| {"year": "2013", "tier": "T2", "problem_label": "1", "problem_type": null, "exam": "Canada_MO", "problem": "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.", "solution": "The answer is $P(x)$ being any constant polynomial and $P(x) \\equiv$ $k x^{2}+k x+c$ for any (nonzero) constant $k$ and constant $c$.\n\nLet $\\Lambda$ be the expression $(x+1) P(x-1)-(x-1) P(x)$, i.e. the expression in the problem statement.\n\nSubstituting $x=-1$ into $\\Lambda$ yields $2 P(-1)$ and substituting $x=1$ into $\\Lambda$ yield $2 P(0)$. Since $(x+1) P(x-1)-(x-1) P(x)$ is a constant polynomial, $2 P(-1)=2 P(0)$. Hence, $P(-1)=P(0)$.\n\nLet $c=P(-1)=P(0)$ and $Q(x)=P(x)-c$. Then $Q(-1)=Q(0)=0$. Hence, $0,-1$ are roots of $Q(x)$. Consequently, $Q(x)=x(x+1) R(x)$ for some polynomial $R$. Then $P(x)-c=x(x+1) R(x)$, or equivalently, $P(x)=x(x+1) R(x)+c$.\n\nSubstituting this into $\\Lambda$ yield\n\n$$\n(x+1)((x-1) x R(x-1)+c)-(x-1)(x(x+1) R(x)+c)\n$$\n\nThis is a constant polynomial and simplifies to\n\n$$\nx(x-1)(x+1)(R(x-1)-R(x))+2 c .\n$$\n\n## Sun $\\mathbb{E}^{\\circ}$\n\n## Life Financial\n\nSince this expression is a constant, so is $x(x-1)(x+1)(R(x-1)-R(x))$. Therefore, $R(x-1)-R(x)=0$ as a polynomial. Therefore, $R(x)=R(x-1)$ for all $x \\in \\mathbb{R}$. Then $R(x)$ is a polynomial that takes on certain values for infinitely values of $x$. Let $k$ be such a value. Then $R(x)-k$ has infinitely many roots, which can occur if and only if $R(x)-k=0$. Therefore, $R(x)$ is identical to a constant $k$. Hence, $Q(x)=k x(x+1)$ for some constant $k$. Therefore, $P(x)=k x(x+1)+c=k x^{2}+k x+c$.\n\nFinally, we verify that all such $P(x)=k x(x+1)+c$ work. Substituting this into $\\Lambda$ yield\n\n$$\n\\begin{aligned}\n& (x+1)(k x(x-1)+c)-(x-1)(k x(x+1)+c) \\\\\n= & k x(x+1)(x-1)+c(x+1)-k x(x+1)(x-1)-c(x-1)=2 c .\n\\end{aligned}\n$$\n\nHence, $P(x)=k x(x+1)+c=k x^{2}+k x+c$ is a solution to the given equation for any constant $k$. Note that this solution also holds for $k=0$. Hence, constant polynomials are also solutions to this equation.", "metadata": {"resource_path": "Canada_MO/segmented/en-sol2013.jsonl", "problem_match": "\n1.", "solution_match": "\nSolution 1:"}} |
| {"year": "2013", "tier": "T2", "problem_label": "1", "problem_type": null, "exam": "Canada_MO", "problem": "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.", "solution": "As in Solution 1, any constant polynomial $P$ satisfies the given property. Hence, we will assume that $P$ is not a constant polynomial.\n\nLet $n$ be the degree of $P$. Since $P$ is not constant, $n \\geq 1$. Let\n\n$$\nP(x)=\\sum_{i=0}^{n} a_{i} x^{i}\n$$\n\nwith $a_{n} \\neq 0$. Then\n\n$$\n(x+1) \\sum_{i=0}^{n} a_{i}(x-1)^{i}-(x-1) \\sum_{i=0}^{n} a_{i} x^{i}=C,\n$$\n\nfor some constant $C$. We will compare the coefficient of $x^{n}$ of the left-hand side of this equation with the right-hand side. Since $C$ is a constant and $n \\geq 1$, the coefficient of $x^{n}$ of the right-hand side is equal to zero. We now determine the coefficient of $x^{n}$ of the left-hand side of this expression.\n\nThe left-hand side of the equation simplifies to\n\n$$\nx \\sum_{i=0}^{n} a_{i}(x-1)^{i}+\\sum_{i=0}^{n} a_{i}(x-1)^{i}-x \\sum_{i=0}^{n} a_{i} x^{i}+\\sum_{i=0}^{n} a_{i} x^{i}\n$$\n\n## Sun\n\n## Life Financial\n\nWe will determine the coefficient $x^{n}$ of each of these four terms.\n\nBy the Binomial Theorem, the coefficient of $x^{n}$ of the first term is equal to that of $x\\left(a_{n-1}(x-1)^{n-1}+a_{n}(x-1)^{n}\\right)=a_{n-1}-\\left(\\begin{array}{c}n \\\\ n-1\\end{array}\\right) a_{n}=a_{n-1}-n a_{n}$.\n\nThe coefficient of $x^{n}$ of the second term is equal to that of $a_{n}(x-1)^{n}$, which is $a_{n}$.\n\nThe coefficient of $x^{n}$ of the third term is equal to $a_{n-1}$ and that of the fourth term is equal to $a_{n}$.\n\nSumming these four coefficients yield $a_{n-1}-n a_{n}+a_{n}-a_{n-1}+a_{n}=(2-n) a_{n}$.\n\nThis expression is equal to 0 . Since $a_{n} \\neq 0, n=2$. Hence, $P$ is a quadratic polynomial.\n\nLet $P(x)=a x^{2}+b x+c$, where $a, b, c$ are real numbers with $a \\neq 0$. Then\n\n$$\n(x+1)\\left(a(x-1)^{2}+b(x-1)+c\\right)-(x-1)\\left(a x^{2}+b x+c\\right)=C .\n$$\n\nSimplifying the left-hand side yields\n\n$$\n(b-a) x+2 c=2 C .\n$$\n\nTherefore, $b-a=0$ and $2 c=2 C$. Hence, $P(x)=a x^{2}+a x+c$. As in Solution 1, this is a valid solution for all $a \\in \\mathbb{R} \\backslash\\{0\\}$.\n\n## Sun\n\n## Life Financial", "metadata": {"resource_path": "Canada_MO/segmented/en-sol2013.jsonl", "problem_match": "\n1.", "solution_match": "\nSolution 2:"}} |
| {"year": "2013", "tier": "T2", "problem_label": "2", "problem_type": null, "exam": "Canada_MO", "problem": "The sequence $a_{1}, a_{2}, \\ldots, a_{n}$ consists of the numbers $1,2, \\ldots, n$ in some order. For which positive integers $n$ is it possible that $0, a_{1}, a_{1}+a_{2}, \\ldots, a_{1}+a_{2}+\\ldots+a_{n}$ all have different remainders when divided by $n+1$ ?", "solution": "It is possible if and only if $n$ is odd.\n\nIf $n$ is even, then $a_{1}+a_{2}+\\ldots+a_{n}=1+2+\\ldots+n=\\frac{n}{2} \\cdot(n+1)$, which is congruent to $0 \\bmod n+1$. Therefore, the task is impossible.\n\nNow suppose $n$ is odd. We will show that we can construct $a_{1}, a_{2}, \\ldots, a_{n}$ that satisfy the conditions given in the problem. Then let $n=2 k+1$ for some non-negative integer $k$. Consider the sequence: $1,2 k, 3,2 k-2,5,2 k-3, \\ldots, 2,2 k+1$, i.e. for each $1 \\leq i \\leq 2 k+1, a_{i}=i$ if $i$ is odd and $a_{i}=2 k+2-i$ if $i$ is even.\n\nWe first show that each term $1,2, \\ldots, 2 k+1$ appears exactly once. Clearly, there are $2 k+1$ terms. For each odd number $m$ in $\\{1,2, \\ldots, 2 k+1\\}, a_{m}=m$. For each even number $m$ in this set, $a_{2 k+2-m}=2 k+2-(2 k+2-m)=m$. Hence, every number appears in $a_{1}, \\ldots, a_{2 k+1}$. Hence, $a_{1}, \\ldots, a_{2 k+1}$ does consist of the numbers $1,2, \\ldots, 2 k+1$ in some order.\n\nWe now determine $a_{1}+a_{2}+\\ldots+a_{m}(\\bmod 2 k+2)$. We will consider the cases when $m$ is odd and when $m$ is even separately. Let $b_{m}=a_{1}+a_{2} \\ldots+a_{m}$.\n\nIf $m$ is odd, note that $a_{1} \\equiv 1(\\bmod 2 k+2), a_{2}+a_{3}=a_{4}+a_{5}=\\ldots=a_{2 k}+$ $a_{2 k+1}=2 k+3 \\equiv 1(\\bmod 2 k+2)$. Therefore, $\\left\\{b_{1}, b_{3}, \\ldots, b_{2 k+1}\\right\\}=\\{1,2,3, \\ldots, k+1\\}$ $(\\bmod 2 k+2)$.\n\nIf $m$ is even, note that $a_{1}+a_{2}=a_{3}+a_{4}=\\ldots=a_{2 k-1}+a_{2 k}=2 k+1 \\equiv-1$ $(\\bmod 2 k+2)$. Therefore, $\\left\\{b_{2}, b_{4}, \\ldots, b_{2 k}\\right\\}=\\{-1,-2, \\ldots,-k\\}(\\bmod 2 k+2) \\equiv$ $\\{2 k+1,2 k, \\ldots, k+2\\}(\\bmod 2 k+2)$.\n\nTherefore, $b_{1}, b_{2}, \\ldots, b_{2 k+1}$ do indeed have different remainders when divided by $2 k+2$. This completes the problem.\n\n## Sun\n\n## Life Financial", "metadata": {"resource_path": "Canada_MO/segmented/en-sol2013.jsonl", "problem_match": "\n2.", "solution_match": "\nSolution:"}} |
| {"year": "2013", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "Canada_MO", "problem": "Let $G$ be the centroid of a right-angled triangle $A B C$ with $\\angle B C A=90^{\\circ}$. Let $P$ be the point on ray $A G$ such that $\\angle C P A=\\angle C A B$, and let $Q$ be the point on ray $B G$ such that $\\angle C Q B=\\angle A B C$. Prove that the circumcircles of triangles $A Q G$ and $B P G$ meet at a point on side $A B$.", "solution": "Since $\\angle C=90^{\\circ}$, the point $C$ lies on the semicircle with diameter $A B$ which implies that, if $M$ is te midpoint of side $A B$, then $M A=M C=M B$. This implies that triangle $A M C$ is isosceles and hence that $\\angle A C M=\\angle A$. By definition, $G$ lies on segment $M$ and it follows that $\\angle A C G=\\angle A C M=\\angle A=\\angle C P A$. This implies that triangles $A P C$ and $A C G$ are similar and hence that $A C^{2}=A G \\cdot A P$. Now if $D$ denotes the foot of the perpendicular from $C$ to $A B$, it follows that triangles $A C D$ and $A B C$ are similar which implies that $A C^{2}=A D \\cdot A B$. Therefore $A G \\cdot A P=$ $A C^{2}=A D \\cdot A B$ and, by power of a point, quadrilateral $D G P B$ is cyclic. This implies that $D$ lies on the circumcircle of triangle $B P G$ and, by a symmetric argument, it follows that $D$ also lies on the circumcircle of triangle $A G Q$. Therefore these two circumcircles meet at the point $D$ on side $A B$.", "metadata": {"resource_path": "Canada_MO/segmented/en-sol2013.jsonl", "problem_match": "\n3.", "solution_match": "\nSolution 1."}} |
| {"year": "2013", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "Canada_MO", "problem": "Let $G$ be the centroid of a right-angled triangle $A B C$ with $\\angle B C A=90^{\\circ}$. Let $P$ be the point on ray $A G$ such that $\\angle C P A=\\angle C A B$, and let $Q$ be the point on ray $B G$ such that $\\angle C Q B=\\angle A B C$. Prove that the circumcircles of triangles $A Q G$ and $B P G$ meet at a point on side $A B$.", "solution": "Define $D$ and $M$ as in Solution 1. Let $R$ be the point on side $A B$ such that $A C=C R$ and triangle $A C R$ is isosceles. Since $\\angle C R A=\\angle A=\\angle C P A$, it follows that $C P R A$ is cyclic and hence that $\\angle G P R=\\angle A P R=\\angle A C R=180^{\\circ}-$ $2 \\angle A$. As in Solution $1, M C=M B$ and hence $\\angle G M R=\\angle C M B=2 \\angle A=180^{\\circ}-$ $\\angle G P R$. Therefore $G P R M$ is cyclic and, by power of a point, $A M \\cdot A R=A G \\cdot A P$. Since $A C R$ is isosceles, $D$ is the midpoint of $A R$ and thus, since $M$ is the midpoint of $A B$, it follows that $A M \\cdot A R=A D \\cdot A B=A G \\cdot A P$. Therefore $D G P B$ is cyclic, implying the result as in Solution 1.\n\n## Sun $\\mathscr{H}^{*}$\n\n## Life Financial", "metadata": {"resource_path": "Canada_MO/segmented/en-sol2013.jsonl", "problem_match": "\n3.", "solution_match": "\nSolution 2."}} |
| {"year": "2013", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "Canada_MO", "problem": "Let $n$ be a positive integer. For any positive integer $j$ and positive real number $r$, define\n\n$$\nf_{j}(r)=\\min (j r, n)+\\min \\left(\\frac{j}{r}, n\\right), \\quad \\text { and } \\quad g_{j}(r)=\\min (\\lceil j r\\rceil, n)+\\min \\left(\\left\\lceil\\frac{j}{r}\\right\\rceil, n\\right) \\text {, }\n$$\n\nwhere $\\lceil x\\rceil$ denotes the smallest integer greater than or equal to $x$. Prove that\n\n$$\n\\sum_{j=1}^{n} f_{j}(r) \\leq n^{2}+n \\leq \\sum_{j=1}^{n} g_{j}(r)\n$$", "solution": "We first prove the left hand side inequality. We begin by drawing an $n \\times n$ board, with corners at $(0,0),(n, 0),(0, n)$ and $(n, n)$ on the Cartesian plane.\n\nConsider the line $\\ell$ with slope $r$ passing through $(0,0)$. For each $j \\in\\{1, \\ldots, n\\}$, consider the point $(j, \\min (j r, n))$. Note that each such point either lies on $\\ell$ or the top edge of the board. In the $j^{\\text {th }}$ column from the left, draw the rectangle of height $\\min (j r, n)$. Note that the sum of the $n$ rectangles is equal to the area of the board under the line $\\ell$ plus $n$ triangles (possibly with area 0 ) each with width at most 1 and whose sum of the heights is at most $n$. Therefore, the sum of the areas of these $n$ triangles is at most $n / 2$. Therefore, $\\sum_{j=1}^{n} \\min (j r, n)$ is at most the area of the square under $\\ell$ plus $n / 2$.\n\nConsider the line with slope $1 / r$. By symmetry about the line $y=x$, the area of the square under the line with slope $1 / r$ is equal to the area of the square above the line $\\ell$. Therefore, using the same reasoning as before, $\\sum_{j=1}^{n} \\min (j / r, n)$ is at most the area of the square above $\\ell$ plus $n / 2$.\n\nTherefore, $\\sum_{j=1}^{n} f_{j}(r)=\\sum_{j=1}^{n}\\left(\\min (j r, n)+\\min \\left(\\frac{j}{r}, n\\right)\\right)$ is at most the area of the board plus $n$, which is $n^{2}+n$. This proves the left hand side inequality.\n\nTo prove the right hand side inequality, we will use the following lemma:\n\nLemma: Consider the line $\\ell$ with slope $s$ passing through $(0,0)$. Then the number of squares on the board that contain an interior point below $\\ell$ is $\\sum_{j=1}^{n} \\min (\\lceil j s\\rceil, n)$.\n\nProof of Lemma: For each $j \\in\\{1, \\ldots, n\\}$, we count the number of squares in the $j^{\\text {th }}$ column (from the left) that contain an interior point lying below the line $\\ell$. The line $x=j$ intersect the line $\\ell$ at $(j, j s)$. Hence, since each column contains $n$ squares\n\n## Sun\n\n## Life Financial\n\ntotal, the number of such squares is $\\min (\\lceil j s\\rceil, n)$. Summing over all $j \\in\\{1,2, \\ldots, n\\}$ proves the lemma. End Proof of Lemma\n\nBy the lemma, the rightmost expression of the inequality is equal to the number of squares containing an interior point below the line with slope $r$ plus the number of squares containing an interior point below the line with slope $1 / r$. By symmetry about the line $y=x$, the latter number is equal to the number of squares containing an interior point above the line with slope $r$. Therefore, the rightmost expression of the inequality is equal to the number of squares of the board plus the number of squares of which $\\ell$ passes through the interior. The former is equal to $n^{2}$. Hence, to prove the inequality, it suffices to show that every line passes through the interior of at least $n$ squares. Since $\\ell$ has positive slope, each $\\ell$ passes through either $n$ rows and/or $n$ columns. In either case, $\\ell$ passes through the interior of at least $n$ squares. Hence, the right inequality holds.", "metadata": {"resource_path": "Canada_MO/segmented/en-sol2013.jsonl", "problem_match": "\n4.", "solution_match": "\nSolution 1:"}} |
| {"year": "2013", "tier": "T2", "problem_label": "4", "problem_type": null, "exam": "Canada_MO", "problem": "Let $n$ be a positive integer. For any positive integer $j$ and positive real number $r$, define\n\n$$\nf_{j}(r)=\\min (j r, n)+\\min \\left(\\frac{j}{r}, n\\right), \\quad \\text { and } \\quad g_{j}(r)=\\min (\\lceil j r\\rceil, n)+\\min \\left(\\left\\lceil\\frac{j}{r}\\right\\rceil, n\\right) \\text {, }\n$$\n\nwhere $\\lceil x\\rceil$ denotes the smallest integer greater than or equal to $x$. Prove that\n\n$$\n\\sum_{j=1}^{n} f_{j}(r) \\leq n^{2}+n \\leq \\sum_{j=1}^{n} g_{j}(r)\n$$", "solution": "We first prove the left inequality. Define the function $f(r)=$ $\\sum_{j=1}^{n} f_{j}(r)$. Note that $f(r)=f(1 / r)$ for all $r>0$. Therefore, we may assume that $r \\geq 1$.\n\nLet $m=\\lfloor n / r\\rfloor$, where $\\lfloor x\\rfloor$ denotes the largest integer less than or equal to $x$. Then $\\min (j r, n)=j r$ for all $j \\in\\{1, \\ldots, m\\}$ and $\\min (j r, n)=n$ for all $j \\in\\{m+1, \\ldots, n\\}$. Note that since $r \\geq 1, \\min (j / r, n) \\leq n$ for all $j \\in\\{1, \\ldots, n\\}$. Therefore,\n\n$$\n\\begin{gathered}\nf(r)=\\sum_{j=1}^{n} f_{j}(r)=(1+2+\\ldots m) r+(n-m) n+(1+2+\\ldots+n) \\cdot \\frac{1}{r} \\\\\n=\\frac{m(m+1)}{2} \\cdot r+\\frac{n(n+1)}{2} \\cdot \\frac{1}{r}+n(n-m)\n\\end{gathered}\n$$\n\nThen by (??), note that $f(r) \\leq n^{2}+n$ if and only if\n\n$$\n\\frac{m(m+1) r}{2}+\\frac{n(n+1)}{2 r} \\leq n(m+1)\n$$\n\nif and only if\n\n$$\nm(m+1) r^{2}+n(n+1) \\leq 2 r n(m+1)\n$$\n\nSince $m=\\lfloor n / r\\rfloor$, there exist a real number $b$ satisfying $0 \\leq b<r$ such that $n=m r+b$. Substituting this into (??) yields\n\n$$\nm(m+1) r^{2}+(m r+b)(m r+b+1) \\leq 2 r(m r+b)(m+1),\n$$\n\n## Sun $\\Psi^{*}$\n\n## Life Financial\n\nif and only if\n\n$$\n2 m^{2} r^{2}+m r^{2}+(2 m b+m) r+b^{2}+b \\leq 2 m^{2} r^{2}+2 m r^{2}+2 m b r+2 b r,\n$$\n\nwhich simplifies to $m r+b^{2}+b \\leq m r^{2}+2 b r \\Leftrightarrow b(b+1-2 r) \\leq m r(r-1) \\Leftrightarrow$ $b((b-r)+(1-r)) \\leq m r(r-1)$. This is true since\n\n$$\nb((b-r)+(1-r)) \\leq 0 \\leq m r(r-1),\n$$\n\nwhich holds since $r \\geq 1$ and $b<r$. Therefore, the left inequality holds.\n\nWe now prove the right inequality. Define the function $g(r)=\\sum_{j=1}^{n}=g_{j}(r)$. Note that $g(r)=g(1 / r)$ for all $r>0$. Therefore, we may assume that $r \\geq 1$. We will consider two cases; $r \\geq n$ and $1 \\leq r<n$.\n\nIf $r \\geq n$, then $\\min (\\lceil j r\\rceil, n)=n$ and $\\min (\\lceil j / r\\rceil, n)=1$ for all $j \\in\\{1, \\ldots, n\\}$. Hence, $g_{j}(r)=n+1$ for all $j \\in\\{1, \\ldots, n\\}$. Therefore, $g(r)=n(n+1)=n^{2}+n$, implying that the inequality is true.\n\nNow we consider the case $1 \\leq r<n$. Let $m=\\lfloor n / r\\rfloor$. Hence, $j r \\leq n$ for all $j \\in\\{1, \\ldots, m\\}$, i.e. $\\min (\\lceil j r\\rceil, n)=,\\lceil j r\\rceil$ and $j r \\geq n$ for all $j \\in\\{m+1, \\ldots, n\\}$, i.e. $\\min (\\lceil j r\\rceil, n)=n$. Therefore,\n\n$$\n\\sum_{j=1}^{n} \\min (\\lceil j r\\rceil, n)=\\sum_{j=1}^{m}\\lceil j r\\rceil+(n-m) n\n$$\n\nWe will now consider the second sum $\\sum_{j=1}^{n} \\min \\{\\lceil j / r\\rceil, n\\}$.\n\nSince $r \\geq 1, \\min (\\lceil j / r\\rceil, n) \\leq \\min (\\lceil n / r\\rceil, n) \\leq n$. Therefore, $\\min (\\lceil j / r\\rceil, n)=$ $\\lceil j / r\\rceil$. Since $m=\\lfloor n / r\\rfloor,\\lceil n / r\\rceil \\leq m+1$. Since $r>1, m<n$, which implies that $m+1 \\leq n$. Therefore, $\\min \\{\\lceil j / r\\rceil, n\\}=\\lceil j / r\\rceil \\leq\\lceil n / r\\rceil \\leq m+1$ for all $j \\in\\{1, \\ldots, n\\}$.\n\nFor each positive integer $k \\in\\{1, \\ldots, m+1\\}$, we now determine the number of positive integers $j \\in\\{1, \\ldots, n\\}$ such that $\\lceil j / r\\rceil=k$. We denote this number by $s_{k}$.\n\nNote that $\\lceil j / r\\rceil=k$ if and only if $k-1<j / r \\leq k$ if and only if $(k-1) r<j \\leq$ $\\min (k r, n)$, since $j \\leq n$. We will handle the cases $k \\in\\{1, \\ldots, m\\}$ and $k=m+1$ separately. If $k \\in\\{1, \\ldots, m\\}$, then $\\min (k r, n)=k r$, since $r \\leq m$ and $m=\\lfloor n / r\\rfloor$.\n\n## Sun 4\n\n## Life Financial\n\nThe set of positive integers $j$ satisfying $(k-1) r<j \\leq k r$ is $\\{\\lfloor(k-1) r\\rfloor+1,\\lfloor(k-$ 1) $r\\rfloor+2, \\ldots,\\lfloor k r\\rfloor\\}$. Hence,\n\n$$\ns_{k}=\\lfloor r k\\rfloor-(\\lfloor r(k-1)\\rfloor+1)+1=\\lfloor r k\\rfloor-\\lfloor r(k-1)\\rfloor\n$$\n\nfor all $k \\in\\{1, \\ldots, m\\}$. If $k=m+1$, then $(k-1) r<j \\leq \\min (k r, n)=n$. The set of positive integers $j$ satisfying $(k-1) r<j \\leq k r$ is $\\{\\lfloor(k-1) r\\rfloor+1, \\ldots, n\\}$. Then $s_{m+1}=n-\\lfloor r(k-1)\\rfloor=n-\\lfloor m r\\rfloor$. Note that this number is non-negative by the definition of $m$. Therefore, by the definition of $s_{k}$, we have\n\n$$\n\\begin{aligned}\n& \\sum_{j=1}^{n} \\min \\left(\\left\\lceil\\frac{j}{r}\\right\\rceil, n\\right)=\\sum_{k=1}^{m+1} k s_{k} \\\\\n& \\quad=\\sum_{k=1}^{m}(k(\\lfloor k r\\rfloor-\\lfloor(k-1) r\\rfloor))+(m+1)(n-\\lfloor r m\\rfloor)=(m+1) n-\\sum_{k=1}^{m}\\lfloor k r\\rfloor .\n\\end{aligned}\n$$\n\nSumming (??) and (??) yields that\n\n$$\ng(r)=n^{2}+n+\\sum_{j=1}^{m}(\\lceil j r\\rceil-\\lfloor j r\\rfloor) \\geq n^{2}+n\n$$\n\nwhich proves the right inequality.\n\n## Sun\n\n## Life Financial", "metadata": {"resource_path": "Canada_MO/segmented/en-sol2013.jsonl", "problem_match": "\n4.", "solution_match": "\nSolution 2:"}} |
| {"year": "2013", "tier": "T2", "problem_label": "5", "problem_type": null, "exam": "Canada_MO", "problem": "Let $O$ denote the circumcentre of an acute-angled triangle $A B C$. A circle $\\Gamma$ passing through vertex $A$ intersects segments $A B$ and $A C$ at points $P$ and $Q$ such that $\\angle B O P=\\angle A B C$ and $\\angle C O Q=\\angle A C B$. Prove that the reflection of $B C$ in the line $P Q$ is tangent to $\\Gamma$.", "solution": "Let the circumcircle of triangle $O B P$ intersect side $B C$ at the points $R$ and $B$ and let $\\angle A, \\angle B$ and $\\angle C$ denote the angles at vertices $A, B$ and $C$, respectively.\n\nNow note that since $\\angle B O P=\\angle B$ and $\\angle C O Q=\\angle C$, it follows that\n\n$\\angle P O Q=360^{\\circ}-\\angle B O P-\\angle C O Q-\\angle B O C=360^{\\circ}-(180-\\angle A)-2 \\angle A=180^{\\circ}-\\angle A$.\n\nThis implies that $A P O Q$ is a cyclic quadrilateral. Since $B P O R$ is cyclic,\n\n$\\angle Q O R=360^{\\circ}-\\angle P O Q-\\angle P O R=360^{\\circ}-\\left(180^{\\circ}-\\angle A\\right)-\\left(180^{\\circ}-\\angle B\\right)=180^{\\circ}-\\angle C$.\n\nThis implies that $C Q O R$ is a cyclic quadrilateral. Since $A P O Q$ and $B P O R$ are cyclic,\n\n$\\angle Q P R=\\angle Q P O+\\angle O P R=\\angle O A Q+\\angle O B R=\\left(90^{\\circ}-\\angle B\\right)+\\left(90^{\\circ}-\\angle A\\right)=\\angle C$.\n\nSince $C Q O R$ is cyclic, $\\angle Q R C=\\angle C O Q=\\angle C=\\angle Q P R$ which implies that the circumcircle of triangle $P Q R$ is tangent to $B C$. Further, since $\\angle P R B=\\angle B O P=$ $\\angle B$,\n\n$$\n\\angle P R Q=180^{\\circ}-\\angle P R B-\\angle Q R C=180^{\\circ}-\\angle B-\\angle C=\\angle A=\\angle P A Q\n$$\n\nThis implies that the circumcircle of $P Q R$ is the reflection of $\\Gamma$ in line $P Q$. By symmetry in line $P Q$, this implies that the reflection of $B C$ in line $P Q$ is tangent to $\\Gamma$.", "metadata": {"resource_path": "Canada_MO/segmented/en-sol2013.jsonl", "problem_match": "\n5.", "solution_match": "\nSolution."}} |
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