{"year": "2010", "tier": "T3", "problem_label": "1", "problem_type": null, "exam": "Benelux_MO", "problem": "A finite set of integers is called bad if its elements add up to 2010. A finite set of integers is a Benelux-set if none of its subsets is bad. Determine the smallest integer $n$ such that the set $\\{502,503,504, \\ldots, 2009\\}$ can be partitioned into $n$ Benelux-sets.\n(A partition of a set $S$ into $n$ subsets is a collection of $n$ pairwise disjoint subsets of $S$, the union of which equals $S$.)", "solution": "As $502+1508=2010$, the set $S=\\{502,503, \\ldots, 2009\\}$ is not a Benelux-set, so $n=1$ does not work. We will prove that $n=2$ does work, i.e. that $S$ can be partitioned into 2 Benelux-sets.\nDefine the following subsets of $S$ :\n\n$$\n\\begin{aligned}\n& A=\\{502,503, \\ldots, 670\\}, \\\\\n& B=\\{671,672, \\ldots, 1005\\}, \\\\\n& C=\\{1006,1007, \\ldots, 1339\\}, \\\\\n& D=\\{1340,1341, \\ldots, 1508\\}, \\\\\n& E=\\{1509,1510, \\ldots, 2009\\} .\n\\end{aligned}\n$$\n\nWe will show that $A \\cup C \\cup E$ and $B \\cup D$ are both Benelux-sets.\nNote that there does not exist a bad subset of $S$ of one element, since that element would have to be 2010. Also, there does not exist a bad subset of $S$ of more than three elements, since the sum of four or more elements would be at least $502+503+504+505=2014>2010$. So any possible bad subset of $S$ contains two or three elements.\nConsider a bad subset of two elements $a$ and $b$. As $a, b \\geq 502$ and $a+b=2010$, we have $a, b \\leq 2010-502=1508$. Furthermore, exactly one of $a$ and $b$ is smaller than 1005 and one is larger than 1005. So one of them, say $a$, is an element of $A \\cup B$, and the other is an element of $C \\cup D$. Suppose $a \\in A$, then $b \\geq 2010-670=1340$, so $b \\in D$. On the other hand, suppose $a \\in B$, then $b \\leq 2010-671=1339$, so $b \\in C$. Hence $\\{a, b\\}$ cannot be a subset of $A \\cup C \\cup E$, nor of $B \\cup D$.\nNow consider a bad subset of three elements $a, b$ and $c$. As $a, b, c \\geq 502, a+b+c=2010$, and the three elements are pairwise distinct, we have $a, b, c \\leq 2010-502-503=1005$. So $a, b, c \\in A \\cup B$. At least one of the elements, say $a$, is smaller than $\\frac{2010}{3}=670$, and at least one of the elements, say $b$, is larger than 670 . So $a \\in A$ and $b \\in B$. We conclude that $\\{a, b, c\\}$ cannot be a subset of $A \\cup C \\cup E$, nor of $B \\cup D$.\nThis proves that $A \\cup C \\cup E$ and $B \\cup D$ are Benelux-sets, and therefore the smallest $n$ for which $S$ can be partitioned into $n$ Benelux-sets is $n=2$.\n\nRemark. Observe that $A \\cup C \\cup E_{1}$ and $B \\cup D \\cup E_{2}$ are also Benelux-sets, where $\\left\\{E_{1}, E_{2}\\right\\}$ is any partition of $E$.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2010-zz.jsonl", "problem_match": "\nProblem 1.", "solution_match": "\nSolution."}}
{"year": "2010", "tier": "T3", "problem_label": "2", "problem_type": null, "exam": "Benelux_MO", "problem": "Find all polynomials $p(x)$ with real coefficients such that\n\n$$\np(a+b-2 c)+p(b+c-2 a)+p(c+a-2 b)=3 p(a-b)+3 p(b-c)+3 p(c-a)\n$$\n\nfor all $a, b, c \\in \\mathbb{R}$.", "solution": ". For $a=b=c$, we have $3 p(0)=9 p(0)$, hence $p(0)=0$. Now set $b=c=0$, then we have\n\n$$\np(a)+p(-2 a)+p(a)=3 p(a)+3 p(-a)\n$$\n\nfor all $a \\in \\mathbb{R}$. So we find a polynomial equation\n\n$$\np(-2 x)=p(x)+3 p(-x)\n$$\n\nNote that the zero polynomial is a solution to this equation. Now suppose that $p$ is not the zero polynomial, and let $n \\geq 0$ be the degree of $p$. Let $a_{n} \\neq 0$ be the coefficient of $x^{n}$ in $p(x)$. At the left-hand side of (1), the coefficient of $x^{n}$ is $(-2)^{n} \\cdot a_{n}$, while at the right-hand side the coefficient of $x^{n}$ is $a_{n}+3 \\cdot(-1)^{n} \\cdot a_{n}$. Hence $(-2)^{n}=1+3 \\cdot(-1)^{n}$. For $n$ even, we find $2^{n}=4$, so $n=2$, and for $n$ odd, we find $-2^{n}=-2$, so $n=1$. As we already know that $p(0)=0$, we must have $p(x)=a_{2} x^{2}+a_{1} x$, where $a_{1}$ and $a_{2}$ are real numbers (possibly zero).\nThe polynomial $p(x)=x$ is a solution to our problem, as\n\n$$\n(a+b-2 c)+(b+c-2 a)+(c+a-2 b)=0=3(a-b)+3(b-c)+3(c-a)\n$$\n\nfor all $a, b, c \\in \\mathbb{R}$. Also, $p(x)=x^{2}$ is a solution, since\n\n$$\n\\begin{aligned}\n(a+b-2 c)^{2}+(b+c-2 a)^{2}+(c+a-2 b)^{2} & =6\\left(a^{2}+b^{2}+c^{2}\\right)-6(a b+b c+c a) \\\\\n& =3(a-b)^{2}+3(b-c)^{2}+3(c-a)^{2}\n\\end{aligned}\n$$\n\nfor all $a, b, c \\in \\mathbb{R}$.\nNow note that if $p(x)$ is a solution to our problem, then so is $\\lambda p(x)$ for all $\\lambda \\in \\mathbb{R}$. Also, if $p(x)$ and $q(x)$ are both solutions, then so is $p(x)+q(x)$. We conclude that for all real numbers $a_{2}$ and $a_{1}$ the polynomial $a_{2} x^{2}+a_{1} x$ is a solution. Since we have already shown that there can be no other solutions, these are the only solutions.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2010-zz.jsonl", "problem_match": "\nProblem 2.", "solution_match": "\nSolution 1"}}
{"year": "2010", "tier": "T3", "problem_label": "2", "problem_type": null, "exam": "Benelux_MO", "problem": "Find all polynomials $p(x)$ with real coefficients such that\n\n$$\np(a+b-2 c)+p(b+c-2 a)+p(c+a-2 b)=3 p(a-b)+3 p(b-c)+3 p(c-a)\n$$\n\nfor all $a, b, c \\in \\mathbb{R}$.", "solution": ". For $a=b=c$, we have $3 p(0)=9 p(0)$, hence $p(0)=0$. Now set $b=c=0$, then we have\n\n$$\np(a)+p(-2 a)+p(a)=3 p(a)+3 p(-a)\n$$\n\nfor all $a \\in \\mathbb{R}$. So we find a polynomial equation\n\n$$\np(-2 x)=p(x)+3 p(-x)\n$$\n\nDefine $q(x)=p(x)+p(-x)$, then we find that\n\n$$\nq(2 x)=p(2 x)+p(-2 x)=(p(-x)+3 p(x))+(p(x)+3 p(-x))=4 q(x)\n$$\n\nNote that the zero polynomial is a solution to this equation. Now suppose that $q$ is not the zero polynomial, and let $m \\geq 0$ be the degree of $q$. Let $b_{m} \\neq 0$ be the coefficient of $x^{m}$ in $q(x)$. At the left-hand side of (3), the coefficient of $x^{m}$ is $2^{m} \\cdot b_{m}$, while at the right-hand side the coefficient of $x^{m}$ is $4 b_{m}$. Hence $m=2$. As $q(x)=p(x)+p(-x)$, the polynomial $q(x)$ does not contain any nonzero terms with odd exponent of $x$. Since also $q(0)=2 p(0)=0$, we conclude that\n\n$$\nq(x)=b_{2} x^{2}\n$$\n\nwhere $b_{2}$ is a real number (possibly zero).\nFrom (2) we now deduce that $p(2 x)=p(-x)+3 p(x)=2 p(x)+q(x)$, so\n\n$$\np(2 x)-2 p(x)=b_{2} x^{2}\n$$\n\nSuppose that that degree $n$ of $p$ is greater than 2 . Let $a_{n} \\neq 0$ be the coefficient of $x^{n}$ in $p(x)$. At the left-hand side of (4), the coefficient of $x^{n}$ is $\\left(2^{n}-2\\right) \\cdot a_{n} \\neq 0$. But the coefficient of $x^{n}$ at the right-hand side vanishes, yielding a contradiction. So the degree of $p$ is at most 2 . As we already know that $p(0)=0$, we must have $p(x)=a_{2} x^{2}+a_{1} x$, where $a_{1}$ and $a_{2}$ are real numbers (possibly zero).\nWe finally check that every polynomial of this form is indeed a solution (see solution 1 ).", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2010-zz.jsonl", "problem_match": "\nProblem 2.", "solution_match": "\nSolution 2"}}
{"year": "2010", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "Benelux_MO", "problem": "On a line $l$ there are three different points $A, B$ and $P$ in that order. Let $a$ be the line through $A$ perpendicular to $l$, and let $b$ be the line through $B$ perpendicular to $l$. A line through $P$, not coinciding with $l$, intersects $a$ in $Q$ and $b$ in $R$. The line through $A$ perpendicular to $B Q$ intersects $B Q$ in $L$ and $B R$ in $T$. The line through $B$ perpendicular to $A R$ intersects $A R$ in $K$ and $A Q$ in $S$.\n(a) Prove that $P, T, S$ are collinear.\n(b) Prove that $P, K, L$ are collinear.\n\n#", "solution": ".\n\n(a) Since $P, R$ and $Q$ are collinear, we have $\\triangle P A Q \\sim \\triangle P B R$, hence\n\n$$\n\\frac{|A Q|}{|B R|}=\\frac{|A P|}{|B P|}\n$$\n\nConversely, $P, T$ and $S$ are collinear if it holds that\n\n$$\n\\frac{|A S|}{|B T|}=\\frac{|A P|}{|B P|}\n$$\n\nSo it suffices to prove\n\n$$\n\\frac{|B T|}{|B R|}=\\frac{|A S|}{|A Q|}\n$$\n\nSince $\\angle A B T=90^{\\circ}=\\angle A L B$ and $\\angle T A B=\\angle B A L$, we have $\\triangle A B T \\sim \\triangle A L B$. And since $\\angle A L B=90^{\\circ}=\\angle Q A B$ and $\\angle L B A=\\angle A B Q$, we have $\\triangle A L B \\sim \\triangle Q A B$. Hence $\\triangle A B T \\sim \\triangle Q A B$, so\n\n$$\n\\frac{|B T|}{|B A|}=\\frac{|A B|}{|A Q|}\n$$\n\nSimilarly, we have $\\triangle A B R \\sim \\triangle A K B \\sim \\triangle S A B$, so\n\n$$\n\\frac{|B R|}{|B A|}=\\frac{|A B|}{|A S|}\n$$\n\nCombining both results, we get\n\n$$\n\\frac{|B T|}{|B R|}=\\frac{|B T| /|B A|}{|B R| /|B A|}=\\frac{|A B| /|A Q|}{|A B| /|A S|}=\\frac{|A S|}{|A Q|}\n$$\n\nwhich had to be proved.\n(b) Let the line $P K$ intersect $B R$ in $B_{1}$ and $A Q$ in $A_{1}$ and let the line $P L$ intersect $B R$ in $B_{2}$ and $A Q$ in $A_{2}$. Consider the points $A_{1}, A$ and $S$ on the line $A Q$, and the points $B_{1}, B$ and $T$ on the line $B R$. As $A Q \\| B R$ and the three lines $A_{1} B_{1}, A B$ and $S T$ are concurrent (in $P$ ), we have\n\n$$\nA_{1} A: A S=B_{1} B: B T\n$$\n\nwhere all lengths are directed. Similarly, as $A_{1} B_{1}, A R$ and $S B$ are concurrent (in $K$ ), we have\n\n$$\nA_{1} A: A S=B_{1} R: R B\n$$\n\nThis gives\n\n$$\n\\frac{B B_{1}}{B T}=\\frac{R B_{1}}{R B}=\\frac{R B+B B_{1}}{R B}=1+\\frac{B B_{1}}{R B}=1-\\frac{B B_{1}}{B R}\n$$\n\nso\n\n$$\nB B_{1}=\\frac{1}{\\frac{1}{B T}+\\frac{1}{B R}} .\n$$\n\nSimilary, using the lines $A_{2} B_{2}, A B$ and $Q R$ (concurrent in $P$ ) and the lines $A_{2} B_{2}$, $A T$ and $Q B$ (concurrent in $L$ ), we find\n\n$$\nB_{2} B: B R=A_{2} A: A Q=B_{2} T: T B\n$$\n\nThis gives\n\n$$\n\\frac{B B_{2}}{B R}=\\frac{T B_{2}}{T B}=\\frac{T B+B B_{2}}{T B}=1+\\frac{B B_{2}}{T B}=1-\\frac{B B_{2}}{B T}\n$$\n\nso\n\n$$\nB B_{2}=\\frac{1}{\\frac{1}{B R}+\\frac{1}{B T}}\n$$\n\nWe conclude that $B_{1}=B_{2}$, which implies that $P, K$ and $L$ are collinear.\n\n#", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2010-zz.jsonl", "problem_match": "\nProblem 3.", "solution_match": "# Solution 1"}}
{"year": "2010", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "Benelux_MO", "problem": "On a line $l$ there are three different points $A, B$ and $P$ in that order. Let $a$ be the line through $A$ perpendicular to $l$, and let $b$ be the line through $B$ perpendicular to $l$. A line through $P$, not coinciding with $l$, intersects $a$ in $Q$ and $b$ in $R$. The line through $A$ perpendicular to $B Q$ intersects $B Q$ in $L$ and $B R$ in $T$. The line through $B$ perpendicular to $A R$ intersects $A R$ in $K$ and $A Q$ in $S$.\n(a) Prove that $P, T, S$ are collinear.\n(b) Prove that $P, K, L$ are collinear.\n\n#", "solution": ".\n\n(a) Define $X$ as the intersection of $A T$ and $B S$, and $Y$ as the intersection of $A R$ and $B Q$. To prove that $P, S$ and $T$ are collinear, we will use Menelaos' theorem in $\\triangle A B X$, so we have to prove\n\n$$\n\\frac{A P}{P B} \\frac{B S}{S X} \\frac{X T}{T A}=-1\n$$\n\nNote that $B$ is between $P$ and $A, X$ is between $S$ and $B$, and $X$ is between $T$ and $A$, so it suffices to prove that\n\n$$\n\\frac{|A P|}{|P B|} \\frac{|B S|}{|S X|} \\frac{|X T|}{|T A|}=1\n$$\n\nBecause $A Q$ and $B R$ are parallel, we have $\\triangle A Q P \\sim \\triangle B R P$, hence\n\n$$\n\\frac{|A P|}{|B P|}=\\frac{|Q A|}{|R B|}\n$$\n\nAlso, since $\\angle A S B=\\angle K B R$ and $\\angle B A S=90^{\\circ}=\\angle B K R$, we have $\\triangle A S B \\sim$ $\\triangle K B R$, hence\n\n$$\n\\frac{|B S|}{|R B|}=\\frac{|A S|}{|K B|}, \\quad \\text { so } \\quad|B S|=\\frac{|A S|}{|K B|}|R B| \\text {. }\n$$\n\nSimilarly, we have $\\triangle A T B \\sim \\triangle Q A L$, hence\n\n$$\n\\frac{|T A|}{|A Q|}=\\frac{|T B|}{|A L|}, \\quad \\text { so } \\quad|T A|=\\frac{|T B|}{|A L|}|A Q| \\text {. }\n$$\n\nAs $\\angle A S X=\\angle A S B=90^{\\circ}-\\angle A B S=90^{\\circ}-\\angle A B K=\\angle K A B=\\angle Y A B$, and $\\angle S A X=90^{\\circ}-\\angle X A B=90^{\\circ}-\\angle L A B=\\angle A B L=\\angle A B Y$, we have $\\triangle S X A \\sim$ $\\triangle A Y B$, hence\n\n$$\n\\frac{|S X|}{|A Y|}=\\frac{|A S|}{|B A|}, \\quad \\text { so } \\quad|S X|=\\frac{|A S|}{|B A|}|A Y| \\text {. }\n$$\n\nSimilarly, we have $\\triangle B X T \\sim \\triangle A Y B$, hence\n\n$$\n\\frac{|X T|}{|Y B|}=\\frac{|B T|}{|A B|}, \\quad \\text { so } \\quad|X T|=\\frac{|B T|}{|A B|}|Y B| \\text {. }\n$$\n\nBy combining (5) - (9), we find\n\n$$\n\\begin{aligned}\n\\frac{|A P|}{|P B|} \\frac{|B S|}{|S X|} \\frac{|X T|}{|T A|} & =\\frac{|Q A|}{|R B|} \\cdot \\frac{|A S|}{|K B|}|R B| \\cdot \\frac{|B A|}{|A S||A Y|} \\cdot \\frac{|B T|}{|A B|}|Y B| \\cdot \\frac{|A L|}{|T B||A Q|} \\\\\n& =\\frac{|A L|}{|K B|} \\frac{|Y B|}{|A Y|} .\n\\end{aligned}\n$$\n\nSince $\\angle Y L A=90^{\\circ}=\\angle Y K B$ and $\\angle A Y L=\\angle B Y K$, we have $\\triangle A Y L \\sim \\triangle B Y K$, hence\n\n$$\n\\frac{|A L|}{|B K|}=\\frac{|A Y|}{|B Y|}, \\quad \\text { so } \\quad \\frac{|A L|}{|B K|} \\frac{|B Y|}{|A Y|}=1\n$$\n\nBy combining (10) and (11), we find\n\n$$\n\\frac{|A P|}{|P B|} \\frac{|B S|}{|S X|} \\frac{|X T|}{|T A|}=1,\n$$\n\nas we wanted to prove.\n(b) Again, we will use Menelaos' theorem in $\\triangle A B X$, so we have to prove\n\n$$\n\\frac{A P}{P B} \\frac{B K}{K X} \\frac{X L}{L A}=-1\n$$\n\nNote that $\\frac{A P}{P B}<0$, and $\\frac{B K}{K X}<0$ if and only if $\\frac{X L}{L A}<0$, so it suffices to prove that\n\n$$\n\\frac{|A P|}{|P B|} \\frac{|B K|}{|K X|} \\frac{|X L|}{|L A|}=1\n$$\n\nAs $\\angle B X L=\\angle A X K$ and $\\angle B L X=90^{\\circ}=\\angle A K X$, we have $\\triangle B L X \\sim \\triangle A K X$, hence\n\n$$\n\\frac{|X L|}{|X K|}=\\frac{|B L|}{|A K|}\n$$\n\nSince $\\angle A L B=90^{\\circ}=\\angle Q A B$, we have $\\triangle A L B \\sim \\triangle Q A B$, hence\n\n$$\n\\frac{|L A|}{|A Q|}=\\frac{|L B|}{|A B|}, \\quad \\text { so } \\quad|L A|=\\frac{|L B|}{|A B|}|A Q| \\text {. }\n$$\n\nSimilarly, we have $\\triangle A K B \\sim \\triangle A B R$, hence\n\n$$\n\\frac{|B K|}{|R B|}=\\frac{|A K|}{|A B|}, \\quad \\text { so } \\quad|B K|=\\frac{|A K|}{|A B|}|R B|\n$$\n\nBy combining (5) and (12) - (14), we find\n\n$$\n\\left.\\frac{|A P|}{|P B|}\\left|\\frac{|B K|}{|K X|} \\frac{|X L|}{|L A|}=\\frac{|Q A|}{|R B|} \\cdot \\frac{|B L|}{|A K|} \\cdot \\frac{|A B|}{|L B||A Q|} \\cdot \\frac{|A K|}{|A B|}\\right| R B \\right\\rvert\\,=1,\n$$\n\nwhich is what we wanted to prove.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2010-zz.jsonl", "problem_match": "\nProblem 3.", "solution_match": "# Solution 2"}}
{"year": "2010", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "Benelux_MO", "problem": "On a line $l$ there are three different points $A, B$ and $P$ in that order. Let $a$ be the line through $A$ perpendicular to $l$, and let $b$ be the line through $B$ perpendicular to $l$. A line through $P$, not coinciding with $l$, intersects $a$ in $Q$ and $b$ in $R$. The line through $A$ perpendicular to $B Q$ intersects $B Q$ in $L$ and $B R$ in $T$. The line through $B$ perpendicular to $A R$ intersects $A R$ in $K$ and $A Q$ in $S$.\n(a) Prove that $P, T, S$ are collinear.\n(b) Prove that $P, K, L$ are collinear.\n\n#", "solution": ". As $\\angle A K B=\\angle A L B=90^{\\circ}$, the points $K$ and $L$ belong to the circle with diameter $A B$. Since $\\angle Q A B=\\angle A B R=90^{\\circ}$, the lines $A Q$ and $B R$ are tangents to this circle.\nApply Pascal's theorem to the points $A, A, K, L, B$ and $B$, all on the same circle. This yields that the intersection $Q$ of the tangent in $A$ and the line $B L$, the intersection $R$ of the tangent in $B$ and the line $A K$, and the intersection of $K L$ and $A B$ are collinear. So $K L$ passes through the intersection of $A B$ and $Q R$, which is point $P$. Hence $P, K$ and $L$ are collinear. This proves part b.\nNow apply Pascal's theorem to the points $A, A, L, K, B$ and $B$. This yields that the intersection $S$ of the tangent in $A$ and the line $B K$, the intersection $T$ of the tangent in $B$ and the line $A L$, and the intersection $P$ of $K L$ and $A B$ are collinear. This proves part a.\n\n#", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2010-zz.jsonl", "problem_match": "\nProblem 3.", "solution_match": "\nSolution 3"}}
{"year": "2010", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "Benelux_MO", "problem": "On a line $l$ there are three different points $A, B$ and $P$ in that order. Let $a$ be the line through $A$ perpendicular to $l$, and let $b$ be the line through $B$ perpendicular to $l$. A line through $P$, not coinciding with $l$, intersects $a$ in $Q$ and $b$ in $R$. The line through $A$ perpendicular to $B Q$ intersects $B Q$ in $L$ and $B R$ in $T$. The line through $B$ perpendicular to $A R$ intersects $A R$ in $K$ and $A Q$ in $S$.\n(a) Prove that $P, T, S$ are collinear.\n(b) Prove that $P, K, L$ are collinear.\n\n#", "solution": ".\n\n(a) W.l.o.g. we may assume that $A=(0,0)$ and $B=(1,0)$ and the line through $P$ is in the upper half plane, so $l$ is the $x$-axis, $a$ is the $y$-axis and $b$ is the line $x=1$. Take $P=(p, 0)(p>1)$ and $Q=(0, q)(q>0)$. Since $P Q$ is given by $\\frac{x}{p}+\\frac{y}{q}=1$, we find $R=\\left(1, \\frac{q(p-1)}{p}\\right)$.\nNow $A R$ is given by $y=\\frac{q(p-1)}{p} x$, hence $B S$, the line perpendicular to $A R$ and passing through $B=(1,0)$, is given by $y=-\\frac{p}{q(p-1)}(x-1)$. We find $S=\\left(0, \\frac{p}{q(p-1)}\\right)$.\nMoreover $B Q$ is given by $y=-q(x-1)$, hence $A T$, the line perpendicular to $B Q$ and passing through $A=(0,0)$, is given by $y=\\frac{1}{q} x$. We find $T=\\left(1, \\frac{1}{q}\\right)$. Since $\\frac{|B T|}{|B P|}=\\frac{1 / q}{p-1}=\\frac{\\frac{p}{q(p-1)}}{p}=\\frac{|A S|}{|A P|}$, we conclude that $P, T$ and $S$ are collinear.\n(b) Point $K$ is the intersection of $A R$ and $B S$. Solving for $x$ yields\n\n$$\n\\begin{aligned}\n\\frac{q(p-1)}{p} x & =-\\frac{p}{q(p-1)}(x-1) \\\\\n\\left(\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}\\right) x & =\\frac{p}{q(p-1)} \\\\\nx & =\\frac{\\frac{p}{q(p-1)}}{\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}}\n\\end{aligned}\n$$\n\nso\n\n$$\nK=\\left(\\frac{\\frac{p}{q(p-1)}}{\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}}, \\frac{1}{\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}}\\right)\n$$\n\nPoint $L$ is the point of intersection of $A T$ and $B Q$. Solving for $x$ yields\n\n$$\n\\begin{aligned}\n\\frac{1}{q} x & =-q(x-1) \\\\\n\\left(\\frac{1}{q}+q\\right) x & =q \\\\\nx & =\\frac{q}{\\frac{1}{q}+q}\n\\end{aligned}\n$$\n\nso\n\n$$\nL=\\left(\\frac{q}{\\frac{1}{q}+q}, \\frac{1}{\\frac{1}{q}+q}\\right)\n$$\n\nLet $K_{0}$ and $L_{0}$ be the projections of $K$ and $L$ on the $x$-axis. We have to show that the following fractions are equal:\n\n$$\n\\frac{\\left|K_{0} K\\right|}{\\left|K_{0} P\\right|}=\\frac{\\frac{1}{\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}}}{p-\\frac{\\frac{p}{q-1)}}{\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}}} \\quad \\text { and } \\quad \\frac{\\left|L_{0} L\\right|}{\\left|L_{0} P\\right|}=\\frac{\\frac{1}{\\frac{1}{q}+q}}{p-\\frac{q}{\\frac{1}{q}+q}}\n$$\n\nWorking out cross products twice, this comes down to\n\n$$\n\\begin{aligned}\n\\frac{1}{\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}} \\cdot\\left(p-\\frac{q}{\\frac{1}{q}+q}\\right) & \\stackrel{?}{=} \\frac{1}{\\frac{1}{q}+q} \\cdot\\left(p-\\frac{\\frac{p}{q(p-1)}}{\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}}\\right) \\\\\n\\left(\\frac{1}{q}+q\\right) \\cdot\\left(p-\\frac{q}{\\frac{1}{q}+q}\\right) & \\stackrel{?}{=}\\left(\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}\\right) \\cdot\\left(p-\\frac{\\frac{p}{q(p-1)}}{\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}}\\right) \\\\\n\\frac{p}{q}+p q-q & \\stackrel{?}{=} q(p-1)+\\frac{p^{2}}{q(p-1)}-\\frac{p}{q(p-1)} \\\\\n\\frac{p}{q}+p q-q & \\stackrel{?}{=} q(p-1)+\\frac{p(p-1)}{q(p-1)}\n\\end{aligned}\n$$\n\nwhich is clearly true.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2010-zz.jsonl", "problem_match": "\nProblem 3.", "solution_match": "# Solution 4"}}
{"year": "2010", "tier": "T3", "problem_label": "4", "problem_type": null, "exam": "Benelux_MO", "problem": "Find all quadruples $(a, b, p, n)$ of positive integers, such that $p$ is a prime and\n\n$$\na^{3}+b^{3}=p^{n}\n$$", "solution": ". Let $(a, b, p, n)$ be a solution. Note that we can write the given equation as\n\n$$\n(a+b)\\left(a^{2}-a b+b^{2}\\right)=p^{n}\n$$\n\nAs $a$ and $b$ are positive integers, we have $a+b \\geq 2$, so $p \\mid a+b$. Furthermore, $a^{2}-a b+b^{2}=$ $(a-b)^{2}+a b$, so either $a=b=1$ or $a^{2}-a b+b^{2} \\geq 2$. Assume that the latter is the case. Then $p$ is a divisor of both $a+b$ and $a^{2}-a b+b^{2}$, hence also of $(a+b)^{2}-\\left(a^{2}-a b+b^{2}\\right)=3 a b$. This means that $p$ either is equal to 3 or is a divisor of $a b$. Since $p$ is a divisor of $a+b$, we have $p|a \\Leftrightarrow p| b$, hence either $p=3$, or $p \\mid a$ and $p \\mid b$. If $p \\mid a$ and $p \\mid b$, then we can write $a=p a^{\\prime}, b=p b^{\\prime}$ with $a^{\\prime}$ and $b^{\\prime}$ positive integers, and we have $\\left(a^{\\prime}\\right)^{3}+\\left(b^{\\prime}\\right)^{3}=p^{n-3}$, so $\\left(a^{\\prime}, b^{\\prime}, p, n-3\\right)$ then is another solution (note that $\\left(a^{\\prime}\\right)^{3}+\\left(b^{\\prime}\\right)^{3}$ is a positive integer greater than 1 , so $n-3$ is positive).\nNow assume that $\\left(a_{0}, b_{0}, p_{0}, n_{0}\\right)$ is a solution such that $p \\nmid a$. From the reasoning above it follows that either $a_{0}=b_{0}=1$, or $p_{0}=3$. After all, if we do not have $a_{0}=b_{0}=1$ and we have $p_{0} \\neq 3$, then $p \\mid a$. Also, given an arbitrary solution $(a, b, p, n)$, we can divide everything by $p$ repeatedly until there are no factors $p$ left in $a$.\nSuppose $a_{0}=b_{0}=1$. Then the solution is $(1,1,2,1)$.\nSuppose $p_{0}=3$. Assume that $3^{2} \\mid\\left(a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}\\right)$. As $3^{2} \\mid\\left(a_{0}+b_{0}\\right)^{2}$, we then have $3^{2} \\mid\\left(a_{0}+b_{0}\\right)^{2}-\\left(a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}\\right)=3 a_{0} b_{0}$, so $3 \\mid a_{0} b_{0}$. But $3 \\nmid a_{0}$ by assumption, and $3 \\mid a_{0}+b_{0}$, so $3 \\nmid b_{0}$, which contradicts $3 \\mid a_{0} b_{0}$. We conclude that $3^{2} \\nmid\\left(a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}\\right)$. As both $a_{0}+b_{0}$ and $a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}$ must be powers of 3 , we have $a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}=3$. Hence $\\left(a_{0}-b_{0}\\right)^{2}+a_{0} b_{0}=3$. We must have $\\left(a_{0}-b_{0}\\right)^{2}=0$ or $\\left(a_{0}-b_{0}\\right)^{2}=1$. The former does not give a solution; the latter gives $a_{0}=2$ and $b_{0}=1$ or $a_{0}=1$ and $b_{0}=2$.\nSo all solutions with $p \\nmid a$ are $(1,1,2,1),(2,1,3,2)$ and $(1,2,3,2)$. From the above it follows that all other solutions are of the form $\\left(p_{0}^{k} a_{0}, p_{0}^{k} b_{0}, p_{0}, n_{0}+3 k\\right)$, where $\\left(a_{0}, b_{0}, p_{0}, n_{0}\\right)$ is one of these three solutions. Hence we find three families of solutions:\n\n- $\\left(2^{k}, 2^{k}, 2,3 k+1\\right)$ with $k \\in \\mathbb{Z}_{\\geq 0}$,\n- $\\left(2 \\cdot 3^{k}, 3^{k}, 3,3 k+2\\right)$ with $k \\in \\mathbb{Z}_{\\geq 0}$,\n- $\\left(3^{k}, 2 \\cdot 3^{k}, 3,3 k+2\\right)$ with $k \\in \\mathbb{Z}_{\\geq 0}$.\n\nIt is easy to check that all these quadruples are indeed solutions.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2010-zz.jsonl", "problem_match": "\nProblem 4.", "solution_match": "\nSolution 1"}}
{"year": "2010", "tier": "T3", "problem_label": "4", "problem_type": null, "exam": "Benelux_MO", "problem": "Find all quadruples $(a, b, p, n)$ of positive integers, such that $p$ is a prime and\n\n$$\na^{3}+b^{3}=p^{n}\n$$", "solution": ". Let $(a, b, p, n)$ be a solution. Note that we can write the given equation as\n\n$$\n(a+b)\\left(a^{2}-a b+b^{2}\\right)=p^{n}\n$$\n\nAs $a$ and $b$ are positive integers, we have $a+b \\geq 2$ and $a^{2}-a b+b^{2}=(a-b)^{2}+a b \\geq 1$. So both factors are positive and therefore must be powers of $p$. Let $k$ be an integer with $1 \\leq k \\leq n$ such that $a+b=p^{k}$. Then $a^{2}-a b+b^{2}=p^{n-k}$. If we substitute $b=p^{k}-a$, we find\n\n$$\np^{n-k}=(a+b)^{2}-3 a b=p^{2 k}-3 a\\left(p^{k}-a\\right) .\n$$\n\nWe can rewrite this as:\n\n$$\n3 a^{2}-3 p^{k} a+p^{2 k}-p^{n-k}=0\n$$\n\nfrom which we see that $a$ is a solution of the following quadratic equation in $x$ :\n\n$$\n3 x^{2}-3 p^{k} x+p^{2 k}-p^{n-k}=0 .\n$$\n\nThe discriminant of (15) is\n\n$$\nD=\\left(-3 p^{k}\\right)^{2}-4 \\cdot 3 \\cdot\\left(p^{2 k}-p^{n-k}\\right)=3 \\cdot\\left(4 p^{n-k}-p^{2 k}\\right)=3 p^{n-k} \\cdot\\left(4-p^{3 k-n}\\right)\n$$\n\nAs $p^{n-k}=(a+b)^{2}-3 a b<(a+b)^{2}=p^{2 k}$, we have $n-k<2 k$, so $3 k-n>0$. Since $a$ is a solution of (15), the discriminant must be nonnegative. Hence $4-p^{3 k-n} \\geq 0$. If $p=2$, this implies $3 k-n=1$ or $3 k-n=2$; if $p=3$, this implies $3 k-n=1$; and if $p>3$, then $p \\geq 5$ so $4 \\geq p^{3 k-n}$ can never be true.\nSuppose $p=2$ and $3 k-n=1$. Then $D=3 \\cdot 2^{2 k-1} \\cdot(4-2)=3 \\cdot 2^{2 k}$. But this is a not a square, so the solutions of (15) will not be integers, which yields a contradiction.\nSuppose $p=2$ and $3 k-n=2$. Then $D=3 \\cdot 2^{2 k-2} \\cdot(4-4)=0$, so the only solution of (15) is $x=\\frac{3 \\cdot 2^{k}}{2 \\cdot 3}=2^{k-1}$. Therefore $a=2^{k-1}$ and $b=2^{k}-a=2^{k-1}$, and this gives a solution for all $k \\geq 1$, namely ( $\\left.2^{k-1}, 2^{k-1}, 2,3 k-2\\right)$.\nSuppose $p=3$ and $3 k-n=1$. Then $D=3 \\cdot 3^{2 k-1} \\cdot(4-3)=3^{2 k}$, so the solutions of (15) are $x=\\frac{3^{k+1} \\pm 3^{k}}{2 \\cdot 3}=\\frac{1}{2}\\left(3^{k} \\pm 3^{k-1}\\right)$. Therefore $a=2 \\cdot 3^{k-1}$ or $a=3^{k-1}$. For all $k \\geq 1$ we find the solutions $\\left(2 \\cdot 3^{k-1}, 3^{k-1}, 3,3 k-1\\right)$ and $\\left(3^{k-1}, 2 \\cdot 3^{k-1}, 3,3 k-1\\right)$.\nWe conclude that there are three families of solutions:\n\n- $\\left(2^{k-1}, 2^{k-1}, 2,3 k-2\\right)$ with $k \\in \\mathbb{Z}_{\\geq 1}$,\n- $\\left(2 \\cdot 3^{k-1}, 3^{k-1}, 3,3 k-1\\right)$ with $k \\in \\mathbb{Z}_{\\geq 1}$,\n- $\\left(3^{k-1}, 2 \\cdot 3^{k-1}, 3,3 k-1\\right)$ with $k \\in \\mathbb{Z}_{\\geq 1}$.\n\nIt is easy to check that all these quadruples are indeed solutions.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2010-zz.jsonl", "problem_match": "\nProblem 4.", "solution_match": "\nSolution 2"}}