| {"year": "2020", "tier": "T3", "problem_label": "1", "problem_type": null, "exam": "Benelux_MO", "problem": "Find all positive integers $d$ with the following property: there exists a polynomial $P$ of degree $d$ with integer coefficients such that $|P(m)|=1$ for at least $d+1$ different integers $m$.\n\n#", "solution": "Note that $P(x)=c$ for a fixed constant has at most $d$ solutions, since the polynomial $P(x)-c$ of degree $d$ cancels at most $d$ times. This implies that there are integers $m$ satisfying $P(m)=1$, as well as integers $m$ such that $P(m)=-1$.\nNext, we prove the following lemma.\nLemma. If $a$ and $b$ are integers such that $P(a)=-1$ and $P(b)=1$, then $|b-a| \\leq 2$.\nProof Since $b-a \\mid P(b)-P(a)$ by a well-known lemma (corollary of $a-b \\mid a^{n}-b^{n}$ for every integer $n \\geq 0$ ), the conclusion follows.\n\nLet us first consider the case that $d \\geq 4$, and assume that there exists a polynomial P with at least $d+1 \\geq 5$ solutions to $|P(m)|=1$. Let $a$ and $b$ be the smallest and largest solution respectively. Since $b-a \\geq 4$, we need $P(a)=P(b)$ by the lemma. Without loss of generality (by switching $P$ with $-P$ if necessary) we can assume $P(a)=P(b)=1$. Take a value $m$ such that $P(m)=-1$. Due to the lemma, we need $b-m$ and $m-a$ to be both at most 2 . Since $b-a \\geq 4$, there is only one possibility left in which case $b-a=4$ and thus $d=4$. By considering $P(x-m)$, we can assume $P( \\pm 2)=P( \\pm 1)=1$ and $P(0)=-1$. The unique fourth degree polynomial satisfying these equalities is $P(x)=-0.5\\left(x^{2}-4\\right)\\left(x^{2}-1\\right)+1$ which is not a polynomial with integer coefficients.\n\nFor $1 \\leq d \\leq 3$, there exist polynomials satisfying the conditions.\nFor $d=1$ we can take $P_{1}(X)=X$ as $P_{1}(-1)=-1$ and $P_{1}(1)=1$.\nFor $d=2, P_{2}(X)=2 X(X-2)+1$ satisfies $P_{2}(0)=1, P_{2}(1)=-1$ and $P_{2}(2)=1$.\nFor $d=3$, the polynomial ; $P_{3}(X)=(X+1) X(X-2)+1$ satisfies $P_{3}(-1)=1=P_{3}(0)=P_{3}(2)=1$ and $P_{3}(1)=-1$. So $\\left|P_{3}(m)\\right|=1$ for $m \\in\\{-1,0,1,2\\}$.\n\nThus the integers with the required property are precisely $d=1,2,3$. This completes the proof.\n\n## BxMO 2020: Problems and Solutions\n\n#", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2020-zz.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution\n"}} |
| {"year": "2020", "tier": "T3", "problem_label": "2", "problem_type": null, "exam": "Benelux_MO", "problem": "Let $N$ be a positive integer. A collection of $4 N^{2}$ unit tiles with two segments drawn on them as shown is assembled into a $2 N \\times 2 N$ board. Tiles can be rotated.\n\n\nThe segments on the tiles define paths on the board. Determine the least possible number and the largest possible number of such paths.\n\n#", "solution": "Let $p$ denote the number of paths. Notice that there are two types of paths: (1) those that start and end at a point on the boundary of the board and (2) closed paths in the interior of the board. Let $p_{1}, p_{2}$ denote the respective numbers of paths of either type. There are $8 N$ points on the boundary of the board, and each of these is the starting point or endpoint of exactly one path, so $p_{1}=4 \\mathrm{~N}$. Trivially, $p_{2} \\geqslant 0$, so $p=p_{1}+p_{2} \\geqslant 4 N$.\n\nThe paths on the board are made up of $8 N^{2}$ segments in total. There are only 4 possible paths of one segment, in the corners of the board. All other paths on the boundary of the board therefore consist of at least 2 segments. Moreover, all closed paths in the interior of the board consist of at least 4 segments. Hence\n\n$$\n8 N^{2} \\geqslant 4 \\cdot 1+\\left(p_{1}-4\\right) \\cdot 2+p_{2} \\cdot 4 \\Longleftrightarrow p_{2} \\leqslant N^{2}+(N-1)^{2}, \\text { so } p=p_{1}+p_{2} \\leqslant N^{2}+(N+1)^{2}\n$$\n\nWe have thus shown that $4 N \\leqslant p \\leqslant N^{2}+(N+1)^{2}$. These minimum and maximum values can indeed be attained, as shown below for $N=3$.\n\n\nThe constructions generalise easily.The constructions clearly attain the required bounds because they satisfy the equalities in our arguments showing the two bounds. Indeed the board on the left has $p_{2}=0$ so give the lower bound. The one on the right has exactly 4 boundary paths with one segment, all other $4 N-4$ boundary paths with 2 segments, and all remaining paths with 4 segments.\nRemark. The same ideas solve the analogous problem for a $(2 N+1) \\times(2 N+1)$ assembled from $(2 N+1)^{2}$ such tiles. For this board, $p \\geqslant p_{1}=2(2 N+1)$. Next, there are $2(2 N+1)^{2}$ segments, so, again,\n\n$$\n2(2 N+1)^{2} \\geqslant 4 \\cdot 1+\\left(p_{1}-4\\right) \\cdot 2+p_{2} \\cdot 4 \\Longleftrightarrow p_{2} \\leqslant 2 N^{2}+\\frac{1}{2}\n$$\n\nBut $p_{2}$ is an integer, so $p_{2} \\leqslant 2 N^{2}$, and hence $p \\leqslant 2(N+1)^{2}$. We have thus shown that $2(2 N+1) \\leqslant p \\leqslant 2(N+1)^{2}$. The construction for the lower bound is the same as for the $2 N \\times 2 N$ board; the construction for the upper bound, shown for $N=3$, is a small modification of that for the $2 N \\times 2 N$ board and again generalises easily.\n\n\n## BxMO 2020: Problems and Solutions\n\n#", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2020-zz.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution\n"}} |
| {"year": "2020", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "Benelux_MO", "problem": "Let $A B C$ be a triangle. The circle $\\omega_{A}$ through $A$ is tangent to line $B C$ at $B$. The circle $\\omega_{C}$ through $C$ is tangent to line $A B$ at $B$. Let $\\omega_{A}$ and $\\omega_{C}$ meet again at $D$. Let $M$ be the midpoint of line segment [ $B C]$, and let $E$ be the intersection of lines $M D$ and $A C$. Show that $E$ lies on $\\omega_{A}$.\n\n#", "solution": "Let $N$ be the midpoint of $[A B]$. By tangential angles, $\\angle C B D=\\angle B A D$ and $\\angle D B A=\\angle D C B$, so triangles $D A B$ and $D B C$ are similar. By definition of the midpoints $M, N$, so are triangles $A N D$ and $B M D$. In particular, $\\angle B N D=180^{\\circ}-\\angle D N A=180^{\\circ}-\\angle D M B$, so $B N D M$ is cyclic. But $M N \\| A C$ by construction, so $\\angle D B A=\\angle D B N=\\angle D M N=\\angle E M N=\\angle M E A=\\angle D E A$, hence $E A D B$ is cyclic, completing the proof.\n\n\n#", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2020-zz.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution 1"}} |
| {"year": "2020", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "Benelux_MO", "problem": "Let $A B C$ be a triangle. The circle $\\omega_{A}$ through $A$ is tangent to line $B C$ at $B$. The circle $\\omega_{C}$ through $C$ is tangent to line $A B$ at $B$. Let $\\omega_{A}$ and $\\omega_{C}$ meet again at $D$. Let $M$ be the midpoint of line segment [ $B C]$, and let $E$ be the intersection of lines $M D$ and $A C$. Show that $E$ lies on $\\omega_{A}$.\n\n#", "solution": "Let $S$ be the point on $\\omega_{C}$ such that $B S$ is parallel to $A C$, and let $E^{\\prime}$ be the reflection of $S$ in $M$ (such that $B S C E^{\\prime}$ is a parallellogram and $E^{\\prime}$ lies on $A C$ ). Now we do some (directed) angle chasing:\n\n$$\n\\begin{aligned}\n\\angle A E^{\\prime} B & =\\angle C E^{\\prime} B\\left(\\text { since } A, E^{\\prime} \\text { and } C \\text { are collineair }\\right) \\\\\n& =\\angle B S C\\left(B S C E^{\\prime} \\text { is a parallellogram }\\right) \\\\\n& =\\angle A B C\\left(\\text { inscribed angles on } \\omega_{C}\\right) \\\\\n& =\\angle A D B\\left(\\text { inscribed angles on } \\omega_{A}\\right) .\n\\end{aligned}\n$$\n\nHence $E^{\\prime}$ lies on $\\omega_{A}$. Further $\\angle A E^{\\prime} D=\\angle A B D=\\angle B S D$, and because $A E^{\\prime}$ is parallel to $B S$ we find that $E^{\\prime} D$ is parallel to $S D$. This means that $E^{\\prime}$ lies on $M D$, so $E^{\\prime}=E$ and $E$ lies on $\\omega_{A}$.\n\n# BxMO 2020: Problems and Solutions \n\n#", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2020-zz.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution 2"}} |
| {"year": "2020", "tier": "T3", "problem_label": "4", "problem_type": null, "exam": "Benelux_MO", "problem": "A divisor $d$ of a positive integer $n$ is said to be a close divisor of $n$ if $\\sqrt{n}<d<2 \\sqrt{n}$. Does there exist a positive integer with exactly 2020 close divisors?\n\n#", "solution": "Let $m$ be an odd integer with exactly 2020 positive divisors and which is (automatically) not a square. For example, $m=3^{2019}$ suffices, but of course there are many alternatives. Now consider $n=2^{k} m$, for some integer $k$ such that $2^{k}>m$. Any divisor of $n$ is then of the form $2^{\\ell} d$ where $d$ is a divisor of $m$. We will now show that for every such divisor $d$, there exists a unique $\\ell$ such that $2^{\\ell} d$ is a close divisor. Because $\\sqrt{n}$ is not an integer, there certainly is a unique integer $a$ such that $\\sqrt{n}<2^{a} d<2 \\sqrt{n}$. Because $2^{k}>m$, we have $m<\\sqrt{n}<2^{k}$. Combining with $1 \\leq d \\leq m$, we find $1<\\frac{\\sqrt{n}}{d}<2^{a}<\\frac{2 \\sqrt{n}}{d}<2 \\cdot 2^{k}$ and so we see that $0<a \\leqslant k$. So $2^{a} d$ is indeed a close divisor. Because $m$ has exactly 2020 divisors, we find that $n$ has exactly 2020 close divisors.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2020-zz.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution\n"}} |
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