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 |
|---|---|---|---|---|---|---|---|---|---|
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 $$ \frac{a}{(b+c-a)^{2}}+\frac{b}{(c+a-b)^{2}}+\frac{c}{(a+b-c)^{2}} \geq \frac{3}{(a b c)^{2}} $$ | As in $$ a^{5}+b^{5}+c^{5} \geq 3 $$ which is weaker than the given one. Due to the symmetry we may assume that $a \geq b \geq c$. In view of (3)), it suffices to prove the inequality $$ \sum \frac{a^{3} b^{2} c^{2}}{(b+c-a)^{2}} \geq \sum a^{5} $$ or, moving all the terms into the left-hand part, $$ \sum \frac{a^{3}}{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 111 | 575 |
2011 | T0 | C1 | Combinatorics | IMO-SL | Let $n>0$ be an integer. We are given a balance and $n$ weights of weight $2^{0}, 2^{1}, \ldots, 2^{n-1}$. In a sequence of $n$ moves we place all weights on the balance. In the first move we choose a weight and put it on the left pan. In each of the following moves we choose one of the remaining weights and we add it ... | Assume $n \geq 2$. We claim $$ f(n)=(2 n-1) f(n-1) . $$ Firstly, note that after the first move the left pan is always at least 1 heavier than the right one. Hence, any valid way of placing the $n$ weights on the scale gives rise, by not considering weight 1 , to a valid way of placing the weights $2,2^{2}, \ldots, 2^{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 219 | 571 |
2011 | T0 | C1 | Combinatorics | IMO-SL | Let $n>0$ be an integer. We are given a balance and $n$ weights of weight $2^{0}, 2^{1}, \ldots, 2^{n-1}$. In a sequence of $n$ moves we place all weights on the balance. In the first move we choose a weight and put it on the left pan. In each of the following moves we choose one of the remaining weights and we add it ... | We present a different way of obtaining (1). Set $f(0)=1$. Firstly, we find a recurrent formula for $f(n)$. Assume $n \geq 1$. Suppose that weight $2^{n-1}$ is placed on the balance in the $i$-th move with $1 \leq i \leq n$. This weight has to be put on the left pan. For the previous moves we have $\left(\begin{array}{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 219 | 618 |
2011 | T0 | C3 | Combinatorics | IMO-SL | Let $\mathcal{S}$ be a finite set of at least two points in the plane. Assume that no three points of $\mathcal{S}$ are collinear. By a windmill we mean a process as follows. Start with a line $\ell$ going through a point $P \in \mathcal{S}$. Rotate $\ell$ clockwise around the pivot $P$ until the line contains another ... | Give the rotating line an orientation and distinguish its sides as the oranje side and the blue side. Notice that whenever the pivot changes from some point $T$ to another point $U$, after the change, $T$ is on the same side as $U$ was before. Therefore, the number of elements of $\mathcal{S}$ on the oranje side and th... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 285 | 1,102 |
2011 | T0 | C3 | Combinatorics | IMO-SL | Let $\mathcal{S}$ be a finite set of at least two points in the plane. Assume that no three points of $\mathcal{S}$ are collinear. By a windmill we mean a process as follows. Start with a line $\ell$ going through a point $P \in \mathcal{S}$. Rotate $\ell$ clockwise around the pivot $P$ until the line contains another ... | There are various examples showing that $k=3$ does indeed have the property under consideration. E.g. one can take $$ \begin{gathered} A_{1}=\{1,2,3\} \cup\{3 m \mid m \geq 4\}, \\ A_{2}=\{4,5,6\} \cup\{3 m-1 \mid m \geq 4\}, \\ A_{3}=\{7,8,9\} \cup\{3 m-2 \mid m \geq 4\} \end{gathered} $$ To check that this partition ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 285 | 929 |
2011 | T0 | C3 | Combinatorics | IMO-SL | Let $\mathcal{S}$ be a finite set of at least two points in the plane. Assume that no three points of $\mathcal{S}$ are collinear. By a windmill we mean a process as follows. Start with a line $\ell$ going through a point $P \in \mathcal{S}$. Rotate $\ell$ clockwise around the pivot $P$ until the line contains another ... | Again we only prove that $k \leq 3$. Assume that $A_{1}, A_{2}, \ldots, A_{k}$ is a partition satisfying the given property. We construct a graph $\mathcal{G}$ on the set $V=\{1,2, \ldots, 18\}$ of vertices as follows. For each $i \in\{1,2, \ldots, k\}$ and each $d \in\{15,16,17,19\}$ we choose one pair of distinct ele... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 285 | 649 |
2011 | T0 | C5 | Combinatorics | IMO-SL | Let $m$ be a positive integer and consider a checkerboard consisting of $m$ by $m$ unit squares. At the midpoints of some of these unit squares there is an ant. At time 0 , each ant starts moving with speed 1 parallel to some edge of the checkerboard. When two ants moving in opposite directions meet, they both turn $90... | For $m=1$ the answer is clearly correct, so assume $m>1$. In the sequel, the word collision will be used to denote meeting of exactly two ants, moving in opposite directions. If at the beginning we place an ant on the southwest corner square facing east and an ant on the southeast corner square facing west, then they w... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 176 | 1,017 |
2011 | T0 | C6 | Combinatorics | IMO-SL | Let $n$ be a positive integer and let $W=\ldots x_{-1} x_{0} x_{1} x_{2} \ldots$ be an infinite periodic word consisting of the letters $a$ and $b$. Suppose that the minimal period $N$ of $W$ is greater than $2^{n}$. A finite nonempty word $U$ is said to appear in $W$ if there exist indices $k \leq \ell$ such that $U=... | Throughout the solution, all the words are nonempty. For any word $R$ of length $m$, we call the number of indices $i \in\{1,2, \ldots, N\}$ for which $R$ coincides with the subword $x_{i+1} x_{i+2} \ldots x_{i+m}$ of $W$ the multiplicity of $R$ and denote it by $\mu(R)$. Thus a word $R$ appears in $W$ if and only if $... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 175 | 1,178 |
2011 | T0 | C7 | Combinatorics | IMO-SL | On a square table of 2011 by 2011 cells we place a finite number of napkins that each cover a square of 52 by 52 cells. In each cell we write the number of napkins covering it, and we record the maximal number $k$ of cells that all contain the same nonzero number. Considering all possible napkin configurations, what is... | Let $m=39$, then $2011=52 m-17$. We begin with an example showing that there can exist 3986729 cells carrying the same positive number. To describe it, we number the columns from the left to the right and the rows from the bottom to the top by $1,2, \ldots, 2011$. We will denote each napkin by the coordinates of its lo... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 89 | 1,347 |
2011 | T0 | C7 | Combinatorics | IMO-SL | On a square table of 2011 by 2011 cells we place a finite number of napkins that each cover a square of 52 by 52 cells. In each cell we write the number of napkins covering it, and we record the maximal number $k$ of cells that all contain the same nonzero number. Considering all possible napkin configurations, what is... | We present a different proof of the estimate which is the hard part of the problem. Let $S=35, H=17, m=39$; so the table size is $2011=S m+H(m-1)$, and the napkin size is $52=S+H$. Fix any positive integer $M$ and call a cell vicious if it contains a number distinct from $M$. We will prove that there are at least $H^{2... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 89 | 2,515 |
2011 | T0 | G1 | Geometry | IMO-SL | Let $A B C$ be an acute triangle. Let $\omega$ be a circle whose center $L$ lies on the side $B C$. Suppose that $\omega$ is tangent to $A B$ at $B^{\prime}$ and to $A C$ at $C^{\prime}$. Suppose also that the circumcenter $O$ of the triangle $A B C$ lies on the shorter $\operatorname{arc} B^{\prime} C^{\prime}$ of $\o... | The point $B^{\prime}$, being the perpendicular foot of $L$, is an interior point of side $A B$. Analogously, $C^{\prime}$ lies in the interior of $A C$. The point $O$ is located inside the triangle $A B^{\prime} C^{\prime}$, hence $\angle C O B<\angle C^{\prime} O B^{\prime}$. Let $\alpha=\angle C A B$. The angles $\a... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 126 | 684 |
2011 | T0 | G2 | Geometry | IMO-SL | Let $A_{1} A_{2} A_{3} A_{4}$ be a non-cyclic quadrilateral. Let $O_{1}$ and $r_{1}$ be the circumcenter and the circumradius of the triangle $A_{2} A_{3} A_{4}$. Define $O_{2}, O_{3}, O_{4}$ and $r_{2}, r_{3}, r_{4}$ in a similar way. Prove that $$ \frac{1}{O_{1} A_{1}^{2}-r_{1}^{2}}+\frac{1}{O_{2} A_{2}^{2}-r_{2}^{2... | Let $M$ be the point of intersection of the diagonals $A_{1} A_{3}$ and $A_{2} A_{4}$. On each diagonal choose a direction and let $x, y, z$, and $w$ be the signed distances from $M$ to the points $A_{1}, A_{2}, A_{3}$, and $A_{4}$, respectively. Let $\omega_{1}$ be the circumcircle of the triangle $A_{2} A_{3} A_{4}$ ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 196 | 504 |
2011 | T0 | G2 | Geometry | IMO-SL | Let $A_{1} A_{2} A_{3} A_{4}$ be a non-cyclic quadrilateral. Let $O_{1}$ and $r_{1}$ be the circumcenter and the circumradius of the triangle $A_{2} A_{3} A_{4}$. Define $O_{2}, O_{3}, O_{4}$ and $r_{2}, r_{3}, r_{4}$ in a similar way. Prove that $$ \frac{1}{O_{1} A_{1}^{2}-r_{1}^{2}}+\frac{1}{O_{2} A_{2}^{2}-r_{2}^{2... | Introduce a Cartesian coordinate system in the plane. Every circle has an equation of the form $p(x, y)=x^{2}+y^{2}+l(x, y)=0$, where $l(x, y)$ is a polynomial of degree at most 1 . For any point $A=\left(x_{A}, y_{A}\right)$ we have $p\left(x_{A}, y_{A}\right)=d^{2}-r^{2}$, where $d$ is the distance from $A$ to the ce... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 196 | 1,353 |
2011 | T0 | G3 | Geometry | IMO-SL | Let $A B C D$ be a convex quadrilateral whose sides $A D$ and $B C$ are not parallel. Suppose that the circles with diameters $A B$ and $C D$ meet at points $E$ and $F$ inside the quadrilateral. Let $\omega_{E}$ be the circle through the feet of the perpendiculars from $E$ to the lines $A B, B C$, and $C D$. Let $\omeg... | Denote by $P, Q, R$, and $S$ the projections of $E$ on the lines $D A, A B, B C$, and $C D$ respectively. The points $P$ and $Q$ lie on the circle with diameter $A E$, so $\angle Q P E=\angle Q A E$; analogously, $\angle Q R E=\angle Q B E$. So $\angle Q P E+\angle Q R E=\angle Q A E+\angle Q B E=90^{\circ}$. By simila... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 162 | 787 |
2011 | T0 | G4 | Geometry | IMO-SL | Let $A B C$ be an acute triangle with circumcircle $\Omega$. Let $B_{0}$ be the midpoint of $A C$ and let $C_{0}$ be the midpoint of $A B$. Let $D$ be the foot of the altitude from $A$, and let $G$ be the centroid of the triangle $A B C$. Let $\omega$ be a circle through $B_{0}$ and $C_{0}$ that is tangent to the circl... | If $A B=A C$, then the statement is trivial. So without loss of generality we may assume $A B<A C$. Denote the tangents to $\Omega$ at points $A$ and $X$ by $a$ and $x$, respectively. Let $\Omega_{1}$ be the circumcircle of triangle $A B_{0} C_{0}$. The circles $\Omega$ and $\Omega_{1}$ are homothetic with center $A$, ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 131 | 640 |
2011 | T0 | G5 | Geometry | IMO-SL | Let $A B C$ be a triangle with incenter $I$ and circumcircle $\omega$. Let $D$ and $E$ be the second intersection points of $\omega$ with the lines $A I$ and $B I$, respectively. The chord $D E$ meets $A C$ at a point $F$, and $B C$ at a point $G$. Let $P$ be the intersection point of the line through $F$ parallel to $... | Since $$ \angle I A F=\angle D A C=\angle B A D=\angle B E D=\angle I E F $$ the quadrilateral $A I F E$ is cyclic. Denote its circumcircle by $\omega_{1}$. Similarly, the quadrilateral $B D G I$ is cyclic; denote its circumcircle by $\omega_{2}$. The line $A E$ is the radical axis of $\omega$ and $\omega_{1}$, and the... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 163 | 517 |
2011 | T0 | G5 | Geometry | IMO-SL | Let $A B C$ be a triangle with incenter $I$ and circumcircle $\omega$. Let $D$ and $E$ be the second intersection points of $\omega$ with the lines $A I$ and $B I$, respectively. The chord $D E$ meets $A C$ at a point $F$, and $B C$ at a point $G$. Let $P$ be the intersection point of the line through $F$ parallel to $... | Let $M$ be the intersection point of the tangents to $\omega$ at $D$ and $E$, and let the lines $A E$ and $B D$ meet at $T$; if $A E$ and $B D$ are parallel, then let $T$ be their common ideal point. It is well-known that the points $K$ and $M$ lie on the line $T I$ (as a consequence of PASCAL's theorem, applied to the... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 163 | 617 |
2011 | T0 | G6 | Geometry | IMO-SL | Let $A B C$ be a triangle with $A B=A C$, and let $D$ be the midpoint of $A C$. The angle bisector of $\angle B A C$ intersects the circle through $D, B$, and $C$ in a point $E$ inside the triangle $A B C$. The line $B D$ intersects the circle through $A, E$, and $B$ in two points $B$ and $F$. The lines $A F$ and $B E$... | Let $D^{\prime}$ be the midpoint of the segment $A B$, and let $M$ be the midpoint of $B C$. By symmetry at line $A M$, the point $D^{\prime}$ has to lie on the circle $B C D$. Since the $\operatorname{arcs} D^{\prime} E$ and $E D$ of that circle are equal, we have $\angle A B I=\angle D^{\prime} B E=\angle E B D=I B K... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 149 | 548 |
2011 | T0 | G6 | Geometry | IMO-SL | Let $A B C$ be a triangle with $A B=A C$, and let $D$ be the midpoint of $A C$. The angle bisector of $\angle B A C$ intersects the circle through $D, B$, and $C$ in a point $E$ inside the triangle $A B C$. The line $B D$ intersects the circle through $A, E$, and $B$ in two points $B$ and $F$. The lines $A F$ and $B E$... | It can be shown in the same way as in the first solution that $I$ lies on the angle bisector of $\angle A B K$. Here we restrict ourselves to proving that $K I$ bisects $\angle A K B$. Denote the circumcircle of triangle $B C D$ and its center by $\omega_{1}$ and by $O_{1}$, respectively. Since the quadrilateral $A B F... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 149 | 531 |
2011 | T0 | G7 | Geometry | IMO-SL | Let $A B C D E F$ be a convex hexagon all of whose sides are tangent to a circle $\omega$ with center $O$. Suppose that the circumcircle of triangle $A C E$ is concentric with $\omega$. Let $J$ be the foot of the perpendicular from $B$ to $C D$. Suppose that the perpendicular from $B$ to $D F$ intersects the line $E O$... | Since $\omega$ and the circumcircle of triangle $A C E$ are concentric, the tangents from $A$, $C$, and $E$ to $\omega$ have equal lengths; that means that $A B=B C, C D=D E$, and $E F=F A$. Moreover, we have $\angle B C D=\angle D E F=\angle F A B$. Consider the rotation around point $D$ mapping $C$ to $E$; let $B^{\p... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 123 | 888 |
2011 | T0 | G7 | Geometry | IMO-SL | Let $A B C D E F$ be a convex hexagon all of whose sides are tangent to a circle $\omega$ with center $O$. Suppose that the circumcircle of triangle $A C E$ is concentric with $\omega$. Let $J$ be the foot of the perpendicular from $B$ to $C D$. Suppose that the perpendicular from $B$ to $D F$ intersects the line $E O$... | Let us denote the points of tangency of $A B, B C, C D, D E, E F$, and $F A$ to $\omega$ by $R, S, T, U, V$, and $W$, respectively. As in the previous solution, we mention that $A R=$ $A W=C S=C T=E U=E V$. The reflection in the line $B O$ maps $R$ to $S$, therefore $A$ to $C$ and thus also $W$ to $T$. Hence, both line... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 123 | 964 |
2011 | T0 | G8 | Geometry | IMO-SL | Let $A B C$ be an acute triangle with circumcircle $\omega$. Let $t$ be a tangent line to $\omega$. Let $t_{a}, t_{b}$, and $t_{c}$ be the lines obtained by reflecting $t$ in the lines $B C, C A$, and $A B$, respectively. Show that the circumcircle of the triangle determined by the lines $t_{a}, t_{b}$, and $t_{c}$ is ... | Denote by $T$ the point of tangency of $t$ and $\omega$. Let $A^{\prime}=t_{b} \cap t_{c}, B^{\prime}=t_{a} \cap t_{c}$, $C^{\prime}=t_{a} \cap t_{b}$. Introduce the point $A^{\prime \prime}$ on $\omega$ such that $T A=A A^{\prime \prime}\left(A^{\prime \prime} \neq T\right.$ unless $T A$ is a diameter). Define the poi... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 107 | 1,381 |
2011 | T0 | G8 | Geometry | IMO-SL | Let $A B C$ be an acute triangle with circumcircle $\omega$. Let $t$ be a tangent line to $\omega$. Let $t_{a}, t_{b}$, and $t_{c}$ be the lines obtained by reflecting $t$ in the lines $B C, C A$, and $A B$, respectively. Show that the circumcircle of the triangle determined by the lines $t_{a}, t_{b}$, and $t_{c}$ is ... | Define the points $T, A^{\prime}, B^{\prime}$, and $C^{\prime}$ in the same way as in the previous solution. Let $X, Y$, and $Z$ be the symmetric images of $T$ about the lines $B C, C A$, and $A B$, respectively. Note that the projections of $T$ on these lines form a Simson line of $T$ with respect to $A B C$, therefor... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 107 | 1,023 |
2011 | T0 | N1 | Number Theory | IMO-SL | For any integer $d>0$, let $f(d)$ be the smallest positive integer that has exactly $d$ positive divisors (so for example we have $f(1)=1, f(5)=16$, and $f(6)=12$ ). Prove that for every integer $k \geq 0$ the number $f\left(2^{k}\right)$ divides $f\left(2^{k+1}\right)$. | For any positive integer $n$, let $d(n)$ be the number of positive divisors of $n$. Let $n=\prod_{p} p^{a(p)}$ be the prime factorization of $n$ where $p$ ranges over the prime numbers, the integers $a(p)$ are nonnegative and all but finitely many $a(p)$ are zero. Then we have $d(n)=\prod_{p}(a(p)+1)$. Thus, $d(n)$ is ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 99 | 515 |
2011 | T0 | N2 | Number Theory | IMO-SL | Consider a polynomial $P(x)=\left(x+d_{1}\right)\left(x+d_{2}\right) \cdot \ldots \cdot\left(x+d_{9}\right)$, where $d_{1}, d_{2}, \ldots, d_{9}$ are nine distinct integers. Prove that there exists an integer $N$ such that for all integers $x \geq N$ the number $P(x)$ is divisible by a prime number greater than 20 . | Given a nonzero integer $N$ as well as a prime number $p$ we write $v_{p}(N)$ for the exponent with which $p$ occurs in the prime factorization of $|N|$. Evidently, if the statement of the problem were not true, then there would exist an infinite sequence $\left(x_{n}\right)$ of positive integers tending to infinity su... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 104 | 689 |
2011 | T0 | N2 | Number Theory | IMO-SL | Consider a polynomial $P(x)=\left(x+d_{1}\right)\left(x+d_{2}\right) \cdot \ldots \cdot\left(x+d_{9}\right)$, where $d_{1}, d_{2}, \ldots, d_{9}$ are nine distinct integers. Prove that there exists an integer $N$ such that for all integers $x \geq N$ the number $P(x)$ is divisible by a prime number greater than 20 . | Obviously, all functions in the answer satisfy the condition of the problem. We will show that there are no other functions satisfying that condition. Let $f$ be a function satisfying the given condition. For each integer $n$, the function $g$ defined by $g(x)=f(x)+n$ also satisfies the same condition. Therefore, by su... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 104 | 683 |
2011 | T0 | N4 | Number Theory | IMO-SL | For each positive integer $k$, let $t(k)$ be the largest odd divisor of $k$. Determine all positive integers $a$ for which there exists a positive integer $n$ such that all the differences $$ t(n+a)-t(n), \quad t(n+a+1)-t(n+1), \quad \ldots, \quad t(n+2 a-1)-t(n+a-1) $$ are divisible by 4 . | A pair $(a, n)$ satisfying the condition of the problem will be called a winning pair. It is straightforward to check that the pairs $(1,1),(3,1)$, and $(5,4)$ are winning pairs. Now suppose that $a$ is a positive integer not equal to 1,3 , and 5 . We will show that there are no winning pairs $(a, n)$ by distinguishing... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 95 | 652 |
2011 | T0 | N5 | Number Theory | IMO-SL | Let $f$ be a function from the set of integers to the set of positive integers. Suppose that for any two integers $m$ and $n$, the difference $f(m)-f(n)$ is divisible by $f(m-n)$. Prove that for all integers $m, n$ with $f(m) \leq f(n)$ the number $f(n)$ is divisible by $f(m)$. | We split the solution into a sequence of claims; in each claim, the letters $m$ and $n$ denote arbitrary integers. Claim 1. $f(n) \mid f(m n)$. Proof. Since trivially $f(n) \mid f(1 \cdot n)$ and $f(n) \mid f((k+1) n)-f(k n)$ for all integers $k$, this is easily seen by using induction on $m$ in both directions. Claim ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 86 | 1,280 |
2011 | T0 | N6 | Number Theory | IMO-SL | Let $P(x)$ and $Q(x)$ be two polynomials with integer coefficients such that no nonconstant polynomial with rational coefficients divides both $P(x)$ and $Q(x)$. Suppose that for every positive integer $n$ the integers $P(n)$ and $Q(n)$ are positive, and $2^{Q(n)}-1$ divides $3^{P(n)}-1$. Prove that $Q(x)$ is a constan... | First we show that there exists an integer $d$ such that for all positive integers $n$ we have $\operatorname{gcd}(P(n), Q(n)) \leq d$. Since $P(x)$ and $Q(x)$ are coprime (over the polynomials with rational coefficients), EuCLID's algorithm provides some polynomials $R_{0}(x), S_{0}(x)$ with rational coefficients such... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 93 | 1,062 |
2011 | T0 | N7 | Number Theory | IMO-SL | Let $p$ be an odd prime number. For every integer $a$, define the number $$ S_{a}=\frac{a}{1}+\frac{a^{2}}{2}+\cdots+\frac{a^{p-1}}{p-1} $$ Let $m$ and $n$ be integers such that $$ S_{3}+S_{4}-3 S_{2}=\frac{m}{n} $$ Prove that $p$ divides $m$. | For rational numbers $p_{1} / q_{1}$ and $p_{2} / q_{2}$ with the denominators $q_{1}, q_{2}$ not divisible by $p$, we write $p_{1} / q_{1} \equiv p_{2} / q_{2}(\bmod p)$ if the numerator $p_{1} q_{2}-p_{2} q_{1}$ of their difference is divisible by $p$. We start with finding an explicit formula for the residue of $S_{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 108 | 751 |
2011 | T0 | N7 | Number Theory | IMO-SL | Let $p$ be an odd prime number. For every integer $a$, define the number $$ S_{a}=\frac{a}{1}+\frac{a^{2}}{2}+\cdots+\frac{a^{p-1}}{p-1} $$ Let $m$ and $n$ be integers such that $$ S_{3}+S_{4}-3 S_{2}=\frac{m}{n} $$ Prove that $p$ divides $m$. | One may solve the problem without finding an explicit formula for $S_{a}$. It is enough to find the following property. Lemma. For every integer $a$, we have $S_{a+1} \equiv S_{-a}(\bmod p)$. Proof. We expand $S_{a+1}$ using the binomial formula as $$ S_{a+1}=\sum_{k=1}^{p-1} \frac{1}{k} \sum_{j=0}^{k}\left(\begin{arra... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 108 | 1,236 |
2011 | T0 | N7 | Number Theory | IMO-SL | Let $p$ be an odd prime number. For every integer $a$, define the number $$ S_{a}=\frac{a}{1}+\frac{a^{2}}{2}+\cdots+\frac{a^{p-1}}{p-1} $$ Let $m$ and $n$ be integers such that $$ S_{3}+S_{4}-3 S_{2}=\frac{m}{n} $$ Prove that $p$ divides $m$. | Let $N=\{1,2, \ldots, n-1\}$. For $a, b \in N$, we say that $b$ follows $a$ if there exists an integer $g$ such that $b \equiv g^{a}(\bmod n)$ and denote this property as $a \rightarrow b$. This way we have a directed graph with $N$ as set of vertices. If $a_{1}, \ldots, a_{n-1}$ is a permutation of $1,2, \ldots, n-1$ ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 108 | 2,854 |
2011 | T0 | A1 | Number Theory | 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
} | 218 | 933 |
2011 | T0 | A2 | Number Theory | 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 $$ x_{1}^{n}+2 x_{2}^{n}+\cdots+2011 x_{2011}^{n}=a^{n+1}+1 . $$ Answer. The only sequence that satisfies the condition is $$ \left(x_{1}, \ldots, x_{2011}\... | Throughout this solution, the set of positive integers will be denoted by $\mathbb{Z}_{+}$. Put $k=2+3+\cdots+2011=2023065$. We have $$ 1^{n}+2 k^{n}+\cdots 2011 k^{n}=1+k \cdot k^{n}=k^{n+1}+1 $$ for all $n$, so $(1, k, \ldots, k)$ is a valid sequence. We shall prove that it is the only one. Let a valid sequence $\lef... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 164 | 966 |
2011 | T0 | A3 | Number Theory | IMO-SL | Determine all pairs $(f, g)$ of functions from the set of real numbers to itself that satisfy $$ g(f(x+y))=f(x)+(2 x+y) g(y) $$ for all real numbers $x$ and $y$. Answer. Either both $f$ and $g$ vanish identically, or there exists a real number $C$ such that $f(x)=x^{2}+C$ and $g(x)=x$ for all real numbers $x$. | Clearly all these pairs of functions satisfy the functional equation in question, so it suffices to verify that there cannot be any further ones. Substituting $-2 x$ for $y$ in the given functional equation we obtain $$ 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
} | 104 | 808 |
2011 | T0 | A4 | Number Theory | 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 $\underbrace{f(f(\ldots f}_{k}(n) \ldots))$. Answer. The only pair $(f, g)$ of functions that satisfies the equatio... | 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
} | 128 | 898 |
2011 | T0 | A7 | Number Theory | 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 $$ \frac{a}{(b+c-a)^{2}}+\frac{b}{(c+a-b)^{2}}+\frac{c}{(a+b-c)^{2}} \geq \frac{3}{(a b c)^{2}} $$ Throughout both solutions, we denote the sums of the form $f(a, b, c)+f(b, c, a)+f(c, a, b)$ ... | 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 $$ \frac{x_{1}^{p+1}}{y_{1}^{p}}+\frac{x_{1}^{p+1}}{y_{1}^{p}}+\ldots+\frac{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 156 | 894 |
2011 | T0 | A7 | Number Theory | IMO-SL | Let $a, b$, and $c$ be positive real numbers satisfying $\min (a+b, b+c, c+a)>\sqrt{2}$ and $a^{2}+b^{2}+c^{2}=3$. Prove that $$ \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}} $$ Throughout both solutions, we denote the sums of the form $f(a, b, c)+f(b, c, a)+f(c, a, b)$ ... | As in $$ a^{5}+b^{5}+c^{5} \geq 3 $$ which is weaker than the given one. Due to the symmetry we may assume that $a \geq b \geq c$. In view of (3)), it suffices to prove the inequality $$ \sum \frac{a^{3} b^{2} c^{2}}{(b+c-a)^{2}} \geq \sum a^{5} $$ or, moving all the terms into the left-hand part, $$ \sum \frac{a^{3}}{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 156 | 575 |
2011 | T0 | C1 | Number Theory | IMO-SL | Let $n>0$ be an integer. We are given a balance and $n$ weights of weight $2^{0}, 2^{1}, \ldots, 2^{n-1}$. In a sequence of $n$ moves we place all weights on the balance. In the first move we choose a weight and put it on the left pan. In each of the following moves we choose one of the remaining weights and we add it ... | Assume $n \geq 2$. We claim $$ f(n)=(2 n-1) f(n-1) . $$ Firstly, note that after the first move the left pan is always at least 1 heavier than the right one. Hence, any valid way of placing the $n$ weights on the scale gives rise, by not considering weight 1 , to a valid way of placing the weights $2,2^{2}, \ldots, 2^{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 210 | 571 |
2011 | T0 | C1 | Number Theory | IMO-SL | Let $n>0$ be an integer. We are given a balance and $n$ weights of weight $2^{0}, 2^{1}, \ldots, 2^{n-1}$. In a sequence of $n$ moves we place all weights on the balance. In the first move we choose a weight and put it on the left pan. In each of the following moves we choose one of the remaining weights and we add it ... | We present a different way of obtaining (1). Set $f(0)=1$. Firstly, we find a recurrent formula for $f(n)$. Assume $n \geq 1$. Suppose that weight $2^{n-1}$ is placed on the balance in the $i$-th move with $1 \leq i \leq n$. This weight has to be put on the left pan. For the previous moves we have $\left(\begin{array}{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 210 | 618 |
2011 | T0 | C2 | Number Theory | IMO-SL | Suppose that 1000 students are standing in a circle. Prove that there exists an integer $k$ with $100 \leq k \leq 300$ such that in this circle there exists a contiguous group of $2 k$ students, for which the first half contains the same number of girls as the second half. | Number the students consecutively from 1 to 1000. Let $a_{i}=1$ if the $i$ th student is a girl, and $a_{i}=0$ otherwise. We expand this notion for all integers $i$ by setting $a_{i+1000}=$ $a_{i-1000}=a_{i}$. Next, let $$ S_{k}(i)=a_{i}+a_{i+1}+\cdots+a_{i+k-1} $$ Now the statement of the problem can be reformulated a... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 75 | 2,468 |
2011 | T0 | C3 | Number Theory | IMO-SL | Let $\mathcal{S}$ be a finite set of at least two points in the plane. Assume that no three points of $\mathcal{S}$ are collinear. By a windmill we mean a process as follows. Start with a line $\ell$ going through a point $P \in \mathcal{S}$. Rotate $\ell$ clockwise around the pivot $P$ until the line contains another ... | Give the rotating line an orientation and distinguish its sides as the oranje side and the blue side. Notice that whenever the pivot changes from some point $T$ to another point $U$, after the change, $T$ is on the same side as $U$ was before. Therefore, the number of elements of $\mathcal{S}$ on the oranje side and th... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 179 | 1,102 |
2011 | T0 | C3 | Number Theory | IMO-SL | Let $\mathcal{S}$ be a finite set of at least two points in the plane. Assume that no three points of $\mathcal{S}$ are collinear. By a windmill we mean a process as follows. Start with a line $\ell$ going through a point $P \in \mathcal{S}$. Rotate $\ell$ clockwise around the pivot $P$ until the line contains another ... | There are various examples showing that $k=3$ does indeed have the property under consideration. E.g. one can take $$ \begin{gathered} A_{1}=\{1,2,3\} \cup\{3 m \mid m \geq 4\}, \\ A_{2}=\{4,5,6\} \cup\{3 m-1 \mid m \geq 4\}, \\ A_{3}=\{7,8,9\} \cup\{3 m-2 \mid m \geq 4\} \end{gathered} $$ To check that this partition ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 179 | 929 |
2011 | T0 | C3 | Number Theory | IMO-SL | Let $\mathcal{S}$ be a finite set of at least two points in the plane. Assume that no three points of $\mathcal{S}$ are collinear. By a windmill we mean a process as follows. Start with a line $\ell$ going through a point $P \in \mathcal{S}$. Rotate $\ell$ clockwise around the pivot $P$ until the line contains another ... | Again we only prove that $k \leq 3$. Assume that $A_{1}, A_{2}, \ldots, A_{k}$ is a partition satisfying the given property. We construct a graph $\mathcal{G}$ on the set $V=\{1,2, \ldots, 18\}$ of vertices as follows. For each $i \in\{1,2, \ldots, k\}$ and each $d \in\{15,16,17,19\}$ we choose one pair of distinct ele... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 179 | 649 |
2011 | T0 | C5 | Number Theory | IMO-SL | Let $m$ be a positive integer and consider a checkerboard consisting of $m$ by $m$ unit squares. At the midpoints of some of these unit squares there is an ant. At time 0 , each ant starts moving with speed 1 parallel to some edge of the checkerboard. When two ants moving in opposite directions meet, they both turn $90... | For $m=1$ the answer is clearly correct, so assume $m>1$. In the sequel, the word collision will be used to denote meeting of exactly two ants, moving in opposite directions. If at the beginning we place an ant on the southwest corner square facing east and an ant on the southeast corner square facing west, then they w... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 202 | 1,017 |
2011 | T0 | C7 | Number Theory | IMO-SL | On a square table of 2011 by 2011 cells we place a finite number of napkins that each cover a square of 52 by 52 cells. In each cell we write the number of napkins covering it, and we record the maximal number $k$ of cells that all contain the same nonzero number. Considering all possible napkin configurations, what is... | Let $m=39$, then $2011=52 m-17$. We begin with an example showing that there can exist 3986729 cells carrying the same positive number. To describe it, we number the columns from the left to the right and the rows from the bottom to the top by $1,2, \ldots, 2011$. We will denote each napkin by the coordinates of its lo... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 152 | 1,347 |
2011 | T0 | C7 | Number Theory | IMO-SL | On a square table of 2011 by 2011 cells we place a finite number of napkins that each cover a square of 52 by 52 cells. In each cell we write the number of napkins covering it, and we record the maximal number $k$ of cells that all contain the same nonzero number. Considering all possible napkin configurations, what is... | We present a different proof of the estimate which is the hard part of the problem. Let $S=35, H=17, m=39$; so the table size is $2011=S m+H(m-1)$, and the napkin size is $52=S+H$. Fix any positive integer $M$ and call a cell vicious if it contains a number distinct from $M$. We will prove that there are at least $H^{2... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 152 | 2,515 |
2011 | T0 | G1 | Number Theory | IMO-SL | Let $A B C$ be an acute triangle. Let $\omega$ be a circle whose center $L$ lies on the side $B C$. Suppose that $\omega$ is tangent to $A B$ at $B^{\prime}$ and to $A C$ at $C^{\prime}$. Suppose also that the circumcenter $O$ of the triangle $A B C$ lies on the shorter arc $B^{\prime} C^{\prime}$ of $\omega$. Prove th... | The point $B^{\prime}$, being the perpendicular foot of $L$, is an interior point of side $A B$. Analogously, $C^{\prime}$ lies in the interior of $A C$. The point $O$ is located inside the triangle $A B^{\prime} C^{\prime}$, hence $\angle C O B<\angle C^{\prime} O B^{\prime}$. Let $\alpha=\angle C A B$. The angles $\a... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 122 | 684 |
2011 | T0 | G8 | Number Theory | IMO-SL | Let $A B C$ be an acute triangle with circumcircle $\omega$. Let $t$ be a tangent line to $\omega$. Let $t_{a}, t_{b}$, and $t_{c}$ be the lines obtained by reflecting $t$ in the lines $B C, C A$, and $A B$, respectively. Show that the circumcircle of the triangle determined by the lines $t_{a}, t_{b}$, and $t_{c}$ is ... | Denote by $T$ the point of tangency of $t$ and $\omega$. Let $A^{\prime}=t_{b} \cap t_{c}, B^{\prime}=t_{a} \cap t_{c}$, $C^{\prime}=t_{a} \cap t_{b}$. Introduce the point $A^{\prime \prime}$ on $\omega$ such that $T A=A A^{\prime \prime}\left(A^{\prime \prime} \neq T\right.$ unless $T A$ is a diameter). Define the poi... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 260 | 1,381 |
2011 | T0 | G8 | Number Theory | IMO-SL | Let $A B C$ be an acute triangle with circumcircle $\omega$. Let $t$ be a tangent line to $\omega$. Let $t_{a}, t_{b}$, and $t_{c}$ be the lines obtained by reflecting $t$ in the lines $B C, C A$, and $A B$, respectively. Show that the circumcircle of the triangle determined by the lines $t_{a}, t_{b}$, and $t_{c}$ is ... | Define the points $T, A^{\prime}, B^{\prime}$, and $C^{\prime}$ in the same way as in the previous solution. Let $X, Y$, and $Z$ be the symmetric images of $T$ about the lines $B C, C A$, and $A B$, respectively. Note that the projections of $T$ on these lines form a Simson line of $T$ with respect to $A B C$, therefor... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 260 | 1,023 |
2011 | T0 | N4 | Number Theory | IMO-SL | For each positive integer $k$, let $t(k)$ be the largest odd divisor of $k$. Determine all positive integers $a$ for which there exists a positive integer $n$ such that all the differences $$ t(n+a)-t(n), \quad t(n+a+1)-t(n+1), \quad \ldots, \quad t(n+2 a-1)-t(n+a-1) $$ are divisible by 4 . Answer. $a=1,3$, or 5 . | A pair $(a, n)$ satisfying the condition of the problem will be called a winning pair. It is straightforward to check that the pairs $(1,1),(3,1)$, and $(5,4)$ are winning pairs. Now suppose that $a$ is a positive integer not equal to 1,3 , and 5 . We will show that there are no winning pairs $(a, n)$ by distinguishing... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2011SL.jsonl",
"solution_match": null
} | 109 | 652 |
2012 | T0 | A1 | Algebra | IMO-SL | Find all the functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $$ f(a)^{2}+f(b)^{2}+f(c)^{2}=2 f(a) f(b)+2 f(b) f(c)+2 f(c) f(a) $$ for all integers $a, b, c$ satisfying $a+b+c=0$. | The substitution $a=b=c=0$ gives $3 f(0)^{2}=6 f(0)^{2}$, hence $$ f(0)=0 \text {. } $$ The substitution $b=-a$ and $c=0$ gives $\left((f(a)-f(-a))^{2}=0\right.$. Hence $f$ is an even function: $$ f(a)=f(-a) \quad \text { for all } a \in \mathbb{Z} $$ Now set $b=a$ and $c=-2 a$ to obtain $2 f(a)^{2}+f(2 a)^{2}=2 f(a)^{... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 86 | 1,788 |
2012 | T0 | A2 | Algebra | IMO-SL | Let $\mathbb{Z}$ and $\mathbb{Q}$ be the sets of integers and rationals respectively. a) Does there exist a partition of $\mathbb{Z}$ into three non-empty subsets $A, B, C$ such that the sets $A+B, B+C, C+A$ are disjoint? b) Does there exist a partition of $\mathbb{Q}$ into three non-empty subsets $A, B, C$ such that... | a) The residue classes modulo 3 yield such a partition: $$ A=\{3 k \mid k \in \mathbb{Z}\}, \quad B=\{3 k+1 \mid k \in \mathbb{Z}\}, \quad C=\{3 k+2 \mid k \in \mathbb{Z}\} $$ b) The answer is no. Suppose that $\mathbb{Q}$ can be partitioned into non-empty subsets $A, B, C$ as stated. Note that for all $a \in A, b \in ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 167 | 648 |
2012 | T0 | A2 | Algebra | IMO-SL | Let $\mathbb{Z}$ and $\mathbb{Q}$ be the sets of integers and rationals respectively. a) Does there exist a partition of $\mathbb{Z}$ into three non-empty subsets $A, B, C$ such that the sets $A+B, B+C, C+A$ are disjoint? b) Does there exist a partition of $\mathbb{Q}$ into three non-empty subsets $A, B, C$ such that... | We prove that the example for $\mathbb{Z}$ from the first solution is unique, and then use this fact to solve part b). Let $\mathbb{Z}=A \cup B \cup C$ be a partition of $\mathbb{Z}$ with $A, B, C \neq \emptyset$ and $A+B, B+C, C+A$ disjoint. We need the relations (1) which clearly hold for $\mathbb{Z}$. Fix two consec... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 167 | 954 |
2012 | T0 | A3 | Algebra | IMO-SL | Let $a_{2}, \ldots, a_{n}$ be $n-1$ positive real numbers, where $n \geq 3$, such that $a_{2} a_{3} \cdots a_{n}=1$. Prove that $$ \left(1+a_{2}\right)^{2}\left(1+a_{3}\right)^{3} \cdots\left(1+a_{n}\right)^{n}>n^{n} . $$ | The substitution $a_{2}=\frac{x_{2}}{x_{1}}, a_{3}=\frac{x_{3}}{x_{2}}, \ldots, a_{n}=\frac{x_{1}}{x_{n-1}}$ transforms the original problem into the inequality $$ \left(x_{1}+x_{2}\right)^{2}\left(x_{2}+x_{3}\right)^{3} \cdots\left(x_{n-1}+x_{1}\right)^{n}>n^{n} x_{1}^{2} x_{2}^{3} \cdots x_{n-1}^{n} $$ for all $x_{1}... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 105 | 731 |
2012 | T0 | A4 | Algebra | IMO-SL | Let $f$ and $g$ be two nonzero polynomials with integer coefficients and $\operatorname{deg} f>\operatorname{deg} g$. Suppose that for infinitely many primes $p$ the polynomial $p f+g$ has a rational root. Prove that $f$ has a rational root. | Since $\operatorname{deg} f>\operatorname{deg} g$, we have $|g(x) / f(x)|<1$ for sufficiently large $x$; more precisely, there is a real number $R$ such that $|g(x) / f(x)|<1$ for all $x$ with $|x|>R$. Then for all such $x$ and all primes $p$ we have $$ |p f(x)+g(x)| \geq|f(x)|\left(p-\frac{|g(x)|}{|f(x)|}\right)>0 $$ ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 65 | 1,086 |
2012 | T0 | A4 | Algebra | IMO-SL | Let $f$ and $g$ be two nonzero polynomials with integer coefficients and $\operatorname{deg} f>\operatorname{deg} g$. Suppose that for infinitely many primes $p$ the polynomial $p f+g$ has a rational root. Prove that $f$ has a rational root. | Like in the first solution, there is a real number $R$ such that the real roots of all polynomials of the form $p f+g$ lie in the interval $[-R, R]$. Let $p_{1}<p_{2}<\cdots$ be an infinite sequence of primes so that for every index $k$ the polynomial $p_{k} f+g$ has a rational root $r_{k}$. The sequence $r_{1}, r_{2},... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 65 | 1,087 |
2012 | T0 | A5 | Algebra | IMO-SL | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy the conditions $$ f(1+x y)-f(x+y)=f(x) f(y) \quad \text { for all } x, y \in \mathbb{R} $$ and $f(-1) \neq 0$. | The only solution is the function $f(x)=x-1, x \in \mathbb{R}$. We set $g(x)=f(x)+1$ and show that $g(x)=x$ for all real $x$. The conditions take the form $$ g(1+x y)-g(x+y)=(g(x)-1)(g(y)-1) \quad \text { for all } x, y \in \mathbb{R} \text { and } g(-1) \neq 1 $$ Denote $C=g(-1)-1 \neq 0$. Setting $y=-1$ in (1) gives ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 75 | 1,234 |
2012 | T0 | A6 | Algebra | IMO-SL | Let $f: \mathbb{N} \rightarrow \mathbb{N}$ be a function, and let $f^{m}$ be $f$ applied $m$ times. Suppose that for every $n \in \mathbb{N}$ there exists a $k \in \mathbb{N}$ such that $f^{2 k}(n)=n+k$, and let $k_{n}$ be the smallest such $k$. Prove that the sequence $k_{1}, k_{2}, \ldots$ is unbounded. | We restrict attention to the set $$ S=\left\{1, f(1), f^{2}(1), \ldots\right\} $$ Observe that $S$ is unbounded because for every number $n$ in $S$ there exists a $k>0$ such that $f^{2 k}(n)=n+k$ is in $S$. Clearly $f$ maps $S$ into itself; moreover $f$ is injective on $S$. Indeed if $f^{i}(1)=f^{j}(1)$ with $i \neq j$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 115 | 1,077 |
2012 | T0 | A7 | Algebra | IMO-SL | We say that a function $f: \mathbb{R}^{k} \rightarrow \mathbb{R}$ is a metapolynomial if, for some positive integers $m$ and $n$, it can be represented in the form $$ f\left(x_{1}, \ldots, x_{k}\right)=\max _{i=1, \ldots, m} \min _{j=1, \ldots, n} P_{i, j}\left(x_{1}, \ldots, x_{k}\right) $$ where $P_{i, j}$ are mult... | We use the notation $f(x)=f\left(x_{1}, \ldots, x_{k}\right)$ for $x=\left(x_{1}, \ldots, x_{k}\right)$ and $[m]=\{1,2, \ldots, m\}$. Observe that if a metapolynomial $f(x)$ admits a representation like the one in the statement for certain positive integers $m$ and $n$, then they can be replaced by any $m^{\prime} \geq... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 151 | 2,605 |
2012 | T0 | C1 | Combinatorics | IMO-SL | Several positive integers are written in a row. Iteratively, Alice chooses two adjacent numbers $x$ and $y$ such that $x>y$ and $x$ is to the left of $y$, and replaces the pair $(x, y)$ by either $(y+1, x)$ or $(x-1, x)$. Prove that she can perform only finitely many such iterations. | Let the current numbers be $a_{1}, a_{2}, \ldots, a_{n}$. Define the score $s_{i}$ of $a_{i}$ as the number of $a_{j}$ 's that are less than $a_{i}$. Call the sequence $s_{1}, s_{2}, \ldots, s_{n}$ the score sequence of $a_{1}, a_{2}, \ldots, a_{n}$. Let us say that a sequence $x_{1}, \ldots, x_{n}$ dominates a sequenc... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 83 | 702 |
2012 | T0 | C2 | Combinatorics | IMO-SL | Let $n \geq 1$ be an integer. What is the maximum number of disjoint pairs of elements of the set $\{1,2, \ldots, n\}$ such that the sums of the different pairs are different integers not exceeding $n$ ? | Consider $x$ such pairs in $\{1,2, \ldots, n\}$. The sum $S$ of the $2 x$ numbers in them is at least $1+2+\cdots+2 x$ since the pairs are disjoint. On the other hand $S \leq n+(n-1)+\cdots+(n-x+1)$ because the sums of the pairs are different and do not exceed $n$. This gives the inequality $$ \frac{2 x(2 x+1)}{2} \leq... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 56 | 1,002 |
2012 | T0 | C3 | Combinatorics | IMO-SL | In a $999 \times 999$ square table some cells are white and the remaining ones are red. Let $T$ be the number of triples $\left(C_{1}, C_{2}, C_{3}\right)$ of cells, the first two in the same row and the last two in the same column, with $C_{1}$ and $C_{3}$ white and $C_{2}$ red. Find the maximum value $T$ can attain. | We prove that in an $n \times n$ square table there are at most $\frac{4 n^{4}}{27}$ such triples. Let row $i$ and column $j$ contain $a_{i}$ and $b_{j}$ white cells respectively, and let $R$ be the set of red cells. For every red cell $(i, j)$ there are $a_{i} b_{j}$ admissible triples $\left(C_{1}, C_{2}, C_{3}\right... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 101 | 782 |
2012 | T0 | C4 | Combinatorics | IMO-SL | Players $A$ and $B$ play a game with $N \geq 2012$ coins and 2012 boxes arranged around a circle. Initially $A$ distributes the coins among the boxes so that there is at least 1 coin in each box. Then the two of them make moves in the order $B, A, B, A, \ldots$ by the following rules: - On every move of his $B$ passes... | We argue for a general $n \geq 7$ instead of 2012 and prove that the required minimum $N$ is $2 n-2$. For $n=2012$ this gives $N_{\min }=4022$. a) If $N=2 n-2$ player $A$ can achieve her goal. Let her start the game with a regular distribution: $n-2$ boxes with 2 coins and 2 boxes with 1 coin. Call the boxes of the two... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 200 | 1,655 |
2012 | T0 | C5 | Combinatorics | IMO-SL | The columns and the rows of a $3 n \times 3 n$ square board are numbered $1,2, \ldots, 3 n$. Every square $(x, y)$ with $1 \leq x, y \leq 3 n$ is colored asparagus, byzantium or citrine according as the modulo 3 remainder of $x+y$ is 0,1 or 2 respectively. One token colored asparagus, byzantium or citrine is placed on ... | Without loss of generality it suffices to prove that the A-tokens can be moved to distinct $\mathrm{A}$-squares in such a way that each $\mathrm{A}$-token is moved to a distance at most $d+2$ from its original place. This means we need a perfect matching between the $3 n^{2} \mathrm{~A}$-squares and the $3 n^{2}$ A-tok... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 231 | 1,023 |
2012 | T0 | C6 | Combinatorics | IMO-SL | Let $k$ and $n$ be fixed positive integers. In the liar's guessing game, Amy chooses integers $x$ and $N$ with $1 \leq x \leq N$. She tells Ben what $N$ is, but not what $x$ is. Ben may then repeatedly ask Amy whether $x \in S$ for arbitrary sets $S$ of integers. Amy will always answer with yes or no, but she might lie... | Consider an answer $A \in\{y e s, n o\}$ to a question of the kind "Is $x$ in the set $S$ ?" We say that $A$ is inconsistent with a number $i$ if $A=y e s$ and $i \notin S$, or if $A=n o$ and $i \in S$. Observe that an answer inconsistent with the target number $x$ is a lie. a) Suppose that Ben has determined a set $T$... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 210 | 1,418 |
2012 | T0 | C7 | Combinatorics | IMO-SL | There are given $2^{500}$ points on a circle labeled $1,2, \ldots, 2^{500}$ in some order. Prove that one can choose 100 pairwise disjoint chords joining some of these points so that the 100 sums of the pairs of numbers at the endpoints of the chosen chords are equal. | The proof is based on the following general fact. Lemma. In a graph $G$ each vertex $v$ has degree $d_{v}$. Then $G$ contains an independent set $S$ of vertices such that $|S| \geq f(G)$ where $$ f(G)=\sum_{v \in G} \frac{1}{d_{v}+1} $$ Proof. Induction on $n=|G|$. The base $n=1$ is clear. For the inductive step choose... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 76 | 1,325 |
2012 | T0 | G1 | Geometry | IMO-SL | In the triangle $A B C$ the point $J$ is the center of the excircle opposite to $A$. This excircle is tangent to the side $B C$ at $M$, and to the lines $A B$ and $A C$ at $K$ and $L$ respectively. The lines $L M$ and $B J$ meet at $F$, and the lines $K M$ and $C J$ meet at $G$. Let $S$ be the point of intersection of ... | Let $\alpha=\angle C A B, \beta=\angle A B C$ and $\gamma=\angle B C A$. The line $A J$ is the bisector of $\angle C A B$, so $\angle J A K=\angle J A L=\frac{\alpha}{2}$. By $\angle A K J=\angle A L J=90^{\circ}$ the points $K$ and $L$ lie on the circle $\omega$ with diameter $A J$. The triangle $K B M$ is isosceles a... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 154 | 602 |
2012 | T0 | G3 | Geometry | IMO-SL | In an acute triangle $A B C$ the points $D, E$ and $F$ are the feet of the altitudes through $A$, $B$ and $C$ respectively. The incenters of the triangles $A E F$ and $B D F$ are $I_{1}$ and $I_{2}$ respectively; the circumcenters of the triangles $A C I_{1}$ and $B C I_{2}$ are $O_{1}$ and $O_{2}$ respectively. Prove ... | Let $\angle C A B=\alpha, \angle A B C=\beta, \angle B C A=\gamma$. We start by showing that $A, B, I_{1}$ and $I_{2}$ are concyclic. Since $A I_{1}$ and $B I_{2}$ bisect $\angle C A B$ and $\angle A B C$, their extensions beyond $I_{1}$ and $I_{2}$ meet at the incenter $I$ of the triangle. The points $E$ and $F$ are o... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 131 | 850 |
2012 | T0 | G6 | Geometry | IMO-SL | Let $A B C$ be a triangle with circumcenter $O$ and incenter $I$. The points $D, E$ and $F$ on the sides $B C, C A$ and $A B$ respectively are such that $B D+B F=C A$ and $C D+C E=A B$. The circumcircles of the triangles $B F D$ and $C D E$ intersect at $P \neq D$. Prove that $O P=O I$. | By Miquel's theorem the circles $(A E F)=\omega_{A},(B F D)=\omega_{B}$ and $(C D E)=\omega_{C}$ have a common point, for arbitrary points $D, E$ and $F$ on $B C, C A$ and $A B$. So $\omega_{A}$ passes through the common point $P \neq D$ of $\omega_{B}$ and $\omega_{C}$. Let $\omega_{A}, \omega_{B}$ and $\omega_{C}$ me... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 105 | 1,979 |
2012 | T0 | G7 | Geometry | IMO-SL | Let $A B C D$ be a convex quadrilateral with non-parallel sides $B C$ and $A D$. Assume that there is a point $E$ on the side $B C$ such that the quadrilaterals $A B E D$ and $A E C D$ are circumscribed. Prove that there is a point $F$ on the side $A D$ such that the quadrilaterals $A B C F$ and $B C D F$ are circumscr... | Let $\omega_{1}$ and $\omega_{2}$ be the incircles and $O_{1}$ and $O_{2}$ the incenters of the quadrilaterals $A B E D$ and $A E C D$ respectively. A point $F$ with the stated property exists only if $\omega_{1}$ and $\omega_{2}$ are also the incircles of the quadrilaterals $A B C F$ and $B C D F$. Let the tangents fr... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 121 | 1,552 |
2012 | T0 | G8 | Geometry | IMO-SL | Let $A B C$ be a triangle with circumcircle $\omega$ and $\ell$ a line without common points with $\omega$. Denote by $P$ the foot of the perpendicular from the center of $\omega$ to $\ell$. The side-lines $B C, C A, A B$ intersect $\ell$ at the points $X, Y, Z$ different from $P$. Prove that the circumcircles of the t... | Let $\omega_{A}, \omega_{B}, \omega_{C}$ and $\omega$ be the circumcircles of triangles $A X P, B Y P, C Z P$ and $A B C$ respectively. The strategy of the proof is to construct a point $Q$ with the same power with respect to the four circles. Then each of $P$ and $Q$ has the same power with respect to $\omega_{A}, \om... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 124 | 2,023 |
2012 | T0 | G8 | Geometry | IMO-SL | Let $A B C$ be a triangle with circumcircle $\omega$ and $\ell$ a line without common points with $\omega$. Denote by $P$ the foot of the perpendicular from the center of $\omega$ to $\ell$. The side-lines $B C, C A, A B$ intersect $\ell$ at the points $X, Y, Z$ different from $P$. Prove that the circumcircles of the t... | First we prove that there is an inversion in space that takes $\ell$ and $\omega$ to parallel circles on a sphere. Let $Q R$ be the diameter of $\omega$ whose extension beyond $Q$ passes through $P$. Let $\Pi$ be the plane carrying our objects. In space, choose a point $O$ such that the line $Q O$ is perpendicular to $... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 124 | 1,104 |
2012 | T0 | N2 | Number Theory | IMO-SL | Find all triples $(x, y, z)$ of positive integers such that $x \leq y \leq z$ and $$ x^{3}\left(y^{3}+z^{3}\right)=2012(x y z+2) \text {. } $$ | First note that $x$ divides $2012 \cdot 2=2^{3} \cdot 503$. If $503 \mid x$ then the right-hand side of the equation is divisible by $503^{3}$, and it follows that $503^{2} \mid x y z+2$. This is false as $503 \mid x$. Hence $x=2^{m}$ with $m \in\{0,1,2,3\}$. If $m \geq 2$ then $2^{6} \mid 2012(x y z+2)$. However the h... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 60 | 1,006 |
2012 | T0 | N3 | Number Theory | IMO-SL | Determine all integers $m \geq 2$ such that every $n$ with $\frac{m}{3} \leq n \leq \frac{m}{2}$ divides the binomial coefficient $\left(\begin{array}{c}n \\ m-2 n\end{array}\right)$. | The integers in question are all prime numbers. First we check that all primes satisfy the condition, and even a stronger one. Namely, if $p$ is a prime then every $n$ with $1 \leq n \leq \frac{p}{2}$ divides $\left(\begin{array}{c}n \\ p-2 n\end{array}\right)$. This is true for $p=2$ where $n=1$ is the only possibilit... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 68 | 609 |
2012 | T0 | N4 | Number Theory | IMO-SL | An integer $a$ is called friendly if the equation $\left(m^{2}+n\right)\left(n^{2}+m\right)=a(m-n)^{3}$ has a solution over the positive integers. a) Prove that there are at least 500 friendly integers in the set $\{1,2, \ldots, 2012\}$. b) Decide whether $a=2$ is friendly. | a) Every $a$ of the form $a=4 k-3$ with $k \geq 2$ is friendly. Indeed the numbers $m=2 k-1>0$ and $n=k-1>0$ satisfy the given equation with $a=4 k-3$ : $$ \left(m^{2}+n\right)\left(n^{2}+m\right)=\left((2 k-1)^{2}+(k-1)\right)\left((k-1)^{2}+(2 k-1)\right)=(4 k-3) k^{3}=a(m-n)^{3} \text {. } $$ Hence $5,9, \ldots, 200... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 97 | 575 |
2012 | T0 | N5 | Number Theory | IMO-SL | For a nonnegative integer $n$ define $\operatorname{rad}(n)=1$ if $n=0$ or $n=1$, and $\operatorname{rad}(n)=p_{1} p_{2} \cdots p_{k}$ where $p_{1}<p_{2}<\cdots<p_{k}$ are all prime factors of $n$. Find all polynomials $f(x)$ with nonnegative integer coefficients such that $\operatorname{rad}(f(n))$ divides $\operatorn... | We are going to prove that $f(x)=a x^{m}$ for some nonnegative integers $a$ and $m$. If $f(x)$ is the zero polynomial we are done, so assume that $f(x)$ has at least one positive coefficient. In particular $f(1)>0$. Let $p$ be a prime number. The condition is that $f(n) \equiv 0(\bmod p)$ implies $$ f\left(n^{\operator... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 136 | 607 |
2012 | T0 | N5 | Number Theory | IMO-SL | For a nonnegative integer $n$ define $\operatorname{rad}(n)=1$ if $n=0$ or $n=1$, and $\operatorname{rad}(n)=p_{1} p_{2} \cdots p_{k}$ where $p_{1}<p_{2}<\cdots<p_{k}$ are all prime factors of $n$. Find all polynomials $f(x)$ with nonnegative integer coefficients such that $\operatorname{rad}(f(n))$ divides $\operatorn... | Let $f(x)$ be a polynomial with integer coefficients (not necessarily nonnegative) such that $\operatorname{rad}(f(n))$ divides $\operatorname{rad}\left(f\left(n^{\operatorname{rad}(n)}\right)\right)$ for any nonnegative integer $n$. We give a complete description of all polynomials with this property. More precisely, ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 136 | 1,398 |
2012 | T0 | N7 | Number Theory | IMO-SL | Find all $n \in \mathbb{N}$ for which there exist nonnegative integers $a_{1}, a_{2}, \ldots, a_{n}$ such that $$ \frac{1}{2^{a_{1}}}+\frac{1}{2^{a_{2}}}+\cdots+\frac{1}{2^{a_{n}}}=\frac{1}{3^{a_{1}}}+\frac{2}{3^{a_{2}}}+\cdots+\frac{n}{3^{a_{n}}}=1 . $$ | Such numbers $a_{1}, a_{2}, \ldots, a_{n}$ exist if and only if $n \equiv 1(\bmod 4)$ or $n \equiv 2(\bmod 4)$. Let $\sum_{k=1}^{n} \frac{k}{3^{a} k}=1$ with $a_{1}, a_{2}, \ldots, a_{n}$ nonnegative integers. Then $1 \cdot x_{1}+2 \cdot x_{2}+\cdots+n \cdot x_{n}=3^{a}$ with $x_{1}, \ldots, x_{n}$ powers of 3 and $a \... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 115 | 1,733 |
2012 | T0 | C5 | Combinatorics | IMO-SL | The columns and the rows of a $3 n \times 3 n$ square board are numbered $1,2, \ldots, 3 n$. Every square $(x, y)$ with $1 \leq x, y \leq 3 n$ is colored asparagus, byzantium or citrine according as the modulo 3 remainder of $x+y$ is 0 , 1 or 2 respectively. One token colored asparagus, byzantium or citrine is placed o... | Without loss of generality it suffices to prove that the A-tokens can be moved to distinct $\mathrm{A}$-squares in such a way that each $\mathrm{A}$-token is moved to a distance at most $d+2$ from its original place. This means we need a perfect matching between the $3 n^{2} \mathrm{~A}$-squares and the $3 n^{2}$ A-tok... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 232 | 1,023 |
2012 | T0 | N5 | Number Theory | IMO-SL | For a nonnegative integer $n$ define $\operatorname{rad}(n)=1$ if $n=0$ or $n=1$, and $\operatorname{rad}(n)=p_{1} p_{2} \cdots p_{k}$ where $p_{1}<p_{2}<\cdots<p_{k}$ are all prime factors of $n$. Find all polynomials $f(x)$ with nonnegative integer coefficients such that $\operatorname{rad}(f(n))$ divides $\operatorn... | We are going to prove that $f(x)=a x^{m}$ for some nonnegative integers $a$ and $m$. If $f(x)$ is the zero polynomial we are done, so assume that $f(x)$ has at least one positive coefficient. In particular $f(1)>0$. Let $p$ be a prime number. The condition is that $f(n) \equiv 0(\bmod p)$ implies $$ f\left(n^{\operator... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 139 | 607 |
2012 | T0 | N5 | Number Theory | IMO-SL | For a nonnegative integer $n$ define $\operatorname{rad}(n)=1$ if $n=0$ or $n=1$, and $\operatorname{rad}(n)=p_{1} p_{2} \cdots p_{k}$ where $p_{1}<p_{2}<\cdots<p_{k}$ are all prime factors of $n$. Find all polynomials $f(x)$ with nonnegative integer coefficients such that $\operatorname{rad}(f(n))$ divides $\operatorn... | Let $f(x)$ be a polynomial with integer coefficients (not necessarily nonnegative) such that $\operatorname{rad}(f(n))$ divides $\operatorname{rad}\left(f\left(n^{\operatorname{rad}(n)}\right)\right)$ for any nonnegative integer $n$. We give a complete description of all polynomials with this property. More precisely, ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2012SL.jsonl",
"solution_match": null
} | 139 | 1,398 |
2013 | T0 | A1 | Algebra | IMO-SL | Let $n$ be a positive integer and let $a_{1}, \ldots, a_{n-1}$ be arbitrary real numbers. Define the sequences $u_{0}, \ldots, u_{n}$ and $v_{0}, \ldots, v_{n}$ inductively by $u_{0}=u_{1}=v_{0}=v_{1}=1$, and $$ u_{k+1}=u_{k}+a_{k} u_{k-1}, \quad v_{k+1}=v_{k}+a_{n-k} v_{k-1} \quad \text { for } k=1, \ldots, n-1 . $$ ... | We prove by induction on $k$ that $$ u_{k}=\sum_{\substack{0<i_{1}<\ldots<i_{t}<k \\ i_{j}+1-i_{j} \geqslant 2}} a_{i_{1}} \ldots a_{i_{t}} $$ Note that we have one trivial summand equal to 1 (which corresponds to $t=0$ and the empty sequence, whose product is 1 ). For $k=0,1$ the sum on the right-hand side only contai... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 169 | 719 |
2013 | T0 | A1 | Algebra | IMO-SL | Let $n$ be a positive integer and let $a_{1}, \ldots, a_{n-1}$ be arbitrary real numbers. Define the sequences $u_{0}, \ldots, u_{n}$ and $v_{0}, \ldots, v_{n}$ inductively by $u_{0}=u_{1}=v_{0}=v_{1}=1$, and $$ u_{k+1}=u_{k}+a_{k} u_{k-1}, \quad v_{k+1}=v_{k}+a_{n-k} v_{k-1} \quad \text { for } k=1, \ldots, n-1 . $$ ... | Define recursively a sequence of multivariate polynomials by $$ P_{0}=P_{1}=1, \quad P_{k+1}\left(x_{1}, \ldots, x_{k}\right)=P_{k}\left(x_{1}, \ldots, x_{k-1}\right)+x_{k} P_{k-1}\left(x_{1}, \ldots, x_{k-2}\right), $$ so $P_{n}$ is a polynomial in $n-1$ variables for each $n \geqslant 1$. Two easy inductive arguments... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 169 | 592 |
2013 | T0 | A1 | Algebra | IMO-SL | Let $n$ be a positive integer and let $a_{1}, \ldots, a_{n-1}$ be arbitrary real numbers. Define the sequences $u_{0}, \ldots, u_{n}$ and $v_{0}, \ldots, v_{n}$ inductively by $u_{0}=u_{1}=v_{0}=v_{1}=1$, and $$ u_{k+1}=u_{k}+a_{k} u_{k-1}, \quad v_{k+1}=v_{k}+a_{n-k} v_{k-1} \quad \text { for } k=1, \ldots, n-1 . $$ ... | Using matrix notation, we can rewrite the recurrence relation as $$ \left(\begin{array}{c} u_{k+1} \\ u_{k+1}-u_{k} \end{array}\right)=\left(\begin{array}{c} u_{k}+a_{k} u_{k-1} \\ a_{k} u_{k-1} \end{array}\right)=\left(\begin{array}{cc} 1+a_{k} & -a_{k} \\ a_{k} & -a_{k} \end{array}\right)\left(\begin{array}{c} u_{k} ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 169 | 1,814 |
2013 | T0 | A2 | Algebra | IMO-SL | Prove that in any set of 2000 distinct real numbers there exist two pairs $a>b$ and $c>d$ with $a \neq c$ or $b \neq d$, such that $$ \left|\frac{a-b}{c-d}-1\right|<\frac{1}{100000} $$ (Lithuania) | For any set $S$ of $n=2000$ distinct real numbers, let $D_{1} \leqslant D_{2} \leqslant \cdots \leqslant D_{m}$ be the distances between them, displayed with their multiplicities. Here $m=n(n-1) / 2$. By rescaling the numbers, we may assume that the smallest distance $D_{1}$ between two elements of $S$ is $D_{1}=1$. Le... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 83 | 994 |
2013 | T0 | A3 | Algebra | IMO-SL | Let $\mathbb{Q}_{>0}$ be the set of positive rational numbers. Let $f: \mathbb{Q}_{>0} \rightarrow \mathbb{R}$ be a function satisfying the conditions $$ f(x) f(y) \geqslant f(x y) \text { and } f(x+y) \geqslant f(x)+f(y) $$ for all $x, y \in \mathbb{Q}_{>0}$. Given that $f(a)=a$ for some rational $a>1$, prove that $... | Denote by $\mathbb{Z}_{>0}$ the set of positive integers. Plugging $x=1, y=a$ into (1) we get $f(1) \geqslant 1$. Next, by an easy induction on $n$ we get from (2) that $$ f(n x) \geqslant n f(x) \text { for all } n \in \mathbb{Z}_{>0} \text { and } x \in \mathbb{Q}_{>0} $$ In particular, we have $$ f(n) \geqslant n f(... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 148 | 876 |
2013 | T0 | A4 | Algebra | IMO-SL | Let $n$ be a positive integer, and consider a sequence $a_{1}, a_{2}, \ldots, a_{n}$ of positive integers. Extend it periodically to an infinite sequence $a_{1}, a_{2}, \ldots$ by defining $a_{n+i}=a_{i}$ for all $i \geqslant 1$. If $$ a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n} \leqslant a_{1}+n $$ and $... | First, we claim that $$ a_{i} \leqslant n+i-1 \quad \text { for } i=1,2, \ldots, n \text {. } $$ Assume contrariwise that $i$ is the smallest counterexample. From $a_{n} \geqslant a_{n-1} \geqslant \cdots \geqslant a_{i} \geqslant n+i$ and $a_{a_{i}} \leqslant n+i-1$, taking into account the periodicity of our sequence... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 192 | 904 |
2013 | T0 | A4 | Algebra | IMO-SL | Let $n$ be a positive integer, and consider a sequence $a_{1}, a_{2}, \ldots, a_{n}$ of positive integers. Extend it periodically to an infinite sequence $a_{1}, a_{2}, \ldots$ by defining $a_{n+i}=a_{i}$ for all $i \geqslant 1$. If $$ a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n} \leqslant a_{1}+n $$ and $... | In the first quadrant of an infinite grid, consider the increasing "staircase" obtained by shading in dark the bottom $a_{i}$ cells of the $i$ th column for $1 \leqslant i \leqslant n$. We will prove that there are at most $n^{2}$ dark cells. To do it, consider the $n \times n$ square $S$ in the first quadrant with a v... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 192 | 705 |
2013 | T0 | A4 | Algebra | IMO-SL | Let $n$ be a positive integer, and consider a sequence $a_{1}, a_{2}, \ldots, a_{n}$ of positive integers. Extend it periodically to an infinite sequence $a_{1}, a_{2}, \ldots$ by defining $a_{n+i}=a_{i}$ for all $i \geqslant 1$. If $$ a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n} \leqslant a_{1}+n $$ and $... | As in For $1 \leqslant i<j \leqslant n$ we have $a_{i} \leqslant a_{j}$ and $i<j$, so $c_{i} \leqslant c_{j}$. Also $a_{n} \leqslant a_{1}+n$ and $n<1+n$ imply $c_{n} \leqslant c_{1}+n$. Finally, the definitions imply that $c_{c_{i}} \in\left\{a_{a_{i}}, a_{i}, a_{i}-n, i\right\}$ so $c_{c_{i}} \leqslant n+i-1$ by (2) ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 192 | 1,410 |
2013 | T0 | A5 | Algebra | IMO-SL | Let $\mathbb{Z}_{\geqslant 0}$ be the set of all nonnegative integers. Find all the functions $f: \mathbb{Z}_{\geqslant 0} \rightarrow \mathbb{Z}_{\geqslant 0}$ satisfying the relation $$ f(f(f(n)))=f(n+1)+1 $$ for all $n \in \mathbb{Z}_{\geqslant 0}$. (Serbia) | To start, we get from (*) that $$ f^{4}(n)=f\left(f^{3}(n)\right)=f(f(n+1)+1) \quad \text { and } \quad f^{4}(n+1)=f^{3}(f(n+1))=f(f(n+1)+1)+1 $$ thus $$ f^{4}(n)+1=f^{4}(n+1) . $$ I. Let us denote by $R_{i}$ the range of $f^{i}$; note that $R_{0}=\mathbb{Z}_{\geqslant 0}$ since $f^{0}$ is the identity function. Obviou... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 106 | 1,772 |
2013 | T0 | A5 | Algebra | IMO-SL | Let $\mathbb{Z}_{\geqslant 0}$ be the set of all nonnegative integers. Find all the functions $f: \mathbb{Z}_{\geqslant 0} \rightarrow \mathbb{Z}_{\geqslant 0}$ satisfying the relation $$ f(f(f(n)))=f(n+1)+1 $$ for all $n \in \mathbb{Z}_{\geqslant 0}$. (Serbia) | I. For convenience, let us introduce the function $g(n)=f(n)+1$. Substituting $f(n)$ instead of $n$ into (*) we obtain $$ f^{4}(n)=f(f(n)+1)+1, \quad \text { or } \quad f^{4}(n)=g^{2}(n) . $$ Applying $f$ to both parts of (*) and using (5) we get $$ f^{4}(n)+1=f(f(n+1)+1)+1=f^{4}(n+1) . $$ Thus, if $g^{2}(0)=f^{4}(0)=c... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 106 | 1,410 |
2013 | T0 | A6 | Algebra | IMO-SL | Let $m \neq 0$ be an integer. Find all polynomials $P(x)$ with real coefficients such that $$ \left(x^{3}-m x^{2}+1\right) P(x+1)+\left(x^{3}+m x^{2}+1\right) P(x-1)=2\left(x^{3}-m x+1\right) P(x) $$ for all real numbers $x$. | Let $P(x)=a_{n} x^{n}+\cdots+a_{0} x^{0}$ with $a_{n} \neq 0$. Comparing the coefficients of $x^{n+1}$ on both sides gives $a_{n}(n-2 m)(n-1)=0$, so $n=1$ or $n=2 m$. If $n=1$, one easily verifies that $P(x)=x$ is a solution, while $P(x)=1$ is not. Since the given condition is linear in $P$, this means that the linear ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 97 | 793 |
2013 | T0 | A6 | Algebra | IMO-SL | Let $m \neq 0$ be an integer. Find all polynomials $P(x)$ with real coefficients such that $$ \left(x^{3}-m x^{2}+1\right) P(x+1)+\left(x^{3}+m x^{2}+1\right) P(x-1)=2\left(x^{3}-m x+1\right) P(x) $$ for all real numbers $x$. | Multiplying (1) by $x$, we rewrite it as $$ x\left(x^{3}-m x^{2}+1\right) P(x+1)+x\left(x^{3}+m x^{2}+1\right) P(x-1)=[(x+1)+(x-1)]\left(x^{3}-m x+1\right) P(x) . $$ After regrouping, it becomes $$ \left(x^{3}-m x^{2}+1\right) Q(x)=\left(x^{3}+m x^{2}+1\right) Q(x-1) \text {, } $$ where $Q(x)=x P(x+1)-(x+1) P(x)$. If $... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 97 | 1,086 |
2013 | T0 | C1 | Combinatorics | IMO-SL | Let $n$ be a positive integer. Find the smallest integer $k$ with the following property: Given any real numbers $a_{1}, \ldots, a_{d}$ such that $a_{1}+a_{2}+\cdots+a_{d}=n$ and $0 \leqslant a_{i} \leqslant 1$ for $i=1,2, \ldots, d$, it is possible to partition these numbers into $k$ groups (some of which may be empty... | We will show that it is even possible to split the sequence $a_{1}, \ldots, a_{d}$ into $2 n-1$ contiguous groups so that the sum of the numbers in each groups does not exceed 1. Consider a segment $S$ of length $n$, and partition it into segments $S_{1}, \ldots, S_{d}$ of lengths $a_{1}, \ldots, a_{d}$, respectively, ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 134 | 645 |
2013 | T0 | C1 | Combinatorics | IMO-SL | Let $n$ be a positive integer. Find the smallest integer $k$ with the following property: Given any real numbers $a_{1}, \ldots, a_{d}$ such that $a_{1}+a_{2}+\cdots+a_{d}=n$ and $0 \leqslant a_{i} \leqslant 1$ for $i=1,2, \ldots, d$, it is possible to partition these numbers into $k$ groups (some of which may be empty... | First put all numbers greater than $\frac{1}{2}$ in their own groups. Then, form the remaining groups as follows: For each group, add new $a_{i} \mathrm{~S}$ one at a time until their sum exceeds $\frac{1}{2}$. Since the last summand is at most $\frac{1}{2}$, this group has sum at most 1 . Continue this procedure until... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 134 | 524 |
2013 | T0 | C2 | Combinatorics | IMO-SL | In the plane, 2013 red points and 2014 blue points are marked so that no three of the marked points are collinear. One needs to draw $k$ lines not passing through the marked points and dividing the plane into several regions. The goal is to do it in such a way that no region contains points of both colors. Find the mi... | Firstly, let us present an example showing that $k \geqslant 2013$. Mark 2013 red and 2013 blue points on some circle alternately, and mark one more blue point somewhere in the plane. The circle is thus split into 4026 arcs, each arc having endpoints of different colors. Thus, if the goal is reached, then each arc shou... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 106 | 510 |
2013 | T0 | C2 | Combinatorics | IMO-SL | In the plane, 2013 red points and 2014 blue points are marked so that no three of the marked points are collinear. One needs to draw $k$ lines not passing through the marked points and dividing the plane into several regions. The goal is to do it in such a way that no region contains points of both colors. Find the mi... | Let us present a different proof of the fact that $k=2013$ suffices. In fact, we will prove a more general statement: If $n$ points in the plane, no three of which are collinear, are colored in red and blue arbitrarily, then it suffices to draw $\lfloor n / 2\rfloor$ lines to reach the goal. We proceed by induction on ... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 106 | 593 |
2013 | T0 | C3 | Combinatorics | IMO-SL | A crazy physicist discovered a new kind of particle which he called an imon, after some of them mysteriously appeared in his lab. Some pairs of imons in the lab can be entangled, and each imon can participate in many entanglement relations. The physicist has found a way to perform the following two kinds of operations ... | Again, we will use the graph language. I. We start with the following observation. Lemma. Assume that a graph $G$ contains an isolated vertex $A$, and a graph $G^{\circ}$ is obtained from $G$ by deleting this vertex. Then, if one can apply a sequence of operations which makes a graph with no edges from $G^{\circ}$, the... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 252 | 1,283 |
2013 | T0 | C4 | Combinatorics | IMO-SL | Let $n$ be a positive integer, and let $A$ be a subset of $\{1, \ldots, n\}$. An $A$-partition of $n$ into $k$ parts is a representation of $n$ as a sum $n=a_{1}+\cdots+a_{k}$, where the parts $a_{1}, \ldots, a_{k}$ belong to $A$ and are not necessarily distinct. The number of different parts in such a partition is the... | If there are no $A$-partitions of $n$, the result is vacuously true. Otherwise, let $k_{\min }$ be the minimum number of parts in an $A$-partition of $n$, and let $n=a_{1}+\cdots+a_{k_{\min }}$ be an optimal partition. Denote by $s$ the number of different parts in this partition, so we can write $S=\left\{a_{1}, \ldot... | {
"problem_match": null,
"resource_path": "IMO_SL/segmented/en-IMO2013SL.jsonl",
"solution_match": null
} | 216 | 1,096 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.