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 |
|---|---|---|---|---|---|---|---|---|---|
2016 | T0 | G6 | Geometry | IMO-SL | Let $A B C D$ be a convex quadrilateral with $\angle A B C=\angle A D C<90^{\circ}$. The internal angle bisectors of $\angle A B C$ and $\angle A D C$ meet $A C$ at $E$ and $F$ respectively, and meet each other at point $P$. Let $M$ be the midpoint of $A C$ and let $\omega$ be the circumcircle of triangle $B P D$. Segm... | We first state two results which will be needed in our proof. - Claim 1. In $\triangle X^{\prime} Y^{\prime} Z^{\prime}$ with $X^{\prime} Y^{\prime} \neq X^{\prime} Z^{\prime}$, let $N^{\prime}$ be the midpoint of $Y^{\prime} Z^{\prime}$ and $W^{\prime}$ be the foot of internal angle bisector from $X^{\prime}$. Then $\... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
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} | 160 | 1,644 |
2016 | T0 | G7 | Geometry | IMO-SL | Let $I$ be the incentre of a non-equilateral triangle $A B C, I_{A}$ be the $A$-excentre, $I_{A}^{\prime}$ be the reflection of $I_{A}$ in $B C$, and $l_{A}$ be the reflection of line $A I_{A}^{\prime}$ in $A I$. Define points $I_{B}, I_{B}^{\prime}$ and line $l_{B}$ analogously. Let $P$ be the intersection point of $l... | (a) Let $A^{\prime}$ be the reflection of $A$ in $B C$ and let $M$ be the second intersection of line $A I$ and the circumcircle $\Gamma$ of triangle $A B C$. As triangles $A B A^{\prime}$ and $A O C$ are isosceles with $\angle A B A^{\prime}=2 \angle A B C=\angle A O C$, they are similar to each other. Also, triangles... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 211 | 765 |
2016 | T0 | G7 | Geometry | IMO-SL | Let $I$ be the incentre of a non-equilateral triangle $A B C, I_{A}$ be the $A$-excentre, $I_{A}^{\prime}$ be the reflection of $I_{A}$ in $B C$, and $l_{A}$ be the reflection of line $A I_{A}^{\prime}$ in $A I$. Define points $I_{B}, I_{B}^{\prime}$ and line $l_{B}$ analogously. Let $P$ be the intersection point of $l... | (a) Note that triangles $A I_{B} C$ and $I_{A} B C$ are similar since their corresponding interior angles are equal. Therefore, the four triangles $A I_{B}^{\prime} C, A I_{B} C, I_{A} B C$ and $I_{A}^{\prime} B C$ are all similar. From $\triangle A I_{B}^{\prime} C \sim \triangle I_{A}^{\prime} B C$, we get $\triangle... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 211 | 917 |
2016 | T0 | G8 | Geometry | IMO-SL | Let $A_{1}, B_{1}$ and $C_{1}$ be points on sides $B C, C A$ and $A B$ of an acute triangle $A B C$ respectively, such that $A A_{1}, B B_{1}$ and $C C_{1}$ are the internal angle bisectors of triangle $A B C$. Let $I$ be the incentre of triangle $A B C$, and $H$ be the orthocentre of triangle $A_{1} B_{1} C_{1}$. Show... | Without loss of generality, assume $\alpha=\angle B A C \leqslant \beta=\angle C B A \leqslant \gamma=\angle A C B$. Denote by $a, b, c$ the lengths of $B C, C A, A B$ respectively. We first show that triangle $A_{1} B_{1} C_{1}$ is acute. Choose points $D$ and $E$ on side $B C$ such that $B_{1} D / / A B$ and $B_{1} E... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
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} | 138 | 1,360 |
2016 | T0 | N1 | Number Theory | IMO-SL | For any positive integer $k$, denote the sum of digits of $k$ in its decimal representation by $S(k)$. Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n \geqslant 2016$, the integer $P(n)$ is positive and $$ S(P(n))=P(S(n)) $$ | We consider three cases according to the degree of $P$. - Case 1. $P(x)$ is a constant polynomial. Let $P(x)=c$ where $c$ is an integer constant. Then (1) becomes $S(c)=c$. This holds if and only if $1 \leqslant c \leqslant 9$. - Case 2. $\operatorname{deg} P=1$. We have the following observation. For any positive inte... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 77 | 841 |
2016 | T0 | N1 | Number Theory | IMO-SL | For any positive integer $k$, denote the sum of digits of $k$ in its decimal representation by $S(k)$. Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n \geqslant 2016$, the integer $P(n)$ is positive and $$ S(P(n))=P(S(n)) $$ | Let $P(x)=a_{d} x^{d}+a_{d-1} x^{d-1}+\cdots+a_{0}$. Clearly $a_{d}>0$. There exists an integer $m \geqslant 1$ such that $\left|a_{i}\right|<10^{m}$ for all $0 \leqslant i \leqslant d$. Consider $n=9 \times 10^{k}$ for a sufficiently large integer $k$ in (1). If there exists an index $0 \leqslant i \leqslant d-1$ such... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 77 | 798 |
2016 | T0 | N2 | Number Theory | IMO-SL | Let $\tau(n)$ be the number of positive divisors of $n$. Let $\tau_{1}(n)$ be the number of positive divisors of $n$ which have remainders 1 when divided by 3 . Find all possible integral values of the fraction $\frac{\tau(10 n)}{\tau_{1}(10 n)}$. | In this solution, we always use $p_{i}$ to denote primes congruent to $1 \bmod 3$, and use $q_{j}$ to denote primes congruent to $2 \bmod 3$. When we express a positive integer $m$ using its prime factorization, we also include the special case $m=1$ by allowing the exponents to be zeros. We first compute $\tau_{1}(m)$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 74 | 1,494 |
2016 | T0 | N3 | Number Theory | IMO-SL | Define $P(n)=n^{2}+n+1$. For any positive integers $a$ and $b$, the set $$ \{P(a), P(a+1), P(a+2), \ldots, P(a+b)\} $$ is said to be fragrant if none of its elements is relatively prime to the product of the other elements. Determine the smallest size of a fragrant set. | We have the following observations. (i) $(P(n), P(n+1))=1$ for any $n$. We have $(P(n), P(n+1))=\left(n^{2}+n+1, n^{2}+3 n+3\right)=\left(n^{2}+n+1,2 n+2\right)$. Noting that $n^{2}+n+1$ is odd and $\left(n^{2}+n+1, n+1\right)=(1, n+1)=1$, the claim follows. (ii) $(P(n), P(n+2))=1$ for $n \not \equiv 2(\bmod 7)$ and $(... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 87 | 1,459 |
2016 | T0 | N4 | Number Theory | IMO-SL | Let $n, m, k$ and $l$ be positive integers with $n \neq 1$ such that $n^{k}+m n^{l}+1$ divides $n^{k+l}-1$. Prove that - $m=1$ and $l=2 k$; or - $l \mid k$ and $m=\frac{n^{k-l}-1}{n^{l}-1}$. | It is given that $$ n^{k}+m n^{l}+1 \mid n^{k+l}-1 $$ This implies $$ n^{k}+m n^{l}+1 \mid\left(n^{k+l}-1\right)+\left(n^{k}+m n^{l}+1\right)=n^{k+l}+n^{k}+m n^{l} . $$ We have two cases to discuss. - Case 1. $l \geqslant k$. Since $\left(n^{k}+m n^{l}+1, n\right)=1$,(2) yields $$ n^{k}+m n^{l}+1 \mid n^{l}+m n^{l-k}+1... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 94 | 708 |
2016 | T0 | N5 | Number Theory | IMO-SL | Let $a$ be a positive integer which is not a square number. Denote by $A$ the set of all positive integers $k$ such that $$ k=\frac{x^{2}-a}{x^{2}-y^{2}} $$ for some integers $x$ and $y$ with $x>\sqrt{a}$. Denote by $B$ the set of all positive integers $k$ such that (1) is satisfied for some integers $x$ and $y$ with... | We first prove the following preliminary result. - Claim. For fixed $k$, let $x, y$ be integers satisfying (1). Then the numbers $x_{1}, y_{1}$ defined by $$ x_{1}=\frac{1}{2}\left(x-y+\frac{(x-y)^{2}-4 a}{x+y}\right), \quad y_{1}=\frac{1}{2}\left(x-y-\frac{(x-y)^{2}-4 a}{x+y}\right) $$ are integers and satisfy (1) (wi... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 127 | 991 |
2016 | T0 | N5 | Number Theory | IMO-SL | Let $a$ be a positive integer which is not a square number. Denote by $A$ the set of all positive integers $k$ such that $$ k=\frac{x^{2}-a}{x^{2}-y^{2}} $$ for some integers $x$ and $y$ with $x>\sqrt{a}$. Denote by $B$ the set of all positive integers $k$ such that (1) is satisfied for some integers $x$ and $y$ with... | The relation (1) is equivalent to $$ k y^{2}-(k-1) x^{2}=a $$ Motivated by Pell's Equation, we prove the following, which is essentially the same as the Claim in Solution 1. - Claim. If $\left(x_{0}, y_{0}\right)$ is a solution to $(2)$, then $\left((2 k-1) x_{0} \pm 2 k y_{0},(2 k-1) y_{0} \pm 2(k-1) x_{0}\right)$ is ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 127 | 744 |
2016 | T0 | N5 | Number Theory | IMO-SL | Let $a$ be a positive integer which is not a square number. Denote by $A$ the set of all positive integers $k$ such that $$ k=\frac{x^{2}-a}{x^{2}-y^{2}} $$ for some integers $x$ and $y$ with $x>\sqrt{a}$. Denote by $B$ the set of all positive integers $k$ such that (1) is satisfied for some integers $x$ and $y$ with... | It suffices to show $A \cup B$ is a subset of $A \cap B$. We take any $k \in A \cup B$, which means there exist integers $x, y$ satisfying (1). Since $a$ is not a square, it follows that $k \neq 1$. As in Without loss of generality, assume $x, y \geqslant 0$. Let $u=x+y$ and $v=x-y$. Then $u \geqslant v$ and (1) become... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 127 | 756 |
2016 | T0 | N6 | Number Theory | IMO-SL | Denote by $\mathbb{N}$ the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for all positive integers $m$ and $n$, the integer $f(m)+f(n)-m n$ is nonzero and divides $m f(m)+n f(n)$. | It is given that $$ f(m)+f(n)-m n \mid m f(m)+n f(n) $$ Taking $m=n=1$ in (1), we have $2 f(1)-1 \mid 2 f(1)$. Then $2 f(1)-1 \mid 2 f(1)-(2 f(1)-1)=1$ and hence $f(1)=1$. Let $p \geqslant 7$ be a prime. Taking $m=p$ and $n=1$ in (1), we have $f(p)-p+1 \mid p f(p)+1$ and hence $$ f(p)-p+1 \mid p f(p)+1-p(f(p)-p+1)=p^{2... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 75 | 886 |
2016 | T0 | N7 | Number Theory | IMO-SL | Let $n$ be an odd positive integer. In the Cartesian plane, a cyclic polygon $P$ with area $S$ is chosen. All its vertices have integral coordinates, and all squares of its side lengths are divisible by $n$. Prove that $2 S$ is an integer divisible by $n$. | Let $P=A_{1} A_{2} \ldots A_{k}$ and let $A_{k+i}=A_{i}$ for $i \geqslant 1$. By the Shoelace Formula, the area of any convex polygon with integral coordinates is half an integer. Therefore, $2 S$ is an integer. We shall prove by induction on $k \geqslant 3$ that $2 S$ is divisible by $n$. Clearly, it suffices to consi... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
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} | 64 | 1,760 |
2016 | T0 | N8 | Number Theory | IMO-SL | Find all polynomials $P(x)$ of odd degree $d$ and with integer coefficients satisfying the following property: for each positive integer $n$, there exist $n$ positive integers $x_{1}, x_{2}, \ldots, x_{n}$ such that $\frac{1}{2}<\frac{P\left(x_{i}\right)}{P\left(x_{j}\right)}<2$ and $\frac{P\left(x_{i}\right)}{P\left(x... | Let $P(x)=a_{d} x^{d}+a_{d-1} x^{d-1}+\cdots+a_{0}$. Consider the substitution $y=d a_{d} x+a_{d-1}$. By defining $Q(y)=P(x)$, we find that $Q$ is a polynomial with rational coefficients without the term $y^{d-1}$. Let $Q(y)=b_{d} y^{d}+b_{d-2} y^{d-2}+b_{d-3} y^{d-3}+\cdots+b_{0}$ and $B=\max _{0 \leqslant i \leqslant... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 154 | 3,041 |
2016 | T0 | A2 | Algebra | IMO-SL | Find the smallest real constant $C$ such that for any positive real numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ (not necessarily distinct), one can always choose distinct subscripts $i, j, k$ and $l$ such that $$ \left|\frac{a_{i}}{a_{j}}-\frac{a_{k}}{a_{l}}\right| \leqslant C . $$ Answer. The smallest $C$ is $\... | We first show that $C \leqslant \frac{1}{2}$. For any positive real numbers $a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant a_{4} \leqslant a_{5}$, consider the five fractions $$ \frac{a_{1}}{a_{2}}, \frac{a_{3}}{a_{4}}, \frac{a_{1}}{a_{5}}, \frac{a_{2}}{a_{3}}, \frac{a_{4}}{a_{5}} . $$ Each of them lies in the interv... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 119 | 649 |
2016 | T0 | A3 | Algebra | IMO-SL | Find all integers $n \geqslant 3$ with the following property: for all real numbers $a_{1}, a_{2}, \ldots, a_{n}$ and $b_{1}, b_{2}, \ldots, b_{n}$ satisfying $\left|a_{k}\right|+\left|b_{k}\right|=1$ for $1 \leqslant k \leqslant n$, there exist $x_{1}, x_{2}, \ldots, x_{n}$, each of which is either -1 or 1 , such that... | For any even integer $n \geqslant 4$, we consider the case $$ a_{1}=a_{2}=\cdots=a_{n-1}=b_{n}=0 \quad \text { and } \quad b_{1}=b_{2}=\cdots=b_{n-1}=a_{n}=1 $$ The condition $\left|a_{k}\right|+\left|b_{k}\right|=1$ is satisfied for each $1 \leqslant k \leqslant n$. No matter how we choose each $x_{k}$, both sums $\su... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 200 | 1,053 |
2016 | T0 | A3 | Algebra | IMO-SL | Find all integers $n \geqslant 3$ with the following property: for all real numbers $a_{1}, a_{2}, \ldots, a_{n}$ and $b_{1}, b_{2}, \ldots, b_{n}$ satisfying $\left|a_{k}\right|+\left|b_{k}\right|=1$ for $1 \leqslant k \leqslant n$, there exist $x_{1}, x_{2}, \ldots, x_{n}$, each of which is either -1 or 1 , such that... | The even case can be handled in the same way as Firstly, for $n=3$, we may assume without loss of generality $a_{1} \geqslant a_{2} \geqslant a_{3} \geqslant 0$ and $b_{1}=a_{1}-1$ (if $b_{1}=1-a_{1}$, we may replace each $b_{k}$ by $-b_{k}$ ). - Case 1. $b_{2}=a_{2}-1$ and $b_{3}=a_{3}-1$, in which case we take $\left... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 200 | 2,218 |
2016 | T0 | A6 | Algebra | IMO-SL | The equation $$ (x-1)(x-2) \cdots(x-2016)=(x-1)(x-2) \cdots(x-2016) $$ is written on the board. One tries to erase some linear factors from both sides so that each side still has at least one factor, and the resulting equation has no real roots. Find the least number of linear factors one needs to erase to achieve th... | Since there are 2016 common linear factors on both sides, we need to erase at least 2016 factors. We claim that the equation has no real roots if we erase all factors $(x-k)$ on the left-hand side with $k \equiv 2,3(\bmod 4)$, and all factors $(x-m)$ on the right-hand side with $m \equiv 0,1(\bmod 4)$. Therefore, it su... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 102 | 1,433 |
2016 | T0 | A8 | Answer. | IMO-SL | Determine the largest real number $a$ such that for all $n \geqslant 1$ and for all real numbers $x_{0}, x_{1}, \ldots, x_{n}$ satisfying $0=x_{0}<x_{1}<x_{2}<\cdots<x_{n}$, we have $$ \frac{1}{x_{1}-x_{0}}+\frac{1}{x_{2}-x_{1}}+\cdots+\frac{1}{x_{n}-x_{n-1}} \geqslant a\left(\frac{2}{x_{1}}+\frac{3}{x_{2}}+\cdots+\fr... | We first show that $a=\frac{4}{9}$ is admissible. For each $2 \leqslant k \leqslant n$, by the CauchySchwarz Inequality, we have $$ \left(x_{k-1}+\left(x_{k}-x_{k-1}\right)\right)\left(\frac{(k-1)^{2}}{x_{k-1}}+\frac{3^{2}}{x_{k}-x_{k-1}}\right) \geqslant(k-1+3)^{2}, $$ which can be rewritten as $$ \frac{9}{x_{k}-x_{k-... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 177 | 677 |
2016 | T0 | A8 | Answer. | IMO-SL | Determine the largest real number $a$ such that for all $n \geqslant 1$ and for all real numbers $x_{0}, x_{1}, \ldots, x_{n}$ satisfying $0=x_{0}<x_{1}<x_{2}<\cdots<x_{n}$, we have $$ \frac{1}{x_{1}-x_{0}}+\frac{1}{x_{2}-x_{1}}+\cdots+\frac{1}{x_{n}-x_{n-1}} \geqslant a\left(\frac{2}{x_{1}}+\frac{3}{x_{2}}+\cdots+\fr... | We shall give an alternative method to establish (1) with $a=\frac{4}{9}$. We define $y_{k}=x_{k}-x_{k-1}>0$ for $1 \leqslant k \leqslant n$. By the Cauchy-Schwarz Inequality, for $1 \leqslant k \leqslant n$, we have $$ \left(y_{1}+y_{2}+\cdots+y_{k}\right)\left(\sum_{j=1}^{k} \frac{1}{y_{j}}\binom{j+1}{2}^{2}\right) \... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 177 | 658 |
2016 | T0 | C2 | Combinatorics | IMO-SL | Find all positive integers $n$ for which all positive divisors of $n$ can be put into the cells of a rectangular table under the following constraints: - each cell contains a distinct divisor; - the sums of all rows are equal; and - the sums of all columns are equal. Answer. 1. | Clearly $n=1$ works. Then we assume $n>1$ and let its prime factorization be $n=p_{1}^{r_{1}} p_{2}^{r_{2}} \cdots p_{t}^{r_{t}}$. Suppose the table has $k$ rows and $l$ columns with $1<k \leqslant l$. Note that $k l$ is the number of positive divisors of $n$ and the sum of all entries is the sum of positive divisors o... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 66 | 771 |
2016 | T0 | C3 | Combinatorics | IMO-SL | Let $n$ be a positive integer relatively prime to 6 . We paint the vertices of a regular $n$-gon with three colours so that there is an odd number of vertices of each colour. Show that there exists an isosceles triangle whose three vertices are of different colours. | For $k=1,2,3$, let $a_{k}$ be the number of isosceles triangles whose vertices contain exactly $k$ colours. Suppose on the contrary that $a_{3}=0$. Let $b, c, d$ be the number of vertices of the three different colours respectively. We now count the number of pairs $(\triangle, E)$ where $\triangle$ is an isosceles tri... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 60 | 585 |
2016 | T0 | C4 | Combinatorics | IMO-SL | Find all positive integers $n$ for which we can fill in the entries of an $n \times n$ table with the following properties: - each entry can be one of $I, M$ and $O$; - in each row and each column, the letters $I, M$ and $O$ occur the same number of times; and - in any diagonal whose number of entries is a multiple of... | We first show that such a table exists when $n$ is a multiple of 9 . Consider the following $9 \times 9$ table. $$ \left(\begin{array}{ccccccccc} I & I & I & M & M & M & O & O & O \\ M & M & M & O & O & O & I & I & I \\ O & O & O & I & I & I & M & M & M \\ I & I & I & M & M & M & O & O & O \\ M & M & M & O & O & O & I ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 120 | 771 |
2016 | T0 | C5 | Combinatorics | IMO-SL | Let $n \geqslant 3$ be a positive integer. Find the maximum number of diagonals of a regular $n$-gon one can select, so that any two of them do not intersect in the interior or they are perpendicular to each other. Answer. $n-2$ if $n$ is even and $n-3$ if $n$ is odd. | We consider two cases according to the parity of $n$. - Case 1. $n$ is odd. We first claim that no pair of diagonals is perpendicular. Suppose $A, B, C, D$ are vertices where $A B$ and $C D$ are perpendicular, and let $E$ be the vertex lying on the perpendicular bisector of $A B$. Let $E^{\prime}$ be the opposite point... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 82 | 960 |
2016 | T0 | C8 | Combinatorics | IMO-SL | Let $n$ be a positive integer. Determine the smallest positive integer $k$ with the following property: it is possible to mark $k$ cells on a $2 n \times 2 n$ board so that there exists a unique partition of the board into $1 \times 2$ and $2 \times 1$ dominoes, none of which contains two marked cells. Answer. $2 n$. | We first construct an example of marking $2 n$ cells satisfying the requirement. Label the rows and columns $1,2, \ldots, 2 n$ and label the cell in the $i$-th row and the $j$-th column $(i, j)$. For $i=1,2, \ldots, n$, we mark the cells $(i, i)$ and $(i, i+1)$. We claim that the required partition exists and is unique... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 88 | 758 |
2016 | T0 | N1 | Number Theory | IMO-SL | For any positive integer $k$, denote the sum of digits of $k$ in its decimal representation by $S(k)$. Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n \geqslant 2016$, the integer $P(n)$ is positive and $$ S(P(n))=P(S(n)) $$ Answer. - $P(x)=c$ where $1 \leqslant c \leqslan... | We consider three cases according to the degree of $P$. - Case 1. $P(x)$ is a constant polynomial. Let $P(x)=c$ where $c$ is an integer constant. Then (1) becomes $S(c)=c$. This holds if and only if $1 \leqslant c \leqslant 9$. - Case 2. $\operatorname{deg} P=1$. We have the following observation. For any positive inte... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 117 | 841 |
2016 | T0 | N1 | Number Theory | IMO-SL | For any positive integer $k$, denote the sum of digits of $k$ in its decimal representation by $S(k)$. Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n \geqslant 2016$, the integer $P(n)$ is positive and $$ S(P(n))=P(S(n)) $$ Answer. - $P(x)=c$ where $1 \leqslant c \leqslan... | Let $P(x)=a_{d} x^{d}+a_{d-1} x^{d-1}+\cdots+a_{0}$. Clearly $a_{d}>0$. There exists an integer $m \geqslant 1$ such that $\left|a_{i}\right|<10^{m}$ for all $0 \leqslant i \leqslant d$. Consider $n=9 \times 10^{k}$ for a sufficiently large integer $k$ in (1). If there exists an index $0 \leqslant i \leqslant d-1$ such... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 117 | 798 |
2016 | T0 | N2 | Number Theory | IMO-SL | Let $\tau(n)$ be the number of positive divisors of $n$. Let $\tau_{1}(n)$ be the number of positive divisors of $n$ which have remainders 1 when divided by 3 . Find all possible integral values of the fraction $\frac{\tau(10 n)}{\tau_{1}(10 n)}$. Answer. All composite numbers together with 2. | In this solution, we always use $p_{i}$ to denote primes congruent to $1 \bmod 3$, and use $q_{j}$ to denote primes congruent to $2 \bmod 3$. When we express a positive integer $m$ using its prime factorization, we also include the special case $m=1$ by allowing the exponents to be zeros. We first compute $\tau_{1}(m)$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 85 | 1,494 |
2016 | T0 | N3 | Number Theory | IMO-SL | Define $P(n)=n^{2}+n+1$. For any positive integers $a$ and $b$, the set $$ \{P(a), P(a+1), P(a+2), \ldots, P(a+b)\} $$ is said to be fragrant if none of its elements is relatively prime to the product of the other elements. Determine the smallest size of a fragrant set. Answer. 6. | We have the following observations. (i) $(P(n), P(n+1))=1$ for any $n$. We have $(P(n), P(n+1))=\left(n^{2}+n+1, n^{2}+3 n+3\right)=\left(n^{2}+n+1,2 n+2\right)$. Noting that $n^{2}+n+1$ is odd and $\left(n^{2}+n+1, n+1\right)=(1, n+1)=1$, the claim follows. (ii) $(P(n), P(n+2))=1$ for $n \not \equiv 2(\bmod 7)$ and $(... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 93 | 1,459 |
2016 | T0 | N6 | Number Theory | IMO-SL | Denote by $\mathbb{N}$ the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for all positive integers $m$ and $n$, the integer $f(m)+f(n)-m n$ is nonzero and divides $m f(m)+n f(n)$. Answer. $f(n)=n^{2}$ for any $n \in \mathbb{N}$. | It is given that $$ f(m)+f(n)-m n \mid m f(m)+n f(n) $$ Taking $m=n=1$ in (1), we have $2 f(1)-1 \mid 2 f(1)$. Then $2 f(1)-1 \mid 2 f(1)-(2 f(1)-1)=1$ and hence $f(1)=1$. Let $p \geqslant 7$ be a prime. Taking $m=p$ and $n=1$ in (1), we have $f(p)-p+1 \mid p f(p)+1$ and hence $$ f(p)-p+1 \mid p f(p)+1-p(f(p)-p+1)=p^{2... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 98 | 886 |
2016 | T0 | N7 | Number Theory | IMO-SL | Let $n$ be an odd positive integer. In the Cartesian plane, a cyclic polygon $P$ with area $S$ is chosen. All its vertices have integral coordinates, and the squares of its side lengths are all divisible by $n$. Prove that $2 S$ is an integer divisible by $n$. | Let $P=A_{1} A_{2} \ldots A_{k}$ and let $A_{k+i}=A_{i}$ for $i \geqslant 1$. By the Shoelace Formula, the area of any convex polygon with integral coordinates is half an integer. Therefore, $2 S$ is an integer. We shall prove by induction on $k \geqslant 3$ that $2 S$ is divisible by $n$. Clearly, it suffices to consi... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 65 | 1,760 |
2016 | T0 | N8 | Number Theory | IMO-SL | Find all polynomials $P(x)$ of odd degree $d$ and with integer coefficients satisfying the following property: for each positive integer $n$, there exist $n$ positive integers $x_{1}, x_{2}, \ldots, x_{n}$ such that $\frac{1}{2}<\frac{P\left(x_{i}\right)}{P\left(x_{j}\right)}<2$ and $\frac{P\left(x_{i}\right)}{P\left(x... | Let $P(x)=a_{d} x^{d}+a_{d-1} x^{d-1}+\cdots+a_{0}$. Consider the substitution $y=d a_{d} x+a_{d-1}$. By defining $Q(y)=P(x)$, we find that $Q$ is a polynomial with rational coefficients without the term $y^{d-1}$. Let $Q(y)=b_{d} y^{d}+b_{d-2} y^{d-2}+b_{d-3} y^{d-3}+\cdots+b_{0}$ and $B=\max _{0 \leqslant i \leqslant... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2016SL.jsonl",
"solution_match": null
} | 205 | 3,041 |
2017 | T0 | A1 | Algebra | IMO-SL | Let $a_{1}, a_{2}, \ldots, a_{n}, k$, and $M$ be positive integers such that $$ \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}=k \quad \text { and } \quad a_{1} a_{2} \ldots a_{n}=M $$ If $M>1$, prove that the polynomial $$ P(x)=M(x+1)^{k}-\left(x+a_{1}\right)\left(x+a_{2}\right) \cdots\left(x+a_{n}\right) $... | We first prove that, for $x>0$, $$ a_{i}(x+1)^{1 / a_{i}} \leqslant x+a_{i}, $$ with equality if and only if $a_{i}=1$. It is clear that equality occurs if $a_{i}=1$. If $a_{i}>1$, the AM-GM inequality applied to a single copy of $x+1$ and $a_{i}-1$ copies of 1 yields $$ \frac{(x+1)+\overbrace{1+1+\cdots+1}^{a_{i}-1 \t... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 156 | 788 |
2017 | T0 | A1 | Algebra | IMO-SL | Let $a_{1}, a_{2}, \ldots, a_{n}, k$, and $M$ be positive integers such that $$ \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}=k \quad \text { and } \quad a_{1} a_{2} \ldots a_{n}=M $$ If $M>1$, prove that the polynomial $$ P(x)=M(x+1)^{k}-\left(x+a_{1}\right)\left(x+a_{2}\right) \cdots\left(x+a_{n}\right) $... | We will prove that, in fact, all coefficients of the polynomial $P(x)$ are non-positive, and at least one of them is negative, which implies that $P(x)<0$ for $x>0$. Indeed, since $a_{j} \geqslant 1$ for all $j$ and $a_{j}>1$ for some $j$ (since $a_{1} a_{2} \ldots a_{n}=M>1$ ), we have $k=\frac{1}{a_{1}}+\frac{1}{a_{2... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 156 | 1,556 |
2017 | T0 | A2 | Algebra | IMO-SL | Let $q$ be a real number. Gugu has a napkin with ten distinct real numbers written on it, and he writes the following three lines of real numbers on the blackboard: - In the first line, Gugu writes down every number of the form $a-b$, where $a$ and $b$ are two (not necessarily distinct) numbers on his napkin. - In the... | Call a number $q$ good if every number in the second line appears in the third line unconditionally. We first show that the numbers 0 and $\pm 2$ are good. The third line necessarily contains 0 , so 0 is good. For any two numbers $a, b$ in the first line, write $a=x-y$ and $b=u-v$, where $x, y, u, v$ are (not necessari... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 215 | 579 |
2017 | T0 | A2 | Algebra | IMO-SL | Let $q$ be a real number. Gugu has a napkin with ten distinct real numbers written on it, and he writes the following three lines of real numbers on the blackboard: - In the first line, Gugu writes down every number of the form $a-b$, where $a$ and $b$ are two (not necessarily distinct) numbers on his napkin. - In the... | Let $q$ be a good number, as defined in the first solution, and define the polynomial $P\left(x_{1}, \ldots, x_{10}\right)$ as $$ \prod_{i<j}\left(x_{i}-x_{j}\right) \prod_{a_{i} \in S}\left(q\left(x_{1}-x_{2}\right)\left(x_{3}-x_{4}\right)-\left(a_{1}-a_{2}\right)^{2}-\left(a_{3}-a_{4}\right)^{2}+\left(a_{5}-a_{6}\rig... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 215 | 1,039 |
2017 | T0 | A3 | Algebra | IMO-SL | Let $S$ be a finite set, and let $\mathcal{A}$ be the set of all functions from $S$ to $S$. Let $f$ be an element of $\mathcal{A}$, and let $T=f(S)$ be the image of $S$ under $f$. Suppose that $f \circ g \circ f \neq g \circ f \circ g$ for every $g$ in $\mathcal{A}$ with $g \neq f$. Show that $f(T)=T$. (India) | For $n \geqslant 1$, denote the $n$-th composition of $f$ with itself by $$ f^{n} \stackrel{\text { def }}{=} \underbrace{f \circ f \circ \cdots \circ f}_{n \text { times }} $$ By hypothesis, if $g \in \mathcal{A}$ satisfies $f \circ g \circ f=g \circ f \circ g$, then $g=f$. A natural idea is to try to plug in $g=f^{n}... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 116 | 1,130 |
2017 | T0 | A4 | Algebra | IMO-SL | A sequence of real numbers $a_{1}, a_{2}, \ldots$ satisfies the relation $$ a_{n}=-\max _{i+j=n}\left(a_{i}+a_{j}\right) \quad \text { for all } n>2017 $$ Prove that this sequence is bounded, i.e., there is a constant $M$ such that $\left|a_{n}\right| \leqslant M$ for all positive integers $n$. | Set $D=2017$. Denote $$ M_{n}=\max _{k<n} a_{k} \quad \text { and } \quad m_{n}=-\min _{k<n} a_{k}=\max _{k<n}\left(-a_{k}\right) . $$ Clearly, the sequences $\left(m_{n}\right)$ and $\left(M_{n}\right)$ are nondecreasing. We need to prove that both are bounded. Consider an arbitrary $n>D$; our first aim is to bound $a... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 107 | 869 |
2017 | T0 | A4 | Algebra | IMO-SL | A sequence of real numbers $a_{1}, a_{2}, \ldots$ satisfies the relation $$ a_{n}=-\max _{i+j=n}\left(a_{i}+a_{j}\right) \quad \text { for all } n>2017 $$ Prove that this sequence is bounded, i.e., there is a constant $M$ such that $\left|a_{n}\right| \leqslant M$ for all positive integers $n$. | As in the previous solution, let $D=2017$. If the sequence is bounded above, say, by $Q$, then we have that $a_{n} \geqslant \min \left\{a_{1}, \ldots, a_{D},-2 Q\right\}$ for all $n$, so the sequence is bounded. Assume for sake of contradiction that the sequence is not bounded above. Let $\ell=\min \left\{a_{1}, \ldot... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 107 | 3,505 |
2017 | T0 | A5 | Algebra | IMO-SL | An integer $n \geqslant 3$ is given. We call an $n$-tuple of real numbers $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ Shiny if for each permutation $y_{1}, y_{2}, \ldots, y_{n}$ of these numbers we have $$ \sum_{i=1}^{n-1} y_{i} y_{i+1}=y_{1} y_{2}+y_{2} y_{3}+y_{3} y_{4}+\cdots+y_{n-1} y_{n} \geqslant-1 $$ Find the l... | First of all, we show that we may not take a larger constant $K$. Let $t$ be a positive number, and take $x_{2}=x_{3}=\cdots=t$ and $x_{1}=-1 /(2 t)$. Then, every product $x_{i} x_{j}(i \neq j)$ is equal to either $t^{2}$ or $-1 / 2$. Hence, for every permutation $y_{i}$ of the $x_{i}$, we have $$ y_{1} y_{2}+\cdots+y_... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 225 | 2,012 |
2017 | T0 | A5 | Algebra | IMO-SL | An integer $n \geqslant 3$ is given. We call an $n$-tuple of real numbers $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ Shiny if for each permutation $y_{1}, y_{2}, \ldots, y_{n}$ of these numbers we have $$ \sum_{i=1}^{n-1} y_{i} y_{i+1}=y_{1} y_{2}+y_{2} y_{3}+y_{3} y_{4}+\cdots+y_{n-1} y_{n} \geqslant-1 $$ Find the l... | We present another proof that $\sum_{i<j} x_{i} x_{j} \geqslant-(n-1) / 2$ for any Shiny $n$-tuple $\left(x_{1}, \ldots, x_{n}\right)$. Assume an ordering of the $x_{i}$ as in (2), and let $\ell=n-k$. Assume without loss of generality that $k \geqslant \ell$. Also assume $k \neq n$, (as otherwise, all of the $x_{i}$ ar... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 225 | 933 |
2017 | T0 | A7 | Algebra | IMO-SL | Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of integers and $b_{0}, b_{1}, b_{2}, \ldots$ be a sequence of positive integers such that $a_{0}=0, a_{1}=1$, and $$ a_{n+1}=\left\{\begin{array}{ll} a_{n} b_{n}+a_{n-1}, & \text { if } b_{n-1}=1 \\ a_{n} b_{n}-a_{n-1}, & \text { if } b_{n-1}>1 \end{array} \quad \text {... | The value of $b_{0}$ is irrelevant since $a_{0}=0$, so we may assume that $b_{0}=1$. Lemma. We have $a_{n} \geqslant 1$ for all $n \geqslant 1$. Proof. Let us suppose otherwise in order to obtain a contradiction. Let $$ n \geqslant 1 \text { be the smallest integer with } a_{n} \leqslant 0 \text {. } $$ Note that $n \g... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 206 | 1,460 |
2017 | T0 | A7 | Algebra | IMO-SL | Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of integers and $b_{0}, b_{1}, b_{2}, \ldots$ be a sequence of positive integers such that $a_{0}=0, a_{1}=1$, and $$ a_{n+1}=\left\{\begin{array}{ll} a_{n} b_{n}+a_{n-1}, & \text { if } b_{n-1}=1 \\ a_{n} b_{n}-a_{n-1}, & \text { if } b_{n-1}>1 \end{array} \quad \text {... | We say that an index $n>1$ is bad if $b_{n-1}=1$ and $b_{n-2}>1$; otherwise $n$ is good. The value of $b_{0}$ is irrelevant to the definition of $\left(a_{n}\right)$ since $a_{0}=0$; so we assume that $b_{0}>1$. Lemma 1. (a) $a_{n} \geqslant 1$ for all $n>0$. (b) If $n>1$ is good, then $a_{n}>a_{n-1}$. Proof. Induction... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 206 | 1,462 |
2017 | T0 | A8 | Algebra | IMO-SL | Assume that a function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies the following condition: For every $x, y \in \mathbb{R}$ such that $(f(x)+y)(f(y)+x)>0$, we have $f(x)+y=f(y)+x$. Prove that $f(x)+y \leqslant f(y)+x$ whenever $x>y$. (Netherlands) | Define $g(x)=x-f(x)$. The condition on $f$ then rewrites as follows: For every $x, y \in \mathbb{R}$ such that $((x+y)-g(x))((x+y)-g(y))>0$, we have $g(x)=g(y)$. This condition may in turn be rewritten in the following form: If $g(x) \neq g(y)$, then the number $x+y$ lies (non-strictly) between $g(x)$ and $g(y)$. Notic... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 95 | 1,472 |
2017 | T0 | A8 | Algebra | IMO-SL | Assume that a function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies the following condition: For every $x, y \in \mathbb{R}$ such that $(f(x)+y)(f(y)+x)>0$, we have $f(x)+y=f(y)+x$. Prove that $f(x)+y \leqslant f(y)+x$ whenever $x>y$. (Netherlands) | As in the previous solution, we pass to the function $g$ satisfying ( $*$ ) and notice that we need to prove the condition (1). We will also make use of the function $g_{1}$. If $g$ is constant, then (1) is clearly satisfied. So, in the sequel we assume that $g$ takes on at least two different values. Now we collect so... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 95 | 2,617 |
2017 | T0 | C2 | Combinatorics | IMO-SL | Let $n$ be a positive integer. Define a chameleon to be any sequence of $3 n$ letters, with exactly $n$ occurrences of each of the letters $a, b$, and $c$. Define a swap to be the transposition of two adjacent letters in a chameleon. Prove that for any chameleon $X$, there exists a chameleon $Y$ such that $X$ cannot be... | To start, notice that the swap of two identical letters does not change a chameleon, so we may assume there are no such swaps. For any two chameleons $X$ and $Y$, define their distance $d(X, Y)$ to be the minimal number of swaps needed to transform $X$ into $Y$ (or vice versa). Clearly, $d(X, Y)+d(Y, Z) \geqslant d(X, ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 109 | 919 |
2017 | T0 | C2 | Combinatorics | IMO-SL | Let $n$ be a positive integer. Define a chameleon to be any sequence of $3 n$ letters, with exactly $n$ occurrences of each of the letters $a, b$, and $c$. Define a swap to be the transposition of two adjacent letters in a chameleon. Prove that for any chameleon $X$, there exists a chameleon $Y$ such that $X$ cannot be... | We use the notion of distance from In any chameleon $X$, we enumerate the positions in it from left to right by $1,2, \ldots, 3 n$. Define $s_{c}(X)$ as the sum of positions occupied by $c$. The value of $s_{c}$ changes by at most 1 on each swap, but this fact alone does not suffice to solve the problem; so we need an ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 109 | 2,861 |
2017 | T0 | C3 | Combinatorics | IMO-SL | Sir Alex plays the following game on a row of 9 cells. Initially, all cells are empty. In each move, Sir Alex is allowed to perform exactly one of the following two operations: (1) Choose any number of the form $2^{j}$, where $j$ is a non-negative integer, and put it into an empty cell. (2) Choose two (not necessarily ... | We will solve a more general problem, replacing the row of 9 cells with a row of $k$ cells, where $k$ is a positive integer. Denote by $m(n, k)$ the maximum possible number of moves Sir Alex can make starting with a row of $k$ empty cells, and ending with one cell containing the number $2^{n}$ and all the other $k-1$ c... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 184 | 2,012 |
2017 | T0 | C3 | Combinatorics | IMO-SL | Sir Alex plays the following game on a row of 9 cells. Initially, all cells are empty. In each move, Sir Alex is allowed to perform exactly one of the following two operations: (1) Choose any number of the form $2^{j}$, where $j$ is a non-negative integer, and put it into an empty cell. (2) Choose two (not necessarily ... | Define merges and insertions as in We will need the following lemma. Lemma. If the binary representation of a positive integer $A$ has $d$ nonzero digits, then $A$ cannot be represented as a sum of fewer than $d$ powers of 2 . Moreover, any representation of $A$ as a sum of $d$ powers of 2 must coincide with its binary... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 184 | 1,093 |
2017 | T0 | C4 | Combinatorics | IMO-SL | Let $N \geqslant 2$ be an integer. $N(N+1)$ soccer players, no two of the same height, stand in a row in some order. Coach Ralph wants to remove $N(N-1)$ people from this row so that in the remaining row of $2 N$ players, no one stands between the two tallest ones, no one stands between the third and the fourth tallest... | Split the row into $N$ blocks with $N+1$ consecutive people each. We will show how to remove $N-1$ people from each block in order to satisfy the coach's wish. First, construct a $(N+1) \times N$ matrix where $x_{i, j}$ is the height of the $i^{\text {th }}$ tallest person of the $j^{\text {th }}$ block-in other words,... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 110 | 1,025 |
2017 | T0 | C4 | Combinatorics | IMO-SL | Let $N \geqslant 2$ be an integer. $N(N+1)$ soccer players, no two of the same height, stand in a row in some order. Coach Ralph wants to remove $N(N-1)$ people from this row so that in the remaining row of $2 N$ players, no one stands between the two tallest ones, no one stands between the third and the fourth tallest... | This is essentially the same as solution 1, but presented inductively. The essence of the argument is the following lemma. Lemma. Assume that we have $N$ disjoint groups of at least $N+1$ people in each, all people have distinct heights. Then one can choose two people from each group so that among the chosen people, th... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 110 | 831 |
2017 | T0 | C5 | Combinatorics | IMO-SL | A hunter and an invisible rabbit play a game in the Euclidean plane. The hunter's starting point $H_{0}$ coincides with the rabbit's starting point $R_{0}$. In the $n^{\text {th }}$ round of the game $(n \geqslant 1)$, the following happens. (1) First the invisible rabbit moves secretly and unobserved from its current ... | If the answer were "yes", the hunter would have a strategy that would "work", no matter how the rabbit moved or where the radar pings $R_{n}^{\prime}$ appeared. We will show the opposite: with bad luck from the radar pings, there is no strategy for the hunter that guarantees that the distance stays below 100 in $10^{9}... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 247 | 1,358 |
2017 | T0 | C6 | Combinatorics | IMO-SL | Let $n>1$ be an integer. An $n \times n \times n$ cube is composed of $n^{3}$ unit cubes. Each unit cube is painted with one color. For each $n \times n \times 1$ box consisting of $n^{2}$ unit cubes (of any of the three possible orientations), we consider the set of the colors present in that box (each color is listed... | Call a $n \times n \times 1$ box an $x$-box, a $y$-box, or a $z$-box, according to the direction of its short side. Let $C$ be the number of colors in a valid configuration. We start with the upper bound for $C$. Let $\mathcal{C}_{1}, \mathcal{C}_{2}$, and $\mathcal{C}_{3}$ be the sets of colors which appear in the big... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 155 | 1,369 |
2017 | T0 | C6 | Combinatorics | IMO-SL | Let $n>1$ be an integer. An $n \times n \times n$ cube is composed of $n^{3}$ unit cubes. Each unit cube is painted with one color. For each $n \times n \times 1$ box consisting of $n^{2}$ unit cubes (of any of the three possible orientations), we consider the set of the colors present in that box (each color is listed... | We will approach a new version of the original problem. In this new version, each cube may have a color, or be invisible (not both). Now we make sets of colors for each $n \times n \times 1$ box as before (where "invisible" is not considered a color) and group them by orientation, also as before. Finally, we require th... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 155 | 629 |
2017 | T0 | C7 | Combinatorics | IMO-SL | For any finite sets $X$ and $Y$ of positive integers, denote by $f_{X}(k)$ the $k^{\text {th }}$ smallest positive integer not in $X$, and let $$ X * Y=X \cup\left\{f_{X}(y): y \in Y\right\} $$ Let $A$ be a set of $a>0$ positive integers, and let $B$ be a set of $b>0$ positive integers. Prove that if $A * B=B * A$, t... | For any function $g: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ and any subset $X \subset \mathbb{Z}_{>0}$, we define $g(X)=$ $\{g(x): x \in X\}$. We have that the image of $f_{X}$ is $f_{X}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash X$. We now show a general lemma about the operation *, with the goal of ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 192 | 1,491 |
2017 | T0 | C7 | Combinatorics | IMO-SL | For any finite sets $X$ and $Y$ of positive integers, denote by $f_{X}(k)$ the $k^{\text {th }}$ smallest positive integer not in $X$, and let $$ X * Y=X \cup\left\{f_{X}(y): y \in Y\right\} $$ Let $A$ be a set of $a>0$ positive integers, and let $B$ be a set of $b>0$ positive integers. Prove that if $A * B=B * A$, t... | We will use Lemma 1 from $$ f_{X}=f_{Y} \Longleftrightarrow f_{X}\left(\mathbb{Z}_{>0}\right)=f_{Y}\left(\mathbb{Z}_{>0}\right) \Longleftrightarrow\left(\mathbb{Z}_{>0} \backslash X\right)=\left(\mathbb{Z}_{>0} \backslash Y\right) \Longleftrightarrow X=Y, $$ where the first equivalence is because $f_{X}$ and $f_{Y}$ ar... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 192 | 1,833 |
2017 | T0 | G1 | Geometry | IMO-SL | Let $A B C D E$ be a convex pentagon such that $A B=B C=C D, \angle E A B=\angle B C D$, and $\angle E D C=\angle C B A$. Prove that the perpendicular line from $E$ to $B C$ and the line segments $A C$ and $B D$ are concurrent. (Italy) | Throughout the solution, we refer to $\angle A, \angle B, \angle C, \angle D$, and $\angle E$ as internal angles of the pentagon $A B C D E$. Let the perpendicular bisectors of $A C$ and $B D$, which pass respectively through $B$ and $C$, meet at point $I$. Then $B D \perp C I$ and, similarly, $A C \perp B I$. Hence $A... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 80 | 538 |
2017 | T0 | G1 | Geometry | IMO-SL | Let $A B C D E$ be a convex pentagon such that $A B=B C=C D, \angle E A B=\angle B C D$, and $\angle E D C=\angle C B A$. Prove that the perpendicular line from $E$ to $B C$ and the line segments $A C$ and $B D$ are concurrent. (Italy) | We present yet another proof that $E I \perp B C$. In pentagon $A B C D E, \angle E<$ $180^{\circ} \Longleftrightarrow \angle A+\angle B+\angle C+\angle D>360^{\circ}$. Then $\angle A+\angle B=\angle C+\angle D>180^{\circ}$, so rays $E A$ and $C B$ meet at a point $P$, and rays $B C$ and $E D$ meet at a point $Q$. Now,... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 80 | 530 |
2017 | T0 | G5 | Geometry | IMO-SL | Let $A B C C_{1} B_{1} A_{1}$ be a convex hexagon such that $A B=B C$, and suppose that the line segments $A A_{1}, B B_{1}$, and $C C_{1}$ have the same perpendicular bisector. Let the diagonals $A C_{1}$ and $A_{1} C$ meet at $D$, and denote by $\omega$ the circle $A B C$. Let $\omega$ intersect the circle $A_{1} B C... | If $A A_{1}=C C_{1}$, then the hexagon is symmetric about the line $B B_{1}$; in particular the circles $A B C$ and $A_{1} B C_{1}$ are tangent to each other. So $A A_{1}$ and $C C_{1}$ must be different. Since the points $A$ and $A_{1}$ can be interchanged with $C$ and $C_{1}$, respectively, we may assume $A A_{1}<C C... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 149 | 626 |
2017 | T0 | G6 | Geometry | IMO-SL | Let $n \geqslant 3$ be an integer. Two regular $n$-gons $\mathcal{A}$ and $\mathcal{B}$ are given in the plane. Prove that the vertices of $\mathcal{A}$ that lie inside $\mathcal{B}$ or on its boundary are consecutive. (That is, prove that there exists a line separating those vertices of $\mathcal{A}$ that lie inside $... | In both solutions, by a polygon we always mean its interior together with its boundary. We start with finding a regular $n$-gon $\mathcal{C}$ which $(i)$ is inscribed into $\mathcal{B}$ (that is, all vertices of $\mathcal{C}$ lie on the perimeter of $\mathcal{B}$ ); and (ii) is either a translation of $\mathcal{A}$, or... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 119 | 2,550 |
2017 | T0 | G6 | Geometry | IMO-SL | Let $n \geqslant 3$ be an integer. Two regular $n$-gons $\mathcal{A}$ and $\mathcal{B}$ are given in the plane. Prove that the vertices of $\mathcal{A}$ that lie inside $\mathcal{B}$ or on its boundary are consecutive. (That is, prove that there exists a line separating those vertices of $\mathcal{A}$ that lie inside $... | Let $O_{A}$ and $O_{B}$ be the centers of $\mathcal{A}$ and $\mathcal{B}$, respectively. Denote $[n]=\{1,2, \ldots, n\}$. We start with introducing appropriate enumerations and notations. Enumerate the sidelines of $\mathcal{B}$ clockwise as $\ell_{1}, \ell_{2}, \ldots, \ell_{n}$. Denote by $\mathcal{H}_{i}$ the half-p... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 119 | 1,833 |
2017 | T0 | G7 | Geometry | IMO-SL | A convex quadrilateral $A B C D$ has an inscribed circle with center $I$. Let $I_{a}, I_{b}, I_{c}$, and $I_{d}$ be the incenters of the triangles $D A B, A B C, B C D$, and $C D A$, respectively. Suppose that the common external tangents of the circles $A I_{b} I_{d}$ and $C I_{b} I_{d}$ meet at $X$, and the common ex... | Denote by $\omega_{a}, \omega_{b}, \omega_{c}$ and $\omega_{d}$ the circles $A I_{b} I_{d}, B I_{a} I_{c}, C I_{b} I_{d}$, and $D I_{a} I_{c}$, let their centers be $O_{a}, O_{b}, O_{c}$ and $O_{d}$, and let their radii be $r_{a}, r_{b}, r_{c}$ and $r_{d}$, respectively. Claim 1. $I_{b} I_{d} \perp A C$ and $I_{a} I_{c... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 161 | 1,653 |
2017 | T0 | G8 | Geometry | IMO-SL | There are 2017 mutually external circles drawn on a blackboard, such that no two are tangent and no three share a common tangent. A tangent segment is a line segment that is a common tangent to two circles, starting at one tangent point and ending at the other one. Luciano is drawing tangent segments on the blackboard,... | First, consider a particular arrangement of circles $C_{1}, C_{2}, \ldots, C_{n}$ where all the centers are aligned and each $C_{i}$ is eclipsed from the other circles by its neighbors - for example, taking $C_{i}$ with center $\left(i^{2}, 0\right)$ and radius $i / 2$ works. Then the only tangent segments that can be ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 117 | 696 |
2017 | T0 | G8 | Geometry | IMO-SL | There are 2017 mutually external circles drawn on a blackboard, such that no two are tangent and no three share a common tangent. A tangent segment is a line segment that is a common tangent to two circles, starting at one tangent point and ending at the other one. Luciano is drawing tangent segments on the blackboard,... | First note that all tangent segments lying on the boundary of the convex hull of the circles are always drawn since they do not intersect anything else. Now in the final picture, aside from the $n$ circles, the blackboard is divided into regions. We can consider the picture as a plane (multi-)graph $G$ in which the cir... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 117 | 1,240 |
2017 | T0 | N1 | Number Theory | IMO-SL | The sequence $a_{0}, a_{1}, a_{2}, \ldots$ of positive integers satisfies $$ a_{n+1}=\left\{\begin{array}{ll} \sqrt{a_{n}}, & \text { if } \sqrt{a_{n}} \text { is an integer } \\ a_{n}+3, & \text { otherwise } \end{array} \quad \text { for every } n \geqslant 0\right. $$ Determine all values of $a_{0}>1$ for which th... | Since the value of $a_{n+1}$ only depends on the value of $a_{n}$, if $a_{n}=a_{m}$ for two different indices $n$ and $m$, then the sequence is eventually periodic. So we look for the values of $a_{0}$ for which the sequence is eventually periodic. Claim 1. If $a_{n} \equiv-1(\bmod 3)$, then, for all $m>n, a_{m}$ is no... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 151 | 1,122 |
2017 | T0 | N2 | Number Theory | IMO-SL | Let $p \geqslant 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1, \ldots, p-1\}$ that was not chosen before by either of the two players and then chooses an element $a_{i}$ of the set $\{0,1,2,3,4,5,6,7,... | We say that a player makes the move $\left(i, a_{i}\right)$ if he chooses the index $i$ and then the element $a_{i}$ of the set $\{0,1,2,3,4,5,6,7,8,9\}$ in this move. If $p=2$ or $p=5$ then Eduardo chooses $i=0$ and $a_{0}=0$ in the first move, and wins, since, independently of the next moves, $M$ will be a multiple o... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 252 | 1,072 |
2017 | T0 | N3 | Number Theory | IMO-SL | Determine all integers $n \geqslant 2$ with the following property: for any integers $a_{1}, a_{2}, \ldots, a_{n}$ whose sum is not divisible by $n$, there exists an index $1 \leqslant i \leqslant n$ such that none of the numbers $$ a_{i}, a_{i}+a_{i+1}, \ldots, a_{i}+a_{i+1}+\cdots+a_{i+n-1} $$ is divisible by $n$. ... | Let us first show that, if $n=a b$, with $a, b \geqslant 2$ integers, then the property in the statement of the problem does not hold. Indeed, in this case, let $a_{k}=a$ for $1 \leqslant k \leqslant n-1$ and $a_{n}=0$. The sum $a_{1}+a_{2}+\cdots+a_{n}=a \cdot(n-1)$ is not divisible by $n$. Let $i$ with $1 \leqslant i... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 146 | 1,019 |
2017 | T0 | N4 | Number Theory | IMO-SL | Call a rational number short if it has finitely many digits in its decimal expansion. For a positive integer $m$, we say that a positive integer $t$ is $m$-tastic if there exists a number $c \in\{1,2,3, \ldots, 2017\}$ such that $\frac{10^{t}-1}{c \cdot m}$ is short, and such that $\frac{10^{k}-1}{c \cdot m}$ is not sh... | First notice that $x \in \mathbb{Q}$ is short if and only if there are exponents $a, b \geqslant 0$ such that $2^{a} \cdot 5^{b} \cdot x \in \mathbb{Z}$. In fact, if $x$ is short, then $x=\frac{n}{10^{k}}$ for some $k$ and we can take $a=b=k$; on the other hand, if $2^{a} \cdot 5^{b} \cdot x=q \in \mathbb{Z}$ then $x=\... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 172 | 1,251 |
2017 | T0 | N5 | Number Theory | IMO-SL | Find all pairs $(p, q)$ of prime numbers with $p>q$ for which the number $$ \frac{(p+q)^{p+q}(p-q)^{p-q}-1}{(p+q)^{p-q}(p-q)^{p+q}-1} $$ is an integer. | Let $M=(p+q)^{p-q}(p-q)^{p+q}-1$, which is relatively prime with both $p+q$ and $p-q$. Denote by $(p-q)^{-1}$ the multiplicative inverse of $(p-q)$ modulo $M$. By eliminating the term -1 in the numerator, $$ \begin{aligned} (p+q)^{p+q}(p-q)^{p-q}-1 & \equiv(p+q)^{p-q}(p-q)^{p+q}-1 \quad(\bmod M) \\ (p+q)^{2 q} & \equiv... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 69 | 1,373 |
2017 | T0 | N6 | Number Theory | IMO-SL | Find the smallest positive integer $n$, or show that no such $n$ exists, with the following property: there are infinitely many distinct $n$-tuples of positive rational numbers ( $a_{1}, a_{2}, \ldots, a_{n}$ ) such that both $$ a_{1}+a_{2}+\cdots+a_{n} \quad \text { and } \quad \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+... | For $n=1, a_{1} \in \mathbb{Z}_{>0}$ and $\frac{1}{a_{1}} \in \mathbb{Z}_{>0}$ if and only if $a_{1}=1$. Next we show that (i) There are finitely many $(x, y) \in \mathbb{Q}_{>0}^{2}$ satisfying $x+y \in \mathbb{Z}$ and $\frac{1}{x}+\frac{1}{y} \in \mathbb{Z}$ Write $x=\frac{a}{b}$ and $y=\frac{c}{d}$ with $a, b, c, d ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 125 | 1,515 |
2017 | T0 | N6 | Number Theory | IMO-SL | Find the smallest positive integer $n$, or show that no such $n$ exists, with the following property: there are infinitely many distinct $n$-tuples of positive rational numbers ( $a_{1}, a_{2}, \ldots, a_{n}$ ) such that both $$ a_{1}+a_{2}+\cdots+a_{n} \quad \text { and } \quad \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+... | Call the $n$-tuples $\left(a_{1}, a_{2}, \ldots, a_{n}\right) \in \mathbb{Q}_{>0}^{n}$ satisfying the conditions of the problem statement good, and those for which $$ f\left(a_{1}, \ldots, a_{n}\right) \stackrel{\text { def }}{=}\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 125 | 1,511 |
2017 | T0 | N7 | Number Theory | IMO-SL | Say that an ordered pair $(x, y)$ of integers is an irreducible lattice point if $x$ and $y$ are relatively prime. For any finite set $S$ of irreducible lattice points, show that there is a homogenous polynomial in two variables, $f(x, y)$, with integer coefficients, of degree at least 1 , such that $f(x, y)=1$ for eac... | First of all, we note that finding a homogenous polynomial $f(x, y)$ such that $f(x, y)= \pm 1$ is enough, because we then have $f^{2}(x, y)=1$. Label the irreducible lattice points $\left(x_{1}, y_{1}\right)$ through $\left(x_{n}, y_{n}\right)$. If any two of these lattice points $\left(x_{i}, y_{i}\right)$ and $\left... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 183 | 1,111 |
2017 | T0 | N7 | Number Theory | IMO-SL | Say that an ordered pair $(x, y)$ of integers is an irreducible lattice point if $x$ and $y$ are relatively prime. For any finite set $S$ of irreducible lattice points, show that there is a homogenous polynomial in two variables, $f(x, y)$, with integer coefficients, of degree at least 1 , such that $f(x, y)=1$ for eac... | As in the previous solution, label the irreducible lattice points $\left(x_{1}, y_{1}\right), \ldots,\left(x_{n}, y_{n}\right)$ and assume without loss of generality that no two of the points are collinear with the origin. We induct on $n$ to construct a homogenous polynomial $f(x, y)$ such that $f\left(x_{i}, y_{i}\ri... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 183 | 1,733 |
2017 | T0 | N8 | Number Theory | IMO-SL | Let $p$ be an odd prime number and $\mathbb{Z}_{>0}$ be the set of positive integers. Suppose that a function $f: \mathbb{Z}_{>0} \times \mathbb{Z}_{>0} \rightarrow\{0,1\}$ satisfies the following properties: - $f(1,1)=0$; - $f(a, b)+f(b, a)=1$ for any pair of relatively prime positive integers $(a, b)$ not both equal... | Denote by $\mathbb{A}$ the set of all pairs of coprime positive integers. Notice that for every $(a, b) \in \mathbb{A}$ there exists a pair $(u, v) \in \mathbb{Z}^{2}$ with $u a+v b=1$. Moreover, if $\left(u_{0}, v_{0}\right)$ is one such pair, then all such pairs are of the form $(u, v)=\left(u_{0}+k b, v_{0}-k a\righ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 182 | 1,149 |
2017 | T0 | N8 | Number Theory | IMO-SL | Let $p$ be an odd prime number and $\mathbb{Z}_{>0}$ be the set of positive integers. Suppose that a function $f: \mathbb{Z}_{>0} \times \mathbb{Z}_{>0} \rightarrow\{0,1\}$ satisfies the following properties: - $f(1,1)=0$; - $f(a, b)+f(b, a)=1$ for any pair of relatively prime positive integers $(a, b)$ not both equal... | We provide a different proof for the Lemma. For this purpose, we use continued fractions to find $g(a, b)=(u, v)$ explicitly. The function $f$ is completely determined on $\mathbb{A}$ by the following Claim. Represent $a / b$ as a continued fraction; that is, let $a_{0}$ be an integer and $a_{1}, \ldots, a_{k}$ be posi... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 182 | 3,833 |
2017 | T0 | A2 | Algebra | IMO-SL | Let $q$ be a real number. Gugu has a napkin with ten distinct real numbers written on it, and he writes the following three lines of real numbers on the blackboard: - In the first line, Gugu writes down every number of the form $a-b$, where $a$ and $b$ are two (not necessarily distinct) numbers on his napkin. - In the... | Call a number $q$ good if every number in the second line appears in the third line unconditionally. We first show that the numbers 0 and $\pm 2$ are good. The third line necessarily contains 0 , so 0 is good. For any two numbers $a, b$ in the first line, write $a=x-y$ and $b=u-v$, where $x, y, u, v$ are (not necessari... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 227 | 579 |
2017 | T0 | A2 | Algebra | IMO-SL | Let $q$ be a real number. Gugu has a napkin with ten distinct real numbers written on it, and he writes the following three lines of real numbers on the blackboard: - In the first line, Gugu writes down every number of the form $a-b$, where $a$ and $b$ are two (not necessarily distinct) numbers on his napkin. - In the... | Let $q$ be a good number, as defined in the first solution, and define the polynomial $P\left(x_{1}, \ldots, x_{10}\right)$ as $$ \prod_{i<j}\left(x_{i}-x_{j}\right) \prod_{a_{i} \in S}\left(q\left(x_{1}-x_{2}\right)\left(x_{3}-x_{4}\right)-\left(a_{1}-a_{2}\right)^{2}-\left(a_{3}-a_{4}\right)^{2}+\left(a_{5}-a_{6}\rig... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 227 | 1,039 |
2017 | T0 | A4 | Algebra | IMO-SL | A sequence of real numbers $a_{1}, a_{2}, \ldots$ satisfies the relation $$ a_{n}=-\max _{i+j=n}\left(a_{i}+a_{j}\right) \quad \text { for all } n>2017 $$ Prove that this sequence is bounded, i.e., there is a constant $M$ such that $\left|a_{n}\right| \leqslant M$ for all positive integers $n$. (Russia) | Set $D=2017$. Denote $$ M_{n}=\max _{k<n} a_{k} \quad \text { and } \quad m_{n}=-\min _{k<n} a_{k}=\max _{k<n}\left(-a_{k}\right) . $$ Clearly, the sequences $\left(m_{n}\right)$ and $\left(M_{n}\right)$ are nondecreasing. We need to prove that both are bounded. Consider an arbitrary $n>D$; our first aim is to bound $a... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 110 | 869 |
2017 | T0 | A4 | Algebra | IMO-SL | A sequence of real numbers $a_{1}, a_{2}, \ldots$ satisfies the relation $$ a_{n}=-\max _{i+j=n}\left(a_{i}+a_{j}\right) \quad \text { for all } n>2017 $$ Prove that this sequence is bounded, i.e., there is a constant $M$ such that $\left|a_{n}\right| \leqslant M$ for all positive integers $n$. (Russia) | As in the previous solution, let $D=2017$. If the sequence is bounded above, say, by $Q$, then we have that $a_{n} \geqslant \min \left\{a_{1}, \ldots, a_{D},-2 Q\right\}$ for all $n$, so the sequence is bounded. Assume for sake of contradiction that the sequence is not bounded above. Let $\ell=\min \left\{a_{1}, \ldot... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 110 | 3,505 |
2017 | T0 | A5 | Algebra | IMO-SL | An integer $n \geqslant 3$ is given. We call an $n$-tuple of real numbers $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ Shiny if for each permutation $y_{1}, y_{2}, \ldots, y_{n}$ of these numbers we have $$ \sum_{i=1}^{n-1} y_{i} y_{i+1}=y_{1} y_{2}+y_{2} y_{3}+y_{3} y_{4}+\cdots+y_{n-1} y_{n} \geqslant-1 $$ Find the l... | First of all, we show that we may not take a larger constant $K$. Let $t$ be a positive number, and take $x_{2}=x_{3}=\cdots=t$ and $x_{1}=-1 /(2 t)$. Then, every product $x_{i} x_{j}(i \neq j)$ is equal to either $t^{2}$ or $-1 / 2$. Hence, for every permutation $y_{i}$ of the $x_{i}$, we have $$ y_{1} y_{2}+\cdots+y_... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 243 | 2,012 |
2017 | T0 | A5 | Algebra | IMO-SL | An integer $n \geqslant 3$ is given. We call an $n$-tuple of real numbers $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ Shiny if for each permutation $y_{1}, y_{2}, \ldots, y_{n}$ of these numbers we have $$ \sum_{i=1}^{n-1} y_{i} y_{i+1}=y_{1} y_{2}+y_{2} y_{3}+y_{3} y_{4}+\cdots+y_{n-1} y_{n} \geqslant-1 $$ Find the l... | We present another proof that $\sum_{i<j} x_{i} x_{j} \geqslant-(n-1) / 2$ for any Shiny $n$-tuple $\left(x_{1}, \ldots, x_{n}\right)$. Assume an ordering of the $x_{i}$ as in (2), and let $\ell=n-k$. Assume without loss of generality that $k \geqslant \ell$. Also assume $k \neq n$, (as otherwise, all of the $x_{i}$ ar... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 243 | 933 |
2017 | T0 | C3 | Combinatorics | IMO-SL | Sir Alex plays the following game on a row of 9 cells. Initially, all cells are empty. In each move, Sir Alex is allowed to perform exactly one of the following two operations: (1) Choose any number of the form $2^{j}$, where $j$ is a non-negative integer, and put it into an empty cell. (2) Choose two (not necessarily ... | We will solve a more general problem, replacing the row of 9 cells with a row of $k$ cells, where $k$ is a positive integer. Denote by $m(n, k)$ the maximum possible number of moves Sir Alex can make starting with a row of $k$ empty cells, and ending with one cell containing the number $2^{n}$ and all the other $k-1$ c... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 206 | 2,012 |
2017 | T0 | C3 | Combinatorics | IMO-SL | Sir Alex plays the following game on a row of 9 cells. Initially, all cells are empty. In each move, Sir Alex is allowed to perform exactly one of the following two operations: (1) Choose any number of the form $2^{j}$, where $j$ is a non-negative integer, and put it into an empty cell. (2) Choose two (not necessarily ... | Define merges and insertions as in We will need the following lemma. Lemma. If the binary representation of a positive integer $A$ has $d$ nonzero digits, then $A$ cannot be represented as a sum of fewer than $d$ powers of 2 . Moreover, any representation of $A$ as a sum of $d$ powers of 2 must coincide with its binary... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 206 | 1,093 |
2017 | T0 | C5 | Combinatorics | IMO-SL | A hunter and an invisible rabbit play a game in the Euclidean plane. The hunter's starting point $H_{0}$ coincides with the rabbit's starting point $R_{0}$. In the $n^{\text {th }}$ round of the game ( $n \geqslant 1$ ), the following happens. (1) First the invisible rabbit moves secretly and unobserved from its curren... | If the answer were "yes", the hunter would have a strategy that would "work", no matter how the rabbit moved or where the radar pings $R_{n}^{\prime}$ appeared. We will show the opposite: with bad luck from the radar pings, there is no strategy for the hunter that guarantees that the distance stays below 100 in $10^{9}... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 268 | 1,358 |
2017 | T0 | C6 | Combinatorics | IMO-SL | Let $n>1$ be an integer. An $n \times n \times n$ cube is composed of $n^{3}$ unit cubes. Each unit cube is painted with one color. For each $n \times n \times 1$ box consisting of $n^{2}$ unit cubes (of any of the three possible orientations), we consider the set of the colors present in that box (each color is listed... | Call a $n \times n \times 1$ box an $x$-box, a $y$-box, or a $z$-box, according to the direction of its short side. Let $C$ be the number of colors in a valid configuration. We start with the upper bound for $C$. Let $\mathcal{C}_{1}, \mathcal{C}_{2}$, and $\mathcal{C}_{3}$ be the sets of colors which appear in the big... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 177 | 1,369 |
2017 | T0 | C6 | Combinatorics | IMO-SL | Let $n>1$ be an integer. An $n \times n \times n$ cube is composed of $n^{3}$ unit cubes. Each unit cube is painted with one color. For each $n \times n \times 1$ box consisting of $n^{2}$ unit cubes (of any of the three possible orientations), we consider the set of the colors present in that box (each color is listed... | We will approach a new version of the original problem. In this new version, each cube may have a color, or be invisible (not both). Now we make sets of colors for each $n \times n \times 1$ box as before (where "invisible" is not considered a color) and group them by orientation, also as before. Finally, we require th... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 177 | 629 |
2017 | T0 | C7 | Combinatorics | IMO-SL | For any finite sets $X$ and $Y$ of positive integers, denote by $f_{X}(k)$ the $k^{\mathrm{th}}$ smallest positive integer not in $X$, and let $$ X * Y=X \cup\left\{f_{X}(y): y \in Y\right\} $$ Let $A$ be a set of $a>0$ positive integers, and let $B$ be a set of $b>0$ positive integers. Prove that if $A * B=B * A$, t... | For any function $g: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ and any subset $X \subset \mathbb{Z}_{>0}$, we define $g(X)=$ $\{g(x): x \in X\}$. We have that the image of $f_{X}$ is $f_{X}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash X$. We now show a general lemma about the operation *, with the goal of ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 192 | 1,491 |
2017 | T0 | C7 | Combinatorics | IMO-SL | For any finite sets $X$ and $Y$ of positive integers, denote by $f_{X}(k)$ the $k^{\mathrm{th}}$ smallest positive integer not in $X$, and let $$ X * Y=X \cup\left\{f_{X}(y): y \in Y\right\} $$ Let $A$ be a set of $a>0$ positive integers, and let $B$ be a set of $b>0$ positive integers. Prove that if $A * B=B * A$, t... | We will use Lemma 1 from $$ f_{X}=f_{Y} \Longleftrightarrow f_{X}\left(\mathbb{Z}_{>0}\right)=f_{Y}\left(\mathbb{Z}_{>0}\right) \Longleftrightarrow\left(\mathbb{Z}_{>0} \backslash X\right)=\left(\mathbb{Z}_{>0} \backslash Y\right) \Longleftrightarrow X=Y, $$ where the first equivalence is because $f_{X}$ and $f_{Y}$ ar... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 192 | 1,833 |
2017 | T0 | G8 | Geometry | IMO-SL | There are 2017 mutually external circles drawn on a blackboard, such that no two are tangent and no three share a common tangent. A tangent segment is a line segment that is a common tangent to two circles, starting at one tangent point and ending at the other one. Luciano is drawing tangent segments on the blackboard,... | First, consider a particular arrangement of circles $C_{1}, C_{2}, \ldots, C_{n}$ where all the centers are aligned and each $C_{i}$ is eclipsed from the other circles by its neighbors - for example, taking $C_{i}$ with center $\left(i^{2}, 0\right)$ and radius $i / 2$ works. Then the only tangent segments that can be ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 163 | 696 |
2017 | T0 | G8 | Geometry | IMO-SL | There are 2017 mutually external circles drawn on a blackboard, such that no two are tangent and no three share a common tangent. A tangent segment is a line segment that is a common tangent to two circles, starting at one tangent point and ending at the other one. Luciano is drawing tangent segments on the blackboard,... | First note that all tangent segments lying on the boundary of the convex hull of the circles are always drawn since they do not intersect anything else. Now in the final picture, aside from the $n$ circles, the blackboard is divided into regions. We can consider the picture as a plane (multi-)graph $G$ in which the cir... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 163 | 1,240 |
2017 | T0 | N1 | Number Theory | IMO-SL | The sequence $a_{0}, a_{1}, a_{2}, \ldots$ of positive integers satisfies $$ a_{n+1}=\left\{\begin{array}{ll} \sqrt{a_{n}}, & \text { if } \sqrt{a_{n}} \text { is an integer } \\ a_{n}+3, & \text { otherwise } \end{array} \quad \text { for every } n \geqslant 0\right. $$ Determine all values of $a_{0}>1$ for which th... | Since the value of $a_{n+1}$ only depends on the value of $a_{n}$, if $a_{n}=a_{m}$ for two different indices $n$ and $m$, then the sequence is eventually periodic. So we look for the values of $a_{0}$ for which the sequence is eventually periodic. Claim 1. If $a_{n} \equiv-1(\bmod 3)$, then, for all $m>n, a_{m}$ is no... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 160 | 1,122 |
2017 | T0 | N3 | Number Theory | IMO-SL | Determine all integers $n \geqslant 2$ with the following property: for any integers $a_{1}, a_{2}, \ldots, a_{n}$ whose sum is not divisible by $n$, there exists an index $1 \leqslant i \leqslant n$ such that none of the numbers $$ a_{i}, a_{i}+a_{i+1}, \ldots, a_{i}+a_{i+1}+\cdots+a_{i+n-1} $$ is divisible by $n$. ... | Let us first show that, if $n=a b$, with $a, b \geqslant 2$ integers, then the property in the statement of the problem does not hold. Indeed, in this case, let $a_{k}=a$ for $1 \leqslant k \leqslant n-1$ and $a_{n}=0$. The sum $a_{1}+a_{2}+\cdots+a_{n}=a \cdot(n-1)$ is not divisible by $n$. Let $i$ with $1 \leqslant i... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 156 | 1,019 |
2017 | T0 | N4 | Number Theory | IMO-SL | Call a rational number short if it has finitely many digits in its decimal expansion. For a positive integer $m$, we say that a positive integer $t$ is $m$-tastic if there exists a number $c \in\{1,2,3, \ldots, 2017\}$ such that $\frac{10^{t}-1}{c \cdot m}$ is short, and such that $\frac{10^{k}-1}{c \cdot m}$ is not sh... | First notice that $x \in \mathbb{Q}$ is short if and only if there are exponents $a, b \geqslant 0$ such that $2^{a} \cdot 5^{b} \cdot x \in \mathbb{Z}$. In fact, if $x$ is short, then $x=\frac{n}{10^{k}}$ for some $k$ and we can take $a=b=k$; on the other hand, if $2^{a} \cdot 5^{b} \cdot x=q \in \mathbb{Z}$ then $x=\... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 179 | 1,251 |
2017 | T0 | N5 | Number Theory | IMO-SL | Find all pairs $(p, q)$ of prime numbers with $p>q$ for which the number $$ \frac{(p+q)^{p+q}(p-q)^{p-q}-1}{(p+q)^{p-q}(p-q)^{p+q}-1} $$ is an integer. (Japan) Answer: The only such pair is $(3,2)$. | Let $M=(p+q)^{p-q}(p-q)^{p+q}-1$, which is relatively prime with both $p+q$ and $p-q$. Denote by $(p-q)^{-1}$ the multiplicative inverse of $(p-q)$ modulo $M$. By eliminating the term -1 in the numerator, $$ \begin{aligned} (p+q)^{p+q}(p-q)^{p-q}-1 & \equiv(p+q)^{p-q}(p-q)^{p+q}-1 \quad(\bmod M) \\ (p+q)^{2 q} & \equiv... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 85 | 1,373 |
2017 | T0 | N6 | Number Theory | IMO-SL | Find the smallest positive integer $n$, or show that no such $n$ exists, with the following property: there are infinitely many distinct $n$-tuples of positive rational numbers $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ such that both $$ a_{1}+a_{2}+\cdots+a_{n} \quad \text { and } \quad \frac{1}{a_{1}}+\frac{1}{a_{2}... | For $n=1, a_{1} \in \mathbb{Z}_{>0}$ and $\frac{1}{a_{1}} \in \mathbb{Z}_{>0}$ if and only if $a_{1}=1$. Next we show that (i) There are finitely many $(x, y) \in \mathbb{Q}_{>0}^{2}$ satisfying $x+y \in \mathbb{Z}$ and $\frac{1}{x}+\frac{1}{y} \in \mathbb{Z}$ Write $x=\frac{a}{b}$ and $y=\frac{c}{d}$ with $a, b, c, d ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 133 | 1,515 |
2017 | T0 | N6 | Number Theory | IMO-SL | Find the smallest positive integer $n$, or show that no such $n$ exists, with the following property: there are infinitely many distinct $n$-tuples of positive rational numbers $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ such that both $$ a_{1}+a_{2}+\cdots+a_{n} \quad \text { and } \quad \frac{1}{a_{1}}+\frac{1}{a_{2}... | Call the $n$-tuples $\left(a_{1}, a_{2}, \ldots, a_{n}\right) \in \mathbb{Q}_{>0}^{n}$ satisfying the conditions of the problem statement good, and those for which $$ f\left(a_{1}, \ldots, a_{n}\right) \stackrel{\text { def }}{=}\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 133 | 1,511 |
2017 | T0 | N8 | Number Theory | IMO-SL | Let $p$ be an odd prime number and $\mathbb{Z}_{>0}$ be the set of positive integers. Suppose that a function $f: \mathbb{Z}_{>0} \times \mathbb{Z}_{>0} \rightarrow\{0,1\}$ satisfies the following properties: - $f(1,1)=0$; - $f(a, b)+f(b, a)=1$ for any pair of relatively prime positive integers $(a, b)$ not both equal... | Denote by $\mathbb{A}$ the set of all pairs of coprime positive integers. Notice that for every $(a, b) \in \mathbb{A}$ there exists a pair $(u, v) \in \mathbb{Z}^{2}$ with $u a+v b=1$. Moreover, if $\left(u_{0}, v_{0}\right)$ is one such pair, then all such pairs are of the form $(u, v)=\left(u_{0}+k b, v_{0}-k a\righ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 186 | 1,149 |
2017 | T0 | N8 | Number Theory | IMO-SL | Let $p$ be an odd prime number and $\mathbb{Z}_{>0}$ be the set of positive integers. Suppose that a function $f: \mathbb{Z}_{>0} \times \mathbb{Z}_{>0} \rightarrow\{0,1\}$ satisfies the following properties: - $f(1,1)=0$; - $f(a, b)+f(b, a)=1$ for any pair of relatively prime positive integers $(a, b)$ not both equal... | We provide a different proof for the Lemma. For this purpose, we use continued fractions to find $g(a, b)=(u, v)$ explicitly. The function $f$ is completely determined on $\mathbb{A}$ by the following Claim. Represent $a / b$ as a continued fraction; that is, let $a_{0}$ be an integer and $a_{1}, \ldots, a_{k}$ be posi... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2017SL.jsonl",
"solution_match": null
} | 186 | 3,833 |
2018 | T0 | A1 | Algebra | IMO-SL | Let $\mathbb{Q}_{>0}$ denote the set of all positive rational numbers. Determine all functions $f: \mathbb{Q}_{>0} \rightarrow \mathbb{Q}_{>0}$ satisfying for all $x, y \in \mathbb{Q}_{>0}$. $$ f\left(x^{2} f(y)^{2}\right)=f(x)^{2} f(y) $$ (Switzerland) | Take any $a, b \in \mathbb{Q}_{>0}$. By substituting $x=f(a), y=b$ and $x=f(b), y=a$ into $(*)$ we get $$ f(f(a))^{2} f(b)=f\left(f(a)^{2} f(b)^{2}\right)=f(f(b))^{2} f(a) $$ which yields $$ \frac{f(f(a))^{2}}{f(a)}=\frac{f(f(b))^{2}}{f(b)} \quad \text { for all } a, b \in \mathbb{Q}_{>0} $$ In other words, this shows ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2018SL.jsonl",
"solution_match": null
} | 97 | 812 |
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