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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
01sh | Let $I$ be the incenter of a triangle $ABC$. Let $A_1$, $B_1$, $C_1$ be the tangency points of the incircle of the triangle $ABC$ with the sides $BC$, $CA$, $AB$, respectively. Let the circumcircle of the triangle $BC_1B$ intersect the line $BC$ at points $B$ and $K$, and the circumcircle of the triangle $CB_1C_1$ inte... | [
"Let $X$ be the intersection point of the lines $LC_1$ and $KB_1$. To prove the problem statement it suffices to show that points $X$, $I$ and $A_1$ lie on the same line. Let $\\angle CAB = \\alpha$, $\\angle ABC = \\beta$, $\\angle BCA = \\gamma$. Since $LC_1B_1C$ is an inscribed quadrilateral, we have\n$$\n\\angl... | Belarus | FINAL ROUND | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
02j6 | Problem:
Resolva geometricamente as equações:
a. $|x-5|=2$
b. $|x+3|=1$
c. $|3x-7|=9$
d. $|x+2|=|x-5|$ | [
"Solution:\n\n## Interpretação geométrica\n$|a-b|=$ distância entre $a$ e $b$\n\na. $|x-5|=2 \\Leftrightarrow$ números cuja distância ao $5$ é $2$.\nLogo as raízes são $3$ e $7$.\n\n\n\nb. $|x+3|=1 \\Leftrightarrow$ números cuja distância ao $-3$ é $1$.\nLogo as raízes são $-4$ e $-2$.\n\n!... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | {'a': [3, 7], 'b': [-4, -2], 'c': [-2.0, 16.0], 'c_simplified': [-0.6666666667, 5.3333333333], 'c_exact': ['-2/3', '16/3'], 'd': '3/2'} | |
02uq | Problem:
O professor Carlão decidiu fazer uma questão de matemática que vale no total 10 pontos e possui três itens: $a, b$ e $c$. Após elaborar os itens, ele ficou na dúvida sobre qual a melhor maneira de distribuir os 10 pontos entre os itens de modo que cada um valha um número inteiro positivo de pontos.
a) Joana,... | [
"Solution:\n\na) Se Carlão seguir a sugestão de Joana o item $c$ valerá 5 pontos e os itens $a$ e $b$ devem somar outros 5 pontos. Teremos então quatro divisões possíveis de itens $(a, b, c): (1,4,5), (2,3,5), (3,2,5)$ e $(4,1,5)$.\n\nb) Uma vez definidas as pontuações dos itens $a$ e $b$, o item $c$ valerá $10-a-b... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a) 4; b) 36 | |
0jeg | Does there exist a pair $(g, h)$ of functions $g, h : \mathbb{R} \to \mathbb{R}$ such that the only function $f : \mathbb{R} \to \mathbb{R}$ satisfying $f(g(x)) = g(f(x))$ and $f(h(x)) = h(f(x))$ for all $x \in \mathbb{R}$ is the identity function $f(x) = x$?
(This problem was suggested by Alexander Betts from the Unit... | [
"**Solution 1.** We claim that $g(x) = 2x$, $h(x) = \\lfloor x \\rfloor - 1$ works. We must show that $f(x) = x$ is the only solution to the system of equations\n$$\n2f(x) = f(2x)\n$$\n$$\nf(\\lfloor x \\rfloor - 1) = \\lfloor f(x) \\rfloor - 1.\n$$\nfor $f(x)$. First, note that $2f(0) = f(0)$, so $f(0) = 0$. Also,... | United States | RMM | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Other"
] | null | proof and answer | Yes. For example, g(x) = 2x and h(x) = floor(x) − 1; then the only f with f(g(x)) = g(f(x)) and f(h(x)) = h(f(x)) for all real x is f(x) = x. | |
0hr1 | Problem:
Show that there exist infinitely many triples of positive integers $x, y, z$ which satisfy $x^{999} + y^{1000} = z^{1001}$. | [
"Solution:\nWe will choose $x$, $y$, and $z$ to be powers of $2$ such that the terms $x^{999}$ and $y^{1000}$ are equal and $z^{1001}$ is their sum. Writing $x = 2^{a}$, $y = 2^{b}$, $z = 2^{c}$, we have the conditions\n$$\n999a = 1000b = 1001c - 1.\n$$\nThe first equation suggests that we try $a = 1000d$, $b = 999... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof only | null | |
0b4e | Problem:
Bryce has $7$ blue socks and $7$ red socks mixed in a drawer. He plays a game with Sean. Blindfolded, Bryce takes two socks from the drawer. Sean looks at the socks, and if they have the same color, Sean gives Bryce $1$ point. Bryce keeps drawing socks until the drawer is empty, at which time the game ends. T... | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 613 | |
0g63 | 正三角形 $XYZ$ 之三頂點 $X, Y, Z$ 分別落在一銳角三角形 $ABC$ 之三邊 $BC, CA$ 與 $AB$ 上。
試證: 三角形 $ABC$ 之內心落在三角形 $XYZ$ 之內部。 | [
"底下證明一個較強的結果：三角形 $ABC$ 之內心 $I$ 落在三角形 $XYZ$ 的內切圓內。記 $d(U, VW)$ 表點 $U$ 至線 $VW$ 之距離。\n令 $O$ 為 $\\triangle XYZ$ 之內心且 $r, r'$ 與 $R', R''$ 分別為 $\\triangle ABC$, $\\triangle XYZ$ 之內切圓半徑與 $\\triangle XYZ$ 之外接圓半徑。則 $R' = 2r'$，且所求等價於證不等式 $OI \\le r'$。假設 $O \\ne I$（若 $O = I$ 顯然成立）。\n令 $\\triangle ABC$ 之內切圓分別與邊 $BC, AC, AB$ 相切... | Taiwan | 二〇一一數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0l93 | Find all polynomials $P(x)$ with integer coefficients such that the polynomial
$$
Q(x) = (x^2 + 6x + 10)[P(x)]^2 - 1
$$
is the square of a polynomial with integer coefficients. | [] | Vietnam | CONTEST FOR THE SELECTION OF VIETNAMESE INTERNATIONAL MATHEMATICAL OLYMPIAD TEAM | [
"Number Theory > Diophantine Equations > Pell's equations"
] | English | proof and answer | Let y = x + 3. The condition is equivalent to finding integer polynomials R, P with R^2 − (y^2 + 1) P^2 = −1. All solutions are obtained from the odd powers of the fundamental solution y + sqrt{y^2 + 1}: for each n ≥ 0,
(y + sqrt{y^2 + 1})^{2n+1} = R_n(y) + P_n(y) sqrt{y^2 + 1},
with R_n, P_n ∈ Z[y]. Then Q(x) = (x^2... | |
06eg | Let $AD$, $BE$ and $CF$ be respectively the altitudes of $\triangle ABC$. Let $P$, $Q$ and $R$ be points on the lines $BC$, $CA$ and $AB$ respectively such that $AP$ is perpendicular to $EF$, $BQ$ is perpendicular to $FD$ and $CR$ is perpendicular to $DE$. Show that $AP$, $BQ$ and $CR$ are concurrent. | [
"By a corollary of Carnot's theorem, the perpendicular lines drawn from $A$, $B$, $C$ to $EF$, $FD$, $DE$ are concurrent if and only if the perpendicular lines drawn from $D$, $E$, $F$ to $BC$, $CA$, $AB$ are concurrent. The latter is clearly true since the perpendicular lines are just $AD$, $BE$, $CF$, which are c... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem"
] | null | proof only | null | |
0fx6 | Problem:
Sei $ABC$ ein Dreieck und $D$ ein Punkt im Innern der Strecke $BC$. Sei $X$ ein weiterer Punkt im Innern der Strecke $BC$ verschieden von $D$ und sei $Y$ der Schnittpunkt von $AX$ mit dem Umkreis von $ABC$. Sei $P$ der zweite Schnittpunkt der Umkreise von $ABC$ und $DXY$. Beweise, dass $P$ unabhängig von der ... | [
"Solution:\n\nSei $A'$ die Spiegelung von $A$ an der Mittelsenkrechten von $BC$. Die Gerade $AA'$ ist dann parallel zu $BC$ und $A'$ liegt auf dem Umkreis von $ABC$. Sei $P'$ der zweite Schnittpunkt von $A'D$ mit dem Umkreis von $ABC$. Nun gilt\n$$\n\\angle XYP' = \\angle AYP' = \\angle AA'P' = \\angle CDP' = 180^\... | Switzerland | IMO Selektion 2008 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0395 | Alexander and Denitza play the following game. Alexander cuts (if possible) a band of positive integer length to three bands of positive integer lengths such that the largest band is unique. Then Denitza cuts (if possible) the largest band in the same way and so on. The winner is the one who makes the last move. Consid... | [
"Consider a band of length $n$. It is clear that no move is possible for $n = 1, 2, 3$. For $4 \\le n \\le 7 = 3+2+2$ Alexander has a winning move. For $n = 8$ and $9$ after the first move of Alexander the largest length is between $4$ and $7$ and then Denitza has a winning move. Similarly, for $10 \\le n \\le 9+8+... | Bulgaria | Winter Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalitie... | English | proof and answer | Denitza has a winning strategy exactly for band lengths n equal to 3^k or 3^k − 1 with k > 1. Among lengths of the form a^b with a ≥ 2 and b ≥ 2, this occurs precisely when either a = 3 and b > 1 (so n = 3^b), or a = 2 and b = 3 (so n = 8). | |
07q3 | Let $k > 3$ be an integer and suppose a doubly infinite sequence of real numbers $\ldots$, $a_{-2}$, $a_{-1}$, $a_0$, $a_1$, $a_2$, $\ldots$ has the property that
$$
a_{n+k} = \frac{1}{k} \sum_{j=0}^{k-1} a_{n+j}, \quad \text{for all integers } n.
$$
If this sequence is bounded, does it follow it is constant? | [] | Ireland | Ireland | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | Yes | |
0kvo | Find all positive integers $n$ for which it is possible to color some cells of an infinite grid of unit squares red, such that each rectangle consisting of exactly $n$ cells (and whose edges lie along the lines of the grid) contains an odd number of red cells. | [
"*Proof.* This can be achieved as follows: assuming the cells are labeled with $(x, y) \\in \\mathbb{Z}^2$, color a cell red if $x \\equiv 0 \\pmod{2^m}$ and $y \\equiv 0 \\pmod{2^{k-m}}$. For example, a $4 \\times 2$ rectangle gets the following coloring:\n\n\n\nA $2^m \\times 2^{k-m}$ rec... | United States | USA TSTST | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof and answer | All n that are powers of two; that is, n = 2^k for some nonnegative integer k. | |
0gej | 設正整數 $n \ge 6$。平面上有 $n$ 個兩兩互斥的圓盤 $D_1, D_2, \dots, D_n$, 其半徑依序為 $R_1 \ge R_2 \ge \dots \ge R_n$。對每一個 $i = 1, 2, \dots, n$, 在 $D_i$ 中都標了一個點 $P_i$。設 $O$ 為平面上的任意點。試證：
$$
\sum_{i=1}^{n} OP_i \ge \sum_{j=6}^{n} R_j
$$
(註：這裡的圓盤都包含其邊界。)
In the plane, there are $n \ge 6$ pairwise disjoint disks $D_1, D_2, \dots, D_n$ with rad... | [
"We will make use of the following lemma.\n\n*Lemma.* Let $D_1, \\dots, D_6$ be disjoint disks in the plane with radii $R_1, \\dots, R_6$. Let $P_i$ be a point in $D_i$, and let $O$ be an arbitrary point. Then there exist indices $i$ and $j$ such that $OP_i \\ge R_j$.\n\n*Proof.* Let $O_i$ be the center of $D_i$. C... | Taiwan | 2021 數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
00yk | Problem:
Let $p > 2$ be a prime number and $1 + \frac{1}{2^{3}} + \frac{1}{3^{3}} + \cdots + \frac{1}{(p-1)^{3}} = \frac{m}{n}$ where $m$ and $n$ are relatively prime. Show that $m$ is a multiple of $p$. | [
"Solution:\n\nThe sum has an even number of terms; they can be joined in pairs in such a way that the sum is the sum of the terms\n$$\n\\frac{1}{k^{3}} + \\frac{1}{(p-k)^{3}} = \\frac{p^{3} - 3 p^{2} k + 3 p k^{2}}{k^{3} (p-k)^{3}}.\n$$\nThe sum of all terms of this type has a denominator in which every prime facto... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | m is a multiple of p | |
0egy | Problem:
Katera izmed navedenih funkcij je odvod funkcije $f(x)=\frac{1-2x}{\sqrt{2x}}$?
(A) $f'(x)=\frac{-\sqrt{2x}(2x-1)}{4x^2}$
(B) $f'(x)=\frac{-\sqrt{2x}(2x-1)}{8x^2}$
(C) $f'(x)=\frac{-\sqrt{2x}(2x+1)}{4x^2}$
(D) $f'(x)=\frac{-\sqrt{2x}(2x+1)}{8x^2}$
(E) $f'(x)=\frac{-\sqrt{2x}(1-2x)}{4x^2}$ | [
"Solution:\nIzračunamo odvod funkcije $f$ po obrazcu $f'(x)=\\frac{-2\\sqrt{2x}-(1-2x)\\cdot\\frac{1}{2}(2x)^{-\\frac{1}{2}}\\cdot 2}{2x}$. Razširimo števec na skupni imenovalec, razrešimo dvojni ulomek in z racionaliziranjem ulomka dobimo $f'(x)=\\frac{-2x\\sqrt{2x}-\\sqrt{2x}}{4x^2}$. Izpostavimo skupni faktor v ... | Slovenia | Državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Calculus > Differential Calculus > Derivatives"
] | null | MCQ | C | |
05wv | Problem:
Soit $(a_{n})$ une suite de réels. On suppose que $a_{0}=1$ et que pour tout $n \geqslant 1$, $a_{n}$ est la plus petite solution strictement positive de
$$
\left(a_{n}-a_{n-1}\right)\left(a_{n}+a_{n-1}-2 \sqrt{n}\right)=2 .
$$
Trouver le plus petit entier $n$ tel que $a_{n} \geqslant 2022$. | [
"Solution:\n\nPour se donner une idée du problème, on calcule les premières valeurs de $a_{n}$ : pour $a_{1}$, l'équation s'écrit\n$$\n\\left(a_{1}-1\\right)\\left(a_{1}-1\\right)=2\n$$\ndonc $a_{1}-1=\\sqrt{2}$ ou $-\\sqrt{2}$. Seul le premier cas donne une solution positive $a_{1}=1+\\sqrt{2}$. Cherchons maintena... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOi 2 : Algèbre | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1022121 | |
0371 | Problem:
The price of a merchandise dropped from March to April by $x\%$, and went up from April to May by $y\%$. It turned out that in the period from March to May the price dropped by $(y-x)\%$. Find $x$ and $y$ if they are positive integers (the price is positive for the whole period). | [
"Solution:\n\nIt follows from the condition of the problem that\n$$\n\\frac{100-x}{100} \\cdot \\frac{100+y}{100} = \\frac{100-y+x}{100}\n$$\nHence $(100-x)(100+y) = 100(100-y+x)$ which can be written as $(200-x)(200+y) = 200^{2}$. Further, $100 < 200-x < 200$, i.e. $200-x$ equals $125$ or $160$.\n\nTherefore the s... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | x=75, y=120; x=40, y=50 | |
0l6v | Problem:
Prove that there exists a real number $\varepsilon > 0$ such that there are infinitely many sequences of integers $0 < a_1 < a_2 < \dots < a_{2025}$ satisfying
$$
\text{gcd}(a_1^2 + 1, a_2^2 + 1, \dots, a_{2025}^2 + 1) > a_{2025}^{1+\varepsilon}.
$$ | [] | United States | TST2025 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Linear Algebra > Determinants",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
... | null | proof only | null | |
0018 | En una escuela militar hay $200$ alumnos. El director decidió que cada día un grupo de $9$ alumnos debe patrullar la escuela y preparó la lista con las patrullas para todo el año, respetando la siguiente regla: si un alumno está en la patrulla cierto día, entonces no puede estar en la patrulla los siguientes $10$ días.... | [] | Argentina | XII Olimpiada Matemática Rióplatense | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | español | proof and answer | Yes | |
010c | Problem:
Let $n$ and $k$ be positive integers. There are $n k$ objects (of the same size) and $k$ boxes, each of which can hold $n$ objects. Each object is coloured in one of $k$ different colours. Show that the objects can be packed in the boxes so that each box holds objects of at most two colours. | [
"Solution:\n\nIf $k=1$, it is obvious how to do the packing. Now assume $k>1$. There are not more than $n$ objects of a certain colour - say, pink - and also not fewer than $n$ objects of some other colour - say, grey. Pack all pink objects into one box; if there is space left, fill the box up with grey objects. Th... | Baltic Way | Baltic Way 1998 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0jqb | Problem:
Let $\mathcal{H}$ be the unit hypercube of dimension $4$ with a vertex at $(x, y, z, w)$ for each choice of $x, y, z, w \in \{0,1\}$. (Note that $\mathcal{H}$ has $2^{4}=16$ vertices.) A bug starts at the vertex $(0,0,0,0)$. In how many ways can the bug move to $(1,1,1,1)$ (the opposite corner of $\mathcal{H}$... | [
"Solution:\nAnswer: $24$\n\nYou may think of this as sequentially adding $1$ to each coordinate of $(0,0,0,0)$. There are $4$ ways to choose the first coordinate, $3$ ways to choose the second, and $2$ ways to choose the third. The product is $24$."
] | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 24 | |
0e0o | Prove the inequality $\frac{9}{4} < \log_2 \pi + \log_4 \pi < \frac{5}{2}$. | [
"Since\n$$\n\\log_2 \\pi + \\log_4 \\pi = \\frac{1}{\\log_\\pi 2} + \\frac{1}{\\log_\\pi 4} = \\frac{3}{2 \\cdot \\log_\\pi 2} = \\frac{3}{2} \\log_2 \\pi.\n$$\nwe have to show that $\\frac{9}{2} < 3 \\log_2 \\pi < 5$.\n\nThe inequality $\\frac{9}{2} < 3 \\log_2 \\pi$ is equivalent to $3 < \\log_2 \\pi^2$ or $2^3 <... | Slovenia | National Math Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof only | null | |
0261 | Problem:
Um corpo em queda livre demora $11$ segundos para tocar o solo. No primeiro segundo ele percorre $4,5\ \mathrm{m}$, em cada segundo que segue, a distância percorrida aumenta de $9,8\ \mathrm{m}$. Qual a altura da queda e quantos metros ele percorreu no último segundo? | [
"Solution:\n\nSeja $d_n$ a distância percorrida no $n$-ésimo segundo.\n\nTemos:\n\n$d_1 = 4,5\\ \\mathrm{m}$\n\ne a cada segundo, a distância aumenta de $9,8\\ \\mathrm{m}$:\n\n$d_2 = d_1 + 9,8$\n$d_3 = d_2 + 9,8 = d_1 + 2 \\times 9,8$\n$\\ldots$\n$d_n = d_1 + (n-1) \\times 9,8$\n\nA altura total $H$ é a soma das d... | Brazil | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | Height: 588.5 m; last second distance: 102.5 m | |
0bmi | Let $ABC$ be a triangle. Let $P_1$ and $P_2$ be points on the side $AB$ such that $P_2$ lies on the segment $BP_1$ and $AP_1 = BP_2$; similarly, let $Q_1$ and $Q_2$ be points on the side $BC$ such that $Q_2$ lies on the segment $BQ_1$ and $BQ_1 = CQ_2$. The segments $P_1Q_2$ and $P_2Q_1$ meet at $R$, and the circles $P... | [
"Throughout the solution, $[XYZ]$ and $d(X, YZ)$ denote the area of the triangle $XYZ$ and the distance from the point $X$ to the line $YZ$, respectively.\n\nSince the quadrilaterals $SRQ_2Q_1$ and $SRP_2P_1$ are cyclic, $\\angle SQ_1R = \\angle SQ_2R$ and $\\angle SP_1R = \\angle SP_2R$ (see Fig. 6), so the triang... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0jfb | Problem:
$$
\left(\begin{array}{lllll}
1 & 1 & 1 & \cdots & 1
\end{array}\right)
$$
is written on a board, with 2013 ones in between the outer parentheses. Between each pair of consecutive ones you may write either "+" or ")(" (you cannot leave the space blank). What is the maximum possible value of the resulting expr... | [
"Solution:\n\nThe answer is $3^{671}$, formed by dividing the ones into groups of three as the answer suggests.\nIt should be obvious that a maximum value exists, since there are only finitely many ways to place the pluses and parentheses and one (or more) of them must give the largest value.\n\nIf there is a facto... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 3^{671} | |
0gg8 | 令 $N$ 與 $s$ 為正整數, 且 $N > s$。台電園區裡有若干棟建物, 其中恰有 $N$ 棟為發電廠, 另有一棟為總部。若干對建物之間有僅能單向送電的電線, 滿足:
(i) 所有連接發電廠的電線都只會把電送出發電廠。
(ii) 對於每個非總部的建物, 都存在唯一的一系列電線, 構成從該建物通向總部的電路。
某建物被稱為 $s$ 級有電, 若且唯若當我們移除園區內的任何一條電線, 該建物仍能從至少 $s$ 個電廠供電。試求 $s$ 級有電建物數量的最大可能值。 | [
"解.\nIn fact, the order of $ABCDE$ on $Ω$ doesn't affect the result. Let $U, V$ be the second intersections of $⊙(ABC)$ and $DS, DT$, respectively, and $R$ be the point on $ST$ such that $∠BAD = ∠RAC$. Since $△ABD \\sim \\triangle ART$, we have $∠ART = ∠ABD = ∠AVD = ∠AVT$, implying that $A, R, T, V$ are concyclic. ... | Taiwan | 2022 數學奧林匹亞競賽第二階段選訓營,國際競賽實作(二) | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | Chinese; English | proof and answer | ⌊(N − 1) / s⌋ | |
0ht9 | Problem:
Let $ABC$ be a triangle such that $\angle A = 90^{\circ}$ and $\angle B < \angle C$. The tangent at $A$ to its circumcircle $\omega$ meets the line $BC$ at $D$. Let $E$ be the reflection of $A$ across $BC$, $X$ the foot of the perpendicular from $A$ to $BE$, and $Y$ the midpoint of $AX$. Let the line $BY$ mee... | [
"Solution:\n\nLet $M$ be the point of intersection of $AE$ and $BC$, and let $N$ be the point on $\\omega$ diametrically opposite $A$. Since $\\angle B < \\angle C$, points $N$ and $B$ are on the same side of $AE$.\n\nFurthermore, $\\angle NAE = \\angle BAX = 90^{\\circ} - \\angle ABE$; hence the triangles $NAE$ an... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ic4 | Problem:
A class of 10 students took a math test. Each problem was solved by exactly 7 of the students. If the first nine students each solved 4 problems, how many problems did the tenth student solve? | [
"Solution:\nSuppose the last student solved $n$ problems, and the total number of problems on the test was $p$. Then the total number of correct solutions written was $7p$ (seven per problem), and also equal to $36 + n$ (the sum of the students' scores), so $p = (36 + n)/7$. The smallest $n \\geq 0$ for which this ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 6 | |
02rs | Problem:
Uma tartaruga corredora anda em linha reta da seguinte maneira. No primeiro trecho do caminho, que mede $\frac{1}{2}~\mathrm{m}$, ela corre à velocidade de $3~\mathrm{m}/\mathrm{s}$. No segundo trecho, que mede $\frac{1}{3}~\mathrm{m}$, ela corre à velocidade de $4~\mathrm{m}/\mathrm{s}$. No terceiro trecho, ... | [
"Solution:\n\na) Como velocidade é a razão entre a distância percorrida e o tempo gasto, temos que, no primeiro trecho,\n$$\n3 = \\frac{\\frac{1}{2}}{t}\n$$\ne daí obtemos $t = \\frac{1}{6}~\\mathrm{s}$. Para escrever $\\frac{1}{6}$ como diferença de duas frações unitárias, basta notar que\n$$\n\\frac{1}{6} = \\fra... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | a) 1/6 s = 1/2 − 1/3; b) 1/12 s = 1/3 − 1/4; c) 1/2 − 1/2015 s | |
0hc0 | Given positive integers $a$, $b$, $c$, $d$ such that $a < b < c < d$. Is it possible for the least common multiple of $a$ and $b$ to be greater than the least common multiple of $c$ and $d$? | [
"Let $a = 8$, $b = 9$, then $[a; b] = 72$. We will choose the other two numbers such that $[c; d] = 50$: $c = 10$, $d = 25$."
] | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof and answer | Yes; for example, a=8, b=9, c=10, d=25 gives lcm(a,b)=72 and lcm(c,d)=50. | |
0d9x | Suppose that 2018 numbers $1$ and $-1$ are written around a circle. For every two adjacent numbers, their product is taken. Suppose that the sum of all 2018 products is negative. Find all possible values of sum of 2018 given numbers. | [
"First, we show that the sum is at most $1008$ and at least $-1008$.\nIndeed, suppose that there are $a$ numbers $+1$ in $2018$ given numbers and $b$ products equal $-1$. Then the sum of $2018$ given numbers is $a + (-1)(2018 - a) = 2a - 2018$, and sum of $2018$ products is\n$$\n(2018 - b) + (-1) b = 2018 - 2b.\n$$... | Saudi Arabia | Team selection tests for BMO 2018 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All even integers from -1008 to 1008 inclusive | |
03v7 | $A$ and $B$ are playing ping-pong, with the agreement that the winner of a game will get $1$ point and the loser $0$ point; the match ends as soon as one of the players is ahead by $2$ points or the number of games reaches $6$. Suppose that the probabilities of $A$ and $B$ winning a game are $\frac{2}{3}$ and $\frac{1}... | [
"**Solution I** It is easy to see that $\\xi$ can only be $2$, $4$ or $6$. We divide the six games into three rounds, each consisting of two consecutive games. If one of the players wins two games in the first round, the match ends and the probability is\n$$\n\\left(\\frac{2}{3}\\right)^2 + \\left(\\frac{1}{3}\\rig... | China | China Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Expected values"
] | English | MCQ | B | |
03xb | For acute triangle $ABC$ with $AB > AC$, let $M$ be the midpoint of side $BC$ and $P$ a point inside $\triangle AMC$ such that $\angle MAB = \angle PAC$. Let $O$, $O_1$ and $O_2$ be the circumcenters of $\triangle ABC$, $\triangle ABP$ and $\triangle ACP$ respectively. Prove that line $AO$ bisects segment $O_1O_2$. (Po... | [
"As shown in Fig. 1, draw the circumcircles of $\\triangle ABC$, $\\triangle ABP$ and $\\triangle ACP$, respectively. Let the extension of $AP$ meet $\\odot O$ at $D$, join $BD$, and draw the line tangent to $\\odot O$ at $A$, intersecting $\\odot O_1$ and $\\odot O_2$ at $E$ and $F$ respectively.\n\nIt is clear th... | China | China National Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | English | proof only | null | |
05ot | Problem:
Dans les cases d'un tableau rectangulaire à $n$ lignes et $m$ colonnes sont écrits des nombres réels. On suppose que pour toute ligne ou colonne, la somme des nombres écrits sur cette ligne ou colonne est un entier. Montrer qu'il est possible de remplacer chaque réel $x$ par l'entier $\lfloor x\rfloor$ ou $\l... | [
"Solution:\n\nQuitte à soustraire sa partie entière à chaque entrée du tableau, on peut supposer que chaque nombre est compris entre $0$ et $1$. On dit qu'une case est « fractionnaire » si son nombre n'est pas entier. Remarquons que, si $C$ est une case fractionnaire, alors il existe nécessairement une autre case f... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
02rx | Problem:
Considere um retângulo $ABCD$ onde os comprimentos dos lados são $\overline{AB}=4$ e $\overline{BC}=8$. Sobre os lados $BC$ e $AD$ se fixam os pontos $M$ e $N$, respectivamente, de modo que o quadrilátero $BMDN$ seja um losango. Calcule a área deste losango. | [
"Solution:\nObserve o seguinte desenho:\n\n\n\nComo $BMDN$ é um losango, então os comprimentos dos segmentos $BM$, $MD$, $DN$ e $NB$ são iguais a um mesmo valor, digamos $x$. Pelo dado do problema, $\\overline{AD}=8$, logo $\\overline{AN}=8-x$.\n\n\n\nPodemos agora ... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 20 | |
0jvz | Problem:
Victor has a drawer with two red socks, two green socks, two blue socks, two magenta socks, two lavender socks, two neon socks, two mauve socks, two wisteria socks, and 2000 copper socks, for a total of 2016 socks. He repeatedly draws two socks at a time from the drawer at random, and stops if the socks are o... | [
"Solution:\n\nThere are $\\binom{2000}{2} + 8\\binom{2}{2} = 1999008$ ways to get socks which are matching colors, and four extra ways to get a red-green pair, hence the answer."
] | United States | HMMT February | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 499752/499753 | |
0kn8 | Problem:
Let $n$ be an integer and
$$
m=(n-1001)(n-2001)(n-2002)(n-3001)(n-3002)(n-3003) .
$$
Given that $m$ is positive, find the minimum number of digits of $m$. | [
"Solution:\nOne can show that if $m>0$, then we must either have $n>3003$ or $n<1001$. If $n<1001$, each term other than $n-1001$ has absolute value at least $1000$, so $m>1000^{5}$, meaning that $m$ has at least $16$ digits. However, if $n>3003$, it is clear that the minimal $m$ is achieved at $n=3004$, which make... | United States | HMMT November 2021 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 11 | |
0bew | a) Consider $f: [0, \infty) \to [0, \infty)$ a differentiable and convex function such that $f(x) \le x$, for $x \ge 0$. Then $f'(x) \le 1$, for all $x \ge 0$.
b) Determine all differentiable and convex functions $f: [0, \infty) \to [0, \infty)$ with the properties $f(0) = 0$ and $f'(x) \cdot f(f(x)) = x$, for all $x ... | [
"a) Suppose the contrary, that is, there exists $a \\ge 0$ such that $f'(a) > 1$. Because $\\lim_{x \\to a} \\frac{f(x) - f(a)}{x - a} > 1$, there is $b > a$ such that $\\frac{f(b) - f(a)}{b - a} > 1$. For any $x > b$, by the convexity of $f$ we get $\\frac{f(x) - f(b)}{x - b} \\ge \\frac{f(b) - f(a)}{b - a} = m > ... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications",
"Precalculus > Functions"
] | null | proof and answer | f(x) = x for all x ≥ 0 | |
0bwe | Let $n$ be a positive integer, coprime to $2$ and $5$. Show that there is a positive integer $k \geq n$ with the property: for any base $10$ positive integer $\overline{a_1a_2...a_k}$, divisible by $k$, with all digits different from $0$, the numbers $\overline{a_2...a_k a_1}$, $\overline{a_3...a_k a_1 a_2}$, ..., $\ov... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof only | null | |
0dex | In each cell of a $10 \times 10$ board one arrow is placed. Each arrow is pointing in one of the four directions $\{\uparrow, \to, \downarrow, \leftarrow\}$. Find the smallest number $n$ with the following property: it is always (regardless of the initial placement of the arrows) possible to remove at most $n$ arrows f... | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 50 | |
09ye | Problem:
Zij $ABC$ een scherphoekige en niet-gelijkbenige driehoek met hoogtepunt $H$. Zij $O$ het middelpunt van de omgeschreven cirkel van driehoek $ABC$ en zij $K$ het middelpunt van de omgeschreven cirkel van driehoek $AHO$. Bewijs dat de spiegeling van $K$ in $OH$ op $BC$ ligt.
 | [
"Solution:\n\nWe bekijken de configuratie zoals in de figuur. Andere configuraties gaan analoog. Noem $D$ het tweede snijpunt van $AH$ met de omgeschreven cirkel van $\\triangle ABC$. Noem $S$ het tweede snijpunt van de omgeschreven cirkel van $ABC$ en de omgeschreven cirkel van $AHO$. (Omdat $\\triangle ABC$ scher... | Netherlands | IMO-selectietoets | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0l2m | Problem:
Let $ABCD$ be a square, and let $\ell$ be a line passing through the midpoint of segment $\overline{AB}$ that intersects segment $\overline{BC}$. Given that the distances from $A$ and $C$ to $\ell$ are $4$ and $7$, respectively, compute the area of $ABCD$. | [
"Solution:\n\n\n\nConsider the line $\\ell'$ through $B$ parallel to $\\ell$, and drop perpendiculars from $A$ to $\\ell'$ and $C$ to $\\ell'$. Note that because $\\ell$ passes through the midpoint of segment $AB$, the distance from $B$ to $\\ell$ is $4$. Thus, the distances from $A$ to $\\... | United States | HMMT February 2024 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 185 | |
0005 | Problem:
Sea $A$ un subconjunto del conjunto $N$ de los números enteros positivos. Diremos que un subconjunto $B$ de $N$ es una base de $A$ si las sumas de los elementos de cada subconjunto no vacío de $B$ son distintas y cada elemento de $A$ es igual a una de estas sumas. Demostrar que para cada $n = 1, 2, 3, \ldots$... | [] | Argentina | XI Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Other",
"Number Theory > Other"
] | español | proof and answer | k(n) = n | |
0cg5 | The non-nil real numbers $a$ and $b$ fulfill $3(a^6 + b^6) = a^2b^2(a^2b^2 + 9)$. Prove that at least one of the numbers $a$ and $b$ is irrational. | [
"The equality can be written $a^4b^4 - 3a^6 - 3b^6 + 9a^2b^2 = 0$, $a^4(b^4 - 3a^2) - 3b^2(b^4 - 3a^2) = 0$, $(a^4 - 3b^2)(b^4 - 3a^2) = 0$. So $a^4 = 3b^2$ or $b^4 = 3a^2$ and, since $a \\neq 0$ and $b \\neq 0$, $\\sqrt{3} = \\pm\\frac{a^2}{b}$ or $\\sqrt{3} = \\pm\\frac{b^2}{a}$.\nIf $a$ and $b$ are rational, the... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof only | null | |
0ex8 | Problem:
Given a lattice of regular hexagons. A bug crawls from vertex $A$ to vertex $B$ along edges of the hexagons, taking the shortest possible path (or one of them). Prove that it travels a distance at least $AB/2$ in one direction. If it travels exactly $AB/2$ in one direction, how many edges does it traverse? | [
"Solution:\n\n\n\nSuppose vertex $A$ is that marked $*$ at the bottom left. Without loss of generality, $B$ is in a $60$ degree sector as shown. Assume the edges have unit length. The vertices can be partitioned into two sets (marked $\\circ$ and $\\cdot$ in the diagram). Each set forms a s... | Soviet Union | 4th ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 3 | |
0c13 | Let $p$ be an integer greater or equal to $2$ and let $(M, \cdot)$ be a finite semi-group, such that $a^p \neq a$, for all $a \in M \setminus \{e\}$, where $e$ is the null element in $M$. Prove that $(M, \cdot)$ is a group.
Traian Preda | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
0kf9 | Problem:
Anne-Marie has a deck of 16 cards, each with a distinct positive factor of $2002$ written on it. She shuffles the deck and begins to draw cards from the deck without replacement. She stops when there exists a nonempty subset of the cards in her hand whose numbers multiply to a perfect square. What is the expe... | [
"Solution:\n\nNote that $2002 = 2 \\cdot 7 \\cdot 11 \\cdot 13$, so that each positive factor of $2002$ is included on exactly one card. Each card can be identified simply by whether or not it is divisible by each of the 4 primes, and we can uniquely achieve all of the $2^{4}$ possibilities. Also, when considering ... | United States | HMMT February 2020 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 837/208 | |
0l59 | There are exactly three positive real numbers $k$ such that the function
$$
f(x) = \frac{(x - 18)(x - 72)(x - 98)(x - k)}{x}
$$
defined over the positive real numbers $x$ achieves its minimum value at exactly two positive real numbers $x$. Find the sum of these three values of $k$. | [] | United States | AIME II | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 42113/98 | |
0i10 | Problem:
Let $ABCD$ be a cyclic quadrilateral (i.e. inscribed in a circle) such that $DC = AD + BC$. Prove that the intersection point of the angle bisectors of angle $A$ and angle $B$ lies on $CD$. | [
"Solution:\nLet $E$ lie on segment $CD$ so that $CE = BC$. Then, $DE = DC - CE = AD$. So, $\\triangle BCE$ and $\\triangle ADE$ are both isosceles. Let $\\alpha = \\angle DAE = \\angle AED$, and let $\\beta = \\angle CEB = \\angle EBC$. Then $\\angle CDA = \\angle EDA = \\pi - 2\\alpha$, but since the quadrilateral... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0at7 | Problem:
Find the only value of $x$ in the open interval $(-\pi / 2, 0)$ that satisfies the equation
$$
\frac{\sqrt{3}}{\sin x} + \frac{1}{\cos x} = 4
$$ | [
"Solution:\n$-\\frac{4\\pi}{9}$"
] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Equations and Inequalities"
] | null | final answer only | -4π/9 | |
00tn | Denote by $\ell(n)$ the largest prime divisor of $n$. Let $a_{n+1} = a_n + \ell(a_n)$ be a recursively defined sequence of integers with $a_1 = 2$. Determine all natural numbers $m$ such that there exists some $i \in \mathbb{N}$ with $a_i = m^2$. | [
"Let $p_1, p_2, \\dots$ be the sequence of prime numbers. We will prove the following:\n\n**Claim:** Assume $a_n = p_i p_{i+1}$. Then for each $k = 1, 2, \\dots, p_{i+2} - p_i$ we have that $a_{n+k} = (p_i + k) p_{i+1}$.\n\n**Proof.** By induction on $k$. Since $\\ell(a_n) = p_{i+1}$, then $a_{n+1} = p_i p_{i+1} + ... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | All prime numbers m (i.e., m is prime). | |
038h | Problem:
An air company operates 36 airlines in a country with 16 airports. Prove that one can make a round trip that includes 4 airports. | [
"Solution:\n\nConsider a graph $G$ whose vertices are the airports in the country. Two vertices form an edge if there is an airline between the corresponding airports. Suppose that a round trip satisfying the conditions of the problem does not exist, i.e. there is no cycle of length 4 in $G$.\n\nIf $x$ is a vertex ... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
00k0 | Points $A$, $B$ and $C$ lie on a line in this order. For every circle $k$ passing through $B$ and $C$, let $D$ be one of the common points of $k$ and the bisector of $BC$. Furthermore, let $E$ be the second common point of the line $AD$ and $k$.
*Prove that the ratio $BE : CE$ is constant for all circles $k$.*
G. Bar... | [
"Let $F$ be the diametrically opposite point to $D$ on $k$. Since $F$ is a point on the bisector of $BC$, we have $FB = FC$, and therefore $\\angle BEF = \\angle CEF$.\n\n\n\nWe see that $EF$ is the internal bisector of $\\angle BEC$, and since $ED \\perp EF$, $ED = AD$ is the external bise... | Austria | AustriaMO2013 | [
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
07xm | In a mathematics class, the teacher shows an apparatus containing $n$ switches, where each switch can be in any of three positions, labelled 1–3, and $n > 1$. Initially, all switches are in position 1, and we need to move the switches so that all are in position 2. The teacher shows one method of doing this that involv... | [
"We provide three methods to show that $c_n = n \\cdot (n+1)!/2 - 1$ for all $n \\in \\mathbb{N}$.\n(We omit the assumption $n > 1$ when proving this.) Let $d_n := c_n + 1$, i.e. $d_n$\nis the number of $(n+1)$-step methods including that of the teacher.\n\n**Method 1.** Consider the first step of any $(n + 1)$-ste... | Ireland | IRL_ABooklet_2025 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0aza | Problem:
Simplify $\sqrt[3]{5 \sqrt{2}+7}-\sqrt[3]{5 \sqrt{2}-7}$ into a rational number. | [
"Solution:\n\nLet $a=\\sqrt[3]{5 \\sqrt{2}+7}$ and let $b=\\sqrt[3]{5 \\sqrt{2}-7}$. Note that\n$$\na^{3}-b^{3}=14 \\text{ and } a b=\\sqrt[3]{50-49}=1\n$$\nNow, $a^{3}-b^{3}=(a-b)\\left(a^{2}+a b+b^{2}\\right)=(a-b)\\left[(a-b)^{2}+3 a b\\right]$. Letting $x=a-b$, we have the resulting equation $x\\left(x^{2}+3\\r... | Philippines | 20th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | 2 | |
0bc2 | Given in the plane equilateral triangles $ABC$ and $BDE$, with $C \in (BD)$ and $A, E$ are on different sides of the line $BD$. Denote $M, N, P$ the midpoints of the segments $(AB)$, $(CD)$, $(BE)$ respectively. Find the measures of the angles of the triangle $MNP$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof and answer | 60°, 60°, 60° | |
0get | For every positive integer $N$, determine the smallest real number $b_N$ such that, for all real $x$,
$$
\sqrt[N]{\frac{x^{2N} + 1}{2}} \le b_N(x - 1)^2 + x.
$$
對每個正整數 $N$，請決定最小的實數 $b_N$ 使得對所有的實數 $x$ 下列不等式恆成立：
$$
\sqrt[N]{\frac{x^{2N} + 1}{2}} \le b_N(x - 1)^2 + x.
$$ | [
"The answer is $b_N = N/2$.\n\n**Solution 1.** First of all, assume that $b_N < N/2$ satisfies the condition. Take $x = 1 + t$ for $t > 0$, we should have\n$$\n\\frac{(1 + t)^{2N} + 1}{2} \\le (1 + t + b_N t^2)^N.\n$$\nExpanding the brackets we get\n$$\n(1 + t + b_N t^2)^N - \\frac{(1 + t)^{2N} + 1}{2} = \\left(Nb_... | Taiwan | 2021 數學奧林匹亞競賽第二階段選訓營, 獨立研究(二) | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | b_N = N/2 | |
0a5x | Problem:
Let $ABCD$ be a square (vertices labelled in clockwise order). Let $Z$ be any point on diagonal $AC$ between $A$ and $C$ such that $AZ > ZC$. Points $X$ and $Y$ exist such that $AXYZ$ is a square (vertices labelled in clockwise order) and point $B$ lies inside $AXYZ$. Let $M$ be the point of intersection of l... | [
"Solution:\n\nSince $Z$ lies on diagonal $AC$, we have $\\angle DAZ = 45^{\\circ}$ and $\\angle ZAB = 45^{\\circ}$. Therefore $B$ lies on diagonal $AY$ of square $AXYZ$ and $\\angle BAX = 45^{\\circ}$.\n\n\n\nSince $AB = AD$ and $AX = AZ$ and $\\angle BAX = 45^{\\circ} = \\angle DAZ$, we ha... | New Zealand | NZMO Round Two | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09s0 | Problem:
Laat $m \geq 3$ en $n$ positieve gehele getallen zijn met $n > m(m-2)$. Vind het grootste positieve gehele getal $d$ zodat $d \mid n!$ en $k \nmid d$ voor alle $k \in \{m, m+1, \ldots, n\}$. | [
"Solution:\nWe gaan bewijzen dat $d = m-1$ de grootste is die voldoet. Merk eerst op dat $m-1 \\mid n!$ en dat voor $k \\geq m$ geldt $k \\nmid m-1$, dus $d = m-1$ voldoet inderdaad.\n\nStel nu dat voor zekere $d$ geldt: $d \\mid n!$ en $k \\nmid d$ voor alle $k \\in \\{m, m+1, \\ldots, n\\}$. We gaan bewijzen dat ... | Netherlands | Selectietoets | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | m-1 | |
06bf | Prove that there are infinitely many primes $p$ such that $N_p = p^2$, where $N_p$ is the total number of solutions mod $p$ to the equation $3x^3 + 4y^3 + 5z^3 - y^4z = 0$. | [
"By Dirichlet's theorem, there are infinitely many primes $p$ of the form $3k + 2$. We claim that $N_p = p^2$ for all such primes $p$.\n\nIndeed, for each of the $p^2$ pairs of $(y, z)$ modulo $p$, we need to solve\n$$\nx^3 \\equiv -3^{-1}(4y^3 + 5z^3 - y^4z) \\pmod{p}.\n$$\nAs $(p-1, 3) = 1$, this equation has a u... | Hong Kong | 1997-2023 IMO HK TST | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof only | null | |
06w2 | Let $n$ and $k$ be positive integers. Prove that for $a_{1}, \ldots, a_{n} \in [1,2^{k}]$ one has
$$
\sum_{i=1}^{n} \frac{a_{i}}{\sqrt{a_{1}^{2}+\ldots+a_{i}^{2}}} \leqslant 4 \sqrt{k n}
$$ | [
"Partition the set of indices $\\{1,2, \\ldots, n\\}$ into disjoint subsets $M_{1}, M_{2}, \\ldots, M_{k}$ so that $a_{\\ell} \\in [2^{j-1}, 2^{j}]$ for $\\ell \\in M_{j}$. Then, if $|M_{j}|=: p_{j}$, we have\n$$\n\\sum_{\\ell \\in M_{j}} \\frac{a_{\\ell}}{\\sqrt{a_{1}^{2}+\\ldots+a_{\\ell}^{2}}} \\leqslant \\sum_{... | IMO | IMO 2020 Shortlisted Problems | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof only | null | |
02cy | Problem:
Seja $ABCD$ um retângulo com $AB = \sqrt{3}$. Se $\angle ACD = 75^{\circ}$, calcule o comprimento de $AC$ e o valor de $\cos 75^{\circ}$.
 | [
"Solution:\n\n\nConsidere os pontos $F$ e $E$ sobre o lado $BC$ de modo que $FC = FA$ e $EF = AE$. Então, pelo Teorema do Ângulo Externo, temos\n$$\n\\angle BEA = 2 \\cdot \\angle EFA = 4 \\cdot \\angle FCA = 4 \\cdot (90^{\\circ} - \\angle ACD) = 60^{\\circ}\n$$\nDaí, $\\frac{AB}{AE} = \\s... | Brazil | NÍVEL 3 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | AC = 2√(6 + 3√3); cos 75° = 1/(2√(2 + √3)) | |
05ho | Problem:
Soit $ABC$ un triangle dont les angles sont aigus, $\omega$ son cercle circonscrit et $O$ le centre de $\omega$. La perpendiculaire issue de $A$ par rapport à $(BC)$ intersecte $(BC)$ en $D$ et $\omega$ en $E$. Soit $F$ un point sur le segment $[AE]$ tel que $2 \cdot FD = AE$. Soit $\ell$ la perpendiculaire à... | [
"Solution:\n\n\n\nOn désire montrer que trois droites sont concourantes. Une stratégie serait donc de considérer le point d'intersection de deux de ces droites et de montrer que ce point appartient à la troisième droite, c'est la stratégie que nous allons adopter ici.\n\nDans le cas de notr... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"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 | |
0k2k | Problem:
There are $2018$ frogs in a pool and there is $1$ frog on the shore. In each time-step thereafter, one random frog moves position. If it was in the pool, it jumps to the shore, and vice versa. Find the expected number of time-steps before all frogs are in the pool for the first time. | [
"Solution:\n\nConsider the general case of $n$ frogs. Let $E_{i}$ be the expected time for all frogs to enter the pool when $i$ frogs are on the shore and $n-i$ frogs are in the pool. We have $E_{0}=0$, $E_{n}=1+E_{n-1}$, and\n$$\nE_{i} = \\frac{i}{n} E_{i-1} + \\frac{n-i}{n} E_{i+1} + 1\n$$\nfor $0 < i < n$. Defin... | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 2^2018 - 1 | |
0g01 | Problem:
Finde alle Paare $(a, b)$ teilerfremder ganzer Zahlen, sodass gilt:
$$
a^{2}+a=b^{3}+b
$$ | [
"Solution:\nComme $a^{2}+a \\geq -\\frac{1}{4}$, on a tout de suite $b^{3}+b \\geq 0$, donc $b \\geq 0$. L'équation quadratique $a^{2}-a-b\\left(b^{2}+1\\right)=0$, si elle possède des solutions réelles, a toujours une racine positive ou nulle et une autre négative ou nulle. Ainsi, sans perte de généralité, prenons... | Switzerland | IMO-Selektion | [
"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), (-2,1), (-1,0), (5,3)] | |
0dg6 | Given a non-isosceles acute angled triangle $\triangle ABC$ where $O$ is the midpoint of $BC$. Let the circle with diameter $BC$, intersects $AB$, $AC$ at $D$, $E$ respectively. Let the angle bisectors of $\angle A$ and $\angle DOE$ intersect at $P$. If the circumcircles of $\triangle BPD$ and $\triangle CPE$ intersect... | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0csm | Let $M$ be a midpoint of the side $AC$ of an acute-angled triangle $ABC$ with $AB > BC$. Let $\Omega$ be the circumcircle of the triangle $ABC$. The tangents to $\Omega$ at points $A$ and $C$ meet at $P$. The segments $BP$ and $AC$ meet at $S$. Let $AD$ be the altitude in the triangle $ABP$. The circumcircle $\omega$ o... | [
"Поскольку $\\angle AMP = \\angle ADP = 90^\\circ$, точки $M$ и $D$ лежат на окружности $\\gamma$ с диаметром $AP$. Поскольку $PA$ — касательная к $\\Omega$, имеем $\\angle KAP = \\angle ACK$. Так как точки $C, K, D$ и $S$ лежат на окружности $\\omega$, имеем $\\angle ACK = \\angle KDP$. Значит, $\\angle KAP = \\an... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hg9 | There are $n \ge 3$ segments, their lengths in centimeters are distinct positive integers. It's known that it's possible to form a nondegenerate triangle from any three of these $n$ segments. Suppose that among these segments there are segments with lengths $5$ cm and $12$ cm. What's the largest value $n$ can attain? | [
"Reorder the segments by their lengths, so that $a_1 < a_2 < \\dots < a_n$. Clearly, any three segments form a triangle if and only if the sum of the lengths of the smallest two segments is larger than the length of the longest segment. So, the smallest segment except from the given two can't have a length smaller ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 6 | |
0378 | Problem:
Consider a set $S$ of 2006 points in the plane. A pair $(A, B) \in S \times S$ is called "isolated" if the disk with diameter $A B$ does not contain other points from $S$. Find the maximum number of "isolated" pairs. | [
"Solution:\nConsider a graph with vertices the given points. Two points form an edge if the corresponding pair of points is \"isolated\". We first prove that the graph is connected. To do this suppose that it has more than one connected component and choose points $A$ and $B$ from different components such that the... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | 2005 | |
095p | Problem:
Găsiți toate valorile parametrului $a \in \mathbb{R}$, pentru care sistemul
$$
\left\{\begin{array}{c}
\sqrt{x^{2}+y^{2}+6 x+10 y+34}+\sqrt{x^{2}+y^{2}+6 x+6 y+18}=2 \\
y^{2}+x^{2}-4 x+10=a
\end{array}\right.
$$ | [
"Solution:\nVom rezolva acest sistem geometric. Soluțiile sistemului sunt punctele de intersecție ale curbelor plane determinate de ecuațiile sistemului. Sistemul inițial este echivalent cu următorul sistem:\n$$\n\\left\\{\\begin{array}{c}\n\\sqrt{(x+3)^{2}+(y+5)^{2}}+\\sqrt{(x+3)^{2}+(y+3)^{2}}=2 \\\\\n(x-2)^{2}+y... | Moldova | A 62 - A OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | a ∈ [40, 56] | |
0elf | Problem:
Dana je enačba $6 z^{10}+a_{9} z^{9}+a_{8} z^{8}+a_{7} z^{7}+\ldots+a_{3} z^{3}+a_{2} z^{2}+a_{1} z+9216=0$, kjer so $a_{1}, a_{2}, \ldots, a_{9}$ realna števila in $z$ neznanka. Poznamo 5 rešitev dane enačbe, in sicer $1+i, 2+2 i, 3+3 i, 4+4 i$ ter $\frac{1}{2}$, kjer je $i$ imaginarna enota. Določi vrednost ... | [
"Solution:\nKer ima polinom na levi strani enačbe realne koeficiente, njegove kompleksne ničle nastopanjo v kompleksnih parih. Poleg danih rešitev enačbe so zato rešitve enačbe tudi $1-i, 2-2 i$, $3-3 i$ in $4-4 i$. Poznamo torej 9 ničel polinoma. Ker je polinom stopnje 10, ima 10 ničel štetih z večkratnostjo. Ozna... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | -125 | |
0721 | Problem:
Consider an acute triangle $A B C$ and let $P$ be an interior point of $A B C$. Suppose the lines $B P$ and $C P$, when produced, meet $A C$ and $A B$ in $E$ and $F$ respectively. Let $D$ be the point where $A P$ intersects the line segment $E F$ and $K$ be the foot of perpendicular from $D$ on to $B C$. Show... | [
"Solution:\n\nProduce $A P$ to meet $B C$ in $Q$. Join $K E$ and $K F$. Draw perpendiculars from $F$ and $E$ on to $B C$ to meet it in $M$ and $L$ respectively. Let us denote $\\angle B K F$ by $\\alpha$ and $\\angle C K E$ by $\\beta$. We show that $\\alpha=\\beta$ by proving $\\tan \\alpha=\\tan \\beta$. This imp... | India | INMO | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0cnt | On 11 sheets of paper 11 following statements are written (each sheet contains exactly one statement):
1) There are no sheets with false statements to the left of this sheet.
2) There is exactly 1 sheet with a false statement to the left of this sheet.
3) There are exactly 2 sheets with false statements to the left of ... | [
"Answer: 6.\n\nIf the sheets are arranged in the following order:\n\n(for brevity, each sheet is marked with the number of false statements to its left), then there will be 6 true statements.\n\nLet us show that there cannot be more. Suppose that two sheets with true statements are placed n... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | 6 | |
01kw | Find all triples of positive integers $(x, y, z)$ satisfying the equality $3^x + 7^y = 4^z$. | [
"$z \\ge 3$, then $2^z - 1 \\vdash 3$, but it is possible only if $z$ is even. Let $z = 2c$ for some nonnegative integer $c$. Then $4^c - 3^a = 1$. If $a = 1$, we find $c = 1$ and obtain $x = 1, z = 2, y = 1$. If $a > 1$, then $4^c = 3^a + 1$ has remainder 1 when divided by 9. The power of 4 has this remainder only... | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (2,1,2) | |
0j3s | Problem:
Let $ABCD$ be a quadrilateral with an inscribed circle centered at $I$. Let $CI$ intersect $AB$ at $E$. If $\angle IDE = 35^{\circ}$, $\angle ABC = 70^{\circ}$, and $\angle BCD = 60^{\circ}$, then what are all possible measures of $\angle CDA$? | [
"Solution:\nArbitrarily defining $B$ and $C$ determines $I$ and $E$ up to reflections across $BC$. $D$ lies on both the circle determined by $\\angle EDI = 35^{\\circ}$ and the line through $C$ tangent to the circle (and on the opposite side of $B$); since the intersection of a line and a circle has at most two poi... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 70°, 160° | |
08li | Problem:
For a fixed triangle $A B C$ we choose a point $M$ on the ray $C A$ (after $A$), a point $N$ on the ray $A B$ (after $B$), and a point $P$ on the ray $B C$ (after $C$) in a way such that $A M - B C = B N - A C = C P - A B$. Prove that the angles of triangle $M N P$ do not depend on the choice of $M, N, P$. | [
"Solution:\n\nConsider the points $M'$ on the ray $B A$ (after $A$), $N'$ on the ray $C B$ (after $B$), and $P'$ on the ray $A C$ (after $C$), so that $A M = A M'$, $B N = B N'$, $C P = C P'$. Since $A M - B C = B N - A C = B N' - A C$, we get $C M = A C + A M = B C + B N' = C N'$. Thus triangle $M C N'$ is isoscel... | JBMO | 2008 Shortlist JBMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
06vd | On a certain social network, there are $2019$ users, some pairs of which are friends, where friendship is a symmetric relation. Initially, there are $1010$ people with $1009$ friends each and $1009$ people with $1010$ friends each. However, the friendships are rather unstable, so events of the following kind may happen... | [
"All of the solutions presented below will use this reformulation.\n\nNote that the given graph is connected, since the total degree of any two vertices is at least $2018$ and hence they are either adjacent or have at least one neighbour in common. Hence the given graph satisfies the following condition:\n\n> Every... | IMO | IMO 2019 Shortlisted Problems | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
09l0 | There were $59$ coins of different weights. Show that you can find the heaviest and second heaviest coins by weighing them a total of $64$ times using a balanced scale. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
05c7 | Find all functions $f$ from the set of all non-negative real numbers to the set of all real numbers such that $f(1) = 1$ and
$$
(f(x+y))^2 \le f(x^2 - 2xy + y^2)
$$
for all real numbers $x$ and $y$ that satisfy the inequality $x + y \ge 0$. | [
"The given inequality can be rewritten as\n$$\n(f(x+y))^2 \\le f((x-y)^2).\n$$\n\nTake $y = 1-x$. Then $1 = 1^2 = (f(1))^2 \\le f((2x-1)^2)$. As $(2x-1)^2$ obtains all non-negative real values, we can conclude that $1 \\le f(z)$ for every non-negative real number $z$.\nNow take $x \\ge -\\frac{1}{2}$ and $y = 1+x$.... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 1 for all non-negative real x | |
0dbk | Each integer is painted one of three colors. Prove that there exist two distinct numbers of the same color, whose difference is a perfect square. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | English | proof only | null | |
0iy2 | Problem:
Five guys are eating hamburgers. Each one puts a top half and a bottom half of a hamburger bun on the grill. When the buns are toasted, each guy randomly takes two pieces of bread off of the grill. What is the probability that each guy gets a top half and a bottom half? | [
"Solution:\n\nSay a guy is content if he gets a top half and a bottom half. Suppose, without loss of generality, that the first guy's first piece of bread is a top. Then there is a $\\frac{5}{9}$ chance that his second piece of bread is a bottom. By the same reasoning, given that the first guy is content, there is ... | United States | 2nd Annual Harvard-MIT November Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 8/63 | |
0aiw | A circle $k$ with center at $O$ and radius $r$ and a line $p$ which doesn't have a common point with $k$ are given. Let $E$ be the foot of the perpendicular from $O$ to $p$. An arbitrary point $M$ different from $E$ is chosen on $p$ and the two tangents are drawn from $M$ to $k$ which touch the circle $k$ at points $A$... | [
"Let $G$ be the intersection of $OM$ and $AB$. Since $\\triangle OGH \\sim \\triangle OEM$ we get $\\overline{OG} = \\overline{OE}$ hence $\\overline{OE} \\cdot \\overline{OH} = \\overline{OM} \\cdot \\overline{OG}$. On the other hand, since $\\triangle AOG \\sim \\triangle MOA$, we have $\\overline{OA} = \\overlin... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
09qf | Problem:
Het polynoom $A(x)=x^{2}+a x+b$ met gehele coëfficiënten heeft de eigenschap dat voor elk priemgetal $p$ er een geheel getal $k$ bestaat zodat $A(k)$ en $A(k+1)$ beide deelbaar zijn door $p$. Bewijs dat er een geheel getal $m$ bestaat zodat $A(m)=A(m+1)=0$. | [
"Solution:\n\nOplossing I. Zij $p$ een priemgetal en zij $k$ zodat $A(k)$ en $A(k+1)$ beide deelbaar zijn door $p$. Het verschil van $A(k)$ en $A(k+1)$ is dan ook deelbaar door $p$ en dit is gelijk aan\n$$\nA(k+1)-A(k)=(k+1)^{2}+a(k+1)+b-\\left(k^{2}+a k+b\\right)=2 k+1+a,\n$$\ndus $2 k \\equiv -1-a \\bmod p$. Omda... | Netherlands | toets | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0ahw | A regular hexagon with side-length $1$ is given. Inside the hexagon $m$ points are given so that no three of them are collinear. The hexagon is split into triangles, so that each of the given $m$ points and each vertex of the hexagon is a vertex of one such triangle. The triangles into which the hexagon is split have n... | [
"First we determine the total number of splitting triangles into which the hexagon is split. Let $A$ be one arbitrary point of the interior $m$ points. The sum of all angles at the point $A$ is $360^\\circ$ (the sum of all angles at $A$ of all triangles having that point as a vertex). On the other hand, the sum of ... | North Macedonia | Junior Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0j1j | Problem:
Convex quadrilateral $BCDE$ lies in the plane. Lines $EB$ and $DC$ intersect at $A$, with $AB = 2$, $AC = 5$, $AD = 200$, $AE = 500$, and $\cos \angle BAC = \frac{7}{9}$. What is the largest number of nonoverlapping circles that can lie in quadrilateral $BCDE$ such that all of them are tangent to both lines $... | [
"Solution:\n\nLet $\\theta = \\angle BAC$, and $\\cos \\theta = \\frac{7}{9}$ implies $\\cos \\frac{\\theta}{2} = \\sqrt{\\frac{1 + \\frac{7}{9}}{2}} = \\frac{2\\sqrt{2}}{3}$; $\\sin \\frac{\\theta}{2} = \\frac{1}{3}$; $BC = \\sqrt{4 + 25 - 2(2)(5) \\frac{7}{9}} = \\frac{11}{3}$.\n\nLet $O_1$ be the excircle of $\\... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Method... | null | proof and answer | 5 | |
00ig | On a circular billiard table, a ball is reflected from a cushion as if it were being reflected from the tangent of the circle in the point of reflection.
A regular hexagon is drawn on a circular billiard table with its vertices on the circle.
A ball (in the form of a point) is placed on a side of the hexagon (not in on... | [
"Whenever a ball is reflected on the perimeter of the circular cushion, the incoming and outgoing angles to the radius must be equal. This means that the incoming and outgoing chords of the circle on the path on the ball must be of equal length. The periodic path must therefore be a regular polygon, and since it mu... | Austria | Austria 2010 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 4 | |
0550 | Masha has an electric carouse in her garden that she rides every day. As she likes order, she always leaves the carouse in the same position after each ride. But every night three bears sneak into the garden and start turning the carouse. Bear dad turns the carouse each time by $\frac{1}{7}$ of the full circle. Bear mu... | [
"As $7 \\cdot 9 \\cdot 32 = 2016$, all turns are integral multiples of $\\frac{1}{2016}$ of the full turn. Thus the carouse can be in at most 2016 distinct positions. It remains to show that all these positions are possible. For that, we show that the bears can turn the carouse by exactly $\\frac{1}{2016}$ of the f... | Estonia | Open Contests | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof and answer | 2016 | |
0eh6 | Problem:
Na krožnici s polmerom $r$ naključno izberemo dve točki. Kolikšna je verjetnost dogodka, da sta izbrani točki oddaljeni za več kot $\sqrt{2} r$ in manj kot $\sqrt{3} r$? | [
"Solution:\n\nPrivzemimo oznake s slike, na kateri je prva izbrana točka na krožnici označena z $A$. Da bo razdalja med izbranima točkama večja od $\\sqrt{2} r$ in manjša od $\\sqrt{3} r$, mora druga izbrana točka $B$ ležati na enem od dveh odebeljenih lokov na sliki. Verjetnost tega dogodk... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 1/6 | |
0kjx | A rectangle with side lengths $1$ and $3$, a square with side length $1$, and a rectangle $R$ are inscribed inside a larger square as shown. The sum of all possible values for the area of $R$ can be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m + n$?
![](attache... | [
"Label the diagram as shown below, where $O$, $P$, and $Q$ are the feet of the perpendicular segments to the respective sides. Let $x = AE$ and $y = AF$.\n\n\n\nBecause\n$$\n\\triangle FAE \\cong \\triangle EOH \\cong \\triangle IOH \\cong \\triangle KCJ \\cong \\triangle FPG,\n$$\nit follo... | United States | Fall 2021 AMC 10 B | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | MCQ | E | |
06kh | Find all positive integers $a_0$, $a_1$, $a_2$, $b_0$, $b_1$, $b_2$ such that
$a_2 b_2 n^2 + a_1 b_1 n + a_0 b_0$ divides $(a_2^{2017n} + b_2)n^2 + (a_1^{2017n} + b_1)n + (a_0^{2017n} + b_0)$ for any positive integer $n$. | [
"Suppose that\n$$\na_2 b_2 n^2 + a_1 b_1 n + a_0 b_0 \\mid (a_2^{2017n} + b_2)n^2 + (a_1^{2017n} + b_1)n + (a_0^{2017n} + b_0) \\quad (1)\n$$\nWe first claim that $a_0 = b_0 = 1$. Let $n$ be a large multiple of $a_0 b_0$. Since $a_0 b_0$ divides the left-hand side of (1), it also divides the right-hand side and hen... | Hong Kong | Pre-IMO 2017 Mock Exam | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization >... | English | proof and answer | a0 = a1 = a2 = b0 = b1 = b2 = 1 | |
09d6 | $ABCD$ гүдгэр 4 өнцөгтийн $\angle BAC = \angle CAD$, $\angle ABC = \angle ACD$ болно. $AD$ ба $BC$ талуудын үргэлжлэл $E$ цэгээр огтлоцөх бол $AB \cdot DE = BC \cdot CE$ гэж батал. | [
"\n$$\n\\angle ACE = \\angle ABC + \\angle BAC \\Rightarrow \\angle ACE = \\angle ACD + \\angle CAD\n$$\n⇒ $\\angle CAD = \\angle DCE$ болж $CED$ ба $AEC$ гурвалжнууд төсөөтэй. Иймд (1) $\\frac{CE}{AE} = \\frac{DE}{CE}$. Мөн $\\angle BAC = \\angle CAE$ гэдгээс биссектрисийн чанараар (2) $\\... | Mongolia | ММО-48 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | Mongolian | proof only | null | |
062u | Problem:
Eine Folge $x_{1}, x_{2}, \ldots$ ist definiert durch $x_{1}=1$ und $x_{2k}=-x_{k}$, $x_{2k-1}=(-1)^{k+1} x_{k}$ für alle $k \geq 1$. Man zeige, dass für alle $n \geq 1$ gilt: $x_{1}+x_{2}+\ldots+x_{n} \geq 0$. | [
"Solution:\n\nWir bezeichnen $S_{n}=x_{1}+\\ldots+x_{n}$ und führen einen Induktionsbeweis. Nachrechnen ergibt $x_{1}=1$, $x_{2}=-1$, $x_{3}=1$, $x_{4}=1$ und damit $S_{n} \\geq 0$ für $1 \\leq n \\leq 4$.\n\nNun nehmen wir an, dass $S_{k} \\geq 0$ für alle natürlichen $k<n$ gilt, und behaupten $S_{n} \\geq 0$.\n\n... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0849 | Problem:
Calcolare l'area dell'intersezione di tre cerchi aventi come rispettivi diametri i tre lati di un triangolo equilatero di lato unitario.
(A) $\frac{\sqrt{3}-\sqrt{2}}{3}$
(B) $\frac{\pi-3}{2}$
(C) $\pi-3$
(D) $\frac{\pi-\sqrt{3}}{4}$
(E) $\frac{\pi-\sqrt{3}}{8}$. | [] | Italy | Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO TRIENNIO | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | E | |
02bk | Problem:
Uma brincadeira - É feita uma brincadeira com quatro números inteiros da seguinte maneira: some três desses números, divida essa soma por $3$ e o resultado some com o quarto número. Existem quatro formas de fazer esta brincadeira, obtendo os seguintes resultados: $17$, $21$, $23$ e $29$. Qual é o maior dos qu... | [
"Solution:\n\nSejam $a$, $b$, $c$ e $d$ os números procurados. São dados os números\n$$\n\\frac{a+b+c}{3} + d, \\quad \\frac{a+b+d}{3} + c, \\quad \\frac{a+c+d}{3} + b, \\quad \\text{e} \\quad \\frac{b+c+d}{3} + a\n$$\nmas não sabemos a ordem deles. Como\n$$\n\\begin{aligned}\n\\frac{a+b+c}{3} + d + \\frac{a+b+d}{3... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 21 | |
0e4n | How many positive integers smaller than $1000$ have the sum of the digits divisible by $7$ and are multiples of $3$?
(A) $7$
(B) $19$
(C) $21$
(D) $28$
(E) $37$ | [
"The sum of the digits of a multiple of $3$ is divisible by $3$. So, we are looking for numbers with the sum of the digits divisible by $21$. The sum of the digits of any number smaller than $1000$ is less than or equal to $9 + 9 + 9 = 27$. So, in our case the sum of the digits is equal to $21$.\n\nThe smallest pos... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | MCQ | D | |
08oz | Problem:
What is the greatest number of integers that can be selected from a set of 2015 consecutive numbers so that no sum of any two selected numbers is divisible by their difference? | [
"Solution:\nWe take any two chosen numbers. If their difference is $1$, it is clear that their sum is divisible by their difference. If their difference is $2$, they will be of the same parity, and their sum is divisible by their difference. Therefore, the difference between any chosen numbers will be at least $3$.... | JBMO | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 672 | |
04fw | An $8 \times 8$ board is initially painted as a standard chessboard in black and white. In each move we choose one row or column and change the colour of each of the eight squares in that row or column from white to black or vice versa. Is it possible to achieve that after a finite sequence of such moves exactly one sq... | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | No | |
0lg3 | Problem:
The area of a convex pentagon $ABCDE$ is $S$, and the circumradii of the triangles $ABC$, $BCD$, $CDE$, $DEA$, $EAB$ are $R_{1}$, $R_{2}$, $R_{3}$, $R_{4}$, $R_{5}$. Prove the inequality
$$
R_{1}^{4} + R_{2}^{4} + R_{3}^{4} + R_{4}^{4} + R_{5}^{4} \geq \frac{4}{5 \sin^{2} 108^{\circ}} S^{2}
$$ | [
"Solution:\n\nLemma 1. The area $S$ of a convex $n$-gon $A_{1}A_{2}\\ldots A_{n}$ satisfies\n$$\n2S \\leq \\frac{1}{2} \\sum_{i=1}^{n} A_{i-1}A_{i+1} \\cdot R_{i} = \\sum_{i=1}^{n} R_{i}^{2} \\sin A_{i}\n$$\nwhere $R_{i}$ is the radius of the circumcircle of $\\triangle A_{i-1}A_{i}A_{i+1}$, and the indices are red... | Zhautykov Olympiad | XI International Zhautykov Olympiad in Sciences | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Jensen/smoothing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Eq... | null | proof only | null | |
0led | Given a positive integer $n > 1$. Let $T$ be the set of all ordered triples $(x, y, z)$ where $x, y$ and $z$ are different positive integers and $1 \le x, y, z \le 2n$. A set $A$ containing ordered pairs $(u, v)$ is called 'connected' to $T$ if for all $(x, y, z) \in T$ then
$$
\{(x, y), (x, z), (y, z)\} \cap A \neq \e... | [
"a) The answer is $|T| = 2n(2n-1)(2n-2)$.\n\nb) Consider the set\n$$\nA = \\{(i, j) : 1 \\le i, j \\le n, i \\ne j\\} \\cup \\{(i, j) : n+1 \\le i, j \\le 2n, i \\ne j\\}.\n$$\nClearly, $|A| = 2 \\cdot A_n^2 = 2n(n-1)$. Furthermore, for all $(x, y, z) \\in T$ there always exist two in three numbers $x, y, z$ belong... | Vietnam | VMO | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) |T| = 2n(2n−1)(2n−2). b) There exists a connected set A with |A| = 2n(n−1). c) Every connected set A has at least 2n(n−1) elements, so the minimum possible size is 2n(n−1). |
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