{"year": "2024", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "APMO", "problem": "Let $A B C$ be an acute triangle. Let $D$ be a point on side $A B$ and $E$ be a point on side $A C$ such that lines $B C$ and $D E$ are parallel. Let $X$ be an interior point of $B C E D$. Suppose rays $D X$ and $E X$ meet side $B C$ at points $P$ and $Q$, respectively such that both $P$ and $Q$ lie between $B$ and $C$. Suppose that the circumcircles of triangles $B Q X$ and $C P X$ intersect at a point $Y \\neq X$. Prove that points $A, X$, and $Y$ are collinear.", "solution": "![](https://cdn.mathpix.com/cropped/2024_11_22_0826db6e0d83838635cfg-01.jpg?height=723&width=1003&top_left_y=678&top_left_x=492)\n\nLet $\\ell$ be the radical axis of circles $B Q X$ and $C P X$. Since $X$ and $Y$ are on $\\ell$, it is sufficient to show that $A$ is on $\\ell$. Let line $A X$ intersect segments $B C$ and $D E$ at $Z$ and $Z^{\\prime}$, respectively. Then it is sufficient to show that $Z$ is on $\\ell$. By $B C \\| D E$, we obtain\n\n$$\n\\frac{B Z}{Z C}=\\frac{D Z^{\\prime}}{Z^{\\prime} E}=\\frac{P Z}{Z Q},\n$$\n\nthus $B Z \\cdot Q Z=C Z \\cdot P Z$, which implies that $Z$ is on $\\ell$.", "metadata": {"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution 1"}}
{"year": "2024", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "APMO", "problem": "Let $A B C$ be an acute triangle. Let $D$ be a point on side $A B$ and $E$ be a point on side $A C$ such that lines $B C$ and $D E$ are parallel. Let $X$ be an interior point of $B C E D$. Suppose rays $D X$ and $E X$ meet side $B C$ at points $P$ and $Q$, respectively such that both $P$ and $Q$ lie between $B$ and $C$. Suppose that the circumcircles of triangles $B Q X$ and $C P X$ intersect at a point $Y \\neq X$. Prove that points $A, X$, and $Y$ are collinear.", "solution": "![](https://cdn.mathpix.com/cropped/2024_11_22_0826db6e0d83838635cfg-01.jpg?height=721&width=998&top_left_y=1995&top_left_x=492)\n\nLet circle $B Q X$ intersect line $A B$ at a point $S$ which is different from $B$. Then $\\angle D E X=$ $\\angle X Q C=\\angle B S X$, thus $S$ is on circle $D E X$. Similarly, let circle $C P X$ intersect line $A C$ at a point $T$ which is different from $C$. Then $T$ is on circle $D E X$. The power of $A$ with respect to the circle $D E X$ is $A S \\cdot A D=A T \\cdot A E$. Since $\\frac{A D}{A B}=\\frac{A E}{A C}, A S \\cdot A B=A T \\cdot A C$. Then $A$ is in the radical axis of circles $B Q X$ and $C P X$, which implies that three points $A, X$ and $Y$ are collinear.", "metadata": {"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution 2"}}
{"year": "2024", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "APMO", "problem": "Let $A B C$ be an acute triangle. Let $D$ be a point on side $A B$ and $E$ be a point on side $A C$ such that lines $B C$ and $D E$ are parallel. Let $X$ be an interior point of $B C E D$. Suppose rays $D X$ and $E X$ meet side $B C$ at points $P$ and $Q$, respectively such that both $P$ and $Q$ lie between $B$ and $C$. Suppose that the circumcircles of triangles $B Q X$ and $C P X$ intersect at a point $Y \\neq X$. Prove that points $A, X$, and $Y$ are collinear.", "solution": "Consider the (direct) homothety that takes triangle $A D E$ to triangle $A B C$, and let $Y^{\\prime}$ be the image of $Y$ under this homothety; in other words, let $Y^{\\prime}$ be the intersection of the line parallel to $B Y$ through $D$ and the line parallel to $C Y$ through $E$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_0826db6e0d83838635cfg-02.jpg?height=732&width=1008&top_left_y=682&top_left_x=484)\n\nThe homothety implies that $A, Y$, and $Y^{\\prime}$ are collinear, and that $\\angle D Y^{\\prime} E=\\angle B Y C$. Since $B Q X Y$ and $C P X Y$ are cyclic,\n$\\angle D Y^{\\prime} E=\\angle B Y C=\\angle B Y X+\\angle X Y C=\\angle X Q P+\\angle X P Q=180^{\\circ}-\\angle P X Q=180^{\\circ}-\\angle D X E$,\nwhich implies that $D Y^{\\prime} E X$ is cyclic. Therefore\n\n$$\n\\angle D Y^{\\prime} X=\\angle D E X=\\angle P Q X=\\angle B Y X\n$$\n\nwhich, combined with $D Y^{\\prime} \\| B Y$, implies $Y^{\\prime} X \\| Y X$. This proves that $X, Y$, and $Y^{\\prime}$ are collinear, which in turn shows that $A, X$, and $Y$ are collinear.", "metadata": {"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution 3"}}
{"year": "2024", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "APMO", "problem": "Consider a $100 \\times 100$ table, and identify the cell in row $a$ and column $b, 1 \\leq a, b \\leq 100$, with 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 to 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 $\\left(x_{0}, y_{0}\\right)=(1,1)$, $\\left(x_{1}, y_{1}\\right),\\left(x_{2}, y_{2}\\right), \\ldots,\\left(x_{n}, y_{n}\\right)$ such that, for all $i=1,2, \\ldots, n, 1 \\leq x_{i}, y_{i} \\leq 100$ and the $k$-knight can move from $\\left(x_{i-1}, y_{i-1}\\right)$ to $\\left(x_{i}, y_{i}\\right)$. In this case, each cell $\\left(x_{i}, y_{i}\\right)$ is said to be reachable. For each $k$, find $L(k)$, the number of reachable cells.\n\nAnswer: $L(k)=\\left\\{\\begin{array}{ll}100^{2}-(2 k-100)^{2} & \\text { if } k \\text { is even } \\\\ \\frac{100^{2}-(2 k-100)^{2}}{2} & \\text { if } k \\text { is odd }\\end{array}\\right.$.", "solution": "Cell $(x, y)$ is directly reachable from another cell if and only if $x-k \\geq 1$ or $x+k \\leq 100$ or $y-k \\geq 1$ or $y+k \\leq 100$, that is, $x \\geq k+1$ 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 $(x, y) \\rightarrow(x+k, y \\pm 1) \\rightarrow(x, y \\pm 2)$ or $(x, y) \\rightarrow$ $(x \\pm 1, y+k) \\rightarrow(x \\pm 2, y)$. In the second case, move $(x, y) \\rightarrow(x-k, y \\pm 1) \\rightarrow(x, y \\pm 2)$ or $(x, y) \\rightarrow(x \\pm 1, y-k) \\rightarrow(x \\pm 2, y)$.\nHence if the table is colored in two colors like a chessboard, if $k \\leq 50$, cells with the same color as $(1,1)$ are reachable. Moreover, 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 then the reachable cells consists of all cells except the center square defined by $101-k \\leq x \\leq k$ and $101-k \\leq y \\leq k$, that is, $L(k)=100^{2}-(2 k-100)^{2}$; if $k$ is odd, then only half of the cells are reachable: the ones with the same color as $(1,1)$, and $L(k)=\\frac{1}{2}\\left(100^{2}-(2 k-100)^{2}\\right)$.", "metadata": {"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution\n\n"}}
{"year": "2024", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "APMO", "problem": "Let $n$ be a positive integer and $a_{1}, a_{2}, \\ldots, a_{n}$ be positive real numbers. Prove that\n\n$$\n\\sum_{i=1}^{n} \\frac{1}{2^{i}}\\left(\\frac{2}{1+a_{i}}\\right)^{2^{i}} \\geq \\frac{2}{1+a_{1} a_{2} \\ldots a_{n}}-\\frac{1}{2^{n}}\n$$", "solution": "We first prove the following lemma:\nLemma 1. For $k$ positive integer and $x, y>0$,\n\n$$\n\\left(\\frac{2}{1+x}\\right)^{2^{k}}+\\left(\\frac{2}{1+y}\\right)^{2^{k}} \\geq 2\\left(\\frac{2}{1+x y}\\right)^{2^{k-1}}\n$$\n\nThe proof goes by induction. For $k=1$, we have\n\n$$\n\\left(\\frac{2}{1+x}\\right)^{2}+\\left(\\frac{2}{1+y}\\right)^{2} \\geq 2\\left(\\frac{2}{1+x y}\\right)\n$$\n\nwhich reduces to\n\n$$\nx y(x-y)^{2}+(x y-1)^{2} \\geq 0 .\n$$\n\nFor $k>1$, by the inequality $2\\left(A^{2}+B^{2}\\right) \\geq(A+B)^{2}$ applied at $A=\\left(\\frac{2}{1+x}\\right)^{2^{k-1}}$ and $B=\\left(\\frac{2}{1+y}\\right)^{2^{k-1}}$ followed by the induction hypothesis\n\n$$\n\\begin{aligned}\n2\\left(\\left(\\frac{2}{1+x}\\right)^{2^{k}}+\\left(\\frac{2}{1+y}\\right)^{2^{k}}\\right) & \\geq\\left(\\left(\\frac{2}{1+x}\\right)^{2^{k-1}}+\\left(\\frac{2}{1+y}\\right)^{2^{k-1}}\\right)^{2} \\\\\n& \\geq\\left(2\\left(\\frac{2}{1+x y}\\right)^{2^{k-2}}\\right)^{2}=4\\left(\\frac{2}{1+x y}\\right)^{2^{k-1}}\n\\end{aligned}\n$$\n\nfrom which the lemma follows.\nThe problem now can be deduced from summing the following applications of the lemma, multiplied by the appropriate factor:\n\n$$\n\\begin{aligned}\n\\frac{1}{2^{n}}\\left(\\frac{2}{1+a_{n}}\\right)^{2^{n}}+\\frac{1}{2^{n}}\\left(\\frac{2}{1+1}\\right)^{2^{n}} & \\geq \\frac{1}{2^{n-1}}\\left(\\frac{2}{1+a_{n} \\cdot 1}\\right)^{2^{n-1}} \\\\\n\\frac{1}{2^{n-1}}\\left(\\frac{2}{1+a_{n-1}}\\right)^{2^{n-1}}+\\frac{1}{2^{n-1}}\\left(\\frac{2}{1+a_{n}}\\right)^{2^{n-1}} & \\geq \\frac{1}{2^{n-2}}\\left(\\frac{2}{1+a_{n-1} a_{n}}\\right)^{2^{n-2}} \\\\\n\\frac{1}{2^{n-2}}\\left(\\frac{2}{1+a_{n-2}}\\right)^{2^{n-2}}+\\frac{1}{2^{n-2}}\\left(\\frac{2}{1+a_{n-1} a_{n}}\\right)^{2^{n-2}} & \\geq \\frac{1}{2^{n-3}}\\left(\\frac{2}{1+a_{n-2} a_{n-1} a_{n}}\\right)^{2^{n-3}} \\\\\n\\ldots & )^{2^{k}} \\\\\n\\frac{1}{2^{k}}\\left(\\frac{2}{1+a_{k}}\\right)^{2^{k}}+\\frac{1}{2^{k}}\\left(\\frac{2}{1+a_{k+1} \\ldots a_{n-1} a_{n}}\\right)^{2^{k-1}} & \\geq \\frac{1}{2^{k-1}}\\left(\\frac{2}{1+a_{k} \\ldots a_{n-2} a_{n-1} a_{n}}\\right)^{2} \\\\\n\\frac{1}{2}\\left(\\frac{2}{1+a_{1}}\\right)^{2}+\\frac{1}{2}\\left(\\frac{2}{1+a_{2} \\ldots a_{n-1} a_{n}}\\right)^{2} & \\geq \\frac{2}{1+a_{1} \\ldots a_{n-2} a_{n-1} a_{n}}\n\\end{aligned}\n$$\n\nComment: Equality occurs if and only if $a_{1}=a_{2}=\\cdots=a_{n}=1$.\n\nComment: The main motivation for the lemma is trying to \"telescope\" the sum\n\n$$\n\\frac{1}{2^{n}}+\\sum_{i=1}^{n} \\frac{1}{2^{i}}\\left(\\frac{2}{1+a_{i}}\\right)^{2^{i}}\n$$\n\nthat is,\n\n$$\n\\frac{1}{2}\\left(\\frac{2}{1+a_{1}}\\right)^{2}+\\cdots+\\frac{1}{2^{n-1}}\\left(\\frac{2}{1+a_{n-1}}\\right)^{2^{n-1}}+\\frac{1}{2^{n}}\\left(\\frac{2}{1+a_{n}}\\right)^{2^{n}}+\\frac{1}{2^{n}}\\left(\\frac{2}{1+1}\\right)^{2^{n}}\n$$\n\nto obtain an expression larger than or equal to\n\n$$\n\\frac{2}{1+a_{1} a_{2} \\ldots a_{n}}\n$$\n\nIt seems reasonable to obtain a inequality that can be applied from right to left, decreases the exponent of the factor $1 / 2^{k}$ by 1 , and multiplies the variables in the denominator. Given that, the lemma is quite natural:\n\n$$\n\\frac{1}{2^{k}}\\left(\\frac{2}{1+x}\\right)^{2^{k}}+\\frac{1}{2^{k}}\\left(\\frac{2}{1+y}\\right)^{2^{k}} \\geq \\frac{1}{2^{k-1}}\\left(\\frac{2}{1+x y}\\right)^{2^{i-1}}\n$$\n\nor\n\n$$\n\\left(\\frac{2}{1+x}\\right)^{2^{k}}+\\left(\\frac{2}{1+y}\\right)^{2^{k}} \\geq 2\\left(\\frac{2}{1+x y}\\right)^{2^{k-1}}\n$$", "metadata": {"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution\n\n"}}
{"year": "2024", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "APMO", "problem": "Prove that for every positive integer $t$ there is a unique permutation $a_{0}, a_{1}, \\ldots, a_{t-1}$ of $0,1, \\ldots, t-$ 1 such that, for every $0 \\leq i \\leq t-1$, the binomial coefficient $\\binom{t+i}{2 a_{i}}$ is odd and $2 a_{i} \\neq t+i$.", "solution": "We constantly make use of Kummer's theorem which, in particular, implies that $\\binom{n}{k}$ is odd if and only if $k$ and $n-k$ have ones in different positions in binary. In other words, if $S(x)$ is the set of positions of the digits 1 of $x$ in binary (in which the digit multiplied by $2^{i}$ is in position $i),\\binom{n}{k}$ is odd if and only if $S(k) \\subseteq S(n)$. Moreover, if we set $k<n, S(k)$ is a proper subset of $S(n)$, that is, $|S(k)|<|S(n)|$.\nWe start with a lemma that guides us how the permutation should be set.\n\n## Lemma 1.\n\n$$\n\\sum_{i=0}^{t-1}|S(t+i)|=t+\\sum_{i=0}^{t-1}|S(2 i)|\n$$\n\nThe proof is just realizing that $S(2 i)=\\{1+x, x \\in S(i)\\}$ and $S(2 i+1)=\\{0\\} \\cup\\{1+x, x \\in S(i)\\}$, because $2 i$ in binary is $i$ followed by a zero and $2 i+1$ in binary is $i$ followed by a one. Therefore\n\n$$\n\\begin{aligned}\n\\sum_{i=0}^{t-1}|S(t+i)| & =\\sum_{i=0}^{2 t-1}|S(i)|-\\sum_{i=0}^{t-1}|S(i)|=\\sum_{i=0}^{t-1}|S(2 i)|+\\sum_{i=0}^{t-1}|S(2 i+1)|-\\sum_{i=0}^{t-1}|S(i)| \\\\\n& =\\sum_{i=0}^{t-1}|S(i)|+\\sum_{i=0}^{t-1}(1+|S(i)|)-\\sum_{i=0}^{t-1}|S(i)|=t+\\sum_{i=0}^{t-1}|S(i)|=t+\\sum_{i=0}^{t-1}|S(2 i)|\n\\end{aligned}\n$$\n\nThe lemma has an immediate corollary: since $t+i>2 a_{i}$ and $\\binom{t+i}{2 a_{i}}$ is odd for all $i, 0 \\leq i \\leq t-1$, $S\\left(2 a_{i}\\right) \\subset S(t+i)$ with $\\left|S\\left(2 a_{i}\\right)\\right| \\leq|S(t+i)|-1$. Since the sum of $\\left|S\\left(2 a_{i}\\right)\\right|$ is $t$ less than the sum of $|S(t+i)|$, and there are $t$ values of $i$, equality must occur, that is, $\\left|S\\left(2 a_{i}\\right)\\right|=|S(t+i)|-1$, which in conjunction with $S\\left(2 a_{i}\\right) \\subset S(t+i)$ means that $t+i-2 a_{i}=2^{k_{i}}$ for every $i, 0 \\leq i \\leq t-1$, $k_{i} \\in S(t+i)$ (more precisely, $\\left\\{k_{i}\\right\\}=S(t+i) \\backslash S\\left(2 a_{i}\\right)$.)\nIn particular, for $t+i$ odd, this means that $t+i-2 a_{i}=1$, because the only odd power of 2 is 1. Then $a_{i}=\\frac{t+i-1}{2}$ for $t+i$ odd, which takes up all the numbers greater than or equal to $\\frac{t-1}{2}$. Now we need to distribute the numbers that are smaller than $\\frac{t-1}{2}$ (call these numbers small). If $t+i$ is even then by Lucas' Theorem $\\binom{t+i}{2 a_{i}} \\equiv\\binom{\\frac{t+i}{2}}{a_{i}}(\\bmod 2)$, so we pair numbers from $\\lceil t / 2\\rceil$ to $t-1$ (call these numbers big) with the small numbers.\nSay that a set $A$ is paired with another set $B$ whenever $|A|=|B|$ and there exists a bijection $\\pi: A \\rightarrow B$ such that $S(a) \\subset S(\\pi(a))$ and $|S(a)|=|S(\\pi(a))|-1$; we also say that $a$ and $\\pi(a)$ are paired. We prove by induction in $t$ that $A_{t}=\\{0,1,2, \\ldots,\\lfloor t / 2\\rfloor-1\\}$ (the set of small numbers) and $B_{t}=\\{\\lceil t / 2\\rceil, \\ldots, t-2, t-1\\}$ (the set of big numbers) can be uniquely paired.\nThe claim is immediate for $t=1$ and $t=2$. For $t>2$, there is exactly one power of two in $B_{t}$, since $t / 2 \\leq 2^{a}<t \\Longleftrightarrow a=\\left\\lceil\\log _{2}(t / 2)\\right\\rceil$. Let $2^{a}$ be this power of two. Then, since $2^{a} \\geq t / 2$, no number in $A_{t}$ has a one in position $a$ in binary. Since for every number $x, 2^{a} \\leq x<t, a \\in S(x)$ and $a \\notin S(y)$ for all $y \\in A_{t}, x$ can only be paired with $x-2^{a}$, since $S(x)$ needs to be stripped of exactly one position. This takes cares of $x \\in B_{t}, 2^{a} \\leq x<t$, and $y \\in A_{t}, 0 \\leq y<t-2^{a}$.\nNow we need to pair the numbers from $A^{\\prime}=\\left\\{t-2^{a}, t-2^{a}+1, \\ldots,\\lfloor t / 2\\rfloor-1\\right\\} \\subset A$ with the numbers from $B^{\\prime}=\\left\\{\\lceil t / 2\\rceil,\\lceil t / 2\\rceil+1, \\ldots, 2^{a}-1\\right\\} \\subset B$. In order to pair these $t-2\\left(t-2^{a}\\right)=$ $2^{a+1}-t<t$ numbers, we use the induction hypothesis and a bijection between $A^{\\prime} \\cup B^{\\prime}$ and $B_{2^{a+1}-t} \\cup A_{2^{a+1}-t}$. Let $S=S\\left(2^{a}-1\\right)=\\{0,1,2, \\ldots, a-1\\}$. Then take a pair $x, y, x \\in A_{2^{a+1}-t}$ and $y \\in B_{2^{a+1}-t}$ and biject it with $2^{a}-1-x \\in B^{\\prime}$ and $2^{a}-1-y \\in A^{\\prime}$. In fact,\n\n$$\n0 \\leq x \\leq\\left\\lfloor\\frac{2^{a+1}-t}{2}\\right\\rfloor-1=2^{a}-\\left\\lceil\\frac{t}{2}\\right\\rceil-1 \\Longleftrightarrow\\left\\lceil\\frac{t}{2}\\right\\rceil \\leq 2^{a}-1-x \\leq 2^{a}-1\n$$\n\nand\n\n$$\n\\left\\lceil\\frac{2^{a+1}-t}{2}\\right\\rceil=2^{a}-\\left\\lfloor\\frac{t}{2}\\right\\rfloor \\leq y \\leq 2^{a+1}-t-1 \\Longleftrightarrow t-2^{a} \\leq 2^{a}-1-y \\leq\\left\\lfloor\\frac{t}{2}\\right\\rfloor-1\n$$\n\nMoreover, $S\\left(2^{a}-1-x\\right)=S \\backslash S(x)$ and $S\\left(2^{a}-1-y\\right)=S \\backslash S(y)$ are complements with respect to $S$, and $S(x) \\subset S(y)$ and $|S(x)|=|S(y)|-1$ implies $S\\left(2^{a}-1-y\\right) \\subset S\\left(2^{a}-1-x\\right)$ and $\\left|S\\left(2^{a}-1-y\\right)\\right|=\\left|S\\left(2^{a}-1-x\\right)\\right|-1$. Therefore a pairing between $A^{\\prime}$ and $B^{\\prime}$ corresponds to a pairing between $A_{2^{a+1}-t}$ and $B_{2^{a+1}-t}$. Since the latter pairing is unique, the former pairing is also unique, and the result follows.\nWe illustrate the bijection by showing the case $t=23$ :\n\n$$\nA_{23}=\\{0,1,2, \\ldots, 10\\}, \\quad B_{23}=\\{12,13,14, \\ldots, 22\\}\n$$\n\nThe pairing is\n\n$$\n\\left(\\begin{array}{ccccccccccc}\n12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 \\\\\n8 & 9 & 10 & 7 & 0 & 1 & 2 & 3 & 4 & 5 & 6\n\\end{array}\\right)\n$$\n\nin which the bijection is between\n\n$$\n\\left(\\begin{array}{cccc}\n12 & 13 & 14 & 15 \\\\\n8 & 9 & 10 & 7\n\\end{array}\\right) \\quad \\text { and } \\quad\\left(\\begin{array}{llll}\n3 & 2 & 1 & 0 \\\\\n7 & 6 & 5 & 8\n\\end{array}\\right) \\rightarrow\\left(\\begin{array}{llll}\n5 & 6 & 7 & 8 \\\\\n1 & 2 & 3 & 0\n\\end{array}\\right) .\n$$", "metadata": {"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution\n\n"}}
{"year": "2024", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "APMO", "problem": "Line $\\ell$ intersects sides $B C$ and $A D$ of cyclic quadrilateral $A B C D$ in its interior points $R$ and $S$ respectively, and intersects ray $D C$ beyond point $C$ at $Q$, and ray $B A$ beyond point $A$ at $P$. Circumcircles of the triangles $Q C R$ and $Q D S$ intersect at $N \\neq Q$, while circumcircles of the triangles $P A S$ and $P B R$ intersect at $M \\neq P$. Let lines $M P$ and $N Q$ meet at point $X$, lines $A B$ and $C D$ meet at point $K$ and lines $B C$ and $A D$ meet at point $L$. Prove that point $X$ lies on line $K L$.", "solution": "We start with the following lemma.\nLemma 1. Points $M, N, P, Q$ are concyclic.\nPoint $M$ is the Miquel point of lines $A P=A B, P S=\\ell, A S=A D$, and $B R=B C$, and point $N$ is the Miquel point of lines $C Q=C D, R C=B C, Q R=\\ell$, and $D S=A D$. Both points $M$ and $N$ are on the circumcircle of the triangle determined by the common lines $A D, \\ell$, and $B C$, which is $L R S$.\nThen, since quadrilaterals $Q N R C, P M A S$, and $A B C D$ are all cyclic, using directed angles (modulo $180^{\\circ}$ )\n\n$$\n\\begin{aligned}\n\\measuredangle N M P & =\\measuredangle N M S+\\measuredangle S M P=\\measuredangle N R S+\\measuredangle S A P=\\measuredangle N R Q+\\measuredangle D A B=\\measuredangle N R Q+\\measuredangle D C B \\\\\n& =\\measuredangle N R Q+\\measuredangle Q C R=\\measuredangle N R Q+\\measuredangle Q N R=\\measuredangle N Q R=\\measuredangle N Q P,\n\\end{aligned}\n$$\n\nwhich implies that $M N Q P$ is a cyclic quadrilateral.\n![](https://cdn.mathpix.com/cropped/2024_11_22_0826db6e0d83838635cfg-08.jpg?height=975&width=1115&top_left_y=1306&top_left_x=439)\n\nLet $E$ be the Miquel point of $A B C D$ (that is, of lines $A B, B C, C D, D A$ ). It is well known that $E$ lies in the line $t$ connecting the intersections of the opposite lines of $A B C D$. Let lines $N Q$ and $t$ meet at $T$. If $T \\neq E$, using directed angles, looking at the circumcircles of $L A B$ (which contains, by definition, $E$ and $M$ ), $A P S$ (which also contains $M$ ), and $M N Q P$,\n\n$$\n\\measuredangle T E M=\\measuredangle L E M=\\measuredangle L A M=\\measuredangle S A M=\\measuredangle S P M=\\measuredangle Q P M=\\measuredangle Q N M=\\measuredangle T N M,\n$$\n\nthat is, $T$ lies in the circumcircle $\\omega$ of $E M N$. If $T=E$, the same computation shows that $\\measuredangle L E M=\\measuredangle E N M$, which means that $t$ is tangent to $\\omega$.\n\nNow let lines $M P$ and $t$ meet at $V$. An analogous computation shows, by looking at the circumcircles of $L C D$ (which contains $E$ and $N$ ), $C Q R$, and $M N Q P$, that $V$ lies in $\\omega$ as well, and that if $V=E$ then $t$ is tangent to $\\omega$.\nTherefore, since $\\omega$ meet $t$ at $T, V$, and $E$, either $T=V$ if both $T \\neq E$ and $V \\neq E$ or $T=V=E$. At any rate, the intersection of lines $M P$ and $N Q$ lies in $t$.", "metadata": {"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution 1"}}
{"year": "2024", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "APMO", "problem": "Line $\\ell$ intersects sides $B C$ and $A D$ of cyclic quadrilateral $A B C D$ in its interior points $R$ and $S$ respectively, and intersects ray $D C$ beyond point $C$ at $Q$, and ray $B A$ beyond point $A$ at $P$. Circumcircles of the triangles $Q C R$ and $Q D S$ intersect at $N \\neq Q$, while circumcircles of the triangles $P A S$ and $P B R$ intersect at $M \\neq P$. Let lines $M P$ and $N Q$ meet at point $X$, lines $A B$ and $C D$ meet at point $K$ and lines $B C$ and $A D$ meet at point $L$. Prove that point $X$ lies on line $K L$.", "solution": "Barycentric coordinates are a viable way to solve the problem, but even the solution we have found had some clever computations. Here is an outline of this solution.\n\nLemma 2. Denote by $\\operatorname{pow}_{\\omega} X$ the power of point $X$ with respect to circle $\\omega$. Let $\\Gamma_{1}$ and $\\Gamma_{2}$ be circles with different centers. Considering $A B C$ as the reference triangle in barycentric coordinates, the radical axis of $\\Gamma_{1}$ and $\\Gamma_{2}$ is given by\n\n$$\n\\left(\\operatorname{pow}_{\\Gamma_{1}} A-\\operatorname{pow}_{\\Gamma_{2}} A\\right) x+\\left(\\operatorname{pow}_{\\Gamma_{1}} B-\\operatorname{pow}_{\\Gamma_{2}} B\\right) y+\\left(\\operatorname{pow}_{\\Gamma_{1}} C-\\operatorname{pow}_{\\Gamma_{2}} C\\right) z=0\n$$\n\nProof: Let $\\Gamma_{i}$ have the equation $\\Gamma_{i}(x, y, z)=-a^{2} y z-b^{2} z x-c^{2} x y+(x+y+z)\\left(r_{i} x+s_{i} y+t_{i} z\\right)$. Then $\\operatorname{pow}_{\\Gamma_{i}} P=\\Gamma_{i}(P)$. In particular, $\\operatorname{pow}_{\\Gamma_{i}} A=\\Gamma_{i}(1,0,0)=r_{i}$ and, similarly, $\\operatorname{pow}_{\\Gamma_{i}} B=s_{i}$ and $\\operatorname{pow}_{\\Gamma_{i}} C=t_{i}$.\nFinally, the radical axis is\n\n$$\n\\begin{aligned}\n& \\operatorname{pow}_{\\Gamma_{1}} P=\\operatorname{pow}_{\\Gamma_{2}} P \\\\\n\\Longleftrightarrow & \\Gamma_{1}(x, y, z)=\\Gamma_{2}(x, y, z) \\\\\n\\Longleftrightarrow & r_{1} x+s_{1} y+t_{1} z=r_{2} x+s_{2} y+t_{2} z \\\\\n\\Longleftrightarrow & \\left(\\operatorname{pow}_{\\Gamma_{1}} A-\\operatorname{pow}_{\\Gamma_{2}} A\\right) x+\\left(\\operatorname{pow}_{\\Gamma_{1}} B-\\operatorname{pow}_{\\Gamma_{2}} B\\right) y+\\left(\\operatorname{pow}_{\\Gamma_{1}} C-\\operatorname{pow}_{\\Gamma_{2}} C\\right) z=0 .\n\\end{aligned}\n$$\n\nWe still use the Miquel point $E$ of $A B C D$. Notice that the problem is equivalent to proving that lines $M P, N Q$, and $E K$ are concurrent. The main idea is writing these three lines as radical axes. In fact, by definition of points $M, N$, and $E$ :\n\n- $M P$ is the radical axis of the circumcircles of $P A S$ and $P B R$;\n- $N Q$ is the radical axis of the circumcircles of $Q C R$ and $Q D S$;\n- $E K$ is the radical axis of the circumcircles of $K B C$ and $K A D$.\n\nLooking at these facts and the diagram, it makes sense to take triangle $K Q P$ the reference triangle. Because of that, we do not really need to draw circles nor even points $M$ and $N$, as all powers can be computed directly from points in lines $K P, K Q$, and $P Q$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_0826db6e0d83838635cfg-09.jpg?height=515&width=1043&top_left_y=2050&top_left_x=478)\n\nAssociate $P$ with the $x$-coordinate, $Q$ with the $y$-coordinate, and $K$ with the $z$-coordinate. Applying the lemma, the equations of lines $P M, Q N$, and $E K$ are\n\n- MP: $(K A \\cdot K P-K B \\cdot K P) x+(Q S \\cdot Q P-Q R \\cdot Q P) y=0$\n- $N Q:(K C \\cdot K Q-K D \\cdot K Q) x+(P R \\cdot P Q-P S \\cdot P Q) z=0$\n- MP: $(-Q C \\cdot Q K+Q D \\cdot Q K) y+(P B \\cdot P K-P A \\cdot P K) z=0$\n\nThese equations simplify to\n\n- $M P:(A B \\cdot K P) x+(P Q \\cdot R S) y=0$\n- $N Q:(-C D \\cdot K Q) x+(P Q \\cdot R S) z=0$\n- $M P:(C D \\cdot K Q) y+(A B \\cdot K P) z=0$\n\nNow, if $u=A B \\cdot K P, v=P Q \\cdot R S$, and $w=C D \\cdot K Q$, it suffices to show that\n\n$$\n\\left|\\begin{array}{ccc}\nu & v & 0 \\\\\n-w & 0 & v \\\\\n0 & w & u\n\\end{array}\\right|=0\n$$\n\nwhich is a straightforward computation.", "metadata": {"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution 2"}}