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 |
|---|---|---|---|---|---|---|---|---|---|
2008 | T0 | N3 | Number Theory | IMO-SL | Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, $\operatorname{gcd}\left(a_{i}, a_{i+1}\right)>a_{i-1}$. Prove that $a_{n} \geq 2^{n}$ for all $n \geq 0$. | Since $a_{i} \geq \operatorname{gcd}\left(a_{i}, a_{i+1}\right)>a_{i-1}$, the sequence is strictly increasing. In particular $a_{0} \geq 1, a_{1} \geq 2$. For each $i \geq 1$ we also have $a_{i+1}-a_{i} \geq \operatorname{gcd}\left(a_{i}, a_{i+1}\right)>a_{i-1}$, and consequently $a_{i+1} \geq a_{i}+a_{i-1}+1$. Hence $... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2008SL.jsonl",
"solution_match": null
} | 97 | 1,668 |
2008 | T0 | N4 | Number Theory | IMO-SL | Let $n$ be a positive integer. Show that the numbers $$ \left(\begin{array}{c} 2^{n}-1 \\ 0 \end{array}\right), \quad\left(\begin{array}{c} 2^{n}-1 \\ 1 \end{array}\right), \quad\left(\begin{array}{c} 2^{n}-1 \\ 2 \end{array}\right), \quad \ldots, \quad\left(\begin{array}{c} 2^{n}-1 \\ 2^{n-1}-1 \end{array}\right) $$ ... | It is well-known that all these numbers are odd. So the assertion that their remainders $\left(\bmod 2^{n}\right)$ make up a permutation of $\left\{1,3, \ldots, 2^{n}-1\right\}$ is equivalent just to saying that these remainders are all distinct. We begin by showing that $$ \left(\begin{array}{c} 2^{n}-1 \\ 2 k \end{ar... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2008SL.jsonl",
"solution_match": null
} | 167 | 1,734 |
2008 | T0 | N4 | Number Theory | IMO-SL | Let $n$ be a positive integer. Show that the numbers $$ \left(\begin{array}{c} 2^{n}-1 \\ 0 \end{array}\right), \quad\left(\begin{array}{c} 2^{n}-1 \\ 1 \end{array}\right), \quad\left(\begin{array}{c} 2^{n}-1 \\ 2 \end{array}\right), \quad \ldots, \quad\left(\begin{array}{c} 2^{n}-1 \\ 2^{n-1}-1 \end{array}\right) $$ ... | We again proceed by induction, writing for brevity $N=2^{n-1}$ and keeping notation $a_{k}=\left(\begin{array}{c}N-1 \\ k\end{array}\right), b_{m}=\left(\begin{array}{c}2 N-1 \\ m\end{array}\right)$. Assume that the result holds for the sequence $\left(a_{0}, a_{1}, a_{2}, \ldots, a_{N / 2-1}\right)$. In view of the sy... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2008SL.jsonl",
"solution_match": null
} | 167 | 1,387 |
2008 | T0 | N5 | Number Theory | IMO-SL | For every $n \in \mathbb{N}$ let $d(n)$ denote the number of (positive) divisors of $n$. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ with the following properties: (i) $d(f(x))=x$ for all $x \in \mathbb{N}$; (ii) $f(x y)$ divides $(x-1) y^{x y-1} f(x)$ for all $x, y \in \mathbb{N}$. | There is a unique solution: the function $f: \mathbb{N} \rightarrow \mathbb{N}$ defined by $f(1)=1$ and $f(n)=p_{1}^{p_{1}^{a_{1}}-1} p_{2}^{p_{2}^{a_{2}}-1} \cdots p_{k}^{p_{k}^{a_{k}}-1}$ where $n=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{k}^{a_{k}}$ is the prime factorization of $n>1$. Direct verification shows that thi... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2008SL.jsonl",
"solution_match": null
} | 120 | 1,739 |
2009 | T0 | A1 | Algebra | IMO-SL | CZE Find the largest possible integer $k$, such that the following statement is true: Let 2009 arbitrary non-degenerated triangles be given. In every triangle the three sides are colored, such that one is blue, one is red and one is white. Now, for every color separately, let us sort the lengths of the sides. We obta... | We will prove that the largest possible number $k$ of indices satisfying the given condition is one. Firstly we prove that $b_{2009}, r_{2009}, w_{2009}$ are always lengths of the sides of a triangle. Without loss of generality we may assume that $w_{2009} \geq r_{2009} \geq b_{2009}$. We show that the inequality $b_{2... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 259 | 638 |
2009 | T0 | A2 | Algebra | IMO-SL | EST Let $a, b, c$ be positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$. Prove that $$ \frac{1}{(2 a+b+c)^{2}}+\frac{1}{(2 b+c+a)^{2}}+\frac{1}{(2 c+a+b)^{2}} \leq \frac{3}{16} $$ | For positive real numbers $x, y, z$, from the arithmetic-geometric-mean inequality, $$ 2 x+y+z=(x+y)+(x+z) \geq 2 \sqrt{(x+y)(x+z)} $$ we obtain $$ \frac{1}{(2 x+y+z)^{2}} \leq \frac{1}{4(x+y)(x+z)} $$ Applying this to the left-hand side terms of the inequality to prove, we get $$ \begin{aligned} \frac{1}{(2 a+b+c)^{2}... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 100 | 651 |
2009 | T0 | A2 | Algebra | IMO-SL | EST Let $a, b, c$ be positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$. Prove that $$ \frac{1}{(2 a+b+c)^{2}}+\frac{1}{(2 b+c+a)^{2}}+\frac{1}{(2 c+a+b)^{2}} \leq \frac{3}{16} $$ | Equivalently, we prove the homogenized inequality $$ \frac{(a+b+c)^{2}}{(2 a+b+c)^{2}}+\frac{(a+b+c)^{2}}{(a+2 b+c)^{2}}+\frac{(a+b+c)^{2}}{(a+b+2 c)^{2}} \leq \frac{3}{16}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) $$ for all positive real numbers $a, b, c$. Without loss of generality we choose $a+b+c=1$. ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 100 | 660 |
2009 | T0 | A3 | Algebra | IMO-SL | FRA Determine all functions $f$ from the set of positive integers into the set of positive integers such that for all $x$ and $y$ there exists a non degenerated triangle with sides of lengths $$ x, \quad f(y) \quad \text { and } \quad f(y+f(x)-1) $$ | The identity function $f(x)=x$ is the only solution of the problem. If $f(x)=x$ for all positive integers $x$, the given three lengths are $x, y=f(y)$ and $z=$ $f(y+f(x)-1)=x+y-1$. Because of $x \geq 1, y \geq 1$ we have $z \geq \max \{x, y\}>|x-y|$ and $z<x+y$. From this it follows that a triangle with these side leng... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 69 | 835 |
2009 | T0 | A5 | Algebra | IMO-SL | BLR Let $f$ be any function that maps the set of real numbers into the set of real numbers. Prove that there exist real numbers $x$ and $y$ such that $$ f(x-f(y))>y f(x)+x $$ | Assume that $$ f(x-f(y)) \leq y f(x)+x \quad \text { for all real } x, y $$ Let $a=f(0)$. Setting $y=0$ in (1) gives $f(x-a) \leq x$ for all real $x$ and, equivalently, $$ f(y) \leq y+a \quad \text { for all real } y $$ Setting $x=f(y)$ in (1) yields in view of (2) $$ a=f(0) \leq y f(f(y))+f(y) \leq y f(f(y))+y+a . $$ ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 54 | 565 |
2009 | T0 | A5 | Algebra | IMO-SL | BLR Let $f$ be any function that maps the set of real numbers into the set of real numbers. Prove that there exist real numbers $x$ and $y$ such that $$ f(x-f(y))>y f(x)+x $$ | Assume that $$ f(x-f(y)) \leq y f(x)+x \quad \text { for all real } x, y $$ Let $a=f(0)$. Setting $y=0$ in (7) gives $f(x-a) \leq x$ for all real $x$ and, equivalently, $$ f(y) \leq y+a \quad \text { for all real } y $$ Now we show that $$ f(z) \geq 0 \quad \text { for all } z \geq 1 $$ Let $z \geq 1$ be fixed, set $b=... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 54 | 753 |
2009 | T0 | A6 | Algebra | IMO-SL | USA Suppose that $s_{1}, s_{2}, s_{3}, \ldots$ is a strictly increasing sequence of positive integers such that the subsequences $$ s_{s_{1}}, s_{s_{2}}, s_{s_{3}}, \ldots \quad \text { and } \quad s_{s_{1}+1}, s_{s_{2}+1}, s_{s_{3}+1}, \ldots $$ are both arithmetic progressions. Prove that $s_{1}, s_{2}, s_{3}, \l... | Let $D$ be the common difference of the progression $s_{s_{1}}, s_{s_{2}}, \ldots$. Let for $n=$ 1, 2, ... $$ d_{n}=s_{n+1}-s_{n} $$ We have to prove that $d_{n}$ is constant. First we show that the numbers $d_{n}$ are bounded. Indeed, by supposition $d_{n} \geq 1$ for all $n$. Thus, we have for all $n$ $$ d_{n}=s_{n+1... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 130 | 925 |
2009 | T0 | A6 | Algebra | IMO-SL | USA Suppose that $s_{1}, s_{2}, s_{3}, \ldots$ is a strictly increasing sequence of positive integers such that the subsequences $$ s_{s_{1}}, s_{s_{2}}, s_{s_{3}}, \ldots \quad \text { and } \quad s_{s_{1}+1}, s_{s_{2}+1}, s_{s_{3}+1}, \ldots $$ are both arithmetic progressions. Prove that $s_{1}, s_{2}, s_{3}, \l... | Let the integers $D$ and $E$ be the common differences of the progressions $s_{s_{1}}, s_{s_{2}}, \ldots$ and $s_{s_{1}+1}, s_{s_{2}+1}, \ldots$, respectively. Let briefly $A=s_{s_{1}}-D$ and $B=s_{s_{1}+1}-E$. Then, for all positive integers $n$, $$ s_{s_{n}}=A+n D, \quad s_{s_{n}+1}=B+n E $$ Since the sequence $s_{1}... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 130 | 837 |
2009 | T0 | A7 | Algebra | IMO-SL | JPN Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all real $x, y$ the identity $$ f(x f(x+y))=f(y f(x))+x^{2} $$ | It is no hard to see that the two functions given by $f(x)=x$ and $f(x)=-x$ for all real $x$ respectively solve the functional equation. In the sequel, we prove that there are no further solutions. Let $f$ be a function satisfying the given equation. It is clear that $f$ cannot be a constant. Let us first show that $f(... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 52 | 1,113 |
2009 | T0 | A7 | Algebra | IMO-SL | JPN Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all real $x, y$ the identity $$ f(x f(x+y))=f(y f(x))+x^{2} $$ | As in Now we prove that $f$ is injective. For this purpose, let us assume that $f(r)=f(s)$ for some $r \neq s$. Then, by (2) $$ r^{2}=f(r f(r))=f(r f(s))=f((s-r) f(r))+r^{2} $$ where the last statement follows from the given functional equation with $x=r$ and $y=s-r$. Hence, $h=(s-r) f(r)$ satisfies $f(h)=0$ which impl... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 52 | 645 |
2009 | T0 | C2 | Combinatorics | IMO-SL | ROU For any integer $n \geq 2$, let $N(n)$ be the maximal number of triples $\left(a_{i}, b_{i}, c_{i}\right), i=1, \ldots, N(n)$, consisting of nonnegative integers $a_{i}, b_{i}$ and $c_{i}$ such that the following two conditions are satisfied: (1) $a_{i}+b_{i}+c_{i}=n$ for all $i=1, \ldots, N(n)$, (2) If $i \neq j... | Let $n \geq 2$ be an integer and let $\left\{T_{1}, \ldots, T_{N}\right\}$ be any set of triples of nonnegative integers satisfying the conditions (1) and (2). Since the $a$-coordinates are pairwise distinct we have $$ \sum_{i=1}^{N} a_{i} \geq \sum_{i=1}^{N}(i-1)=\frac{N(N-1)}{2} $$ Analogously, $$ \sum_{i=1}^{N} b_{i... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 308 | 1,192 |
2009 | T0 | C3 | Combinatorics | IMO-SL | RUS Let $n$ be a positive integer. Given a sequence $\varepsilon_{1}, \ldots, \varepsilon_{n-1}$ with $\varepsilon_{i}=0$ or $\varepsilon_{i}=1$ for each $i=1, \ldots, n-1$, the sequences $a_{0}, \ldots, a_{n}$ and $b_{0}, \ldots, b_{n}$ are constructed by the following rules: $$ \begin{gathered} a_{0}=b_{0}=1, \qua... | For a binary word $w=\sigma_{1} \ldots \sigma_{n}$ of length $n$ and a letter $\sigma \in\{0,1\}$ let $w \sigma=$ $\sigma_{1} \ldots \sigma_{n} \sigma$ and $\sigma w=\sigma \sigma_{1} \ldots \sigma_{n}$. Moreover let $\bar{w}=\sigma_{n} \ldots \sigma_{1}$ and let $\emptyset$ be the empty word (of length 0 and with $\ba... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 360 | 1,208 |
2009 | T0 | C4 | Combinatorics | IMO-SL | NLD For an integer $m \geq 1$, we consider partitions of a $2^{m} \times 2^{m}$ chessboard into rectangles consisting of cells of the chessboard, in which each of the $2^{m}$ cells along one diagonal forms a separate rectangle of side length 1 . Determine the smallest possible sum of rectangle perimeters in such a pa... | For a $k \times k$ chessboard, we introduce in a standard way coordinates of the vertices of the cells and assume that the cell $C_{i j}$ in row $i$ and column $j$ has vertices $(i-1, j-1),(i-$ $1, j),(i, j-1),(i, j)$, where $i, j \in\{1, \ldots, k\}$. Without loss of generality assume that the cells $C_{i i}$, $i=1, \... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 81 | 1,295 |
2009 | T0 | C4 | Combinatorics | IMO-SL | NLD For an integer $m \geq 1$, we consider partitions of a $2^{m} \times 2^{m}$ chessboard into rectangles consisting of cells of the chessboard, in which each of the $2^{m}$ cells along one diagonal forms a separate rectangle of side length 1 . Determine the smallest possible sum of rectangle perimeters in such a pa... | We start as in $$ p=2\left(\sum_{i=1}^{M} r_{i}+\sum_{i=1}^{M} c_{i}\right) . $$ No rectangle can simultaneously cover cells from row $i$ and from column $i$ since otherwise it would also cover the cell $C_{i i}$. We classify subsets $S$ of rectangles of the partition as follows. We say that $S$ is of type $i, 1 \leq i... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 81 | 506 |
2009 | T0 | C5 | Combinatorics | IMO-SL | NLD Five identical empty buckets of 2-liter capacity stand at the vertices of a regular pentagon. Cinderella and her wicked Stepmother go through a sequence of rounds: At the beginning of every round, the Stepmother takes one liter of water from the nearby river and distributes it arbitrarily over the five buckets. T... | No, the Stepmother cannot enforce a bucket overflow and Cinderella can keep playing forever. Throughout we denote the five buckets by $B_{0}, B_{1}, B_{2}, B_{3}$, and $B_{4}$, where $B_{k}$ is adjacent to bucket $B_{k-1}$ and $B_{k+1}(k=0,1,2,3,4)$ and all indices are taken modulo 5 . Cinderella enforces that the foll... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 129 | 613 |
2009 | T0 | C5 | Combinatorics | IMO-SL | NLD Five identical empty buckets of 2-liter capacity stand at the vertices of a regular pentagon. Cinderella and her wicked Stepmother go through a sequence of rounds: At the beginning of every round, the Stepmother takes one liter of water from the nearby river and distributes it arbitrarily over the five buckets. T... | We prove that Cinderella can maintain the following two conditions and hence she can prevent the buckets from overflow: (1') Every two non-adjacent buckets contain a total of at most 1. (2') The total contents of all five buckets is at most $\frac{3}{2}$. We use the same notations as in the first solution. The two cond... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 129 | 2,212 |
2009 | T0 | C6 | Combinatorics | IMO-SL | BGR On a $999 \times 999$ board a limp rook can move in the following way: From any square it can move to any of its adjacent squares, i.e. a square having a common side with it, and every move must be a turn, i.e. the directions of any two consecutive moves must be perpendicular. A nonintersecting route of the limp ... | The answer is $998^{2}-4=4 \cdot\left(499^{2}-1\right)$ squares. First we show that this number is an upper bound for the number of cells a limp rook can visit. To do this we color the cells with four colors $A, B, C$ and $D$ in the following way: for $(i, j) \equiv(0,0) \bmod 2$ use $A$, for $(i, j) \equiv(0,1) \bmod ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 172 | 1,009 |
2009 | T0 | C8 | Combinatorics | IMO-SL | AUT For any integer $n \geq 2$, we compute the integer $h(n)$ by applying the following procedure to its decimal representation. Let $r$ be the rightmost digit of $n$. (1) If $r=0$, then the decimal representation of $h(n)$ results from the decimal representation of $n$ by removing this rightmost digit 0 . (2) If $1 ... | We identify integers $n \geq 2$ with the digit-strings, briefly strings, of their decimal representation and extend the definition of $h$ to all non-empty strings with digits from 0 to 9. We recursively define ten functions $f_{0}, \ldots, f_{9}$ that map some strings into integers for $k=$ $9,8, \ldots, 1,0$. The func... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 290 | 1,367 |
2009 | T0 | C8 | Combinatorics | IMO-SL | AUT For any integer $n \geq 2$, we compute the integer $h(n)$ by applying the following procedure to its decimal representation. Let $r$ be the rightmost digit of $n$. (1) If $r=0$, then the decimal representation of $h(n)$ results from the decimal representation of $n$ by removing this rightmost digit 0 . (2) If $1 ... | We identify integers $n \geq 2$ with the digit-strings, briefly strings, of their decimal representation and extend the definition of $h$ to all non-empty strings with digits from 0 to 9. Moreover, let us define that the empty string, $\varepsilon$, is being mapped to the empty string. In the following all functions ma... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 290 | 2,143 |
2009 | T0 | C8 | Combinatorics | IMO-SL | AUT For any integer $n \geq 2$, we compute the integer $h(n)$ by applying the following procedure to its decimal representation. Let $r$ be the rightmost digit of $n$. (1) If $r=0$, then the decimal representation of $h(n)$ results from the decimal representation of $n$ by removing this rightmost digit 0 . (2) If $1 ... | We commence by introducing some terminology. Instead of integers, we will consider the set $S$ of all strings consisting of the digits $0,1, \ldots, 9$, including the empty string $\epsilon$. If $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ is a nonempty string, we let $\rho(a)=a_{n}$ denote the terminal digit of $a$ and ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 290 | 3,674 |
2009 | T0 | G1 | Geometry | IMO-SL | BEL Let $A B C$ be a triangle with $A B=A C$. The angle bisectors of $A$ and $B$ meet the sides $B C$ and $A C$ in $D$ and $E$, respectively. Let $K$ be the incenter of triangle $A D C$. Suppose that $\angle B E K=45^{\circ}$. Find all possible values of $\angle B A C$. | Answer: $\angle B A C=60^{\circ}$ or $\angle B A C=90^{\circ}$ are possible values and the only possible values. Let $I$ be the incenter of triangle $A B C$, then $K$ lies on the line $C I$. Let $F$ be the point, where the incircle of triangle $A B C$ touches the side $A C$; then the segments $I F$ and $I D$ have the s... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 93 | 1,041 |
2009 | T0 | G1 | Geometry | IMO-SL | BEL Let $A B C$ be a triangle with $A B=A C$. The angle bisectors of $A$ and $B$ meet the sides $B C$ and $A C$ in $D$ and $E$, respectively. Let $K$ be the incenter of triangle $A D C$. Suppose that $\angle B E K=45^{\circ}$. Find all possible values of $\angle B A C$. | Denote angles at $A, B$ and $C$ as usual by $\alpha, \beta$ and $\gamma$. Since triangle $A B C$ is isosceles, we have $\beta=\gamma=90^{\circ}-\frac{\alpha}{2}<90^{\circ}$, so $\angle E C K=45^{\circ}-\frac{\alpha}{4}=\angle K C D$. Since $K$ is the incenter of triangle $A D C$, we have $\angle C D K=\angle K D A=45^{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 93 | 1,230 |
2009 | T0 | G2 | Geometry | IMO-SL | RUS Let $A B C$ be a triangle with circumcenter $O$. The points $P$ and $Q$ are interior points of the sides $C A$ and $A B$, respectively. The circle $k$ passes through the midpoints of the segments $B P$, $C Q$, and $P Q$. Prove that if the line $P Q$ is tangent to circle $k$ then $O P=O Q$. | Let $K, L, M, B^{\prime}, C^{\prime}$ be the midpoints of $B P, C Q, P Q, C A$, and $A B$, respectively (see Figure 1). Since $C A \| L M$, we have $\angle L M P=\angle Q P A$. Since $k$ touches the segment $P Q$ at $M$, we find $\angle L M P=\angle L K M$. Thus $\angle Q P A=\angle L K M$. Similarly it follows from $A... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 95 | 535 |
2009 | T0 | G2 | Geometry | IMO-SL | RUS Let $A B C$ be a triangle with circumcenter $O$. The points $P$ and $Q$ are interior points of the sides $C A$ and $A B$, respectively. The circle $k$ passes through the midpoints of the segments $B P$, $C Q$, and $P Q$. Prove that if the line $P Q$ is tangent to circle $k$ then $O P=O Q$. | Again, denote by $K, L, M$ the midpoints of segments $B P, C Q$, and $P Q$, respectively. Let $O, S, T$ be the circumcenters of triangles $A B C, K L M$, and $A P Q$, respectively (see Figure 2). Note that $M K$ and $L M$ are the midlines in triangles $B P Q$ and $C P Q$, respectively, so $\overrightarrow{M K}=\frac{1}... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 95 | 511 |
2009 | T0 | G3 | Geometry | IMO-SL | IRN Let $A B C$ be a triangle. The incircle of $A B C$ touches the sides $A B$ and $A C$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $B Y$ and $C Z$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $B C Y R$ and $B C S Z$ are parallelograms. Prove that $G R=G S$. | Denote $x=A Z=A Y, y=B Z=B X, z=C X=C Y, p=Z G, q=G C$. Several lengthy calculations (Menelaos' theorem in triangle $A Z C$, law of Cosines in triangles $A B C$ and $A Z C$ and Stewart's theorem in triangle $Z C S$ ) give four equations for $p, q, \cos \alpha$ and $G S$ in terms of $x, y$, and $z$ that can be resolved ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 112 | 1,171 |
2009 | T0 | G4 | Geometry | IMO-SL | UNK Given a cyclic quadrilateral $A B C D$, let the diagonals $A C$ and $B D$ meet at $E$ and the lines $A D$ and $B C$ meet at $F$. The midpoints of $A B$ and $C D$ are $G$ and $H$, respectively. Show that $E F$ is tangent at $E$ to the circle through the points $E, G$, and $H$. | It suffices to show that $\angle H E F=\angle H G E$ (see Figure 1), since in circle $E G H$ the angle over the chord $E H$ at $G$ equals the angle between the tangent at $E$ and $E H$. First, $\angle B A D=180^{\circ}-\angle D C B=\angle F C D$. Since triangles $F A B$ and $F C D$ have also a common interior angle at ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 98 | 573 |
2009 | T0 | G4 | Geometry | IMO-SL | UNK Given a cyclic quadrilateral $A B C D$, let the diagonals $A C$ and $B D$ meet at $E$ and the lines $A D$ and $B C$ meet at $F$. The midpoints of $A B$ and $C D$ are $G$ and $H$, respectively. Show that $E F$ is tangent at $E$ to the circle through the points $E, G$, and $H$. | We use the following Lemma (Gauß). Let $A B C D$ be a quadrilateral. Let $A B$ and $C D$ intersect at $P$, and $B C$ and $D A$ intersect at $Q$. Then the midpoints $K, L, M$ of $A C, B D$, and $P Q$, respectively, are collinear. Proof: Let us consider the points $Z$ that fulfill the equation $$ (A B Z)+(C D Z)=(B C Z)+... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 98 | 607 |
2009 | T0 | G4 | Geometry | IMO-SL | UNK Given a cyclic quadrilateral $A B C D$, let the diagonals $A C$ and $B D$ meet at $E$ and the lines $A D$ and $B C$ meet at $F$. The midpoints of $A B$ and $C D$ are $G$ and $H$, respectively. Show that $E F$ is tangent at $E$ to the circle through the points $E, G$, and $H$. | As in $$ \left(\frac{\boldsymbol{e}-\boldsymbol{f}}{2}\right)^{2}=\left(\frac{\boldsymbol{e}+\boldsymbol{f}}{2}-\frac{\boldsymbol{a}+\boldsymbol{b}}{2}\right)\left(\frac{\boldsymbol{e}+\boldsymbol{f}}{2}-\frac{\boldsymbol{c}+\boldsymbol{d}}{2}\right) $$ or equivalently $$ 4 \boldsymbol{e f}-(\boldsymbol{e}+\boldsymbol{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 98 | 614 |
2009 | T0 | G5 | Geometry | IMO-SL | POL Let $P$ be a polygon that is convex and symmetric to some point $O$. Prove that for some parallelogram $R$ satisfying $P \subset R$ we have $$ \frac{|R|}{|P|} \leq \sqrt{2} $$ where $|R|$ and $|P|$ denote the area of the sets $R$ and $P$, respectively. | We will construct two parallelograms $R_{1}$ and $R_{3}$, each of them containing $P$, and prove that at least one of the inequalities $\left|R_{1}\right| \leq \sqrt{2}|P|$ and $\left|R_{3}\right| \leq \sqrt{2}|P|$ holds (see Figure 1). First we will construct a parallelogram $R_{1} \supseteq P$ with the property that ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 87 | 1,455 |
2009 | T0 | G5 | Geometry | IMO-SL | POL Let $P$ be a polygon that is convex and symmetric to some point $O$. Prove that for some parallelogram $R$ satisfying $P \subset R$ we have $$ \frac{|R|}{|P|} \leq \sqrt{2} $$ where $|R|$ and $|P|$ denote the area of the sets $R$ and $P$, respectively. | We construct the parallelograms $R_{1}, R_{2}$ and $R_{3}$ in the same way as in Figure 2 Recall that affine one-to-one maps of the plane preserve the ratio of areas of subsets of the plane. On the other hand, every parallelogram can be transformed with an affine map onto a square. It follows that without loss of gener... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 87 | 645 |
2009 | T0 | G6 | Geometry | IMO-SL | UKR Let the sides $A D$ and $B C$ of the quadrilateral $A B C D$ (such that $A B$ is not parallel to $C D$ ) intersect at point $P$. Points $O_{1}$ and $O_{2}$ are the circumcenters and points $H_{1}$ and $H_{2}$ are the orthocenters of triangles $A B P$ and $D C P$, respectively. Denote the midpoints of segments $O_... | We keep triangle $A B P$ fixed and move the line $C D$ parallel to itself uniformly, i.e. linearly dependent on a single parameter $\lambda$ (see Figure 1). Then the points $C$ and $D$ also move uniformly. Hence, the points $O_{2}, H_{2}$ and $E_{2}$ move uniformly, too. Therefore also the perpendicular from $E_{2}$ on... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 182 | 1,259 |
2009 | T0 | G6 | Geometry | IMO-SL | UKR Let the sides $A D$ and $B C$ of the quadrilateral $A B C D$ (such that $A B$ is not parallel to $C D$ ) intersect at point $P$. Points $O_{1}$ and $O_{2}$ are the circumcenters and points $H_{1}$ and $H_{2}$ are the orthocenters of triangles $A B P$ and $D C P$, respectively. Denote the midpoints of segments $O_... | Let the perpendicular from $E_{1}$ on $C D$ meet $P H_{1}$ at $X$, and the perpendicular from $E_{2}$ on $A B$ meet $P H_{2}$ at $Y$ (see Figure 3). Let $\varphi$ be the intersection angle of $A B$ and $C D$. Denote by $M, N$ the midpoints of $P H_{1}, P H_{2}$ respectively. Figure 3 We will prove now that triangles $E... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 182 | 739 |
2009 | T0 | G7 | Geometry | IMO-SL | IRN Let $A B C$ be a triangle with incenter $I$ and let $X, Y$ and $Z$ be the incenters of the triangles $B I C, C I A$ and $A I B$, respectively. Let the triangle $X Y Z$ be equilateral. Prove that $A B C$ is equilateral too. | $A Z, A I$ and $A Y$ divide $\angle B A C$ into four equal angles; denote them by $\alpha$. In the same way we have four equal angles $\beta$ at $B$ and four equal angles $\gamma$ at $C$. Obviously $\alpha+\beta+\gamma=\frac{180^{\circ}}{4}=45^{\circ}$; and $0^{\circ}<\alpha, \beta, \gamma<45^{\circ}$. Easy calculation... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 78 | 1,531 |
2009 | T0 | G8 | Geometry | IMO-SL | BGR Let $A B C D$ be a circumscribed quadrilateral. Let $g$ be a line through $A$ which meets the segment $B C$ in $M$ and the line $C D$ in $N$. Denote by $I_{1}, I_{2}$, and $I_{3}$ the incenters of $\triangle A B M$, $\triangle M N C$, and $\triangle N D A$, respectively. Show that the orthocenter of $\triangle I_... | Let $k_{1}, k_{2}$ and $k_{3}$ be the incircles of triangles $A B M, M N C$, and $N D A$, respectively (see Figure 1). We shall show that the tangent $h$ from $C$ to $k_{1}$ which is different from $C B$ is also tangent to $k_{3}$. Figure 1 To this end, let $X$ denote the point of intersection of $g$ and $h$. Then $A B... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 122 | 764 |
2009 | T0 | G8 | Geometry | IMO-SL | BGR Let $A B C D$ be a circumscribed quadrilateral. Let $g$ be a line through $A$ which meets the segment $B C$ in $M$ and the line $C D$ in $N$. Denote by $I_{1}, I_{2}$, and $I_{3}$ the incenters of $\triangle A B M$, $\triangle M N C$, and $\triangle N D A$, respectively. Show that the orthocenter of $\triangle I_... | We start by proving that $C, I_{1}, I_{2}$, and $I_{3}$ are concyclic. Figure 2 To this end, notice first that $I_{2}, M, I_{1}$ are collinear, as are $N, I_{2}, I_{3}$ (see Figure 2). Denote by $\alpha, \beta, \gamma, \delta$ the internal angles of $A B C D$. By considerations in triangle $C M N$, it follows that $\an... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 122 | 891 |
2009 | T0 | N2 | Number Theory | IMO-SL | PER A positive integer $N$ is called balanced, if $N=1$ or if $N$ can be written as a product of an even number of not necessarily distinct primes. Given positive integers $a$ and $b$, consider the polynomial $P$ defined by $P(x)=(x+a)(x+b)$. (a) Prove that there exist distinct positive integers $a$ and $b$ such that... | Define a function $f$ on the set of positive integers by $f(n)=0$ if $n$ is balanced and $f(n)=1$ otherwise. Clearly, $f(n m) \equiv f(n)+f(m) \bmod 2$ for all positive integers $n, m$. (a) Now for each positive integer $n$ consider the binary sequence $(f(n+1), f(n+2), \ldots, f(n+$ $50)$ ). As there are only $2^{50}$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 138 | 599 |
2009 | T0 | N3 | Number Theory | IMO-SL | EST Let $f$ be a non-constant function from the set of positive integers into the set of positive integers, such that $a-b$ divides $f(a)-f(b)$ for all distinct positive integers $a, b$. Prove that there exist infinitely many primes $p$ such that $p$ divides $f(c)$ for some positive integer $c$. | Assume that there are only finitely many primes $p_{1}, p_{2}, \ldots, p_{m}$ that divide some function value of $f$. Since $f$ is not identically 1 , we must have $m \geq 1$. Then there exist non-negative integers $\alpha_{1}, \ldots, \alpha_{m}$ such that $$ f(1)=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{m}^{\a... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 77 | 575 |
2009 | T0 | N4 | Number Theory | IMO-SL | PRK Find all positive integers $n$ such that there exists a sequence of positive integers $a_{1}, a_{2}, \ldots, a_{n}$ satisfying $$ a_{k+1}=\frac{a_{k}^{2}+1}{a_{k-1}+1}-1 $$ for every $k$ with $2 \leq k \leq n-1$. | Such a sequence exists for $n=1,2,3,4$ and no other $n$. Since the existence of such a sequence for some $n$ implies the existence of such a sequence for all smaller $n$, it suffices to prove that $n=5$ is not possible and $n=4$ is possible. Assume first that for $n=5$ there exists a sequence of positive integers $a_{1... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 89 | 1,537 |
2009 | T0 | N4 | Number Theory | IMO-SL | PRK Find all positive integers $n$ such that there exists a sequence of positive integers $a_{1}, a_{2}, \ldots, a_{n}$ satisfying $$ a_{k+1}=\frac{a_{k}^{2}+1}{a_{k-1}+1}-1 $$ for every $k$ with $2 \leq k \leq n-1$. | It is easy to check that for $n=4$ the sequence $a_{1}=4, a_{2}=33, a_{3}=217$ and $a_{4}=1384$ is possible. Now assume there is a sequence with $n \geq 5$. Then we have in particular $$ \begin{aligned} & a_{2}^{2}+1=\left(a_{1}+1\right)\left(a_{3}+1\right), \\ & a_{3}^{2}+1=\left(a_{2}+1\right)\left(a_{4}+1\right), \\... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 89 | 1,647 |
2009 | T0 | N5 | Number Theory | IMO-SL | HUN Let $P(x)$ be a non-constant polynomial with integer coefficients. Prove that there is no function $T$ from the set of integers into the set of integers such that the number of integers $x$ with $T^{n}(x)=x$ is equal to $P(n)$ for every $n \geq 1$, where $T^{n}$ denotes the $n$-fold application of $T$. | Assume there is a polynomial $P$ of degree at least 1 with the desired property for a given function $T$. Let $A(n)$ denote the set of all $x \in \mathbb{Z}$ such that $T^{n}(x)=x$ and let $B(n)$ denote the set of all $x \in \mathbb{Z}$ for which $T^{n}(x)=x$ and $T^{k}(x) \neq x$ for all $1 \leq k<n$. Both sets are fi... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 92 | 1,255 |
2009 | T0 | N6 | Number Theory | IMO-SL | TUR Let $k$ be a positive integer. Show that if there exists a sequence $a_{0}, a_{1}, \ldots$ of integers satisfying the condition $$ a_{n}=\frac{a_{n-1}+n^{k}}{n} \quad \text { for all } n \geq 1 $$ then $k-2$ is divisible by 3 . | Part $A$. For each positive integer $k$, there exists a polynomial $P_{k}$ of degree $k-1$ with integer coefficients, i. e., $P_{k} \in \mathbb{Z}[x]$, and an integer $q_{k}$ such that the polynomial identity $$ x P_{k}(x)=x^{k}+P_{k}(x-1)+q_{k} $$ is satisfied. To prove this, for fixed $k$ we write $$ P_{k}(x)=b_{k-1}... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 88 | 843 |
2009 | T0 | N6 | Number Theory | IMO-SL | TUR Let $k$ be a positive integer. Show that if there exists a sequence $a_{0}, a_{1}, \ldots$ of integers satisfying the condition $$ a_{n}=\frac{a_{n-1}+n^{k}}{n} \quad \text { for all } n \geq 1 $$ then $k-2$ is divisible by 3 . | Part $A$ and $B$. Let $k$ be a positive integer, and suppose there is a sequence $a_{0}, a_{1}, \ldots$ as required. We prove: There exists a polynomial $P \in \mathbb{Z}[x]$, i. e., with integer coefficients, such that $a_{n}=P(n), n=0,1, \ldots$, and $\quad x P(x)=x^{k}+P(x-1)$. To prove this, we write $\quad P(x)=b_... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 88 | 1,754 |
2009 | T0 | N7 | Number Theory | IMO-SL | MNG Let $a$ and $b$ be distinct integers greater than 1 . Prove that there exists a positive integer $n$ such that $\left(a^{n}-1\right)\left(b^{n}-1\right)$ is not a perfect square. | At first we notice that $$ \begin{aligned} (1-\alpha)^{\frac{1}{2}}(1-\beta)^{\frac{1}{2}} & =\left(1-\frac{1}{2} \cdot \alpha-\frac{1}{8} \cdot \alpha^{2}-\cdots\right)\left(1-\frac{1}{2} \cdot \beta-\frac{1}{8} \cdot \beta^{2}-\cdots\right) \\ & =\sum_{k, \ell \geq 0} c_{k, \ell} \cdot \alpha^{k} \beta^{\ell} \quad \... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 57 | 1,932 |
2009 | T0 | N7 | Number Theory | IMO-SL | MNG Let $a$ and $b$ be distinct integers greater than 1 . Prove that there exists a positive integer $n$ such that $\left(a^{n}-1\right)\left(b^{n}-1\right)$ is not a perfect square. | We set $a^{2}=A, b^{2}=B$, and $z_{n}=\sqrt{\left(A^{n}-1\right)\left(B^{n}-1\right)}$. Let us assume that $z_{n}$ is an integer for $n=1,2, \ldots$. Without loss of generality, we may suppose that $b<a$. We determine an integer $k \geq 2$ such that $b^{k-1} \leq a<b^{k}$, and define a sequence $\gamma_{1}, \gamma_{2},... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 57 | 1,575 |
2009 | T0 | N7 | Number Theory | IMO-SL | MNG Let $a$ and $b$ be distinct integers greater than 1 . Prove that there exists a positive integer $n$ such that $\left(a^{n}-1\right)\left(b^{n}-1\right)$ is not a perfect square. | Lemma. Let $c$ be a positive integer, which is not a perfect square. Then there exists an odd prime $p$ such that $c$ is not a quadratic residue modulo $p$. Proof. Denoting the square-free part of $c$ by $c^{\prime}$, we have the equality $\left(\frac{c^{\prime}}{p}\right)=\left(\frac{c}{p}\right)$ of the corresponding... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 57 | 2,620 |
2009 | T0 | A1 | Algebra | IMO-SL | CZE Find the largest possible integer $k$, such that the following statement is true: Let 2009 arbitrary non-degenerated triangles be given. In every triangle the three sides are colored, such that one is blue, one is red and one is white. Now, for every color separately, let us sort the lengths of the sides. We obta... | We will prove that the largest possible number $k$ of indices satisfying the given condition is one. Firstly we prove that $b_{2009}, r_{2009}, w_{2009}$ are always lengths of the sides of a triangle. Without loss of generality we may assume that $w_{2009} \geq r_{2009} \geq b_{2009}$. We show that the inequality $b_{2... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 260 | 638 |
2009 | T0 | A3 | Algebra | IMO-SL | FRA Determine all functions $f$ from the set of positive integers into the set of positive integers such that for all $x$ and $y$ there exists a non degenerated triangle with sides of lengths $$ x, \quad f(y) \quad \text { and } \quad f(y+f(x)-1) . $$ | The identity function $f(x)=x$ is the only solution of the problem. If $f(x)=x$ for all positive integers $x$, the given three lengths are $x, y=f(y)$ and $z=$ $f(y+f(x)-1)=x+y-1$. Because of $x \geq 1, y \geq 1$ we have $z \geq \max \{x, y\}>|x-y|$ and $z<x+y$. From this it follows that a triangle with these side leng... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 70 | 835 |
2009 | T0 | N2 | Number Theory | IMO-SL | PER A positive integer $N$ is called balanced, if $N=1$ or if $N$ can be written as a product of an even number of not necessarily distinct primes. Given positive integers $a$ and $b$, consider the polynomial $P$ defined by $P(x)=(x+a)(x+b)$. (a) Prove that there exist distinct positive integers $a$ and $b$ such that... | Define a function $f$ on the set of positive integers by $f(n)=0$ if $n$ is balanced and $f(n)=1$ otherwise. Clearly, $f(n m) \equiv f(n)+f(m) \bmod 2$ for all positive integers $n, m$. (a) Now for each positive integer $n$ consider the binary sequence $(f(n+1), f(n+2), \ldots, f(n+$ $50)$ ). As there are only $2^{50}$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2009SL.jsonl",
"solution_match": null
} | 136 | 599 |
2010 | T0 | A2 | Algebra | IMO-SL | Let the real numbers $a, b, c, d$ satisfy the relations $a+b+c+d=6$ and $a^{2}+b^{2}+c^{2}+d^{2}=12$. Prove that $$ 36 \leq 4\left(a^{3}+b^{3}+c^{3}+d^{3}\right)-\left(a^{4}+b^{4}+c^{4}+d^{4}\right) \leq 48 $$ (Ukraine) | Observe that $$ \begin{gathered} 4\left(a^{3}+b^{3}+c^{3}+d^{3}\right)-\left(a^{4}+b^{4}+c^{4}+d^{4}\right)=-\left((a-1)^{4}+(b-1)^{4}+(c-1)^{4}+(d-1)^{4}\right) \\ +6\left(a^{2}+b^{2}+c^{2}+d^{2}\right)-4(a+b+c+d)+4 \\ =-\left((a-1)^{4}+(b-1)^{4}+(c-1)^{4}+(d-1)^{4}\right)+52 \end{gathered} $$ Now, introducing $x=a-1,... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 119 | 604 |
2010 | T0 | A2 | Algebra | IMO-SL | Let the real numbers $a, b, c, d$ satisfy the relations $a+b+c+d=6$ and $a^{2}+b^{2}+c^{2}+d^{2}=12$. Prove that $$ 36 \leq 4\left(a^{3}+b^{3}+c^{3}+d^{3}\right)-\left(a^{4}+b^{4}+c^{4}+d^{4}\right) \leq 48 $$ (Ukraine) | First, we claim that $0 \leq a, b, c, d \leq 3$. Actually, we have $$ a+b+c=6-d, \quad a^{2}+b^{2}+c^{2}=12-d^{2} $$ hence the power mean inequality $$ a^{2}+b^{2}+c^{2} \geq \frac{(a+b+c)^{2}}{3} $$ rewrites as $$ 12-d^{2} \geq \frac{(6-d)^{2}}{3} \quad \Longleftrightarrow \quad 2 d(d-3) \leq 0 $$ which implies the de... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 119 | 636 |
2010 | T0 | A2 | Algebra | IMO-SL | Let the real numbers $a, b, c, d$ satisfy the relations $a+b+c+d=6$ and $a^{2}+b^{2}+c^{2}+d^{2}=12$. Prove that $$ 36 \leq 4\left(a^{3}+b^{3}+c^{3}+d^{3}\right)-\left(a^{4}+b^{4}+c^{4}+d^{4}\right) \leq 48 $$ (Ukraine) | First, expanding $48=4\left(a^{2}+b^{2}+c^{2}+d^{2}\right)$ and applying the AM-GM inequality, we have $$ \begin{aligned} a^{4}+b^{4}+c^{4}+d^{4}+48 & =\left(a^{4}+4 a^{2}\right)+\left(b^{4}+4 b^{2}\right)+\left(c^{4}+4 c^{2}\right)+\left(d^{4}+4 d^{2}\right) \\ & \geq 2\left(\sqrt{a^{4} \cdot 4 a^{2}}+\sqrt{b^{4} \cdo... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 119 | 990 |
2010 | T0 | A3 | Algebra | IMO-SL | Let $x_{1}, \ldots, x_{100}$ be nonnegative real numbers such that $x_{i}+x_{i+1}+x_{i+2} \leq 1$ for all $i=1, \ldots, 100$ (we put $x_{101}=x_{1}, x_{102}=x_{2}$ ). Find the maximal possible value of the sum $$ S=\sum_{i=1}^{100} x_{i} x_{i+2} $$ (Russia) Answer. $\frac{25}{2}$. | Let $x_{2 i}=0, x_{2 i-1}=\frac{1}{2}$ for all $i=1, \ldots, 50$. Then we have $S=50 \cdot\left(\frac{1}{2}\right)^{2}=\frac{25}{2}$. So, we are left to show that $S \leq \frac{25}{2}$ for all values of $x_{i}$ 's satisfying the problem conditions. Consider any $1 \leq i \leq 50$. By the problem condition, we get $x_{2... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 138 | 662 |
2010 | T0 | A3 | Algebra | IMO-SL | Let $x_{1}, \ldots, x_{100}$ be nonnegative real numbers such that $x_{i}+x_{i+1}+x_{i+2} \leq 1$ for all $i=1, \ldots, 100$ (we put $x_{101}=x_{1}, x_{102}=x_{2}$ ). Find the maximal possible value of the sum $$ S=\sum_{i=1}^{100} x_{i} x_{i+2} $$ (Russia) Answer. $\frac{25}{2}$. | We present another proof of the estimate. From the problem condition, we get $$ \begin{aligned} S=\sum_{i=1}^{100} x_{i} x_{i+2} \leq \sum_{i=1}^{100} x_{i}\left(1-x_{i}-x_{i+1}\right) & =\sum_{i=1}^{100} x_{i}-\sum_{i=1}^{100} x_{i}^{2}-\sum_{i=1}^{100} x_{i} x_{i+1} \\ & =\sum_{i=1}^{100} x_{i}-\frac{1}{2} \sum_{i=1}... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 138 | 640 |
2010 | T0 | A4 | Algebra | IMO-SL | A sequence $x_{1}, x_{2}, \ldots$ is defined by $x_{1}=1$ and $x_{2 k}=-x_{k}, x_{2 k-1}=(-1)^{k+1} x_{k}$ for all $k \geq 1$. Prove that $x_{1}+x_{2}+\cdots+x_{n} \geq 0$ for all $n \geq 1$. (Austria) | We start with some observations. First, from the definition of $x_{i}$ it follows that for each positive integer $k$ we have $$ x_{4 k-3}=x_{2 k-1}=-x_{4 k-2} \quad \text { and } \quad x_{4 k-1}=x_{4 k}=-x_{2 k}=x_{k} $$ Hence, denoting $S_{n}=\sum_{i=1}^{n} x_{i}$, we have $$ \begin{gathered} S_{4 k}=\sum_{i=1}^{k}\le... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 106 | 679 |
2010 | T0 | A4 | Algebra | IMO-SL | A sequence $x_{1}, x_{2}, \ldots$ is defined by $x_{1}=1$ and $x_{2 k}=-x_{k}, x_{2 k-1}=(-1)^{k+1} x_{k}$ for all $k \geq 1$. Prove that $x_{1}+x_{2}+\cdots+x_{n} \geq 0$ for all $n \geq 1$. (Austria) | We will use the notation of $S_{n}$ and the relations (1)-(3) from the previous solution. Assume the contrary and consider the minimal $n$ such that $S_{n+1}<0$; surely $n \geq 1$, and from $S_{n} \geq 0$ we get $S_{n}=0, x_{n+1}=-1$. Hence, we are especially interested in the set $M=\left\{n: S_{n}=0\right\}$; our aim... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 106 | 531 |
2010 | T0 | A5 | Algebra | IMO-SL | Denote by $\mathbb{Q}^{+}$the set of all positive rational numbers. Determine all functions $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$ which satisfy the following equation for all $x, y \in \mathbb{Q}^{+}$: $$ f\left(f(x)^{2} y\right)=x^{3} f(x y) $$ (Switzerland) Answer. The only such function is $f(x)=\frac{1}{... | By substituting $y=1$, we get $$ f\left(f(x)^{2}\right)=x^{3} f(x) $$ Then, whenever $f(x)=f(y)$, we have $$ x^{3}=\frac{f\left(f(x)^{2}\right)}{f(x)}=\frac{f\left(f(y)^{2}\right)}{f(y)}=y^{3} $$ which implies $x=y$, so the function $f$ is injective. Now replace $x$ by $x y$ in (2), and apply (1) twice, second time to ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 117 | 983 |
2010 | T0 | A6 | Algebra | IMO-SL | Suppose that $f$ and $g$ are two functions defined on the set of positive integers and taking positive integer values. Suppose also that the equations $f(g(n))=f(n)+1$ and $g(f(n))=$ $g(n)+1$ hold for all positive integers. Prove that $f(n)=g(n)$ for all positive integer $n$. (Germany) | Throughout the solution, by $\mathbb{N}$ we denote the set of all positive integers. For any function $h: \mathbb{N} \rightarrow \mathbb{N}$ and for any positive integer $k$, define $h^{k}(x)=\underbrace{h(h(\ldots h}_{k}(x) \ldots)$ ) (in particular, $\left.h^{0}(x)=x\right)$. Observe that $f\left(g^{k}(x)\right)=f\le... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 81 | 1,453 |
2010 | T0 | A6 | Algebra | IMO-SL | Suppose that $f$ and $g$ are two functions defined on the set of positive integers and taking positive integer values. Suppose also that the equations $f(g(n))=f(n)+1$ and $g(f(n))=$ $g(n)+1$ hold for all positive integers. Prove that $f(n)=g(n)$ for all positive integer $n$. (Germany) | We start with the same observations, introducing the relation $\sim$ and proving Claim 1 from the previous solution. Note that $f(a)>a$ since otherwise we have $f(a)=a$ and hence $g(a)=g(f(a))=g(a)+1$, which is false. Claim 2'. $a=b$. Proof. We can assume that $a \leq b$. Since $f(a) \geq a+1$, there exists some $x \in... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 81 | 684 |
2010 | T0 | A7 | Algebra | IMO-SL | Let $a_{1}, \ldots, a_{r}$ be positive real numbers. For $n>r$, we inductively define $$ a_{n}=\max _{1 \leq k \leq n-1}\left(a_{k}+a_{n-k}\right) $$ Prove that there exist positive integers $\ell \leq r$ and $N$ such that $a_{n}=a_{n-\ell}+a_{\ell}$ for all $n \geq N$. (Iran) | First, from the problem conditions we have that each $a_{n}(n>r)$ can be expressed as $a_{n}=a_{j_{1}}+a_{j_{2}}$ with $j_{1}, j_{2}<n, j_{1}+j_{2}=n$. If, say, $j_{1}>r$ then we can proceed in the same way with $a_{j_{1}}$, and so on. Finally, we represent $a_{n}$ in a form $$ \begin{gathered} a_{n}=a_{i_{1}}+\cdots+a... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 115 | 1,186 |
2010 | T0 | A7 | Algebra | IMO-SL | Let $a_{1}, \ldots, a_{r}$ be positive real numbers. For $n>r$, we inductively define $$ a_{n}=\max _{1 \leq k \leq n-1}\left(a_{k}+a_{n-k}\right) $$ Prove that there exist positive integers $\ell \leq r$ and $N$ such that $a_{n}=a_{n-\ell}+a_{\ell}$ for all $n \geq N$. (Iran) | As in the previous solution, we involve the expansion (2), (3), and we fix some index $1 \leq \ell \leq r$ such that $$ \frac{a_{\ell}}{\ell}=s=\max _{1 \leq i \leq r} \frac{a_{i}}{i} $$ Now, we introduce the sequence $\left(b_{n}\right)$ as $b_{n}=a_{n}-s n$; then $b_{\ell}=0$. We prove by induction on $n$ that $b_{n}... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 115 | 982 |
2010 | T0 | A8 | Algebra | IMO-SL | Given six positive numbers $a, b, c, d, e, f$ such that $a<b<c<d<e<f$. Let $a+c+e=S$ and $b+d+f=T$. Prove that $$ 2 S T>\sqrt{3(S+T)(S(b d+b f+d f)+T(a c+a e+c e))} $$ (South Korea) | We define also $\sigma=a c+c e+a e, \tau=b d+b f+d f$. The idea of the solution is to interpret (1) as a natural inequality on the roots of an appropriate polynomial. Actually, consider the polynomial $$ \begin{aligned} & P(x)=(b+d+f)(x-a)(x-c)(x-e)+(a+c+e)(x-b)(x-d)(x-f) \\ &=T\left(x^{3}-S x^{2}+\sigma x-a c e\right)... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 83 | 856 |
2010 | T0 | A8 | Algebra | IMO-SL | Given six positive numbers $a, b, c, d, e, f$ such that $a<b<c<d<e<f$. Let $a+c+e=S$ and $b+d+f=T$. Prove that $$ 2 S T>\sqrt{3(S+T)(S(b d+b f+d f)+T(a c+a e+c e))} $$ (South Korea) | Let $$ U=\frac{1}{2}\left((e-a)^{2}+(c-a)^{2}+(e-c)^{2}\right)=S^{2}-3(a c+a e+c e) $$ and $$ V=\frac{1}{2}\left((f-b)^{2}+(f-d)^{2}+(d-b)^{2}\right)=T^{2}-3(b d+b f+d f) $$ Then $$ \begin{aligned} & \text { (L.H.S. })^{2}-(\text { R.H.S. })^{2}=(2 S T)^{2}-(S+T)(S \cdot 3(b d+b f+d f)+T \cdot 3(a c+a e+c e))= \\ & \qu... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 83 | 541 |
2010 | T0 | A8 | Algebra | IMO-SL | Given six positive numbers $a, b, c, d, e, f$ such that $a<b<c<d<e<f$. Let $a+c+e=S$ and $b+d+f=T$. Prove that $$ 2 S T>\sqrt{3(S+T)(S(b d+b f+d f)+T(a c+a e+c e))} $$ (South Korea) | We introduce the expressions $\sigma$ and $\tau$ as in the previous solutions. The idea of the solution is to change the values of variables $a, \ldots, f$ keeping the left-hand side unchanged and increasing the right-hand side; it will lead to a simpler inequality which can be proved in a direct way. Namely, we change... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 83 | 1,377 |
2010 | T0 | C1 | Combinatorics | IMO-SL | In a concert, 20 singers will perform. For each singer, there is a (possibly empty) set of other singers such that he wishes to perform later than all the singers from that set. Can it happen that there are exactly 2010 orders of the singers such that all their wishes are satisfied? (Austria) Answer. Yes, such an examp... | We say that an order of singers is good if it satisfied all their wishes. Next, we say that a number $N$ is realizable by $k$ singers (or $k$-realizable) if for some set of wishes of these singers there are exactly $N$ good orders. Thus, we have to prove that a number 2010 is 20-realizable. We start with the following ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 78 | 1,514 |
2010 | T0 | C2 | Combinatorics | IMO-SL | On some planet, there are $2^{N}$ countries $(N \geq 4)$. Each country has a flag $N$ units wide and one unit high composed of $N$ fields of size $1 \times 1$, each field being either yellow or blue. No two countries have the same flag. We say that a set of $N$ flags is diverse if these flags can be arranged into an $... | When speaking about the diagonal of a square, we will always mean the main diagonal. Let $M_{N}$ be the smallest positive integer satisfying the problem condition. First, we show that $M_{N}>2^{N-2}$. Consider the collection of all $2^{N-2}$ flags having yellow first squares and blue second ones. Obviously, both colors... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 159 | 1,194 |
2010 | T0 | C2 | Combinatorics | IMO-SL | On some planet, there are $2^{N}$ countries $(N \geq 4)$. Each country has a flag $N$ units wide and one unit high composed of $N$ fields of size $1 \times 1$, each field being either yellow or blue. No two countries have the same flag. We say that a set of $N$ flags is diverse if these flags can be arranged into an $... | We present a different proof of the estimate $M_{N} \leq 2^{N-2}+1$. We do not use the induction, involving Hall's lemma on matchings instead. Consider arbitrary $2^{N-2}+1$ distinct flags and arrange them into a large $\left(2^{N-2}+1\right) \times N$ flag. Construct two bipartite graphs $G_{\mathrm{y}}=\left(V \cup V... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 159 | 1,128 |
2010 | T0 | C3 | Combinatorics | IMO-SL | 2500 chess kings have to be placed on a $100 \times 100$ chessboard so that (i) no king can capture any other one (i.e. no two kings are placed in two squares sharing a common vertex); (ii) each row and each column contains exactly 25 kings. Find the number of such arrangements. (Two arrangements differing by rotation... | Suppose that we have an arrangement satisfying the problem conditions. Divide the board into $2 \times 2$ pieces; we call these pieces blocks. Each block can contain not more than one king (otherwise these two kings would attack each other); hence, by the pigeonhole principle each block must contain exactly one king. N... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 102 | 1,102 |
2010 | T0 | C4 | Combinatorics | IMO-SL | Six stacks $S_{1}, \ldots, S_{6}$ of coins are standing in a row. In the beginning every stack contains a single coin. There are two types of allowed moves: Move 1: If stack $S_{k}$ with $1 \leq k \leq 5$ contains at least one coin, you may remove one coin from $S_{k}$ and add two coins to $S_{k+1}$. Move 2: If stack $... | Denote by $\left(a_{1}, a_{2}, \ldots, a_{n}\right) \rightarrow\left(a_{1}^{\prime}, a_{2}^{\prime}, \ldots, a_{n}^{\prime}\right)$ the following: if some consecutive stacks contain $a_{1}, \ldots, a_{n}$ coins, then it is possible to perform several allowed moves such that the stacks contain $a_{1}^{\prime}, \ldots, a... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 272 | 2,097 |
2010 | T0 | C5 | Combinatorics | IMO-SL | $n \geq 4$ players participated in a tennis tournament. Any two players have played exactly one game, and there was no tie game. We call a company of four players bad if one player was defeated by the other three players, and each of these three players won a game and lost another game among themselves. Suppose that th... | For any tournament $T$ satisfying the problem condition, denote by $S(T)$ sum under consideration, namely $$ S(T)=\sum_{i=1}^{n}\left(w_{i}-\ell_{i}\right)^{3} $$ First, we show that the statement holds if a tournament $T$ has only 4 players. Actually, let $A=\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ be the number of wi... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 145 | 1,309 |
2010 | T0 | C5 | Combinatorics | IMO-SL | $n \geq 4$ players participated in a tennis tournament. Any two players have played exactly one game, and there was no tie game. We call a company of four players bad if one player was defeated by the other three players, and each of these three players won a game and lost another game among themselves. Suppose that th... | Similarly to the first solution, we call the subsets of players as companies, and the $k$-element subsets will be called as $k$-companies. In any company of the players, call a player the local champion of the company if he defeated all other members of the company. Similarly, if a player lost all his games against the... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 145 | 1,097 |
2010 | T0 | C6 | Combinatorics | IMO-SL | Given a positive integer $k$ and other two integers $b>w>1$. There are two strings of pearls, a string of $b$ black pearls and a string of $w$ white pearls. The length of a string is the number of pearls on it. One cuts these strings in some steps by the following rules. In each step: (i) The strings are ordered by th... | Denote the situation after the $i$ th step by $A_{i}$; hence $A_{0}$ is the initial situation, and $A_{i-1} \rightarrow A_{i}$ is the $i$ th step. We call a string containing $m$ pearls an $m$-string; it is an $m$-w-string or a $m$-b-string if it is white or black, respectively. We continue the process until every stri... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 293 | 1,459 |
2010 | T0 | C6 | Combinatorics | IMO-SL | Given a positive integer $k$ and other two integers $b>w>1$. There are two strings of pearls, a string of $b$ black pearls and a string of $w$ white pearls. The length of a string is the number of pearls on it. One cuts these strings in some steps by the following rules. In each step: (i) The strings are ordered by th... | We use the same notations as introduced in the first paragraph of the previous solution. We claim that at every stage, there exist a $u$-b-string and a $v$-w-string such that either (i) $u>v \geq 1$, or (ii) $2 \leq u \leq v<2 u$, and there also exist $k-1$ of ( $>v / 2$ )-strings other than considered above. First, we... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 293 | 1,523 |
2010 | T0 | C7 | Combinatorics | IMO-SL | Let $P_{1}, \ldots, P_{s}$ be arithmetic progressions of integers, the following conditions being satisfied: (i) each integer belongs to at least one of them; (ii) each progression contains a number which does not belong to other progressions. Denote by $n$ the least common multiple of steps of these progressions; let... | First, we prove the key lemma, and then we show how to apply it to finish the solution. Let $n_{1}, \ldots, n_{k}$ be positive integers. By an $n_{1} \times n_{2} \times \cdots \times n_{k}$ grid we mean the set $N=$ $\left\{\left(a_{1}, \ldots, a_{k}\right): a_{i} \in \mathbb{Z}, 0 \leq a_{i} \leq n_{i}-1\right\}$; th... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 148 | 3,223 |
2010 | T0 | C7 | Combinatorics | IMO-SL | Let $P_{1}, \ldots, P_{s}$ be arithmetic progressions of integers, the following conditions being satisfied: (i) each integer belongs to at least one of them; (ii) each progression contains a number which does not belong to other progressions. Denote by $n$ the least common multiple of steps of these progressions; let... | We start with introducing some notation. For positive integer $r$, we denote $[r]=\{1,2, \ldots, r\}$. Next, we say that a set of progressions $\mathcal{P}=\left\{P_{1}, \ldots, P_{s}\right\}$ cover $\mathbb{Z}$ if each integer belongs to some of them; we say that this covering is minimal if no proper subset of $\mathc... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 148 | 3,762 |
2010 | T0 | G1 | Geometry | IMO-SL | Let $A B C$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $B C, C A, A B$ respectively. One of the intersection points of the line $E F$ and the circumcircle is $P$. The lines $B P$ and $D F$ meet at point $Q$. Prove that $A P=A Q$. (United Kingdom) | The line $E F$ intersects the circumcircle at two points. Depending on the choice of $P$, there are two different cases to consider. Case 1: The point $P$ lies on the ray $E F$ (see Fig. 1). Let $\angle C A B=\alpha, \angle A B C=\beta$ and $\angle B C A=\gamma$. The quadrilaterals $B C E F$ and $C A F D$ are cyclic du... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 88 | 636 |
2010 | T0 | G2 | Geometry | IMO-SL | Point $P$ lies inside triangle $A B C$. Lines $A P, B P, C P$ meet the circumcircle of $A B C$ again at points $K, L, M$, respectively. The tangent to the circumcircle at $C$ meets line $A B$ at $S$. Prove that $S C=S P$ if and only if $M K=M L$. | As in the previous solution, we assume that $S$ lies on the ray $A B$. 1. Let $P$ be an arbitrary point inside both the circumcircle $\omega$ of the triangle $A B C$ and the angle $A S C$, the points $K, L, M$ defined as in the problem. We claim that $S P=S C$ implies $M K=M L$. Let $E$ and $F$ be the points of interse... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 84 | 758 |
2010 | T0 | G2 | Geometry | IMO-SL | Point $P$ lies inside triangle $A B C$. Lines $A P, B P, C P$ meet the circumcircle of $A B C$ again at points $K, L, M$, respectively. The tangent to the circumcircle at $C$ meets line $A B$ at $S$. Prove that $S C=S P$ if and only if $M K=M L$. | We present a different proof of the converse direction, that is, $M K=M L \Rightarrow$ $S P=S C$. As in the previous solutions we assume that $C A>C B$, and the line $S P$ meets $\omega$ at $E$ and $F$. From $M L=M K$ we get $\widehat{M E K}=\widehat{M F L}$. Now we claim that $\widehat{M E}=\widehat{M F}$ and $\wideha... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 84 | 583 |
2010 | T0 | G3 | Geometry | IMO-SL | Let $A_{1} A_{2} \ldots A_{n}$ be a convex polygon. Point $P$ inside this polygon is chosen so that its projections $P_{1}, \ldots, P_{n}$ onto lines $A_{1} A_{2}, \ldots, A_{n} A_{1}$ respectively lie on the sides of the polygon. Prove that for arbitrary points $X_{1}, \ldots, X_{n}$ on sides $A_{1} A_{2}, \ldots, A_{... | Denote $P_{n+1}=P_{1}, X_{n+1}=X_{1}, A_{n+1}=A_{1}$. Lemma. Let point $Q$ lies inside $A_{1} A_{2} \ldots A_{n}$. Then it is contained in at least one of the circumcircles of triangles $X_{1} A_{2} X_{2}, \ldots, X_{n} A_{1} X_{1}$. Proof. If $Q$ lies in one of the triangles $X_{1} A_{2} X_{2}, \ldots, X_{n} A_{1} X_{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 191 | 684 |
2010 | T0 | G3 | Geometry | IMO-SL | Let $A_{1} A_{2} \ldots A_{n}$ be a convex polygon. Point $P$ inside this polygon is chosen so that its projections $P_{1}, \ldots, P_{n}$ onto lines $A_{1} A_{2}, \ldots, A_{n} A_{1}$ respectively lie on the sides of the polygon. Prove that for arbitrary points $X_{1}, \ldots, X_{n}$ on sides $A_{1} A_{2}, \ldots, A_{... | As in We will prove a bit stronger inequality, namely $$ \max \left\{\frac{X_{1} X_{2}}{P_{1} P_{2}} \cos \alpha_{1}, \ldots, \frac{X_{n} X_{1}}{P_{n} P_{1}} \cos \alpha_{n}\right\} \geq 1 $$ where $\alpha_{i}(1 \leq i \leq n)$ is the angle between lines $X_{i} X_{i+1}$ and $P_{i} P_{i+1}$. We denote $\beta_{i}=\angle ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 191 | 907 |
2010 | T0 | G4 | Geometry | IMO-SL | Let $I$ be the incenter of a triangle $A B C$ and $\Gamma$ be its circumcircle. Let the line $A I$ intersect $\Gamma$ at a point $D \neq A$. Let $F$ and $E$ be points on side $B C$ and $\operatorname{arc} B D C$ respectively such that $\angle B A F=\angle C A E<\frac{1}{2} \angle B A C$. Finally, let $G$ be the midpoin... | Let $X$ be the second point of intersection of line $E I$ with $\Gamma$, and $L$ be the foot of the bisector of angle $B A C$. Let $G^{\prime}$ and $T$ be the points of intersection of segment $D X$ with lines $I F$ and $A F$, respectively. We are to prove that $G=G^{\prime}$, or $I G^{\prime}=G^{\prime} F$. By the Men... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 139 | 701 |
2010 | T0 | G4 | Geometry | IMO-SL | Let $I$ be the incenter of a triangle $A B C$ and $\Gamma$ be its circumcircle. Let the line $A I$ intersect $\Gamma$ at a point $D \neq A$. Let $F$ and $E$ be points on side $B C$ and $\operatorname{arc} B D C$ respectively such that $\angle B A F=\angle C A E<\frac{1}{2} \angle B A C$. Finally, let $G$ be the midpoin... | As in the previous solution, we introduce the points $X, T$ and $K$ and note that it suffice to prove the equality $$ \frac{T F}{A T}=\frac{D I}{A D} \quad \Longleftrightarrow \quad \frac{T F+A T}{A T}=\frac{D I+A D}{A D} \quad \Longleftrightarrow \quad \frac{A T}{A D}=\frac{A F}{D I+A D} $$ Since $\angle F A D=\angle ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 139 | 553 |
2010 | T0 | G5 | Geometry | IMO-SL | Let $A B C D E$ be a convex pentagon such that $B C \| A E, A B=B C+A E$, and $\angle A B C=$ $\angle C D E$. Let $M$ be the midpoint of $C E$, and let $O$ be the circumcenter of triangle $B C D$. Given that $\angle D M O=90^{\circ}$, prove that $2 \angle B D A=\angle C D E$. (Ukraine) | Let $\angle C B D=\alpha, \angle B D C=\beta, \angle A D E=\gamma$, and $\angle A B C=\angle C D E=2 \varphi$. Then we have $\angle A D B=2 \varphi-\beta-\gamma, \angle B C D=180^{\circ}-\alpha-\beta, \angle A E D=360^{\circ}-\angle B C D-\angle C D E=$ $180^{\circ}-2 \varphi+\alpha+\beta$, and finally $\angle D A E=18... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 104 | 1,111 |
2010 | T0 | G7 | Geometry | IMO-SL | Three circular arcs $\gamma_{1}, \gamma_{2}$, and $\gamma_{3}$ connect the points $A$ and $C$. These arcs lie in the same half-plane defined by line $A C$ in such a way that arc $\gamma_{2}$ lies between the arcs $\gamma_{1}$ and $\gamma_{3}$. Point $B$ lies on the segment $A C$. Let $h_{1}, h_{2}$, and $h_{3}$ be thre... | Denote by $O_{i}$ and $R_{i}$ the center and the radius of $\gamma_{i}$, respectively. Denote also by $H$ the half-plane defined by $A C$ which contains the whole configuration. For every point $P$ in the half-plane $H$, denote by $d(P)$ the distance between $P$ and line $A C$. Furthermore, for any $r>0$, denote by $\O... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 508 | 3,136 |
2010 | T0 | N2 | Number Theory | IMO-SL | Find all pairs $(m, n)$ of nonnegative integers for which $$ m^{2}+2 \cdot 3^{n}=m\left(2^{n+1}-1\right) $$ (Australia) Answer. $(6,3),(9,3),(9,5),(54,5)$. | For fixed values of $n$, the equation (1) is a simple quadratic equation in $m$. For $n \leq 5$ the solutions are listed in the following table. | case | equation | discriminant | integer roots | | :--- | :--- | :--- | :--- | | $n=0$ | $m^{2}-m+2=0$ | -7 | none | | $n=1$ | $m^{2}-3 m+6=0$ | -15 | none | | $n=2$ | $m^{2... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 69 | 1,186 |
2010 | T0 | N3 | Number Theory | IMO-SL | Find the smallest number $n$ such that there exist polynomials $f_{1}, f_{2}, \ldots, f_{n}$ with rational coefficients satisfying $$ x^{2}+7=f_{1}(x)^{2}+f_{2}(x)^{2}+\cdots+f_{n}(x)^{2} . $$ (Poland) Answer. The smallest $n$ is 5 . | The equality $x^{2}+7=x^{2}+2^{2}+1^{2}+1^{2}+1^{2}$ shows that $n \leq 5$. It remains to show that $x^{2}+7$ is not a sum of four (or less) squares of polynomials with rational coefficients. Suppose by way of contradiction that $x^{2}+7=f_{1}(x)^{2}+f_{2}(x)^{2}+f_{3}(x)^{2}+f_{4}(x)^{2}$, where the coefficients of po... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 91 | 1,059 |
2010 | T0 | N3 | Number Theory | IMO-SL | Find the smallest number $n$ such that there exist polynomials $f_{1}, f_{2}, \ldots, f_{n}$ with rational coefficients satisfying $$ x^{2}+7=f_{1}(x)^{2}+f_{2}(x)^{2}+\cdots+f_{n}(x)^{2} . $$ (Poland) Answer. The smallest $n$ is 5 . | We prove that $n \leq 4$ is impossible. Define the numbers $a_{i}, b_{i}$ for $i=1,2,3,4$ as in the previous solution. By Euler's identity we have $$ \begin{aligned} \left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}+b_{4}^{2}\right) & =\left(a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}+a_{4... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 91 | 721 |
2010 | T0 | N4 | Number Theory | IMO-SL | Let $a, b$ be integers, and let $P(x)=a x^{3}+b x$. For any positive integer $n$ we say that the pair $(a, b)$ is $n$-good if $n \mid P(m)-P(k)$ implies $n \mid m-k$ for all integers $m, k$. We say that $(a, b)$ is very good if $(a, b)$ is $n$-good for infinitely many positive integers $n$. (a) Find a pair $(a, b)$ whi... | (a) We show that the pair $\left(1,-51^{2}\right)$ is good but not very good. Let $P(x)=x^{3}-51^{2} x$. Since $P(51)=P(0)$, the pair $\left(1,-51^{2}\right)$ is not $n$-good for any positive integer that does not divide 51 . Therefore, $\left(1,-51^{2}\right)$ is not very good. On the other hand, if $P(m) \equiv P(k)(... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 149 | 1,570 |
2010 | T0 | N5 | Number Theory | IMO-SL | Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that the number $(f(m)+n)(m+f(n))$ is a square for all $m, n \in \mathbb{N}$. Answer. All functions of the form $f(n)=n+c$, where $c \in \mathbb{N} \cup\{0\}$. | First, it is clear that all functions of the form $f(n)=n+c$ with a constant nonnegative integer $c$ satisfy the problem conditions since $(f(m)+n)(f(n)+m)=(n+m+c)^{2}$ is a square. We are left to prove that there are no other functions. We start with the following Lemma. Suppose that $p \mid f(k)-f(\ell)$ for some pri... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 102 | 895 |
2010 | T0 | N6 | Number Theory | IMO-SL | The rows and columns of a $2^{n} \times 2^{n}$ table are numbered from 0 to $2^{n}-1$. The cells of the table have been colored with the following property being satisfied: for each $0 \leq i, j \leq 2^{n}-1$, the $j$ th cell in the $i$ th row and the $ Prove that the maximal possible number of colors is $2^{n}$. | Throughout the solution we denote the cells of the table by coordinate pairs; $(i, j)$ refers to the $j$ th cell in the $i$ th row. Consider the directed graph, whose vertices are the cells of the board, and the edges are the arrows $(i, j) \rightarrow(j, i+j)$ for all $0 \leq i, j \leq 2^{n}-1$. From each vertex $(i, ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2010SL.jsonl",
"solution_match": null
} | 100 | 3,884 |
2011 | T0 | A1 | Algebra | IMO-SL | For any set $A=\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}$ of four distinct positive integers with sum $s_{A}=a_{1}+a_{2}+a_{3}+a_{4}$, let $p_{A}$ denote the number of pairs $(i, j)$ with $1 \leq i<j \leq 4$ for which $a_{i}+a_{j}$ divides $s_{A}$. Among all sets of four distinct positive integers, determine those sets... | Firstly, we will prove that the maximum value of $p_{A}$ is at most 4 . Without loss of generality, we may assume that $a_{1}<a_{2}<a_{3}<a_{4}$. We observe that for each pair of indices $(i, j)$ with $1 \leq i<j \leq 4$, the sum $a_{i}+a_{j}$ divides $s_{A}$ if and only if $a_{i}+a_{j}$ divides $s_{A}-\left(a_{i}+a_{j... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 134 | 933 |
2011 | T0 | A2 | Algebra | IMO-SL | Determine all sequences $\left(x_{1}, x_{2}, \ldots, x_{2011}\right)$ of positive integers such that for every positive integer $n$ there is an integer $a$ with $$ x_{1}^{n}+2 x_{2}^{n}+\cdots+2011 x_{2011}^{n}=a^{n+1}+1 . $$ | Throughout this solution, the set of positive integers will be denoted by $\mathbb{Z}_{+}$. Put $k=2+3+\cdots+2011=2023065$. We have $$ 1^{n}+2 k^{n}+\cdots 2011 k^{n}=1+k \cdot k^{n}=k^{n+1}+1 $$ for all $n$, so $(1, k, \ldots, k)$ is a valid sequence. We shall prove that it is the only one. Let a valid sequence $\lef... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 93 | 966 |
2011 | T0 | A3 | Algebra | IMO-SL | Determine all pairs $(f, g)$ of functions from the set of real numbers to itself that satisfy $$ g(f(x+y))=f(x)+(2 x+y) g(y) $$ for all real numbers $x$ and $y$. | Clearly all these pairs of functions satisfy the functional equation in question, so it suffices to verify that there cannot be any further ones. Substituting $-2 x$ for $y$ in the given functional equation we obtain $$ g(f(-x))=f(x) . $$ Using this equation for $-x-y$ in place of $x$ we obtain $$ f(-x-y)=g(f(x+y))=f(x... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 52 | 808 |
2011 | T0 | A4 | Algebra | IMO-SL | Determine all pairs $(f, g)$ of functions from the set of positive integers to itself that satisfy $$ f^{g(n)+1}(n)+g^{f(n)}(n)=f(n+1)-g(n+1)+1 $$ for every positive integer $n$. Here, $f^{k}(n)$ means $\underbrace{f(f(\ldots f}_{k}(n) \ldots))$. | The given relation implies $$ f\left(f^{g(n)}(n)\right)<f(n+1) \text { for all } n $$ which will turn out to be sufficient to determine $f$. Let $y_{1}<y_{2}<\ldots$ be all the values attained by $f$ (this sequence might be either finite or infinite). We will prove that for every positive $n$ the function $f$ attains a... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 90 | 898 |
2011 | T0 | A5 | Algebra | IMO-SL | Prove that for every positive integer $n$, the set $\{2,3,4, \ldots, 3 n+1\}$ can be partitioned into $n$ triples in such a way that the numbers from each triple are the lengths of the sides of some obtuse triangle. | Throughout the solution, we denote by $[a, b]$ the set $\{a, a+1, \ldots, b\}$. We say that $\{a, b, c\}$ is an obtuse triple if $a, b, c$ are the sides of some obtuse triangle. We prove by induction on $n$ that there exists a partition of $[2,3 n+1]$ into $n$ obtuse triples $A_{i}$ $(2 \leq i \leq n+1)$ having the for... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 62 | 807 |
2011 | T0 | A6 | Algebra | IMO-SL | Let $f$ be a function from the set of real numbers to itself that satisfies $$ f(x+y) \leq y f(x)+f(f(x)) $$ for all real numbers $x$ and $y$. Prove that $f(x)=0$ for all $x \leq 0$. | We will also use the condition of the problem in form (2)). For clarity we divide the argument into four steps. Step 1. We begin by proving that $f$ attains nonpositive values only. Assume that there exist some real number $z$ with $f(z)>0$. Substituting $x=z$ into (2) and setting $A=f(z)$, $B=-z f(z)-f(f(z))$ we get $... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 66 | 1,259 |
2011 | T0 | A7 | Algebra | IMO-SL | Let $a, b$, and $c$ be positive real numbers satisfying $\min (a+b, b+c, c+a)>\sqrt{2}$ and $a^{2}+b^{2}+c^{2}=3$. Prove that $$ \frac{a}{(b+c-a)^{2}}+\frac{b}{(c+a-b)^{2}}+\frac{c}{(a+b-c)^{2}} \geq \frac{3}{(a b c)^{2}} $$ | The condition $b+c>\sqrt{2}$ implies $b^{2}+c^{2}>1$, so $a^{2}=3-\left(b^{2}+c^{2}\right)<2$, i.e. $a<\sqrt{2}<b+c$. Hence we have $b+c-a>0$, and also $c+a-b>0$ and $a+b-c>0$ for similar reasons. We will use the variant of HÖLDER's inequality $$ \frac{x_{1}^{p+1}}{y_{1}^{p}}+\frac{x_{1}^{p+1}}{y_{1}^{p}}+\ldots+\frac{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 111 | 894 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.