{"year": "2016", "tier": "T3", "problem_label": "1", "problem_type": null, "exam": "Benelux_MO", "problem": "Find the greatest positive integer $N$ with the following property: there exist integers $x_{1}, \\ldots, x_{N}$ such that $x_{i}^{2}-x_{i} x_{j}$ is not divisible by 1111 for any $i \\neq j$.", "solution": "We prove that the greatest $N$ with the required property is $N=1000$. First note that $x_{i}^{2}-x_{i} x_{j}=x_{i}\\left(x_{i}-x_{j}\\right)$, and that the prime factorisation of 1111 is $11 \\cdot 101$.\n\nWe first show that we can find 1000 integers $x_{1}, x_{2}, \\ldots, x_{1000}$ such that $x_{i}^{2}-x_{i} x_{j}$ is not divisible by 1111 for any $i \\neq j$. Consider the set $\\{1,2, \\ldots, 1110\\}$. This set contains 10 integers divisible by 101 , and it contains 100 integers divisible by 11 . None of the integers in the set are divisible by both 11 and 101. If we delete all of these $10+100$ integers from the set, we are left with 1000 integers. Call these $x_{1}, x_{2}, \\ldots, x_{1000}$. Now we have $11 \\nmid x_{i}$ and $101 \\nmid x_{i}$ for all $i$. Suppose there are $i \\neq j$ with $1111 \\mid x_{i}\\left(x_{i}-x_{j}\\right)$, then we must have $1111 \\mid x_{i}-x_{j}$, which is a contradiction, since $x_{i}, x_{j} \\in\\{1,2, \\ldots, 1110\\}$. So this set satisfies the requirement.\n\nWe now prove that given 1001 (or more) integers $x_{1}, x_{2}, \\ldots, x_{1001}$ there are $i \\neq j$ with $1111 \\mid x_{i}\\left(x_{i}-x_{j}\\right)$. Suppose for a contradiction that for all indices $i \\neq j$, we have that $x_{i}\\left(x_{i}-x_{j}\\right)$ is not divisible by 1111, and write $X=\\left\\{x_{1}, \\ldots, x_{1001}\\right\\}$. We may (after reducing modulo 1111 if necessary) assume that $x_{i} \\in\\{0,1, \\ldots, 1110\\}$ for all $i$. Then we know that $x_{i} \\neq 0$ for all $i$, and $x_{i} \\neq x_{j}$ for all $i \\neq j$. Suppose for some $i$ we have $11 \\mid x_{i}$. (Since $x_{i} \\neq 0$, we know that $101 \\nmid x_{i}$.) Then any integer $a \\neq x_{i}$ with $a \\equiv x_{i} \\bmod 101$ cannot be an element of $X$, since $1111 \\mid x_{i}\\left(x_{i}-a\\right)$. In $\\{1,2, \\ldots, 1110\\}$ there are 10 such integers, all of them coprime with $11 \\cdot 101$. If there are exactly $k$ different values of $i$ such that $11 \\mid x_{i}$, there are $10 k$ different integers from $\\{1,2, \\ldots, 1110\\}$ that cannot be elements of $X$, all of them coprime with $11 \\cdot 101$. Similarly, if there are exactly $m$ different values of $i$ such that $101 \\mid x_{i}$, then there are 100 m different integers from $\\{1,2, \\ldots, 1110\\}$ that cannot be elements of $X$, all of them coprime with $11 \\cdot 101$. (Note that those $10 k$ and 100 m integers can overlap.)\n\nIn $\\{1,2, \\ldots, 1110\\}$ there are 100 multiples of 11 , there are 10 multiples of 101 and there is no multiple of $11 \\cdot 101$, so there are 1000 integers that are coprime with $11 \\cdot 101$. In $X$ we have $1001-k-m$ integers that are coprime with $11 \\cdot 101$, so exactly $k+m-1$ of the coprime integers in $\\{1,2, \\ldots, 1110\\}$ are not in $X$. This implies that $10 k \\leqslant k+m-1$ and $100 m \\leqslant k+m-1$. Adding these two inequalities we find $8 k+98 m \\leqslant-2$, a clear contradiction. So $N<1001$.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2016-zz.jsonl", "problem_match": "\nProblem 1.", "solution_match": "\nSolution."}}
{"year": "2016", "tier": "T3", "problem_label": "2", "problem_type": null, "exam": "Benelux_MO", "problem": "Let $n$ be a positive integer. Suppose that its positive divisors can be partitioned into pairs (i.e. can be split in groups of two) in such a way that the sum of each pair is a prime number. Prove that these prime numbers are distinct and that none of these are a divisor of $n$.", "solution": "Let $d_{1}$ and $d_{2}$ be positive divisors of $n$ that form a pair as given in the problem. If $d_{1}$ and $d_{2}$ have a non-trivial prime divisor $p$ in common, then $p \\mid d_{1}+d_{2}$ and $p \\leqslant d_{1}<d_{1}+d_{2}$, so $d_{1}+d_{2}$ cannot be prime. Hence $\\operatorname{gcd}\\left(d_{1}, d_{2}\\right)=1$, which implies that $d_{1} d_{2} \\mid n$. Suppose the number of positive divisors of $n$ is $2 t$ (it is even since the divisors can be split into pairs). If we now multiply all divisors, then on one hand we have the product of all $d_{1} d_{2}$ where $\\left\\{d_{1}, d_{2}\\right\\}$ is a pair, so that product is at most $n^{t}$. On the other hand for every divisor $d$ there is another divisor $\\frac{n}{d}$ (since the number of divisors is even, the case $d=\\frac{n}{d}$ does not occur), and the product of all those is equal to $n^{t}$. Hence there must be equality in every inequality $d_{1} d_{2} \\leqslant n$. So the pairs of divisors given in the problem are all of the form $\\left\\{d, \\frac{n}{d}\\right\\}$.\n\nNow we prove the two statements in the problem. Suppose that $d, d^{\\prime}$ are positive divisors of $n$ such that $d+\\frac{n}{d}=d^{\\prime}+\\frac{n}{d^{\\prime}}$. Then $d^{2} d^{\\prime}+n d^{\\prime}=d\\left(d^{\\prime}\\right)^{2}+n d$, so $d d^{\\prime}\\left(d-d^{\\prime}\\right)=n\\left(d-d^{\\prime}\\right)$ and hence $\\left(d d^{\\prime}-n\\right)\\left(d-d^{\\prime}\\right)=0$. Therefore either $d=d^{\\prime}$ or $d d^{\\prime}=n$, which implies that $\\left\\{d, \\frac{n}{d}\\right\\}=\\left\\{d^{\\prime}, \\frac{n}{d^{\\prime}}\\right\\}$, as required.\n\nNow let $d$ be a positive divisor of $n$. Every prime divisor $p$ of $n$ divides precisely one of $d$ and $\\frac{n}{d}$, since $d \\frac{n}{d}=n$ and $\\operatorname{gcd}\\left(d, \\frac{n}{d}\\right)=1$; so $p \\nmid d+\\frac{n}{d}$. Therefore $\\operatorname{gcd}\\left(n, d+\\frac{n}{d}\\right)=1$. Since $d+\\frac{n}{d}>1$ we conclude that $d+\\frac{n}{d}$ cannot be a divisor of $n$.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2016-zz.jsonl", "problem_match": "\nProblem 2.", "solution_match": "\nSolution."}}
{"year": "2016", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "Benelux_MO", "problem": "Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{Z}$ such that\n\n$$\n(f(f(y)-x))^{2}+f(x)^{2}+f(y)^{2}=f(y) \\cdot(1+2 f(f(y)))\n$$\n\nfor all $x, y \\in \\mathbb{R}$.", "solution": "Take $x=y=0$ and write $c=f(0)$, then we find $f(c)^{2}+c^{2}+c^{2}=c+2 c f(c)$, so $(f(c)-c)^{2}=c-c^{2}$. The left-hand side is non-negative, so the right-hand side must be non-negative as well, hence $c-c^{2} \\geqslant 0$, so $c(1-c) \\geqslant 0$. This implies $0 \\leqslant c \\leqslant 1$, and since $c \\in \\mathbb{Z}$, we get $c=0$ or $c=1$. In both cases we have $c-c^{2}=0$, so $f(c)-c=0$, hence $f(c)=c$. Now taking $y=0$ we find for all $x \\in \\mathbb{R}$ that\n\n$$\n(f(c-x))^{2}+f(x)^{2}+c^{2}=c+2 c^{2} .\n$$\n\nIf $c=0$, then this equation reduces to $f(-x)^{2}+f(x)^{2}=0$, and since the left-hand side consists of the sum of two squares, both squares must be zero. Therefore $f(x)=0$ for all $x$. This function is indeed a solution of the given functional equation.\n\nNow we consider the other case: $c=1$. Then (1) reduces to $f(1-x)^{2}+f(x)^{2}=2$. The left-hand side consists of two squares of integers, so they must both be 1 . Therefore for all $x$ we have $f(x)=1$ or $f(x)=-1$.\n\nSuppose there is an $a$ with $f(a)=-1$. We take $y=a$ in the functional equation; using that $f(w)^{2}=1$ for any $w$, we find $1+1+1=-1 \\cdot(1+2 f(-1))$. Both with $f(-1)=1$ and with $f(-1)=-1$ this gives a contradiction. So there exists no such $a$, and we conclude that $f(x)=1$ for all $x \\in \\mathbb{R}$. This is also a solution of the given functional equation.\n\nWe conclude that there are two solutions: $f(x)=0$ for all $x$, and $f(x)=1$ for all $x$.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2016-zz.jsonl", "problem_match": "\nProblem 3.", "solution_match": "\nSolution I."}}
{"year": "2016", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "Benelux_MO", "problem": "Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{Z}$ such that\n\n$$\n(f(f(y)-x))^{2}+f(x)^{2}+f(y)^{2}=f(y) \\cdot(1+2 f(f(y)))\n$$\n\nfor all $x, y \\in \\mathbb{R}$.", "solution": "We proceed as in the first solution up to the point of deriving $f(x)= \\pm 1$ for all $x \\in \\mathbb{R}$. Now we take $x=0$ in the original functional equation, and we find\n\n$$\n(f(f(y)))^{2}+f(0)^{2}+f(y)^{2}=f(y) \\cdot(1+2 f(f(y)))\n$$\n\nWe can rewrite this as\n\n$$\n(f(y)-f(f(y)))^{2}+f(0)^{2}=f(y)\n$$\n\nThe left-hand side is non-negative, so $f(y) \\geqslant 0$ for all $y$. If we combine this with $f(x)= \\pm 1$ for all $x$, we can conclude that $f(x)=1$ for all $x \\in \\mathbb{R}$. And this is a solution.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2016-zz.jsonl", "problem_match": "\nProblem 3.", "solution_match": "\nSolution II."}}
{"year": "2016", "tier": "T3", "problem_label": "4", "problem_type": null, "exam": "Benelux_MO", "problem": "A circle $\\omega$ passes through the two vertices $B$ and $C$ of a triangle $A B C$. Furthermore, $\\omega$ intersects segment $A C$ in $D \\neq C$ and segment $A B$ in $E \\neq B$. On the ray from $B$ through $D$ lies a point $K$ such that $|B K|=|A C|$, and on the ray from $C$ through $E$ lies a point $L$ such that $|C L|=|A B|$. Show that the circumcentre $O$ of triangle $A K L$ lies on $\\omega$.", "solution": "Let $M$ be the midpoint of the arc $B C$ of $\\omega$ that is on the same side of $B C$ as $A$. Then $|B M|=|C M|$. We also have $|B A|=|C L|$ and $\\angle A B M=\\angle E B M=\\angle E C M=$ $\\angle L C M$. Hence $\\triangle A B M \\cong \\triangle L C M$. So $|A M|=|L M|$. (In case $M=E$ the triangles $A B M$ and $L C M$ are degenerate, but then the proof still works, since $|A M|=|A B|-|B M|=$ $|L C|-|C M|=|L M|$.) Similarly, we prove that $|A M|=|K M|$, so $M$ is the circumcentre of $\\triangle A K L$. This means that $M=O$, and hence we are done as $M$ was defined to be on $\\omega$.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2016-zz.jsonl", "problem_match": "\nProblem 4.", "solution_match": "\nSolution I."}}
{"year": "2016", "tier": "T3", "problem_label": "4", "problem_type": null, "exam": "Benelux_MO", "problem": "A circle $\\omega$ passes through the two vertices $B$ and $C$ of a triangle $A B C$. Furthermore, $\\omega$ intersects segment $A C$ in $D \\neq C$ and segment $A B$ in $E \\neq B$. On the ray from $B$ through $D$ lies a point $K$ such that $|B K|=|A C|$, and on the ray from $C$ through $E$ lies a point $L$ such that $|C L|=|A B|$. Show that the circumcentre $O$ of triangle $A K L$ lies on $\\omega$.", "solution": "We consider the configuration where $O$ is in the interior of triangle $A B C$. The proof for other configurations is similar. Furthermore, we exclude the case that $O=D$ or $O=E$; in those cases it is immediate that $O$ is on $\\omega$.\n\nWe have $\\angle A B K=\\angle E B D=\\angle E C D=\\angle L C A$. Together with $|A B|=|C L|$ and $|A C|=$ $|B K|$ this implies $\\triangle A B K \\cong \\triangle L C A$. Hence $|A K|=|A L|, \\angle A K B=\\angle L A C$ (denote this angle by $\\alpha$ ) and $\\angle B A K=\\angle C L A$ (denote this angle by $\\beta$ ). Furthermore, let $\\gamma=\\angle B E C=$ $\\angle B D C$. Then $\\angle B A C=180^{\\circ}-\\angle B D A-\\angle A B D=\\angle B D C-\\angle A B D=\\gamma-\\angle A B D$. Therefore $\\angle K A L=\\angle B A K+\\angle L A C-\\angle B A C=\\alpha+\\beta-(\\gamma-\\angle A B D)=\\alpha+\\beta+\\angle A B D-\\gamma$. Note that in triangle $A B K$ we have $\\alpha+\\beta+\\angle A B D=180^{\\circ}$, hence $\\angle K A L=180^{\\circ}-\\gamma$.\n\nWe have $\\angle K O L=2\\left(180^{\\circ}-\\angle K A L\\right)=2 \\gamma$. Since $|A K|=|A L|$, this implies $\\angle A O L=\\gamma$. As we also have $\\angle A E L=\\angle B E C=\\gamma$, the quadrilateral $O E L A$ is cyclic. Analogously, $O D K A$ is cyclic. Now we have\n\n$$\n\\begin{gathered}\n\\angle D O E=360^{\\circ}-\\angle D O A-\\angle A O E=180^{\\circ}-\\angle D O A+180^{\\circ}-\\angle A O E \\\\\n=\\angle D K A+\\angle A L E=\\angle B K A+\\angle A L C=\\alpha+\\beta\n\\end{gathered}\n$$\n\nOn the other hand\n\n$$\n\\angle D B E=\\angle K B A=180^{\\circ}-\\angle B A K-\\angle A K B=180^{\\circ}-\\alpha-\\beta .\n$$\n\nHence $\\angle D O E+\\angle D B E=180^{\\circ}$, so $O$ is on the circle containing $D, B$ and $E$, which is $\\omega$.", "metadata": {"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2016-zz.jsonl", "problem_match": "\nProblem 4.", "solution_match": "\nSolution II."}}