{"year": "2013", "tier": "T3", "problem_label": "1", "problem_type": null, "exam": "Benelux_MO", "problem": "Let $n \\geqslant 3$ be an integer. A frog is to jump along the real axis, starting at the point 0 and making $n$ jumps: one of length 1 , one of length $2, \\ldots$, one of length $n$. It may perform these $n$ jumps in any order. If at some point the frog is sitting on a number $a \\leqslant 0$, its next jump must be to the right (towards the positive numbers). If at some point the frog is sitting on a number $a>0$, its next jump must be to the left (towards the negative numbers). Find the largest positive integer $k$ for which the frog can perform its jumps in such an order that it never lands on any of the numbers $1,2, \\ldots, k$.", "solution": "We claim that the largest positive integer $k$ with the given property is $\\left\\lfloor\\frac{n-1}{2}\\right\\rfloor$, where $\\lfloor x\\rfloor$ is by definition the largest integer not exceeding $x$.\n\nConsider a sequence of $n$ jumps of length $1,2, \\ldots n$ such that the frog never lands on any of the numbers $1,2, \\ldots, k$, where $k \\geqslant 1$. Note that we must have $k<n$ in order for the frog to be able to make its first jump. As the frog jumps to the right only if it is in a number $a \\leqslant 0$, and the largest jump has length $n$, it is impossible to reach numbers greater than $n$. On the other hand, suppose the frog is in a number $a>0$, then it must even be in a number $a \\geqslant k+1$, since it is not allowed to hit the numbers $1,2, \\ldots, k$. So the frog jumps to the left only if it is in a number $a \\geqslant k+1$, and therefore it is impossible to reach numbers less than $(k+1)-n=k-n+1$. This means the frog only possibly lands on the numbers $i$ satisfying\n\n$$\nk-n+1 \\leqslant i \\leqslant 0 \\quad \\text { or } \\quad k+1 \\leqslant i \\leqslant n \\text {. }\n$$\n\nWhen performing a jump of length $k$, the frog has to remain at either side of the numbers $1,2, \\ldots, k$. Indeed, jumping over $1,2, \\ldots, k$ requires a jump of at least length $k+1$. In case it starts at a number $a>0$ (in fact $k+1 \\leqslant a \\leqslant n$ ), it lands in $a-k$ and we must also have $a-k \\geqslant k+1$. So $2 k+1 \\leqslant a \\leqslant n$, therefore $2 k+1 \\leqslant n$. In case it starts at a number $a \\leqslant 0$ (in fact $k-n+1 \\leqslant a \\leqslant 0$ ), it lands in $a+k$ and we must also have $a+k \\leqslant 0$. Adding $k$ to both sides of $k-n+1 \\leqslant a$, we obtain $2 k-n+1 \\leqslant a+k \\leqslant 0$, so in this case we have $2 k+1 \\leqslant n$ as well. We conclude that $k \\leqslant \\frac{n-1}{2}$. Since $k$ is integer, we even have $k \\leqslant\\left\\lfloor\\frac{n-1}{2}\\right\\rfloor$.\n\nNext we prove that this upperbound is sharp: for $k=\\left\\lfloor\\frac{n-1}{2}\\right\\rfloor$ the frog really can perform its jumps in such an order that it never lands on any of the numbers $1,2, \\ldots, k$.\n\nSuppose $n$ is odd, then $\\frac{n-1}{2}$ is an integer and we have $k=\\frac{n-1}{2}$, so $n=2 k+1$. We claim that when the frog performs the jumps of length $1, \\ldots, 2 k+1$ in the following order, it does never land on $1,2, \\ldots, k$ : it starts with a jump of length $k+1$, then it performs two jumps, one of length $k+2$ followed by one of length 1 , next two jumps of length $k+3$ and $2, \\ldots$, next two jumps of length $k+(i+1)$ and $i, \\ldots$, and finally two jumps of length $k+(k+1)$ and $k$. In this order of the jumps every length between 1 and $n=2 k+1$ does occur: it performs a pair of jumps for $1 \\leqslant i \\leqslant k$, which are the jumps of length $1,2, \\ldots$, $k$ and the jumps of length $k+2, k+3, \\ldots, 2 k+1$, and it starts with the jump of length $k+1$.\n\nWe now prove the correctness of this jumping scheme. After the first jump the frog lands in $k+1>k$. Now suppose the frog is in 0 or $k+1$ and is about to perform the pair of jumps of length $k+(i+1)$ and $i$. Starting from 0 , it lands in $k+(i+1)>k$, after which it lands in $(k+i+1)-i=k+1>k$. If on the contrary it starts in $k+1$, it lands in $(k+1)-(k+(i+1))=-i<1$, after which it lands in $(-i)+i=0$. We see that, starting in 0 , the frog lands in $k+1$ after the pair of jumps, while starting in $k+1$ the frog lands in 0 , while in both cases the jumps do not touch $1,2, \\ldots k$. This proves the correctness of its series of jumps. As the frog (after its first jump) alters between $k+1$ and 0 exactly $k$ times, for odd $k$ it will end up in 0 , while for even $k$ it will end up in $k+1$.\n\nSuppose $n$ is even, then $\\frac{n-1}{2}$ is not an integer and we have $k=\\frac{n-1}{2}-\\frac{1}{2}=\\frac{n-2}{2}$, so $n=2 k+2$. Let the frog firstly perform the same series of jumps as in the previous case; they still do not touch $1,2, \\ldots, k$. Now let the frog make a final extra jump of length $2 k+2$. It will land in $0+(2 k+2)=2 k+2>k$ if $k$ is odd, or in $(k+1)-(2 k+2)=-k-1<1$ if $k$ is even, and its series of jumps is correct again.\n\nWe conclude that the largest positive integer $k$ with the given property is $\\left\\lfloor\\frac{n-1}{2}\\right\\rfloor$.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2013-zz.jsonl", "problem_match": "\nProblem 1.", "solution_match": "\nSolution."}}
{"year": "2013", "tier": "T3", "problem_label": "2", "problem_type": null, "exam": "Benelux_MO", "problem": "Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that\n\n$$\nf(x+y)+y \\leqslant f(f(f(x)))\n$$\n\nholds for all $x, y \\in \\mathbb{R}$.", "solution": "Let $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ be a function satisfying the given inequality (2). Writing $z$ for $x+y$, we find that $f(z)+(z-x) \\leqslant f(f(f(x)))$, or equivalently\n\n$$\nf(z)+z \\leqslant f(f(f(x)))+x\n$$\n\nfor all $x, z \\in \\mathbb{R}$. Substituting $z=f(f(x))$ yields $f(f(f(x)))+f(f(x)) \\leqslant f(f(f(x)))+x$, from which we see that\n\n$$\nf(f(x)) \\leqslant x\n$$\n\nfor all $x \\in \\mathbb{R}$. Substituting $f(x)$ for $x$ we get $f(f(f(x))) \\leqslant f(x)$, which combined with (3) gives $f(z)+z \\leqslant f(f(f(x)))+x \\leqslant f(x)+x$. So\n\n$$\nf(z)+z \\leqslant f(x)+x\n$$\n\nfor all $x, z \\in \\mathbb{R}$. By symmetry we see that we also have $f(x)+x \\leqslant f(z)+z$, from which we conclude that in fact we even have\n\n$$\nf(z)+z=f(x)+x\n$$\n\nfor all $x, z \\in \\mathbb{R}$. So $f(z)+z=f(0)+0$ for all $z \\in \\mathbb{R}$, and we conclude that $f(z)=c-z$ for some $c \\in \\mathbb{R}$.\n\nNow we check whether all functions of this form satisfy the given inequality. Let $c \\in \\mathbb{R}$ be given and consider the function $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ defined by $f(z)=c-z$ for all $z \\in \\mathbb{R}$. Note that $f(f(z))=c-(c-z)=z$ for all $z \\in \\mathbb{R}$. For the lefthand side of (2) we find\n\n$$\nf(x+y)+y=(c-(x+y))+y=c-x,\n$$\n\nwhile the righthand side reads\n\n$$\nf(f(f(x)))=f(x)=c-x .\n$$\n\nWe see that inequality (2) holds; in fact we even have equality here.\nWe conclude that the solutions to (2) are given by the functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ defined by $f(z)=c-z$ for all $z \\in \\mathbb{R}$, where $c$ is an arbitrary real constant.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2013-zz.jsonl", "problem_match": "\nProblem 2.", "solution_match": "\nSolution."}}
{"year": "2013", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "Benelux_MO", "problem": "Let $\\triangle A B C$ be a triangle with circumcircle $\\Gamma$, and let $I$ be the center of the incircle of $\\triangle A B C$. The lines $A I, B I$ and $C I$ intersect $\\Gamma$ in $D \\neq A, E \\neq B$ and $F \\neq C$. The tangent lines to $\\Gamma$ in $F, D$ and $E$ intersect the lines $A I, B I$ and $C I$ in $R, S$ and $T$, respectively. Prove that\n\n$$\n|A R| \\cdot|B S| \\cdot|C T|=|I D| \\cdot|I E| \\cdot|I F|\n$$", "solution": "We first prove that $|D B|=|D I|$. (This may also be claimed by referring to the lemma that $D$ is the centre of the circumcircle of $B I C I_{a}$.) By the constant angle theorem and the fact that $A D$ and $B E$ are angle bisectors of triangle $A B C$, we see that\n\n$$\n\\angle D B I=\\angle D B C+\\angle C B I=\\angle D A C+\\angle C B E=\\angle D A B+\\angle A B E,\n$$\n\nwhile\n\n$$\n\\angle D I B=180^{\\circ}-\\angle A I B=\\angle I A B+\\angle A B I=\\angle D A B+\\angle A B E .\n$$\n\nSo $\\triangle B D I$ has equal angles $\\angle D B I=\\angle D I B$, so $|D B|=|D I|$. This proves our claim. We similarly deduce that $|E C|=|E I|$ and $|F A|=|F I|$.\n\nRewriting (7) into $\\frac{|A R|}{|I F|} \\cdot \\frac{|B S|}{|I D|} \\cdot \\frac{|C T|}{|I E|}=1$, we see that it suffices to prove that\n\n$$\n\\frac{|A R|}{|A F|} \\cdot \\frac{|B S|}{|B D|} \\cdot \\frac{|C T|}{|C E|}=1\n$$\n\nWe now prove by angle chasing that $\\triangle R F A \\sim \\triangle A C I$. As $R F$ is tangent to the circumcircle of $\\triangle A F C$, we obtain (using also that $C F$ is angle bisector of $\\angle A C B$ )\n\n$$\n\\angle R F A=\\angle F C A=\\angle I C A .\n$$\n\nMoreover, from $|F A|=|F I|$ we deduce that $\\angle F A I=\\angle F I A$, so\n\n$$\n\\angle F A R=180^{\\circ}-\\angle F A I=180^{\\circ}-\\angle F I A=\\angle C I A\n$$\n\nThis proves our similarity, which entails that $\\frac{|A R|}{|A F|}=\\frac{|I A|}{|I C|}$. In the same way we deduce that $\\frac{|B S|}{|B D|}=\\frac{|I B|}{|I A|}$ and $\\frac{|C T|}{|C E|}=\\frac{|I C|}{|I B|}$. By these equal ratios we know that\n\n$$\n\\frac{|A R|}{|A F|} \\cdot \\frac{|B S|}{|B D|} \\cdot \\frac{|C T|}{|C E|}=\\frac{|I A|}{|I C|} \\cdot \\frac{|I B|}{|I A|} \\cdot \\frac{|I C|}{|I B|}=1\n$$\n\nwhich proves (8), as required.\n\n#", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2013-zz.jsonl", "problem_match": "\nProblem 3.", "solution_match": "\nSolution."}}
{"year": "2013", "tier": "T3", "problem_label": "4", "problem_type": null, "exam": "Benelux_MO", "problem": "a) Find all positive integers $g$ with the following property: for each odd prime number $p$ there exists a positive integer $n$ such that $p$ divides the two integers\n\n$$\ng^{n}-n \\quad \\text { and } \\quad g^{n+1}-(n+1)\n$$\n\nb) Find all positive integers $g$ with the following property: for each odd prime number $p$ there exists a positive integer $n$ such that $p$ divides the two integers\n\n$$\ng^{n}-n^{2} \\quad \\text { and } \\quad g^{n+1}-(n+1)^{2}\n$$\n\n#", "solution": "a) Let $g$ be a positive integer with the given property. So for each odd prime number $p$ there exists a positive integer $n$ such that $p \\mid g^{n}-n$ and $p \\mid g^{n+1}-(n+1)$.\nIf $g$ has an odd prime factor $p$, then from $p \\mid g^{n}-n$ it follows that $p \\mid n$, while from $p \\mid g^{n+1}-(n+1)$ we deduce that $p \\mid n+1$. But $p$ cannot divide both $n$ and $n+1$; contradiction. So $g$ is a power of $2: g=2^{k}$ for some $k \\geqslant 0$.\nIf $g=2^{0}=1$, then $p \\mid 1-n$ and $p \\mid 1-(n+1)$, which is again a contradiction.\nSuppose $k \\geqslant 2$. Then $g-1$ has an odd prime factor $p$, therefore $g \\equiv 1(\\bmod p)$ so $0 \\equiv g^{n}-n \\equiv 1-n(\\bmod p)$ and $0 \\equiv g^{n+1}-(n+1) \\equiv 1-(n+1)(\\bmod p)$, which is again a contradiction.\nNow we prove that $g=2^{1}=2$ does satisfy the condition. Let a prime $p>2$ be given. Choose $n=(p-1)^{2}$, then we have $n \\equiv(-1)^{2}=1(\\bmod p)$. By Fermat's little theorem (using $\\operatorname{gcd}(2, p)=1)$ we know that $2^{p-1} \\equiv 1(\\bmod p)$, so\n\n$$\n2^{n}=2^{(p-1)^{2}}=\\left(2^{p-1}\\right)^{p-1} \\equiv 1 \\equiv n \\quad(\\bmod p)\n$$\n\nMultiplying both sides by 2 , we see that also\n\n$$\n2^{n+1} \\equiv 2 n=n+n \\equiv n+1 \\quad(\\bmod p)\n$$\n\nWe conclude that only $g=2$ has the given property.\nb) Let $g$ be a positive integer with the given property. So for each odd prime number $p$ there exists a positive integer $n$ such that $p \\mid g^{n}-n^{2}$ and $p \\mid g^{n+1}-(n+1)^{2}$.\nIf $g$ has an odd prime factor $p$, then from $p \\mid g^{n}-n^{2}$ it follows that $p \\mid n^{2}$, so also $p \\mid n$, while from $p \\mid g^{n+1}-(n+1)^{2}$ we deduce that $p \\mid(n+1)^{2}$, so also $p \\mid n+1$.\n\nBut $p$ cannot divide both $n$ and $n+1$; contradiction. So $g$ is a power of $2: g=2^{k}$ for some $k \\geqslant 0$.\nIf $g=2^{0}=1$, then for any odd prime $p$ we have $p \\mid 1-n^{2}=(1-n)(1+n)$ and $p \\mid 1-(n+1)^{2}=(1-(n+1))(1+(n+1))$. Now take $p=5$. The first statement says that $n \\equiv 1$ or $n \\equiv-1 \\equiv 4(\\bmod 5)$, and the second that $n \\equiv 0$ or $n \\equiv-2 \\equiv 3$ $(\\bmod 5)$. But this yields a contradiction.\nIf $g=2^{1}=2$, then for any odd prime $p$ we have $p \\mid 2^{n}-n^{2}$ and $p \\mid 2^{n+1}-(n+1)^{2}$. Now take $p=3$. As $3 \\nmid 2^{n}$ and $3 \\nmid 2^{n+1}$, we know that $3 \\nmid n^{2}$ and $3 \\nmid(n+1)^{2}$. So these two squares must be 1 modulo 3 (as 2 can never be a square modulo 3 ). Therefore also $2^{n}$ and $2^{n+1}$ must be 1 modulo 3 , which gives $2 \\cdot 1 \\equiv 2 \\cdot 2^{n}=2^{n+1} \\equiv 1(\\bmod 3)$; contradiction.\nNow suppose $k \\geqslant 2$. Then $g-1$ has an odd prime factor $p$, therefore $g \\equiv 1(\\bmod p)$ so $0 \\equiv g^{n}-n^{2} \\equiv 1-n^{2}=(1-n)(1+n)(\\bmod p)$ and $0 \\equiv g^{n+1}-(n+1)^{2} \\equiv$ $1-(n+1)^{2}=(1-(n+1))(1+(n+1))(\\bmod p)$. Suppose $p \\geqslant 5$. The first statement says that $n \\equiv 1$ or $n \\equiv-1(\\bmod p)$, and the second that $n \\equiv 0$ or $n \\equiv-2(\\bmod p)$. But $n$ can only be congruent to at most one of the numbers $-2,-1,0$ and 1 , since $p \\geqslant 5$; contradiction. We conclude that $p=3$, so $g-1$ contains only prime factors 3 . Hence $2^{k}-1=3^{\\ell}$ for some $\\ell>0$. We see that $2^{k}-1 \\equiv(-1)^{k}-1(\\bmod 3)$, while $3^{\\ell} \\equiv 0(\\bmod 3)$. So $k$ has to be even, say $k=2 m$, and our equation becomes $2^{2 m}-1=3^{\\ell}$, or equivalently $\\left(2^{m}-1\\right)\\left(2^{m}+1\\right)=3^{\\ell}$. Not both factors on the left-hand side can be divisible by 3 , so $2^{m}-1=1$ and $2^{m}+1=3^{\\ell}$, so $m=1$. Hence $g=2^{2}=4$.\nNow we show that $g=4$ does have the given property. For this we use that $g=2$ is a solution to part (a): for any odd prime $p$ there exists a positive integer $n$ such that\n\n$$\nn \\equiv 2^{n} \\quad(\\bmod p) \\quad \\text { and } \\quad n+1 \\equiv 2^{n+1} \\quad(\\bmod p) .\n$$\n\nTaking the square of both congruences, we obtain\n\n$$\nn^{2} \\equiv\\left(2^{n}\\right)^{2}=\\left(2^{2}\\right)^{n}=4^{n} \\quad(\\bmod p)\n$$\n\nand\n\n$$\n(n+1)^{2} \\equiv\\left(2^{n+1}\\right)^{2}=\\left(2^{2}\\right)^{n+1}=4^{n+1} \\quad(\\bmod p)\n$$\n\nas desired.\nWe conclude that only $g=4$ has the given property.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2013-zz.jsonl", "problem_match": "# Problem 4.", "solution_match": "# Solution."}}