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 |
|---|---|---|---|---|---|---|---|---|---|
2013 | T0 | C4 | Combinatorics | IMO-SL | Let $n$ be a positive integer, and let $A$ be a subset of $\{1, \ldots, n\}$. An $A$-partition of $n$ into $k$ parts is a representation of $n$ as a sum $n=a_{1}+\cdots+a_{k}$, where the parts $a_{1}, \ldots, a_{k}$ belong to $A$ and are not necessarily distinct. The number of different parts in such a partition is the... | Assume, to the contrary, that the statement is false, and choose the minimum number $n$ for which it fails. So there exists a set $A \subseteq\{1, \ldots, n\}$ together with an optimal $A$ partition $n=a_{1}+\cdots+a_{k_{\min }}$ of $n$ refuting our statement, where, of course, $k_{\min }$ is the minimum number of part... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 216 | 1,354 |
2013 | T0 | C5 | Combinatorics | IMO-SL | Let $r$ be a positive integer, and let $a_{0}, a_{1}, \ldots$ be an infinite sequence of real numbers. Assume that for all nonnegative integers $m$ and $s$ there exists a positive integer $n \in[m+1, m+r]$ such that $$ a_{m}+a_{m+1}+\cdots+a_{m+s}=a_{n}+a_{n+1}+\cdots+a_{n+s} $$ Prove that the sequence is periodic, i... | For every indices $m \leqslant n$ we will denote $S(m, n)=a_{m}+a_{m+1}+\cdots+a_{n-1}$; thus $S(n, n)=0$. Let us start with the following lemma. Lemma. Let $b_{0}, b_{1}, \ldots$ be an infinite sequence. Assume that for every nonnegative integer $m$ there exists a nonnegative integer $n \in[m+1, m+r]$ such that $b_{m}... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 155 | 1,332 |
2013 | T0 | C6 | Combinatorics | IMO-SL | In some country several pairs of cities are connected by direct two-way flights. It is possible to go from any city to any other by a sequence of flights. The distance between two cities is defined to be the least possible number of flights required to go from one of them to the other. It is known that for any city the... | Let us denote by $d(a, b)$ the distance between the cities $a$ and $b$, and by $$ S_{i}(a)=\{c: d(a, c)=i\} $$ the set of cities at distance exactly $i$ from city $a$. Assume that for some city $x$ the set $D=S_{4}(x)$ has size at least 2551 . Let $A=S_{1}(x)$. A subset $A^{\prime}$ of $A$ is said to be substantial, if... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 110 | 1,568 |
2013 | T0 | C7 | Combinatorics | IMO-SL | Let $n \geqslant 2$ be an integer. Consider all circular arrangements of the numbers $0,1, \ldots, n$; the $n+1$ rotations of an arrangement are considered to be equal. A circular arrangement is called beautiful if, for any four distinct numbers $0 \leqslant a, b, c, d \leqslant n$ with $a+c=b+d$, the chord joining num... | Given a circular arrangement of $[0, n]=\{0,1, \ldots, n\}$, we define a $k$-chord to be a (possibly degenerate) chord whose (possibly equal) endpoints add up to $k$. We say that three chords of a circle are aligned if one of them separates the other two. Say that $m \geqslant 3$ chords are aligned if any three of them... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 195 | 1,839 |
2013 | T0 | C7 | Combinatorics | IMO-SL | Let $n \geqslant 2$ be an integer. Consider all circular arrangements of the numbers $0,1, \ldots, n$; the $n+1$ rotations of an arrangement are considered to be equal. A circular arrangement is called beautiful if, for any four distinct numbers $0 \leqslant a, b, c, d \leqslant n$ with $a+c=b+d$, the chord joining num... | Notice that there are exactly $N$ irreducible fractions $f_{1}<\cdots<f_{N}$ in $(0,1)$ whose denominator is at most $n$, since the pair $(x, y)$ with $x+y \leqslant n$ and $(x, y)=1$ corresponds to the fraction $x /(x+y)$. Write $f_{i}=\frac{a_{i}}{b_{i}}$ for $1 \leqslant i \leqslant N$. We begin by constructing $N+1... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
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} | 195 | 2,097 |
2013 | T0 | C8 | Combinatorics | IMO-SL | Players $A$ and $B$ play a paintful game on the real line. Player $A$ has a pot of paint with four units of black ink. A quantity $p$ of this ink suffices to blacken a (closed) real interval of length $p$. In every round, player $A$ picks some positive integer $m$ and provides $1 / 2^{m}$ units of ink from the pot. Pla... | We will present a strategy for player $B$ that guarantees that the interval $[0,1]$ is completely blackened, once the paint pot has become empty. At the beginning of round $r$, let $x_{r}$ denote the largest real number for which the interval between 0 and $x_{r}$ has already been blackened; for completeness we define ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 198 | 2,011 |
2013 | T0 | G1 | Geometry | IMO-SL | Let $A B C$ be an acute-angled triangle with orthocenter $H$, and let $W$ be a point on side $B C$. Denote by $M$ and $N$ the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_{1}$ the circumcircle of $B W N$, and let $X$ be the point on $\omega_{1}$ which is diametrically opposite to $W$. Analogo... | Let $L$ be the foot of the altitude from $A$, and let $Z$ be the second intersection point of circles $\omega_{1}$ and $\omega_{2}$, other than $W$. We show that $X, Y, Z$ and $H$ lie on the same line. Due to $\angle B N C=\angle B M C=90^{\circ}$, the points $B, C, N$ and $M$ are concyclic; denote their circle by $\om... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 165 | 556 |
2013 | T0 | G2 | Geometry | IMO-SL | Let $\omega$ be the circumcircle of a triangle $A B C$. Denote by $M$ and $N$ the midpoints of the sides $A B$ and $A C$, respectively, and denote by $T$ the midpoint of the $\operatorname{arc} B C$ of $\omega$ not containing $A$. The circumcircles of the triangles $A M T$ and $A N T$ intersect the perpendicular bisect... | Let $L$ be the second common point of the line $A C$ with the circumcircle $\gamma$ of the triangle $A M T$. From the cyclic quadrilaterals $A B T C$ and $A M T L$ we get $\angle B T C=180^{\circ}-$ $\angle B A C=\angle M T L$, which implies $\angle B T M=\angle C T L$. Since $A T$ is an angle bisector in these quadril... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 163 | 756 |
2013 | T0 | G3 | Geometry | IMO-SL | In a triangle $A B C$, let $D$ and $E$ be the feet of the angle bisectors of angles $A$ and $B$, respectively. A rhombus is inscribed into the quadrilateral $A E D B$ (all vertices of the rhombus lie on different sides of $A E D B$ ). Let $\varphi$ be the non-obtuse angle of the rhombus. Prove that $\varphi \leqslant \... | Let $K, L, M$, and $N$ be the vertices of the rhombus lying on the sides $A E, E D, D B$, and $B A$, respectively. Denote by $d(X, Y Z)$ the distance from a point $X$ to a line $Y Z$. Since $D$ and $E$ are the feet of the bisectors, we have $d(D, A B)=d(D, A C), d(E, A B)=d(E, B C)$, and $d(D, B C)=d(E, A C)=0$, which ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 123 | 667 |
2013 | T0 | G4 | Geometry | IMO-SL | Let $A B C$ be a triangle with $\angle B>\angle C$. Let $P$ and $Q$ be two different points on line $A C$ such that $\angle P B A=\angle Q B A=\angle A C B$ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $B Q$ for which $P D=P B$. Let the ray $A D$ intersect the circl... | Again, denote by $\omega$ the circumcircle of the triangle $A B C$. Denote $\angle A C B=\gamma$. Since $\angle P B A=\gamma$, the line $P B$ is tangent to $\omega$. Let $E$ be the second intersection point of $B Q$ with $\omega$. If $V^{\prime}$ is any point on the ray $C E$ beyond $E$, then $\angle B E V^{\prime}=180... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 128 | 662 |
2013 | T0 | G4 | Geometry | IMO-SL | Let $A B C$ be a triangle with $\angle B>\angle C$. Let $P$ and $Q$ be two different points on line $A C$ such that $\angle P B A=\angle Q B A=\angle A C B$ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $B Q$ for which $P D=P B$. Let the ray $A D$ intersect the circl... | Denote by $\omega$ and $O$ the circumcircle of the triangle $A B C$ and its center, respectively. From the condition $\angle P B A=\angle B C A$ we know that $B P$ is tangent to $\omega$. Let $E$ be the second point of intersection of $\omega$ and $B D$. Due to the isosceles triangle $B D P$, the tangent of $\omega$ at... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 128 | 852 |
2013 | T0 | G5 | Geometry | IMO-SL | Let $A B C D E F$ be a convex hexagon with $A B=D E, B C=E F, C D=F A$, and $\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$. Prove that the diagonals $A D, B E$, and $C F$ are concurrent. (Ukraine) | Let $x=A B=D E, y=C D=F A, z=E F=B C$. Consider the points $P, Q$, and $R$ such that the quadrilaterals $C D E P, E F A Q$, and $A B C R$ are parallelograms. We compute $$ \begin{aligned} \angle P E Q & =\angle F E Q+\angle D E P-\angle E=\left(180^{\circ}-\angle F\right)+\left(180^{\circ}-\angle D\right)-\angle E \\ &... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 77 | 639 |
2013 | T0 | G5 | Geometry | IMO-SL | Let $A B C D E F$ be a convex hexagon with $A B=D E, B C=E F, C D=F A$, and $\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$. Prove that the diagonals $A D, B E$, and $C F$ are concurrent. (Ukraine) | Let $X=C D \cap E F, Y=E F \cap A B, Z=A B \cap C D, X^{\prime}=F A \cap B C, Y^{\prime}=$ $B C \cap D E$, and $Z^{\prime}=D E \cap F A$. From $\angle A+\angle B+\angle C=360^{\circ}+\theta / 2$ we get $\angle A+\angle B>180^{\circ}$ and $\angle B+\angle C>180^{\circ}$, so $Z$ and $X^{\prime}$ are respectively on the o... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 77 | 1,295 |
2013 | T0 | G5 | Geometry | IMO-SL | Let $A B C D E F$ be a convex hexagon with $A B=D E, B C=E F, C D=F A$, and $\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$. Prove that the diagonals $A D, B E$, and $C F$ are concurrent. (Ukraine) | Place the hexagon on the complex plane, with $A$ at the origin and vertices labelled clockwise. Now $A, B, C, D, E, F$ represent the corresponding complex numbers. Also consider the complex numbers $a, b, c, a^{\prime}, b^{\prime}, c^{\prime}$ given by $B-A=a, D-C=b, F-E=c, E-D=a^{\prime}$, $A-F=b^{\prime}$, and $C-B=c... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 77 | 829 |
2013 | T0 | G6 | Geometry | IMO-SL | Let the excircle of the triangle $A B C$ lying opposite to $A$ touch its side $B C$ at the point $A_{1}$. Define the points $B_{1}$ and $C_{1}$ analogously. Suppose that the circumcentre of the triangle $A_{1} B_{1} C_{1}$ lies on the circumcircle of the triangle $A B C$. Prove that the triangle $A B C$ is right-angled... | Denote the circumcircles of the triangles $A B C$ and $A_{1} B_{1} C_{1}$ by $\Omega$ and $\Gamma$, respectively. Denote the midpoint of the arc $C B$ of $\Omega$ containing $A$ by $A_{0}$, and define $B_{0}$ as well as $C_{0}$ analogously. By our hypothesis the centre $Q$ of $\Gamma$ lies on $\Omega$. Lemma. One has $... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 104 | 1,044 |
2013 | T0 | G6 | Geometry | IMO-SL | Let the excircle of the triangle $A B C$ lying opposite to $A$ touch its side $B C$ at the point $A_{1}$. Define the points $B_{1}$ and $C_{1}$ analogously. Suppose that the circumcentre of the triangle $A_{1} B_{1} C_{1}$ lies on the circumcircle of the triangle $A B C$. Prove that the triangle $A B C$ is right-angled... | Let $Q$ again denote the centre of the circumcircle of the triangle $A_{1} B_{1} C_{1}$, that lies on the circumcircle $\Omega$ of the triangle $A B C$. We first consider the case where $Q$ coincides with one of the vertices of $A B C$, say $Q=B$. Then $B C_{1}=B A_{1}$ and consequently the triangle $A B C$ is isoscele... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 104 | 1,509 |
2013 | T0 | N2 | Number Theory | IMO-SL | Prove that for any pair of positive integers $k$ and $n$ there exist $k$ positive integers $m_{1}, m_{2}, \ldots, m_{k}$ such that $$ 1+\frac{2^{k}-1}{n}=\left(1+\frac{1}{m_{1}}\right)\left(1+\frac{1}{m_{2}}\right) \cdots\left(1+\frac{1}{m_{k}}\right) . $$ (Japan) | Consider the base 2 expansions of the residues of $n-1$ and $-n$ modulo $2^{k}$ : $$ \begin{aligned} n-1 & \equiv 2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{r}}\left(\bmod 2^{k}\right) & & \text { where } 0 \leqslant a_{1}<a_{2}<\ldots<a_{r} \leqslant k-1 \\ -n & \equiv 2^{b_{1}}+2^{b_{2}}+\cdots+2^{b_{s}}\left(\bmod 2^{k}\right... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 113 | 943 |
2013 | T0 | N3 | Number Theory | IMO-SL | Prove that there exist infinitely many positive integers $n$ such that the largest prime divisor of $n^{4}+n^{2}+1$ is equal to the largest prime divisor of $(n+1)^{4}+(n+1)^{2}+1$. (Belgium) | Let $p_{n}$ be the largest prime divisor of $n^{4}+n^{2}+1$ and let $q_{n}$ be the largest prime divisor of $n^{2}+n+1$. Then $p_{n}=q_{n^{2}}$, and from $$ n^{4}+n^{2}+1=\left(n^{2}+1\right)^{2}-n^{2}=\left(n^{2}-n+1\right)\left(n^{2}+n+1\right)=\left((n-1)^{2}+(n-1)+1\right)\left(n^{2}+n+1\right) $$ it follows that $... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 64 | 1,143 |
2013 | T0 | N4 | Number Theory | IMO-SL | Determine whether there exists an infinite sequence of nonzero digits $a_{1}, a_{2}, a_{3}, \ldots$ and a positive integer $N$ such that for every integer $k>N$, the number $\overline{a_{k} a_{k-1} \ldots a_{1}}$ is a perfect square. (Iran) | Assume that $a_{1}, a_{2}, a_{3}, \ldots$ is such a sequence. For each positive integer $k$, let $y_{k}=$ $\overline{a_{k} a_{k-1} \ldots a_{1}}$. By the assumption, for each $k>N$ there exists a positive integer $x_{k}$ such that $y_{k}=x_{k}^{2}$. I. For every $n$, let $5^{\gamma_{n}}$ be the greatest power of 5 divi... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 77 | 1,447 |
2013 | T0 | N4 | Number Theory | IMO-SL | Determine whether there exists an infinite sequence of nonzero digits $a_{1}, a_{2}, a_{3}, \ldots$ and a positive integer $N$ such that for every integer $k>N$, the number $\overline{a_{k} a_{k-1} \ldots a_{1}}$ is a perfect square. (Iran) | Again, we assume that a sequence $a_{1}, a_{2}, a_{3}, \ldots$ satisfies the problem conditions, introduce the numbers $x_{k}$ and $y_{k}$ as in the previous solution, and notice that $$ y_{k+1}-y_{k}=\left(x_{k+1}-x_{k}\right)\left(x_{k+1}+x_{k}\right)=10^{k} a_{k+1} $$ for all $k>N$. Consider any such $k$. Since $a_{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 77 | 1,644 |
2013 | T0 | N5 | Number Theory | IMO-SL | Fix an integer $k \geqslant 2$. Two players, called Ana and Banana, play the following game of numbers: Initially, some integer $n \geqslant k$ gets written on the blackboard. Then they take moves in turn, with Ana beginning. A player making a move erases the number $m$ just written on the blackboard and replaces it by... | Let us first observe that the number appearing on the blackboard decreases after every move; so the game necessarily ends after at most $n$ steps, and consequently there always has to be some player possessing a winning strategy. So if some $n \geqslant k$ is bad, then Ana has a winning strategy in the game with starti... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 233 | 1,382 |
2013 | T0 | N5 | Number Theory | IMO-SL | Fix an integer $k \geqslant 2$. Two players, called Ana and Banana, play the following game of numbers: Initially, some integer $n \geqslant k$ gets written on the blackboard. Then they take moves in turn, with Ana beginning. A player making a move erases the number $m$ just written on the blackboard and replaces it by... | We use the same analysis of the game of numbers as in the first five paragraphs of the first solution. Let us call a prime number $p$ small in case $p \leqslant k$ and big otherwise. We again call two integers similar if their sets of small prime factors coincide. Claim 4. For each integer $b \geqslant k$ having some s... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 233 | 1,594 |
2013 | T0 | N6 | Number Theory | IMO-SL | Determine all functions $f: \mathbb{Q} \longrightarrow \mathbb{Z}$ satisfying $$ f\left(\frac{f(x)+a}{b}\right)=f\left(\frac{x+a}{b}\right) $$ for all $x \in \mathbb{Q}, a \in \mathbb{Z}$, and $b \in \mathbb{Z}_{>0}$. (Here, $\mathbb{Z}_{>0}$ denotes the set of positive integers.) (Israel) | I. We start by verifying that these functions do indeed satisfy (1). This is clear for all constant functions. Now consider any triple $(x, a, b) \in \mathbb{Q} \times \mathbb{Z} \times \mathbb{Z}_{>0}$ and set $$ q=\left\lfloor\frac{x+a}{b}\right\rfloor . $$ This means that $q$ is an integer and $b q \leqslant x+a<b(q... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 113 | 2,009 |
2013 | T0 | N6 | Number Theory | IMO-SL | Determine all functions $f: \mathbb{Q} \longrightarrow \mathbb{Z}$ satisfying $$ f\left(\frac{f(x)+a}{b}\right)=f\left(\frac{x+a}{b}\right) $$ for all $x \in \mathbb{Q}, a \in \mathbb{Z}$, and $b \in \mathbb{Z}_{>0}$. (Here, $\mathbb{Z}_{>0}$ denotes the set of positive integers.) (Israel) | Here we just give another argument for the second case of the above solution. Again we use equation (2). It follows that the set $S$ of all zeros of $f$ contains for each $x \in \mathbb{Q}$ exactly one term from the infinite sequence $\ldots, x-2, x-1, x, x+1, x+2, \ldots$. Next we claim that $$ \text { if }(p, q) \in ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 113 | 1,342 |
2013 | T0 | N7 | Number Theory | IMO-SL | Let $\nu$ be an irrational positive number, and let $m$ be a positive integer. A pair $(a, b)$ of positive integers is called good if $$ a\lceil b \nu\rceil-b\lfloor a \nu\rfloor=m . $$ A good pair $(a, b)$ is called excellent if neither of the pairs $(a-b, b)$ and $(a, b-a)$ is good. (As usual, by $\lfloor x\rfloor$... | For positive integers $a$ and $b$, let us denote $$ f(a, b)=a\lceil b \nu\rceil-b\lfloor a \nu\rfloor . $$ We will deal with various values of $m$; thus it is convenient to say that a pair $(a, b)$ is $m$-good or $m$-excellent if the corresponding conditions are satisfied. To start, let us investigate how the values $f... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 180 | 3,803 |
2013 | T0 | A3 | Algebra | IMO-SL | Let $\mathbb{Q}_{>0}$ be the set of positive rational numbers. Let $f: \mathbb{Q}_{>0} \rightarrow \mathbb{R}$ be a function satisfying the conditions $$ \begin{aligned} & f(x) f(y) \geqslant f(x y) \\ & f(x+y) \geqslant f(x)+f(y) \end{aligned} $$ for all $x, y \in \mathbb{Q}_{>0}$. Given that $f(a)=a$ for some ratio... | Denote by $\mathbb{Z}_{>0}$ the set of positive integers. Plugging $x=1, y=a$ into (1) we get $f(1) \geqslant 1$. Next, by an easy induction on $n$ we get from (2) that $$ f(n x) \geqslant n f(x) \text { for all } n \in \mathbb{Z}_{>0} \text { and } x \in \mathbb{Q}_{>0} $$ In particular, we have $$ f(n) \geqslant n f(... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 156 | 876 |
2013 | T0 | A5 | Algebra | IMO-SL | Let $\mathbb{Z}_{\geqslant 0}$ be the set of all nonnegative integers. Find all the functions $f: \mathbb{Z}_{\geqslant 0} \rightarrow \mathbb{Z}_{\geqslant 0}$ satisfying the relation $$ f(f(f(n)))=f(n+1)+1 $$ for all $n \in \mathbb{Z}_{\geqslant 0}$. (Serbia) Answer. There are two such functions: $f(n)=n+1$ for a... | To start, we get from (*) that $$ f^{4}(n)=f\left(f^{3}(n)\right)=f(f(n+1)+1) \quad \text { and } \quad f^{4}(n+1)=f^{3}(f(n+1))=f(f(n+1)+1)+1 $$ thus $$ f^{4}(n)+1=f^{4}(n+1) . $$ I. Let us denote by $R_{i}$ the range of $f^{i}$; note that $R_{0}=\mathbb{Z}_{\geqslant 0}$ since $f^{0}$ is the identity function. Obviou... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 342 | 1,772 |
2013 | T0 | A5 | Algebra | IMO-SL | Let $\mathbb{Z}_{\geqslant 0}$ be the set of all nonnegative integers. Find all the functions $f: \mathbb{Z}_{\geqslant 0} \rightarrow \mathbb{Z}_{\geqslant 0}$ satisfying the relation $$ f(f(f(n)))=f(n+1)+1 $$ for all $n \in \mathbb{Z}_{\geqslant 0}$. (Serbia) Answer. There are two such functions: $f(n)=n+1$ for a... | I. For convenience, let us introduce the function $g(n)=f(n)+1$. Substituting $f(n)$ instead of $n$ into (*) we obtain $$ f^{4}(n)=f(f(n)+1)+1, \quad \text { or } \quad f^{4}(n)=g^{2}(n) . $$ Applying $f$ to both parts of (*) and using (5) we get $$ f^{4}(n)+1=f(f(n+1)+1)+1=f^{4}(n+1) . $$ Thus, if $g^{2}(0)=f^{4}(0)=c... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 342 | 1,410 |
2013 | T0 | A6 | Algebra | IMO-SL | Let $m \neq 0$ be an integer. Find all polynomials $P(x)$ with real coefficients such that $$ \left(x^{3}-m x^{2}+1\right) P(x+1)+\left(x^{3}+m x^{2}+1\right) P(x-1)=2\left(x^{3}-m x+1\right) P(x) $$ for all real numbers $x$. (Serbia) Answer. $P(x)=t x$ for any real number $t$. | Let $P(x)=a_{n} x^{n}+\cdots+a_{0} x^{0}$ with $a_{n} \neq 0$. Comparing the coefficients of $x^{n+1}$ on both sides gives $a_{n}(n-2 m)(n-1)=0$, so $n=1$ or $n=2 m$. If $n=1$, one easily verifies that $P(x)=x$ is a solution, while $P(x)=1$ is not. Since the given condition is linear in $P$, this means that the linear ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 119 | 793 |
2013 | T0 | A6 | Algebra | IMO-SL | Let $m \neq 0$ be an integer. Find all polynomials $P(x)$ with real coefficients such that $$ \left(x^{3}-m x^{2}+1\right) P(x+1)+\left(x^{3}+m x^{2}+1\right) P(x-1)=2\left(x^{3}-m x+1\right) P(x) $$ for all real numbers $x$. (Serbia) Answer. $P(x)=t x$ for any real number $t$. | Multiplying (1) by $x$, we rewrite it as $$ x\left(x^{3}-m x^{2}+1\right) P(x+1)+x\left(x^{3}+m x^{2}+1\right) P(x-1)=[(x+1)+(x-1)]\left(x^{3}-m x+1\right) P(x) . $$ After regrouping, it becomes $$ \left(x^{3}-m x^{2}+1\right) Q(x)=\left(x^{3}+m x^{2}+1\right) Q(x-1) \text {, } $$ where $Q(x)=x P(x+1)-(x+1) P(x)$. If $... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 119 | 1,086 |
2013 | T0 | C1 | Combinatorics | IMO-SL | Let $n$ be a positive integer. Find the smallest integer $k$ with the following property: Given any real numbers $a_{1}, \ldots, a_{d}$ such that $a_{1}+a_{2}+\cdots+a_{d}=n$ and $0 \leqslant a_{i} \leqslant 1$ for $i=1,2, \ldots, d$, it is possible to partition these numbers into $k$ groups (some of which may be empty... | We will show that it is even possible to split the sequence $a_{1}, \ldots, a_{d}$ into $2 n-1$ contiguous groups so that the sum of the numbers in each groups does not exceed 1. Consider a segment $S$ of length $n$, and partition it into segments $S_{1}, \ldots, S_{d}$ of lengths $a_{1}, \ldots, a_{d}$, respectively, ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 145 | 645 |
2013 | T0 | C1 | Combinatorics | IMO-SL | Let $n$ be a positive integer. Find the smallest integer $k$ with the following property: Given any real numbers $a_{1}, \ldots, a_{d}$ such that $a_{1}+a_{2}+\cdots+a_{d}=n$ and $0 \leqslant a_{i} \leqslant 1$ for $i=1,2, \ldots, d$, it is possible to partition these numbers into $k$ groups (some of which may be empty... | First put all numbers greater than $\frac{1}{2}$ in their own groups. Then, form the remaining groups as follows: For each group, add new $a_{i} \mathrm{~S}$ one at a time until their sum exceeds $\frac{1}{2}$. Since the last summand is at most $\frac{1}{2}$, this group has sum at most 1 . Continue this procedure until... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 145 | 524 |
2013 | T0 | C2 | Combinatorics | IMO-SL | In the plane, 2013 red points and 2014 blue points are marked so that no three of the marked points are collinear. One needs to draw $k$ lines not passing through the marked points and dividing the plane into several regions. The goal is to do it in such a way that no region contains points of both colors. Find the mi... | Firstly, let us present an example showing that $k \geqslant 2013$. Mark 2013 red and 2013 blue points on some circle alternately, and mark one more blue point somewhere in the plane. The circle is thus split into 4026 arcs, each arc having endpoints of different colors. Thus, if the goal is reached, then each arc shou... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 117 | 510 |
2013 | T0 | C2 | Combinatorics | IMO-SL | In the plane, 2013 red points and 2014 blue points are marked so that no three of the marked points are collinear. One needs to draw $k$ lines not passing through the marked points and dividing the plane into several regions. The goal is to do it in such a way that no region contains points of both colors. Find the mi... | Let us present a different proof of the fact that $k=2013$ suffices. In fact, we will prove a more general statement: If $n$ points in the plane, no three of which are collinear, are colored in red and blue arbitrarily, then it suffices to draw $\lfloor n / 2\rfloor$ lines to reach the goal. We proceed by induction on ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 117 | 593 |
2013 | T0 | C5 | Combinatorics | IMO-SL | Let $r$ be a positive integer, and let $a_{0}, a_{1}, \ldots$ be an infinite sequence of real numbers. Assume that for all nonnegative integers $m$ and $s$ there exists a positive integer $n \in[m+1, m+r]$ such that $$ a_{m}+a_{m+1}+\cdots+a_{m+s}=a_{n}+a_{n+1}+\cdots+a_{n+s} $$ Prove that the sequence is periodic, i... | For every indices $m \leqslant n$ we will denote $S(m, n)=a_{m}+a_{m+1}+\cdots+a_{n-1}$; thus $S(n, n)=0$. Let us start with the following lemma. Lemma. Let $b_{0}, b_{1}, \ldots$ be an infinite sequence. Assume that for every nonnegative integer $m$ there exists a nonnegative integer $n \in[m+1, m+r]$ such that $b_{m}... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 159 | 1,332 |
2013 | T0 | C8 | Combinatorics | IMO-SL | Players $A$ and $B$ play a paintful game on the real line. Player $A$ has a pot of paint with four units of black ink. A quantity $p$ of this ink suffices to blacken a (closed) real interval of length p. In every round, player $A$ picks some positive integer $m$ and provides $1 / 2^{m}$ units of ink from the pot. Playe... | We will present a strategy for player $B$ that guarantees that the interval $[0,1]$ is completely blackened, once the paint pot has become empty. At the beginning of round $r$, let $x_{r}$ denote the largest real number for which the interval between 0 and $x_{r}$ has already been blackened; for completeness we define ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 214 | 2,011 |
2013 | T0 | G5 | Geometry | IMO-SL | Let $A B C D E F$ be a convex hexagon with $A B=D E, B C=E F, C D=F A$, and $\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$. Prove that the diagonals $A D, B E$, and $C F$ are concurrent. (Ukraine) In all three solutions, we denote $\theta=\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$ and assume with... | Let $x=A B=D E, y=C D=F A, z=E F=B C$. Consider the points $P, Q$, and $R$ such that the quadrilaterals $C D E P, E F A Q$, and $A B C R$ are parallelograms. We compute $$ \begin{aligned} \angle P E Q & =\angle F E Q+\angle D E P-\angle E=\left(180^{\circ}-\angle F\right)+\left(180^{\circ}-\angle D\right)-\angle E \\ &... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 124 | 639 |
2013 | T0 | G5 | Geometry | IMO-SL | Let $A B C D E F$ be a convex hexagon with $A B=D E, B C=E F, C D=F A$, and $\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$. Prove that the diagonals $A D, B E$, and $C F$ are concurrent. (Ukraine) In all three solutions, we denote $\theta=\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$ and assume with... | Let $X=C D \cap E F, Y=E F \cap A B, Z=A B \cap C D, X^{\prime}=F A \cap B C, Y^{\prime}=$ $B C \cap D E$, and $Z^{\prime}=D E \cap F A$. From $\angle A+\angle B+\angle C=360^{\circ}+\theta / 2$ we get $\angle A+\angle B>180^{\circ}$ and $\angle B+\angle C>180^{\circ}$, so $Z$ and $X^{\prime}$ are respectively on the o... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 124 | 1,295 |
2013 | T0 | G5 | Geometry | IMO-SL | Let $A B C D E F$ be a convex hexagon with $A B=D E, B C=E F, C D=F A$, and $\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$. Prove that the diagonals $A D, B E$, and $C F$ are concurrent. (Ukraine) In all three solutions, we denote $\theta=\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$ and assume with... | Place the hexagon on the complex plane, with $A$ at the origin and vertices labelled clockwise. Now $A, B, C, D, E, F$ represent the corresponding complex numbers. Also consider the complex numbers $a, b, c, a^{\prime}, b^{\prime}, c^{\prime}$ given by $B-A=a, D-C=b, F-E=c, E-D=a^{\prime}$, $A-F=b^{\prime}$, and $C-B=c... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 124 | 829 |
2013 | T0 | N4 | Number Theory | IMO-SL | Determine whether there exists an infinite sequence of nonzero digits $a_{1}, a_{2}, a_{3}, \ldots$ and a positive integer $N$ such that for every integer $k>N$, the number $\overline{a_{k} a_{k-1} \ldots a_{1}}$ is a perfect square. (Iran) Answer. No. | Assume that $a_{1}, a_{2}, a_{3}, \ldots$ is such a sequence. For each positive integer $k$, let $y_{k}=$ $\overline{a_{k} a_{k-1} \ldots a_{1}}$. By the assumption, for each $k>N$ there exists a positive integer $x_{k}$ such that $y_{k}=x_{k}^{2}$. I. For every $n$, let $5^{\gamma_{n}}$ be the greatest power of 5 divi... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 82 | 1,447 |
2013 | T0 | N4 | Number Theory | IMO-SL | Determine whether there exists an infinite sequence of nonzero digits $a_{1}, a_{2}, a_{3}, \ldots$ and a positive integer $N$ such that for every integer $k>N$, the number $\overline{a_{k} a_{k-1} \ldots a_{1}}$ is a perfect square. (Iran) Answer. No. | Again, we assume that a sequence $a_{1}, a_{2}, a_{3}, \ldots$ satisfies the problem conditions, introduce the numbers $x_{k}$ and $y_{k}$ as in the previous solution, and notice that $$ y_{k+1}-y_{k}=\left(x_{k+1}-x_{k}\right)\left(x_{k+1}+x_{k}\right)=10^{k} a_{k+1} $$ for all $k>N$. Consider any such $k$. Since $a_{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 82 | 1,644 |
2013 | T0 | N6 | Number Theory | IMO-SL | Determine all functions $f: \mathbb{Q} \longrightarrow \mathbb{Z}$ satisfying $$ f\left(\frac{f(x)+a}{b}\right)=f\left(\frac{x+a}{b}\right) $$ for all $x \in \mathbb{Q}, a \in \mathbb{Z}$, and $b \in \mathbb{Z}_{>0}$. (Here, $\mathbb{Z}_{>0}$ denotes the set of positive integers.) (Israel) Answer. There are three k... | I. We start by verifying that these functions do indeed satisfy (1). This is clear for all constant functions. Now consider any triple $(x, a, b) \in \mathbb{Q} \times \mathbb{Z} \times \mathbb{Z}_{>0}$ and set $$ q=\left\lfloor\frac{x+a}{b}\right\rfloor . $$ This means that $q$ is an integer and $b q \leqslant x+a<b(q... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 140 | 2,009 |
2013 | T0 | N6 | Number Theory | IMO-SL | Determine all functions $f: \mathbb{Q} \longrightarrow \mathbb{Z}$ satisfying $$ f\left(\frac{f(x)+a}{b}\right)=f\left(\frac{x+a}{b}\right) $$ for all $x \in \mathbb{Q}, a \in \mathbb{Z}$, and $b \in \mathbb{Z}_{>0}$. (Here, $\mathbb{Z}_{>0}$ denotes the set of positive integers.) (Israel) Answer. There are three k... | Here we just give another argument for the second case of the above solution. Again we use equation (2). It follows that the set $S$ of all zeros of $f$ contains for each $x \in \mathbb{Q}$ exactly one term from the infinite sequence $\ldots, x-2, x-1, x, x+1, x+2, \ldots$. Next we claim that $$ \text { if }(p, q) \in ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 140 | 1,342 |
2013 | T0 | N7 | Number Theory | IMO-SL | Let $\nu$ be an irrational positive number, and let $m$ be a positive integer. A pair $(a, b)$ of positive integers is called good if $$ a\lceil b \nu\rceil-b\lfloor a \nu\rfloor=m $$ A good pair $(a, b)$ is called excellent if neither of the pairs $(a-b, b)$ and $(a, b-a)$ is good. (As usual, by $\lfloor x\rfloor$ a... | For positive integers $a$ and $b$, let us denote $$ f(a, b)=a\lceil b \nu\rceil-b\lfloor a \nu\rfloor . $$ We will deal with various values of $m$; thus it is convenient to say that a pair $(a, b)$ is $m$-good or $m$-excellent if the corresponding conditions are satisfied. To start, let us investigate how the values $f... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 179 | 3,803 |
2014 | T0 | A1 | Algebra | IMO-SL | Let $z_{0}<z_{1}<z_{2}<\cdots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n \geqslant 1$ such that $$ z_{n}<\frac{z_{0}+z_{1}+\cdots+z_{n}}{n} \leqslant z_{n+1} $$ (Austria) | For $n=1,2, \ldots$ define $$ d_{n}=\left(z_{0}+z_{1}+\cdots+z_{n}\right)-n z_{n} $$ The sign of $d_{n}$ indicates whether the first inequality in (1) holds; i.e., it is satisfied if and only if $d_{n}>0$. Notice that $$ n z_{n+1}-\left(z_{0}+z_{1}+\cdots+z_{n}\right)=(n+1) z_{n+1}-\left(z_{0}+z_{1}+\cdots+z_{n}+z_{n+1... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 91 | 556 |
2014 | T0 | A2 | Algebra | IMO-SL | Define the function $f:(0,1) \rightarrow(0,1)$ by $$ f(x)= \begin{cases}x+\frac{1}{2} & \text { if } x<\frac{1}{2} \\ x^{2} & \text { if } x \geqslant \frac{1}{2}\end{cases} $$ Let $a$ and $b$ be two real numbers such that $0<a<b<1$. We define the sequences $a_{n}$ and $b_{n}$ by $a_{0}=a, b_{0}=b$, and $a_{n}=f\left... | Note that $$ f(x)-x=\frac{1}{2}>0 $$ if $x<\frac{1}{2}$ and $$ f(x)-x=x^{2}-x<0 $$ if $x \geqslant \frac{1}{2}$. So if we consider $(0,1)$ as being divided into the two subintervals $I_{1}=\left(0, \frac{1}{2}\right)$ and $I_{2}=\left[\frac{1}{2}, 1\right)$, the inequality $$ \left(a_{n}-a_{n-1}\right)\left(b_{n}-b_{n-... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 218 | 978 |
2014 | T0 | A3 | Algebra | IMO-SL | For a sequence $x_{1}, x_{2}, \ldots, x_{n}$ of real numbers, we define its price as $$ \max _{1 \leqslant i \leqslant n}\left|x_{1}+\cdots+x_{i}\right| $$ Given $n$ real numbers, Dave and George want to arrange them into a sequence with a low price. Diligent Dave checks all possible ways and finds the minimum possib... | If the initial numbers are $1,-1,2$, and -2 , then Dave may arrange them as $1,-2,2,-1$, while George may get the sequence $1,-1,2,-2$, resulting in $D=1$ and $G=2$. So we obtain $c \geqslant 2$. Therefore, it remains to prove that $G \leqslant 2 D$. Let $x_{1}, x_{2}, \ldots, x_{n}$ be the numbers Dave and George have... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 310 | 1,122 |
2014 | T0 | A4 | Algebra | IMO-SL | Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfying $$ f(f(m)+n)+f(m)=f(n)+f(3 m)+2014 $$ for all integers $m$ and $n$. | Let $f$ be a function satisfying (1). Set $C=1007$ and define the function $g: \mathbb{Z} \rightarrow \mathbb{Z}$ by $g(m)=f(3 m)-f(m)+2 C$ for all $m \in \mathbb{Z}$; in particular, $g(0)=2 C$. Now (1) rewrites as $$ f(f(m)+n)=g(m)+f(n) $$ for all $m, n \in \mathbb{Z}$. By induction in both directions it follows that ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 57 | 1,890 |
2014 | T0 | A5 | Algebra | IMO-SL | Consider all polynomials $P(x)$ with real coefficients that have the following property: for any two real numbers $x$ and $y$ one has $$ \left|y^{2}-P(x)\right| \leqslant 2|x| \quad \text { if and only if } \quad\left|x^{2}-P(y)\right| \leqslant 2|y| $$ Determine all possible values of $P(0)$. | Part I. We begin by verifying that these numbers are indeed possible values of $P(0)$. To see that each negative real number $-C$ can be $P(0)$, it suffices to check that for every $C>0$ the polynomial $P(x)=-\left(\frac{2 x^{2}}{C}+C\right)$ has the property described in the statement of the problem. Due to symmetry i... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 100 | 2,796 |
2014 | T0 | A6 | Algebra | IMO-SL | Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $$ n^{2}+4 f(n)=f(f(n))^{2} $$ for all $n \in \mathbb{Z}$. | Part I. Let us first check that each of the functions above really satisfies the given functional equation. If $f(n)=n+1$ for all $n$, then we have $$ n^{2}+4 f(n)=n^{2}+4 n+4=(n+2)^{2}=f(n+1)^{2}=f(f(n))^{2} . $$ If $f(n)=n+1$ for $n>-a$ and $f(n)=-n+1$ otherwise, then we have the same identity for $n>-a$ and $$ n^{2}... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 55 | 2,959 |
2014 | T0 | A6 | Algebra | IMO-SL | Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $$ n^{2}+4 f(n)=f(f(n))^{2} $$ for all $n \in \mathbb{Z}$. | Let us provide an alternative proof for Part II, which also proceeds in several steps. Step 1. Let $a$ be an arbitrary integer and $b=f(a)$. We first concentrate on the case where $|a|$ is sufficiently large. 1. If $b=0$, then (1) applied to $a$ yields $a^{2}=f(f(a))^{2}$, thus $$ f(a)=0 \quad \Rightarrow \quad a= \pm ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 55 | 3,445 |
2014 | T0 | C1 | Combinatorics | IMO-SL | Let $n$ points be given inside a rectangle $R$ such that no two of them lie on a line parallel to one of the sides of $R$. The rectangle $R$ is to be dissected into smaller rectangles with sides parallel to the sides of $R$ in such a way that none of these rectangles contains any of the given points in its interior. Pr... | Let $k$ denote the number of rectangles. In the following, we refer to the directions of the sides of $R$ as 'horizontal' and 'vertical' respectively. Our goal is to prove the inequality $k \geqslant n+1$ for fixed $n$. Equivalently, we can prove the inequality $n \leqslant k-1$ for each $k$, which will be done by indu... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 100 | 592 |
2014 | T0 | C2 | Combinatorics | IMO-SL | We have $2^{m}$ sheets of paper, with the number 1 written on each of them. We perform the following operation. In every step we choose two distinct sheets; if the numbers on the two sheets are $a$ and $b$, then we erase these numbers and write the number $a+b$ on both sheets. Prove that after $m 2^{m-1}$ steps, the su... | Let $P_{k}$ be the product of the numbers on the sheets after $k$ steps. Suppose that in the $(k+1)^{\text {th }}$ step the numbers $a$ and $b$ are replaced by $a+b$. In the product, the number $a b$ is replaced by $(a+b)^{2}$, and the other factors do not change. Since $(a+b)^{2} \geqslant 4 a b$, we see that $P_{k+1}... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 105 | 2,657 |
2014 | T0 | C3 | Combinatorics | IMO-SL | Let $n \geqslant 2$ be an integer. Consider an $n \times n$ chessboard divided into $n^{2}$ unit squares. We call a configuration of $n$ rooks on this board happy if every row and every column contains exactly one rook. Find the greatest positive integer $k$ such that for every happy configuration of rooks, we can find... | Let $\ell$ be a positive integer. We will show that (i) if $n>\ell^{2}$ then each happy configuration contains an empty $\ell \times \ell$ square, but (ii) if $n \leqslant \ell^{2}$ then there exists a happy configuration not containing such a square. These two statements together yield the answer. (i). Assume that $n>... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 111 | 1,134 |
2014 | T0 | C4 | Combinatorics | IMO-SL | Construct a tetromino by attaching two $2 \times 1$ dominoes along their longer sides such that the midpoint of the longer side of one domino is a corner of the other domino. This construction yields two kinds of tetrominoes with opposite orientations. Let us call them Sand Z-tetrominoes, respectively. ^{j}$ into the squ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 293 | 802 |
2014 | T0 | C5 | Combinatorics | IMO-SL | Consider $n \geqslant 3$ lines in the plane such that no two lines are parallel and no three have a common point. These lines divide the plane into polygonal regions; let $\mathcal{F}$ be the set of regions having finite area. Prove that it is possible to colour $\lceil\sqrt{n / 2}\rceil$ of the lines blue in such a wa... | Let $L$ be the given set of lines. Choose a maximal (by inclusion) subset $B \subseteq L$ such that when we colour the lines of $B$ blue, no region in $\mathcal{F}$ has a completely blue boundary. Let $|B|=k$. We claim that $k \geqslant\lceil\sqrt{n / 2}\rceil$. Let us colour all the lines of $L \backslash B$ red. Call... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 101 | 1,293 |
2014 | T0 | C6 | Combinatorics | IMO-SL | We are given an infinite deck of cards, each with a real number on it. For every real number $x$, there is exactly one card in the deck that has $x$ written on it. Now two players draw disjoint sets $A$ and $B$ of 100 cards each from this deck. We would like to define a rule that declares one of them a winner. This rul... | We prove a more general statement for sets of cardinality $n$ (the problem being the special case $n=100$, then the answer is $n$ ). In the following, we write $A>B$ or $B<A$ for " $A$ beats $B$ ". Part I. Let us first define $n$ different rules that satisfy the conditions. To this end, fix an index $k \in\{1,2, \ldots... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 347 | 933 |
2014 | T0 | C6 | Combinatorics | IMO-SL | We are given an infinite deck of cards, each with a real number on it. For every real number $x$, there is exactly one card in the deck that has $x$ written on it. Now two players draw disjoint sets $A$ and $B$ of 100 cards each from this deck. We would like to define a rule that declares one of them a winner. This rul... | Another possible approach to Part II of this problem is induction on $n$. For $n=1$, there is trivially only one rule in view of the second condition. In the following, we assume that our claim (namely, that there are no possible rules other than those given in Part I) holds for $n-1$ in place of $n$. We start with the... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 347 | 2,319 |
2014 | T0 | C7 | Combinatorics | IMO-SL | Let $M$ be a set of $n \geqslant 4$ points in the plane, no three of which are collinear. Initially these points are connected with $n$ segments so that each point in $M$ is the endpoint of exactly two segments. Then, at each step, one may choose two segments $A B$ and $C D$ sharing a common interior point and replace ... | A line is said to be red if it contains two points of $M$. As no three points of $M$ are collinear, each red line determines a unique pair of points of $M$. Moreover, there are precisely $\binom{n}{2}<\frac{n^{2}}{2}$ red lines. By the value of a segment we mean the number of red lines intersecting it in its interior, ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 132 | 1,267 |
2014 | T0 | C8 | Combinatorics | IMO-SL | A card deck consists of 1024 cards. On each card, a set of distinct decimal digits is written in such a way that no two of these sets coincide (thus, one of the cards is empty). Two players alternately take cards from the deck, one card per turn. After the deck is empty, each player checks if he can throw out one of hi... | Let us identify each card with the set of digits written on it. For any collection of cards $C_{1}, C_{2}, \ldots, C_{k}$ denote by their sum the set $C_{1} \triangle C_{2} \triangle \cdots \triangle C_{k}$ consisting of all elements belonging to an odd number of the $C_{i}$ 's. Denote the first and the second player b... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 142 | 1,963 |
2014 | T0 | C9 | Combinatorics | IMO-SL | There are $n$ circles drawn on a piece of paper in such a way that any two circles intersect in two points, and no three circles pass through the same point. Turbo the snail slides along the circles in the following fashion. Initially he moves on one of the circles in clockwise direction. Turbo always keeps sliding alo... | Replace every cross (i.e. intersection of two circles) by two small circle arcs that indicate the direction in which the snail should leave the cross (see Figure 1.1). Notice that the placement of the small arcs does not depend on the direction of moving on the curves; no matter which direction the snail is moving on t... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 128 | 1,181 |
2014 | T0 | C9 | Combinatorics | IMO-SL | There are $n$ circles drawn on a piece of paper in such a way that any two circles intersect in two points, and no three circles pass through the same point. Turbo the snail slides along the circles in the following fashion. Initially he moves on one of the circles in clockwise direction. Turbo always keeps sliding alo... | We present a different proof of (*). We perform a sequence of small modification steps on the configuration of the circles in such a way that at the end they have no intersection at all (see Figure 6.1). We use two kinds of local changes to the structure of the orbits (see Figure 6.2): - Type-1 step: An arc of a circle... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 128 | 966 |
2014 | T0 | C9 | Combinatorics | IMO-SL | There are $n$ circles drawn on a piece of paper in such a way that any two circles intersect in two points, and no three circles pass through the same point. Turbo the snail slides along the circles in the following fashion. Initially he moves on one of the circles in clockwise direction. Turbo always keeps sliding alo... | Like in the previous solutions, we do not need all circle pairs to intersect but we assume that the circles form a connected set. Denote by $\mathcal{C}$ and $\mathcal{P}$ the sets of circles and their intersection points, respectively. The circles divide the plane into several simply connected, bounded regions and one... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 128 | 1,676 |
2014 | T0 | G2 | Geometry | IMO-SL | Let $A B C$ be a triangle. The points $K, L$, and $M$ lie on the segments $B C, C A$, and $A B$, respectively, such that the lines $A K, B L$, and $C M$ intersect in a common point. Prove that it is possible to choose two of the triangles $A L M, B M K$, and $C K L$ whose inradii sum up to at least the inradius of the ... | Denote $$ a=\frac{B K}{K C}, \quad b=\frac{C L}{L A}, \quad c=\frac{A M}{M B} . $$ By Ceva's theorem, $a b c=1$, so we may, without loss of generality, assume that $a \geqslant 1$. Then at least one of the numbers $b$ or $c$ is not greater than 1 . Therefore at least one of the pairs $(a, b)$, $(b, c)$ has its first co... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 113 | 787 |
2014 | T0 | G4 | Geometry | IMO-SL | Consider a fixed circle $\Gamma$ with three fixed points $A, B$, and $C$ on it. Also, let us fix a real number $\lambda \in(0,1)$. For a variable point $P \notin\{A, B, C\}$ on $\Gamma$, let $M$ be the point on the segment $C P$ such that $C M=\lambda \cdot C P$. Let $Q$ be the second point of intersection of the circu... | Throughout the solution, we denote by $\Varangle(a, b)$ the directed angle between the lines $a$ and $b$. Let $D$ be the point on the segment $A B$ such that $B D=\lambda \cdot B A$. We will show that either $Q=D$, or $\Varangle(D Q, Q B)=\Varangle(A B, B C)$; this would mean that the point $Q$ varies over the constant... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 141 | 763 |
2014 | T0 | G4 | Geometry | IMO-SL | Consider a fixed circle $\Gamma$ with three fixed points $A, B$, and $C$ on it. Also, let us fix a real number $\lambda \in(0,1)$. For a variable point $P \notin\{A, B, C\}$ on $\Gamma$, let $M$ be the point on the segment $C P$ such that $C M=\lambda \cdot C P$. Let $Q$ be the second point of intersection of the circu... | As in the previous solution, we introduce the radical centre $X=A P \cap B C \cap M Q$ of the circles $\omega_{A}, \omega_{B}$, and $\Gamma$. Next, we also notice that the points $A, Q, B$, and $X$ lie on a common circle $\Omega$. If the point $P$ lies on the arc $B A C$ of $\Gamma$, then the point $X$ is outside $\Gam... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 141 | 617 |
2014 | T0 | G4 | Geometry | IMO-SL | Consider a fixed circle $\Gamma$ with three fixed points $A, B$, and $C$ on it. Also, let us fix a real number $\lambda \in(0,1)$. For a variable point $P \notin\{A, B, C\}$ on $\Gamma$, let $M$ be the point on the segment $C P$ such that $C M=\lambda \cdot C P$. Let $Q$ be the second point of intersection of the circu... | Let us perform an inversion centred at $C$. Denote by $X^{\prime}$ the image of a point $X$ under this inversion. The circle $\Gamma$ maps to the line $\Gamma^{\prime}$ passing through the constant points $A^{\prime}$ and $B^{\prime}$, and containing the variable point $P^{\prime}$. By the problem condition, the point ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 141 | 827 |
2014 | T0 | G5 | Geometry | IMO-SL | Let $A B C D$ be a convex quadrilateral with $\angle B=\angle D=90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $B D$. The points $S$ and $T$ are chosen on the sides $A B$ and $A D$, respectively, in such a way that $H$ lies inside triangle $S C T$ and $$ \angle S H C-\angle B S C=90^{\circ}, \quad... | Let the line passing through $C$ and perpendicular to the line $S C$ intersect the line $A B$ at $Q$ (see Figure 1). Then $$ \angle S Q C=90^{\circ}-\angle B S C=180^{\circ}-\angle S H C, $$ which implies that the points $C, H, S$, and $Q$ lie on a common circle. Moreover, since $S Q$ is a diameter of this circle, we i... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 154 | 1,001 |
2014 | T0 | G6 | Geometry | IMO-SL | Let $A B C$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $A C$ and $A B$, respectively, and let $M$ be the midpoint of $E F$. Let the perpendicular bisector of $E F$ intersect the line $B C$ at $K$, and let the perpendicular bisector of $M K$ intersect the lines $A C$ and $A B$ ... | For any interesting pair $(E, F)$, we will say that the corresponding triangle $E F K$ is also interesting. Let $E F K$ be an interesting triangle. Firstly, we prove that $\angle K E F=\angle K F E=\angle A$, which also means that the circumcircle $\omega_{1}$ of the triangle $A E F$ is tangent to the lines $K E$ and $... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 200 | 2,876 |
2014 | T0 | G6 | Geometry | IMO-SL | Let $A B C$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $A C$ and $A B$, respectively, and let $M$ be the midpoint of $E F$. Let the perpendicular bisector of $E F$ intersect the line $B C$ at $K$, and let the perpendicular bisector of $M K$ intersect the lines $A C$ and $A B$ ... | Let $(E, F)$ be an interesting pair. This time we prove that $$ \frac{A M}{A K}=\cos \angle A $$ As in Solution 1, we introduce the circle $\omega$ passing through the points $K, S$, $A$, and $T$, together with the points $N$ and $L$ at which the line $A M$ intersect the line $S T$ and the circle $\omega$ for the secon... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 200 | 1,023 |
2014 | T0 | G7 | Geometry | IMO-SL | Let $A B C$ be a triangle with circumcircle $\Omega$ and incentre $I$. Let the line passing through $I$ and perpendicular to $C I$ intersect the segment $B C$ and the $\operatorname{arc} B C($ not containing $A)$ of $\Omega$ at points $U$ and $V$, respectively. Let the line passing through $U$ and parallel to $A I$ int... | We start with some general observations. Set $\alpha=\angle A / 2, \beta=\angle B / 2, \gamma=\angle C / 2$. Then obviously $\alpha+\beta+\gamma=90^{\circ}$. Since $\angle U I C=90^{\circ}$, we obtain $\angle I U C=\alpha+\beta$. Therefore $\angle B I V=\angle I U C-\angle I B C=\alpha=\angle B A I=\angle B Y V$, which... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 183 | 777 |
2014 | T0 | G7 | Geometry | IMO-SL | Let $A B C$ be a triangle with circumcircle $\Omega$ and incentre $I$. Let the line passing through $I$ and perpendicular to $C I$ intersect the segment $B C$ and the $\operatorname{arc} B C($ not containing $A)$ of $\Omega$ at points $U$ and $V$, respectively. Let the line passing through $U$ and parallel to $A I$ int... | As in $$ \angle B A V=\angle C A E $$ Proof. Let $\rho$ be the composition of the inversion with centre $A$ and radius $\sqrt{A B \cdot A C}$, and the symmetry with respect to $A I$. Clearly, $\rho$ interchanges $B$ and $C$. Let $J$ be the excentre of the triangle $A B C$ opposite to $A$ (see Figure 2). Then we have $\... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 183 | 1,909 |
2014 | T0 | N1 | Number Theory | IMO-SL | Let $n \geqslant 2$ be an integer, and let $A_{n}$ be the set $$ A_{n}=\left\{2^{n}-2^{k} \mid k \in \mathbb{Z}, 0 \leqslant k<n\right\} . $$ Determine the largest positive integer that cannot be written as the sum of one or more (not necessarily distinct) elements of $A_{n}$. (Serbia) | Part I. First we show that every integer greater than $(n-2) 2^{n}+1$ can be represented as such a sum. This is achieved by induction on $n$. For $n=2$, the set $A_{n}$ consists of the two elements 2 and 3 . Every positive integer $m$ except for 1 can be represented as the sum of elements of $A_{n}$ in this case: as $m... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 103 | 1,337 |
2014 | T0 | N1 | Number Theory | IMO-SL | Let $n \geqslant 2$ be an integer, and let $A_{n}$ be the set $$ A_{n}=\left\{2^{n}-2^{k} \mid k \in \mathbb{Z}, 0 \leqslant k<n\right\} . $$ Determine the largest positive integer that cannot be written as the sum of one or more (not necessarily distinct) elements of $A_{n}$. (Serbia) | The fact that $m=(n-2) 2^{n}+1$ cannot be represented as a sum of elements of $A_{n}$ can also be shown in other ways. We prove the following statement by induction on $n$ : Claim. If $a, b$ are integers with $a \geqslant 0, b \geqslant 1$, and $a+b<n$, then $a 2^{n}+b$ cannot be written as a sum of elements of $A_{n}$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 103 | 744 |
2014 | T0 | N1 | Number Theory | IMO-SL | Let $n \geqslant 2$ be an integer, and let $A_{n}$ be the set $$ A_{n}=\left\{2^{n}-2^{k} \mid k \in \mathbb{Z}, 0 \leqslant k<n\right\} . $$ Determine the largest positive integer that cannot be written as the sum of one or more (not necessarily distinct) elements of $A_{n}$. (Serbia) | Denote by $B_{n}$ the set of all positive integers that can be written as a sum of elements of $A_{n}$. In this solution, we explicitly describe all the numbers in $B_{n}$ by an argument similar to the first solution. For a positive integer $n$, we denote by $\sigma_{2}(n)$ the sum of its digits in the binary represent... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 103 | 1,224 |
2014 | T0 | N3 | Number Theory | IMO-SL | A coin is called a Cape Town coin if its value is $1 / n$ for some positive integer $n$. Given a collection of Cape Town coins of total value at most $99+\frac{1}{2}$, prove that it is possible to split this collection into at most 100 groups each of total value at most 1. (Luxembourg) | We will show that for every positive integer $N$ any collection of Cape Town coins of total value at most $N-\frac{1}{2}$ can be split into $N$ groups each of total value at most 1 . The problem statement is a particular case for $N=100$. We start with some preparations. If several given coins together have a total val... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 77 | 730 |
2014 | T0 | N4 | Number Theory | IMO-SL | Let $n>1$ be a given integer. Prove that infinitely many terms of the sequence $\left(a_{k}\right)_{k \geqslant 1}$, defined by $$ a_{k}=\left\lfloor\frac{n^{k}}{k}\right\rfloor $$ are odd. (For a real number $x,\lfloor x\rfloor$ denotes the largest integer not exceeding $x$.) (Hong Kong) | If $n$ is odd, let $k=n^{m}$ for $m=1,2, \ldots$. Then $a_{k}=n^{n^{m}-m}$, which is odd for each $m$. Henceforth, assume that $n$ is even, say $n=2 t$ for some integer $t \geqslant 1$. Then, for any $m \geqslant 2$, the integer $n^{2^{m}}-2^{m}=2^{m}\left(2^{2^{m}-m} \cdot t^{2^{m}}-1\right)$ has an odd prime divisor ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 99 | 505 |
2014 | T0 | N4 | Number Theory | IMO-SL | Let $n>1$ be a given integer. Prove that infinitely many terms of the sequence $\left(a_{k}\right)_{k \geqslant 1}$, defined by $$ a_{k}=\left\lfloor\frac{n^{k}}{k}\right\rfloor $$ are odd. (For a real number $x,\lfloor x\rfloor$ denotes the largest integer not exceeding $x$.) (Hong Kong) | Treat the (trivial) case when $n$ is odd as in Now assume that $n$ is even and $n>2$. Let $p$ be a prime divisor of $n-1$. Proceed by induction on $i$ to prove that $p^{i+1}$ is a divisor of $n^{p^{i}}-1$ for every $i \geqslant 0$. The case $i=0$ is true by the way in which $p$ is chosen. Suppose the result is true for... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 99 | 920 |
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) | 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
} | 61 | 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) | 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
} | 61 | 1,433 |
2014 | T0 | N6 | Number Theory | IMO-SL | Let $a_{1}<a_{2}<\cdots<a_{n}$ be pairwise coprime positive integers with $a_{1}$ being prime and $a_{1} \geqslant n+2$. On the segment $I=\left[0, a_{1} a_{2} \cdots a_{n}\right]$ of the real line, mark all integers that are divisible by at least one of the numbers $a_{1}, \ldots, a_{n}$. These points split $I$ into a... | Let $A=a_{1} \cdots a_{n}$. Throughout the solution, all intervals will be nonempty and have integer end-points. For any interval $X$, the length of $X$ will be denoted by $|X|$. Define the following two families of intervals: $$ \begin{aligned} \mathcal{S} & =\{[x, y]: x<y \text { are consecutive marked points }\} \\ ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 146 | 1,807 |
2014 | T0 | N6 | Number Theory | IMO-SL | Let $a_{1}<a_{2}<\cdots<a_{n}$ be pairwise coprime positive integers with $a_{1}$ being prime and $a_{1} \geqslant n+2$. On the segment $I=\left[0, a_{1} a_{2} \cdots a_{n}\right]$ of the real line, mark all integers that are divisible by at least one of the numbers $a_{1}, \ldots, a_{n}$. These points split $I$ into a... | The conventions from the first paragraph of the first solution are still in force. We shall prove the following more general statement: ( $\boxplus$ ) Let $p$ denote a prime number, let $p=a_{1}<a_{2}<\cdots<a_{n}$ be $n$ pairwise coprime positive integers, and let $d$ be an integer with $1 \leqslant d \leqslant p-n$. ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 146 | 1,398 |
2014 | T0 | N7 | Number Theory | IMO-SL | Let $c \geqslant 1$ be an integer. Define a sequence of positive integers by $a_{1}=c$ and $$ a_{n+1}=a_{n}^{3}-4 c \cdot a_{n}^{2}+5 c^{2} \cdot a_{n}+c $$ for all $n \geqslant 1$. Prove that for each integer $n \geqslant 2$ there exists a prime number $p$ dividing $a_{n}$ but none of the numbers $a_{1}, \ldots, a_{... | Let us define $x_{0}=0$ and $x_{n}=a_{n} / c$ for all integers $n \geqslant 1$. It is easy to see that the sequence $\left(x_{n}\right)$ thus obtained obeys the recursive law $$ x_{n+1}=c^{2}\left(x_{n}^{3}-4 x_{n}^{2}+5 x_{n}\right)+1 $$ for all integers $n \geqslant 0$. In particular, all of its terms are positive in... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 139 | 1,718 |
2014 | T0 | N8 | Number Theory | IMO-SL | For every real number $x$, let $\|x\|$ denote the distance between $x$ and the nearest integer. Prove that for every pair $(a, b)$ of positive integers there exist an odd prime $p$ and a positive integer $k$ satisfying $$ \left\|\frac{a}{p^{k}}\right\|+\left\|\frac{b}{p^{k}}\right\|+\left\|\frac{a+b}{p^{k}}\right\|=1 ... | Notice first that $\left\lfloor x+\frac{1}{2}\right\rfloor$ is an integer nearest to $x$, so $\|x\|=\left\lfloor\left.\left\lfloor x+\frac{1}{2}\right\rfloor-x \right\rvert\,\right.$. Thus we have $$ \left\lfloor x+\frac{1}{2}\right\rfloor=x \pm\|x\| . $$ For every rational number $r$ and every prime number $p$, denote... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 114 | 2,190 |
2014 | T0 | A1 | Algebra | IMO-SL | Let $z_{0}<z_{1}<z_{2}<\cdots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n \geqslant 1$ such that $$ z_{n}<\frac{z_{0}+z_{1}+\cdots+z_{n}}{n} \leqslant z_{n+1} . $$ (Austria) | For $n=1,2, \ldots$ define $$ d_{n}=\left(z_{0}+z_{1}+\cdots+z_{n}\right)-n z_{n} $$ The sign of $d_{n}$ indicates whether the first inequality in (1) holds; i.e., it is satisfied if and only if $d_{n}>0$. Notice that $$ n z_{n+1}-\left(z_{0}+z_{1}+\cdots+z_{n}\right)=(n+1) z_{n+1}-\left(z_{0}+z_{1}+\cdots+z_{n}+z_{n+1... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 92 | 556 |
2014 | T0 | A3 | Algebra | IMO-SL | For a sequence $x_{1}, x_{2}, \ldots, x_{n}$ of real numbers, we define its price as $$ \max _{1 \leqslant i \leqslant n}\left|x_{1}+\cdots+x_{i}\right| $$ Given $n$ real numbers, Dave and George want to arrange them into a sequence with a low price. Diligent Dave checks all possible ways and finds the minimum possib... | If the initial numbers are $1,-1,2$, and -2 , then Dave may arrange them as $1,-2,2,-1$, while George may get the sequence $1,-1,2,-2$, resulting in $D=1$ and $G=2$. So we obtain $c \geqslant 2$. Therefore, it remains to prove that $G \leqslant 2 D$. Let $x_{1}, x_{2}, \ldots, x_{n}$ be the numbers Dave and George have... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 317 | 1,122 |
2014 | T0 | A4 | Algebra | IMO-SL | Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfying $$ f(f(m)+n)+f(m)=f(n)+f(3 m)+2014 $$ for all integers $m$ and $n$. (Netherlands) Answer. There is only one such function, namely $n \longmapsto 2 n+1007$. | Let $f$ be a function satisfying (1). Set $C=1007$ and define the function $g: \mathbb{Z} \rightarrow \mathbb{Z}$ by $g(m)=f(3 m)-f(m)+2 C$ for all $m \in \mathbb{Z}$; in particular, $g(0)=2 C$. Now (1) rewrites as $$ f(f(m)+n)=g(m)+f(n) $$ for all $m, n \in \mathbb{Z}$. By induction in both directions it follows that ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 86 | 1,890 |
2014 | T0 | A5 | Algebra | IMO-SL | Consider all polynomials $P(x)$ with real coefficients that have the following property: for any two real numbers $x$ and $y$ one has $$ \left|y^{2}-P(x)\right| \leqslant 2|x| \quad \text { if and only if } \quad\left|x^{2}-P(y)\right| \leqslant 2|y| $$ Determine all possible values of $P(0)$. (Belgium) Answer. The s... | Part I. We begin by verifying that these numbers are indeed possible values of $P(0)$. To see that each negative real number $-C$ can be $P(0)$, it suffices to check that for every $C>0$ the polynomial $P(x)=-\left(\frac{2 x^{2}}{C}+C\right)$ has the property described in the statement of the problem. Due to symmetry i... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 135 | 2,796 |
2014 | T0 | A6 | Algebra | IMO-SL | Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $$ n^{2}+4 f(n)=f(f(n))^{2} $$ for all $n \in \mathbb{Z}$. (United Kingdom) Answer. The possibilities are: - $f(n)=n+1$ for all $n$; - or, for some $a \geqslant 1, \quad f(n)= \begin{cases}n+1, & n>-a, \\ -n+1, & n \leqslant-a ;\end{cases}$ - or $f(... | Part I. Let us first check that each of the functions above really satisfies the given functional equation. If $f(n)=n+1$ for all $n$, then we have $$ n^{2}+4 f(n)=n^{2}+4 n+4=(n+2)^{2}=f(n+1)^{2}=f(f(n))^{2} . $$ If $f(n)=n+1$ for $n>-a$ and $f(n)=-n+1$ otherwise, then we have the same identity for $n>-a$ and $$ n^{2}... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 180 | 2,959 |
2014 | T0 | A6 | Algebra | IMO-SL | Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $$ n^{2}+4 f(n)=f(f(n))^{2} $$ for all $n \in \mathbb{Z}$. (United Kingdom) Answer. The possibilities are: - $f(n)=n+1$ for all $n$; - or, for some $a \geqslant 1, \quad f(n)= \begin{cases}n+1, & n>-a, \\ -n+1, & n \leqslant-a ;\end{cases}$ - or $f(... | Let us provide an alternative proof for Part II, which also proceeds in several steps. Step 1. Let $a$ be an arbitrary integer and $b=f(a)$. We first concentrate on the case where $|a|$ is sufficiently large. 1. If $b=0$, then (1) applied to $a$ yields $a^{2}=f(f(a))^{2}$, thus $$ f(a)=0 \quad \Rightarrow \quad a= \pm ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 180 | 3,445 |
2014 | T0 | C3 | Combinatorics | IMO-SL | Let $n \geqslant 2$ be an integer. Consider an $n \times n$ chessboard divided into $n^{2}$ unit squares. We call a configuration of $n$ rooks on this board happy if every row and every column contains exactly one rook. Find the greatest positive integer $k$ such that for every happy configuration of rooks, we can find... | Let $\ell$ be a positive integer. We will show that (i) if $n>\ell^{2}$ then each happy configuration contains an empty $\ell \times \ell$ square, but (ii) if $n \leqslant \ell^{2}$ then there exists a happy configuration not containing such a square. These two statements together yield the answer. (i). Assume that $n>... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 125 | 1,134 |
2014 | T0 | C4 | Combinatorics | IMO-SL | Construct a tetromino by attaching two $2 \times 1$ dominoes along their longer sides such that the midpoint of the longer side of one domino is a corner of the other domino. This construction yields two kinds of tetrominoes with opposite orientations. Let us call them Sand Z-tetrominoes, respectively. ^{j}$ into the squ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 204 | 802 |
2014 | T0 | C5 | Combinatorics | IMO-SL | Consider $n \geqslant 3$ lines in the plane such that no two lines are parallel and no three have a common point. These lines divide the plane into polygonal regions; let $\mathcal{F}$ be the set of regions having finite area. Prove that it is possible to colour $\lceil\sqrt{n / 2}\rceil$ of the lines blue in such a wa... | Let $L$ be the given set of lines. Choose a maximal (by inclusion) subset $B \subseteq L$ such that when we colour the lines of $B$ blue, no region in $\mathcal{F}$ has a completely blue boundary. Let $|B|=k$. We claim that $k \geqslant\lceil\sqrt{n / 2}\rceil$. Let us colour all the lines of $L \backslash B$ red. Call... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 106 | 1,293 |
2014 | T0 | C6 | Combinatorics | IMO-SL | We are given an infinite deck of cards, each with a real number on it. For every real number $x$, there is exactly one card in the deck that has $x$ written on it. Now two players draw disjoint sets $A$ and $B$ of 100 cards each from this deck. We would like to define a rule that declares one of them a winner. This rul... | We prove a more general statement for sets of cardinality $n$ (the problem being the special case $n=100$, then the answer is $n$ ). In the following, we write $A>B$ or $B<A$ for " $A$ beats $B$ ". Part I. Let us first define $n$ different rules that satisfy the conditions. To this end, fix an index $k \in\{1,2, \ldots... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 354 | 933 |
2014 | T0 | C6 | Combinatorics | IMO-SL | We are given an infinite deck of cards, each with a real number on it. For every real number $x$, there is exactly one card in the deck that has $x$ written on it. Now two players draw disjoint sets $A$ and $B$ of 100 cards each from this deck. We would like to define a rule that declares one of them a winner. This rul... | Another possible approach to Part II of this problem is induction on $n$. For $n=1$, there is trivially only one rule in view of the second condition. In the following, we assume that our claim (namely, that there are no possible rules other than those given in Part I) holds for $n-1$ in place of $n$. We start with the... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 354 | 2,319 |
2014 | T0 | C8 | Combinatorics | IMO-SL | A card deck consists of 1024 cards. On each card, a set of distinct decimal digits is written in such a way that no two of these sets coincide (thus, one of the cards is empty). Two players alternately take cards from the deck, one card per turn. After the deck is empty, each player checks if he can throw out one of hi... | Let us identify each card with the set of digits written on it. For any collection of cards $C_{1}, C_{2}, \ldots, C_{k}$ denote by their sum the set $C_{1} \triangle C_{2} \triangle \cdots \triangle C_{k}$ consisting of all elements belonging to an odd number of the $C_{i}$ 's. Denote the first and the second player b... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 154 | 1,963 |
2014 | T0 | C9 | Combinatorics | IMO-SL | There are $n$ circles drawn on a piece of paper in such a way that any two circles intersect in two points, and no three circles pass through the same point. Turbo the snail slides along the circles in the following fashion. Initially he moves on one of the circles in clockwise direction. Turbo always keeps sliding alo... | Replace every cross (i.e. intersection of two circles) by two small circle arcs that indicate the direction in which the snail should leave the cross (see Figure 1.1). Notice that the placement of the small arcs does not depend on the direction of moving on the curves; no matter which direction the snail is moving on t... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 131 | 1,181 |
2014 | T0 | C9 | Combinatorics | IMO-SL | There are $n$ circles drawn on a piece of paper in such a way that any two circles intersect in two points, and no three circles pass through the same point. Turbo the snail slides along the circles in the following fashion. Initially he moves on one of the circles in clockwise direction. Turbo always keeps sliding alo... | We present a different proof of (*). We perform a sequence of small modification steps on the configuration of the circles in such a way that at the end they have no intersection at all (see Figure 6.1). We use two kinds of local changes to the structure of the orbits (see Figure 6.2): - Type-1 step: An arc of a circle... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 131 | 966 |
2014 | T0 | C9 | Combinatorics | IMO-SL | There are $n$ circles drawn on a piece of paper in such a way that any two circles intersect in two points, and no three circles pass through the same point. Turbo the snail slides along the circles in the following fashion. Initially he moves on one of the circles in clockwise direction. Turbo always keeps sliding alo... | Like in the previous solutions, we do not need all circle pairs to intersect but we assume that the circles form a connected set. Denote by $\mathcal{C}$ and $\mathcal{P}$ the sets of circles and their intersection points, respectively. The circles divide the plane into several simply connected, bounded regions and one... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 131 | 1,676 |
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 $A B$ ... | For any interesting pair $(E, F)$, we will say that the corresponding triangle $E F K$ is also interesting. Let $E F K$ be an interesting triangle. Firstly, we prove that $\angle K E F=\angle K F E=\angle A$, which also means that the circumcircle $\omega_{1}$ of the triangle $A E F$ is tangent to the lines $K E$ and $... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2014SL.jsonl",
"solution_match": null
} | 203 | 2,876 |
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