id stringlengths 4 4 | problem_markdown stringlengths 36 3.59k | solutions_markdown listlengths 0 10 | images images listlengths 0 15 | country stringclasses 58
values | competition stringlengths 3 108 ⌀ | topics_flat listlengths 0 12 | language stringclasses 18
values | problem_type stringclasses 4
values | final_answer stringlengths 1 1.22k ⌀ |
|---|---|---|---|---|---|---|---|---|---|
07ky | Let $S$ be a set of $n$ different points in the plane so that no three lie on a line. Show that there are at least $\frac{n(n-2)}{3}$ triangles whose vertices are among the points of $S$ and such that the triangles do not contain any other points of $S$. | [
"Choose two points $X$ and $Y$ from $S$, and consider the half plane $h$ with boundary the line $XY$, and containing at least one other point of $S$. Now select a point $Z$ of $S$ in the interior of $h$ whose distance from the line $XY$ is a minimum. If there are several such points, choose an arbitrary one.\n\nNow... | Ireland | Irska | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
053x | Find all positive real solutions of the system of equations
$$x + \frac{1}{x} - w = 2,$$
$$y + \frac{1}{y} - w = 2,$$
$$z + \frac{1}{z} + w = 2,$$
$$y + \frac{1}{z} + w = 2.$$ | [
"Subtracting the second equation from the first and multiplying by $xy$ we get $x^2y + y - xy^2 - x = 0$, which gives either $x = y$ or $x = \\frac{1}{y}$.\nIf $x = y$, then subtracting the fourth equation from the third and multiplying by $xz$, we similarly get either $x = z$ or $x = -\\frac{1}{z}$. If $x = z$, th... | Estonia | Estonian Math Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (x, y, z, w) = (2, 1/2, 1, 1/2) | |
00mu | Let $n \ge 2$ be an integer.
Ariane and Bérénice play a game on the set of residue classes modulo $n$. In the beginning, the residue class $1$ is written on a piece of paper. In each move, the player whose turn it is replaces the current residue class $x$ with either $x+1$ or $2x$. The two players alternate with Ariane... | [
"**Answer.** Ariane wins for $n = 2, 4$ and $8$, for all other $n \\ge 2$ Bérénice wins.\n\nWe observe: If Ariane can win for a certain $n$, she will also win for all divisors of $n$, and conversely, if Bérénice can win for a certain $n$, she will also win for all multiples of $n$ because a residue $0$ modulo $n$ i... | Austria | AUT_ABooklet_2020 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Other"
] | English | proof and answer | Ariane wins for n = 2, 4, 8; for all other integers n ≥ 2, Bérénice wins. | |
0kqj | Problem:
Unit segments $A B$ and $C D$ are given such that $A B \sharp C D$. If $M$ is the midpoint of $A C$, $N$ the midpoint of $B D$, and $X$ the intersection of lines $A B$ and $C D$, given $M \neq N$ and $A \neq X \neq C$, show that the angle bisector of $\angle A X C$ is either parallel or perpendicular to $M N$... | [
"Solution:\n\nLet $E$ be such that $A B E D$ is a parallelogram. $E D$ is then a unit segment parallel to $A B$, so the angle bisector of $\\angle C D E$ is parallel either to the internal or to the external angle bisector of $\\angle A X C$, the angle between lines $A B$ and $C D$. Furthermore, the midpoint of $B ... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
01ux | The altitudes $AA_1$, $BB_1$ and $CC_1$ are drawn in the acute triangle $ABC$. The bisector of the angle $AA_1C$ intersects the segments $CC_1$ and $CA$ at $E$ and $D$ respectively. The bisector of the angle $AA_1B$ intersects the segments $BB_1$ and $BA$ at $F$ and $G$ respectively. The circumcircles of the triangles ... | [
"Since $\\angle BB_1C = \\angle CC_1B = 90^\\circ$, it is sufficient to prove that the points $B$, $C_1$, $X$ and $B_1$ lie on a circle. Since $A_1D$ and $A_1G$ are the bisectors of the angles $\\angle AA_1C$ and $\\angle AA_1B$ respectively, $\\angle GA_1D = 90^\\circ$. Since $\\angle FB_1D = \\angle FA_1D = 90^\\... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0j8m | Let $n \ge 1$ be an integer, and let $S$ be a set of integer pairs $(a, b)$ with $1 \le a < b \le 2^n$. Assume $|S| > n \cdot 2^{n+1}$. Prove that there exist four integers $a < b < c < d$ such that $S$ contains all three pairs $(a, c)$, $(b, d)$ and $(a, d)$. | [
"Let $p$ and $q$ be integers with $1 \\le p \\le 2^n$ and $0 \\le q \\le n-1$. We say that a pair $(a, b)$ has type $A(p, q)$ if $a = p$ and $2^q \\le b - a < 2^{q+1}$, and we say that it has type $B(p, q)$ if $b = p$ and $2^q \\le b - a < 2^{q+1}$. Because there are $n2^{n+1}$ total types and $|S| > n2^{n+1}$, we ... | United States | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
098l | Problem:
Fie funcția $f: \mathbf{R} \rightarrow (0,+\infty)$ derivabilă, strict descrescătoare, cu $f(0)=1$ și $f^{\prime}(0)=-1$. Fie șirul $\left(a_{n}\right)_{n=1}^{\infty}$, definit prin: $a_{1}=1$ şi $a_{n+1}=a_{n} f\left(a_{n}\right), \forall n \geq 1$. Arătați că există limita $\lim _{n \rightarrow \infty}\left... | [
"Solution:\n\nDeoarece $a_{1}=1>0$ și $a_{n+1}=a_{n} f\\left(a_{n}\\right), \\forall n \\geq 1$, prin inducție matematică se arată că $a_{n}>0, \\forall n \\geq 1$.\nDin faptul că funcția $f$ este strict descrescătoare, rezultă inegalitățile: $0<a_{n+1}=a_{n} f\\left(a_{n}\\right)<a_{n} f(0)=a_{n}, \\forall n \\geq... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 1 | |
002r | Sea $p > 3$ un número primo. Si
$$
\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{(p-1)^2} = \frac{n}{m}
$$
donde el máximo común divisor de $n$ y $m$ es $1$, demuestre que $p^3$ divide a $n$. | [] | Argentina | XX Olimpiada Iberoamericana de Matemáticas | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | Español | proof only | null | |
0iw5 | Problem:
Let $n \geq 3$ be a positive integer. A triangulation of a convex $n$-gon is a set of $n-3$ of its diagonals which do not intersect in the interior of the polygon. Along with the $n$ sides, these diagonals separate the polygon into $n-2$ disjoint triangles. Any triangulation can be viewed as a graph: the verti... | [
"Solution:\nWe will show that all triangulations have chromatic number $3$, by induction on $n$.\n\nAs a base case, if $n=3$, a triangle has chromatic number $3$.\n\nNow, given a triangulation of an $n$-gon for $n>3$, every edge is either a side or a diagonal of the polygon. There are $n$ sides and only $n-3$ diago... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0gq3 | Find all pairs $(m, n)$ of positive integers satisfying $m^6 = n^{n+1} + n - 1$. | [
"If $n = 1$, the only solution is $(m, n) = (1, 1)$. Let $n > 1$.\n\n$n$ is not odd, otherwise since $2n^{\\frac{n+1}{2}} + 1 > n^{\\frac{n+1}{2}} > n > n-1 > 0$ we get\n$$\n(n^{\\frac{n+1}{2}})^2 < (m^3)^2 = n^{n+1} + n - 1 < (n^{\\frac{n+1}{2}} + 1)^2\n$$\n\nand $(m^3)^2$ is strictly between two consecutive squar... | Turkey | Team Selection Test for IMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof and answer | (1, 1) | |
07p9 | Suppose $ABCD$ is a simple quadrilateral, with side lengths
$$
a = |BC|,\ b = |CD|,\ c = |DA|,\ d = |AB|.
$$
Let $p = |AC|$. Prove that
$$
\max(|d-a|,\ |b-c|) < p < \min(d+a,\ b+c).
$$
Conversely, if five positive numbers $a, b, c, d, p$ satisfy this condition, prove that $a, b, c, d$ are the side lengths of a simple q... | [
"Suppose $ABCD$ is a quadrilateral. The numbers $d, a, p$ are then the side lengths of the triangle $ABC$. Hence\n$$\n||AB| - |BC|| < |AC| < |AB| + |BC| \\iff |d - a| < p < d + a.\n$$\nSimilarly, $b, c, p$ are the side lengths of the triangle $CDA$, and so $|b - c| < p < b + c$, whence $p < \\min(d + a, b + c)$ and... | Ireland | Irska 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
00a5 | Let $N$ denote the number of ordered 9-tuples $(x_1, x_2, \dots, x_9)$ of positive integers such that
$$
\frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_9} = 1.
$$
Decide if $N$ is even or odd. Justify your answer. | [
"There are an even number of solutions $(x_1, x_2, \\dots, x_9)$ in which $x_1 \\neq x_2$. Indeed they can be divided into pairs such that the two solutions in a pair are obtained from one another by swapping $x_1$ and $x_2$. Hence, so far as the parity of $N$ is concerned, one may assume $x_1 = x_2$. Likewise ther... | Argentina | Argentine National Olympiad 2015 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | even | |
081x | Problem:
Un sottoinsieme $A$ dei numeri naturali compresi fra 1 e 100 è tale che la somma di due suoi elementi qualsiasi è divisibile per 6. Quanti elementi può avere, al massimo, il sottoinsieme $A$?
(A) 11
(B) 16
(C) 17
(D) 25
(E) 33. | [
"Solution:\n\nLa risposta è (C). I numeri debbono essere o tutti pari o tutti dispari, poiché le somme a 2 a 2 sono tutte pari.\n\nAnalogamente, i numeri debbono essere tutti divisibili per 3, perché se in $A$ ci fosse un numero non divisibile per 3 (con resto $r$, diciamo), allora ce ne potrebbe essere solo un alt... | Italy | Progetto Olimpiadi di Matematica | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | MCQ | C | |
0da3 | 1. For integer $n \geq 3$, suppose that $0 < a_{1} < a_{2} < \ldots < a_{n}$ is an arithmetic sequence and $0 < b_{1} < b_{2} < \ldots < b_{n}$ is a geometric sequence with $a_{1} = b_{1}$, $a_{n} = b_{n}$. Prove that $a_{k} > b_{k}$ for all $k = 2, 3, \ldots, n-1$.
2. Prove that for every positive integer $n \geq 3$,... | [
"1) Put $a_{1} = b_{1} = x$ and $a_{n} = b_{n} = y$, then $0 < x < y$. Denote $d$ as the difference and the ratio of two consecutive terms of $(a_{n})$, $(b_{n})$ respectively.\nWe have $a_{n} = a_{1} + (n-1)d$ and $a_{k} = a_{1} + (k-1)d$ for any $k \\in \\{2, 3, \\ldots, n-1\\}$. Hence,\n$$\na_{k} = \\frac{(n-k)a... | Saudi Arabia | Team selection tests for IMO 2018 | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
0j88 | Problem:
In a one-player game, the player begins with $4m$ fair coins. On each of $m$ turns, the player takes 4 unused coins, flips 3 of them randomly to heads or tails, and then selects whether the 4th one is heads or tails (these four coins are then considered used). After $m$ turns, when the sides of all $4m$ coins... | [
"Solution:\n\nLet $q_{n}(k)$ be the probability that, with $n$ turns left and $|H-T|=k$, the player wins playing the optimal strategy. Note that $k$ is always even. We prove by induction on $n$ that with $n$ turns left in the game, this is the optimal strategy, and that $q_{n}(k) \\geq q_{n}(k+2)$.\n\nBase Case: $n... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
05m4 | Problem:
Soit $ABCD$ un quadrilatère convexe (c'est-à-dire que ses diagonales sont à l'intérieur de $ABCD$), et $P$ l'intersection de ses diagonales $[AC]$ et $[BD]$. On note $O_1, O_2, O_3$ et $O_4$ les centres des cercles circonscrits à $ABP$, $BCP$, $CDP$ et $DAP$.
Montrer que $O_1O_2O_3O_4$ est un parallélogramme.
... | [
"Solution:\n$O_1$ et $O_2$ sont sur la médiatrice de $[PB]$, donc $O_1O_2$ est la médiatrice de $[PB]$. De même, $(O_3O_4)$ est la médiatrice de $[PD]$, donc $(O_1O_2)$ et $(O_3O_4)$ sont toutes deux perpendiculaires à $(BD)$, donc elles sont parallèles. De même, $(O_2O_3)$ et $(O_4O_1)$ sont toutes deux perpendicu... | France | French Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0gre | Each square of an $n \times n$ chessboard either contains a rook or is empty. Suppose that for any two rooks not threatening each other there is an empty square which is threatened by each of these rooks. Find the maximal possible number of rooks on the chessboard.
*Note: A rook is threatening a square if they lie on ... | [
"Answer: $\\lfloor 3n/2 \\rfloor$.\n\nEach column contains at most two rooks, otherwise the uppermost and lowermost rooks on this column do not satisfy the conditions. Let us consider two columns each containing two rooks. Let $a_1, a_2$ and $b_1, b_2$ be rooks on the first and second columns, respectively ($a_1$ i... | Turkey | 23rd Junior Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | ⌊3n/2⌋ | |
0kpk | Problem:
A trapezoid has height $12$ and diagonals of length $13$ and $15$. What is its area? | [
"Solution:\n\nPlace trapezoid $ABCD$ in the coordinate plane so that $A = (a, 12)$, $B = (b, 12)$. If $AC = 13$, by the Pythagorean theorem we then have $C = (a + 5, 0)$, $D = (b - 9, 0)$. Thus, the average length of the bases is\n$$\n\\frac{(b - a) + ((a + 5) - (b - 9))}{2} = 7,\n$$\nso the area of $ABCD$ is $7 \\... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 84 | |
07cm | Let $k > 1$ be an integer. The sequence $\{a_i\}_{i=1}^\infty$ is defined as $a_1 = 1$, $a_2 = k$, and for all $n > 1$ we have
$$
a_{n+1} - (k+1)a_n + a_{n-1} = 0.
$$
Find all positive integers $n$ such that $a_n$ is a power of $k$. | [
"We are going to find the general term of the sequence. Note that since $k > 1$, the characteristic polynomial $P(x) = x^2 - (k+1)x + 1$ has two different roots\n$$\nr_1 = \\frac{k+1+\\sqrt{(k-1)(k+3)}}{2}, \\quad r_2 = \\frac{k+1-\\sqrt{(k-1)(k+3)}}{2}.\n$$\nSo $a_n$ has a general form as\n$$\na_n = c_1 r_1^n + c_... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | n = 1 and n = 2 | |
00bo | Let $n$ be a positive integer. Prove that $n(2^n - 1)$ can be expressed as a sum of $n$ distinct powers of $2$. | [
"First, we observe that $n < 2^n$ for every positive integer $n$ (it can be proved easily by induction).\nConsider a positive integer $n \\ge 2$ (for $n = 1$ we have $1 \\cdot (2^1 - 1) = 2^0$). Let $k$ be the number of ones in the binary representation of $n-1$. Then,\n$$\nn-1 = 2^{m_1} + \\dots + 2^{m_k},\n$$\nwh... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof only | null | |
0b04 | Problem:
A triangle has sides of lengths $20$ and $19$. If the triangle is not acute, how many possible integer lengths can the third side have? | [
"Solution:\nWe begin by determining the possible integer lengths of the third side $s$. By the triangle inequality, $2 \\leq s \\leq 38$.\n\nNow, we determine when the triangle is not acute. This means that the triangle is either a right triangle or an obtuse triangle. Since we are only considering integer lengths,... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof and answer | 16 | |
0df1 | You are given $n$ different primes $p_1, p_2, \dots, p_n$. Consider the polynomial
$$
x^n + a_1x^{n-1} + a_2x^{n-2} + \dots + a_{n-1}x + a_n,
$$
where $a_i$ is the product of the first $i$ given prime numbers. For what $n$ can it have an integer root? | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | n = 1 | |
06t8 | For every real number $x$, let $\|x\|$ denote the distance between $x$ and the nearest integer. Prove that for every pair $(a, b)$ of positive integers there exist an odd prime $p$ and a positive integer $k$ satisfying
$$
\left\|\frac{a}{p^{k}}\right\|+\left\|\frac{b}{p^{k}}\right\|+\left\|\frac{a+b}{p^{k}}\right\|=1 .... | [
"Notice first that $\\left\\lfloor x+\\frac{1}{2}\\right\\rfloor$ is an integer nearest to $x$, so $\\|x\\|=\\left|\\left\\lfloor x+\\frac{1}{2}\\right\\rfloor-x\\right|$. Thus we have\n$$\n\\left\\lfloor x+\\frac{1}{2}\\right\\rfloor=x \\pm\\|x\\| .\n$$\nFor every rational number $r$ and every prime number $p$, de... | IMO | 55th International Mathematical Olympiad Shortlist | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0c7p | Prove that, if positive real numbers $a_1, a_2, \dots, a_n$ have the product $1$, then
$$
\left(\frac{a_1}{a_2}\right)^{n-1} + \left(\frac{a_2}{a_3}\right)^{n-1} + \dots + \left(\frac{a_{n-1}}{a_n}\right)^{n-1} + \left(\frac{a_n}{a_1}\right)^{n-1} \geq a_1^2 + a_2^2 + \dots + a_n^2.
$$ | [
"We prove that, for all $a_1, a_2, \\dots, a_n > 0$, the following inequality holds\n$$\n\\left(\\frac{a_1}{a_2}\\right)^{n-1} + \\left(\\frac{a_2}{a_3}\\right)^{n-1} + \\dots + \\left(\\frac{a_{n-1}}{a_n}\\right)^{n-1} + \\left(\\frac{a_n}{a_1}\\right)^{n-1} \\geq \\frac{a_1^2 + a_2^2 + \\dots + a_n^2}{\\sqrt[n]{a... | Romania | Stars of Mathematics Competition | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
01ie | Let $n \ge 2$ be an integer. Let $a_1, a_2, \dots, a_n \in \{1, 2, 3, \dots, 2n\}$ be such that $\text{lcm}(a_i, a_j) > 2n$ for any $1 \le i < j \le 2n$. Prove that
$$
a_1 a_2 \dots a_n \mid (n+1)(n+2) \dots (2n-1)(2n).
$$ | [
"For every $i = 1, 2, \\dots, n$ let $a_i = b_i \\cdot 2^{c_i}$ where $b_i$ is odd.\nNote that the $b_i$'s are pairwise distinct. Indeed, if $b_i = b_j$ for some $i < j$ then one of the numbers $a_i, a_j$ divides the other one, so $\\text{lcm}(a_i, a_j) = \\max(a_i, a_j) \\le 2n$ which is a contradiction.\nAlso, it... | Baltic Way | Baltic Way 2021 Shortlist | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
08d6 | Problem:
Jacopo ha a disposizione 6 colori (tra cui il bianco) per colorare tutti i numeri interi. Vuole rispettare però queste condizioni: $n$ e $n+5$ devono avere lo stesso colore per ogni $n$ intero e inoltre se $ab$ è bianco, allora almeno uno tra $a$ e $b$ deve essere bianco. In quanti modi Jacopo può colorare gl... | [
"Solution:\n\nLa risposta è $\\mathbf{(E)}$. La condizione che $n$ e $n+5$ hanno lo stesso colore è equivalente a restringere il problema alle 5 classi di congruenza modulo 5 (che continueremo a chiamare $0,1,2,3,4$).\n\nStudiamo ora la seconda condizione; scriviamo tutti i possibili modi per ottenere un numero com... | Italy | Gara di Febbraio | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | MCQ | E | |
0lcn | Let $P(x)$ be a polynomial with real coefficients satisfying the condition: there are infinitely many pairs of integers $m, n$ for which $P(m) + P(n) = 0$. Prove that the graph of function $y = P(x)$ has a center of symmetry. | [
"If $\\deg P(x) = 0$ then $P(x) = 0$, and the statement of the problem is obviously true. Consider the case $\\deg P(x) > 0$. Without loss of generality, we can assume that the leading coefficient of $P(x)$ equals $1$. If the degree of $P(x)$ is even then for $x$ with absolute value big enough, we will have $P(x) >... | Vietnam | Vietnamese Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
04rv | Find all real $p$ such that the inequality
$$
\sqrt{a^2 + p b^2} + \sqrt{b^2 + p a^2} \ge a + b + (p-1)\sqrt{ab}
$$
holds for any real $a$ and $b$. | [
"If $a = b = 1$ the parameter $p > 0$ has to satisfy:\n$$\n\\begin{cases} 2\\sqrt{p+1} \\ge p+1, \\\\ 2 \\ge \\sqrt{p+1}, \\\\ p \\le 3. \\end{cases}\n$$\nWe show that for $p \\in (0, 3)$ the inequality holds for any real $a$ and $b$. If $p \\in (0, 1)$ the inequality holds trivially:\n$$\n\\sqrt{a^2 + p b^2} > a, ... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof and answer | 0 ≤ p ≤ 3 | |
0ayx | Problem:
The number $\sqrt{49+6 \sqrt{6}+12 \sqrt{14}+4 \sqrt{21}}$ can be expressed as $a \sqrt{2}+b \sqrt{3}+c \sqrt{7}$ for some integers $a, b, c$. Find $a+b+c$. | [
"Solution:\n\nWe seek for integers $a, b, c$ such that $\\sqrt{49+6 \\sqrt{6}+12 \\sqrt{14}+4 \\sqrt{21}}=a \\sqrt{2}+b \\sqrt{3}+c \\sqrt{7}$. The latter equation is equivalent to\n$$\n\\begin{aligned}\n49+6 \\sqrt{6}+12 \\sqrt{14}+4 \\sqrt{21} & =(a \\sqrt{2}+b \\sqrt{3}+c \\sqrt{7})^{2} \\\\\n& =2 a^{2}+3 b^{2}+... | Philippines | 20th Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Other"
] | null | final answer only | 6 | |
04mf | A positive integer $N$ is called interesting if it is divisible by $36$, and if there exists a positive integer $k$ smaller than $10$ such that $1, 2, \dots, k$ are, in some order, the digits of the decimal representation of $N$. Find the smallest interesting positive integer. (Lucija Validžić) | [] | Croatia | Croatia_2018 | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 12345768 | |
0dbe | Let $I$ be the incenter of triangle $A B C$ and $J$ the excenter of the side $B C$. Let $M$ be the midpoint of $C B$ and $N$ the midpoint of arc $B A C$ of circle $(A B C)$. If $T$ is the symmetric of the point $N$ by the point $A$, prove that the quadrilateral $J M I T$ is cyclic. | [
"Suppose that $A I$ cuts $B C$ at $D$ and cuts $(A B C)$ again at $P$. We have known that $B, C, I, J$ lie on circle $(P, \\frac{I J}{2})$.\n\nSuppose that $A N$ meets $B C$ at $E$, then $A E$ is the external angle bisector of $\\angle B A C$, so $(D E, B C) = -1$. We have\n$$\n\\overline{D E} \\cdot \\overline{D M... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Co... | English | proof only | null | |
07v5 | The numbers $1$ to $9$ are arranged in a $3 \times 3$ square. Such a square is *row-balanced* if the sum of the numbers in each row is the same.
How many of the $9!$ possible squares are row-balanced? | [
"The target row sum is $9 \\cdot 10/2 \\cdot 3 = 15$. We first note that the small numbers $1$, $2$, and $3$ must be in separate rows since if two of them were in the same row, the sum would be too small: we would get at most $2 + 3 + 9 = 14$. In a similar fashion, the large numbers $7$, $8$, and $9$ must be in sep... | Ireland | IRL_ABooklet | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2592 | |
07jj | Prove that for every positive integer $n$, one can find three pairwise coprime positive integers $a, b, c > n$ such that the set of prime divisors of $a + b + c$ and $ab + ac + bc$ are the same. | [
"We seek for large enough positive integers $a, b, c$ such that $\\frac{ab+ac+bc}{a+b+c} = 3^N$ for some positive integer $N$. Indeed, one can write $\\frac{ab+ac+bc}{a+b+c} = a+b-\\frac{a^2+ab+b^2}{a+b+c}$. Now, choose $c$ such that $\\frac{a^2+ab+b^2}{a+b+c} = 1$ that is, $c = a^2+b^2+ab-a-b$ then we need to find... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
083z | Problem:
Una lumaca si arrampica su una colonna cilindrica alta $8$ metri, la cui circonferenza di base è lunga $3$ metri. Sapendo che partendo dalla base raggiunge la cima facendo due giri intorno alla colonna e che, arrivata in cima, si trova esattamente sopra il punto da cui era partita, quanto è lunga la strada pi... | [] | Italy | Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO BIENNIO | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | B | |
0b7o | In the Cartesian plane are given $51$ points of integer coordinates, with pairwise integer distances. Prove that more than $49\%$ of the distances are even integers. | [
"Among the given $51$ points there cannot exist simultaneously a point with odd coordinates and a point with even coordinates. Indeed, that be the case, let $(2a, 2b)$ and $(2c+1, 2d+1)$ be such points. The square of the distance between them is $(2a - 2c - 1)^2 + (2b - 2d - 1)^2 = M4 + 2$, which is not the square ... | Romania | NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
04y4 | Let $ABC$ be an acute-angled triangle with $AB < AC$. Tangent to its circumcircle $\Omega$ at $A$ intersects the line $BC$ at $D$. Let $G$ be the centroid of $ABC$ and let $AG$ meet $\Omega$ again at $H \neq A$. Suppose the line $DG$ intersects the lines $AB$ and $AC$ at $E$ and $F$, respectively. Prove that $\angle EH... | [
"Let $p$ be the line parallel to $BC$ that passes through $A$ and let $P = p \\cap DG$. We denote the midpoint of $BC$ by $T$.\nWe project the harmonic ratio $(BTC_\\infty) = -1$ from $A$ onto the line $DG$ and learn that $(EGFP) = -1$. Therefore by a well-known Apollonian property of harmonic ratios it suffices to... | Czech-Polish-Slovak Mathematical Match | Czech-Polish-Slovak Match | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, E... | English | proof only | null | |
0avh | Problem:
Find all triples of positive real numbers $(x, y, z)$ which satisfy the system
$$
\begin{cases}
\sqrt[3]{x} - \sqrt[3]{y} - \sqrt[3]{z} = 64 \\
\sqrt[4]{x} - \sqrt[4]{y} - \sqrt[4]{z} = 32 \\
\sqrt[6]{x} - \sqrt[6]{y} - \sqrt[6]{z} = 8
\end{cases}
$$ | [
"Solution:\nUsing the first and the third equations, we find that\n$$\n(8 + \\sqrt[6]{y} + \\sqrt[6]{z})^2 = 64 + \\sqrt[3]{y} + \\sqrt[3]{z}\n$$\nSimplifying, we obtain the following equation\n$$\n8 \\sqrt[6]{y} + 8 \\sqrt[6]{z} + \\sqrt[6]{y z} = 0\n$$\nwhich has no solution."
] | Philippines | 18th PMO National Stage Oral Phase | [
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | no solution | |
0000 | Determinar todos los pares $(a, b)$ de enteros positivos para los cuales
$$
\frac{a^2 b + b}{ab^2 + 9}
$$
es un número entero. | [] | Argentina | XI Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | español | proof and answer | All pairs are (a, b) = (9 t^2, 9 t) for any positive integer t, together with the exceptional pairs (32, 1), (73, 1), and (22, 2). | |
019c | Find the smallest value of $k$ with the following property: Among any $k$ consecutive positive integers there exists a number $n$ such that the sum of all positive divisors of $n$ is even. | [
"We claim that $k = 3$. Obviously $k = 2$ is not enough: the numbers $1$, $2$ or $8$, $9$ have odd divisor sums.\n\nNow suppose that the sum $\\sigma(n)$ of all positive divisors of $n$ is odd. Consider a decomposition $n = 2^r \\cdot m$, where $r \\ge 0$ and $m$ is an odd positive integer. Then the set of odd divi... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 3 | |
0105 | Problem:
If we add $1996$ and $1997$, we first add the unit digits $6$ and $7$. Obtaining $13$, we write down $3$ and "carry" $1$ to the next column. Thus we make a carry. Continuing, we see that we are to make three carries in total:
$$
\begin{array}{r}
111 \\
1996 \\
+1997 \\\hline
3993
\end{array}
$$
Does there exi... | [
"Solution:\n\nAnswer: yes.\nThe key to the proof is noting that if we add two positive integers and the result is an integer consisting only of digits $9$ then the process of addition must have gone without any carries. Therefore it is enough to prove that there exists an integer $k$ such that $3993 k$ is of the fo... | Baltic Way | Baltic Way 1997 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | yes | |
053n | There are $125$ distinct positive integers in a row in such a way that among every three consecutive numbers the second one is larger than the arithmetic mean of the first and the third one. Find the largest number in the row, given that it is as small as possible under such conditions. | [
"Let the numbers in a row be $a_1, a_2, \\dots, a_{125}$. By conditions, we have $a_{i+1} > \\frac{a_i + a_{i+2}}{2}$ for every $i = 1, 2, \\dots, 123$, which is equivalent to $a_{i+1} - a_i > a_{i+2} - a_{i+1}$. Denoting $d_i = a_{i+1} - a_i$, we have $d_1 > d_2 > \\dots > d_{124}$.\n\nLet $a_m$ be the largest amo... | Estonia | Estonian Math Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 2016 | |
07cf | Let $P$ and $P'$ be two unequal regular $n$-gons and $A$ and $A'$ two points inside $P$ and $P'$, respectively. Suppose $\{d_1, d_2, \dots, d_n\}$ are the distances from $A$ to the vertices of $P$ and $\{d'_1, d'_2, \dots, d'_n\}$ are the distances from $A'$ to the vertices of $P'$. Is it possible for $\{d'_1, d'_2, \d... | [
"Suppose that $P = P_1P_2 \\dots P_n$, $P' = P'_1P'_2 \\dots P'_n$ and that $AP_i$'s are a permutation of $A'P'_i$'s. Let $O$ and $O'$ be the centers of $P$ and $P'$, respectively. Without loss of generality it can be assumed that $A$ is inside or on the perimeter of triangle $OP_1P_2$ and $AP_1 \\le AP_2$; $A'$ is... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof only | null | |
01em | $AD$ is a bisector of the triangle $ABC$. Line $AD$ intersects a second time the circumcircle of $\triangle ABC$ at point $E$. Let $K, L, M$ and $N$ be the midpoints of the segments $AB, BD, CD$ and $AC$ respectively, $P$ be the circumcenter of the triangle $EKL$, $Q$ be the circumcenter of the triangle $EMN$. Prove th... | [
"Triangles $AEB$ and $BED$ are similar since $\\angle BAE = \\angle EAC = \\angle DBE$. Hence $\\angle AEK = \\angle BEL$ as the angles between a median and a side in similar triangles. Denote these angles by $\\varphi$. Then $\\angle EKL = \\varphi$ since $KL$ is a midline of $\\triangle ABD$.\n\nAnalogously, let ... | Baltic Way | Baltic Way shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06jl | Find the largest possible positive integer $n$, so that there exist $n$ distinct positive real numbers $x_1, x_2, \dots, x_n$ satisfying the following inequality: for any $1 \le i, j \le n$,
$$
(3x_i - x_j)(x_i - 3x_j) \ge (1 - x_ix_j)^2.
$$ | [
"The largest possible $n$ is $3$.\nNote that\n$$\n\\begin{aligned}\n& (3x_i - x_j)(x_i - 3x_j) \\ge (1 - x_ix_j)^2 \\\\\n\\Leftrightarrow & 3x_i^2 - 10x_ix_j + 3x_j^2 \\ge 1 - 2x_ix_j + x_i^2x_j^2 \\\\\n\\Leftrightarrow & 3x_i^2 - 6x_ix_j + 3x_j^2 \\ge 1 + 2x_ix_j + x_i^2x_j^2 \\\\\n\\Leftrightarrow & 3(x_i - x_j)^... | Hong Kong | Year 2016 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 3 | |
00hh | Let $ABC$ be a scalene triangle with circumcircle $\Gamma$. Let $M$ be the midpoint of $BC$. A variable point $P$ is selected in the line segment $AM$. The circumcircles of triangles $BPM$ and $CPM$ intersect $\Gamma$ again at points $D$ and $E$, respectively. The lines $DP$ and $EP$ intersect (a second time) the circu... | [
"Let $N$ be the radical center of the circumcircles of triangles $ABC$, $BMP$ and $CMP$. The pairwise radical axes of these circles are $BD$, $CE$ and $PM$, and hence they concur at $N$. Now, note that in directed angles:\n$$\n\\angle MCE = \\angle MPE = \\angle MPY = \\angle MBY.\n$$\n\nIt... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
02kq | Problem:
Frações inteiras - Quantos números inteiros positivos $n$ existem tais que $\frac{2 n^{2}+4 n+18}{3 n+3}$ é um inteiro? | [
"Solution:\n\nComo\n$$\n\\frac{2 n^{2}+4 n+18}{3 n+3} = \\frac{2}{3} \\frac{\\left(n^{2}+2 n+1\\right)+8}{n+1} = \\frac{1}{3}\\left(2 n+2+\\frac{16}{n+1}\\right)\n$$\nsegue que $n+1$ tem que dividir $16$. Assim, $n$ tem que pertencer ao conjunto $\\{1,3,7,15\\}$. Em cada um destes casos temos\n\n| $n$ | $\\frac{2 n... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 4 | |
0926 | Problem:
The numbers from $1$ to $2013^{2}$ are written row by row into a table consisting of $2013 \times 2013$ cells. Afterwards, all columns and all rows containing at least one of the perfect squares $1, 4, 9, \ldots, 2013^{2}$ are simultaneously deleted.
How many cells remain? | [
"Solution:\nLet $m = 503$ and $n = 4m + 1 = 2013$.\nNote that\n$$\n(m-1)n = (m-1)(4m+1) < m \\cdot 4m = (2m)^2 < m(4m+1) = mn\n$$\nso the perfect square $(2m)^2$ is in the $m$-th row.\nNote that $(k+1)^2 - k^2 = 2k + 1$ is at most $n$ if $k \\leq 2m$ and is at least $n$ if $k \\geq 2m$. Therefore, on the one hand, ... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 825423 | |
0gnm | A circle $\Gamma$ and a line $\ell$, which does not intersect the circle, are given in the plane. Determine the intersection of all circles with diameters $AB$, where $\{A, B\}$ is any pair of points on the line $\ell$ for which points $P, Q, R, S$ on the circle $\Gamma$ satisfying $PQ \cap RS = \{A\}$ and $PS \cap QR ... | [
"Let $O$ be the center and $r$ be the radius of the circle $\\Gamma$. If $PQ \\cap RS = \\{A\\}$ and $PS \\cap QR = \\{B\\}$ where the points $P$, $Q$, $R$, $S$ lie on the circle $\\Gamma$; then by Miquel's theorem for the triangle $ABQ$, the circumcircles of the triangles $APS$ and $BRS$ intersect at a point $K$ l... | Turkey | 16th Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | Let O be the center of Γ, r its radius, and K the foot of the perpendicular from O to the line ℓ. Then all the circles described pass through exactly two fixed points: the two points on the line OK whose distance from ℓ equals √(OK² − r²). | |
03x8 | For positive integer $n$, $a_n = n\sqrt{5} - \lfloor n\sqrt{5} \rfloor$. Compute the maximum value and the minimum value of $a_1, a_2, \dots, a_{2009}$. (For real number $x$, $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.) (Posed by Wang Zhixiong) | [
"Let $b_0 = 0$, $b_1 = 1$, $b_n = 4b_{n-2} + b_{n-1}$ ($n \\ge 2$). Then\n$$\nb_n = \\frac{(2+\\sqrt{5})^n - (2-\\sqrt{5})^n}{2\\sqrt{5}}.\n$$\nIn particular, $b_6 = 1292$, $b_7 = 5473$.\nFor every $k = 1, 2, \\dots, 5473$, there are unique integers $x_k, y_k$ such that $1292k = x_k + 5473y_k$, and $1 \\le x_k \\le... | China | China Girls' Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | Maximum value: 1 − (2 − √5)^6 (attained at n = 1292). Minimum value: 1 − 5470(2 − √5)^6 + 1291(2 − √5)^7 (attained at n = 1597). | |
0g34 | Problem:
Déterminer toutes les paires d'entiers strictement positifs $(a, b)$ telles que
$$
a b+2=a^{3}+2 b
$$ | [
"Solution:\nEn passant tous les $b$ à gauche, on obtient\n$$\nb(a-2)=a^{3}-2\n$$\nC'est-à-dire, $a-2$ est un diviseur de $a^{3}-2$ et $b=\\left(a^{3}-2\\right)/(a-2)$. Quand est-ce que $a-2$ divise $a^{3}-2$ ? L'idée est de se débarrasser du $a^{3}$ pour y voir plus clair. On obtient\n$$\n\\begin{aligned}\n& a-2 \\... | Switzerland | Vorrunde 2019 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | (1, 1), (3, 25), (4, 31), (5, 41), (8, 85) | |
0j2f | Problem:
A sphere is the set of points at a fixed positive distance $r$ from its center. Let $\mathcal{S}$ be a set of 2010-dimensional spheres. Suppose that the number of points lying on every element of $\mathcal{S}$ is a finite number $n$. Find the maximum possible value of $n$. | [
"Solution:\nThe answer is $2$ for any number of dimensions. We prove this by induction on the dimension.\n\nNote that 1-dimensional spheres are pairs of points, and 2-dimensional spheres are circles.\n\nBase case, $d=2$:\nThe intersection of two circles is either a circle (if the original circles are identical and ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2 | |
09md | We consider numbers written in digits $0$ and $1$. We say that $M$ is a *subnumber* of $N$ if it is possible to obtain $M$ from $N$ by deleting some digits from the beginning and some digits from the end of $N$.
Fix $M = 10101$ and let $g(N)$ denote the number of times $M$ appears as a subnumber of $N$. For example, $... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | (n - 4) 2^{n - 5} | |
04ei | Prove that among any four numbers from the interval $\langle 0, \frac{\pi}{2} \rangle$ one can choose two numbers, namely $x$ and $y$, such that
$$
8 \cos x \cos y \cos(x - y) + 1 > 4 (\cos^2 x + \cos^2 y).
$$ | [
"The given inequality is successively equivalent to\n$$8 \\cos x \\cos y (\\cos x \\cos y + \\sin x \\sin y) + 1 > 4 \\cos^2 x + 4 \\cos^2 y,$$\n$$8 \\cos^2 x \\cos^2 y - 4 \\cos^2 x - 4 \\cos^2 y + 2 \\sin 2x \\sin 2y + 1 > 0,$$\n$$2 (2 \\cos^2 x - 1) (2 \\cos^2 y - 1) + 2 \\sin 2x \\sin 2y - 1 > 0,$$\n$$\\cos 2x ... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0a8u | Problem:
Let $a$, $b$, and $c$ be the side lengths of a triangle and let $R$ be its circumradius. Show that
$$
\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a} \geq \frac{1}{R^{2}}
$$ | [
"Solution:\nBy the well-known (Euler) theorem, the inradius $r$ and circumradius $R$ of any triangle satisfy $2 r \\leq R$. (In fact, $R(R-2 r)=d^{2}$, where $d$ is the distance between the incenter and circumcenter.) The area $S$ of a triangle can be written as\n$$\nA=\\frac{r}{2}(a+b+c)\n$$\nand, by the sine theo... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 18 | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0l8i | Alice the architect and Bob the builder play a game. First, Alice chooses two points $P$ and $Q$ in the plane and a subset $S$ of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each othe... | [] | United States | USAMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | Bob | |
09an | Three problems $A$, $B$ and $C$ were given in a mathematical olympiad and there were $25$ participants. Every participant solved at least $1$ problem. Among the participants who didn't solve problem $A$, the number of participants who solved $B$ is twice as much as the number of participants who solved problem $C$. The... | [
"Let $A$, $B$ and $C$ be the numbers of participants who solved just one problem $A$, $B$, $C$ respectively. And let $AB$ be the number of participants who solved only $A$ and $B$ and etc. Then\n$$\n\\left\\{ \n\\begin{array}{l}\nA + B + C + AB + AC + BC + ABC = 25 \\quad (1) \\\\\nB + BC = 2(C + BC) \\quad (2) \\\... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 6 | |
065j | We consider the numbers:
$$
A = \frac{1}{4} \cdot \frac{3}{6} \cdot \frac{5}{8} \cdots \frac{595}{598} \cdot \frac{597}{600} \text{ and } B = \frac{2}{5} \cdot \frac{4}{7} \cdot \frac{6}{9} \cdots \frac{596}{599} \cdot \frac{598}{601}.
$$
Prove that: (α) $A < B$, (β) $A < \frac{1}{5990}$. | [
"(α) To each fraction of $A$ of the form $\\frac{2\\nu - 1}{2\\nu + 2}$, $\\nu = 1, 2, \\dots, 299$, corresponds a fraction from $B$ of the form $\\frac{2\\nu}{2\\nu + 3}$, $\\nu = 1, 2, \\dots, 299$. Since\n$$\n0 < \\frac{2\\nu - 1}{2\\nu + 2} < \\frac{2\\nu}{2\\nu + 3} \\quad \\text{for every } \\nu \\in \\mathbb... | Greece | 26th Hellenic Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof only | null | |
0116 | Problem:
Let $K$ be a point inside the triangle $A B C$. Let $M$ and $N$ be points such that $M$ and $K$ are on opposite sides of the line $A B$, and $N$ and $K$ are on opposite sides of the line $B C$. Assume that
$$
\angle M A B=\angle M B A=\angle N B C=\angle N C B=\angle K A C=\angle K C A .
$$
Show that $M B N K... | [
"Solution:\n\nDenote $\\angle M A B=\\angle M B A=\\cdots=\\alpha$. Then\n$$\n\\angle M A K=\\alpha+(\\angle B A C-\\alpha)=\\angle B A C,\n$$\n$\\frac{|A M|}{|A B|}=\\frac{1}{2 \\cos \\alpha}$ and $\\frac{|A K|}{|A C|}=\\frac{1}{2 \\cos \\alpha}$ (see Figure 1). Hence the triangles $M A K$ and $B A C$ are similar,... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kw3 | Problem:
Let
$$
P=\prod_{i=0}^{2016}\left(i^{3}-i-1\right)^{2}
$$
The remainder when $P$ is divided by the prime $2017$ is not zero. Compute this remainder. | [
"Solution:\nLet $Q(x)=x^{3}-x-1=(x-a)(x-b)(x-c)$, for $a, b, c \\in \\mathbb{F}_{p^{3}}$. Then, we can write\n$$\nP=\\prod_{i=0}^{2016}(i-a)(i-b)(i-c)\n$$\nIf we consider each root separately, then\n$$\nP=-\\left(a^{2017}-a\\right)\\left(b^{2017}-b\\right)\\left(c^{2017}-c\\right)\n$$\nThe key observation is that $... | United States | HMMT February 2023 | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Abstract Algebra > Field Theory"
] | null | final answer only | 1994 | |
0axb | Problem:
In a tournament, $2017$ teams participate. To save time, in each round, instead of a one-on-one battle, three teams (randomly chosen) participate; the game for each round is designed so that exactly one of the three teams survives in the tournament (unless of course it happens that only two teams are remainin... | [
"Solution:\n\nAfter each battle, $2$ teams get eliminated. After $1007$ battles, $2017-2(1007)=3$ teams remain, who will fight it out for one more battle.\n\nAlternatively, there are a total of\n$$\n\\left\\lfloor\\frac{2017}{3}\\right\\rfloor+\\left\\lfloor\\frac{673}{3}\\right\\rfloor+\\left\\lfloor\\frac{225}{3}... | Philippines | Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 1008 | |
07p3 | Suppose that $\cos(m^\circ) = \sqrt{2} - 1$. Prove that $m$ is not an integer. | [
"Let $x = \\sqrt{2}-1$, $t = m^\\circ$. Then $x$ is irrational and $x^2 = 3-2\\sqrt{2} = 1-2x$. We would like to establish that all the numbers $\\cos(2nt)$ are distinct for $n \\ge 1$. This will follow from the following claim.\n\nClaim: The sequences of integers $(a_n)$ and $(b_n)$ given recursively by\n$$\n\\beg... | Ireland | Irska 2014 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0cqw | Дана функция $f$, определённая на множестве действительных чисел и принимающая действительные значения. Известно, что для любых $x$ и $y$ таких, что $x > y$, верно неравенство $(f(x))^2 \le f(y)$. Докажите, что множество значений функции содержится в промежутке $[0, 1]$.
(А. Храбров) | [
"По условию $f(y) \\ge (f(y+1))^2 \\ge 0$ для любого $y$, поэтому все значения функции неотрицательны.\n\nПусть теперь $f(x_0) = 1 + a > 1$ для некоторого $x_0$. Докажем индукцией по $n$, что для любого $y < x_0$ верно неравенство $f(y) > 1 + 2^n a$.\n\nПри $n = 1$ имеем $f(y) \\ge (f(x_0))^2 = 1 + 2a + a^2 > 1 + 2... | Russia | XL Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0k09 | Problem:
Consider all ordered pairs of integers $(a, b)$ such that $1 \leq a \leq b \leq 100$ and
$$
\frac{(a+b)(a+b+1)}{a b}
$$
is an integer.
Among these pairs, find the one with largest value of $b$. If multiple pairs have this maximal value of $b$, choose the one with largest $a$. For example choose $(3,85)$ over ... | [
"Solution:\n\nFirstly note that $\\frac{(a+b)(a+b+1)}{a b}=2+\\frac{a^{2}+b^{2}+a+b}{a b}$. Let $c$ be this fraction so that $(a+b)(a+b+1)=a b(c+2)$ for some integers $a, b, c$. Suppose $(a, b)$ with $a \\geq b$ is a solution for some $c$. Consider the quadratic\n$$\nx^{2}-(b c-1) x+b^{2}+b=0\n$$\nIt has one root $... | United States | February 2017 | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (35, 90) | |
0iv4 | Problem:
Prove that the sum of the $2009$th powers of the first $2009$ positive integers is divisible by $2009$. | [
"Solution:\nUsing the factoring formula\n$$\nx^{2009} + y^{2009} = (x + y)\\left(x^{2008} - x^{2007}y + x^{2006}y^{2} - \\cdots - x y^{2007} + y^{2008}\\right),\n$$\nwe find that each of the numbers\n$$\n0^{2009} + 2009^{2009},\\ 1^{2009} + 2008^{2009},\\ 2^{2009} + 2007^{2009},\\ \\ldots,\\ 1004^{2009} + 1005^{200... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
026v | Problem:
Numa misteriosa ilha havia 13 camaleões cinza, 15 camaleões marrons e 17 camaleões vermelhos. Quando dois camaleões de cores diferentes se encontram, os dois tomam a terceira cor. Por exemplo, se um cinza se encontra com um vermelho, então os dois ficam marrons. Por causa de uma tempestade, ocorreram 2 encontr... | [
"Solution:\n16 cinzas, 18 marrons e 11 vermelhos"
] | Brazil | null | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | final answer only | 16 gray, 18 brown, 11 red | |
059y | Does there exist a positive integer $n$ such that across all representations $n = ab$, the digits $0, 1, \dots, 9$ are all present as the final digit of $a^b$ at least once? | [
"**Answer:** No.\n\nIf $n$ is odd, both $a$ and $b$ will always be odd, but then $a^b$ will never end with a $0$. If $n$ is even, then $a$ or $b$ will be even, so $a^b$ will be even or a perfect square. In either case, it will never end in a $7$."
] | Estonia | Estonian Math Competitions | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | No | |
09pv | Problem:
Laat $\triangle ABC$ een rechthoekige driehoek zijn met $\angle B = 90^\circ$ en $|AB| > |BC|$; en zij $\Gamma$ de halve cirkel met middellijn $AB$ aan de kant van $AB$ waar ook $C$ ligt. Zij punt $P$ op $\Gamma$ zo dat $|BP| = |BC|$ en zij $Q$ op lijnstuk $AB$ zo dat $|AP| = |AQ|$. Bewijs dat het midden van ... | [] | Netherlands | TOETS TRAININGSKAMP | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0522 | Can $2013$ be represented as the difference of two cubes of integers? | [
"*Answer:* No.\n\nSuppose that $2013 = x^3 - y^3$ where $x$ and $y$ are integers. Note that\n$$\nx^3 - y^3 = (x - y)^3 + 3x^2y - 3xy^2 = (x - y)^3 + 3xy(x - y).\n$$\nAs $2013$ is divisible by $3$ and so is $3xy(x - y)$, the difference $(x - y)^3$ must be divisible by $3$. Thus also $x - y$ is divisible by $3$ as $3... | Estonia | Final Round of National Olympiad | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities... | English | proof and answer | No | |
0foj | Consideramos un número primo $p$. Debemos diseñar un torneo de $p$-parchís sujeto a las siguientes reglas.
* En el torneo participan $p^2$ jugadores.
* En cada partida juegan $p$ jugadores.
* El torneo se divide en rondas. Las rondas se dividen en partidas. Cada jugador juega una, o ninguna, partida en cada ronda.
* A... | [
"El número de partidas que disputa cada jugador es\n$$\n\\frac{\\text{número de jugadores a los que se enfrenta}}{\\text{número de jugadores a los que se enfrenta en cada partida}} = \\frac{p^2 - 1}{p - 1}.\n$$\nO sea, cada jugador juega $p+1$ partidas. Por tanto el número de rondas es, al menos, $p+1$, y es exacta... | Spain | L Olimpiada Matemática Española | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Other"
] | Spanish | proof and answer | Yes, it is possible; the minimum number of rounds is p+1. | |
0f3u | Problem:
A six digit (decimal) number has six different digits, none of them $0$, and is divisible by $37$. Show that you can obtain at least $23$ other numbers which are divisible by $37$ by permuting the digits. | [
"Solution:\n\nSuppose the digits are $a_1, a_2, a_3, a_4, a_5, a_6$ and that $a_1a_2a_3a_4a_5a_6$ is divisible by $37$. We claim that $a_2a_3a_4a_5a_6a_1$ is also divisible by $37$. Put $n = a_2a_3a_4a_5a_6$ and $m = a_1$. The original number is $10^{5}m + n$ and the derived number is $10n + m$. But $37$ divides $1... | Soviet Union | ASU | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0f34 | Problem:
A convex quadrilateral is divided by its diagonals into four triangles. The incircles of each of the four are equal. Show that the quadrilateral has all its sides equal. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscel... | null | proof only | null | |
06ua | 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$. | [
"Define $g(x)=x-f(x)$. The condition on $f$ then rewrites as follows:\nFor every $x, y \\in \\mathbb{R}$ such that $((x+y)-g(x))((x+y)-g(y))>0$, we have $g(x)=g(y)$.\nThis condition may in turn be rewritten in the following form:\nIf $g(x) \\neq g(y)$, then the number $x+y$ lies (non-strictly) between $g(x)$ and $g... | IMO | International Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof only | null | |
05wr | Problem:
Un terrible groupe de bandits se prépare à se partager un butin. Chaque bandit vise deux autres bandits avec ses pistolets. Les bandits sont appelés dans un certain ordre. Lorsqu'un bandit est appelé, s'il est encore en vie, il tire sur les bandits qu'il visait. Après que tous les bandits ont été appelés, il ... | [
"Solution:\n\nSupposons par l'absurde qu'il existe un ordre qui cause 9 victimes ou moins. Il y a alors trois catégories de bandits :\n- Ceux qui sont morts avant de tirer\n- Ceux qui ont pu tirer puis sont morts\n- Ceux qui sont encore en vie à la fin\n\nLes bandits de 1ère et 2ème catégorie sont au total au plus ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 10 | |
0ase | Problem:
What is the probability that a randomly chosen positive divisor of $2010$ has two digits? | [
"$\\frac{1}{4}$"
] | Philippines | Philippines Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | final answer only | 1/4 | |
0b1z | Problem:
Consider an equilateral triangle with side $700$. Suppose that one move consists of changing the length of any of the sides of a triangle such that the result will still be a triangle. Find the minimum number of moves to change the given triangle to an equilateral triangle with side $2$. | [
"Solution:\n\nWork backwards by starting from $(2,2,2)$. The fastest way to do this is to lengthen the shortest side, making it as long as the sum of the other two sides. Denote $k^{-}$ to be a real number arbitrarily close to $k$, but less than $k$.\n\nFirst move: $(2,2,4^{-})$\n\nSecond move: $(2,4^{-}, 6^{-})$\n... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 14 | |
07ei | Point $P$ lies inside of parallelogram $ABCD$. Perpendicular lines to $PA, PB, PC$ and $PD$ through $A, B, C$ and $D$ construct convex quadrilateral $XYZT$. Prove that the area of $XYZT$ is not less than twice the area of $ABCD$. | [
"At first we shall prove following lemma:\n**Lemma.** If point $A'$ is antipode of vertex $A$ in circumcircle of triangle $ABC$, then\n$$\nS_{ABC} - S_{A'BC} = \\frac{1}{2} |BC|^2 \\cot \\angle A,\n$$\nwhere $S_{XYZ}$ denotes the area of triangle $XYZ$.\n*Proof.* Let $H$ be the orthocenter of triangle $ABC$. Notice... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > ... | English | proof only | null | |
016f | Determine all pairs $(m, n)$ of positive integers satisfying
$$
2^m = 3^n + 5.
$$ | [
"It is easy to find that $(3,1)$ and $(5,3)$ are solutions and that there are no solutions with $m = 1, 2, 4$.\n\nNow let $m \\ge 6$. We will prove that there is no solution.\nAssume that there is a solution $(m, n)$ with $m \\ge 6$. Then $64|2^m$. Hence, $3^n \\equiv 59 \\pmod{64}$. $3^n$ runs through the remainde... | Baltic Way | Baltic Way SHL | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | Pairs (m, n) are (3, 1) and (5, 3). | |
073f | Define $f, g, h$ on $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}$ as follows:
$$
f(x, y, z) = (3x + 2y + 2z, 2x + 2y + z, 2x + y + 2z),
$$
$$
g(x, y, z) = (3x + 2y - 2z, 2x + 2y - z, 2x + y - 2z),
$$
$$
h(x, y, z) = (3x - 2y + 2z, 2x - y + 2z, 2x - 2y + z).
$$
Given a primitive Pythagorean triplet $(x, y, z)$, with ... | [
"Let\n$$\nu = u(x, y, z) = 3x - 2y - 2z,$$\n$$v = v(x, y, z) = -2x + 2y + z,$$\n$$w = w(x, y, z) = -2x + y + 2z.$$\nFirst of all, if $(x, y, z) \\neq (5, 4, 3)$ is a primitive Pythagorean triplet with $x > y > z > 0$, then, it is easy to check that $u^2 = v^2 + w^2$. Further, $x^2 + (2y - 2z)^2 > 0 \\Rightarrow 9x^... | India | Indija TS 2007 | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Algebra > Linear Algebra > Linear transformations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
06f5 | In a school there are $2007$ male and $2007$ female students. Each student joins not more than $100$ clubs in the school. It is known that any two students of opposite genders have joined at least one common club. Show that there is a club with at least $11$ male and $11$ female members. | [
"Suppose on the contrary that no club contains at least $11$ male and $11$ female members. We shall count the number of triples $(M, F, C)$ where $M$ is a male student, $F$ is a female student and $C$ is a club joined by these two students.\n\nFirstly, for each of the $2007$ choices of $M$ and each of the $2007$ ch... | Hong Kong | CHKMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0a85 | Problem:
a. For which positive numbers $n$ does there exist a sequence $x_{1}, x_{2}, \ldots, x_{n}$, which contains each of the numbers $1,2, \ldots, n$ exactly once and for which $x_{1}+x_{2}+\cdots+x_{k}$ is divisible by $k$ for each $k=1,2, \ldots, n$?
b. Does there exist an infinite sequence $x_{1}, x_{2}, x_{3}... | [
"Solution:\n\n**a.**\nWe assume that $x_{1}, \\ldots, x_{n}$ is the sequence required in the problem. Then $x_{1}+x_{2}+\\cdots+x_{n}=\\frac{n(n+1)}{2}$. This sum should be divisible by $n$. If $n$ is odd, this is possible, since $\\frac{(n+1)}{2}$ is an integer. If, on the other hand, $n=2m$, then $\\frac{n(n+1)}{... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 12 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | a: n equals 1 or 3. b: Yes, such an infinite sequence exists. | |
0ako | Let $ABC$ be an acute triangle, $A'$, $B'$ and $C'$ be the reflections of the vertices $A$, $B$ and $C$ with respect to $BC$, $CA$ and $AB$, respectively, and let the circumcircles of triangles $ABB'$ and $ACC'$ meet again at $A_1$. Points $B_1$ and $C_1$ are defined similarly. Prove that the lines $AA_1$, $BB_1$ and $... | [
"Let $O_1$, $O_2$ and $O$ be the circumcenters of triangles $ABB'$, $ACC'$ and $ABC$ respectively. As $AB$ is the perpendicular bisector of the line segment $CC'$, $O_2$ is the intersection of the perpendicular bisector of $AC$ with $AB$. Similarly, $O_1$ is the intersection of the perpendicular bisector of $AB$ wi... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | English | proof only | null | |
0gvr | Prove for every integer $n \ge 2$, that such a set $A_n$ of $n$ pairwise distinct positive integers can be collected that for any different $i, j \in A_n$ the sum of their squares $i^2 + j^2$ is divisible by their difference $i - j$. | [
"$$\nA_n = \\{1 \\cdot n!; 2 \\cdot n!; \\dots; n \\cdot n!\\}\n$$\nsatisfies the condition of the problem. Indeed, take any two distinct numbers from the set $A_n$: $k \\cdot n!$ and $m \\cdot n!$, where $1 \\le k < m \\le n$. Since $m-k \\le n$, it follows that $m-k|n!$. Therefore, obviously, $(k^2 + m^2) \\cdot ... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof only | null | |
0hcw | Each player - Andriy and Olesya - has a set of 2019 cards, which contains numbers $1, 2, \ldots, 2019$ (each number is exactly one time for each player). The game follows the following rules. At the beginning of the game there is a card with a number $k \in \{1, 2, \ldots, 2019\}$ on the table. After that, the players ... | [
"For each time Andriy plays a card $n > k$, Olesya plays a card $n - 1 < n$. Since Andriy has a set of cards $\\{1, 2, \\ldots, k, k, k+1, \\ldots, n-1, n+1, n+2, \\ldots, 2018\\}$, he must put one of the cards that is greater or equal to $n+1$. Olesya plays the card $n$ back. Thus, after each move, Andriy should p... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Olesya | |
071p | Problem:
Let $ABC$ be a triangle in which no angle is $90^{\circ}$. For any point $P$ in the plane of the triangle, let $A_{1}, B_{1}, C_{1}$ denote the reflections of $P$ in the sides $BC, CA, AB$ respectively. Prove the following statements:
a. If $P$ is the incentre or an excentre of $ABC$, then $P$ is the circumce... | [
"Solution:\n\na.\n\nIf $P=I$ is the incentre of triangle $ABC$, and $r$ its inradius, then it is clear that $A_{1}I = B_{1}I = C_{1}I = 2r$. It follows that $I$ is the circumcentre of $A_{1}B_{1}C_{1}$. On the other hand if $P=I_{1}$ is the excentre of $ABC$ opposite $A$ and $r_{1}$ the cor... | India | INMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
0e9y | Problem:
Naj bo $n$ naravno število. Poenostavi izraz
$$
\left(\frac{a+a^{-1} b^{2}}{a-a^{-1} b^{2}}-1\right)^{-1}\left(b^{n}(a-b)^{-1}-b^{n}(a+b)^{-1}\right)
$$
Za katere vrednosti $a$ in $b$ izraz nima pomena? | [] | Slovenia | 14. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | Simplified expression: b^{n-1}. Undefined when a = 0, or b = 0, or a^2 = b^2 (i.e., a = b or a = −b). | |
0c82 | Find the least positive real number $\lambda$ such that, for any real numbers $a_1, a_2, a_3 \in [0, \frac{1}{2}]$ and $b_1, b_2, b_3 \in (0, \infty)$ with $\sum_{i=1}^3 a_i = \sum_{i=1}^3 b_i = 1$, the following inequality holds
$$
b_1b_2b_3 \le \lambda(a_1b_1 + a_2b_2 + a_3b_3).
$$ | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 1/8 | |
0flb | Problem:
Calcula las soluciones reales de la ecuación:
$$
\sqrt[4]{97-X}+\sqrt[4]{X}=5
$$ | [
"Solution:\nSi llamamos $a=\\sqrt[4]{97-x}$ y $b=\\sqrt[4]{x}$, se verifica:\n$$\n\\left.\\begin{array}{l}\na+b=5 \\\\\na^{4}+b^{4}=97\n\\end{array}\\right\\}\n$$\nPara resolver este sistema procedemos como sigue:\n$$\n25=(a+b)^{2}=a^{2}+2 a b+b^{2}\n$$\ny de aquí se tiene $a^{2}+b^{2}=25-2 a b$. Por otro lado\n$$\... | Spain | Spain | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 16, 81 | |
0ed6 | Let $P(x)$ and $Q(x)$ be quadratic polynomials with rational coefficients and assume that the roots of $P(x)$ are irrational numbers $a$ and $b$. If $a + 2016$ is one of the roots of $Q(x)$, find all possible values for the other root. | [
"Let $c$ denote the other root of $Q$. Since $P$ and $Q$ have rational coefficients the Vieta's formulas imply\n$$\n\\begin{aligned}\na+b &= p_1 \\in \\mathbb{Q} \\quad \\text{and} \\quad ab = p_2 \\in \\mathbb{Q}, \\\\\n(a+2016)+c &= q_1 \\in \\mathbb{Q} \\quad \\text{and} \\quad (a+2016)c = q_2 \\in \\mathbb{Q}.\... | Slovenia | Slovenija 2016 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | b + 2016 | |
0d0f | For a positive integer $n$, let $\phi(n)$ be the number of positive integers less than $n$ and relatively prime to $n$ (by convention $\phi(1) = 1$). Let $\tau(n)$ be the number of positive integers that are divisors of $n$. Find all positive integers $n$ such that
$$
\phi(n) + \tau(n) > n.
$$ | [
"The answer is $n = 1$, $n = 4$, or $n$ is any prime number. Let $S$ be the set of integers less than $n$ and relatively prime to $n$, and let $T$ be the set of integers that are positive divisors of $n$. Then\n$$\n(i) \\quad |S| = \\phi(n) \\text{ and } |T| = \\tau(n),\n$$\n$$\n(ii) \\quad S, T \\subseteq \\{1, 2,... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Divisibil... | English | proof and answer | n = 1, n = 4, or n is prime | |
0cwi | On a plane, $10^6$ points are marked no three of which are collinear; and all segments between these points are drawn. On each drawn segment Grisha placed a real number whose absolute value does not exceed $1$. Then for each six marked points, he calculated the sum of the numbers on all $15$ segments connecting these p... | [
"Consider a graph whose vertices are the marked points and edges are the drawn segments.\n\n*Upper bound.* We prove $C \\le \\frac{15}{4}$. The condition states that in our complete graph there exist both $6$-vertex subsets where the sum of edge weights is positive, and $6$-vertex subsets where it's negative. There... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Russian | proof and answer | 15/4 | |
0e7d | Problem:
Poišči vsa praštevila $p, q$ in $r$, za katera velja $p+q^{2}=r^{4}$. | [
"Solution:\n\nEnačbo preoblikujemo v $p = r^{4} - q^{2} = (r^{2} - q)(r^{2} + q)$. Ker je $p$ praštevilo, mora biti $r^{2} - q = 1$ in $r^{2} + q = p$. Prvo enačbo preoblikujemo v $q = r^{2} - 1 = (r - 1)(r + 1)$. Ker je $q$ praštevilo, mora biti $r - 1 = 1$. Torej je $r = 2$ in $q = 3$. Iz enačbe $r^{2} + q = p$ d... | Slovenia | 57. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | p=7, q=3, r=2 | |
0iky | Problem:
A regular 12-sided polygon is inscribed in a circle of radius $1$. How many chords of the circle that join two of the vertices of the 12-gon have lengths whose squares are rational? (No proof is necessary.) | [
"Solution:\n\nThe chords joining vertices subtend minor arcs of $30^{\\circ}, 60^{\\circ}, 90^{\\circ}, 120^{\\circ}, 150^{\\circ}$, or $180^{\\circ}$. There are $12$ chords of each of the first five kinds and $6$ diameters. For a chord with central angle $\\theta$, we can draw radii from the two endpoints of the c... | United States | Harvard-MIT Mathematics Tournament, Team Round B | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | 42 | |
05v2 | Problem:
On dit que deux entiers $a, b \in \mathbb{N}^{\star}$ sont reliés s'il existe un nombre premier $p$ tel que $a = p b$ ou $b = p a$.
Trouver tous les entiers $n \geqslant 1$ ayant la propriété suivante : on peut écrire tous les diviseurs positifs de $n$ (1 et $n$ compris) exactement une fois sur un cercle de ... | [
"Solution:\n\nTraitons d'abord le cas où $n$ n'a qu'un seul facteur premier, donc $n = p^{\\alpha}$. Si $\\alpha \\leqslant 1$, n'importe quelle configuration marche. En revanche, si $\\alpha \\geqslant 2$, $1$ doit être placé à côté de deux nombres différents, et n'est relié qu'à un seul autre diviseur de $p^{\\al... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | Exactly those n that are neither perfect squares nor prime powers with exponent at least two; equivalently, all primes and all integers with at least two distinct prime factors that are not perfect squares. | |
0idb | Problem:
A best-of-9 series is to be played between two teams; that is, the first team to win $5$ games is the winner. The Mathletes have a chance of $2/3$ of winning any given game. What is the probability that exactly $7$ games will need to be played to determine a winner? | [
"Solution:\nIf the Mathletes are to win, they must win exactly $5$ out of the $7$ games. One of the $5$ games they win must be the $7$th game, because otherwise they would win the tournament before $7$ games are completed. Thus, in the first $6$ games, the Mathletes must win $4$ games and lose $2$. The probability ... | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 20/81 | |
0gj2 | 令 $ABC$ 是一個銳角三角形，且其外接圓為 $\omega$。$\Gamma$ 內切 $\omega$ 於點 $A$，同時切 $BC$ 於點 $D$。令 $AB$ 和 $AC$ 分別與 $\Gamma$ 相交於 $P$ 和 $Q$。令 $M$ 和 $N$ 為在直線 $BC$ 上的點，滿足 $B$ 是 $DM$ 的中點，$C$ 是 $DN$ 的中點。直線 $MP$ 和 $NQ$ 交於 $K$，並且再次與 $\Gamma$ 分別於 $I$ 和 $J$ 相交。已知射線 $KA$ 與三角形 $IJK$ 的外接圓交於 $X \neq K$。證明 $\angle BXP = \angle CXQ$。
Let $ABC$ be an acu... | [
"Let $MP$ and $NQ$ intersect $AD$ at $K_1$ and $K_2$ respectively. By applying Menelaus' theorem to triangle $ABD$ and line $MPK_1$, we have\n$$\n\\frac{AK_1}{K_1D} = \\frac{AP}{PB} \\cdot \\frac{BM}{MD} = \\frac{AP}{2PB}\n$$\nand similarly $\\frac{AK_2}{K_2D} = \\frac{AQ}{2QC}$. A homothety at $A$ takes $\\Gamma \... | Taiwan | IMO 2J, Mock Exam 2 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic... | Chinese; English | proof only | null | |
07gj | Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for all positive integers $n$
i. $f(n) < f(n + 1) < f(n) + 2020$;
ii. $S(f(n)) = f(S(n))$.
(Where $S(n)$ is the sum of digits of $n$ in base 10.) | [
"It is clear that $f(n) = n$ is a solution. Assume that there is another solution. Since $f$ is strictly monotone it follows that $n < f(n) < 2020n + C$ for all large $n$. Let $n = 10^t - 1$ for some large $t$ it follows that $S(f(10^t - 1)) = f(9t) > 9t$. On the other hand $f(10^t - 1) \\ge 10^t$ so we have\n$$\nf... | Iran | 38th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Other"
] | null | proof and answer | f(n) = n | |
04tw | Find the largest positive integer $n$ such that
$$
\lfloor\sqrt{1}\rfloor + \lfloor\sqrt{2}\rfloor + \lfloor\sqrt{3}\rfloor + \dots + \lfloor\sqrt{n}\rfloor
$$
is a prime ($\lfloor x \rfloor$ denotes the largest integer not exceeding $x$). (Patrik Bak) | [
"Consider the infinite sequence $\\{a_n\\}_{n=1}^{\\infty}$ defined by $a_n = \\lfloor\\sqrt{n}\\rfloor$. This sequence is clearly non-decreasing and since\n$$\nk = \\sqrt{k^2} < \\sqrt{k^2 + 1} < \\dots < \\sqrt{k^2 + 2k} < \\sqrt{k^2 + 2k + 1} = k + 1,\n$$\nit contains every integer $k$ precisely $(2k + 1)$-times... | Czech Republic | 67th Czech and Slovak Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 47 | |
0ced | Let $n \ge 2022$ be an integer and $A$ be a finite set so that $\{S(B) \mid B \subset A\} = \{1, 2, \dots, n\}$, where $S(X)$ is the sum of the elements of the set $X$. Find the smallest and the largest possible value of the cardinal of the set $A$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Smallest |A| = ceil(log2(n + 1)); Largest |A| = floor((sqrt(8n + 1) − 1) / 2). | |
01yb | Let $a \neq b$ be coprime positive integers such that for each integer $k$ from $1$ to $10$ the numbers $a + k$ and $b + k$ are not coprime.
Find the minimal possible value of $|a - b|$. | [
"Answer: $2310$. Suppose $a$ and $b$ satisfy the problem conditions. For each integer $k$ from $1$ to $10$ denote by $p_k$ the minimal prime divisor of $\\gcd(a+k, b+k)$. Since the $\\gcd$ of numbers divides their difference, among $10$ successive integers at most $\\lfloor \\frac{10}{p_k} \\rfloor$ can be a multip... | Belarus | SELECTION and TRAINING SESSION | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 2310 | |
0269 | Problem:
Todos com o 2 - Qual operação devemos fazer com todos os 5 números
$$
418, 244, 816, 426, 24
$$
para obter 5 números que tenham todos o algarismo 2?
(a) dividir 2;
(b) somar 4;
(c) dividir por 6;
(d) subtrair 5;
(e) multiplicar por 3. | [
"Solution:\nTodos com o 2 - Resposta: multiplicar por 3."
] | Brazil | null | [
"Math Word Problems"
] | null | MCQ | e |
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