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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0ipp | Problem:
Evaluate the infinite sum $\sum_{n=1}^{\infty} \frac{n}{n^{4}+4}$. | [
"Solution:\n\n$\\displaystyle\n\\begin{aligned}\n\\sum_{n=1}^{\\infty} \\frac{n}{n^{4}+4} & =\\sum_{n=1}^{\\infty} \\frac{n}{\\left(n^{2}+2 n+2\\right)\\left(n^{2}-2 n+2\\right)} \\\\\n& =\\frac{1}{4} \\sum_{n=1}^{\\infty}\\left(\\frac{1}{n^{2}-2 n+2}-\\frac{1}{n^{2}+2 n+2}\\right) \\\\\n& =\\frac{1}{4} \\sum_{n=1}... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 3/8 | |
08bg | Problem:
Alberto e Barbara giocano a biliardino. Prima di iniziare, decidono che la partita finirà non appena uno dei due avrà fatto 3 gol più dell'altro. Sapendo che, per ogni pallina giocata, sia Alberto che Barbara hanno il $50\%$ di probabilità di segnare, qual è la probabilità che la partita non termini prima del... | [
"Solution:\n\nLa risposta è (A). Indichiamo con $P_{n}$ la probabilità che la partita non termini con l'$n$-simo gol (né prima). Supponiamo che dopo un numero dispari $(2k-1)$ di gol la partita non sia ancora finita: la differenza tra i gol segnati dai due giocatori è allora uguale a $1$ in valore assoluto, dal mom... | Italy | Gara di Febbraio | [
"Statistics > Probability > Counting Methods > Other"
] | null | MCQ | A | |
0c3u | Problem:
Fie $ABC$ un triunghi ascuţitunghic în care $AB < BC$, $O$ centrul cercului său circumscris şi $B'$ piciorul înălţimii din $B$. Paralela prin $B'$ la $CO$ intersectează dreapta $BO$ în punctul $X$. Arătaţi că $X$ şi mijloacele segmentelor $[AB]$ şi $[AC]$ sunt coliniare. | [
"Solution:\n\nFie $M$ mijlocul laturii $[AB]$. Atunci $m(\\angle OBC) = m(\\angle OCB) = 90^\\circ - m(\\angle A)$ şi $MX \\parallel BC$ revine la $m(\\angle MXB) = 90^\\circ - m(\\angle A)$. Dar $m(\\angle B'XB) = m(\\angle XOC) = 2 m(\\angle OBC) = 180^\\circ - 2 m(\\angle A)$. În triunghiul $ABB'$, avem $MA = MB... | Romania | Al doilea test de selectie pentru OBMJ | [
"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 | |
0emk | Let $a$ be a positive integer and $a > 1$. Prove that, for every positive integer $n$, the number
$$
n(2n + 1)(3n + 1)\dots(an + 1)
$$
is divisible by all prime numbers smaller than $a$. | [
"Let $p < a$ be a prime. If $p \\nmid n$, the claim holds.\nSuppose then that $n$ is not divisible by $p$. Then the numbers $2n + 1, 3n + 1, \\dots, (p+1)n + 1$ give different remainders when divided by $p$, which are denoted, respectively, by $r_1, r_2, \\dots, r_p$. Indeed, if $r_i = r_j$, then\n$$\np|((i+1)n + 1... | South Africa | South-Afrika 2011-2013 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0i6q | Problem:
A man, standing on a lawn, is wearing a circular sombrero of radius $3$ feet. Unfortunately, the hat blocks the sunlight so effectively that the grass directly under it dies instantly. If the man walks in a circle of radius $5$ feet, what area of dead grass will result? | [
"Solution:\n$60\\pi\\ \\mathrm{ft}^2$\n\nLet $O$ be the center of the man's circular trajectory. The sombrero kills all the grass that is within $3$ feet of any point that is $5$ feet away from $O$—i.e., all the grass at points $P$ with $2 \\leq OP \\leq 8$. The area of this annulus is then\n$$\n\\pi\\left(8^2 - 2^... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 60π | |
0ga3 | 令所有正實數所成的集合為 $\mathbb{R}^+$. 求所有函數 $f : \mathbb{R}^+ \to \mathbb{R}^+$, 滿足:
$$
f(x + y + f(y)) = 4030x - f(x) + f(2016y),
$$
對所有正實數 $x, y$ 都成立。
Let $\mathbb{R}^+$ be the set of all positive real numbers. Determine all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ satisfying
$$
f(x + y + f(y)) = 4030x - f(x) + f(2016y)... | [
"It's trivial that if $f$ maps $x$ to $2015x$, then the functional equation holds. And we'll show that it is the only function that satisfies our requirement. As usual, there are some parts in our proof. For simplicity, let $k = 2015$, then the functional equation becomes\n$$\nf(x + y + f(y)) = 2kx - f(x) + f((k+1)... | Taiwan | 二〇一六數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 2015x | |
06go | Some of the lattice points $(x, y)$, with $1 \le x \le 101$ and $1 \le y \le 101$ are marked so that no 4 marked points form the vertices of an isosceles trapezoid with bases parallel to the $x$-axis or the $y$-axis (a rectangle is counted as an isosceles trapezoid). Determine the maximum number of marked points. (A la... | [
"The answer is $251$.\n\nConsider pairs of marked points with the same $y$-coordinates. If there exist two pairs of marked points with the same sum of $x$-coordinates and different $y$-coordinates, then the 4 points in these pairs are the vertices of an isosceles trapezoid.\n\nSuppose there are $a_1, a_2, \\dots, a... | Hong Kong | CHKMO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | 251 | |
0ld6 | Find the smallest positive integer $n$ such that there exist $n$ real numbers $a_1, a_2, \dots, a_n$ satisfying the following conditions:
i) $a_1 + \dots + a_n > 0$;
ii) $a_1^3 + \dots + a_n^3 < 0$; and
iii) $a_1^5 + \dots + a_n^5 > 0$. | [
"For an $n$-tuple $\\mathbf{a} = (a_1, a_2, \\dots, a_n)$ and a nonnegative integer $k$, let\n$$\nS_k(\\mathbf{a}) = a_1^k + \\dots + a_n^k.\n$$\n\nWe need to find the smallest positive integer $n$ such that there exist an $n$-tuple $\\mathbf{a} = (a_1, \\dots, a_n)$ of real numbers such that $S_1(\\mathbf{a}) > 0$... | Vietnam | IMO 2015 Team Selection Tests | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 5 | |
0a6j | Problem:
Let $a$ and $b$ be positive integers with no common factor greater than $1$. What are the possible values for the greatest common divisor of $(a + b)$ and $(a - b)$? | [
"Solution:\nLet $d$ be a common divisor of both $(a + b)$ and $(a - b)$, and therefore divides linear combinations of $(a + b)$ and $(a - b)$. In particular,\n$$\nd \\mid [(a + b) + (a - b)] = 2a,\\quad d \\mid [(a + b) - (a - b)] = 2b.\n$$\nAs $a$ and $b$ don't share any common factors, the only common factors $2a... | New Zealand | NZMO Round One | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 1 and 2 | |
0628 | Problem:
Es seien $n$ eine positive ganze Zahl größer als Eins und $B=\{1,2, \ldots, 2^{n}\}$. Eine Teilmenge $A$ von $B$ heiße ulkig, wenn sie von je zwei verschiedenen Elementen $x, y$ von $B$, deren Summe eine Zweierpotenz ist, genau eines enthält. Wie viele ulkige Teilmengen hat $B$? | [
"Solution:\n\nDie Anzahl ist $2^{n+1}$. Wir argumentieren mit vollständiger Induktion nach $n$ und setzen hierzu $B_{n}=\\{1,2, \\ldots, 2^{n}\\}$. \n\nInduktionsanfang, $n=1$. Dieser Fall ist klar: alle vier Teilmengen von $B_{1}=\\{1,2\\}$ sind ulkig.\n\nInduktionsschritt, $n \\longrightarrow n+1$. Es sei also be... | Germany | 1. IMO-Auswahlklausur | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2^{n+1} | |
0h35 | Prove that there exist a point $A$ on the graph of $f(x) = x^4$ and a point $B$ on the graph of $g(x) = x^4 + x^2 + x + 1$ such that the distance between $A$ and $B$ is less than $\frac{1}{100}$. | [
"Розглянемо точку $B(100, 100010101)$, яка лежить на графіку функції $g$, і точку $A(\\sqrt[4]{100010101}, 100010101)$, яка лежить на графіку функції $f$. Відстань між цими точками дорівнює\n$$\n\\sqrt[4]{100010101} - 100 = \\frac{100010101 - 100^4}{(\\sqrt[4]{100010101} + 100)(\\sqrt{100010101} + 100^2)} < \\frac{... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0b3k | Problem:
Let $m$ and $n$ be relatively prime positive integers. If $m^{3} n^{5}$ has 209 positive divisors, then how many positive divisors does $m^{5} n^{3}$ have? | [
"Solution:\n\nLet $d(N)$ denote the number of positive divisors of an integer $N$. Suppose that the prime factorizations of $m$ and $n$ are $\\prod\\left(p_{i}^{a_{i}}\\right)$ and $\\prod\\left(q_{i}^{b_{i}}\\right)$ respectively. Observe that 209 has four positive divisors: $1, 11, 19, 209$.\n\nIf $m=1$, then $n^... | Philippines | 24th Philippine Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 217 | |
0kfp | Problem:
Alice and Bob take turns removing balls from a bag containing 10 black balls and 10 white balls, with Alice going first. Alice always removes a black ball if there is one, while Bob removes one of the remaining balls uniformly at random. Once all balls have been removed, the expected number of black balls whi... | [
"Solution:\n\nSuppose $a$ is the number of black balls and $b$ is the number of white balls, and let $E_{a, b}$ denote the expected number of black balls Bob has once all the balls are removed with Alice going first. Then we want to find $E_{10,10}$. It is evident that if $E_{0, b}=0$. Also, since Bob chooses a bla... | United States | HMMO | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 4519 | |
01zj | Find the smallest number $n$ with the following property: if among the numbers from $1$ to $1000$ we choose $n$ numbers such that no two of them are divisible by the square of the same prime number, then at least one of these numbers is necessarily the square of a prime number. | [
"Answer: $617$.\n\nTo begin with, let's give an example of $616$ numbers from $1$ to $1000$, among which there are no prime squares and no two of which are divisible by the square of a prime. To do this, take $8$ numbers of the form $2p^2$ not exceeding $500$, where $p$ is a prime number:\n$$\n2 \\cdot 2^2,\\ 2 \\c... | Belarus | SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 617 | |
0e3t | At a volleyball tournament any two of the $n$ teams played against each other exactly once. For any two different teams $A$ and $B$ there were exactly $k$ teams that lost against both of them. Prove that $n = 4k + 3$. | [] | Slovenia | Selection Examinations for the IMO | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
03ti | A positive integer $m$ is called good, if there is a positive integer $n$ such that $m$ is the quotient of $n$ over the number of positive integer divisors of $n$ (including 1 and $n$ itself). Prove that $1, 2, \ldots, 17$ are good numbers and that $18$ is not a good number. | [
"For positive integer $n$, let $d(n)$ denote the number of positive divisors of $n$ (including $1$ and $n$ itself).\n\nFirstly, note that $1$ and $2$ are good, since $1 = \\frac{2}{d(2)}$ and\n$$\n2 = \\frac{8}{d(8)}.\n$$\n\nSecondly, we note that if $p$ is an odd prime, then $p$ is good. This is because $d(8p) = 8... | China | China Girls' Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
05rt | Problem:
Soit $ABC$ un triangle, et soit $M$ le pied de la médiane issue de $A$. Soit également $\ell_{\mathrm{b}}$ la bissectrice de $\widehat{AMB}$ et $\ell_{c}$ la bissectrice de $\widehat{AMC}$. Enfin, soit $B'$ le projeté orthogonal de $B$ sur $\ell_{b}$, soit $C'$ le projeté orthogonal de $C$ sur $\ell_{c}$, et ... | [
"Solution:\n\nSoit $\\beta$ l'angle $\\widehat{AMB'}$ et $\\gamma$ l'angle $\\widehat{AMC'}$. Par construction, on sait que $2(\\beta+\\gamma)=180^\\circ$, donc que $\\widehat{B'MC'}=\\beta+\\gamma=90^\\circ$. Ainsi, la droite $\\ell_{\\mathrm{b}} = (MB')$ est perpendiculaire aux droites $(BB')$ et $\\ell_{c} = (MC... | France | Préparation Olympique Française de Mathématiques - Test du 15 Mai 2019 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
02kd | Problem:
Se $3$ e $\frac{1}{3}$ são as raízes da equação $a x^{2}-6 x+c=0$, qual o valor de $a+c$?
A) $1$
B) $0$
C) $-\frac{9}{5}$
D) $\frac{18}{5}$
E) $-5$ | [
"Solution:\n\nSolução 1 - Como $3$ e $\\frac{1}{3}$ são raízes da equação $a x^{2}-6 x+c=0$ temos:\n$$\n9a - 18 + c = 0 \\Rightarrow 9a + c = 18 \\text{ e } \\frac{a}{9} - 2 + c = 0 \\Rightarrow \\frac{a}{9} + c = 2\n$$\nResolvendo o sistema\n$$\n\\left\\{\\begin{array}{l}\n9a + c = 18 \\\\\n\\frac{a}{9} + c = 2\n\... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | D | |
053o | There are 8 white pawns on the squares at one edge of an $8 \times 8$ chessboard and 8 black pawns on the squares at the opposite edge. On each move, a player shifts one of his pawns by one or more squares forward (toward the opponent's piece) or backward, but moving a pawn to a square containing the opponent's pawn or... | [
"Black can use the following strategy. If white moves his $k$th pawn counting from his left, by $n$ squares forward, black moves his $k$th pawn counting from his left, by $n$ squares forward. If white moves his pawn by $n$ squares backward, black moves his pawn on the same file by $n$ squares forward. After each mo... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Black | |
02eb | Given points $A_1 = (x_1, y_1, z_1)$, $A_2 = (x_2, y_2, z_2)$, ..., $A_n = (x_n, y_n, z_n)$ let $P = (x, y, z)$ be the point which minimizes $\sum_{i=1}^n (|x - x_i| + |y - y_i| + |z - z_i|)$. Give an example (for each $n > 4$) of points $A_i$ for which the point $P$ lies outside the convex hull of the points $A_i$. | [
"Suppose $n$ is $3m$. Take the points to be\n$$\n(1, 0, 0), (2, 0, 0), \\dots, (m, 0, 0), \\\\\n(0, 1, 0), (0, 2, 0), \\dots, (0, m, 0), \\\\\n(0, 0, 1), (0, 0, 2), \\dots, (0, 0, m)\n$$\nTake $P$ to be $(x, y, z)$. Then the sum is $2m|x| + |x-1| + |x-2| + \\dots + |x-m|$ plus similar terms in $y$ and $z$. Evidentl... | Brazil | IX OBM | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof only | null | |
0hhq | Prove that there are no natural numbers $n$ and $k$ that satisfy the equation:
$$
n^n + (n + 1)^{n+1} + (n + 2)^{n+2} = 2023^k.
$$ | [
"Consider the given equation modulo $3$. The right-hand side, $2023^k = 1 \\pmod{3}$. Among any three consecutive natural numbers $n$, $n+1$, $n+2$, one is divisible by $3$, one leaves a remainder of $1$ when divided by $3$, and one leaves a remainder of $2$.\n\nIf $x$ is divisible by $3$, then $x^x$ is also divisi... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0fhv | Problem:
El ángulo $A$ de un triángulo isósceles $ABC$ mide $2/5$ de recto, siendo los ángulos $B$ y $C$ iguales. La bisectriz del ángulo $C$ corta al lado opuesto en el punto $D$. Calcular las medidas de los ángulos del triángulo $BCD$. Expresar la medida $a$ del lado $BC$ en función de la medida $b$ del lado $AC$, s... | [
"Solution:\n\nCon los datos del enunciado tenemos\n\nen el triángulo $ABC$, $\\widehat{BAC} = 36^\\circ$; $\\widehat{ABC} = \\widehat{ACB} = 72^\\circ$,\nen el triángulo $CBD$, $\\widehat{BCD} = 36^\\circ$; $\\widehat{CDB} = \\widehat{BDC} = 72^\\circ$\nen el triángulo $ADC$, $\\widehat{DAC} = \\widehat{ACD} = 72^\... | Spain | OME 30 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | Angles of triangle BCD: 36°, 72°, 72°. Relation between sides: a = ((√5 − 1)/2) · b. | |
003i | Se tienen varios enteros positivos menores que $10^6$ tales que el producto de cada pareja de dos números distintos no es divisible por $2006$. ¿Cuál es la mayor cantidad de números que se pueden tener? | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Español | proof and answer | 983050 | |
0dpc | Positive real numbers $a$, $b$, $c$, $d$ satisfy:
$$
a(c^2 - 1) = b(b^2 + c^2) \text{ and } d \le 1.
$$
Prove that
$$
d(a\sqrt{1-d^2} + b^2\sqrt{1+d^2}) \le \frac{(a+b)c}{2}.
$$ | [] | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0jjw | Problem:
Determine the number of sequences of sets $S_{1}, S_{2}, \ldots, S_{999}$ such that
$$
S_{1} \subseteq S_{2} \subseteq \cdots \subseteq S_{999} \subseteq \{1,2, \ldots, 999\}
$$
Here $A \subseteq B$ means that all elements of $A$ are also elements of $B$. | [
"Solution:\n$10^{2997}$ OR $1000^{999}$ The idea is to look at each element individually, rather than each subset. For each $k \\in \\{1,2, \\ldots, 999\\}$, there are $1000$ choices for the first subset in the chain that contains $k$. This count includes the possibility that $k$ doesn't appear in any of the subset... | United States | HMMT November 2014 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 1000^{999} | |
0957 | Problem:
Fie polinomul $P(X) = a X^{2} + b X + c$, unde $a, b, c \in \mathbb{R}$, $a \neq 0$ şi $x_{1}, x_{2}$ rădăcinile acestui polinom. Să se arate că dacă $a$ se află între $c$ şi $b-c$, atunci în intervalul $(-1,1)$ se poate conţine cel mult una dintre aceste rădăcini. | [
"Solution:\n\nPresupunem contrariul: fie că polinomul are ambele rădăcini $x_{1}, x_{2} \\in (-1,1)$. Atunci $|x_{1}| < 1$ şi $|x_{2}| < 1$, de unde $|x_{1} x_{2}| < 1$, adică $-1 < x_{1} x_{2} < 1$. Avem sistemul:\n$$\n\\begin{cases}\n1 - x_{1} x_{2} > 0, \\\\\n1 + x_{1} > 0, \\\\\n1 + x_{2} > 0.\n\\end{cases}\n$$... | Moldova | Olimpiada de Matematică a Republicii Moldova | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0kn0 | Problem:
Compute the number of ways to fill each cell in a $8 \times 8$ square grid with one of the letters $H, M$, or $T$ such that every $2 \times 2$ square in the grid contains the letters $H, M, M, T$ in some order. | [
"Solution:\n\nWe solve the problem for general $n \\times n$ boards where $n$ even. Let the cell in the $i$-th row and $j$-th column be $a_{i, j}$.\n\nClaim: In any valid configuration, either the rows (or columns) alternate between ( $\\cdots, H, M, H, M, \\cdots$ ) and $(\\cdots, T, M, T, M, \\cdots)$ or $(\\cdot... | United States | HMMT Spring 2021 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1076 | |
0exd | Problem:
Given two relatively prime natural numbers $r$ and $s$, call an integer good if it can be represented as $mr + ns$ with $m, n$ non-negative integers and bad otherwise. Prove that we can find an integer $c$, such that just one of $k$, $c - k$ is good for any $k$. How many bad numbers are there? | [
"Solution:\n\nNotice that $0$ is good and all negative numbers are bad. Take $c = rs - r - s$. First, $c$ is bad. For suppose otherwise: $c = mr + ns$. Then $mr + ns = (s - 1)r - s$. Hence $(s - 1 - m)r = (n + 1)s$, so $r$ divides $n + 1$. Say $n + 1 = kr$, and then $s - 1 - m = ks$, so $m = (1 - k)s - 1$. But $n +... | Soviet Union | 5th ASU | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | Choose c = rs − r − s. The number of bad nonnegative integers is (rs − r − s + 1)/2. | |
0l8w | In the plane, let be given an isosceles triangle $ABC$ ($AB = AC$). A variable circle $(O)$ with center $O$ on the line $BC$, passes through $A$ but does not touch the lines $AB$, $AC$. Let $M$, $N$ be respectively the second points of intersection of the circle $(O)$ with the lines $AB$, $AC$. Find the locus of the or... | [
"1st case: $\\angle A = 90^\\circ$: the locus is the singleton $\\{A\\}$.\n\n2nd case: $\\angle A \\neq 90^\\circ$: let $D$ be the point symmetric to $A$ with respect to $BC$, $K$ be the point symmetric to $D$ with respect to $MN$ then the line $HK$ is the image of the line $BC$ under the homothety with center $D$ ... | Vietnam | THE 2002 VIETNAMESE MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Con... | English | proof and answer | If the angle at A is a right angle, the locus of the orthocenter of triangle AMN is the single point A.
If the angle at A is not a right angle: let D be the reflection of A across BC. Let d be the image of the line BC under the homothety centered at D with ratio 4·sin^2(A/2). Then the locus of the orthocenter H of tria... | |
0hfg | Monica and Bogdan are playing a game that depends on two positive integers $n$ and $k$. First, Monica chooses and writes $k$ positive numbers. Bogdan wins if he manages to mark $n$ points on the plane so that for each number $m$ written by Monica, there are two marked points at a distance precisely $m$, otherwise Monic... | [
"If $k < n$, Bogdan can choose $n$ points on a line so that the distance between the first and second points is equal to the first written number, between the second and third to the second written number, and so on.\n\nIf $k \\ge n$, Monica can choose numbers $2^0, 2^1, \\dots, 2^{k-1}$. Suppose that Bogdan can wi... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Discrete Mathematics > Other"
] | English | proof and answer | Bogdan wins if k < n; Monica wins if k >= n. | |
0ko2 | Problem:
Is
$$
\prod_{k=0}^{\infty}\left(1-\frac{1}{2022^{k!}}\right)
$$
rational? | [
"Solution:\nIt suffices to prove that the product $A=\\prod_{k=1}^{\\infty}\\left(1-\\frac{1}{2022^{k!}}\\right)$ is irrational. Suppose for the sake of contradiction that $A$ is rational. Note that for each non-negative integer $n$, there exists at most one subset $S$ of $\\{1!, 2!, 3!, \\ldots\\}$ such that the s... | United States | HMIC | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | proof and answer | irrational | |
0bt0 | The positive numbers $a$, $b$, $c$ are such that
$$
\frac{a}{b+c+1} + \frac{b}{a+c+1} + \frac{c}{a+b+1} \le 1.
$$
Prove that:
$$
\frac{1}{b+c+1} + \frac{1}{a+c+1} + \frac{1}{a+b+1} \ge 1.
$$ | [
"Denote by $S = \\sum \\frac{1}{b+c+1}$. Using the inequality from the hypothesis we obtain\n$$\n\\sum \\left( \\frac{a}{b+c+1} + 1 \\right) \\le 4,\n$$\nand thus $(a+b+c+1)S \\le 4$.\n\nUsing arithmetic mean – harmonic mean inequality we get\n$$\nS \\ge \\frac{9}{2(a + b + c + 1) + 1},\n$$\nhence $S \\ge 9 - 2(a +... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
000j | Sean $1 = a_1 \le a_2 \le \dots \le a_n \le \dots$ números enteros tales que existen infinitos enteros positivos $k$ con $k = \frac{i}{a_i}$ para algún $i$.
Demostrar que para cada entero positivo $n$, existe un entero positivo $j$ tal que $n = \frac{j}{a_j}$. | [] | Argentina | XI Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series"
] | español | proof only | null | |
07b5 | The vertices of an $n$-vertex tree are labeled by numbers $1$, $2$, $\ldots$, $n$. At each stage one can choose an edge that has not been selected before, and switch the labels of its two ends, until all of the edges have been selected. Show that the final permutation of the labels is a complete cycle of $n$ letters. | [
"We will prove the statement by induction on $n$. Let $e$ be the last edge, with $i$, $j$ as the labels of its vertices. Deleting $e$ divides the tree into two trees with vertices $I = \\{i = i_1, i_2, \\dots, i_k\\}$, $J = \\{j = j_1, \\dots, j_l\\}$ (we denote every vertex by its initial label). By the induction ... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof only | null | |
03f5 | a) Find all values of $a$ for which the inequality
$$
x \log_{\frac{1}{2}} a^4 - x^2 > 3 + 2 \log_2 a^2
$$
has a solution.
b) Calculate the limit
$$
\lim_{a \to -\infty} \left( \sqrt{a^2 - a + 1} + a \right).
$$ | [
"a) Since $\\log_{\\frac{1}{2}}(a^4) = -2 \\log_2(a^2)$, then by putting $2 \\log_2(a^2) = b$, we get the inequality $x^2 + b \\cdot x + 3 + b < 0$. For this inequality to have at least one solution, it is necessary and sufficient that $D = b^2 - 4b - 12 > 0$ whose solutions are $b < -2$ or $b > 6$, whence $\\log_2... | Bulgaria | 3 Bulgarian Spring Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof and answer | Part (a): a ∈ (-∞, -2√2) ∪ (-√2/2, 0) ∪ (0, √2/2) ∪ (2√2, ∞). Part (b): 1/2. | |
0384 | Problem:
Let $A$ and $B$ be given points on a circle $k$. For an arbitrary point $L$ on $k$ denote by $M$ the point on the line $A L$ such that $L M = L B$ and $L$ is between $A$ and $M$. Find the locus of the points $M$. | [
"Solution:\nLet $C D$ be the diameter of $k$ such that $C D \\perp A B$ and let $L \\in A \\widehat{C} B$. Then $\\angle A L B = 2 \\alpha$ is constant and we have $\\angle A M B = \\alpha$ since $\\triangle M L B$ is isosceles. Therefore $M$ belongs to an arc of the circle $k_{1}$, from which the segment $A B$ is ... | Bulgaria | Spring Mathematical Competition | [
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Let α be half of the inscribed angle under which the segment AB is seen from points on the arc of the given circle containing the points between A and the diameter perpendicular to AB. Then the locus of M is the union of two arcs: the arc of the circle from which AB is seen at angle α and the arc of the circle from whi... | |
0h5p | Two circles $w_1, w_2$ are externally tangent at a point $Q$. A common external tangent line to these circles (that doesn't pass through $Q$) is tangent to $w_1$ at a point $B$, and $BA$ is a diameter of this circle. The point $A$ belongs to the line tangent to the circle $w_2$ at a point $C$ such that $B$ and $C$ are ... | [
"Let $K$ be the point where the common tangent touches the circle $w_2$ (Fig. 6). Consider the common tangent line to the two circles that passes through the point $Q$. Suppose it intersects the line $BK$ at a point $P$. By the properties of lines tangent to circles,\n$$\nPB = PQ = PK.\n$$\nTherefore, $\\angle BQK ... | Ukraine | Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0cd6 | A $3 \times 3 \times 3$ cube is divided into 27 unit-cubes. Call a *strip* any $1 \times 1 \times 3$ rectangular cuboid (block) consisting of three unit-cubes.
A positive integer is written inside each unit-cube such that any number $n$, strictly greater than 1, written in a unit-cube, is the sum of the numbers written... | [
"If all the 27 numbers are equal to 1, we have nothing to prove. Assume the cube contains some numbers greater than 1 and suppose there is an even number between them. If $n$ is the smallest even number written inside a cube, then $n$ should be the sum of three odd numbers, which is impossible due to parity reasons... | Romania | THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - FIRST SELECTION TEST | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0j77 | Problem:
Find the number of sequences consisting of $100$ $R$'s and $2011$ $S$'s that satisfy the property that among the first $k$ letters, the number of $S$'s is strictly more than $20$ times the number of $R$'s for all $1 \leq k \leq 2111$. | [
"Solution:\n\nAnswer: $\\frac{11}{2111}\\binom{2111}{100}$\n\nGiven positive integers $r$ and $s$ such that $s \\geq 20 r$, let $N(s, r)$ denote the number of sequences of $s$ copies of $S$ and $r$ copies of $R$ such that for all $1 \\leq k \\leq r+s-1$, among the first $k$ letters, the number of $S$'s is strictly ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | final answer only | 11/2111 * C(2111, 100) | |
04v6 | Alice and Ben play the game on a board with $72$ cells around a circle. First, Ben chooses some cells and places one chip on each of them. Each round, Alice first chooses one empty cell and then Ben moves a chip from one of the adjacent cells onto the chosen one. If Ben fails to do so, the game ends; otherwise, another... | [
"We show that the smallest possible number of chips is $36$.\n\nIn the first part, we describe the strategy of Ben in which he can ensure that the game will never end. At the beginning, Ben places $36$ chips on even cells of the game board and the odd cells he lets empty. Moreover, he firmly divides all $72$ cells ... | Czech Republic | 72nd Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 36 | |
04be | Azra thought of four real numbers and wrote on the blackboard the sums of all pairs of imagined numbers, and then she deleted one of the sums. There were numbers $-2$, $1$, $2$, $3$ and $6$ left on the blackboard. What numbers did Azra think of? (M. Bašić, M. Bombardelli) | [
"Let $a$, $b$, $c$ and $d$ be the numbers Azra thought of. Without loss of generality, we can assume that the deleted sum is $c+d$. Then there are numbers $a+b$, $a+c$, $a+d$, $b+c$ and $b+d$ written on the blackboard, i.e.\n$$\n\\{-2, 1, 2, 3, 6\\} = \\{a+b, a+c, a+d, b+c, b+d\\}.\n$$\nSince\n$$\n(a+c) + (b+d) = (... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | −3/2, −1/2, 5/2, 7/2 | |
07ld | Find all pairs $(a, b)$ of positive integers, such that $(ab)^2 - 4(a+b)$ is the square of an integer. | [
"If $(ab)^2 - 4(a+b) = x^2$ with positive integers $a, b$ and an integer $x \\ge 0$, we have $x < ab$. As $(ab)^2 - (ab-1)^2 = 2ab - 1$ is odd, we even have $x \\le ab - 2$. This implies $(ab)^2 - 4(a+b) \\le (ab-2)^2 = (ab)^2 - 4ab + 4$, from which we obtain\n$$\nab \\le a + b + 1. \\qquad (5)\n$$\nAfter swapping ... | Ireland | Irska | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (1,5), (5,1), (2,2), (2,3), (3,2) | |
020k | Problem:
A circle $\omega$ passes through the two vertices $B$ and $C$ of a triangle $A B C$. Furthermore, $\omega$ intersects segment $A C$ in $D \neq C$ and segment $A B$ in $E \neq B$. On the ray from $B$ through $D$ lies a point $K$ such that $|B K|=|A C|$, and on the ray from $C$ through $E$ lies a point $L$ such ... | [
"Solution:\nLet $M$ be the midpoint of the arc $B C$ of $\\omega$ that is on the same side of $B C$ as $A$. Then $|B M|=|C M|$. We also have $|B A|=|C L|$ and $\\angle A B M=\\angle E B M=\\angle E C M=\\angle L C M$. Hence $\\triangle A B M \\cong \\triangle L C M$. So $|A M|=|L M|$. (In case $M=E$ the triangles $... | Benelux Mathematical Olympiad | 8th Benelux 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"
] | null | proof only | null | |
081m | Problem:
Siano $a < b < c$ interi positivi tali che $a^{2} + b^{2} + c^{2}$ ha lo stesso numero di cifre di $a + b + c$. Qual è il massimo valore che può assumere $c$? | [
"Solution:\n\nLa risposta è $9$. La condizione che $a + b + c$ abbia lo stesso numero di cifre di $a^{2} + b^{2} + c^{2}$ implica che\n$$\n\\frac{a^{2} + b^{2} + c^{2}}{a + b + c} < 10\n$$\nPertanto si deve avere che\n$$\na^{2} - 10a + b^{2} - 10b + c^{2} - 10c < 0\n$$\nil che è equivalente ad affermare che\n$$\n(a... | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 9 | |
03oy | Find all pairs of positive integers (*x*, *y*) satisfying $x^y = y^{x-y}$. | [
"If $x = 1$, then $y = 1$. If $y = 1$, then $x = 1$.\nIf $x = y$, then $x^y = 1$, so $x = y = 1$.\n\nWe will discuss the circumstances when $x > y \\ge 2$ below. By assumption\n$$\n1 < \\left(\\frac{x}{y}\\right)^y = y^{x-2y},\n$$\nso\n$x > 2y$, and $y \\mid x$.\n\nAssume $x = k y$ then $k \\ge 3$, and\n$$\nk y = y... | China | China Girls' Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | (1, 1), (8, 2), (9, 3) | |
0kr2 | Problem:
In last year's HMMT Spring competition, 557 students submitted at least one answer to each of the three individual tests. Let $S$ be the set of these students, and let $P$ be the set containing the 30 problems on the individual tests. Estimate $A$, the number of subsets $R \subseteq P$ for which some student ... | [] | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | final answer only | null | |
0l9r | Let be given a convex quadrilateral **ABCD**. A point **M** moves on the line **AB** but does not coincide with **A** and **B**. Let **N** be the second point of meeting (distinct from **M**) of the circles (**MAC**) and (**MBD**). Prove that
i) $N$ moves on a fixed circle,
ii) The line $MN$ passes through a fixed po... | [
"Let $I$ be the point of intersection of the two diagonals of the quadrilateral $ABCD$ (figure 1).\n\ni) The quadrilateral $DCIN$ is cyclic because $\\widehat{ICN} = \\widehat{IDN}$ (as both angles are equal to $\\widehat{AMN}$). Consequently, $N$ moves on the fixed circle.\n\nii) Draw the line $l$ passing through ... | Vietnam | Vijetnam 2006 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0kx9 | Problem:
There are five people in a room. They each simultaneously pick two of the other people in the room independently and uniformly at random and point at them. Compute the probability that there exists a group of three people such that each of them is pointing at the other two in the group. | [
"Solution:\nThe desired probability is the number of ways to pick the two isolated people times the probability that the remaining three point at each other. So,\n$$\nP = \\binom{5}{2} \\cdot \\left(\\frac{\\binom{2}{2}}{\\binom{4}{2}}\\right)^3 = 10 \\cdot \\left(\\frac{1}{6}\\right)^3 = \\frac{5}{108}\n$$\nis the... | United States | HMMT November 2023 | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 5/108 | |
03g4 | Given a scalene triangle $ABC$. On the rays $AC \rightarrow$ and $BC \rightarrow$, the points $C_a$ and $C_b$ are chosen, respectively, such that $AC_a = BC_b = AB$. We denote by $O_c$ the center of the circumcircle about $\triangle CC_a C_b$. Analogously, we define the points $O_a$ and $O_b$. Prove that the lines $AO_... | [
"Let $O$ and $I$ be the centers of the circumscribed and inscribed circle of $ABC$, and $A_1$, $B_1$, $C_1$ be the centers of the arcs $\\widehat{AB}$, $\\widehat{AC}$, $\\widehat{BC}$ (not containing the third vertices) of the circumscribed circle. By symmetry with respect to $AA_1$ we have $A_1C_a = A_1B = A_1C$ ... | Bulgaria | 6 TST for BMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0168 | Determine the smallest number $ab + c$, which can be obtained from six different positive integers $a, b, c, d, e, f$, which fulfill $ab + c = de + f$. | [
"Let's call a 3-set of positive integers a *party* and $ab + c$ a *product-sum* in the party $\\{a, b, c\\}$. Let $m(A)$ denote the smallest product-sum of the party $A$. If $a < b < c$ then $ab + c < ac + b < bc + a$ and hence $m(\\{a, b, c\\}) = ab + c$. Let's also write $\\{a, b, c\\} \\le \\{d, e, f\\}$ if $a <... | Baltic Way | Baltic Way SHL | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | null | proof and answer | 11 | |
08px | Problem:
Let $A$ and $B$ be two non-empty subsets of $X=\{1,2, \ldots, 11\}$ with $A \cup B = X$. Let $P_{A}$ be the product of all elements of $A$ and let $P_{B}$ be the product of all elements of $B$. Find the minimum and maximum possible value of $P_{A} + P_{B}$ and find all possible equality cases. | [
"Solution:\nFor the maximum, we use the fact that $(P_{A} - 1)(P_{B} - 1) \\geqslant 0$, to get that $P_{A} + P_{B} \\leqslant P_{A} P_{B} + 1 = 11! + 1$. Equality holds if and only if $A = \\{1\\}$ or $B = \\{1\\}$.\n\nFor the minimum observe, first that $P_{A} \\cdot P_{B} = 11! = c$. Without loss of generality l... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | Maximum value: 11! + 1, attained exactly when one of the sets is {1}. Minimum value: 12636, attained when the products are 6300 and 6336. All equality cases for the minimum occur with A equal to one of {2,5,7,9,10}, {1,2,5,7,9,10}, {3,5,6,7,10}, {1,3,5,6,7,10}, and B being the complement of A in {1,2,3,4,5,6,7,8,9,10,1... | |
03rj | A natural number $a$ is called a "lucky number" if the sum of its digits is $7$. Arrange all "lucky numbers" in ascending order, and we get a sequence $a_1, a_2, \dots$. If $a_n = 2005$, then $a_{5n} = \_\_\_\_\_\_$. | [
"Since the number of non-negative integer solutions of the equation $x_1 + x_2 + \\dots + x_k = m$ is $C_{m+k-1}^k$, the number of integer solutions, when $x_1 \\ge 1$ and $x_i \\ge 0\\ (i \\ge 2)$, is $C_{m+k-2}^{m-1}$. Let $m=7$, the number of lucky numbers with $k$ digits is $p(k) = C_{k+5}^6$.\n\nSince $2005$ i... | China | China Mathematical Competition (Jiangxi) | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | final answer only | 52000 | |
0dj3 | On a line, 200 points are marked and numbered $1, 2, 3, \ldots, 200$ from left to right. Various crickets jump around the line. Each starts at point $1$, jumping on the marked points and ending up at point $200$. In addition, each cricket jumps from a marked point to another marked point with a greater number. When all... | [
"For every pair $(i, j)$ where $1 \\le i \\le 100$ and $101 \\le j \\le 200$ there is a cricket that jumped from $i$ to $j$ and no cricket can do two such jumps. Therefore there are at least $100^2 = 10000$ crickets.\n\nConsider the following paths of crickets:\n\n1. $1 \\to 200$;\n2. $1 \\to a \\to 200$ where $a \... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 10000 | |
02pv | Problem:
Considere o conjunto $A=\{1,2,3, \ldots, 2011\}$. Quantos subconjuntos de $A$ existem de modo que a soma de seus elementos seja 2023060? | [
"Solution:\n\nObserve que a soma $1+2+\\cdots+2011=\\frac{2011 \\times 2012}{2}=2023066$. Logo, para obtermos um subconjunto de $A$ que tenha para soma de seus elementos $2023060$, basta retirarmos de $A$ os elementos cuja soma é $6$. Os possíveis casos são:\n\n- Subconjuntos com um elemento: $\\{6\\}$.\n- Subconju... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 4 | |
0eoy | The sums of three out of four numbers (omitting each of the four numbers in turn) are $20$, $22$, $24$ and $27$, respectively. What is the sum of the four numbers? | [
"Since each of the given sums involves exactly three of the numbers, it follows that the sum of the four sums is three times the sum of the four numbers. Thus the required sum is equal to $(20 + 22 + 24 + 27)/3 = 93/3 = 31$."
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | final answer only | 31 | |
0gkm | Let $f: \mathbb{R} \to \mathbb{R}$ be such that for all $x, y \in \mathbb{R}$,
$$
|f(x + y)| = |f(x) + f(y)|.
$$
Prove that $f(x + y) = f(x) + f(y)$ for all $x, y \in \mathbb{R}$. | [
"Suppose that there are $a, b \\in \\mathbb{R}$ such that $f(a + b) \\neq f(a) + f(b)$. By the assumption we have $f(a + b) = -f(a) - f(b)$. If $f(a + b) = 0$ then $f(a) = -f(b)$.\n$$\n\\begin{align*}\n|f(2a + 2b)| &= |f(a + (a + b + b))| \\\\\n&= |f(a) + f((a + b) + b)| \\\\\n&= |f(a) + f(a + b) + f(b)| \\text{ or... | Thailand | Thailand Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof only | null | |
03vn | Let $n$ and $k$ be positive integers with $k \ge n$ and $k - n$ an even number. Let $2n$ lamps labelled $1, 2, \dots, 2n$ be given, each of which can be either on or off. Initially all the lamps are off. We consider sequences of steps: at each step one of the lamps is switched from on to off or from off to on.
Let $N$ ... | [
"**Lemma** For any positive integer $t$, call a $t$-element array $(a_1, a_2, \\dots, a_t)$ which consists of $0, 1$ ($a_1, a_2, \\dots, a_t \\in \\{0, 1\\}$) \"good\" if there are odd '0's in it. Prove that there are $2^{t-1}$ \"good\" arrays.\n\n**Proof:** In fact, for the same $a_1, a_2, \\dots, a_t$, when $a_t$... | China | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 2^(k-n) | |
0ans | Problem:
Let $a$, $b$, and $c$ be three consecutive even numbers such that $a > b > c$. What is the value of $a^{2} + b^{2} + c^{2} - ab - bc - ac$? | [] | Philippines | 18th PMO Area Stage | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 12 | |
0h3z | Let $SABC$ be a tetrahedron with $MA > 1$, $MB > 1$, $MC > 1$, where $M$ is the centroid of $ABC$. Prove that
$$
SA + SB + SC > 3.
$$ | [
"Розглянемо вектори $\\overrightarrow{MA} = \\vec{a}$, $\\overrightarrow{MB} = \\vec{b}$, $\\overrightarrow{MC} = \\vec{c}$, $\\overrightarrow{MS} = \\vec{s}$, $\\overrightarrow{SA} = \\vec{a} - \\vec{s}$, $\\overrightarrow{SB} = \\vec{b} - \\vec{s}$ і $\\overrightarrow{SC} = \\vec{c} - \\vec{s}$. За умовою задачі,... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle tr... | English | proof only | null | |
0fqy | Problem:
Consideramos el polinomio
$$
p(x) = (x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a)
$$
Demostrar que $p(x) \geq 0$ para todo $x \in \mathbb{R}$ si y solamente si $a = b = c$. | [
"Solution:\nEn primer lugar, observamos que cuando $a = b = c$ se tiene que $p(x) = 3(x-a)^2$, que claramente satisface $p(x) \\geq 0$ para todo $x \\in \\mathbb{R}$.\n\nSupongamos ahora que $p(x) \\geq 0$ para todo $x \\in \\mathbb{R}$. Desarrollando la expresión como un polinomio cuadrático, obtenemos que\n$$\np(... | Spain | FASE LOCAL DE LA OLIMPIADA MATEMÁTICA ESPAÑOLA. | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
00h8 | A sequence of real numbers $a_{0}, a_{1}, \ldots$ is said to be $\operatorname{good}$ if the following three conditions hold.
(i) The value of $a_{0}$ is a positive integer.
(ii) For each non-negative integer $i$ we have $a_{i+1}=2 a_{i}+1$ or $a_{i+1}=\frac{a_{i}}{a_{i}+2}$.
(iii) There exists a positive integer $k$ s... | [
"Note that\n$$\na_{i+1}+1=2\\left(a_{i}+1\\right) \\text{ or } a_{i+1}+1=\\frac{a_{i}+a_{i}+2}{a_{i}+2}=\\frac{2\\left(a_{i}+1\\right)}{a_{i}+2} .\n$$\nHence\n$$\n\\frac{1}{a_{i+1}+1}=\\frac{1}{2} \\cdot \\frac{1}{a_{i}+1} \\text{ or } \\frac{1}{a_{i+1}+1}=\\frac{a_{i}+2}{2\\left(a_{i}+1\\right)}=\\frac{1}{2} \\cdo... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 60 | |
0blx | Find all triples $(a, b, c)$ of non-zero complex numbers with equal absolute values, for which
$$
\frac{a}{b} + \frac{b}{c} + \frac{c}{a} + 1 = 0.
$$ | [
"Observe that $\\overline{\\left(\\frac{a}{b}\\right)} = \\frac{b}{a}$, hence, taking conjugates yields\n$$\n\\frac{b}{a} + \\frac{c}{b} + \\frac{a}{c} + 1 = 0.\n$$\nClearing denominators and adding up gives\n$$\na^2 b + b^2 c + c^2 a + ab^2 + bc^2 + ca^2 + 2abc = 0,\n$$\nwhich factors as $(a+b)(b+c)(c+a) = 0$, the... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | All triples that are permutations of t, t, minus t where t is any nonzero complex number. | |
030s | Problem:
Se consideră două cercuri $\mathcal{C}_1$ și $\mathcal{C}_2$ tangente interior în punctul $P$ (cercul $\mathcal{C}_2$ este interior cercului $\mathcal{C}_1$).
O coardă $AB$ din cercul $\mathcal{C}_1$ este tangentă cercului $\mathcal{C}_2$ în punctul $C$. Fie $D$ al doilea punct de intersectie dintre dreapta $... | [] | Brazil | Al treilea baraj de selecție pentru OBMJ | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fox | Determine all integers $n \ge 1$ for which the number $n^8 + n^6 + n^4 + 4$ is prime. | [
"We have\n$$\nn^8 + n^6 + n^4 + 4 = (n^4 - n^3 + n^2 - 2n + 2)(n^4 + n^3 + n^2 + 2n + 3) = f(n)g(n)\n$$\nThe first factor $f(n)$ satisfies\n$$\nf(n) = n^4 - n^3 + n^2 - 2n + 2 = n^3(n-1) + (n-1)^2 + 1\n$$\nand hence $f(n) \\ge 2$ for all $n \\ge 2$. The second factor $g(n) = n^4 + n^3 + n^2 + 2n + 3$ is strictly gr... | Spain | MEDITERRANEAN MATHEMATICAL COMPETITION | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Spanish | proof and answer | n = 1 | |
0fxz | Problem:
Finde alle injektiven Funktionen $f: \mathbb{N} \rightarrow \mathbb{N}$, sodass für alle natürlichen Zahlen $n$ gilt
$$
f(f(n)) \leq \frac{f(n)+n}{2}
$$ | [
"Solution:\nWir zeigen, dass die Identität $f(n)=n$ die einzige solche Funktion ist. Nehme an, es gelte $f(n)<n$ für eine natürliche Zahl $n$. Dann folgt\n$$\nf(f(n)) \\leq \\frac{f(n)+n}{2}<n\n$$\nund eine einfache Induktion zeigt\n$$\nf^{k}(n)<n \\quad \\forall n, k \\in \\mathbb{N}\n$$\ndabei bezeichet $f^{k}=f ... | Switzerland | SMO Finalrunde | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(n) = n for all n | |
06j6 | Let $n \ge 2$ be an integer. There are $n$ distinct circles on the plane such that any two circles have two distinct intersections and no three circles have a common intersection. Initially there is a coin on each of the intersection points of the circles. Starting from $X$, players $X$ and $Y$ alternately take away a ... | [
"$Y$ has a winning strategy if and only if $n \\ge 4$.\nFor $n = 2, 3$, after $X$ takes away any coin, $Y$ cannot take away any other coin. Therefore, $Y$ does not have a winning strategy.\nFor $n \\ge 4$, we claim that $Y$ has a winning strategy. Since the total number $2\\binom{n}{2}$ of coins is even, it suffice... | Hong Kong | CHKMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Geometry > Plane Geometry > Circles"
] | null | proof and answer | n ≥ 4 | |
0d69 | Let $ABC$ be an acute, non-isosceles triangle which is inscribed in a circle $(O)$. A point $I$ belongs to the segment $BC$. Denote by $H$ and $K$ the projections of $I$ on $AB$ and $AC$, respectively. Suppose that the line $HK$ intersects $(O)$ at $M, N$ ($H$ is between $M, K$ and $K$ is between $H, N$). Let $X, Y$ be... | [
"1) We will use the inversion to solve this problem.\nNote that $\\triangle AHI \\sim \\triangle AIB$ and that $\\triangle AKI \\sim \\triangle AIC$, hence\n$$\nAI^{2} = AH \\cdot AB = AK \\cdot AC =: k.\n$$\nLet $f$ be the inversion with center $A$ and power $k$. Then\n$$\nf(I) = I,\\ f(H) = B,\\ f(B) = H,\\ f(K) ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0k8a | Problem:
You are trying to cross a 400 foot wide river. You can jump at most 4 feet, but you have many stones you can throw into the river. You will stop throwing stones and cross the river once you have placed enough stones to be able to do so. You can throw straight, but you can't judge distance very well, so each s... | [
"$$\n\\frac{100}{100}+\\frac{100}{99}+\\cdots+\\frac{100}{1} \\approx 100 \\ln 100\n$$\nstone throws (it takes $\\frac{100}{100-k}$ moves on average to get a stone into a new section if $k$ sections already have a stone). So the answer is at least $100 \\ln 100 \\approx 450$.\n\nOn the other hand, if we divide the ... | United States | HMMT November 2019 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | final answer only | approximately 712.811 | |
0gks | Let $\omega$ be the incircle of a triangle $ABC$ with $\omega$ tangent to $BC$ and $AC$ at the points $D$ and $E$, respectively. Draw a line perpendicular to $BC$ at $D$ to meet the circle $\omega$ at a point $P$ closer to $A$. The line $AP$ meets $BC$ at $M$. Let $N$ be a point on the segment $AC$ such that $AE = CN$,... | [
"From the point $P$, draw a line parallel to $BC$ meeting the lines $AB, AC$ at the points $B', C'$, respectively.\n\n\nClearly, the line $B'C'$ is externally tangent to the circle $\\omega$ at the point $P$. Let $Y, Z$ be two points on the line $BC$ and let $Y', Z'$ be the two intersecting... | Thailand | The 10th Thailand Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Transformatio... | null | proof only | null | |
06t4 | Let $ABC$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $AC$ and $AB$, respectively, and let $M$ be the midpoint of $EF$. Let the perpendicular bisector of $EF$ intersect the line $BC$ at $K$, and let the perpendicular bisector of $MK$ intersect the lines $AC$ and $AB$ at $S$ and... | [
"Solution 1. For any interesting pair $(E, F)$, we will say that the corresponding triangle $EFK$ is also interesting.\nLet $EFK$ be an interesting triangle. Firstly, we prove that $\\angle KEF = \\angle KFE = \\angle A$, which also means that the circumcircle $\\omega_1$ of the triangle $AEF$ is tangent to the lin... | IMO | 55th International Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > C... | null | proof only | null | |
03f2 | Find all pairs $(x, y)$ of real numbers for which
$$
4y^4 + x^4 + 12y^3 + 5x^2(y^2 + 1) + y^2 + 4 = 12y.
$$ | [
"We have the inequalities $x^4 \\geq 0$, $5x^2(y^2+1) \\geq 0$ and $4y^4 + 12y^3 + y^2 - 12y + 4 = (2y-1)^2(y+2)^2 \\geq 0$. The sum of the left sides is 0 if and only if each of them is equal to 0. The first two lead to $x = 0$, and the third to $y = -2$ or $y = \\frac{1}{2}$. $\\square$"
] | Bulgaria | Bulgarian Spring Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | (0, -2) and (0, 1/2) | |
0jba | Problem:
Let $\pi$ be a permutation of the numbers from $2$ through $2012$. Find the largest possible value of $\log_{2} \pi(2) \cdot \log_{3} \pi(3) \cdots \log_{2012} \pi(2012)$. | [
"Solution:\nNote that\n$$\n\\begin{aligned}\n\\prod_{i=2}^{2012} \\log_{i} \\pi(i) &= \\prod_{i=2}^{2012} \\frac{\\log \\pi(i)}{\\log i} \\\\\n&= \\frac{\\prod_{i=2}^{2012} \\log \\pi(i)}{\\prod_{i=2}^{2012} \\log i} \\\\\n&= 1\n\\end{aligned}\n$$\nwhere the last equality holds since $\\pi$ is a permutation of the ... | United States | HMMT November | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 1 | |
0gnd | Let $A_1, B_1, C_1$ be the midpoints of the sides $BC, CA, AB$, respectively, of an acute triangle $ABC$ with orthocenter $H$ and circumcenter $O$. The rays $HA_1, HB_1, HC_1$ cut the circumcircle at the points $A_0, B_0, C_0$, respectively. Show that $O, H,$ and $H_0$ are collinear where $H_0$ is the orthocenter of $A... | [
"Since $H$ is the orthocenter, $\\angle BHC = 180^\\circ - \\angle BAC = \\angle BA_0C$. As $A_1$ is the midpoint of $BC$, this is possible only if $BHCA_0$ is a parallelogram. Then $\\angle ACA_0 = 90^\\circ$, and $AA_0$ is a diameter of the circumcircle of $ABC$. Therefore, the reflection across $O$ takes $ABC$ t... | Turkey | 16th Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and Collinea... | English | proof only | null | |
061j | Problem:
Man beweise: Sind $x$, $y$, $z$ die Längen der Winkelhalbierenden eines Dreiecks mit dem Umfang $6$, dann gilt
$$
\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}} \geq 1
$$ | [
"Solution:\n\nMan beachte nebenstehende Figur.\nDP und DQ sind die Parallelen durch $D$ zu $AB$ und $AC$. Da $AD = x$ die Winkelhalbierende von $\\alpha$ ist, ist $AQDP$ eine Raute, deren Seitenlänge mit $u$ bezeichnet wurde.\nAus der Ähnlichkeit der Dreiecke $PDC$ mit $QBD$ folgt $\\frac{u}{c-u} = \\frac{b-u}{u}$,... | Germany | Auswahlwettbewerb zur IMO 2002 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0g0o | Problem:
Déterminer tous les entiers naturels $n$ tels que pour chaque diviseur positif $d$ de $n$ on ait
$$
d+1 \mid n+1
$$ | [
"Solution:\n\nPour tout nombre naturel $n$, $1$ est un diviseur de $n$ donc nous avons $2 \\mid n+1$. Ainsi $n$ doit être impair.\n\nSi $n=1$, $d=1$ est le seul diviseur de $n$ et clairement $2 \\mid 2$ donc $1$ est solution.\n\nSinon soit $p$ le plus petit facteur premier de $n$. Comme $p$ divise $n$ nous avons au... | Switzerland | SMO - Vorrunde | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | n = 1 or n is an odd prime | |
05gd | Problem:
Soit $ABC$ un triangle isocèle en $A$. Soient $M$ et $N$ deux points de $[BC]$. Les droites $[AM]$ et $[AN]$ recoupent le cercle circonscrit à $ABC$ en $P$ et $Q$.
1) Montrer que $M, N, P, Q$ sont cocycliques.
2) Soient $R_{1}$ et $R_{2}$ les rayons des cercles circonscrits à $BMP$ et $CMP$. Calculer $R_{1}... | [
"Solution:\n\n\n\n1) $\\widehat{PMN}=\\widehat{PMC}=\\pi-\\widehat{MCP}-\\widehat{CPM}=\\pi-\\widehat{BCP}-\\widehat{CPA}=\\pi-\\widehat{BCP}-\\widehat{CBA}$, et $\\widehat{NQP}=\\widehat{AQP}=\\widehat{ACP}=\\widehat{ACB}+\\widehat{BCP}=\\widehat{CBA}+\\widehat{BCP}$,\ndonc $\\widehat{PMN}... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | R1 + R2 = (AB × BC) / (2 √(AB^2 − (BC/2)^2)) | |
0kiv | Problem:
Let $X_{0}$ be the interior of a triangle with side lengths $3$, $4$, and $5$. For all positive integers $n$, define $X_{n}$ to be the set of points within $1$ unit of some point in $X_{n-1}$. The area of the region outside $X_{20}$ but inside $X_{21}$ can be written as $a \pi + b$, for integers $a$ and $b$. C... | [
"Solution:\n\n$X_{n}$ is the set of points within $n$ units of some point in $X_{0}$. The diagram above shows $X_{0}$, $X_{1}$, $X_{2}$, and $X_{3}$. As seen above it can be verified that $X_{n}$ is the union of\n- $X_{0}$,\n- three rectangles of height $n$ with the sides of $X_{0}$ as base... | United States | HMMT Spring 2021 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 4112 | |
08iu | Problem:
Let $n \geq 1$ be a positive integer. A square table of dimensions $n \times n$ is filled arbitrarily with the numbers $1, 2, \ldots, n^2$, so that every number appears exactly once in the table. From each row, select the smallest number, and let the greatest of them be denoted by $x$. From each column, selec... | [] | JBMO | The third selection test for IMO 2003 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | n^2 · ((n−1)!)^2 · ((n−1)^2)! · C(n^2, 2n−1) | |
00nj | Let $ABCD$ be an inscribed convex quadrilateral with diagonals $AC$ and $BD$. Each of the four vertices is reflected on the diagonal it does not lie on.
Prove that the resulting four points lie on a common circle or a common line.
a. Investigate when the four resulting points lie on a common line and give a simple equ... | [
"a. We denote the reflections of $A$, $B$, $C$ and $D$ with $A'$, $B'$, $C'$ resp. $D'$ and we denote the intersection of the diagonals with $S$. Since the points $A$ and $C$ are reflected in the same line $BD$ and the point $S$ remains invariant under this reflection, the whole line $ASC$ becomes $A'SC'$ after ref... | Austria | Austrian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Mi... | English | proof and answer | The four reflected points are collinear if and only if the angle between the diagonals is sixty degrees; in all other cases the four points are concyclic. | |
0fdw | Problem:
En una reunión entre cuatro países de la ONU, digamos $A$, $B$, $C$ y $D$, el país $A$ tiene el doble de representantes que el $B$, el triple que el $C$, y el cuádruple que el $D$. Se pretende distribuir a los representantes en mesas con el mismo número de personas en cada una. Sólo hay una condición: en cada... | [
"Solution:\n\nLa respuesta es $25$. Veamos la demostración. Sean $a$, $b$, $c$ y $d$ el número de representantes de cada país. Como $a$ debe ser múltiplo de $3$ y de $4$, también debe ser múltiplo de $12$. Por tanto, existe un número $k$ tal que $a=12k$, luego $b=6k$, $c=4k$ y $d=3k$. El número total de representan... | Spain | XLVII Olimpiada Matemática Española Primera Fase | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 25 | |
08ac | Problem:
Sia $ABC$ un triangolo acutangolo. Siano $AM$, $BN$ e $CL$ le mediane, che si intersecano nel baricentro $G$. Siano $M'$, $N'$ e $L'$ i punti medi di $AG$, $BG$ e $CG$, rispettivamente. Mostrare che i sei punti $M$, $M'$, $N$, $N'$, $L$, $L'$ giacciono su una circonferenza se e solo se $ABC$ è equilatero. | [
"Solution:\n\nOsserviamo che prendendo in considerazione la mediana $AM$, abbiamo che i tre segmenti $AM'$, $M'G$ e $GM$ sono uguali tra loro, poiché il baricentro divide la mediana in due segmenti uno il doppio dell'altro ed $M'$ è per costruzione il punto medio del segmento più lungo dei due, $AG$. Un analogo ris... | Italy | Progetto Olimpiadi della Matematica - Gara di Febbraio | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
05gb | Problem:
Soit $ABC$ un triangle dont $O$ est le centre du cercle circonscrit. On note $\ell$ une droite perpendiculaire à la droite $(AO)$. La droite $(\ell)$ intersecte les côtés $(AB)$ et $(AC)$ en les points $D$ et $E$. Montrer que les points $B$, $C$, $E$ et $D$ sont cocycliques. | [
"Solution:\n\n\n\nPosons $X$ et $Y$ les projetés orthogonaux de $O$ sur $(AC)$ et $(DE)$ respectivement. Notons que $X$ est le milieu du segment $[AC]$ et aussi le pied de la bissectrice de l'angle $\\widehat{COA}$ dans le triangle $AOC$ isocèle en $O$.\n\nComme $\\widehat{EXO} = 90^{\\circ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08pd | Problem:
Let $[AB]$ be a chord of a circle $(c)$ centered at $O$, and let $K$ be a point on the segment $(AB)$ such that $AK < BK$. Two circles through $K$, internally tangent to $(c)$ at $A$ and $B$, respectively, meet again at $L$. Let $P$ be one of the points of intersection of the line $KL$ and the circle $(c)$, a... | [
"Solution:\n\nLet $(c_1)$ and $(c_2)$ be circles through $K$, internally tangent to $(c)$ at $A$ and $B$, respectively, and meeting again at $L$, and let the common tangent to $(c_1)$ and $(c)$ meet the common tangent to $(c_2)$ and $(c)$ at $Q$. Then the point $Q$ is the radical center of the circles $(c_1)$, $(c_... | JBMO | Junior Balkan Mathematics Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Mis... | null | proof only | null | |
01r0 | Consider all polynomials $P(x)$ with real coefficients that have the following property: for all real $x$ and $y$ one has
$$
|y^2 - P(x)| \le 2x \iff |x^2 - P(y)| \le 2|y|.
$$
Determine all possible values of $P(0)$. | [
"3. See IMO-2014 Shortlist, Problem A5."
] | Belarus | SELECTION and TRAINING SESSION | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (-∞, 0) | |
0i6y | Problem:
The real numbers $x, y, z, w$ satisfy
\begin{aligned}
2x + y + z + w &= 1 \\
x + 3y + z + w &= 2 \\
x + y + 4z + w &= 3 \\
x + y + z + 5w &= 25.
\end{aligned}
Find the value of $w$. | [
"Solution:\n\n$11/2$. Multiplying the four equations by $12, 6, 4, 3$ respectively, we get\n\n\\begin{aligned}\n24x + 12y + 12z + 12w &= 12 \\\\\n6x + 18y + 6z + 6w &= 12 \\\\\n4x + 4y + 16z + 4w &= 12 \\\\\n3x + 3y + 3z + 15w &= 75\n\\end{aligned}\n\nAdding these yields $37x + 37y + 37z + 37w = 111$, or $x + y + z... | United States | Harvard-MIT Math Tournament | [
"Algebra > Linear Algebra > Vectors",
"Algebra > Linear Algebra > Matrices"
] | null | final answer only | 11/2 | |
0h3m | Two players play the following game. They start with a pile of $2012$ pebbles and take some amounts of pebbles by turns. The player starts can take $1$ or $4$ pebbles each turn at his discretion, another player can take $1$ or $3$ each turn. The player unable to make his turn loses. Which player has a winning strategy? | [
"Перший гравець може забезпечити собі перемогу, узявши першим ходом чотири камінці, а всіма наступними — по одному камінцю. Тоді кількість узятих загалом камінців буде парною лише після ходів першого гравця, а перемагає той гравець, після ходу якого ця величина стане рівною $2012$.\n\n*Відповідь:* Перший гравець."
... | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | First player | |
0dfu | Let the sequence $a_1, a_2, \dots, a_n$ is such that $a_1 = 0$, $|a_2| = |a_1 + 1|$, $|a_3| = |a_2 + 1|$, $\dots$, $|a_n| = |a_{n-1} + 1|$. Prove that
$$
\frac{a_1 + a_2 + \dots + a_n}{n} \geq -\frac{1}{2}.
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
04ww | Find all polynomials $P$ with real coefficients for which the equality
$$
P(x^2) = P(x) \cdot P(x+2)
$$
holds for every real number $x$. | [
"The constant polynomial $P(x) = c$ is a solution if and only if $c = c^2$, thus the polynomials $P(x) = 0$ and $P(x) = 1$ are solutions of the problem.\nWe claim that the only polynomial of a positive degree $n$ which solves the equation is of the form $P(x) = (x-1)^n$. In view of the identity $(x^2 - 1)^n = (x-1)... | Czech-Polish-Slovak Mathematical Match | Czech-Slovak-Polish Match | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | P(x) = 0, P(x) = 1, or P(x) = (x - 1)^n for any positive integer n | |
0hq4 | Problem:
The cafeteria in a certain laboratory is open from noon until 2 in the afternoon every Monday for lunch. Two professors eat 15 minute lunches sometime between noon and 2. What is the probability that they are in the cafeteria simultaneously on any given Monday? | [
"Solution:\n\n$\\frac{15}{64}$."
] | United States | null | [
"Statistics > Probability > Counting Methods > Other",
"Math Word Problems"
] | null | final answer only | 15/64 | |
0gym | In triangle $ABC$ $\angle ABC = 120^\circ$. The bisector of this angle intersects side $AC$ at $M$, and the bisector of angle adjacent to $\angle BCA$, intersects line $AB$ at $P$. Segment $MP$ intersects side $BC$ in $K$. Prove that $\angle AKM = \angle KPC$. | [
"Let us prove that $AK$ is the bisector of $\\angle BAM$. Point $P$ is equidistant from the lines $BC$ and $AC$, and also from the lines $BM$ and $BC$, since $\\angle ABM = \\angle MBC = \\angle CBP = 60^\\circ$ (fig. 13). Thus $P$ is equidistant from the lines $BM$ and $MC$, which means that $P$ belongs to the bis... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0a78 | Problem:
Let $m$, $n$, and $p$ be odd positive integers. Prove that the number
$$
\sum_{k=1}^{(n-1)^{p}} k^{m}
$$
is divisible by $n$. | [
"Solution:\nSince $n$ is odd, the sum has an even number of terms. So we can write it as\n$$\n\\sum_{k=1}^{\\frac{1}{2}(n-1)^{p}}\\left(k^{m}+\\left((n-1)^{p}-k+1\\right)^{m}\\right)\n$$\nBecause $m$ is odd, each term in the sum has $k+(n-1)^{p}-k+1=(n-1)^{p}+1$ as a factor. As $p$ is odd, too, $(n-1)^{p}+1=(n-1)^{... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 4 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0l5z | Let $A_1A_2\cdots A_{2025}$ be a convex 2025-gon, and let $A_i = A_{i+2025}$ for all integers $i$. Distinct points $P$ and $Q$ lie in its interior such that $\angle A_{i-1}A_iP = \angle QA_iA_{i+1}$ for all $i$. Define points $P_i^j$ and $Q_i^j$ for integers $i$ and positive integers $j$ as follows:
* For all $i$, $P_i... | [] | United States | TST2025 | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"G... | null | proof only | null | |
0c0b | For every integer $n \ge 2$, let $B_n$ denote the set of all binary $n$-tuples of zeroes and ones, and split $B_n$ into equivalence classes by letting two $n$-tuples be *equivalent* if one is obtained from the other by a cyclic permutation of the entries. Determine the integers $n \ge 2$ for which $B_n$ splits into an ... | [
"Only $B_2$ splits into an odd number of equivalence classes, namely, three: $\\{(0,0)\\}$, $\\{(0,1), (1,0)\\}$ and $\\{(1,1)\\}$. If $n > 2$, then $B_n$ always splits into an even number of classes, as we are presently going to show.\n\nCall two $n$-tuples of $B_n$ conjugate if one is obtained from the other by r... | Romania | 69th NMO Selection Tests for BMO and IMO | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | n = 2 only | |
0jbm | Problem:
Mr. Canada chooses a positive real $a$ uniformly at random from $(0,1]$, chooses a positive real $b$ uniformly at random from $(0,1]$, and then sets $c = a / (a + b)$. What is the probability that $c$ lies between $1/4$ and $3/4$? | [
"Solution:\n\nAnswer: $2/3$\n\nFrom $c \\geq 1/4$ we get\n$$\n\\frac{a}{a+b} \\geq \\frac{1}{4} \\Longleftrightarrow b \\leq 3a\n$$\nand similarly $c \\leq 3/4$ gives\n$$\n\\frac{a}{a+b} \\leq \\frac{3}{4} \\Longleftrightarrow a \\leq 3b\n$$\n\nChoosing $a$ and $b$ randomly from $[0,1]$ is equivalent to choosing a ... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 2/3 | |
02zw | Problem:
O quadrado $A B C D$ de lado $1~\mathrm{cm}$ está inscrito em uma circunferência de centro $O$. O ponto $M$ está sobre o arco $B C$, o segmento $A M$ encontra $B D$ no ponto $P$, o segmento $D M$ encontra $A C$ no ponto $Q$.
a) Verifique que $\angle A Q D=\angle P A D$.
b) Encontre a... | [
"Solution:\n\na) Como $\\angle A M D=45^{\\circ}=\\angle O A D=\\angle O D A$, temos\n$$\n\\begin{aligned}\n\\angle A Q D & =\\angle A M D+\\angle M A Q \\\\\n& =45^{\\circ}+\\angle M A Q \\\\\n& =\\angle O A D+\\angle M A Q \\\\\n& =\\angle P A D\n\\end{aligned}\n$$\n\nb) De modo semelhante ao item anterior, podem... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterillaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 1/2 cm^2 | |
08m6 | Problem:
A parallelogram $ABCD$ with obtuse angle $\angle ABC$ is given. After rotating the triangle $ACD$ around the vertex $C$, we get a triangle $CD' A'$, such that points $B$, $C$ and $D'$ are collinear. The extension of the median of triangle $CD' A'$ that passes through $D'$ intersects the straight line $BD$ at p... | [
"Solution:\nLet $AC \\cap BD = \\{X\\}$ and $PD' \\cap CA' = \\{Y\\}$. Because $AX = CX$ and $CY = YA'$, we deduce:\n$$\n\\triangle ABC \\cong \\triangle CDA \\cong \\triangle CD' A' \\Rightarrow \\triangle ABX \\cong \\triangle CD' Y, \\triangle BCX \\cong \\triangle D'A' Y\n$$\nIt follows that\n$$\n\\angle ABX = ... | JBMO | 2009 Shortlist JBMO | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0iau | Problem:
How many ordered pairs of integers $(a, b)$ satisfy all of the following inequalities?
$$
\begin{aligned}
a^{2}+b^{2} & <16 \\
a^{2}+b^{2} & <8a \\
a^{2}+b^{2} & <8b
\end{aligned}
$$ | [
"Solution:\nThis is easiest to see by simply graphing the inequalities. They correspond to the (strict) interiors of circles of radius $4$ and centers at $(0,0)$, $(4,0)$, $(0,4)$, respectively. So we can see that there are $6$ lattice points in their intersection (circled in the figure).\n... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 6 | |
0gk3 | Let $n$ be a positive integer for which $5n + 1$ is a perfect square. Show that $n + 1$ is a sum of 5 perfect squares. | [
"Let $5n + 1 = m^2 \\equiv 1 \\pmod{5}$. Thus $m = 5k \\pm 1$ for some integer $k$. We have\n$$\nn + 1 = \\frac{(5k \\pm 1)^2 + 4}{5} = 5k^2 \\pm 2k + 1 = 4k^2 + (k \\pm 1)^2\n$$\nwhich can be written as a sum of 5 perfect squares as desired."
] | Thailand | Thai Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0e6t | We multiplied a number whose cube is equal to $2012^{12}$ by the square of $2012^{11}$. Which number did we get?
(A) $2012^{58}$ (B) $2012^{26}$ (C) $2012^{88}$ (D) $2012^{15}$ (E) $2012^{12}$ | [
"The number $2012^4$ was multiplied by $(2012^{11})^2$, hence we got $2012^4 \\cdot 2012^{22} = 2012^{26}$."
] | Slovenia | National Math Olympiad 2012 | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | MCQ | B | |
0c89 | A nonempty, finite set $A$ of positive integers is called *quadratic* if the sum of the elements of $A$ equals the square of the number of elements of $A$.
For example $A = \{1, 3, 4, 8\}$ is *quadratic* since $1 + 3 + 4 + 8 = 4^2$.
a) Give an example of a *quadratic* set with 20 elements.
b) Prove that every *quadra... | [
"a) An example is: $A = \\{1, 2, 3, \\ldots, 19\\} \\cup \\{210\\}$.\nThe sum of elements is $1 + 2 + \\ldots + 19 + 210 = \\frac{19 \\cdot 20}{2} + 210 = 400 = 20^2$.\n\nb) We argue by contradiction. Suppose $A$ is a *quadratic* set containing $n$ even positive integers. The sum of the elements of $A$ is at least ... | Romania | RMC 2020 | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | {1, 2, 3, ..., 19, 210} |
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