year stringdate 1961-01-01 00:00:00 2025-01-01 00:00:00 ⌀ | tier stringclasses 5
values | problem_label stringclasses 119
values | problem_type stringclasses 13
values | exam stringclasses 28
values | problem stringlengths 87 2.77k | solution stringlengths 834 13k | metadata dict | problem_tokens int64 50 903 | solution_tokens int64 500 3.93k |
|---|---|---|---|---|---|---|---|---|---|
2014 | T0 | G6 | Geometry | IMO-SL | Let $A B C$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $A C$ and $A B$, respectively, and let $M$ be the midpoint of $E F$. Let the perpendicular bisector of $E F$ intersect the line $B C$ at $K$, and let the perpendicular bisector of $M K$ intersect the lines $A C$ and $A B$ ... | Let $(E, F)$ be an interesting pair. This time we prove that $$ \frac{A M}{A K}=\cos \angle A $$ As in Solution 1, we introduce the circle $\omega$ passing through the points $K, S$, $A$, and $T$, together with the points $N$ and $L$ at which the line $A M$ intersect the line $S T$ and the circle $\omega$ for the secon... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 203 | 1,023 |
2014 | T0 | G7 | Geometry | IMO-SL | Let $A B C$ be a triangle with circumcircle $\Omega$ and incentre $I$. Let the line passing through $I$ and perpendicular to $C I$ intersect the segment $B C$ and the $\operatorname{arc} B C$ (not containing $A$ ) of $\Omega$ at points $U$ and $V$, respectively. Let the line passing through $U$ and parallel to $A I$ in... | We start with some general observations. Set $\alpha=\angle A / 2, \beta=\angle B / 2, \gamma=\angle C / 2$. Then obviously $\alpha+\beta+\gamma=90^{\circ}$. Since $\angle U I C=90^{\circ}$, we obtain $\angle I U C=\alpha+\beta$. Therefore $\angle B I V=\angle I U C-\angle I B C=\alpha=\angle B A I=\angle B Y V$, which... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 185 | 777 |
2014 | T0 | G7 | Geometry | IMO-SL | Let $A B C$ be a triangle with circumcircle $\Omega$ and incentre $I$. Let the line passing through $I$ and perpendicular to $C I$ intersect the segment $B C$ and the $\operatorname{arc} B C$ (not containing $A$ ) of $\Omega$ at points $U$ and $V$, respectively. Let the line passing through $U$ and parallel to $A I$ in... | As in $$ \angle B A V=\angle C A E $$ Proof. Let $\rho$ be the composition of the inversion with centre $A$ and radius $\sqrt{A B \cdot A C}$, and the symmetry with respect to $A I$. Clearly, $\rho$ interchanges $B$ and $C$. Let $J$ be the excentre of the triangle $A B C$ opposite to $A$ (see Figure 2). Then we have $\... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 185 | 1,909 |
2014 | T0 | N1 | Number Theory | IMO-SL | Let $n \geqslant 2$ be an integer, and let $A_{n}$ be the set $$ A_{n}=\left\{2^{n}-2^{k} \mid k \in \mathbb{Z}, 0 \leqslant k<n\right\} . $$ Determine the largest positive integer that cannot be written as the sum of one or more (not necessarily distinct) elements of $A_{n}$. (Serbia) Answer. $(n-2) 2^{n}+1$. | Part I. First we show that every integer greater than $(n-2) 2^{n}+1$ can be represented as such a sum. This is achieved by induction on $n$. For $n=2$, the set $A_{n}$ consists of the two elements 2 and 3 . Every positive integer $m$ except for 1 can be represented as the sum of elements of $A_{n}$ in this case: as $m... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 118 | 1,337 |
2014 | T0 | N1 | Number Theory | IMO-SL | Let $n \geqslant 2$ be an integer, and let $A_{n}$ be the set $$ A_{n}=\left\{2^{n}-2^{k} \mid k \in \mathbb{Z}, 0 \leqslant k<n\right\} . $$ Determine the largest positive integer that cannot be written as the sum of one or more (not necessarily distinct) elements of $A_{n}$. (Serbia) Answer. $(n-2) 2^{n}+1$. | The fact that $m=(n-2) 2^{n}+1$ cannot be represented as a sum of elements of $A_{n}$ can also be shown in other ways. We prove the following statement by induction on $n$ : Claim. If $a, b$ are integers with $a \geqslant 0, b \geqslant 1$, and $a+b<n$, then $a 2^{n}+b$ cannot be written as a sum of elements of $A_{n}$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 118 | 744 |
2014 | T0 | N1 | Number Theory | IMO-SL | Let $n \geqslant 2$ be an integer, and let $A_{n}$ be the set $$ A_{n}=\left\{2^{n}-2^{k} \mid k \in \mathbb{Z}, 0 \leqslant k<n\right\} . $$ Determine the largest positive integer that cannot be written as the sum of one or more (not necessarily distinct) elements of $A_{n}$. (Serbia) Answer. $(n-2) 2^{n}+1$. | Denote by $B_{n}$ the set of all positive integers that can be written as a sum of elements of $A_{n}$. In this solution, we explicitly describe all the numbers in $B_{n}$ by an argument similar to the first solution. For a positive integer $n$, we denote by $\sigma_{2}(n)$ the sum of its digits in the binary represent... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 118 | 1,224 |
2014 | T0 | N2 | Number Theory | IMO-SL | Determine all pairs $(x, y)$ of positive integers such that $$ \sqrt[3]{7 x^{2}-13 x y+7 y^{2}}=|x-y|+1 $$ Answer. Either $(x, y)=(1,1)$ or $\{x, y\}=\left\{m^{3}+m^{2}-2 m-1, m^{3}+2 m^{2}-m-1\right\}$ for some positive integer $m \geqslant 2$. | Let $(x, y)$ be any pair of positive integers solving (1). We shall prove that it appears in the list displayed above. The converse assertion that all these pairs do actually satisfy (1) either may be checked directly by means of a somewhat laborious calculation, or it can be seen by going in reverse order through the ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 114 | 827 |
2014 | T0 | N5 | Number Theory | IMO-SL | Find all triples $(p, x, y)$ consisting of a prime number $p$ and two positive integers $x$ and $y$ such that $x^{p-1}+y$ and $x+y^{p-1}$ are both powers of $p$. (Belgium) Answer. $(p, x, y) \in\{(3,2,5),(3,5,2)\} \cup\left\{\left(2, n, 2^{k}-n\right) \mid 0<n<2^{k}\right\}$. | For $p=2$, clearly all pairs of two positive integers $x$ and $y$ whose sum is a power of 2 satisfy the condition. Thus we assume in the following that $p>2$, and we let $a$ and $b$ be positive integers such that $x^{p-1}+y=p^{a}$ and $x+y^{p-1}=p^{b}$. Assume further, without loss of generality, that $x \leqslant y$, ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 122 | 1,594 |
2014 | T0 | N5 | Number Theory | IMO-SL | Find all triples $(p, x, y)$ consisting of a prime number $p$ and two positive integers $x$ and $y$ such that $x^{p-1}+y$ and $x+y^{p-1}$ are both powers of $p$. (Belgium) Answer. $(p, x, y) \in\{(3,2,5),(3,5,2)\} \cup\left\{\left(2, n, 2^{k}-n\right) \mid 0<n<2^{k}\right\}$. | Again, we can focus on the case that $p>2$. If $p \mid x$, then also $p \mid y$. In this case, let $p^{k}$ and $p^{\ell}$ be the highest powers of $p$ that divide $x$ and $y$ respectively, and assume without loss of generality that $k \leqslant \ell$. Then $p^{k}$ divides $x+y^{p-1}$ while $p^{k+1}$ does not, but $p^{k... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 122 | 1,433 |
2015 | T0 | A1 | Algebra | IMO-SL | Suppose that a sequence $a_{1}, a_{2}, \ldots$ of positive real numbers satisfies $$ a_{k+1} \geqslant \frac{k a_{k}}{a_{k}^{2}+(k-1)} $$ for every positive integer $k$. Prove that $a_{1}+a_{2}+\cdots+a_{n} \geqslant n$ for every $n \geqslant 2$. | From the constraint (1), it can be seen that $$ \frac{k}{a_{k+1}} \leqslant \frac{a_{k}^{2}+(k-1)}{a_{k}}=a_{k}+\frac{k-1}{a_{k}} $$ and so $$ a_{k} \geqslant \frac{k}{a_{k+1}}-\frac{k-1}{a_{k}} . $$ Summing up the above inequality for $k=1, \ldots, m$, we obtain $$ a_{1}+a_{2}+\cdots+a_{m} \geqslant\left(\frac{1}{a_{2... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 104 | 876 |
2015 | T0 | A2 | Algebra | IMO-SL | Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ with the property that $$ f(x-f(y))=f(f(x))-f(y)-1 $$ holds for all $x, y \in \mathbb{Z}$. (Croatia) | It is immediately checked that both functions mentioned in the answer are as desired. Now let $f$ denote any function satisfying (1) for all $x, y \in \mathbb{Z}$. Substituting $x=0$ and $y=f(0)$ into (1) we learn that the number $z=-f(f(0))$ satisfies $f(z)=-1$. So by plugging $y=z$ into (1) we deduce that $$ f(x+1)=f... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 64 | 693 |
2015 | T0 | A2 | Algebra | IMO-SL | Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ with the property that $$ f(x-f(y))=f(f(x))-f(y)-1 $$ holds for all $x, y \in \mathbb{Z}$. (Croatia) | Set $d=f(0)$. By plugging $x=f(y)$ into (1) we obtain $$ f^{3}(y)=f(y)+d+1 $$ for all $y \in \mathbb{Z}$, where the left-hand side abbreviates $f(f(f(y)))$. When we replace $x$ in (1) by $f(x)$ we obtain $f(f(x)-f(y))=f^{3}(x)-f(y)-1$ and as a consequence of (4) this simplifies to $$ f(f(x)-f(y))=f(x)-f(y)+d $$ Now we ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 64 | 818 |
2015 | T0 | A3 | Algebra | IMO-SL | Let $n$ be a fixed positive integer. Find the maximum possible value of $$ \sum_{1 \leqslant r<s \leqslant 2 n}(s-r-n) x_{r} x_{s} $$ where $-1 \leqslant x_{i} \leqslant 1$ for all $i=1,2, \ldots, 2 n$. | Let $Z$ be the expression to be maximized. Since this expression is linear in every variable $x_{i}$ and $-1 \leqslant x_{i} \leqslant 1$, the maximum of $Z$ will be achieved when $x_{i}=-1$ or 1 . Therefore, it suffices to consider only the case when $x_{i} \in\{-1,1\}$ for all $i=1,2, \ldots, 2 n$. For $i=1,2, \ldots... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 90 | 1,553 |
2015 | T0 | A3 | Algebra | IMO-SL | Let $n$ be a fixed positive integer. Find the maximum possible value of $$ \sum_{1 \leqslant r<s \leqslant 2 n}(s-r-n) x_{r} x_{s} $$ where $-1 \leqslant x_{i} \leqslant 1$ for all $i=1,2, \ldots, 2 n$. | We present a different method of obtaining the bound $Z \leqslant n(n-1)$. As in the previous solution, we reduce the problem to the case $x_{i} \in\{-1,1\}$. For brevity, we use the notation $[2 n]=\{1,2, \ldots, 2 n\}$. Consider any $x_{1}, x_{2}, \ldots, x_{2 n} \in\{-1,1\}$. Let $$ A=\left\{i \in[2 n]: x_{i}=1\righ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 90 | 1,224 |
2015 | T0 | A4 | Algebra | IMO-SL | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying the equation $$ f(x+f(x+y))+f(x y)=x+f(x+y)+y f(x) $$ for all real numbers $x$ and $y$. | Clearly, each of the functions $x \mapsto x$ and $x \mapsto 2-x$ satisfies (1). It suffices now to show that they are the only solutions to the problem. Suppose that $f$ is any function satisfying (1). Then setting $y=1$ in (1), we obtain $$ f(x+f(x+1))=x+f(x+1) $$ in other words, $x+f(x+1)$ is a fixed point of $f$ for... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 57 | 753 |
2015 | T0 | A5 | Algebra | IMO-SL | Let $2 \mathbb{Z}+1$ denote the set of odd integers. Find all functions $f: \mathbb{Z} \rightarrow 2 \mathbb{Z}+1$ satisfying $$ f(x+f(x)+y)+f(x-f(x)-y)=f(x+y)+f(x-y) $$ for every $x, y \in \mathbb{Z}$. | Throughout the solution, all functions are assumed to map integers to integers. For any function $g$ and any nonzero integer $t$, define $$ \Delta_{t} g(x)=g(x+t)-g(x) $$ For any nonzero integers $a$ and $b$, notice that $\Delta_{a} \Delta_{b} g=\Delta_{b} \Delta_{a} g$. Moreover, if $\Delta_{a} g=0$ and $\Delta_{b} g=... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 86 | 1,566 |
2015 | T0 | A6 | Algebra | IMO-SL | Let $n$ be a fixed integer with $n \geqslant 2$. We say that two polynomials $P$ and $Q$ with real coefficients are block-similar if for each $i \in\{1,2, \ldots, n\}$ the sequences $$ \begin{aligned} & P(2015 i), P(2015 i-1), \ldots, P(2015 i-2014) \quad \text { and } \\ & Q(2015 i), Q(2015 i-1), \ldots, Q(2015 i-201... | For convenience, we set $k=2015=2 \ell+1$. Part (a). Consider the following polynomials of degree $n+1$ : $$ P(x)=\prod_{i=0}^{n}(x-i k) \quad \text { and } \quad Q(x)=\prod_{i=0}^{n}(x-i k-1) $$ Since $Q(x)=P(x-1)$ and $P(0)=P(k)=P(2 k)=\cdots=P(n k)$, these polynomials are block-similar (and distinct). Part (b). For ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 203 | 2,020 |
2015 | T0 | C1 | Combinatorics | IMO-SL | In Lineland there are $n \geqslant 1$ towns, arranged along a road running from left to right. Each town has a left bulldozer (put to the left of the town and facing left) and a right bulldozer (put to the right of the town and facing right). The sizes of the $2 n$ bulldozers are distinct. Every time when a right and a... | We start with the same enumeration and the same observation as in Clearly, there is no town which can sweep $T_{n}$ away from the right. Then we may choose the leftmost town $T_{k}$ which cannot be swept away from the right. One can observe now that no town $T_{i}$ with $i>k$ may sweep away some town $T_{j}$ with $j<k$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 284 | 1,090 |
2015 | T0 | C1 | Combinatorics | IMO-SL | In Lineland there are $n \geqslant 1$ towns, arranged along a road running from left to right. Each town has a left bulldozer (put to the left of the town and facing left) and a right bulldozer (put to the right of the town and facing right). The sizes of the $2 n$ bulldozers are distinct. Every time when a right and a... | We separately prove that $(i)$ there exists a town which cannot be swept away, and that (ii) there is at most one such town. We also make use of the two observations from the previous solutions. To prove ( $i$ ), assume contrariwise that every town can be swept away. Let $t_{1}$ be the leftmost town; next, for every $k... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 284 | 732 |
2015 | T0 | C2 | Combinatorics | IMO-SL | Let $\mathcal{V}$ be a finite set of points in the plane. We say that $\mathcal{V}$ is balanced if for any two distinct points $A, B \in \mathcal{V}$, there exists a point $C \in \mathcal{V}$ such that $A C=B C$. We say that $\mathcal{V}$ is center-free if for any distinct points $A, B, C \in \mathcal{V}$, there does n... | Part ( $\boldsymbol{a}$ ). Assume that $n$ is odd. Consider a regular $n$-gon. Label the vertices of the $n$-gon as $A_{1}, A_{2}, \ldots, A_{n}$ in counter-clockwise order, and set $\mathcal{V}=\left\{A_{1}, \ldots, A_{n}\right\}$. We check that $\mathcal{V}$ is balanced. For any two distinct vertices $A_{i}$ and $A_{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 191 | 1,309 |
2015 | T0 | C3 | Combinatorics | IMO-SL | For a finite set $A$ of positive integers, we call a partition of $A$ into two disjoint nonempty subsets $A_{1}$ and $A_{2}$ good if the least common multiple of the elements in $A_{1}$ is equal to the greatest common divisor of the elements in $A_{2}$. Determine the minimum value of $n$ such that there exists a set of... | Let $A=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$, where $a_{1}<a_{2}<\cdots<a_{n}$. For a finite nonempty set $B$ of positive integers, denote by $\operatorname{lcm} B$ and $\operatorname{gcd} B$ the least common multiple and the greatest common divisor of the elements in $B$, respectively. Consider any good partitio... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 102 | 1,540 |
2015 | T0 | C4 | Combinatorics | IMO-SL | Let $n$ be a positive integer. Two players $A$ and $B$ play a game in which they take turns choosing positive integers $k \leqslant n$. The rules of the game are: (i) A player cannot choose a number that has been chosen by either player on any previous turn. (ii) A player cannot choose a number consecutive to any of th... | For brevity, we denote by $[n]$ the set $\{1,2, \ldots, n\}$. Firstly, we show that $B$ wins whenever $n \neq 1,2,4,6$. For this purpose, we provide a strategy which guarantees that $B$ can always make a move after $A$ 's move, and also guarantees that the game does not end in a draw. We begin with an important observa... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 148 | 1,773 |
2015 | T0 | C5 | Combinatorics | IMO-SL | Consider an infinite sequence $a_{1}, a_{2}, \ldots$ of positive integers with $a_{i} \leqslant 2015$ for all $i \geqslant 1$. Suppose that for any two distinct indices $i$ and $j$ we have $i+a_{i} \neq j+a_{j}$. Prove that there exist two positive integers $b$ and $N$ such that $$ \left|\sum_{i=m+1}^{n}\left(a_{i}-b... | We visualize the set of positive integers as a sequence of points. For each $n$ we draw an arrow emerging from $n$ that points to $n+a_{n}$; so the length of this arrow is $a_{n}$. Due to the condition that $m+a_{m} \neq n+a_{n}$ for $m \neq n$, each positive integer receives at most one arrow. There are some positive ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 150 | 1,039 |
2015 | T0 | C5 | Combinatorics | IMO-SL | Consider an infinite sequence $a_{1}, a_{2}, \ldots$ of positive integers with $a_{i} \leqslant 2015$ for all $i \geqslant 1$. Suppose that for any two distinct indices $i$ and $j$ we have $i+a_{i} \neq j+a_{j}$. Prove that there exist two positive integers $b$ and $N$ such that $$ \left|\sum_{i=m+1}^{n}\left(a_{i}-b... | Set $s_{n}=n+a_{n}$ for all positive integers $n$. By our assumptions, we have $$ n+1 \leqslant s_{n} \leqslant n+2015 $$ for all $n \in \mathbb{Z}_{>0}$. The members of the sequence $s_{1}, s_{2}, \ldots$ are distinct. We shall investigate the set $$ M=\mathbb{Z}_{>0} \backslash\left\{s_{1}, s_{2}, \ldots\right\} $$ C... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 150 | 1,166 |
2015 | T0 | C6 | Combinatorics | IMO-SL | Let $S$ be a nonempty set of positive integers. We say that a positive integer $n$ is clean if it has a unique representation as a sum of an odd number of distinct elements from $S$. Prove that there exist infinitely many positive integers that are not clean. | Define an odd (respectively, even) representation of $n$ to be a representation of $n$ as a sum of an odd (respectively, even) number of distinct elements of $S$. Let $\mathbb{Z}_{>0}$ denote the set of all positive integers. Suppose, to the contrary, that there exist only finitely many positive integers that are not c... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 58 | 1,429 |
2015 | T0 | C6 | Combinatorics | IMO-SL | Let $S$ be a nonempty set of positive integers. We say that a positive integer $n$ is clean if it has a unique representation as a sum of an odd number of distinct elements from $S$. Prove that there exist infinitely many positive integers that are not clean. | We will also use Property 1 from We first define some terminology and notations used in this solution. Let $\mathbb{Z}_{\geqslant 0}$ denote the set of all nonnegative integers. All sums mentioned are regarded as sums of distinct elements of $S$. Moreover, a sum is called even or odd depending on the parity of the numb... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 58 | 2,531 |
2015 | T0 | C7 | Combinatorics | IMO-SL | In a company of people some pairs are enemies. A group of people is called unsociable if the number of members in the group is odd and at least 3 , and it is possible to arrange all its members around a round table so that every two neighbors are enemies. Given that there are at most 2015 unsociable groups, prove that ... | Let $G=(V, E)$ be a graph where $V$ is the set of people in the company and $E$ is the set of the enemy pairs - the edges of the graph. In this language, partitioning into 11 disjoint enemy-free subsets means properly coloring the vertices of this graph with 11 colors. We will prove the following more general statement... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 97 | 1,422 |
2015 | T0 | C7 | Combinatorics | IMO-SL | In a company of people some pairs are enemies. A group of people is called unsociable if the number of members in the group is odd and at least 3 , and it is possible to arrange all its members around a round table so that every two neighbors are enemies. Given that there are at most 2015 unsociable groups, prove that ... | We provide a different proof of the claim from the previous solution. We say that a graph is critical if deleting any vertex from the graph decreases the graph's chromatic number. Obviously every graph contains a critical induced subgraph with the same chromatic number. Lemma 2. Suppose that $G=(V, E)$ is a critical gr... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 97 | 1,646 |
2015 | T0 | G2 | Geometry | IMO-SL | Let $A B C$ be a triangle inscribed into a circle $\Omega$ with center $O$. A circle $\Gamma$ with center $A$ meets the side $B C$ at points $D$ and $E$ such that $D$ lies between $B$ and $E$. Moreover, let $F$ and $G$ be the common points of $\Gamma$ and $\Omega$. We assume that $F$ lies on the arc $A B$ of $\Omega$ n... | Again, we denote the circumcircle of $B D K F$ by $\omega_{B}$. In addition, we set $\alpha=$ $\angle B A C, \varphi=\angle A B F$, and $\psi=\angle E D A=\angle A E D$ (see Figure 2). Notice that $A F=A G$ entails $\varphi=\angle G C A$, so all three of $\alpha, \varphi$, and $\psi$ respect the "symmetry" between $B$ ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 212 | 963 |
2015 | T0 | G3 | Geometry | IMO-SL | Let $A B C$ be a triangle with $\angle C=90^{\circ}$, and let $H$ be the foot of the altitude from $C$. A point $D$ is chosen inside the triangle $C B H$ so that $C H$ bisects $A D$. Let $P$ be the intersection point of the lines $B D$ and $C H$. Let $\omega$ be the semicircle with diameter $B D$ that meets the segment... | Let $\Gamma$ be the circumcircle of $A B C$, and let $A D$ meet $\omega$ at $T$. Then $\angle A T B=$ $\angle A C B=90^{\circ}$, so $T$ lies on $\Gamma$ as well. As in the previous solution, let $K$ be the projection of $D$ onto $A B$; then $A H=H K$ (see Figure 2). Our goal now is to prove that the points $C, Q$, and ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 150 | 872 |
2015 | T0 | G4 | Geometry | IMO-SL | Let $A B C$ be an acute triangle, and let $M$ be the midpoint of $A C$. A circle $\omega$ passing through $B$ and $M$ meets the sides $A B$ and $B C$ again at $P$ and $Q$, respectively. Let $T$ be the point such that the quadrilateral $B P T Q$ is a parallelogram. Suppose that $T$ lies on the circumcircle of the triang... | Let $S$ be the center of the parallelogram $B P T Q$, and let $B^{\prime} \neq B$ be the point on the ray $B M$ such that $B M=M B^{\prime}$ (see Figure 1). It follows that $A B C B^{\prime}$ is a parallelogram. Then, $\angle A B B^{\prime}=\angle P Q M$ and $\angle B B^{\prime} A=\angle B^{\prime} B C=\angle M P Q$, a... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 118 | 772 |
2015 | T0 | G5 | Geometry | IMO-SL | Let $A B C$ be a triangle with $C A \neq C B$. Let $D, F$, and $G$ be the midpoints of the sides $A B, A C$, and $B C$, respectively. A circle $\Gamma$ passing through $C$ and tangent to $A B$ at $D$ meets the segments $A F$ and $B G$ at $H$ and $I$, respectively. The points $H^{\prime}$ and $I^{\prime}$ are symmetric ... | We may assume that $C A>C B$. Observe that $H^{\prime}$ and $I^{\prime}$ lie inside the segments $C F$ and $C G$, respectively. Therefore, $M$ lies outside $\triangle A B C$ (see Figure 1). Due to the powers of points $A$ and $B$ with respect to the circle $\Gamma$, we have $$ C H^{\prime} \cdot C A=A H \cdot A C=A D^{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 190 | 815 |
2015 | T0 | G5 | Geometry | IMO-SL | Let $A B C$ be a triangle with $C A \neq C B$. Let $D, F$, and $G$ be the midpoints of the sides $A B, A C$, and $B C$, respectively. A circle $\Gamma$ passing through $C$ and tangent to $A B$ at $D$ meets the segments $A F$ and $B G$ at $H$ and $I$, respectively. The points $H^{\prime}$ and $I^{\prime}$ are symmetric ... | Let $X=H I \cap A B$, and let the tangent to $\Gamma$ at $C$ meet $A B$ at $Y$. Let $X C$ meet $\Gamma$ again at $X^{\prime}$ (see Figure 3). Projecting from $C, X$, and $C$ again, we have $(X, A ; D, B)=$ $\left(X^{\prime}, H ; D, I\right)=(C, I ; D, H)=(Y, B ; D, A)$. Since $A$ and $B$ are symmetric about $D$, it fol... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 190 | 737 |
2015 | T0 | G6 | Geometry | IMO-SL | Let $A B C$ be an acute triangle with $A B>A C$, and let $\Gamma$ be its circumcircle. Let $H$, $M$, and $F$ be the orthocenter of the triangle, the midpoint of $B C$, and the foot of the altitude from $A$, respectively. Let $Q$ and $K$ be the two points on $\Gamma$ that satisfy $\angle A Q H=90^{\circ}$ and $\angle Q ... | Let $A^{\prime}$ be the point diametrically opposite to $A$ on $\Gamma$. Since $\angle A Q A^{\prime}=90^{\circ}$ and $\angle A Q H=90^{\circ}$, the points $Q, H$, and $A^{\prime}$ are collinear. Similarly, if $Q^{\prime}$ denotes the point on $\Gamma$ diametrically opposite to $Q$, then $K, H$, and $Q^{\prime}$ are co... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 141 | 612 |
2015 | T0 | G7 | Geometry | IMO-SL | Let $A B C D$ be a convex quadrilateral, and let $P, Q, R$, and $S$ be points on the sides $A B, B C, C D$, and $D A$, respectively. Let the line segments $P R$ and $Q S$ meet at $O$. Suppose that each of the quadrilaterals $A P O S, B Q O P, C R O Q$, and $D S O R$ has an incircle. Prove that the lines $A C, P Q$, and... | Denote by $\gamma_{A}, \gamma_{B}, \gamma_{C}$, and $\gamma_{D}$ the incircles of the quadrilaterals $A P O S, B Q O P$, $C R O Q$, and $D S O R$, respectively. We start with proving that the quadrilateral $A B C D$ also has an incircle which will be referred to as $\Omega$. Denote the points of tangency as in Figure 1... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 133 | 939 |
2015 | T0 | G7 | Geometry | IMO-SL | Let $A B C D$ be a convex quadrilateral, and let $P, Q, R$, and $S$ be points on the sides $A B, B C, C D$, and $D A$, respectively. Let the line segments $P R$ and $Q S$ meet at $O$. Suppose that each of the quadrilaterals $A P O S, B Q O P, C R O Q$, and $D S O R$ has an incircle. Prove that the lines $A C, P Q$, and... | Applying Menelaus' theorem to $\triangle A B C$ with the line $P Q$ and to $\triangle A C D$ with the line $R S$, we see that the line $A C$ meets $P Q$ and $R S$ at the same point (possibly at infinity) if and only if $$ \frac{A P}{P B} \cdot \frac{B Q}{Q C} \cdot \frac{C R}{R D} \cdot \frac{D S}{S A}=1 $$ So, it suff... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 133 | 1,146 |
2015 | T0 | G7 | Geometry | IMO-SL | Let $A B C D$ be a convex quadrilateral, and let $P, Q, R$, and $S$ be points on the sides $A B, B C, C D$, and $D A$, respectively. Let the line segments $P R$ and $Q S$ meet at $O$. Suppose that each of the quadrilaterals $A P O S, B Q O P, C R O Q$, and $D S O R$ has an incircle. Prove that the lines $A C, P Q$, and... | We present another approach for showing (1) from Lemma 2. Let $E F G H$ and $E^{\prime} F^{\prime} G^{\prime} H^{\prime}$ be circumscribed quadrilaterals such that $\angle E+\angle E^{\prime}=$ $\angle F+\angle F^{\prime}=\angle G+\angle G^{\prime}=\angle H+\angle H^{\prime}=180^{\circ}$. Then $$ \frac{E F \cdot G H}{F... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 133 | 1,638 |
2015 | T0 | G8 | Geometry | IMO-SL | A triangulation of a convex polygon $\Pi$ is a partitioning of $\Pi$ into triangles by diagonals having no common points other than the vertices of the polygon. We say that a triangulation is a Thaiangulation if all triangles in it have the same area. Prove that any two different Thaiangulations of a convex polygon $\... | We denote by [S] the area of a polygon $S$. Recall that each triangulation of a convex $n$-gon has exactly $n-2$ triangles. This means that all triangles in any two Thaiangulations of a convex polygon $\Pi$ have the same area. Let $\mathcal{T}$ be a triangulation of a convex polygon $\Pi$. If four vertices $A, B, C$, a... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 122 | 2,133 |
2015 | T0 | G8 | Geometry | IMO-SL | A triangulation of a convex polygon $\Pi$ is a partitioning of $\Pi$ into triangles by diagonals having no common points other than the vertices of the polygon. We say that a triangulation is a Thaiangulation if all triangles in it have the same area. Prove that any two different Thaiangulations of a convex polygon $\... | We will make use of the preliminary observations from Arguing indirectly, we choose a convex polygon $\Pi$ with the least possible number of sides such that some two Thaiangulations $\mathcal{T}_{1}$ and $\mathcal{T}_{2}$ of $\Pi$ violate the statement (thus $\Pi$ has at least five sides). Assume that $\mathcal{T}_{1}$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 122 | 1,599 |
2015 | T0 | N1 | Number Theory | IMO-SL | Determine all positive integers $M$ for which the sequence $a_{0}, a_{1}, a_{2}, \ldots$, defined by $a_{0}=\frac{2 M+1}{2}$ and $a_{k+1}=a_{k}\left\lfloor a_{k}\right\rfloor$ for $k=0,1,2, \ldots$, contains at least one integer term. (Luxembourg) | Define $b_{k}=2 a_{k}$ for all $k \geqslant 0$. Then $$ b_{k+1}=2 a_{k+1}=2 a_{k}\left\lfloor a_{k}\right\rfloor=b_{k}\left\lfloor\frac{b_{k}}{2}\right\rfloor . $$ Since $b_{0}$ is an integer, it follows that $b_{k}$ is an integer for all $k \geqslant 0$. Suppose that the sequence $a_{0}, a_{1}, a_{2}, \ldots$ does not... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 96 | 543 |
2015 | T0 | N2 | Number Theory | IMO-SL | Let $a$ and $b$ be positive integers such that $a!b$ ! is a multiple of $a!+b!$. Prove that $3 a \geqslant 2 b+2$. (United Kingdom) | If $a>b$, we immediately get $3 a \geqslant 2 b+2$. In the case $a=b$, the required inequality is equivalent to $a \geqslant 2$, which can be checked easily since $(a, b)=(1,1)$ does not satisfy $a!+b!\mid a!b!$. We now assume $a<b$ and denote $c=b-a$. The required inequality becomes $a \geqslant 2 c+2$. Suppose, to th... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 51 | 935 |
2015 | T0 | N2 | Number Theory | IMO-SL | Let $a$ and $b$ be positive integers such that $a!b$ ! is a multiple of $a!+b!$. Prove that $3 a \geqslant 2 b+2$. (United Kingdom) | As in $$ N=1+(a+1)(a+2) \cdots(a+c) \mid(a+c)! $$ which implies that all prime factors of $N$ are at most $a+c$. Let $p$ be a prime factor of $N$. If $p \leqslant c$ or $p \geqslant a+1$, then $p$ divides one of $a+1, \ldots, a+c$ which is impossible. Hence $a \geqslant p \geqslant c+1$. Furthermore, we must have $2 p>... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 51 | 532 |
2015 | T0 | N3 | Number Theory | IMO-SL | Let $m$ and $n$ be positive integers such that $m>n$. Define $x_{k}=(m+k) /(n+k)$ for $k=$ $1,2, \ldots, n+1$. Prove that if all the numbers $x_{1}, x_{2}, \ldots, x_{n+1}$ are integers, then $x_{1} x_{2} \cdots x_{n+1}-1$ is divisible by an odd prime. (Austria) | Assume that $x_{1}, x_{2}, \ldots, x_{n+1}$ are integers. Define the integers $$ a_{k}=x_{k}-1=\frac{m+k}{n+k}-1=\frac{m-n}{n+k}>0 $$ for $k=1,2, \ldots, n+1$. Let $P=x_{1} x_{2} \cdots x_{n+1}-1$. We need to prove that $P$ is divisible by an odd prime, or in other words, that $P$ is not a power of 2 . To this end, we ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 109 | 1,034 |
2015 | T0 | N4 | Number Theory | IMO-SL | Suppose that $a_{0}, a_{1}, \ldots$ and $b_{0}, b_{1}, \ldots$ are two sequences of positive integers satisfying $a_{0}, b_{0} \geqslant 2$ and $$ a_{n+1}=\operatorname{gcd}\left(a_{n}, b_{n}\right)+1, \quad b_{n+1}=\operatorname{lcm}\left(a_{n}, b_{n}\right)-1 $$ for all $n \geqslant 0$. Prove that the sequence $\le... | Let $s_{n}=a_{n}+b_{n}$. Notice that if $a_{n} \mid b_{n}$, then $a_{n+1}=a_{n}+1, b_{n+1}=b_{n}-1$ and $s_{n+1}=s_{n}$. So, $a_{n}$ increases by 1 and $s_{n}$ does not change until the first index is reached with $a_{n} \nmid s_{n}$. Define $$ W_{n}=\left\{m \in \mathbb{Z}_{>0}: m \geqslant a_{n} \text { and } m \nmid... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 188 | 1,080 |
2015 | T0 | N4 | Number Theory | IMO-SL | Suppose that $a_{0}, a_{1}, \ldots$ and $b_{0}, b_{1}, \ldots$ are two sequences of positive integers satisfying $a_{0}, b_{0} \geqslant 2$ and $$ a_{n+1}=\operatorname{gcd}\left(a_{n}, b_{n}\right)+1, \quad b_{n+1}=\operatorname{lcm}\left(a_{n}, b_{n}\right)-1 $$ for all $n \geqslant 0$. Prove that the sequence $\le... | By Claim 1 in the first solution, we have $a_{n} \leqslant w_{n} \leqslant w_{0}$, so the sequence $\left(a_{n}\right)$ is bounded, and hence it has only finitely many values. Let $M=\operatorname{lcm}\left(a_{1}, a_{2}, \ldots\right)$, and consider the sequence $b_{n}$ modulo $M$. Let $r_{n}$ be the remainder of $b_{n... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 188 | 1,414 |
2015 | T0 | N5 | Number Theory | IMO-SL | Determine all triples $(a, b, c)$ of positive integers for which $a b-c, b c-a$, and $c a-b$ are powers of 2 . Explanation: A power of 2 is an integer of the form $2^{n}$, where $n$ denotes some nonnegative integer. (Serbia) | It can easily be verified that these sixteen triples are as required. Now let ( $a, b, c$ ) be any triple with the desired property. If we would have $a=1$, then both $b-c$ and $c-b$ were powers of 2 , which is impossible since their sum is zero; because of symmetry, this argument shows $a, b, c \geqslant 2$. Case 1. A... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 71 | 1,165 |
2015 | T0 | N5 | Number Theory | IMO-SL | Determine all triples $(a, b, c)$ of positive integers for which $a b-c, b c-a$, and $c a-b$ are powers of 2 . Explanation: A power of 2 is an integer of the form $2^{n}$, where $n$ denotes some nonnegative integer. (Serbia) | As in the beginning of the first solution, we observe that $a, b, c \geqslant 2$. Depending on the parities of $a, b$, and $c$ we distinguish three cases. Case 1. The numbers $a, b$, and $c$ are even. Let $2^{A}, 2^{B}$, and $2^{C}$ be the largest powers of 2 dividing $a, b$, and $c$ respectively. We may assume without... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 71 | 2,120 |
2015 | T0 | N6 | Number Theory | IMO-SL | Let $\mathbb{Z}_{>0}$ denote the set of positive integers. Consider a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$. For any $m, n \in \mathbb{Z}_{>0}$ we write $f^{n}(m)=\underbrace{f(f(\ldots f}_{n}(m) \ldots))$. Suppose that $f$ has the following two properties: (i) If $m, n \in \mathbb{Z}_{>0}$, then $\... | We split the solution into three steps. In the first of them, we show that the function $f$ is injective and explain how this leads to a useful visualization of $f$. Then comes the second step, in which most of the work happens: its goal is to show that for any $n \in \mathbb{Z}_{>0}$ the sequence $n, f(n), f^{2}(n), \... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 226 | 2,461 |
2015 | T0 | N7 | Number Theory | IMO-SL | Let $\mathbb{Z}_{>0}$ denote the set of positive integers. For any positive integer $k$, a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ is called $k$-good if $\operatorname{gcd}(f(m)+n, f(n)+m) \leqslant k$ for all $m \neq n$. Find all $k$ such that there exists a $k$-good function. (Canada) | For any function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$, let $G_{f}(m, n)=\operatorname{gcd}(f(m)+n, f(n)+m)$. Note that a $k$-good function is also $(k+1)$-good for any positive integer $k$. Hence, it suffices to show that there does not exist a 1-good function and that there exists a 2-good function. We fir... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 109 | 736 |
2015 | T0 | N7 | Number Theory | IMO-SL | Let $\mathbb{Z}_{>0}$ denote the set of positive integers. For any positive integer $k$, a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ is called $k$-good if $\operatorname{gcd}(f(m)+n, f(n)+m) \leqslant k$ for all $m \neq n$. Find all $k$ such that there exists a $k$-good function. (Canada) | We provide an alternative construction of a 2 -good function $f$. Let $\mathcal{P}$ be the set consisting of 4 and all odd primes. For every $p \in \mathcal{P}$, we say that a number $a \in\{0,1, \ldots, p-1\}$ is $p$-useful if $a \not \equiv-a(\bmod p)$. Note that a residue modulo $p$ which is neither 0 nor 2 is $p$-u... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 109 | 2,336 |
2015 | T0 | N8 | Number Theory | IMO-SL | For every positive integer $n$ with prime factorization $n=\prod_{i=1}^{k} p_{i}^{\alpha_{i}}$, define $$ \mho(n)=\sum_{i: p_{i}>10^{100}} \alpha_{i} . $$ That is, $\mho(n)$ is the number of prime factors of $n$ greater than $10^{100}$, counted with multiplicity. Find all strictly increasing functions $f: \mathbb{Z} ... | A straightforward check shows that all the functions listed in the answer satisfy the problem condition. It remains to show the converse. Assume that $f$ is a function satisfying the problem condition. Notice that the function $g(x)=f(x)-f(0)$ also satisfies this condition. Replacing $f$ by $g$, we assume from now on t... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 171 | 3,565 |
2015 | T0 | A2 | Algebra | IMO-SL | Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ with the property that $$ f(x-f(y))=f(f(x))-f(y)-1 $$ holds for all $x, y \in \mathbb{Z}$. (Croatia) Answer. There are two such functions, namely the constant function $x \mapsto-1$ and the successor function $x \mapsto x+1$. | It is immediately checked that both functions mentioned in the answer are as desired. Now let $f$ denote any function satisfying (1) for all $x, y \in \mathbb{Z}$. Substituting $x=0$ and $y=f(0)$ into (1) we learn that the number $z=-f(f(0))$ satisfies $f(z)=-1$. So by plugging $y=z$ into (1) we deduce that $$ f(x+1)=f... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 97 | 693 |
2015 | T0 | A2 | Algebra | IMO-SL | Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ with the property that $$ f(x-f(y))=f(f(x))-f(y)-1 $$ holds for all $x, y \in \mathbb{Z}$. (Croatia) Answer. There are two such functions, namely the constant function $x \mapsto-1$ and the successor function $x \mapsto x+1$. | Set $d=f(0)$. By plugging $x=f(y)$ into (1) we obtain $$ f^{3}(y)=f(y)+d+1 $$ for all $y \in \mathbb{Z}$, where the left-hand side abbreviates $f(f(f(y)))$. When we replace $x$ in (1) by $f(x)$ we obtain $f(f(x)-f(y))=f^{3}(x)-f(y)-1$ and as a consequence of (4) this simplifies to $$ f(f(x)-f(y))=f(x)-f(y)+d $$ Now we ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 97 | 818 |
2015 | T0 | A3 | Algebra | IMO-SL | Let $n$ be a fixed positive integer. Find the maximum possible value of $$ \sum_{1 \leqslant r<s \leqslant 2 n}(s-r-n) x_{r} x_{s}, $$ where $-1 \leqslant x_{i} \leqslant 1$ for all $i=1,2, \ldots, 2 n$. (Austria) Answer. $n(n-1)$. | Let $Z$ be the expression to be maximized. Since this expression is linear in every variable $x_{i}$ and $-1 \leqslant x_{i} \leqslant 1$, the maximum of $Z$ will be achieved when $x_{i}=-1$ or 1 . Therefore, it suffices to consider only the case when $x_{i} \in\{-1,1\}$ for all $i=1,2, \ldots, 2 n$. For $i=1,2, \ldots... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 103 | 1,553 |
2015 | T0 | A3 | Algebra | IMO-SL | Let $n$ be a fixed positive integer. Find the maximum possible value of $$ \sum_{1 \leqslant r<s \leqslant 2 n}(s-r-n) x_{r} x_{s}, $$ where $-1 \leqslant x_{i} \leqslant 1$ for all $i=1,2, \ldots, 2 n$. (Austria) Answer. $n(n-1)$. | We present a different method of obtaining the bound $Z \leqslant n(n-1)$. As in the previous solution, we reduce the problem to the case $x_{i} \in\{-1,1\}$. For brevity, we use the notation $[2 n]=\{1,2, \ldots, 2 n\}$. Consider any $x_{1}, x_{2}, \ldots, x_{2 n} \in\{-1,1\}$. Let $$ A=\left\{i \in[2 n]: x_{i}=1\righ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 103 | 1,224 |
2015 | T0 | A4 | Algebra | IMO-SL | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying the equation $$ f(x+f(x+y))+f(x y)=x+f(x+y)+y f(x) $$ for all real numbers $x$ and $y$. (Albania) Answer. There are two such functions, namely the identity function and $x \mapsto 2-x$. | Clearly, each of the functions $x \mapsto x$ and $x \mapsto 2-x$ satisfies (1). It suffices now to show that they are the only solutions to the problem. Suppose that $f$ is any function satisfying (1). Then setting $y=1$ in (1), we obtain $$ f(x+f(x+1))=x+f(x+1) $$ in other words, $x+f(x+1)$ is a fixed point of $f$ for... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 84 | 753 |
2015 | T0 | A5 | Algebra | IMO-SL | Let $2 \mathbb{Z}+1$ denote the set of odd integers. Find all functions $f: \mathbb{Z} \rightarrow 2 \mathbb{Z}+1$ satisfying $$ f(x+f(x)+y)+f(x-f(x)-y)=f(x+y)+f(x-y) $$ for every $x, y \in \mathbb{Z}$. Answer. Fix an odd positive integer $d$, an integer $k$, and odd integers $\ell_{0}, \ell_{1}, \ldots, \ell_{d-1}$... | Throughout the solution, all functions are assumed to map integers to integers. For any function $g$ and any nonzero integer $t$, define $$ \Delta_{t} g(x)=g(x+t)-g(x) $$ For any nonzero integers $a$ and $b$, notice that $\Delta_{a} \Delta_{b} g=\Delta_{b} \Delta_{a} g$. Moreover, if $\Delta_{a} g=0$ and $\Delta_{b} g=... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 190 | 1,566 |
2015 | T0 | A6 | Step 3. We describe all functions $f$. | IMO-SL | Let $n$ be a fixed integer with $n \geqslant 2$. We say that two polynomials $P$ and $Q$ with real coefficients are block-similar if for each $i \in\{1,2, \ldots, n\}$ the sequences $$ \begin{aligned} & P(2015 i), P(2015 i-1), \ldots, P(2015 i-2014) \quad \text { and } \\ & Q(2015 i), Q(2015 i-1), \ldots, Q(2015 i-201... | For convenience, we set $k=2015=2 \ell+1$. Part (a). Consider the following polynomials of degree $n+1$ : $$ P(x)=\prod_{i=0}^{n}(x-i k) \quad \text { and } \quad Q(x)=\prod_{i=0}^{n}(x-i k-1) $$ Since $Q(x)=P(x-1)$ and $P(0)=P(k)=P(2 k)=\cdots=P(n k)$, these polynomials are block-similar (and distinct). Part (b). For ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 206 | 2,020 |
2015 | T0 | C2 | Combinatorics | IMO-SL | Let $\mathcal{V}$ be a finite set of points in the plane. We say that $\mathcal{V}$ is balanced if for any two distinct points $A, B \in \mathcal{V}$, there exists a point $C \in \mathcal{V}$ such that $A C=B C$. We say that $\mathcal{V}$ is center-free if for any distinct points $A, B, C \in \mathcal{V}$, there does n... | Part ( $\boldsymbol{a}$ ). Assume that $n$ is odd. Consider a regular $n$-gon. Label the vertices of the $n$-gon as $A_{1}, A_{2}, \ldots, A_{n}$ in counter-clockwise order, and set $\mathcal{V}=\left\{A_{1}, \ldots, A_{n}\right\}$. We check that $\mathcal{V}$ is balanced. For any two distinct vertices $A_{i}$ and $A_{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 210 | 1,309 |
2015 | T0 | C3 | Combinatorics | IMO-SL | For a finite set $A$ of positive integers, we call a partition of $A$ into two disjoint nonempty subsets $A_{1}$ and $A_{2}$ good if the least common multiple of the elements in $A_{1}$ is equal to the greatest common divisor of the elements in $A_{2}$. Determine the minimum value of $n$ such that there exists a set of... | Let $A=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$, where $a_{1}<a_{2}<\cdots<a_{n}$. For a finite nonempty set $B$ of positive integers, denote by $\operatorname{lcm} B$ and $\operatorname{gcd} B$ the least common multiple and the greatest common divisor of the elements in $B$, respectively. Consider any good partitio... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 110 | 1,540 |
2015 | T0 | C4 | Combinatorics | IMO-SL | Let $n$ be a positive integer. Two players $A$ and $B$ play a game in which they take turns choosing positive integers $k \leqslant n$. The rules of the game are: (i) A player cannot choose a number that has been chosen by either player on any previous turn. (ii) A player cannot choose a number consecutive to any of th... | For brevity, we denote by $[n]$ the set $\{1,2, \ldots, n\}$. Firstly, we show that $B$ wins whenever $n \neq 1,2,4,6$. For this purpose, we provide a strategy which guarantees that $B$ can always make a move after $A$ 's move, and also guarantees that the game does not end in a draw. We begin with an important observa... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
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} | 175 | 1,773 |
2015 | T0 | C5 | Combinatorics | IMO-SL | Consider an infinite sequence $a_{1}, a_{2}, \ldots$ of positive integers with $a_{i} \leqslant 2015$ for all $i \geqslant 1$. Suppose that for any two distinct indices $i$ and $j$ we have $i+a_{i} \neq j+a_{j}$. Prove that there exist two positive integers $b$ and $N$ such that $$ \left|\sum_{i=m+1}^{n}\left(a_{i}-b... | We visualize the set of positive integers as a sequence of points. For each $n$ we draw an arrow emerging from $n$ that points to $n+a_{n}$; so the length of this arrow is $a_{n}$. Due to the condition that $m+a_{m} \neq n+a_{n}$ for $m \neq n$, each positive integer receives at most one arrow. There are some positive ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 153 | 1,039 |
2015 | T0 | C5 | Combinatorics | IMO-SL | Consider an infinite sequence $a_{1}, a_{2}, \ldots$ of positive integers with $a_{i} \leqslant 2015$ for all $i \geqslant 1$. Suppose that for any two distinct indices $i$ and $j$ we have $i+a_{i} \neq j+a_{j}$. Prove that there exist two positive integers $b$ and $N$ such that $$ \left|\sum_{i=m+1}^{n}\left(a_{i}-b... | Set $s_{n}=n+a_{n}$ for all positive integers $n$. By our assumptions, we have $$ n+1 \leqslant s_{n} \leqslant n+2015 $$ for all $n \in \mathbb{Z}_{>0}$. The members of the sequence $s_{1}, s_{2}, \ldots$ are distinct. We shall investigate the set $$ M=\mathbb{Z}_{>0} \backslash\left\{s_{1}, s_{2}, \ldots\right\} $$ C... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 153 | 1,166 |
2015 | T0 | C7 | Combinatorics | IMO-SL | In a company of people some pairs are enemies. A group of people is called unsociable if the number of members in the group is odd and at least 3, and it is possible to arrange all its members around a round table so that every two neighbors are enemies. Given that there are at most 2015 unsociable groups, prove that i... | Let $G=(V, E)$ be a graph where $V$ is the set of people in the company and $E$ is the set of the enemy pairs - the edges of the graph. In this language, partitioning into 11 disjoint enemy-free subsets means properly coloring the vertices of this graph with 11 colors. We will prove the following more general statement... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 100 | 1,422 |
2015 | T0 | C7 | Combinatorics | IMO-SL | In a company of people some pairs are enemies. A group of people is called unsociable if the number of members in the group is odd and at least 3, and it is possible to arrange all its members around a round table so that every two neighbors are enemies. Given that there are at most 2015 unsociable groups, prove that i... | We provide a different proof of the claim from the previous solution. We say that a graph is critical if deleting any vertex from the graph decreases the graph's chromatic number. Obviously every graph contains a critical induced subgraph with the same chromatic number. Lemma 2. Suppose that $G=(V, E)$ is a critical gr... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 100 | 1,646 |
2015 | T0 | G4 | Geometry | IMO-SL | Let $A B C$ be an acute triangle, and let $M$ be the midpoint of $A C$. A circle $\omega$ passing through $B$ and $M$ meets the sides $A B$ and $B C$ again at $P$ and $Q$, respectively. Let $T$ be the point such that the quadrilateral $B P T Q$ is a parallelogram. Suppose that $T$ lies on the circumcircle of the triang... | Let $S$ be the center of the parallelogram $B P T Q$, and let $B^{\prime} \neq B$ be the point on the ray $B M$ such that $B M=M B^{\prime}$ (see Figure 1). It follows that $A B C B^{\prime}$ is a parallelogram. Then, $\angle A B B^{\prime}=\angle P Q M$ and $\angle B B^{\prime} A=\angle B^{\prime} B C=\angle M P Q$, a... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 126 | 772 |
2015 | T0 | N1 | Number Theory | IMO-SL | Determine all positive integers $M$ for which the sequence $a_{0}, a_{1}, a_{2}, \ldots$, defined by $a_{0}=\frac{2 M+1}{2}$ and $a_{k+1}=a_{k}\left\lfloor a_{k}\right\rfloor$ for $k=0,1,2, \ldots$, contains at least one integer term. (Luxembourg) Answer. All integers $M \geqslant 2$. | Define $b_{k}=2 a_{k}$ for all $k \geqslant 0$. Then $$ b_{k+1}=2 a_{k+1}=2 a_{k}\left\lfloor a_{k}\right\rfloor=b_{k}\left\lfloor\frac{b_{k}}{2}\right\rfloor . $$ Since $b_{0}$ is an integer, it follows that $b_{k}$ is an integer for all $k \geqslant 0$. Suppose that the sequence $a_{0}, a_{1}, a_{2}, \ldots$ does not... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 110 | 543 |
2015 | T0 | N5 | Number Theory | IMO-SL | Determine all triples $(a, b, c)$ of positive integers for which $a b-c, b c-a$, and $c a-b$ are powers of 2 . Explanation: A power of 2 is an integer of the form $2^{n}$, where $n$ denotes some nonnegative integer. (Serbia) Answer. There are sixteen such triples, namely $(2,2,2)$, the three permutations of $(2,2,3)$,... | It can easily be verified that these sixteen triples are as required. Now let ( $a, b, c$ ) be any triple with the desired property. If we would have $a=1$, then both $b-c$ and $c-b$ were powers of 2 , which is impossible since their sum is zero; because of symmetry, this argument shows $a, b, c \geqslant 2$. Case 1. A... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 124 | 1,165 |
2015 | T0 | N5 | Number Theory | IMO-SL | Determine all triples $(a, b, c)$ of positive integers for which $a b-c, b c-a$, and $c a-b$ are powers of 2 . Explanation: A power of 2 is an integer of the form $2^{n}$, where $n$ denotes some nonnegative integer. (Serbia) Answer. There are sixteen such triples, namely $(2,2,2)$, the three permutations of $(2,2,3)$,... | As in the beginning of the first solution, we observe that $a, b, c \geqslant 2$. Depending on the parities of $a, b$, and $c$ we distinguish three cases. Case 1. The numbers $a, b$, and $c$ are even. Let $2^{A}, 2^{B}$, and $2^{C}$ be the largest powers of 2 dividing $a, b$, and $c$ respectively. We may assume without... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 124 | 2,120 |
2015 | T0 | N7 | Number Theory | IMO-SL | Let $\mathbb{Z}_{>0}$ denote the set of positive integers. For any positive integer $k$, a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ is called $k$-good if $\operatorname{gcd}(f(m)+n, f(n)+m) \leqslant k$ for all $m \neq n$. Find all $k$ such that there exists a $k$-good function. (Canada) Answer. $k \ge... | For any function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$, let $G_{f}(m, n)=\operatorname{gcd}(f(m)+n, f(n)+m)$. Note that a $k$-good function is also $(k+1)$-good for any positive integer $k$. Hence, it suffices to show that there does not exist a 1-good function and that there exists a 2-good function. We fir... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 121 | 736 |
2015 | T0 | N7 | Number Theory | IMO-SL | Let $\mathbb{Z}_{>0}$ denote the set of positive integers. For any positive integer $k$, a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ is called $k$-good if $\operatorname{gcd}(f(m)+n, f(n)+m) \leqslant k$ for all $m \neq n$. Find all $k$ such that there exists a $k$-good function. (Canada) Answer. $k \ge... | We provide an alternative construction of a 2 -good function $f$. Let $\mathcal{P}$ be the set consisting of 4 and all odd primes. For every $p \in \mathcal{P}$, we say that a number $a \in\{0,1, \ldots, p-1\}$ is $p$-useful if $a \not \equiv-a(\bmod p)$. Note that a residue modulo $p$ which is neither 0 nor 2 is $p$-u... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 121 | 2,336 |
2015 | T0 | N8 | Number Theory | IMO-SL | For every positive integer $n$ with prime factorization $n=\prod_{i=1}^{k} p_{i}^{\alpha_{i}}$, define $$ \mathcal{W}(n)=\sum_{i: p_{i}>10^{100}} \alpha_{i} . $$ That is, $\mho(n)$ is the number of prime factors of $n$ greater than $10^{100}$, counted with multiplicity. Find all strictly increasing functions $f: \mat... | A straightforward check shows that all the functions listed in the answer satisfy the problem condition. It remains to show the converse. Assume that $f$ is a function satisfying the problem condition. Notice that the function $g(x)=f(x)-f(0)$ also satisfies this condition. Replacing $f$ by $g$, we assume from now on t... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2015SL.jsonl",
"solution_match": null
} | 214 | 3,565 |
2016 | T0 | A1 | Algebra | IMO-SL | Let $a, b$ and $c$ be positive real numbers such that $\min \{a b, b c, c a\} \geqslant 1$. Prove that $$ \sqrt[3]{\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)} \leqslant\left(\frac{a+b+c}{3}\right)^{2}+1 $$ | We first show the following. - Claim. For any positive real numbers $x, y$ with $x y \geqslant 1$, we have $$ \left(x^{2}+1\right)\left(y^{2}+1\right) \leqslant\left(\left(\frac{x+y}{2}\right)^{2}+1\right)^{2} $$ Proof. Note that $x y \geqslant 1$ implies $\left(\frac{x+y}{2}\right)^{2}-1 \geqslant x y-1 \geqslant 0$. ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 103 | 829 |
2016 | T0 | A2 | Algebra | IMO-SL | Find the smallest real constant $C$ such that for any positive real numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ (not necessarily distinct), one can always choose distinct subscripts $i, j, k$ and $l$ such that $$ \left|\frac{a_{i}}{a_{j}}-\frac{a_{k}}{a_{l}}\right| \leqslant C $$ | We first show that $C \leqslant \frac{1}{2}$. For any positive real numbers $a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant a_{4} \leqslant a_{5}$, consider the five fractions $$ \frac{a_{1}}{a_{2}}, \frac{a_{3}}{a_{4}}, \frac{a_{1}}{a_{5}}, \frac{a_{2}}{a_{3}}, \frac{a_{4}}{a_{5}} . $$ Each of them lies in the interv... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 101 | 649 |
2016 | T0 | A3 | Algebra | IMO-SL | Find all integers $n \geqslant 3$ with the following property: for all real numbers $a_{1}, a_{2}, \ldots, a_{n}$ and $b_{1}, b_{2}, \ldots, b_{n}$ satisfying $\left|a_{k}\right|+\left|b_{k}\right|=1$ for $1 \leqslant k \leqslant n$, there exist $x_{1}, x_{2}, \ldots, x_{n}$, each of which is either -1 or 1 , such that... | For any even integer $n \geqslant 4$, we consider the case $$ a_{1}=a_{2}=\cdots=a_{n-1}=b_{n}=0 \quad \text { and } \quad b_{1}=b_{2}=\cdots=b_{n-1}=a_{n}=1 $$ The condition $\left|a_{k}\right|+\left|b_{k}\right|=1$ is satisfied for each $1 \leqslant k \leqslant n$. No matter how we choose each $x_{k}$, both sums $\su... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 181 | 1,053 |
2016 | T0 | A3 | Algebra | IMO-SL | Find all integers $n \geqslant 3$ with the following property: for all real numbers $a_{1}, a_{2}, \ldots, a_{n}$ and $b_{1}, b_{2}, \ldots, b_{n}$ satisfying $\left|a_{k}\right|+\left|b_{k}\right|=1$ for $1 \leqslant k \leqslant n$, there exist $x_{1}, x_{2}, \ldots, x_{n}$, each of which is either -1 or 1 , such that... | The even case can be handled in the same way as Firstly, for $n=3$, we may assume without loss of generality $a_{1} \geqslant a_{2} \geqslant a_{3} \geqslant 0$ and $b_{1}=a_{1}-1$ (if $b_{1}=1-a_{1}$, we may replace each $b_{k}$ by $-b_{k}$ ). - Case 1. $b_{2}=a_{2}-1$ and $b_{3}=a_{3}-1$, in which case we take $\left... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 181 | 2,218 |
2016 | T0 | A4 | Algebra | IMO-SL | Denote by $\mathbb{R}^{+}$the set of all positive real numbers. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$such that $$ x f\left(x^{2}\right) f(f(y))+f(y f(x))=f(x y)\left(f\left(f\left(x^{2}\right)\right)+f\left(f\left(y^{2}\right)\right)\right) $$ for all positive real numbers $x$ and $y$. | Taking $x=y=1$ in (1), we get $f(1) f(f(1))+f(f(1))=2 f(1) f(f(1))$ and hence $f(1)=1$. Putting $x=1$ in (1), we have $f(f(y))+f(y)=f(y)\left(1+f\left(f\left(y^{2}\right)\right)\right)$ so that $$ f(f(y))=f(y) f\left(f\left(y^{2}\right)\right) $$ Putting $y=1$ in (1), we get $x f\left(x^{2}\right)+f(f(x))=f(x)\left(f\l... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 119 | 843 |
2016 | T0 | A5 | Algebra | IMO-SL | (a) Prove that for every positive integer $n$, there exists a fraction $\frac{a}{b}$ where $a$ and $b$ are integers satisfying $0<b \leqslant \sqrt{n}+1$ and $\sqrt{n} \leqslant \frac{a}{b} \leqslant \sqrt{n+1}$. (b) Prove that there are infinitely many positive integers $n$ such that there is no fraction $\frac{a}{b}$... | (a) Let $r$ be the unique positive integer for which $r^{2} \leqslant n<(r+1)^{2}$. Write $n=r^{2}+s$. Then we have $0 \leqslant s \leqslant 2 r$. We discuss in two cases according to the parity of $s$. - Case 1. $s$ is even. Consider the number $\left(r+\frac{s}{2 r}\right)^{2}=r^{2}+s+\left(\frac{s}{2 r}\right)^{2}$.... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 158 | 838 |
2016 | T0 | A6 | Algebra | IMO-SL | The equation $$ (x-1)(x-2) \cdots(x-2016)=(x-1)(x-2) \cdots(x-2016) $$ is written on the board. One tries to erase some linear factors from both sides so that each side still has at least one factor, and the resulting equation has no real roots. Find the least number of linear factors one needs to erase to achieve th... | Since there are 2016 common linear factors on both sides, we need to erase at least 2016 factors. We claim that the equation has no real roots if we erase all factors $(x-k)$ on the left-hand side with $k \equiv 2,3(\bmod 4)$, and all factors $(x-m)$ on the right-hand side with $m \equiv 0,1(\bmod 4)$. Therefore, it su... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 94 | 1,433 |
2016 | T0 | A7 | Algebra | IMO-SL | Denote by $\mathbb{R}$ the set of all real numbers. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(0) \neq 0$ and $$ f(x+y)^{2}=2 f(x) f(y)+\max \left\{f\left(x^{2}\right)+f\left(y^{2}\right), f\left(x^{2}+y^{2}\right)\right\} $$ for all real numbers $x$ and $y$. | Taking $x=y=0$ in (1), we get $f(0)^{2}=2 f(0)^{2}+\max \{2 f(0), f(0)\}$. If $f(0)>0$, then $f(0)^{2}+2 f(0)=0$ gives no positive solution. If $f(0)<0$, then $f(0)^{2}+f(0)=0$ gives $f(0)=-1$. Putting $y=0$ in (1), we have $f(x)^{2}=-2 f(x)+f\left(x^{2}\right)$, which is the same as $(f(x)+1)^{2}=f\left(x^{2}\right)+1... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 120 | 2,752 |
2016 | T0 | A7 | Algebra | IMO-SL | Denote by $\mathbb{R}$ the set of all real numbers. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(0) \neq 0$ and $$ f(x+y)^{2}=2 f(x) f(y)+\max \left\{f\left(x^{2}\right)+f\left(y^{2}\right), f\left(x^{2}+y^{2}\right)\right\} $$ for all real numbers $x$ and $y$. | Taking $x=y=0$ in (1), we get $f(0)^{2}=2 f(0)^{2}+\max \{2 f(0), f(0)\}$. If $f(0)>0$, then $f(0)^{2}+2 f(0)=0$ gives no positive solution. If $f(0)<0$, then $f(0)^{2}+f(0)=0$ gives $f(0)=-1$. Putting $y=0$ in (1), we have $$ f(x)^{2}=-2 f(x)+f\left(x^{2}\right) $$ Replace $x$ by $-x$ in (5) and compare with (5) again... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 120 | 1,218 |
2016 | T0 | A7 | Algebra | IMO-SL | Denote by $\mathbb{R}$ the set of all real numbers. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(0) \neq 0$ and $$ f(x+y)^{2}=2 f(x) f(y)+\max \left\{f\left(x^{2}\right)+f\left(y^{2}\right), f\left(x^{2}+y^{2}\right)\right\} $$ for all real numbers $x$ and $y$. | As in $$ (f(x)+1)^{2}=f\left(x^{2}\right)+1 $$ and $$ f(x)=f(-x) \quad \text { or } \quad f(x)+f(-x)=-2 $$ for any $x \in \mathbb{R}$. We shall show that one of the statements in (9) holds for all $x \in \mathbb{R}$. Suppose $f(a)=f(-a)$ but $f(a)+f(-a) \neq-2$, while $f(b) \neq f(-b)$ but $f(b)+f(-b)=-2$. Clearly, $a,... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 120 | 949 |
2016 | T0 | A8 | Algebra | IMO-SL | Determine the largest real number $a$ such that for all $n \geqslant 1$ and for all real numbers $x_{0}, x_{1}, \ldots, x_{n}$ satisfying $0=x_{0}<x_{1}<x_{2}<\cdots<x_{n}$, we have $$ \frac{1}{x_{1}-x_{0}}+\frac{1}{x_{2}-x_{1}}+\cdots+\frac{1}{x_{n}-x_{n-1}} \geqslant a\left(\frac{2}{x_{1}}+\frac{3}{x_{2}}+\cdots+\fr... | We first show that $a=\frac{4}{9}$ is admissible. For each $2 \leqslant k \leqslant n$, by the CauchySchwarz Inequality, we have $$ \left(x_{k-1}+\left(x_{k}-x_{k-1}\right)\right)\left(\frac{(k-1)^{2}}{x_{k-1}}+\frac{3^{2}}{x_{k}-x_{k-1}}\right) \geqslant(k-1+3)^{2}, $$ which can be rewritten as $$ \frac{9}{x_{k}-x_{k-... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 160 | 677 |
2016 | T0 | A8 | Algebra | IMO-SL | Determine the largest real number $a$ such that for all $n \geqslant 1$ and for all real numbers $x_{0}, x_{1}, \ldots, x_{n}$ satisfying $0=x_{0}<x_{1}<x_{2}<\cdots<x_{n}$, we have $$ \frac{1}{x_{1}-x_{0}}+\frac{1}{x_{2}-x_{1}}+\cdots+\frac{1}{x_{n}-x_{n-1}} \geqslant a\left(\frac{2}{x_{1}}+\frac{3}{x_{2}}+\cdots+\fr... | We shall give an alternative method to establish (1) with $a=\frac{4}{9}$. We define $y_{k}=x_{k}-x_{k-1}>0$ for $1 \leqslant k \leqslant n$. By the Cauchy-Schwarz Inequality, for $1 \leqslant k \leqslant n$, we have $$ \left(y_{1}+y_{2}+\cdots+y_{k}\right)\left(\sum_{j=1}^{k} \frac{1}{y_{j}}\binom{j+1}{2}^{2}\right) \... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 160 | 658 |
2016 | T0 | C2 | Combinatorics | IMO-SL | Find all positive integers $n$ for which all positive divisors of $n$ can be put into the cells of a rectangular table under the following constraints: - each cell contains a distinct divisor; - the sums of all rows are equal; and - the sums of all columns are equal. | Clearly $n=1$ works. Then we assume $n>1$ and let its prime factorization be $n=p_{1}^{r_{1}} p_{2}^{r_{2}} \cdots p_{t}^{r_{t}}$. Suppose the table has $k$ rows and $l$ columns with $1<k \leqslant l$. Note that $k l$ is the number of positive divisors of $n$ and the sum of all entries is the sum of positive divisors o... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 60 | 771 |
2016 | T0 | C3 | Combinatorics | IMO-SL | Let $n$ be a positive integer relatively prime to 6 . We paint the vertices of a regular $n$-gon with three colours so that there is an odd number of vertices of each colour. Show that there exists an isosceles triangle whose three vertices are of different colours. $\mathbf{C 4}$. Find all positive integers $n$ for wh... | For $k=1,2,3$, let $a_{k}$ be the number of isosceles triangles whose vertices contain exactly $k$ colours. Suppose on the contrary that $a_{3}=0$. Let $b, c, d$ be the number of vertices of the three different colours respectively. We now count the number of pairs $(\triangle, E)$ where $\triangle$ is an isosceles tri... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 175 | 585 |
2016 | T0 | C5 | Combinatorics | IMO-SL | Let $n \geqslant 3$ be a positive integer. Find the maximum number of diagonals of a regular $n$-gon one can select, so that any two of them do not intersect in the interior or they are perpendicular to each other. | We consider two cases according to the parity of $n$. - Case 1. $n$ is odd. We first claim that no pair of diagonals is perpendicular. Suppose $A, B, C, D$ are vertices where $A B$ and $C D$ are perpendicular, and let $E$ be the vertex lying on the perpendicular bisector of $A B$. Let $E^{\prime}$ be the opposite point... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 55 | 960 |
2016 | T0 | C6 | Combinatorics | IMO-SL | There are $n \geqslant 3$ islands in a city. Initially, the ferry company offers some routes between some pairs of islands so that it is impossible to divide the islands into two groups such that no two islands in different groups are connected by a ferry route. After each year, the ferry company will close a ferry ro... | Initially, we pick any pair of islands $A$ and $B$ which are connected by a ferry route and put $A$ in set $\mathcal{A}$ and $B$ in set $\mathcal{B}$. From the condition, without loss of generality there must be another island which is connected to $A$. We put such an island $C$ in set $\mathcal{B}$. We say that two se... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 212 | 938 |
2016 | T0 | C7 | Combinatorics | IMO-SL | Let $n \geqslant 2$ be an integer. In the plane, there are $n$ segments given in such a way that any two segments have an intersection point in the interior, and no three segments intersect at a single point. Jeff places a snail at one of the endpoints of each of the segments and claps his hands $n-1$ times. Each time ... | We consider a big disk which contains all the segments. We extend each segment to a line $l_{i}$ so that each of them cuts the disk at two distinct points $A_{i}, B_{i}$. (a) For odd $n$, we travel along the circumference of the disk and mark each of the points $A_{i}$ or $B_{i}$ 'in' and 'out' alternately. Since every... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 216 | 690 |
2016 | T0 | C8 | Combinatorics | IMO-SL | Let $n$ be a positive integer. Determine the smallest positive integer $k$ with the following property: it is possible to mark $k$ cells on a $2 n \times 2 n$ board so that there exists a unique partition of the board into $1 \times 2$ and $2 \times 1$ dominoes, none of which contains two marked cells. | We first construct an example of marking $2 n$ cells satisfying the requirement. Label the rows and columns $1,2, \ldots, 2 n$ and label the cell in the $i$-th row and the $j$-th column $(i, j)$. For $i=1,2, \ldots, n$, we mark the cells $(i, i)$ and $(i, i+1)$. We claim that the required partition exists and is unique... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 81 | 758 |
2016 | T0 | G1 | Geometry | IMO-SL | In a convex pentagon $A B C D E$, let $F$ be a point on $A C$ such that $\angle F B C=90^{\circ}$. Suppose triangles $A B F, A C D$ and $A D E$ are similar isosceles triangles with $$ \angle F A B=\angle F B A=\angle D A C=\angle D C A=\angle E A D=\angle E D A $$ Let $M$ be the midpoint of $C F$. Point $X$ is chosen... | From $\angle C A D=\angle E D A$, we have $A C / / E D$. Together with $A C / / E X$, we know that $E, D, X$ are collinear. Denote the common angle in (1) by $\theta$. From $\triangle A B F \sim \triangle A C D$, we get $\frac{A B}{A C}=\frac{A F}{A D}$ so that $\triangle A B C \sim \triangle A F D$. This yields $\angl... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 147 | 805 |
2016 | T0 | G1 | Geometry | IMO-SL | In a convex pentagon $A B C D E$, let $F$ be a point on $A C$ such that $\angle F B C=90^{\circ}$. Suppose triangles $A B F, A C D$ and $A D E$ are similar isosceles triangles with $$ \angle F A B=\angle F B A=\angle D A C=\angle D C A=\angle E A D=\angle E D A $$ Let $M$ be the midpoint of $C F$. Point $X$ is chosen... | Let the common angle in (1) be $\theta$. From $\triangle A B F \sim \triangle A C D$, we have $\frac{A B}{A C}=\frac{A F}{A D}$ so that $\triangle A B C \sim \triangle A F D$. Then $\angle A D F=\angle A C B=90^{\circ}-2 \theta=90^{\circ}-\angle B A D$ and hence $D F \perp A B$. As $F A=F B$, this implies $\triangle D ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 147 | 579 |
2016 | T0 | G2 | Geometry | IMO-SL | Let $A B C$ be a triangle with circumcircle $\Gamma$ and incentre $I$. Let $M$ be the midpoint of side $B C$. Denote by $D$ the foot of perpendicular from $I$ to side $B C$. The line through $I$ perpendicular to $A I$ meets sides $A B$ and $A C$ at $F$ and $E$ respectively. Suppose the circumcircle of triangle $A E F$ ... | Let $A M$ meet $\Gamma$ again at $Y$ and $X Y$ meet $B C$ at $D^{\prime}$. It suffices to show $D^{\prime}=D$. We shall apply the following fact. - Claim. For any cyclic quadrilateral $P Q R S$ whose diagonals meet at $T$, we have $$ \frac{Q T}{T S}=\frac{P Q \cdot Q R}{P S \cdot S R} $$ Proof. We use $\left[W_{1} W_{2... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 131 | 1,387 |
2016 | T0 | G2 | Geometry | IMO-SL | Let $A B C$ be a triangle with circumcircle $\Gamma$ and incentre $I$. Let $M$ be the midpoint of side $B C$. Denote by $D$ the foot of perpendicular from $I$ to side $B C$. The line through $I$ perpendicular to $A I$ meets sides $A B$ and $A C$ at $F$ and $E$ respectively. Suppose the circumcircle of triangle $A E F$ ... | Let $\omega_{A}$ be the $A$-mixtilinear incircle of triangle $A B C$. From the properties of mixtilinear incircles, $\omega_{A}$ touches sides $A B$ and $A C$ at $F$ and $E$ respectively. Suppose $\omega_{A}$ is tangent to $\Gamma$ at $T$. Let $A M$ meet $\Gamma$ again at $Y$, and let $D_{1}, T_{1}$ be the reflections ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 131 | 677 |
2016 | T0 | G3 | Geometry | IMO-SL | Let $B=(-1,0)$ and $C=(1,0)$ be fixed points on the coordinate plane. A nonempty, bounded subset $S$ of the plane is said to be nice if (i) there is a point $T$ in $S$ such that for every point $Q$ in $S$, the segment $T Q$ lies entirely in $S$; and (ii) for any triangle $P_{1} P_{2} P_{3}$, there exists a unique point... | If in the similarity of $\triangle A B C$ and $\triangle P_{\sigma(1)} P_{\sigma(2)} P_{\sigma(3)}, B C$ corresponds to the longest side of $\triangle P_{1} P_{2} P_{3}$, then we have $B C \geqslant A B \geqslant A C$. The condition $B C \geqslant A B$ is equivalent to $(x+1)^{2}+y^{2} \leqslant 4$, while $A B \geqslan... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 283 | 885 |
2016 | T0 | G4 | Geometry | IMO-SL | Let $A B C$ be a triangle with $A B=A C \neq B C$ and let $I$ be its incentre. The line $B I$ meets $A C$ at $D$, and the line through $D$ perpendicular to $A C$ meets $A I$ at $E$. Prove that the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$. | Let $I^{\prime}$ be the reflection of $I$ in $A C$. Denote by $T$ and $M$ the projections from $I$ to sides $A B$ and $B C$ respectively. Since $B I$ is the perpendicular bisector of $T M$, we have $$ D T=D M $$ Since $\angle A D E=\angle A T I=90^{\circ}$ and $\angle D A E=\angle T A I$, we have $\triangle A D E \sim ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 94 | 888 |
2016 | T0 | G5 | Geometry | IMO-SL | Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $A B C$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $A B$ and $A C$ at $X$ and $Y$ respectively. Let $P$ be the foot of ... | Denote the orthocentre and circumcentre of triangle $A B C$ by $H$ and $O$ respectively. Let $Q$ be the midpoint of $A H$ and $N$ be the nine-point centre of triangle $A B C$. It is known that $Q$ lies on the nine-point circle of triangle $A B C, N$ is the midpoint of $Q M$ and that $Q M$ is parallel to $A O$. Let the ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 140 | 1,192 |
2016 | T0 | G5 | Geometry | IMO-SL | Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $A B C$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $A B$ and $A C$ at $X$ and $Y$ respectively. Let $P$ be the foot of ... | Denote the orthocentre and circumcentre of triangle $A B C$ by $H$ and $O$ respectively. Let $O_{1}$ be the circumcentre of triangle $X S Y$. Consider two other possible positions of $S$. We name them $S^{\prime}$ and $S^{\prime \prime}$ and define the analogous points $X^{\prime}, Y^{\prime}, O_{1}^{\prime}, X^{\prime... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 140 | 817 |
2016 | T0 | G6 | Geometry | IMO-SL | Let $A B C D$ be a convex quadrilateral with $\angle A B C=\angle A D C<90^{\circ}$. The internal angle bisectors of $\angle A B C$ and $\angle A D C$ meet $A C$ at $E$ and $F$ respectively, and meet each other at point $P$. Let $M$ be the midpoint of $A C$ and let $\omega$ be the circumcircle of triangle $B P D$. Segm... | Let $\omega_{1}$ be the circumcircle of triangle $A B C$. We first prove that $Y$ lies on $\omega_{1}$. Let $Y^{\prime}$ be the point on ray $M D$ such that $M Y^{\prime} \cdot M D=M A^{2}$. Then triangles $M A Y^{\prime}$ and $M D A$ are oppositely similar. Since $M C^{2}=M A^{2}=M Y^{\prime} \cdot M D$, triangles $M ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 160 | 887 |
2016 | T0 | G6 | Geometry | IMO-SL | Let $A B C D$ be a convex quadrilateral with $\angle A B C=\angle A D C<90^{\circ}$. The internal angle bisectors of $\angle A B C$ and $\angle A D C$ meet $A C$ at $E$ and $F$ respectively, and meet each other at point $P$. Let $M$ be the midpoint of $A C$ and let $\omega$ be the circumcircle of triangle $B P D$. Segm... | Denote by $\omega_{1}$ and $\omega_{2}$ the circumcircles of triangles $A B C$ and $A D C$ respectively. Since $\angle A B C=\angle A D C$, we know that $\omega_{1}$ and $\omega_{2}$ are symmetric with respect to the midpoint $M$ of $A C$. Firstly, we show that $X$ lies on $\omega_{2}$. Let $X_{1}$ be the second inters... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 160 | 1,108 |
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