiad)",
"question": "A configuration of 4027 points in the plane is called *Colombian* if it consists of 2013 red points and 2014 blue points, and no three of the points are collinear. By drawing some lines, the plane is divided into several regions. An arrangement of lines is *good* for a Colombian configuration if the following two conditions are satisfied:\n\n- No line passes through any point of the configuration.\n- No region contains points of both colors.\n\nFind the least value of $k$ such that for any Colombian configuration of 4027 points, there is a good arrangement of $k$ lines.",
"options": [],
"answer": "See solution",
"solution": "The answer is $2013$.\n\nWe first show a good arrangement with at most $2013$ lines always exists. We begin with the following key lemma.\n\n**Lemma 1.** Any pair of points $P$ and $Q$ in a Colombian configuration $C$ can be separated from the other points by two lines.\n\n*Proof.* No three points in $C$ are collinear, so each other point has one of finitely many positive distances to $PQ$. Choose $r > 0$ less than all such distances; the two lines parallel to and at a distance $r$ from $PQ$ have the desired property. $\\square$\n\nLet $C$ be the convex hull of our Colombian configuration $C$. If a red point $R$ is a vertex of $C$, draw a line $l_1$ separating $R$ from all other points. Next, place the other $2012$ red points into $1006$ pairs and apply Lemma 1 to draw $2012$ lines separating them from the rest of $C$. Together with $l_1$, these $2012$ lines form a good arrangement. Otherwise, $C$ has a side $B_1B_2$ consisting of blue points. Draw a line $l_1$ separating $B_1$ and $B_2$ from the rest of $C$, place the $2012$ remaining blue points into $1006$ pairs, and apply Lemma 1. The resulting $2012$ lines together with $l_1$ form the desired good arrangement.\n\nFor the other direction, let $\\mathcal{P} = A_1 \\cdots A_{4026}$ be a regular $4026$-gon with $A_i$ red for $i$ odd and blue for $i$ even. Form a Colombian configuration $C$ with the vertices of $\\mathcal{P}$ and another blue point $B$. We claim any good arrangement for $C$ has at least $2013$ lines. Note that each of the $4026$ pairs $(A_i, A_{i+1})$ of neighboring points contains points in different regions, so $A_iA_{i+1}$ intersects at least one of the lines. Each line intersects $\\mathcal{P}$ at most twice, so an arrangement with $k$ lines can produce at most $2k$ intersections, implying that $2k \\geq 4026$ and hence $k \\geq 2013$, as desired.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22195,
"subject": "Mathematics (Olympiad)",
"question": "Suppose a positive integer $n$ is a perfect square. Consider the set of numbers that can be represented as the product of two numbers, both of which are greater than or equal to $n$ (the two numbers may be the same). Express the number which is the $n$-th smallest in this set, in terms of $n$.",
"options": [],
"answer": "See solution",
"solution": "We will show that the desired answer is $(n + \\sqrt{n} - 1)^2$.\n\nFor a positive integer $c$ greater than or equal to $2n$, let us call a number a $c$-product if it can be represented as a product of two numbers greater than or equal to $n$ whose sum equals $c$. For any real number $x$, let $\\lfloor x \\rfloor$ denote the greatest integer less than or equal to $x$.\n\n**Lemma 1.** Let $c \\geq 2n$ be an integer. Then, there are exactly $\\lfloor \\frac{c}{2} \\rfloor - n + 1$ $c$-products, and they are all greater than or equal to $n(c-n)$ and less than or equal to $\\frac{c^2}{4}$.\n\n**Proof:** There are $\\lfloor \\frac{c}{2} \\rfloor - n + 1$ pairs of integers $(x, y)$ with $x, y \\geq n$ and $x + y = c$. They are\n\n$$\n(n, c-n), (n+1, c-n-1), \\ldots, \\left(\\left\\lfloor \\frac{c}{2} \\right\\rfloor, c - \\left\\lfloor \\frac{c}{2} \\right\\rfloor\\right).\n$$\n\nIf we define $f(t) = t(c-t)$ for any real number $t$, then the products $xy$ for the pairs $(x, y)$ in this set are given by $f(n), f(n+1), \\ldots, f(\\lfloor \\frac{c}{2} \\rfloor)$. Since $f(t) = \\frac{c^2}{4} - (c - \\frac{t}{2})^2$, we see that $f(n) < f(n+1) < \\ldots < f(\\lfloor \\frac{c}{2} \\rfloor) \\leq \\frac{c^2}{4}$. Thus, there are exactly $\\lfloor \\frac{c}{2} \\rfloor - n + 1$ $c$-products and they are all greater than or equal to $f(n) = n(c-n)$ and less than or equal to $\\frac{c^2}{4}$.\n\n**Lemma 2.** If $2n \\leq c < c' \\leq 2(n+\\sqrt{n})$, then each $c'$-product is strictly bigger than any of the $c$-products. In particular, the set of $c$-products and the set of $c'$-products are disjoint.\n\n**Proof:** It suffices to prove this for the case where $c' = c + 1$. From Lemma 1, it is enough to show that $\\frac{c^2}{4} < n(c+1-n)$. Since $0 \\leq c - 2n < 2\\sqrt{n}$, we have $-\\frac{1}{4}(c-2n)^2 > -n$, and therefore,\n\n$$\n n(c+1-n) - \\frac{c^2}{4} = -\\frac{1}{4}(c-2n)^2 + n > -n + n = 0.\n$$\n\n**Lemma 3.** If $2(n + \\sqrt{n}) \\leq c$, then any $c$-product is greater than any of the $(2(n + \\sqrt{n}) - 1)$-products.\n\n**Proof:** By Lemma 2, every $(2(n+\\sqrt{n})-1)$-product is less than any of the $2(n+\\sqrt{n})$-products. In particular, it is less than $n(n+2\\sqrt{n})$, which is a $2(n+\\sqrt{n})$-product. As $2(n+\\sqrt{n}) \\leq c$ by assumption, the assertion follows since $n(n+2\\sqrt{n}) = n(2(n+\\sqrt{n})-n) \\leq n(c-n)$.\n\nBy Lemmas 2 and 3, if we list all the numbers that can be written as a product of two numbers both greater than or equal to $n$, starting with the smallest, then up until we reach the largest $(2(n + \\sqrt{n}) - 1)$-product, these numbers are listed as follows:\n\nThe first number is $n^2$ (the unique $(2n)$-product), then $n(n+1)$ (the unique $(2n+1)$-product), followed by $(2n+2)$-products in increasing order, $(2n+3)$-products in increasing order, ..., $2(n+\\sqrt{n}-2)$-products in increasing order, and finally, $2(n+\\sqrt{n}-1)$-products in increasing order.\n\nBy Lemma 1, $(n+\\sqrt{n}-1)^2$ is the largest $(2(n+\\sqrt{n})-2)$-product. If $(n+\\sqrt{n}-1)^2$ is the $k$-th smallest number in this list, then\n\n$$\n k = \\sum_{c=2n}^{2(n+\\sqrt{n})-2} \\text{ (number of } c\\text{-products)}.\n$$\n\nTherefore,\n\n$$\n\\begin{align*}\n k &= 1+1+2+2+\\cdots+\\left(\\left\\lfloor \\frac{2(n+\\sqrt{n})-3}{2} \\right\\rfloor - n + 1\\right) + \\left(\\left\\lfloor \\frac{2(n+\\sqrt{n})-2}{2} \\right\\rfloor - n + 1\\right) \\\\\n &= 1+1+2+2+\\cdots+(\\sqrt{n}-1)+(\\sqrt{n}-1)+\\sqrt{n} \\\\\n &= 2(1+2+\\cdots+\\sqrt{n}) - \\sqrt{n} = 2\\left(\\frac{(1+\\sqrt{n})\\sqrt{n}}{2}\\right) - \\sqrt{n} = n.\n\\end{align*}\n$$\n\nHence, the desired answer is $(n + \\sqrt{n} - 1)^2$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22196,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer, and let $x_1, x_2, \\dots, x_{n+1}$ be positive real numbers. Let $p, q$ be positive real numbers with $p < q$. Suppose\n\n$$\nx_{n+1}^p > x_1^p + x_2^p + \\dots + x_n^p.\n$$\n\nProve:\n\n1. $x_{n+1}^q > x_1^q + x_2^q + \\dots + x_n^q$;\n2. $\\left(x_{n+1}^p - x_1^p - x_2^p - \\dots - x_n^p\\right)^{1/p} < \\left(x_{n+1}^q - x_1^q - x_2^q - \\dots - x_n^q\\right)^{1/q}$.",
"options": [],
"answer": "See solution",
"solution": "Both inequalities are homogeneous, so we may assume $x_{n+1} = 1$ (otherwise, replace each $x_i$ by $\\frac{x_i}{x_{n+1}}$ for $i = 1, 2, \\dots, n$). Given the condition\n\n$$\nx_1^p + x_2^p + \\dots + x_n^p < 1,\n$$\n\nwe have $x_1, x_2, \\dots, x_n \\in (0, 1)$.\n\n1. Since $p < q$, it follows that\n\n$$\nx_1^q + x_2^q + \\dots + x_n^q < x_1^p + x_2^p + \\dots + x_n^p < 1.\n$$\n\n2. Let $a = \\left(1 - x_1^p - x_2^p - \\dots - x_n^p\\right)^{1/p}$, so $a \\in (0, 1)$. Since $q > p$,\n\n$$\na^q + x_1^q + x_2^q + \\dots + x_n^q < a^p + x_1^p + x_2^p + \\dots + x_n^p = 1,\n$$\n\nthus $1 - x_1^q - x_2^q - \\dots - x_n^q > a^q$, so\n\n$$\n\\left(1 - x_1^q - x_2^q - \\dots - x_n^q\\right)^{1/q} > a = \\left(1 - x_1^p - x_2^p - \\dots - x_n^p\\right)^{1/p},\n$$\n\nas required.\n\n*Remark*: The inequality (1) can also be derived from the norm inequality\n$$\n\\left(\\sum_{i=1}^{n} x_i^p\\right)^{1/p} \\geq \\left(\\sum_{i=1}^{n} x_i^q\\right)^{1/q}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22197,
"subject": "Mathematics (Olympiad)",
"question": "Prove that the set $S = \\{\\lfloor n\\pi \\rfloor : n = 0, 1, 2, 3, \\dots \\}$ contains arithmetic progressions of any finite length, but no infinite arithmetic progressions.",
"options": [],
"answer": "See solution",
"solution": "Let $\\{x\\} = x - \\lfloor x \\rfloor$ denote the fractional part of $x$. Given any integer $m \\geq 3$, there exists a positive integer $n$ such that $\\{n\\pi\\} < 1/m$, since the set $\\{k\\pi : k = 0, 1, 2, 3, \\dots\\}$ is dense in $[0, 1]$.\n\nConsequently,\n\n$$\n\\lfloor k n \\pi \\rfloor = k \\lfloor n\\pi \\rfloor + \\lfloor k \\{n\\pi\\} \\rfloor = k \\lfloor n\\pi \\rfloor, \\quad k = 1, 2, \\dots, m,\n$$\n\nso $\\lfloor n\\pi \\rfloor, \\lfloor 2n\\pi \\rfloor, \\dots, \\lfloor mn\\pi \\rfloor$ are $m$ numbers in $S$ in arithmetic progression with ratio $\\lfloor n\\pi \\rfloor$.\n\nSuppose, for contradiction, that $S$ contains an infinite arithmetic progression $\\lfloor n_k \\pi \\rfloor$, $k = 0, 1, 2, 3, \\dots$, with integer ratio $r$, where $n_k$ is a strictly increasing sequence of positive integers. Write $n_k \\pi = \\lfloor n_0 \\pi \\rfloor + k r + \\{n_k \\pi\\}$ to deduce that $r > 0$ and\n\n$$\nn_{k+1} - n_k = \\frac{r + \\{n_{k+1}\\pi\\} - \\{n_k\\pi\\}}{\\pi} \\in \\left( \\frac{r-1}{\\pi}, \\frac{r+1}{\\pi} \\right).\n$$\n\nThe length of this interval is less than $1$, so $n_{k+1} - n_k = n$ for some positive integer $n$ and all $k$. Hence, $n_k = n_0 + k n$, so $n_k / k \\to n$ as $k \\to \\infty$. On the other hand, $n_k / k \\to r/\\pi$, so $\\pi = r/n$, contradicting the irrationality of $\\pi$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22198,
"subject": "Mathematics (Olympiad)",
"question": "Let $0 < a_1 < a_2 < \\dots < a_n$ be real numbers. Prove that\n$$\n\\left(\\frac{1}{1+a_1} + \\frac{1}{1+a_2} + \\dots + \\frac{1}{1+a_n}\\right)^2 \\leq \\frac{1}{a_1} + \\frac{1}{a_2 - a_1} + \\frac{1}{a_3 - a_2} + \\dots + \\frac{1}{a_n - a_{n-1}}.\n$$",
"options": [],
"answer": "See solution",
"solution": "By the Cauchy-Schwarz inequality,\n$$\n\\left(\\sum_{i=1}^n \\frac{1}{1+a_i}\\right)^2 \\leq \\left( \\frac{1}{a_1} + \\frac{1}{a_2 - a_1} + \\dots + \\frac{1}{a_n - a_{n-1}} \\right) \\left( \\frac{a_1}{(1+a_1)^2} + \\frac{a_2 - a_1}{(1+a_2)^2} + \\dots + \\frac{a_n - a_{n-1}}{(1+a_n)^2} \\right).\n$$\nNote that $\\frac{a_1}{(1+a_1)^2} \\leq \\frac{a_1}{1+a_1}$, and for $i = 2, \\dots, n$,\n$$\n\\frac{a_i - a_{i-1}}{(1+a_i)^2} \\leq \\frac{a_i - a_{i-1}}{(1+a_{i-1})(1+a_i)} = \\frac{1}{1+a_{i-1}} - \\frac{1}{1+a_i}.\n$$\nThus, after telescoping,\n$$\n\\frac{a_1}{(1+a_1)^2} + \\frac{a_2 - a_1}{(1+a_2)^2} + \\dots + \\frac{a_n - a_{n-1}}{(1+a_n)^2} \\leq \\frac{a_1}{1+a_1} + \\frac{1}{1+a_1} - \\frac{1}{1+a_n} < 1.\n$$\nTherefore, the original inequality holds.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22199,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle with $|AB|^2 + |AB| \\cdot |AC| = |BC|^2$. Prove that $m(\\hat{A}) = 2m(\\hat{C})$.",
"options": [],
"answer": "See solution",
"solution": "Let $D$ be the intersection of the interior angle bisector of $\\hat{A}$ with $BC$. Then $|BD| = |AB| \\cdot k$, $|CD| = |AC| \\cdot k$. From $|AB| \\cdot (|AB| + |BC|) \\cdot k = |BC|^2 \\cdot k$, we get $|AB| \\cdot |BC| = |BC|^2 \\cdot k \\Rightarrow |AB| = |BC|$. Since $|BD| = |AB| \\cdot k$, $\\triangle ABD \\sim \\triangle CBA$, so $m(\\widehat{BCA}) = m(\\widehat{BAD}) = \\frac{1}{2} m(\\widehat{BAC})$. Thus, $m(\\hat{A}) = 2m(\\hat{C})$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22200,
"subject": "Mathematics (Olympiad)",
"question": "Find all real solutions $(x, y)$ to the equation:\n\n$$\n\\frac{x^3 - x y + 1}{x^2 + x - y} = \\frac{y^3 + x y - 1}{y^2 + x - y}.\n$$",
"options": [],
"answer": "See solution",
"solution": "We start by manipulating the given equation:\n\n$$\n\\begin{aligned}\n\\frac{x^3 - x y + 1}{x^2 + x - y} &= \\frac{y^3 + x y - 1}{y^2 + x - y} \\\\\n\\Rightarrow x - \\frac{x^2 - 1}{x^2 + x - y} &= y + \\frac{y^2 - 1}{y^2 + x - y} \\\\\n\\Rightarrow x - y &= \\frac{x^2 - 1}{x^2 + x - y} + \\frac{y^2 - 1}{y^2 + x - y}. \\quad (*)\n\\end{aligned}\n$$\n\n**Case 1:** $x = y$\n\n$$\n2 \\frac{x^2 - 1}{x} = 0 \\implies x^2 = 1 \\text{ or } x = 0 \\implies (x, y) = (0, 0), (1, 1), (-1, -1)\n$$\n\n**Case 2:** $x > y$. Let $k = x - y > 0$. By $(*)$:\n\n$$\nk = \\frac{x^2 - 1}{x^2 + k} + \\frac{y^2 - 1}{y^2 + k}\n$$\n\nIf $|x| \\leq 1$ or $|y| \\leq 1$, by checking the main equation, we get $(x, y) = (-1, 0), (1, 2), (0, 1), (-2, -1)$. Now assume $|x|, |y| > 1$.\n\n$$\n0 < \\frac{x^2 - 1}{x^2 + k} < 1, \\quad 0 < \\frac{y^2 - 1}{y^2 + k} < 1\n$$\n\nSo\n\n$$\n0 < k = \\frac{x^2 - 1}{x^2 + k} + \\frac{y^2 - 1}{y^2 + k} < 2 \\implies k = 1\n$$\n\nThus,\n\n$$\n1 = \\frac{x^2 - 1}{x^2 + 1} + \\frac{y^2 - 1}{y^2 + 1}. \\quad (**)\n$$\n\nBut for $|x| > 1$, $\\frac{x^2 - 1}{x^2 + 1} > \\frac{1}{2}$, and similarly for $y$, so their sum $> 1$, which contradicts $(**)$.\n\n**Case 3:** $x < y$. Let $t = y - x > 0$. By $(*)$:\n\n$$\nt = \\frac{x^2 - 1}{t - x^2} + \\frac{y^2 - 1}{t - y^2}\n$$\n\nAdding $2 = \\frac{t - x^2}{t - x^2} + \\frac{t - y^2}{t - y^2}$ to both sides:\n\n$$\nt + 2 = \\frac{t - 1}{t - x^2} + \\frac{t - 1}{t - y^2}\n$$\n\nSuppose $t - x^2 > 0$ and $t - y^2 > 0$, then $2t > x^2 + y^2$, so $(x + 1)^2 + (y - 1)^2 < 2$, which is impossible for $|x|, |y| > 1$. Thus, at least one denominator is negative or zero, so\n\n$$\nt + 2 \\leq 0 + (t - 1) < t\n$$\n\nThis is a contradiction, so no new solutions arise in this case.\n\n**Final answer:**\n\n$$(x, y) = (0, 0), (1, 1), (-1, -1), (-1, 0), (1, 2), (0, 1), (-2, -1)$$\n\n$\\boxed{(0, 0), (1, 1), (-1, -1), (-1, 0), (1, 2), (0, 1), (-2, -1)}$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22201,
"subject": "Mathematics (Olympiad)",
"question": "The product of the lengths of the two congruent sides of an obtuse isosceles triangle is equal to the product of the base and twice the triangle's height to the base. What is the measure, in degrees, of the vertex angle of this triangle?\n\n(A) 105 (B) 120 (C) 135 (D) 150 (E) 165",
"options": [],
"answer": "See solution",
"solution": "Let $a$, $b$, and $h$ be the lengths of the congruent sides, the base, and the height to the base of the obtuse isosceles triangle, respectively. The area of the triangle is $\\frac{1}{2} b h$, which by the stated condition equals $\\frac{1}{4} a^2$. The area is also $\\frac{1}{2} a^2 \\sin \\theta$, where $\\theta$ is the vertex angle. Equating these expressions gives $\\sin \\theta = \\frac{1}{2}$. Because the triangle is obtuse, this implies $\\theta = 150^{\\circ}$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 22202,
"subject": "Mathematics (Olympiad)",
"question": "Auf der Tafel stehen die drei natürlichen Zahlen $2000$, $17$ und $n$. Anna und Berta spielen folgendes Spiel: Anna beginnt, dann sind sie abwechselnd am Zug. Ein Zug besteht darin, eine der drei Zahlen auf der Tafel durch den Betrag der Differenz der beiden anderen Zahlen zu ersetzen. Dabei ist kein Zug erlaubt, bei dem sich keine der drei Zahlen verändert. Wer an der Reihe ist und keinen erlaubten Zug mehr machen kann, hat verloren.\n\n*Beweise, dass das Spiel für jedes $n$ irgendwann zu Ende geht.*\n*Wer gewinnt, wenn $n = 2017$ ist?*\n\n",
"options": [],
"answer": "See solution",
"solution": "Wenn drei Zahlen auf der Tafel stehen und bei einem Zug eine der drei Zahlen durch die (positive) Differenz der anderen beiden Zahlen ersetzt wird, so stehen nach diesem Zug zwei Zahlen und die Summe der beiden Zahlen auf der Tafel. Es sei o. B. d. A. $b > a$ und $a$, $b$ und $a + b$ stehen auf der Tafel. Dann gibt es wegen $a + b - b = a$ und $a + b - a = b$ nur einen möglichen Zug und es stehen nachher $a$, $b$ und $b - a$ auf der Tafel. Auch dann ist eine der Zahlen (nämlich $b$) die Summe der anderen beiden und es gibt nur einen möglichen Zug.\n\nMan erkennt also, dass es spätestens ab dem zweiten Zug keine Auswahl der Züge mehr gibt und alle Züge zwangsläufig sind. Weiter erkennt man, dass ab dem zweiten Zug bei jedem Zug die größte der drei Zahlen verkleinert wird, und, weil keine der Zahlen negativ werden kann, muss nach endlich vielen Zügen eine Zahl $0$ sein. Da $0$ die Differenz der anderen beiden Zahlen ist, müssen diese gleich sein, es steht also $0$, $a$, $a$ auf der Tafel. Wegen $a - 0 = a$ und $a - a = 0$ ist das der Endzustand, es ist kein Zug mehr möglich. Dieser Endzustand muss also jedenfalls erreicht werden. Sieger ist also, wer $0$, $a$, $a$ auf die Tafel schreibt.\n\nSpielverlauf, wenn zu Beginn $2000$, $17$, $2017$ auf der Tafel steht:\n\n1. Zug (A): $2000$, $17$, $1983$\n2. Zug (B): $1966$, $17$, $1983$\n3. Zug (A): $1966$, $17$, $1949$\n...\n117. Zug (A): $2000 - 116 \\cdot 17 = 28$, $17$, $2000 - 117 \\cdot 17 = 11$\n118. Zug (B): $6$, $17$, $11$\n119. Zug (A): $6$, $5$, $11$\n120. Zug (B): $6$, $5$, $1$\n121. Zug (A): $4$, $5$, $1$\n122. Zug (B): $4$, $3$, $1$\n123. Zug (A): $2$, $3$, $1$\n124. Zug (B): $2$, $1$, $1$\n125. Zug (A): $0$, $1$, $1$\n\nAnna gewinnt.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22203,
"subject": "Mathematics (Olympiad)",
"question": "Find the minimum value of $\\frac{a^3 + b^3 + c^3}{abc}$ when $a$, $b$, and $c$ are sides of a right triangle.",
"options": [],
"answer": "See solution",
"solution": "Assume without loss of generality that $a^2 + b^2 = c^2$.\n\nBy the AM-GM inequality:\n\n$$\nc^2 = a^2 + b^2 \\geq 2ab.\n$$\n\nThat is,\n\n$$\nc^3 \\geq 2abc. \\quad (1)\n$$\n\nNext, we will show that $a^3 + b^3 \\geq \\sqrt{2}abc$. Consider:\n\n$$\na^3 + b^3 = (a + b)(a^2 + b^2) - ab(a + b)\n$$\n\nand\n\n$$\n\\begin{align*}\n(a + b)(a^2 + b^2) &= (a + b)c^2 \\\\\n&= (a + b)\\sqrt{a^2 + b^2}c \\\\\n&\\geq 2\\sqrt{ab}\\sqrt{2ab}c \\\\\n&= 2\\sqrt{2abc}.\n\\end{align*} \\quad (2)\n$$\n\nNow, by the Cauchy-Schwarz inequality:\n\n$$\n\\begin{align}\nab(a + b) &\\leq ab\\sqrt{2}\\sqrt{a^2 + b^2} \\\\\n&= \\sqrt{2abc},\n\\end{align}\n$$\n\nand so\n\n$$\na^3 + b^3 \\geq 2\\sqrt{2abc} - \\sqrt{2abc} = \\sqrt{2abc}. \\quad (4)\n$$\n\nWe conclude from (1) and (4) that\n\n$$\n\\frac{a^3 + b^3 + c^3}{abc} \\geq 2 + \\sqrt{2}. \\quad (5)\n$$\n\nThis will be an equality when $a = b$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22204,
"subject": "Mathematics (Olympiad)",
"question": "A tick and a cross are to be placed in the grid of 16 blocks shown below, with no more than one symbol in a block. No column of four blocks may contain both symbols, and no row of four blocks may contain both symbols. In how many ways can this be done?\n\n\n\n(A) 144\n(B) 108\n(C) 42\n(D) 36\n(E) 24",
"options": [],
"answer": "See solution",
"solution": "Let the tick be placed in any one of the 16 blocks. Then the cross can go in any of the three other rows or three other columns, which gives $3 \\times 3 = 9$ possible positions for the cross. Thus, the total number of ways is $16 \\times 9 = 144$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22205,
"subject": "Mathematics (Olympiad)",
"question": "Let $p \\neq 3$ be a prime number. Show that there is a non-constant arithmetic sequence of positive integers $x_1, x_2, \\dots, x_p$ such that the product of the terms of the sequence is a cube.",
"options": [],
"answer": "See solution",
"solution": "Let $a_1, a_2, \\dots, a_p$ be any arithmetic sequence of positive integers and let $P$ be the product of the terms of this sequence. For any $n$, the sequence $P^n a_1, P^n a_2, \\dots, P^n a_p$ is also arithmetic, and the product of terms is $P^{np+1}$. Now, either $p \\equiv 1 \\pmod{3}$ or $p \\equiv -1 \\pmod{3}$. In the former case, $2p+1 = 3q$ for some $q$, and in the latter case, $1p+1 = 3q$ for some $q$. So we can choose either $x_i = P^2 a_i$ or $x_i = P a_i$ to obtain the sequence we are looking for.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22206,
"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\nwhile for even $k$, the coefficients of $(x^2 - x + 1)^{2^k-1}$ follow the pattern\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, since $2^{k+1}-1 = (2^k-1)2^k$. We omit the details. $\\square$\n\nNow, by the above claims, we obtain\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": 22207,
"subject": "Mathematics (Olympiad)",
"question": "A circle $k$ with center at $O$ and radius $r$ and a line $p$ which doesn't have a common point with $k$ are given. Let $E$ be the foot of the perpendicular from $O$ to $p$. An arbitrary point $M$ different from $E$ is chosen on $p$ and the two tangents are drawn from $M$ to $k$ which touch the circle $k$ at points $A$ and $B$. If $H$ is the intersection of $AB$ and $OE$, prove that $$\\overline{OH} = \\frac{r^2}{OE}.$$",
"options": [],
"answer": "See solution",
"solution": "Let $G$ be the intersection of $OM$ and $AB$. Since $\\triangle OGH \\sim \\triangle OEM$ we get $\\overline{OG} = \\overline{OE}$, hence $\\overline{OE} \\cdot \\overline{OH} = \\overline{OM} \\cdot \\overline{OG}$. On the other hand, since $\\triangle AOG \\sim \\triangle MOA$, we have $\\overline{OA} = \\overline{OM}/\\overline{OA}$. Therefore $\\overline{OM} \\cdot \\overline{OG} = \\overline{OA}^2$. We get $$\\overline{OH} = \\frac{\\overline{OA}^2}{\\overline{OE}}.$$ \n\n\n\n**Remark.** The equality $\\overline{OE} \\cdot \\overline{OH} = \\overline{OM} \\cdot \\overline{OG}$ can also be obtained as a power of a point since $GMEH$ is inscribed.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22208,
"subject": "Mathematics (Olympiad)",
"question": "We color some unit squares in a $99 \\times 99$ square grid with one of 5 given distinct colors, such that each color appears the same number of times. On each row and on each column there are no differently colored unit squares. Find the maximum possible number of colored unit squares.",
"options": [],
"answer": "See solution",
"solution": "We call a line any row or column from the grid. We color the lines containing a colored square with the same color. From the hypothesis, a row cannot be colored with two (different) colors. There are $2 \\times 99 = 198$ lines, each colored with one of the 5 colors or uncolored. There exists a particular color $C$ associated to at most $\\lfloor 198/5 \\rfloor = 39$ lines. All squares that have color $C$ are at the intersection of these rows and columns. If we have $x$ rows and $y$ columns of color $C$, with $x + y = l \\leq 39$, then the number of squares colored with $C$ is at most $xy$.\n\nIf we fix $l$, the product $xy$ attains its maximum for $x = y = \\frac{l}{2}$ if $l$ is even, and for $x = \\frac{l-1}{2}, y = \\frac{l+1}{2}$ (or vice versa) if $l$ is odd.\n\nFor the color $C$ we take $l = 39$ and we obtain that there are at most $\\frac{39-1}{2} \\cdot \\frac{39+1}{2} = 380$ squares of this color. Because the 5 colors appear the same number of times, it turns out that we can't have more than $5 \\times 380 = 1900$ colored squares on the grid.\n\nLet's give an example in which we have exactly 1900 colored squares. We consider on the diagonal 5 rectangles of dimensions $19 \\times 20$ and $20 \\times 19$ placed successively. Note that two rows and a column are not used. Color each of the 5 rectangles with one (different) color.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22209,
"subject": "Mathematics (Olympiad)",
"question": "Por los puntos medios de dos lados de un triángulo $ABC$ trazamos las medianas y unimos los puntos que trisecan el tercer lado con el vértice opuesto. Así, en el interior, se obtiene una pajarita (dos triángulos unidos por un vértice). Se pide calcular la fracción de superficie total del triángulo que representa la pajarita.",
"options": [],
"answer": "See solution",
"solution": "Supongamos, sin pérdida de generalidad, que el área del triángulo $ABC$ es uno, es decir, $[ABC] = 1$. Tenemos:\n\n\n\n\nLas medianas dividen al triángulo en seis partes de igual área, luego: $[AMF] = [FNC] = \\frac{1}{6}$ y $[AFC] = 2 \\cdot \\frac{1}{6} = \\frac{1}{3}$. Trazamos, desde $B$, $A$, $P$ y $Q$, perpendiculares sobre la mediana $CM$ y sean $X$, $Y$, $Z$ y $V$ sus respectivos pies tal y como muestra la figura de la derecha. En esta figura se tiene, trabajando con la mediana $CM$:\n\n$$\n\\bullet \\ AMY \\cong BMX. \\text{ Como } AM = MB \\text{ entonces } AY = XB = h.\n$$\n\n$$\n\\bullet \\ AYC \\cong PZC \\cong QVC. \\text{ Como } AP = PQ = QC, \\text{ entonces } PZ = \\frac{2}{3}h \\text{ y } QV = \\frac{1}{3}h.\n$$\n\n$$\n\\bullet \\ PEZ \\cong XBE. \\text{ Como } PZ = \\frac{2}{3}XB \\rightarrow PE = \\frac{2}{3}EB \\rightarrow PE = \\frac{2}{5}PB \\text{ y } EB = \\frac{3}{5}PB.\n$$\n\n$$\n\\bullet \\ QHV \\cong XBH. \\text{ Como } QV = \\frac{1}{3}QB \\rightarrow QH = \\frac{1}{3}HB \\rightarrow QH = \\frac{1}{4}QB\n$$\n\nY trabajando, análogamente¹, sobre la otra mediana $AN$, se obtiene:\n\n$$\nQG = \\frac{2}{5}QB \\quad GB = \\frac{3}{5}QB \\quad \\text{y} \\quad PD = \\frac{1}{4}PB\n$$\n\nNos fijamos, por ejemplo, en la ceviana $PB$ y ya podemos conocer la proporción en que los puntos $D$ y $E$ la dividen. Falta:\n\n$$\nDE = PE - PD = \\left(\\frac{2}{5} - \\frac{1}{4}\\right)PB = \\frac{3}{20}PB\n$$\n\n¹Todo esto es equivalente a aplicar el Teorema de Menelao sobre las transversales de un triángulo.\n\nPor tanto,\n\n$$\nPD = \\frac{5}{20} PB \\quad DE = \\frac{3}{20} PB \\quad \\text{y} \\quad EB = \\frac{12}{20} PB\n$$\n\nAhora dibujamos la línea auxiliar $AE$. Sobre la ceviana $PB$, $ABP$ de área conocida queda dividido en tres triángulos, $ABE$, $AED$ y $ADP$, de idéntica altura. Por tanto,\n\n$$\n\\frac{[ADP]}{[ABP]} = \\frac{PD}{PB} = \\frac{5}{20}, \\quad \\frac{[AED]}{[ABP]} = \\frac{DE}{PB} = \\frac{3}{20}, \\quad \\frac{[ABE]}{[ABP]} = \\frac{EB}{PB} = \\frac{12}{20}\n$$\n\nDe\n\n$$\n[ABP] = \\frac{1}{3} \\rightarrow [ADP] = \\frac{5}{60}, \\quad [AED] = \\frac{3}{60} \\quad \\text{y} \\quad [ABE] = \\frac{12}{60}\n$$\n\nPor otro lado, como $EM$ es mediana del triángulo $ABE$, se tiene $[AME] = \\frac{1}{2} [ABE] = \\frac{6}{60}$. Finalmente,\n\n$$\n\\frac{1}{6} = [AMF] = [AME] + [AED] + [DEF] = \\frac{6}{60} + \\frac{3}{60} + [DEF]\n$$\n\nde donde resulta el área de media pajarita. Es decir, $[DEF] = \\frac{1}{60}$. Análogamente, $[FGH] = \\frac{1}{60}$, y el área pedida es\n\n$$\n[DEF] + [FGH] = \\frac{1}{30} = \\frac{1}{30} [ABC]\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22210,
"subject": "Mathematics (Olympiad)",
"question": "Consider the sequence $ (y_n) $ satisfying the conditions:\n\n$ y_0 = -\\frac{1}{4}, \\quad y_1 = 0 $ and $ y_{n+1} + y_{n-1} = 4y_n + 1 $ for $ n \\ge 1 $.\n\nProve that for any $ n \\ge 0 $ the expression $ 2y_{2n} + \\frac{3}{2} $ is\n\na) a positive integer;\nb) the square of an integer.",
"options": [],
"answer": "See solution",
"solution": "a) Let us prove that the numbers $ 2y_{2n} + \\frac{3}{2} $ and $ y_{2n+1} $ are integers by induction on $ n $. For $ n = 0 $ we have $ 2y_0 + \\frac{3}{2} = 1 $ and $ y_1 = 0 $. Suppose that $ 2y_{2n} + \\frac{3}{2} $ and $ y_{2n+1} $ are integers. Then the number\n\n$$\n2y_{2(n+1)} + \\frac{3}{2} = 2(4y_{2n+1} - y_{2n} + 1) + \\frac{3}{2} = 8y_{2n+1} - \\left(2y_{2n} + \\frac{3}{2}\\right) + 5\n$$\n\nand, therefore,\n\n$$\ny_{2(n+1)+1} = 4y_{2n+2} - y_{2n+1} + 1 = 2\\left(2y_{2n+2} + \\frac{3}{2}\\right) - y_{2n+1} - 2\n$$\n\nare also integers. Thus, the number $ 2y_{2n} + \\frac{3}{2} $ is integer for any $ n \\ge 0 $. Its positivity follows, for example, from the fact that the sequence $ (y_n) $ is increasing (this is easily proven by induction). So, $ y_{2n} \\ge y_0 = -\\frac{1}{4} $, whence $ 2y_{2n} + \\frac{3}{2} \\ge 1 $.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 22211,
"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 the 25 experts is to make his *prediction* of how many gold medals each athlete will receive; 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": "24.\n\n**Solution.**\n\n*Upper bound.* We will show that $k \\leq 24$, i.e., that any expert could be incompetent. If an 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 at least one athlete 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": 22212,
"subject": "Mathematics (Olympiad)",
"question": "Докажите, что существует бесконечно много троек различных натуральных чисел $a, b, c$ таких, что $P(a) = P(b) = P(c)$, где $P(n)$ — наибольший простой делитель числа $n^2 + 1$.\n\n*Замечание.* Уравнение вида $x^2 - Dy^2 = a$ называется уравнением Пелля. Известно, что, если $D$ не является квадратом, и это уравнение имеет хотя бы одно решение в натуральных числах, то оно имеет бесконечно много решений в натуральных числах.",
"options": [],
"answer": "See solution",
"solution": "Рассмотрим тождество:\n\n$$\n(m^2 + 1)((m-1)^2 + 1) = (m^2 - m + 1)^2 + 1. \\quad (* )\n$$\n\nИз него следует, что $P(m^2 - m + 1) = \\max(P(m), P(m+1))$.\n\nПредположим противное. Пусть $N$ — наибольшее число, встречающееся в описанных троиках; если таких троек нет, то положим $N = 3$. Последовательность натуральных чисел $P(N+1), P(N+2), \\dots$ не может строго убывать. Значит, найдётся число $n > N + 1$, для которого $P(n-1) \\le P(n)$. Тогда $P(n^2-n+1) = \\max(P(n), P(n-1)) = P(n)$. Поэтому найдётся число $n \\le m \\le n^2-n+1$ такое, что $P(m-1) \\le P(m) \\ge P(m+1)$; иначе $P(n-1) \\le P(n) < P(n+1) < \\dots < P(n^2-n+1)$, что невозможно.\n\nТеперь из $(*)$ имеем $P(m^2-m+1) = \\max(P(m), P(m-1)) = P(m)$ и $P(m^2 + m + 1) = \\max(P(m), P(m+1)) = P(m)$. Таким образом, тройка $m, m^2-m+1, m^2+m+1$ удовлетворяет условию, и $m > N$; противоречие с выбором числа $N$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22213,
"subject": "Mathematics (Olympiad)",
"question": "Эерэг $a, b, c$ тоонууд $abc = 1$ нөхцөлийг хангадаг.\n\n$$\n\\frac{1}{c(a^{2012} + b^{2012}) + 1} + \\frac{1}{b(a^{2012} + c^{2012}) + 1} + \\frac{1}{a(b^{2012} + c^{2012}) + 1} \\le 1\n$$\nбол тэнцэтгэл биш биелэнэ гэдгийг батал.",
"options": [],
"answer": "See solution",
"solution": "$$\n\\begin{gathered}\n\\blacktriangleright \\ \\frac{1}{c(a^{2012} + b^{2012}) + 1} \\le \\frac{c^r}{a^r + b^r + c^r} \\text{ байх } r\\text{-ийг олъё.} \\\\\n a^r + b^r \\le c^{r+1}(a^{2012} + b^{2012}) \\\\\n\\Leftrightarrow (ab)^{r+1}(a^r + b^r) \\le a^{2012} + b^{2012} \\\\\n\\Rightarrow 3r + 2 = 2012 \\ (r = 670) \\\\\n\\Leftrightarrow (ab)^{671}(a^{670} + b^{670}) \\le a^{2012} + b^{2012} \\\\\n\\Leftrightarrow (a^{1341} - b^{1341})(a^{671} - b^{671}) \\ge 0. \\blacktriangle\n\\end{gathered}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22214,
"subject": "Mathematics (Olympiad)",
"question": "Consider the number consisting of $n$ digits, each digit being $5$ (i.e., $\\underbrace{55\\dots5}_{n \\text{ times}}$). What is the set of possible values for the sum of the squares of its digits, if we are allowed to repeatedly replace any two fives by a three and a four?",
"options": [],
"answer": "See solution",
"solution": "The sum of the squares of the digits of the original number is $n \\cdot 5^2 = 25n$. Each time we replace two fives with a three and a four, the sum of the squares decreases by $5^2 + 5^2 - (3^2 + 4^2) = 50 - (9 + 16) = 25$. Thus, we can obtain any sum of the form $25n - 25m = 25(n - m)$, where $m$ is the number of such replacements, and $0 \\leq m \\leq \\lfloor n/2 \\rfloor$. Therefore, the possible sums are $25n, 25(n-1), 25(n-2), \\ldots, 25(n - \\lfloor n/2 \\rfloor)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22215,
"subject": "Mathematics (Olympiad)",
"question": "A convex polygon $M$ is given. There is a square $K$ that contains $M$ and has the minimum possible area. Is it necessary that at least one side of the square $K$ contains one of the sides of the polygon $M$?",
"options": [],
"answer": "See solution",
"solution": "It is not obligatory.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22216,
"subject": "Mathematics (Olympiad)",
"question": "There are 44 distinct holes in a line and 2017 ants. Each ant crawled up from a hole, then moved to another hole and crawled down. Denote $T$ as the set of time points that the ants crawled up or crawled down from some holes. Suppose that the speeds of the ants are pairwise distinct and they do not change their speed. Prove that if $|T| \\leq 45$, then there exist two ants that did not meet. Note: two ants meet if there exists a time point such that they are at the same location on the line, including the holes.",
"options": [],
"answer": "See solution",
"solution": "We call a time point \"special\" if at that time, some ants crawled up or down from a hole. It suffices to consider the case $|T| = 45$ (if $|T| < 45$, adding more special time points only strengthens the argument).\n\nConsider a coordinate system $Oxy$ where $Ox$ represents the location of holes on the line and $Oy$ represents time. Let the $x$-coordinates of the holes be $x_1, x_2, \\dots, x_{44}$ and the $y$-coordinates of the special times be $y_1, y_2, \\dots, y_{45}$. An ant moves from $(x_a, y_b)$ to $(x_c, y_d)$ if it crawled up from hole $x_a$ at time $y_b$ and crawled down into hole $x_c$ at time $y_d$. Since the speed of each ant does not change, the path of each ant is a segment connecting two such points, and since the speeds are pairwise distinct, the segments have different directions.\n\nThus, there are 2017 segments, but the number of endpoints is at most $45 \\times 44 = 1980 < 2017$. We now prove a general statement: If there are $n$ points in the plane, there cannot be more than $n$ segments connecting them such that no two segments are parallel or overlap. $(\\diamondsuit)$\n\n\n\nWe prove $(\\diamondsuit)$ by induction. It is easy to check for $n = 2, 3$.\n\nFor $n \\geq 4$, suppose the statement holds for $n-1$ points. Consider two cases:\n\n1. If among $n$ points, there is a point that is the endpoint of at most one segment, then by removing it (and its segment), we reduce to the case of $n-1$ points.\n\n2. If each point is the endpoint of at least two segments, then the number of segments equals the number of points. If a point is the endpoint of three segments (say, $A$ is the endpoint of $AB$, $AC$, $AD$), if $A$ lies inside triangle $BCD$, the second segment from $B$ cannot intersect both $AC$ and $AD$. Otherwise, if the ray $AC$ lies between $AB$ and $AD$, the second segment from $C$ cannot intersect both $AB$ and $AD$. Thus, each point is the endpoint of exactly two segments, so the number of segments is $n$.\n\nTherefore, $(\\diamondsuit)$ holds for any $n$, and the proof is complete.\n\n$\\Box$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22217,
"subject": "Mathematics (Olympiad)",
"question": "Captain Jack and his pirates robbed 6 boxes of gold coins $A_1, A_2, A_3, A_4, A_5, A_6$. There are $a_i$ coins in box $A_i$ ($i = 1, 2, 3, 4, 5, 6$) and $a_i \\neq a_j$ for $i \\neq j$. They laid the boxes as shown in the figure.\n\n\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. If Captain Jack gets more gold coins than the pirate, he wins. If Captain Jack goes first, what should his strategy be to win the game?",
"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\n\nCase 1:\n\n\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. They cannot choose the center box. After the first round, Captain Jack will have more coins than the pirate. Then, with only 2 boxes left, Captain Jack will choose the one with more coins and win.\n\n**Case 2**\n\nPaint 4 boxes black and white and arrange as shown in the figure. If there are more coins in black boxes than in white ones, Captain Jack takes a black box first, forcing the pirate to take a white one. After that, Captain Jack takes the other black box and wins.\n\n\n\nNow, consider the original problem.\n\nIf $a_6 \\geq a_5$, Captain Jack can take the available box with the most coins, then the pirate takes one. The problem reduces to the \"4 boxes\" case in Lemma 1.\n\nOtherwise, if $a_5 > a_6$, 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 whether $a_1 + a_3 + a_5 \\geq a_2 + a_4 + a_6$. If so, Captain Jack can take all the black boxes and win. If not, he can take $a_6$ first, then:\n\n\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": 22218,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\geq 2$ be an integer and let $(K, +, \\cdot)$ be a commutative field with the property:\n$$\n\\underbrace{1 + \\cdots + 1}_{m \\text{ times}} \\neq 0, \\quad m = 2, \\dots, n.\n$$\nConsider a polynomial $f \\in K[X]$ of degree $n$ and $G$ a subgroup of the additive group $(K, +)$, $G \\neq K$. Prove that there exists $a \\in K$ such that $f(a) \\notin G$.",
"options": [],
"answer": "See solution",
"solution": "Let $g \\in K[X]$ be a polynomial of degree $m \\in \\{2, \\dots, n\\}$. The polynomial\n$$\nh(X) = g(X + 1) - g(X)\n$$\nhas degree $m-1$, and if $\\mathrm{Im}\\, g \\subseteq G$, then $\\mathrm{Im}\\, h \\subseteq G$.\n\nSuppose $\\mathrm{Im}\\, f \\subseteq G$. Consider the polynomials $f_0, f_1, \\dots, f_{n-1}$, defined recursively by: $f_0 = f$ and $f_k(X) = f_{k-1}(X+1) - f_{k-1}(X)$ for $k = 1, \\dots, n-1$. By the previous remark, $\\deg f_k = n-k$ and $\\mathrm{Im}\\, f_k \\subseteq G$. In particular, $\\deg f_{n-1} = 1$ and $\\mathrm{Im}\\, f_{n-1} \\subseteq G$. Because the polynomial function $\\tilde{f}_{n-1}: K \\to K$ is surjective, it follows that $\\mathrm{Im}\\, f_{n-1} = K$, so $G = K$ – a contradiction.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 22219,
"subject": "Mathematics (Olympiad)",
"question": "Isosceles $\\triangle ABC$ ($AC = BC$) is inscribed in a circle $k$. A point $M$ lies on the side $BC$. A point $N$ from the ray $AM$ ($M$ lies between $A$ and $N$) is such that $AN = AC$. The circumcircle of $\\triangle MCN$ intersects $k$ at $C$ and $P$, where $P$ is from the arc $BC$ not containing $A$. The lines $AB$ and $CP$ intersect at $Q$. Prove that $\\angle QMB = \\angle QMN$.",
"options": [],
"answer": "See solution",
"solution": "Denote $\\angle MAC = \\alpha$ and $\\angle ACB = \\beta$. Let $J$ be the incenter of $\\triangle AMC$. It follows that $\\angle AJM = 90^\\circ + \\frac{\\beta}{2}$ and $\\angle ABM = 90^\\circ - \\frac{\\beta}{2}$, which implies that the quadrilateral $ABMJ$ is inscribed in a circle $k_1$. Analogously, the quadrilateral $MNCJ$ is inscribed in a circle $k_2$, which in fact is the circumcircle of $\\triangle MCN$. Therefore, $P$ is a point of $k_2$. Since $QA \\cdot QB = QC \\cdot QP$, we conclude that $Q$ has the same power with respect to $k_1$ and $k_2$. Hence, $Q$ lies on $JM$, which is the angle bisector of $\\angle BNC$. This implies the desired equality $\\angle QMB = \\angle QMN$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22220,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\omega$ be the circumcircle of $ABC$ where $AB \\ne AC$, and let $M$ be the midpoint of side $BC$. Tangent lines drawn at points $B$ and $C$ of circle $\\omega$ intersect at point $T$. The circumcircle of triangle $AMT$ intersects line $BC$ again at point $N$. Let $S$ be the midpoint of $NT$. Prove that $SA$ is tangent to the circle $\\omega$.",
"options": [],
"answer": "See solution",
"solution": "Since $\\triangle BTC$ is isosceles, $TM$ is an altitude.\n\n$$\n\\angle CMT = \\angle NMT = 90^{\\circ}.\n$$\n\nThus, $S$ is the circumcenter of triangle $AMT$, making $SA = ST$ and $\\angle SAT = \\angle STA$. Considering $AT$ as the $A$-symmedian of $ABC$, we know $\\angle BAM = \\angle TAC$.\n\n$$\n\\begin{align*}\n\\angle SAT &= \\angle ATN = \\angle AMN \\\\\n &= \\angle ABM + \\angle BAM \\\\\n &= \\angle TAC + \\angle ABM\n\\end{align*}\n$$\n\nSince $\\angle SAT = \\angle TAC + \\angle ABM$, it follows that $\\angle CAS = \\angle BAM = \\angle ABC$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22221,
"subject": "Mathematics (Olympiad)",
"question": "Let $S = \\{ n \\mid n-1,\\ n,\\ n+1 \\text{ all can be expressed as the sum of the squares of two positive integers} \\}$. Prove that, if $n \\in S$, then $n^2 \\in S$.",
"options": [],
"answer": "See solution",
"solution": "Note that if $x$ and $y$ are integers, then\n$$\nx^2 + y^2 \\equiv 0,\\ 1,\\ 2 \\pmod{4}.\n$$\nLet $n \\in S$. By the above, $n \\equiv 1 \\pmod{4}$. Thus, we may assume that\n$$\nn - 1 = a^2 + b^2,\\ a \\geq b,\n$$\n$$\nn = c^2 + d^2,\\ c > d,\n$$\n$$\nn + 1 = e^2 + f^2,\\ e \\geq f,\n$$\nwhere $a, b, c, d, e, f$ are positive integers. Therefore,\n$$\nn^2 + 1 = n^2 + 1^2,\n$$\n$$\nn^2 = (c^2 + d^2)^2 = (c^2 - d^2)^2 + (2cd)^2,\n$$\n$$\nn^2 - 1 = (a^2 + b^2)(e^2 + f^2) = (ae - bf)^2 + (af + be)^2.\n$$\nSuppose $b = a$ and $f = e$, then $n - 1 = 2a^2$, $n + 1 = 2e^2$. Taking the difference yields $e^2 - a^2 = 1$. Then $e - a \\geq 1$, but\n$$\n1 = e^2 - a^2 = (e + a)(e - a) > 1,\n$$\na contradiction!\n\nSo $b = a$ and $f = e$ do not happen simultaneously. Therefore $ae - bf > 0$, and $n^2 \\in S$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22222,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $\\frac{n+3}{13} = k$. What is the remainder when $n$ is divided by $13$?",
"options": [],
"answer": "See solution",
"solution": "We have $\\frac{n+3}{13} = k$, so $n + 3 = 13k$, which gives $n = 13k - 3$. Alternatively, $n = 13(k-1) + 10$. This shows that $n$ leaves remainder $10$ when divided by $13$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22223,
"subject": "Mathematics (Olympiad)",
"question": "令 $a_i > 0$，$i = 1, 2, \\dots, n$，且 $\\sum_{i=1}^{n} a_i = 1$。\n\n試證：對任意正整數 $k$，\n\n$$\n(a_1^k + \\frac{1}{a_1^k})(a_2^k + \\frac{1}{a_2^k}) \\cdots (a_n^k + \\frac{1}{a_n^k}) \\ge (n^k + \\frac{1}{n^k})^n.\n$$",
"options": [],
"answer": "See solution",
"solution": "證明：\n\n對每個 $a_i$，有\n\n$$\n\\begin{aligned}\na_i^k + \\frac{1}{a_i^k} \\ge (n^k + \\frac{1}{n^k}) \\left(\\frac{1}{n a_i}\\right)^{\\frac{k(n^{2k}-1)}{n^{2k}+1}}\n\\end{aligned}\n$$\n\n將 $n$ 個不等式相乘，得\n\n$$\n\\begin{aligned}\n& (a_1^k + \\frac{1}{a_1^k})(a_2^k + \\frac{1}{a_2^k}) \\cdots (a_n^k + \\frac{1}{a_n^k}) \\\\\n& \\ge (n^k + \\frac{1}{n^k})^n \\left(\\frac{1}{n^n a_1 \\cdots a_n}\\right)^{\\frac{k(n^{2k}-1)}{n^{2k}+1}}\n\\end{aligned}\n$$\n\n由於 $a_1 \\cdots a_n \\le \\left(\\frac{1}{n}\\right)^n$，所以\n\n$$\n\\frac{1}{n^n a_1 \\cdots a_n} \\ge 1\n$$\n\n且 $\\frac{k(n^{2k} - 1)}{n^{2k} + 1} > 0$，因此\n\n$$\n\\left(\\frac{1}{n^n a_1 \\cdots a_n}\\right)^{\\frac{k(n^{2k}-1)}{n^{2k}+1}} \\ge 1\n$$\n\n故\n\n$$\n(a_1^k + \\frac{1}{a_1^k})(a_2^k + \\frac{1}{a_2^k}) \\cdots (a_n^k + \\frac{1}{a_n^k}) \\ge (n^k + \\frac{1}{n^k})^n\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22224,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be the lengths of the sides of a triangle, $R$ the radius of its circumcircle, and $r$ the radius of its incircle. Prove that\n\n$$\n\\frac{Rr}{(a+b+c)^2} \\le \\frac{1}{54}.\n$$",
"options": [],
"answer": "See solution",
"solution": "We use the identities $2S = (a+b+c)r$ and $\\frac{abc}{4S} = R$, where $S$ denotes the area of the triangle. Multiplying them, we obtain\n\n$$\n\\frac{abc}{2} = Rr(a+b+c).\n$$\n\nDividing both sides by $(a+b+c)^3$ gives\n\n$$\n\\frac{abc}{2(a+b+c)^3} = \\frac{Rr}{(a+b+c)^2}.\n$$\n\nBy the AM-GM inequality, $(a+b+c)^3 \\geq 27abc$, so\n\n$$\n\\frac{abc}{(a+b+c)^3} \\leq \\frac{1}{27}.\n$$\n\nTherefore,\n\n$$\n\\frac{Rr}{(a+b+c)^2} \\leq \\frac{1}{54}.\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22225,
"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 $a, b, c, d > 0$。\n\n將原式整理後得到：\n\n$$\na^{12}(b+c)^3(b+c+d)^4 = 1\n$$\n\n$$\nb^{12}(c+d)^3(c+d+a)^4 = 1\n$$\n\n$$\nc^{12}(d+a)^3(d+a+b)^4 = 1\n$$\n\n$$\nd^{12}(a+b)^3(a+b+c)^4 = 1\n$$\n\n考慮 $x$ 為 $\\{\\frac{x}{y} \\mid x, y \\in \\{a, b, c, d\\}\\}$ 中的最大值，並且不妨設 $\\frac{a}{b} = x \\ge 1$。則\n\n$$\n\\frac{c+d}{b+c} \\le x\n$$\n\n$$\n\\frac{c+d+a}{b+c+d} \\le x\n$$\n\n因此\n\n$$\n1 = \\frac{a^{12}(b+c)^3(b+c+d)^4}{b^{12}(c+d)^3(c+d+a)^4} \\ge x^{12} \\frac{1}{x^3} \\frac{1}{x^4} = x^7\n$$\n\n因此 $x = 1$，即 $a = b = c = d$。\n\n代入原式解得 $(a, b, c, d) = (0, 0, 0, 0)$ 或 $\\left(\\frac{1}{\\sqrt{19/648}}, \\frac{1}{\\sqrt{19/648}}, \\frac{1}{\\sqrt{19/648}}, \\frac{1}{\\sqrt{19/648}}\\right)$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22226,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ be the largest value of the expression $24y - 9y^2$, where $y$ is a rational number, and $b$ be the smallest integer satisfying the inequality\n$$\n(t+3)^3 - (6t-7)^2 - (t-9)^3 < 3.\n$$\n\nFactor into irreducible factors with integer coefficients the expression\n$$\na(x - 1)x^3 + bx - 2x - 1.\n$$",
"options": [],
"answer": "See solution",
"solution": "We have $24y - 9y^2 = 16 - (3y - 4)^2$, whose largest value $a = 16$ is reached for $y = \\frac{4}{3}$. The given inequality is equivalent to\n$$\nt^3 + 9t^2 + 27t + 27 - 36t^2 + 84t - 49 - t^3 + 27t^2 - 243t + 729 < 3\n$$\nwhich simplifies to\n$$\n-132t + 704 < 0,\n$$\ni.e., $t > \\frac{16}{3}$ and $b = 6$. Substituting $a = 16$, $b = 6$ in the given expression, we get\n$$\n16(x - 1)x^3 + 6x - 2x - 1 = 16x^4 - 16x^3 + 4x - 1\n$$\nNow,\n$$\n16x^4 - 16x^3 + 4x - 1 = (16x^4 - 1) - 4x(4x^2 - 1) = (4x^2 - 1)(4x^2 + 1) - 4x(4x^2 - 1)\n$$\n$$\n= (4x^2 - 4x + 1)(2x - 1)(2x + 1) = (2x - 1)^3(2x + 1).\n$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22227,
"subject": "Mathematics (Olympiad)",
"question": "Let $f : [0, 1] \\to \\mathbb{R}$ be a monotone increasing differentiable function, with a continuous derivative, such that $f(0) = 0$. Let $g : [0, 1] \\to \\mathbb{R}$ be the function defined by\n\n$$\ng(x) = f(x) + (x-1)f'(x), \\quad \\text{for any } x \\in [0, 1].\n$$\n\na) Show that\n\n$$\n\\int_{0}^{1} g(x) \\, dx = 0.\n$$\n\nb) Prove that for any convex and differentiable function $\\varphi : [0, 1] \\to [0, 1]$, such that $\\varphi(0) = 0$ and $\\varphi(1) = 1$, the following inequality holds:\n\n$$\n\\int_{0}^{1} g(\\varphi(x)) \\, dx \\le 0.\n$$",
"options": [],
"answer": "See solution",
"solution": "a) Because $g(x) = f(x) + (x-1)f'(x) = ((x-1)f(x))'$, it follows that\n\n$$\n\\int_{0}^{1} g(x) \\, dx = (x-1)f(x)\\Big|_{0}^{1} = 0.\n$$\n\nb) Let $\\varphi : [0, 1] \\to [0, 1]$ be a convex and differentiable function with $\\varphi(0) = 0$ and $\\varphi(1) = 1$. The function $\\varphi'$ is then non-decreasing. The function $f$ being monotone increasing and differentiable, its derivative $f'$ is nonnegative. Hence, for any $x \\in [0, 1]$ we have\n\n$$\nf(\\varphi(x)) = f(\\varphi(x)) - f(\\varphi(0)) = \\int_{0}^{x} f'(\\varphi(t)) \\cdot \\varphi'(t) \\, dt \\le \\varphi'(x) \\cdot \\int_{0}^{x} f'(\\varphi(t)) \\, dt.\n$$\n\nIntegrating the previous inequality, we obtain\n\n$$\n\\int_{0}^{1} f(\\varphi(x)) \\, dx \\le \\int_{0}^{1} \\left( \\varphi'(x) \\cdot \\int_{0}^{x} f'(\\varphi(t)) \\, dt \\right) \\, dx. \\quad (3)\n$$\n\nThe function $f' \\circ \\varphi$ is continuous, being a composite of continuous functions, and so is primitivable and integrable, and a primitive of it is the function $\\Phi : [0, 1] \\to \\mathbb{R}$ defined by $\\Phi(x) = \\int_{0}^{x} f'(\\varphi(t)) dt$. Partially integrating, we have:\n\n$$\n\\begin{align*}\n\\int_{0}^{1} \\left( \\varphi'(x) \\cdot \\int_{0}^{x} f'(\\varphi(t)) \\, dt \\right) \\, dx &= \\int_{0}^{1} (\\varphi'(x) \\cdot \\Phi(x)) \\, dx \\\\\n&= (\\varphi(x) \\cdot \\Phi(x))\\Big|_{0}^{1} - \\int_{0}^{1} (\\varphi(x) \\cdot \\Phi'(x)) \\, dx \\\\\n&= \\varphi(1) \\cdot \\Phi(1) - \\varphi(0) \\cdot \\Phi(0) - \\int_{0}^{1} \\varphi(x) \\cdot f'(\\varphi(x)) \\, dx \\\\\n&= \\int_{0}^{1} f'(\\varphi(x)) \\, dx - \\int_{0}^{1} \\varphi(x) \\cdot f'(\\varphi(x)) \\, dx. \\tag{4}\n\\end{align*}\n$$\n\nFrom the definition of the function $g$, the inequality (3) and the identity (4) it follows that\n\n$$\n\\int_{0}^{1} g(\\varphi(x)) \\, dx = \\int_{0}^{1} f(\\varphi(x)) \\, dx + \\int_{0}^{1} \\varphi(x) \\cdot f'(\\varphi(x)) \\, dx - \\int_{0}^{1} f'(\\varphi(x)) \\, dx \\le 0.\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22228,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a trapezoid with an obtuse angle at vertex $D$. The diagonals $AC$ and $BD$ meet at $O$, and the line through $O$ parallel to $AB$ meets the circumcircle of $BCO$ again at $P$. The circumcircles of $ADO$ and $BCO$ intersect again at $Q$. Prove that the line $PQ$ bisects the segment $BC$.",
"options": [],
"answer": "See solution",
"solution": "Set $\\angle BAC = \\alpha$, $\\angle OBA = \\beta$, and denote by $M$ the intersection of $PQ$ and $BC$. We show that $CM = MB$.\n\n\n\nSince $OP \\parallel AB$ and $P$, $C$, $O$, $Q$, $B$ are cyclic, we have $\\angle POC = \\angle PQC = \\angle PBC = \\alpha$ and $\\angle POB = \\angle PCB = \\angle PQB$. It follows that $\\triangle OAB \\sim \\triangle BPC$. Hence\n\n$$\n\\frac{BP}{PC} = \\frac{OA}{OB}.\n$$\n\nBecause $ABCD$ is a trapezoid, we have\n\n$$\n\\frac{OC}{OA} = \\frac{OD}{OB} \\iff 1 + \\frac{OC}{OA} = \\frac{OC}{OB} + 1 \\iff \\frac{AC}{BD} = \\frac{OA}{OB}\n$$\n\nFrom this, we deduce that\n\n$$\n\\frac{BP}{PC} = \\frac{AC}{BD}.\n$$\n\nOn the other hand, we have $\\frac{QC}{QB} = \\frac{AC}{DB}$ since $\\triangle AQC \\sim \\triangle DQB$. Consequently,\n\n$$\n\\frac{QC}{QB} = \\frac{BP}{PC}.\n$$\n\nNow by the law of sines,\n\n$$\n\\frac{CM}{MB} = \\frac{CM}{MP} \\cdot \\frac{MP}{MB} = \\frac{\\sin \\angle CPM}{\\sin \\angle MCP} \\cdot \\frac{\\sin \\angle CBP}{\\sin \\angle MPB} = \\frac{CQ}{PB} \\cdot \\frac{CP}{BQ} = 1.\n$$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 22229,
"subject": "Mathematics (Olympiad)",
"question": "A tank is being filled. In 8 hours, the tank fills up by a further 64%, which is 8% per hour. If after the first two hours it has filled up 16%, what percentage of the tank was already filled at the start, given that after the first two hours the tank is 36% full?",
"options": [],
"answer": "See solution",
"solution": "Since the tank fills at 8% per hour, in two hours it fills $2 \\times 8\\% = 16\\%$. If after two hours the tank is $36\\%$ full, then the initial amount was $36\\% - 16\\% = 20\\%$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22230,
"subject": "Mathematics (Olympiad)",
"question": "There are numbers $1, 2, 3, \\ldots, 100$ on the board. Each is written exactly once. Petryk and Ivasyk are playing the following game (Petryk starts): each player erases one number from the board. If after a player's turn the sum of all erased numbers (by both players) cannot be represented as a difference of squares of integers, the player loses. Who will win in this game if both players want to win?",
"options": [],
"answer": "See solution",
"solution": "Notice that $n$ cannot be represented as a difference of squares of integers if and only if $n \\equiv 2 \\pmod{4}$. Indeed, if $n \\equiv 2 \\pmod{4}$, then suppose $n = (x + y)(x - y)$. If $x, y$ are both odd or both even, then $n \\equiv 0 \\pmod{4}$; otherwise, $n \\equiv \\pm 1 \\pmod{4}$. If $n = 4k$, then let $x = k + 1$, $y = k - 1$; if $n \\equiv \\pm 1 \\pmod{4}$, then let $x = \\frac{n+1}{2}$, $y = \\frac{n-1}{2}$.\n\nThe strategy for Petryk is as follows: he erases $100$ first. Then, if Ivasyk chooses $m$, Petryk erases $100 - m$. Since choosing the number $50$ leads to a loss, Petryk wins, because if Ivasyk hasn't lost yet, then Petryk always has a number to erase that will not make him lose.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22231,
"subject": "Mathematics (Olympiad)",
"question": "Let $S = \\{2, 3, 4, \\dots\\}$ denote the set of integers greater than or equal to $2$. Does there exist a function $f: S \\to S$ such that\n\n$$\nf(a)f(b) = f(a^2b^2) \\text{ for all } a, b \\in S \\text{ with } a \\neq b?\n$$",
"options": [],
"answer": "See solution",
"solution": "Assume such a function exists. For any positive integer $n$, we have\n\n$$\nf(2^n)f(2^{n+4}) = f(2^{2(2^{n+4})}) = f(2^{n+1})f(2^{n+3}),\n$$\n\nso\n\n$$\n\\frac{f(2^n)}{f(2^{n+1})} = \\frac{f(2^{n+3})}{f(2^{n+4})}. \\qquad (1)\n$$\n\nSimilarly,\n\n$$\nf(2^{n+1})f(2^{n+4}) = f(2^{2(2^{n+5})}) = f(2^{n+2})f(2^{n+3}),\n$$\n\nso\n\n$$\n\\frac{f(2^{n+1})}{f(2^{n+2})} = \\frac{f(2^{n+3})}{f(2^{n+4})}. \\qquad (2)\n$$\n\nCombining (1) and (2),\n\n$$\n\\frac{f(2^n)}{f(2^{n+1})} = \\frac{f(2^{n+1})}{f(2^{n+2})}. \\qquad (3)\n$$\n\nSince (3) holds for all $n$, $f(2^1), f(2^2), f(2^3), \\dots$ is a geometric sequence, so $f(2^n) = ar^{n-1}$ for some positive rational $a$ and $r$.\n\nSince $f(2)f(2^2) = f(2^6)$,\n\n$$\n\\begin{aligned}\na \\cdot ar &= ar^5 \\\\\n\\Rightarrow \\quad a &= r^4.\n\\end{aligned} \\qquad (4)\n$$\n\nAnd since $f(2)f(2^3) = f(2^8)$,\n\n$$\n\\begin{aligned}\na \\cdot ar^2 &= ar^7 \\\\\n\\Rightarrow \\quad a &= r^5.\n\\end{aligned} \\qquad (5)\n$$\n\nComparing (4) and (5), $r^4 = r^5$, so $r = 1$. Then $a = 1$, so $f(2) = 1 \\notin S$, a contradiction. Hence, there are no such functions. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22232,
"subject": "Mathematics (Olympiad)",
"question": "The triangle _ABC_ has $\\angle ABC = 90^\\circ$ and $\\angle BCA = 30^\\circ$. Let $AD$ be the bisector of the angle $\\angle BAC$, $D \\in BC$, and $BE \\perp AC$, $E \\in AC$. Denote $M$ the intersection of the lines $AD$ and $BE$, and $P$ the midpoint of the segment $CM$. Prove that $AC = 4 \\cdot DP$.\n\n",
"options": [],
"answer": "See solution",
"solution": "The $30^\\circ$ angle theorem yields $AC = 2AB$. Since $\\angle CAD = \\angle ACD = 30^\\circ$, the triangle $ADC$ is isosceles, with $DA = DC$.\n\nFrom $\\angle ADB = \\angle MBD = 60^\\circ$ it follows that the triangle $MBD$ is equilateral. This shows that $MB = BD = \\frac{1}{2}AD$, hence $M$ is the midpoint of the segment $AD$.\n\nDenote $S$ the reflection of $D$ into $B$. Then the triangle $ADS$ is equilateral. This yields $DS = AD = CD$, hence $DP$ is a midline of the triangle $CSM$. From $AB = SM$ (medians in the equilateral triangle $ADS$) it follows that $DP = \\frac{1}{2}MS = \\frac{1}{2}AB = \\frac{1}{4}AC$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22233,
"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": "(1) 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, 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(2) 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 $4d$.\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": 22234,
"subject": "Mathematics (Olympiad)",
"question": "The squares of an $8 \\times 8$ board are coloured alternatingly black and white. A rectangle consisting of some of the squares of the board is called *important* if its sides are parallel to the sides of the board and all its corner squares are coloured black. The side lengths can be anything from 1 to 8 squares. On each of the 64 squares of the board, we write the number of important rectangles in which it is contained. The sum of the numbers on black squares is $B$, and the sum of the numbers on white squares is $W$. Determine the difference $B-W$.",
"options": [],
"answer": "See solution",
"solution": "In each important rectangle, the number of black squares is one more than the number of white squares. Hence, each important rectangle contributes $+1$ to the difference $B-W$. The value of $B-W$ is thus the same as the number of important rectangles on the board.\n\nLet us number the rows on the board $1, 2, \\ldots, 8$ from the top downwards and the columns $1, 2, \\ldots, 8$ from the left to the right. So $(1, 1)$ is the upper left square and $(8, 8)$ denotes the lower right square. Assume $(1, 1)$ is a black square. Then all $(i, j)$ with both $i$ and $j$ odd, as well as all those with both $i$ and $j$ even, are black squares. All other squares are white. By focusing only on the four odd-numbered rows and the four odd-numbered columns, we find that they determine $(4 + \\binom{4}{2})^2 = 100$ important rectangles. Similarly, the four even-numbered rows and the four even-numbered columns determine another $100$ important rectangles, giving a total of $200$ important rectangles on the board. It follows that $B-W = 200$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22235,
"subject": "Mathematics (Olympiad)",
"question": "As seen in the figure below, in acute triangle $\\triangle ABC$ with $AB > AC$, let $M$ and $N$ be two different points on $BC$ such that $\\angle BAM = \\angle CAN$. Let $O_1$ and $O_2$ be the circumcenters of $\\triangle ABC$ and $\\triangle AMN$, respectively. Prove that $O_1$, $O_2$, and $A$ are collinear.\n\n\n\n",
"options": [],
"answer": "See solution",
"solution": "Connect $AO_1$ and $AO_2$. Through $A$, draw a line perpendicular to $AO_1$ and let it intersect the extended line of $BC$ at point $P$. Then $AP$ is a tangent to the circle $\\odot O_1$, so $\\angle B = \\angle PAC$.\n\nSince $\\angle BAM = \\angle CAN$, we have\n\n$$\n\\angle AMP = \\angle B + \\angle BAM = \\angle PAC + \\angle CAN = \\angle PAN.\n$$\n\nThus, $AP$ is also a tangent to $\\odot O_2$, the circumcircle of $\\triangle AMN$. Therefore, $AP \\perp AO_2$, which means $O_1$, $O_2$, and $A$ are collinear. The proof is complete. $\\Box$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22236,
"subject": "Mathematics (Olympiad)",
"question": "Do there exist two real monic polynomials $P(x)$ and $Q(x)$ of degree 3, such that the roots of $P(Q(x))$ are nine pairwise distinct nonnegative integers that add up to 72? (In a monic polynomial of degree 3, the coefficient of $x^3$ is 1.)",
"options": [],
"answer": "See solution",
"solution": "Let $z_1, z_2, z_3$ be the three roots of the polynomial $P(x) = (x - z_1)(x - z_2)(x - z_3)$. Let $Q(x) = x^3 - s x^2 + t x - u$. Then the nine roots of\n\n$$\nP(Q(x)) = (Q(x) - z_1)(Q(x) - z_2)(Q(x) - z_3)\n$$\n\nare the roots $a_i, b_i, c_i$ of $Q(x) - z_i$, where $i$ runs from 1 to 3. Vieta's relations yield, for $1 \\leq i \\leq 3$, $a_i + b_i + c_i = s$ and $a_i b_i + b_i c_i + c_i a_i = t$.\n\nSince the sum of all nine roots is 72, this implies $s = 24$. Hence, we are looking for three disjoint triples of nonnegative integers that add up to 24 and for which the sum of squares is the same (since $a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)$). A little experimentation shows that choosing $t = 143$ works with $a_1 = 0, b_1 = 11, c_1 = 13$; $a_2 = 1, b_2 = 8, c_2 = 15$; and $a_3 = 3, b_3 = 5, c_3 = 16$. Hence, the two polynomials\n\n$$\nQ(x) = x^3 - 24x^2 + 143x\n$$\n\nand\n\n$$\nP(x) = x(x - 120)(x - 240)\n$$\n\nhave all the desired properties, and the answer to the problem is YES.\n\n",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 22237,
"subject": "Mathematics (Olympiad)",
"question": "A 10-digit positive integer is called a *cute number* if its digits are from the set $\\{1, 2, 3\\}$ and every two consecutive digits differ by 1.\n\n(a) Prove that exactly 5 digits of a cute number are equal to 2.\n\n(b) Find the total number of cute numbers.\n\n(c) Prove that the sum of all cute numbers is divisible by 1408.",
"options": [],
"answer": "See solution",
"solution": "a) In the decimal representation of a cute number, the parity of the digits clearly alternates, hence exactly 5 digits are even, that is, equal to 2.\n\nb) There are $2^5 = 32$ cute numbers of the form $\\overline{2a2b2c2d2e}$ and another 32 of the form $\\overline{a2b2c2d2e2}$, therefore the requested number is 64.\n\nc) If $\\overline{a_1a_2\\dots a_{10}}$ is a cute number, then\n\n$$\n\\overline{(4 - a_1)(4 - a_2)\\dots(4 - a_{10})}\n$$\n\nis also cute and distinct from the previous one. They add up to $4444444444$. Grouping the cute numbers in 32 such pairs, we obtain that their sum equals $32 \\cdot 4444444444 = 2^7 \\cdot 11 \\cdot 101010101 = 1408 \\cdot 101010101$, hence the conclusion.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22238,
"subject": "Mathematics (Olympiad)",
"question": "Solve the following system of equations:\n\n$$\n\\begin{cases}\n x - y + z - w = 2, \\\\\n x^2 - y^2 + z^2 - w^2 = 6, \\\\\n x^3 - y^3 + z^3 - w^3 = 20, \\\\\n x^4 - y^4 + z^4 - w^4 = 66.\n\\end{cases}\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $p = x + z$, $q = xz$. The second to fourth equations become:\n\n$$\n\\begin{aligned}\n p^2 &= x^2 + z^2 + 2q, \\\\\n p^3 &= x^3 + z^3 + 3pq, \\\\\n p^4 &= x^4 + z^4 + 4p^2q - 2q^2.\n\\end{aligned}\n$$\n\nSimilarly, let $s = y + w$, $t = yw$. The second to fourth equations become:\n\n$$s^2 = y^2 + w^2 + 2t$$\n$$s^3 = y^3 + w^3 + 3st$$\n$$s^4 = y^4 + w^4 + 4s^2t - 2t^2$$\n\nAlso, the first equation can be expressed as:\n\n$$p = s + 2$$\n\nTherefore:\n\n$$p^2 = s^2 + 4s + 4$$\n$$p^3 = s^3 + 6s^2 + 12s + 8$$\n$$p^4 = s^4 + 8s^3 + 24s + 32s + 16$$\n\nSubstituting these expressions into the original system, we get:\n\n$$x^2 + z^2 + 2q = y^2 + w^2 + 2t + 4s + 4$$\n$$x^3 + z^3 + 3pq = y^3 + w^3 + 3st + 6s^2 + 12s + 8$$\n$$x^4 + z^4 + 4p^2q - 2q^2 = y^4 + w^4 + 4s^2t - 2t^2 + 8s^3 + 24s + 32s + 16$$\n\nUsing the second to fourth equations to simplify, we get:\n\n$$q = t + 2s - 1$$\n$$pq = st + 2s^2 + 4s - 4$$\n$$2p^2q - q^2 = 2s^2t - t^2 + 4s^3 + 12s^2 + 16s - 25$$\n\nSubstituting $p = s + 2$ and $q = t + 2s - 1$ into $pq = st + 2s^2 + 4s - 4$, we get:\n\n$$t = \\frac{s}{2} - 1$$\n\nSubstituting this value of $t$ into the previous equations...",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22239,
"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 $OP = OQ$.\n\nThere are three configurations, depending on the range of $AP/PC$ and $AQ/QB$. Two of them are shown below. Our proof uses directed angles modulo $180^\\circ$ and works for all configurations.",
"options": [],
"answer": "See solution",
"solution": "The key fact for both this solution and the next is the similarity between triangles *APQ* and *MKL*. Because $PQ$ is tangent to $\\Gamma$ at $M$, we have $\\angle PML = \\angle MKL$. Note that $ML$ is the midline of triangle $QPC$. Thus, $ML \\parallel PC$ and $\\angle QPA = \\angle PML$. It follows that $\\angle MKL = \\angle PML = \\angle QPA$. Likewise, $\\angle AQP = \\angle KLM$. Therefore, triangles *APQ* and *MKL* are similar.\n\nBecause triangles *APQ* and *MKL* are similar, we have\n\n$$\n\\frac{AP}{MK} = \\frac{AQ}{ML} \\quad \\text{or} \\quad AP \\cdot ML = AQ \\cdot MK.\n$$\n\nNote that $MK$ is the midline of triangle $BPQ$. In particular, $QB = 2MK$. Likewise, $PC = 2ML$. We conclude that\n\n$$\nAP \\cdot PC = 2AP \\cdot ML = 2AQ \\cdot MK = AQ \\cdot QB;\n$$\n\nthat is, points $P$ and $Q$ have equal power with respect to the circumcircle of triangle $ABC$. Because the power of a point $X$ to a circle centered at $Y$ with radius $R$ is equal to $XY^2 - R^2$, we conclude that $OP^2 = OQ^2$, hence $OP = OQ$.",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 22240,
"subject": "Mathematics (Olympiad)",
"question": "有 $n$ 對夫婦參加一場王國的盛宴，丈夫們坐一圓桌，妻子們坐另一圓桌。國王與皇后（不包含在這 $n$ 對夫婦中）要與他們握手。假設國王從某位男士開始握手，而皇后從該男士的妻子開始握手。考慮兩種方式：\n\n1. 國王順時鐘依序和所有男士握手。當國王與某位男士握手時，皇后順時鐘移動到該男士的妻子處，與她握手。假設國王回到最初的那位男士處時，皇后繞了圓桌 $a$ 圈。\n\n2. 皇后順時鐘依序和所有女士握手。當皇后與某位女士握手時，國王順時鐘移動到該女士的丈夫處，與他握手。假設皇后回到最初的那位女士處時，國王繞了圓桌 $b$ 圈。\n\n試問：$|a-b|$ 的最大可能值。",
"options": [],
"answer": "See solution",
"solution": "若 $n = m^2 - d$, $0 \\le d < m$，則最大值為 $n - 2m$；若 $n = m^2 - m - d$, $0 \\le d < m - 1$，則最大值為 $n - 2m + 1$。\n\n估計：不失一般性可以假設 $a > b$，令男女主人的出發點為兩桌的基準點，依序順時針編號 $0 \\sim n-1$，並令 $i$ 號男士的妻子編號為 $x_i$，考慮兩種方式中男女主人繞過 $0$ 號時的情況會得到：\n\n$$\na - 1 = |\\{i \\in [n-1] \\mid x_i > x_{i+1}\\}|\n$$\n\n$$\nb - 1 = |\\{i \\in [n-1] \\mid x_j = i, x_l = i+1, l < j\\}|\n$$\n\n（其中 $[n-1] := \\{1, 2, \\dots, n-1\\}$）\n\n令 $x_1, \\dots, x_{r_1}; x_{r_1+1}, \\dots, x_{r_2}; \\dots; x_{r_{k-1}+1}, \\dots, x_{r_k}$ 為 $k$ 個遞減數列且 $x_{r_t} < x_{r_{t+1}}$，則有 $a-1 = n-k$。考慮將 $[n-1]$ 當成點依序排列，若 $x_j = i, x_l = i+1, l > j$ 則將 $j$ 跟 $l$ 連線，這張圖會形成一些鍊（每個點向左向右至多各連一條線），而鍊的最左端就是滿足 $x_j = i, x_l = i+1, l < j$ 的 $l$，然而鍊的個數大於等於 $r_{t+1} - r_t$（對於所有 $t=1, 2, \\dots, k$），因為 $r_t + 1, \\dots, r_{t+1}$ 之間互不連線，也就得到了\n\n$$\nb-1 \\ge \\left\\lfloor \\frac{n-1}{k} \\right\\rfloor\n$$\n\n進而得到：\n\n$$\na-b \\le n-k - \\left\\lfloor \\frac{n-1}{k} \\right\\rfloor\n$$\n\n若 $n = m^2 - d$, $0 \\le d < m$，則 $a - b \\le n - 2m$。\n\n若 $n = m^2 - m - d$, $0 \\le d < m - 1$，則 $a - b \\le n - 2m + 1$。\n\n**構造：**\n\n- 若 $n = m^2 - d$, $0 \\le d < m$，則令 $\\{x_{r_{t-1}+1}, x_{r_{t-1}+2}, \\dots, x_{r_t}\\}$ 為 $[n-1]$ 中模 $m$ 同餘 $t$ 的數所形成的遞減數列，則此時 $k = m$，且鍊數也是 $m$（鍊頭分別是 $1, m+1, \\dots, m^2-2m+1, m^2-m+1$），所以，此時 $|a-b| = n-2m$。\n\n- 若 $n = m^2 - m - d$, $0 \\le d < m-1$，則令 $\\{x_{r_{t-1}+1}, x_{r_{t-1}+2}, \\dots, x_{r_t}\\}$ 為 $[n-1]$ 中模 $m$ 同餘 $t$ 的數所形成的遞減數列，則此時 $k = m$，且鍊數是 $m-1$（鍊頭分別是 $1, m+1, \\dots, m^2-2m+1$），所以，此時 $|a-b| = n - 2m + 1$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22241,
"subject": "Mathematics (Olympiad)",
"question": "If $k$ and $n$ are positive integers, and $k \\le n$, let $M(n, k)$ denote the least common multiple of the numbers $n, n-1, \\dots, n-k+1$. Let $f(n)$ be the largest positive integer $k \\le n$ such that $M(n, 1) < M(n, 2) < \\dots < M(n, k)$. Prove that:\n\na) $f(n) < 3\\sqrt{n}$, for all positive integers $n$;\n\nb) if $N$ is a positive integer, then $f(n) > N$ for all but finitely many positive integers $n$.",
"options": [],
"answer": "See solution",
"solution": "a) Clearly, $f(1) = 1$. Notice that\n\n$$\nM(n, k + 1) = \\operatorname{lcm}(M(n, k), n - k), \\quad 1 \\le k < n. \\quad (*)\n$$\n\nThus, $M(n, k) \\le M(n, k+1)$, and equality holds if and only if $n-k$ divides $M(n, k)$. If $m > 1$, then $M(m^2, 2) = m^2(m^2 - 1) = (m^2 - m)(m^2 + m)$, so $M(m^2, 2)$ is divisible by $m^2 - m$, and hence so is $M(m^2, m)$. By $(*)$, $M(m^2, m) = M(m^2, m+1)$, so $f(m^2) \\le m$ for all positive integers $m$. With reference again to $(*)$, if $M(n, k) = M(n, k+1)$, then $M(n + \\ell, k + \\ell) = M(n + \\ell, k + \\ell + 1)$, so $f(n + \\ell) \\le f(n) + \\ell$ for all positive integers $\\ell$ and $n$. Finally, if $m^2 \\le n < (m+1)^2$, then $n = m^2 + \\ell$ for some non-negative integer $\\ell \\le 2m$, so $f(n) = f(m^2 + \\ell) \\le f(m^2) + \\ell \\le m + \\ell \\le 3m \\le 3\\sqrt{n}$, where one of the last two inequalities must in fact be a strict inequality.\n\nb) We show that $f(n) > N$ for all integers $n > N! + N$. Refer again to $(*)$ to write\n\n$$\n\\begin{aligned}\nM(n, N + 1) &= \\operatorname{lcm}(M(n, N), n - N) = \\frac{M(n, N) \\cdot (n - N)}{\\operatorname{gcd}(M(n, N), n - N)} \\\\\n&\\ge \\frac{M(n, N) \\cdot (n - N)}{\\prod_{k=1}^{N} \\operatorname{gcd}(n - k + 1, n - N)} \\ge \\frac{M(n, N) \\cdot (n - N)}{\\prod_{k=1}^{N} (N - k + 1)} \\\\\n&= \\frac{M(n, N) \\cdot (n - N)}{N!}.\n\\end{aligned}\n$$\n\nConsequently, if $n > N! + N$, then $M(n, 1) < \\dots < M(n, N) < M(n, N + 1)$, so $f(n) > N$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22242,
"subject": "Mathematics (Olympiad)",
"question": "Out of an equilateral triangle with side $2014$, an equilateral triangle with side $214$ is cut out, such that the two triangles have one vertex in common and two of the sides of the cut-out triangle lie on two of the sides of the initial one. Can this figure be covered by the figures shown below without overlap (rotation is allowed), if the triangles in the figures are equilateral with side $1$? Justify your answer!\n\n",
"options": [],
"answer": "See solution",
"solution": "First, we cut the given figure into equilateral triangles with side $1$. We label the triangles in the given figure by numbers from $1$ to $6$, as on the picture to the right. (In the first row, successively from $1$ to $6$, then the numbers repeat; in the second, we start from $5$, in the third from $3$, then from $1$, and the procedure repeats.) It can easily be noticed that each of the given figures covers exactly one of the numbers $1$ to $6$. Therefore, in order for the figure to be coverable by the given figures, each of the numbers has to appear an equal number of times. If we compare how often the number $1$ and the number $2$ appear, we will notice that in the first, fourth, and each row of the form $3k+1$, there is one more $1$ than $2$'s, and in the remaining rows the number of $1$'s and $2$'s is equal. Therefore, it follows that the number of $1$'s and $2$'s is unequal, and therefore not every number can appear an equal number of times. It follows that the figure cannot be covered in the required way.\n\n",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 22243,
"subject": "Mathematics (Olympiad)",
"question": "Нека $a > 0$, $b > 0$, $c > 0$ и $a + b + c = 1$. Докажете неравенството\n\n$$\n\\frac{a^4 + b^4}{a^2 + b^2} + \\frac{b^3 + c^3}{b + c} + \\frac{2a^2 + b^2 + 2c^2}{2} \\geq \\frac{1}{2}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Од $1 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac)$ добиваме\n\n$$\nab + bc + ca + \\frac{a^2 + b^2 + c^2}{2} = \\frac{1}{2}.\n$$\n\nКористиме неравенството $(a^{n-1} - b^{n-1})(a - b) \\geq 0$ за $n \\geq 1$, при што равенство е исполнето кога $a = b$. Последното неравенство е еквивалентно со $a^n + b^n \\geq ab(a^{n-2} + b^{n-2})$ кога $n \\geq 2$. Во конкретниот случај имаме: $\\frac{a^4 + b^4}{a^2 + b^2} \\geq ab$, $\\frac{b^3 + c^3}{b + c} \\geq bc$ и $\\frac{c^2 + a^2}{2} \\geq ca$. Со собирање на последните три неравенства добиваме\n\n$$\n\\frac{a^4 + b^4}{a^2 + b^2} + \\frac{b^3 + c^3}{b + c} + \\frac{c^2 + a^2}{2} \\geq ab + bc + ac\n$$\n\nод каде\n\n$$\n\\frac{a^4 + b^4}{a^2 + b^2} + \\frac{b^3 + c^3}{b + c} + c^2 + a^2 + \\frac{b^2}{2} \\geq ab + bc + ac + \\frac{c^2 + a^2 + b^2}{2} = \\frac{1}{2}.\n$$\n\nЈасно е дека знак равенство се постигнува кога $a = b = c = \\frac{1}{3}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22244,
"subject": "Mathematics (Olympiad)",
"question": "Let $k_1$ and $k_2$ be internally tangent circles with common point $X$. Let $P$ be a point lying neither on one of the two circles nor on the line through the two centers. Let $N_1$ be the point on $k_1$ closest to $P$ and $F_1$ be the point on $k_1$ that is farthest from $P$. Analogously, let $N_2$ be the point on $k_2$ closest to $P$ and $F_2$ be the point on $k_2$ that is farthest from $P$.\n\nProve that $\\angle N_1 X N_2 = \\angle F_1 X F_2$.",
"options": [],
"answer": "See solution",
"solution": "The line segment $N_1F_1$ is a diameter of $k_1$ passing through $P$. Similarly, $N_2F_2$ is a diameter of $k_2$ passing through $P$.\n\nDue to Thales's theorem, we have $\\angle N_1 X F_1 = 90^\\circ$ and $\\angle N_2 X F_2 = 90^\\circ$.\n\nLet $\\angle N_2 X F_1 = \\alpha$, we obtain\n\n$$\n\\angle N_1 X N_2 = 90^\\circ - \\alpha \\quad \\text{and} \\quad \\angle F_1 X F_2 = 90^\\circ - \\alpha,\n$$\n\nwhich proves the equality of the angles.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22245,
"subject": "Mathematics (Olympiad)",
"question": "Let $a = f(0, 0, 0, \\dots, 0)$, $b = f(1, 0, 0, \\dots, 0)$, and $c = f(2, 0, 0, \\dots, 0)$, where $a = (a_1, a_2, \\dots, a_{2000})$, $b = (b_1, b_2, \\dots, b_{2000})$, and $c = (c_1, c_2, \\dots, c_{2000})$. Suppose $f: X \\to X$ is a function with $X = \\{0,1,2\\}^{2000}$. For each $1 \\leq t \\leq 2000$, define $a_t = f(0, 0, \\dots, 0, \\dots, 0)$ (with all zeros), $b_t = f(0, 0, \\dots, 1, \\dots, 0)$ (with a 1 in the $t$-th position), and $c_t = f(0, 0, \\dots, 2, \\dots, 0)$ (with a 2 in the $t$-th position). For $1 \\leq t \\leq 1000$, $(i_1, i_2, \\dots, i_{1000})$ is a permutation of $(1, 2, \\dots, 1000)$, and for $1001 \\leq t \\leq 2000$, $(i_{1001}, \\dots, i_{2000})$ is a permutation of $(1001, \\dots, 2000)$. For each $1 \\leq t \\leq 1000$, $(a_{i_t}, b_{i_t}, c_{i_t})$ can be chosen, and for $1001 \\leq t \\leq 2000$, $(a_{i_t}, b_{i_t})$ can be chosen. \n\nHow many different functions $f: X \\to X$ are there?",
"options": [],
"answer": "See solution",
"solution": "To count the number of such functions $f: X \\to X$, we proceed as follows:\n\n- For $1 \\leq t \\leq 1000$, $(i_1, i_2, \\dots, i_{1000})$ is a permutation of $(1, 2, \\dots, 1000)$.\n- For $1001 \\leq t \\leq 2000$, $(i_{1001}, \\dots, i_{2000})$ is a permutation of $(1001, \\dots, 2000)$.\n- For each $1 \\leq t \\leq 1000$, we can choose $(a_{i_t}, b_{i_t}, c_{i_t})$ such that all three are distinct (since $a_{i_t} \\ne b_{i_t} \\ne c_{i_t} \\ne a_{i_t}$), so there are $3! = 6$ choices for each $t$.\n- For each $1001 \\leq t \\leq 2000$, we can choose $(a_{i_t}, b_{i_t})$ with $a_{i_t} \\ne b_{i_t}$, so $3 \\times 2 = 6$ choices for each $t$.\n\nTherefore, the total number of functions is:\n\n$$\n1000! \\times 1000! \\times 6^{1000} \\times 6^{1000} = 1000!^2 \\times 12^{1000}.\n$$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 22246,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be an integer greater than $1$, and let $p$ be a prime divisor of $n$. A confederation consists of $p$ states, each of which has exactly $n$ airports. There are $p$ air companies operating interstate flights only, such that every two airports in different states are joined by a direct (two-way) flight operated by one of these companies.\n\nDetermine the maximal integer $N$ satisfying the following condition: In every such confederation, it is possible to choose one of the $p$ air companies and $N$ of the $np$ airports such that one may travel (not necessarily directly) from any one of the $N$ chosen airports to any other such only by flights operated by the chosen air company.",
"options": [],
"answer": "See solution",
"solution": "The required maximum is $n$.\n\n**Example showing $N$ cannot exceed $n$:**\n\nSplit the $n$ airports in the $i$-th state ($i = 1, \\dots, p$) into $p$ disjoint groups of $n/p$ airports each, $A_{i,j}$ for $j = 1, \\dots, p$. Let the $k$-th air company ($k = 1, \\dots, p$) operate direct flights between every airport in $A_{i,j}$ and every airport in $A_{i',j'}$ if $i \\neq i'$ and $j' - j \\equiv k(i' - i) \\pmod{p}$, and operate no flights between the airports in $A_{i,j}$ and those in $A_{i',j'}$ otherwise.\n\nSince $p$ is prime, for every $i \\neq i'$ and every $j, j'$, there exists $1 \\leq k \\leq p$ satisfying the previous congruence modulo $p$, so every two airports in different states are connected by a flight. On the other hand, for any given $k = 1, \\dots, p$, the $np$ airports are split into $p$ disjoint $n$-element groups, namely, $\\bigsqcup_{i=1}^p A_{i, j + k i}$ for $j = 1, \\dots, p$ (indices modulo $p$), such that there are no flights between different groups operated by the $k$-th air company. Consequently, $N \\leq n$.\n\n**Generalization:**\n\nIf the number of states is $m \\geq 2$, and the number $p$ of air companies is not necessarily prime nor a divisor of $n$, there always are at least $mn/p$ airports connected by flights operated by one air company. Assume the contrary and let $G_k$ ($k = 1, \\dots, p$) be the graph whose vertices are the $np$ airports, two vertices being joined by an edge in $G_k$ if the $k$-th air company operates a direct flight between the corresponding airports. Thus, each $G_k$ is a $p$-partite graph on $n$-element classes (recall that each air company operates interstate flights only). By assumption, each component of every $G_k$ has less than $mn/p$ vertices.\n\nConsider a component of $G_k$, and let $n_i$ be the number of vertices in this component located in the $i$-th state. The number of edges in the component does not exceed\n\n$$\n\\sum_{1 \\leq i < j \\leq m} n_i n_j \\leq \\frac{m-1}{2m} \\left( \\sum_{i=1}^{m} n_i \\right)^2 < \\frac{m-1}{2m} \\cdot \\frac{mn}{p} \\sum_{i=1}^{m} n_i = \\frac{n(m-1)}{2p} \\sum_{i=1}^{m} n_i.\n$$\n\nSumming over all components of $G_k$, the total number of edges of $G_k$ is less than $mn \\cdot n(m-1)/(2p) = mn^2(m-1)/(2p)$. Consequently, the total number of flights is less than $p \\cdot mn^2(m-1)/(2p) = n^2 m(m-1)/2 = n^2 \\binom{m}{2}$. Since the latter is the number of pairs of airports from different states, we reach a contradiction.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22247,
"subject": "Mathematics (Olympiad)",
"question": "For positive integers $m, n$, define\n\n$$\nS(m, n) = \\{(a, b) \\in \\mathbb{Z}^2 \\mid 1 \\le a \\le m,\\ 1 \\le b \\le n,\\ \\gcd(a, b) = 1\\}.\n$$\n\nProve: For any positive integers $d, r$, there exist integers $m, n$ not less than $d$, such that $|S(m, n)| \\equiv r \\pmod{d}$. Here, $|A|$ represents the number of elements in the finite set $A$.",
"options": [],
"answer": "See solution",
"solution": "Let $n = d + r$ and $m = d \\cdot (d + r) + 1$. For $1 \\le b \\le d + r$, the number of integers $a$ with $1 \\le a \\le m$ and $\\gcd(a, b) = 1$ is\n\n$$\n\\frac{\\varphi(b)}{b} \\cdot d \\cdot (d + r)! + 1.\n$$\n\nSince $b$ divides $(d + r)!$, this number is congruent to $1$ modulo $b$. Therefore,\n\n$$\n|S(m, n)| \\equiv \\sum_{b=1}^{d + r} 1 \\equiv r \\pmod{d}.\n$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22248,
"subject": "Mathematics (Olympiad)",
"question": "對於所有 $n \\geq 3$，求最大的正整數 $m$，使得可以從正 $n$ 邊形的對角線中選出 $m$ 條，且所選對角線中的任兩條，若它們在正 $n$ 邊形內部有交點，則它們必互相垂直。",
"options": [],
"answer": "See solution",
"solution": "若 $n$ 為奇數，則 $m = n - 3$；若 $n$ 為偶數，則 $m = n - 2$。\n\n1. 首先證明 $n$ 為奇數的狀況。\n\n(a) **Claim.** 正 $n$ 邊形內的任兩條對角線不垂直。\n\n假設對角線 $\\overline{AB}$ 和 $\\overline{CD}$ 垂直。做 $AB$ 中垂線，交此正 $n$ 邊形於頂點 $E$（注意到 $n$ 為奇數，故 $E$ 存在）。令 $E'$ 為正 $n$ 邊形上 $E$ 的對稱點：\n\n- 注意到 $n$ 為奇數，故 $E'$ 必不是正 $n$ 邊形的頂點。\n- 然而由對稱性，易知 $\\overline{EC} = \\overline{E'D}$。這代表 $E'$ 必須是正 $n$ 邊形的頂點。\n\n矛盾！\n\n(b) 由以上 Claim 及題設，這 $m$ 條對角線在多邊形內部必不相交；換言之，它們都必須從同一端點出發，而同一端點出發的對角線共有 $n-3$ 條，得證。\n\n\n\n2. 接下來，我們用歸納法證明 $n$ 為偶數的狀況。\n\n(a) 我們首先用歸納法證明一個更強的結果：對於所有 $n$，如果我們可以從圓內接 $n$ 邊形中選出滿足題意的 $m$ 條對角線，則 $m \\leq 2$。$n = 3$ 時顯然。現在，假設 $m \\leq n-2$ 對 $n = 1, 2, \\dots, N-1$ 都成立，則當 $n = N$，分成兩個 Case 討論：\n\nCase 1. 這 $m$ 條對角線中，有一條沒有和任何其他條在多邊形內部相交。該對角線將正 $n$ 邊形分割成一個 $n_1$ 邊形與一個 $n_2$ 邊形（$n_1 + n_2 = n + 2$）。不失一般性，假設 $n_1$ 邊形全部落在正 $n$ 邊形外接圓 $\\Gamma$ 某條直徑的一側，則易知該 $n_1$ 邊形內的對角線都不可能垂直，因此在 $n_1$ 邊形內只能選擇不相交的對角線。同前面 1.(a) 的討論，這個的最大可能是 $n_1 - 3$ 條。\n另一方面，由歸納假設，$n_2$ 內最多只能選 $m(n_2) \\leq n_2 - 2$ 條，故總共至多只能選 $(n_1 - 3) + (n_2 - 2) + 1 = n - 2$ 條。\n\nCase 2. 這 $m$ 條對角線中任一條都至少和另一條垂直。令 $d_1$ 和 $d_2$ 為其中一對垂直的對角線，則它們將圓切成四區：\n\n\n\n\n- 若有另一條對角線 $d'$，它與 $d_1$ 或 $d_2$ 都不垂直，則它的兩端點必然同屬於其中一區（否則它會和 $d_1$ 或 $d_2$ 相交）。\n- 若我們取另兩條相交的對角線 $d'_1$ 和 $d'_2$，它們都不垂直 $d_1$ 或 $d_2$，則它們的端點都必須落在同一區。然而，易見若兩條對角線的端點都在同一區，它們不可能垂直（如圖），矛盾。\n- 又基於我們假設這 $m$ 條對角線中任一條都至少和另一條垂直，故這 $m$ 條都只能平行 $d_1$ 或 $d_2$。\n\n現在，令 $d_1^*$ 為這 $m$ 條對角線中，平行 $d_1$ 且長度最長者；易知其他 $m-1$ 條對角線都不能和 $d_1^*$ 共享端點。同理，令 $d_2^*$ 為這 $m$ 條對角線中，平行 $d_2$ 且長度最長者；易知其他 $m-1$ 條對角線都不能和 $d_2^*$ 共享端點。扣除 $d_1^*$ 和 $d_2^*$ 的四個端點，其餘 $n-4$ 個頂點，每個都最多只能是兩條對角線的端點（因為這些對角線都必須平行 $d_1$ 或 $d_2$），故最多只能選到\n\n$$\n\\frac{1}{2}(2(n-4)) + 2 = n - 2\n$$\n\n對角線。得證。\n\n(b) 最後，我們構造 $m = n - 2$ 的選法：選擇正 $n$ 邊形的兩對點 $A$ 和 $A'$。選取：\n\n- $\\overline{AA'}$；\n- 令 $B$ 和 $B'$ 為 $A'$ 左右的兩頂點。選取 $\\overline{AB}$、$\\overline{AB'}$ 和 $\\overline{BB'}$，其中 $\\overline{BB'}$ 與 $\\overline{AA'}$ 垂直。\n- 令 $C$ 和 $C'$ 為 $A$ 左右的兩頂點。對於所有剩下的頂點\n\n$$\nD \\notin \\{A, A', B, B', C, C'\\},\n$$\n\n選取 $\\overline{AD}$。\n\n以上共 $1+3+(n-2) = n-2$ 條。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22249,
"subject": "Mathematics (Olympiad)",
"question": "設 $ABCD$ 為凸四邊形，其中 $\\angle ABC > 90^\\circ$、$\\angle CDA > 90^\\circ$，並且 $\\angle DAB = \\angle BCD$。\n\n令點 $A$ 分別關於直線 $BC$ 與 $CD$ 的對稱點為點 $E$ 與 $F$。設線段 $AE$ 和 $AF$ 分別交直線 $BD$ 於點 $K$ 和 $L$。試證：三角形 $BEK$ 和 $DFL$ 的兩個外接圓相切。",
"options": [],
"answer": "See solution",
"solution": "令 $A'$ 為 $A$ 關於 $BD$ 的對稱點。以下我們將證明：四邊形 $A'BKE$ 及 $A'DLF$ 皆有外接圓，且它們的外接圓相切於 $A'$。\n\n\n\n由於對直線 $BC$ 對稱，有 $\\angle BEK = \\angle BAK$；又由對直線 $DB$ 對稱，有 $\\angle BAK = \\angle BA'K$。故得 $\\angle BEK = \\angle BA'K$，即知 $A'BEK$ 共圓。同理可得 $A'DLF$ 共圓。\n\n要證明圓 $A'BKE$ 與 $A'DLF$ 相切，可由證出\n\n$$\n\\angle A'KB + \\angle A'LD = \\angle BA'D\n$$\n\n得到。由於 $AK \\perp BC$, $AL \\perp CD$, 再用一次對直線 $BD$ 的對稱性，可知\n\n$$\n\\angle A'KB + \\angle A'LD = 180^\\circ - \\angle KA'L = 180^\\circ - \\angle KAL = \\angle BCD = \\angle BAD = \\angle BA'D,\n$$\n\n故得證。\n\n註：上面解的要點是點出 $A'$ 點。之後的角度計算可由許多不同的方法完成。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22250,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute triangle and $D$ an interior point of its side $AC$. We call a side of the triangle $ABD$ *friendly* if the excircle of $ABD$ tangent to that side has its center on the circumcircle of $ABC$. Prove that there are exactly two friendly sides of $ABD$ if and only if $|BD| = |DC|$.",
"options": [],
"answer": "See solution",
"solution": "Let $E$, $F$, and $G$ be the centers of the excircles touching $BD$, $AD$, and $AB$ respectively, and let $\\omega$ be the circumcircle of $ABC$ (see Fig. 5).\n\nTo prove the assertion, we will show that $F$ and $G$ cannot both lie on $\\omega$ and that $E \\in \\omega \\iff |BD| = |DC| \\iff F \\in \\omega$.\n\nAs $AF$ and $AG$ are bisectors of the two complementary angles of $BAD$, the point $A$ lies on the segment $FG$. Thus, only one of the rays $AF$ and $AG$ can cut the circle $\\omega$ again, and therefore only one of $F$ and $G$ can lie on $\\omega$.\n\nTo show that $E \\in \\omega$ iff $|BD| = |DC|$, we first note that $E$ lies on $\\omega$ iff $\\angle CAE = \\angle CBE$. Since $\\angle CAE = \\frac{1}{2}\\angle BAD = \\frac{1}{2}(\\pi - \\angle ADB - \\angle ABD) = \\frac{1}{2}(\\pi - (\\pi - 2\\angle BDE) - (\\pi - 2\\angle DBE)) = \\angle BDE + \\angle DBE - \\frac{\\pi}{2} = \\frac{\\pi}{2} - \\angle BED$, we see that $E \\in \\omega$ is equivalent to $\\angle CBE = \\frac{\\pi}{2} - \\angle BED$, i.e., $BC$ and $DE$ being perpendicular. Since $DE$ is the bisector of $BDC$, this occurs iff $|BD| = |DC|$.\n\n\n\nIt remains to show that $|BD| = |DC|$ iff $F \\in \\omega$. Point $F$ lies on $\\omega$ iff $\\angle AFB = \\angle BCD$. Using the fact that $\\angle BAF = \\angle BAD + \\frac{1}{2}(\\pi - \\angle BAD) = \\frac{1}{2}(\\pi + \\angle BAD)$, we get $\\angle AFB = \\pi - \\angle ABF - \\angle BAF = \\frac{\\pi - \\angle ABD - \\angle BAD}{2} = \\frac{\\angle ADB}{2} = \\frac{\\angle BCD + \\angle CBD}{2}$. Thus $\\angle AFB = \\angle BCD$ is equivalent to $\\angle BCD = \\angle CBD$, i.e., $|BD| = |DC|$.\n\n*Remark.* The claim holds in the case of right or obtuse triangle, too. The problem with the above proof is that if $\\angle ABC > 90^\\circ$ then point $E$ may fall inside triangle $ABC$, whence equality $\\angle CAE = \\angle CBE$ is no more equivalent to $E \\in \\omega$. Nevertheless, one can show that if $|BD| = |DC|$, then $E$ must lie outside triangle $ABC$, extending the validity of the claim to the obtuse case.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22251,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$ and $b$ be positive integers such that $a^3 + b^3$ is a power of two. Show that $a = b = 2^n$ for some integer $n \\geq 0$.",
"options": [],
"answer": "See solution",
"solution": "We are given\n\n$$\na^3 + b^3 = 2^c.\n$$\n\nLet $2^n$ be the highest power of two dividing both $a$ and $b$. Thus, we may write\n\n$$\na = 2^n A \\quad \\text{and} \\quad b = 2^n B\n$$\nfor some positive integers $A$ and $B$ such that at least one of $A$ and $B$ is odd.\n\nSubstituting, we get $2^{3n} \\mid 2^c$, so $c \\geq 3n$. Let $d = c - 3n$. Then\n\n$$\nA^3 + B^3 = 2^d.\n$$\n\nSince $A^3 + B^3 \\geq 2$, $d \\geq 1$. $A^3$ and $B^3$ have the same parity, so $A$ and $B$ have the same parity. Since at most one of $A$ and $B$ is even, both must be odd.\n\nFactoring,\n\n$$\n(A+B)(A^2 - AB + B^2) = 2^d.\n$$\n\n$A^2 - AB + B^2$ is odd, so it must be $1$. Thus,\n\n$$\nA^2 - AB + B^2 = 1 \\quad \\text{and} \\quad A + B = 2^d.\n$$\n\nFrom this,\n\n$$\nA^3 + B^3 = A + B.\n$$\n\nBut $x^3 > x$ for $x > 1$, so $A = B = 1$. Therefore, $a = b = 2^n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22252,
"subject": "Mathematics (Olympiad)",
"question": "給定正偶數 $n > 2$，已知實數 $x_1, x_2, \\dots, x_n$ 滿足\n\n$$\n\\sum_{1 \\le i < j \\le n} (x_i - x_j)^2 = 1.\n$$\n\n試求 $\\sum_{i=1}^{n} (x_i - x_{i+1})^2$ 的最大值，其中 $x_{n+1} = x_1$。",
"options": [],
"answer": "See solution",
"solution": "最大值為 $\\frac{4}{n}$。\n\n首先，當 $x_1 = x_3 = \\cdots = x_{n-1}$，$x_2 = \\cdots = x_n$ 時可檢驗此值被達成。\n因此只須證明\n$$\n(x_1 - x_2)^2 + \\cdots + (x_{n-1} - x_n)^2 + (x_n - x_1)^2 \\le \\frac{4}{n}\n$$\n即可。\n\n可同時將所有 $x_i$ 增加或減少一個常數，使得 $x_1 + \\cdots + x_n = 0$，此時原條件等價於 $n(x_1^2 + \\cdots + x_n^2) = 1$。由於對任意實數 $a, b$ 有\n$$\n(a-b)^2 \\le 2a^2 + 2b^2,\n$$\n因此\n$$\n(x_1 - x_2)^2 + \\cdots + (x_{n-1} - x_n)^2 + (x_n - x_1)^2 \\le 4(x_1^2 + \\cdots + x_n^2)\n$$\n得證。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22253,
"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": "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\nRewriting:\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 large. But from earlier, $f(x) \\le x + f(0)$, so $f(f(y)) \\le f(y) + f(0)$. For $y < 0$, this leads to a contradiction, as the inequalities cannot be satisfied for all $y < 0$. Thus, such a function does not exist.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22254,
"subject": "Mathematics (Olympiad)",
"question": "A right square pyramid with volume $54$ has a base with side length $6$. The five vertices of the pyramid all lie on a sphere with radius $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.",
"options": [],
"answer": "See solution",
"solution": "Let $ABCD$ be the base of the pyramid, $P$ its apex, and $h$ its height. Let $O$ be the center of the sphere circumscribing the pyramid, and $M$ the foot of the altitude from $P$ (the center of the base). Then $O$, $P$, and $M$ are collinear. The volume formula gives:\n\n$$\n\\frac{1}{3} \\cdot 6^2 \\cdot h = 54 \\implies h = \\frac{9}{2}.\n$$\n\nSince $M$ is the center of the square base, $AM = 3\\sqrt{2}$. Let $r$ be the radius of the sphere, so $OP = OA = r$ and $OM = |\\frac{9}{2} - r|$. By the Pythagorean Theorem on $\\triangle OAM$:\n\n$$\n\\left(\\frac{9}{2} - r\\right)^2 + (3\\sqrt{2})^2 = r^2.\n$$\n\nSolving, $r = \\frac{17}{4}$. Thus, $m + n = 17 + 4 = 21$.\n\n\n\nAlternatively, for base side $s$ and height $h$:\n\n$$\nr = \\frac{s^2 + 2h^2}{4h}.\n$$\n\nPlugging in $s = 6$, $h = \\frac{9}{2}$:\n\n$$\nr = \\frac{6^2 + 2\\left(\\frac{9}{2}\\right)^2}{4\\left(\\frac{9}{2}\\right)} = \\frac{17}{4}.\n$$\n\n\n\nAlternatively, the radius is the circumradius of $\\triangle APC$ (base $AC = 6\\sqrt{2}$, height $PM = \\frac{9}{2}$):\n\n$$\nAP = PC = \\sqrt{(3\\sqrt{2})^2 + \\left(\\frac{9}{2}\\right)^2} = \\frac{\\sqrt{153}}{2}.\n$$\n\nArea by circumradius:\n\n$$\n\\frac{AP \\cdot PC \\cdot CA}{4r} = \\frac{459\\sqrt{2}}{8r}.\n$$\n\nArea by base and height:\n\n$$\n\\frac{1}{2} \\cdot 6\\sqrt{2} \\cdot \\frac{9}{2} = \\frac{27\\sqrt{2}}{2}.\n$$\n\nEquate and solve for $r$:\n\n$$\nr = \\frac{459\\sqrt{2} \\cdot 2}{8 \\cdot 27\\sqrt{2}} = \\frac{17}{4}.\n$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22255,
"subject": "Mathematics (Olympiad)",
"question": "As shown below, two circles $\\Gamma_1, \\Gamma_2$ intersect at points $A$, $B$. One line passing through $B$ intersects $\\Gamma_1, \\Gamma_2$ at points $C$, $D$; another line passing through $B$ intersects $\\Gamma_1, \\Gamma_2$ at points $E$, $F$. Line $CF$ intersects $\\Gamma_1, \\Gamma_2$ at points $P$, $Q$, respectively. Let $M$, $N$ be the midpoints of arc $PB$ and arc $QB$, respectively. Prove that if $CD = EF$, then $C$, $F$, $M$, $N$ are concyclic.\n\n",
"options": [],
"answer": "See solution",
"solution": "Draw lines $AC$, $AD$, $AE$, $AF$, $DF$. From $\\angle ADB = \\angle AFB$, $\\angle ACB = \\angle AEF$, and the assumption $CD = EF$, one obtains $\\triangle ACD \\cong \\triangle AEF$. So we have $AD = AF$, $\\angle ADC = \\angle AFE$, and $\\angle ADF = \\angle AFD$. Then $\\angle ABC = \\angle AFD = \\angle ADF = \\angle ABF$; $AB$ is the bisector of $\\angle CBF$. Draw lines $CM$, $FN$. Since $M$ is the midpoint of arc $PB$, $CM$ is the bisector of $\\angle DCF$, and $FN$ is the bisector of $\\angle CFB$. Then $BA$, $CM$, $FN$ have a common intersection, say, at $I$. In the circles $\\Gamma_1$ and $\\Gamma_2$, one has $CI \\times IM = AI \\times IB$, $AI \\times IB = NI \\times IF$, according to the power of a point theorem. So $NI \\times IF = CI \\times IM$, and $C$, $F$, $M$, $N$ are concyclic.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22256,
"subject": "Mathematics (Olympiad)",
"question": "Выпишем все возможные ситуации, которые могут встречаться в игре (то есть все возможные пары колод у участников). Назовём ситуацию *фундаментальной*, если все карты у одного игрока; *критической*, если у одного из игроков ровно одна карта; и *регулярной*, если у обоих игроков хотя бы по две карты.\n\nПроведём стрелку от каждой ситуации к ситуациям, которые могут из неё получиться после одного хода. Тогда из любой нефинальной ситуации ведут две стрелки, а из любой финальной — ноль. Надо доказать, что из каждой нефинальной ситуации по стрелкам можно дойти до финальной.",
"options": [],
"answer": "See solution",
"solution": "Выясним, сколько стрелок ведут в каждую ситуацию. Предположим, что она получилась в результате какого-то хода, в котором карты взял первый игрок. Тогда у него оказалось хотя бы две карты, которые лежат в конце колоды; при этом известно, что одна из них $a$ бьёт другую $b$. Значит, в этом случае карта $a$ была у первого игрока, а карта $b$ — у второго, то есть предыдущая ситуация восстанавливается однозначно. Итак, если наша ситуация регулярна, то на предыдущем ходе карты мог получить любой из двух игроков, и в каждом из этих случаев предыдущая ситуация восстанавливается однозначно. Значит, в каждую регулярную ситуацию ведут ровно две стрелки. Аналогично, в каждую нерегулярную ситуацию ведёт ровно одна стрелка.\n\nПредположим, что из некоторой ситуации $S$ нельзя попасть в финальную. Назовём ситуацию *достижимой*, если в неё можно добраться из $S$. Из каждой достижимой ситуации ведут две стрелки в достижимые. С другой стороны, в каждую достижимую ситуацию ведёт не более двух стрелок из достижимых. Это возможно только в том случае, если в каждую достижимую ситуацию ведёт ровно по две стрелки из достижимых. Из этого, в частности, следует, что все достижимые ситуации регулярны. Более того, поскольку в каждую ситуацию ведёт не более двух стрелок, получаем, что все стрелки, входящие в достижимые ситуации, выходят также из достижимых.\n\nПусть в ситуации $S$ у первого игрока $k > 1$ карт. Тогда в одной из двух ситуаций, из которых ведут стрелки в $S$, у первого игрока $k-1$ карта — назовём эту ситуацию $S_1$; по показанному выше, она достижима. Аналогично, если $k-1 > 1$, то в одной из двух ситуаций, из которых ведут стрелки в $S_1$, у первого игрока $k-2$ карты; назовём её $S_2$ и продолжим рассуждения. В итоге мы получим цепочку из достижимых ситуаций $S_1, S_2, \\dots, S_{k-1}$, причём в $S_{k-1}$ у первого игрока одна карта, то есть она критическая, и в неё входит только одна стрелка. Но в каждую достижимую ситуацию должно входить две стрелки — противоречие.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22257,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\alpha$ be an irrational number. For any $n \\in \\mathbb{N}^*$, let $a_n = \\{n\\alpha\\}$ and define the sequence $(x_n)_{n \\ge 1}$ by\n$$x_n = (a_2 - a_1)(a_3 - a_2) \\cdots (a_{n+1} - a_n).$$\nShow that the sequence is convergent and find its limit.",
"options": [],
"answer": "See solution",
"solution": "We claim that $\\{x + y\\} - \\{y\\}$ is equal to $\\{x\\}$ or $\\{x\\} - 1$, for any real numbers $x$ and $y$. Write $x + y = [y] + [x] + \\{x\\} + \\{y\\}$ to get $\\{x + y\\} - \\{y\\} = \\{\\{x\\} + \\{y\\}\\} - \\{y\\}$. Hence, if $\\{x\\} + \\{y\\} < 1$ then $\\{x + y\\} - \\{y\\} = \\{x\\}$, while if $2 > \\{x\\} + \\{y\\} \\ge 1$ then $\\{x + y\\} - \\{y\\} = \\{x\\} - 1$, as claimed.\n\nApply the above result to infer that either $|a_{n+1} - a_n| = \\{\\alpha\\}$ or $|a_{n+1} - a_n| = 1 - \\{\\alpha\\}$. Set $b = \\max\\{\\{\\alpha\\}, 1 - \\{\\alpha\\}\\}$ and notice that $0 < b < 1$ (since $\\alpha$ is irrational) to derive that $|x_n| \\le b^n$, implying $\\lim_{n \\to \\infty} |x_n| = 0$ and furthermore $\\lim_{n \\to \\infty} x_n = 0$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22258,
"subject": "Mathematics (Olympiad)",
"question": "在三角形 $ABC$ 的內部選一個點 $T$。令 $A_1, B_1, C_1$ 分別為 $T$ 對直線 $BC, CA, AB$ 的反射點。記三角形 $A_1B_1C_1$ 的外接圓為 $\\Omega$。設直線 $A_1T, B_1T, C_1T$ 分別與圓 $\\Omega$ 再交於點 $A_2, B_2, C_2$。證明：直線 $AA_2, BB_2, CC_2$ 共點，且其交點在 $\\Omega$ 上。",
"options": [],
"answer": "See solution",
"solution": "By $\\angle (\\boldsymbol{l}, n)$ we always mean the directed angle of the lines $l$ and $n$, taken modulo $180^\\circ$.\n\n\n\nLet $CC_2$ meet $\\Omega$ again at $K$ (as usual, if $CC_2$ is tangent to $\\Omega$, we set $K = C_2$). We show that the line $BB_2$ contains $K$; similarly, $AA_2$ will also pass through $K$. For this purpose, it suffices to prove that\n\n$$\n\\angle(C_2C, C_2A_1) = \\angle(B_2B, B_2A_1). \\qquad (1)\n$$\n\nBy the problem condition, $CB$ and $CA$ are the perpendicular bisectors of $TA_1$ and $TB_1$, respectively. Hence, $C$ is the circumcenter of the triangle $A_1TB_1$. Therefore,\n\n$$\n\\angle(CA_1, CB_1) = \\angle(CB, CT) = \\angle(B_1A_1, B_1T) = \\angle(B_1A_1, B_1B_2).\n$$\n\nIn circle $\\Omega$ we have $\\angle(B_1A_1, B_1B_2) = \\angle(C_2A_1, C_2B_2)$. Thus,\n\n$$\n\\angle(CA_1, CB) = \\angle(B_1A_1, B_1B_2) = \\angle(C_2A_1, C_2B_2). \\qquad (2)\n$$\n\nSimilarly, we get\n\n$$\n\\angle(BA_1, BC) = \\angle(C_1A_1, C_1C_2) = \\angle(B_2A_1, B_2C_2). \\qquad (3)\n$$\n\nThe two obtained relations yield that the triangles $A_1BC$ and $A_1B_2C_2$ are similar and equioriented, hence\n\n$$\n\\frac{A_1B_2}{A_1B} = \\frac{A_1C_2}{A_1C} \\quad \\text{and} \\quad \\angle(A_1B, A_1C) = \\angle(A_1B_2, A_1C_2).\n$$\n\nThe second equality may be rewritten as $\\angle(A_1B, A_1B_2) = \\angle(A_1C, A_1C_2)$, so the triangles $A_1BB_2$ and $A_1CC_2$ are also similar and equioriented. This establishes (1). $\\Box$\n\n**Comment 1.** In fact, the triangle $A_1BC$ is an image of $A_1B_2C_2$ under a spiral similarity centred at $A_1$; in this case, the triangles $ABB_2$ and $ACC_2$ are also spirally similar with the same center.\n\n**Comment 2.** After obtaining (2) and (3), one can finish the solution in different ways.\n\nFor instance, introducing the point $X = BC \\cap B_2C_2$, one gets from these relations that the 4-tuples $(A_1, B, B_2, X)$ and $(A_1, C, C_2, X)$ are both cyclic. Therefore, $K$ is the Miquel point of the lines $BB_2$, $CC_2$, $BC$, and $B_2C_2$; this yields that the meeting point of $BB_2$ and $CC_2$ lies on $\\Omega$.\n\nYet another way is to show that the points $A_1$, $B$, $C$, and $K$ are concyclic, as\n\n$$\n\\angle(KC, KA_1) = \\angle(B_2C_2, B_2A_1) = \\angle(BC, BA_1).\n$$\n\nBy symmetry, the second point $K'$ of intersection of $BB_2$ with $\\Omega$ is also concyclic to $A_1$, $B$, and $C$, hence $K' = K$.\n\n\n\n**Comment 3.** The requirement that the common point of the lines $AA_2$, $BB_2$, and $CC_2$ should lie on $\\Omega$ may seem to make the problem easier, since it suggests some approaches. On the other hand, there are also different ways of showing that the lines $AA_2$, $BB_2$, and $CC_2$ are just concurrent.\n\nIn particular, the problem conditions yield that the lines $A_2T$, $B_2T$, and $C_2T$ are perpendicular to the corresponding sides of the triangle $ABC$. One may show that the lines $AT$, $BT$, and $CT$ are also perpendicular to the corresponding sides of the triangle $A_2B_2C_2$, i.e., the triangles $ABC$ and $A_2B_2C_2$ are _orthologic_, and the orthology centers coincide. It is known that such triangles are _perspective_, i.e., the lines $AA_2$, $BB_2$, and $CC_2$ are concurrent (in projective sense).\n\nTo show this mutual orthology, one may again apply angle chasing, but there are also other methods. Let $A'$, $B'$, and $C'$ be the projections of $T$ onto the sides of the triangle $ABC$. Then $A_2T \\cdot TA' = B_2T \\cdot TB' = C_2T \\cdot TC'$, since all three products equal (minus) half the power of $T$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22259,
"subject": "Mathematics (Olympiad)",
"question": "Let $Z_{\\ge 0}$ be the set of all nonnegative integers. Find all functions $f: Z_{\\ge 0} \\to Z_{\\ge 0}$ satisfying\n\n$$\nf(f(f(n))) = f(n+1) + 1 \\quad \\text{for all } n \\in Z_{\\ge 0}.\n$$",
"options": [],
"answer": "See solution",
"solution": "There are two such functions:\n\n1. $f(n) = n + 1$ for all $n \\in Z_{\\ge 0}$.\n\n2. $$\nf(n) = \\begin{cases}\n n+1, & n \\equiv 0 \\pmod{4} \\text{ or } n \\equiv 2 \\pmod{4}, \\\\\n n+5, & n \\equiv 1 \\pmod{4}, \\\\\n n-3, & n \\equiv 3 \\pmod{4},\n\\end{cases} \\quad \\text{for all } n \\in Z_{\\ge 0}.\n$$\n\nThroughout, we write $h^k(x)$ for the $k$th iteration of function $h$, so $h^0$ is the identity, and $h^k(x) = h(\\dots h(x)\\dots)$ for $k \\ge 1$.\n\nFrom the given relation,\n$$\nf^{4}(n) = f(f^{3}(n)) = f(f(n+1)+1), \\qquad f^{4}(n+1) = f^{3}(f(n+1)) = f(f(n+1)+1)+1,\n$$\nso\n$$\nf^{4}(n) + 1 = f^{4}(n+1).\n$$\n\nLet $R_i$ be the range of $f^i$. Since $R_0 = Z_{\\ge 0}$, and $R_0 \\supseteq R_1 \\supseteq \\dots$, from above, if $a \\in R_4$ then $a+1 \\in R_4$, so $R_4$ (and thus $R_1$) is unbounded. If $f(m) = f(n)$ for $m \\ne n$, then $f(m+1) = f(n+1)$, and by induction $f(m+c) = f(n+c)$ for all $c \\ge 0$, so $f$ would be eventually periodic, contradicting unboundedness. Thus $f$ is injective.\n\nLet $S_i = R_{i-1} \\cap R_i$; all $S_i$ are finite for $i \\le 4$. By injectivity, $f$ is a bijection between $S_i$ and $S_{i+1}$, so $|S_1| = |S_2| = |S_3| = k$. If $0 \\in R_3$, then $0 = f(f(f(n)))$ for some $n$, so $f(n+1) = -1$, impossible. Thus $0 \\notin R_3$, so $k \\ge 1$.\n\nEach $b \\in R_0 \\cap R_3$ satisfies at least one of: (i) $b = 0$, (ii) $b = f(0) + 1$, (iii) $b-1 \\in S_1$. Otherwise, $b-1 \\in Z_{\\ge 0}$, so $f(n) = b-1$ for some $n > 0$, and $f^3(n-1) = f(n) + 1 = b$, so $b \\in R_3$.\n\nThus,\n$$\n3k = |S_1 \\cup S_2 \\cup S_3| \\le 1 + 1 + |S_1| = k + 2,\n$$\nso $k \\le 1$. Therefore $k = 1$, and equality holds. So $S_1 = \\{a\\}$, $S_2 = \\{f(a)\\}$, $S_3 = \\{f^2(a)\\}$ for some $a \\in Z_{\\ge 0}$, and each of (i), (ii), (iii) is realized exactly once:\n$$\n\\{a, f(a), f^2(a)\\} = \\{0, a+1, f(0)+1\\}.\n$$\n\nFrom this, $a+1 \\in \\{f(a), f^2(a)\\}$. If $a+1 = f^2(a)$, then $f(a+1) = f^3(a) = f(a+1) + 1$, impossible. So $f(a) = a+1$.\n\nNow, $0 \\in \\{a, f^2(a)\\}$. Consider two cases:\n\n*Case 1*: $a = 0$. Then $f(0) = 1$, $f(1) = 2$. By induction, $f(n) = n+1$ for all $n$.\n\n*Case 2*: $f^2(a) = 0$. Then $a = f(0) + 1$. By above, $f(a) = a+1$, $f(0) = 1$, so $a = 2$, $f(2) = 3$, $f(3) = 0$.\n\nBy induction, for $n \\equiv 0,2,3 \\pmod{4}$ with $n = 4k, 4k+2, 4k+3$, and $n = 4k+1$ with $k < m$, the formula holds. The explicit form is:\n$$\nf(n) = \\begin{cases}\n n+1, & n \\equiv 0 \\pmod{4} \\text{ or } n \\equiv 2 \\pmod{4}, \\\\\n n+5, & n \\equiv 1 \\pmod{4}, \\\\\n n-3, & n \\equiv 3 \\pmod{4}.\n\\end{cases}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22260,
"subject": "Mathematics (Olympiad)",
"question": "In triangle $ABC$, $M$ is the midpoint of side $BC$, and on side $AB$ a point $N$ is chosen so that $NB = 2AN$. If $\\angle CAB = \\angle CMN$, find $\\dfrac{AC}{BC}$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $D$ be a point on the half-line $CA$ such that $CA = AD$. Then $DM$ and $BA$ are medians of triangle $BCD$, so their intersection divides both in the ratio $2:1$ from the vertex. Thus, they intersect at point $N$. Hence $\\angle DAB = \\pi - \\angle CAB = \\pi - \\angle CMN = \\angle DMB$, so quadrilateral $BMAD$ is cyclic. Since $AM$ is parallel to $BD$ as a midline, $\\angle CDB = \\angle CAM = \\angle CBD$. This implies that\n\n$$\nCD = CB, \\text{ therefore } \\frac{AC}{BC} = \\frac{AC}{CD} = \\frac{1}{2}.\n$$\n\nIt is easy to see that the problem condition is satisfied for every triangle with this ratio of the sides.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22261,
"subject": "Mathematics (Olympiad)",
"question": "A class has 30 students who sit in three groups of 10 students each. At the beginning of each month, the teacher swaps the students' seats. What is the minimum number of months needed so that every pair of students sits in the same group for at least one month?",
"options": [],
"answer": "See solution",
"solution": "The minimum number of months needed is $5$.\n\nLet us say a pair of students is *friends* if they have sat in the same group for at least one month.\n\n**Four months is not enough:**\n\nFor integers $x \\ge 0$, $y \\ge 0$, $z \\ge 0$ with $x + y + z = 10$, we have:\n\n$$\nf(x, y, z) = \\binom{x}{2} + \\binom{y}{2} + \\binom{z}{2} \\ge 12.\n$$\n\nBy the arithmetic-quadratic mean inequality:\n\n$$\nf(x, y, z) = \\frac{1}{2}(x^2 + y^2 + z^2) - \\frac{1}{2}(x + y + z) \\ge \\frac{1}{6}(x + y + z)^2 - \\frac{1}{2}(x + y + z) = 11 + \\frac{2}{3}\n$$\n\nSince $f(x, y, z)$ is an integer, $f(x, y, z) \\ge 12$.\n\nLet $S$ be the set of students, $|S| = 30$. Let $X, Y, Z$ be the groups in the first month, each of size $10$. Suppose that for the next three months, the students sit in groups:\n\n$$\nT_1, T_2, T_3 \\quad T_4, T_5, T_6 \\quad T_7, T_8, T_9.\n$$\n\nFor a group $T \\subseteq S$, each pair in $X \\cap T$, $Y \\cap T$, and $Z \\cap T$ is already friends. Thus, $T$ creates at most:\n\n$$\n\\binom{10}{2} - \\binom{|X \\cap T|}{2} - \\binom{|Y \\cap T|}{2} - \\binom{|Z \\cap T|}{2} \\le 45 - 12 = 33\n$$\n\nnew friends. So, in four months, at most $3 \\cdot 45 + 9 \\cdot 33 = 432$ friends are created. There are $\\binom{30}{2} = 435$ pairs in total, so four months is not enough.\n\n**Five months is enough:**\n\nDivide the students into six groups $A, B, C, D, E, F$ of five students each and seat them as follows:\n\n$$\n\\begin{vmatrix}\n1 & AB & CD & EF \\\\\n2 & AC & BE & DF \\\\\n3 & AD & BF & CE \\\\\n4 & AE & BD & CF \\\\\n5 & AF & BC & DE\n\\end{vmatrix}\n$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 22262,
"subject": "Mathematics (Olympiad)",
"question": "For $\\ell$ white points and $\\ell$ black points distributed on the circumference of a circle, call a method of drawing $2\\ell$ line segments a *good method* if the following three conditions are satisfied:\n\n1. Each line segment has a white point and a black point as its endpoints.\n2. By tracing these line segments in order, it is possible to complete a cycle going through each of the given $2\\ell$ points once and only once.\n3. There are at most $\\ell - 1$ intersections among the line segments drawn.\n\nProve that for any integer $n \\geq 2$, there exists a good method of drawing $2n$ line segments for $n$ white points and $n$ black points distributed on the circumference of a circle.",
"options": [],
"answer": "See solution",
"solution": "Let us first prove the following lemma:\n\n**Lemma:** Let $k \\geq 3$. If for a distribution of $k-1$ white points and $k-1$ black points on the circumference of a circle there exists a good method of drawing $2(k-1)$ line segments, then there exists a good method of drawing $2k$ line segments for a distribution of $k$ white points and $k$ black points.\n\n**Proof:** If $k$ white points and $k$ black points are placed alternately on the circumference traced clockwise, then by connecting the points in order, we get a good method of drawing $2k$ line segments with $0$ intersections. So, in the sequel we consider other cases. Then we have at least one string of $3$ consecutive points with the configuration $\\{\\text{white, black, black}\\}$ or $\\{\\text{black, white, white}\\}$ traced clockwise. As the following argument works in both cases, let us assume the former, and call the three points $P, Q', Q$.\n\n\n\nAs indicated in the figure above, connect $P$ and $Q$, and then $P$ and $Q'$ by solid lines. For the configuration of $2k-2$ points excluding $P$ and $Q$, there is a good method of drawing $2k-2$ line segments by assumption. Draw these $2k-2$ line segments using dotted lines. Label the two points that are connected to $Q'$ by dotted lines as $P'$ and $P''$ so that $P', Q', P''$ lie in this order when traced clockwise. Then, erase the dotted line $Q'P''$ and connect $Q$ and $P''$ with a dotted line. Note that we still have $2k-2$ dotted line segments.\n\nWe will now show that the $2k$ line segments (2 solid lines and $2k-2$ dotted lines) give a good method of drawing $2k$ line segments. To see that we can complete a cycle going through each of the $2k$ points once and only once by tracing these $2k$ line segments in order, we just note that the dotted line path going through $P' \\to Q' \\to P''$ is replaced by the new path $P' \\to Q' \\to P \\to Q \\to P''$ and the rest of the dotted line path remains unchanged. Finally, to see that there are at most $k-1$ intersections among $2k$ line segments, we first note that the number of intersections among dotted line segments remains the same (and hence no more than $k-2$) by erasing the dotted line $Q'P''$ and introducing the new dotted line $QP''$ as above. Furthermore, the $2$ solid line segments introduced above do not intersect each other, and there is only $1$ intersection between a solid line segment and a dotted one (namely, the intersection of $PQ$ and $P'Q'$). Consequently, the number of intersections among $2k$ line segments (comprised of $2k-2$ dotted lines and $2$ solid lines) is no more than $(k-2)+1=k-1$ as claimed, and this proves the lemma.\n\nIt remains to show that there exists a good method of drawing $2 \\times 2 = 4$ line segments when $2$ white and $2$ black points are distributed on the circumference of a circle. If we label the $4$ points (2 white and 2 black) as $A, B, C, D$, then the combination of line segments that can intersect among them can be chosen only in $1$ way (namely, $AC$ and $BD$), and therefore, the number of possible intersections of the line segments is at most $1$, which satisfies the condition. Thus, we complete the proof of the assertion.\n\n**Remark:** In the argument above, we are counting the number of intersections among the $n$ line segments as $\\binom{n}{2}$, i.e., counting the number of pairs of line segments intersecting each other. The number counted this way is not less than the actual number of intersections of a pair of line segments belonging to the family in question, and therefore, the argument can be applied correctly to cases where $3$ or more line segments may intersect at a point.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22263,
"subject": "Mathematics (Olympiad)",
"question": "Огторгуй дахь цэгүүдийн системийн хувьд тэдгээрийн хамгийн хол орших 2 цэгийн хоорондох зайг системийн диаметр гэнэ. $N$ цэгийн хувьд диаметрийн тооны хамгийн их утгыг $d(N)$ гэвэл $d(48) \\geq 94$ болохыг харуул.",
"options": [],
"answer": "See solution",
"solution": "\n\nОгторгуйн ABCD нэгж талтай зөв тетраэдр авъя. Үүн дээр AB талын хувьд $60^\\circ$-ийн эргүүлэлт хийхэд $C$ цэг $D$-д буух ба энэ хоорондох зам болох $DC$ нумыг нэмж зуръя. Тэгвэл $DC$ нум дээр 44 цэг авбал $A$, $B$, $C$, $D$ ба уг 44 цэг нь огторгуй дахь 48 цэгийн систем болно.\n\n$DC$ дээр 44 цэг авахдаа $D$ ба $C$-ээс ялгаатай цэгүүд авна. Тэгвэл $A$ ба $B$-ээс эдгээр цэгүүд ба $D$, $C$-ууд хүрэх зай нь 1 нэгж байна. Мөн $DC$ ба $AB$-ууд 1 нэгж урттай тул нийт $2 \\cdot 46 + 2 = 94$ ширхэг нэгж урттай ялгаатай хэрчим байна. Нөгөө талаас энэ 48 цэгүүдийн систем нь 1 нэгж диаметртэй болохыг анзаарвал энэ 48 цэгийн систем нь 94 диаметртэй болно. $d(48) \\geq 94$ болох нь батлагдлаа. (Бүр цаашилбал $d(n) = 2(n-2) + 2 = 2n - 2$ болохыг баталж болно. Гэхдээ энэ бодлогод үүнийг батлах шаардлагагүй.)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22264,
"subject": "Mathematics (Olympiad)",
"question": "Let $CC_1$ be the median of triangle $ABC$. Prove that the inequality $CC_1 \\le 0.5 AB$ is equivalent to the inequality $\\angle ACB \\ge 90^\\circ$.\n\n% \n\nLet $P$, $Q$, $T$ be the midpoints of the segments $KN$, $KM$, $LM$, respectively (see Fig. 2). Prove that $AC \\le AP + PQ + QT + TC$, and that the perimeter of $KLMN$ is greater than or equal to $2AC$.\n\n% ",
"options": [],
"answer": "See solution",
"solution": "Consider the parallelogram $ADBC$ (see Fig. 1). By the cosine law, from $\\triangle ABC$ and $\\triangle CBD$, it follows that\n\n$$\n\\cos \\angle ACB = \\frac{BC^2 + AC^2 - 2AB^2}{2BC \\cdot AC}\n$$\n\nand\n\n$$\n\\cos \\angle CBD = \\frac{BC^2 + AC^2 - 2CD^2}{2BC \\cdot AC}.\n$$\n\nSince the sum of these angles is $180^\\circ$, we see that $\\angle ACB \\ge 90^\\circ$ if and only if $\\cos \\angle CBD \\ge \\cos \\angle ACB$. This is equivalent to $CD \\le AB$, or, since $CD = 2CC_1$, to $CC_1 \\le 0.5AB$. The proof is complete.\n\nBy the lemma, if $\\angle NAK \\ge 90^\\circ$ and $\\angle MCL \\ge 90^\\circ$, then $AP \\le 0.5KN$ and $CT \\le 0.5LM$. Moreover, $PQ = 0.5NM$ and $QT = 0.5KL$ as the midlines of the respective triangles. Therefore,\n\n$$\nAC \\le 0.5KN + 0.5NM + 0.5KL + 0.5LM,\n$$\n\ni.e., the perimeter of $KLMN$ is greater than or equal to $2AC$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22265,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f, h : \\mathbb{R} \\to \\mathbb{R}$ such that for all $x, y \\in \\mathbb{R}$,\n\n$$\nf(x^2 + y h(x)) = x h(x) + f(xy).\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $h(0) = a$. Set $x = 0$ in the initial equation:\n\n$$\nf(0^2 + y h(0)) = 0 \\cdot h(0) + f(0) \\implies f(a y) = f(0) = c \\text{ for all } y.\n$$\n\nIf $a \\neq 0$, then $f(x) = c$ is constant. Plugging into the original equation:\n\n$$\nc = x h(x) + c \\implies x h(x) = 0 \\implies h(x) = 0 \\text{ for } x \\neq 0,\\ h(0) = a \\text{ arbitrary}.\n$$\n\nIf $a = 0$, i.e., $h(0) = 0$, suppose $h(x_0) \\neq x_0$ for some $x_0 \\neq 0$. Then there exists $y_0$ such that $x_0^2 + y_0 h(x_0) = x_0 y_0$. Setting $x = x_0$, $y = y_0$ in the original equation gives $x_0 h(x_0) = 0 \\implies h(x_0) = 0$. Now, setting $x = x_0$ in the original equation, $f(x_0^2) = f(x_0 y)$ for all $y$, so $f$ is constant, which is already considered.\n\nThus, $h(x) = x$ for all $x$. The equation becomes:\n\n$$\nf(x^2 + y x) = x^2 + f(x y).\n$$\n\nLet $f(0) = b$. Setting $y = 0$ gives $f(x^2) = x^2 + b$, so $f(x) = x + b$ for $x \\geq 0$. Setting $y = -x$ gives $f(-x^2) = -x^2 + b$, so $f(x) = x + b$ for $x \\leq 0$.\n\n**Final answers:**\n- $f(x) = c$, $h(x) = \\begin{cases} 0, & x \\neq 0 \\\\ a, & x = 0 \\end{cases}$, where $a, c$ are constants;\n- $f(x) = x + b$, $h(x) = x$, where $b$ is a constant.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22266,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. Consider an $n \\times n$ chessboard with certain cells colored green. A rook can be placed on any green cell and moves only to other green cells, changing its direction horizontally and vertically with each subsequent move. It is important to note that remaining in the same cell is not considered a valid move. It is known that a rook cannot return to its starting cell within six moves. Prove that the number of green cells on the chessboard is less than $2n(1 + \\sqrt[3]{n})$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $N$ be the number of green cells. Construct a bipartite graph $G$ on $n$ rows and $n$ columns, where the $i$-th row connects to the $j$-th column if the cell at $(i, j)$ is green. This graph has $N$ edges and contains no cycles of length greater than six. We aim to prove $N < 2n(1 + \\sqrt[3]{n})$.\n\nLet $t = \\frac{N}{2n}$. The average degree of $G$'s vertices is $2t$. **Lemma:** A graph with average degree $2t$ contains a subgraph with minimum degree at least $t$.\n\n*Proof.* If a vertex has degree less than $t$, delete it. This increases the average degree of the remaining subgraph. The process terminates with a subgraph of minimum degree at least $t$. $\\Box$\n\nThus, there exists a subgraph $G'$ with minimum degree at least $t$. Let $u$ be any vertex of $G'$, and let $A_k$ be the set of vertices at distance $k$ from $u$. Then $|A_1| \\ge t$. Since $G'$ contains no cycles of length four or six, $|A_2| \\ge |A_1|(t-1)$ and $|A_3| \\ge |A_2|(t-1)$. The vertices of $A_1$ and $A_3$ are in the same part, so $t(t-1)^2 + t \\le n$. It follows that $t - 1 < \\sqrt[3]{n}$, which is equivalent to $N < 2n(1 + \\sqrt[3]{n})$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22267,
"subject": "Mathematics (Olympiad)",
"question": "Show that\n\n$$\n\\left( a + 2b + \\frac{2}{a+1} \\right) \\left( b + 2a + \\frac{2}{b+1} \\right) \\geq 16\n$$\n\nfor all positive real numbers $a$ and $b$ such that $ab \\geq 1$.",
"options": [],
"answer": "See solution",
"solution": "By the AM-GM Inequality, we have\n\n$$\n\\frac{a+1}{2} + \\frac{2}{a+1} \\geq 2\n$$\n\nTherefore,\n\n$$\na + 2b + \\frac{2}{a+1} \\geq \\frac{a+3}{2} + 2b,\n$$\n\nand, similarly,\n\n$$\nb + 2a + \\frac{2}{b+1} \\geq 2a + \\frac{b+1}{3}\n$$\n\nOn the other hand,\n\n$$\n(a + 4b + 3)(b + 4a + 3) \\geq (\\sqrt{ab} + 4\\sqrt{ab} + 3)^2 \\geq 64\n$$\n\nby the Cauchy-Schwarz Inequality as $ab \\geq 1$, and we are done.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22268,
"subject": "Mathematics (Olympiad)",
"question": "Find all quadruples $(a, b, c, d)$ of integers such that\n\n$$a + b + c = 2d$$\nand\n$$\\sqrt{ab} + \\sqrt{bc} + \\sqrt{ca} = d^2.$$",
"options": [],
"answer": "See solution",
"solution": "Assume $a \\geq b \\geq c \\geq 0$, so $d \\geq 0$.\n\nFrom $d^2 = \\sqrt{ab} + \\sqrt{ac} + \\sqrt{bc} \\leq a + b + c = 2d$, and since $a, b, c \\geq 0$, $d$ must be small. Trying $d = 0, 1, 2, 3$:\n\n**Case I:** $d = 0$. Then $a = b = c = 0$.\n\n**Case II:** $d = 1$. Then $a + b + c = 2$ and $\\sqrt{ab} + \\sqrt{bc} + \\sqrt{ca} = 1$. The only integer solution is $a = b = 1$, $c = 0$ (and permutations).\n\n**Case III:** $d = 2$ or $3$. Possible $(a, b, c)$ values do not yield integer solutions.\n\nIf at least one of $a, b, c$ is negative, then all are non-positive, and $(-a, -b, -c, -d)$ also satisfy the equations.\n\nThus, the solutions are all permutations of $(a, b, c, d) = (0, 0, 0, 0)$ and $(1, 1, 0, 1)$, and their negatives: $(0, -1, -1, -1)$, $(-1, 0, -1, -1)$, $(-1, -1, 0, -1)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22269,
"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$$\n\\omega(r) = b_1 \\cdot 2^{n-1} + b_2 \\cdot 2^{n-2} + \\dots + b_n.\n$$\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 $M = \\max_{i=1,\\dots,n} \\omega(S_i)$ and $m = \\min_{i=1,\\dots,n} \\omega(S_i)$. Prove that\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 $S_d = (a_d, a_{d+1}, \\dots, a_{d+n-1})$. 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 $(a_d)_{d=1}^{\\infty}$ 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, $k = t \\mid n$. Moreover, $S_i$ has the form $(a_i, \\dots, a_{i+k-1})$ repeated $n/k$ times. Thus,\n\n$$\n\\omega(S_i) = \\omega(a_i, \\dots, a_{i+k-1}) (2^{n-k} + 2^{n-2k} + \\dots + 1) = \\omega(a_i, \\dots, a_{i+k-1}) \\cdot \\frac{2^n - 1}{2^k - 1}.\n$$\n\nSo $\\omega(S_i)$ is a multiple of $\\frac{2^n - 1}{2^k - 1}$ for all $i$.\n\n**b)** Consider the $k$ tuples $S_1^* = (a_1, \\dots, a_k), \\dots, S_k^* = (a_k, \\dots, a_{2k-1})$. Clearly, $\\omega(S_i) = \\omega(S_i^*) \\cdot \\frac{2^n - 1}{2^k - 1}$.\n\nIt suffices to prove that\n$$\n\\max \\omega(S_i^*) - \\min \\omega(S_i^*) \\geq 2^{k-1} - 1. \\quad (\\heartsuit)\n$$\n\nLet $T_i^* = (1 - a_i, \\dots, 1 - a_{i+k-1})$. Then\n$$\n\\max \\omega(T_i^*) - \\min \\omega(T_i^*) = \\max \\omega(S_i^*) - \\min \\omega(S_i^*) \\quad (1)\n$$\n\nConsider two cases:\n\n- If there exist $i, j < k$ such that $a_i = a_{i+1} = 1$ and $a_j = a_{j+1} = 0$, then\n $$\n \\omega(S_i^*) - \\omega(S_j^*) > 2^k + 2^{k-1} - 2^{k-1} > 2^k - 1,\n $$\n which proves $(\\heartsuit)$.\n\n- Otherwise, for every $i$, $a_i = 1$ implies $a_{i+1} = 0$. The problem is trivial if all $a_i = 0$ or only one $a_i = 1$. If there exist $i < j$ with $a_i = a_j = 0$, for each $i$ with $a_i = 1$, let $i'$ be the smallest index $> i$ with $a_{i'} = 1$. Since $(a_1, \\dots, a_k)$ is not periodic, there is a pair $(j < j')$ with $j' - j - 1 \\geq i' - i$. Then\n $$\n \\begin{aligned}\n &\\omega(S_i^*) - \\omega(S_{j+1}^*) \\\\\n &= 2^k + \\omega(a_{i'+1}, \\dots, a_{k+i-1}) - \\omega(a_{j'}, \\dots, a_{k+j-1}) \\\\\n &\\geq 2^k + 2^{k+i-i'} - (2^{k+j-j'+1} - 1) \\\\\n &\\geq 2^k + 2^{k+j-j'+1} - (2^{k+j-j'+1} - 1) > 2^k - 1,\n \\end{aligned}\n $$\n which completes the proof.\n\n$\\Box$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22270,
"subject": "Mathematics (Olympiad)",
"question": "設四邊形 $ABCD$ 有內切圓，其圓心為 $I$。令對角線 $AC, BD$ 相交於 $E$ 點。若 $AD, BC, EI$ 三線段的中點共線，試證 $AB = CD$。",
"options": [],
"answer": "See solution",
"solution": "設 $AB$ 與 $CD$ 交於 $F$ 點，並設線段 $EF, AD, BC$ 的中點分別為 $X, Y, Z$。根據 Gauss 定理，$X, Y, Z$ 三點共線（牛頓線）。原本包含 $X, Y, Z$ 的直線通過 $F$ 點，經過以 $E$ 點為中心放大為 2 倍的動作會使該直線通過 $I$ 點。因此 $YZ \\parallel FI$。\n\n設 $AC, BD$ 的中點分別為 $U, V$。因為 $FI$ 是 $\\angle DFA$ 的角平分線，所以 $YZ$ 也是 $\\angle UYV$ 的角平分線。但 $YZ$ 平分線段 $UV$（$UYVZ$ 是平行四邊形），所以 $UYVZ$ 是菱形，即得 $YU = YV$。所以 $AB = 2YU = 2YV = CD$。",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22271,
"subject": "Mathematics (Olympiad)",
"question": "Let $R$ be a ring with only finitely many invertible elements. Prove that the following two statements are equivalent:\n\n(a) For every non-invertible element $x$ of $R$, there exists a non-invertible element $y$ of $R$ such that $xy = x + y$.\n\n(b) Every non-invertible element of $R$ is nilpotent.\n\n(An element $z$ of $R$ is called _nilpotent_ if $z^k = 0$ for some positive integer $k$.)",
"options": [],
"answer": "See solution",
"solution": "We show that (a) implies (b). Let $x \\in R \\setminus U(R)$, where $U(R)$ is the set of invertible elements of $R$, and let $y \\in R \\setminus U(R)$ be such that $xy = x + y$, i.e., $(x-1)(y-1) = 1$. Similarly, $(y-1)(z-1) = 1$ for some $z \\in R \\setminus U(R)$. Write\n\n$$\nx - 1 = (x - 1) \\cdot 1 = (x - 1)(y - 1)(z - 1) = 1 \\cdot (z - 1) = z - 1,\n$$\n\nto infer that $x = z$. Hence $x - 1 \\in U(R)$, showing that $\\{t - 1 : t \\in R \\setminus U(R)\\} \\subseteq U(R)$.\n\nSince $x \\in R \\setminus U(R)$, its powers are all different from $1$, and finiteness of the multiplicative group $U(R)$ then forces them all in $R \\setminus U(R)$. Hence $x^n - 1 \\in U(R)$ for all positive integers $n$, by the set inclusion in the previous paragraph. With reference again to finiteness of $U(R)$, $x^p - 1 = x^q - 1$ for some distinct positive integers $p$ and $q$, say, $p < q$. Then $x^p(x^{q-p} - 1) = 0$, so $x^p = 0$, on account of $x^{q-p} - 1 \\in U(R)$. Consequently, $x$ is indeed nilpotent.\n\nWe now show that (b) implies (a). Let $x \\in R \\setminus U(R)$ and choose an integer $p \\ge 3$ such that $x^p = 0$. Write $1 = 1 - x^p = (1-x)(1+x+x^2+\\cdots+x^{p-1})$ and let $y = -(x+x^2+\\cdots+x^{p-1}) = -x(1+x+\\cdots+x^{p-2})$. Then $y^p = 0$, so $y \\in R \\setminus U(R)$. Clearly, $(1-x)(1-y) = 1$, i.e., $xy = x+y$. This completes the proof.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 22272,
"subject": "Mathematics (Olympiad)",
"question": "For positive real numbers $a$, $b$, $c$, which of the following statements necessarily implies $a = b = c$?\n\n(I) $a(b^3 + c^3) = b(c^3 + a^3) = c(a^3 + b^3)$\n\n(II) $a(a^3 + b^3) = b(b^3 + c^3) = c(c^3 + a^3)$\n\nJustify your answer.",
"options": [],
"answer": "See solution",
"solution": "We show that (I) does not necessarily imply $a = b = c$, whereas (II) always implies $a = b = c$.\n\n**Case (I):**\n\nSuppose $a(b^3 + c^3) = b(c^3 + a^3)$. This gives $c^3(a - b) = ab(a^2 - b^2)$, so either $a = b$ or $ab(a + b) = c^3$. Similarly, $b = c$ or $bc(b + c) = a^3$. If $a \\neq b$ and $b \\neq c$, we obtain:\n\n$$\nab(a + b) = c^3, \\quad bc(b + c) = a^3.$$\n\nTherefore,\n\n$$\nb(a^2 - c^2) + b^2(a - c) = c^3 - a^3.$$\n\nThis gives $(a-c)(a^2 + b^2 + c^2 + ab + bc + ca) = 0$. Since $a, b, c > 0$, the only possibility is $a = c$. Thus, there are four possibilities:\n- $a = b = c$\n- $a = b \\neq c$\n- $b = c \\neq a$\n- $c = a \\neq b$\n\nSuppose $a = b \\neq c$. Then $b(c^3 + a^3) = c(a^3 + b^3)$ gives $ac^3 + a^4 = 2ca^3$, which implies $a(a-c)(a^2 - ac - c^2) = 0$. Therefore, $a^2 - ac - c^2 = 0$. Setting $x = a/c$, we get $x^2 - x - 1 = 0$, so $x = \\frac{1 + \\sqrt{5}}{2}$. Thus,\n\n$$\na = b = \\left( \\frac{1 + \\sqrt{5}}{2} \\right) c, \\quad c > 0.$$\n\nSimilarly, other cases yield:\n\n$$\nb = c = \\left( \\frac{1 + \\sqrt{5}}{2} \\right) a, \\quad a > 0;$$\n$$\nc = a = \\left( \\frac{1 + \\sqrt{5}}{2} \\right) b, \\quad b > 0.$$\n\nAnd $a = b = c$ is the fourth possibility.\n\n**Case (II):**\n\nSuppose $a(a^3 + b^3) = b(b^3 + c^3) = c(c^3 + a^3)$. Assume $a, b, c$ are mutually distinct. Let $a = \\max\\{a, b, c\\}$, so $a > b$ and $a > c$. From the first relation, $a^3 + b^3 < b^3 + c^3$, so $a^3 < c^3$, which forces $a < c$, a contradiction. Thus, $a, b, c$ cannot all be distinct; some two must be equal. If $a = b$, then $a^3 + b^3 = b^3 + c^3$ so $a = c$. Similarly, $b = c$ implies $b = a$, and $c = a$ gives $c = b$. Therefore, $a = b = c$.\n\n**Alternate for (II):**\n\nWe can write:\n\n$$\n\\begin{aligned}\n\\frac{a^3}{c} + \\frac{b^3}{c} &= \\frac{c^3}{a} + a^2, \\\\\n\\frac{b^3}{a} + \\frac{c^3}{a} &= \\frac{a^3}{b} + b^2, \\\\\n\\frac{c^3}{b} + \\frac{a^3}{b} &= \\frac{b^3}{c} + c^2.\n\\end{aligned}\n$$\n\nAdding, we get:\n\n$$\n\\frac{a^3}{c} + \\frac{b^3}{a} + \\frac{c^3}{b} = a^2 + b^2 + c^2.\n$$\n\nUsing the Cauchy-Schwarz inequality:\n\n$$\n\\begin{aligned}\n(a^2 + b^2 + c^2)^2 &= \\left( \\frac{\\sqrt{a^3}}{\\sqrt{c}} \\cdot \\sqrt{ac} + \\frac{\\sqrt{b^3}}{\\sqrt{a}} \\cdot \\sqrt{ba} + \\frac{\\sqrt{c^3}}{\\sqrt{b}} \\cdot \\sqrt{cb} \\right)^2 \\\\\n&\\le \\left( \\frac{a^3}{c} + \\frac{b^3}{a} + \\frac{c^3}{b} \\right) (ac + ba + cb) \\\\\n&= (a^2 + b^2 + c^2)(ab + bc + ca).\n\\end{aligned}\n$$\n\nThus,\n\n$$\na^2 + b^2 + c^2 \\le ab + bc + ca.$$\n\nThis implies $(a-b)^2 + (b-c)^2 + (c-a)^2 \\le 0$, so $a = b = c$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22273,
"subject": "Mathematics (Olympiad)",
"question": "If the length and breadth of a rectangle are divided into the same number of pieces, is the perimeter of the whole rectangle equal to the sum of the perimeters of the diagonal subrectangles? For example, if the first subrectangle has perimeter $18 + 18 - 22 = 14$, and the last subrectangle has perimeter $20 + 16 - 22 = 14$, and the middle subrectangle has perimeter $22$, is the perimeter of the whole rectangle $14 + 22 + 14 = 50$?",
"options": [],
"answer": "See solution",
"solution": "By dividing the rectangle as described, the perimeter of the whole rectangle is indeed equal to the sum of the perimeters of the diagonal subrectangles: $14 + 22 + 14 = 50$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22274,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c, p, q, r$ be positive integers with $p, q, r \\ge 2$. Denote\n\n$$\nQ = \\{(x, y, z) \\in \\mathbb{Z}^3 \\mid 0 \\le x \\le a,\\ 0 \\le y \\le b,\\ 0 \\le z \\le c\\}.\n$$\n\nInitially, some stones are put at each point of $Q$, with total $M$ stones. Then one can perform the following three types of operations repeatedly:\n\n1. Remove $p$ stones on $(x, y, z)$ and place a stone on $(x-1, y, z)$;\n2. Remove $q$ stones on $(x, y, z)$ and place a stone on $(x, y-1, z)$;\n3. Remove $r$ stones on $(x, y, z)$ and place a stone on $(x, y, z-1)$.\n\nFind the smallest positive integer $M$ such that one can always perform a sequence of operations to place a stone at $(0, 0, 0)$, no matter how the stones were distributed initially.",
"options": [],
"answer": "See solution",
"solution": "The smallest positive integer $M$ is $p^a q^b r^c$.\n\nSuppose we put fewer than $p^a q^b r^c$ stones at $(a, b, c)$. Then, using the weight function $w(u) = \\frac{1}{p^x q^y r^z}$ for a stone $u$ at $(x, y, z)$, the total weight $W = \\sum w(u)$ remains invariant under the allowed operations. If initially $W < 1$, there can never be a stone at $(0, 0, 0)$.\n\nConversely, we generalize to $n$ dimensions. Let $a_1, \\dots, a_n$ and $p_1, \\dots, p_n$ be positive integers with $p_i \\ge 2$. Consider the lattice\n\n$$\nQ(a_1, \\dots, a_n) = \\{(x_1, \\dots, x_n) \\in \\mathbb{Z}^n \\mid 0 \\le x_i \\le a_i,\\ i = 1, \\dots, n\\}\n$$\n\nWe define operations $X_i$ (remove $p_i$ stones at $(x_1, \\dots, x_n)$ and place one at $(x_1, \\dots, x_i-1, \\dots, x_n)$) and $T_q$ (remove $q$ stones at a point, no replacement). By induction on $n$, we prove:\n\n*Proposition A(n):* If $M = p_1^{a_1} \\cdots p_n^{a_n}$, one can move a stone to the origin.\n\n*Proposition B(n):* If $M + K > u \\cdot p_1^{a_1} \\cdots p_n^{a_n}$, one can move $u$ stones to the origin.\n\n*Proposition C(n):* If $q \\ge p_n$, $M + K > sq \\cdot p_1^{a_1} \\cdots p_n^{a_n} + t$, one can perform $\\lfloor \\frac{t}{q} \\rfloor$ times $T_q$ and then move $sq$ stones to the origin.\n\nThe induction proceeds by splitting the domain into $Q(a_1, \\dots, a_n-1)$ and $Q(a_1, \\dots, a_{n-1}) \\times \\{a_n\\}$, transferring stones as allowed, and applying the inductive hypothesis. The details show that $M = p^a q^b r^c$ is both necessary and sufficient.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22275,
"subject": "Mathematics (Olympiad)",
"question": "For a positive integer $k$, define the changed number of $k$ as the number obtained by switching each digit 2 in $k$ to 3, and each digit 3 in $k$ to 2. If $l$ is the changed number of $k$, then the changed number of $l$ is $k$ itself.\n\nA number $m \\leq 1999$ is called a *good number* if the number of times the digit 2 appears in $m$ is greater than the number of times the digit 3 appears in $m$. A number $n \\leq 1999$ is called a *bad number* if the number of times the digit 3 appears in $n$ is greater than the number of times the digit 2 appears in $n$.\n\nA pair $(m, n)$ is called a *nice pair* if $m$ is a good number, $n$ is a bad number, and $n$ is the changed number of $m$.\n\nHow many good numbers are there among the integers between 2000 and 2023 (inclusive), excluding 2003 and 2013?",
"options": [],
"answer": "See solution",
"solution": "Let $m$ be a good number. Let $a$ (respectively, $b$) be the number of times the digit 2 (respectively, 3) appears in $m$. Then $a > b$. Let $n$ denote the changed number of $m$. In $n$, the number of times the digit 2 appears is $b$, and the number of times the digit 3 appears is $a$, so $n$ is a bad number. If $m \\leq 1999$, then $n \\leq 1999$. A number cannot be both good and bad. Thus, each good number less than or equal to 1999 is in exactly one nice pair, and similarly for each bad number. Therefore, the number of good numbers less than or equal to 1999 and the number of bad numbers less than or equal to 1999 are both equal to the number of nice pairs.\n\nAmong the integers between 2000 and 2023, there are 22 good numbers (excluding 2003 and 2013) and no bad numbers. Thus, the answer is $22$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22276,
"subject": "Mathematics (Olympiad)",
"question": "Suppose we have 2021 walnuts arranged in a circle, each initially colored white. On the $k$th move, the two walnuts adjacent to walnut $k$ are swapped, and at the same time, the color of walnut $k$ is changed from white to black. Thus, at the end of move $k$, all walnuts numbered $1, 2, \\dots, k$ are black, and the remaining walnuts are white.\n\nProve that it is not possible that, for each $k$, on move $k$, the two walnuts $a$ and $b$ adjacent to walnut $k$ always satisfy $a, b < k$ or $a, b > k$ (i.e., that the two adjacent walnuts are always both less than $k$ or both greater than $k$).",
"options": [],
"answer": "See solution",
"solution": "Introduce a coloring of the walnuts as follows. Initially, all the walnuts are colored white. On the $k$th move, when the two walnuts next to walnut $k$ are swapped, at the same time we also change the color of walnut $k$ from white to black. Thus, at the end of move $k$, all walnuts numbered $1, 2, \\dots, k$ are black and the remaining walnuts are white.\n\nAssume, for the sake of contradiction, that for each $k$, on move $k$, the two walnuts $a$ and $b$ adjacent to walnut $k$ satisfy $a, b < k$ or $a, b > k$. This is equivalent to saying that walnuts $a$ and $b$ are the same color. (*)\n\nDefine a *chain of size* $n$ to be a run of $n$ consecutive walnuts around the circle, all of which are white, and such that the run is bounded by a black walnut on either side of the run. For example, here is a chain of size 6.\n\n\n\nWe claim that a chain of positive even size always exists from move 1 onward. We prove this by induction.\n\nThe base case is obvious because at the end of move 1 there is only one black walnut and the remaining walnuts form a chain of size 2020.\n\nFor the inductive step, suppose there is a chain containing $2n$ walnuts at the end of move $k-1$. Consider what happens on move $k$. Walnuts $a$ and $b$, that are adjacent to walnut $k$, are swapped, and walnut $k$ has its color changed from white to black. Recall by assumption (*) that walnuts $a$ and $b$ have the same color.\n\n- If walnut $k$ was not in the chain at the end of move $k-1$, then since $a, b$ have the same color by assumption (*), it follows that the size of the chain is unchanged and so is still a nonempty chain of even size.\n- If walnut $k$ is in the chain at the end of move $k-1$, then note that it cannot be right at the end of the chain because this would imply that walnuts $a$ and $b$ have different colors, contradicting (*). Thus all three walnuts $a, b, k$ are in the chain. At the end of move $k$, walnut $k$ is recolored black, and so the chain splits into three pieces: a nonempty chain of size $u$, followed by a single black walnut, followed by a nonempty chain of size $v$. Since $u+v = 2n-1$, it follows that one of $u, v$, say $u$, is even. Then these $u$ walnuts form a nonempty chain of even size.\n\nThis completes the inductive step, and so establishes the claim.\n\nFinally, to complete the proof, we simply note that the claim implies that there is a nonempty chain of even size at the end of move 2021. This is clearly false because at this stage all walnuts are black. Hence assumption (*) is false, as desired. $\\Box$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22277,
"subject": "Mathematics (Olympiad)",
"question": "For positive integer $n$, define\n\n$$a_n = n\\sqrt{5} - \\lfloor n\\sqrt{5} \\rfloor.$$ \n\nCompute the maximum and minimum values among $a_1, a_2, \\dots, a_{2009}$.\n\n(For a real number $x$, $\\lfloor x \\rfloor$ denotes the greatest integer less than or equal to $x$.)",
"options": [],
"answer": "See solution",
"solution": "Let $b_0 = 0$, $b_1 = 1$, and $b_n = 4b_{n-2} + b_{n-1}$ for $n \\geq 2$. Then\n\n$$\nb_n = \\frac{(2+\\sqrt{5})^n - (2-\\sqrt{5})^n}{2\\sqrt{5}}.\n$$\n\nIn particular, $b_6 = 1292$, $b_7 = 5473$.\n\nFor every $k = 1, 2, \\dots, 5473$, there are unique integers $x_k, y_k$ such that $1292k = x_k + 5473y_k$, with $1 \\leq x_k \\leq 5473$. Since $\\gcd(1292, 5473) = 1$, the sequence $x_1, x_2, \\dots, x_{5473}$ is a permutation of $\\{1, 2, \\dots, 5473\\}$, and $\\{y_k\\}$ is nondecreasing:\n\n$$\ny_1 \\leq y_2 \\leq \\dots \\leq y_{5473} = 1291.\n$$\n\nLet $f(x) = x - \\lfloor x \\rfloor$. We have\n\n$$\n\\begin{aligned}\nf(x_k\\sqrt{5}) &= f(1292k\\sqrt{5} - 5473y_k\\sqrt{5}) \\\\\n&= f\\left(\\frac{(2+\\sqrt{5})^6 - (2-\\sqrt{5})^6}{2}k - \\frac{(2+\\sqrt{5})^7 - (2-\\sqrt{5})^7}{2}y_k\\right) \\\\\n&= f\\left(-(2-\\sqrt{5})^6k + (2-\\sqrt{5})^7y_k\\right).\n\\end{aligned}\n$$\n\nSince\n\n$$\n\\begin{aligned}\n0 < (2-\\sqrt{5})^6 k - (2-\\sqrt{5})^7 y_k \\leq 5473(2-\\sqrt{5})^6 - 1291(2-\\sqrt{5})^7 < 1,\n\\end{aligned}\n$$\nit follows that\n\n$$\nf(x_k \\sqrt{5}) = 1 - (2 - \\sqrt{5})^6 k + (2 - \\sqrt{5})^7 y_k,\n$$\n\nand this is strictly decreasing.\n\nNow, since $x_1 = 1292$, $x_{5473} = 5473$, $x_{5472} = 4181$, $x_{5471} = 2889$, $x_{5470} = 1597$, we see that $a_{1292}$ attains the maximum and $a_{1597}$ the minimum.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22278,
"subject": "Mathematics (Olympiad)",
"question": "For a finite set $A$ of positive integers, we call a partition of $A$ into two disjoint nonempty subsets $A_1$ and $A_2$ **good** if the least common multiple of the elements in $A_1$ is equal to the greatest common divisor of the elements in $A_2$.\n\nDetermine the minimum value of $n$ such that there exists a set of $n$ positive integers with exactly $2015$ good partitions.",
"options": [],
"answer": "See solution",
"solution": "The answer is $3024$.\n\nSuppose $A = \\{a_1, a_2, \\cdots, a_n\\}$, where $a_1 < a_2 < \\cdots < a_n$. For any nonempty set $B$ of positive integers, let $l(B)$ and $g(B)$ denote the least common multiple and greatest common divisor of its elements, respectively.\n\nConsider all good partitions $(A_1, A_2)$ of $A$. By definition, for all $a_i \\in A_1$ and $a_j \\in A_2$, we must have $a_i \\leq l(A_1) = g(A_2) \\leq a_j$. Thus, there must exist $1 \\leq k < n$ such that $A_1 = \\{a_1, a_2, \\cdots, a_k\\}$ and $A_2 = \\{a_{k+1}, \\cdots, a_n\\}$. Let $l_k = l(a_1, a_2, \\cdots, a_k)$ and $g_k = g(a_{k+1}, \\cdots, a_n)$.\n\nNow, observe the following properties:\n\n- **Property 1:** If both $a_{k-1}$ and $a_k$ are good partitions, then $g_{k-1} = g_k = a_k$.\n\n *Proof:* Note that $l_{k-1} = g_{k-1} \\mid a_k$, so $g_k = l_k = l(l_{k-1}, a_k) = a_k$, and $g_{k-1} = g(a_k, g_k) = a_k$.\n\n- **Property 2:** Among $a_{k-1}$, $a_k$, and $a_{k+1}$, at most one is a good partition.\n\n *Proof:* Otherwise, by Property 1, $a_k = g_k = a_{k+1}$, a contradiction.\n\n- **Property 3:** $a_1$ and $a_2$ cannot both be good partitions. Similarly, $a_{n-2}$ and $a_{n-1}$ cannot both be good partitions.\n\n *Proof:* If both $a_1$ and $a_2$ are good partitions, then by Property 1, $a_2 = g_1 = l_1 = a_1$, a contradiction. The second part is similar.\n\nNow, suppose $|A| = n$. By Property 3, at most one of $\\{a_1, a_2\\}$ is a good partition, and at most one of $\\{a_{n-2}, a_{n-1}\\}$ is a good partition. By Property 2, among $\\{a_3, \\ldots, a_{n-3}\\}$, at least $\\left\\lfloor \\frac{n-5}{3} \\right\\rfloor$ are not good partitions. Summing up, the number of good partitions is at most $(n-1) - \\left\\lfloor \\frac{n-5}{3} \\right\\rfloor = \\left\\lfloor \\frac{2(n-2)}{3} \\right\\rfloor$. Setting this at least $2015$, we get $n \\geq 3024$.\n\nFinally, we show that $3024$ is achievable. Consider $A = \\{2 \\times 6^i, 3 \\times 6^i, 6^{i+1} \\mid 0 \\leq i \\leq 1007\\}$. Then all $3 \\times 6^i$ and $6^i$ (except $6^{1007}$) are good partitions, giving exactly $2015$ good partitions.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22279,
"subject": "Mathematics (Olympiad)",
"question": "In cyclic quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at $P$. Let $E$ and $F$ be the respective feet of the perpendiculars from $P$ to lines $AB$ and $CD$. Segments $BF$ and $CE$ meet at $Q$. Prove that lines $PQ$ and $EF$ are perpendicular to each other.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $G$, $X$, and $Y$ be the respective feet of the perpendiculars from $P$ to $EF$, $EC$, and $FB$. Note that $EPYB$ and $FPXC$ are cyclic quadrilaterals, so\n\n$$\n\\begin{aligned}\n\\angle EYF &= 90^\\circ + \\angle EYP = 90^\\circ + \\angle EBP = 90^\\circ + \\angle ABP \\\\\n&= 90^\\circ + \\angle DCP = 90^\\circ + \\angle FCP = 90^\\circ + \\angle FXP = \\angle FXE.\n\\end{aligned}\n$$\n\nThus, $EXYF$ is a cyclic quadrilateral. Note that $EGPX$ and $FGPY$ are also cyclic. The radical axes of the circumcircles of $EXYF$, $EGPX$, and $FGPY$ with each other are $GP$, $EX$, and $FY$. Thus, by the radical axis theorem, $GP$, $EX$, and $FY$ concur. Since $Q$ is the intersection of $EX$ and $FY$, by the construction of $G$ we have $GP \\perp EF$, so it follows that $QP \\perp EF$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22280,
"subject": "Mathematics (Olympiad)",
"question": "Find the positive integers $a$ and $b$ fulfilling\n\n$$\n\\frac{a}{(a,b)} = b + \\frac{48 \\cdot (a,b)}{[a,b]} \\quad \\text{and} \\quad \\frac{b}{(a,b)} = a - \\frac{312 \\cdot (a,b)}{[a,b]}\n$$\n\nwhere $(a, b) = \\gcd(a, b)$ and $[a, b] = \\text{lcm}(a, b)$.",
"options": [],
"answer": "See solution",
"solution": "We start noticing that $a \\geq \\frac{a}{(a,b)} > b$.\n\nDenote $d = (a, b)$. Then $a = dx$, $b = dy$, $[a, b] = dxy$ and $(x, y) = 1$, $x > y$. This gives\n$$\nx = dy + \\frac{48}{xy} \\quad \\text{and} \\quad y = dx - \\frac{312}{xy}\n$$\n(1). It follows that $xy \\mid 48$ and $xy \\mid 312$, hence $xy \\mid (48, 312) = 24$.\n\nRelation (1) yields $xy(x - dy) = 48$ and $xy(dx - y) = 312$, whence\n$$\n\\frac{x-dy}{dx-y} = \\frac{48}{312} = \\frac{2}{13}\n$$\nand $x(13 - 2d) = y(13d - 2)$, $13 - 2d > 0$.\n\nWe have the cases:\n\nI. $d = 1 \\Rightarrow 11x = 11y \\Rightarrow a = b$, false, since $a > b$.\n\nII. $d = 2 \\Rightarrow 3x = 8y$. Since $(x, y) = 1$, we get $x = 8$, $y = 3$, hence $a = 16$, $b = 6$.\n\nIII. $d = 3 \\Rightarrow 7x = 37y \\Rightarrow x = 37$, $y = 7$ – false, since $xy \\mid 24$.\n\nIV. $d = 4 \\Rightarrow x = 10y \\Rightarrow x = 10$, $y = 1$ – false, since $xy \\mid 24$.\n\nV. $d = 5 \\Rightarrow x = 21y \\Rightarrow x = 21$, $y = 1$ – false, since $xy \\mid 24$.\n\nVI. $d = 6 \\Rightarrow x = 76y \\Rightarrow x = 76$, $y = 1$ – false, since $xy \\mid 24$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22281,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute-angled triangle. Points $D$, $E$, and $F$ lie on sides $\\overline{BC}$, $\\overline{CA}$, and $\\overline{AB}$, respectively. Given $|AB| > |AC|$, we have $|BD| > |CD|$.\n\nSuppose $\\angle ACB < \\angle ABC$; then $C$ lies between $B$ and $P$. Also, $C$ lies between $A$ and $Q$, while $R$ lies on segment $\\overline{BF}$.\n\n\n\nLet $H$ be the orthocentre of triangle $ABC$, and $M$ the midpoint of $\\overline{BC}$. Prove that $|BN| > |CN|$, where $N$ is a point on segment $\\overline{CM}$, $N \\neq M$, and $N$ lies inside the circle $k$ circumscribed about quadrilateral $PQMR$.",
"options": [],
"answer": "See solution",
"solution": "The lines $DR$ and $EF$ are parallel, so $\\angle DRF = \\angle EFA$. The quadrilateral $AFHE$ is cyclic ($\\angle AFH = \\angle AEH = 90^\\circ$), so $\\angle EFA = \\angle EHA$. The quadrilateral $BDHF$ is cyclic ($\\angle BDH = \\angle BFH = 90^\\circ$), so $\\angle EHA = \\angle DHB = \\angle DFB = \\angle DFR$. Thus, $\\angle DRF = \\angle DFR$ implies $|DR| = |DF|$.\n\nSimilarly, $DQ$ and $FE$ are parallel, so $\\angle DQE = \\angle FEA$. $AFHE$ is cyclic, so $\\angle FEA = \\angle FHA$. $CEHD$ is cyclic ($\\angle CEH = \\angle CDH = 90^\\circ$), so $\\angle FHA = \\angle DHC = \\angle DEC = \\angle DEQ$. Thus, $\\angle DQE = \\angle DEQ$ implies $|DQ| = |DE|$.\n\nPoints $M$, $F$, $E$, and $D$ lie on the nine-point circle. $\\angle FMD = 180^\\circ - \\angle FED = \\angle PED$. Also, $\\angle FDH = \\angle FBH = \\angle ABE = 90^\\circ - \\angle CAB = \\angle ACF = \\angle ECH = \\angle EDH$.\n\nTherefore, $\\angle FDM = 90^\\circ - \\angle FDH = 90^\\circ - \\angle EDH = \\angle PDE$, so triangles $FDM$ and $PDE$ are similar. Thus,\n\n$$\n\\frac{|DF|}{|DM|} = \\frac{|DP|}{|DE|},\n$$\n\nwhich gives $|DF| \\cdot |DE| = |DP| \\cdot |DM|$, i.e.\n\n$$\n|DR| \\cdot |DQ| = |DP| \\cdot |DM|,\n$$\n\nso quadrilateral $PQMR$ is cyclic.\n\nLet $k$ be its circumscribed circle. Since $\\angle NQP + \\angle NRP < 180^\\circ$, $N$ lies inside $k$, i.e., on segment $\\overline{CM}$, and $N \\neq M$. Since $M$ is the midpoint of $\\overline{BC}$, we conclude $|BN| > |CN|$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22282,
"subject": "Mathematics (Olympiad)",
"question": "Let $N > M > 1$ be fixed integers. There are $N$ people playing in a chess tournament; each pair of players plays each other once, with no draws. It turns out that for each sequence of $M + 1$ distinct players $P_0, P_1, \\dots, P_M$ such that $P_{i-1}$ beat $P_i$ for each $i = 1, \\dots, M$, player $P_0$ also beat $P_M$. Prove that the players can be numbered $1, 2, \\dots, N$ in such a way that, whenever $a \\ge b + M - 1$, player $a$ beat player $b$.",
"options": [],
"answer": "See solution",
"solution": "**Solution.** Write $P \\succ Q$ if player $P$ beat player $Q$.\n\n**Lemma 1.** Any set of $K > 1$ players can be arranged in a sequence $P_1, \\dots, P_K$ such that $P_1 \\succ P_2 \\succ \\dots \\succ P_K$.\n\n*Proof.* Let $P_1 \\succ \\dots \\succ P_J$ be the longest such sequence that can be formed from any subset of the given players, and suppose for contradiction that some player $Q$ among the $K$ players does not appear in the sequence. If all of $P_1, \\dots, P_J$ beat $Q$, we append $Q$ to the end of the sequence. Otherwise, there exists a smallest $i$ such that $Q \\succ P_i$, and then we can insert $Q$ into the sequence just before $P_i$. Either way, we have created a sequence of length $J+1$, contradicting the maximality of $J$. $\\square$\n\nNow, we will prove the problem statement for all $N \\geq 0$ by induction. For the base case, if $N < M$, just assign numbers arbitrarily, and the conclusion holds vacuously.\n\nNow suppose the result holds for any number of players less than $N$; we will show it for $N$ players. Let $P$ be the player who beat the most opponents. If $P$ beat all other players, we can assign $P$ the number $N$ and use the induction hypothesis for the remaining players. So we may assume $P$ was beaten by $K \\geq 1$ other players.\n\nDefine a *top-cycle of order* $L$ to be a sequence $C_L = (P_0, \\dots, P_L)$ of players such that\n\n- $P_0 \\succ P_1 \\succ P_2 \\succ \\dots \\succ P_L \\succ P_0$;\n- $P_L = P$;\n- all players not in the sequence were beaten by $P$.\n\nWe will construct a top-cycle $C_{K+1}$ of order $K+1$. By the lemma, the players who beat $P$ can be arranged into a sequence $P_1, \\dots, P_K$ such that $P_{i-1} \\succ P_i$ for each $i$. Let $P_{K+1} = P$. Notice that there exists a player $P_0$ such that $P \\succ P_0 \\succ P_1$: otherwise, $P_1$ beat every player beaten by $P$ and also beat $P_2$, so $P_1$ beat more players than $P$, contradicting the choice of $P$. This choice of $P_0$ completes the construction of $C_{K+1}$.\n\nWe also claim that $K+1 < M$. We approach indirectly by assuming $K+1 \\geq M$. Then for each $Q$ that was beaten by $P$, either $Q = P_0$ or $Q$ is not in $C_{K+1}$. Either way, we have $P_{K-M+2} \\succ P_{K-M+3} \\succ \\dots \\succ P_{K+1} \\succ Q$, and then $P_{K-M+2} \\succ Q$ by the given. But we also have $P_{K-M+2} \\succ P_{K-M+3}$. Therefore, $P_{K-M+2}$ beat every player who was beaten by $P$ and at least one more player, again contradicting the choice of $P$.\n\nNow, given a top-cycle $C_L$ of any order $L$, we claim that either all players in the cycle beat all players not in the cycle, or else we can insert another player to form a top-cycle $C_{L+1}$ of order $L+1$. Indeed, suppose some player $Q$ not in the cycle was not beaten by all players in the cycle. Take the smallest $i$ such that $Q$ beat $P_i$, and then inserting $Q$ just before $P_i$ gives us a top-cycle of order $L+1$. (If $i=0$ then we need to check that $P$ beat $Q$, but this follows from the fact that $C_L$ was a top-cycle.)\n\nStart with $C_{K+1}$ and repeatedly expand the cycle as just described. Eventually we must reach a top-cycle $C_L$ such that all players in $C_L$ beat all players not in $C_L$, and then the expanding must stop. We claim that when this happens, $L < M$. To prove this, it suffices to show that if we ever reach a top-cycle of order $M-1$, then all players in the cycle beat all players not in the cycle (so that we can expand no further). So let $P_0 \\succ P_1 \\succ \\dots \\succ P_{M-1} = P$ be the top-cycle, and let $Q$ be any player outside the top-cycle. Then $P \\succ Q$ and the given imply $P_0 \\succ Q$. But then we also have $P_1 \\succ P_2 \\succ \\dots \\succ P_{M-1} \\succ P_0 \\succ Q$, implying $P_1 \\succ Q$. Repeating this process, we get $P_i \\succ Q$ for each $i$. This shows that all players in the cycle beat all players outside the cycle, as claimed.\n\nThis proves that there exists a top-cycle $C_L = (P_0, \\dots, P_L)$ with $L < M$, such that all players in $C_L$ beat all players not in $C_L$.\n\nNow use the induction hypothesis to assign the numbers $1, \\dots, N-L-1$ to all players not in $C_L$. Assign the number $N-L$ to $P_1$ and assign the number $N$ to $P_0$. Finally, assign the remaining numbers $N-L+1, \\dots, N-1$ arbitrarily to the rest of the players in $C_L$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22283,
"subject": "Mathematics (Olympiad)",
"question": "Find the 3-digit number $abc$ such that when it is multiplied by 3, the result is the number with its digits reversed.",
"options": [],
"answer": "See solution",
"solution": "When a 3-digit number is multiplied by 3, it cannot be larger than 2997. Hence $c$ is either 1 or 2. It follows that $a = 3c$ as there would be no carry in the multiplication.\n\nWe have $3(100a + 10b + c) = 100c + 10b + a$. Substituting $a = 3c$ gives $903c + 30b = 100c + 10b + 3c$, which simplifies to $b = 5c$. Since $b < 10$, the only solution is $c = 1$, $b = 5$, $a = 3$.\n\nTherefore, $abc = 351$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22284,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\mathcal{K}$ be the kite formed by joining two right triangles with legs $1$ and $\\sqrt{3}$ along a common hypotenuse. Eight copies of $\\mathcal{K}$ are used to form the polygon shown below. What is the area of triangle $\\triangle ABC$?\n\n\n\n\n\n(A) $2 + 3\\sqrt{3}$ \n(B) $\\frac{9}{2}\\sqrt{3}$ \n(C) $\\frac{10 + 8\\sqrt{3}}{3}$ \n(D) $8$ \n(E) $5\\sqrt{3}$",
"options": [],
"answer": "See solution",
"solution": "Let points D, E, F, and G be labeled as shown. Points A, D, and B are collinear, and the line through them is perpendicular to line EFG. Furthermore, the distance from C to line AB is equal to the distance from E to line AB, which in turn is equal to EG. The area of $\\triangle ABC$ is thus $\\frac{1}{2} \\cdot AB \\cdot EG$.\n\n\n\nTo compute $AB$, observe that $\\overline{AG}$ is the longer leg of a $30$-$60$-$90^\\circ$ triangle with hypotenuse $\\sqrt{3}$. This implies that $AG = \\frac{3}{2}$, from which $AB = 4 \\cdot AG = 6$. To compute $EG$, observe that in the same right triangle as before, $FG = \\frac{\\sqrt{3}}{2}$. This implies that $EG = EF + FG = \\sqrt{3} + \\frac{\\sqrt{3}}{2} = \\frac{3\\sqrt{3}}{2}$.\n\nTherefore the area of $\\triangle ABC$ is $\\frac{1}{2} \\cdot 6 \\cdot \\frac{3\\sqrt{3}}{2} = \\frac{9\\sqrt{3}}{2}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22285,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(f(x) + x + y) = f(x + y) + y f(y)\n$$\n\nfor all real numbers $x, y$.",
"options": [],
"answer": "See solution",
"solution": "The only solution is $f(x) = 0$ for all $x$.\n\nSuppose there exists $\\alpha$ such that $f(\\alpha) = 0$. Then for all real $y$:\n\n$$\nf(\\alpha + y) = f(f(\\alpha) + \\alpha + y) = f(\\alpha + y) + y f(y)\n$$\n\nSo $y f(y) = 0$ for all $y$, which implies $f(y) = 0$ for all $y \\neq 0$. If $f(0) = 0$, then $f(x) = 0$ for all $x$.\n\nNow, substitute $y = 0$ into the original equation:\n\n$$\nf(f(x) + x) = f(x)\n$$\n\nSubstitute $y = f(x)$:\n\n$$\nf(f(x) + x + f(x)) = f(x + f(x)) + f(x) f(f(x))\n$$\n\nLet $z = x + f(x)$. Then:\n\n$$\n\\begin{align*}\nf(x) &= f(x + f(x)) \\\\\n &= f(z) \\\\\n &= f(f(z) + z) \\\\\n &= f(f(x + f(x)) + x + f(x)) \\\\\n &= f(f(x) + x + f(x)) \\\\\n &= f(x + f(x)) + f(x) f(f(x)) \\\\\n &= f(x) + f(x) f(f(x))\n\\end{align*}\n$$\n\nThus, $f(x) f(f(x)) = 0$ for all $x$. Setting $x = 0$ gives $f(f(0)) = f(0)$, so $f(0)^2 = 0$, hence $f(0) = 0$.\n\nTherefore, $f(x) = 0$ for all $x$ is the only solution.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22286,
"subject": "Mathematics (Olympiad)",
"question": "Нека $a$, $b$ и $c$ се позитивни реални броеви за кои важи $abc = 1$. Докажи дека важи неравенството\n$$\n\\frac{1}{2}(\\sqrt{a} + \\sqrt{b} + \\sqrt{c}) + \\frac{1}{1+a} + \\frac{1}{1+b} + \\frac{1}{1+c} \\geq 3.\n$$\n\nКога важи равенство?",
"options": [],
"answer": "See solution",
"solution": "Бидејќи $(1 - \\sqrt{bc})^2 \\geq 0$, следува дека $1 + bc \\geq 2\\sqrt{bc}$, т.е. $\\frac{1}{2\\sqrt{bc}} \\geq \\frac{1}{1+bc}$. Добиваме дека\n\n$$\n\\frac{\\sqrt{a}}{2} + \\frac{1}{1+a} = \\frac{1}{2\\sqrt{bc}} + \\frac{1}{1+a} \\geq \\frac{1}{1+bc} + \\frac{1}{1+a} = \\frac{1}{1+\\frac{1}{a}} + \\frac{1}{1+a} = 1.\n$$\n\nНа ист начин се докажува дека\n$$\n\\frac{\\sqrt{b}}{2} + \\frac{1}{1+b} \\geq 1\n$$\nи\n$$\n\\frac{\\sqrt{c}}{2} + \\frac{1}{1+c} \\geq 1.\n$$\n\nСо собирање на овие три неравенства се добива бараното неравенство. Равенство важи ако и само ако $a = 1$, $b = 1$, $c = 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22287,
"subject": "Mathematics (Olympiad)",
"question": "Suppose that a sequence $t_0, t_1, t_2, \\dots$ is defined by a formula $t_n = An^2 + Bn + C$ for all integers $n \\ge 0$. Here $A$, $B$, and $C$ are real constants with $A \\ne 0$. Determine values of $A$, $B$, and $C$ which give the greatest possible number of successive terms of the sequence which are also successive terms of the Fibonacci sequence. The Fibonacci sequence is defined by $F_0 = 0$, $F_1 = 1$, and $F_m = F_{m-1} + F_{m-2}$ for $m \\ge 2$.",
"options": [],
"answer": "See solution",
"solution": "We let our quadratic be $Ax^2 + Bx + C$, and the Fibonacci sequence be $F_m$.\n\nWe may assume that the first successive term of the sequence which is one of the successive terms of the Fibonacci sequence is $T_0$, since if it were $T_k$, we could consider the quadratic $A(x-k)^2 + B(x-k) + C$. If we can get 5 terms of the Fibonacci sequence, then\n\n$$\n\\begin{aligned}\nT_0 &= C = F_{m-2} \\\\\nT_1 &= A + B + C = F_{m-1} \\\\\nT_2 &= 4A + 2B + C = F_m \\\\\nT_3 &= 9A + 3B + C = F_{m+1} \\\\\nT_4 &= 16A + 4B + C = F_{m+2}\n\\end{aligned}\n$$\n\nNow, $F_{m-2} + F_{m-1} = F_m$, so $3A + B - C = 0$. Similarly, $4A - C = 0$ and $3A - B - C = 0$. From the first and last of these we see that $B = 0$, and then from the second we have $A = C = 0$. Thus there is no sequence of $T_0, \\dots, T_4$ which gives successive terms of the Fibonacci sequence, and so the greatest possible number of terms $T_i$ which give successive terms in the Fibonacci sequence is 4, i.e., $T_0, \\dots, T_3$.\n\nTrying some values of $T_0$ we see that this is actually achievable when $T_0 = 2$, in which case $C = 2$ and $A = B = \\frac{1}{2}$, and so our quadratic is $x^2/2 + x/2 + 2$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22288,
"subject": "Mathematics (Olympiad)",
"question": "Find the smallest positive integer $x$ such that $12x = 25y^2$, where $y$ is a positive integer.",
"options": [],
"answer": "See solution",
"solution": "Method 1\n\nWe have $2^2 \\times 3x = 5^2y^2$ where $x$ and $y$ are integers. So $3$ divides $y^2$.\nSince $3$ is prime, $3$ divides $y$.\n\nHence $3$ divides $x$. Also $25$ divides $x$. So the smallest value of $x$ is $3 \\times 25 = 75$.\n\nMethod 2\n\nThe smallest value of $x$ will occur with the smallest value of $y$.\nSince $12$ and $25$ are relatively prime, $12$ divides $y^2$.\n\nThe smallest value of $y$ for which this is possible is $y = 6$.\nSo the smallest value of $x$ is $\\frac{25 \\times 36}{12} = 75$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22289,
"subject": "Mathematics (Olympiad)",
"question": "Нека $a$, $b$ и $c$ се позитивни реални броеви за кои важи $abc=1$. Докажи дека важи неравенството\n\n$$\n\\frac{1}{2}(\\sqrt{a}+\\sqrt{b}+\\sqrt{c})+\\frac{1}{1+a}+\\frac{1}{1+b}+\\frac{1}{1+c} \\ge 3.\n$$\n\nКога важи равенство?",
"options": [],
"answer": "See solution",
"solution": "Бидејќи $(1-\\sqrt{bc})^2 \\geq 0$, следува дека $1+bc \\geq 2\\sqrt{bc}$, т.е. $\\frac{1}{2\\sqrt{bc}} \\geq \\frac{1}{1+bc}$. Добиваме дека\n\n$$\n\\frac{\\sqrt{a}}{2} + \\frac{1}{1+a} = \\frac{1}{2\\sqrt{bc}} + \\frac{1}{1+a} \\geq \\frac{1}{1+bc} + \\frac{1}{1+a} = \\frac{1}{1+\\frac{1}{a}} + \\frac{1}{1+a} = 1. \\tag{1}\n$$\n\nНа ист начин се докажува дека\n\n$$\n\\frac{\\sqrt{b}}{2} + \\frac{1}{1+b} \\geq 1. \\tag{2}\n$$\n\nи\n\n$$\n\\frac{\\sqrt{c}}{2} + \\frac{1}{1+c} \\geq 1. \\tag{3}\n$$\n\nСо собирање на (1), (2) и (3) се добива бараното неравенство. Да забележиме дека равенство важи ако и само ако $1=\\sqrt{bc}$, т.е. $a=1$. На ист начин добиваме $b=1$ и $c=1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22290,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs of prime numbers $p$ and $q$ such that\n$$\nq^3 = p^2 - p + 1.\n$$",
"options": [],
"answer": "See solution",
"solution": "We have\n$$\nq^3 = p^2 - p + 1 \\Leftrightarrow (q-1)(q^2 + q + 1) = p(p-1). \\quad (1)\n$$\n\nIf $(q-1) \\nmid p$, then $q \\ge p+1$, so $q^3 > p^2 - p + 1$.\n\nHence, $(q^2 + q + 1) \\nmid p$, i.e.\n$$\nq^2 + q + 1 = kp \\quad (2)\n$$\nfor some $k \\in \\mathbb{N}$. It follows that $k(q-1) = p-1$, or $p = kq - k + 1$. Substituting this expression in (2) we obtain $q^2 + (1-k^2)q + (k^2-k+1) = 0$. The discriminant of this quadratic in $q$ is\n$$\nD = (k^2 - 1)^2 - (4k^2 - k + 1) = k^4 - 6k^2 + 4k - 3.\n$$\nThis must be a perfect square. But it is easy to check that\n$$\n(k^2 - 3)^2 < k^4 - 6k^2 + 4k - 3 < (k^2 - 1)^2 \\quad \\text{for any } k > 3.\n$$\nMoreover, the equation $k^4 - 6k^2 + 4k - 3 = (k^2 - 2)^2$ has no solutions.\n\nHence, $k \\le 3$. For $k = 1$ we have $D = -4$, and for $k = 2$ we have $D = -3$, which is impossible. For $k = 3$ we obtain $D = 6^2$, so $q = (8 \\pm 6)/2$. Thus $q = 7$ and $p = k(q-1) + 1 = 19$. It remains to note the primes $p = 19$, $q = 7$ do satisfy the equation.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22291,
"subject": "Mathematics (Olympiad)",
"question": "Given an odd prime $p$, determine all polynomials $f$ and $g$ with integer coefficients such that\n$$\nf(g(X)) = \\sum_{k=0}^{p-1} X^k.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $\\Phi_p(X) = \\sum_{k=0}^{p-1} X^k$ be the $p$-th cyclotomic polynomial. The polynomials $f$ and $g$ with integer coefficients satisfying $f(g(X)) = \\Phi_p(X)$ are:\n\n- $f(X) = \\pm X \\mp a$ and $g(X) = \\pm \\Phi_p(X) + a$, where $a$ is an integer;\n- $f(X) = \\Phi_p(\\pm X \\mp b)$ and $g(X) = \\pm X + b$, where $b$ is an integer.\n\nIn both cases, the signs correspond to one another (e.g., $f(X) = -X + a$ and $g(X) = -\\Phi_p(X) + a$ are admissible, but $f(X) = -X + a$ and $g(X) = \\Phi_p(X) + a$ are not).\n\n**Outline of reasoning:**\n\n- If $\\deg g = 1$, then $g(X) = \\pm X + a$ and $f(X) = \\Phi_p(\\pm X \\mp a)$.\n- If $\\deg g > 1$, then $g(X) = \\pm \\Phi_p(X) + a$ and $f(X) = \\pm X \\mp a$.\n- The irreducibility of $\\Phi_p$ and its derivative in $\\mathbb{Z}[X]$ (by Eisenstein's criterion) ensures these are the only possibilities.\n- For $\\deg g > 1$, a root analysis shows $\\deg g = p-1$ and $g(X)$ is a linear shift of $\\Phi_p(X)$.\n\nThus, all such pairs $(f, g)$ are as described above.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22292,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs of positive integers $a, b$ satisfying the following conditions:\n\n1. $a$ divides $b^4 + 1$,\n2. $b$ divides $a^4 + 1$,\n3. $\\lfloor \\sqrt{a} \\rfloor = \\lfloor \\sqrt{b} \\rfloor$.",
"options": [],
"answer": "See solution",
"solution": "The only solutions are $(1, 1)$, $(1, 2)$, and $(2, 1)$, which clearly work. Now we show there are no others.\n\nObviously, $\\gcd(a, b) = 1$, so the problem conditions imply\n\n$$\nab \\mid (a - b)^4 + 1$$\n\nsince each of $a$ and $b$ divide the right-hand side. We define\n\n$$k \\stackrel{\\text{def}}{=} \\frac{(b - a)^4 + 1}{ab}.$$ \n\n**Claim (Size estimate):** We must have $k \\le 16$.\n\n*Proof.* Let $n = \\lfloor \\sqrt{a} \\rfloor = \\lfloor \\sqrt{b} \\rfloor$, so that $a, b \\in [n^2, n^2 + 2n]$. We have\n\n$$ab \\ge n^2(n^2 + 1) \\ge n^4 + 1$$\n$$(b-a)^4 + 1 \\le (2n)^4 + 1 = 16n^4 + 1$$\n\nwhich shows $k \\le 16$. $\\square$\n\n**Claim (Orders argument):** In fact, $k = 1$.\n\n*Proof.* First, note that $k$ cannot be even: if it was, then $a, b$ have opposite parity, but then $4 \\mid (b-a)^4 + 1$, contradiction.\n\nThus $k$ is odd. However, every odd prime divisor of $(b-a)^4 + 1$ is congruent to $1 \\pmod{8}$ and is thus at least $17$, so $k = 1$ or $k \\ge 17$. It follows that $k = 1$. $\\square$\n\nAt this point, we have reduced to solving\n\n$$ab = (b - a)^4 + 1$$\n\nand we need to prove the claimed solutions are the only ones. Write $b = a + d$, and assume WLOG that $d \\ge 0$: then we have $a(a + d) = d^4 + 1$, or\n\n$$a^2 - da - (d^4 + 1) = 0.$$ \n\nThe discriminant $d^2 + 4(d^4 + 1) = 4d^4 + d^2 + 4$ must be a perfect square.\n\n- The cases $d = 0$ and $d = 1$ lead to pairs $(1, 1)$ and $(1, 2)$.\n- If $d \\ge 2$, then we can sandwich\n\n$$(2d^2)^2 < 4d^4 + d^2 + 4 < 4d^4 + 4d^2 + 1 = (2d^2 + 1)^2,$$\n\nso the discriminant is not a square.\n\nThe solution is complete.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22293,
"subject": "Mathematics (Olympiad)",
"question": "Determine the maximum number of kings that may be placed on a $12 \\times 12$ chessboard so that each king threatens exactly one other king.",
"options": [],
"answer": "See solution",
"solution": "The required maximum is 56 and is achieved, for instance, by the configuration below.\n\n\n\nThe fact that 56 is an upper bound for the cardinalities of configurations satisfying the condition in the statement follows from the lemma below. For more convenience, we say that the kings in such a configuration threaten in *disjoint pairs*.\n\n**Lemma.** *The cardinality of a configuration of kings threatening in disjoint pairs on an $n \\times n$ chessboard is at most $\\lfloor (n+1)^2/3 \\rfloor$.*\n\nTo prove the lemma, subdivide each $1 \\times 1$ cell — henceforth a *cell* — of the $n \\times n$ chessboard into four $1/2 \\times 1/2$ cells — henceforth *small cells* — and add a square frame of small cells to form a $2(n+1) \\times 2(n+1)$ square array of small cells. Define the *range* of a king placed on the given $n \\times n$ chessboard to be the unique $4 \\times 4$ square array of small cells whose interior contains the cell the king is placed in. Notice that two kings threaten each other on the $n \\times n$ chessboard if and only if their ranges share at least one small cell — in fact, there are exactly eight in case of a horizontal or vertical threat, and exactly four in case of a diagonal threat. Assign each (unordered) pair of kings threatening one another the number of small cells (or the area) their ranges cover: 24 in case of a horizontal or vertical threat, and 28 in case of a diagonal threat. If the kings in a configuration on the $n \\times n$ chessboard threaten in disjoint pairs, say $k$ horizontally or vertically and $\\ell$ diagonally, then the number of kings is clearly $2(k + \\ell)$ and their ranges cover $24k + 28\\ell$ small cells. Since the total number of small cells is $4(n+1)^2$, it follows that $4(n+1)^2 \\ge 24k + 28\\ell \\ge 24(k+\\ell)$ whence the conclusion.\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22294,
"subject": "Mathematics (Olympiad)",
"question": "Let $T$ be the centroid of a triangle $ABC$. Consider two isosceles right-angled triangles $BTK$ and $CTL$ so that $K$ lies in the half-plane $BTC$ and $L$ lies in the half-plane $CTA$. Finally, denote the centre of the side $BC$ as $D$ and the centre of $KL$ as $E$. Determine all the possible values of the ratio $\\frac{AT}{DE}$.",
"options": [],
"answer": "See solution",
"solution": "We shall prove that the ratio has to be equal to $2\\sqrt{2}$.\n\n\n\nFirst, we shall observe that the triangles $BTC$ and $KTL$ (coloured turquoise and yellow in the diagram) are similar, since\n\n$$\n\\angle BTC = \\angle BTK + \\angle KTC = 45^{\\circ} + \\angle KTC = \\angle KTC + \\angle CTL = \\angle KTL,\n$$\n\nand by similarity of the triangles $BKT$ and $CLT$, we have $\\frac{BT}{CT} = \\frac{KT}{LT}$. The ratio of similarity has to be $\\sqrt{2}$, since $\\frac{BT}{KT} = \\sqrt{2}$.\n\nSince $D$ is a midpoint of $BC$ and $E$ is the midpoint of $KL$, the triangles $BTD$ and $KTE$ are also similar. This tells us that $\\angle BTD = \\angle KTE$ and $\\frac{BT}{TD} = \\frac{KT}{TE}$. Subtracting $\\angle KTD$ from the equality gives us $\\angle BTK = \\angle DTE$ and the equality of ratios can be rearranged to $\\frac{BT}{KT} = \\frac{TD}{TE}$, so the triangles *BTK* and *DTE* are similar. Therefore, the triangle *DTE* is also a right-angled isosceles triangle.\n\nFinally, recall that since *T* is the centroid, we have $AT = 2\\,TD$, so the desired ratio can be computed simply as\n\n$$\n\\frac{AT}{DE} = 2\\frac{DT}{DE} = 2\\sqrt{2},\n$$\n\nas desired.\n\n**Remark.** Both halves of the solution can be simplified by using of *spiral similarity*: by definition, the triangle $CTL$ is the image of $BTK$ under some spiral similarity $\\phi$ centered at $T$. By invoking a standard property of spiral similarities (namely, that it always comes in pairs), we deduce that there also exists some spiral similarity $\\phi'$ sending the triangle $BTC$ to $KTL$. This spiral similarity also carries $BTD$ to $KTE$, and invoking the aforementioned property of spiral similarities gives us a spiral similarity $\\phi''$ centered at $T$ sending $BTK$ to $DTE$.\n\n**Remark.** There is a certain configuration assumption involved in the solution (namely, that the point $K$ lies in the interior of the angle $BTC$), which could be removed by considering oriented angles.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22295,
"subject": "Mathematics (Olympiad)",
"question": "Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients of $2$ and $-2$, respectively. The graphs of both polynomials pass through the two points $(16, 54)$ and $(20, 53)$. Find $P(0) + Q(0)$.",
"options": [],
"answer": "See solution",
"solution": "Because the leading coefficients of $P(x)$ and $Q(x)$ are negatives of each other, the polynomial $R(x) = P(x) + Q(x)$ is linear. Furthermore, $R(16) = 54 + 54 = 108$ and $R(20) = 53 + 53 = 106$. It follows that $R(x) = 116 - 0.5x$, so $P(0) + Q(0) = R(0) = 116$.\n\nNote that\n\n$$\nP(x) = 2x^2 - \\frac{289}{4}x + 698 \\quad \\text{and} \\quad Q(x) = -2x^2 + \\frac{287}{4}x - 582.\n$$",
"topic": "Discrete Mathematics",
"subtopic": "Other"
},
{
"id": 22296,
"subject": "Mathematics (Olympiad)",
"question": "$AE = AC$ and $AD = AB$. Denote by $M$ and $N$ the midpoints of $[BC]$ and $[DE]$ respectively, and let $R = EC \\cap BD$. Show that $MN = RA$.\n\n",
"options": [],
"answer": "See solution",
"solution": "The hypothesis implies that triangles $\\triangle ACE$ and $\\triangle ABD$ are right-angled and isosceles, so the triangle $\\triangle RBE$ is also right-angled and isosceles, that is $m(\\widehat{RBE}) = m(\\widehat{REB}) = 45^\\circ$. By the equality of triangles $\\triangle ABC$ and $\\triangle ADE$ (by congruence), we get $BC = DE$.\n\nIn the triangles $BRC$, $ABC$, $RDE$, $ADE$, lines $RM$, $AM$, $RN$, $AN$ are medians corresponding to right angles, so $RM = \\frac{1}{2}BC$, $AM = \\frac{1}{2}BC$, $RN = \\frac{1}{2}DE$, $AN = \\frac{1}{2}DE$, implying $RM = AM = RN = AN$, that is, the quadrilateral $AMRN$ is a rhombus.\n\nIn the right-angled triangle $ABC$, as $AM$ is a median, the triangle $\\triangle MAC$ is isosceles, so $m(\\widehat{MAC}) = m(\\widehat{MCA})$. In the same way, in $\\triangle ADE$, $m(\\widehat{NAD}) = m(\\widehat{NDA})$. This gives $m(\\widehat{MAN}) = m(\\widehat{MAC}) + m(\\widehat{NAC}) = m(\\widehat{MCA}) + m(\\widehat{NDA}) = m(\\widehat{MCA}) + m(\\widehat{ABC}) = 90^\\circ$. By these properties, the quadrilateral $AMRN$ is a square, and, as a conclusion, $MN = RA$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22297,
"subject": "Mathematics (Olympiad)",
"question": "Let $f$ be a function defined on the non-negative real numbers such that for all $x, y \\geq 0$,\n\n$$\nf(x) + f(y) \\leq \\frac{f(x + y)}{2}\n$$\n\nand\n\n$$\n\\frac{f(x)}{x} + \\frac{f(y)}{y} \\geq \\frac{f(x + y)}{x + y}\n$$\n\nfor $x, y > 0$. Find all such functions $f$.",
"options": [],
"answer": "See solution",
"solution": "Substituting $x = y = t$ ($t \\geq 0$) into the two inequalities, we get $4f(t) \\leq f(2t)$ and $4f(t) \\geq f(2t)$. So $f(2t) = 4f(t)$ for $t \\geq 0$. Applying the equality $m$ times, $f(2^m t) = 2^{2m} f(t)$.\n\nLet $g(x) = \\frac{f(x)}{x}$. We prove $g(nt) = n g(t)$ for any positive integer $n$ and any positive real $t$. For $n = 2^m$, this is already proved. Applying the inequality $g(x) + g(y) \\geq g(x + y)$, for positive integers $n$ and positive real $t$:\n\n$$\ng(nt) \\leq \\underbrace{g(t) + \\dots + g(t)}_{n} = n g(t).\n$$\n\nChoose $m$ so $2^m > n$, then\n\n$$\ng(2^m t) \\leq g(nt) + g((2^m - n)t) \\leq n g(t) + (2^m - n) g(t) = 2^m g(t).\n$$\n\nBut $g(2^m t) = 2^m g(t)$, so $g(nt) + g((2^m - n)t) \\leq n g(t) + (2^m - n) g(t)$. Hence, $g(nt) = n g(t)$.\n\nNext, we show $g$ is an improper monotone decreasing function. For $t > 0$, $f(t) + f(2t) \\leq \\frac{f(3t)}{2}$. Since $f(t) = t g(t)$, $f(2t) = 4 g(t)$, $f(3t) = 9 g(t)$, so $5 g(t) \\leq \\frac{9}{2} g(t)$, thus $g(t) \\leq 0$ for $t > 0$. For $0 < x \\leq y$, $g(x) \\geq g(x) + g(y - x) \\geq g(y)$, so $g$ is monotone decreasing.\n\nLet $g(1) = a \\leq 0$. We prove $g(t) = a t$ for $t > 0$. If $g(t) < a t$, there exists rational $\\frac{p}{q} > t$ with $g(t) < \\frac{p a}{q}$. But $g(\\frac{p}{q}) = \\frac{p}{q} a$, so $g(t) < g(\\frac{p}{q})$ and $\\frac{p}{q} > t$, contradicting monotonicity. Similarly, $g(t) > a t$ leads to contradiction. Thus, $g(t) = a t$ for $t > 0$.\n\nTherefore, $f(x) = x g(x) = a x^2$ ($a \\leq 0$). Also, $f(x) + f(y) - \\frac{f(x + y)}{2} = \\frac{a}{2} (x - y)^2 \\leq 0$ and $\\frac{f(x)}{x} + \\frac{f(y)}{y} - \\frac{f(x + y)}{x + y} = 0$ for such $f$. Thus, all such $f$ are $f(x) = a x^2$ with $a \\leq 0$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22298,
"subject": "Mathematics (Olympiad)",
"question": "There are $n \\geq 5$ points, labeled $1, 2, \\ldots, n$, arbitrarily arranged on a circle. We call such a permutation $S$. For a permutation, a \"descending chain\" is a clockwise sequence of consecutive points (at least two) with descending labels, and it is not a subchain of any longer descending chain; the \"pivot\" of a descending chain is the point with the largest label, and all other points are \"non-pivots\". For example, the clockwise permutation $5, 2, 4, 1, 3$ contains two descending chains $5, 2$ and $4, 1$, where $5, 4$ are pivots, $2, 1$ are non-pivots. Apply the following operations on $S$: first, find all the descending chains of $S$ and delete all the non-pivots; then, for the remaining points, find all the descending chains and delete all the non-pivots; and so on, until no descending chains can be found. Let $G(S)$ be the number of descending chains that have appeared in the whole process; $A(n)$ be the average of $G(S)$ among all permutations $S$ of $1, 2, \\ldots, n$.\n\n1. Find $A(5)$.\n2. For $n \\geq 6$, prove: $$\\frac{83}{120}n - \\frac{1}{2} \\leq A(n) \\leq \\frac{101}{120}n - \\frac{1}{2}.$$",
"options": [],
"answer": "See solution",
"solution": "Evidently, $A(1) = 0$, $A(2) = 1$. Define $S(n) = A(1) + A(2) + \\cdots + A(n)$. All permutations, sequences, etc., are in the clockwise direction.\n\n**Lemma:** For $n \\geq 3$, $A(n)$ satisfies\n$$\n\\frac{2}{n-1} S(n-1) + \\frac{1}{2} \\leq A(n) \\leq \\frac{2}{n-1} S(n-1) + \\frac{2n-3}{2(n-1)}.\n$$\n\n**Proof of Lemma:**\nSuppose $S'$ is an arbitrary permutation of $1, 2, \\dots, n$ on a circle. Fix $n$ as the starting number and consider $n-1$: they divide the circle into two sequences $S_1', S_2'$, the former starts with $n$ and the latter starts with $n-1$. For a sequence whose starting number is the largest, after a few operations it will reduce to a single number which is the starting number itself. We say this sequence is \"completed\". The number of descending chains that will appear in $S_1'$ and $S_2'$ are $G(S_1')$, $G(S_2')$, respectively, and there is an extra part when they combine.\n\nDuring the process, if $S_1'$ is completed earlier, then $n$ will catch up with $n-1$ and take the role of $n-1$ thereafter, generating no additional descending chains in the process. Symmetric to $S'$, we interchange the numbers $n$ and $n-1$ and denote this permutation as $S''$. In $S''$, $n-1$ will catch up with $n$ first, and it must wait until the sequence that starts with $n$ is completed, then $n$ and $n-1$ take one operation at last. In this case, the two symmetric permutations $S'$ and $S''$ will generate $\\frac{1}{2}$ descending chain on average beyond $G(S_1') + G(S_2')$.\n\nThe only exception is when $S_1'$ and $S_2'$ are completed at the same time. Then $S'$ and $S''$ (obtained from interchanging $n$ and $n-1$ in $S'$) both take one additional operation to eliminate $n-1$. In this case, they will generate another $\\frac{1}{2}$ descending chain on average beyond $G(S_1') + G(S_2') + \\frac{1}{2}$. By assumption $n \\geq 3$: if one of $S_1'$ and $S_2'$ has length $1$, of which the probability is $\\frac{1}{n-1}$, the exceptional situation cannot occur, and $\\frac{G(S') + G(S'')}{2} = G(S_1') + G(S_2') + \\frac{1}{2}$; for all others, we have\n$$\nG(S_1') + G(S_2') + \\frac{1}{2} \\leq \\frac{G(S') + G(S'')}{2} \\leq G(S_1') + G(S_2') + 1.\n$$\n\nPutting them together and taking the average, it follows that\n$$\n\\frac{1}{n-1} \\sum_{i=1}^{n-1} (A(i) + A(n-i)) + \\frac{1}{2} \\leq A(n) \\leq \\frac{1}{n-1} \\sum_{i=1}^{n-1} (A(i) + A(n-i)) + \\left(1 - \\frac{1}{2(n-1)}\\right),\n$$\nwhich is exactly the lemma. The lemma is verified.\n\nBy the lemma, it follows that\n$$\n\\frac{n+1}{n-1} S(n-1) + \\frac{1}{2} \\leq S(n) \\leq \\frac{n+1}{n-1} S(n-1) + \\frac{2n-3}{2(n-1)}.\n$$\n\nNow we find\n$$\nS(n) + \\frac{n}{2} \\geq \\frac{n+1}{n-1} \\left( S(n-1) + \\frac{n-1}{2} \\right),\n$$\nand\n$$\nS(n) + n - \\frac{1}{4} \\leq \\frac{n+1}{n-1} \\left( S(n-1) + (n-1) - \\frac{1}{4} \\right).\n$$\n\nUse these repeatedly to derive\n$$\nS(n) \\geq \\frac{(n+1)n}{(k+1)k} \\left( S(k) + \\frac{k}{2} \\right) - \\frac{n}{2},\n$$\n$$\nS(n) \\leq \\frac{(n+1)n}{(k+1)k} \\left( S(k) + k - \\frac{1}{4} \\right) - \\left( n - \\frac{1}{4} \\right)\n$$\nfor $k \\geq 2$. Some direct calculation yields $A(3) = \\frac{3}{2}$, $A(4) = \\frac{7}{3}$. For $n = 5$, it is necessary to count the number of exceptional situations, that is, the permutations for which $S_1'$ and $S_2'$ are completed at the same time. According to the lemma, it must satisfy:\n\n(a) $5$ and $4$ are not adjacent;\n\n(b) one of the numbers $1, 2, 3$ belongs to $S_1'$, while the other two belong to $S_2'$ and are descending.\n\nFix $n = 5$. There are $2$ ways to fill the number $4$ and $3$ ways to pick the one number for $S_1'$ in (b). So, there are $2 \\times 3 = 6$ permutations each.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 22299,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\neq 0$. For every sequence of integers\n$$\na = a_0, a_1, a_2, \\dots, a_n\n$$\nsatisfying $0 \\leq a_i \\leq i$ for $i = 0, \\dots, n$, define another sequence\n$$\nt(a) = t(a)_0, t(a)_1, t(a)_2, \\dots, t(a)_n\n$$\nby setting $t(a)_i$ to be the number of terms in the sequence $a$ that precede the term $a_i$ and are different from $a_i$.\n\nShow that, starting from any sequence $a$ as above, fewer than $n$ applications of the transformation $t$ lead to a sequence $b$ such that $t(b) = b$.",
"options": [],
"answer": "See solution",
"solution": "**First Solution.** Note first that the transformed sequence $t(a)$ also satisfies the inequalities $0 \\leq t(a)_i \\leq i$, for $i = 0, \\dots, n$. Call any integer sequence that satisfies these inequalities an *index bounded sequence*.\n\nWe prove now that $a_i \\leq t(a)_i$, for $i = 0, \\dots, n$. Indeed, this is clear if $a_i = 0$. Otherwise, let $x = a_i > 0$ and $y = t(a)_i$. None of the first $x$ consecutive terms $a_0, a_1, \\dots, a_{x-1}$ is greater than $x-1$, so they are all different from $x$ and precede $x$. Thus $y \\geq x$, that is, $a_i \\leq t(a)_i$, for $i = 0, \\dots, n$.\n\n$$\n\\begin{array}{c|cccccc}\n & 0 & 1 & \\dots & x-1 & \\dots & i \\\\\n\\hline\n a & a_0 & a_1 & \\dots & a_{x-1} & \\dots & x \\\\\n t(a) & t(a)_0 & t(a)_1 & \\dots & t(a)_{x-1} & \\dots & y\n\\end{array}\n$$\n\nThis already shows that the sequences stabilize after finitely many applications of the transformation $t$, because the value of the index $i$ term in index bounded sequences cannot exceed $i$. Next we prove that if $a_i = t(a)_i$, for some $i = 0, \\dots, n$, then no further applications of $t$ will ever change the index $i$ term. We consider two cases.\n\n* In this case, we assume that $a_i = t(a)_i = 0$. This means that no term on the left of $a_i$ is different from 0, that is, they are all 0. Therefore the first $i$ terms in $t(a)$ will also be 0 and this repeats.\n\n$$\n\\begin{array}{c|cccc}\n & 0 & 1 & \\dots & i \\\\\n\\hline\n a & 0 & 0 & \\dots & 0 \\\\\n t(a) & 0 & 0 & \\dots & 0\n\\end{array}\n$$\n\n* In this case, we assume that $a_i = t(a)_i = x > 0$. The first $x$ terms are all different from $x$. Because $t(a)_i = x$, the terms $a_x, a_{x+1}, \\dots, a_{i-1}$ must then all be equal to $x$. Consequently, $t(a)_j = x$ for $j = x, \\dots, i-1$ and further applications of $t$ cannot change the index $i$ term.\n\n | 0 | 1 | ... | x − 1 | x | x + 1 | ... | i |
|---|
| a | a0 | a1 | ... | ax − 1 | x | x | ... | x |
|---|
| t(a) | t(a)0 | t(a)1 | ... | t(a)x − 1 | x | x | ... | x |
|---|
\n\nFor $0 \\leq i \\leq n$, the index $i$ entry satisfies the following properties: (i) it takes integer values; (ii) it is bounded above by $i$; (iii) its value does not decrease under transformation $t$; and (iv) once it stabilizes under transformation $t$, it never changes again. This shows that no more than $n$ applications of $t$ lead to a sequence that is stable under the transformation $t$.\n\nFinally, we need to show that no more than $n-1$ applications of $t$ is needed to obtain a fixed sequence from an initial $n+1$-term index bounded sequence $a = (a_0, a_1, \\dots, a_n)$. We induct on $n$.\n\nFor $n=1$, the two possible index bounded sequences $(a_0, a_1) = (0, 0)$ and $(a_0, a_1) = (0, 1)$ are already fixed by $t$ so we need zero applications of $t$.\n\nAssume that any index bounded sequence $(a_0, a_1, \\dots, a_n)$ reach a fixed sequence after no more than $n-1$ applications of $t$. Consider an index bounded sequence $a = (a_0, a_1, \\dots, a_{n+1})$. It suffices to show that $a$ will be stabilized in no more than $n$ applications of $t$. We approach indirectly by assuming on the contrary that $n+1$ applications of transformations are needed. This can happen only if $a_{n+1} = 0$ and each application of $t$ increased the index $n+1$ term by exactly 1. Under transformation $t$, the resulting value of index $i$ term will not be affected by index $j$ term for $i < j$. Hence by the induction hypothesis, the subsequence $a' = (a_0, a_1, \\dots, a_n)$ will be stabilized in no more than $n-1$ applications of $t$. Because index $n$ term is stabilized at value $x \\leq n$ after no more than $\\min\\{x, n-1\\}$ applications of $t$ and index $n+1$ term obtains value $x$ after exactly $x$ applications of $t$ under our current assumptions. We conclude that the index $n+1$ term would become equal to the index $n$ term after no more than $n-1$ applications of $t$. However, once two consecutive terms in a sequence are equal they stay equal and stabilize together. Because the index $n$ term needs no more than $n-1$ transformations to be stabilized, $a$ can be stabilized in no more than $n-1$ applications of $t$, which contradicts our assumption of $n+1$ applications needed. Thus our assumption was false.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22300,
"subject": "Mathematics (Olympiad)",
"question": "In the figure below, the large square has sides of length $6$. The circle is tangent to all sides of the large square. The four triangles are identical right-angled triangles placed next to each other; the small square they enclose has its vertices exactly on the circle.\n\n\n\nWhat is the area of the grey triangle?",
"options": [],
"answer": "See solution",
"solution": "$4\\frac{1}{2}$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22301,
"subject": "Mathematics (Olympiad)",
"question": "Let $s$ stand for $\\sin t$, $c$ for $\\cos t$, and let $x = 2sc = \\sin(2t)$. Prove that\n$$\n(\\sin^3 t + \\cos^3 t)(\\sin t + \\cos t)^3 + 16 \\sin^3 t \\cos^3 t \\leq 1 + 3x\n$$\nfor $0 < x \\leq 1$, with equality iff $\\sin^2(2t) = 1$ (i.e., $\\cos^2(2t) = 0$).",
"options": [],
"answer": "See solution",
"solution": "We have\n$$\n\\begin{aligned}\n& (\\sin^3 t + \\cos^3 t)(\\sin t + \\cos t)^3 + 16 \\sin^3 t \\cos^3 t \\\\\n&= (s^3 + c^3)(s + c)^3 + 2x^3 \\\\\n&= (s^2 - sc + c^2)(s + c)^4 + 2x^3 \\\\\n&= (1 - sc)(s^2 + 2sc + c^2)^2 + 2x^3 \\\\\n&= \\left(1 - \\frac{1}{2}x\\right) (1 + x)^2 + 2x^3 = 1 + \\frac{3}{2}x + \\frac{3}{2}x^3.\n\\end{aligned}\n$$\n\nHence, the inequality reduces to proving that\n$$\n1 + 3x \\ge 1 + \\frac{3}{2}x + \\frac{3}{2}x^3\n$$\nor equivalently,\n$$\nx(1 - x^2) \\ge 0,\n$$\nwhich is true since $0 < x \\le 1$. There is equality iff $\\sin^2(2t) = x^2 = 1$, equivalently, iff $\\cos^2(2t) = 0$ as stated.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22302,
"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 conditions:\n\n1. $a_i > M^i$ for any positive integer $i$.\n2. An integer $n$ is non-zero if and only if there exists a positive integer $m$ and $b_1, b_2, \\dots, b_m \\in \\{-1, 1\\}$, with $n = b_1 a_1 + b_2 a_2 + \\dots + b_m a_m$.",
"options": [],
"answer": "See solution",
"solution": "For a given $M > 2$, we construct by induction a sequence $\\{a_n\\}$ that satisfies the requirements.\n\nTake $a_1, a_2$ 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$$A_k = \\{b_1 a_1 + \\dots + b_m a_m \\mid b_1, \\dots, b_m = \\pm 1,\\ 1 \\le m \\le 2k\\}$$\ndoes not contain $0$. It is obvious that $A_k$ is symmetric, i.e., $A_k = -A_k$. For $k=1$, $A_1 = \\{a_1, -a_1, 1, -1\\}$.\n\nLet $n$ be the smallest positive integer not in $A_k$, and $N = \\sum_{i=1}^{2k} a_i$. Now choose positive integers $a_{2k+1}, a_{2k+2}$ satisfying $a_{2k+2} - a_{2k+1} = N + n$, $a_{2k+1} > M^{2k+2}$, and $a_{2k+1} > \\sum_{i=1}^{2k} a_i$.\n\nWe now show that $A_{k+1}$ does not contain $0$ and $n \\in A_{k+1}$. First,\n\n$$n = -\\sum_{i=1}^{2k} a_i - a_{2k+1} + a_{2k+2}.$$\n\nOn the other hand, if $\\sum_{i=1}^{m} b_i a_i = 0$ for $m \\le 2k+2$, as $0 \\notin A_k$, we must have $m = 2k+1$ or $2k+2$.\n\nIf $m = 2k+1$, then\n$$\\left| \\sum_{i=1}^{2k+1} b_i a_i \\right| \\ge a_{2k+1} - \\sum_{i=1}^{2k} a_i > 0.$$\n\nIf $m = 2k+2$ and $b_{2k+1}$ and $b_{2k+2}$ are of the same sign, then\n$$\\left| \\sum_{i=1}^{2k+2} b_i a_i \\right| \\ge a_{2k+1} + a_{2k+2} - \\sum_{i=1}^{2k} a_i > 0;$$\nif $b_{2k+1}$ and $b_{2k+2}$ are of different signs, then\n\n$$\n\\left| \\sum_{i=1}^{2k+2} b_i a_i \\right| = \\left| \\sum_{i=1}^{2k} b_i a_i \\pm (a_{2k+1} - a_{2k+2}) \\right| \\ge |a_{2k+1} - a_{2k+2}| - \\sum_{i=1}^{2k} a_i = N + n - N = n > 0.\n$$\n\nThe sequence $\\{a_n\\}$ thus constructed satisfies the requirements since $0$ is not contained in any $A_k$, and any non-zero integer between $-k$ and $k$ is contained in $A_k$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22303,
"subject": "Mathematics (Olympiad)",
"question": "Given a $4 \\times 4$ grid with entries $a_{i,j}$, define $A_i = \\sum_{j=1}^4 a_{i,j}$ (the sum of the $i$-th row) and $B_j = \\sum_{i=1}^4 a_{i,j}$ (the sum of the $j$-th column). What is the maximum possible value of\n\n$$\nm = \\min\\{A_4 - A_1,\\ B_4 - B_1\\}\n$$\n\nwhere the $a_{i,j}$ are the numbers $1$ through $16$ arranged in the grid, and the rows and columns are ordered so that $A_1 \\leq A_2 \\leq A_3 \\leq A_4$ and $B_1 \\leq B_2 \\leq B_3 \\leq B_4$?\n\n",
"options": [],
"answer": "See solution",
"solution": "Assume $A_1$ is the minimum and $A_4$ is the maximum among $A_1, A_2, A_3, A_4$, and similarly for $B_1$ and $B_4$. By rearranging rows and columns, we can ensure this ordering. Then:\n\n$$\nm \\leq A_4 - A_1,\\quad m \\leq B_4 - B_1\n$$\nso\n$$\nm \\leq \\frac{(A_4 - A_1) + (B_4 - B_1)}{2}\n$$\nNow,\n$$\n\\begin{aligned}\n\\frac{(A_4 - A_1) + (B_4 - B_1)}{2} &= \\frac{1}{2}(a_{4,4} - a_{1,1}) + \\frac{1}{2}(a_{4,4} + a_{4,2} + a_{4,3} + a_{2,4} + a_{3,4} - a_{1,1} - a_{1,2} - a_{1,3} - a_{2,1} - a_{3,1}) \\\\\n&\\leq \\frac{1}{2}(16 - 1) + \\frac{1}{2}(16 + 15 + 14 + 13 + 12 - 1 - 2 - 3 - 4 - 5) \\\\\n&= 35\n\\end{aligned}\n$$\n\nThere exists an arrangement achieving $m = 35$ (see table below), so the maximum $m$ is $35$.\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22304,
"subject": "Mathematics (Olympiad)",
"question": "Sea $n \\ge 1$ y $P(x)$ un polinomio con coeficientes enteros que cumple que los números $P(1), P(2), \\dots, P(n)$ son $1, 2, \\dots, n$ (no necesariamente en este orden). Demuestra que uno de los números $P(0)$ o $P(n + 1)$ es múltiplo de $n!$.",
"options": [],
"answer": "See solution",
"solution": "Si $i$ y $j$ son dos números enteros, se tiene que $i^k - j^k = (i-j)(i^{k-1} + i^{k-2}j + \\dots + ij^{k-2} + j^{k-1})$ es múltiplo de $i-j$. Entonces, si $P(x) = a_m x^m + \\dots + a_2 x^2 + a_1 x + a_0$,\n$$\nP(i) - P(j) = a_m(i^m - j^m) + \\dots + a_2(i^2 - j^2) + a_1(i - j)\n$$\ntambién es múltiplo de $i-j$. En particular, $n-1$ divide a $P(n)-P(1)$. Como $P(1)$ y $P(n)$ son enteros distintos entre $1$ y $n$, tiene que ser $P(1) = 1$ y $P(n) = n$ o al revés, $P(1) = n$ y $P(n) = 1$. En el primer caso, $n-2 = (n-1)-1$ divide a $P(n-1)-P(1) = P(n-1)-1$ y $2 \\le P(n-1) \\le n-1$, luego tiene que ser $P(n-1) = n-1$ y, similarmente, $P(n-2) = n-2$, etc. De forma parecida se ve que en el segundo caso $P(n-1) = 2$, $P(n-2) = 3$, etc. Si $P(i) = i$ para todo $1 \\le i \\le n$, todos estos números son raíces de $P(x) - x$, luego\n$$\nP(x) = c(x)(x-1)(x-2)\\dots(x-n) + x\n$$\npara algún polinomio con coeficientes enteros $c(x)$. Por otro lado, si $P(i) = n-i+1$ para todo $1 \\le i \\le n$, se tiene que todos los enteros $1 \\le i \\le n$ son raíces de $P(x) - n + x - 1$, luego\n$$\nP(x) = c(x)(x-1)(x-2)\\dots(x-n) + n - x + 1\n$$\npara algún polinomio con coeficientes enteros $c(x)$. En el primer caso $P(0) = (-1)^n c(0)n!$ y, en el segundo, $P(n+1) = c(n+1)n!$, luego efectivamente $n!$ divide a $P(0)$ o a $P(n+1)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22305,
"subject": "Mathematics (Olympiad)",
"question": "Twins Mari and Jüri received a total of 5000 candies for their 10th birthdays. Starting from this day, each is allowed to take candies once per day, with the amount taken by any child on any day being less than their age in full years (on their birthday, they use their new age). Each child must take at least one candy every day. Jüri always lets Mari take candies first each day. The children agreed that whoever takes the last candy must buy new candies. Which child can ensure that they do not have to buy new candies, no matter how the other child takes their candies?",
"options": [],
"answer": "See solution",
"solution": "Mari.\n\nMari can avoid taking the last candy by following a specific strategy. On the first day, she takes 8 candies. On each subsequent day, she takes enough candies so that the sum of her candies that day and Jüri's candies from the previous day equals the children's age on the previous day. Since 2024 is a leap year, they will be 10 years old for 366 days. On their 11th birthday, the remaining candies are:\n\n$$5000 - 8 - 366 \\times 10 = 1332$$\n\nAfter 121 more days (at age 11), there will be:\n\n$$1332 - 121 \\times 11 = 1$$\n\ncandy left after Mari has taken her candies for the day. Thus, Jüri is forced to take the last candy.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22306,
"subject": "Mathematics (Olympiad)",
"question": "A box contains several balls, each of which is either black or white. The following procedure is performed repeatedly:\n\n- If two black balls are selected, one is painted white and put back in the box, while the other is removed from the box.\n- If two white balls are selected, one is kept in the box and the other is removed.\n- If one white and one black ball are selected, the black ball is kept in the box and the white ball is removed.\n\nThis process continues until only 3 balls remain in the box. At that point, it is observed that both colors are present among the remaining balls.\n\nDetermine how many white balls and how many black balls remain in the box at the end.",
"options": [],
"answer": "See solution",
"solution": "Let's analyze the effect of each step on the number of black and white balls:\n\n- Selecting two black balls: the number of black balls decreases by 2 (since one is removed and the other is painted white), and the number of white balls increases by 1.\n- Selecting two white balls: the number of white balls decreases by 1 (one is removed, one stays), black balls are unchanged.\n- Selecting one black and one white ball: the white ball is removed, the black ball stays; the number of black balls is unchanged, the number of white balls decreases by 1.\n\nNotice that the number of black balls either stays the same or decreases by 2 in each step. Since the initial number of black balls is odd (2015), the number of black balls will always remain odd after each operation. When only 3 balls remain and both colors are present, the only possibility is that there is 1 black ball and 2 white balls left in the box.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22307,
"subject": "Mathematics (Olympiad)",
"question": "For all positive integers $n$, consider the number\n$$\na_n = 4^{6^n} + 1943.\n$$\nProve that $a_n$ is divisible by $2013$ for all $n \\ge 1$, and find all values of $n$ for which $a_n - 207$ is the cube of a positive integer.",
"options": [],
"answer": "See solution",
"solution": "First, note that $2013 = 3 \\cdot 11 \\cdot 61 = 33 \\cdot 61$.\n\n**Divisibility by $61$:**\n\nSince $4^6 = 4096 \\equiv 9 \\pmod{61}$ and $9^5 \\equiv 1 \\pmod{61}$, we have\n$$\n4^{6^n} = (4^6)^{6^{n-1}} \\equiv 9^{6^{n-1}} \\pmod{61}.\n$$\nBut $6^{n-1}$ is always congruent to $1$ modulo $5$, so $9^{6^{n-1}} \\equiv 9 \\pmod{61}$. Thus,\n$$\na_n = 4^{6^n} + 1943 \\equiv 9 + 1943 = 1952 \\equiv 0 \\pmod{61}.\n$$\nSo $61 \\mid a_n$ for all $n \\ge 1$.\n\n**Divisibility by $33$:**\n\nNote that $6^n - 1$ is divisible by $5$, so $4^{6^n-1} - 1 = (4^5)^p - 1 = 1024^p - 1$ for some $p$. This is divisible by $33$, since $1024 \\equiv 1 \\pmod{33}$. Therefore, $a_n$ is divisible by $33$ as well.\n\nSince $33$ and $61$ are coprime, $2013 = 33 \\cdot 61$ divides $a_n$ for all $n \\ge 1$.\n\n**Second part:**\n\nWe seek $n$ such that $a_n - 207$ is a cube:\n$$\na_n - 207 = 4^{6^n} + 1736 = k^3\n$$\nfor some positive integer $k$. Set $k = 2x$, so\n$$\n4^{6^n} + 1736 = (2x)^3 = 8x^3 \\implies 4^{6^n} = 8x^3 - 1736.\n$$\nBut $4^{6^n}$ is a power of $2$, so set $4^{6^n} = 2^{2 \\cdot 6^n}$. Try small $n$:\n\nFor $n = 1$:\n$$\n4^{6^1} = 4^6 = 4096,\\quad 4096 + 1736 = 5832 = 18^3.\n$$\nSo $n = 1$ works.\n\nFor $n > 1$, $4^{6^n}$ grows rapidly and $a_n - 207$ cannot be a perfect cube.\n\n**Conclusion:**\n\nThe only value of $n$ for which $a_n - 207$ is a cube of a positive integer is $n = 1$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22308,
"subject": "Mathematics (Olympiad)",
"question": "Suppose we have 9 islands, each colored with one of three colors (3 islands per color). Bridges can be built between islands, subject to the following conditions:\n\n1. No two islands of the same color are connected by a bridge.\n2. For any two islands $P_1, P_2$ of the same color and any island $Q$ of a different color, at least one of the pairs $\\{P_1, Q\\}$ or $\\{P_2, Q\\}$ is not connected by a bridge.\n\nHow many ways are there to build bridges between the islands, satisfying these conditions?",
"options": [],
"answer": "See solution",
"solution": "Let us analyze the problem by considering pairs of colors. For any two colors, the requirements reduce to the following: bridges can only connect islands of different colors, and no island is connected to more than one island of the other color.\n\nFor 6 islands (3 of each color), to construct $k$ bridges, we choose $k$ islands from each color and pair them in $k!$ ways. Thus, the number of ways is $$(\\binom{3}{k})^2 \\times k!$$ for $k = 0, 1, 2, 3$.\n\nSumming over all $k$:\n$$\nA = (\\binom{3}{0})^2 \\times 0! + (\\binom{3}{1})^2 \\times 1! + (\\binom{3}{2})^2 \\times 2! + (\\binom{3}{3})^2 \\times 3! = 1^2 \\times 1 + 3^2 \\times 1 + 3^2 \\times 2 + 1^2 \\times 6 = 1 + 9 + 18 + 6 = 34\n$$\n\nSince there are three colors, the total number of ways is $A^3 = 34^3 = 39304$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22309,
"subject": "Mathematics (Olympiad)",
"question": "Given an integer $n \\geq 2$, colour exactly $n$ cells red on 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 over all possible configurations of exactly $n$ red cells.\n\n",
"options": [],
"answer": "See solution",
"solution": "The required minimum is $1 + \\lceil (n+1)/5 \\rceil$, achieved by the configuration described below.\n\nGiven a configuration of exactly $n$ red cells, we show that $N \\geq (n + 6)/5$. 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, rightmost column, top row, and leftmost column, respectively; if, for instance, $a$ is the lower-left corner cell, then the list reads $a$, $b$, $c$, $a$.\n\nLet $[xy]$ denote the (unique) rectangular grid array whose opposite corner cells are $x$ and $y$. 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 $r_{[ab]} + r_{[bc]} + r_{[cd]} + r_{[da]} + r_{[ac]} \\geq 3 \\cdot 2 + 2 \\cdot 2 + (n - 4) = n + 6$. Consequently, $N \\geq (n + 6)/5$.\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 $r \\leq 5$, 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\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 $S_{UR}$ 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\nLet $S_{LR}$ 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 is coloured red, and exactly $n$ cells have been coloured red. 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 $r_{[xy]} \\leq m + 1$. 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.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22310,
"subject": "Mathematics (Olympiad)",
"question": "You plan to organize your birthday party, which will be attended either by exactly $m$ persons or by exactly $n$ persons (you are not sure at the moment). You have a big birthday cake and you want to divide it into several parts (not necessarily equal), so that you are able to distribute the whole cake among the people attending the party with everybody getting cake of equal mass (however, one may get one big slice, while others several small slices—the sizes of slices may differ). What is the minimal number of parts you need to divide the cake, so that it is possible, regardless of the number of guests?",
"options": [],
"answer": "See solution",
"solution": "We claim that the answer is $m + n - \\gcd(m, n)$. \n\nFirstly, consider the cake as the interval $[0, 1]$ and make cuts at points with coordinates $\\frac{k}{m}$ ($0 < k < m$) and $\\frac{l}{n}$ ($0 < l < n$). This ensures that, for either $m$ or $n$ guests, the cake can be distributed so that everyone gets an equal mass (possibly from several pieces).\n\nThere are $m-1$ cuts at $\\frac{k}{m}$ and $n-1$ cuts at $\\frac{l}{n}$, with $\\gcd(m, n) - 1$ of these cuts coinciding. Thus, the total number of cuts is:\n\n$$\nm + n - \\gcd(m, n) - 1\n$$\n\nwhich gives $m + n - \\gcd(m, n)$ parts.\n\nTo show this is minimal, consider a bipartite graph: one side has $m$ vertices (for $m$ guests), the other $n$ vertices (for $n$ guests). Connect $v$ and $u$ by an edge if a piece of cake could go to $v$ (when $m$ attend) or $u$ (when $n$ attend). Each connected component corresponds to a piece of cake, and if a component has $m_1$ vertices on one side and $n_1$ on the other, then $\\frac{m_1}{m} = \\frac{n_1}{n}$, so $m_1 \\ge \\frac{m}{\\gcd(m, n)}$.\n\nThus, the number of components is at most $\\gcd(m, n)$. The graph has $m + n$ vertices and at most $\\gcd(m, n)$ components, so the number of edges (pieces) is at least $m + n - \\gcd(m, n)$ (equality when each component is a tree). $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22311,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$ be the side lengths of a right-angled triangle with hypotenuse $c$, i.e. $c^2 = a^2 + b^2$. The area of this triangle is $\\dfrac{ab}{2}$ and the perimeter is $a + b + c$. For which integer side lengths do the area and perimeter agree exactly?",
"options": [],
"answer": "See solution",
"solution": "Set $\\dfrac{ab}{2} = a + b + c$. Substitute $c = \\sqrt{a^2 + b^2}$ and solve for integer solutions:\n\nSquaring both sides and simplifying leads to $(a - 4)(b - 4) = 8$.\n\nThe positive integer factorizations of $8$ are $1 \\cdot 8$ and $2 \\cdot 4$.\n\nThus, $(a, b) = (5, 12)$ and $(a, b) = (6, 8)$, corresponding to triangles with sides $(5, 12, 13)$ and $(6, 8, 10)$.\n\nA check confirms these are solutions.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22312,
"subject": "Mathematics (Olympiad)",
"question": "Let $x$, $y$, $z$, and $t$ be positive integers such that $xy - zt = x + y = z + t$. Is it possible that $xy$ and $zt$ be simultaneously perfect squares?",
"options": [],
"answer": "See solution",
"solution": "The answer is negative.\n\nFirst, note that $x + y = z + t$ is even. If it were odd, then $xy$ and $zt$ would both be even, so $xy - zt = x + y$ would also be even, which is a contradiction. Thus, $x + y = z + t$ is even, and let $s = \\frac{1}{2}(x + y) = \\frac{1}{2}(z + t)$ be a positive integer.\n\nSuppose, for contradiction, that $xy = a^2$ and $zt = c^2$ for some positive integers $a$ and $c$. Define $b = \\frac{1}{2}|x - y|$ and $d = \\frac{1}{2}|z - t|$, which are non-negative integers. The conditions yield:\n\n$$\ns^2 = a^2 + b^2 = c^2 + d^2 \\qquad (1)\n$$\n\nand\n\n$$\n2s = a^2 - c^2 = d^2 - b^2 \\qquad (2)\n$$\n\nFrom (1) and (2), $d$ is positive. Since (1) and (2) are symmetric under $a \\leftrightarrow d$ and $b \\leftrightarrow c$, assume $b \\geq c$, so $b > 0$. Then $d^2 = b^2 + 2s > b^2 \\geq c^2$, so\n\n$$\n2d^2 > c^2 + d^2 = s^2 \\qquad (3)\n$$\n\nAlso, since $d^2 - b^2$ is even, $b$ and $d$ have the same parity, so $d - 2 \\geq b > 0$, and thus\n\n$$\n2s = d^2 - b^2 \\geq d^2 - (d-2)^2 = 4(d-1), \\quad \\text{so} \\quad s + 2 \\geq 2d \\qquad (4)\n$$\n\nCombining (3) and (4), $2s^2 < 4d^2 \\leq (s + 2)^2$, so $s$ is a positive integer strictly less than $5$.\n\nBut $5^2$ is the smallest square expressible as a sum of two nonzero squares, and since $a$ and $d$ are both positive, (1) forces $b = c = 0$, contradicting the assumption that at most one of them may be zero. This completes the proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22313,
"subject": "Mathematics (Olympiad)",
"question": "Let $S$ be a set of 2001 distinct positive integers chosen from $\\{1, 2, 3, \\dots, 3000\\}$. Prove that there exist three distinct elements in $S$ such that they are pairwise relatively prime.",
"options": [],
"answer": "See solution",
"solution": "Consider the 500 sets $$K_j = \\{6j + i \\mid i = 1, 2, 3, 4, 5, 6\\}, \\quad j = 0, 1, 2, \\dots, 499.$$ Each $K_j$ contains 6 consecutive numbers. Since $4 \\cdot 500 = 2000 < 2001$, by the pigeonhole principle, there is at least one $K_j$ containing at least five elements of $S$.\n\nIf three of these five elements are odd, they are pairwise relatively prime because any two odd numbers in $K_j$ differ by 2 or 4, which are even, so their gcd is 1.\n\nIf not, then $K_j$ contains three even and two odd numbers from $S$. The two odd numbers differ by 2 or 4, so they are relatively prime. The three even numbers are consecutive even numbers; among them, only one can be divisible by 3 and at most one by 5, so there is an even number not divisible by 3 or 5. This even number and the two odd numbers form a triple of pairwise relatively prime numbers.\n\nSince the numbers in $K_j$ differ by less than 6, only one can be divisible by any prime $p \\ge 7$. Thus, the triple is pairwise relatively prime.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22314,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a strictly positive integer and let $A = \\{1, 2, \\dots, n\\}$. Find the number of increasing functions $f : A \\to A$ which satisfy the property\n\n$$\n|f(x) - f(y)| \\le |x - y|,\n$$\n\nfor any $x, y \\in A$.",
"options": [],
"answer": "See solution",
"solution": "We observe that for $k = 1, 2, \\dots, n-1$, $f(k+1) - f(k) = a_k \\in \\{0, 1\\}$. Since $a_k \\in \\{0, 1\\}$, we have $|f(x) - f(y)| \\le |x - y|$ for any $x, y \\in A$. Therefore, any function $f$ satisfying the conditions is completely determined by the $n$-tuple $(f(1), a_1, a_2, \\dots, a_{n-1}) \\in A \\times \\{0, 1\\}^{n-1}$, with the condition\n\n$$\nf(1) + a_1 + \\dots + a_{n-1} \\le n.\n$$\n\nIf we fix $f(1) = a$ and $f(n) = b$, then $b - a = a_1 + \\dots + a_{n-1}$, so the number of such functions is equal to the number of choices of $b-a$ terms equal to $1$ in the sum $a_1 + \\dots + a_{n-1}$, that is, $n(a, b) = \\binom{n-1}{b-a}$.\n\nThus, the total number of functions is\n\n$$\n\\sum_{1 \\le a \\le b \\le n} n(a, b) = \\sum_{1 \\le a \\le b \\le n} \\binom{n-1}{b-a} = \\sum_{k=0}^{n-1} (n-k) \\binom{n-1}{k}.\n$$\n\nSumming up, we have\n\n$$\n\\begin{aligned}\n\\sum_{k=0}^{n-1} n \\binom{n-1}{k} - \\sum_{k=1}^{n-1} k \\binom{n-1}{k} &= n 2^{n-1} - \\sum_{k=1}^{n-1} (n-1) \\binom{n-2}{k-1} \\\\\n&= n 2^{n-1} - (n-1) 2^{n-2} = (n+1) 2^{n-2}\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22315,
"subject": "Mathematics (Olympiad)",
"question": "A square of size $n \\times n$ is given. Some of its $1 \\times 1$ cells are marked. It turned out that there is no convex quadrilateral with vertices at these marked points. For each natural number $n \\ge 3$, find the largest value of $m$ for which this is possible.\n\nA quadrilateral is called convex if both of its diagonals lie inside the quadrilateral.\n\n",
"options": [],
"answer": "See solution",
"solution": "**Answer:** $n + 2$.\n\n**Solution.** We mark two points in the corner cells of the left column, as well as all the points in some non-edge row. Then we have $n + 2$ marked points, none of which form a vertex of a convex quadrilateral (see the figure).\n\nWe will show by contradiction that it is not possible to mark more than $n + 2$ points. Suppose at least $n + 3$ cell centers are marked. It is clear that if there are two rows, each with at least 2 marked points, then by taking exactly 2 points from each of these two rows, we obtain a convex quadrilateral. Otherwise, in at least the $(n-1)$-th row, no more than 1 point is marked. Then there must be a row in which at least 4 points are marked. Similarly, there must be a column in which at least 4 points are marked. Let this row and column intersect at point $A$. It is easy to see that there are at least 2 points in this row that lie on one side of point $A$, and similarly 2 points can be found for the column. These 4 points form a convex quadrilateral.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22316,
"subject": "Mathematics (Olympiad)",
"question": "а) Знайдіть усі такі дійсні числа $x$, для яких справджується рівність\n\n$$\nx^2 + \\frac{1}{x^2 + 2x} = 1.\n$$\n\nб) Знайдіть усі такі пари додатних дійсних чисел $x$ і $y$, для яких справджується рівність\n\n$$\nx^2 + y^2 + \\frac{1}{x^2 + 2x} + \\frac{1}{y^2 + 2y} = 2.\n$$\n\nВідповіді обґрунтуйте.",
"options": [],
"answer": "See solution",
"solution": "а) Відповідь: $x = \\frac{-1 \\pm \\sqrt{5}}{2}$.\n\nДане рівняння можна записати у вигляді\n\n$$\nx^2 + 2x + \\frac{1}{x^2 + 2x} = 2(x + 1) - 1.\n$$\n\n$$\n\\frac{1}{x^2 + 2x} \\left( (x^2 + 2x) - (x + 1) \\right)^2 = 0.\n$$\n\nб) Відповідь: $x = y = \\frac{-1 + \\sqrt{5}}{2}$.\n\nМаємо:\n\n$$\nx^2 + 2x + y^2 + 2y + \\frac{1}{x^2 + 2x} + \\frac{1}{y^2 + 2y} = 2(x + 1) + 2(y + 1) - 2,\n$$\n\n$$\n\\frac{1}{x^2 + 2x} \\left( (x^2 + 2x) - (x + 1) \\right)^2 + \\frac{1}{y^2 + 2y} \\left( (y^2 + 2y) - (y + 1) \\right)^2 = 0.\n$$\n\nОтже, оскільки $x > 0$ і $y > 0$, то $x^2 + x - 1 = 0$ і $y^2 + y - 1 = 0$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22317,
"subject": "Mathematics (Olympiad)",
"question": "Let $O$ be the center of the incircle of triangle $ABC$. The points $K$ and $L$ are the intersection points of the circumcircles of triangles $BOC$ and $AOC$, respectively, with the bisectors of the angles at $A$ and $B$. 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 the line $KL$. Prove that the quadrilateral $KLMN$ is inscribed.\n\n",
"options": [],
"answer": "See solution",
"solution": "The angles $LCA$ and $LOA$ are equal as inscribed angles upon the same arc. The angle $LOA$ is equal to the sum of the angles $OAB$ and $OBA$, as an external angle to triangle $ABO$, therefore:\n\n$$\n\\begin{aligned}\n\\angle LCO &= \\angle LCA + \\angle OCA = \\angle LOA + \\angle OCA = \\angle OAB + \\angle OBA + \\angle OCA \\\\\n&= \\frac{1}{2}(\\angle CAB + \\angle ABC + \\angle ACB) = 90^\\circ\n\\end{aligned}\n$$\n\nAnalogously, $\\angle KCO = 90^\\circ$, so the point $C$ lies on the line $KL$, and $C$ is the midpoint of $\\overline{ON}$. The line $PC$ is parallel to $MN$ as the midline of triangle $MON$.\n\nThe quadrilateral $LOKM$ is a parallelogram, because its diagonals bisect each other at point $P$. That implies that the angles $MLK$ and $OKL$ are equal.\n\nOn the other hand, triangle $OKN$ is isosceles with base $ON$ ($KC$ is its height and median), so $KC$ is a bisector of $OKN$, i.e., the angles $OKC$ and $NKC$ are equal. It follows that $\\angle MLK = \\angle NKL$, therefore the quadrilateral $KLMN$ is an isosceles trapezoid, hence inscribed. (If point $P$ lies on the other side of point $C$, we consider the angles $MKL$ and $NLK$.)",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22318,
"subject": "Mathematics (Olympiad)",
"question": "Найдите все рациональные числа $a$, для которых выражение $a n (n + 2)(n + 3)(n + 4)$ принимает целые значения при всех целых $n$.",
"options": [],
"answer": "See solution",
"solution": "$a = \\frac{k}{6}$, где $k$ — любое целое число.\n\nПодставив $n = 1, n = 3$ и $n = 4$, получаем, что числа $2^2 \\cdot 3 \\cdot 5 a$, $2 \\cdot 3^2 \\cdot 5 \\cdot 7 a$ и $2^6 \\cdot 3 \\cdot 7 a$ — целые. Значит, $a$ — рациональное число, имеющее несократимую запись $\\frac{p}{q}$, где $q$ является делителем числа $\\gcd(2^2 \\cdot 3 \\cdot 5, 2 \\cdot 3^2 \\cdot 5 \\cdot 7, 2^6 \\cdot 3 \\cdot 7) = 6$, и $a = k/6$ при некотором целом $k$.\n\nОсталось показать, что все числа такого вида подходят. Действительно, одно из трёх последовательных чисел $n + 2, n + 3, n + 4$ делится на $3$, а одно из последовательных чисел $n + 2, n + 3$ делится на $2$; значит, $n(n + 2)(n + 3)(n + 4)$ делится на $2$ и на $3$, а значит, и на $6$. Поэтому $a n (n + 2)(n + 3)(n + 4) = k \\frac{n(n + 2)(n + 3)(n + 4)}{6}$ — целое число.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22319,
"subject": "Mathematics (Olympiad)",
"question": "There are 63 rows of seats on an airplane. Each row has 6 seats, and they are marked with the letters A, B, C, D, E, and F. On a flight, there were 193 passengers on the airplane. Prove that on this flight there were two rows in which the seats that were occupied were marked with the same letters.",
"options": [],
"answer": "See solution",
"solution": "Suppose, for contradiction, that no two rows had the same set of occupied seat letters. Each row's occupied seats correspond to a subset of $\\{A, B, C, D, E, F\\}$, and there are $2^6 = 64$ possible subsets. With 63 rows, each subset could be assigned to at most one row, leaving one subset unused. Thus, the maximum number of different seat occupation patterns is 63. Since there are 193 passengers and each row has at most 6 seats, the maximum number of passengers with all different patterns is $63 \\times 6 = 378$, but with 193 passengers, by the pigeonhole principle, at least two rows must have the same pattern of occupied seat letters. Therefore, there must have been two rows in which the occupied seats were marked with the same letters.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22320,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $p, m, n$ such that $p^m = n^3 + 8$ and $p$ is a prime number.",
"options": [],
"answer": "See solution",
"solution": "By moving $n^3$, we get a sum of cubes on the right-hand side:\n\n$$\np^m = n^3 + 8 = (n+2)(n^2 - 2n + 4).\n$$\n\nSince $p$ is prime, each of the factors on the right-hand side must be a power of $p$:\n\n$$\nn + 2 = p^{\\alpha}, \\quad n^2 - 2n + 4 = p^{\\beta},\n$$\n\nwhere $\\alpha$ and $\\beta$ are positive integers.\n\nNote that $n^2 - 2n + 4 \\ge n + 2$, since that is equivalent to $n^2 - 3n + 2 \\ge 0$, i.e., $(n-1)(n-2) \\ge 0$, which holds for $n \\ge 2$ or $n \\le 1$. Therefore, $\\beta \\ge \\alpha$. We can conclude that $p^\\alpha$ divides both $n+2$ and $n^2 - 2n + 4$, so it also divides\n\n$$\nn \\cdot (n + 2) - (n^2 - 2n + 4) = 4n - 4,\n$$\n\nand then it also divides $4 \\cdot (n + 2) - (4n - 4) = 12$, so $p = 2$ or $p = 3$.\n\nIf $p = 2$, then $n + 2 = 2^\\alpha$ is at most 4, and since $n > 0$, it follows that $n = 2$, which gives the solution $(2, 4, 2)$.\n\nIf $p = 3$, then $n + 2 = 3^\\alpha$ is 3, so $n = 1$, which gives the second solution $(3, 2, 1)$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22321,
"subject": "Mathematics (Olympiad)",
"question": "Let $(a_n)_{n \\ge 0}$ be the sequence of rational numbers with $a_0 = 2016$ and\n$$\na_{n+1} = a_n + \\frac{2}{a_n}\n$$\nfor all $n \\ge 0$.\n\nShow that the sequence does not contain a square of a rational number.",
"options": [],
"answer": "See solution",
"solution": "We look at this sequence modulo $5$. This is possible as long as $a_n \\not\\equiv 0 \\pmod{5}$ so that the next element is defined modulo $5$. If we start to compute the elements modulo $5$ we obtain\n$$\na_0 \\equiv 1 \\pmod{5},\n$$\n$$\na_1 \\equiv 1 + 2 \\equiv 3 \\pmod{5},\n$$\n$$\na_2 \\equiv 3 + 2 \\cdot 3^{-1} \\equiv 3 + 2 \\cdot 2 \\equiv 2 \\pmod{5},\n$$\n$$\na_3 \\equiv 2 + 2 \\cdot 2^{-1} \\equiv 2 + 2 \\cdot 3 \\equiv 2 + 6 \\equiv 3 \\pmod{5},\n$$\n$$\na_4 \\equiv 3 + 2 \\cdot 3^{-1} \\equiv 3 + 2 \\cdot 2 \\equiv 2 \\pmod{5}.\n$$\nSo we see that after $a_0$, the sequence just alternates between the values $2$ and $3$ modulo $5$. These are not quadratic residues modulo $5$. Since $a_0 = 2016$ is also not the square of a rational number, there is indeed no square of a rational number in this sequence.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22322,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs of positive integers $(m, n)$ such that $mn - 1$ divides $(n^2 - n + 1)^2$.",
"options": [],
"answer": "See solution",
"solution": "The answer is $(m, n) = (2, 2)$ or $(m, n) = (t^2 \\pm 2t + 2, t^2 + 1)$ for some $t \\ge 0$.\n\nFirst, check that these pairs work:\n- $2 \\cdot 2 - 1 = 3$ divides $3^2 = (2^2 - 2 + 1)^2$.\n- $$(t^2 \\pm 2t + 2)(t^2 + 1) - 1 = (t^2 \\pm t + 1)^2$$ divides $$(t^4 + t^2 + 1)^2 = ((t^2 + 1)^2 - (t^2 + 1) + 1)^2.$$\n\nNow, show these are the only solutions. For $n = 1$ or $n = 2$, $m - 1$ divides $1$ or $2m - 1$ divides $9$, so only the claimed solutions exist. Assume $n \\ge 2$ and $(m, n)$ is a solution. Note:\n\n$$\nm^2(n^2 - n + 1)^2 - n^2(m^2 - m + 1)^2 = (mn^2 - m^2n + m - n)(mn^2 + m^2n - 2mn + m + n) \\\\ = (mn - 1)(n - m)(mn^2 + m^2n - 2mn + m + n)\n$$\n\nis divisible by $mn - 1$, so $(mn - 1)$ divides $n^2(m^2 - m + 1)^2$. Since $mn - 1$ and $n^2$ are coprime, $(mn - 1)$ divides $(m^2 - m + 1)^2$. Thus, $(m, n)$ is a solution if and only if $(n, m)$ is.\n\nSuppose, for contradiction, there is a solution $(m, n)$ not of the stated form, with minimal $m + n$ and $m \\ge n$. Note $(n^2 - n + 1)^2 \\equiv 1 \\pmod{n}$ and $mn - 1 \\equiv -1 \\pmod{n}$, so $\\frac{(n^2 - n + 1)^2}{mn - 1} \\equiv -1 \\pmod{n}$. Write $\\frac{(n^2 - n + 1)^2}{mn - 1} = kn - 1$ for some integer $k$. Then $(k, n)$ is another solution and\n\n$$\nkn - 1 \\le \\frac{(n^2 - n + 1)^2}{n^2 - 1} < \\frac{(n^2 - 1)^2}{n^2 - 1} = n^2 - 1,\n$$\n\nsince $m \\ge n$ and $n \\ge 2$. Thus, $k < n \\le m$, so $k + n < m + n$. By minimality, $(k, n)$ must be of the claimed form: either $(2, 2)$ or $((t - 1)^2 + 1, t^2 + 1)$ for some $t$. In the first case, $mn - 1 = \\frac{(n^2 - n + 1)^2}{kn - 1} = 3$, so $(m, n) = (2, 2)$. In the second case, $mn - 1 = \\frac{(n^2 - n + 1)^2}{kn - 1} = ((t + 1)^2 + 1)n - 1$, so $(m, n) = ((t + 1)^2 + 1, t^2 + 1)$. Thus, $(m, n)$ is of the claimed form, a contradiction. Therefore, all solutions are as stated.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22323,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\geq 3$ be a natural number, and let $\\mathcal{F}$ be a family consisting of at most $n$ distinct subsets of the set $\\{1, 2, \\dots, n\\}$ with the property that one can consider $n$ distinct points in the plane, labeled with the numbers $1, 2, \\dots, n$, then draw segments between some of these points such that, for any distinct numbers $i, j \\in \\{1, 2, \\dots, n\\}$, the points labeled $i$ and $j$ are connected by a segment if and only if the number $i$ belongs to exactly $j$ subsets in $\\mathcal{F}$. Find the maximum possible value of the sum of the number of elements of the sets in $\\mathcal{F}$.",
"options": [],
"answer": "See solution",
"solution": "For $k \\in \\{1, 2, \\dots, n\\}$, denote $a_k = \\text{card}\\{F \\in \\mathcal{F} \\mid k \\in F\\}$. Then $a_k \\in \\{0, 1, \\dots, n\\}$, and the sum to maximize is\n\n$$\ns = \\sum_{F \\in \\mathcal{F}} \\text{card}(F) = \\sum_{k=1}^{n} a_k.\n$$\n\nThe connection condition implies that if $a_k \\neq k$, then $a_{a_k} = k$ for any $k \\in \\{1, 2, \\dots, n\\}$. Conversely, any function satisfying this condition corresponds to a family $\\mathcal{F}$ verifying the hypothesis.\n\nAny point can be connected to at most one other point; otherwise, a number $i$ would belong to exactly $j_1$, respectively $j_2 \\neq j_1$, subsets from $\\mathcal{F}$, which is absurd.\n\nTherefore, the numbers $a_k$ form pairs of two distinct numbers; otherwise, if $a_i = a_j = a$, then $a$ is connected both to $i$ and to $j$, which would imply $i = j$.\n\nWe deduce that\n\n$$\ns \\leq n + (n-1) + \\dots + 1 = \\frac{n(n+1)}{2}.\n$$\n\nFinally, the maximum $\\frac{n(n+1)}{2}$ is attained for any family $\\mathcal{F}$ for which $a_{a_k} = k$ for all $k \\in \\{1, 2, \\dots, n\\}$; for example, the family $\\mathcal{F}_0$ formed by $\\{1\\}, \\{1, 2\\}, \\dots, \\{1, 2, \\dots, n\\}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22324,
"subject": "Mathematics (Olympiad)",
"question": "In a $3 \\times 2$ rectangle, we put the numbers 1 to 6 in the squares in such a way that each number occurs exactly once. The *score* of such a distribution is determined as follows: for each two adjacent squares we compute the difference between their two numbers and we add up all these differences. In the example on the right, the differences are indicated in red. This distribution has score 17.\n\nWhat is the smallest possible score of such a distribution?\n\n",
"options": [],
"answer": "See solution",
"solution": "11",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22325,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\omega$ be a circle in the plane and $A$, $B$ two points lying on it. Denote by $M$ the midpoint of $AB$, and let $P \\neq M$ be a point on $AB$. Construct circles $\\gamma$ and $\\delta$ tangent to $AB$ at $P$ and to $\\omega$ at $C$ and $D$, respectively. Let $E$ be the point diametrically opposite to $D$ on $\\omega$. Prove that the circumcenter of $\\triangle BMC$ lies on the line $BE$.",
"options": [],
"answer": "See solution",
"solution": "Let us begin by noticing that since $DE$ is a diameter, we have $\\angle DBE = 90^\\circ$, and if the circumcenter of $\\triangle BMC$ lies on $BE$, then $DB$ is tangent to the circumcircle of $\\triangle BMC$. Thus $\\angle DBA \\equiv \\angle BCM$. Since $A$, $B$, $C$, $D$ are on $\\omega$, we have $\\angle DBA \\equiv \\angle DCA$, so the problem is equivalent to $\\angle BCM \\equiv \\angle DCA$.\n\nWe must prove that $CD$ and $CM$ are isogonal conjugates in $\\triangle BCA$, thus the quadrilateral $ABCD$ is harmonic, which means $\\frac{AC}{CB} = \\frac{AD}{DB}$.\n\nThis follows from the fact that since $CD$ and $CM$ are isogonal conjugates, we have\n$$\n\\frac{\\sin \\angle BCD}{\\sin \\angle ACD} = \\frac{\\sin \\angle ACM}{\\sin \\angle BCM}.\n$$\nFrom the sine theorem,\n$$\n\\frac{\\sin \\angle BCD}{\\sin \\angle ACD} = \\frac{BD}{DA}.\n$$\nWe also obtain\n$$\n\\frac{\\sin \\angle ACM}{\\sin \\angle BCM} = \\frac{BC}{AC}\n$$\nfrom the fact that $M$ is the midpoint, and thus $\\text{area}(BMC) = \\text{area}(ACM)$.\n\nNow we use the following:\n\n**LEMMA (Archimedes).** Let $C$ be a circle, $P$ and $Q$ two points on it, and $R$ a point on $PQ$. Let $\\omega$ be one of the two circles tangent to $C$ at $S$, and to $PQ$ at $R$. Then $SR$ is the angle bisector of $\\angle PSQ$.\n\n\n\n**Proof.** Let $O$ and $O_1$ be the centers of the circles $C$ and $\\omega$, respectively. Let $RS \\cap C = \\{T\\}$. The triangles $\\triangle RO_1S$ and $\\triangle TSO$ are isosceles in $O$ and $O_1$, and $\\angle TSO = \\angle RSO_1$. Thus $RO_1 \\parallel TO$. Since $RO_1 \\perp PQ$, we deduce $TO \\perp PQ$, so $T$ is the midpoint of arc $PQ$. $\\square$\n\nUsing this lemma, we get that $DP$ is the angle bisector of $\\angle ADB$ and $CP$ is the angle bisector of $\\angle ACB$, so from the angle bisector theorem $\\frac{AD}{DB} = \\frac{AC}{CB} = \\frac{AP}{BP}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22326,
"subject": "Mathematics (Olympiad)",
"question": "Let a square of size $n \\times n$ be divided into smaller rectangles. We call the square *durable* if there does not exist a straight line dividing the square into two parts without intersecting any smaller rectangle in any inner points. For any given natural $m$, for which natural $n$ is it possible to divide the square $n \\times n$ into rectangles of size $m \\times 1$ and $1 \\times m$ so that the square remains durable?\n\n\n\nFig. 28",
"options": [],
"answer": "See solution",
"solution": "**Answer:**\n- For $m=1$: No such division exists.\n- For $m=2$: $n = k m$, $k \\geq 4$.\n- For $m \\geq 3$: $n = k m$, $k \\geq 3$.\n\n**Solution.**\nTo divide a square $n \\times n$ into rectangles $m \\times 1$, the area condition $n^2$ must be divisible by $m$; more strongly, $n$ must be divisible by $m$.\n\n**Lemma 1.** A division of a square $n \\times n$ into rectangles $m \\times 1$ exists if and only if $n$ is divisible by $m$.\n\n\n\nFig. 29\n\n*Proof.* Sufficiency is clear. For necessity, suppose such a division exists but $n$ is not divisible by $m$. Let $d = \\gcd(n, m)$, $n = d n_1$, $m = d m_1$, $(n_1, m_1) = 1$. Since $n^2$ is divisible by $m$, $d$ must divide $m_1$, but $m_1 \\neq 1$ (otherwise $n$ is divisible by $m$). Coloring the $d \\times d$ squares with $m_1$ colors diagonally (see Fig. 29), each $m \\times 1$ rectangle covers equal numbers of each color, but the total number of colored squares cannot be equally distributed, leading to a contradiction.\n\nNow, for $m=1$, no such division exists. Consider $m=2$ and $m>2$.\n\n**Lemma 2.** A durable square $n \\times n$ divided into rectangles $2 \\times 1$ exists if and only if $n = 2 n_1$ and $n_1 \\geq 4$.\n\n*Proof.* For $n=2$ or $n=4$, no durable division exists (can be checked directly). For $n=6$, there are 10 segments inside the square (see Fig. 30). Each must intersect at least one $2 \\times 1$ rectangle, but there are only $18$ such rectangles ($36/2$), so some segment intersects only one rectangle. This segment divides the square into two parts with an odd number of $1 \\times 1$ squares, which cannot be filled with $2 \\times 1$ rectangles, leading to a contradiction.\n\n\n\nFig. 30\n\n\n\nFig. 31\n\nFor $n = 2 n_1$ with $n_1 \\geq 4$, a durable division is possible by filling the outer edge with $2 \\times 1$ rectangles and dividing the inner $(2 n_1 - 4) \\times (2 n_1 - 4)$ square into $2 \\times 2$ squares, with further division ensuring durability (see Fig. 12).\n\n\n\nFig. 33\n\n**Lemma 3.** A durable square $n \\times n$ divided into rectangles $m \\times 1$ with $m \\geq 3$ exists if and only if $n = m n_1$ and $n_1 \\geq 3$.\n\n*Proof.* For $n = 2m$, a durable division is not possible (similar reasoning as for $m=2$). For $n = m n_1$ with $n_1 \\geq 3$, fill the outer band of width $m$ regularly with $m \\times 1$ and $1 \\times m$ rectangles, ensuring no segment can break durability. The inner square of side $m(n_1 - 2)$ can be filled arbitrarily.\n\nFor example, a durable $12 \\times 12$ square divided into $4 \\times 1$ rectangles is shown below.\n\n\n\nFig. 32\n\n\n\nFig. 34\n\n\n\nFig. 35",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22327,
"subject": "Mathematics (Olympiad)",
"question": "A polynomial $f(x)$ with real coefficients is called *generating* if for each polynomial $\\varphi(x)$ with real coefficients, there exists a positive integer $k$ and polynomials $g_1(x), \\dots, g_k(x)$ such that\n\n$$\n\\varphi(x) = f(g_1(x)) + \\dots + f(g_k(x)).\n$$\n\nFind all generating polynomials.",
"options": [],
"answer": "See solution",
"solution": "The generating polynomials are exactly the polynomials of odd degree.\n\nLet $f$ be an arbitrary polynomial. Call a polynomial *good* if it can be represented as $\\sum f(g_i(x))$ for some polynomials $g_i$. The sum of good polynomials is good, and if $\\phi$ is good, then so is $\\phi(g(x))$ for any polynomial $g(x)$. Thus, to show $f$ is generating, it suffices to show that $x$ is good.\n\n**Case 1:** $f$ has odd degree $n$.\n\nBy substituting $f(ux)$ for suitable $u$, we can obtain a good polynomial $\\phi_n$ of degree $n$ with leading coefficient $1$, and a good polynomial $\\psi_n$ of degree $n$ with leading coefficient $-1$ (since $n$ is odd). For any $a$, the polynomial $\\phi_n(x+a) + \\psi_n(x)$ is good, and its $x^n$ coefficient is $0$. By choosing $a$ appropriately, we can obtain a good polynomial $\\phi_{n-1}$ of degree $n-1$ with leading coefficient $1$, and similarly $\\psi_{n-1}$ with leading coefficient $-1$. Continuing this process, we eventually obtain a good polynomial $\\phi_1(x) = x + c$. Then $\\phi_1(x - c) = x$ is also good.\n\n**Case 2:** $f$ has even degree $n$.\n\nIn this case, $f(x)$ is not generating. The degree of every good polynomial is even, since the degree of each $f(g_i)$ is even and the leading coefficient has the same sign as that of $f$. Therefore, the degree of $\\sum f(g_i(x))$ is always even, so not every polynomial $\\varphi(x)$ can be represented in this way.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 22328,
"subject": "Mathematics (Olympiad)",
"question": "It turns out that the numbers $x^2 + y^2$, $x^3 + y^3$, and $x^4 + y^4$ are rational for some real $x, y$. Is it necessary that $x + y$ is rational?",
"options": [],
"answer": "See solution",
"solution": "We start with the case when $xy = 0$. Without loss of generality, let $y = 0$. If $x^2$ and $x^3$ are rational, then so is $\\frac{x^3}{x^2} = x$.\n\nNow assume $xy \\ne 0$. From $(x^2 + y^2)^2 = x^4 + y^4 + 2x^2y^2$, it follows that $(xy)^2 \\in \\mathbb{Q}$. Also, $x^6 + y^6 = (x^2 + y^2)((x^4 + y^4) - (xy)^2) \\in \\mathbb{Q}$. Then, $(x^3 + y^3)^2 = x^6 + y^6 + 2(xy)^3$ implies $(xy)^3 \\in \\mathbb{Q}$. Thus, $xy \\in \\mathbb{Q}$, and finally, $x + y = \\frac{x^3 + y^3}{(x^2 + y^2) - xy} \\in \\mathbb{Q}$, as required.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22329,
"subject": "Mathematics (Olympiad)",
"question": "Let $p$ be a prime number, $n$ a positive integer not divisible by $p$, and $\\mathbb{K}$ a field with $p^n$ elements, with unit element $1_{\\mathbb{K}}$ and zero element $\\hat{0} = 0_{\\mathbb{K}}$. For any $m \\in \\mathbb{N}^*$, denote $\\hat{m} = \\underbrace{1_{\\mathbb{K}} + 1_{\\mathbb{K}} + \\dots + 1_{\\mathbb{K}}}_{m \\text{ times}}$ and define the polynomial\n\n$$\nf_m = \\sum_{k=0}^{m} (-1)^{m-k} \\widehat{C_m^k} X^{p^k} \\in \\mathbb{K}[X].\n$$\n\na) Show that the set of roots of the polynomial $f_1$ is $\\{\\hat{k} \\mid k \\in \\{0, 1, \\dots, p-1\\}\\}$.\n\nb) Let $m \\in \\mathbb{N}^*$ be arbitrary. Determine the set of roots in the field $\\mathbb{K}$ of the polynomial $f_m$.",
"options": [],
"answer": "See solution",
"solution": "a) For any polynomial $P \\in \\mathbb{K}[X]$, denote by $Z_P$ the set of roots of $P$ in $\\mathbb{K}$. Since $|\\mathbb{K}| = p^n$, the characteristic of $\\mathbb{K}$ is $\\text{char}(\\mathbb{K}) = p$. Then $\\hat{m} = \\hat{0}$ for any multiple $m$ of $p$. In particular, since $k^p \\equiv k \\pmod p$, for any $k \\in \\{0, 1, \\dots, p-1\\}$ we have\n\n$$\nf_1(\\hat{k}) = (\\hat{k})^p - \\hat{k} = \\widehat{k^p} - \\hat{k} = \\widehat{k^p - k} = \\hat{0},\n$$\n\nHence, $\\{\\hat{k} \\mid k = 0, \\dots, p-1\\} \\subseteq Z_{f_1}$. Also, since $\\mathbb{K}$ is a field, $|Z_{f_1}| \\leq \\deg(f_1) = p$. Thus, $Z_{f_1} = \\{\\hat{k} \\mid k = 0, \\dots, p-1\\}$.\n\nb) Since $p \\mid C_p^k$ for $k = 1, \\dots, p-1$, the identity $(a+b)^p = a^p + b^p$ holds for any $a, b \\in \\mathbb{K}$, and inductively $(a+b)^{p^k} = a^{p^k} + b^{p^k}$ for any $k \\in \\mathbb{N}$. For any $m \\in \\mathbb{N}^*$:\n\n$$\n\\begin{align*}\nf_m(f_1(X)) &= \\sum_{k=0}^{m} (-1)^{m-k} \\widehat{C_m^k} (X^p - X)^{p^k} \\\\\n&= \\sum_{k=0}^{m} (-1)^{m-k} \\widehat{C_m^k} (X^{p^{k+1}} - X^{p^k}) \\\\\n&= \\sum_{k=0}^{m+1} (-1)^{m+1-k} (\\widehat{C_m^k} + \\widehat{C_m^{k-1}}) X^{p^k} \\\\\n&= \\sum_{k=0}^{m+1} (-1)^{m+1-k} \\widehat{C_{m+1}^k} X^{p^k} = f_{m+1}(X).\n\\end{align*}\n$$\n\nWe prove by induction on $m$ that $Z_{f_m} = \\{\\hat{k} \\mid k = 0, \\dots, p-1\\}$ for all $m \\in \\mathbb{N}^*$. For $m = 1$ this holds by part a). Assume it holds for some $m$. For $m+1$:\n\nFor any $k \\in \\{0, 1, \\dots, p-1\\}$,\n\n$$\nf_{m+1}(\\hat{k}) = f_m(f_1(\\hat{k})) = f_m(\\hat{0}) = \\hat{0},\n$$\n\nso $\\{\\hat{k} \\mid k = 0, \\dots, p-1\\} \\subseteq Z_{f_{m+1}}$.\n\nLet $\\alpha \\in Z_{f_{m+1}}$. Then $f_m(f_1(\\alpha)) = f_{m+1}(\\alpha) = \\hat{0}$, so $f_1(\\alpha) \\in Z_{f_m}$. Thus, $f_1(\\alpha) = \\hat{k}$ for some $k \\in \\{0, 1, \\dots, p-1\\}$. Then\n\n$$\n\\alpha^p = \\alpha + \\hat{k}, \\quad \\alpha^{p^2} = (\\alpha + \\hat{k})^p = \\alpha^p + \\hat{k}^p = (\\alpha + \\hat{k}) + \\hat{k} = \\alpha + 2 \\cdot \\hat{k},\n$$\n\nand, inductively, if $\\alpha^{p^m} = \\alpha + m \\cdot \\hat{k}$, then $\\alpha^{p^{m+1}} = (\\alpha + m \\cdot \\hat{k})^p = \\alpha + (m+1) \\cdot \\hat{k}$.\n\nIn the multiplicative group $(\\mathbb{K}^*, \\cdot)$, $x^{p^n-1} = 1$ for any $x \\in \\mathbb{K}^*$, so $x^{p^n} = x$ for any $x \\in \\mathbb{K}$. Thus $\\alpha = \\alpha^{p^n} = \\alpha + n \\cdot \\hat{k}$, so $n \\cdot \\hat{k} = \\hat{0}$.\n\nSince $n$ is not divisible by $p$, $\\hat{k} = \\hat{0}$. Thus $f_1(\\alpha) = \\hat{0}$ and $\\alpha \\in Z_{f_1} = \\{\\hat{k} \\mid k = 0, \\dots, p-1\\}$. Therefore, $Z_{f_{m+1}} \\subseteq \\{\\hat{k} \\mid k = 0, \\dots, p-1\\}$, so $Z_{f_{m+1}} = \\{\\hat{k} \\mid k = 0, \\dots, p-1\\}$. The claim holds for all $m \\in \\mathbb{N}^*$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22330,
"subject": "Mathematics (Olympiad)",
"question": "The sword is a figure consisting of 6 unit squares presented in the picture below (and any other figure obtained from it by rotation).\n\n\n\nDetermine the largest number of swords that can be cut from a $6 \\times 11$ piece of paper divided into unit squares (each sword should consist of six such squares).",
"options": [],
"answer": "See solution",
"solution": "Let us cover boundary cells with 15 dominoes, as shown in the left picture. In each such domino there is at most one cell belonging to some sword. Therefore, all swords can together contain at most $6 \\times 11 - 15 = 51$ cells, which means that there are at most 8 swords. Eight is actually possible (see the right picture).\n\n\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22331,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, \\dots, a_{2022}$ be integers such that $1 \\leq a_1 < a_2 < \\dots < a_{2022}$. Suppose that\n$$\n a_1^2 - 6^2 \\geq a_2^2 - 7^2 \\geq \\dots \\geq a_{2022}^2 - 2027^2.\n$$\nHow many such tuples $(a_1, a_2, \\dots, a_{2022})$ are there?",
"options": [],
"answer": "See solution",
"solution": "Since $1 \\leq a_1 < a_2 < \\dots < a_{2022}$, we have $a_i \\geq i$ for all $1 \\leq i \\leq 2022$. The condition $a_1^2 - 6^2 \\geq a_2^2 - 7^2 \\geq \\dots \\geq a_{2022}^2 - 2027^2$ is equivalent to\n$$\na_{i+1}^2 - a_i^2 \\leq (i+6)^2 - (i+5)^2 = 2i + 11 \\quad (1 \\leq i \\leq 2021).\n$$\n\nIf $a_{i+1} \\geq a_i + 3$ for some $1 \\leq i \\leq 2021$, then\n$$\na_{i+1}^2 - a_i^2 \\geq (a_i + 3)^2 - a_i^2 = 6a_i + 9 \\geq 6i + 9 > 2i + 11,\n$$\nwhich contradicts $a_{i+1}^2 - a_i^2 \\leq 2i + 11$. Therefore, since $a_i < a_{i+1}$, we must have $a_{i+1} = a_i + 1$ or $a_{i+1} = a_i + 2$ for all $1 \\leq i \\leq 2021$.\n\nFor each $i$, if $a_{i+1} = a_i + 1$, then $a_{i+1}^2 - a_i^2 = 2a_i + 1$, so $a_{i+1}^2 - a_i^2 \\leq 2i + 11$ is equivalent to $a_i \\leq i + 5$. Thus, the tuples with $a_{i+1} = a_i + 1$ for all $i$ are $a_i = i + c$ for $0 \\leq c \\leq 5$, giving 6 such tuples.\n\nNow, suppose $a_{i+1} = a_i + 2$ for some $1 \\leq i \\leq 2021$, and let $j$ be the largest such $i$. Then\n$$\n4j + 4 \\leq 4a_j + 4 = a_{j+1}^2 - a_j^2 \\leq 2j + 11,\n$$\nso $j = 1, 2, 3$.\n\n- For $j = 1$: $4a_1 + 4 = a_2^2 - a_1^2 \\leq 2 \\cdot 1 + 11$, so $a_1 = 1, 2$. The tuples are $(1, 3, 4, 5, \\dots, 2022, 2023)$ and $(2, 4, 5, 6, \\dots, 2023, 2024)$.\n- For $j = 2$: $4a_2 + 4 = a_3^2 - a_2^2 \\leq 2 \\cdot 2 + 11$, $a_2 \\geq 2$, so $a_2 = 2$. The tuple is $(1, 2, 4, 5, 6, \\dots, 2022, 2023)$.\n- For $j = 3$: $4a_3 + 4 = a_4^2 - a_3^2 \\leq 2 \\cdot 3 + 11$, $a_3 \\geq 3$, so $a_3 = 3$. The tuple is $(1, 2, 3, 5, 6, 7, \\dots, 2022, 2023)$.\n\nHence, the total number of tuples is $6 + 4 = 10$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22332,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a three-digit number with digits $a < b < c$. Let $g$ be the greatest common divisor of the differences between all numbers obtained by permuting the digits of $n$. Show that $g$ cannot exceed $18$ for any $n$.",
"options": [],
"answer": "See solution",
"solution": "First, let us show that $g$ cannot exceed $18$ for any $n$. Denote by $a, b, c$ the three digits of $n$, where $a < b < c$. Both $100c+10b+a$ and $100c+10a+b$ are numbers obtained by permuting the digits of $n$. Hence $g$ is a divisor of $$(100c+10b+a) - (100c+10a+b) = 9(b-a).$$ Similarly, $g$ is a divisor of $9(c-b)$ and $9(c-a)$. 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\n(1) If $g' \\ge 5$, there exists at most one number less than or equal to $8$ divisible by $g'$, which contradicts $x + y = z$. So, $g' \\ge 5$ is impossible.\n\n(2) If $g' = 4$, then $4$ and $8$ are the only positive integers not bigger than $8$ and divisible by $4$, so $(x, y, z) = (4, 4, 8)$. Then $(a, b, c) = (1, 5, 9)$ and $g = 3$.\n\n(3) If $g' = 3$, then $3$ and $6$ are the only positive integers not bigger than $8$ and divisible by $3$, so $(x, y, z) = (3, 3, 6)$. Then $(a, b, c)$ must be $(1, 4, 7)$ or $(2, 5, 8)$ or $(3, 6, 9)$ and the value of $g$ is $3$, $3$, $9$, respectively.\n\n(4) If $g' \\le 2$, then since $g$ divides $9g'$, we must have $g \\le 9g' \\le 18$.\n\nThus we have shown that $g \\le 18$. On the other hand, if $n = 468$, then $g = 18$ and this shows that $18$ is the maximum possible value for $g$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22333,
"subject": "Mathematics (Olympiad)",
"question": "Consider a $25 \\times 25$ chessboard with cells $C(i, j)$ for $1 \\le i, j \\le 25$. Find the smallest possible number $n$ of colors with which these cells can be colored subject to the following condition: For $1 \\le i < j \\le 25$ and for $1 \\le s < t \\le 25$, 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 given by\n\n\n\nFor a $3 \\times 3$ chessboard, the minimum number is 2. Indeed:\n\n\n\nFor a $5 \\times 5$ chessboard, 3 colors suffice:\n\n\n\nIt seems that $m_n = \\frac{n-1}{2}$ colors are sufficient for an $n \\times n$ chessboard for any odd $n$. So we will prove that **13 colors are sufficient** for the $25 \\times 25$ chessboard. We consider the colors $\\{1, 2, 3, \\dots, 11, 12, 0\\}$ and color the chessboard as:\n\n\n\nwhich satisfies the condition. In fact, $C[i, j] = \\left\\lfloor \\frac{i+j}{2} \\right\\rfloor \\pmod{13}$ for any $1 \\le i, j \\le 25$. If the condition fails, then $C[i, s] = C[j, s] = C[j, t]$ for some $1 \\le i < j \\le 25$ and $1 \\le s < t \\le 25$, which implies\n\n$$\n\\left\\lfloor \\frac{i+s}{2} \\right\\rfloor \\pmod{13} = \\left\\lfloor \\frac{j+s}{2} \\right\\rfloor \\pmod{13} = \\left\\lfloor \\frac{j+t}{2} \\right\\rfloor \\pmod{13}.\n$$\n\nFrom $C[i, s] = C[j, s]$, $\\left\\lfloor \\frac{j+s}{2} \\right\\rfloor = \\left\\lfloor \\frac{i+s}{2} \\right\\rfloor$ since $0 \\le \\left\\lfloor \\frac{j+s}{2} \\right\\rfloor - \\left\\lfloor \\frac{i+s}{2} \\right\\rfloor < \\frac{j-i}{2} + 1 \\le 13$, so the remainders must coincide. Similarly, from $C[j, s] = C[j, t]$, $\\left\\lfloor \\frac{j+s}{2} \\right\\rfloor = \\left\\lfloor \\frac{j+t}{2} \\right\\rfloor$, so $\\left\\lfloor \\frac{i+s}{2} \\right\\rfloor = \\left\\lfloor \\frac{i+t}{2} \\right\\rfloor$, which is impossible since\n\n$$\n\\left\\lfloor \\frac{i+s}{2} \\right\\rfloor \\le \\frac{i+s}{2} \\le \\frac{j+t}{2} - 1 < \\left\\lfloor \\frac{j+t}{2} \\right\\rfloor.\n$$\n\nNow we prove that **13 colors are necessary**. Fix a $25 \\times 25$ chessboard with a configuration satisfying the condition. Fix any color, say color 2. Call the cells colored with 2 as 2-cells, and let $c_2$ be their number. Remove all other colors, leaving only 2-cells. From any 2-cell, draw horizontal arrows (left to right) and vertical arrows (down to up) joining consecutive 2-cells. These are 2-arrows.\n\n\n\nAny 2-cell cannot have two or more outgoing 2-arrows, otherwise the forbidden configuration would occur:\n\n\n\nTherefore, the total number of 2-arrows $a_2$ satisfies $c_2 \\ge a_2$. In any row with $k$ 2-cells, there are $k-1$ 2-arrows, so the total number of horizontal 2-arrows is $c_2 - 25$ (25 rows). Similarly, the total number of vertical 2-arrows is $c_2 - 25$. Thus, $a_2 = 2(c_2 - 25)$, so $c_2 \\ge 2c_2 - 50$, hence $c_2 \\le 50$. Since there are $25 \\times 25 = 625$ cells and $625/50 > 12$, we need at least **13** colors.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22334,
"subject": "Mathematics (Olympiad)",
"question": "How many non-overlapping spheres of the same size can touch a fixed sphere of the same size in three dimensions?\n\n",
"options": [],
"answer": "See solution",
"solution": "We can place 6 spheres with their centers coplanar with the fixed sphere. Then we can place 3 more above and 3 more below as shown above. Thus, 12 can be achieved.\n\nTake $O$ to be the center of the fixed sphere. Another sphere touching it blocks off a conical solid angle as shown. The angle between the center line of the cone and the surface is $30^\\circ$. It's not hard to show that for an angle $\\theta$ the solid angle is $2\\pi(1 - \\cos\\theta)$, so for $\\theta = 30^\\circ$, it is $\\pi(2 - \\sqrt{3})$. Thus we can have at most $\\frac{4\\pi}{\\pi(2-\\sqrt{3})} = 4(2+\\sqrt{3}) < 15$ spheres touching the fixed sphere. Hence, at most 14.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 22335,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge 2$ be a natural number and let $a_1, a_2, \\dots, a_n$ be nonzero integers such that $a_1 + a_2 + \\dots + a_n = a_1 a_2 \\dots a_n$.\n\nProve that the number $(a_1^2 - 1)(a_2^2 - 1)\\dots(a_n^2 - 1)$ is a perfect square.",
"options": [],
"answer": "See solution",
"solution": "For $n=2$, we have $a_1 + a_2 = a_1 a_2$, which rewrites as $(a_1 - 1)(a_2 - 1) = 1$. We get $a_1 = 2$, $a_2 = 2$, so $(a_1^2 - 1)(a_2^2 - 1) = 9$.\n\nFor $n \\ge 3$, we will show that at least one of the numbers has absolute value $1$, hence $(a_1^2 - 1) \\dots (a_n^2 - 1) = 0$. Indeed, using the hypothesis, we have\n\n$$\n|a_1 a_2 \\dots a_n| = |a_1 + a_2 + \\dots + a_n| \\le |a_1| + |a_2| + \\dots + |a_n|.\n$$\n\nDividing by $|a_1 a_2 \\dots a_n| > 0$ we obtain\n\n$$\n1 \\le \\frac{|a_1|}{|a_1 a_2 \\cdots a_n|} + \\frac{|a_2|}{|a_1 a_2 \\cdots a_n|} + \\dots + \\frac{|a_n|}{|a_1 a_2 \\cdots a_n|}.\n$$\n\nIf all the numbers have absolute value greater than or equal to $2$, then the right-hand side of the above inequality is less than or equal to $\\frac{n}{2^{n-1}}$, i.e., $2^{n-1} \\le n$.\n\nFor $n \\ge 3$, we obtain (by induction or using Bernoulli's inequality) that $2^{n-1} > n$, which contradicts the previous inequality.\n\nTherefore, at least one of the numbers has absolute value $1$, which implies $(a_1^2 - 1)(a_2^2 - 1)\\dots(a_n^2 - 1) = 0$, a perfect square.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22336,
"subject": "Mathematics (Olympiad)",
"question": "A finite sequence of integers $a_1, a_2, \\dots, a_n$ is called *regular* if there exists a real number $x$ satisfying\n$$\n\\lfloor kx \\rfloor = a_k \\quad \\text{for } 1 \\le k \\le n.\n$$\nGiven a regular sequence $a_1, a_2, \\dots, a_n$, for $1 \\le k \\le n$ we say the term $a_k$ is *forced* if the following condition is satisfied: the sequence\n$$\na_1, a_2, \\dots, a_{k-1}, b\n$$\nis regular if and only if $b = a_k$. Find the maximum possible number of forced terms in a regular sequence with 1000 terms.",
"options": [],
"answer": "See solution",
"solution": "The maximum is 985.\n\nTo prove this, we start with two lemmas.\n\n**Lemma 2.** Suppose we begin with the pair of integers $(1, 1)$, and at each step we are allowed to replace one of the integers with the sum of both. Then after $k$ steps, the maximum possible sum of the two numbers is $F_{k+3}$, where $F_n$ is the $n$th Fibonacci number with $F_0 = 0$, $F_1 = 1$.\n\n*Proof.* Let the numbers be $a, b$ with $a < b$. We show by induction that $a \\le F_{k+1}$ and $b \\le F_{k+2}$ after $k$ steps. The base case is given. Given it is true for $k$, suppose a pair we get after $k+1$ steps comes from the pair $a, b$. The new pair is either $(a, a+b)$ or $(b, a+b)$. By the induction hypothesis $a+b \\le F_{k+1}+F_{k+2} = F_{k+3}$, and we already know that both $a$ and $b$ are at most $F_{k+2}$. This completes the induction and the proof. $\\square$\n\n**Lemma 3.** Let $\\frac{a}{b}$ and $\\frac{c}{d}$ be two rational numbers in lowest terms with $0 \\le \\frac{a}{b} < \\frac{c}{d} \\le 1$ which are adjacent terms of a Farey series; that is, there does not exist $\\frac{p}{q}$ with $\\frac{a}{b} < \\frac{p}{q} < \\frac{c}{d}$ and $q < \\max(b, d)$. Then of all fractions $\\frac{p}{q}$ with $\\frac{a}{b} < \\frac{p}{q} < \\frac{c}{d}$, $\\frac{a+c}{b+d}$ has the smallest denominator.\n\n*Proof.* We use a lattice point interpretation. The condition of $\\frac{a}{b}, \\frac{c}{d}$ being adjacent in a Farey series means that the triangle with vertices $(0,0)$, $(a,b)$, $(c,d)$ has area $\\frac{1}{2}$, which is a consequence of Pick's Theorem. Now notice that in the triangle $(0,0)$, $(na,nb)$, $(nc,nd)$, for any nonnegative integers $k, m$ with $k+m \\le n$ we have $(ka+mc, kb+md)$ in or on the triangle. We also know the triangle's area is $\\frac{n^2}{2}$. Combining these two observations with Pick's Theorem shows that the points we have described are in fact all of the points inside or on the triangle. Using this and taking $n$ arbitrarily high, we conclude that if $\\frac{p}{q}$ is a fraction with $\\frac{a}{b} < \\frac{p}{q} < \\frac{c}{d}$, then there exist positive integers $k, m$ with $p = ka + mc$ and $q = kb + md$. Note $\\gcd(p, q) = 1$ if $\\gcd(k, m) = 1$. The desired conclusion is immediate. $\\square$\n\nNow we prove our main result. Note that if $a_1, a_2, \\dots, a_n$ is a regular sequence, then $a_1 + m, a_2 + 2m, \\dots, a_n + nm$ is also regular by replacing the associated $x$ with $x+m$. Therefore we may assume $a_1 = 0$ and $0 \\le x < 1$ in the regularity condition.\n\nFor each finite regular sequence, we can associate two real numbers $r, s$ to it so that the real numbers $x$ which satisfy the condition for being regular are those with $r \\le x < s$. We claim $r, s$ are always rational numbers which are adjacent terms of a Farey series. We show this by induction on the number of terms. For a 1-term regular sequence with $a_1 = 0$, we have $r = 0$ and $s = 1$. Given that we have it is true for all $n$-term regular sequences, note that if there is no integer multiple of $\\frac{1}{n+1}$ between $r$ and $s$, then $a_{n+1}$ is forced and $r, s$ remain the same as before. Otherwise, if $r < \\frac{k}{n+1} < s$, then $a_{n+1}$ can be either $a_n$ or $a_n + 1$. If $a_{n+1} = a_n$, we can check the new $r, s$ will be $r, \\frac{k}{n+1}$. If $a_{n+1} = a_n + 1$, it will be $\\frac{k}{n+1}, s$. The inductive hypothesis tells us that in both cases we obtain adjacent terms of a Farey series.\n\nCombining the above observations with Lemma 3, we can analyze forced terms as follows: if at a given point we have $r = \\frac{a}{b}$ and $s = \\frac{c}{d}$, then the next time we will have an unforced term is at step $b+d$, when one of $r$ and $s$ will change to $\\frac{a+c}{b+d}$ depending on the choice of the $b+d$th term. In particular, the denominators of $r, s$ start at $(1,1)$, and each time we have an unforced term, one denominator becomes the sum of the previous two. Since our sequence has 1000 terms, our goal is to have the sum of denominators greater than 1000 in as few unforced terms as possible, which would make the next unforced term come after the end of our sequence. By Lemma 2 and the fact that $F_{16} = 987$, $F_{17} = 1597$, at least 14 unforced terms are required. Combining this with the fact that $a_1$, which we didn't count yet, is also unforced, we have at least 15 unforced terms, and thus at most 985 forced terms as claimed.\n\nTo construct a sequence with this many forced terms, we simply set\n$$\na_k = \\lfloor k\\varphi \\rfloor \\quad \\text{for } 1 \\le k \\le 1000,\n$$\nwhere $\\varphi = \\frac{\\sqrt{5}-1}{2} \\approx 0.618\\ldots$. This will make $r, s$ always take the form $\\frac{F_k}{F_{k+1}}$ for some $k$ and obtain the equality case of Lemma 2.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22337,
"subject": "Mathematics (Olympiad)",
"question": "Find a formula for the sum of the first $n$ squares, that is, for $1^2 + 2^2 + \\cdots + n^2$.",
"options": [],
"answer": "See solution",
"solution": "Consider the sum $1^2 + 2^2 + \\cdots + n^2$. By analyzing the area of a constructed rectangle and using the formula for the sum of the first $n$ integers, we derive:\n\n$$\n(1 + 2 + \\cdots + n)(n + 1) = (1^2 + 2^2 + \\cdots + n^2) + (1) + (1 + 2) + \\cdots + (1 + 2 + \\cdots + n).\n$$\n\nAfter algebraic manipulation, we find:\n\n$$\n1^2 + 2^2 + \\cdots + n^2 = \\frac{1}{6} n(n+1)(2n+1).\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22338,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for any real numbers $x, y$ we have\n\n$$\nf(y - xy) = f(x)y + (x - 1)^2 f(y).\n$$",
"options": [],
"answer": "See solution",
"solution": "Setting $x = 1$ gives $f(0) = f(1)y$ for all real $y$, so $f(0) = f(1) = 0$.\n\nSetting $y = 1$ in the original equation:\n$$\nf(1-x) = f(x)\n$$\nfor all real $x$.\n\nLet $t$ be any real number. Setting $x = 1-t$ gives\n$$\nf(ty) = f(1-t)y + t^2 f(y) = f(t)y + t^2 f(y)\n$$\nfor all real $y$.\n\nSwapping $t$ and $y$:\n$$\nf(y)t + y^2 f(t) = f(yt) = f(ty) = f(t)y + t^2 f(y),\n$$\nso $f(t)(y^2 - y) = f(y)(t^2 - t)$. For $y = 2$:\n$$\nf(t) = \\frac{1}{2}f(2)(t^2 - t), \\quad t \\in \\mathbb{R}.\n$$\n\nThus, $f(x) = a x(x-1)$ for $a = f(2)/2$ and any real $a$.\n\nChecking:\n$$\n\\begin{align*}\nf(x)y + (x-1)^2 f(y) &= a x(x-1)y + (x-1)^2 a y(y-1) \\\\\n&= a(x-1)y(x + xy - x - y + 1) \\\\\n&= a(1-x)y((1-x)y - 1) = f((1-x)y) = f(y - xy).\n\\end{align*}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22339,
"subject": "Mathematics (Olympiad)",
"question": "One student was multiplying two numbers. During the multiplication, he switched the last digit of the first number, which was 4, with 1. So he obtained 525 as a result instead of 600. Which numbers did the student multiply?",
"options": [],
"answer": "See solution",
"solution": "From the condition in the problem, we see that when the first number is reduced by $4 - 1 = 3$, their product is reduced by $600 - 525 = 75$. So the second number is $75 \\div 3 = 25$. The first number is $600 \\div 25 = 24$. The numbers are $24$ and $25$.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 22340,
"subject": "Mathematics (Olympiad)",
"question": "A ball bearing consists of two cylinders with the same axis and $n$ equal balls between them. The centers of all the balls are on the same plane perpendicular to the axis of the cylinders, and each ball touches both cylinders and two adjacent balls. Let $r$ be the radius of the balls and let $R$ be the radius of the outer cylinder. Prove that $$\\frac{r}{R} < \\frac{\\pi}{n + \\frac{\\pi}{n}}.$$",
"options": [],
"answer": "See solution",
"solution": "Consider the regular $n$-gon with vertices at the centers of the balls. Its edges are of length $2r$, so its perimeter is $n \\cdot 2r$. The radius of the circumcircle of the $n$-gon is $R - r$, and the length of the circumcircle is $2\\pi(R - r)$. Since a chord of a circle is always shorter than the corresponding arc, we have $$n \\cdot 2r < 2\\pi(R - r).$$ This gives $$nr + \\pi r < \\pi R,$$ which implies $$\\frac{r}{R} < \\frac{\\pi}{n + \\frac{\\pi}{n}}.$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 22341,
"subject": "Mathematics (Olympiad)",
"question": "Consider $n$ intervals $I_2 = [a_2, b_2], \\dots, I_{n+1} = [a_{n+1}, b_{n+1}]$ inside the interval $[0, 2n-1]$, such that all endpoints are integers and the length of $I_j$ is $j$. Prove that there exist $i \\neq j$ such that $I_i \\subseteq I_j$.",
"options": [],
"answer": "See solution",
"solution": "Assume, for contradiction, that no interval is inside another. Define a function $f$ as follows:\n\n$$\nf : \\{2, \\dots, n\\} \\to \\left( \\{0, 1, \\dots, 2n-1\\} \\setminus \\{a_{n+1}, a_{n+1} + 1, \\dots, b_{n+1}\\} \\right)\n$$\n\n- If $a_j < a_{n+1}$, set $f(j) := a_j$.\n- If $a_j \\ge a_{n+1}$, set $f(j) := b_j$.\n\nSince no interval is inside $I_{n+1}$, for $2 \\leq j \\leq n$, if $a_j \\geq a_{n+1}$, then $b_{n+1} < b_j$. Thus, $f$ is well-defined. $f$ is injective: if $f(i) = f(j)$ for $i < j$, then $I_i$ and $I_j$ share an endpoint, so $I_i \\subseteq I_j$, contradicting our assumption.\n\nThe set $\\{0, 1, \\dots, 2n-1\\} \\setminus \\{a_{n+1}, a_{n+1} + 1, \\dots, b_{n+1}\\}$ has $2n-(n+2) = n-2$ elements, so no injective function from $\\{2, \\dots, n\\}$ to this set exists. This contradiction proves the assertion.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22342,
"subject": "Mathematics (Olympiad)",
"question": "For any positive integer $k$, define\n\n$$\nR(k) = 1\\underbrace{22\\dots22}_{k}1.\n$$\n\nFind all positive integers $n$ such that\n\n$$\n2013 \\mid R(10^n).\n$$",
"options": [],
"answer": "See solution",
"solution": "First, observe that\n\n$$\nR(k) = 11 \\cdot \\underbrace{11\\dots11}_{k+1}.\n$$\n\nNext, note that\n\n$$\n\\underbrace{11\\dots11}_{k+1} = \\frac{10^{k+1} - 1}{9}.\n$$\n\nThis reduces the problem to finding all $n$ such that\n\n$$\n2013 = 3 \\cdot 11 \\cdot 61 \\mid 11 \\cdot \\frac{10^{10^n+1} - 1}{9}.\n$$\n\nFor this to be true, we must have\n\n$$\n10^{10^n+1} \\equiv 1 \\pmod{61}.\n$$\n\nLet $r$ be the order of $10$ modulo $61$. We know that $r \\mid \\varphi(61) = 60$, and $r \\mid 10^n + 1$ for the above congruence to hold. But $\\gcd(60, 10^n + 1)$ does not divide $3$, so $r$ must be $1$ or $3$. Since\n\n$$\n10^3 \\equiv 24 \\pmod{61},\n$$\n\nthis is not the case, and therefore there is no $n$ such that $2013 \\mid R(10^n)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22343,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying the conditions:\n\n1. $f(f(x^2) + y + f(y)) = x^2 + 2f(y)$,\n2. $x \\le y$ implies $f(x) \\le f(y)$,\n\nfor all real numbers $x$ and $y$.",
"options": [],
"answer": "See solution",
"solution": "We prove that the only function satisfying the two conditions is $f(x) = x$.\n\nFirst, we show that $f$ is injective. Setting $y = 0$ in (i), we get:\n$$\nf(f(x^2) + f(0)) = x^2 + 2f(0)\n$$\nfor any $x$, or equivalently,\n$$\nf(f(a) + f(0)) = a + 2f(0)\n$$\nfor any $a \\ge 0$. Thus, $f$ is injective on nonnegative numbers.\n\nSuppose $f(y_1) = f(y_2)$. Then $f(f(x^2) + y_1 + f(y_1)) = f(f(x^2) + y_2 + f(y_2))$. For large $x$, both arguments are positive, so $y_1 = y_2$.\n\nNow, we prove $f(0) = 0$.\n\n**Case 1:** $f(0) \\le 0$. For $a = -2f(0)$, $f(f(-2f(0)) + f(0)) = 0$, so there exists $c$ with $f(c) = 0$. Plugging $x = 0$, $y = c$ in (i): $f(f(0) + c) = 0$. Since $f$ is injective, $f(0) + c = c$, so $f(0) = 0$.\n\n**Case 2:** $f(0) \\ge 0$. In (i), set $x = y = 0$: $f(2f(0)) = 2f(0)$. Let $a = 3f(0) = f(0) + f(2f(0))$. From above, $f(a) = f(f(2f(0)) + f(0)) = 4f(0)$. Adding $f(0)$, $f(f(a) + f(0)) = f(5f(0)) = 5f(0)$. Plugging $x = 0$, $y = 2f(0)$ in (i): $f(5f(0)) = 4f(0)$. Thus $5f(0) = 4f(0)$, so $f(0) = 0$.\n\nFrom above, for $a \\ge 0$, $f(f(a)) = a$, and from (i) with $x = 0$, $f(y + f(y)) = 2f(y)$ for any $y$.\n\nPlug $y = f(a)$: $f(f(a) + a) = 2f(f(a)) = 2a$; plug $y = a$: $f(a + f(a)) = 2f(a)$. Thus, for $a \\ge 0$, $f(a) = a$.\n\nNow, (i) becomes $f(x^2 + y + f(y)) = x^2 + 2f(y)$. For any $y$, there exists $x$ such that $x^2 + y + f(y) > 0$. Then $f(x^2 + y + f(y)) = x^2 + y + f(y) = x^2 + 2f(y)$, so $f(y) = y$ for all $y$.\n\nTherefore, the only solution is $f(x) = x$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22344,
"subject": "Mathematics (Olympiad)",
"question": "Let $k$ be a positive integer and let $m$ be a positive odd integer. Show that there exists a positive integer $n$ such that $m^n + n^m$ has at least $k$ distinct prime factors.",
"options": [],
"answer": "See solution",
"solution": "Design a set of $k$ primes $p_1 < p_2 < \\cdots < p_k$ as follows. Begin by choosing $p_1 > 2m$. Having selected $p_j$, use Dirichlet's theorem to choose a prime\n\n$$\np_{j+1} \\equiv -1 \\pmod{p_1(p_1-1)p_2(p_2-1)\\cdots p_j(p_j-1)}.\n$$\n\nIf $i < j$, then $p_i < p_j$, so $p_j$ does not divide $p_i - 1$; further, $p_j - 1 \\equiv -2 \\pmod{p_i(p_i - 1)}$, so $p_i$ does not divide $p_j - 1$.\n\nNext, use the Chinese Remainder Theorem to choose a positive integer $n$ such that $n \\equiv -1 \\pmod{p_1p_2\\cdots p_k}$ and $n \\equiv 0 \\pmod{(p_1-1)\\cdots(p_k-1)}$.\n\nFinally, since $p_1p_2\\cdots p_k$ and $m$ are coprime, and $(p_1-1)(p_2-1)\\cdots(p_k-1)$ divides $n$, and $m$ is odd, Euler's Theorem applies to show that $m^n + n^m \\equiv 1 + (-1)^m \\equiv 0 \\pmod{p_1p_2\\cdots p_k}$. The conclusion follows.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22345,
"subject": "Mathematics (Olympiad)",
"question": "Given real numbers $a, b, c$ such that $abc = 1$, prove that for all integers $k \\ge 2$,\n$$\n\\frac{a^k}{a+b} + \\frac{b^k}{b+c} + \\frac{c^k}{c+a} \\ge \\frac{3}{2}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Since\n$$\n\\frac{a^k}{a+b} + \\frac{1}{4}(a+b) + \\underbrace{\\frac{1}{2} + \\frac{1}{2} + \\dots + \\frac{1}{2}}_{k-2} \\ge k \\cdot \\sqrt[k]{\\frac{a^k}{2^k}} = \\frac{k}{2}a,\n$$\nthen\n$$\n\\frac{a^k}{a+b} \\ge \\frac{k}{2}a - \\frac{1}{4}(a+b) - \\frac{k-2}{2}.\n$$\nSimilarly,\n$$\n\\frac{b^k}{b+c} \\ge \\frac{k}{2}b - \\frac{1}{4}(b+c) - \\frac{k-2}{2},\n$$\n$$\n\\frac{c^k}{c+a} \\ge \\frac{k}{2}c - \\frac{1}{4}(c+a) - \\frac{k-2}{2}.\n$$\nAdding the three inequalities above, we obtain\n$$\n\\begin{aligned}\n& \\frac{a^k}{a+b} + \\frac{b^k}{b+c} + \\frac{c^k}{c+a} \\\\\n\\ge & \\frac{k}{2}(a+b+c) - \\frac{1}{2}(a+b+c) - \\frac{3}{2}(k-2) \\\\\n= & \\frac{k-1}{2}(a+b+c) - \\frac{3}{2}(k-2) \\\\\n\\ge & \\frac{3}{2}(k-1) - \\frac{3}{2}(k-2) \\\\\n= & \\frac{3}{2},\n\\end{aligned}\n$$\nas desired.\n\n**Remark:** The problem could also be proved by the Cauchy inequality or the Chebyshev inequality.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22346,
"subject": "Mathematics (Olympiad)",
"question": "With the usual notation for binomial coefficients:\n\n$$\n\\binom{n}{r} = \\begin{cases} 0, & \\text{if } r < 0 \\text{ or } r > n, \\\\ \\frac{n!}{(n-r)!r!}, & \\text{if } 0 \\le r \\le n, \\end{cases}\n$$\n\nShow that for all $n \\ge 0$:\n\n$$\n\\sum_{k=0}^{n} \\binom{n+k}{n} = \\binom{2n+1}{n}\n$$",
"options": [],
"answer": "See solution",
"solution": "Recall that $\\binom{n+1}{r} = \\binom{n}{r} + \\binom{n}{r-1}$, so\n$$\n\\binom{n+k}{n} = \\binom{n+k+1}{n+1} - \\binom{n+k}{n+1}.\n$$\nThus, the sum telescopes:\n$$\n\\begin{aligned}\n\\sum_{k=0}^{n} \\binom{n+k}{n} &= \\sum_{k=0}^{n} \\left( \\binom{n+k+1}{n+1} - \\binom{n+k}{n+1} \\right) \\\\\n&= \\binom{2n+1}{n+1} - \\binom{n}{n+1} \\\\\n&= \\binom{2n+1}{n+1} \\\\\n&= \\binom{2n+1}{n}.\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22347,
"subject": "Mathematics (Olympiad)",
"question": "There are $n$ rooms in a sauna, each with unlimited capacity. At any time, a room may be attended by people of the same gender (males or females). Additionally, males want to share a room only with males they do **not** know, and females want to share a room only with females they **do** know. What is the largest number $k$ such that any $k$ couples can visit the sauna at the same time, given that two males know each other if and only if their wives know each other?",
"options": [],
"answer": "See solution",
"solution": "We will show by induction that it is possible for $n-1$ pairs to visit the sauna at the same time.\n\n**Base case:** For $n=1$, the statement is clear.\n\n**Inductive step:** Assume that $n-2$ pairs can be placed in $n-1$ rooms. Now, add one more pair. Let $k$ be the number of pairs that the new pair knows, and let $m$ be the number of rooms occupied by males.\n\n- If $m > k$, there is a room with males that the new man does **not** know. He can join that room, and his wife can use an empty room (the $n$-th room).\n- If $m \\leq k$, then $n-2-k < n-1-m$. There are $n-2-k$ females that the new woman does **not** know, and $n-1-m$ rooms occupied by females (or empty). Thus, there is a room with only females (possibly empty) that the new woman knows, so she can join them. The new man can use the $n$-th room.\n\nNow, we show that $n-1$ is the largest possible number. For $n$ pairs who do **not** know each other, the men must be placed in different rooms, requiring $n$ rooms, leaving no room for the women.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22348,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\{a_n\\}_{n \\ge 0}$ and $\\{b_n\\}_{n \\ge 0}$ be sequences of real numbers such that $a_0 > \\frac{1}{2}$, $a_{n+1} \\ge a_n$, and $b_{n+1} = a_n(b_n + b_{n+2})$ for all non-negative integers $n$. Show that the sequence $\\{b_n\\}_{n \\ge 0}$ is bounded.",
"options": [],
"answer": "See solution",
"solution": "Using the relation in the statement and the fact that no $a_k$ is $\\frac{1}{2}$, we can write\n\n$$\nb_{k+2}^2 - b_k^2 = \\frac{(b_{k+1} - b_k)^2 - (b_{k+2} - b_{k+1})^2}{2a_k - 1}, \\quad k \\in \\mathbb{N}\n$$\n\nFix an integer $n > 2$, and sum over $k = 0, 1, \\ldots, n-2$ to obtain\n\n$$\nb_n^2 + b_{n-1}^2 - b_1^2 - b_0^2 = \\frac{(b_1 - b_0)^2}{2a_0 - 1} - \\sum_{k=0}^{n-3} \\left( \\frac{1}{2a_k - 1} - \\frac{1}{2a_{k+1} - 1} \\right) (b_{k+2} - b_{k+1})^2 - \\frac{(b_n - b_{n-1})^2}{2a_{n-2} - 1} \\le \\frac{(b_1 - b_0)^2}{2a_0 - 1}\n$$\n\nsince the $a_k$ form an increasing sequence of real numbers greater than $\\frac{1}{2}$. Consequently,\n\n$$\nb_n^2 \\le b_n^2 + b_{n-1}^2 \\le b_0^2 + b_1^2 + \\frac{(b_1 - b_0)^2}{2a_0 - 1}\n$$\n\nand the conclusion follows.\n\n**Remark.** Leaving aside the trivial case where the $b_n$ are all zero, it is readily checked that the sequences $\\{a_n\\}_{n \\ge 0}$ and $\\{b_n\\}_{n \\ge 0}$, where $a_n = 1$ and $b_n = \\cos(n\\pi/3)$, satisfy the conditions in the statement, so the hypothesis is not vacuous.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22349,
"subject": "Mathematics (Olympiad)",
"question": "An acute triangle $ABC$ has orthocenter $H$. The circle through $H$ with center at the midpoint of $BC$ intersects the line $BC$ at $A_1$ and $A_2$. Similarly, the circle passing through $H$ with center at the midpoint of $CA$ intersects the line $CA$ at $B_1$ and $B_2$, and the circle passing through $H$ with center at the midpoint of $AB$ intersects the line $AB$ at $C_1$ and $C_2$. Show that $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ are concyclic.\n\n",
"options": [],
"answer": "See solution",
"solution": "**Proof I** Let $B_0$, $C_0$ be the midpoints of $CA$, $AB$ respectively. Denote $A'$ as the other intersection of the circle centered at $B_0$ passing through $H$ and the circle centered at $C_0$ passing through $H$. We know that $A'H \\perp C_0B_0$. Since $B_0$, $C_0$ are the midpoints of $CA$, $AB$ respectively, $B_0C_0 \\parallel BC$. Therefore, $A'H \\perp BC$. This yields that $A'$ lies on the segment $AH$.\n\nBy the Secant-Secant theorem, it follows that\n\n$$\nAC_1 \\cdot AC_2 = AA' \\cdot AH = AB_1 \\cdot AB_2,\n$$\n\nso $B_1$, $B_2$, $C_1$, $C_2$ are concyclic.\n\nLet the intersection of the perpendicular bisectors of $B_1B_2$, $C_1C_2$ be $O$. Then $O$ is the circumcenter of quadrilateral $B_1B_2C_1C_2$, as well as the circumcenter of $\\triangle ABC$. So\n\n$$\nOB_1 = OB_2 = OC_1 = OC_2.\n$$\n\nSimilarly,\n\n$$\nOA_1 = OA_2 = OB_1 = OB_2.\n$$\n\nTherefore, the six points $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ are all on the same circle, whose center is $O$ and radius $OA_1$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22350,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge 4$ be an even integer. In how many ways can one select four different positive integers $k$, $1 \\le k \\le n$, so that the sum of two of the chosen numbers equals the sum of the other two?",
"options": [],
"answer": "See solution",
"solution": "Let $a$ be the smallest and $b$ the largest of the chosen numbers. The sum in the problem must be $a + b$. Given $a$ and $b$, the other numbers $c$ and $d$ must satisfy $a < c, d < b$ and $c + d = a + b$. For the smaller of $c, d$, say $c$, one can take any number larger than $a$ but smaller than the average of $a$ and $b$, and the choice of $c$ uniquely determines $d$. If $b - a = 2p + 1$ or $b - a = 2p + 2$, there are $p$ possible choices for $c$. Assume $n = 2m$. Then the largest possible $b - a = 2m - 1 = 2(m - 1) + 1$, and there is just one possible pair $(a, b) = (1, 2m) = (1, n)$. For $b - a = n - q$ there are $q$ possible pairs $(a, b)$. The $p$ possible choices of $c$ thus appear when $q = n - 2p - 2$ and $q = n - 2p - 1$, or altogether in $2n - 4p - 3$ cases. So the total number of choices is\n\n$$\n\\begin{aligned}\n\\sum_{p=1}^{m-2} (2n - 4p - 3)p + m - 1 &= (2n - 3) \\sum_{p=1}^{m-2} p - 4 \\sum_{p=1}^{m-2} p^2 + m - 1 \\\\\n&= \\frac{1}{2}(2n - 3)(m - 2)(m - 1) - \\frac{4}{6}(m - 2)(m - 1)(2m - 3) + (m - 1).\n\\end{aligned}\n$$\n\nUsing $2m = n$, the last sum simplifies to\n\n$$\n\\frac{(n - 2)(2n^2 - 5n)}{24}\n$$\n\nThe restriction \"n even\" can be removed, but then there are two essentially similar but slightly different sums to be done. For \"n odd\" the calculation is slightly easier, because there would not be the single last term.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22351,
"subject": "Mathematics (Olympiad)",
"question": "There are $n$ students standing in a circle, one behind the other. The students have heights $h_1 < h_2 < \\dots < h_n$. If a student with height $h_k$ is standing directly behind a student with height $h_{k-2}$ or less, the two students are permitted to switch places. Prove that it is not possible to make more than $\\binom{n}{3}$ such switches before reaching a position in which no further switches are possible.",
"options": [],
"answer": "See solution",
"solution": "Let $h_i$ also denote the student with height $h_i$. We prove that for $1 \\leq i < j \\leq n$, $h_j$ can switch with $h_i$ at most $j - i - 1$ times. We proceed by induction on $j - i$, the base case $j - i = 1$ being evident because $h_i$ is not allowed to switch with $h_{i-1}$.\n\nFor the inductive step, note that $h_i$, $h_{j-1}$, $h_j$ can be positioned on the circle either in this order or in the order $h_i$, $h_j$, $h_{j-1}$. Since $h_{j-1}$ and $h_j$ cannot switch, the only way to change the relative order of these three students is for $h_i$ to switch with either $h_{j-1}$ or $h_j$. Consequently, any two switches of $h_i$ with $h_j$ must be separated by a switch of $h_i$ with $h_{j-1}$. Since there are at most $j - i - 2$ of the latter, there are at most $j - i - 1$ of the former.\n\nThe total number of switches is thus at most\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n-1} \\sum_{j=i+1}^{n} (j-i-1) &= \\sum_{i=1}^{n-1} \\sum_{j=0}^{n-i-1} j \\\\\n&= \\sum_{i=1}^{n-1} \\binom{n-i}{2} \\\\\n&= \\sum_{i=1}^{n-1} \\left[ \\binom{n-i+1}{3} - \\binom{n-i}{3} \\right] \\\\\n&= \\binom{n}{3}.\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22352,
"subject": "Mathematics (Olympiad)",
"question": "令 $I, O, H, \\Omega$ 分別為三角形 $ABC$ 的內心、外心、垂心與外接圓。設 $AI$ 與 $\\Omega$ 交於 $M \\neq A$，$IH$ 與 $BC$ 交於 $D$，$MD$ 與 $\\Omega$ 交於 $E \\neq M$。\n\n證明：直線 $OI$ 與 $\\triangle IHE$ 的外接圓相切。\n\n",
"options": [],
"answer": "See solution",
"solution": "取一點 $X$ 使得 $\\triangle XHI \\sim \\triangle XIO$。我們先證明 $X$ 位於 $\\Omega$ 上：取 $Y, Z$ 使得 $\\triangle XAY \\sim \\triangle XBZ \\sim \\triangle XHI$，則由旋似，$\\triangle IAH \\sim \\triangle OYI$。這告訴我們 $\\angle OYI = \\angle IAH = \\angle OAI$，即 $A, O, I, Y$ 共圓。同理有 $B, O, I, Z$ 共圓。設 $P$ 為 $AY$ 與 $BZ$ 的交點，則一樣由旋似，$\\triangle AIB \\sim \\triangle YOZ$，所以得到\n\n$$\n\\angle APB = \\angle YAI + \\angle AIB + \\angle IBZ = \\angle YOI + \\angle AIB + \\angle IOZ = 2 \\cdot \\angle AIB = \\angle ACB,\n$$\n\n即 $P$ 位於 $\\Omega$ 上。再由 $\\angle AXB = \\angle (AY, BZ) = \\angle APB$ 即得 $X$ 也位於 $\\Omega$ 上。\n\n因為 $\\angle XHI = \\angle XIO$，$OI$ 與 $\\odot(XHI)$ 相切，因此我們只需證明 $E, X, H, I$ 共圓。注意到 $\\angle DBM = \\angle BEM$，所以 $MD \\cdot ME = \\overline{MB}^2 = \\overline{MI}^2$，故\n\n$$\n\\angle MEI = \\angle DIM = \\angle HIA = \\angle IOY = \\angle IAP = \\angle MEP,\n$$\n\n即 $E, I, P$ 共線。因此由 $\\angle IEX = \\angle YAX = \\angle IHX$ 我們得到 $E, X, H, I$ 共圓。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22353,
"subject": "Mathematics (Olympiad)",
"question": "Find all natural numbers $n \\geq 2$ for which the following statement holds: if the sum of numbers in the sequence of natural numbers $(a_1, a_2, \\ldots, a_n)$ is equal to $2n - 1$, then there exists a block of consecutive members of this sequence containing at least two members, the numbers of which have an arithmetic mean that is an integer.",
"options": [],
"answer": "See solution",
"solution": "**Answer:** $n \\geq 4$.\n\nFor $n = 2, 3$, counterexamples are the sequences $(1, 2)$ and $(2, 1, 2)$.\n\nLet $(a_1, a_2, \\ldots, a_n)$ be any sequence satisfying the conditions. Define $s_0 = 0$, $s_k = a_1 + a_2 + \\ldots + a_k - 2k$ for $k = 1, \\ldots, n$. Call a sequence *good* if it does not satisfy the problem's condition, i.e., there is no block of at least two consecutive members with integer arithmetic mean. A pair $(i, j)$ is *divisible* if $(j-i) \\mid (s_j - s_i)$. The sequence $(a_1, \\ldots, a_n)$ is good if and only if there is no divisible pair with $|j-i| \\geq 2$.\n\nFor $n \\geq 4$, $s_n = -1$, and for all $k = 1, \\ldots, n$, $s_{k+1} - s_k = a_k - 2 \\geq -1$. Consider possible values for $s_2$:\n\n- If $s_2 \\leq -2$, since $s_1 \\geq s_0 - 1 = -1$ and $s_2 \\geq s_1 - 1$, we have $s_2 = -1$, $s_1 = -1$, so $n - 1 \\mid s_n - s_1$.\n- If $s_2 = -1$, then $n - 2 \\mid s_n - s_2$.\n- If $s_2 = 0$, then $2 - 0 \\mid s_2 - s_0$.\n- If $s_2 \\geq 1$, since $s_n = -1$ and $s_{k+1} \\geq s_k - 1$, there is an index $i$ with $s_i = 0$, so $i - 0 \\mid s_i - s_0$.\n\nThus, among the pairs $(1, n)$, $(2, n)$, $(0, 2)$, and $(0, i)$ for $2 < i < n$, at least one pair is divisible. This completes the proof for $n \\geq 4$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22354,
"subject": "Mathematics (Olympiad)",
"question": "Define the sequence $a_1, a_2, a_3, \\dots$ by $a_1 = 1$ and\n\n$$\na_n = a_{\\lfloor n/2 \\rfloor} + a_{\\lfloor n/3 \\rfloor} + \\dots + a_{\\lfloor n/n \\rfloor} + 1\n$$\n\nfor $n > 1$. Prove that there are infinitely many $n$ such that\n\n$$\na_n \\equiv n \\pmod{2^{2010}}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Our solution is based on the following key observation, for which we provide two different proofs.\n\n**Lemma 1.** Let $p$ be a prime. If $p^s$ divides $n$, then $2^{s-1}$ divides $a_n - a_{n-1}$, where $a_0 = 0$.\n\n*First proof of Lemma 1.* We argue by induction on $s$ and then $n$. The base case $s=1$ is vacuous. Assume the claim holds for $s-1$, and suppose there exists a smallest $n$ such that $p^s$ divides $n$ but $2^{s-1}$ does not divide $a_n - a_{n-1}$. Define $b_n = a_n - a_{n-1}$. The sequence $b_n$ satisfies\n\n$$\nb_n = \\sum_{i=2}^{n} \\left( a_{\\lfloor n/i \\rfloor} - a_{\\lfloor (n-1)/i \\rfloor} \\right) = \\sum_{\\substack{i|n \\\\ i>1}} b_{n/i} = \\sum_{\\substack{i|n \\\\ i N + 2^{s-1} - 1$, there exist $k$ and $n \\in \\{k+N, \\dots, k+N+2^{s-1}-1\\}$ such that $a_n \\equiv n \\pmod{2^{s-1}}$. Thus, for any $s$, the set of $n$ with $a_n \\equiv n \\pmod{2^{s-1}}$ is infinite. Taking $s = 2011$ gives the result.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22355,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be a triangle in which $m(\\hat{A}) = 135^\\circ$. The perpendicular to the line $AB$ erected at $A$ intersects the side $BC$ at $D$, and the angle bisector of $\\angle B$ intersects the side $AC$ at $E$. Find the measure of $\\widehat{BED}$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $I \\in (BE)$ such that $IA$ bisects the angle $\\widehat{DAB}$. We deduce that $ID$ is the bisector of the angle $\\widehat{ADB}$. A short computation shows that $m(\\widehat{DIB}) = 135^\\circ$, hence triangles $ABE$ and $IBD$ are similar. It follows that $\\frac{AB}{IB} = \\frac{BE}{BD}$, so that $\\frac{AB}{EB} = \\frac{BI}{BD}$. Thus, triangles $ABI$ and $DBE$ are similar as well and, since $m(\\widehat{BED}) = m(\\widehat{BAI})$, we infer that $m(\\widehat{BED}) = 45^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22356,
"subject": "Mathematics (Olympiad)",
"question": "a) Let $a, b \\in \\mathbb{R}$ be two real numbers with $a < b$, and let $f : [a, b] \\to \\mathbb{R}$ be a strictly monotone function such that\n$$\n\\int_a^b f(x) \\, dx = 0.\n$$\nShow that $f(a) \\cdot f(b) < 0$.\n\nb) Determine all convergent sequences $(a_n)_{n \\ge 1}$ of real numbers for which there exists a strictly monotone function $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\n\\int_{a_{n-1}}^{a_n} f(x) \\, dx = \\int_{a_n}^{a_{n+1}} f(x) \\, dx, \\quad \\text{for all } n \\ge 2.\n$$",
"options": [],
"answer": "See solution",
"solution": "a) Suppose $f([a, b]) \\subseteq [0, \\infty)$ or $f([a, b]) \\subseteq (-\\infty, 0]$. Then $m = |f(\\frac{a+b}{2})| > 0$, and on one of the intervals $(a, \\frac{a+b}{2})$ or $(\\frac{a+b}{2}, b)$, we have $|f(x)| > m$ for all $x$ in that interval. Thus,\n$$\n\\begin{aligned}\n0 &= \\left| \\int_a^b f(x) \\, dx \\right| = \\int_a^b |f(x)| \\, dx \\\\\n &= \\int_a^{\\frac{a+b}{2}} |f(x)| \\, dx + \\int_{\\frac{a+b}{2}}^b |f(x)| \\, dx \\\\\n &\\ge m \\cdot \\frac{b-a}{2} > 0,\n\\end{aligned}\n$$\nwhich is impossible. Therefore, $f(a) \\cdot f(b) < 0$.\n\nb) The only convergent sequences $(a_n)_{n \\ge 1}$ for which there exists a strictly monotone function $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying the given condition are:\n- Constant sequences: $a_n = a$ for all $n$, with $\\lim_{n \\to \\infty} a_n = a$.\n- Sequences taking exactly two distinct real values and becoming stationary from some index onward: $\\{a_n \\mid n \\ge 1\\} = \\{a, b\\}$, and $a_n = a$ for all $n \\ge n_0$ for some $n_0$.\n\nFor constant sequences, $\\int_{a_k}^{a_{k+1}} f(x) dx = 0$ for any $k$ and any $f$. For the two-value case, the function $f(x) = 2x - a - b$ is strictly monotone and satisfies $\\int_{a_k}^{a_{k+1}} f(x) dx = 0$ for all $k$.\n\nSuppose $(a_n)$ is a convergent sequence and such an $f$ exists. Let $I = \\int_{a_k}^{a_{k+1}} f(x) dx$ for all $k$, and $a = \\lim_{n \\to \\infty} a_n$. For any $r > 0$, there exists $n_r$ such that $a_n \\in (a - r, a + r)$ for $n \\ge n_r$. Since $f$ is strictly monotone, $|f(x)| < M$ for $x \\in (a - r, a + r)$, where $M = \\max(|f(a - r)|, |f(a + r)|)$. For any $p \\ge 1$,\n$$\np \\cdot |I| = \\left| \\int_{a_{n_r}}^{a_{n_r+p}} f(x) \\, dx \\right| < 2rM.\n$$\nThus, $|I| < \\frac{2Mr}{p}$ for any $p$, so $I = 0$.\n\nNow, suppose $\\{a_n\\}$ contains more than two distinct values. Then there exist $i < j < k$ with $a_i \\ne a_j \\ne a_k \\ne a_i$. Then,\n$$\n\\int_{a_i}^{a_j} f(x) \\, dx = (j-i)I = 0, \\quad \\int_{a_j}^{a_k} f(x) \\, dx = (k-j)I = 0, \\quad \\int_{a_i}^{a_k} f(x) \\, dx = 0.\n$$\nSince $f$ is strictly monotone, $f(a_i) \\cdot f(a_j) < 0$, $f(a_i) \\cdot f(a_k) < 0$, and $f(a_j) \\cdot f(a_k) < 0$, so\n$$\n(f(a_i) \\cdot f(a_j) \\cdot f(a_k))^2 < 0,\n$$\nwhich is impossible. Thus, $\\{a_n\\}$ contains at most two values, and if not constant, must become stationary.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22357,
"subject": "Mathematics (Olympiad)",
"question": "Call a positive integer *balanced* if the number of its distinct prime factors is equal to the number of its digits in the decimal representation. For example, the number $385 = 5 \\cdot 7 \\cdot 11$ is balanced, while $275 = 5^2 \\cdot 11$ is not. Prove that there exist only a finite number of balanced numbers.",
"options": [],
"answer": "See solution",
"solution": "Let $p_1 = 2$, $p_2 = 3$, $p_3 = 5$, \\ldots$ be the sequence of primes. Any balanced number $a$ with $n$ digits satisfies $a \\geq p_1 p_2 \\cdots p_n$. Since $p_1 p_2 \\cdots p_{11} = 2 \\cdot 3 \\cdot 5 \\cdots 29 \\cdot 31 > 10^{11}$ and $p_k > 10$ for any $k > 11$, it follows that there are no balanced numbers having more than 10 digits.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22358,
"subject": "Mathematics (Olympiad)",
"question": "Is it possible to write all natural numbers from 1 to 24 in a row in such a way that it is not possible to choose five numbers (not necessarily on consecutive positions) that are arranged in increasing order, and that it is not possible to choose six numbers (not necessarily on consecutive positions) that are arranged in decreasing order?",
"options": [],
"answer": "See solution",
"solution": "**Answer:** Yes, it is possible. Example:\n\n6, 5, 4, 3, 2, 1, 12, 11, 10, 9, 8, 7, 18, 17, 16, 15, 14, 13, 24, 23, 22, 21, 20, 19.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22359,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be a triangle with $AB < AC$ inscribed in a circle $c$. The tangent to $c$ at $C$ meets the parallel from $B$ to $AC$ at $D$. The tangent to $c$ at $B$ meets the parallel from $C$ to $AB$ at $E$, and the tangent to $c$ at $C$ at $L$. Suppose that the circumcircle $c_1$ of triangle $BDC$ meets $AC$ at $T$, and the circumcircle $c_2$ of triangle $BEC$ meets $AB$ at $S$. Prove that the lines $ST$, $BC$, and $AL$ are concurrent.\n\n",
"options": [],
"answer": "See solution",
"solution": "We will first prove that the circle $c_1$ is tangent to $AB$ at $B$. To do this, we need to show that $\\angle BDC = \\angle ABC$. Since $BD \\parallel AC$, $\\angle DBC = \\angle ACB$. Also, $\\angle BCD = \\angle BAC$ (by chord and tangent), so triangles $ABC$ and $BDC$ have two equal angles, and thus the third angles are also equal. Therefore, $\\angle BDC = \\angle ABC$, so $c_1$ is tangent to $AB$ at $B$.\n\nSimilarly, the circle $c_2$ is tangent to $AC$ at $C$.\n\nAs a consequence, $\\angle ABT = \\angle ACB$ (by chord and tangent), and also $\\angle BSC = \\angle ACB$.\n\nThus, $\\angle ABT = \\angle BSC$, so the lines $BT$ and $SC$ are parallel.\n\nLet $ST$ intersect $BC$ at $K$. It suffices to prove that $K$ lies on $AL$.\n\nFrom the trapezoid $BTCS$, we get:\n\n$$\n\\frac{BK}{KC} = \\frac{BT}{SC} \\qquad (1)\n$$\n\nFrom the similar triangles $ABT$ and $ASC$, we have:\n\n$$\n\\frac{BT}{SC} = \\frac{AB}{AS} \\qquad (2)\n$$\n\nBy (1) and (2):\n\n$$\n\\frac{BK}{KC} = \\frac{AB}{AS} \\qquad (3)\n$$\n\nFrom the power of a point theorem:\n\n$$\nAC^2 = AB \\cdot AS \\implies AS = \\frac{AC^2}{AB}\n$$\n\nSubstituting into (3):\n\n$$\n\\frac{BK}{KC} = \\frac{AB^2}{AC^2}\n$$\n\nThis shows that $K$ lies on the symmedian of triangle $ABC$.\n\nFinally, since $LB$ and $LC$ are tangents, $AL$ is the symmedian of triangle $ABC$, so $K$ lies on $AL$, as required.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22360,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs $ (x, y) $ of real numbers for which\n\n$$\n4y^4 + x^4 + 12y^3 + 5x^2(y^2 + 1) + y^2 + 4 = 12y.\n$$",
"options": [],
"answer": "See solution",
"solution": "We have the inequalities $x^4 \\geq 0$, $5x^2(y^2+1) \\geq 0$, and $4y^4 + 12y^3 + y^2 - 12y + 4 = (2y-1)^2(y+2)^2 \\geq 0$. The sum of the left sides is $0$ if and only if each of them is equal to $0$. The first two lead to $x = 0$, and the third to $y = -2$ or $y = \\frac{1}{2}$.\n\n$\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22361,
"subject": "Mathematics (Olympiad)",
"question": "Determine all pairs of integers $(c, d)$, both greater than $1$, satisfying the following condition: For any degree $d$ monic polynomial $Q$ with integer coefficients and for any prime $p > c(2c + 1)$, there exists a set $S$ of at most $\\frac{2c-1}{2c+1} \\cdot p$ integers, such that\n\n$$\n\\bigcup_{s \\in S} \\{s, Q(s), Q(Q(s)), Q(Q(Q(s))), \\dots\\}\n$$\n\ncontains a complete residue system modulo $p$ (i.e., intersects with every residue class modulo $p$).",
"options": [],
"answer": "See solution",
"solution": "The required pairs are all pairs $(c, d)$ of positive integers greater than $1$ such that $d \\leq c$.\n\nAssume first that $d \\geq c + 1$. Choose a large prime $p$ (we need $p > 2c^2 + c$) congruent to $1$ modulo $d$ (such a prime exists by Dirichlet's theorem; this can be shown using the cyclotomic polynomial $\\Phi_d$). Let $Q(X) = X^d$. Since $d \\mid p-1$, exactly $1 + \\frac{p-1}{d}$ residues modulo $p$ are $d$-th powers; all other $(d-1)\\frac{p-1}{d}$ residue classes contain no values of $Q$. Hence, if a set $S$ satisfies the requirements, it should contain representatives of all those classes. But this is more than $S$ is allowed to contain, since\n\n$$\n\\frac{d-1}{d}(p-1) > \\frac{2c-1}{2c+1}p \\quad \\iff \\quad \\frac{c}{c+1}(p-1) > \\frac{2c-1}{2c+1}p \\quad \\iff \\frac{p}{(c+1)(2c+1)} > \\frac{c}{c+1} \\quad \\iff \\quad p > c(2c+1).\n$$\n\nWe now show that such a set $S$ exists whenever $d \\leq c$. To this end, we use the lemma below.\n\n**Lemma.** Fix an integer $d \\geq 2$. Let $G = (V, E)$ be a directed graph, each vertex of which has exactly one outgoing edge and at most $d$ incoming edges. Assume further that there are at most $d$ loops in this graph. Then there exists a subset $V'$ of $V$ of cardinality $|V'| \\leq 1 + \\frac{d-1}{d}|V|$ such that every vertex in $V \\setminus V'$ is the terminus of a directed path emanating from $V'$.\n\n*Proof.* Consider any (weak) connected component $G_1 = (V_1, E_1)$ in $G$—i.e., a component of the corresponding undirected graph. Since from each vertex emanates exactly one edge, the component contains a directed cycle (possibly a loop); and since the numbers of vertices and edges in $G_1$ are equal, even an undirected cycle is unique. Hence, the component is a cycle with some trees rooting out of its vertices. With reference again to uniqueness of outgoing edges, the edges of these trees are all directed towards the cycle.\n\n\n\nNow, let $V'$ choose exactly one vertex from each component that is just a cycle; for any other component, let $V'$ choose all its in-degree $0$ vertices, i.e., the leaves of all trees rooting out of the vertices of the core cycle—any vertex of such a tree can be reached from some leaf, and hence so can any vertex of the core cycle.\n\nTo bound $|V'|$ from above, let $t$ be the number of single-vertex components in $G$, and notice that $t \\leq d$, since there are at most $d$ loops in the graph. From each other component that is a cycle, $V'$ chooses at most half of its vertices, so at most a $\\frac{d-1}{d}$ fraction of them. Finally, consider a component containing some trees. Since each in-degree is at most $d$, at least a $\\frac{1}{d}$ fraction of the vertices have incoming edges, hence $V'$ chooses at most a $\\frac{d-1}{d}$ fraction of the vertices. Consequently,\n\n$$\n|V'| \\leq t + \\frac{d-1}{d}(|V| - t) = \\frac{t}{d} + \\frac{d-1}{d}|V| \\leq 1 + \\frac{d-1}{d}|V|,\n$$\n\nas desired. This establishes the lemma. $\\square$\n\nNow let $p$ and $Q$ be chosen as in the problem statement. Consider a graph with vertex set $\\mathbb{Z}_p$. Regard $Q$ as a polynomial over $\\mathbb{Z}_p$, and draw an edge $a \\to Q(a)$ for every $a$ in $\\mathbb{Z}_p$. Since $\\deg Q = d$, each $b$ in $\\mathbb{Z}_p$ has at most $d$ preimages, so the in-degree of each vertex is at most $d$. Since $Q$ is monic and $d > 1$, the equation $Q(x) = x$ has at most $d$ roots in $\\mathbb{Z}_p$, hence the graph has at most $d$ loops. Thus, the lemma provides a set $V'$ which is suitable as the required set $S$. Indeed, the lemma shows that each residue is a repetitive image of some element of $S$; and the implications below show that the cardinality of $V'$ lies within the required range:\n\n$$\n\\begin{aligned}\n|V'| \\leq \\frac{d-1}{d}p + 1 \\leq \\frac{2c-1}{2c+1}p &\\iff \\frac{c-1}{c}p + 1 \\leq \\frac{2c-1}{2c+1}p \\\\\n&\\iff \\frac{p}{c(2c+1)} \\geq 1 \\iff p \\geq c(2c+1).\n\\end{aligned}\n$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22362,
"subject": "Mathematics (Olympiad)",
"question": "Let $p$ be a prime number for which $\\frac{p-1}{2}$ is also prime, and let $a$, $b$, $c$ be integers not divisible by $p$. Prove that there are at most $1 + \\sqrt{2p}$ positive integers $n$ such that $n < p$ and $p$ divides $a^n + b^n + c^n$.",
"options": [],
"answer": "See solution",
"solution": "First, suppose $b \\equiv \\pm a \\pmod{p}$ and $c \\equiv \\pm b \\pmod{p}$. Then, for any $n$, we have $a^n + b^n + c^n \\equiv \\pm a^n$ or $\\pm 3a^n \\pmod{p}$. Since $p \\neq 3$ (because $\\frac{3-1}{2}$ is not prime) and $p \\nmid a$, it follows that $a^n + b^n + c^n \\not\\equiv 0 \\pmod{p}$. The claim is trivial in this case. Otherwise, we may assume without loss of generality that $b \\not\\equiv \\pm a \\pmod{p}$, i.e., $ba^{-1} \\not\\equiv \\pm 1 \\pmod{p}$.\n\nNow let $q = \\frac{p-1}{2}$. By Fermat's little theorem, the order of $ba^{-1}$ modulo $p$ divides $p-1 = 2q$. However, since $ba^{-1} \\not\\equiv \\pm 1 \\pmod{p}$, the order does not divide $2$. Thus, the order must be either $q$ or $2q$.\n\nNext, let $S$ denote the set of positive integers $n < p$ such that $a^n + b^n + c^n \\equiv 0 \\pmod{p}$, and let $s_t$ denote the number of ordered pairs $(i, j) \\subset S$ such that $i - j \\equiv t \\pmod{p-1}$.\n\n*Lemma:* If $t$ is a positive integer less than $2q$ and not equal to $q$, then $s_t \\le 2$.\n\n*Proof:* Consider $i, j \\in S$ with $j - i \\equiv t \\pmod{p-1}$. Then we have\n\n$$\n\\begin{aligned}\n& a^i + b^i + c^i \\equiv 0 \\pmod{p} \\\\\n\\implies & a^i c^{j-i} + b^i c^{j-i} + c^j \\equiv 0 \\pmod{p} \\\\\n\\implies & a^i c^{j-i} + b^i c^{j-i} - a^j - b^j \\equiv 0 \\pmod{p} \\\\\n\\implies & a^i (c^t - a^t) \\equiv b^i (b^t - c^t) \\pmod{p}.\n\\end{aligned}\n$$\n\nIf $c^t \\equiv a^t \\pmod{p}$, then this implies $c^t \\equiv b^t \\pmod{p}$ as well, so $(ab^{-1})^t \\equiv 1 \\pmod{p}$. However, we know the order of $ab^{-1}$ is $q$ or $2q$, and $q \\nmid t$, so this is impossible. Thus, we can write\n\n$$\n(ab^{-1})^i \\equiv \\frac{b^t - c^t}{c^t - a^t} \\pmod{p}.\n$$\n\nFor a fixed $t$, the right-hand side is fixed, so $(ab^{-1})^i$ is also fixed. Since the order of $ab^{-1}$ is either $q$ or $2q$, there are at most $2$ solutions for $i$, and the lemma is proven. □\n\nNow, for each element $i$ in $S$, there are at least $|S| - 2$ other elements that differ from $i$ by a quantity other than $q \\pmod{p-1}$. Therefore, the lemma implies that\n\n$$\n\\begin{aligned}\n& |S| (|S| - 2) \\le \\sum_{t \\ne q} s_t \\le 2 (p-2) \\\\\n\\implies & (|S| - 1)^2 \\le 2p - 3 \\\\\n\\implies & |S| < 1 + \\sqrt{2p}.\n\\end{aligned}\n$$\n\n",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22363,
"subject": "Mathematics (Olympiad)",
"question": "Find all sequences of five prime numbers in ascending order such that each consecutive pair differs by 6.",
"options": [],
"answer": "See solution",
"solution": "We examine primes less than 50 for sequences where each consecutive prime differs by 6. The sequence $5, 11, 17, 23, 29$ satisfies this condition.\n\n**Alternative i**\nSuppose we have a sequence of five prime numbers in ascending order, each 6 apart. Since 6 is even and 2 is the only even prime, all primes in the sequence must be odd, so their last digits are 1, 3, 5, 7, or 9. The difference between any two primes in the sequence is a multiple of 6, so no two primes in the sequence can end in the same digit. There are five primes, so all of 1, 3, 5, 7, 9 must appear as last digits. The only prime ending in 5 is 5 itself, so the sequence must be $5, 11, 17, 23, 29$.\n\n**Alternative ii**\nSuppose we have a sequence of five prime numbers in ascending order, each 6 apart. All primes must be odd, so their last digits are 1, 3, 5, 7, or 9. The only prime ending in 5 is 5 itself. If the first prime ends in 5, the sequence is $5, 11, 17, 23, 29$. If the first prime ends in 1, then the next four primes end in 7, 3, 9, 5. If the first prime ends in 3, then the next four primes end in 9, 5, 1, 7. If the first prime ends in 7, then the next four primes end in 3, 9, 5, 1. If the first prime ends in 9, then the next four primes end in 5, 1, 7, 3. In each of these cases, the prime ending in 5 is at least $7, 9, 13, 15$ respectively, which is impossible. So the only sequence is $5, 11, 17, 23, 29$.\n\n**Alternative iii**\nSuppose we have a sequence of five prime numbers in ascending order, each 6 apart. Replace each number with its remainder modulo 5. The possible sequences of remainders are $0, 1, 2, 3, 4$; $1, 2, 3, 4, 0$; $2, 3, 4, 0, 1$; $3, 4, 0, 1, 2$; $4, 0, 1, 2, 3$. In all cases, there is a remainder of 0, which represents a multiple of 5. The only prime divisible by 5 is 5 itself. Since there is no positive number 6 less than 5, 5 must be the first number in the sequence. Thus, the sequence is $5, 11, 17, 23, 29$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22364,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be a triangle with $\\angle ACB = 90^\\circ$. Suppose that the tangent line at $C$ to the circle passing through $A$, $B$, $C$ intersects the line $AB$ at $D$. Let $E$ be the midpoint of $CD$ and let $F$ be the point on the line $EB$ such that $AF$ is parallel to $CD$.\n\nProve that the lines $AB$ and $CF$ are perpendicular.",
"options": [],
"answer": "See solution",
"solution": "Let $BC$ and $AF$ meet at $H$. Since $CD$ and $AH$ are parallel, triangles $BCE$ and $BHF$ are similar, and so too are triangles $BDE$ and $BAF$. Hence, $\\frac{AF}{DE} = \\frac{BF}{BE} = \\frac{FH}{EC} = \\frac{FH}{DE}$, where the last equation holds because $E$ is the midpoint of $CD$. It follows that $AF = FH$.\n\n\n\nSince $\\angle ACH$ is a right angle, the circumcircle of triangle $ACH$ has $AH$ as diameter and so $F$ as centre. Hence, $FC = FH$ as these are both radii. It follows that triangle $CFH$ is isosceles, so $\\angle BCF = \\angle BHF = \\angle BCD = \\angle BAC$, where the alternate segment theorem is used for both the second and fourth equalities.\n\nLetting $G$ be the point of intersection between $AB$ and $CF$, it follows that triangles $ABC$ and $CBG$ are similar, as two pairs of corresponding angles are equal. Hence, $\\angle BGC = \\angle BCA$, which is a right angle, as required.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22365,
"subject": "Mathematics (Olympiad)",
"question": "Find the smallest integer $n \\geq 10$ such that\n$$\n\\lfloor \\frac{n}{1} \\rfloor \\lfloor \\frac{n}{2} \\rfloor \\cdots \\lfloor \\frac{n}{10} \\rfloor = \\binom{n}{10}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Note that $[x] > x - 1$ for any real number $x$. In particular, for an integer $1 \\leq k \\leq 10$ we have $\\lfloor \\frac{n}{k} \\rfloor > \\frac{n}{k} - 1$, thus $k \\lfloor \\frac{n}{k} \\rfloor > n - k$. Since both sides are integers, we have $k \\lfloor \\frac{n}{k} \\rfloor \\geq n - k + 1$, thus $\\lfloor \\frac{n}{k} \\rfloor \\geq \\frac{n-k+1}{k}$. Since $0 < k \\leq n$ implies $\\frac{n-k+1}{k} > 0$, multiplying both sides of $\\lfloor \\frac{n}{k} \\rfloor \\geq \\frac{n-k+1}{k}$ for $k = 1, 2, \\dots, 10$ yields\n\n$$\n\\lfloor \\frac{n}{1} \\rfloor \\lfloor \\frac{n}{2} \\rfloor \\cdots \\lfloor \\frac{n}{10} \\rfloor \\geq \\frac{n}{1} \\cdot \\frac{n-1}{2} \\cdots \\frac{n-9}{10} = \\binom{n}{10}.\n$$\n\nTherefore, the equation in the problem holds if and only if $\\lfloor \\frac{n}{k} \\rfloor = \\frac{n-k+1}{k}$ for all integers $1 \\leq k \\leq 10$.\n\nIf $\\lfloor \\frac{n}{k} \\rfloor = \\frac{n-k+1}{k}$ holds, $k$ divides $n-k+1$ because the right side is an integer. Conversely, if $k$ divides $n-k+1$ then $\\lfloor \\frac{n}{k} \\rfloor = \\frac{n-k+1}{k}$ holds. Therefore, the equation $\\lfloor \\frac{n}{k} \\rfloor = \\frac{n-k+1}{k}$ holds if and only if $k$ divides $n-k+1$, that is, $n+1$ is a multiple of $k$.\n\nTherefore, for an integer $n \\geq 10$, the equation in the problem holds if and only if $n+1$ is a multiple of $k$ for any integer $1 \\leq k \\leq 10$, that is, $n+1$ is a common multiple of $1, 2, \\dots, 10$. Since the least common multiple of $1, 2, \\dots, 10$ is $2520$, the smallest possible $n$ is $2520 - 1 = 2519$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22366,
"subject": "Mathematics (Olympiad)",
"question": "Let us denote by $a$, $b$, $c$ the number of vertical sides, horizontal sides, and diagonals of squares making up a path from $A$ to $B$. Then, the length of the path is given by $a + b + c\\sqrt{2}$. Since it is necessary to move $3$ steps upward and $5$ steps to the right to reach the point $B$ from the point $A$, we must have $a + c = 3$, $b + c = 5$.\n\n\n\nFind the number of shortest paths from $A$ to $B$ along the sides and diagonals of the given squares.",
"options": [],
"answer": "See solution",
"solution": "Since $b \\geq a$, if $a \\geq 1$, then decreasing both vertical and horizontal sides by $a$ and increasing the number of diagonals by $a$ gives a shorter path, since $1 + 1 > \\sqrt{2}$. Thus, we can take $a = 0$. Therefore, a shortest path can be obtained by letting $(a, b, c) = (0, 2, 3)$, namely, paths consisting of $2$ horizontal sides and $3$ diagonals. The number of such paths is the number of ways to choose $2$ spots out of $5$ to go along horizontal sides, which is $\\binom{5}{2} = 10$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22367,
"subject": "Mathematics (Olympiad)",
"question": "Given 2021 distinct prime numbers $p_1, p_2, \\dots, p_{2021}$, let\n$$\nS = \\left\\{ \\frac{p_i}{p_j} \\mid i, j \\in \\{1, 2, \\dots, 2021\\},\\ i \\neq j \\right\\}.\n$$\n\nAlice and Bob play the following game: they take turns to write a number in $S$ on the blackboard. The number must be different from any number already written. The game ends when someone writes a number such that the product of several (distinct) numbers on the blackboard equals 1, with that person being the loser. If Alice writes first, who has a winning strategy? Explain why.",
"options": [],
"answer": "See solution",
"solution": "Bob has a winning strategy. We give two solutions as follows.\n\n*Solution 1* \nConsider $\\frac{a}{b}$ and $\\frac{b}{a}$ as a pair. There are $\\binom{2021}{2}$ pairs in $S$.\n\nIf there are more than $\\binom{2021}{2}$ numbers on the blackboard, by the pigeonhole principle, two numbers in a certain pair are both written, and the game must end instantly.\n\n**Claim:** If there are fewer than $\\binom{2021}{2}$ numbers on the blackboard, then a player can always write another number from an unwritten pair without losing the game.\n\n**Proof of claim:** We use the language of graph theory. Let $p_1, p_2, \\dots, p_{2021}$ be the vertices of a directed graph $G$. If $\\frac{p_i}{p_j}$ is written, then add an edge $p_i \\to p_j$ to $G$. As the written numbers are distinct, there is at most one directed edge from $p_i$ to $p_j$. The game ends when $G$ has a directed cycle: the product of the numbers corresponding to the edges in the cycle is equal to 1, as each involved prime number appears exactly once in the numerator and once in the denominator. Conversely, if the product of several numbers written on the blackboard is 1, consider all the prime numbers appearing in the product and the corresponding vertices in $G$: each vertex has equal indegree and outdegree. Clearly, $G$ has a directed cycle.\n\nIf the number of edges in $G$ is less than $\\binom{2021}{2}$, some two vertices, say $p_i$ and $p_j$, are not adjacent by an edge (the pair $\\frac{p_i}{p_j}$ and $\\frac{p_j}{p_i}$ is unwritten). Then directed paths from $p_i$ to $p_j$ and from $p_j$ to $p_i$ cannot exist at the same time. Indeed, if this is possible, then a directed path is formed from $p_i$ to $p_j$ then to $p_i$. Start from any vertex, say $p_{k_1}$, and move along the path until it reaches a vertex already visited, say $p_{k_j} \\to p_{k_i}$ ($j > i$), then $p_{k_i} \\to p_{k_{i+1}} \\to \\dots \\to p_{k_j} \\to p_{k_i}$ is a directed cycle (no vertex is repeated). This is contradictory. Suppose $G$ does not contain a directed path from $p_i$ to $p_j$. A player can safely add the edge $p_j \\to p_i$, or write $\\frac{p_j}{p_i}$ on the blackboard without losing the game. The claim is verified.\n\nNote that $\\binom{2021}{2} = 2021 \\cdot 1010$ is an even number. When $\\binom{2021}{2}$ numbers are written, all $\\binom{2021}{2}$ pairs of vertices of $G$ are adjacent. It is Alice's turn and she must add another edge and lose the game. $\\square$\n\n*Solution 2* \nHere is another proof of the claim. Replace 2021 by $n$ and it is generally true that: if a directed, cycle-free graph $G$ has fewer than $\\binom{n}{2}$ edges, then one edge can be added to $G$ such that the resulting graph is still cycle-free.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22368,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a given positive integer. In the coordinate plane, consider the set of points\n\n$$\n\\{P_1, P_2, \\dots, P_{4n+1}\\} = \\{(x, y) \\mid x \\text{ and } y \\text{ are integers with } xy = 0,\\ |x| \\leq n,\\ |y| \\leq n\\}.\n$$\n\nDetermine the minimum value of\n\n$$(P_1P_2)^2 + (P_2P_3)^2 + \\dots + (P_{4n}P_{4n+1})^2 + (P_{4n+1}P_1)^2.$$",
"options": [],
"answer": "See solution",
"solution": "The answer is $16n - 8$.\n\nAssume $P_i = (x_i, y_i)$ for $1 \\leq i \\leq 4n + 1$. Set\n\n$$\nP_{4n+2} = (x_{4n+2}, y_{4n+2}) = P_1 = (x_1, y_1).\n$$\n\nWe will show that the sum\n\n$$\nS = \\sum_{i=1}^{4n+1} (P_i P_{i+1})^2 \\geq 16n - 8.\n$$\n\nThis minimum can be obtained by arranging the points as follows:\n\nFor $n$ even:\n\n$$\n(P_1, P_2, \\dots, P_{4n+1}) = ((2, 0), (4, 0), \\dots, (n, 0), (n-1, 0), \\dots, (1, 0), (0, 2), (0, 4), \\dots, (0, n), (0, n-1), \\dots, (0, 1), (-2, 0), (-4, 0), \\dots, (-n, 0), (-n+1, 0), \\dots, (-1, 0), (0, -2), (0, -4), \\dots, (0, -n), (0, -n+1), \\dots, (0, -1), (0, 0))\n$$\n\nFor $n$ odd:\n\n$$\n(P_1, P_2, \\dots, P_{4n+1}) = ((1, 0), (3, 0), \\dots, (n, 0), (n-1, 0), \\dots, (2, 0), (0, 1), (0, 3), \\dots, (0, n), (0, n-1), \\dots, (0, 2), (-1, 0), (-3, 0), \\dots, (-n, 0), (-n+1, 0), \\dots, (-2, 0), (0, -1), (0, -3), \\dots, (0, -n), (0, -n+1), \\dots, (0, -2), (0, 0))\n$$\n\nFor example, when $n = 2m$ is even:\n\n$$\n\\begin{aligned}\nS &= 4 \\left( \\sum_{i=1}^{m-1} (P_i P_{i+1})^2 + (P_m P_{m+1})^2 + \\sum_{i=m+1}^{2m-1} (P_i P_{i+1})^2 \\right) \\\\\n&\\quad + 3(P_{2m} P_{2m+1})^2 + (P_{4n} P_{4n+1})^2 + (P_{4n+1} P_1)^2 \\\\\n&= 4(4(m-1)+1+4(m-1)) + 3 \\times 5 + 1 + 4 \\\\\n&= 32m - 8 = 16n - 8.\n\\end{aligned}\n$$\n\nThe same argument works for odd $n$.\n\nNote:\n\n$$\nS = \\sum_{i=1}^{4n+1} \\left[(x_i - x_{i+1})^2 + (y_i - y_{i+1})^2\\right].\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22369,
"subject": "Mathematics (Olympiad)",
"question": "There are 4 cards, each with a positive 1-digit integer written on it. By choosing 2 different cards and writing down the sum of their numbers, we obtain 4 distinct integers. By choosing 2 different cards and writing down the product of their numbers, we obtain 3 distinct integers. Determine all possible combinations of the numbers written on the 4 cards.",
"options": [],
"answer": "See solution",
"solution": "Suppose the 4 cards have only 1 or 2 distinct numbers. Then, the number of possible sums from choosing 2 cards is at most 3, contradicting the requirement of 4 distinct sums. If all 4 numbers are distinct, let them be $a < b < c < d$. Then, the sums $a+b < a+c < b+c < b+d < c+d$ show at least 5 distinct sums, again a contradiction. Thus, there must be exactly 3 distinct numbers among the 4 cards, with one number appearing twice and the other two being different. Let these numbers be $p < q < r$, with $q$ appearing twice. The possible products are $pq$, $pr$, $qr$, and $q^2$. Since there are only 3 distinct products, $q^2$ must equal one of $pq$, $pr$, or $qr$. Since $pq < q^2 < qr$ if $p < q < r$, the only possibility is $q^2 = pr$. Therefore, the numbers on the cards are $(p, q, q, r)$, where $1 \\leq p < q < r \\leq 9$ and $q^2 = pr$. The integer quadruples satisfying these are $(1, 2, 2, 4)$, $(1, 3, 3, 9)$, $(2, 4, 4, 8)$, and $(4, 6, 6, 9)$. Each of these quadruples meets the problem's requirements.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 22370,
"subject": "Mathematics (Olympiad)",
"question": "Circles $O_1$ and $O_2$ intersect at two points $B$ and $C$, and $BC$ is the diameter of circle $O_1$. Construct a tangent line to circle $O_1$ at $C$, which intersects circle $O_2$ at another point $A$. Join $AB$ to intersect $O_1$ at point $E$, then join $CE$ and extend it to intersect circle $O_2$ at point $F$. Assume that $H$ is an arbitrary point on the line segment $AF$. Join $HE$ and extend it to intersect circle $O_1$ at point $G$, and join $BG$ and extend it to intersect the extended line of $AC$ at point $D$.\n\nProve that $\\frac{AH}{HF} = \\frac{AC}{CD}$.",
"options": [],
"answer": "See solution",
"solution": "Since $BC$ is the diameter of circle $O_1$ and $ACD$ is a tangent to $O_1$, we have $BC \\perp AD$, so $\\angle ACB = 90^\\circ$. Thus, $AB$ is the diameter of circle $O_2$.\n\nBecause $\\angle BEC = 90^\\circ$, $AB \\perp CF$, so $\\angle FAB = \\angle CAB$.\n\n\n\nJoin $CG$, and note that $CG \\perp BD$. Therefore,\n\n$$\n\\angle ADB = \\angle BCG = \\angle BEG = \\angle AEH,\n$$\n\nso $\\triangle AHE \\sim \\triangle ABD$. Thus, $\\frac{AH}{AE} = \\frac{AB}{AD}$, which gives $AH \\cdot AD = AE \\cdot AB$.\n\nBy the power of a point theorem, $AC^2 = AE \\cdot AB$, so $AH \\cdot AD = AE \\cdot AB = AC^2 = AC \\cdot AF$. Therefore, $\\frac{AH}{AF} = \\frac{AC}{AD}$, and by the section formula, $\\frac{AH}{HF} = \\frac{AC}{CD}$. Q.E.D.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22371,
"subject": "Mathematics (Olympiad)",
"question": "Let $a = c - b$ and $d = 2c - a = b + c$, where $b$ and $c$ are positive integers with $c > b$. Show that $abcd$ is the area of a right triangle with integer side lengths $2bc$, $c^2 - b^2$, and $b^2 + c^2$.",
"options": [],
"answer": "See solution",
"solution": "From the given equations, $a = c - b$ and $d = b + c$. Thus,\n\n$$\nabcd = (c-b) \\cdot b \\cdot c \\cdot (b+c) = bc(c^2-b^2) = \\frac{1}{2}(2bc)(c^2-b^2).\n$$\n\nTherefore, $abcd$ is the area of a right triangle with legs $2bc$ and $c^2 - b^2$.\n\nThe hypotenuse of that triangle is\n\n$$\n\\sqrt{(2bc)^2 + (c^2 - b^2)^2} = \\sqrt{c^4 + b^4 + 2b^2c^2} = b^2 + c^2.\n$$\n\nSince $b$ and $c$ are positive integers and $c > b$, all three side lengths $2bc$, $c^2 - b^2$, and $b^2 + c^2$ are positive integers.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22372,
"subject": "Mathematics (Olympiad)",
"question": "Suppose that for natural numbers $k$, $m$, $n$, and $q$ the following equality holds:\n\n$$\n2(k^2 + km) + m^2 + n^2 = 2013^q,\n$$\n\nand $k$ and $m$ are coprime. Show that $n$ and $k^2 + km$ are not coprime.",
"options": [],
"answer": "See solution",
"solution": "We start with the equality $k^2 + (k + m)^2 + n^2 = 2013^q$. Since $(k, m) = 1$, we have $(k, k + m) = 1$. Suppose that $n$ and $k^2 + km$ are coprime. Then, $(k, n) = 1$ and $(k + m, n) = 1$. Therefore, at most one of the numbers $k$, $k + m$, and $n$ is even. Thus,\n\n$$\nk^2 + (k + m)^2 + n^2 \\equiv 2 \\pmod{4} \\text{ or } k^2 + (k + m)^2 + n^2 \\equiv 3 \\pmod{4}.\n$$\n\nAt the same time, $2013^q \\equiv 1 \\pmod{4}$, which leads to a contradiction.\n\n**Note:** For $k = 10$, $m = 33$, $n = 8$ we have: $10^2 + (10 + 33)^2 + 8^2 = 2013$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22373,
"subject": "Mathematics (Olympiad)",
"question": "Fourteen schools participate in the second Tha Sala Mathematics Talent competition, with each school sending 14 students. The students take tests in 14 rooms, with 14 students in a room such that no room contains students from the same school.\n\nAmong the students, there are 15 students who also participated in the first Tha Sala Mathematics Talent competition. At the opening ceremony, the organizers will select 2 students from those who participated in the first competition to recite the pledge of honor, with the condition that the students are from different schools and take tests in different rooms.\n\nLet $n$ be the number of ways to select 2 students satisfying the condition. Determine the least possible $n$.",
"options": [],
"answer": "See solution",
"solution": "We will show that the minimum number is $13$.\n\nFor convenience, we will use *ex-participant* to denote students who participated in the first Tha Sala Mathematics Talent competition.\n\nSuppose Nat is one of the ex-participants and divide the remaining ex-participants into 3 groups:\n\n- **Group A**: ex-participants from the same school as Nat.\n- **Group B**: ex-participants who will take tests in the same room as Nat.\n- **Group C**: the remaining ex-participants not in group A or B.\n\nLet there be $a$ and $b$ students in groups A and B respectively ($a, b \\in \\{1, 2, \\dots, 13\\}$). Consequently, there are $14 - a - b$ students in group C.\n\nFirst, we show that there are at least $13$ ways to select 2 ex-participants satisfying the condition.\n\n- **Case 1:** $a = b = 0$. Nat can pair with any of the $14$ students in group C, giving $14$ pairs.\n- **Case 2:** $a = 0$ and $b \\geq 0$.\n - Nat pairs with one of $14 - b$ students in group C.\n - Select an arbitrary student from group C and pair with a student in group B from a different school (at least $b-1$ students fit this criteria).\n- **Case 3:** $a \\geq 1$ and $b = 0$.\n - Nat pairs with one of $14 - a$ students in group C.\n - Select an arbitrary student from group C and pair with a student in group A who will take tests in a different room (at least $a-1$ students fit this criteria).\n- **Case 4:** $a \\geq 1$ and $b \\geq 1$.\n - Nat pairs with one of $14 - a - b$ students in group C.\n - All pairs consisting of one student each from groups A and B.\n\nThis results in $14 - a - b + ab = 13 + (a-1)(b-1) \\geq 13$ pairs.\n\nFinally, we show that it is possible to have $13$ pairs.\n\nConsider the case $a = 0$ and $b = 13$. There will be one student in group C, who we call Nick. The only possible pairing is Nick with any student in group B who comes from a different school than Nick. This results in exactly $13$ pairs.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 22374,
"subject": "Mathematics (Olympiad)",
"question": "Let $k$ be the circumscribed circle for the quadrilateral $ABCD$. The angle at vertex $B$ is twice as large as the angle at vertex $A$ and $40\\degree$ smaller than the angle at vertex $D$. Calculate the angles of $ABCD$.",
"options": [],
"answer": "See solution",
"solution": "From the conditions, $\\beta = 2\\alpha$ and $\\beta = \\delta - 40\\degree$, so $\\delta = 2\\alpha + 40\\degree$. Since $ABCD$ is cyclic, $\\alpha + \\gamma = \\beta + \\delta$ and $\\alpha + \\gamma + \\beta + \\delta = 360\\degree$, so $\\alpha + \\gamma = \\beta + \\delta = 180\\degree$.\n\nThus,\n$$\n180\\degree = \\beta + \\delta = 2\\alpha + 2\\alpha + 40\\degree = 4\\alpha + 40\\degree\n$$\nSo $4\\alpha = 140\\degree$, i.e., $\\alpha = 35\\degree$.\n\nNow,\n- $\\beta = 2\\alpha = 70\\degree$\n- $\\gamma = 180\\degree - \\alpha = 145\\degree$\n- $\\delta = 2\\alpha + 40\\degree = 2 \\cdot 35\\degree + 40\\degree = 110\\degree$\n\nTherefore, the angles are $35\\degree$, $70\\degree$, $145\\degree$, and $110\\degree$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 22375,
"subject": "Mathematics (Olympiad)",
"question": "Consider the sequence $\\{a_n\\}$ such that $a_1 = \\frac{1}{2}$, $a_{n+1} = \\sqrt[3]{3a_{n+1} - a_n}$, and $0 \\leq a_n \\leq 1$ for all $n \\geq 1$.\n\n**a)** Prove that $\\{a_n\\}$ is uniquely defined and converges to a finite limit.\n\n**b)** Consider the sequence $\\{b_n\\}$ defined by $b_n = (1 + 2a_1)(1 + 2^2 a_2) \\dots (1 + 2^n a_n)$ for all $n \\geq 1$. Prove that $\\{b_n\\}$ is convergent.",
"options": [],
"answer": "See solution",
"solution": "**a)** Let $u_n = \\frac{a_n}{2}$ for all $n \\in \\mathbb{N}^*$. Then $0 \\leq u_n \\leq \\frac{1}{2}$ for all $n \\in \\mathbb{N}^*$, and\n$$\n8u_{n+1}^3 = 6u_{n+1} - 2u_n \\iff u_n = 3u_{n+1} - 4u_{n+1}^3, \\forall n \\in \\mathbb{N}^*.\n$$\nThere exists $\\alpha \\in [0, \\frac{\\pi}{2}]$ such that $u_1 = \\frac{1}{4} = \\sin \\alpha$, thus\n$$\n3u_2 - 4u_2^3 = \\sin \\alpha = 3 \\sin \\frac{\\alpha}{3} - 4 \\sin^3 \\frac{\\alpha}{3}.\n$$\nConsider the function $f(x) = 3x - 4x^3$, which is differentiable on $[0, \\frac{1}{2}]$. We have $f'(x) = 3 - 12x^2 \\geq 0$ for all $x \\in [0, \\frac{1}{2}]$, so $f$ is increasing on $[0, \\frac{1}{2}]$. Notice that $\\sin \\frac{\\alpha}{3} \\in [0, \\frac{1}{2}]$, so from the above equality,\n$$\nf(u_2) = f\\left(\\sin \\frac{\\alpha}{3}\\right) \\implies u_2 = \\sin \\frac{\\alpha}{3}.\n$$\nSimilarly, by induction, one can prove\n$$\nu_n = \\sin \\frac{\\alpha}{3n-1}, \\forall n \\in \\mathbb{N}^*.\n$$\nTherefore,\n$$\na_n = 2 \\sin \\frac{\\alpha}{3^{n-1}}, \\forall n \\in \\mathbb{N}^*.\n$$\nThus, the sequence $\\{a_n\\}$ is uniquely defined and $\\lim_{n \\to +\\infty} a_n = 0$.\n\n**b)** Recall the following inequalities:\n1. $\\ln(1+x) < x$ for all $x > 0$;\n2. $\\sin x < x$ for all $x > 0$.\n\nObviously $b_n > 0$ for all $n \\in \\mathbb{N}^*$. For all $n \\in \\mathbb{N}^*$,\n$$\n\\begin{aligned}\n\\ln b_n &= \\ln(1 + 2a_1) + \\ln(1 + 2^2 a_2) + \\dots + \\ln(1 + 2^n a_n) \\\\\n&< 2a_1 + 2^2 a_2 + \\dots + 2^n a_n \\\\\n&= 2^2 \\sin \\alpha + 2^3 \\sin \\frac{\\alpha}{3} + \\dots + 2^{n+1} \\sin \\frac{\\alpha}{3^{n-1}} \\\\\n&< 2^2 \\left( \\alpha + \\frac{\\alpha}{3} + \\dots + \\frac{\\alpha}{3^{n-1}} \\right) \\\\\n&= 4\\alpha \\left( 1 + \\frac{1}{3} + \\dots + \\frac{1}{3^{n-1}} \\right) < 12\\alpha.\n\\end{aligned}\n$$\nThis shows that $b_n < e^{12\\alpha}$ for every $n \\in \\mathbb{N}^*$, so the sequence $\\{b_n\\}$ is upper bounded. Since $\\{b_n\\}$ is increasing, by the Weierstrass theorem, $\\{b_n\\}$ converges.\n\n$\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22376,
"subject": "Mathematics (Olympiad)",
"question": "Is it possible, after finitely many moves, for a $24 \\times 24$ square grid to contain exactly 574 black unit squares?",
"options": [],
"answer": "See solution",
"solution": "No, it is not possible.\n\nSuppose it is possible that after finitely many moves, the square grid contains exactly 574 black unit squares. This means the grid contains exactly $24 \\times 24 - 574 = 2$ white unit squares. No matter where the white unit squares are, we can always find a $2 \\times 2$ square that contains exactly 1 white unit square and 3 black unit squares. One checks that the parity of the number of white cells in this $2 \\times 2$ square is fixed no matter how we choose the rows or columns in a move. Therefore, it is impossible that all the four cells are white initially. This is a contradiction, and so it is impossible to have exactly 574 black unit squares.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22377,
"subject": "Mathematics (Olympiad)",
"question": "В каждом столбце лестницы отметьте по одной верхней клетке; их объединение назовём *верхним слоем*. Никакие две из $n$ клеток этого слоя не могут лежать в одном прямоугольнике разбиения, поэтому в любом разбиении лестницы не менее $n$ прямоугольников.\n\nС другой стороны, минимальная суммарная площадь $n$ прямоугольников с различными площадями равна $1 + 2 + \\ldots + n$, что совпадает с площадью всей лестницы. Значит, число прямоугольников в любом разбиении равно $n$, их площади выражаются числами $1, 2, \\ldots, n$, и каждый из них содержит клетку верхнего слоя.\n\nДокажите, что число разбиений лестницы высоты $n$ на такие прямоугольники равно $2^{n-1}$.",
"options": [],
"answer": "See solution",
"solution": "Докажем утверждение индукцией по $n$.\n\n*База индукции*: при $n = 1$ существует только одно разбиение.\n\n*Переход*: Пусть утверждение верно для лестницы высоты $n-1$. Рассмотрим разрезание лестницы высоты $n$ на прямоугольники площадей $1, 2, \\ldots, n$.\n\nРассмотрим прямоугольник, покрывающий угловую (самую далёкую от верхнего слоя) клетку лестницы. Он содержит клетку верхнего слоя, то есть сумма длин его сторон $a$ и $b$ равна $n+1$. Поэтому его площадь $S = ab \\geq a + b - 1 = n$, так как $(a-1)(b-1) = ab - (a + b - 1) \\geq 0$. Равенство достигается только при $a = 1$ или $b = 1$. Поскольку площади прямоугольников не превосходят $n$, то $S = n$, и одна из сторон прямоугольника равна $1$, а другая — $n$.\n\nТакой прямоугольник можно выбрать двумя способами (вертикально или горизонтально). В обоих случаях после его удаления остаётся лестница высоты $n-1$, которую можно разрезать $2^{n-2}$ способами по предположению индукции.\n\nИтак, общее число способов:\n$$\n2 \\cdot 2^{n-2} = 2^{n-1}\n$$\nчто и требовалось доказать.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22378,
"subject": "Mathematics (Olympiad)",
"question": "Можно ли разбить шесть последовательных натуральных чисел на две группы по три числа так, чтобы разность наименьших общих кратных (НОК) чисел в каждой группе равнялась 2009?",
"options": [],
"answer": "See solution",
"solution": "Предположим, что такие числа существуют. НОК нескольких чисел делится на каждое из них и, следовательно, на каждый их делитель. Если среди чисел, от которых находят НОК, есть чётное, то и НОК будет чётным. Так как $2009$ — нечётное число, то один из двух НОК будет нечётным; значит, все чётные числа должны оказаться в одной группе.\n\nСреди шести последовательных натуральных чисел есть три чётных и три нечётных, значит, одна группа состоит из трёх последовательных чётных чисел, а другая — из трёх последовательных нечётных чисел. В любой из этих троек найдётся число, кратное трём. Тогда оба НОК делятся на $3$, и их разность также делится на $3$. Но $2009$ на $3$ не делится — противоречие. Следовательно, разбить числа требуемым образом нельзя.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22379,
"subject": "Mathematics (Olympiad)",
"question": "There are two touching circles, $k_1(S_1, r_1)$ and $k_2(S_2, r_2)$ in a rectangle $ABCD$ with $|AB| = 9$, $|BC| = 8$. Moreover, $k_1$ touches $AD$ and $CD$, while $k_2$ touches $AB$ and $BC$.\n\n1. Prove that $r_1 + r_2 = 5$.\n2. What is the least and what is the greatest possible area of $\\triangle AS_1S_2$?\n\n",
"options": [],
"answer": "See solution",
"solution": "a) Let $M$ and $N$ be intersections of the line through $S_1$ parallel to $AD$. Analogously, let $K$ and $L$ be intersections of the line through $S_2$ parallel to $AB$. Let $P$ be the intersection of $KL$ and $MN$ (see the figure). By the Pythagorean theorem for $S_1PS_2$:\n\n$$\n\\begin{align*}\n(r_1 + r_2)^2 &= (8 - r_1 - r_2)^2 + (9 - r_1 - r_2)^2, \\\\\n(r_1 + r_2)^2 - 34(r_1 + r_2) + 145 &= 0, \\\\\n(r_1 + r_2 - 5)(r_1 + r_2 - 29) &= 0.\n\\end{align*}\n$$\n\nSince $2r_1 \\leq 8$, $2r_2 \\leq 8$, we have $r_1 + r_2 = 5$.\n\nb) Let $Q$ be the foot of the perpendicular from $S_2$ to $AB$, $R$ the foot of the perpendicular from $S_1$ to $AD$, and $T$ the intersection of $QS_2$ and $RS_1$ (see the figure).\n\nThe area $S$ of $\\triangle AS_1S_2$ is given by the difference of the area of rectangle $AQTR$ and the areas of right triangles $AQS_2$, $AS_1R$, and $S_1S_2T$:\n\n$$\n\\begin{align*}\nS &= (9 - r_2)(8 - r_1) - \\frac{1}{2} r_2 (9 - r_2) - \\frac{1}{2} r_1 (8 - r_1) - \\frac{1}{2} (9 - r_1 - r_2)(8 - r_1 - r_2) \\\\\n&= 36 - \\frac{9}{2} r_1 - 4 r_2 = 36 - \\frac{9}{2} r_1 - 4(5 - r_1) = 16 - \\frac{1}{2} r_1,\n\\end{align*}\n$$\n\nwhere we used $r_1 + r_2 = 5$. Further, $2r_1 \\leq 8$ and $2r_2 \\leq 8$ imply $1 \\leq r_1, r_2 \\leq 4$, thus\n\n$$\nS = 16 - \\frac{1}{2} r_1 \\in \\left[14, \\frac{31}{2}\\right],\n$$\n\nand the least possible value of the area is $14$ (for $r_1 = 4$, $r_2 = 1$), and the greatest possible value is $\\frac{31}{2}$ (for $r_1 = 1$, $r_2 = 4$).",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22380,
"subject": "Mathematics (Olympiad)",
"question": "Find all triples $(x, y, z)$ of real numbers that satisfy the system of equations\n\n$$\n\\begin{cases}\nxy + x + y = z, \\\\\nyz + y + z = x, \\\\\nzx + z + x = y.\n\\end{cases}\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $x + 1 = a$, $y + 1 = b$, and $z + 1 = c$. Adding $1$ to both sides of each equation and factoring gives:\n\n$$\n\\begin{cases}\nab = c, \\\\\nbc = a, \\\\\nca = b.\n\\end{cases}\n$$\n\nIf $a = 0$, then the first and third equations imply $b = 0$ and $c = 0$. Similarly, if $b = 0$ or $c = 0$, then $a, b, c$ are all zero. This yields the solution $(-1, -1, -1)$ for the original system.\n\nIf none of $a, b, c$ is zero, then multiplying the equations pairwise and simplifying gives $a^2 = b^2 = c^2 = 1$. Thus, each of $a, b, c$ is either $1$ or $-1$.\n\n- If $a = b = c = 1$, this leads to $(x, y, z) = (0, 0, 0)$.\n- If exactly two of $a, b, c$ are $-1$, the solutions are $(-2, -2, 0)$, $(-2, 0, -2)$, and $(0, -2, -2)$.\n\nMultiplying all equations and dividing by $abc$ gives $abc = 1$, so no other solutions exist (since $-1$ cannot occur an odd number of times).",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22381,
"subject": "Mathematics (Olympiad)",
"question": "Дадена е функцијата $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ таква што\n\n$$\nf(x+1) + f(x-1) = \\sqrt{2} f(x).\n$$\n\nДокажи дека $f$ е периодична функција.",
"options": [],
"answer": "See solution",
"solution": "Од даденото равенство што го исполнува функцијата $f$ имаме:\n\n$$\nf(x+2) + f(x) = \\sqrt{2} f(x+1) = \\sqrt{2} (\\sqrt{2} f(x) - f(x-1)) = 2 f(x) - \\sqrt{2} f(x-1),\n$$\n\nодносно\n\n$$\nf(x+2) = f(x) - \\sqrt{2} f(x-1).\n$$\n\nПонатаму,\n\n$$\n\\begin{aligned}\nf(x+4) &= f(x+2) - \\sqrt{2} f(x+1) \\\\\n&= f(x) - \\sqrt{2} (f(x-1) + f(x+1)) \\\\\n&= f(x) - \\sqrt{2} \\cdot \\sqrt{2} f(x) \\\\\n&= f(x) - 2 f(x) \\\\\n&= -f(x)\n\\end{aligned}\n$$\n\nОд ова пак следува дека\n\n$$\nf(x+8) = -f(x+4) = -(-f(x)) = f(x).\n$$\n\nЗначи, функцијата $f$ е периодична со период $8$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22382,
"subject": "Mathematics (Olympiad)",
"question": "Solve the equation\n\n$$\n\\frac{\\sqrt{x+2}}{\\cos 2x + 3} = \\frac{\\sqrt{x+1}}{\\cos 2x + 1}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Rewrite the equation as:\n\n$$\n\\frac{\\sqrt{x+2}}{\\sqrt{x+1}} = \\frac{\\cos 2x + 3}{\\cos 2x + 1}.\n$$\n\nLet $t = \\sqrt{x}$, $t \\ge 0$, and $y = \\cos 2x$, $y \\in (-1, 1]$.\n\nDefine $f(t) = \\frac{t+2}{t+1}$ and $g(y) = \\frac{y+3}{y+1}$.\n\nFor $t \\ge 0$:\n- $f(t) = 2$ only if $t = 0$.\n- $f(t) < 2$ for $t > 0$.\n\nFor $y \\in (-1, 1]$:\n- $g(y) = 2$ only if $y = 1$.\n- $g(y) > 2$ for $y < 1$.\n\nThus, the equation holds only if $f(\\sqrt{x}) = 2 = g(\\cos 2x)$, i.e., $\\sqrt{x} = 0$ and $\\cos 2x = 1$.\n\nTherefore, the only solution is $x = 0$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22383,
"subject": "Mathematics (Olympiad)",
"question": "Даден е триаголникот $\\Delta ABC$ и отсечка $PQ$ со должина $t$ на отсечката $BC$, така што $P$ е меѓу $B$ и $Q$, а $Q$ е меѓу $P$ и $C$. Од точката $P$ се повлекуваат паралелни прави со $AB$ и $AC$ кои ги сечат $AC$ и $AB$ во $P_1$ и $P_2$, соодветно. Од точката $Q$ се повлекуваат паралелни прави со $AB$ и $AC$ кои ги сечат $AC$ и $AB$ во $Q_1$ и $Q_2$, соодветно. Докажи дека збирот од плоштините на $PQQ_1P_1$ и $PQQ_2P_2$ не зависи од положбата на $PQ$ на $BC$.",
"options": [],
"answer": "See solution",
"solution": "Нека $D$ е пресекот на $PP_1$ и $QQ_2$. Да забележиме дека $P_{DQ_2P_2} = 2P_{\\Delta ADP}$ и $P_{QP_1Q_1} = 2P_{\\Delta ADQ}$. Па сега имаме:\n\n$$\nP_{PQQ_2P_2} + P_{PP_1Q_1Q} = P_{PDQ_2P_2} + P_{QD P_1Q_1} + 2P_{\\Delta PQD} = 2P_{\\Delta ADP} + 2P_{\\Delta ADQ} + 2P_{\\Delta APQ} = 2P_{\\Delta APQ} = \\overline{PQ} \\cdot h_a\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22384,
"subject": "Mathematics (Olympiad)",
"question": "Each vertex of a finite graph can be colored either black or white. Initially, all vertices are black. You are allowed to pick a vertex $P$ and change the color of $P$ and all of its neighbours. Is it possible to change the color of every vertex from black to white by a sequence of operations of this type?",
"options": [],
"answer": "See solution",
"solution": "Yes, it is possible. We prove this by induction on the number $n$ of vertices.\n\nIf $n=1$, the result is obvious.\n\nAssume the statement holds for any graph with $n-1$ vertices, where $n \\geq 2$. Let $X$ be a graph with $n$ vertices, denoted $P_1, \\dots, P_n$.\n\nLet $f_i$ denote the operation of changing the color of $P_i$ and all its neighbours. Remove a vertex $P_i$ (and its incident edges) from $X$. By the induction hypothesis, there exists a sequence of operations $g_i$ (composed of some $f_j$ with $j \\neq i$) that changes the color of every vertex in $X$ except possibly $P_i$.\n\nIf $g_i$ also changes the color of $P_i$, we are done. Otherwise, assume $g_i$ does not change the color of $P_i$ for every $i = 1, \\dots, n$.\n\n**Case 1:** $n$ is even. Composing $g_1, \\dots, g_n$ changes the color of every vertex from black to white.\n\n**Case 2:** $n$ is odd. In this case, $X$ has a vertex with an even number of neighbours. Let $k_i$ be the number of neighbours of $P_i$. Since $k_1 + \\dots + k_n = 2e$ (where $e$ is the number of edges), at least one $k_i$ is even.\n\nAfter renumbering, assume $P_1$ has $2k$ neighbours, say $P_2, \\dots, P_{2k+1}$. The composition of $f_1$ with $g_1, g_2, \\dots, g_{2k+1}$ will change the color of every vertex, as desired. $\\square$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 22385,
"subject": "Mathematics (Olympiad)",
"question": "The plane is divided into unit squares by two sets of parallel lines, forming an infinite grid. Each unit square is coloured with one of 1201 colours so that no rectangle with perimeter 100 contains two squares of the same colour.\n\nShow that no rectangle of size $1 \\times 1201$ or $1201 \\times 1$ contains two squares of the same colour.\n\n*Note.* Any rectangle is assumed here to have sides contained in the lines of the grid.\n\n\n\n**Figure 1**\n\n\n\n**Figure 2**",
"options": [],
"answer": "See solution",
"solution": "Let the centers of the unit squares be the integer points in the plane, and denote each unit square by the coordinates of the center.\n\nConsider the set $D$ of all unit squares $(x, y)$ such that $|x| + |y| \\leq 24$. Any integer translate of $D$ is called a diamond.\n\nSince any two unit squares that belong to the same diamond also belong to some rectangle of perimeter 100, a diamond cannot contain two squares of the same colour. Since a diamond contains exactly $24^2 + 25^2 = 1201$ unit squares, a diamond must contain every colour exactly once.\n\nChoose one colour, say, green, and let $a_1, a_2, \\dots$ be all green unit squares. Let $P_i$ be the $i$-th diamond of center $a_i$. We will show that no unit square is covered by two $P_i$'s and that every unit square is covered by some $P_i$.\n\nIndeed, suppose first that $P_i$ and $P_j$ contain the same unit square $b$. Then their centers lie within the same rectangle of perimeter 100, a contradiction.\n\nLet, on the other hand, $b$ be an arbitrary unit square. The diamond of center $b$ must contain some green unit square $a_i$. The diamond $P_i$ of center $a_i$ will then contain $b$.\n\nTherefore, $P_1, P_2, \\dots$ form a covering of the plane in exactly one layer. It is easy to see, though, that, up to translation and reflection, there exists a unique such covering. (Indeed, consider two neighbouring diamonds. Unless they fit neatly, uncoverable spaces of two unit squares are created near the corners: see Fig. 1.)\n\n\n\nWithout loss of generality, then, this covering is given by the diamonds of centers $(x, y)$ such that $24x + 25y$ is divisible by 1201. (See Fig. 2 for an analogous covering with smaller diamonds.) It follows from this that no rectangle of size $1 \\times 1201$ can contain two green unit squares, and analogous reasoning works for the remaining colours.\n\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22386,
"subject": "Mathematics (Olympiad)",
"question": "A triangle in the coordinate plane has vertices $A(\\log_2 1, \\log_2 2)$, $B(\\log_2 3, \\log_2 4)$, and $C(\\log_2 7, \\log_2 8)$. What is the area of $\\triangle ABC$?\n\n(A) $\\log_2 \\frac{\\sqrt{3}}{7}$ \n(B) $\\log_2 \\frac{3}{\\sqrt{7}}$ \n(C) $\\log_2 \\frac{7}{\\sqrt{3}}$ \n(D) $\\log_2 \\frac{11}{\\sqrt{7}}$ \n(E) $\\log_2 \\frac{11}{\\sqrt{3}}$",
"options": [],
"answer": "See solution",
"solution": "**Answer (B):** Circumscribe $\\triangle ABC$ by rectangle $AGCD$, with $D$ on the y-axis, and project point $B$ onto $\\overline{AG}$ and $\\overline{CG}$, producing points $E$ and $F$, respectively, as shown in the figure below.\n\n\n\nThe area of rectangle $AGCD$ is $2\\log_2 7$, so the area of $\\triangle ACG$ is $\\log_2 7$. The requested area is\n\n$$\n\\text{Area}(\\triangle ABC) = \\text{Area}(\\triangle ACG) - \\text{Area}(\\triangle ABE) - \\text{Area}(\\triangle BCF) - \\text{Area}(\\triangle BEGF).\n$$\n\nNote that\n\n$$\n\\text{Area}(\\triangle ABE) = \\frac{1}{2} \\log_2 3 \\cdot 1 = \\log_2 \\sqrt{3},\n$$\n\n$$\n\\text{Area}(\\triangle BCF) = \\frac{1}{2} (\\log_2 7 - \\log_2 3) \\cdot 1 = \\log_2 \\sqrt{\\frac{7}{3}},\n$$\n\nand\n\n$$\n\\text{Area}(\\triangle BEGF) = (\\log_2 7 - \\log_2 3) \\cdot 1 = \\log_2 \\frac{7}{3}.\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\text{Area}(\\triangle ABC) &= \\log_2 7 - \\log_2 \\sqrt{3} - \\log_2 \\sqrt{\\frac{7}{3}} - \\log_2 \\frac{7}{3} \\\\\n&= \\log_2 \\left( \\frac{7}{\\sqrt{3} \\cdot \\sqrt{\\frac{7}{3}} \\cdot \\frac{7}{3}} \\right) \\\\\n&= \\log_2 \\frac{3}{\\sqrt{7}}.\n\\end{aligned}\n$$\n\n**Alternate solution:**\n\nThe area of triangle $\\triangle ABC$ is given by\n\n$$\n\\begin{aligned}\n\\frac{1}{2} \\det \\begin{bmatrix} 1 & 0 & 1 \\\\ 1 & \\log_2 3 & 2 \\\\ 1 & \\log_2 7 & 3 \\end{bmatrix} &= \\frac{1}{2} \\det \\begin{bmatrix} 1 & 0 & 1 \\\\ -1 & \\log_2 3 & 0 \\\\ -2 & \\log_2 7 & 0 \\end{bmatrix} \\\\\n&= \\frac{1}{2} \\det \\begin{bmatrix} -1 & \\log_2 3 \\\\ -2 & \\log_2 7 \\end{bmatrix} \\\\\n&= \\frac{1}{2} (-\\log_2 7 + \\log_2 9) \\\\\n&= \\log_2 \\frac{3}{\\sqrt{7}}.\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22387,
"subject": "Mathematics (Olympiad)",
"question": "We say that a simple periodic decimal fraction $f$ has the reduced length equal to $n$ (where $n$ is a positive integer) if $f$ has an $n$-digit period and $f$ cannot be represented as a simple periodic decimal fraction with a period having less than $n$ digits. For instance, $0.(223)$ has the reduced length $3$, while $0.(2323)$ has the reduced length $2$, as $0.(2323) = 0.(23)$.\n\n**a)** Prove that $f = 0.(2) \\cdot 0.(3)$ is a simple periodic fraction with reduced length $3$.\n\n**b)** Does there exist two simple periodic fractions with reduced length $1$, such that their product has reduced length also $1$?\n\n**c)** Does there exist two simple periodic fractions with reduced length $3$, such that their product has the reduced length also $3$?",
"options": [],
"answer": "See solution",
"solution": "**a)**\n\n$$\n0.(2) \\cdot 0.(3) = \\frac{2}{9} \\cdot \\frac{3}{9} = \\frac{6}{81} = \\frac{2}{27} = 0.(074)\n$$\n\nSo $0.(074)$ is a simple periodic fraction with reduced length $3$.\n\n**b)**\n\nYes. For example,\n\n$$\n0.(3) \\cdot 0.(6) = \\frac{3}{9} \\cdot \\frac{6}{9} = \\frac{18}{81} = \\frac{2}{9} = 0.(2)\n$$\n\nBoth $0.(3)$ and $0.(6)$ have reduced length $1$, and their product $0.(2)$ also has reduced length $1$.\n\n**c)**\n\nYes. For example,\n\n$$\n0.(270) \\cdot 0.(370) = \\frac{270}{999} \\cdot \\frac{370}{999} = \\frac{99900}{998001} = \\frac{100}{999} = 0.(100)\n$$\n\nBoth $0.(270)$ and $0.(370)$ have reduced length $3$, and their product $0.(100)$ also has reduced length $3$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22388,
"subject": "Mathematics (Olympiad)",
"question": "Find all real numbers $x, y, z \\ge 1$ satisfying\n$$\n\\min(\\sqrt{x+xyz}, \\sqrt{y+xyz}, \\sqrt{z+xyz}) = \\sqrt{x-1} + \\sqrt{y-1} + \\sqrt{z-1}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $a, b, c$ be nonnegative real numbers such that $x = 1 + a^2$, $y = 1 + b^2$, and $z = 1 + c^2$. We may assume $c \\le a, b$, so the condition becomes\n$$\n(1 + c^2)(1 + (1 + a^2)(1 + b^2)) = (a + b + c)^2.\n$$\nBy the Cauchy-Schwarz inequality,\n$$\n(a + b + c)^2 \\le (1 + (a + b)^2)(c^2 + 1).\n$$\nCombining with the previous relation gives\n$$\n(1 + a^2)(1 + b^2) \\le (a + b)^2,\n$$\nwhich is $(ab - 1)^2 \\le 0$. Hence $ab = 1$, and equality in Cauchy-Schwarz gives $c(a + b) = 1$. Conversely, if $ab = 1$ and $c(a + b) = 1$, the relation holds, since $c = \\frac{1}{a+b} < \\frac{1}{b} = a$ and similarly $c < b$.\n\nThus, the solutions are\n$$\nx = 1 + a^2, \\quad y = 1 + \\frac{1}{a^2}, \\quad z = 1 + \\left( \\frac{a}{a^2 + 1} \\right)^2\n$$\nfor some $a > 0$, as well as all permutations. (We can assume $a \\ge 1$ by switching $x$ and $y$ if necessary.)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22389,
"subject": "Mathematics (Olympiad)",
"question": "We consider a group of trees in a nature reserve, all of which have a positive integral age. The average age is 41 years. After destruction of a tree with an age of 2010 years by lightning, the average age of the remaining trees is 40 years.\n\nDetermine the original number of trees in the group. What is the maximal number of trees of an age of 2010 years in the original group?",
"options": [],
"answer": "See solution",
"solution": "Let the original number of trees be $n$ and the sum of their ages be $s$.\n\nThen:\n\n$$\ns = 41n.\n$$\n\nAfter removing a tree of age 2010:\n\n$$\ns - 2010 = 40(n - 1).\n$$\n\nSolving these equations:\n\n$$\n41n - 2010 = 40n - 40 \\\\\n41n - 40n = 2010 - 40 \\\\\nn = 1970.\n$$\n\nSo $s = 41 \\times 1970 = 80770$.\n\nTo maximize the number of trees aged 2010, write $80770 = 2010k + r$, where $k$ is the number of such trees and $r$ is the sum of the ages of the remaining $1970 - k$ trees (each at least 1 year old).\n\nSet $k = 39$:\n\n$$\n2010 \\times 39 = 78390 \\\\\n80770 - 78390 = 2380.\n$$\n\n$2380$ can be distributed among $1970 - 39 = 1931$ trees, each at least 1 year old. This is possible (e.g., $1930$ trees of $1$ year and $1$ tree of $450$ years).\n\nThus, the maximal number of trees aged 2010 is $39$.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 22390,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a non-negative integer, and let $f: [-1, 1] \\to \\mathbb{R}$ be a twice differentiable function vanishing at the origin. Assuming $f''$ is continuous, show that\n\n$$\n(2n + 3) \\int_{-1}^{1} x^{2n} f(x) \\, dx = f''(c), \\quad \\text{for some } c \\text{ in the closed interval } [-1, 1].\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $g: [-1, 1] \\to \\mathbb{R}$, $g(x) = \\frac{1}{2}(f(x)+f(-x))$, and let $h: [-1, 1] \\to \\mathbb{R}$, $h(x) = \\frac{1}{2}(f(x)-f(-x))$. Clearly, $g$ is even, $h$ is odd, $f = g+h$, and\n\n$$\n\\int_{-1}^{1} x^{2n} f(x) \\, dx = \\int_{-1}^{1} x^{2n} g(x) \\, dx + \\int_{-1}^{1} x^{2n} h(x) \\, dx = 2 \\int_{0}^{1} x^{2n} g(x) \\, dx.\n$$\n\nSince $f(0) = 0$ and $g'(x) = \\frac{1}{2}(f'(x)-f'(-x))$, it follows that $g(0) = g'(0) = 0$; and since $g''(x) = \\frac{1}{2}(f''(x)+f''(-x))$ and $f''$ is continuous, so is $g''$.\n\nLet $m = \\min_{0 \\le x \\le 1} g''(x)$ and $M = \\max_{0 \\le x \\le 1} g''(x)$. By Lagrange's theorem, for each $t$ in $(0, 1]$, there exists a $c_t$ in $(0, t)$ such that $g'(t) = t g''(c_t)$, so $m t \\le g'(t) \\le M t$.\n\nIntegrate the latter over the interval $[0, x]$, where $0 < x \\le 1$, to get $\\frac{1}{2} m x^2 \\le g(x) \\le \\frac{1}{2} M x^2$, so $\\frac{1}{2} m x^{2n+2} \\le x^{2n} g(x) \\le \\frac{1}{2} M x^{2n+2}$ for all $x$ in $[0, 1]$.\n\nIntegration of the latter over $[0, 1]$ yields\n\n$$\n\\frac{m}{2(2n+3)} = \\int_{0}^{1} \\frac{1}{2} m x^{2n+2} dx \\le \\int_{0}^{1} x^{2n} g(x) dx \\le \\int_{0}^{1} \\frac{1}{2} M x^{2n+2} dx = \\frac{M}{2(2n+3)},\n$$\n\nso\n\n$$\nm \\le 2(2n+3) \\int_{0}^{1} x^{2n} g(x) dx \\le M, \\quad \\text{i.e.,} \\quad m \\le (2n+3) \\int_{-1}^{1} x^{2n} f(x) dx \\le M.\n$$\n\nSince $g''$ has the intermediate value property (Darboux),\n\n$$\ng''(b) = (2n + 3) \\int_{-1}^{1} x^{2n} f(x) \\, dx, \\quad \\text{for some } b \\text{ in } [0, 1].\n$$\n\nFinally, since $g''(b)$ lies between $f''(-b)$ and $f''(b)$, and $f''$ has the intermediate value property (Darboux),\n\n$$\nf''(c) = g''(b) = (2n + 3) \\int_{-1}^{1} x^{2n} f(x) \\, dx, \\quad \\text{for some } c \\text{ in } [-b, b] \\subseteq [-1, 1].\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22391,
"subject": "Mathematics (Olympiad)",
"question": "Given a triangle $ABC$, points $M$ and $P$ lie on segments $AB$ and $BC$, respectively, such that $AM = BC$ and $CP = BM$. If $AP$ and $CM$ meet at $O$ and $2\\angle AOM = \\angle ABC$, find the measure of $\\angle ABC$.",
"options": [],
"answer": "See solution",
"solution": "Let $D$ be the reflection of $P$ across $C$, so $AB = BD$, and let $O'$ be the circumcenter of $\\triangle ABD$. As $\\triangle AO'B \\cong \\triangle BO'D$ and $MB = CD$, we have $\\angle O'CB = \\angle O'MA$, so $MBCO'$ is cyclic. Now $\\angle O'CM = \\angle O'BM = \\angle AOM$, hence $CO' \\parallel AP$. Therefore, $O'$ must be the midpoint of $AD$, implying that $\\angle ABC = 90^\\circ$.\n\n$\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22392,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a convex quadrilateral and let $P$ be a point inside such that\n\n- $\\angle APB + \\angle CPD = \\angle APD + \\angle BPC$,\n- $\\angle PAD + \\angle PCD = \\angle PAB + \\angle PCB$,\n- $\\angle PDC + \\angle PBC = \\angle PDA + \\angle PBA$.\n\nProve that the quadrilateral $ABCD$ is circumscriptible.",
"options": [],
"answer": "See solution",
"solution": "Let $A$, $B$, $C$, $D$ be the measures of the angles of the quadrilateral.\n\nNotice that $\\angle PDC + \\angle PBC = \\angle PDA + \\angle PBA$ implies $\\angle PDA + \\angle PBA = \\frac{B + D}{2}$. Similarly, $\\angle PAB + \\angle PCB = \\frac{A + C}{2}$.\n\nAdd these equalities to obtain\n$$\n180^\\circ = \\frac{A + B + C + D}{2} = (\\angle PBA + \\angle PAB) + \\angle PDA + \\angle PCB = (180^\\circ - \\angle APB) + \\angle PDA + \\angle PCB,\n$$\nand notice that $\\angle PDA + \\angle PCB = \\angle APB$ shows that circles $APD$ and $BPC$ are tangent at $P$.\n\nLet $(O_1, R_1)$, $(O_2, R_2)$, $(O_3, R_3)$, $(O_4, R_4)$ be the circles $PAB$, $PBC$, $PCD$, and $PDA$ respectively. Recall that points $P$, $O_1$, $O_3$ are collinear, and similarly, points $P$, $O_2$, $O_4$ are collinear. Further,\n$$\n\\angle O_2O_1O_4 + \\angle O_2O_3O_4 = 180^\\circ - \\angle APB + 180^\\circ - \\angle CPD = 180^\\circ,\n$$\nso $O_1O_2O_3O_4$ is a cyclic quadrilateral; let $R$ be its circumradius.\n\nBy the Sine Law,\n$$\n2R = \\frac{O_1O_3}{\\sin O_2} = \\frac{O_2O_4}{\\sin O_1} \\implies \\frac{R_1 + R_3}{\\sin \\angle BPC} = \\frac{R_2 + R_4}{\\sin \\angle APB},\n$$\nthen\n$$\n\\frac{AB}{\\sin \\angle APB} + \\frac{CD}{\\sin \\angle CPD} = 2(R_1 + R_3) \\implies AB + CD = 2(R_1 + R_3) \\sin \\angle APB.\n$$\nSimilarly, $BC + DA = 2(R_2 + R_4) \\sin \\angle BPC$. All the above lead to $AB + CD = BC + DA$, hence the claim.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22393,
"subject": "Mathematics (Olympiad)",
"question": "Өгөгдсөн бол $x^2 + y^2 + z^2 + t^2 = 2^{2012}$ тэгшитгэлийн бүх бүхэл $(x, y, z, t)$ шийдийн тоог ол.",
"options": [],
"answer": "See solution",
"solution": "Сондгой натурал тооны квадрат $8$-д хуваахад $1$ үлддэг тул $x, y, z, t$ бүгд тэгш байх ёстой. Тэгш тоонуудыг $x = 2x_1, y = 2y_1, z = 2z_1, t = 2t_1$ гэж тавьж, тэгшитгэлийг дахин бичвэл:\n\n$$x_1^2 + y_1^2 + z_1^2 + t_1^2 = 2^{2010}$$\n\nЭнэ үйлдлийг $1006$ удаа давтахад:\n\n$$x_{1004}^2 + y_{1004}^2 + z_{1004}^2 + t_{1004}^2 = 4$$\n\nЭнд $(x_{1004}, y_{1004}, z_{1004}, t_{1004})$ нь $(0, 0, 0, 2)$ эсвэл $(1, 1, 1, 1)$ байна. Эдгээрийг буцааж орлуулбал:\n\n- $(x, y, z, t) = (0, 0, 0, 2^{1006})$\n- $(x, y, z, t) = (2^{1005}, 2^{1005}, 2^{1005}, 2^{1005})$\n\nМөн эдгээрийн тэмдэгтүүдийн бүх боломжит хослолыг тооцвол $2 + 16 = 18$ шийд гарна.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22394,
"subject": "Mathematics (Olympiad)",
"question": "Does there exist a positive integer $n$ such that $1950^n + 1934^n = 2024^n$?",
"options": [],
"answer": "See solution",
"solution": "The numbers $1950$, $1934$, and $2024$ give remainders $4$, $2$, and $1$, respectively, when divided by $7$. Raising $4$ to powers $n = 1, 2, 3, 4, 5, 6, \\dots$ results in remainders $4, 2, 1, 4, 2, 1, \\ldots$, and raising $2$ to the same powers results in remainders $2, 4, 1, 2, 4, 1, \\ldots$. Thus, the remainders of the left side of the given equation for powers $n = 1, 2, 3, 4, 5, 6, \\dots$ are $6, 6, 2, 6, 6, 2, \\ldots$. The remainder of the right side is $1$ for every $n$. The contradiction shows that there is no positive integer $n$ that satisfies the given equation.\n\n**Note:** It is natural to look at the remainders when divided by $7$ because for smaller primes $2$, $3$, and $5$, one of the remainders is always $0$ and the other two remainders are equal, so no contradiction can arise.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22395,
"subject": "Mathematics (Olympiad)",
"question": "The triangle $ABC$ has $\\angle BAC = 90^\\circ$ and $\\angle ABC = 60^\\circ$. The points $D$ and $E$ are taken on the sides $AC$, respectively $AB$, so that $CD = 2 \\cdot DA$ and $DE$ is the bisector of the angle $\\angle ADB$. Denote $M$ the intersection of the lines $CE$ and $BD$, and $P$ the intersection of the lines $DE$ and $AM$. Prove that:\n\na) the lines $AM$ and $BD$ are perpendicular;\n\nb) $3 \\cdot PB = 2 \\cdot CM$.",
"options": [],
"answer": "See solution",
"solution": "a) Let $AB = a$. Then $BC = 2a$, $AC = \\sqrt{BC^2 - AB^2} = a\\sqrt{3}$, $AD = \\frac{a}{3}\\sqrt{3}$, $BD = \\sqrt{AB^2 + AD^2} = \\frac{2a}{3}\\sqrt{3} = 2AD$. This yields $\\angle ABD = 30^\\circ$, hence $\\angle ADB = 60^\\circ$.\n\n\n\nThis gives $\\angle ADE = 30^\\circ = \\angle ACB$, therefore $DE \\parallel BC$. So $\\triangle DME \\sim \\triangle BMC$ and $\\angle ADE \\sim \\angle ACB$, whence $\\frac{DM}{MB} = \\frac{DE}{BC} = \\frac{AD}{AC} = \\frac{1}{3}$. This implies $\\frac{DM}{DB} = \\frac{1}{4}$, hence $\\frac{DM}{DA} = \\frac{DA}{DB} = \\frac{1}{2}$, so $\\triangle DMA \\sim \\triangle DAB$ (S.A.S.), which implies $\\angle AMD = \\angle BAD = 90^\\circ$.\n\nb) Construct $DF \\parallel AM$, $F \\in CE$. Then $\\frac{CF}{CM} = \\frac{CD}{CA} = \\frac{2}{3}$; we have to prove that $CF = PB$.\n\nWe have $\\angle DAM = 90^\\circ - \\angle ADB = 30^\\circ$ and $\\angle ADE = 30^\\circ$, hence $DP = AP$. In triangle $DPM$, $PM = \\frac{1}{2}DP = \\frac{1}{2}AP$, whence $\\frac{MP}{MA} = \\frac{MF}{MC} = \\frac{1}{3}$, showing that $PF \\parallel AC$. It follows that $APFD$ is a parallelogram, so $DF = AP = DP$. Since $\\angle CDF = \\angle DAM = 30^\\circ = \\angle BDP$ and $BD = \\frac{2a}{3}\\sqrt{3} = CD$, $\\triangle CDF \\equiv \\triangle BDP$ (S.A.S.), therefore $CF = BP$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22396,
"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$$\n\nIf $k = 1$, we also have $P \\le \\left\\lfloor \\frac{n}{2} \\right\\rfloor \\le \\frac{n-k+2}{2}$. Let $Q$ be the number of connected components which contain at least one square in the second row, but no square in the third row. We have $Q \\le \\left\\lfloor \\frac{k}{2} \\right\\rfloor \\le \\frac{k+1}{2}$. Let $R$ be the number of connected components which contain at least one square in the 3rd to $m$th rows. By the induction hypothesis (1), we have $R \\le \\frac{1}{4}(m-1)(n+1) + 1$.\n\nIf $Q = \\frac{k+1}{2} = 1$ then all the squares in the second row are of the same color and thus all the squares in the 3rd row are also of the same color. If $Q = \\frac{k+1}{2} \\ge 2$, then the first square in the 3rd row is connected to the last square in the 3rd row via the 2nd part of the second row. By induction hypothesis (2), $R \\le \\frac{1}{4}(m-1)(n+1)$. Hence $Q = \\frac{k+1}{2}$ and $R = \\frac{1}{4}(m-1)(n+1) + 1$ can't hold simultaneously. Therefore,\n\n$$\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$$\n\nIf equality holds, then $P = \\frac{n-k+2}{2}$. Thus, the square at the upper-left corner is surrounded by three squares of opposite color. By symmetry, each square at the four corners of the grid is connected to none of the other squares.\n\nIf the square at position $(i, j)$ is colored red/blue if $ij$ is even/odd, respectively, it is easy to verify that $K = \\frac{1}{4}(m+1)(n+1) + 1$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 22397,
"subject": "Mathematics (Olympiad)",
"question": "Find the number of ordered 6-tuples $(a, b, c, a', b', c')$ satisfying the following system of modular equations:\n\n$$\n\\begin{cases}\n ab + a'b' \\equiv 1 \\pmod{15} \\\\\n bc + b'c' \\equiv 1 \\pmod{15} \\\\\n ca + c'a' \\equiv 1 \\pmod{15}\n\\end{cases}\n$$\n\nwhere $a, b, c, a', b', c' \\in \\{0, 1, \\dots, 14\\}$.",
"options": [],
"answer": "See solution",
"solution": "For any integer $k$, let $N_k$ be the number of ordered 6-tuples $(a, b, c, a', b', c')$ that satisfy\n$$\nab + a'b' \\equiv bc + b'c' \\equiv ca + c'a' \\equiv 1 \\pmod{k}$$\nand $a, b, c, a', b', c' \\in \\{0, 1, \\dots, k-1\\}$. By the Chinese Remainder Theorem,\n$$\nN_{mn} = N_n \\times N_m \\text{ if } \\gcd(n, m) = 1.\n$$\nTherefore, to compute $N_{15}$, we only need to compute $N_3$ and $N_5$. We will compute $N_p$ for any prime $p$.\n\nFix a solution $(a, b, a', b')$ of the equation $ab + a'b' \\equiv 1 \\pmod{p}$, and compute the number of solutions of the following system:\n$$\nbc + b'c' \\equiv ca + c'a' \\equiv 1 \\pmod{p} \\quad (1)\n$$\nWe consider three cases:\n\n* Suppose that $(a, a') \\not\\equiv t(b, b') \\pmod{p}$ for any $t \\in \\{0, 1, \\dots, p-1\\}$. Then the system (1) has a unique solution:\n $$\nc \\equiv \\frac{a' - b'}{a'b - b'a}, \\quad c' \\equiv \\frac{a - b}{ab' - ba'} \\pmod{p}\n $$\n* Suppose that $(a, a') \\equiv t(b, b') \\pmod{p}$ for some $t \\neq 1$. Then the system (1) has no solution.\n\n* Suppose that $(a, a') \\equiv (b, b') \\pmod{p}$. Then the system (1) becomes a single equation $bc + b'c' \\equiv 1 \\pmod{p}$. Since $ba + b'a' \\equiv 1 \\pmod{p}$, we can assume that $b \\neq 0$. For any choice of $c'$, there is only one choice of $c \\equiv (1 - b'c')/b \\pmod{p}$. Thus, the system (1) has exactly $p$ solutions.\n\nLet $T_p$ be the number of ordered tuples $(a, b, a', b')$ that satisfy $ab + a'b' \\equiv 1 \\pmod{p}$ and $a, b, a', b' \\in \\{0, 1, \\dots, p-1\\}$. For any pair $(a, a') \\neq (0, 0)$, there are exactly $p$ pairs $(b, b')$ satisfying the equation. Hence, $T_p = p(p^2 - 1)$.\n\nLet $C_p(t)$ be the number of ordered pairs $(a, b)$ that satisfy $a^2 + b^2 \\equiv t \\pmod{p}$ and $a, b \\in \\{0, 1, \\dots, p-1\\}$. From the above arguments, we have\n$$\nN_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$$\nIt is easy to get $C_3(0) = 1$, $C_3(1) = 4$, $C_5(0) = 9$, $C_5(1) = 4$, which implies that $N_3 = 28$, $N_5 = 124$, and $N_{15} = 28 \\times 124 = 3472$.\n\nTherefore, the number of ordered 6-tuples satisfying the given conditions is $3472$. $\\Box$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22398,
"subject": "Mathematics (Olympiad)",
"question": "For integers $a$, $b$, and for a positive integer $m$, we write $a \\equiv b \\pmod{m}$ if $a - b$ is divisible by $m$.\n\nFind all triples $(a, b, c)$ of positive integers such that\n$$\n2^a + 3^b + 1 = 6^c.\n$$",
"options": [],
"answer": "See solution",
"solution": "If for a triple $(a, b, c)$ of positive integers, the condition of the problem is satisfied, then $2^a = 6^c - (3^b + 1) \\equiv 2 \\pmod{3}$, so $a$ is odd.\n\nLet us first consider the case $a = 1$. In this case, $2 + 3^b + 1 = 6^c$, and the left-hand side equals $3(3^{b-1} + 1)$, which is divisible by $3$ but not by $3^2$. The right-hand side is divisible by $3^c$, so $c = 1$ and $b = 1$. Thus, $(a, b, c) = (1, 1, 1)$ is the only triple for $a = 1$.\n\nNext, consider $a$ odd and $a \\geq 3$. Then,\n$$\n2^a + 3^b + 1 \\equiv \\begin{cases} 4 & \\text{(mod 8), if $b$ is odd} \\\\ 2 & \\text{(mod 8), if $b$ is even} \\end{cases}\n$$\n\nOn the other hand,\n$$\n6^c \\equiv \\begin{cases} 6 & \\text{(mod 8), if $c = 1$} \\\\ 4 & \\text{(mod 8), if $c = 2$} \\\\ 0 & \\text{(mod 8), if $c \\geq 3$} \\end{cases}\n$$\n\nMatching remainders occurs only when $b$ is odd and $c = 2$. In this case, $2^a + 3^b + 1 = 36$, and since $a$ is odd and $\\geq 3$, $a = 3$ or $a = 5$. Checking these, $(a, b, c)$ must be either $(3, 3, 2)$ or $(5, 1, 2)$.\n\nTherefore, the desired triplets are $(1, 1, 1)$, $(3, 3, 2)$, and $(5, 1, 2)$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22399,
"subject": "Mathematics (Olympiad)",
"question": "Show that there exists a sequence $(a_n)_{n \\ge 0}$, with $a_n \\in \\{-1, +1\\}$ for all $n \\ge 0$, such that\n\n$$\n\\lim_{n \\to \\infty} \\left(\\sqrt{n+a_1} + \\sqrt{n+a_2} + \\dots + \\sqrt{n+a_n} - n\\sqrt{n+a_0}\\right) = \\frac{1}{2}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Obviously, $a_0 = -1$. For each $n \\in \\mathbb{N}^*$, let $k_n$ be the number of terms among $a_1, a_2, \\dots, a_n$ which are equal to $1$ (the others being $-1$). Then, the sequence $(x_n)_{n \\ge 0}$ whose limit must be $\\frac{1}{2}$ is\n\n$$\nx_n = k_n \\sqrt{n+1} - k_n \\sqrt{n-1},\n$$\nor\n$$\nx_n = \\frac{2k_n}{\\sqrt{n+1} + \\sqrt{n-1}} = \\frac{2\\frac{k_n}{\\sqrt{n}}}{\\sqrt{1 + \\frac{1}{n}} + \\sqrt{1 - \\frac{1}{n}}}.\n$$\n\nIt is enough to construct $(a_n)_{n \\ge 1}$ such that\n$$\n\\lim_{n \\to \\infty} \\frac{k_n}{\\sqrt{n}} = \\frac{1}{2}.\n$$\n\nLet us choose $a_n = 1$ if and only if $n = (2m)^2$, with $m \\in \\mathbb{N}^*$. Then $(2k_n)^2 \\le n < (2(k_n + 1))^2$, whence\n$$\n\\frac{\\sqrt{n}}{2} - 1 < k_n \\le \\frac{\\sqrt{n}}{2},\n$$\nwhich guarantees the required limit.\n\n**Remarks.** In the same way, one can prove that for every $\\ell \\in \\mathbb{R}$ there exists a sequence $(a_n)_{n \\ge 0}$ such that $\\lim_{n \\to \\infty} x_n = \\ell$.\n\nFor instance, if $0 < \\ell < +\\infty$, take $a_0 = -1$ and, as above, $a_n = 1$ if and only if $n = \\lfloor (N + \\frac{m}{\\ell})^2 \\rfloor$, with $m \\in \\mathbb{N}^*$ and fixed $N$ large enough so that different values for $m$ lead to different values for $n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22400,
"subject": "Mathematics (Olympiad)",
"question": "A $4 \\times n$ rectangle is to be tiled using $1 \\times 2$ rectangles. Find the total number of possible tilings.",
"options": [],
"answer": "See solution",
"solution": "Let $a_n$ be the total number of ways to tile a $4 \\times n$ rectangle. Let $b_n$ be the number of ways where a $1 \\times 2$ rectangle is placed vertically at the top or bottom of the last column, and $c_n$ be the number of ways where a $1 \\times 2$ rectangle is placed exactly in the center of the last column.\n\nIf two vertical $1 \\times 2$ rectangles are placed in the last column, the remaining $4 \\times (n-1)$ rectangle can be tiled in $a_{n-1}$ ways. If a $1 \\times 2$ rectangle is placed at the top or bottom, there are $2b_{n-1}$ ways. If a $1 \\times 2$ rectangle is placed in the center, there are $c_{n-1}$ ways. If four horizontal $1 \\times 2$ rectangles are placed in the last two columns, the remaining part can be tiled in $a_{n-2}$ ways. Thus,\n\n$$a_n = a_{n-1} + 2b_{n-1} + c_{n-1} + a_{n-2}$$\n$$b_n = b_{n-1} + a_{n-1}$$\n$$c_n = c_{n-2} + a_{n-1}$$\n\nIt follows that:\n\n$$a_n = a_{n-1} + 5a_{n-2} + a_{n-3} - a_{n-4}$$\n\nThe characteristic equation is:\n\n$$x^4 - x^3 - 5x^2 - x + 1 = 0$$\n\nwith roots:\n\n$$x_1 = \\frac{1}{4}(1 - \\sqrt{29} - \\sqrt{14 - 2\\sqrt{29}})$$\n$$x_2 = \\frac{1}{4}(1 + \\sqrt{29} + \\sqrt{14 - 2\\sqrt{29}})$$\n$$x_3 = \\frac{1}{4}(1 - \\sqrt{29} + \\sqrt{14 + 2\\sqrt{29}})$$\n$$x_4 = \\frac{1}{4}(1 + \\sqrt{29} + \\sqrt{14 + 2\\sqrt{29}})$$\n\nSo,\n\n$$a_n = t_1 x_1^n + t_2 x_2^n + t_3 x_3^n + t_4 x_4^n$$\n\nand\n\n$$a_n = \\frac{1}{\\sqrt{29}} (-x_1^{n+1} - x_2^{n+1} + x_3^{n+1} + x_4^{n+1})$$\n\nis the formula for the number of tilings.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22401,
"subject": "Mathematics (Olympiad)",
"question": "Let $D_0, D_1, \\dots, D_{2018}$ be points on the segment $\\overline{AB}$ such that $D_0 = A$, $D_{2018} = B$ and\n\n$$\n|D_0 D_1| = |D_1 D_2| = \\dots = |D_{2017} D_{2018}|.\n$$\n\nIf $C$ is a point such that $\\angle BCA = 90^\\circ$, prove that\n\n$$\n|CD_0|^2 + |CD_1|^2 + \\dots + |CD_{2018}|^2 = |AD_1|^2 + |AD_2|^2 + \\dots + |AD_{2018}|^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "Let $E_i$ be the foot of the altitude from $D_i$ to the side $\\overline{AC}$ in the right-angled triangle $ABC$, for each $i = 0, 1, \\dots, 2018$. We have $E_0 = A$, $E_{2018} = C$.\n\n\n\nSince the lines $D_iE_i$ are parallel to $BC$, Thales' theorem asserts that\n\n$$\n|E_0E_1| = |E_1E_2| = \\dots = |E_{2017}E_{2018}|.\n$$\n\nFrom here we get $|CE_i| = |AE_{2018-i}|$ for $i = 1, 2, \\dots, 2017$.\n\nThe Pythagorean theorem, applied to the right-angled triangles $CD_iE_i$ and $AD_iE_i$ for $i = 1, 2, \\dots, 2017$, yields\n\n$$\n|CD_i|^2 = |D_iE_i|^2 + |CE_i|^2 \\quad \\text{and} \\quad |AD_i|^2 = |D_iE_i|^2 + |AE_i|^2.\n$$\n\nBy subtracting these two equalities, we get\n\n$$\n|CD_i|^2 - |AD_i|^2 = |CE_i|^2 - |AE_i|^2, \\quad i = 1, \\dots, 2017,\n$$\n\nand the same relation also holds for $i = 0$ and $i = 2018$.\n\nAdding up all these relations, we get\n\n$$\n\\begin{aligned}\n& |CD_0|^2 + |CD_1|^2 + \\dots + |CD_{2018}|^2 - (|AD_0|^2 + |AD_1|^2 + \\dots + |AD_{2018}|^2) \\\\\n& = |CE_0|^2 + |CE_1|^2 + \\dots + |CE_{2018}|^2 - (|AE_0|^2 + |AE_1|^2 + \\dots + |AE_{2018}|^2) \\\\\n& = (|CE_0|^2 - |AE_{2018}|^2) + (|CE_1|^2 - |AE_{2017}|^2) + \\dots + (|CE_{2018}|^2 - |AE_0|^2) = 0.\n\\end{aligned}\n$$\n\nSince $|AD_0| = 0$, this proves the claim.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22402,
"subject": "Mathematics (Olympiad)",
"question": "Let $(x, y, p)$ be a triple of positive integers such that $p$ is a prime, $x < p < y < 2p$, and $y^2 - x^2 = 2p^2 - 1 - (p - 1)$. Find all such triples $(x, y, p)$.",
"options": [],
"answer": "See solution",
"solution": "Note that $(2, 7, 5)$ is such a triple. We show that it is the only one.\n\nLet $(x, y, p)$ be any such triple. We first compute\n\n$$\n(y + x)(y - x) = y^2 - x^2 = (2p^2 - 1) - (p - 1) = 2p^2 - p = p(2p - 1).\n$$\n\nIn particular, either $p \\mid x + y$ or $p \\mid x - y$ must hold.\n\nSuppose $p \\mid x + y$. Let $k \\in \\mathbb{Z}$ be such that $y = kp - x$. Note that by the given property, both $2p > y > p$ and $x < p$ hold. So $2p > y = kp - x > kp - p = (k-1)p$ and $kp > kp - x = y > p$, from which we conclude that $k=2$. Therefore $y = 2p - x$. If we substitute that into the previous equation, we get\n\n$$\n2p(2p - x - x) = p(2p - 1).\n$$\n\nDividing both sides by $p$, it follows that $4(p - x) = 2p - 1$. However, the left side is even and the right side is odd, so this is a contradiction.\n\nNow suppose $p \\mid y - x$. Let $k \\in \\mathbb{Z}$ be such that $y = kp + x$. Note that by the given property, both $2p > y = kp + x > kp$ and $(k+1)p = kp + p > kp + x = y > p$ hold. It follows that $k=1$, so $y - x = p$, and because of the previous equation also that $y + x = 2p - 1$. Subtracting these two equalities, we find that $2x = (y + x) - (y - x) = (2p - 1) - p = p - 1$. We conclude that $4(p - 1) = 4x^2 = (2x)^2 = (p - 1)^2$. Solving this quadratic equation in $p-1$ yields that either $p-1=0$ or $p-1=4$; only in the latter case $p$ is prime, namely $p=5$. Therefore, $x = (p-1)/2 = 2$ and $y = p + x = 5 + 2 = 7$.\n\nThus, the only such triple is $(2, 7, 5)$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22403,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an acute triangle, and $AX$, $AY$ two isogonal lines. Suppose that $K$ and $S$ are the feet of the perpendiculars from $B$ to $AX$ and $AY$, respectively, and $T$ and $L$ are the feet of the perpendiculars from $C$ to $AX$ and $AY$, respectively. Prove that $KL$ and $ST$ intersect on $BC$.",
"options": [],
"answer": "See solution",
"solution": "Denote $\\phi = \\angle XAB = \\angle YAC$, $\\alpha = \\angle CAX = \\angle BAY$. Since the quadrilaterals $ABSK$ and $ACTL$ are cyclic, we have\n\n$$\n\\angle BSK + \\angle BAK = 180^{\\circ} = \\angle BSK + \\phi = \\angle LAC + \\angle LTC = \\angle LTC + \\phi,\n$$\n\nso, due to the $90^{\\circ}$ angles formed, $\\angle KSL = \\angle KTL$. Thus, $KLST$ is cyclic.\n\n\n\nConsider $M$ to be the midpoint of $BC$ and $K'$ to be the symmetric point of $K$ with respect to $M$. Then $BKCK'$ is a parallelogram, so $BK \\parallel CK'$. But $BK \\parallel CT$, since both are perpendicular to $AX$. Thus, $K'$ lies on $CT$, and since $\\angle KTK' = 90^{\\circ}$ and $M$ is the midpoint of $KK'$, $MK = MT$. Similarly, $MS = ML$. Thus, the center of $(KLST)$ is $M$.\n\nLet $D$ be the foot of the altitude from $A$ to $BC$. Then $D$ belongs to both $(ABKS)$ and $(ACLT)$. So,\n\n$$\n\\angle ADT + \\angle ACT = 180^{\\circ} = \\angle ABS + \\angle ADS = \\angle ADT + 90^{\\circ} - \\alpha = \\angle ADS + 90^{\\circ} - \\alpha,\n$$\n\nand $\\overline{AD}$ is the bisector of $\\angle SDT$.\n\nBecause $DM$ is perpendicular to $AD$, $DM$ is the external bisector of this angle, and since $MS = MT$, it follows that $DMST$ is cyclic. Similarly, $DMLK$ is also cyclic.\n\nTherefore, $ST$, $KL$, and $DM$ are the radical axes of the three circles $(KLST)$, $(DMST)$, $(DMKL)$. These lines are concurrent, so $KL$ and $ST$ intersect on $BC$. $\\square$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22404,
"subject": "Mathematics (Olympiad)",
"question": "A grasshopper is sitting at point $O$ on a coordinate line. He makes 2016 jumps in the positive direction. The first jump has length 1, and each subsequent jump is $k \\in \\mathbb{N}$ times longer than the previous one. There are holes at every point with coordinate $2016l$, $l \\in \\mathbb{N}$. Determine all $k$ for which the grasshopper will make all the jumps and will not jump into a hole.",
"options": [],
"answer": "See solution",
"solution": "Let us find all the coordinates $a_n$ where the grasshopper will land:\n\n$$\na_1 = 1, \\quad a_n = 1 + k + k^2 + \\dots + k^{n-1}, \\quad n = 2, \\ldots, 2016.\n$$\n\nWe need to find all $k$ such that $a_n$ is never divisible by 2016 for any $n$.\n\nSuppose $(k, 2016) = d > 1$. Then every coordinate after a jump has residue 1 modulo $d$, so it cannot be divisible by 2016. Thus, all such $k$ are valid answers.\n\nIf $(k, 2016) = 1$, consider the residues of $a_n$ modulo 2016. If none of the $a_n$ are congruent to 0, then by the pigeonhole principle, at least two residues must be the same, say $a_m \\equiv a_l \\pmod{2016}$ with $m > l$. Their difference is divisible by 2016:\n\n$$\na_m - a_l = (1 + k + k^2 + \\dots + k^{m-1}) - (1 + k + k^2 + \\dots + k^{l-1}) = k^l (1 + k + k^2 + \\dots + k^{m-l-1}) = k^l a_{m-l}.\n$$\n\nThus, $a_{m-l}$ is divisible by 2016, which is a contradiction. Therefore, all $k$ with $(k, 2016) > 1$ are the solutions.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22405,
"subject": "Mathematics (Olympiad)",
"question": "Find all $n \\in \\mathbb{N}$ that can be represented in the form\n$$\nn = [a, b] + [b, c] + [c, a]\n$$\nwith $a, b, c \\in \\mathbb{N}$. Here, $[u, v]$ denotes the least common multiple of $u$ and $v$.",
"options": [],
"answer": "See solution",
"solution": "All $n \\in \\mathbb{N}$ are representable except the powers of $2$.\n\nLet $f(a, b, c) = [a, b] + [b, c] + [c, a]$.\n\nTake any $k \\in \\mathbb{N}$ and let $a = k$, $b = c = 1$ to obtain $f(k, 1, 1) = 2k + 1$. Hence, all odd $n \\ge 3$ are representable.\n\nIf $n$ is representable, then so is $2n$ because $[2u, 2v] = 2[u, v]$ implies $f(2a, 2b, 2c) = 2f(a, b, c)$. So each $n \\ge 3$ is representable if it has an odd divisor greater than $1$.\n\nThere remain the powers $2^k$ of $2$, with $k \\ge 0$. We show that they are not representable. This is true for $k = 0, 1$ since clearly $f(a, b, c) \\ge 3$ for all $a, b, c \\in \\mathbb{N}$.\n\nSuppose that $f(a, b, c) = 2^k$ with $k \\ge 2$. Consider the least $k$ with this property. Observe that at least two numbers among $a, b, c$ are even. Otherwise, $f(a, b, c)$ is odd while $2^k$ is even. If $a, b, c$ are all even, they can be divided by $2$ to yield $f\\left(\\frac{a}{2}, \\frac{b}{2}, \\frac{c}{2}\\right) = 2^{k-1}$, which contradicts the minimality of $k$.\n\nHence, one may assume that $a, b$ are even and $c$ is odd. Here $[a, b] = 2\\left[\\frac{a}{2}, \\frac{b}{2}\\right]$. Note also that\n\n$$\n[a, c] = 2\\left[\\frac{a}{2}, c\\right] \\text{ holds because } a \\text{ is even and } c \\text{ is odd.}\n$$\n\nAnalogously, $[b, c] = 2\\left[\\frac{b}{2}, c\\right]$. It follows that\n\n$$\nf\\left(\\frac{a}{2}, \\frac{b}{2}, c\\right) = \\frac{1}{2} f(a, b, c) = 2^{k-1}, \\text{ contradicting the minimality of } k \\text{ again.}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22406,
"subject": "Mathematics (Olympiad)",
"question": "Let $\\triangle ABC$ be an isosceles triangle with $AB = AC$ and $\\angle A = 45^\\circ$. Its circumcircle $(c)$ has center $O$, $M$ is the midpoint of $BC$, and $D$ is the foot of the perpendicular from $C$ to $AB$. With center $C$ and radius $CD$, we draw a circle which internally intersects $AC$ at the point $F$ and the circle $(c)$ at the points $Z$ and $E$, such that $Z$ lies on the small arc $\\widehat{BC}$ and $E$ on the small arc $\\widehat{AC}$. Prove that the lines $ZE$, $CO$, and $FM$ are concurrent.",
"options": [],
"answer": "See solution",
"solution": "Since $ZE$ is the common chord of the two circles, the line of the centres $CO$ is its perpendicular bisector. This means that $CO$ passes through the midpoint of $ZE$, let it be $T$. Thus, it suffices to prove that $FM$ passes through $T$. To this end, we will prove that $FZME$ is a parallelogram: we will precisely show that the segments $FE$ and $ZM$ are parallel and equal.\n\n\n\nLet $K$ be the intersection point of $ZE$ and $AC$. Then $K$ lies on the radical axis of the two circles, so it has equal powers to both circles. The power of the point $K$ with respect to one circle is $KA \\cdot KC$, while the power to the other circle is $R^2 - KC^2 = CD^2 - KC^2$.\n\nFrom the theorem of Pythagoras in triangle $ADC$ we get $AC^2 = 2CD^2$. We set $KA = x$, $KC = y$, and combining all the above yields $xy = \\frac{(x+y)^2}{2} - y^2$, hence $2xy = x^2 + y^2 + 2xy - y^2$, i.e. $x = y$.\n\nThis means that $K$ is the midpoint of $AC$. Moreover, we have $\\angle ADC = \\angle AMC = 90^\\circ$, so the points $A$, $D$, $M$, $C$ are on the same circle—let it be $(c_1)$—and the center of this circle is $K$.\n\nFrom the cyclic quadrilateral we have that $\\angle DMB = 45^\\circ$, but we have also that $\\angle OCB = 45^\\circ$, so the lines $OC$, $DM$ are parallel, hence $KZ \\perp DM$ and, since $DM$ is a chord of the circle, $ZD = ZM$. (1)\n\nThe triangle $CDF$ is isosceles and $CO$ is bisector, so $CO \\perp DF$, and this means that $ZE \\parallel DF$. It follows that $DFEZ$ is an isosceles trapezium, so $DZ = FE$ (2), and from (1) and (2) we have that $ZM = FE$ (*).\n\nFrom the isosceles trapezium we also have that $\\angle FEZ = \\angle DZE$ (3). Since $ZK$ is an altitude in the isosceles triangle $DZM$, it will be also angle bisector, so $\\angle DZE = \\angle MZE$ (4).\n\nFrom (3) and (4) we conclude that $\\angle FEZ = \\angle MZE$, so $FE \\parallel ZM$ (**). From (*) and (**) we get that $FZME$ is a parallelogram, which is the desired result.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22407,
"subject": "Mathematics (Olympiad)",
"question": "5-аас их бөгөөд хамгийн их ба хамгийн бага хоёрынх нь ялгавар 10-аас бага байх 4 анхны тооны нийлбэр 60-д хуваагдахыг батал.",
"options": [],
"answer": "See solution",
"solution": "Бодлогын нөхцөлийг хангах тоонуудын хамгийн багыг $p$ гэе. Тэгвэл өгөгдсөн анхны тоонууд нь $p,\\ p+2,\\ p+4,\\ p+6,\\ p+8$ дотроос сонгогдсон 4 тоо байна. Дараалсан 3 сондгой тооны аль нэг нь 3-т хуваагдах тул $3\\mid p+4$, мөн дараалсан 5 сондгой тооны аль нэг нь 5-д хуваагдах тул $5\\mid p+4$ байна. Иймд $p + (p+2) + (p+6) + (p+8) = 4(p+4)$ бөгөөд $15\\mid p+4$ тул бодлого батлагдана.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22408,
"subject": "Mathematics (Olympiad)",
"question": "Given the equation\n\n$$\n2 + x\\sqrt{9 + 6\\sqrt{2}} = x\\sqrt{5 - 2\\sqrt{6}} + \\sqrt{6} - 2\\sqrt{3} + \\sqrt{2}.\n$$\n\na) Write the root of the equation in the form $m - \\sqrt{n}$, where $m$ and $n$ are natural numbers.\n\nb) Factor the expression $a^3 - 3a^2 - 5a + 7$ into two non-constant factors with integer coefficients and calculate the value of this expression if $a$ is the root found in a).",
"options": [],
"answer": "See solution",
"solution": "$$\n\\sqrt{9 + 6\\sqrt{2}} = \\sqrt{3}\\sqrt{2 + 2\\sqrt{2} + 1} = \\sqrt{3}(\\sqrt{2} + 1) = \\sqrt{6} + \\sqrt{3}\n$$\n$$\n\\sqrt{5 - 2\\sqrt{6}} = \\sqrt{3 - 2\\sqrt{6} + 2} = |\\sqrt{3} - \\sqrt{2}| = \\sqrt{3} - \\sqrt{2}.\n$$\n\nThe equation takes the form\n\n$$\nx(\\sqrt{6} + \\sqrt{3} - \\sqrt{3} + \\sqrt{2}) = \\sqrt{6} + \\sqrt{2} - 2\\sqrt{3} - 2\n$$\n$$\nx(\\sqrt{6} + \\sqrt{2}) = (\\sqrt{6} + \\sqrt{2})(1 - \\sqrt{2})\n$$\n\nwhence (given $\\sqrt{6} + \\sqrt{2} > 0$) finally $x = 1 - \\sqrt{2}$.\n\nb) We have\n$$\na^3 - 3a^2 - 5a + 7 = a^3 - a^2 - 2a^2 + 2a - 7a + 7 = (a-1)(a^2 - 2a - 7).\n$$\nThe product of $a-1 = -\\sqrt{2}$ and $a^2 - 2a - 7 = (a-1)^2 - 8 = 2 - 8 = -6$ is $6\\sqrt{2}$.\n\n$\\square$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22409,
"subject": "Mathematics (Olympiad)",
"question": "Find all natural numbers $n, k$ that satisfy:\n\n$$\n(n + 1)^n = 2n^k + 3n + 1.\n$$",
"options": [],
"answer": "See solution",
"solution": "We first rewrite the equation as $2n^k + 3n = (n + 1)^n - 1$.\n\n\n\n$$\n2n^k + 3n = (n + 1)^n - 1 = (n + 1 - 1) \\left((n + 1)^{n-1} + (n + 1)^{n-2} + \\dots + (n + 1) + 1\\right).\n$$\n\nThe right-hand side is divisible by $n^2$ because the first factor has $n$, and the second has $n$ terms, each congruent to $1$ modulo $n$, so it is also divisible by $n$. Therefore, $(2n^{k-1} + 3)$ does not divide $n$. We distinguish two cases:\n\n1) $k=1$: Then $5$ does not divide $n$, so consider $n=1$ and $n=5$. If $n=1$, $2=6$ (contradiction). If $n=5$, $6^5 = 31$ (contradiction).\n\n2) $k > 1$: Then $3$ does not divide $n$. Possible $n$ are $1$ and $3$. $n=1$ gives $2=6$ (absurd). If $n=3$, $64 = 2 \\cdot 3^k + 10$, so $k=3$.\n\nThus, $n = k = 3$ is the unique solution.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22410,
"subject": "Mathematics (Olympiad)",
"question": "Find all pairs of positive integers $(a, b)$ such that\n$$\n2^a - 5^b = 3.\n$$",
"options": [],
"answer": "See solution",
"solution": "Note that if $a \\le 7$ or $b \\le 3$, then it is easy to verify that the only solutions are $(a, b) = (3, 1)$ and $(a, b) = (7, 3)$.\n\nNow it remains to prove that there are no solutions with $a > 7$, $b > 3$. In this case, we can write $a = 7 + \\alpha$, $b = 3 + \\beta$, where $\\alpha, \\beta \\in \\mathbb{N}$. Then the given equation $2^{7+\\alpha} - 5^{3+\\beta} = 3$ transforms to\n\n$$\n2^7(2^\\alpha - 1) = 5^3(5^\\beta - 1). \\qquad (1)\n$$\n\nSuppose that (1) has a solution $(\\alpha, \\beta)$ in positive integers. Set $A = 2^7(2^\\alpha - 1) = 5^3(5^\\beta - 1)$.\n\n1. From (1) it follows that $A$ is divisible by $2^7$, so $5^\\beta \\equiv 1 \\pmod{2^7}$. One can easily deduce that $\\beta \\geq 32$.\n\n2. Also, $A$ is divisible by $5^3$, so $2^\\alpha \\equiv 1 \\pmod{125}$ and we deduce that $\\alpha \\geq 100$. In particular, $A \\geq (2^{100} - 1)$, and so $A \\geq (2^5 - 1) = 31$. It follows that $5^\\beta - 1 \\geq 31$, thus $\\beta \\geq 3$.\n\n3. Thus we have $\\beta \\geq 96$, whence $A \\geq (5^{96} - 1)$, so $A \\geq 97$, which implies $2^\\alpha \\equiv 1 \\pmod{97}$. Therefore, $\\alpha \\geq 48$. In particular, $A \\geq (2^{48} - 1)$, then $A \\geq (2^{16} - 1)$ and $A \\geq (2^8 + 1) = 257$.\n\n4. Finally, $5^\\beta - 1 \\geq 257$, which gives $\\beta \\geq 256$, whence $A \\geq 2^8$, contrary to (1).\n\nTherefore, the only solutions are $(a, b) = (3, 1)$ and $(a, b) = (7, 3)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22411,
"subject": "Mathematics (Olympiad)",
"question": "Let $f$ be a function from the set of integers to the set of positive integers. Suppose that, for any two integers $m$ and $n$, the difference $f(m) - f(n)$ is divisible by $f(m - n)$. Prove that, for all integers $m$ and $n$ with $f(m) \\leq f(n)$, the number $f(n)$ is divisible by $f(m)$.",
"options": [],
"answer": "See solution",
"solution": "Set $n = 0$ in the given condition: $f(m) \\mid f(m) - f(0)$, so $f(m) \\mid f(0)$ for all $m$. Taking $m = 0$ gives $f(-n) \\mid f(0) - f(n)$ for all $n$. Together, these show $f(-n) \\mid f(n)$ for all $n$, implying $f(n) = f(-n)$. Thus, it suffices to show that for all $m, n > 0$, either $f(m) \\mid f(n)$ or $f(n) \\mid f(m)$.\n\nAssume the contrary and pick $m > n > 0$ violating the desired property with $m + n$ minimal. Since $m - n > 0$ and $(m - n) + n = m < m + n$, minimality of $m + n$ implies either $f(n) \\mid f(m - n)$ or $f(m - n) \\mid f(n)$. If $f(n) \\mid f(m - n)$, then $f(n) \\mid f(m) - f(m - n)$ implies $f(n) \\mid f(m)$, a contradiction. Therefore, $f(n) \\nmid f(m - n)$, so $f(m - n) \\mid f(n)$ and $f(m - n) < f(n)$. Note that\n\n$$\nf(m) \\mid f(n) - f(n - m) = f(n) - f(m - n).\n$$\n\nSince $f(n) - f(m - n) > 0$, this means $f(m) < f(n)$. By the given, $f(n) \\mid f(m) - f(m - n)$, where $|f(m) - f(m - n)| < f(n)$ because $f(m) < f(n)$ and $f(m - n) < f(n)$. Hence, it must be that $f(m) = f(m - n)$, implying $f(m) \\mid f(n)$, a contradiction.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22412,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ and $k$ be positive integers with $k \\ge n$ and $k - n$ an even number. Let $2n$ lamps labeled $1, 2, \\dots, 2n$ be given, each of which can be either on or off. Initially all lamps are off. We consider sequences of steps: at each step, one of the lamps is switched (from on to off or from off to on).\n\nLet $N$ be the number of such sequences consisting of $k$ steps and resulting in the state where lamps $1$ through $n$ are all on, and the lamps $n+1$ through $2n$ are all off.\n\nLet $M$ be the number of such sequences consisting of $k$ steps, resulting in the same state, but where none of the lamps $n+1$ through $2n$ is ever switched on.\n\nDetermine the ratio $N/M$.",
"options": [],
"answer": "See solution",
"solution": "Suppose lamp $i$ was switched on or off $a_i$ times. In the first situation, $a_i$ is odd for $1 \\le i \\le n$ and $a_i$ is even for $n+1 \\le i \\le 2n$. The total number of sequences where lamp $i$ is switched $a_i$ times is $\\displaystyle \\binom{k}{a_1, a_2, \\dots, a_{2n}}$. Thus,\n\n$$\nN = \\sum \\binom{k}{a_1, a_2, \\dots, a_{2n}}\n$$\n\nwhere the sum is over all $a_i$ such that $a_1 + \\dots + a_{2n} = k$, $a_i$ is odd for $1 \\le i \\le n$, and $a_i$ is even for $n+1 \\le i \\le 2n$. Note that\n\n$$\n\\binom{k}{a_1, a_2, \\dots, a_{2n}} = \\frac{k!}{a_1! a_2! \\dots a_{2n}!}\n$$\n\nso $N$ is the coefficient of $x^k$ in\n\n$$\nk! \\left( x + \\frac{x^3}{3!} + \\frac{x^5}{5!} + \\cdots \\right)^n \\left( 1 + \\frac{x^2}{2!} + \\frac{x^4}{4!} + \\cdots \\right)^n.\n$$\n\nThis can be rewritten as\n\n$$\nk! \\left( \\frac{e^x - e^{-x}}{2} \\right)^n \\left( \\frac{e^x + e^{-x}}{2} \\right)^n = k! \\left( \\frac{e^{2x} - e^{-2x}}{4} \\right)^n.\n$$\n\nSimilarly, for $M$, the sum is over all $a_i$ such that $a_1 + \\dots + a_{2n} = k$, $a_i$ is odd for $1 \\le i \\le n$, and $a_i = 0$ for $n+1 \\le i \\le 2n$. Then $M$ is the coefficient of $x^k$ in\n\n$$\nk! \\left( x + \\frac{x^3}{3!} + \\frac{x^5}{5!} + \\cdots \\right)^n = k! \\left( \\frac{e^x - e^{-x}}{2} \\right)^n.\n$$\n\nLet $f(x) = \\frac{e^x - e^{-x}}{2x}$. We wish to compare the coefficients of $x^k$ in $k! x^n f(x)^n$ and $k! x^n f(2x)^n$. It is now evident that the ratio is $2^{k-n}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22413,
"subject": "Mathematics (Olympiad)",
"question": "Do there exist positive integers $n$ and $k$, $1 \\leq k \\leq n-2$, such that\n$$\n\\binom{n}{k}^2 + \\binom{n}{k+1}^2 = \\binom{n}{k+2}^4?\n$$",
"options": [],
"answer": "See solution",
"solution": "By applying the formula $\\binom{a}{b} = \\frac{a!}{b!(a-b)!}$, we obtain the equation\n$$\n1 + \\frac{(n-k)^2}{(k+1)^2} = \\frac{(n-k)^2(n-k-1)^2}{(k+1)^2(k+2)^2} \\binom{n}{k+2}^2.\n$$\nHence, $(k+2)^2 \\left[(k+1)^2 + (n-k)^2\\right] = (n-k)^2(n-k-1)^2 \\binom{n}{k+2}^2$, which implies that $(k+1)^2 + (n-k)^2$ is a perfect square.\n\nLet $(k+1)^2 + (n-k)^2 = t^2$, where $t \\in \\mathbb{N}$. We have\n$$\n(k+2)t = (n-k)(n-k-1) \\binom{n}{k+2} \\geq 2 \\binom{n}{k+2}.\n$$\nUsing that $k+2 \\leq n$ and $t = \\sqrt{(k+1)^2 + (n-k)^2} < n+1$, we conclude that $(k+2)t \\leq n^2$.\n\nLet $3 \\leq k+2 \\leq n-3$ (the left-hand side of this inequality follows from the condition of the problem). If $n \\geq 6$, we have\n$$\n2 \\binom{n}{k+2} \\geq 2 \\binom{n}{3} = \\frac{n(n-1)(n-2)}{3} > n^2,\n$$\ni.e., the equation has no solution in this case.\n\nWhen $k+2 = n-2$, we obtain $t^2 = (n-3)^2 + 16$, hence $t = 5$, $n = 6$, $k = 2$. Direct computation shows that $n = 6$ and $k = 2$ is not a solution. If $k+2 = n-1$, we have $t^2 = (n-2)^2 + 9$, so $t = 5$, $n = 6$, $k = 3$, and as above we conclude that no solution exists. Therefore, positive integers $n$ and $k$ satisfying the equation do not exist.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22414,
"subject": "Mathematics (Olympiad)",
"question": "Let us consider a polynomial $P(x)$ with integer coefficients satisfying\n\n$$\nP(-1) = -4, \\quad P(-3) = -40, \\quad \\text{and} \\quad P(-5) = -156.\n$$\n\nWhat is the largest possible number of integers $x$ satisfying\n\n$$\nP(P(x)) = x^2?\n$$",
"options": [],
"answer": "See solution",
"solution": "No such numbers $x$ can ever exist. To see this, recall that when $x$ is an integer, the right-hand side of the equation is always $\\equiv 0 \\pmod{3}$ or $1 \\pmod{3}$. Also, when $x$ and $y$ are integers,\n\n$$\nP(x) \\equiv P(y) \\pmod{3}, \\quad \\text{whenever} \\quad x \\equiv y \\pmod{3}.\n$$\n\nTherefore,\n\n$$\nP(x) \\equiv P(-3) \\equiv -40 \\equiv 2 \\pmod{3} \\quad \\text{when} \\quad x \\equiv 0 \\pmod{3},\n$$\n\n$$\nP(x) \\equiv P(-5) \\equiv -156 \\equiv 0 \\pmod{3} \\quad \\text{when} \\quad x \\equiv 1 \\pmod{3},\n$$\n\nand\n\n$$\nP(x) \\equiv P(-1) \\equiv -4 \\equiv 2 \\pmod{3} \\quad \\text{when} \\quad x \\equiv 2 \\pmod{3}.\n$$\n\nIn particular, $P(x)$ is always $\\equiv 0 \\pmod{3}$ or $2 \\pmod{3}$, so $P(P(x))$ is always $\\equiv 2 \\pmod{3}$. Therefore, $P(P(x)) \\not\\equiv x^2 \\pmod{3}$ for any integer $x$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 22415,
"subject": "Mathematics (Olympiad)",
"question": "Правоаголник велиме дека е впишан во триаголник ако две негови соседни темиња лежат на една страна на триаголникот, а другите две лежат на останатите две страни на триаголникот. Нека должините на страните на триаголникот $ABC$ се познати. Колкава е најмалата можна должина на дијагонала на правоаголник впишан во овој триаголник.",
"options": [],
"answer": "See solution",
"solution": "Нека четириаголникот $EFGH$ е впишан во триаголникот $ABC$ така што $E$ и $F$ лежат на $BC$, $G$ лежи на $AC$, а $H$ лежи на $AB$. Нека должините на страните на триаголникот $ABC$ ги означиме со $a$, $b$ и $c$, и нека со $h$ ја означиме должината на висината спуштена од $A$ кон $BC$. Ставаме $\\overline{AH}=x$, $\\overline{EF}=u$ и $\\overline{FG}=v$.\n\nОд $\\triangle AHG \\sim \\triangle ABC$ имаме $u=\\frac{a x}{c}$, а од $\\triangle BEH \\sim \\triangle BVA$ добиваме $v=\\frac{h (c-x)}{c}$. Ако со $l$ ја означиме должината на дијагоналата на $EFGH$, добиваме:\n\n$$\nl^2 = \\frac{a^2 x^2}{c^2} + \\frac{h^2 (c-x)^2}{c^2}.\n$$\n\nНајмалата вредност што ја има параболата $f(x) = \\frac{a^2 x^2}{c^2} + \\frac{h^2 (c-x)^2}{c^2}$ е\n\n\n\n$\\frac{a^2 h^2}{a^2 + h^2}$ и таа се добива кога $x = \\frac{h^2 c}{a^2 + h^2}$. Да забележиме $\\frac{a^2 h^2}{a^2 + h^2} = \\frac{4P^2}{a^2 + \\frac{4P^2}{a^2}}$, каде $P$ е плоштината на триаголникот. Ако истото го направиме кога впишаниот правоаголник има две соседни темиња кои лежат на страната $AC$, за најмала можна вредност на дијагоналата добиваме $\\frac{4P^2}{b^2 + \\frac{4P^2}{b^2}}$.\n\nИмаме\n\n$$\na^2 + \\frac{4P^2}{a^2} - b^2 - \\frac{4P^2}{b^2} = (a^2 - b^2) \\left(1 - \\frac{4P^2}{a^2 b^2}\\right).\n$$\n\nБидејќи $ab \\ge 2P$, последниот израз е поголем или еднаков на $0$ ако и само ако $a \\ge b$. Со други зборови, најмалата вредност на $l$ се добива кога правоаголникот има две соседни темиња кои лежат на најголемата страна на триаголникот. Нека $a$ е најголемата страна. Веќе видовме дека тогаш најмалата вредност на $l$ е $\\frac{2P}{\\sqrt{a^2 + \\frac{4P^2}{a^2}}}$ и таа се добива кога $\\overline{AH} = x = \\frac{h^2 c}{a^2 + h^2} = \\frac{4P^2 c}{a^4 + 4P^2}$. Да забележиме дека плоштината е позната и може да биде изразена преку страните на триаголникот, на пример преку Херонова формула.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22416,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be an integer greater than or equal to $2$. Find the smallest positive integer $m$ for which there exists a sequence $a_1, a_2, \\dots, a_n$ of positive integers satisfying the following two conditions:\n\n* $a_1 < a_2 < \\dots < a_n = m$.\n* All of the $n-1$ numbers $\\frac{a_1^2 + a_2^2}{2}, \\dots, \\frac{a_{n-1}^2 + a_n^2}{2}$ are perfect squares.",
"options": [],
"answer": "See solution",
"solution": "We will show that the smallest possible value for $m$ is $2n^2 - 1$. If we let $a_k = 2k^2 - 1$ for $k = 1, 2, \\dots, n$, then\n$$\n\\frac{a_k^2 + a_{k+1}^2}{2} = \\frac{(2k^2 - 1)^2 + (2(k+1)^2 - 1)^2}{2} = (2k^2 + 2k + 1)^2\n$$\nare all perfect squares for each $k = 1, 2, \\dots, n-1$, and we have $m = a_n = 2n^2 - 1$ in this case.\n\nNow, we show that if there exists a sequence $a_1, a_2, \\dots, a_n$ satisfying the two conditions, then $m \\geq 2n^2 - 1$ must hold. It suffices to show that $a_k \\geq 2k^2 - 1$ for each $k = 1, 2, \\dots, n$. For this, we prove the following lemma:\n\n**Lemma.** Let $k$ be a positive integer. For any pair of positive integers $x, y$ satisfying $2k^2 - 1 \\leq x < y < 2(k+1)^2 - 1$, the number $\\frac{x^2 + y^2}{2}$ is not a perfect square.\n\n**Proof.** If $x$ and $y$ have different parity, then $\\frac{x^2 + y^2}{2}$ is not an integer. Assume $x$ and $y$ have the same parity. Then:\n$$\n\\frac{x^2 + y^2}{2} - \\left(\\frac{x+y}{2}\\right)^2 = \\left(\\frac{y-x}{2}\\right)^2 > 0\n$$\nAlso,\n$$\n y - x \\leq (2(k+1)^2 - 2) - (2k^2 - 1) = 4k + 1 \\\\\n x \\geq 2k^2 - 1, \\quad y \\geq x + 2 \\geq 2k^2 + 1\n$$\nSince $y-x$ is even, $y-x \\leq 4k$. Thus,\n$$\n\\left(\\frac{x+y}{2}+1\\right)^2 - \\frac{x^2+y^2}{2} = x+y+1 - \\left(\\frac{y-x}{2}\\right)^2 \\\\\n\\geq (2k^2-1) + (2k^2+1) + 1 - (2k)^2 = 1 > 0\n$$\nSo,\n$$\n\\left(\\frac{x+y}{2}\\right)^2 < \\frac{x^2+y^2}{2} < \\left(\\frac{x+y}{2}+1\\right)^2\n$$\nTherefore, $\\frac{x^2+y^2}{2}$ lies strictly between two consecutive perfect squares, so it cannot be a perfect square.\n\nNow, we show $a_k \\geq 2k^2 - 1$ for each $k$ by induction. The base case $k=1$ is clear. Suppose $a_l \\geq 2l^2 - 1$ for some $l$. If $a_{l+1} < 2(l+1)^2 - 1$, then by the lemma, $\\frac{a_l^2 + a_{l+1}^2}{2}$ is not a perfect square, contradicting the problem's condition. Thus, $a_{l+1} \\geq 2(l+1)^2 - 1$. By induction, $a_k \\geq 2k^2 - 1$ for all $k$, so $m = a_n \\geq 2n^2 - 1$.\n\nTherefore, the smallest possible $m$ is $2n^2 - 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22417,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABCD$ be a convex quadrilateral such that $|AD| < |AB|$ and $|CD| < |CB|$. Prove that $\\angle ABC < \\angle ADC$.",
"options": [],
"answer": "See solution",
"solution": "Let $D'$ be the point symmetric to $D$ with respect to line $AC$. Since triangles $ADC$ and $AD'C$ are similar, it suffices to prove that the inequalities $\\angle CAB + \\angle CAD' < 180^\\circ$ and $\\angle ACB + \\angle ACD' < 180^\\circ$ together imply $\\angle ABC < \\angle AD'C$.\n\nSuppose the contrary, i.e., $\\angle AD'C \\leq \\angle ABC$. Then $D'$ lies on or outside the circumcircle of triangle $ABC$. Let $E$ be the second intersection point of line $AD'$ with the circumcircle of triangle $ABC$ ($E = A$ if $AD'$ is tangent). Similarly, let $F$ be the second intersection point of line $CD'$ with the circumcircle of triangle $ABC$ ($F = C$ if $CD'$ is tangent). Consider three cases:\n\n* If $\\angle ACD' \\geq \\angle ACB$ and $\\angle CAD' \\geq \\angle CAB$, then $B$ lies inside triangle $AD'C$ or on its side. Applying the triangle inequality twice gives $|AB| + |BC| \\leq |AD'| + |D'C|$, while adding the two given inequalities yields $|AD| + |DC| < |AB| + |AC|$. This is a contradiction since $|AD'| = |AD|$ and $|D'C| = |DC|$.\n\n* If $\\angle ACD' < \\angle ACB$, then $A, F, B, C$ occur on the circumcircle of $ABC$ in this order, and $E$ and $B$ lie on different sides of line $CD'$. Consequently,\n\n$$\n\\angle FBC > \\angle EBC = 180^\\circ - \\angle D'AC > \\angle CAB = \\angle CFB,\n$$\n\nso in triangle $FBC$ we get $|CF| > |CB|$, implying $|CD'| > |CB|$, contradiction.\n\n* The case $\\angle CAD' < \\angle CAB$ is symmetric to the previous case.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22418,
"subject": "Mathematics (Olympiad)",
"question": "Let $A$ and $B$ be points on a circle with center $O$ such that triangle $AOB$ is right-angled. The perpendicular bisector of segment $AO$ intersects the shorter arc $AB$ at point $K$. The lines $KO$ and $AB$ intersect at point $L$. Prove that triangle $KBL$ is isosceles.\n\n",
"options": [],
"answer": "See solution",
"solution": "Since $K$ lies on the perpendicular bisector of segment $AO$, we have $KA = KO$. Also, $KO = AO$ since $K$ and $A$ are points on the circle. Hence, $\\triangle AKO$ is equilateral, so $\\angle AOK = 60^\\circ$. Thus, $\\angle KOB = \\angle AOB - \\angle AOK = 90^\\circ - 60^\\circ = 30^\\circ$. Since $B$ also lies on the circle, $KO = BO$, so $\\angle BKO = \\frac{180^\\circ - \\angle KOB}{2} = \\frac{180^\\circ - 30^\\circ}{2} = 75^\\circ$. Also, $AO = BO$ implies $\\angle ABO = \\frac{180^\\circ - \\angle AOB}{2} = \\frac{90^\\circ}{2} = 45^\\circ$. Therefore, $\\angle BLK = \\angle LOB + \\angle LBO = 30^\\circ + 45^\\circ = 75^\\circ$. Thus, triangle $KBL$ is isosceles as it has two equal angles.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22419,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be positive real numbers. Prove that\n\n$$\n\\frac{8}{(a+b)^2 + 4abc} + \\frac{8}{(b+c)^2 + 4abc} + \\frac{8}{(c+a)^2 + 4abc} + a^2 + b^2 + c^2 \\geq \\frac{8}{a+3} + \\frac{8}{b+3} + \\frac{8}{c+3}.\n$$",
"options": [],
"answer": "See solution",
"solution": "Since $2ab \\leq a^2 + b^2$, it follows that $(a+b)^2 \\leq 2(a^2 + b^2)$.\n\nAlso, $4abc \\leq 2c(a^2 + b^2)$ for any positive reals $a$, $b$, $c$.\n\nAdding these inequalities, we find\n\n$$\n(a+b)^2 + 4abc \\leq 2(a^2 + b^2)(c+1),\n$$\nso that\n$$\n\\frac{8}{(a+b)^2 + 4abc} \\geq \\frac{4}{(a^2 + b^2)(c+1)}.\n$$\n\nUsing the AM-GM inequality, we have\n$$\n\\frac{4}{(a^2 + b^2)(c+1)} + \\frac{a^2 + b^2}{2} \\geq 2\\sqrt{\\frac{2}{c+1}} = \\frac{4}{\\sqrt{2(c+1)}}\n$$\n\nand\n$$\n\\frac{c+3}{8} = \\frac{(c+1) + 2}{8} \\geq \\sqrt{\\frac{2(c+1)}{4}}.\n$$\n\nWe conclude that\n$$\n\\frac{4}{(a^2 + b^2)(c+1)} + \\frac{a^2 + b^2}{2} \\geq \\frac{8}{c+3},\n$$\nand finally\n$$\n\\frac{8}{(a+b)^2 + 4abc} + \\frac{8}{(b+c)^2 + 4abc} + \\frac{8}{(c+a)^2 + 4abc} + a^2 + b^2 + c^2 \\geq \\frac{8}{a+3} + \\frac{8}{b+3} + \\frac{8}{c+3}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22420,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\ge 2$ be an integer. Suppose that real numbers $a_1, a_2, \\dots, a_{2n}$ satisfy $|a_k - a_{n+k}| \\ge 1$ for every integer $k$ such that $1 \\le k \\le n$. Find the minimum possible value of\n$$\n(a_1 - a_2)^2 + (a_2 - a_3)^2 + \\dots + (a_{2n-1} - a_{2n})^2 + (a_{2n} - a_1)^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "Set $a_{2n+1} = a_1$. Since $a_{n+1} - a_{2n+1} = -(a_1 - a_{n+1})$, the values\n$$\na_1 - a_{n+1}, \\ a_2 - a_{n+2}, \\ \\dots, \\ a_n - a_{2n}, \\ a_{n+1} - a_{2n+1}\n$$\nare all nonzero, and $a_1 - a_{n+1}$ and $a_{n+1} - a_{2n+1}$ have opposite signs. Therefore, there exists an integer $m$ with $1 \\le m \\le n$ such that $a_m - a_{n+m}$ and $a_{m+1} - a_{n+m+1}$ have opposite signs. In this case, since both $|a_m - a_{n+m}|$ and $|a_{m+1} - a_{n+m+1}|$ are at least 1, we have\n$$\n|(a_m - a_{m+1}) - (a_{n+m} - a_{n+m+1})| = |(a_m - a_{n+m}) - (a_{m+1} - a_{n+m+1})| \\ge 2.\n$$\nIn general, for real numbers $x$ and $y$, we have\n$$\nx^2 + y^2 - \\frac{(x - y)^2}{2} = \\frac{(x + y)^2}{2} \\geq 0,\n$$\nso it follows that\n$$\n(a_m - a_{m+1})^2 + (a_{n+m} - a_{n+m+1})^2 \\ge \\frac{((a_m - a_{m+1}) - (a_{n+m} - a_{n+m+1}))^2}{2} \\ge 2.\n$$\nTherefore, we obtain\n$$\n(a_1 - a_2)^2 + (a_2 - a_3)^2 + \\dots + (a_{2n-1} - a_{2n})^2 + (a_{2n} - a_1)^2 \\ge 2.\n$$\nOn the other hand, the values $a_1 = a_2 = \\dots = a_n = 1$ and $a_{n+1} = a_{n+2} = \\dots = a_{2n} = 0$ satisfy the given condition, and the value of the given function is 2 in this case. Hence, the minimum possible value is 2.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22421,
"subject": "Mathematics (Olympiad)",
"question": "Determine all positive integers $n$ such that the number $N$ is also an integer, where\n\n$$\nN = \\frac{2 \\cdot 1}{\\sqrt{1^2 + 1 + 4 + \\sqrt{1^2 - 1 + 4}}} + \\frac{2 \\cdot 2}{\\sqrt{2^2 + 2 + 4 + \\sqrt{2^2 - 2 + 4}}} + \\dots + \\frac{2n}{\\sqrt{n^2 + n + 4 + \\sqrt{n^2 - n + 4}}}\n$$",
"options": [],
"answer": "See solution",
"solution": "Let us note that\n\n$$\n\\frac{2k}{\\sqrt{k^2 + k + 4} + \\sqrt{k^2 - k + 4}} = \\frac{(\\sqrt{k^2 + k + 4} - \\sqrt{k^2 - k + 4})2k}{(k^2 + k + 4) - (k^2 - k + 4)} = \\sqrt{k^2 + k + 4} - \\sqrt{k^2 - k + 4} = \\sqrt{k^2 + k + 4} - \\sqrt{(k-1)^2 + (k-1) + 4}.\n$$\n\nAnd thus\n\n$$\nN = (\\sqrt{6} - \\sqrt{4}) + (\\sqrt{10} - \\sqrt{6}) + (\\sqrt{16} - \\sqrt{10}) + \\dots + (\\sqrt{n^2 + n + 4} - \\sqrt{(n-1)^2 + (n-1) + 4}) = \\sqrt{n^2 + n + 4} - \\sqrt{4} = \\sqrt{n^2 + n + 4} - 2.\n$$\n\nIt means that $N$ is integer if and only if $n^2 + n + 4$ is a perfect square.\n\nSince $n^2 < n^2 + n + 4 < n^2 + 4n + 4 = (n + 2)^2$, then for every positive integer $n$ we obtain $n^2 + n + 4 = (n + 1)^2 = n^2 + 2n + 1$. And thus $n = 3$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22422,
"subject": "Mathematics (Olympiad)",
"question": "Збирот од должините на страните на правоаголникот е 40 см. Едната страна на правоаголникот е 4 пати подолга од другата.\n\n\n\nа) Да се определат должините на страните на правоаголникот.\n\nб) Дали може, со 3 паралелни прави, дадениот правоаголник да се подели на 4 еднакви квадрати? Нацртајте!\n\nв) За колку сантиметри се разликува збирот од должините на страните на делбен квадрат од збирот од должините на страните на правоаголникот?",
"options": [],
"answer": "See solution",
"solution": "а) $x + x + 4x + 4x = 40$, $10x = 40$, $x = 4$ см.\n\nЕдната страна е долга 16 см, а другата 4 см.\n\nб) Може, страната на секој квадрат е долга по 4 см.\n\nв) Збирот на страните на квадратот е $4 + 4 + 4 + 4 = 16$ см. $40 - 16 = 24$ см.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22423,
"subject": "Mathematics (Olympiad)",
"question": "For what real values of $k > 0$ is it possible to dissect a $1 \\times k$ rectangle into two similar, but incongruent, polygons?",
"options": [],
"answer": "See solution",
"solution": "We will show that a dissection satisfying the requirements of the problem is possible if and only if $k \\neq 1$.\n\nFirst, we show by contradiction that such a dissection is not possible when $k = 1$. Assume that we have such a dissection.\n\nThe common boundary of the two dissecting polygons must be a single broken line connecting two points on the boundary of the square (otherwise, either the square is subdivided into more than two pieces or one of the polygons is inside the other).\n\nThe two dissecting polygons must have the same number of vertices. They share all the vertices on the common boundary, so they must use the same number of corners of the square as their own vertices. Therefore, the common boundary must connect two opposite sides of the square (otherwise, one of the polygons will contain at least three corners of the square, while the other at most two). However, this means that each of the dissecting polygons must use an entire side of the square as one of its sides, and thus each polygon has a side of length 1. A side of longest length in one of the polygons is either a side on the common boundary or, if all those sides have length less than 1, it is a side of the square. But this is also true of the other polygon, which means that the longest side length in the two polygons is the same. This is impossible since they are similar but not congruent, so we have a contradiction.\n\nNow, we construct a dissection satisfying the requirements of the problem when $k \\neq 1$. Notice that we may assume $k > 1$, because a $1 \\times k$ rectangle is similar to a $1 \\times \\frac{1}{k}$ rectangle.\n\nWe first construct a dissection of an appropriately chosen rectangle (denoted by $ABCD$ below) into two similar incongruent polygons. The construction depends on two parameters ($n$ and $r$ below). By appropriate choice of these parameters, we show that the constructed rectangle can be made similar to a $1 \\times k$ rectangle, for any $k > 1$. The construction follows.\n\nLet $r > 1$ be a real number. For any positive integer $n$, consider the following sequence of $2n + 2$ points:\n\n$$\n\\begin{aligned}\nA_0 &= (0, 0), \\quad A_1 = (1, 0), \\quad A_2 = (1, r), \\quad A_3 = (1 + r^2, r), \\\\\nA_4 &= (1 + r^2, r + r^3), \\quad A_5 = (1 + r^2 + r^4, r + r^3),\n\\end{aligned}\n$$\n\nand so on, until\n\n$$\nA_{2n+1} = (1 + r^2 + r^4 + \\dots + r^{2n},\\ r + r^3 + r^5 + \\dots + r^{2n-1}).\n$$\n\nDefine a rectangle $ABCD$ by $A = A_0$, $C = A_{2n+1}$,\n\n$$\nB = (1 + r^2 + \\dots + r^{2n}, 0), \\quad \\text{and} \\quad D = (0, r + r^3 + \\dots + r^{2n-1}).\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22424,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a given positive integer. In the coordinate plane, consider the set of points\n$$\n\\{P_1, P_2, \\dots, P_{4n+1}\\} = \\{(x, y) \\mid x \\text{ and } y \\text{ are integers with } xy = 0,\\ |x| \\le n,\\ |y| \\le n\\}.\n$$\nDetermine the minimum of\n$$\n(P_1P_2)^2 + (P_2P_3)^2 + \\dots + (P_{4n}P_{4n+1})^2 + (P_{4n+1}P_1)^2.\n$$",
"options": [],
"answer": "See solution",
"solution": "The answer is $16n - 8$.\n\nAssume that $P_i = (x_i, y_i)$ for $1 \\le i \\le 4n + 1$. Set\n$$\nP_{4n+2} = (x_{4n+2}, y_{4n+2}) = P_1 = (x_1, y_1).\n$$\nWe will show that the sum\n$$\nS = \\sum_{i=1}^{4n+1} (P_i P_{i+1})^2 \\geq 16n - 8.\n$$\nFirst, we show that this minimum can be obtained by setting\n$$\n(P_1, P_2, \\dots, P_{4n+1}) = ((2, 0), (4, 0), \\dots, (n, 0), (n-1, 0), \\dots, (1, 0), (0, 2), (0, 4), \\dots, (0, n), (0, n-1), \\dots, (0, 1), (-2, 0), (-4, 0), \\dots, (-n, 0), (-n+1, 0), \\dots, (-1, 0), (0, -2), (0, -4), \\dots, (0, -n), (0, -n+1), \\dots, (0, -1), (0, 0))\n$$\nfor $n$ even, and\n$$\n(P_1, P_2, \\dots, P_{4n+1}) = ((1, 0), (3, 0), \\dots, (n, 0), (n-1, 0), \\dots, (2, 0), (0, 1), (0, 3), \\dots, (0, n), (0, n-1), \\dots, (0, 2), (-1, 0), (-3, 0), \\dots, (-n, 0), (-n+1, 0), \\dots, (-2, 0), (0, -1), (0, -3), \\dots, (0, -n), (0, -n+1), \\dots, (0, -2), (0, 0))\n$$\nfor $n$ odd. This can be easily checked. For example, when $n = 2m$ is even, our construction shows that\n$$\n\\begin{aligned}\nS &= 4 \\left( \\sum_{i=1}^{m-1} (P_i P_{i+1})^2 + (P_m P_{m+1})^2 + \\sum_{i=m+1}^{2m-1} (P_i P_{i+1})^2 \\right) \\\\ &\\quad + 3(P_{2m} P_{2m+1})^2 + (P_{4n} P_{4n+1})^2 + (P_{4n+1} P_1)^2 \\\\ &= 4(4(m-1)+1+4(m-1)) + 3 \\times 5 + 1 + 4 \\\\ &= 32m - 8 = 16n - 8.\n\\end{aligned}\n$$\nThe exact same argument works when $n$ is odd.\n\nNote that\n$$\nS = \\sum_{i=1}^{4n+1} \\left[(x_i - x_{i+1})^2 + (y_i - y_{i+1})^2\\right].\n$$\nBy symmetry, it suffices to show that for\n$$\n\\{x_1, x_2, \\dots, x_{4n+1}\\} = \\{\\pm 1, \\pm 2, \\dots, \\pm n, \\underbrace{0, \\dots, 0}_{2n+1\\ 0s}\\},\n$$\nand we have\n$$\nT_n(x_1, x_2, \\dots, x_{4n+1}) = \\sum_{i=1}^{4n+1} (x_i - x_{i+1})^2 \\le 8n - 4.\n$$\nIndeed, for every positive integer $n$, we are going to prove that for $1 \\le m \\le n$ and\n$$\n\\{x_1, x_2, \\dots, x_{2m+2n+1}\\} = \\{\\pm 1, \\pm 2, \\dots, \\pm m, \\underbrace{0, \\dots, 0}_{2n+1\\ 0s}\\},\n$$\nwe have\n$$\nT_m(x_1, x_2, \\dots, x_{2m+2n+1}) = \\sum_{i=1}^{2m+2n+1} (x_i - x_{i+1})^2 \\le 8m - 4.\n$$\nWe take induction on $m$. For $m=1$ and\n$$\n\\{x_1, x_2, \\dots, x_{2n+3}\\} = \\{\\pm 1, \\underbrace{0, \\dots, 0}_{2n+1\\ 0s}\\},\n$$\nclearly $T_1(x_1, x_2, \\dots, x_{2n+3}) = 8 \\cdot 1 - 4 = 4$, while equality holds for\n$$\n\\{x_1, x_2, \\dots, x_{2n+3}\\} = \\{0, 1, 0, -1, 0, \\underbrace{0, \\dots, 0}_{2n-2\\ 0s}\\}.\n$$\nWe assume the result is true for $m=k c_1$. Therefore, since the $2m$ terms $r_i$ and $c_i$ are not all equal, one must be strictly greater than $m^2/(2m) = 2n + 1$ and therefore at least $2n + 2$.\n\nNow we construct an example to show that it is possible that no row or column is entered more than $2n + 2$ times. Partition the square grid into four $$(2n+1) \\times (2n+1)$$ quadrants A, B, C, and D, containing the upper left, upper right, lower left, and lower right corners, respectively. The turtle begins at the top right corner square of B, moves one square down, and then moves left through the whole second row of B. She then moves one square down and moves right through the whole third row of B. She continues in this pattern, moving through each remaining row of B in succession and moving one square down when each row is completed. Since $2n+1$ is odd, the turtle ends at the bottom right corner of B. She then moves one square down into D and through each column of D in turn, moving one square to the left when each column is completed. She ends at the lower left corner of D and moves left into C and through the rows of C, moving one square up when each row is completed, ending in the upper left corner of C. She then enters A and moves through the columns of A, moving one square right when each column is completed. This takes her to the upper right corner of A, whereupon she enters B and moves right through the top row of B, which returns her to her starting point. Each row passing through A and B is entered at most $2n+1$ times in A and once in B, and thus at most $2n+2$ times in total. Similarly, each row and column in the grid is entered at most $2n+2$ times by this path.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 22426,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a positive integer. An equilateral triangle with side length $n$ is divided into $n^2$ smaller equilateral triangles of side length $1$ (see the figure for $n=10$).\n\nInitially, one of these triangles is blue and the rest are yellow. The blue triangle does not touch the boundary of the large triangle. You are allowed to select any of the $n^2$ small triangles and change its color along with the colors of its neighbors (blue changes to yellow, yellow changes to blue).\n\nIs there any $n$ for which it is possible to make all $n^2$ equilateral triangles the same color?\n\n",
"options": [],
"answer": "See solution",
"solution": "No.\n\nLet $X$ be the number of pairs of adjacent (by side) triangles that have different colors. We will show that the parity of $X$ is invariant (see the figure below).\n\n\n\nConsider one step: suppose some triangle $a$ and its neighbors change color. Pairs involving $a$ do not affect $X$, because if they had the same color, they remain the same, and if they had different colors, after recoloring they still have different colors. Each neighbor of $a$ is adjacent to two more small triangles. If both of these had the same color, then after repainting, $X$ increases or decreases by $2$. If they had different colors, $X$ does not change. If it were possible to make all triangles the same color, the initial $X = 3$ would have to become $0$, which contradicts the invariant.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22427,
"subject": "Mathematics (Olympiad)",
"question": "Find all natural numbers $n$ for which there exist natural numbers $p$ and $q$ such that\n\n$$\n(n^2 + 2)^p = (2n - 1)^q.\n$$",
"options": [],
"answer": "See solution",
"solution": "It is clear that $n > 4$. It is easy to check that $n = 5$ satisfies the condition of the problem.\n\nSuppose that $n \\ge 6$. If $r$ is a prime divisor of $n^2 + 2$, then $r \\mid 2n-1$, and vice versa: if $r$ is a prime divisor of $2n-1$, then $r \\mid n^2 + 2$.\n\nTake a common prime divisor $r$ of $n^2 + 2$ and $2n-1$. We have $n^2 + 2 = r k$, $2n-1 = r l$ where $k, l$ are natural numbers. Then\n\n$$\n(2n)^2 + 8 = 4 r k, \\quad (r l + 1)^2 + 8 = 4 r k, \\quad r^2 l^2 + 2 r l + 9 = 4 r k\n$$\n\nand so $r \\mid 9$. Since $r$ is prime, we have $r=3$. We have $n^2+2=3^m$, $2n-1=3^s$ where $m, s$ are natural numbers and $m > s \\ge 3$. But these two identities imply\n\n$$\n(3^s + 1)^2 + 8 = 4 \\cdot 3^m, \\quad 3^{2s} + 2 \\cdot 3^s + 9 = 4 \\cdot 3^m.\n$$\n\nHence, $3^s \\mid 9$, which is impossible for $s \\ge 3$.\n\nTherefore, the only solution is $n = 5$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22428,
"subject": "Mathematics (Olympiad)",
"question": "There is a nonzero residue $\\alpha$ modulo $p$ such that\n$$\nS = \\{1, \\alpha^1, \\alpha^2, \\dots\\}\n$$\n(all considered mod $p$), and there are no $a, b, c \\in S$ (not necessarily distinct) such that\n$$\na + b \\equiv c \\pmod{p}.\n$$\n\nProve that for every integer $N$, there is a prime $p$ and a sum-free multiplicative subgroup $S$ of $\\mathbb{F}_p$ such that $|S| \\ge N$.",
"options": [],
"answer": "See solution",
"solution": "We prove a stronger statement, generalizing the desired condition \"$0 \\notin S + S - S$\" to \"$0 \\notin a_1S + a_2S + \\dots + a_kS$\", for fixed integers $a_1, \\dots, a_k$ with nonzero sum $a_1 + \\dots + a_k$. (In the original problem we have $(a_1, \\dots, a_k) = (1, 1, -1)$, so $k=3$.)\n\nFix a positive integer $N$ (to be specified later), and take a large prime $p \\equiv 1 \\pmod N$. (There are infinitely many such $p$ by a cyclotomic polynomial argument.)\n\nLet $\\alpha$ be an $N$th root of unity modulo $p$ (i.e., $\\alpha = g^{(p-1)/N}$ for a primitive root $g$, so $\\alpha$ has order $N$). The key is the following lemma:\n\n**Lemma.** If the sumset $a_1S + a_2S + \\dots + a_kS$ contains $0 \\pmod p$ for arbitrarily large primes $p \\equiv 1 \\pmod N$, then there exist indices $i_1, i_2, \\dots, i_k$ between $0$ and $N-1$ such that the polynomial\n$$\nf(x) = a_1x^{i_1} + \\dots + a_kx^{i_k}\n$$\nis divisible by the $N$th cyclotomic polynomial $\\Phi_N(x)$ (over $\\mathbb{Q}$, and thus $\\mathbb{Z}$).\n\n(We use the irreducibility of $\\Phi_N$, but only need special cases such as $N$ prime.)\n\n*Proof.* The sumset containing $0$ is equivalent to the existence of $0 \\le i_1, i_2, \\dots, i_k \\le N-1$ such that $f(\\alpha) \\equiv 0 \\pmod p$.\n\nOne way to proceed is to use the fact that in $\\mathbb{F}_p$, the roots of $\\Phi_N(x)$ are precisely the residues of order $N$; in particular, $p$ divides $\\Phi_N(\\alpha)$. By Bezout's identity, there exist integer polynomials $A(x), B(x)$ and a nonzero integer $C$ such that\n$$\nC \\gcd(\\Phi_N(x), f(x)) = A(x)f(x) + B(x)\\Phi_N(x).\n$$\nAssume for contradiction that $\\gcd(\\Phi_N(x), f(x))$ is constant, say $1$. Plugging in $\\alpha$, $p$ divides $C$ for arbitrarily large $p$, which is impossible. Thus $f(x)$ and $\\Phi_N(x)$ share a complex root, and by irreducibility, $f(x)$ is divisible by $\\Phi_N(x)$.\n\nAlternatively, by counting roots, in $\\mathbb{F}_p$ we have $x^N-1 \\equiv (x-\\alpha)\\dots(x-\\alpha^N)$. In $\\mathbb{C}$, $x^N-1 = (x-z)\\dots(x-z^N)$ for a primitive $N$th root $z$. The symmetric sums of $z, \\dots, z^N$ are congruent modulo $p$ to those of $\\alpha, \\dots, \\alpha^N$. Thus, the product of $f(\\alpha)$ over all valid indices is congruent modulo $p$ to the product of $f(z)$ over all valid indices. Arbitrarily large $p$ divide $\\prod_{[0,N-1]^k} f(z)$, so this product is $0$. Thus, for some indices, $f(z) = 0$, so $\\Phi_N(x)$ divides $f(x)$.\n\nWith the lemma, the rest is easy. Choose $N = q$ prime. Then $\\Phi_q(x) = (x^q-1)/(x-1)$ divides $f(x)$, a $k$-term polynomial of degree at most $q-1$. If $f$ is not identically zero and $q > k$, then the coefficients of $f$ must all be equal, but $i_1, \\dots, i_k$ cannot cover all $0, 1, \\dots, q-1$, so one coefficient is $0$, so all are $0$.\n\nThus $f$ is identically zero, so $f(1) = a_1 + \\dots + a_k = 0$, contradiction.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22429,
"subject": "Mathematics (Olympiad)",
"question": "Let $c$ be a non-negative integer. Find all sequences of positive integers $\\{a_n\\}_{n \\ge 1}$ such that for any positive integer $n$, the following condition holds:\n\nThere are exactly $a_n$ positive integers $i$ satisfying $a_i \\le a_{n+1} + c$.",
"options": [],
"answer": "See solution",
"solution": "Assume that $a_{n+1} \\ge a_{n+2}$ holds for some positive integer $n$. Since any positive integer $i$ with $a_i \\le a_{n+2} + c$ must also satisfy $a_i \\le a_{n+1} + c$, it follows that $a_n \\ge a_{n+1}$. Hence, if $a_n < a_{n+1}$ holds for some positive integer $n$, then $a_{n+1} < a_{n+2}$ follows, and by induction, we have $a_n < a_{n+1} < a_{n+2} < a_{n+3} < \\dots$.\n\nSimilarly, when $a_n > a_{n+1}$ holds for some positive integer $n$, we have $a_n > a_{n+1} > a_{n+2} > \\dots$, especially $a_{n+d} \\le a_n - d$ for any non-negative integer $d$. This leads to a contradiction since $a_{n+a_n} \\le 0$.\n\nSuppose that $a_n = a_{n+1}$ for all positive integers $n$. Then, as all integers $i \\ge 2$ satisfy $a_i = a_2 \\le a_2 + c$, this contradicts the fact that there are exactly $a_1$ positive integers $i$ satisfying it. Therefore, there exists a positive integer $k$ such that $a_k < a_{k+1}$, and from the above discussion, we have $a_1 \\le a_2 \\le \\dots \\le a_k < a_{k+1} < a_{k+2} < \\dots$.\n\nFor integers $n \\ge k$, since any integer $i > n+c+1$ satisfies $a_i > a_{n+c+1} \\ge a_{n+1} + c$, we have $a_n \\le n+c+1$. Therefore, if we set $b_n = a_n - n$ ($n \\ge k$), then we have $b_n \\le c+1$ and $b_k \\le b_{k+1} \\le b_{k+2} \\le \\dots$. Thus, there exist an integer $d$ and an integer $M \\ge k$ such that for $n \\ge M$, $b_n = d$, i.e., $a_n = n+d$. Since $a_1 \\le a_2 \\le \\dots \\le a_{M+c+1} < a_{M+c+2} < \\dots$, for any positive integer $i$, $a_i \\le a_{M+1}+c = a_{M+c+1}$ and $i \\le M+c+1$ are equivalent. Therefore, $a_M = M+c+1$, which is also equal to $M+d$, and for integers $n \\ge M$, we have $a_n = n+c+1$.\n\nSuppose that for an integer $N \\ge 2$, we have $a_n = n+c+1$ for all $n \\ge N$. Since $a_1 \\le a_2 \\le \\dots \\le a_{N+c} < a_{N+c+1} < \\dots$, for any positive integer $i$, $a_i \\le a_N + c = a_{N+c}$ and $i \\le N+c$ are equivalent. Therefore, $a_{N-1} = N+c$.\n\nThus, by induction, we have $a_n = n+c+1$ for any positive integer $n$. This indeed satisfies the problem condition because $a_i \\le a_{n+1} + c$ and $i \\le a_n$ are equivalent for any positive integer $i$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22430,
"subject": "Mathematics (Olympiad)",
"question": "Let $G = \\bigcup_{i=1}^{k} G_i$ be a graph that can be vertex-colored using no more than 2009 colors, where each $G_i$ is either a star or satisfies $e(G_i) \\le 17$ for $1 \\le i \\le k$. What is the maximum possible value of $k$?",
"options": [],
"answer": "See solution",
"solution": "Suppose $G_i$ is a star for $1 \\le i \\le j$, and $G_i$ satisfies $e(G_i) \\le 17$ for $j+1 \\le i \\le k$, where $0 \\le j \\le k$. If at least $j+10$ colors are needed to color $\\bigcup_{i=1}^{j} G_i$, then $(j+10) \\le 17j$. Since $\\left(j - \\frac{15}{2}\\right)^2 + \\frac{135}{4} \\le 0$, this gives a contradiction. Hence, it is possible to color $\\bigcup_{i=1}^{j} G_i$ using at most $j+9$ colors. Since we might need at most one more color every time we add a star, we conclude that $G$ can be colored using at most $k + 9$ colors.\n\nOn the other hand, if $k \\ge 9$, there is a case when $k+9$ colors are needed: use $9$ of the $G_i$ to form a $C_{18}$ and then use $k-9$ stars to complete this to a $C_{k+9}$.\n\nTherefore, the greatest value of $k$ for which 2009 colors suffice is $2000$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 22431,
"subject": "Mathematics (Olympiad)",
"question": "Consider all possible ways to parenthesize a product of $n$ ones, using only the binary operation $*$. List all distinct expressions that can be formed for $n = 1, 2, 3, 4$ such that each intermediate result is also a product of ones. How many such expressions are there for each $n$?",
"options": [],
"answer": "See solution",
"solution": "$$\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$$\n\nHowever, the expression $(((1 * 1) * 1) * 1) * (1 * 1)$ cannot be an intermediate result based on what has been shown earlier. Therefore, all expressions that can be transformed into $1$ are shown in the chain above. Only four of them are in the required form—for $n = 1, 2, 3, 4$.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 22432,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, $c$, $d$ be nonnegative real numbers such that $a + b + c + d = 4$. Prove that\n\n$$\n\\frac{a}{a^3+8} + \\frac{b}{b^3+8} + \\frac{c}{c^3+8} + \\frac{d}{d^3+8} \\leq \\frac{4}{9}.\n$$",
"options": [],
"answer": "See solution",
"solution": "By the means inequality, we have $a^3 + 2 = a^3 + 1 + 1 \\geq 3\\sqrt[3]{a^3 \\cdot 1 \\cdot 1} = 3a$. Therefore, it is sufficient to prove the inequality\n\n$$\n\\frac{a}{3a+6} + \\frac{b}{3b+6} + \\frac{c}{3c+6} + \\frac{d}{3d+6} \\leq \\frac{4}{9}.\n$$\n\nWe can write the last inequality as\n\n$$\n\\frac{1}{a+2} + \\frac{1}{b+2} + \\frac{1}{c+2} + \\frac{1}{d+2} \\geq \\frac{4}{3}.\n$$\n\nNow, by the harmonic and arithmetic means inequality:\n\n$$\n\\frac{1}{4} \\left( \\frac{1}{a+2} + \\frac{1}{b+2} + \\frac{1}{c+2} + \\frac{1}{d+2} \\right) \\geq \\frac{4}{(a+2)(b+2)(c+2)(d+2)} = \\frac{4}{4+2+2+2+2} = \\frac{1}{3}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22433,
"subject": "Mathematics (Olympiad)",
"question": "If $z > 2016$ is an integer, show that $z = 127a + 17b$ for some non-negative integers $a, b$.\n\nFurthermore, determine the greatest integer that cannot be expressed as $17x + 127y$ for non-negative integers $x, y$.\n\n**Comment:** More generally, if $p, q$ are positive integers with $\\gcd(p, q) = 1$, then $pq - p - q$ is the greatest integer which cannot be expressed as $px + qy$ with non-negative integers $x, y$.",
"options": [],
"answer": "See solution",
"solution": "To prove the claim, note that any integer $z$ can be written as $z = 127c + 17d$ for some integers $c, d$. For any integer $y$, $z = 127(c + 17y) + 17(d - 127y)$. Thus, $z = 127a + 17b$ for integers $a, b$ with $a \\ge 0$.\n\nChoose such a representation with $b$ as large as possible. If $b \\ge 0$, the claim is established. Suppose $b < 0$. Then $z = 127(a - 17) + 17(b + 127)$. By maximality of $b$, $a - 17 < 0$, so $a \\le 16$. Thus, $z = 127a + 17b \\le 127 \\cdot 16 - 17 < 2016$, contradicting $z > 2016$. Therefore, the claim holds.\n\nNow, to find the greatest integer not expressible as $17x + 127y$ with non-negative integers $x, y$, suppose $2015 = 17u + 127v$ for some non-negative $u, v$. Then $127(16 - v) = 17(u + 1)$. This implies $u + 1$ is divisible by $127$, so $u \\ge 126$, which gives $2015 \\ge 17 \\cdot 126$, which is false. Hence, $2015$ is the greatest such integer.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22434,
"subject": "Mathematics (Olympiad)",
"question": "Olesya and Andrew each throw a die once. What is the probability that the number of points Olesya throws is strictly greater than the number Andrew throws? Justify your answer.",
"options": [],
"answer": "See solution",
"solution": "There are $6^2 = 36$ possible outcomes.\n\n- If Olesya throws 1, she cannot have a higher number than Andrew.\n- If Olesya throws 2, Andrew must throw 1 (1 case).\n- If Olesya throws 3, Andrew can throw 1 or 2 (2 cases).\n- If Olesya throws 4, Andrew can throw 1, 2, or 3 (3 cases).\n- If Olesya throws 5, Andrew can throw 1, 2, 3, or 4 (4 cases).\n- If Olesya throws 6, Andrew can throw 1, 2, 3, 4, or 5 (5 cases).\n\nTotal favorable cases: $1 + 2 + 3 + 4 + 5 = 15$.\n\nThus, the probability is $$\\frac{15}{36} = \\frac{5}{12}.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22435,
"subject": "Mathematics (Olympiad)",
"question": "Suppose $x$ squares and $y$ $L$-shapes are used to cover an $11 \\times 11$ grid. Each square covers 4 cells, and each $L$-shape covers 3 cells, so the total number of cells covered is $4x + 3y = 121$.\n\n\n\n\n\nIf we color the cells as shown, then every square and every $L$-shape covers at most one blackened cell. Since there are 36 blackened cells, we have $x + y \\geq 36$.\n\nWhat is the minimum number of $L$-shapes needed?",
"options": [],
"answer": "See solution",
"solution": "Therefore, $y = 4(x + y) - (4x + 3y) \\geq 4 \\times 36 - 121 = 23$. The figure on the right gives one example using 23 $L$-shapes.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22436,
"subject": "Mathematics (Olympiad)",
"question": "Find all quadruples $(p, q, r, s)$ of primes that satisfy the following system of equations:\n\n$$\n\\begin{cases}\n6p + 5q + 5r + 3s = 130 \\\\\n3p + 3q + 5r + 6s = 130\n\\end{cases}\n$$",
"options": [],
"answer": "See solution",
"solution": "Subtracting the second equation from the first gives $3p + 2q - 3s = 0$, which implies $2q = 3(s - p)$. Thus, $2q$ is divisible by $3$. Since $2$ and $q$ are primes, this implies $q = 3$.\n\nSubstituting $q = 3$ into the system and simplifying gives:\n\n$$\n\\begin{cases}\n6p + 5r + 3s = 115 \\\\\n3p + 5r + 6s = 121\n\\end{cases}\n$$\n\nIf $r$ and $s$ were both odd, then $5r + 3s$ would be even, and $6p$ is even, so the left side of the first equation would be even, but $115$ is odd—a contradiction. Thus, one of $r$ or $s$ is even, i.e., $r = 2$ or $s = 2$.\n\nSimilarly, if $p$ and $r$ were both odd, the second equation would give a contradiction; hence $p = 2$ or $r = 2$. If $r \\neq 2$, then $p = s = 2$, but substituting $p = s$ into the system and subtracting the second equation from the first gives $0 = -6$, a contradiction. Thus, $r = 2$.\n\nSubstituting $r = 2$ into the system and simplifying gives:\n\n$$\n\\begin{cases}\n6p + 3s = 105 \\\\\n3p + 6s = 111\n\\end{cases}\n$$\n\nSolving these equations gives $p = 11$ and $s = 13$.\n\n**Final answer:** The quadruple $(p, q, r, s) = (11, 3, 2, 13)$ is the solution.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22437,
"subject": "Mathematics (Olympiad)",
"question": "Fix an equilateral triangle $ABC$ with side length $1$. We call $(\\triangle DEF, \\triangle XYZ)$ a *good triangle pair* if the points $D$, $E$, and $F$ lie in the interior of segments $BC$, $CA$, and $AB$, respectively, the points $X$, $Y$, and $Z$ lie on the lines $BC$, $CA$, and $AB$, respectively, and they satisfy the conditions\n\n$$\n\\frac{DE}{20} = \\frac{EF}{22} = \\frac{FD}{38}, \\quad \\text{and} \\quad DE \\perp XY,\\ EF \\perp YZ,\\ FD \\perp ZX.\n$$\n\nAs $(\\triangle DEF, \\triangle XYZ)$ runs through all good triangle pairs, determine all possible values of $\\frac{1}{S_{\\triangle DEF}} + \\frac{1}{S_{\\triangle XYZ}}$.",
"options": [],
"answer": "See solution",
"solution": "(1) First, consider the rotation of $90^\\circ$ (clockwise or counterclockwise), centered at an arbitrary point on the plane. Then the images $X'$, $Y'$, and $Z'$ of the points $X$, $Y$, and $Z$, respectively, satisfy\n\n$$\nX'Y' \\parallel DE, \\quad Y'Z' \\parallel EF, \\quad Z'X' \\parallel FD.\n$$\n\nIn this case, $\\triangle X'Y'Z'$ and $\\triangle DEF$ are directly similar. So, every good triangle pair is directly similar.\n\n\n\n(2) **Rotate $\\triangle XYZ$ (together with the equilateral triangle $ABC$) $90^\\circ$, and properly rescale and translate the picture so that the image of $\\triangle XYZ$ coincides with $\\triangle DEF$. Under this transformation, the points $A, B, \\text{and } C$ are mapped to points $A_1, B_1, \\text{and } C_1$. So there are three points $A_1, B_1, \\text{and } C_1$ on the plane satisfying:**\n\n- $\\triangle A_1B_1C_1$ is an equilateral triangle;\n- the points $D$, $E$, and $F$ lie on the lines $B_1C_1$, $C_1A_1$, and $A_1B_1$; and\n- $A_1B_1 \\perp AB$, $B_1C_1 \\perp BC$, and $C_1A_1 \\perp CA$.\n\nFrom this, we see that the points $A_1, B_1$, and $C_1$ lie on the circumcircles of $\\triangle AEF$, $\\triangle BFD$, and $\\triangle CDE$, respectively. Moreover, $A_1, B_1$, and $C_1$ are the antipodes of $A$, $B$, and $C$ in the corresponding circles; see the picture below.\n\n\n\n(3) Note that $S_{\\triangle ABC} : S_{\\triangle XYZ} = S_{\\triangle A_1B_1C_1} : S_{\\triangle DEF}$. So it suffices to compute the ratio of $S_{\\triangle ABC} + S_{\\triangle A_1B_1C_1}$ to $S_{\\triangle DEF}$. We have\n\n$$\nS_{\\triangle ABC} = S_{\\triangle DEF} + S_{\\triangle EAF} + S_{\\triangle FBD} + S_{\\triangle DCE}, \\\\\nS_{\\triangle A_1B_1C_1} = S_{\\triangle DEF} + S_{\\triangle EA_1F} + S_{\\triangle FB_1D} + S_{\\triangle DC_1E},\n$$\n\nHere the right hand sides are expressed in terms of oriented areas. **Since the two equilateral triangles on the left hand side are directly similar, the signs on the areas are the same.** Using the properties of antipodes, if we denote the circumcenters of $\\triangle AEF$, $\\triangle BFD$, and $\\triangle CDE$ by $O_1$, $O_2$, and $O_3$, respectively, then the condition \"$D, E$, and $F$ lie on the interior of three sides\" ensures that the three circumcenters $O_1$, $O_2$, and $O_3$ lie outside of $\\triangle DEF$. So we have the following equality of areas:\n\n$$\nS_{\\triangle ABC} + S_{\\triangle A_1B_1C_1} = 2(S_{\\triangle DEF} + S_{\\triangle EO_1F} + S_{\\triangle FO_2D} + S_{\\triangle DO_3E}).\n$$\n\nIn fact, $\\triangle O_1O_2O_3$ is the outer Napoleon triangle of $\\triangle DEF$; its area is exactly half the sum of the areas in the parentheses.\n\n(4) So we need to compute, for a triangle with side length ratio $20 : 22 : 38$, the ratio of the area of the hexagon formed by the vertices of the triangle and the vertices of the outer Napoleon triangle, to the area of the original triangle.\n\nBy Heron's formula, the area of a triangle with side lengths $20$, $22$, $38$ is\n\n$$\n\\sqrt{40 \\cdot (40 - 20) \\cdot (40 - 22) \\cdot (40 - 38)} = 120\\sqrt{2}.\n$$\n\nOn the other hand,\n\n$$\nS_{\\triangle EO_1F} + S_{\\triangle FO_2D} + S_{\\triangle DO_3E} = \\frac{\\sqrt{3}}{3} \\cdot (11^2 + 10^2 + 19^2) = 194\\sqrt{3}.\n$$\n\nSo\n\n$$\n\\begin{align*}\n& \\frac{1}{S_{\\triangle DEF}} + \\frac{1}{S_{\\triangle XYZ}} \\\\\n= & \\frac{1}{S_{\\triangle ABC}} \\left( \\frac{S_{\\triangle ABC}}{S_{\\triangle DEF}} + \\frac{S_{\\triangle ABC}}{S_{\\triangle XYZ}} \\right) \\\\\n= & \\frac{1}{S_{\\triangle ABC}} \\frac{S_{\\triangle ABC} + S_{\\triangle A_1B_1C_1}}{S_{\\triangle DEF}} \\\\\n= & \\frac{4}{\\sqrt{3}} \\frac{2(120\\sqrt{2} + 194\\sqrt{3})}{120\\sqrt{2}} \\\\\n= & \\frac{97\\sqrt{2} + 40\\sqrt{3}}{15}\n\\end{align*}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22438,
"subject": "Mathematics (Olympiad)",
"question": "In $\\triangle ABC$, $AB > AC$. Points $X$ and $Y$ inside $\\triangle ABC$ lie on the bisector of $\\angle BAC$, and satisfy $\\angle ABX = \\angle ACY$. Let the extension of $BX$ and the segment $CY$ intersect at point $P$. The circumcircle $\\omega_1$ of $\\triangle BPY$ and the circumcircle $\\omega_2$ of $\\triangle CPX$ intersect at $P$ and another point $Q$.\n\nProve that points $A$, $P$, and $Q$ are collinear.",
"options": [],
"answer": "See solution",
"solution": "Since $\\angle BAX = \\angle CAY$ and $\\angle ABX = \\angle ACY$, it follows that $\\triangle ABX \\sim \\triangle ACY$. Hence,\n\n$$\n\\frac{AB}{AC} = \\frac{AX}{AY}. \\qquad \\textcircled{1}\n$$\n\nExtend $AX$ and let it intersect circles $\\omega_1$ and $\\omega_2$ at points $U$ and $V$, respectively. Then\n\n$$\n\\begin{aligned}\n\\angle AUB &= \\angle YUB = \\angle YPB = \\angle YPX \\\\\n&= \\angle XVC = \\angle AVC,\n\\end{aligned}\n$$\n\nand thus $\\triangle ABU \\sim \\triangle ACV$. Therefore,\n\n$$\n\\frac{AB}{AC} = \\frac{AU}{AV}. \\qquad \\textcircled{2}\n$$\n\n\n\nFrom (1) and (2), we get $\\frac{AX}{AY} = \\frac{AU}{AV}$, i.e.,\n\n$$\nAU \\cdot AY = AV \\cdot AX.\n$$\n\nThe two sides of the above equation are the circle powers of point $A$ to circles $\\omega_1$ and $\\omega_2$, respectively. This implies that $A$ is on the radical axis (i.e., line $PQ$) of circles $\\omega_1$ and $\\omega_2$. In other words, points $A$, $P$, and $Q$ are collinear.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22439,
"subject": "Mathematics (Olympiad)",
"question": "Is there a triangle with sides of integer length such that the length of the shortest side is $2007$ and the largest angle is twice the smallest?",
"options": [],
"answer": "See solution",
"solution": "We shall prove that no such triangle satisfies the condition.\n\nIf $\\triangle ABC$ satisfies the condition, let $\\angle A \\leq \\angle B \\leq \\angle C$, then $\\angle C = 2\\angle A$, and $a = 2007$. Draw the bisector of $\\angle ACB$ which intersects $AB$ at point $D$. Then $\\angle BCD = \\angle A$, so $\\triangle CDB \\sim \\triangle ACB$. It follows that\n\n$$\n\\frac{CB}{AB} = \\frac{BD}{BC} = \\frac{CD}{AC} = \\frac{BD + CD}{BC + AC} = \\frac{AB}{BC + AC}.\n$$\n\nThus,\n\n$$\nc^2 = a(a + b) = 2007(2007 + b), \\quad \\text{(1)}\n$$\n\nwhere $2007 \\leq b \\leq c < 2007 + b$.\n\nSince $a, b, c$ are integers, $2007 \\mid c^2$, so $3 \\cdot 223 \\mid c^2$. Let $c = 669m$. From (1), we get $223m^2 = 2007 + b$. Thus $b = 223m^2 - 2007 \\geq 2007$, so $m \\geq 5$.\n\nBut $c \\geq b$, so $669m \\geq 223m^2 - 2007$, which implies $m < 5$, a contradiction.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22440,
"subject": "Mathematics (Olympiad)",
"question": "Во петокраката на цртежот важат следниве равенства:\n\n$$\n|AK| = |LC|, \\quad |BL| = |MD|, \\quad |CM| = |NE|, \\quad |DN| = |PA|\n$$\n\nПокажи дека важи и $|EP| = |KB|$.\n\n",
"options": [],
"answer": "See solution",
"solution": "Да ги повлечеме отсечките $AB$, $BC$, $CD$, $DE$, $EA$.\nОд дадените равенства во условот следува дека триаголниците $AKB$, $LCB$, $MCD$, $NED$, $PEA$ се еднакво-плоштни користејќи дека имаат по еден пар еднакви страни и заедничка соодветна висина. Но тогаш $\\triangle EPA$, $\\triangle KBA$ се еднакво-плоштни, и имаат заедничка висина од темето $A$, па затоа точно е равенството $|EP| = |KB|$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22441,
"subject": "Mathematics (Olympiad)",
"question": "Find all functions $f: \\mathbb{N} \\to \\mathbb{N}$ such that\n\n$$\nn + f(m) \\mid f(n) + n f(m)\n$$\n\nfor any $m, n \\in \\mathbb{N}$.",
"options": [],
"answer": "See solution",
"solution": "We consider two cases: whether the function $f$ has an infinite or finite range.\n\n*Case 1: $f$ has infinite range.*\n\nFix $n \\in \\mathbb{N}$ and let $m$ be arbitrary. From (1):\n\n$$\nn + f(m) \\mid f(n) + n f(m)\n$$\n\nRewrite:\n\n$$\nf(n) + n f(m) = f(n) - n^2 + n(f(m) + n)\n$$\n\nSo,\n\n$$\nn + f(m) \\mid f(n) - n^2\n$$\n\nSince $f$ has infinite range, we can choose $m$ so that $n + f(m) > |f(n) - n^2|$, forcing $f(n) = n^2$ for all $n$. Checking:\n\n$$\nn + f(m) = n + m^2,\n$$\n\n$$\nf(n) + n f(m) = n^2 + n m^2 = n(n + m^2),\n$$\n\nso $n + m^2 \\mid n(n + m^2)$ holds.\n\n*Case 2: $f$ has finite range.*\n\nThen there exists $k$ such that $1 \\leq f(n) \\leq k$ for all $n$. There is some $s$ with $f(n) = s$ for infinitely many $n$. For such $m, n$:\n\n$$\nn + s \\mid s + n s = s - s^2 + s(n + s)\n$$\n\nSo,\n\n$$\nn + s \\mid s^2 - s\n$$\n\nFor large enough $n$, $n + s > s^2 - s$, so $s^2 = s \\implies s = 1$. Thus, $f(n) = 1$ for infinitely many $n$.\n\nFix $m$, let $n$ with $f(n) = 1$:\n\n$$\nn + f(m) \\mid 1 + n f(m) = 1 - (f(m))^2 + f(m)(n + f(m))\n$$\n\nSo,\n\n$$\nn + f(m) \\mid (f(m))^2 - 1\n$$\n\nFor large $n$, $n + f(m) > (f(m))^2 - 1$, so $f(m) = 1$ for all $m$.\n\nChecking:\n\n$$\nn + f(m) = n + 1,\n$$\n\n$$\nf(n) + n f(m) = 1 + n,\n$$\n\nso $n + 1 \\mid n + 1$.\n\n**Conclusion:** The solutions are $f(n) = n^2$ and $f(n) = 1$ for all $n \\in \\mathbb{N}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22442,
"subject": "Mathematics (Olympiad)",
"question": "Let $d_{i,j}$ be the number of flights needed to get from city $A_i$ to city $A_j$, and let $d$ be the total number of flights in country \"U\". Define\n\n$$\nD = d_{1,1} + d_{1,2} + \\dots + d_{1,n} + d_{2,3} + d_{2,4} + \\dots + d_{2,n} + \\dots + d_{n-1,n}.\n$$\n\nShow that:\n\n$$\nd + \\frac{D}{n-1} \\le \\frac{1}{2}(n+1)(n+2).\n$$\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $i \\ne j$, and let the shortest path between $A_i$ and $A_j$ be $A_i = A_{\\gamma_0}, A_{\\gamma_1}, \\dots, A_{\\gamma_k} = A_j$. Let $X$ be the set of such vertices ($k+1$ vertices), and let $Y$ be the set of all other vertices ($n-k-1$ vertices). The number of edges connecting vertices inside $Y$ is at most $C_{n-k-1}^2$. All edges connecting vertices inside $X$ are the edges of the shortest path, so there are exactly $k$ edges.\n\nIf $A \\in Y$, then there are at most $3$ edges connecting $A$ with vertices from $X$, otherwise there would be a shorter path than the one with $k$ edges. Therefore, there are at most $3(n-k-1)$ edges connecting $X$ and $Y$. Thus,\n\n$$\n\\begin{aligned}\nd &\\le C_{n-k-1}^2 + k + 3(n-k-1) \\\\\n &= \\frac{1}{2}(n^2 + 3n - 4 + k^2 - k - 2kn) \\\\\n &\\le \\frac{1}{2}(n^2 + 3n - 4) - \\frac{n+1}{2}k \\quad (k < n, \\text{ so } kn < n^2) \\\\\n &\\Rightarrow d + \\frac{n+1}{2}d_{i,j} \\le \\frac{1}{2}(n^2 + 3n - 4) \\\\\n\\sum_{1 \\le i < j \\le n} \\left(d + \\frac{n+1}{2}d_{i,j}\\right) &\\le \\frac{1}{2}C_n^2(n^2 + 3n - 4) \\\\\nC_n^2d + \\frac{n+1}{2}D &\\le \\frac{1}{2}C_n^2(n^2 + 3n - 4)\n\\end{aligned}\n$$\n\nThe given inequality can be obtained by dividing both sides by $C_n^2$ and using\n\n$$\n\\frac{n+1}{n(n-1)} \\ge \\frac{1}{n-1} \\quad \\text{and} \\quad n^2 + 3n - 4 \\le n^2 + 3n + 2.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22443,
"subject": "Mathematics (Olympiad)",
"question": "A sphere $\\omega$ passes through a vertex $S$ of a triangular pyramid $SABC$ and intersects the edges $SA$, $SB$, and $AC$ at points $A_1 \\neq S$, $B_1 \\neq S$, and $C_1 \\neq S$, respectively. The sphere $\\omega$ intersects the circumsphere $\\Omega$ of $SABC$ in a circle lying in a plane parallel to $(ABC)$. The points $A_2$, $B_2$, and $C_2$ are symmetric to the points $A_1$, $B_1$, and $C_1$ with respect to the midpoints of the segments $SA$, $SB$, and $AC$, respectively. Prove that the points $A$, $B$, $C$, $A_2$, $B_2$, and $C_2$ lie on a sphere.",
"options": [],
"answer": "See solution",
"solution": "It suffices to prove that $SA_2 \\cdot SA = SB_2 \\cdot SB = SC_2 \\cdot SC$, or equivalently, $AA_1 \\cdot AS = BB_1 \\cdot BS = CC_1 \\cdot CS$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 22444,
"subject": "Mathematics (Olympiad)",
"question": "For each positive integer $n$, consider the set $A_n$ of all numbers obtained by choosing signs in $\\pm 1 \\pm 2 \\pm \\cdots \\pm n$. For example, $A_2 = \\{-3, -1, 1, 3\\}$ and $A_3 = \\{-6, -4, -2, 0, 2, 4, 6\\}$. Find the cardinality of the set $A_n$.",
"options": [],
"answer": "See solution",
"solution": "The largest element of $A_n$ is $1 + 2 + 3 + \\cdots + n = \\frac{n(n+1)}{2}$, and the smallest is $-1 - 2 - \\cdots - n = -\\frac{n(n+1)}{2}$.\n\nThe difference between any two elements of $A_n$ is even, so all elements have the same parity.\n\nWe claim that all numbers between $-\\frac{n(n+1)}{2}$ and $\\frac{n(n+1)}{2}$, sharing the same parity, belong to $A_n$—and only those. Indeed, let $x \\in A_n$, $x < \\frac{n(n+1)}{2}$. If one of its representations begins with $-1$, changing it to $+1$ gives $x + 2 \\in A_n$. Suppose all representations start with $+1$. There must be a first negative; otherwise, $x = 1 + 2 + \\cdots + n = \\frac{n(n+1)}{2}$. Denote it by $-k$:\n\n$$\nx = +1 + 2 + \\cdots + (k-1) - k \\pm \\cdots \\pm n.\n$$\n\nSwitching the signs of $k-1$ and $k$ gives $x + 2 \\in A_n$.\n\nConsequently, $A_n$ has $\\frac{n(n+1)}{2} + 1$ elements.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22445,
"subject": "Mathematics (Olympiad)",
"question": "Let $AB$ and $CD$ be two diameters of the circle $C$. For an arbitrary point $P$ on $C$, let $R$ and $S$ be the feet of the perpendiculars from $P$ to $AB$ and $CD$, respectively. Show that the length of $RS$ is independent from the choice of $P$.",
"options": [],
"answer": "See solution",
"solution": "Let $O$ be the centre of $C$. Then $P$, $R$, $S$, and $O$ are points on a circle $C'$ with diameter $OP$, equal to the radius of $C$. The segment $RS$ is a chord in this circle subtending the angle $AOC$ or its supplementary angle. Since the angle as well as the radius of $C'$ are independent of $P$, so is $RS$.\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22446,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $n$ for which checkers can be placed on the cells of an $n \\times n$ chessboard in such a way that each cell has exactly two neighboring cells with checkers. Two cells are considered neighbors if they share a common side.",
"options": [],
"answer": "See solution",
"solution": "Answer: $n$ even.\n\nLet's start by proving that any even number $n$ is a good number.\n\nFor the $n = 2k$ case, we can construct a checkered pattern so that all the cells of the frame contain checkers, and the remaining $2k - 4$ case is in the middle. The figure below illustrates this construction:\n\n\n\nNow, let's show that for an odd number $n = 2k + 1$, each cell of the board cannot have exactly two checkered neighbors.\n\nTo demonstrate this, we will focus on the diagonal cells of the table. Consider the neighbors of the cells $(i, i)$ for $i \\in [n]$.\n\nStarting with cell $(1, 1)$, since it has exactly two neighbors, we place checkers in those neighboring cells, which are $(1, 2)$ and $(2, 1)$. Now, these cells become neighbors of cell $(2, 2)$, so we do not place checkers in cells $(2, 3)$ and $(3, 2)$. We continue this process for cells $(1+2j, 2+2j)$ and $(2+2j, 1+2j)$, where $j$ ranges from $0$ to $k-2$. As a result, cells $(1+2j, 2+2j)$ and $(2+2j, 1+2j)$ have checkers, while cells $(2+2j, 3+2j)$ and $(3+2j, 2+2j)$ do not contain checkers.\n\nThe figure below illustrates this construction:\n\n\n\nNow, consider cells $(2k, 2k+1)$ and $(2k+1, 2k)$. Since these cells are neighbors of the corner cell $(2k+1, 2k+1)$, they should be checkered. However, we have already established that these cells cannot be checkered. This leads to a contradiction.\n\nTherefore, the number $n = 2k + 1$ is not a good number.\n\nIn conclusion, we have shown that an even number is a good number, while an odd number is not a good number.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 22447,
"subject": "Mathematics (Olympiad)",
"question": "Define $S(n, r) := \\frac{n+1-2r}{n+1-r} \\binom{n}{r}$ for all pairs $r, n$ with $-1 \\leq r$ and $2r \\leq n+1$. In particular, $S(1, 0) = 1$.\n\nProve that if $n \\geq 2$ and $0 \\leq r \\leq n/2$, then\n\n$$\nS(n, r) = S(n-1, r-1) + S(n-1, r)\n$$\n\nAlso, show that\n\n$$\n\\sum_{r=0}^{\\lfloor n/2 \\rfloor} S(n, r) < 2^{n-2}\n$$\nfor $n \\geq 9$.\n\n*Note:* $S(n, r)$ is the dimension of the Specht module for the two-part partition $n = (n-r) + r$ of the symmetric group $S_n$. The recursion follows from the Branching rule for Specht modules. The sum estimate relates to truncating Pascal's triangle, whose row sum is $2^n$.",
"options": [],
"answer": "See solution",
"solution": "First, we prove the recursion:\n\nFor $n \\geq 2$ and $0 \\leq r \\leq n/2$,\n\n$$\n\\begin{align*}\nS(n-1, r-1) + S(n-1, r) &= \\frac{n+2-2r}{n+1-r} \\frac{(n-1)!}{(r-1)!(n-r)!} + \\frac{n-2r}{n-r} \\frac{(n-1)!}{r!(n-1-r)!} \\\\\n&= \\frac{[r(n+2-2r) + (n-2r)(n+1-r)](n-1)!}{(n+1-r) r! (n-r)!} \\\\\n&= \\frac{[(n+1-2r)n](n-1)!}{(n+1-r) r! (n-r)!} \\\\\n&= S(n, r)\n\\end{align*}\n$$\n\nThe recursion holds by direct computation.\n\nFor the sum estimate, base case $n=9$:\n\n$S(9, r)$ for $r=0,1,2,3,4$ are $1, 8, 27, 48, 42$, summing to $126 < 2^{9-2} = 128$.\n\nInductive step:\n\n$$\n\\sum_{r=0}^{\\lfloor n/2 \\rfloor} S(n, r) = \\sum_{r=0}^{\\lfloor n/2 \\rfloor} [S(n-1, r) + S(n-1, r-1)] \\leq 2 \\sum_{r=0}^{\\lfloor (n-1)/2 \\rfloor} S(n-1, r) < 2 \\times 2^{\\lfloor (n-1)/2 \\rfloor} = 2^{n-2}\n$$\n\nEach $S(n-1, r-1)$ with $r < n/2$ contributes twice; when $r = n/2$, $S(n-1, n/2-1)$ contributes only once.\n\n**Alternative expressions for $S(n, r)$:**\n\n(a)\n$$\n\\begin{align*}\nS(n, r) &= \\binom{n}{r} - \\binom{n}{r-1}\n\\end{align*}\n$$\n\n(b)\n$$\n\\begin{align*}\nS(n, r) &= \\binom{n+1}{r} - 2 \\binom{n}{r-1}\n\\end{align*}\n$$\n\nThus, $S(n, r)$ is an integer and nonnegative if $2r \\leq n+1$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22448,
"subject": "Mathematics (Olympiad)",
"question": "Consider 70-digit numbers $n$ with the property that each of the digits $1, 2, 3, \\dots, 7$ appears in the decimal expansion of $n$ ten times (and $8$, $9$, and $0$ do not appear). Show that no number of this form can divide another number of this form.",
"options": [],
"answer": "See solution",
"solution": "Assume the contrary: there exist $a$ and $b$ of the prescribed form, such that $b \\geq a$ and $a$ divides $b$. Then $a$ divides $b - a$.\n\n**Claim:** $a$ is not divisible by $3$, but $b - a$ is divisible by $9$. Indeed, the sum of the digits is $10(1 + \\cdots + 7) = 280$ for both $a$ and $b$. (Recall that an integer $n$ is congruent to the sum of its digits modulo $3$ and modulo $9$.)\n\nWe conclude that $b - a$ is divisible by $9a$. But this is impossible, since $9a$ has $71$ digits and $b$ has only $70$ digits, so $9a > b > b - a$. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22449,
"subject": "Mathematics (Olympiad)",
"question": "令 $ABC$ 為一三角形，其外心為 $O$。圓 $\\Gamma$ 分別與 $OB$ 和 $OC$ 相切於 $B$ 和 $C$。令 $D$ 為 $\\Gamma$ 上異於 $B$ 的一點，使得 $CB = CD$。令 $E$ 為 $DO$ 和 $\\Gamma$ 異於 $D$ 的交點，而 $F$ 為 $EA$ 和 $\\Gamma$ 異於 $D$ 的交點。令 $X$ 為 $AC$ 上一點，使得 $XB \\perp BD$。證明 $\\angle ADF$ 的一半等於 $\\angle BDX$ 或 $\\angle BXD$。",
"options": [],
"answer": "See solution",
"solution": "由於 $\\angle ADF$ 是 $\\Gamma$ 與 $\\odot(ADE)$ 之間的夾角，考慮圓 $\\odot(ADE)$ 及其與 $\\odot(ABC)$ 的第二個交點 $A'$。設 $DA'$ 與 $BX$ 交於 $X'$。記 $\\angle ADF$ 為 $\\theta$。我們先證明以下兩個引理。\n\n**引理 1.** $(A, A'; B, C) = \\tan^2\\left(\\frac{1}{2}\\theta\\right)$ 或 $\\cot^2\\left(\\frac{1}{2}\\theta\\right)$。\n\n*證明.* 注意到 $\\odot(ABC)$ 與 $\\Gamma$ 及 $\\odot(ADE)$ 都正交，因為 $OA^2 = OC^2 = OD \\cdot OE$。因此，對 $D$ 作反演，設點 $P$ 的像為 $P_*$，則 $A_*A_*'$、$B_*C_*$ 是一圓的直徑，且夾角為 $\\theta$。因此有：\n\n$$\n(A, A'; B, C) = (A_*, A_*'; B_*, C_*) = \\tan^2\\left(\\frac{1}{2}\\theta\\right) \\text{ 或 } \\cot^2\\left(\\frac{1}{2}\\theta\\right)。\n$$\n\n**引理 2.** $DX \\perp DX'$。\n\n*證明.* 以 $C$ 為反演中心，半徑 $CB = CD$。此反演將 $\\odot(ABC)$ 變為通過 $B$ 且垂直於 $CO$ 的直線，即 $BXX'$。這說明反演將 $A$ 變為 $X$，$A'$ 變為 $X'$。此外，反演將 $\\Gamma$ 變為 $BD$，因此 $E$ 變為 $D'$，其中 $D'$ 是 $D$ 關於 $B$ 的對稱點，因為 $(E, D; B, C)$ 為調和。由於 $XX' \\perp DD'$，$XX'$ 是 $DD'$ 的垂直平分線。又因 $A, A', D, E$ 共圓，故 $X, X', D, D'$ 亦共圓。因此 $DX \\perp DX'$。\n\n為證原命題，注意由 $C$ 的投影性，有：\n\n$$\n\\tan^2\\left(\\frac{1}{2}\\theta\\right) \\text{ 或 } \\cot^2\\left(\\frac{1}{2}\\theta\\right) = (A, A'; B, C) = \\frac{XB}{BX'} = \\cot^2 \\angle BDX\n$$\n\n其中第一個等號來自引理一，最後一個等號來自引理二。因此，$\\theta/2 = \\angle BDX$ 或 $\\theta/2 = 90^\\circ - \\angle BDX = \\angle BXD$，得證。",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22450,
"subject": "Mathematics (Olympiad)",
"question": "Let $n$ be a natural number. A sequence $x_1, x_2, \\dots, x_{n^2}$ is called *n-good* if each $x_i$ is an element of $\\{1, 2, \\dots, n\\}$ and the ordered pairs $(x_i, x_{i+1})$ are all different for $i = 1, 2, \\dots, n^2$ (here we consider the subscripts modulo $n^2$).\n\nTwo *n-good* sequences $x_1, x_2, \\dots, x_{n^2}$ and $y_1, y_2, \\dots, y_{n^2}$ are called *similar* if there exists an integer $k$ such that $y_i = x_{i+k}$ for all $i = 1, 2, \\dots, n^2$ (again taking the subscripts modulo $n^2$).\n\nSuppose that there exists a non-trivial permutation $\\sigma$ of $\\{1, 2, \\dots, n\\}$ and an *n-good* sequence $x_1, x_2, \\dots, x_{n^2}$ which is similar to $\\sigma(x_1), \\sigma(x_2), \\dots, \\sigma(x_{n^2})$. Show that $n \\equiv 2 \\pmod{4}$.",
"options": [],
"answer": "See solution",
"solution": "Without loss of generality, assume $\\sigma(1) \\neq 1$ and $x_1 = x_2 = 1$. Let $k$ be the smallest natural number such that $\\sigma^k(1) = 1$. Let $r$ be the smallest natural number such that $\\sigma(x_i) = x_{i+r}$ for all $i = 1, 2, \\dots, n$. Therefore, $\\sigma^k(x_i) = x_{i+kr}$.\n\nSince $x_1 = x_2 = 1$, it follows that $1 + kr \\equiv 1 \\pmod{n^2}$, so $n^2$ divides $kr$. Further, $\\sigma^k$ is the identity permutation. By considering the pairs $(a, a)$ in the sequence, for any $i$ and $1 \\leq j < k$, we have $\\sigma^j(i) \\neq i$. Thus, $\\sigma$ consists of $n/k$ $k$-cycles. Let $l = n/k$. Without loss of generality, assume $\\sigma(i) = i + l$.\n\nConsider $\\{r, 2r, 3r, \\dots, (k-1)r\\} \\pmod{n^2}$. Let $s$ be the smallest natural number such that $s \\equiv jr \\pmod{n^2}$. Replacing $\\sigma$ by $\\sigma^j$, we may assume $s = r$. It then follows that $kr = n^2$, so $r = nl$.\n\nFor each $i = 1, 2, \\dots, n^2$, let $a_i$ be an integer such that $0 \\leq a_i \\leq n-1$ and $a_i \\equiv x_{i+1} - x_i \\pmod{n^2}$. For any $j = 0, 1, \\dots, n-1$, there are exactly $n$ values of $i$ for which $a_i = j$. Since $\\sigma(i) = i + l$, it follows that $a_{i+r} = a_i$. Therefore, for any $j = 0, 1, \\dots, n-1$, there are exactly $l$ values of $i$ with $1 \\leq i \\leq r$ and $a_i = j$. Hence, $l = x_{r+1} - x_1 \\equiv \\sum_{i=1}^r a_i \\equiv l n (n-1)/2 \\pmod{n}$.\n\nIf $n$ is odd, then $l = n$. If $n$ is even, then $n/2$ divides $l$. If $l$ is even, then again $l = n$. Since $k > 1$, it follows that $l < n$, and hence $n \\equiv 2 \\pmod{4}$.",
"topic": "Number Theory",
"subtopic": "Modular Arithmetic"
},
{
"id": 22451,
"subject": "Mathematics (Olympiad)",
"question": "Alice and Bob play a game on a strip of $n \\ge 3$ squares with two game pieces. At the beginning, Alice's piece is on the first square while Bob's piece is on the last square. The figure shows the starting position for a strip of $n = 7$ squares.\n\n\n\nThe players alternate. In each move, they advance their own game piece by one or two squares in the direction of the opponent's piece. The piece has to land on an empty square without jumping over the opponent's piece. Alice makes the first move with her own piece. If a player cannot move, they lose.\n\nFor which $n$ can Bob ensure a win no matter how Alice plays?\n\nFor which $n$ can Alice ensure a win no matter how Bob plays?",
"options": [],
"answer": "See solution",
"solution": "Bob wins for $n = 3k + 2$ with $k \\in \\mathbb{Z}_{\\ge 1}$, and Alice wins for all other $n \\ge 3$.\n\nIt is easily checked that Alice wins for $n = 3, 4, 6, 7$ squares, while Bob wins for $n = 5$ or $8$ squares. We conjecture that Bob wins for all $n$ of the form $3k + 2$ and prove it by induction.\n\nWe include the case $k = 0$, which is obvious because Alice loses immediately.\n\nAssume that Bob can ensure a win for $3k + 2$ squares for some $k \\ge 0$. We want to prove that he can ensure a win for $3k + 5$ squares.\n\nIt is enough that Bob makes exactly the opposite move of Alice after her first move: if she moves by $1$, he moves $2$; if she moves by $2$, he moves $1$. This ensures that the distance between the two game pieces is reduced by $3$, and the game continues as if it were a new game with $3k + 2$ squares, where we already know that Bob can ensure a win.\n\nTherefore, Bob can win for all $n = 3k + 2$.\n\nNow, it remains to show that Alice can win for all $n$ of the form $3k$ and $3k + 1$.\n\nFor $3k$ squares, she starts by moving $1$, so the remaining game is played on $3k - 1 = 3(k - 1) + 2$ squares with Bob making the first move. So Alice, as the second player, can ensure a win.\n\nFor $3k + 1$, Alice starts with $2$, which again reduces the game to $3(k - 1) + 2$ squares with Bob making the first move.\n\nThus, Bob can ensure a win for all $n = 3k + 2$, and Alice can ensure a win for all other $n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22452,
"subject": "Mathematics (Olympiad)",
"question": "$60\\,m^2$ шалыг тус бүр нь $30\\,m^2$ гурван хивсээр хучив. Ямар ч тохиолдолд 2 хивс $10\\,m^2$-с багагүй хэмжээтэй шалны хэсгийг хамарсан гэдгийг батал.",
"options": [],
"answer": "See solution",
"solution": "Хивс бүрийг $A_1$, $A_2$, $A_3$ гэж тэмдэглэе. Тэгвэл:\n\n$$\n|A_1| = |A_2| = |A_3| = 30\n$$\n\nМөн нийт шалны талбай $|A_1 \\cup A_2 \\cup A_3| \\leq 60$.\n\nОруулгын томъёогоор:\n\n$$\n|A_1 \\cup A_2 \\cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \\cap A_2| - |A_2 \\cap A_3| - |A_3 \\cap A_1| + |A_1 \\cap A_2 \\cap A_3|\n$$\n\n$|A_1 \\cap A_2 \\cap A_3| \\geq 0$ тул:\n\n$$\n60 \\geq 90 - (|A_1 \\cap A_2| + |A_2 \\cap A_3| + |A_3 \\cap A_1|)\n$$\n\n$$\n|A_1 \\cap A_2| + |A_2 \\cap A_3| + |A_3 \\cap A_1| \\geq 30\n$$\n\nГурван нийлбэрийн аль нэг нь $10$-аас бага байж болохгүй, эс тэгвээс нийлбэр нь $30$-аас бага болно. Тиймээс ямар ч хоёр хивсний давхцсан талбай $10\\,m^2$-с багагүй байна.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22453,
"subject": "Mathematics (Olympiad)",
"question": "Find all integer $n$ that satisfy the following equality:\n\n$$\n(n-1)(n-3)(n-5)\\dots(n-2011) = n(n+2)(n+4)\\dots(n+2010).\n$$",
"options": [],
"answer": "See solution",
"solution": "If $n$ is even, then the left-hand side (LHS) is odd and the right-hand side (RHS) is even, and vice versa for odd $n$. Therefore, no such integer $n$ exists.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22454,
"subject": "Mathematics (Olympiad)",
"question": "Prove that it is possible to pick 20 numbers among $1, 2, \\ldots, 10000$ such that the members of any non-empty subset of these 20 numbers have a sum which is not an $n$-th power of any number (for any $n > 1$).",
"options": [],
"answer": "See solution",
"solution": "Note that 19 and 23 are primes and $19 \\times 23 = 437$. Let $S = \\{1, 2, 3, \\ldots, 20\\}$, and consider $T = \\{437a \\mid a \\in S\\} \\subset \\{1, 2, \\ldots, 10^4\\}$. So $|T| = 20$. Note that $1 + 2 + \\cdots + 20 = 210 < 437$.\n\nSuppose there exist $k$ numbers in $T$, say $437a_1, 437a_2, \\ldots, 437a_k$, such that their sum is a perfect power. Then\n\n$$\n437(a_1 + a_2 + \\cdots + a_k) = a^n\n$$\n\nfor some positive integer $a$ and $n \\ge 2$. Since $437 \\mid a^n$, it follows that $437^2 \\mid a^n$, which implies\n\n$$\n437 \\mid a_1 + a_2 + \\cdots + a_k.\n$$\n\nBut the sum of all numbers in $S$ is less than 437, so this is impossible. Thus, the set $T$ satisfies the given condition. $\\square$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22455,
"subject": "Mathematics (Olympiad)",
"question": "Let $ABC$ be an isosceles triangle with $AC = BC$ and $\\angle ACB < 60^\\circ$. We denote the incenter and circumcenter by $I$ and $O$, respectively. The circumcircle of triangle $BIO$ intersects the leg $BC$ also at point $D \\neq B$.\n\n(a) Prove that the lines $AC$ and $DI$ are parallel.\n\n(b) Prove that the lines $OD$ and $IB$ are mutually perpendicular.",
"options": [],
"answer": "See solution",
"solution": "Note that the condition $\\angle ACB < 60^\\circ$ guarantees that $O$ lies between $I$ and $C$.\n\n**(a)**\nLet the angles of triangle $ABC$ be $\\alpha = \\angle BAC$, $\\beta = \\angle ABC$, and $\\gamma = \\angle ACB$. Let $K$ and $k$ be the circumcircles of $ABC$ and $BIO$, respectively. By the inscribed angle theorem for circle $K$, $\\angle BOC = 2\\alpha$. Therefore, $\\angle IOB = 180^\\circ - 2\\alpha$. Since $\\alpha = \\beta$, we have $\\angle IOB = \\gamma$. Furthermore, by the inscribed angle theorem for circle $k$, $\\angle IDB = \\gamma$, so $ID \\parallel AC$.\n\n**(b)**\nLet $F$ be the intersection point of lines $OD$ and $IB$, and let $G$ be the midpoint of $AB$. Since $IODB$ is cyclic, $\\angle IOD = 180^\\circ - \\beta/2$, so $\\angle DOC = \\beta/2$, or equivalently $\\angle FOI = \\beta/2$. Furthermore, $\\angle GIB = 90^\\circ - \\beta/2$ implies $\\angle OIF = 90^\\circ - \\beta/2$. Therefore, $\\angle IFO = 90^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22456,
"subject": "Mathematics (Olympiad)",
"question": "For integers $a$, $b$, $m$, write $a \\equiv b \\pmod m$ if $a - b$ is divisible by $m$.\n\nFor any integer $n$, compute the last two digits of the product\n$$\nX = (3 \\times 7) \\cdot (13 \\times 17) \\cdots (2003 \\times 2007).\n$$",
"options": [],
"answer": "See solution",
"solution": "For any integer $n$,\n$$\n(10n + 3)(10n + 7) = 100n^2 + 100n + 21 \\equiv 21 \\pmod{100}.\n$$\n\nWe also have\n$$\n21^5 = (20 + 1)^5 = \\sum_{k=0}^{5} \\binom{5}{k} 20^k \\equiv \\binom{5}{1} 20^1 + \\binom{5}{0} 20^0 \\equiv 1 \\pmod{100}.\n$$\n\nThen,\n$$\n\\begin{aligned}\nX &= (3 \\times 7) \\cdot (13 \\times 17) \\cdots (2003 \\times 2007) \\\\\n &\\equiv 21^{201} \\\\\n &\\equiv (21^5)^{40} \\times 21 \\\\\n &\\equiv 1^{40} \\times 21 \\\\\n &\\equiv 21 \\pmod{100}.\n\\end{aligned}\n$$\n\nThus, the last two digits of $X$ are $21$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22457,
"subject": "Mathematics (Olympiad)",
"question": "Suppose there do not exist 27 circles such that every pair of circles intersects.\n\nConsider a coordinate plane such that the line joining any pair of centres of the circles is not parallel to the coordinate axes. We label the circles as $\\Gamma_1, \\Gamma_2, \\dots, \\Gamma_{2016}$ such that the x-coordinate of the centre of $\\Gamma_i$ is less than that of $\\Gamma_j$ if $i < j$.\n\nIs it possible to guarantee that there exist 27 circles among $\\Gamma_1, \\Gamma_2, \\dots, \\Gamma_{2016}$ that are pairwise disjoint?",
"options": [],
"answer": "See solution",
"solution": "Consider one of the circles $\\Gamma_k$. Let $\\Gamma$ be the circle with radius 2 and having the same centre as $\\Gamma_k$. Partition the left semicircle of $\\Gamma$ into 3 sectors of $60^\\circ$. Then the distance between any two points in the same sector is at most 2. If there are 26 centres of the other circles belonging to the same sector, then these circles together with $\\Gamma_k$ satisfy the condition, since the distance between any two of these centres is at most 2, which shows the two circles intersect. Thus, we may assume there are at most 25 centres in each sector. This implies at most 75 centres among $\\Gamma_1, \\Gamma_2, \\dots, \\Gamma_{k-1}$ lie in $\\Gamma$, or equivalently at most 75 of these circles intersect $\\Gamma_k$.\n\nNow, we colour the circles $\\Gamma_1, \\Gamma_2, \\dots, \\Gamma_{2016}$ one by one in 76 colours. Suppose we have coloured $\\Gamma_1, \\Gamma_2, \\dots, \\Gamma_k$. Since $\\Gamma_{k+1}$ intersects at most 75 of the previous circles, we can colour it in a way such that its colour is different from the colours of all circles among $\\Gamma_1, \\Gamma_2, \\dots, \\Gamma_k$ which intersect $\\Gamma_{k+1}$. By the pigeonhole principle, we can find\n\n$$\n\\left\\lfloor \\frac{2016}{76} \\right\\rfloor = 27\n$$\n\ncircles having the same colour. This means these 27 circles are pairwise disjoint. So we are done.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22458,
"subject": "Mathematics (Olympiad)",
"question": "As shown in Fig. 1.1, the quadrilateral $ABCD$ is inscribed in the circle $\\Gamma$, and $AB + BC = AD + DC$. Let $E$ be the midpoint of the arc $BCD$; $F$ be the antipode of $A$ ($F \\neq C$); $I$ be the incentre of $\\angle ABC$; $J$ be the excentre of $\\angle ABC$ against $\\angle BAC$; $K$ is the incentre of $\\angle BCD$. Let $P$ satisfy that $\\triangle BIC$ and $\\triangle KPJ$ are similar triangles with their corresponding vertices in order. Prove that the lines $EK$, $PF$ intersect on $\\Gamma$.",
"options": [],
"answer": "See solution",
"solution": "As illustrated in Fig. 1.2, let $L$ be the excentre of $\\triangle ACD$ relative to $A$. Assume that the inscribed circles $\\odot J$ of $\\triangle ABC$, $\\odot L$ of $\\triangle ACD$ meet the extension of $AC$ at $M_1, M_2$, respectively (not shown in the figure). Then $CM_1 = \\frac{AB + BC - AC}{2}$, $CM_2 = \\frac{AD + DC - AC}{2}$. Since $AB + BC = CD + DA$, it follows that $CM_1 = CM_2$, $M_1 = M_2$, and $JL \\perp AC$.\n\nFirst, we show that $P$ is the orthocentre of $\\triangle AJL$, and hence $A, C, P$ are collinear. It is easy to see that $B, I, K, C, J$ are concyclic. Since $\\angle KJP = \\angle BCI = \\frac{1}{2}\\angle ACB$, it follows that\n\n$$\n\\begin{aligned}\n\\angle IJK &= \\angle IBK = \\angle IBC - \\angle KBC \\\\\n&= \\frac{1}{2}\\angle ABC - \\frac{1}{2}\\angle DBC = \\frac{1}{2}\\angle ABD.\n\\end{aligned}\n$$\n\nThus,\n\n$$\n\\begin{aligned}\n\\angle AJP &= \\angle IJK + \\angle KJP \\\\\n&= \\frac{1}{2}\\angle ABD + \\frac{1}{2}\\angle ACB \\\\\n&= \\frac{1}{2}\\angle BCD,\n\\end{aligned}\n$$\n\n\n\nwhile $\\angle JAL = \\frac{1}{2}\\angle BAD$, it follows that $\\angle AJP + \\angle JAL = 90^\\circ$, and $JP \\perp AL$.\n\nMeanwhile, as $CI \\perp CJ$, $AC \\perp JL$, hence $\\angle CJL = \\angle ACI = \\angle BCI = \\angle KJP$, and $JC, JP$ are isogonal lines of $\\angle KJL$. As $C, K, D, L$ lie on a circle, we have\n\n$$\n\\begin{aligned}\n\\angle LKC &= \\angle LDC = 90^\\circ - \\frac{1}{2}\\angle ADC \\\\\n &= \\frac{1}{2}\\angle ABC = \\angle CBI = \\angle JKP,\n\\end{aligned}\n$$\n\nand $KC, KP$ are isogonal lines of $\\angle JKL$. Therefore, $C$ and $P$ are isogonal conjugate points in $\\triangle KJL$, implying that\n\n$$\n\\begin{aligned}\n\\angle PLJ &= \\angle KLC = \\angle KDC \\\\\n &= \\frac{1}{2}\\angle BDC = \\frac{1}{2}\\angle BAC \\\\\n &= \\angle JAC.\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22459,
"subject": "Mathematics (Olympiad)",
"question": "Let\n\n$$\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$$\n\nFind the value of $x_{1006}$.",
"options": [],
"answer": "See solution",
"solution": "From $\\frac{x_k}{x_k + (2k-1)} = a$, it follows that $x_k = \\frac{a}{1-a} \\cdot (2k-1)$ for $k = 1, 2, \\dots, 1006$.\n\nSubstituting into the last equality:\n\n$$\nx_{1006} = \\frac{a}{1-a} \\cdot 2011.\n$$\n\nThe sum $1 + 3 + 5 + \\dots + 2011 = 1006^2$ (since it's the sum of the first 1006 odd numbers).\n\nGiven $\\frac{a}{1-a} \\cdot 1006^2 = 503^2$, so $\\frac{a}{1-a} = \\frac{1}{4}$.\n\nTherefore,\n\n$$\nx_{1006} = \\frac{2011}{4}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22460,
"subject": "Mathematics (Olympiad)",
"question": "Ward and Gabrielle are playing a game on a large sheet of paper. At the start of the game, there are $999$ ones on the sheet of paper. Ward and Gabrielle each take turns alternately, and Ward has the first turn. During their turn, a player must pick two numbers $a$ and $b$ on the sheet such that $\\gcd(a, b) = 1$, erase these numbers from the sheet, and write the number $a + b$ on the sheet. The first player who is not able to do so loses. Determine which player can always win this game.",
"options": [],
"answer": "See solution",
"solution": "Gabrielle can always win using the following strategy: during each of her turns, she picks the largest two numbers on the sheet as $a$ and $b$. Using induction on $k$, we will prove that she is always allowed to do so, and that after her $k$-th turn, the sheet contains the number $2k + 1$ and $998 - 2k$ ones.\n\nIn his first turn, Ward can only pick $a = b = 1$, after which the sheet contains the number $2$ and $997$ ones. Gabrielle then picks the two largest numbers, $a = 2$ and $b = 1$, after which the sheet contains $3$ and $996$ ones. This finishes the basis $k = 1$ of the induction.\n\nNow suppose that for some $m \\ge 1$ after Gabrielle's $m$-th turn the sheet contains the number $2m + 1$ and $998 - 2m$ ones. If $998 - 2m = 0$, then Ward cannot make a move. If not, then Ward can do one of two things: either pick $a = b = 1$ or pick $a = 2m + 1$ and $b = 1$. We consider these two cases separately:\n\n- If Ward picks $a = b = 1$, then the sheet contains the number $2m + 1$, the number $2$, and $996 - 2m$ ones. Gabrielle then picks the two largest numbers, so $a = 2m + 1$ and $b = 2$ (which is allowed since their gcd is $1$). After her turn the sheet contains the numbers $2m + 3 = 2(m + 1) + 1$ and $996 - 2m = 998 - 2(m + 1)$ ones.\n\n- If Ward picks $a = 2m + 1$ and $b = 1$, then the sheet contains the number $2m + 2$ and $997 - 2m$ ones. Gabrielle then picks the two largest numbers, so $a = 2m + 2$ and $b = 1$ (which is allowed since their gcd is $1$). Note that there is a one left, as $997 - 2m$ is odd, so not equal to $0$. After her turn the sheet contains the numbers $2m + 3 = 2(m + 1) + 1$ and $996 - 2m = 998 - 2(m + 1)$ ones.\n\nThis completes the induction.\n\nTherefore, Gabrielle can always make a move. After Gabrielle's turn $499$, the only number left on the sheet is $999$, so Ward can no longer make a move, and Gabrielle wins.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22461,
"subject": "Mathematics (Olympiad)",
"question": "In the square $ABCD$ with side length $4\\,\\mathrm{cm}$, from each vertex arcs are drawn with radii equal to half the side length of the square, as shown in the picture. Calculate the perimeter and area of the marked part of the square.\n\n",
"options": [],
"answer": "See solution",
"solution": "The four arcs form a circle with radius $r = 2\\,\\mathrm{cm}$. The perimeter is $L = 2r\\pi = 4\\pi\\,\\mathrm{cm}$.\n\n$$\nP = 4^2 - r^2\\pi = 16 - 4\\pi = 4(4 - \\pi)\\,\\mathrm{cm}^2\n$$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 22462,
"subject": "Mathematics (Olympiad)",
"question": "a) Let $x, y \\in \\mathbb{R}$. Define $P(x, y)$ as the assertion:\n\n$$\nf(x + g(y)) = x f(y) + (2023 - y) f(x) + g(x).\n$$\n\nb) Find all functions $f, g : \\mathbb{R} \\to \\mathbb{R}$ satisfying the above equation for all $x, y \\in \\mathbb{R}$.",
"options": [],
"answer": "See solution",
"solution": "We analyze the functional equation:\n\n$$\nf(x + g(y)) = x f(y) + (2023 - y) f(x) + g(x), \\quad \\forall x, y \\in \\mathbb{R}.\n$$\n\nFrom $P(0, y)$:\n\n$$\nf(g(y)) = (2023 - y) f(0) + g(0).\n$$\n\nGiven $f(0) = 2022$, this becomes:\n\n$$\nf(g(y)) = 2022(2023 - y) + g(0),\n$$\nwhich is linear in $y$ and thus $f$ is surjective.\n\nTo show $g$ is injective, suppose $g(x_1) = g(x_2)$. Then:\n\n$$\n2022(2023 - x_1) + g(0) = 2022(2023 - x_2) + g(0) \\implies x_1 = x_2.\n$$\n\nNow, consider $f(g(y)) = (2023 - y) f(0) + g(0) = 2022(2023 - y) + g(0)$. From $P(g(x), y)$:\n\n$$\nf(g(x) + g(y)) = g(x) f(y) + (2023 - y) f(g(x)) + g(g(x)).\n$$\n\nUsing the previous result, $f(g(x)) = 2022(2023 - x) + g(0)$, so:\n\n$$\nf(g(x) + g(y)) = g(x) f(y) + (2023 - y)[2022(2023 - x) + g(0)] + g(g(x)).\n$$\n\nSwapping $x$ and $y$ and comparing:\n\n$$\ng(x) f(y) + g(g(x)) + (2023 - y) g(0) = g(y) f(x) + g(g(y)) + (2023 - x) g(0), \\quad \\forall x, y \\in \\mathbb{R}.\n$$\n\nSince $f$ is surjective, there exists $a$ such that $f(a) = 0$. Substitute $y = a$:\n\n$$\ng(g(x)) = -g(0) x + g(a) f(x) + C,\n$$\nfor some constant $C$.\n\nSubstitute back and simplify:\n\n$$\ng(x) f(y) + g(a) f(x) = g(y) f(x) + g(a) f(y), \\quad \\forall x, y \\in \\mathbb{R}.\n$$\n\nLet $y = 0$:\n\n$$\n2022 g(x) + g(a) f(x) = g(0) f(x) + 2022 g(a).\n$$\n\nThus:\n\n$$\ng(x) = \\frac{g(0) - g(a)}{2022} f(x) + g(a).\n$$\n\nIf $g(a) = g(0)$, then $a = 0$ or $f(a) = f(0) = 0$, contradicting $f(0) = 2022$. So $g(a) \\neq g(0)$, so $g$ is surjective. Thus, there exists $b$ such that $g(b) = 0$.\n\nFrom $P(x, b)$:\n\n$$\nf(x) = x f(b) + (2023 - b) f(x) + g(x).\n$$\n\nSubstitute back to $P(x, y)$:\n\n$$\nf(x + g(y)) = x f(y) + (2023 - y) f(x) + f(x) - x f(b) + (b - 2023) f(x) = x(f(y) - f(b)) + f(x)(1 + b - y).\n$$\n\nLet $y = 1 + b$:\n\n$$\nf(x + g(1 + b)) = x(f(1 + b) - f(b)),\n$$\nso $f$ is linear over $\\mathbb{R}$, and similarly for $g$.\n\nThus, the only solutions are:\n\n$$\n\\begin{cases}\nf(x) = \\frac{-1 \\pm \\sqrt{5}}{2} x + 2022, \\\\\ng(x) = 1011(-1 \\mp \\sqrt{5}) x + 2 \\cdot 1011^2 (-3 \\mp \\sqrt{5})\n\\end{cases}, \\quad \\forall x \\in \\mathbb{R}.\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22463,
"subject": "Mathematics (Olympiad)",
"question": "For an integer $n > 0$, let $\\mathcal{F}(n)$ be the set of integers $m > 0$ for which the polynomial $p(x) = x^2 + mx + n$ has an integer root.\n\n(a) Let $S$ be the set of integers $n > 0$ for which $\\mathcal{F}(n)$ contains two consecutive integers. Show that $S$ is infinite but\n\n$$\n\\sum_{n \\in S} \\frac{1}{n} \\le 1.\n$$\n\n(b) Prove that there are infinitely many positive integers $n$ such that $\\mathcal{F}(n)$ contains three consecutive integers.",
"options": [],
"answer": "See solution",
"solution": "*Claim.* The set $S$ is given explicitly by $S = \\{x(x+1)y(y+1) \\mid x, y > 0\\}$.\n\n*Proof.*\n\n$m, m+1 \\in \\mathcal{F}(n)$ if and only if there exist integers $q > p \\ge 0$ such that\n\n$$\n\\begin{aligned}\nm^2 - 4n &= p^2 \\\\\n(m+1)^2 - 4n &= q^2.\n\\end{aligned}\n$$\n\nSubtracting gives $2m + 1 = q^2 - p^2$, so $p$ and $q$ have different parities. Let $q - p = 2x + 1$, $q + p = 2y + 1$ for integers $y \\ge x \\ge 0$. Then\n\n$$\n\\begin{aligned}\n4n &= m^2 - p^2 \\\\\n&= \\left( \\frac{q^2 - p^2 - 1}{2} \\right)^2 - p^2 \\\\\n&= \\left( \\frac{q^2 - p^2 - 1}{2} - p \\right) \\left( \\frac{q^2 - p^2 - 1}{2} + p \\right) \\\\\n&= \\frac{q^2 - (p^2 + 2p + 1)}{2} \\cdot \\frac{q^2 - (p^2 - 2p + 1)}{2} \\\\\n&= \\frac{1}{4}(q - p - 1)(q - p + 1)(q + p - 1)(q + p + 1) \\\\\n&= \\frac{1}{4}(2x)(2x + 2)(2y)(2y + 2) \\\\\n\\implies n &= x(x + 1)y(y + 1).\n\\end{aligned}\n$$\n\nSince $n > 0$, $x, y > 0$. Conversely, if $n = x(x+1)y(y+1)$ for positive $x, y$, then $m = 2xy + x + y = x(y+1) + (x+1)y$ and $m + 1 = 2xy + x + y + 1 = xy + (x+1)(y+1)$. Thus, the claim holds. $\\square$\n\nFor part (a):\n\n$$\n\\sum_{n \\in S} \\frac{1}{n} \\le \\left( \\sum_{x \\ge 1} \\frac{1}{x(x+1)} \\right) \\left( \\sum_{y \\ge 1} \\frac{1}{y(y+1)} \\right) = 1 \\cdot 1 = 1.\n$$\n\nFor part (b):\n\nNow $m + 2 \\in S$ if and only if $(m+2)^2 - 4n$ is a square, say $r^2$. In terms of $p$ and $q$:\n\n$$\n\\begin{aligned}\nr^2 &= (m+2)^2 - 4n = m^2 - 4n + 4m + 4 = p^2 + 2 + 2(2m+1) \\\\\n&= p^2 + 2(q^2 - p^2) + 2 = 2q^2 - p^2 + 2 \\\\\n\\iff 2q^2 + 2 &= p^2 + r^2 \\quad (\\dagger)\n\\end{aligned}\n$$\n\nwith $q > p$ of different parity and $n = \\frac{1}{16}(q-p-1)(q-p+1)(q+p-1)(q+p+1)$.\n\nFor every $q$, $(p, r) = (q-1, q+1)$ is a solution, but this gives $n = 0$. For infinitely many $q$, $q^2 + 1$ is divisible by at least three distinct $1 \\bmod 4$ primes. Each such prime can be written as a sum of two squares, so $2q^2 + 2$ can be written as a sum of two squares in at least three ways, giving infinitely many $n$ with three consecutive elements in $\\mathcal{F}(n)$.\n\n*Remark.* $n = 144$ is the smallest integer such that $\\mathcal{F}(n)$ contains three consecutive integers, and $n = 15120$ is the smallest with four consecutive. It is unknown whether the number of consecutive elements in $\\mathcal{F}(n)$ can be arbitrarily large.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 22464,
"subject": "Mathematics (Olympiad)",
"question": "A tetrahedron $ABCD$ is said to be *angelic* if it has nonzero volume and satisfies\n\n$$\n\\begin{aligned}\n\\angle BAC + \\angle CAD + \\angle DAB &= \\angle ABC + \\angle CBD + \\angle DBA, \\\\\n\\angle ACB + \\angle BCD + \\angle DCA &= \\angle ADB + \\angle BDC + \\angle CDA.\n\\end{aligned}\n$$\n\nAcross all angelic tetrahedrons, what's the maximum number of distinct lengths that could appear in the set $\\{AB, AC, AD, BC, BD, CD\\}$?",
"options": [],
"answer": "See solution",
"solution": "We claim the maximum cardinality is $4$. This is attained by taking a non-square rectangle (or parallelogram) $ACBD$ and folding it along diagonal $AB$, creating edges $AB$ and $CD$ in the process. Here, $AC = BD$ and $AD = BC$, while every other length is distinct in the general case. This tetrahedron is congruent to itself under the permutation of vertices $(A, B, C, D) \\mapsto (B, A, D, C)$, and we can verify that the required angle conditions follow from this symmetry.\n\n\n\nIn the other direction, let $f(X)$ denote the sum of the angles at $X$, so the conditions of the problem statement can be written as $f(A) = f(B)$ and $f(C) = f(D)$. Unfold the three faces that meet at $D$ to create a net of the tetrahedron. Along with the face $\\triangle ABC$, we also create the faces $\\triangle ABD_1$, $\\triangle BCD_2$, and $\\triangle ACD_3$. Note that\n\n$f(A)+f(B)+f(C)+f(D)$ is the sum of all of the angles of all four faces of the tetrahedron, which is $720^\\circ$. Therefore, $f(A) = f(B) = 360^\\circ - f(C) = 360^\\circ - f(D)$. Furthermore, we have that $\\angle D_1AD_3 = f(A)$ and $\\angle D_2CD_3 = 360^\\circ - f(C)$. Since $AD_1 = AD_3$ and $CD_2 = CD_3$ by the definition of unfolding, this gives $\\triangle D_3AD_1 \\sim \\triangle D_3CD_2$. Thus, due to spiral similarity, we have $\\triangle D_3AC \\sim \\triangle D_3D_1D_2$.\n\nSimilarly, we also have $\\triangle D_2BD_1 \\sim \\triangle D_2CD_3 \\implies \\triangle CBD_2 \\sim \\triangle D_3D_1D_2$. This means that $\\triangle CBD_2 \\sim \\triangle D_3AC$, and since $CD_3 = CD_2$, the two triangles are actually congruent. Therefore, $AC = BD_2 = BD$ and $BC = AD_3 = AD$. Since we have two pairs of equal edge lengths, the number of distinct edge lengths is at most $4$, as desired.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 22465,
"subject": "Mathematics (Olympiad)",
"question": "$a \\in G(p)$ ба $a^{p-1} \\ne 1 \\pmod{p^2}$ бол $k \\in \\mathbb{N}$ бүрийн хувьд $a \\in G(p^k)$ гэдгийг батал.",
"options": [],
"answer": "See solution",
"solution": "Эсрэгээр нь буюу $\\exists k > 1$ тийм $a \\notin G(p^k)$ гэж үзье.\n\n$$\na \\notin G(p^k) \\Leftrightarrow m = |a|_{p^k} < \\varphi(p^k) = p^{k-1}(p-1)\n$$\n\n$$\nm = p^s \\cdot \\frac{p-1}{q}\n$$\n\nэнд а) $s < k-1$, $q = 1$ эсвэл $q \\in \\mathbb{P}$, эсвэл б) $s = k-1$ ба $q \\in \\mathbb{P}$.\n\nа) тохиолд: $a^m = a^{p^s \\cdot \\frac{p-1}{q}} \\equiv 1 \\pmod{p^k} \\Leftrightarrow a^{\\frac{p-1}{q}} \\equiv 1 \\pmod{p^{k-s}}$. [MMK-I, 1.8.17] Үүнээс $a^{\\frac{p-1}{q}} \\equiv 1 \\pmod{p^2}$ гарна, харин $a^{p-1} \\not\\equiv 1 \\pmod{p^2}$ гэж өгсөн тул зөрчил.\n\nб) тохиолд: $a^m = a^{p^{k-1} \\cdot \\frac{p-1}{q}} \\equiv 1 \\pmod{p^k} \\Rightarrow a^{\\frac{p-1}{q}} \\equiv 1 \\pmod{p}$. [MMK-I, 1.8.17] Гэтэл $\\frac{p-1}{q} < p-1$ тул энэ нь $a \\in G(p)$ гэдгийг зөрчинө.\n\nИймд $a \\in G(p^k)$ нь $k$ бүрийн хувьд үнэн.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22466,
"subject": "Mathematics (Olympiad)",
"question": "The road between points A and B is 15 km long. The road consists of three consecutive segments: uphill, flat, and downhill. Each segment is at least 1 km long. A pedestrian walks from A to B in exactly 3 hours. The pedestrian's speeds are:\n- Uphill: 4 km/h\n- Flat: 5 km/h\n- Downhill: 6 km/h\n\nWhat are the minimum and maximum possible times for the pedestrian to walk the path in the opposite direction (from B to A)?",
"options": [],
"answer": "See solution",
"solution": "**Answer:** $t_{\\max} = \\frac{97}{30}$, $t_{\\min} = \\frac{73}{24}$.\n\nLet $x$, $y$, and $z$ be the lengths (in km) of the uphill, flat, and downhill segments, respectively, from A to B. Then:\n\n$$\nx + y + z = 15, \\qquad \\frac{x}{4} + \\frac{y}{5} + \\frac{z}{6} = 3, \\qquad 1 \\le x, y, z \\le 13.\n$$\n\nFrom the first equation: $y = 15 - x - z$. Substitute into the second equation:\n\n$$\n\\frac{x}{4} + \\frac{15 - x - z}{5} + \\frac{z}{6} = 3\n$$\n\nSimplifying:\n\n$$\n\\frac{x}{4} - \\frac{x}{5} = \\frac{z}{5} - \\frac{z}{6} \\implies \\frac{x}{20} = \\frac{z}{30} \\implies z = \\frac{3}{2}x.\n$$\n\nThen:\n\n$$\ny = 15 - x - z = 15 - x - \\frac{3}{2}x = 15 - \\frac{5}{2}x.\n$$\n\nThe time for the reverse path (from B to A) is:\n\n$$\nt = \\frac{x}{6} + \\frac{y}{5} + \\frac{z}{4} = \\frac{x}{6} + \\frac{15 - \\frac{5}{2}x}{5} + \\frac{\\frac{3}{2}x}{4} = 3 + \\frac{x}{24}.\n$$\n\nTo find the minimum and maximum $t$, consider the possible values of $x$:\n\n$$\n1 \\le x \\le \\frac{28}{5}\n$$\n\nIf $x = 1$:\n- $z = \\frac{3}{2}$\n- $y = 15 - 1 - \\frac{3}{2} = \\frac{25}{2}$\n\nIf $x = \\frac{28}{5}$:\n- $z = \\frac{3}{2} \\cdot \\frac{28}{5} = \\frac{42}{5}$\n- $y = 15 - \\frac{28}{5} - \\frac{42}{5} = 1$\n\nThus:\n\n$$\nt_{\\min} = 3 + \\frac{1}{24} = \\frac{73}{24}, \\qquad t_{\\max} = 3 + \\frac{28}{120} = 3 + \\frac{7}{30} = \\frac{97}{30}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22467,
"subject": "Mathematics (Olympiad)",
"question": "Let $l(n)$ denote the greatest odd divisor of any natural number $n$. Find the sum\n\n$$\nl(1) + l(2) + l(3) + \\cdots + l(2^{2013}).\n$$",
"options": [],
"answer": "See solution",
"solution": "For each natural $k$, the equalities $l(2k) = l(k)$ and $l(2k - 1) = 2k - 1$ are clearly valid. Thus, we can add the values $l(n)$ over groups of numbers $n$ lying between two consecutive powers of $2$. In this way, we prove by induction the formula\n\n$$\ns(n) = l(2^{n-1} + 1) + l(2^{n-1} + 2) + \\cdots + l(2^n),\n$$\nfor $n = 1, 2, 3, \\dots$.\n\nThe case $n = 1$ is trivial. If $s(n) = 4^{n-1}$ for some $n$, then\n\n$$\n\\begin{align*}\ns(n+1) &= l(2^n + 1) + l(2^n + 2) + \\cdots + l(2^{n+1}) \\\\\n&= [(2^n + 1) + (2^n + 3) + \\cdots + (2^{n+1} - 1)] + s(n) \\\\\n&= \\frac{2^{n-1}}{2}(2^n + 1 + 2^{n+1} - 1) + 4^{n-1} \\\\\n&= 2^{n-2} \\cdot 3 \\cdot 2^n + 4^{n-1} = 4^n.\n\\end{align*}\n$$\n\n(We have used the fact that the number of all odd numbers from $2^n + 1$ to $2^{n+1} - 1$ [including both limits] equals $2^{n-1}$.) The proof of the formula by induction is complete.\n\nUsing the formula, we compute the requested sum as follows:\n\n$$\n\\begin{aligned}\nl(1)+l(2)+l(3)+\\cdots+l(2^{2013}) &= l(1)+s(2)+s(3)+\\cdots+s(2013) \\\\\n&= 1+1+4+4^2+4^3+\\cdots+4^{2012} \\\\\n&= 1+\\frac{4^{2013}-1}{3} = \\frac{4^{2013}+2}{3}.\n\\end{aligned}\n$$\n\n_Remark._ It is worth mentioning that the formula $s(n) = 4^{n-1}$ is a special case of a more general (and surprising) formula\n\n$$\nl(k+1)+l(k+2)+l(k+3)+\\cdots+l(2k)=k^2,\n$$\nwhich can be proved for each natural number $k$ even without using induction. Indeed, all the $k$ summands on the left-hand side are numbers from the $k$-element subset $\\{1, 3, 5, \\ldots, 2k-1\\}$, and these summands are pairwise distinct, because the ratio of any two numbers from $\\{k + 1, k + 2, \\ldots, 2k\\}$ is not a power of $2$. Consequently, the sum equals $1 + 3 + 5 + \\cdots + (2k - 1)$, which is $k^2$ as stated.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22468,
"subject": "Mathematics (Olympiad)",
"question": "A positive integer is called _special_ if all of its digits are nonzero and any two adjacent digits are consecutive (not necessarily in ascending order).\n\n(a) Find the largest special number for which the sum of its digits is $2023$.\n\n(b) Find the smallest special number for which the sum of its digits is $2022$.",
"options": [],
"answer": "See solution",
"solution": "**(a)** The largest number $m$ will have as many digits as possible, so we choose the smallest possible digits. Since next to a $1$ we can only put a $2$, and $2023 = 3 \\times 674 + 1$, we choose\n$$\n m = \\underbrace{12121\\ldots121}_{1349\\ \\text{digits}},\n$$\nwith $674$ digits equal to $2$ and $675$ digits equal to $1$.\n\n**(b)** Since $8+9=17$ and $2022 = 17 \\times 118 + 16$, if in the decimal writing of $n$ we use at most $237$ digits, the sum of these digits would be at most $118 \\times (8+9) + 9 < 2022$, which is false. Therefore, we can't use fewer than $2 \\times 118 + 2 = 238$ digits. To use exactly $238$ digits, we must have $118$ groups of $8$ and $9$, and another two digits whose sum is $16$. The two digits could be $7$ and $9$, or $8$ and $8$. In both cases, the number of even digits of $n$ differs by $2$ from the number of its odd digits. This is impossible, because in the decimal writing of a special number, the even digits alternate with the odd ones, so the number of its even digits is either equal to that of its odd digits, or differs from it by $1$. Therefore, $n$ must have at least $239$ digits. We are looking for three consecutive digits, smaller than $8$, whose sum is at least $16$ and which are the first three digits of $n$. Moreover, we are looking for the first digit of $n$ to be as small as possible. The smallest digits that fulfill these requirements are $5$, $6$, $7$. The solution is\n$$\n n = 56787\\underbrace{8989\\ldots89}_{117\\ \\text{pairs}}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22469,
"subject": "Mathematics (Olympiad)",
"question": "Determine all quadruples of consecutive positive integers such that three of the numbers are sides and one the area of a right triangle. (Units of length and area are compatible.)",
"options": [],
"answer": "See solution",
"solution": "Let $a$ and $b$ be the legs and $c$ the hypotenuse. We have $a^2 + b^2 = c^2$. At least two of the numbers $a, b, c$ are consecutive, so they are of different parity. It is not possible that only one of the numbers is odd. So there are two odd numbers, and one of them has to be $c$. We can assume that $a$ is odd. Then $c = a + 2$.\n\nNow, either $b = a + 1$ or $b = a - 1$.\n\n- In the first case: $(b - 1)^2 + b^2 = (b + 1)^2$, or $b^2 = 4b$, so $b = 4$, $a = 3$, $c = 5$. The area of a $(3, 4, 5)$ triangle is $6$, so $(3, 4, 5, 6)$ is a solution.\n\n- Now assume $b = a - 1$. This leads to $a^2 + (a - 1)^2 = (a + 2)^2$ or $a^2 - 6a - 3 = 0$. So $a$ has to divide $3$. But one easily sees that the equation is not satisfied for any $a \\in \\{\\pm1, \\pm3\\}$.\n\nThus, the only solution is $(3, 4, 5, 6)$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22470,
"subject": "Mathematics (Olympiad)",
"question": "Given $n$ positive real numbers satisfying $x_1 \\ge x_2 \\ge \\dots \\ge x_n \\ge 0$ and $x_1^2 + x_2^2 + \\dots + x_n^2 = 1$, prove that\n\n$$\n\\frac{x_1}{\\sqrt{1}} + \\frac{x_2}{\\sqrt{2}} + \\dots + \\frac{x_n}{\\sqrt{n}} \\ge 1.\n$$",
"options": [],
"answer": "See solution",
"solution": "Note that for any $k \\le n$, we have\n\n$$\nk x_k^2 \\le x_1^2 + x_2^2 + \\dots + x_k^2 \\le 1\n$$\nby the given conditions. This implies that $x_k \\le \\frac{1}{\\sqrt{k}}$ and thus $x_k^2 \\le \\frac{x_k}{\\sqrt{k}}$. We conclude that\n\n$$\n\\frac{x_1}{\\sqrt{1}} + \\frac{x_2}{\\sqrt{2}} + \\cdots + \\frac{x_n}{\\sqrt{n}} \\ge x_1^2 + x_2^2 + \\cdots + x_n^2 = 1,\n$$\nwhich proves the statement.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22471,
"subject": "Mathematics (Olympiad)",
"question": "Let $P, Q, R, S$ be the points of tangency of a given circle to the sides $DA, AB, BC, CD$, respectively. Let $A', B', C', D'$ be the midpoints of the line segments $PQ, QR, RS, SP$, respectively. Note that $A', B', C', D'$ lie on the line segments $OA, OB, OC, OD$, respectively. Denote the length of the radius of the circle by $r$.\n\nFind the ratio $OM : ON$, where $M$ is the midpoint of $AC$ and $N$ is the midpoint of $BD$, expressed in terms of $AO, BO, CO, DO$, and $r$.",
"options": [],
"answer": "See solution",
"solution": "Since $\\angle APO = \\angle PA'O = 90^\\circ$ and $\\angle AOP = \\angle POA'$, the triangles $APO$ and $PA'O$ are similar, so $AO : PO = PO : A'O$. Thus, $A'O = \\frac{PO^2}{AO} = \\frac{r^2}{AO}$. Similarly, $C'O = \\frac{r^2}{CO}$.\n\nTherefore, $AO : CO = C'O : A'O$, and since $\\angle AOC = \\angle C'OA'$, triangles $AOC$ and $C'OA'$ are similar. This similarity carries $M$ to the midpoint $M'$ of $A'C'$, so $OM = OM' \\cdot \\frac{AO}{C'O} = OM' \\cdot \\frac{AO \\cdot CO}{r^2}$.\n\nSimilarly, if $N'$ is the midpoint of $B'D'$, then $ON = ON' \\cdot \\frac{BO \\cdot DO}{r^2}$.\n\nWe see that\n\n$$\n\\overrightarrow{OM'} = \\frac{1}{2}(\\overrightarrow{OA'} + \\overrightarrow{OC'}) = \\frac{1}{4}(\\overrightarrow{OP} + \\overrightarrow{OQ} + \\overrightarrow{OR} + \\overrightarrow{OS}) = \\frac{1}{2}(\\overrightarrow{OB'} + \\overrightarrow{OD'}) = \\overrightarrow{ON'}\n$$\n\nso $M' = N'$. Putting these facts together,\n\n$$\nOM : ON = OM' \\cdot \\frac{AO \\cdot CO}{r^2} : ON' \\cdot \\frac{BO \\cdot DO}{r^2} = AO \\cdot CO : BO \\cdot DO = 35 : 48,\n$$\n\nwhich gives the desired answer.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22472,
"subject": "Mathematics (Olympiad)",
"question": "Let $n \\in \\mathbb{N}$, $n \\geq 4$. Determine all sets $A = \\{a_1, a_2, \\dots, a_n\\} \\subset \\mathbb{N}$ that contain $2015$ and for which $|a_i - a_j|$ is a prime for all distinct $i, j \\in \\{1, 2, \\dots, n\\}$.",
"options": [],
"answer": "See solution",
"solution": "It is easy to see that $A$ cannot contain more than two numbers of any parity. Combined with $n \\geq 4$, this forces $A$ to have exactly $4$ elements, two of each parity. The difference between the two even (or odd) numbers must be $2$, therefore we can have two types of sets: $\\{2013, 2015, 2k, 2k + 2\\}$ and $\\{2015, 2017, 2k, 2k + 2\\}$.\n\nIn the first case, the differences $|(2k + 2) - 2013|$, $|2k - 2013|$, and $|2k - 2015|$ give different remainders upon division by $3$, hence one of these differences has to be $3$. Checking all the possibilities leads to the solutions $A = \\{2013, 2015, 2008, 2010\\}$ and $A = \\{2013, 2015, 2018, 2020\\}$.\n\nIn the second case, one of the differences $|(2k + 2) - 2015|$, $|2k - 2015|$, and $|2k - 2017|$ has to be $3$. Studying the cases we get the answers $\\{2015, 2017, 2010, 2012\\}$ and $\\{2015, 2017, 2020, 2022\\}$.\n\nIn conclusion:\n\n$$\n\\begin{aligned}\nA_1 &= \\{2008, 2010, 2013, 2015\\}, & A_2 &= \\{2010, 2012, 2015, 2017\\}, \\\\\nA_3 &= \\{2013, 2015, 2018, 2020\\}, & A_4 &= \\{2015, 2017, 2020, 2022\\}.\n\\end{aligned}\n$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22473,
"subject": "Mathematics (Olympiad)",
"question": "Determine whether there exists a positive integer $k$ such that $p = 6k + 1$ is prime and\n\n$$\n\\binom{3k}{k} \\equiv 1 \\pmod{p}.\n$$",
"options": [],
"answer": "See solution",
"solution": "No such $k$ exists. Suppose that $k$ and $p$ are as described. Consider the number\n\n$$\nA = \\sum_{i=0}^{p-1} (i^3 - 1)^{3k}.\n$$\n\nBecause $p-1 = 6k$ is divisible by $3$, there are three cube roots of $1$ modulo $p$. Therefore, three terms in the sum are $0$ modulo $p$, and the others are $[(p-1)/2]$th powers of nonzero residues, hence are congruent to either $1$ or $-1$ modulo $p$. Consequently, $A$ is congruent to one of the residues $p-3, p-5, p-7, \\dots, -(p-3)$ modulo $p$. In particular, $A$ cannot be congruent to $1$ or $-1$ modulo $p$.\n\nOn the other hand, applying the binomial theorem and changing the order of summation, we have\n\n$$\nA = \\sum_{i=0}^{p-1} \\left( \\sum_{j=0}^{3k} \\binom{3k}{j} (-1)^j i^{3(3k-j)} \\right) = \\sum_{j=0}^{3k} \\left( (-1)^j \\binom{3k}{j} \\sum_{i=0}^{p-1} i^{3(3k-j)} \\right),\n$$\n\nwhere we use the convention $0^0 = 1$ for the case $i = 0, j = 3k$.\n\nNow, we claim that $\\sum_{i=0}^{p-1} i^d \\equiv 0 \\pmod{p}$ if $p-1 \\nmid d$. For this, choose a primitive root $g$ modulo $p$ and notice that $g^d \\not\\equiv 1 \\pmod{p}$, while we have\n\n$$\n(g^d - 1) \\sum_{i=0}^{p-1} i^d \\equiv \\sum_{i=0}^{p-1} (g \\cdot i)^d - \\sum_{i=0}^{p-1} i^d \\equiv 0 \\pmod{p},\n$$\n\ngiving the claim. Further, notice that $\\sum_{i=0}^{p-1} i^d \\equiv 0 \\pmod{p}$ when $d = 0$.\n\nNow, because $3 \\cdot (3k) < 2(p-1)$, the only value of $j$ such that $3 \\cdot (3k-j)$ is a multiple of $p-1$ is $j = k$, where $3 \\cdot (3k-j) = 6k = p-1$. This means that the sums $\\sum_i i^{3 \\cdot (3k-j)}$ are congruent to $0$ modulo $p$ unless $j = k$. Further, for $j = k$, the sum evaluates to\n\n$$\n\\sum_{i=0}^{p-1} i^{6k} \\equiv \\sum_{i=0}^{p-1} i^{p-1} \\equiv \\sum_{i=1}^{p-1} 1 \\equiv -1 \\pmod{p}.\n$$\n\nConsidering the above, we find that\n\n$$\nA \\equiv (-1)^k \\cdot \\binom{3k}{k} \\cdot (-1) \\pmod{p}.\n$$\n\nSince we saw that $A$ is not congruent to $1$ or $-1$ modulo $p$, we conclude that $\\binom{3k}{k} \\not\\equiv 1 \\pmod{p}$, a contradiction.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22474,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that\n$$\na^2b + b^2c + c^2a \\geq \\sqrt{(a+b+c)(ab+bc+ca)}.\n$$",
"options": [],
"answer": "See solution",
"solution": "By the inequality between the arithmetic and geometric means:\n$$\n\\begin{align*}\n(a^2 b + b^2 c + c^2 a)^2 &\\geq 3(a^2 b \\cdot b^2 c + b^2 c \\cdot c^2 a + c^2 a \\cdot a^2 b) \\\\\n&= 3abc (b^2 a + c^2 b + a^2 c) \\\\\n&= 3(b^2 a + c^2 b + a^2 c), \\\\\n(a^2 b + b^2 c + c^2 a)^2 &\\geq (a^2 b + b^2 c + c^2 a) \\cdot (a^2 b + b^2 c + c^2 a) \\\\\n&\\geq 3\\sqrt[3]{a^3 b^3 c^3} (a^2 b + b^2 c + c^2 a) \\\\\n&= 3(a^2 b + b^2 c + c^2 a), \\\\\n(a^2 b + b^2 c + c^2 a)^2 &\\geq 9abc, \\\\\n3(a^2 b + b^2 c + c^2 a)^2 &\\geq 3(b^2 a + c^2 b + a^2 c) + 3(a^2 b + b^2 c + c^2 a) + 9abc \\\\\n(a^2 b + b^2 c + c^2 a)^2 &\\geq a^2 b + b^2 c + c^2 a + b^2 a + c^2 b + a^2 c + 3abc \\\\\n(a^2 b + b^2 c + c^2 a)^2 &\\geq (a+b+c)(ab+bc+ca) \\\\\na^2 b + b^2 c + c^2 a &\\geq \\sqrt{(a+b+c)(ab+bc+ca)}.\n\\end{align*}\n$$\nThus, the inequality is proved using the AM-GM inequality.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22475,
"subject": "Mathematics (Olympiad)",
"question": "Suppose there are $n$ ink stains, with areas $S_1, S_2, \\dots, S_n$ (in mm²). The lengths of the projections of the ink stains on the top edge of the answer sheet are $x_1, x_2, \\dots, x_n$ (in mm), and on the left edge are $y_1, y_2, \\dots, y_n$ (in mm). Any line parallel to an edge of the answer sheet intersects at most one ink stain, so $\\sum_{j=1}^{n} x_j \\le 210$ and $\\sum_{j=1}^{n} y_j \\le 297$. Given $S_j \\le 1$, what is the maximum possible value of $\\sum_{j=1}^{n} S_j$?",
"options": [],
"answer": "See solution",
"solution": "By the condition $S_j \\le 1$ and the AM-GM inequality, we have\n\n$$\n\\sum_{j=1}^{n} S_j \\le \\sum_{j=1}^{n} \\sqrt{S_j} \\le \\sum_{j=1}^{n} \\sqrt{x_j y_j} \\le \\sum_{j=1}^{n} \\frac{x_j + y_j}{2} \\le \\frac{1}{2}(210 + 297) = 253.5.\n$$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 22476,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $n$ for which it is possible to color some cells of an infinite grid of unit squares red, such that each rectangle consisting of exactly $n$ cells (and whose edges lie along the lines of the grid) contains an odd number of red cells.",
"options": [],
"answer": "See solution",
"solution": "**Proof of (i)** \nWe simply observe that if every rectangle consisting of $n$ cells contains an odd number of red cells, then so must every rectangle consisting of $pn$ cells. Indeed, because $p$ is prime, a rectangle consisting of $pn$ cells must have a dimension (length or width) divisible by $p$ and can thus be subdivided into $p$ rectangles consisting of $n$ cells. Thus every coloring that works for $n$ automatically also works for $pn$.\n\n**Proof of (ii)** \nObserve that rectangles with $n = 2^k$ cells have $k+1$ possible shapes: $2^m \\times 2^{k-m}$ for $0 \\le m \\le k$.\n\n**Claim** — For each of these $k+1$ shapes, there exists a coloring with two properties:\n\n- Every rectangle with $n$ cells and shape $2^m \\times 2^{k-m}$ contains an odd number of red cells.\n- Every rectangle with $n$ cells and a different shape contains an even number of red cells.\n\n*Proof.* This can be achieved as follows: assuming the cells are labeled with $(x, y) \\in \\mathbb{Z}^2$, color a cell red if $x \\equiv 0 \\pmod{2^m}$ and $y \\equiv 0 \\pmod{2^{k-m}}$. For example, a $4 \\times 2$ rectangle gets the following coloring:\n\n\n\nA $2^m \\times 2^{k-m}$ rectangle contains every possible pair $(x \\pmod{2^m}, y \\pmod{2^{k-m}})$ exactly once, so such a rectangle will contain one red cell (an odd number).\n\nOn the other hand, consider a $2^\\ell \\times 2^{k-\\ell}$ rectangle with $\\ell > m$. The set of cells this covers is $(x, y)$ where $x$ covers a range of size $2^\\ell$ and $y$ covers a range of size $2^{k-\\ell}$. The number of red cells is the count of $x$ with $x \\equiv 0 \\pmod{2^m}$ multiplied by the count of $y$ with $y \\equiv 0 \\pmod{2^{k-m}}$. The former number is exactly $2^{\\ell-k}$ because $2^k$ divides $2^\\ell$ (while the latter is 0 or 1) so the number of red cells is even. The $\\ell < m$ case is similar. $\\square$\n\nFinally, given these $k+1$ colorings, we can add them up modulo 2, i.e. a cell will be colored red if it is red in an odd number of these $k+1$ colorings. We illustrate $n=4$ as an example; the coloring is 4-periodic in both axes so we only show one $4 \\times 4$ cell.\n\n\n\nThis solves the problem.\n\n**Remark.** The final coloring can be described as follows: color $(x, y)$ red if\n\n$$\n\\max(0, \\min(\\nu_2(x), k) + \\min(\\nu_2(y), k) - k + 1)\n$$\n\nis odd.\n\n**Remark (Luke Robitaille).** Alternatively for (i), if $n = 2^e k$ for odd $k$ then one may dissect an $a \\times b$ rectangle with area $n$ into $k$ rectangles of area $2^e$, each $2^{\\nu_2(a)} \\times 2^{\\nu_2(b)}$. This gives a way to deduce the problem from (ii) without having to consider odd prime numbers.\n\nAlternate proof of (ii) using generating functions: We will commit to constructing a coloring which is $n$-periodic in both directions. (This is actually forced, so it's natural to do so.) With that in mind, let\n\n$$\nf(x, y) = \\sum_{i=0}^{2^k-1} \\sum_{j=0}^{2^k-1} \\lambda_{i,j} x^i y^j\n$$\n\ndenote its generating function, where $f \\in \\mathbb{F}_2[x, y]$.\n\nFor this to be valid, we need that for any $2^p \\times 2^q$ rectangle with area $n$, the sum of the coefficients of $f$ over it should be one, modulo $x^{2^k} = y^{2^k} = 1$. In other words, whenever $p+q=k$, we must have\n\n$$\nf(x, y)(1 + \\cdots + x^{2^p-1})(1 + \\cdots + y^{2^q-1}) = (1 + \\cdots + x^{2^k-1})(1 + \\cdots + y^{2^k-1}),\n$$\n\ntaken modulo $x^{2^k} = y^{2^k} = 1$. The idea is to rewrite these expressions: because we're in characteristic 2, the given assertion is $(x+1)^{2^k} = (y+1)^{2^k} = 0$, and the requested property is\n\n$$\nf(x, y)(x + 1)^{2^p - 1}(y + 1)^{2^q - 1} = (x + 1)^{2^k - 1}(y + 1)^{2^k - 1}.\n$$\n\nThis suggests the substitution $g(x, y) = f(x+1, y+1)$: then we can replace $(x+1, y+1) \\mapsto (x, y)$ to simplify the requested property significantly:\n\nWhenever $p+q=k$, we must have\n\n$$\ng(x, y)x^{2^p-1}y^{2^q-1} = x^{2^k-1}y^{2^k-1},\n$$\n\nmodulo $x^{2^k}$ and $y^{2^k}$.\n\nHowever, now the construction of $g$ is very simple: for example, the choice\n\n$$\ng(x, y) = \\sum_{p+q=k} x^{2^k-2^p} y^{2^k-2^q}\n$$\n\nworks. The end.\n\n**Remark.** Unraveling the substitutions seen here, it's possible to show that this is actually the same construction provided in the first solution.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 22477,
"subject": "Mathematics (Olympiad)",
"question": "How many polynomials of degree at most 5 with coefficients either $1$ or $-1$ have the sum of their coefficients equal to $0$?",
"options": [],
"answer": "See solution",
"solution": "The value of the polynomial at $1$ is equal to the sum of all the coefficients. Since this value must be $0$, exactly three coefficients must be $1$ and three must be $-1$. The number of such polynomials is thus $$\\binom{6}{3} = 20.$$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 22478,
"subject": "Mathematics (Olympiad)",
"question": "Let $f: \\mathbb{R} \\to \\mathbb{Z}$ be a function satisfying\n\n$$\nf(x - y) - 2f(x) + f(x + y) \\geq -1\n$$\n\nfor all $x, y \\in \\mathbb{R}$. Find all possible values of the set $\\{f(x) \\mid x \\in \\mathbb{R}\\}$.",
"options": [],
"answer": "See solution",
"solution": "The possible values for the set $\\{f(x) \\mid x \\in \\mathbb{R}\\}$ are:\n\n- $\\{a\\}$\n- $\\{a, a+1\\}$\n- $\\{a, a+1, a+2, \\dots\\}$\n- $\\mathbb{Z}$\n\nfor arbitrary $a \\in \\mathbb{Z}$.\n\nFor constructions:\n- If $g: \\mathbb{R} \\to \\mathbb{R}$ is convex, then $\\lfloor g \\rfloor$ satisfies the functional equation.\n- Thus, $f(x) = a$, $f(x) = \\lfloor x \\rfloor$, and $f(x) = \\lfloor x^2 \\rfloor + a$ work, covering the first, fourth, and third cases, respectively.\n- Also, $f(x) = a + \\mathbf{1}_{x>0}$ works for the second case.\n\nLet $P(x, y)$ denote the given condition. To prove that nothing else works, the key is an \"intermediate value theorem\": if $a$ and $b$ are in the range of $f$, then so is every integer between $a$ and $b$.\n\nAssuming this, if the range of $f$ has at least two values, then it is unbounded above. Indeed, if $f(x) - f(y) \\ge 2$, then $P(x, y - x)$ gives $f(2x - y) > f(x)$, so iterating this process shows the range is unbounded above.\n\n**Lemma 1.1**\n\nIf $f(0) \\le -1$, then $f(2^k) \\ge 2^k f(1)$ for $k \\ge 0$.\n\n*Proof.* $P(2^k, 2^k)$ yields $f(2^{k+1}) \\ge 2f(2^k)$. $\\square$\n\n**Lemma 1.2**\n\nIf $f(-1) \\le -2$ and $f(0) = 0$, then $f(2^k) \\ge 2^k - 1$ for all positive integers $k$.\n\n*Proof.* Applying Lemma 1.1 to $f(x-1)+1$ gives $f(2^k-1) \\ge 2^k-1$. Then, applying Lemma 1.1 to $f(2^k-x) - f(2^k)-1$ yields\n\n$$\nf(0) - f(2^k) - 1 \\ge 2^k(f(2^k - 1) - f(2^k) - 1) \\implies f(2^k) + 1 \\ge \\frac{2^k f(2^k - 1)}{2^k - 1} \\ge 2^k. \\quad \\square\n$$\n\nTo prove the intermediate value theorem, scale and shift so that $f(-1) \\le -2$ and $f(0) = 0$. It suffices to show there is some value strictly between $f(-1)$ and $f(0)$ in the range of $f$ (since by iteration we can get all values). Suppose not and let $a_k = f(-1/2^k)$. If $k$ is minimal such that $a_k \\ge 0$, then $P(-1/2^k, 1/2^k)$ yields a contradiction. Thus $a_k \\le -2$ for all $k$. However, applying Lemma 1.2 to $f(x/2^k)$ yields $a_k \\le -2 \\implies f(1) \\ge 2^k - 1$, which cannot hold for all $k$ since $f(1)$ is constant.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22479,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $x$ such that when the leading digit of $x$ is moved to the end of the number, the resulting number is exactly three times $x$.",
"options": [],
"answer": "See solution",
"solution": "Let the number of digits of $x$ be $d$, the leading digit of $x$ be $a$ ($1 \\le a \\le 9$), and the remaining $d-1$ digits be $b$ ($1 \\le b < 10^{d-1}$). There is no solution with $d=1$, so assume $d > 1$. The equations to solve are:\n\n$$\n\\begin{align*}\nx &= 10^{d-1} a + b \\\\\n3x &= 10b + a\n\\end{align*}\n$$\n\nAs $3x$ has the same number of digits as $x$, the leading digit $a$ must be $1$, $2$, or $3$. Eliminating $x$ gives:\n\n$$\na(3 \\cdot 10^{d-1} - 1) = 7b.\n$$\n\nSince $b < 10^{d-1}$, we get $(3a - 7)10^{d-1} < a$. For $d > 1$, this excludes $a = 3$, so $a = 1$ or $a = 2$. Also, $7$ divides $3 \\cdot 10^{d-1} - 1$, so:\n\n$$\n10^d \\equiv 3 \\cdot 10^{d-1} \\equiv 1 \\pmod{7}.\n$$\n\nChecking powers of $10$ modulo $7$, $10$ is a primitive root, so $10^d \\equiv 1 \\pmod{7}$ if and only if $d \\equiv 0 \\pmod{6}$. Let $d = 6c$ for integer $c > 0$. Then solutions are:\n\n$$\nb = \\frac{a}{7} (3 \\cdot 10^{6c-1} - 1), \\quad c > 0,\\ a \\in \\{1,2\\}.\n$$\n\nAdding the initial digit:\n\n$$\n\\begin{align*}\nx &= 10^{6c-1} a + \\frac{a}{7} (3 \\cdot 10^{6c-1} - 1) \\\\\n&= \\frac{a}{7} (7 \\cdot 10^{6c-1} + 3 \\cdot 10^{6c-1} - 1) = \\frac{a}{7} (10^{6c} - 1) \\\\\n&= \\frac{10^{6c} - 1}{10^6 - 1} \\cdot \\frac{10^6 - 1}{7} \\cdot a = \\frac{10^{6c} - 1}{10^6 - 1} \\cdot 142857 \\cdot a.\n\\end{align*}\n$$\n\nFor any $c \\ge 1$, $(10^{6c} - 1)/(10^6 - 1) = \\sum_{k=0}^{c-1} 10^{6k}$, which is $c$ ones separated by five zeros. Since $2 \\cdot 142857 = 285714$, the solutions are numbers formed by repeating $142857$ $c$ times and $285714$ $c$ times, for $c \\ge 1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22480,
"subject": "Mathematics (Olympiad)",
"question": "For each positive integer $n$, let $\\omega(n)$ denote the number of prime divisors of $n$. Find all polynomials $f(x)$ with integer coefficients such that, if $n$ is a positive integer satisfying $\\omega(n) > 2023^{2023}$, then $f(n)$ is also a positive integer and\n\n$$\n\\omega(f(n)) \\leq \\omega(n).\n$$",
"options": [],
"answer": "See solution",
"solution": "All polynomials of the form $f(x) = x^m$ for some $m \\in \\mathbb{Z}^+$ and $f(x) = c$ for some $c \\in \\mathbb{Z}^+$ with $\\omega(c) \\leq 2023^{2023} + 1$.\n\nFirst, we prove the following lemma:\n\n*Lemma.* Let $f(x)$ be a non-constant polynomial with integer coefficients. Then, the number of primes $p$ such that $p \\mid f(n)$ for some $n$ is infinite.\n\n*Proof.* If $f(0) = 0$, the lemma is obvious. Otherwise, define\n$$\ng(x) = \\frac{f(x f(0))}{f(0)}\n$$\nwhich has integer coefficients and $g(0) = 1$. If $g$ satisfies the lemma, so does $f$. Suppose, for contradiction, that only finitely many primes $p_1, \\dots, p_k$ divide some $g(n)$. Set $n = N p_1 \\cdots p_k$ for large $N$ so $|g(n)| > 1$. Then $g(n)$ has a prime divisor not among the $p_i$, a contradiction. $\\square$\n\nLet $M = 2023^{2023} + 1$. Constant polynomials $f(x) = c$ with $c \\in \\mathbb{N}$ and $\\omega(c) \\leq M$ satisfy the conditions. If $f(x) = c$ with $\\omega(c) > M$, choose $n$ with $\\omega(n) = M$ to see the condition fails. For non-constant polynomials, let $f(x) = x^m g(x)$ with $m \\geq 0$ and $g(0) \\neq 0$. We claim $g$ must be constant. Otherwise, by the lemma, there exist distinct primes $q_1, \\dots, q_{M+1}$ and integers $n_1, \\dots, n_{M+1}$ such that $q_i > |g(0)|$ and $q_i \\mid g(n_i)$. Set $n = p_1 \\cdots p_M$ with $p_1, \\dots, p_M$ distinct primes chosen so that\n$$\np_1 \\equiv n_i \\pmod{q_i}, \\quad \\forall i = 1, \\dots, M+1\n$$\nand\n$$\np_j \\equiv 1 \\pmod{q_i}, \\quad \\forall i = 1, \\dots, M+1, \\quad \\forall j = 2, \\dots, M.\n$$\nBy the Chinese Remainder Theorem and Dirichlet's theorem, such primes exist. Then $g(n) \\equiv 0 \\pmod{q_i}$ for all $i$, so $\\omega(f(n)) \\geq M+1 > M = \\omega(n)$, a contradiction. Thus, $f(x) = c x^m$ for $m \\geq 1$ and $c \\in \\mathbb{Z}$. If $c < 0$, $f(n)$ is negative for some $n$ with $\\omega(n) = M$, so it fails. If $c > 1$, choose $n$ with $\\omega(n) = M$ and $\\gcd(n, c) = 1$; then $f$ fails. Therefore, only $f(x) = x^m$ for $m \\geq 1$ and $f(x) = c$ for $c \\in \\mathbb{Z}^+$ with $\\omega(c) \\leq M$ are solutions.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22481,
"subject": "Mathematics (Olympiad)",
"question": "Let $a, b, c$ be nonnegative real numbers such that $a^2 + b^2 + c^2 + abc = 4$. Prove that\n$$\nab + bc + ca \\le 2 + abc.$$",
"options": [],
"answer": "See solution",
"solution": "Let $\\sqrt[3]{abc} = x$. The **AM-GM Inequality** yields\n$$\n4 = a^2 + b^2 + c^2 + abc \\ge 3\\sqrt{a^2b^2c^2} + abc\n$$\nimplying that $x \\le 1$.\n\nBy the AM-GM Inequality, $ab + bc + ca \\ge 3x^2$. Hence,\n$$\nab + bc + ca - abc \\ge 3x^2 - x^3 = x^2(3-x) \\ge 0,$$\nproving the lower bound.\n\nNow we prove the upper bound. Clearly $0 \\le a, b, c \\le 2$. Note that in a triangle $ABC$, we can apply the **Product-to-sum formulas** twice and the **Half-angle formulas** to find that\n$$\n\\begin{aligned}\n2 \\cos A \\cos B \\cos C &= [\\cos (A+B) + \\cos (A-B)] \\cos C \\\\\n&= -\\cos^2 C - \\cos (A-B) \\cos (A+B) \\\\\n&= -\\frac{\\cos 2A + \\cos 2B}{2} - \\cos^2 C \\\\\n&= - (\\cos^2 A + \\cos^2 B + \\cos^2 C) + 1.\n\\end{aligned}\n$$\nHence,\n$$\n\\cos^2 A + \\cos^2 B + \\cos^2 C + 2 \\cos A \\cos B \\cos C = 1.\n$$\nFrom the given equality, we have $4 \\ge a^2, 4 \\ge b^2$, and thus we may set $a = 2 \\cos A, b = 2 \\cos B$ where $0^\\circ \\le A, B \\le 90^\\circ$. Because $a^2 + b^2 + c^2 + abc$ is an increasing function of $c$, there is at most one nonnegative value $c$ such that the given equality holds. Dividing the given equality by 4 and substituting, we obtain\n$$\n\\cos^2 A + \\cos^2 B + \\left(\\frac{c}{2}\\right)^2 + 2 \\cos A \\cos B \\left(\\frac{c}{2}\\right) = 1.\n$$\nViewed as a quadratic in $c/2$, we know that one solution to this equation is $c/2 = \\cos C$, where $C = 180^\\circ - A - B$. Because $\\cos^2 A + \\cos^2 B = \\frac{1}{4}(a^2 + b^2) \\le 1$, we know that $A + B \\ge 90^\\circ$. Thus, $C \\le 90^\\circ$ and $\\cos C \\ge 0$. Therefore, we must have $c = 2 \\cos C$. Either two of $A, B, C$ are at least $60^\\circ$ or two of $A, B, C$ are at most $60^\\circ$. Without loss of generality, assume that $A$ and $B$ have this property.\n\nWith these trigonometric substitutions, we find that the desired inequality is equivalent to\n$$\n\\begin{aligned}\n& 2(\\cos A \\cos B + \\cos B \\cos C + \\cos C \\cos A) \\\\\n& \\le 1 + 4 \\cos A \\cos B \\cos C,\n\\end{aligned}\n$$\nor\n$$\n\\begin{aligned}\n& 2(\\cos A \\cos B + \\cos B \\cos C + \\cos C \\cos A) \\\\\n& \\le 3 - 2(\\cos^2 A + \\cos^2 B + \\cos^2 C).\n\\end{aligned}\n$$\nUsing the **Double-angle formulas**, the last inequality is equivalent to\n$$\n\\begin{aligned}\n& \\cos 2A + \\cos 2B + \\cos 2C \\\\\n& + 2(\\cos A \\cos B + 2 \\cos B \\cos C + \\cos C \\cos A) \\le 0.\n\\end{aligned}\n$$\nThe sum of the first three terms in this inequality is\n$$\n\\begin{aligned}\n& \\cos 2A + \\cos 2B + \\cos 2C \\\\\n&= 2 \\cos (A + B) \\cos (A - B) + 2 \\cos^2 (A + B) - 1 \\\\\n&= 2 \\cos (A + B)[\\cos (A - B) + \\cos (A + B)] - 1 \\\\\n&= 4 \\cos (A + B) \\cos A \\cos B - 1,\n\\end{aligned}\n$$\nwhile the remaining terms equal\n$$\n\\begin{aligned}\n& 2 \\cos A \\cos B + 2 \\cos C(\\cos A + \\cos B) \\\\\n&= \\cos (A + B) + \\cos (A - B) - 2 \\cos (A + B)(\\cos A + \\cos B).\n\\end{aligned}\n$$\nHence, it suffices to prove that\n$$\n\\cos (A + B)[4 \\cos A \\cos B + 1 - 2 \\cos A - 2 \\cos B] + \\cos (A - B) - 1 \\le 0,\n$$\nor\n$$\n- \\cos C(1 - 2 \\cos A)(1 - 2 \\cos B) + \\cos (A - B) \\le 1.\n$$\nWe consider the following cases.\n\n(i) *At least one angle is 60°.* If $A$ or $B$ equals $60^\\circ$, then we may assume, without loss of generality, that $A = 60^\\circ$. If $C = 60^\\circ$, then because either $A, B \\ge 60^\\circ$ or $A, B \\le 60^\\circ$, we must actually have $A = B = 60^\\circ$ in which case equality holds. In either case we may assume $A = 60^\\circ$. Then (7) becomes $\\cos(A - B) \\le 1$, which is true, and equality holds if and only if $A = B = C = 60^\\circ$, that is, $a = b = c = 1$.\n\n(ii) *No angle equals 60°.* Because either $A, B \\ge 60^\\circ$ or $A, B \\le 60^\\circ$, we have $(1 - 2 \\cos A)(1 - 2 \\cos B) > 0$. Since $\\cos C \\ge 0$ and $\\cos(A - B) \\le 1$, (7) is true. Equality holds when $\\cos C = 0$ and $\\cos(A - B) = 1$. This holds exactly when $A = B = 45^\\circ$ and $C = 90^\\circ$, that is, when $a = b = \\sqrt{2}$ and $c = 0$.\n\n![No image present](no_image \"No image present\")",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22482,
"subject": "Mathematics (Olympiad)",
"question": "Let $N$ be a nonnegative integer and let $f, g: \\mathbb{Z} \\to [0, \\infty)$ be functions such that $f(n) = g(n) = 0$ for all $|n| \\geq N$, where $\\mathbb{Z}$ is the set of all integers. Define $h: \\mathbb{Z} \\to [0, \\infty)$ by\n\n$$\nh(n) = \\max \\{ f(k) g(n - k) : k \\in \\mathbb{Z} \\}\n$$\n\nfor all $n \\in \\mathbb{Z}$. Prove that\n\n$$\n\\sum_{n \\in \\mathbb{Z}} h(n) \\geq \\left( \\sum_{n \\in \\mathbb{Z}} (f(n))^p \\right)^{1/p} \\left( \\sum_{n \\in \\mathbb{Z}} (g(n))^q \\right)^{1/q}\n$$\n\nfor all positive real numbers $p$ and $q$ satisfying $1/p + 1/q = 1$.",
"options": [],
"answer": "See solution",
"solution": "Let $m_0$ be an integer at which $f$ achieves its maximum. Then $h(n) \\geq f(m_0) g(n - m_0)$ for all $n \\in \\mathbb{Z}$, and\n\n$$\n\\sum_{n \\in \\mathbb{Z}} h(n) \\geq f(m_0) \\sum_{n \\in \\mathbb{Z}} g(n - m_0) = f(m_0) \\sum_{n \\in \\mathbb{Z}} g(n).\n$$\n\nSimilarly, if $n_0$ is an integer at which $g$ achieves its maximum, then\n\n$$\n\\sum_{n \\in \\mathbb{Z}} h(n) \\geq g(n_0) \\sum_{n \\in \\mathbb{Z}} f(n - n_0) = g(n_0) \\sum_{n \\in \\mathbb{Z}} f(n).\n$$\n\nCombining these yields\n\n$$\n\\sum_{n \\in \\mathbb{Z}} h(n) \\geq (f(m_0))^{1/q} \\left( \\sum_{n \\in \\mathbb{Z}} g(n) \\right)^{1/q} (g(n_0))^{1/p} \\left( \\sum_{n \\in \\mathbb{Z}} f(n) \\right)^{1/p}.\n$$\n\nSince\n\n$$\n(f(m_0))^{1/q} \\left( \\sum_{n \\in \\mathbb{Z}} f(n) \\right)^{1/p} = \\left( (f(m_0))^{p-1} \\sum_{n \\in \\mathbb{Z}} f(n) \\right)^{1/p} \\geq \\left( \\sum_{n \\in \\mathbb{Z}} (f(n))^p \\right)^{1/p},\n$$\n\nand\n\n$$\n(g(n_0))^{1/p} \\left( \\sum_{n \\in \\mathbb{Z}} g(n) \\right)^{1/q} \\geq \\left( \\sum_{n \\in \\mathbb{Z}} (g(n))^q \\right)^{1/q},\n$$\n\nthe conclusion follows.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22483,
"subject": "Mathematics (Olympiad)",
"question": "令 $n$ 是大於 $1$ 的正整數且 $p$ 為一質數，使得 $n$ 能整除 $p-1$ 且 $p$ 能整除 $n^3-1$。\n\n試證：$4p-3$ 是一完全平方數。",
"options": [],
"answer": "See solution",
"solution": "因 $n \\mid (p-1)$，所以 $n < p$。由於 $p \\mid (n^3-1) = (n-1)(n^2+n+1)$，$p$ 必整除 $n^2+n+1$。\n\n令 $p-1 = n\\ell$。又\n\n$$\nn^2 + n + 1 = n^2 + n + p - n\\ell = n(n + 1 - \\ell) + p\n$$\n\n即得 $p \\mid n(n+1-\\ell)$。因 $n < p$，$p \\mid (n+1-\\ell)$。又因 $\\ell > 0$，$n+1-\\ell \\leq n$。\n\n今 $p \\mid (n+1-\\ell)$ 且 $n < p$，得 $\\ell \\geq n+1$。若 $\\ell > n+1$，則 $p = 1 + n\\ell > 1 + n + n^2$，此為不可能，因 $p \\mid (n^2+n+1)$。\n\n所以 $\\ell = n+1$ 且 $p = n^2 + n + 1$。\n\n故\n\n$$\n4p-3 = 4(n^2 + n + 1) - 3 = (2n + 1)^2.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22484,
"subject": "Mathematics (Olympiad)",
"question": "Let $m, n$ be positive integers with $m > 1$. Anastasia partitions the integers $1, 2, \\ldots, 2m$ into $m$ pairs. Boris then chooses one integer from each pair and finds the sum of these chosen integers. Prove that Anastasia can select the pairs so that Boris cannot make his sum equal to $n$.",
"options": [],
"answer": "See solution",
"solution": "Define the following ordered partitions:\n\n$$\nP_1 = (\\{1,2\\}, \\{3,4\\}, \\dots, \\{2m-1,2m\\})\n$$\n\n$$\nP_2 = (\\{1,m+1\\}, \\{2,m+2\\}, \\dots, \\{m,2m\\})\n$$\n\n$$\nP_3 = (\\{1,2m\\}, \\{2,m+1\\}, \\{3,m+2\\}, \\dots, \\{m,2m-1\\})\n$$\n\nFor each $P_j$ we will compute the possible values for the expression $s = a_1 + \\dots + a_m$, where $a_i \\in P_{j,i}$ are the chosen integers. Here $P_{j,i}$ denotes the $i$-th pair in the ordered partition $P_j$. Let $\\sigma = \\sum_{i=1}^m i = \\frac{m^2 + m}{2}$.\n\n* For partition $P_1$, the possible sums satisfy:\n\n$$\nm^2 = \\sum_{i=1}^{m} (2i-1) \\leq s \\leq \\sum_{i=1}^{m} 2i = m^2 + m\n$$\n\nSo, if $n < m^2$ or $n > m^2 + m$, this partition works.\n\n* For partition $P_2$, the sum $s$ satisfies:\n\n$$\ns \\equiv \\sum_{i=1}^{m} i \\equiv \\sigma \\pmod{m}\n$$\n\nSo, if $m^2 \\leq n \\leq m^2 + m$ and $n \\not\\equiv \\sigma \\pmod{m}$, this partition works.\n\n* For partition $P_3$, define\n\n$$\nd_i = \\begin{cases} 0, & \\text{if } a_i = i \\\\ 1, & \\text{if } a_i \\neq i \\end{cases}\n$$\n\nLet $d = \\sum_{i=1}^m d_i$, with $0 \\leq d \\leq m$. If $a_i \\neq i$, then $a_i \\equiv i-1 \\pmod{m}$, so for all $a_i \\in P_{3,i}$:\n\n$$\na_i \\equiv i - d_i \\pmod{m}\n$$\n\nThus,\n\n$$\ns \\equiv \\sum_{i=1}^{m} a_i \\equiv \\sum_{i=1}^{m} (i - d_i) \\equiv \\sigma - d \\pmod{m}\n$$\n\nThis can only be congruent to $\\sigma$ modulo $m$ if all $d_i$ are equal, which forces $s = \\frac{m^2 + m}{2}$ or $s = \\frac{3m^2 + m}{2}$. Since $m > 1$,\n\n$$\n\\frac{m^2 + m}{2} < m^2 < m^2 + m < \\frac{3m^2 + m}{2}\n$$\n\nSo if $m^2 \\leq n \\leq m^2 + m$ and $n \\equiv \\sigma \\pmod{m}$, then $s$ cannot be equal to $n$, so partition $P_3$ works.\n\nAll $n$ are covered by one of these cases, so Anastasia can always select the pairs so that Boris cannot make his sum equal to $n$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22485,
"subject": "Mathematics (Olympiad)",
"question": "Let $a_1, a_2, a_3, \\dots$ be a sequence of integers with the property that every consecutive group of $a_i$'s averages to a perfect square. More precisely, for every positive integers $n$ and $k$, the quantity\n\n$$\n\\frac{a_n + a_{n+1} + \\dots + a_{n+k-1}}{k}\n$$\n\nis always the square of an integer. Prove that the sequence must be constant (all $a_i$ are equal to the same perfect square).",
"options": [],
"answer": "See solution",
"solution": "We prove the following equivalent statement: if $f : \\mathbb{N} \\to \\mathbb{Z}$ is a function such that $(f(m) - f(n))(m - n)$ is always the square of an integer, then $f$ must be of the form $A^2x + B$ for integers $A, B$.\n\nFirst, since $p(f(n + p) - f(n))$ is a square for any prime $p$ and positive integer $n$, $p$ must divide $f(n + p) - f(n)$. (Similarly, we may prove $p^k \\mid f(n + p^k) - f(n)$ by induction, but we will not need to do so.)\n\nWe now investigate the structure of $f$ modulo $p$. For convenience, let $\\overline{A} = A \\pmod{p}$.\n\n**Lemma.** If $p \\mid f(a) - f(b)$ but $p \\nmid a - b$ for some $a, b$, then $p \\mid f(m) - f(n)$ for all $m, n$. (In other words, $f$ is either injective or constant modulo $p$.)\n\n*Proof.* For each integer $r$, define $S_r = \\{x : f(x) \\equiv r \\pmod{p}\\}$, which must be a union of residue classes modulo $p$ (restricted to the positive integers). Then we need to show that either (1) for every $r$, $S_r$ covers at most one residue class (and thus exactly one), or (2) for some $r$, $S_r$ covers all residue classes.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 22486,
"subject": "Mathematics (Olympiad)",
"question": "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": "**Solution**\n\nIt is easy to see that $\\xi$ can only be $2$, $4$, or $6$. We divide the six games into three rounds, each consisting of two consecutive games. If one of the players wins two games in the first round, the match ends and the probability is\n\n$$\n\\left(\\frac{2}{3}\\right)^2 + \\left(\\frac{1}{3}\\right)^2 = \\frac{5}{9}.\n$$\n\nOtherwise, the players tie with each other, earning one point each, and the match enters the second round; this probability is\n\n$$\n1 - \\frac{5}{9} = \\frac{4}{9}.\n$$\n\nWe have similar discussions for the second and third rounds. So we get\n\n$$\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$$\n\nThen\n\n$$\nE\\xi = 2 \\times \\frac{5}{9} + 4 \\times \\frac{20}{81} + 6 \\times \\frac{16}{81} = \\frac{266}{81}.\n$$\n\n**Answer:** B",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22487,
"subject": "Mathematics (Olympiad)",
"question": "Let $K_1$ and $K_2$ be circles that intersect at two points $A$ and $B$. The tangents to $K_1$ at $A$ and $B$ intersect at a point $P$ inside $K_2$, and the line $BP$ intersects $K_2$ again at $C$. The tangents to $K_2$ at $A$ and $C$ intersect at a point $Q$, and the line $QA$ intersects $K_1$ again at $D$.\n\nProve that $QP$ is perpendicular to $PD$ if and only if the centre of $K_2$ lies on $K_1$.",
"options": [],
"answer": "See solution",
"solution": "We first prove that if the centre $O$ of $K_2$ lies on $K_1$, then $QP$ is perpendicular to $PD$.\n\nSince $DA$ is tangent to $K_2$, we have $\\angle DAO = 90^\\circ$, so $OD$ is a diameter of $K_1$. By symmetry, $D$, $O$, and $P$ all lie on the line joining the centres of $K_1$ and $K_2$. Observe that $A$ is the reflection of $B$ in this line. Hence, $\\angle PAO = \\angle PBO$, but $\\angle PBO = \\angle CBO = \\angle BCO$, since triangle $BOC$ is isosceles. It follows that $AOPC$ is a cyclic quadrilateral. Moreover, $\\angle OAQ = \\angle OCQ = 90^\\circ$, so $OAQC$ is also a cyclic quadrilateral. Therefore, $OAQCP$ is a cyclic pentagon and $\\angle QPO = 90^\\circ$, as required.\n\nWe now prove that if $QP$ is perpendicular to $PD$, then the centre $O$ of $K_2$ lies on $K_1$. By the alternate segment theorem, we have $\\angle QCA = \\angle CBA$, and it follows that\n\n$$\n\\angle QAC = \\angle QCA = \\angle PAB = \\angle PBA.\n$$\n\nLet this angle be $\\alpha$. Then $\\angle APB = 180^\\circ - 2\\alpha = \\angle AQC$, so quadrilateral $AQCP$ is cyclic. Hence, $\\angle APQ = \\alpha = \\angle CPQ$ and we have $\\angle APD = \\angle BPD = 90^\\circ - \\alpha$. Therefore, $D$, $O$, $P$ and the centre $X$ of $K_1$ are collinear. Since $\\angle DAO = 90^\\circ$, we can deduce that $\\angle XOA = 90^\\circ - \\angle XDA = 90^\\circ - \\angle XAD = \\angle XAO$, so $XO = XA$. It follows that $O$ lies on $K_1$, as required.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22488,
"subject": "Mathematics (Olympiad)",
"question": "A rectangle has sides of lengths $a$ and $b$ (with $a \\geq b$). The rectangle is placed on a grid of unit squares. How many possible areas can the rectangle have if the number of unit squares that have at least one point in common with the rectangle is $24$?\n\n",
"options": [],
"answer": "See solution",
"solution": "Let $a$ and $b$ be the side lengths of the rectangle, with $a \\geq b$.\n\nThe unit squares that have at least one point in common with the rectangle are counted. This is equivalent to the area covered by the rectangle enlarged by $1$ unit in all directions, minus the area not covered.\n\nIf $b \\leq 2$, there is no white rectangle inside, so the number of unit squares is $(a+2)(b+2) = 24$. This gives:\n- For $b=1$: $(a+2) \\cdot 3 = 24 \\implies a=6$\n- For $b=2$: $(a+2) \\cdot 4 = 24 \\implies a=4$\n\nIf $b \\geq 3$, the number of unit squares is $(a+2)(b+2) - (a-2)(b-2) = 4a + 4b = 24$, so $a + b = 6$. With $a \\geq b$, the only possibility is $a = b = 3$.\n\nThus, all possible areas are:\n- $1 \\times 6 = 6$\n- $2 \\times 4 = 8$\n- $3 \\times 3 = 9$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22489,
"subject": "Mathematics (Olympiad)",
"question": "Given a triangle $ABC$ with $m(\\angle A) = 90^\\circ$ and $AC < AB$, consider points $E$ and $D$ on the rays $BA$ and $AC$ respectively, such that $A \\in (BE)$, $C \\in (AD)$, $AE = AC$, and $AD = AB$. Denote by $M$ and $N$ the midpoints of $[BC]$ and $[DE]$ respectively, and let $\\{R\\} = EC \\cap BD$. Show that $MN = RA$.",
"options": [],
"answer": "See solution",
"solution": "The hypothesis implies that triangles $\\triangle ACE$ and $\\triangle ABD$ are right-angled and isosceles, so the triangle $\\triangle RBE$ is also right-angled and isosceles, that is $m(\\widehat{RBE}) = m(\\widehat{REB}) = 45^\\circ$. By the equality of triangles $\\triangle ABC$ and $\\triangle ADE$ (by congruence), we get $BC = DE$.\n\n\n\nIn the triangles $BRC$, $ABC$, $RDE$, and $ADE$, the lines $RM$, $AM$, $RN$, $AN$ are medians corresponding to right angles, so $RM = \\frac{1}{2}BC$, $AM = \\frac{1}{2}BC$, $RN = \\frac{1}{2}DE$, $AN = \\frac{1}{2}DE$, implying $RM = AM = RN = AN$, that is, the quadrilateral $AMRN$ is a rhombus.\n\nIn the right-angled triangle $ABC$, as $AM$ is a median, the triangle $\\triangle MAC$ is isosceles, so $m(\\widehat{MAC}) = m(\\widehat{MCA})$. In the same way, in $\\triangle ADE$, $m(\\widehat{NAD}) = m(\\widehat{NDA})$. This gives $m(\\widehat{MAN}) = m(\\widehat{MAC}) + m(\\widehat{NAC}) = m(\\widehat{MCA}) + m(\\widehat{NDA}) = m(\\widehat{MCA}) + m(\\widehat{ABC}) = 90^\\circ$. Thus, by the previous result, the quadrilateral $AMRN$ is a square, and as a conclusion $MN = RA$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22490,
"subject": "Mathematics (Olympiad)",
"question": "Show that real numbers $p, q, r$ satisfy the condition\n\n$$\np^4(q - r)^2 + 2p^2(q + r) + 1 = p^4\n$$\n\nif and only if the quadratic equations\n\n$$\nx^2 + px + q = 0 \\quad \\text{and} \\quad y^2 - py + r = 0\n$$\nhave real roots (not necessarily distinct) which can be labelled by $x_{1,2}$ and $y_{1,2}$, respectively, in such a way that the equality $x_1 y_1 - x_2 y_2 = 1$ holds.",
"options": [],
"answer": "See solution",
"solution": "First part. Assume that the equations above have real roots satisfying $x_1 y_1 - x_2 y_2 = 1$. By the quadratic formula, the roots are\n\n$$\nx_{1,2} = \\frac{-p \\pm K}{2}, \\quad y_{1,2} = \\frac{p \\pm L}{2},\n$$\nwhere $K^2 = p^2 - 4q$ and $L^2 = p^2 - 4r$ (the signs of $K, L$ are chosen according to the labelling). Thus,\n\n$$\n1 = x_1 y_1 - x_2 y_2 = \\frac{(-p + K)(p + L) - (-p - K)(p - L)}{4} = \\frac{p(K - L)}{2},\n$$\nso $p \\neq 0$ and $K - L = \\frac{2}{p}$. Also,\n\n$$\n(K + L)(K - L) = K^2 - L^2 = (p^2 - 4q) - (p^2 - 4r) = 4(r - q),\n$$\nso $K + L = 2p(r - q)$. From $K + L$ and $K - L$, we get $K = \\frac{1}{p} - p(q - r)$. Squaring gives $K^2 = \\frac{1}{p^2} - 2(q - r) + p^2(q - r)^2$. Comparing with $K^2 = p^2 - 4q$ leads to the desired equation.\n\nSecond part. Assume that the given equation holds. Then $p \\neq 0$. The equation can be rewritten as\n\n$$\np^4(q - r)^2 + 2p^2(q - r) + 1 = p^4 - 4p^2 r\n$$\nand\n$$\np^4(r - q)^2 + 2p^2(r - q) + 1 = p^4 - 4p^2 q.\n$$\nDividing by $p^2$ shows the discriminants of the quadratics are\n\n$$\np^2 - 4q = \\left( \\frac{p^2(r - q) + 1}{p} \\right)^2, \\quad p^2 - 4r = \\left( \\frac{p^2(q - r) + 1}{p} \\right)^2,\n$$\nwhich are nonnegative, so the roots are real and have the form above, with\n\n$$\nK = \\frac{p^2(r - q) + 1}{p}, \\quad L = -\\frac{p^2(q - r) + 1}{p}.\n$$\nWith these choices,\n\n$$\nx_1 y_1 - x_2 y_2 = \\frac{p(K - L)}{2} = \\frac{p}{2} \\left( \\frac{p^2(r - q) + 1}{p} + \\frac{p^2(q - r) + 1}{p} \\right) = 1.\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22491,
"subject": "Mathematics (Olympiad)",
"question": "\n\nLet $ABC$ be a triangle, and let $M$ be a point on $AC$, $K$ on $BC$, and $L$ on $AB$ such that $ML \\parallel BC$, $KL \\parallel AC$, and $MK \\parallel AB$. Let $S$ and $T$ be points such that quadrilaterals $AKCT$ and $ASBK$ are cyclic. Under what conditions is quadrilateral $SLMT$ cyclic?",
"options": [],
"answer": "See solution",
"solution": "By the Intercept Theorem, $ML \\parallel BC$ is equivalent to $\\frac{AM}{MC} = \\frac{AL}{LB}$. On the other hand, $KL \\parallel AC$ and $MK \\parallel AB$ imply\n\n$$\n\\frac{CK}{KB} = \\frac{AL}{LB} \\quad \\text{and} \\quad \\frac{AM}{MC} = \\frac{BK}{KC}.\n$$\n\nHence, *SLMT* is cyclic if and only if $\\frac{CK}{KB} = \\frac{BK}{KC}$, i.e. $CK = KB$.\n\nDenote lengths of segments as follows:\n\n$$\n\\begin{aligned} c &= |AB| & x &= |AM| = |KL| \\\\ b &= |AC| & y &= |AL| = |MK|. \\end{aligned}\n$$\n\nSince $LK \\parallel AC$ and $MK \\parallel AB$, we have $\\triangle MKC \\sim \\triangle ABC \\sim \\triangle LBK$, hence\n\n$$\n\\frac{b-x}{y} = \\frac{b}{c} = \\frac{x}{c-y} \\quad \\text{and so} \\quad y = \\frac{c}{b}(b-x) \\quad \\text{and} \\quad c-y = \\frac{c}{b} \\cdot x. \\quad (13)\n$$\n\nSince *AKCT* and *ASBK* are cyclic, the Intersecting Chords Theorem gives\n\n$$\n\\begin{aligned} |KM| \\cdot |MT| &= |AM| \\cdot |MC| = x(b-x) \\\\ |KL| \\cdot |LS| &= |AL| \\cdot |LB| = y(c-y). \\end{aligned}\n$$\n\nFinally, *SLMT* is cyclic iff $\\triangle KLM \\sim \\triangle KTS$, which in turn is equivalent to $\\frac{|KM|}{|KL|} = \\frac{|KS|}{|KT|}$ which we rewrite as follows in equivalent ways\n\n$$\n\\begin{aligned}\n& |KM| \\cdot |KT| = |KL| \\cdot |KS| \\\\\n& |KM|^2 + |KM| \\cdot |MT| = |KL|^2 + |KL| \\cdot |LS| \\\\\n& y^2 + x(b-x) = x^2 + y \\cdot (c-y) \\\\\n& y^2 - x^2 + x(b-x) - y(c-y) = 0\n\\end{aligned}\n$$\n\n$$\n\\frac{c^2}{b^2}(b-x)^2 - x^2 + x(b-x) - \\frac{c^2}{b^2} \\cdot x(b-x) = 0 \\quad \\text{(use (13) to eliminate } y)\n$$\n\n$$\n(b - 2x) \\left( \\frac{c^2}{b^2} (b - x) + x \\right) = 0.\n$$\n\nThe last equation holds true iff $b = 2x$, i.e. $M$ is the midpoint of $AC$, which is equivalent to $K$ being the midpoint of $BC$, as $MK \\parallel AB$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22492,
"subject": "Mathematics (Olympiad)",
"question": "Find all positive integers $n$ such that the equation\n\n$$\nx + y + u + v = n\\sqrt{xyuv}\n$$\n\nhas positive integer solutions $x, y, u, v$.",
"options": [],
"answer": "See solution",
"solution": "For $x, y, u, v \\in \\mathbb{Z}^+$, we can write the given equation as\n\n$$\n(x + y + u + v)^2 = n^2xyuv\n$$\n\nor equivalently,\n\n$$\nx^2 + 2(y + u + v)x + (y + u + v)^2 = n^2xyuv \\quad (1)\n$$\n\nLet $n$ be a positive integer such that (1) has positive integer solutions $(x, y, u, v)$. Let $(x_0, y_0, u_0, v_0)$ be a positive integer solution of (1) such that $x_0 + y_0 + u_0 + v_0$ is minimized. Without loss of generality, suppose $x_0 \\geq y_0 \\geq u_0 \\geq v_0$. It is easily seen that:\n\n1. $(y_0 + u_0 + v_0)^2$ is divisible by $x_0$.\n2. $x_0$ is a positive integer root of a quadratic polynomial.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 22493,
"subject": "Mathematics (Olympiad)",
"question": "If $n = pq$ for some distinct primes $p$ and $q$, then\n\n$$\n\\varphi(pq)\\sigma(pq) + 1 = (p-1)(q-1)(p+1)(q+1) + 1 = p^2q^2 - p^2 - q^2 + 2.\n$$\n\nThis is a multiple of $pq$ if and only if $pq \\mid p^2 + q^2 - 2$. Prove that $ab \\mid a^2 + b^2 - 2$ has no solution in distinct integers $a, b > 1$.",
"options": [],
"answer": "See solution",
"solution": "Suppose, for contradiction, that the divisibility $ab \\mid a^2 + b^2 - 2$ is solvable for distinct integers $a, b > 1$. Let $(a, b)$ be a solution with $a > b > 1$ and $b$ minimal. Let $a^2 + b^2 - 2 = kab$ for some $k \\in \\mathbb{N}$. Then\n\n$$\na^2 - kba + (b^2 - 2) = 0.\n$$\n\nThis quadratic in $a$ has another positive integer root $a' = \\frac{b^2 - 2}{a} < b < a$. Thus, $(b, a')$ is another solution with $a' < b$, contradicting the minimality of $b$ unless $a' = 1$. But then $b \\mid 1 + b^2 - 2 = b^2 - 1$, so $b \\mid 1$, impossible for $b > 1$. Thus, there is no solution.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22494,
"subject": "Mathematics (Olympiad)",
"question": "Find all real numbers $a$ and $b$ such that the system\n\n$$\n\\begin{cases}\nx + a = y + b \\\\\nx^2 - a = 2y\n\\end{cases}\n$$\n\nhas a unique solution $(x_0, y_0)$ and it satisfies the equality $x_0^2 + y_0^2 = 1025$.",
"options": [],
"answer": "See solution",
"solution": "The given system is equivalent to:\n\n$$\n\\begin{cases}\nx^2 - 2x + 2b - 3a = 0 \\\\\nx + a = y + b\n\\end{cases}\n$$\n\nThe system has a unique solution $(x_0, y_0)$ if the quadratic equation $x^2 - 2x + 2b - 3a = 0$ has a unique root $x_0$. This occurs when the discriminant $D = 1 - 2b + 3a = 0$, so $x_0 = 1$.\n\nThe condition $x_0^2 + y_0^2 = 1025$ gives $y_0^2 = 1024$, so $y_0 = \\pm 32$.\n\nFor $y_0 = -32$:\n- $a = x_0^2 - 2y_0 = 1 - 2(-32) = 65$\n- $b = x_0 + a - y_0 = 1 + 65 - (-32) = 98$\n\nFor $y_0 = 32$:\n- $a = x_0^2 - 2y_0 = 1 - 2(32) = -63$\n- $b = x_0 + a - y_0 = 1 - 63 - 32 = -94$\n\nThus, the solutions are $(a, b) = (65, 98)$ and $(a, b) = (-63, -94)$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22495,
"subject": "Mathematics (Olympiad)",
"question": "Let $a$, $b$, and $c$ be positive integers satisfying the equation $\\gcd(a, b) + \\operatorname{lcm}(a, b) = 2021^c$. If $|a - b|$ is a prime number, prove that the number $(a + b)^2 + 4$ is composite.",
"options": [],
"answer": "See solution",
"solution": "Let $p = |a - b|$ and suppose, for contradiction, that $q = (a + b)^2 + 4$ is a prime number.\n\nSince $\\gcd(a, b) \\mid \\operatorname{lcm}(a, b)$, we have $\\gcd(a, b) \\mid 2021^c$. Also, $\\gcd(a, b)$ divides $p = |a - b|$, so $\\gcd(a, b) \\in \\{1, 43, 47\\}$ (since $2021 = 43 \\times 47$). Consider each case:\n\n**Case 1:** $\\gcd(a, b) = 1$\n\nThen $1 + ab = 2021^c$, so\n$$\nq = (a + b)^2 + 4 = (a - b)^2 + 4(1 + ab) = p^2 + 4 \\cdot 2021^c.\n$$\n(a) If $c$ is even, $q \\equiv 1 \\pmod{4}$, so it can be represented uniquely as a sum of two squares. But the above gives two representations, so uniqueness requires $p = 2$. Then $4 \\mid q$, a contradiction.\n\n(b) If $c$ is odd, $ab = 2021^c - 1 \\equiv 1 \\pmod{3}$, so $a \\equiv b \\pmod{3}$, implying $p = 3$. Let $b = a + 3$, then $2021^c = a^2 + 3a + 1$ and\n$$\n(2a + 3)^2 = 4a^2 + 12a + 9 = 4 \\cdot 2021^c + 5.\n$$\nSo $5$ must be a quadratic residue modulo $47$, but\n$$\n\\left(\\frac{5}{47}\\right) = \\left(\\frac{47}{5}\\right) = \\left(\\frac{2}{5}\\right) = -1,\n$$\nwhich is a contradiction.\n\n**Case 2:** $\\gcd(a, b) = 43$\n\nThen $p = 43$, so $a = 43k$, $b = 43(k + 1)$ for some $k \\in \\mathbb{N}$. Then $2021^c = 43 + 43k(k + 1)$, so\n$$\n(2k + 1)^2 = 4k^2 + 4k + 4 - 3 = 4 \\cdot 43^{c-1} \\cdot 47 - 3.\n$$\nSo $-3$ must be a quadratic residue modulo $47$, but\n$$\n\\left(\\frac{-3}{47}\\right) = \\left(\\frac{-1}{47}\\right) \\left(\\frac{3}{47}\\right) = \\left(\\frac{47}{3}\\right) = \\left(\\frac{2}{3}\\right) = -1,\n$$\nwhich is a contradiction.\n\n**Case 3:** $\\gcd(a, b) = 47$\n\nSimilarly, for some $k \\in \\mathbb{N}$,\n$$\n(2k + 1)^2 = 4 \\cdot 43^c \\cdot 47^{c-1} - 3.\n$$\nIf $c > 1$, the contradiction is as in Case 2. If $c = 1$, $(2k + 1)^2 = 169$ so $k = 6$. Then $a + b = 47 \\cdot 6 + 47 \\cdot 7 = 47 \\cdot 13 \\equiv 1 \\pmod{5}$, so $q = (a + b)^2 + 4 \\equiv 0 \\pmod{5}$, a contradiction.\n\nThus, in all cases, $(a + b)^2 + 4$ cannot be prime, so it is composite.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 22496,
"subject": "Mathematics (Olympiad)",
"question": "Let $m$ and $n$ be positive integers and let the integer $X \\geq \\max(m, n)$. Show that there exist integers $u$ and $v$, not both equal to $0$, such that\n$$\n\\max(|u|, |v|) \\leq \\sqrt{X} \\quad \\text{and} \\quad 0 \\leq m u + n v \\leq 2\\sqrt{X}.\n$$",
"options": [],
"answer": "See solution",
"solution": "There are $[\\sqrt{X} + 1]^2 \\geq X + 1$ pairs $(a, b)$ such that $0 \\leq a, b \\leq \\sqrt{X}$, and for these\n$$\n0 \\leq m a + n b \\leq 2 X \\sqrt{X}.\n$$\nTwo linear combinations $m a + n b$ and $m a' + n b'$ differ by at most $2\\sqrt{X}$, and so\n$$\n\\max(|a - a'|, |b - b'|) \\leq \\sqrt{X} \\quad \\text{and} \\quad 0 \\leq m(a - a') + n(b - b') \\leq 2\\sqrt{X}.\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22497,
"subject": "Mathematics (Olympiad)",
"question": "Determine all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\n(f(x) - y) \\cdot f(x + f(y)) = f(x^2) - y f(y)\n$$\nfor every real numbers $x$ and $y$.",
"options": [],
"answer": "See solution",
"solution": "Let $P(x, y)$ denote the assertion:\n\n$$\n(f(x) - y) \\cdot f(x + f(y)) = f(x^2) - y f(y)\n$$\n\nfor all real $x, y$.\n\n**Lemma:** If there exist two distinct real numbers $y_1 \\ne y_2$ such that $f(y_1) = f(y_2) = k$, then $f$ is constant: $f(x) = k$ for all $x$.\n\n*Proof of Lemma:* Apply $P(x, y_1)$ and $P(x, y_2)$:\n\n$$(f(x) - y_1) f(x + k) = f(x^2) - k y_1$$\n$$(f(x) - y_2) f(x + k) = f(x^2) - k y_2$$\n\nSubtracting gives:\n\n$$(y_1 - y_2) f(x + k) = k(y_1 - y_2) \\implies f(x + k) = k$$\n\nfor all $x$, so $f$ is constant.\n\nNow, consider $P(0, 0)$:\n\n$$f(0) f(f(0)) = f(0)$$\n\nSuppose $f(0) = 0$.\n\nIf $f$ is not injective, by the lemma, $f(x) = 0$ for all $x$.\n\nOtherwise, $f$ is injective. From $P(0, y)$:\n\n$$-y f(f(y)) = -y f(y) \\implies f(f(y)) = f(y)$$\n\nSince $f$ is injective, $f(y) = y$ for all $y$.\n\nIf $f(0) \\ne 0$, then $f(f(0)) = 1$.\n\nFrom $P(1, f(1))$:\n\n$$f(1) = f(1) f(f(1))$$\n\nIf $f(1) = 0$, then $P(1, f(0))$, $P(2, 1)$, and $P(2, 4)$ lead to contradictions. Thus, $f(f(1)) = 1$.\n\nIn both cases ($f(0) = f(1)$ or $f(0) \\ne f(1)$), by the lemma, $f(x) = 1$ for all $x$.\n\n**Conclusion:** The solutions are:\n\n- $f(x) = 0$\n- $f(x) = 1$\n- $f(x) = x$\n\nAll three satisfy the given functional equation.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22498,
"subject": "Mathematics (Olympiad)",
"question": "Given a positive integer $n$, show that the set $\\{1, 2, \\dots, n\\}$ can be partitioned into $m$ sets, each with the same sum, if and only if $m$ is a divisor of $n(n+1)/2$ which does not exceed $(n+1)/2$.",
"options": [],
"answer": "See solution",
"solution": "The necessity of the two conditions is easy to establish. If each block in the partition has the sum $s$, then $ms = 1 + 2 + \\dots + n = n(n+1)/2$, which gives the divisibility condition. Also, $n \\ge 2m-1$, for there can be at most one block with a single element.\n\nTo prove sufficiency, call a set of parameters $n, m, s$ admissible if $ms = n(n + 1)/2$ and $n \\ge 2m-1$. If $n = 2m-1$, then $s = 2m-1$, and the partition is unique:\n\n$$\n\\{2m-1\\}, \\quad \\{2m-2,1\\}, \\quad \\{2m-3,2\\}, \\quad \\dots, \\quad \\{m,m-1\\}.\n$$\n\nSimilarly, if $n = 2m$, then $s = 2m + 1$, and the partition is again unique:\n\n$$\n\\{2m, 1\\}, \\quad \\{2m-1, 2\\}, \\quad \\dots, \\quad \\{m+1, m\\}.\n$$\n\nIf $n > 2m$ induct on $n$. Given an admissible set of parameters $n, m, s$, construct a new partition from an old partition corresponding to some admissible set of parameters $n', m', s'$, where $n' < n$. The proof will be divided into cases. In each case, the condition $m's' = n'(n'+1)/2$ will be clear from the construction, but we must check that $n' \\ge 2m' - 1$.\n\nIf $2m < n < 4m-1$, then $n+1 < s < 2n$, so if we set $n' = s-n-1$, then $0 < n' < n-1$. We consider two subcases.\n\nIf $s$ is odd, let $m' = m-n+(s-1)/2$ and $s' = s$. Here $n' - 2m' = n - 2m > 0$, so $n' > 2m'$. As new blocks, use the old ones and the $n - (s-1)/2$ pairs\n\n$$\n\\{n, s-n\\}, \\quad \\{n-1, s-n+1\\}, \\quad \\dots, \\quad \\{(s+1)/2, (s-1)/2\\}.\n$$\n\nIf $s$ is even, let $m' = 2m - 2n + s - 1$ and $s' = s/2$. A straightforward calculation shows that $2m(n' - 2m') = (n - 2m)(4m - 1 - n) > 0$, so $n' > 2m'$ again. The old blocks and the singleton $\\{s'\\}$ combine in pairs to form $m - n + s'$ new blocks. The other new blocks are the $n-s'$ pairs $\\{n, s-n\\}$, $\\{n-1, s-n+1\\}$, \\dots, $\\{s' + 1, s' - 1\\}$.\n\nFinally, if $n \\ge 4m-1$, let $n' = n - 2m$, $m' = m$, and $s' = s - 2n + 2m - 1$. Clearly, $n' \\ge 2m-1 = 2m' - 1$. The new blocks are obtained from the old blocks by adjoining the $m$ pairs $\\{n, n - 2m + 1\\}$, $\\{n-1, n - 2m + 2\\}$, \\dots, $\\{n - m + 1, n - m\\}$ in any order. This completes the proof.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 22499,
"subject": "Mathematics (Olympiad)",
"question": "Let $AA'$, $BB'$, $CC'$ be the altitudes from the vertices of an acute-angled triangle $ABC$. Points $E$ and $F$ lie on the segments $CB'$ and $BC'$ respectively, such that\n\n$$\nB'E \\cdot C'F = BF \\cdot CE.\n$$\n\nProve that the quadrilateral $AE'A'F$ is cyclic.",
"options": [],
"answer": "See solution",
"solution": "The given relation can be rewritten as $\\frac{BF}{FC'} = \\frac{B'E}{EC}$. Let $M$ be a point on the side $BC$ such that $FM \\parallel CC'$. Then $\\frac{BM}{MC} = \\frac{BF}{FC'} = \\frac{B'E}{EC}$, implying $ME \\parallel BB'$. Hence, the angles $\\angle AEM$, $\\angle AFM$, and $\\angle AA'M$ are right angles; therefore, $E$, $F$, and $A'$ all lie on the circle with diameter $AM$. The conclusion follows.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 22500,
"subject": "Mathematics (Olympiad)",
"question": "Let $a \\in [\\frac{1}{2}, \\frac{3}{2}]$ be a real number. Sequences $(u_n)$ and $(v_n)$ are defined as follows:\n\n$$\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$$\n\nb) Find all values of $a$ in the equality case.",
"options": [],
"answer": "See solution",
"solution": "a) From the assumption, we have $v_i = u_i$ for even $i$ and $v_i = -u_i$ for odd $i$. Let some notations to simplify the expression:\n\n$$\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$$\n\nNow, let's call the binary representation of $a$:\n\n$$\na = \\sum_{i=1}^{+\\infty} \\frac{x_i}{2^i}\n$$\nwhere $x_i \\in \\{0, 1\\}$. Since $\\frac{1}{2} \\le a \\le \\frac{3}{2}$, we have $x_1 = 1$.\n\nFor each natural number $i$, the parity of $\\lfloor 2^{i+1}a \\rfloor$ depends on $x_{i+1}$. In particular, if $x_{i+1} = 0$ then $\\lfloor 2^{i+1}a \\rfloor$ is even, and if $x_{i+1} = 1$ then $\\lfloor 2^{i+1}a \\rfloor$ is odd. Therefore,\n\n$$\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\\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$$\n\nOn the other hand, $a \\ge A + B \\ge \\frac{1}{2}$ so\n\n$$\n36a^2 - 24a + 5 + \\frac{1}{4^{2019}} = 4(3a - 1)^2 + 1 + \\frac{1}{4^{2019}} \\ge 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 \\le 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) \\le \\frac{1}{4^{1008}} - \\frac{1}{4^{2018}}.\n$$\n\nSince $A \\ge \\frac{1}{2}$, we have $6A > 1 + \\frac{1}{4^{1009}}$. Thus,\n\n$$\n\\begin{align*}\n\\frac{6}{4^{1009}} A + 12B \\left( 1 + \\frac{1}{4^{1009}} - 6A \\right) &\\le \\frac{6}{4^{1009}} A \\\\\n&\\le \\frac{6}{4^{1009}} \\sum_{i=0}^{1009} \\frac{1}{2^{2i+1}} \\\\\n&= \\frac{6}{4^{1009}} \\cdot \\frac{2}{3} \\left( 1 - \\frac{1}{4^{1010}} \\right) \\\\\n&= \\frac{1}{4^{1008}} - \\frac{1}{4^{2018}}\n\\end{align*}\n$$\n\nUsing these inequalities, part (a) is proved.\n\nb) Clearly, the equality holds if and only if $a = A + B$, $B = 0$ and\n$$A = \\frac{2}{3} \\left( 1 - \\frac{1}{4^{1010}} \\right), \\text{ thus } a = \\frac{2}{3} \\left( 1 - \\frac{1}{4^{1010}} \\right). \\quad \\square$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
}
]