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 |
|---|---|---|---|---|---|---|---|---|---|
2022 | T0 | A2 | Algebra | IMO-SL | Let $k \geqslant 2$ be an integer. Find the smallest integer $n \geqslant k+1$ with the property that there exists a set of $n$ distinct real numbers such that each of its elements can be written as a sum of $k$ other distinct elements of the set. (Slovakia) | First we show that $n \geqslant k+4$. Suppose that there exists such a set with $n$ numbers and denote them by $a_{1}<a_{2}<\cdots<a_{n}$. Note that in order to express $a_{1}$ as a sum of $k$ distinct elements of the set, we must have $a_{1} \geqslant a_{2}+\cdots+a_{k+1}$ and, similarly for $a_{n}$, we must have $a_{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 73 | 759 |
2022 | T0 | A3 | Algebra | IMO-SL | Let $\mathbb{R}_{>0}$ be the set of positive real numbers. Find all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that, for every $x \in \mathbb{R}_{>0}$, there exists a unique $y \in \mathbb{R}_{>0}$ satisfying $$ x f(y)+y f(x) \leqslant 2 $$ (Netherlands) | First we prove that the function $f(x)=1 / x$ satisfies the condition of the problem statement. The AM-GM inequality gives $$ \frac{x}{y}+\frac{y}{x} \geqslant 2 $$ for every $x, y>0$, with equality if and only if $x=y$. This means that, for every $x>0$, there exists a unique $y>0$ such that $$ \frac{x}{y}+\frac{y}{x} ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 104 | 712 |
2022 | T0 | A3 | Algebra | IMO-SL | Let $\mathbb{R}_{>0}$ be the set of positive real numbers. Find all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that, for every $x \in \mathbb{R}_{>0}$, there exists a unique $y \in \mathbb{R}_{>0}$ satisfying $$ x f(y)+y f(x) \leqslant 2 $$ (Netherlands) | 0. As in the other solutions we verify that the function $f(x)=1 / x$ is a solution. We first want to prove the following lemma: Lemma 3. For all $x \in \mathbb{R}_{>0}$ we actually have $x f(g(x))+g(x) f(x)=2$ (that is: the inequality is actually an equality). Proof. We proceed by contradiction: Assume there exists so... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 104 | 840 |
2022 | T0 | A3 | Algebra | IMO-SL | Let $\mathbb{R}_{>0}$ be the set of positive real numbers. Find all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that, for every $x \in \mathbb{R}_{>0}$, there exists a unique $y \in \mathbb{R}_{>0}$ satisfying $$ x f(y)+y f(x) \leqslant 2 $$ (Netherlands) | Again, let us prove that $f(x)=1 / x$ is the only solution. Let again $g(x)$ be the unique positive real number such that $(x, g(x))$ is a good pair. Lemma 4. The function $f$ is strictly convex. Proof. Consider the function $q_{s}(x)=f(x)+s x$ for some real number $s$. If $f$ is not strictly convex, then there exist $... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 104 | 506 |
2022 | T0 | A4 | Algebra | IMO-SL | Let $n \geqslant 3$ be an integer, and let $x_{1}, x_{2}, \ldots, x_{n}$ be real numbers in the interval $[0,1]$. Let $s=x_{1}+x_{2}+\ldots+x_{n}$, and assume that $s \geqslant 3$. Prove that there exist integers $i$ and $j$ with $1 \leqslant i<j \leqslant n$ such that $$ 2^{j-i} x_{i} x_{j}>2^{s-3} $$ (Trinidad and ... | Let $1 \leqslant a<b \leqslant n$ be such that $2^{b-a} x_{a} x_{b}$ is maximal. This choice of $a$ and $b$ implies that $x_{a+t} \leqslant 2^{t} x_{a}$ for all $1-a \leqslant t \leqslant b-a-1$, and similarly $x_{b-t} \leqslant 2^{t} x_{b}$ for all $b-n \leqslant t \leqslant b-a+1$. Now, suppose that $x_{a} \in\left(\... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 143 | 1,262 |
2022 | T0 | A5 | Algebra | IMO-SL | Find all positive integers $n \geqslant 2$ for which there exist $n$ real numbers $a_{1}<\cdots<a_{n}$ and a real number $r>0$ such that the $\frac{1}{2} n(n-1)$ differences $a_{j}-a_{i}$ for $1 \leqslant i<j \leqslant n$ are equal, in some order, to the numbers $r^{1}, r^{2}, \ldots, r^{\frac{1}{2} n(n-1)}$. (Czech Re... | We first show a solution for each $n \in\{2,3,4\}$. We will later show the impossibility of finding such a solution for $n \geqslant 5$. For $n=2$, take for example $\left(a_{1}, a_{2}\right)=(1,3)$ and $r=2$. For $n=3$, take the root $r>1$ of $x^{2}-x-1=0$ (the golden ratio) and set $\left(a_{1}, a_{2}, a_{3}\right)=$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 129 | 2,131 |
2022 | T0 | A6 | Algebra | IMO-SL | Let $\mathbb{R}$ be the set of real numbers. We denote by $\mathcal{F}$ the set of all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$ f(x+f(y))=f(x)+f(y) $$ for every $x, y \in \mathbb{R}$. Find all rational numbers $q$ such that for every function $f \in \mathcal{F}$, there exists some $z \in \mathbb{... | Let $Z$ be the set of all rational numbers $q$ such that for every function $f \in \mathcal{F}$, there exists some $z \in \mathbb{R}$ satisfying $f(z)=q z$. Let further $$ S=\left\{\frac{n+1}{n}: n \in \mathbb{Z}, n \neq 0\right\} $$ We prove that $Z=S$ by showing the two inclusions: $S \subseteq Z$ and $Z \subseteq S$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 126 | 1,090 |
2022 | T0 | A7 | Algebra | IMO-SL | For a positive integer $n$ we denote by $s(n)$ the sum of the digits of $n$. Let $P(x)=$ $x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$ be a polynomial, where $n \geqslant 2$ and $a_{i}$ is a positive integer for all $0 \leqslant i \leqslant n-1$. Could it be the case that, for all positive integers $k, s(k)$ and $s(P(k)... | With the notation above, we begin by choosing a positive integer $t$ such that $$ 10^{t}>\max \left\{\frac{100^{n-1} a_{n-1}}{\left(10^{\frac{1}{n-1}}-9^{\frac{1}{n-1}}\right)^{n-1}}, \frac{a_{n-1}}{9} 10^{n-1}, \frac{a_{n-1}}{9}\left(10 a_{n-1}\right)^{n-1}, \ldots, \frac{a_{n-1}}{9}\left(10 a_{0}\right)^{n-1}\right\}... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 136 | 1,482 |
2022 | T0 | A8 | Algebra | IMO-SL | For a positive integer $n$, an $n$-sequence is a sequence $\left(a_{0}, \ldots, a_{n}\right)$ of non-negative integers satisfying the following condition: if $i$ and $j$ are non-negative integers with $i+j \leqslant n$, then $a_{i}+a_{j} \leqslant n$ and $a_{a_{i}+a_{j}}=a_{i+j}$. Let $f(n)$ be the number of $n$-seque... | In order to solve this, we will give a complete classification of $n$-sequences. Let $k=\lfloor n / 2\rfloor$. We will say that an $n$-sequence is large if $a_{i}>k$ for some $i$, and small if no such $i$ exists. For now we will assume that $\left(a_{i}\right)$ is not the identity sequence (in other words, that $a_{i} ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 173 | 3,522 |
2022 | T0 | C1 | Combinatorics | IMO-SL | A $\pm 1$-sequence is a sequence of 2022 numbers $a_{1}, \ldots, a_{2022}$, each equal to either +1 or -1 . Determine the largest $C$ so that, for any $\pm 1$-sequence, there exists an integer $k$ and indices $1 \leqslant t_{1}<\ldots<t_{k} \leqslant 2022$ so that $t_{i+1}-t_{i} \leqslant 2$ for all $i$, and $$ \left|... | First, we prove that this can always be achieved. Without loss of generality, suppose at least $\frac{2022}{2}=1011$ terms of the $\pm 1$-sequence are +1 . Define a subsequence as follows: starting at $t=0$, if $a_{t}=+1$ we always include $a_{t}$ in the subsequence. Otherwise, we skip $a_{t}$ if we can (i.e. if we inc... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 165 | 818 |
2022 | T0 | C2 | Combinatorics | IMO-SL | The Bank of Oslo issues coins made out of two types of metal: aluminium (denoted A) and copper (denoted $C$ ). Morgane has $n$ aluminium coins, and $n$ copper coins, and arranges her $2 n$ coins in a row in some arbitrary initial order. Given a fixed positive integer $k \leqslant 2 n$, she repeatedly performs the follo... | Define a block to be a maximal subsequence of consecutive coins made out of the same metal, and let $M^{b}$ denote a block of $b$ coins of metal $M$. The property that there is at most one aluminium coin adjacent to a copper coin is clearly equivalent to the configuration having two blocks, one consisting of all $A$-s ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 265 | 846 |
2022 | T0 | C3 | Combinatorics | IMO-SL | In each square of a garden shaped like a $2022 \times 2022$ board, there is initially a tree of height 0 . A gardener and a lumberjack alternate turns playing the following game, with the gardener taking the first turn: - The gardener chooses a square in the garden. Each tree on that square and all the surrounding squ... | We solve the problem for a general $3 N \times 3 N$ board. First, we prove that the lumberjack has a strategy to ensure there are never more than $5 N^{2}$ majestic trees. Giving the squares of the board coordinates in the natural manner, colour each square where at least one of its coordinates are divisible by 3 , sho... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 181 | 796 |
2022 | T0 | C4 | Combinatorics | IMO-SL | Let $n>3$ be a positive integer. Suppose that $n$ children are arranged in a circle, and $n$ coins are distributed between them (some children may have no coins). At every step, a child with at least 2 coins may give 1 coin to each of their immediate neighbours on the right and left. Determine all initial distributions... | Number the children $1, \ldots, n$, and denote the number of coins the $i$-th child has by $c_{i}$. A step of this process consists of reducing some $c_{i}$ by 2 , and increasing $c_{i-1}, c_{i+1}$ by 1 . (Indices are considered $(\bmod n)$.) Because $(i-1)-2 i+(i+1)=0$, the quantity $\sum_{i=1}^{n} i c_{i}(\bmod n)$ w... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 96 | 1,964 |
2022 | T0 | C5 | Combinatorics | IMO-SL | Let $m, n \geqslant 2$ be integers, let $X$ be a set with $n$ elements, and let $X_{1}, X_{2}, \ldots, X_{m}$ be pairwise distinct non-empty, not necessary disjoint subsets of $X$. A function $f: X \rightarrow$ $\{1,2, \ldots, n+1\}$ is called nice if there exists an index $k$ such that $$ \sum_{x \in X_{k}} f(x)>\sum... | For a subset $Y \subseteq X$, we write $f(Y)$ for $\sum_{y \in Y} f(y)$. Note that a function $f: X \rightarrow$ $\{1, \ldots, n+1\}$ is nice, if and only if $f\left(X_{i}\right)$ is maximized by a unique index $i \in\{1, \ldots, m\}$. We will first investigate the set $\mathcal{F}$ of functions $f: X \rightarrow\{1, \... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 160 | 731 |
2022 | T0 | C6 | Combinatorics | IMO-SL | Let $n$ be a positive integer. We start with $n$ piles of pebbles, each initially containing a single pebble. One can perform moves of the following form: choose two piles, take an equal number of pebbles from each pile and form a new pile out of these pebbles. For each positive integer $n$, find the smallest number of... | The solution we describe is simple, but not the most effective one. We can combine two piles of $2^{k-1}$ pebbles to make one pile of $2^{k}$ pebbles. In particular, given $2^{k}$ piles of one pebble, we may combine them as follows: $$ \begin{array}{lcc} 2^{k} \text { piles of } 1 \text { pebble } & \rightarrow & 2^{k-... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 100 | 1,247 |
2022 | T0 | C6 | Combinatorics | IMO-SL | Let $n$ be a positive integer. We start with $n$ piles of pebbles, each initially containing a single pebble. One can perform moves of the following form: choose two piles, take an equal number of pebbles from each pile and form a new pile out of these pebbles. For each positive integer $n$, find the smallest number of... | Throughout the solution, we will consider the moves in reverse order. Namely, imagine we have some piles of pebbles, and we are allowed to perform moves as follows: take a pile with an even number of pebbles, split it into two equal halves and add the pebbles from each half to a different pile, possibly forming new pil... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 100 | 2,230 |
2022 | T0 | C7 | Combinatorics | IMO-SL | Lucy starts by writing $s$ integer-valued 2022-tuples on a blackboard. After doing that, she can take any two (not necessarily distinct) tuples $\mathbf{v}=\left(v_{1}, \ldots, v_{2022}\right)$ and $\mathbf{w}=\left(w_{1}, \ldots, w_{2022}\right)$ that she has already written, and apply one of the following operations ... | We solve the problem for $n$-tuples for any $n \geqslant 3$ : we will show that the answer is $s=3$, regardless of the value of $n$. First, let us briefly introduce some notation. For an $n$-tuple $\mathbf{v}$, we will write $\mathbf{v}_{i}$ for its $i$-th coordinate (where $1 \leqslant i \leqslant n$ ). For a positive... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 291 | 1,932 |
2022 | T0 | C8 | Combinatorics | IMO-SL | Alice fills the fields of an $n \times n$ board with numbers from 1 to $n^{2}$, each number being used exactly once. She then counts the total number of good paths on the board. A good path is a sequence of fields of arbitrary length (including 1) such that: (i) The first field in the sequence is one that is only adjac... | We will call any field that is only adjacent to fields with larger numbers a well. Other fields will be called non-wells. Let us make a second $n \times n$ board $B$ where in each field we will write the number of good sequences which end on the corresponding field in the original board $A$. We will thus look for the m... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 154 | 898 |
2022 | T0 | C9 | Combinatorics | IMO-SL | Let $\mathbb{Z}_{\geqslant 0}$ be the set of non-negative integers, and let $f: \mathbb{Z}_{\geqslant 0} \times \mathbb{Z}_{\geqslant 0} \rightarrow \mathbb{Z}_{\geqslant 0}$ be a bijection such that whenever $f\left(x_{1}, y_{1}\right)>f\left(x_{2}, y_{2}\right)$, we have $f\left(x_{1}+1, y_{1}\right)>f\left(x_{2}+1, ... | We defer the constructions to the end of the solution. Instead, we begin by characterizing all such functions $f$, prove a formula and key property for such functions, and then solve the problem, providing constructions. Characterization Suppose $f$ satisfies the given relation. The condition can be written more strong... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 235 | 2,143 |
2022 | T0 | G5 | Geometry | IMO-SL | Let $A B C$ be a triangle, and let $\ell_{1}$ and $\ell_{2}$ be two parallel lines. For $i=1,2$, let $\ell_{i}$ meet the lines $B C, C A$, and $A B$ at $X_{i}, Y_{i}$, and $Z_{i}$, respectively. Suppose that the line through $X_{i}$ perpendicular to $B C$, the line through $Y_{i}$ perpendicular to $C A$, and finally th... | Throughout the solutions, $\Varangle(p, q)$ will denote the directed angle between lines $p$ and $q$, taken modulo $180^{\circ}$. Let the vertices of $\Delta_{i}$ be $D_{i}, E_{i}, F_{i}$, such that lines $E_{i} F_{i}, F_{i} D_{i}$ and $D_{i} E_{i}$ are the perpendiculars through $X, Y$ and $Z$, respectively, and denot... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 164 | 1,070 |
2022 | T0 | G5 | Geometry | IMO-SL | Let $A B C$ be a triangle, and let $\ell_{1}$ and $\ell_{2}$ be two parallel lines. For $i=1,2$, let $\ell_{i}$ meet the lines $B C, C A$, and $A B$ at $X_{i}, Y_{i}$, and $Z_{i}$, respectively. Suppose that the line through $X_{i}$ perpendicular to $B C$, the line through $Y_{i}$ perpendicular to $C A$, and finally th... | As in the first solution, let the vertices of $\Delta_{i}$ be $D_{i}, E_{i}, F_{i}$, such that $E_{i} F_{i}, F_{i} D_{i}$ and $D_{i} E_{i}$ are the perpendiculars through $X_{i}, Y_{i}$ and $Z_{i}$, respectively. In the same way we conclude that $\left(A, D_{1}, D_{2}\right),\left(B, E_{1}, E_{2}\right)$ and $\left(C, ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 164 | 1,548 |
2022 | T0 | G6 | Geometry | IMO-SL | In an acute-angled triangle $A B C$, point $H$ is the foot of the altitude from $A$. Let $P$ be a moving point such that the bisectors $k$ and $\ell$ of angles $P B C$ and $P C B$, respectively, intersect each other on the line segment $A H$. Let $k$ and $A C$ meet at $E$, let $\ell$ and $A B$ meet at $F$, and let $E F... | Let the reflections of the line $B C$ with respect to the lines $A B$ and $A C$ intersect at point $K$. We will prove that $P, Q$ and $K$ are collinear, so $K$ is the common point of the varying line $P Q$. Let lines $B E$ and $C F$ intersect at $I$. For every point $O$ and $d>0$, denote by $(O, d)$ the circle centred ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 140 | 1,012 |
2022 | T0 | G6 | Geometry | IMO-SL | In an acute-angled triangle $A B C$, point $H$ is the foot of the altitude from $A$. Let $P$ be a moving point such that the bisectors $k$ and $\ell$ of angles $P B C$ and $P C B$, respectively, intersect each other on the line segment $A H$. Let $k$ and $A C$ meet at $E$, let $\ell$ and $A B$ meet at $F$, and let $E F... | Again, let $B E$ and $C F$ meet at $I$, that is the incentre in triangle $B C P$; then $P I$ is the third angle bisector. From the tangent segments of the incircle we have $B P-C P=$ $B H-C H$; hence, the possible points $P$ lie on a branch of a hyperbola $\mathcal{H}$ with foci $B, C$, and $H$ is a vertex of $\mathcal... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 140 | 545 |
2022 | T0 | G7 | Geometry | IMO-SL | Let $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$ be two triangles having the same circumcircle $\omega$, and the same orthocentre $H$. Let $\Omega$ be the circumcircle of the triangle determined by the lines $A A^{\prime}, B B^{\prime}$ and $C C^{\prime}$. Prove that $H$, the centre of $\omega$, and the centre of $\O... | In what follows, $\Varangle(p, q)$ will denote the directed angle between lines $p$ and $q$, taken modulo $180^{\circ}$. Denote by $O$ the centre of $\omega$. In any triangle, the homothety with ratio $-\frac{1}{2}$ centred at the centroid of the triangle takes the vertices to the midpoints of the opposite sides and it... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 107 | 2,126 |
2022 | T0 | G8 | Geometry | IMO-SL | Let $A A^{\prime} B C C^{\prime} B^{\prime}$ be a convex cyclic hexagon such that $A C$ is tangent to the incircle of the triangle $A^{\prime} B^{\prime} C^{\prime}$, and $A^{\prime} C^{\prime}$ is tangent to the incircle of the triangle $A B C$. Let the lines $A B$ and $A^{\prime} B^{\prime}$ meet at $X$ and let the l... | Denote by $\omega$ and $\omega^{\prime}$ the incircles of $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ and let $I$ and $I^{\prime}$ be the centres of these circles. Let $N$ and $N^{\prime}$ be the second intersections of $B I$ and $B^{\prime} I^{\prime}$ with $\Omega$, the circumcircle of $A^{\pri... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 160 | 1,170 |
2022 | T0 | N1 | Number Theory | IMO-SL | A number is called Norwegian if it has three distinct positive divisors whose sum is equal to 2022. Determine the smallest Norwegian number. (Note: The total number of positive divisors of a Norwegian number is allowed to be larger than 3.) (Cyprus) | Observe that 1344 is a Norwegian number as 6, 672 and 1344 are three distinct divisors of 1344 and $6+672+1344=2022$. It remains to show that this is the smallest such number. Assume for contradiction that $N<1344$ is Norwegian and let $N / a, N / b$ and $N / c$ be the three distinct divisors of $N$, with $a<b<c$. Then... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 57 | 858 |
2022 | T0 | N2 | Number Theory | IMO-SL | Find all positive integers $n>2$ such that $$ n!\mid \prod_{\substack{p<q \leq n, p, q \text { primes }}}(p+q) $$ (Nigeria) | Assume that $n$ satisfies $n!\mid \prod_{p<q \leqslant n}(p+q)$ and let $2=p_{1}<p_{2}<\cdots<p_{m} \leqslant n$ be the primes in $\{1,2, \ldots, n\}$. Each such prime divides $n$ !. In particular, $p_{m} \mid p_{i}+p_{j}$ for some $p_{i}<p_{j} \leqslant n$. But $$ 0<\frac{p_{i}+p_{j}}{p_{m}}<\frac{p_{m}+p_{m}}{p_{m}}=... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 50 | 619 |
2022 | T0 | N3 | Number Theory | IMO-SL | Let $a>1$ be a positive integer, and let $d>1$ be a positive integer coprime to $a$. Let $x_{1}=1$ and, for $k \geqslant 1$, define $$ x_{k+1}= \begin{cases}x_{k}+d & \text { if } a \text { doesn't divide } x_{k} \\ x_{k} / a & \text { if } a \text { divides } x_{k}\end{cases} $$ Find the greatest positive integer $n... | Like in the first solution, $x_{k}$ is relatively prime to $d$ and $x_{k}<a d$ for all $k$. We wish to prove that there are $n$ consecutive decreasing indices. Let $m_{0}=0$ and $m_{k}$ be the $k$-th smallest decreasing index (an index $k$ is decreasing if $x_{k}=x_{k-1} / a$ ) and define a sequence $\left(y_{k}\right)... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 155 | 695 |
2022 | T0 | N5 | Number Theory | IMO-SL | For each $1 \leqslant i \leqslant 9$ and $T \in \mathbb{N}$, define $d_{i}(T)$ to be the total number of times the digit $i$ appears when all the multiples of 1829 between 1 and $T$ inclusive are written out in base 10. Show that there are infinitely many $T \in \mathbb{N}$ such that there are precisely two distinct v... | Let $n:=1829$. First, we choose some $k$ such that $n \mid 10^{k}-1$. For instance, any multiple of $\varphi(n)$ would work since $n$ is coprime to 10 . We will show that either $T=10^{k}-1$ or $T=10^{k}-2$ has the desired property, which completes the proof since $k$ can be taken to be arbitrary large. For this it suf... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 132 | 888 |
2022 | T0 | N6 | Number Theory | IMO-SL | Let $Q$ be a set of prime numbers, not necessarily finite. For a positive integer $n$ consider its prime factorisation; define $p(n)$ to be the sum of all the exponents and $q(n)$ to be the sum of the exponents corresponding only to primes in $Q$. A positive integer $n$ is called special if $p(n)+p(n+1)$ and $q(n)+q(n+... | Let us call two positive integers $m, n$ friends if $p(m)+p(n)$ and $q(m)+q(n)$ are both even integers. We start by noting that the pairs $(p(k), q(k))$ modulo 2 can take at most 4 different values; thus, among any five different positive integers there are two which are friends. In addition, both functions $p$ and $q$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 217 | 1,067 |
2022 | T0 | A2 | Algebra | IMO-SL | Let $k \geqslant 2$ be an integer. Find the smallest integer $n \geqslant k+1$ with the property that there exists a set of $n$ distinct real numbers such that each of its elements can be written as a sum of $k$ other distinct elements of the set. (Slovakia) Answer: $n=k+4$. | First we show that $n \geqslant k+4$. Suppose that there exists such a set with $n$ numbers and denote them by $a_{1}<a_{2}<\cdots<a_{n}$. Note that in order to express $a_{1}$ as a sum of $k$ distinct elements of the set, we must have $a_{1} \geqslant a_{2}+\cdots+a_{k+1}$ and, similarly for $a_{n}$, we must have $a_{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 81 | 759 |
2022 | T0 | A3 | Algebra | IMO-SL | Let $\mathbb{R}_{>0}$ be the set of positive real numbers. Find all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that, for every $x \in \mathbb{R}_{>0}$, there exists a unique $y \in \mathbb{R}_{>0}$ satisfying $$ x f(y)+y f(x) \leqslant 2 . $$ (Netherlands) Answer: The function $f(x)=1 / x$ is the... | First we prove that the function $f(x)=1 / x$ satisfies the condition of the problem statement. The AM-GM inequality gives $$ \frac{x}{y}+\frac{y}{x} \geqslant 2 $$ for every $x, y>0$, with equality if and only if $x=y$. This means that, for every $x>0$, there exists a unique $y>0$ such that $$ \frac{x}{y}+\frac{y}{x} ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 122 | 712 |
2022 | T0 | A3 | Algebra | IMO-SL | Let $\mathbb{R}_{>0}$ be the set of positive real numbers. Find all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that, for every $x \in \mathbb{R}_{>0}$, there exists a unique $y \in \mathbb{R}_{>0}$ satisfying $$ x f(y)+y f(x) \leqslant 2 . $$ (Netherlands) Answer: The function $f(x)=1 / x$ is the... | 0. As in the other solutions we verify that the function $f(x)=1 / x$ is a solution. We first want to prove the following lemma: Lemma 3. For all $x \in \mathbb{R}_{>0}$ we actually have $x f(g(x))+g(x) f(x)=2$ (that is: the inequality is actually an equality). Proof. We proceed by contradiction: Assume there exists so... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 122 | 840 |
2022 | T0 | A3 | Algebra | IMO-SL | Let $\mathbb{R}_{>0}$ be the set of positive real numbers. Find all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that, for every $x \in \mathbb{R}_{>0}$, there exists a unique $y \in \mathbb{R}_{>0}$ satisfying $$ x f(y)+y f(x) \leqslant 2 . $$ (Netherlands) Answer: The function $f(x)=1 / x$ is the... | Again, let us prove that $f(x)=1 / x$ is the only solution. Let again $g(x)$ be the unique positive real number such that $(x, g(x))$ is a good pair. Lemma 4. The function $f$ is strictly convex. Proof. Consider the function $q_{s}(x)=f(x)+s x$ for some real number $s$. If $f$ is not strictly convex, then there exist $... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 122 | 506 |
2022 | T0 | A5 | Algebra | IMO-SL | Find all positive integers $n \geqslant 2$ for which there exist $n$ real numbers $a_{1}<\cdots<a_{n}$ and a real number $r>0$ such that the $\frac{1}{2} n(n-1)$ differences $a_{j}-a_{i}$ for $1 \leqslant i<j \leqslant n$ are equal, in some order, to the numbers $r^{1}, r^{2}, \ldots, r^{\frac{1}{2} n(n-1)}$. (Czech Re... | We first show a solution for each $n \in\{2,3,4\}$. We will later show the impossibility of finding such a solution for $n \geqslant 5$. For $n=2$, take for example $\left(a_{1}, a_{2}\right)=(1,3)$ and $r=2$. For $n=3$, take the root $r>1$ of $x^{2}-x-1=0$ (the golden ratio) and set $\left(a_{1}, a_{2}, a_{3}\right)=$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 145 | 2,131 |
2022 | T0 | A6 | Algebra | IMO-SL | Let $\mathbb{R}$ be the set of real numbers. We denote by $\mathcal{F}$ the set of all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$ f(x+f(y))=f(x)+f(y) $$ for every $x, y \in \mathbb{R}$. Find all rational numbers $q$ such that for every function $f \in \mathcal{F}$, there exists some $z \in \mathbb{... | Let $Z$ be the set of all rational numbers $q$ such that for every function $f \in \mathcal{F}$, there exists some $z \in \mathbb{R}$ satisfying $f(z)=q z$. Let further $$ S=\left\{\frac{n+1}{n}: n \in \mathbb{Z}, n \neq 0\right\} $$ We prove that $Z=S$ by showing the two inclusions: $S \subseteq Z$ and $Z \subseteq S$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 166 | 1,090 |
2022 | T0 | A7 | Algebra | IMO-SL | For a positive integer $n$ we denote by $s(n)$ the sum of the digits of $n$. Let $P(x)=$ $x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$ be a polynomial, where $n \geqslant 2$ and $a_{i}$ is a positive integer for all $0 \leqslant i \leqslant n-1$. Could it be the case that, for all positive integers $k, s(k)$ and $s(P(k)... | With the notation above, we begin by choosing a positive integer $t$ such that $$ 10^{t}>\max \left\{\frac{100^{n-1} a_{n-1}}{\left(10^{\frac{1}{n-1}}-9^{\frac{1}{n-1}}\right)^{n-1}}, \frac{a_{n-1}}{9} 10^{n-1}, \frac{a_{n-1}}{9}\left(10 a_{n-1}\right)^{n-1}, \ldots, \frac{a_{n-1}}{9}\left(10 a_{0}\right)^{n-1}\right\}... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 170 | 1,482 |
2022 | T0 | A8 | Algebra | IMO-SL | For a positive integer $n$, an $n$-sequence is a sequence $\left(a_{0}, \ldots, a_{n}\right)$ of non-negative integers satisfying the following condition: if $i$ and $j$ are non-negative integers with $i+j \leqslant n$, then $a_{i}+a_{j} \leqslant n$ and $a_{a_{i}+a_{j}}=a_{i+j}$. Let $f(n)$ be the number of $n$-seque... | In order to solve this, we will give a complete classification of $n$-sequences. Let $k=\lfloor n / 2\rfloor$. We will say that an $n$-sequence is large if $a_{i}>k$ for some $i$, and small if no such $i$ exists. For now we will assume that $\left(a_{i}\right)$ is not the identity sequence (in other words, that $a_{i} ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 215 | 3,522 |
2022 | T0 | C1 | Combinatorics | IMO-SL | A $\pm 1$-sequence is a sequence of 2022 numbers $a_{1}, \ldots, a_{2022}$, each equal to either +1 or -1 . Determine the largest $C$ so that, for any $\pm 1$-sequence, there exists an integer $k$ and indices $1 \leqslant t_{1}<\ldots<t_{k} \leqslant 2022$ so that $t_{i+1}-t_{i} \leqslant 2$ for all $i$, and $$ \left|... | First, we prove that this can always be achieved. Without loss of generality, suppose at least $\frac{2022}{2}=1011$ terms of the $\pm 1$-sequence are +1 . Define a subsequence as follows: starting at $t=0$, if $a_{t}=+1$ we always include $a_{t}$ in the subsequence. Otherwise, we skip $a_{t}$ if we can (i.e. if we inc... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 177 | 818 |
2022 | T0 | C2 | Combinatorics | IMO-SL | The Bank of Oslo issues coins made out of two types of metal: aluminium (denoted $A$ ) and copper (denoted $C$ ). Morgane has $n$ aluminium coins, and $n$ copper coins, and arranges her $2 n$ coins in a row in some arbitrary initial order. Given a fixed positive integer $k \leqslant 2 n$, she repeatedly performs the fo... | Define a block to be a maximal subsequence of consecutive coins made out of the same metal, and let $M^{b}$ denote a block of $b$ coins of metal $M$. The property that there is at most one aluminium coin adjacent to a copper coin is clearly equivalent to the configuration having two blocks, one consisting of all $A$-s ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 303 | 846 |
2022 | T0 | C3 | Combinatorics | IMO-SL | In each square of a garden shaped like a $2022 \times 2022$ board, there is initially a tree of height 0 . A gardener and a lumberjack alternate turns playing the following game, with the gardener taking the first turn: - The gardener chooses a square in the garden. Each tree on that square and all the surrounding squ... | We solve the problem for a general $3 N \times 3 N$ board. First, we prove that the lumberjack has a strategy to ensure there are never more than $5 N^{2}$ majestic trees. Giving the squares of the board coordinates in the natural manner, colour each square where at least one of its coordinates are divisible by 3 , sho... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 235 | 796 |
2022 | T0 | C4 | Combinatorics | IMO-SL | Let $n>3$ be a positive integer. Suppose that $n$ children are arranged in a circle, and $n$ coins are distributed between them (some children may have no coins). At every step, a child with at least 2 coins may give 1 coin to each of their immediate neighbours on the right and left. Determine all initial distributions... | Number the children $1, \ldots, n$, and denote the number of coins the $i$-th child has by $c_{i}$. A step of this process consists of reducing some $c_{i}$ by 2 , and increasing $c_{i-1}, c_{i+1}$ by 1 . (Indices are considered $(\bmod n)$.) Because $(i-1)-2 i+(i+1)=0$, the quantity $\sum_{i=1}^{n} i c_{i}(\bmod n)$ w... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 156 | 1,964 |
2022 | T0 | C6 | Combinatorics | IMO-SL | Let $n$ be a positive integer. We start with $n$ piles of pebbles, each initially containing a single pebble. One can perform moves of the following form: choose two piles, take an equal number of pebbles from each pile and form a new pile out of these pebbles. For each positive integer $n$, find the smallest number of... | The solution we describe is simple, but not the most effective one. We can combine two piles of $2^{k-1}$ pebbles to make one pile of $2^{k}$ pebbles. In particular, given $2^{k}$ piles of one pebble, we may combine them as follows: $$ \begin{array}{lcc} 2^{k} \text { piles of } 1 \text { pebble } & \rightarrow & 2^{k-... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 119 | 1,247 |
2022 | T0 | C6 | Combinatorics | IMO-SL | Let $n$ be a positive integer. We start with $n$ piles of pebbles, each initially containing a single pebble. One can perform moves of the following form: choose two piles, take an equal number of pebbles from each pile and form a new pile out of these pebbles. For each positive integer $n$, find the smallest number of... | Throughout the solution, we will consider the moves in reverse order. Namely, imagine we have some piles of pebbles, and we are allowed to perform moves as follows: take a pile with an even number of pebbles, split it into two equal halves and add the pebbles from each half to a different pile, possibly forming new pil... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 119 | 2,230 |
2022 | T0 | C7 | Combinatorics | IMO-SL | Lucy starts by writing $s$ integer-valued 2022-tuples on a blackboard. After doing that, she can take any two (not necessarily distinct) tuples $\mathbf{v}=\left(v_{1}, \ldots, v_{2022}\right)$ and $\mathbf{w}=\left(w_{1}, \ldots, w_{2022}\right)$ that she has already written, and apply one of the following operations ... | We solve the problem for $n$-tuples for any $n \geqslant 3$ : we will show that the answer is $s=3$, regardless of the value of $n$. First, let us briefly introduce some notation. For an $n$-tuple $\mathbf{v}$, we will write $\mathbf{v}_{i}$ for its $i$-th coordinate (where $1 \leqslant i \leqslant n$ ). For a positive... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 303 | 1,932 |
2022 | T0 | C8 | Common remarks. | IMO-SL | Alice fills the fields of an $n \times n$ board with numbers from 1 to $n^{2}$, each number being used exactly once. She then counts the total number of good paths on the board. A good path is a sequence of fields of arbitrary length (including 1) such that: (i) The first field in the sequence is one that is only adjac... | We will call any field that is only adjacent to fields with larger numbers a well. Other fields will be called non-wells. Let us make a second $n \times n$ board $B$ where in each field we will write the number of good sequences which end on the corresponding field in the original board $A$. We will thus look for the m... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 167 | 898 |
2022 | T0 | C9 | Common remarks. | IMO-SL | Let $\mathbb{Z}_{\geqslant 0}$ be the set of non-negative integers, and let $f: \mathbb{Z}_{\geqslant 0} \times \mathbb{Z}_{\geqslant 0} \rightarrow \mathbb{Z}_{\geqslant 0}$ be a bijection such that whenever $f\left(x_{1}, y_{1}\right)>f\left(x_{2}, y_{2}\right)$, we have $f\left(x_{1}+1, y_{1}\right)>f\left(x_{2}+1, ... | We defer the constructions to the end of the solution. Instead, we begin by characterizing all such functions $f$, prove a formula and key property for such functions, and then solve the problem, providing constructions. Characterization Suppose $f$ satisfies the given relation. The condition can be written more strong... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 264 | 2,143 |
2022 | T0 | G8 | Geometry | IMO-SL | Let $A A^{\prime} B C C^{\prime} B^{\prime}$ be a convex cyclic hexagon such that $A C$ is tangent to the incircle of the triangle $A^{\prime} B^{\prime} C^{\prime}$, and $A^{\prime} C^{\prime}$ is tangent to the incircle of the triangle $A B C$. Let the lines $A B$ and $A^{\prime} B^{\prime}$ meet at $X$ and let the l... | Denote by $\omega$ and $\omega^{\prime}$ the incircles of $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ and let $I$ and $I^{\prime}$ be the centres of these circles. Let $N$ and $N^{\prime}$ be the second intersections of $B I$ and $B^{\prime} I^{\prime}$ with $\Omega$, the circumcircle of $A^{\pri... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 163 | 1,170 |
2022 | T0 | N1 | Number Theory | IMO-SL | A number is called Norwegian if it has three distinct positive divisors whose sum is equal to 2022. Determine the smallest Norwegian number. (Note: The total number of positive divisors of a Norwegian number is allowed to be larger than 3.) (Cyprus) Answer: 1344 | Observe that 1344 is a Norwegian number as 6, 672 and 1344 are three distinct divisors of 1344 and $6+672+1344=2022$. It remains to show that this is the smallest such number. Assume for contradiction that $N<1344$ is Norwegian and let $N / a, N / b$ and $N / c$ be the three distinct divisors of $N$, with $a<b<c$. Then... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 64 | 858 |
2022 | T0 | N2 | Number Theory | IMO-SL | Find all positive integers $n>2$ such that $$ n!\mid \prod_{\substack{p<q \leqslant n, p, q \text { primes }}}(p+q) . $$ (Nigeria) Answer: This only holds for $n=7$. | Assume that $n$ satisfies $n!\mid \prod_{p<q \leqslant n}(p+q)$ and let $2=p_{1}<p_{2}<\cdots<p_{m} \leqslant n$ be the primes in $\{1,2, \ldots, n\}$. Each such prime divides $n$ !. In particular, $p_{m} \mid p_{i}+p_{j}$ for some $p_{i}<p_{j} \leqslant n$. But $$ 0<\frac{p_{i}+p_{j}}{p_{m}}<\frac{p_{m}+p_{m}}{p_{m}}=... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 64 | 619 |
2022 | T0 | N3 | Number Theory | IMO-SL | Let $a>1$ be a positive integer, and let $d>1$ be a positive integer coprime to $a$. Let $x_{1}=1$ and, for $k \geqslant 1$, define $$ x_{k+1}= \begin{cases}x_{k}+d & \text { if } a \text { doesn't divide } x_{k} \\ x_{k} / a & \text { if } a \text { divides } x_{k}\end{cases} $$ Find the greatest positive integer $n... | Like in the first solution, $x_{k}$ is relatively prime to $d$ and $x_{k}<a d$ for all $k$. We wish to prove that there are $n$ consecutive decreasing indices. Let $m_{0}=0$ and $m_{k}$ be the $k$-th smallest decreasing index (an index $k$ is decreasing if $x_{k}=x_{k-1} / a$ ) and define a sequence $\left(y_{k}\right)... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 174 | 695 |
2022 | T0 | N4 | Number Theory | IMO-SL | Find all triples of positive integers $(a, b, p)$ with $p$ prime and $$ a^{p}=b!+p $$ (Belgium) Answer: $(2,2,2)$ and $(3,4,3)$. | Clearly, $a>1$. We consider three cases. Case 1: We have $a<p$. Then we either have $a \leqslant b$ which implies $a \mid a^{p}-b!=p$ leading to a contradiction, or $a>b$ which is also impossible since in this case we have $b!\leqslant a!<a^{p}-p$, where the last inequality is true for any $p>a>1$. Case 2: We have $a>p... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 53 | 693 |
2022 | T0 | N4 | Number Theory | IMO-SL | Find all triples of positive integers $(a, b, p)$ with $p$ prime and $$ a^{p}=b!+p $$ (Belgium) Answer: $(2,2,2)$ and $(3,4,3)$. | The cases $a \neq p$ are covered as in solution 1 , as are $p=2,3$. Also $b>p$, as $p^{p}>p!+p$ for $p>2$. The cases $p=5,7,11$ are also checked manually, so assume $p \geqslant 13$. Let $q \mid p+1$ be an odd prime. By LTE $$ v_{q}\left(p^{p}-p\right)=v_{q}\left(\left(p^{2}\right)^{\frac{p-1}{2}}-1\right)=v_{q}\left(p... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 53 | 590 |
2022 | T0 | N6 | Number Theory | IMO-SL | Let $Q$ be a set of prime numbers, not necessarily finite. For a positive integer $n$ consider its prime factorisation; define $p(n)$ to be the sum of all the exponents and $q(n)$ to be the sum of the exponents corresponding only to primes in $Q$. A positive integer $n$ is called special if $p(n)+p(n+1)$ and $q(n)+q(n+... | Let us call two positive integers $m, n$ friends if $p(m)+p(n)$ and $q(m)+q(n)$ are both even integers. We start by noting that the pairs $(p(k), q(k))$ modulo 2 can take at most 4 different values; thus, among any five different positive integers there are two which are friends. In addition, both functions $p$ and $q$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2022SL.jsonl",
"solution_match": null
} | 218 | 1,067 |
2023 | T0 | A1 | Algebra | IMO-SL | Professor Oak is feeding his 100 Pokémon. Each Pokémon has a bowl whose capacity is a positive real number of kilograms. These capacities are known to Professor Oak. The total capacity of all the bowls is 100 kilograms. Professor Oak distributes 100 kilograms of food in such a way that each Pokémon receives a non-negat... | First, consider the situation where 99 bowls have a capacity of 0.5 kilograms and the last bowl has a capacity of 50.5 kilograms. No matter how Professor Oak distributes the food, the dissatisfaction level of every Pokémon will be at least 0.5 . This amounts to a total dissatisfaction level of at least 50 , proving tha... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 175 | 848 |
2023 | T0 | A1 | Algebra | IMO-SL | Professor Oak is feeding his 100 Pokémon. Each Pokémon has a bowl whose capacity is a positive real number of kilograms. These capacities are known to Professor Oak. The total capacity of all the bowls is 100 kilograms. Professor Oak distributes 100 kilograms of food in such a way that each Pokémon receives a non-negat... | We adopt the same notation as in This solution uses the probabilistic method. We consider all distributions in which each Pokémon receives $\left\lfloor C_{i}\right\rfloor+\varepsilon_{i}$ kilograms of food, where $\varepsilon_{i} \in\{0,1\}$ and $\varepsilon_{1}+\varepsilon_{2}+\cdots+\varepsilon_{100}=R$. There are $... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 175 | 525 |
2023 | T0 | A2 | Algebra | IMO-SL | Let $\mathbb{R}$ be the set of real numbers. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that $$ f(x+y) f(x-y) \geqslant f(x)^{2}-f(y)^{2} $$ for every $x, y \in \mathbb{R}$. Assume that the inequality is strict for some $x_{0}, y_{0} \in \mathbb{R}$. Prove that $f(x) \geqslant 0$ for every $x \in \... | We prove the contrapositive of the problem statement. Assume that there exist $a, b \in \mathbb{R}$ such that $f(a)<0$ and $f(b)>0$. We want to prove that the inequality is actually an equality, i.e. it is never strict. Lemma 1. The function $f$ is odd, i.e. $f(x)+f(-x)=0$ for every $x \in \mathbb{R}$. Proof. We plug i... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 164 | 527 |
2023 | T0 | A2 | Algebra | IMO-SL | Let $\mathbb{R}$ be the set of real numbers. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that $$ f(x+y) f(x-y) \geqslant f(x)^{2}-f(y)^{2} $$ for every $x, y \in \mathbb{R}$. Assume that the inequality is strict for some $x_{0}, y_{0} \in \mathbb{R}$. Prove that $f(x) \geqslant 0$ for every $x \in \... | As in In this solution, we construct an argument by multiplying inequalities, rather than adding them as in Solutions 1-3. Lemma 2. $f(b) f(-b)<0$. Proof. Let $x_{1}:=\frac{a+b}{2}$ and $y_{1}:=\frac{a-b}{2}$ so that $a=x_{1}+y_{1}$ and $b=x_{1}-y_{1}$. Plugging in $x=x_{1}$ and $y=y_{1}$, we obtain $$ 0>f(a) f(b)=f\le... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 164 | 932 |
2023 | T0 | A3 | Algebra | IMO-SL | Let $x_{1}, x_{2}, \ldots, x_{2023}$ be distinct real positive numbers such that $$ a_{n}=\sqrt{\left(x_{1}+x_{2}+\cdots+x_{n}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}\right)} $$ is an integer for every $n=1,2, \ldots, 2023$. Prove that $a_{2023} \geqslant 3034$. (Netherlands) | We start with some basic observations. First note that the sequence $a_{1}, a_{2}, \ldots, a_{2023}$ is increasing and thus, since all elements are integers, $a_{n+1}-a_{n} \geqslant 1$. We also observe that $a_{1}=1$ and $$ a_{2}=\sqrt{\left(x_{1}+x_{2}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}\right)}>2 $$ by Cauch... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 140 | 1,017 |
2023 | T0 | A3 | Algebra | IMO-SL | Let $x_{1}, x_{2}, \ldots, x_{2023}$ be distinct real positive numbers such that $$ a_{n}=\sqrt{\left(x_{1}+x_{2}+\cdots+x_{n}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}\right)} $$ is an integer for every $n=1,2, \ldots, 2023$. Prove that $a_{2023} \geqslant 3034$. (Netherlands) | The trick is to compare $a_{n+2}$ and $a_{n}$. Observe that $$ \begin{aligned} a_{n+2}^{2}= & \left(x_{1}+\cdots+x_{n+2}\right)\left(\frac{1}{x_{1}}+\cdots+\frac{1}{x_{n+2}}\right) \\ = & \left(x_{1}+\cdots+x_{n}\right)\left(\frac{1}{x_{1}}+\cdots+\frac{1}{x_{n}}\right)+\left(x_{n+1}+x_{n+2}\right)\left(\frac{1}{x_{n+1... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 140 | 2,339 |
2023 | T0 | A4 | Algebra | IMO-SL | Let $\mathbb{R}_{>0}$ be the set of positive real numbers. Determine all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that $$ x(f(x)+f(y)) \geqslant(f(f(x))+y) f(y) $$ for every $x, y \in \mathbb{R}_{>0}$. (Belgium) | Let $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ be a function that satisfies the inequality of the problem statement. We will write $f^{k}(x)=f(f(\cdots f(x) \cdots))$ for the composition of $f$ with itself $k$ times, with the convention that $f^{0}(x)=x$. Substituting $y=x$ gives $$ x \geqslant f^{2}(x) . $$ Subs... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 93 | 663 |
2023 | T0 | A5 | Algebra | IMO-SL | Let $a_{1}, a_{2}, \ldots, a_{2023}$ be positive integers such that - $a_{1}, a_{2}, \ldots, a_{2023}$ is a permutation of $1,2, \ldots, 2023$, and - $\left|a_{1}-a_{2}\right|,\left|a_{2}-a_{3}\right|, \ldots,\left|a_{2022}-a_{2023}\right|$ is a permutation of $1,2, \ldots, 2022$. Prove that $\max \left(a_{1}, a_{202... | For the sake of clarity, we consider and prove the following generalisation of the original problem (which is the case $N=1012$ ): Let $N$ be a positive integer and $a_{1}, a_{2}, \ldots, a_{2 N-1}$ be positive integers such that - $a_{1}, a_{2}, \ldots, a_{2 N-1}$ is a permutation of $1,2, \ldots, 2 N-1$, and - $\left... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 168 | 1,553 |
2023 | T0 | A6 | Algebra | IMO-SL | Let $k \geqslant 2$ be an integer. Determine all sequences of positive integers $a_{1}, a_{2}, \ldots$ for which there exists a monic polynomial $P$ of degree $k$ with non-negative integer coefficients such that $$ P\left(a_{n}\right)=a_{n+1} a_{n+2} \cdots a_{n+k} $$ for every integer $n \geqslant 1$. (Malaysia) | We assume that $\left(a_{n}\right)$ is not eventually constant. Step 1. The first goal is to show that the sequence must be increasing, i.e. $a_{n}<a_{n+1}$ for all $n \geqslant 1$. First, by comparing the two equalities $$ \begin{aligned} P\left(a_{n}\right) & =a_{n+1} a_{n+2} \cdots a_{n+k}, \\ P\left(a_{n+1}\right) ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 107 | 2,095 |
2023 | T0 | A6 | Algebra | IMO-SL | Let $k \geqslant 2$ be an integer. Determine all sequences of positive integers $a_{1}, a_{2}, \ldots$ for which there exists a monic polynomial $P$ of degree $k$ with non-negative integer coefficients such that $$ P\left(a_{n}\right)=a_{n+1} a_{n+2} \cdots a_{n+k} $$ for every integer $n \geqslant 1$. (Malaysia) | We assume that $\left(a_{n}\right)$ is not eventually constant. In this solution, we first prove an alternative version of Claim 1. Claim 3. There exist infinitely many $n \geqslant 1$ with $$ a_{n} \leqslant \min \left\{a_{n+1}, \ldots, a_{n+k}\right\} $$ Proof. Suppose not, then for all but finitely many $n \geqslant... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 107 | 2,962 |
2023 | T0 | A7 | Algebra | IMO-SL | Let $N$ be a positive integer. Prove that there exist three permutations $a_{1}, a_{2}, \ldots, a_{N}$; $b_{1}, b_{2}, \ldots, b_{N}$; and $c_{1}, c_{2}, \ldots, c_{N}$ of $1,2, \ldots, N$ such that $$ \left|\sqrt{a_{k}}+\sqrt{b_{k}}+\sqrt{c_{k}}-2 \sqrt{N}\right|<2023 $$ for every $k=1,2, \ldots, N$. | The idea is to approximate the numbers $\sqrt{1}, \sqrt{2}, \ldots, \sqrt{N}$ by the nearest integer with errors $<0.5$. This gives the following sequence $$ 1,1,2,2,2,2,3,3,3,3,3,3,4, \ldots $$ More precisely, for each $k \geqslant 1$, we round $\sqrt{k^{2}-k+1}, \ldots, \sqrt{k^{2}+k}$ to $k$, so that there are $2 k$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 140 | 2,238 |
2023 | T0 | A7 | Algebra | IMO-SL | Let $N$ be a positive integer. Prove that there exist three permutations $a_{1}, a_{2}, \ldots, a_{N}$; $b_{1}, b_{2}, \ldots, b_{N}$; and $c_{1}, c_{2}, \ldots, c_{N}$ of $1,2, \ldots, N$ such that $$ \left|\sqrt{a_{k}}+\sqrt{b_{k}}+\sqrt{c_{k}}-2 \sqrt{N}\right|<2023 $$ for every $k=1,2, \ldots, N$. | This is a variation of Let $n$ be an integer satisfying $0 \leqslant n \leqslant m+1$ and define the multiset $T_{m, n}$ by $$ T_{m, n}:=\left\{1_{\times 1}, 2_{\times 2}, 3_{\times 3}, \ldots, m_{\times m},(m+1)_{\times n}\right\} . $$ In other words, $T_{m, 0}=T_{m}, T_{m, n}=T_{m} \sqcup\left\{(m+1)_{\times n}\right... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 140 | 1,543 |
2023 | T0 | A7 | Algebra | IMO-SL | Let $N$ be a positive integer. Prove that there exist three permutations $a_{1}, a_{2}, \ldots, a_{N}$; $b_{1}, b_{2}, \ldots, b_{N}$; and $c_{1}, c_{2}, \ldots, c_{N}$ of $1,2, \ldots, N$ such that $$ \left|\sqrt{a_{k}}+\sqrt{b_{k}}+\sqrt{c_{k}}-2 \sqrt{N}\right|<2023 $$ for every $k=1,2, \ldots, N$. | This solution is based on the geometrical insight of equilateral triangles. Step 1. We first consider the easier case of triangle numbers $$ N=\frac{m(m+1)}{2} $$ As shown in the following picture, consider the triangular shaped lattice points inside an equilateral triangle $A B C$ with a total of $N$ points. The latti... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 140 | 2,550 |
2023 | T0 | C2 | Combinatorics | IMO-SL | Determine the maximal length $L$ of a sequence $a_{1}, \ldots, a_{L}$ of positive integers satisfying both the following properties: - every term in the sequence is less than or equal to $2^{2023}$, and - there does not exist a consecutive subsequence $a_{i}, a_{i+1}, \ldots, a_{j}$ (where $1 \leqslant i \leqslant j \... | We prove more generally that the answer is $2^{k+1}-1$ when $2^{2023}$ is replaced by $2^{k}$ for an arbitrary positive integer $k$. Write $n=2^{k}$. We first show that there exists a sequence of length $L=2 n-1$ satisfying the properties. For a positive integer $x$, denote by $v_{2}(x)$ the maximal nonnegative integer... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 185 | 1,376 |
2023 | T0 | C3 | Combinatorics | IMO-SL | Let $n$ be a positive integer. We arrange $1+2+\cdots+n$ circles in a triangle with $n$ rows, such that the $i^{\text {th }}$ row contains exactly $i$ circles. The following figure shows the case $n=6$. ![](https://cdn.mathpix.com/cropped/2025_01_09_9538693ed865d47d2de5g-07.jpg?height=407&width=392&top_left_y=1604&top_... | We give an alternative proof that there exists a ninja-path passing through at least $N+1$ red circles. Assign numbers to circles as in the previous solution, but we only focus on the numbers assigned to red circles. For each positive integer $i$, denote by $e_{i}$ the number of red circles with number $i$. Claim 2. If... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 216 | 644 |
2023 | T0 | C3 | Combinatorics | IMO-SL | Let $n$ be a positive integer. We arrange $1+2+\cdots+n$ circles in a triangle with $n$ rows, such that the $i^{\text {th }}$ row contains exactly $i$ circles. The following figure shows the case $n=6$. . Denote by $f_{i}$ the number of red circles with number $i$. Note that if a red ci... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 216 | 810 |
2023 | T0 | C4 | Combinatorics | IMO-SL | Let $n \geqslant 2$ be a positive integer. Paul has a $1 \times n^{2}$ rectangular strip consisting of $n^{2}$ unit squares, where the $i^{\text {th }}$ square is labelled with $i$ for all $1 \leqslant i \leqslant n^{2}$. He wishes to cut the strip into several pieces, where each piece consists of a number of consecuti... | For the entirety of the solution, we shall view the labels as taking values in $\mathbb{Z} / n \mathbb{Z}$, as only their values modulo $n$ play a role. Here are two possible constructions consisting of $2 n-1$ pieces. 1. Cut into pieces of sizes $n, 1, n, 1, \ldots, n, 1,1$, and glue the pieces of size 1 to obtain the... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 198 | 610 |
2023 | T0 | C4 | Combinatorics | IMO-SL | Let $n \geqslant 2$ be a positive integer. Paul has a $1 \times n^{2}$ rectangular strip consisting of $n^{2}$ unit squares, where the $i^{\text {th }}$ square is labelled with $i$ for all $1 \leqslant i \leqslant n^{2}$. He wishes to cut the strip into several pieces, where each piece consists of a number of consecuti... | We provide an alternative proof that at least $2 n-1$ pieces are needed. Instead of having a linear strip, we work with a number of circular strips, each having length a multiple of $n$ and labelled as $$ 1,2, \ldots, n, 1,2, \ldots, n, \ldots, 1,2, \ldots, n $$ where there are $n^{2}$ cells in total across all circula... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 198 | 649 |
2023 | T0 | C4 | Combinatorics | IMO-SL | Let $n \geqslant 2$ be a positive integer. Paul has a $1 \times n^{2}$ rectangular strip consisting of $n^{2}$ unit squares, where the $i^{\text {th }}$ square is labelled with $i$ for all $1 \leqslant i \leqslant n^{2}$. He wishes to cut the strip into several pieces, where each piece consists of a number of consecuti... | As with the previous solution, we again work with circular strips. In particular, we start out with $k$ circular strips, each having length a multiple of $n$ and labelled as $$ 1,2, \ldots, n, 1,2, \ldots, n, \ldots, 1,2, \ldots, n $$ where there are $n^{2}$ cells in total across all $k$ circular strips. The goal is st... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 198 | 1,184 |
2023 | T0 | C5 | Combinatorics | IMO-SL | Elisa has 2023 treasure chests, all of which are unlocked and empty at first. Each day, Elisa adds a new gem to one of the unlocked chests of her choice, and afterwards, a fairy acts according to the following rules: - if more than one chests are unlocked, it locks one of them, or - if there is only one unlocked chest... | We will prove that such a constant $C$ exists when there are $n$ chests for $n$ an odd positive integer. In fact we can take $C=n-1$. Elisa's strategy is simple: place a gem in the chest with the fewest gems (in case there are more than one such chests, pick one arbitrarily). For each integer $t \geqslant 0$, let $a_{1... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 142 | 2,549 |
2023 | T0 | C5 | Combinatorics | IMO-SL | Elisa has 2023 treasure chests, all of which are unlocked and empty at first. Each day, Elisa adds a new gem to one of the unlocked chests of her choice, and afterwards, a fairy acts according to the following rules: - if more than one chests are unlocked, it locks one of them, or - if there is only one unlocked chest... | We solve the problem when 2023 is replaced with an arbitrary integer $n$. We assume that Elisa uses the following strategy: At the beginning of the $(n t+1)^{\text {th }}$ day, Elisa first labels her chests as $C_{1}^{t}, \ldots, C_{n}^{t}$ so that before she adds in the gem, the number of gems in $C_{i}^{t}$ is less t... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 142 | 1,727 |
2023 | T0 | C6 | Combinatorics | IMO-SL | Let $N$ be a positive integer, and consider an $N \times N$ grid. A right-down path is a sequence of grid cells such that each cell is either one cell to the right of or one cell below the previous cell in the sequence. A right-up path is a sequence of grid cells such that each cell is either one cell to the right of o... | We define a good parallelogram to be a parallelogram composed of two isosceles right-angled triangles glued together as shown below. Given any partition into $k$ right-down or right-up paths, we can find a corresponding packing of good parallelograms that leaves an area of $k$ empty. Thus, it suffices to prove that we ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 217 | 580 |
2023 | T0 | C6 | Combinatorics | IMO-SL | Let $N$ be a positive integer, and consider an $N \times N$ grid. A right-down path is a sequence of grid cells such that each cell is either one cell to the right of or one cell below the previous cell in the sequence. A right-up path is a sequence of grid cells such that each cell is either one cell to the right of o... | We apply an induction on $N$. The base case $N=1$ is trivial. Suppose that the claim holds for $N-1$ and prove it for $N \geqslant 2$. Let us denote the path containing the upper left corner by $P$. If $P$ is right-up, then every cell in $P$ is in the top row or in the leftmost column. By the induction hypothesis, ther... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 217 | 959 |
2023 | T0 | C7 | Combinatorics | IMO-SL | The Imomi archipelago consists of $n \geqslant 2$ islands. Between each pair of distinct islands is a unique ferry line that runs in both directions, and each ferry line is operated by one of $k$ companies. It is known that if any one of the $k$ companies closes all its ferry lines, then it becomes impossible for a tra... | We reformulate the problem using graph theory. We have a complete graph $K_{n}$ on $n$ nodes (corresponding to islands), and we want to colour the edges (corresponding to ferry lines) with $k$ colours (corresponding to companies), so that every Hamiltonian path contains all $k$ different colours. For a fixed set of $k$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 127 | 3,327 |
2023 | T0 | G3 | Geometry | IMO-SL | Let $A B C D$ be a cyclic quadrilateral with $\angle B A D<\angle A D C$. Let $M$ be the midpoint of the arc $C D$ not containing $A$. Suppose there is a point $P$ inside $A B C D$ such that $\angle A D B=$ $\angle C P D$ and $\angle A D P=\angle P C B$. Prove that lines $A D, P M, B C$ are concurrent. | Let $A D$ and $B C$ meet at $T$. Denote by $p_{a}, p_{b}, m_{a}$ and $m_{b}$ the distances between line $T A$ and $P, T B$ and $P, T A$ and $M$ and between $T B$ and $M$ respectively. Our goal is to prove $p_{a}: p_{b}=m_{a}: m_{b}$ which is equivalent to the collinearity of $T, P$ and $M$. Let $\angle B A C=\angle B D... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 103 | 706 |
2023 | T0 | G7 | Geometry | IMO-SL | Let $A B C$ be an acute, scalene triangle with orthocentre $H$. Let $\ell_{a}$ be the line through the reflection of $B$ with respect to $C H$ and the reflection of $C$ with respect to $B H$. Lines $\ell_{b}$ and $\ell_{c}$ are defined similarly. Suppose lines $\ell_{a}, \ell_{b}$, and $\ell_{c}$ determine a triangle $... | We write $\triangle P_{1} P_{2} P_{3} \stackrel{ \pm}{\sim} \triangle Q_{1} Q_{2} Q_{3}$ (resp. $\triangle P_{1} P_{2} P_{3} \approx \triangle Q_{1} Q_{2} Q_{3}$ ) to indicate that two triangles are directly (resp. oppositely) similar. We use directed angles throughout denoted with $\Varangle$. Denote by $A_{b}, A_{c}$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 133 | 1,078 |
2023 | T0 | G7 | Geometry | IMO-SL | Let $A B C$ be an acute, scalene triangle with orthocentre $H$. Let $\ell_{a}$ be the line through the reflection of $B$ with respect to $C H$ and the reflection of $C$ with respect to $B H$. Lines $\ell_{b}$ and $\ell_{c}$ are defined similarly. Suppose lines $\ell_{a}, \ell_{b}$, and $\ell_{c}$ determine a triangle $... | We use the same notation $A_{b}, A_{c}, B_{c}, B_{a}, C_{a}, C_{b}$ and $A_{1}, B_{1}, C_{1}, O_{1}, H_{1}$ as in Let $O$ be the circumcentre of $\triangle A B C$ and $A_{2}$ be the reflection of $A_{1}$ in $A H$. As $\odot A A_{b} A_{c}$ is centred at $H, A_{2}$ also lies on this circle. By Claim 2, $\Varangle B_{a} A... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 133 | 1,272 |
2023 | T0 | G8 | Geometry | IMO-SL | Let $A B C$ be an equilateral triangle. Points $A_{1}, B_{1}, C_{1}$ lie inside triangle $A B C$ such that triangle $A_{1} B_{1} C_{1}$ is scalene, $B A_{1}=A_{1} C, C B_{1}=B_{1} A, A C_{1}=C_{1} B$ and $$ \angle B A_{1} C+\angle C B_{1} A+\angle A C_{1} B=480^{\circ} . $$ Lines $B C_{1}$ and $C B_{1}$ intersect at ... | Let $\delta_{A}, \delta_{B}, \delta_{C}$ be the circumcircles of $\triangle A A_{1} A_{2}, \triangle B B_{1} B_{2}, \triangle C C_{1} C_{2}$. The general strategy of the solution is to find two different points having equal power with respect to $\delta_{A}, \delta_{B}, \delta_{C}$. Claim. $A_{1}$ is the circumcentre o... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 241 | 1,919 |
2023 | T0 | N3 | Number Theory | IMO-SL | For positive integers $n$ and $k \geqslant 2$ define $E_{k}(n)$ as the greatest exponent $r$ such that $k^{r}$ divides $n$ !. Prove that there are infinitely many $n$ such that $E_{10}(n)>E_{9}(n)$ and infinitely many $m$ such that $E_{10}(m)<E_{9}(m)$. (Brazil) $\mathbf{N 4 .}$ Let $a_{1}, a_{2}, \ldots, a_{n}, b_{1},... | We let $v_{p}(m)$ denote the $p$-adic valuation of $m$. By Legendre's Formula we know, for $p$ prime, that $v_{p}(n!)=\lfloor n / p\rfloor+\left\lfloor n / p^{2}\right\rfloor+\cdots$. We can see that $E_{9}(n)=\left\lfloor\frac{v_{3}(n!)}{2}\right\rfloor$. Since $v_{5}(n!) \leqslant v_{2}(n!)$ and $E_{10}(n)=\min \left... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 285 | 853 |
2023 | T0 | N5 | Number Theory | IMO-SL | Let $a_{1}<a_{2}<a_{3}<\cdots$ be positive integers such that $a_{k+1}$ divides $2\left(a_{1}+a_{2}+\cdots+a_{k}\right)$ for every $k \geqslant 1$. Suppose that for infinitely many primes $p$, there exists $k$ such that $p$ divides $a_{k}$. Prove that for every positive integer $n$, there exists $k$ such that $n$ divid... | For every $k \geqslant 2$ define the quotient $b_{k}=2\left(a_{1}+\cdots+a_{k-1}\right) / a_{k}$, which must be a positive integer. We first prove the following properties of the sequence $\left(b_{k}\right)$ : Claim 1. We have $b_{k+1} \leqslant b_{k}+1$ for all $k \geqslant 2$. Proof. By subtracting $b_{k} a_{k}=2\le... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 122 | 718 |
2023 | T0 | N6 | Number Theory | IMO-SL | A sequence of integers $a_{0}, a_{1}, a_{2}, \ldots$ is called kawaii, if $a_{0}=0, a_{1}=1$, and, for any positive integer $n$, we have $$ \left(a_{n+1}-3 a_{n}+2 a_{n-1}\right)\left(a_{n+1}-4 a_{n}+3 a_{n-1}\right)=0 . $$ An integer is called kawaii if it belongs to a kawaii sequence. Suppose that two consecutive p... | We start by rewriting the condition in the problem as: $$ a_{n+1}=3 a_{n}-2 a_{n-1}, \text { or } a_{n+1}=4 a_{n}-3 a_{n-1} . $$ We have $a_{n+1} \equiv a_{n}$ or $a_{n-1}(\bmod 2)$ and $a_{n+1} \equiv a_{n-1}$ or $a_{n}(\bmod 3)$ for all $n \geqslant 1$. Now, since $a_{0}=0$ and $a_{1}=1$, we have that $a_{n} \equiv 0... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 171 | 719 |
2023 | T0 | N6 | Number Theory | IMO-SL | A sequence of integers $a_{0}, a_{1}, a_{2}, \ldots$ is called kawaii, if $a_{0}=0, a_{1}=1$, and, for any positive integer $n$, we have $$ \left(a_{n+1}-3 a_{n}+2 a_{n-1}\right)\left(a_{n+1}-4 a_{n}+3 a_{n-1}\right)=0 . $$ An integer is called kawaii if it belongs to a kawaii sequence. Suppose that two consecutive p... | We start by proving the following: Claim 1. We have $a_{n} \equiv 0,1 \bmod 3$ for all $n \geqslant 0$. Proof. We have $a_{n+1}=3 a_{n}-2 a_{n-1}=3\left(a_{n}-a_{n-1}\right)+a_{n-1}$ or $a_{n+1}=4 a_{n}-3 a_{n-1}=3\left(a_{n}-\right.$ $\left.a_{n-1}\right)+a_{n}$, so $a_{n+1} \equiv a_{n}$ or $a_{n-1} \bmod 3$, and sin... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 171 | 978 |
2023 | T0 | N6 | Number Theory | IMO-SL | A sequence of integers $a_{0}, a_{1}, a_{2}, \ldots$ is called kawaii, if $a_{0}=0, a_{1}=1$, and, for any positive integer $n$, we have $$ \left(a_{n+1}-3 a_{n}+2 a_{n-1}\right)\left(a_{n+1}-4 a_{n}+3 a_{n-1}\right)=0 . $$ An integer is called kawaii if it belongs to a kawaii sequence. Suppose that two consecutive p... | (This solution is just a different combination of the ideas in For $n \geqslant 1$, put $b_{n}=a_{n}-a_{n-1}$. We have $$ a_{t}=a_{0}+\sum_{k=1}^{t}\left(a_{k}-a_{k-1}\right)=\sum_{k=1}^{t} b_{k} . $$ Note that $$ a_{n+1}-3 a_{n}+2 a_{n-1}=b_{n+1}-2 b_{n} \quad \text { and } \quad a_{n+1}-4 a_{n}+3 a_{n-1}=b_{n+1}-3 b_... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 171 | 1,340 |
2023 | T0 | N7 | Number Theory | IMO-SL | Let $a, b, c, d$ be positive integers satisfying $$ \frac{a b}{a+b}+\frac{c d}{c+d}=\frac{(a+b)(c+d)}{a+b+c+d} $$ Determine all possible values of $a+b+c+d$. (Netherlands) | First, note that if we take $a=\ell, b=k \ell, c=k \ell, d=k^{2} \ell$ for some positive integers $k$ and $\ell$, then we have $$ \frac{a b}{a+b}+\frac{c d}{c+d}=\frac{k \ell^{2}}{\ell+k \ell}+\frac{k^{3} \ell^{2}}{k \ell+k^{2} \ell}=\frac{k \ell}{k+1}+\frac{k^{2} \ell}{k+1}=k \ell $$ and $$ \frac{(a+b)(c+d)}{a+b+c+d}=... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 66 | 762 |
2023 | T0 | N8 | Number Theory | IMO-SL | Let $\mathbb{Z}_{>0}$ be the set of positive integers. Determine all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that $$ f^{b f(a)}(a+1)=(a+1) f(b) $$ holds for all $a, b \in \mathbb{Z}_{>0}$, where $f^{k}(n)=f(f(\cdots f(n) \cdots))$ denotes the composition of $f$ with itself $k$ times. | We divide the solution into 5 steps. Step 1. ( $f$ is injective) Claim 1. For any $a \geqslant 2$, the set $\left\{f^{n}(a) \mid n \in \mathbb{Z}_{>0}\right\}$ is infinite. Proof. First, we have $f^{f(a)}(a+1) \stackrel{P(a, 1)}{=}(a+1) f(1)$. Varying $a$, we see that $f\left(\mathbb{Z}_{>0}\right)$ is infinite. Next, ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 121 | 2,423 |
2023 | T0 | N8 | Number Theory | IMO-SL | Let $\mathbb{Z}_{>0}$ be the set of positive integers. Determine all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that $$ f^{b f(a)}(a+1)=(a+1) f(b) $$ holds for all $a, b \in \mathbb{Z}_{>0}$, where $f^{k}(n)=f(f(\cdots f(n) \cdots))$ denotes the composition of $f$ with itself $k$ times. | In the same way as Steps 1-2 of We first note that Claim 2 in Solution 1 is also true for $a=1$. Claim 2'. For any $a, n \in \mathbb{Z}_{>0}$, we have $f^{n}(a) \neq a$. Proof. If $a \geqslant 2$, the assertion was proved in Claim 2 in Solution 1. If $a=1$, we have that 1 is not in the range of $f$ by Claim 3 in Soluti... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 121 | 916 |
2023 | T0 | N8 | Number Theory | IMO-SL | Let $\mathbb{Z}_{>0}$ be the set of positive integers. Determine all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that $$ f^{b f(a)}(a+1)=(a+1) f(b) $$ holds for all $a, b \in \mathbb{Z}_{>0}$, where $f^{k}(n)=f(f(\cdots f(n) \cdots))$ denotes the composition of $f$ with itself $k$ times. | The following is another way of finishing $$ \begin{aligned} f^{b g(a)}(f(a)) & =f^{b f(f(a)-1)}(f(a)) \stackrel{P(f(a)-1, b)}{=} f(a) f(b) \stackrel{P(f(b)-1, a)}{=} f^{a f(f(b)-1)}(f(b)) \\ & =f^{a g(b)+k}(f(a)) \end{aligned} $$ By Claim 2' in Solution 2, we have $b g(a)=a g(b)+k$, so $f^{k}(a) \cdot g(a)=a \cdot g\l... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 121 | 747 |
2023 | T0 | N8 | Number Theory | IMO-SL | Let $\mathbb{Z}_{>0}$ be the set of positive integers. Determine all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that $$ f^{b f(a)}(a+1)=(a+1) f(b) $$ holds for all $a, b \in \mathbb{Z}_{>0}$, where $f^{k}(n)=f(f(\cdots f(n) \cdots))$ denotes the composition of $f$ with itself $k$ times. | We provide yet another (more technical) solution assuming Step 1 and Step 2 of Solution 1. By Claim 5 in Solution 1, every $a \geqslant 2$ is a descendant of 1 . Let $g$ and $h$ be the functions on $\mathbb{Z}_{\geqslant 2}$ such that $f^{g(a)}(1)=a$ and $h(a)=f(a-1)$. Then, $g: \mathbb{Z}_{\geqslant 2} \rightarrow \ma... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 121 | 2,099 |
2023 | T0 | A1 | Algebra | IMO-SL | Professor Oak is feeding his 100 Pokémon. Each Pokémon has a bowl whose capacity is a positive real number of kilograms. These capacities are known to Professor Oak. The total capacity of all the bowls is 100 kilograms. Professor Oak distributes 100 kilograms of food in such a way that each Pokémon receives a non-negat... | First, consider the situation where 99 bowls have a capacity of 0.5 kilograms and the last bowl has a capacity of 50.5 kilograms. No matter how Professor Oak distributes the food, the dissatisfaction level of every Pokémon will be at least 0.5 . This amounts to a total dissatisfaction level of at least 50 , proving tha... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 186 | 848 |
2023 | T0 | A1 | Algebra | IMO-SL | Professor Oak is feeding his 100 Pokémon. Each Pokémon has a bowl whose capacity is a positive real number of kilograms. These capacities are known to Professor Oak. The total capacity of all the bowls is 100 kilograms. Professor Oak distributes 100 kilograms of food in such a way that each Pokémon receives a non-negat... | We adopt the same notation as in This solution uses the probabilistic method. We consider all distributions in which each Pokémon receives $\left\lfloor C_{i}\right\rfloor+\varepsilon_{i}$ kilograms of food, where $\varepsilon_{i} \in\{0,1\}$ and $\varepsilon_{1}+\varepsilon_{2}+\cdots+\varepsilon_{100}=R$. There are $... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 186 | 525 |
2023 | T0 | A2 | Algebra | IMO-SL | Let $\mathbb{R}$ be the set of real numbers. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that $$ f(x+y) f(x-y) \geqslant f(x)^{2}-f(y)^{2} $$ for every $x, y \in \mathbb{R}$. Assume that the inequality is strict for some $x_{0}, y_{0} \in \mathbb{R}$. Prove that $f(x) \geqslant 0$ for every $x \in \... | We prove the contrapositive of the problem statement. Assume that there exist $a, b \in \mathbb{R}$ such that $f(a)<0$ and $f(b)>0$. We want to prove that the inequality is actually an equality, i.e. it is never strict. Lemma 1. The function $f$ is odd, i.e. $f(x)+f(-x)=0$ for every $x \in \mathbb{R}$. Proof. We plug i... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 189 | 527 |
2023 | T0 | A2 | Algebra | IMO-SL | Let $\mathbb{R}$ be the set of real numbers. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that $$ f(x+y) f(x-y) \geqslant f(x)^{2}-f(y)^{2} $$ for every $x, y \in \mathbb{R}$. Assume that the inequality is strict for some $x_{0}, y_{0} \in \mathbb{R}$. Prove that $f(x) \geqslant 0$ for every $x \in \... | As in In this solution, we construct an argument by multiplying inequalities, rather than adding them as in Solutions 1-3. Lemma 2. $f(b) f(-b)<0$. Proof. Let $x_{1}:=\frac{a+b}{2}$ and $y_{1}:=\frac{a-b}{2}$ so that $a=x_{1}+y_{1}$ and $b=x_{1}-y_{1}$. Plugging in $x=x_{1}$ and $y=y_{1}$, we obtain $$ 0>f(a) f(b)=f\le... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 189 | 932 |
2023 | T0 | A4 | Algebra | IMO-SL | Let $\mathbb{R}_{>0}$ be the set of positive real numbers. Determine all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that $$ x(f(x)+f(y)) \geqslant(f(f(x))+y) f(y) $$ for every $x, y \in \mathbb{R}_{>0}$. (Belgium) Answer: All functions $f(x)=\frac{c}{x}$ for some $c>0$. | Let $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ be a function that satisfies the inequality of the problem statement. We will write $f^{k}(x)=f(f(\cdots f(x) \cdots))$ for the composition of $f$ with itself $k$ times, with the convention that $f^{0}(x)=x$. Substituting $y=x$ gives $$ x \geqslant f^{2}(x) . $$ Subs... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 115 | 663 |
2023 | T0 | A5 | Algebra | IMO-SL | Let $a_{1}, a_{2}, \ldots, a_{2023}$ be positive integers such that - $a_{1}, a_{2}, \ldots, a_{2023}$ is a permutation of $1,2, \ldots, 2023$, and - $\left|a_{1}-a_{2}\right|,\left|a_{2}-a_{3}\right|, \ldots,\left|a_{2022}-a_{2023}\right|$ is a permutation of $1,2, \ldots, 2022$. Prove that $\max \left(a_{1}, a_{202... | For the sake of clarity, we consider and prove the following generalisation of the original problem (which is the case $N=1012$ ): Let $N$ be a positive integer and $a_{1}, a_{2}, \ldots, a_{2 N-1}$ be positive integers such that - $a_{1}, a_{2}, \ldots, a_{2 N-1}$ is a permutation of $1,2, \ldots, 2 N-1$, and - $\left... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 171 | 1,553 |
2023 | T0 | A6 | Algebra | IMO-SL | Let $k \geqslant 2$ be an integer. Determine all sequences of positive integers $a_{1}, a_{2}, \ldots$ for which there exists a monic polynomial $P$ of degree $k$ with non-negative integer coefficients such that $$ P\left(a_{n}\right)=a_{n+1} a_{n+2} \cdots a_{n+k} $$ for every integer $n \geqslant 1$. (Malaysia) Ans... | We assume that $\left(a_{n}\right)$ is not eventually constant. Step 1. The first goal is to show that the sequence must be increasing, i.e. $a_{n}<a_{n+1}$ for all $n \geqslant 1$. First, by comparing the two equalities $$ \begin{aligned} P\left(a_{n}\right) & =a_{n+1} a_{n+2} \cdots a_{n+k}, \\ P\left(a_{n+1}\right) ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 463 | 2,095 |
2023 | T0 | A6 | Algebra | IMO-SL | Let $k \geqslant 2$ be an integer. Determine all sequences of positive integers $a_{1}, a_{2}, \ldots$ for which there exists a monic polynomial $P$ of degree $k$ with non-negative integer coefficients such that $$ P\left(a_{n}\right)=a_{n+1} a_{n+2} \cdots a_{n+k} $$ for every integer $n \geqslant 1$. (Malaysia) Ans... | We assume that $\left(a_{n}\right)$ is not eventually constant. In this solution, we first prove an alternative version of Claim 1. Claim 3. There exist infinitely many $n \geqslant 1$ with $$ a_{n} \leqslant \min \left\{a_{n+1}, \ldots, a_{n+k}\right\} $$ Proof. Suppose not, then for all but finitely many $n \geqslant... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2023SL.jsonl",
"solution_match": null
} | 463 | 2,962 |
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