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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0dk1 | On the board, there are written numbers $1, 2, 3, \dots, 2023^{2024}$. At each step, it is allowed to erase any two numbers $a, b$ on the board and then replace them with $\gcd(2024ab, a^2 + 254ab + b^2)$, keep doing this until there is only one number $x$ left. Find all possible values of $x$.
On the board, there are... | [
"Let $f(a, b) = \\text{gcd}(2024ab, a^2 + 254ab + b^2)$ then for 2 numbers $a, b$ on the board, the new number generated will be $f(a, b)$. We consider some cases:\n* If $a, b$ are same parity then $f(a, b)$ is even.\n* If $a, b$ are different from parity then $f(a, b)$ is odd.\nThis means that the number of odd nu... | Saudi Arabia | Saudi Arabia booklet 2024 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 1 | |
07oz | Let $n$ be a positive integer and $a_1, \dots, a_n$ be positive real numbers. Let $g(x)$ denote the product $(x + a_1) \cdots (x + a_n)$. Let $a_0$ be a real number and let
$$
f(x) = (x - a_0)g(x) = x^{n+1} + b_1 x^n + b_2 x^{n-1} + \dots + b_n x + b_{n+1}.
$$
Prove that all the coefficients $b_1, b_2, \dots, b_{n+1}$ ... | [
"Note first that $b_1 = (a_1 + a_2 + \\dots + a_n) - a_0$, so the condition\n$$\na_0 > a_1 + a_2 + \\dots + a_n\n$$\nis necessary for $b_1 < 0$.\nBelow it will be useful to know that in the case $a_0 = a_1 + a_2 + \\dots + a_n$, the coefficients $b_i$ are all non-positive. First note that $a_0 = a_1 + a_2 + \\dots ... | Ireland | Irska 2014 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
00sx | A strategical video game consists of a map of finitely many towns. In each town there are $k$ directions, labelled from 1 through $k$. One of the towns is designated as initial, and one – as terminal. Starting from the initial town the hero of the game makes a finite sequence of moves. At each move the hero selects a d... | [
"Without loss of generality we may assume that the set of directions is $D = \\{1, 2, \\dots, k\\}$. Let us enumerate the towns in the first game from 1 through $N$ and the towns in the second game from $N+1$ through $N+M$. Without loss of generality we may assume that the initial and terminal towns in the first ga... | Balkan Mathematical Olympiad | BMO Short List | [
"Discrete Mathematics > Algorithms",
"Algebra > Linear Algebra > Matrices",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Other"
] | English | proof only | null | |
0ivd | Problem:
A positive integer is called jubilant if the number of 1's in its binary representation is even. For example, $6=110_{2}$ is a jubilant number. What is the 2009th smallest jubilant number? | [
"Solution:\n\nNotice that for each pair of consecutive positive integers $2k$ and $2k+1$, their binary representation differs by exactly one 1 (in the units digit), so exactly one of $2$ and $3$ is jubilant, exactly one of $4$ and $5$ is jubilant, etc. It follows that there are exactly $2009$ jubilant numbers less ... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | final answer only | 4018 | |
0dvr | Problem:
V jami pod Krimom spi grozna pošast. Ko postane lačna, se zbudi in požre toliko ovc, kolikor je vsota števk tistega leta. Potem spet zaspi za toliko let, kolikor ovc je pojedla. Vemo, da se je zbudila 12. aprila leta 354. Ali je pošast lahko pred vrati? | [
"Solution:\n\nKer je število $354$ deljivo s $3$, je tudi vsota njegovih števk in zato število požrtih ovc deljivo s $3$. Tudi vsako naslednje leto, ko se pošast zbudi, je torej deljivo s $3$. Pošast se leta $2003$ ne more zbuditi, saj $2003$ ni deljivo s $3$; torej tudi danes, $12.$ aprila $2003$, ne more biti pre... | Slovenia | 47. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | No | |
05iq | Problem:
Soient un entier $n$ et des réels $0 < u_{1} < u_{2} < \ldots < u_{n}$ tels que
$$
u_{1} + u_{2} + \ldots + u_{n} = \frac{1}{u_{1}^{2}} + \frac{1}{u_{2}^{2}} + \ldots + \frac{1}{u_{n}^{2}}$$
Montrer que, pour tout entier $k$ inférieur ou égal à $n$, il existe $k$ réels parmi $u_{1}, u_{2}, \ldots, u_{n}$ dont... | [
"Solution:\n\n$\\triangleright$ Commençons par montrer le résultat pour $k = n$. Notons $a = \\frac{u_{1} + \\ldots + u_{n}}{n}$ et $g = (u_{1} \\ldots u_{n})^{\\frac{1}{n}}$ les moyennes arithmétique et géométrique de $u_{1}, \\ldots, u_{n}$. Alors, l'inégalité classique entre moyenne arithmétique-géométrique-harm... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
02z7 | Problem:
O jogo "Dado Geográfico" consiste em cada participante jogar um dado uma quantidade $n$ de vezes e anotar a sequência, sendo que o primeiro lançamento indica a distância, em metros, que o participante andará para o Norte, o segundo para o Leste, o terceiro para o Sul, o quarto para o Oeste, o quinto para o No... | [
"Solution:\n\na) Márcia andou $2+6=8$ m para o Norte, $1+5=6$ m para o Leste, $4$ m para o Sul e $3$ m para o Oeste. Na direção Norte-Sul, Márcia andou $8-4=4$ m para o Norte e na direção Leste-Oeste, Márcia andou $6-3=3$ m para o Leste. Como estas direções são ortogonais, basta aplicarmos o Teorema de Pitágoras pa... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | a) 5 m; b) 36; c) 90 | |
02mm | Problem:
Círculos dentro do quadrado - Dentro de um quadrado são colocados círculos, dois a dois disjuntos ou, então, tangentes externamente. Se o lado do quadrado mede $1~\mathrm{cm}$, será possível colocar tantos desses círculos de tal modo que a soma de seus raios, em centímetros, seja maior do que $2008$? | [] | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | No | |
0i0o | Problem:
Three grasshoppers are on a straight line. Every second one grasshopper jumps. It jumps across one (but not across two) of the other grasshoppers. Prove that after 1999 seconds the grasshoppers cannot be in the initial position. | [
"Solution:\n\nAssume that the grasshoppers are initially in the order $(A, B, C)$ on the line. Call $(A, B, C)$, $(B, C, A)$, $(C, A, B)$ \"even\" orders and $(A, C, B)$, $(B, A, C)$, $(C, B, A)$ \"odd\" orders. We claim that if the grasshoppers are in an even order, they will be in an odd order the next second, an... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof only | null | |
029u | Problem:
O preço do feijão - A tabela e o gráfico, dados a seguir, mostram a evolução do preço médio de três tipos de feijão, $A$, $B$ e $C$, na bolsa de alimentos durante os primeiros quatro meses de certo ano: Desses 3 tipos, os que apresentam, respectivamente, o maior e o menor crescimento percentual no preço nesse... | [] | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | a | |
05u4 | Problem:
On définit la suite $a_{1}, a_{2}, a_{3} \ldots$ de la façon suivante : $a_{1}=63$ et, pour tout entier $n \geqslant 2$, $a_{n}$ est le plus petit multiple de $n$ qui soit supérieur ou égal à $a_{n-1}$. Démontrer que les termes de notre suite sont deux à deux distincts. | [
"Solution:\n\nIl est clair que la suite $\\left(a_{n}\\right)_{n \\geqslant 1}$ est croissante. Il s'agit donc de démontrer qu'elle est strictement croissante.\n\nOn calcule donc les premiers termes de la suite : $a_{1}=63, a_{2}=64, a_{3}=66, a_{4}=68, a_{5}=70$, $a_{6}=72, a_{7}=77, a_{8}=80, a_{9}=81, a_{10}=90,... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
09a6 | Let $M$ be a point on the circumscribed circle of an acute triangle $ABC$ and distinct from the vertices of $ABC$. $MP$ and $MQ$ are perpendicular lines from $M$ to $AB$ and $AC$ respectively and $H$ is orthocenter of triangle $ABC$. Prove that $PQ$ intersects $HM$ at the midpoint of $HM$. | [
"Let us denote the feet of the perpendicular from $M$ to $AC$ by $R$. Then $P$, $Q$ and $R$ are collinear (Simson's line).\n\n\n\nIf $\\angle CAM = \\varphi_1$, $\\angle ACM = \\varphi_2$ and $\\angle ABC = \\beta$, then $\\varphi_1 + \\varphi_2 = \\beta$. We first show that $S_{PHQ} = S_{P... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
09m2 | $a$, $b$ and $c$ are non-zero real numbers such that $\frac{a+b}{c} = \frac{b+c}{a} = \frac{c+a}{b}$.
(1) Prove that $a^3 + b^3 + c^3 \neq 0$.
(2) Determine all possible values of the expression $\frac{(a+b)(b+c)(c+a)}{a^3 + b^3 + c^3}$. | [
"Answer: $F = -1/3, 8/3$.\nLet us denote $s = a + b + c$. Then, under the given condition, we have $\\frac{s}{c} = \\frac{s}{a} = \\frac{s}{b}$.\n(1) If $s = 0$, then $a^3+b^3+c^3 = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc = 3abc \\neq 0$. If $s \\neq 0$, then $a = b = c$. Hence $a^3 + b^3 + c^3 = 3a^3 \\neq 0$. This com... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | -1/3, 8/3 | |
042m | For positive integer $k > 1$, let $f(k)$ be the number of ways of factoring $k$ into product of positive integers greater than 1 (The order of factors are not counted, for example $f(12) = 4$, as $12$ can be factored in these 4 ways: $12$, $2 \times 6$, $3 \times 4$, $2 \times 2 \times 3$).
**Prove:** If $n$ is a posi... | [
"Let $P(n)$ stand for the biggest prime divisor of $n$, and define $P(1) = f(1) = 1$. We first prove two lemmas.\n\nLemma 1: For positive integer $n$ and prime $p \\mid n$, we have $f(n) \\le \\sum_{d|\\frac{n}{p}} f(d)$.\n\n**Proof of Lemma 1:** For any factoring of $n$, write $n = n_1 n_2 \\cdots n_k$. Since $p \... | China | China Team Selection Test | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)"
] | English | proof only | null | |
0l8v | Find all triangles $ABC$ such that the angle $\angle BCA$ is acute, the perpendicular bisector of the segment $BC$ cuts the two rays dividing the angle $\angle BAC$ into three equal angles ($\angle BAx = \angle xAy = \angle yAC$) at the points $N$ and $P$ satisfying the conditions $AB = NP = 2HM$, where $H$ is the orth... | [] | Vietnam | CONTEST FOR THE SELECTION OF VIETNAMESE INTERNATIONAL MATHEMATICAL OLYMPIAD TEAM | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | All right isosceles triangles with right angle at B, i.e., AB ⟂ BC and AB = BC (so angle C is forty-five degrees and acute). | |
06ow | In a triangle $A B C$, let $M_{a}, M_{b}, M_{c}$ be respectively the midpoints of the sides $B C, C A, A B$ and $T_{a}, T_{b}, T_{c}$ be the midpoints of the arcs $B C, C A, A B$ of the circumcircle of $A B C$, not containing the opposite vertices. For $i \in \{a, b, c\}$, let $\omega_{i}$ be the circle with $M_{i} T_{... | [
"Let $T_{a} T_{b}$ intersect circle $\\omega_{b}$ at $T_{b}$ and $U$, and let $T_{a} T_{c}$ intersect circle $\\omega_{c}$ at $T_{c}$ and $V$. Further, let $U X$ be the tangent to $\\omega_{b}$ at $U$, with $X$ on $A C$, and let $V Y$ be the tangent to $\\omega_{c}$ at $V$, with $Y$ on $A B$. The homothety with cen... | IMO | IMO 2006 Shortlisted Problems | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 1/4 | |
0b81 | For each positive integer $n$, find the largest real number $C_n$ with the following property. Given any $n$ real-valued functions $f_1(x), f_2(x), \dots, f_n(x)$ defined on the closed interval $0 \le x \le 1$, one can find numbers $x_1, x_2, \dots, x_n$, such that $0 \le x_i \le 1$, satisfying
$$
|f_1(x_1) + f_2(x_2) ... | [
"First we will prove that $C_n \\ge \\frac{n-1}{2n}$, i.e. that for any $n$ functions $f_1, f_2, \\dots, f_n : [0, 1] \\to \\mathbb{R}$, there exist numbers $x_1, x_2, \\dots, x_n$ in $[0, 1]$ such that\n$$\n|f_1(x_1) + f_2(x_2) + \\dots + f_n(x_n) - x_1x_2\\dots x_n| \\ge \\frac{n-1}{2n}.\n$$\nFor $n=1$ this is tr... | Romania | Local Mathematical Competitions | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | (n-1)/(2n) | |
0k1f | Problem:
$\triangle P N R$ has side lengths $P N = 20$, $N R = 18$, and $P R = 19$. Consider a point $A$ on $P N$. $\triangle N R A$ is rotated about $R$ to $\triangle N' R A'$, so that $R$, $N'$, and $P$ lie on the same line and $A A'$ is perpendicular to $P R$. Find $\frac{P A}{A N}$. | [
"Solution:\n\nDenote the intersection of $P R$ and $A A'$ be $D$. Note $R A' = R A$, so $D$, being the altitude of an isosceles triangle, is the midpoint of $A A'$. Thus,\n$$\n\\angle A R D = \\angle A' R D = \\angle N R A\n$$\nso $R A$ is the angle bisector of $P N R$ through $R$. By the angle bisector theorem, we... | United States | HMMT February 2018 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 19/18 | |
07mi | There are 14 boys in a class. Each boy is asked how many other boys in the class have his first name, and how many have his last name. It turns out that each number from 0 to 6 occurs among the answers.
Prove that there are two boys in the class with the same first name and the same last name. | [
"Consider groups of students with the same first name – these groups partition the 14 students. Also consider groups of students with the same last name – these groups also partition the 14 students. Each student belongs to two groups, and by assumption there are groups of size 1, 2, 3, 4, 5, 6 and 7; but these num... | Ireland | Irish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
00nz | Written on a blackboard are the $2023$ numbers
$2023, 2023, \ldots, 2023$.
The numbers on the blackboard are now modified, in a sequence of moves. In each move, two numbers on the blackboard—call them $x$ and $y$—are chosen, deleted, and replaced by the single number $\frac{x+y}{4}$. Such moves are carried out until th... | [
"The expression $\\frac{x+y}{4}$ reminds us of the arithmetic mean. By the AM-HM inequality, we have\n$$\n\\frac{x+y}{2} \\geq \\frac{2}{\\frac{1}{x} + \\frac{1}{y}}\n$$\nor\n$$\n\\frac{1}{x} + \\frac{1}{y} \\geq \\frac{1}{(x+y)/4}\n$$\nThis inequality leads us to consider an argument concerning the reciprocals of ... | Austria | AUT_ABooklet_2023 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
00mq | On a circle 2018 points are marked.
Each of these points is labeled with an integer. Let each number be larger than the sum of the preceding two numbers in clockwise order.
Determine the maximal number of positive integers that can occur in such a configuration of 2018 integers. | [
"Let the points be labeled $a_0, a_1, \\dots, a_{2017}$ clockwise with cyclical notation, i.e., $a_{k+2018} = a_k$ for all integers $k$.\n\nLemma. In a valid configuration, no two neighbouring numbers can be both non-negative.\n\nProof. Assume that there exist neighbouring numbers $a_{k-1}$ and $a_k$ which are both... | Austria | 49th Austrian Mathematical Olympiad, National Competition (Final Round, part 2) | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 1008 | |
01t9 | For a finite set $A$ of positive integers, we call a partition of $A$ into two disjoint nonempty subsets $A_1$ and $A_2$ good if the least common multiple of the elements in $A_1$ is equal to the greatest common divisor of the elements in $A_2$.
Determine the minimum value of $n$ such that there exists a set of $n$ po... | [] | Belarus | 66th Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | null | |
0ji2 | Problem:
Let $x$, $y$ be complex numbers such that $\frac{x^{2}+y^{2}}{x+y}=4$ and $\frac{x^{4}+y^{4}}{x^{3}+y^{3}}=2$. Find all possible values of $\frac{x^{6}+y^{6}}{x^{5}+y^{5}}$. | [
"Solution:\n\nAnswer: $10 \\pm 2 \\sqrt{17}$\n\nLet $A=\\frac{1}{x}+\\frac{1}{y}$ and let $B=\\frac{x}{y}+\\frac{y}{x}$. Then\n$$\n\\frac{B}{A}=\\frac{x^{2}+y^{2}}{x+y}=4\n$$\nso $B=4A$. Next, note that\n$$\nB^{2}-2=\\frac{x^{4}+y^{4}}{x^{2} y^{2}} \\text{ and } AB-A=\\frac{x^{3}+y^{3}}{x^{2} y^{2}}\n$$\nso\n$$\n\\... | United States | HMMT | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 10 ± 2√17 | |
0khd | A square with side length $3$ is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length $2$ has two vertices on the other square and the other two on sides of the triangle, as shown. What is the area of the triangle?
(A) $19\fra... | [
"Label the vertices as shown in the diagram.\n\nThen $\\triangle ABC$, $\\triangle CDE$, and $\\triangle EFG$ are similar. Because $CD = 2$ and $DE = \\frac{3-2}{2} = \\frac{1}{2}$, the lengths of the legs of each of these triangles are in the ratio of $4$ to $1$. It follows that $FG = \\fr... | United States | Fall 2021 AMC 10 B | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | MCQ | B | |
09fw | Let $x$, $y$ and $z$ be real numbers in the interval $(-1, 1)$. Prove that
$$
\frac{1}{(1-x^2)(1-y^2)(1-z^2)} + \frac{2}{(1-xy)(1-yz)(1-zx)} \geq \\
\frac{1}{(1-x^2)(1-yz)^2} + \frac{1}{(1-y^2)(1-zx)^2} + \frac{1}{(1-z^2)(1-xy)^2}.
$$ | [
"**Lemma.** Let $a, b, c, d, e$ and $f$ be real numbers. If for all real numbers $\\lambda$ and $\\mu$, we have\n$$\na + b\\lambda^2 + c\\mu^2 + 2(d\\lambda + e\\lambda\\mu + f\\mu) \\ge 0, \\qquad (1)\n$$\nthen we have\n$$\nabc + 2def \\ge ae^2 + bf^2 + cd^2. \\qquad (2)\n$$\n\n*Proof.* Clearly, $b \\ge 0$. If $b=... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
07nl | Let $I$ be the incentre of $\triangle ABC$. Prove that the circumcentre of $\triangle BIC$ lies on the circumcircle of $\triangle ABC$. | [
"Let the bisector of $\\angle BAC$ meet the circumcircle of $\\triangle ABC$ at $D$.\n\nBecause $\\angle BAD = \\angle DAC$, we have $|DB| = |DC|$. On the other hand, $\\angle BID = \\angle BAI + \\angle ABI = \\angle CAI + \\angle CBI = \\angle CBD + \\angle CBI = \\angle DBI$ and so $|DB|... | Ireland | Ireland | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hzj | Problem:
Evaluate $\frac{d}{d x}\left(\sin x-\frac{4}{3} \sin ^{3} x\right)$ when $x=15$. | [
"Solution:\n\nOf course this problem can be done by brute force, differentiating and then using the half angle formula to find $\\sin$ and $\\cos$ of $15$, but there is a quicker way. $e^{i x}=\\cos x+i \\sin x$, so $\\sin (3 x)$ is the imaginary part of $(\\cos x+i \\sin x)^{3}$, which is $3 \\cos ^{2} x \\sin x-\... | United States | Harvard-MIT Math Tournament | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | sqrt(2)/2 | |
0k4b | Problem:
A sequence $a_{1}, a_{2}, \ldots$ of positive integers satisfies $a_{1}=1$ and
$$
a_{n+1}=2^{a_{n}}+a_{n}
$$
for $n \geq 1$. Prove that $a_{1}, a_{2}, \ldots, a_{243}$ leave distinct remainders when divided by $243$. | [
"Solution:\nI'll prove by induction on $k \\geq 1$ that any $3^{k}$ consecutive values of $a_{n}$ produce distinct residues modulo $3^{k}$. The base case $k=1$ is easily checked ($a_{n}$ is always odd, hence $a_{n}$ cycles $1,0,2 \\bmod 3$).\n\nFor the inductive step, assume it's true up to $k$. Since $2^{*} (\\bmo... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
08dk | Problem:
Alla Scuola Normale, gli immatricolati di quest'anno nella classe di Scienze sono di quattro tipi: Matematici, Fisici, Chimici e Biologi. A mensa si ritrovano seduti tutti assieme intorno ad un tavolo rotondo; ognuno di essi ha esattamente una persona seduta di fronte a sé, ed inoltre per ogni studente l'insie... | [
"Solution:\nLa risposta è (B). Siano $m$ il numero di Matematici e $n$ il numero totale di immatricolati.\nSe per ogni matricola consideriamo il gruppo costituito da questa, il suo vicino destro, il suo vicino sinistro e la persona di fronte, abbiamo $n$ gruppi, ciascuno dei quali contiene esattamente un Matematico... | Italy | Progetto Olimpiadi della Matematica | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Modular Arithmetic"
] | null | MCQ | B | |
091g | Problem:
The incircle of the triangle $ABC$ touches the sides $BC$, $CA$, and $AB$ in the points $D$, $E$, and $F$, respectively. Let $K$ be the point symmetric to $D$ with respect to the incenter. The lines $DE$ and $FK$ intersect at $S$. Prove that $AS$ is parallel to $BC$. | [
"Solution:\n\nLet $S'$ be the intersection point of the line $FK$ and the line parallel to $BC$ passing through $A$. We need to show that $S'$, $D$ and $E$ are collinear. Let the tangent line to the given circle at the point $K$ intersect $AB$ at $Q$ (it is parallel to $BC$). Then $\\angle AS'F = \\angle QKF = \\an... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"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 > Analytic / Coordinate M... | null | proof only | null | |
0gd3 | 找出所有的正整數 $n$, 存在某個整數 $m$, 使得
$$
\frac{1}{n} \sum_{k=m}^{m+n-1} k^2
$$
也是完全平方數。例如當 $n=7$ 時, 可取 $m=-3$。 | [
"計算並配方得知\n$$\n\\begin{aligned} \\frac{1}{n} \\sum_{k=m}^{m+n-1} k^2 &= m^2 + (n-1)m + \\frac{(n-1)(2n-1)}{6} \\\\ &= \\left(m + \\frac{n-1}{2}\\right)^2 + \\frac{n^2-1}{12} \\end{aligned}\n$$\n為一完全平方數。\n\n因為 $n^2 - 1$ 對於 8 的餘數是 0 或奇數, 所以 $\\frac{n^2-1}{12}$ 必是偶數。同樣由上式知道 $\\frac{n^2-1}{12}$ 是兩完全平方數的差, 又因為是偶數, 所以必然是 ... | Taiwan | 二〇一九數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | null | proof and answer | All positive integers n with n ≡ 1, 7, 17, or 23 modulo 24 (i.e., n = 24p ± 1 or 24p ± 7). | |
0k4f | Problem:
Two triangles $A B C$ and $X Y Z$ have a common circumcircle. Suppose the nine-point circle $\gamma$ of $\triangle A B C$ passes through the midpoints of $\overline{X Y}$ and $\overline{X Z}$. Prove that $\gamma$ also passes through the midpoint of $\overline{Y Z}$. | [
"Solution:\n\nLet $R$ be the circumradius of $\\triangle A B C$. Note $\\gamma$ has radius $R / 2$ and passes through the midpoints $M$ and $N$ of $\\overline{X Y}$ and $\\overline{X Z}$. There are only two circles with these properties: one of them is $\\triangle X M N$ (by homothety) and the other is the nine-poi... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof only | null | |
0j8d | Problem:
Does there exist a row of Pascal's Triangle containing four distinct elements $a$, $b$, $c$ and $d$ such that $b=2a$ and $d=2c$?
Note that the values must be distinct, so $a$, $b$, $c$, $d$ must be four different numbers.
Recall that Pascal's triangle is the pattern of numbers that begins as follows
![](at... | [
"Solution:\n\nYes, there are infinitely many such rows. For example,\n$$\n\\binom{203}{68} = 2\\binom{203}{67} \\text{ and } \\binom{203}{85} = 2\\binom{203}{83}\n$$\nThere are infinitely many rows having two adjacent elements in a 1:2 ratio, for\n$$\n2\\binom{n}{k} = \\binom{n}{k+1}\n$$\nreduces to $2(k+1) = n - k... | United States | 13th Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Diophantine Equations > Pell's equations"
] | null | proof and answer | Yes; in fact there are infinitely many such rows (for example, the row with index two hundred three). | |
0jdh | For a positive integer $a$, define a sequence of integers $x_1, x_2, \dots$ by letting $x_1 = a$ and $x_{n+1} = 2x_n + 1$ for $n \ge 1$. Let $y_n = 2^{x_n} - 1$. Determine the largest possible $k$ such that, for some positive integer $a$, the numbers $y_1, \dots, y_k$ are all prime.
(This problem was suggested by Vale... | [
"The answer is $k=2$. If $a=1$, then $y_1=1$ is not prime. If $a=2$, then $y_1=3$ and $y_2=31$ are prime, while $y_3 = 2^{11}-1 = 23 \\cdot 89$ is composite. It now remains only to check that $y_1, y_2$, and $y_3$ cannot all be prime for $a \\ge 3$.\n\nSuppose otherwise for the sake of contradiction that $y_1, y_2$... | United States | RMM | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 2 | |
00qq | Let $a$, $b$, $c$ be the lengths of the sides of a given triangle and $m_a$, $m_b$, $m_c$ be the lengths of the corresponding medians. Prove that:
$$
m_a \left(\frac{b}{a} - 1\right) \left(\frac{c}{a} - 1\right) + m_b \left(\frac{a}{b} - 1\right) \left(\frac{c}{b} - 1\right) + m_c \left(\frac{a}{c} - 1\right) \left(\fr... | [
"The given inequality is equivalent to\n$$\n\\frac{m_a}{a^2}(a-b)(a-c) + \\frac{m_b}{b^2}(b-a)(b-c) + \\frac{m_c}{c^2}(c-a)(c-b) \\geq 0.\n$$\nThe last one is symmetric with respect to the variables, so without loss of generality, we can assume that $a \\geq b \\geq c$. Then we have that\n$$\n\\frac{m_a}{a^2}(a-b)(... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0cyo | Let $ABCD$ be a square of center $O$. The parallel to $AD$ through $O$ intersects $AB$ and $CD$ at $M$ and $N$ and a parallel to $AB$ intersects diagonal $AC$ at $P$. Prove that
$$
OP^{4} + \left(\frac{MN}{2}\right)^{4} = MP^{2} \cdot NP^{2}
$$ | [
"Let $AB = 2a$ and let $Q$ be the intersection point of the parallel to $AB$ with $MN$. Let $x = OQ$. Then $PQ = x$. We have\n$$\n\\begin{gathered}\nOP^{2} = 2x^{2}, \\quad MP^{2} = x^{2} + (a - x)^{2} \\\\\nNP^{2} = x^{2} + (a + x)^{2}, \\quad MN = 2a\n\\end{gathered}\n$$\n\n\n\nThe relati... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0ejk | Problem:
Nad stranico $CD$ kvadrata $ABCD$ z zunanje stani narišemo pravokotni trikotnik $DCE$ s pravim kotom pri $E$. Dokaži, da simetrala kota $\angle DEC$ razdeli kvadrat $ABCD$ na dva ploščinsko enaka dela. | [
"Solution:\n\n\n\nOznačimo z $G$ presečišče diagonal kvadrata $ABCD$. Ker je $\\angle DEC=90^\\circ=\\angle CGD$, po Talesovem izreku točke $G, C, E$ in $D$ ležijo na isti krožnici $\\mathcal{K}$. Točka $G$ zaradi simetrije razpolavlja lok $\\overparen{DC}$ krožnice $\\mathcal{K}$. Ker tudi... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
06ia | Let $f(x) = ax + b$ where $a$ and $b$ are integers. If $f(f(0)) = 0$ and $f(f(f(4))) = 9$, find the value of $f(f(f(f(1)))) + f(f(f(f(2)))) + \cdots + f(f(f(f(2014))))$.
設 $f(x)=ax+b$，其中 $a$、$b$ 為整數。若 $f(f(0))=0$ 而 $f(f(f(4)))=9$，求 $f(f(f(f(1))))+f(f(f(f(2))))+\cdots+f(f(f(f(2014))))$ 的值。 | [] | Hong Kong | HONG KONG PRELIMINARY SELECTION CONTEST | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English; Chinese | proof and answer | 2029105 | |
02wt | Problem:
Considere um torneio de xadrez envolvendo terráqueos e alienígenas em que cada jogador joga contra todos os outros exatamente uma vez. Ao final do torneio, cada jogador obteve metade dos pontos que conquistou jogando contra terráqueos e metade jogando contra alienígenas. Sendo $t$ e $a$ os números de terráqueo... | [
"Solution:\na) Sendo $n = t + a$ o total de enxadristas presentes, com cada dupla se enfrentando em partidas únicas, o torneio teve\n$$\n\\frac{n(n-1)}{2} = \\frac{(t+a)(t+a-1)}{2} \\text{ partidas. }\n$$\n\nb) Se somarmos a pontuação que cada atleta recebeu por jogo (independentemente do resultado) obteremos sempr... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) (t + a)(t + a − 1) / 2.
b) t(t − 1).
c) t + a = (a − t)^2, so the total number of players is a perfect square. | |
0gcl | 平面上給定三角形 $ABC$ 及一點 $P$。令 $\triangle ABC$, $\triangle BPC$, $\triangle CPA$, $\triangle APB$ 的外接圓圓心分別為點 $O$, $D$, $E$, $F$。設直線 $BC$ 與 $EF$ 交於點 $T$，而點 $O$ 對直線 $EF$ 的對稱點為 $X$。證明：$PT \perp DX$。 | [
"Let $X$ be the reflection of $O$ in $EF$ and $Y$, $Z$ be the reflections of $P$ in $EX$, $FX$, respectively. Since $A$, $B$, $C$ are the reflections of $P$ in $EF$, $FD$, $DE$, respectively, so $O$, $P$ are isogonal conjugate with respect to $\\triangle DEF$ and hence\n$$\n\\angle FEX = \\angle OEF = \\angle DEP, ... | Taiwan | 二〇一九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 > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
... | null | proof only | null | |
02d1 | Problem:
No cabeleireiro - Três clientes estão no cabeleireiro pagando cada um a sua conta no caixa.
- o primeiro cliente paga o mesmo montante que há no caixa e retira 10 reais de troco;
- o segundo cliente efetua a mesma operação que o primeiro;
- o terceiro cliente efetua a mesma operação que os dois primeiros.
Enc... | [
"Solution:\n\nSeja $x$ o montante inicial no caixa. Esse montante mais o que os 3 clientes pagaram nos dará o caixa zerado.\n\n- O 1º cliente paga $x-10$. Depois do primeiro cliente, há $x + (x-10) = 2x - 10$ reais no caixa.\n\n- O 2º cliente paga $(2x-10) - 10 = 2x - 20$. Depois do 2º cliente, há $(2x-10) + (2x-20... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 8.75 reais | |
0bvl | What values can assume $[\sqrt{m} + \frac{1}{m+n}] + [\sqrt{n} + \frac{1}{m+n}] - [\sqrt{m}] - [\sqrt{n}]$, when $m$ and $n$ are integers $\ge 9$? | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 0 | |
0co2 | 100 pairwise distinct real numbers are arranged in a circle. Prove that there exist four successive numbers among them such that the sum of the two middle ones is strictly less than the sum of the other two.
По окружности расставлено 100 попарно различных чисел. Докажите, что можно выбрать 4 подряд стоящих числа таким... | [
"See the solution to problem 9.2."
] | Russia | Final round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof only | null | |
07zi | Problem:
Sia $ABCD$ un trapezio con base maggiore $AB$ tale che le diagonali $AC$ e $BD$ siano perpendicolari. Sia $O$ il centro della circonferenza circoscritta al triangolo $ABC$ e sia $E$ il punto di intersezione tra la retta $OB$ e la retta $CD$. Dimostrare che
$$
\overline{BC}^2 = \overline{CD} \cdot \overline{CE... | [
"Solution:\n\nSia $F$ l'intersezione delle due diagonali del trapezio e sia $M$ il punto medio di $AB$. Poiché il triangolo $AOB$ è isoscele e $OM$ ne è la mediana rispetto alla base, i triangoli $AOM$ e $BOM$ sono uguali e, in particolare, $\\widehat{AOM} = \\widehat{BOM} = \\frac{1}{2} \\widehat{AOB}$. Inoltre, c... | Italy | GARA NAZIONALE di MATEMATICA | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jqv | Problem:
Let $ABCDE$ be a square pyramid of height $\frac{1}{2}$ with square base $ABCD$ of side length $AB=12$ (so $E$ is the vertex of the pyramid, and the foot of the altitude from $E$ to $ABCD$ is the center of square $ABCD$). The faces $ADE$ and $CDE$ meet at an acute angle of measure $\alpha$ (so that $0^{\circ}... | [
"Solution:\n\n$\\boxed{\\frac{17}{144}}$\n\nLet $X$ be the projection of $A$ onto $DE$. Let $b=AB=12$.\n\nThe key fact in this computation is that if $Y$ is the projection of $A$ onto face $CDE$, then the projection of $Y$ onto line $DE$ coincides with the projection of $A$ onto line $DE$ (i.e., $X$ as defined abov... | United States | HMMT February 2015 | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 17/144 | |
0fyk | Problem:
Finde alle positiven reellen Lösungen $(a, b, c, d)$ der Gleichung
$$
\frac{a^{2}-b d}{b+2 c+d}+\frac{b^{2}-c a}{c+2 d+a}+\frac{c^{2}-d b}{d+2 a+b}+\frac{d^{2}-a c}{a+2 b+c}=0
$$ | [
"Solution:\nDie gesuchten Lösungen $(a, b, c, d)$ sind genau diejenigen mit $a=c$ und $b=d$. Sei $A$ die linke Seite der Gleichung, wir werden beweisen, dass für alle $a, b, c, d>0$ die Ungleichung $A \\geq 0$ gilt mit Gleichheit genau dann, wenn $a=c$ und $b=d$. Dass dies wirklich Gleichungsfälle sind, ist leicht ... | Switzerland | IMO Selektion | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | a = c and b = d | |
00ik | We are given a triangle $ABC$ and a point $P$ in its interior. The lines through $P$ and parallel to the sides of the triangle divide the triangle into three parallelograms and three triangles.
a) If $P$ is the incenter of $ABC$, show that the perimeter of each of the three small triangles is equal to the length of th... | [
"a) Let $I$ be the incenter of $ABC$. Let $X$ be the common point of $AB$ with the line through $I$ parallel to $CA$, and $Y$ be the common point of $CA$ with the line through $I$ parallel to $AB$. $AXIY$ is a parallelogram, and since $I$ is the incenter of $ABC$, we have $\\angle IAX = \\angle IAY$. Since $\\angle... | Austria | Austria 2010 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities ... | English | proof and answer | a) When the interior point is the incenter, each small triangle’s perimeter equals the length of the adjacent side. b) The only interior point with this property is the incenter. c) The sum of the areas is minimized at the centroid. | |
03to | Let $ABC$ be an acute triangle. Points $D$, $E$ and $F$ lie on segments $BC$, $CA$ and $AB$ respectively, and each of the three segments $AD$, $BE$ and $CF$ contains the circumcenter of $ABC$. Prove that if any two of the ratios
$$
\frac{BD}{DC}, \frac{CE}{EA}, \frac{AF}{FB}, \frac{BF}{FA}, \frac{AE}{EC}, \frac{CD}{DB}... | [
"**Proof I** Note that there are $\\binom{6}{2} = 15$ possible pairs of ratios among the six given in the problem statement. These pairs are of two types: (i) Three of these pairs are reciprocal pairs involving segments from just one side of triangle $ABC$. (ii) The other 12 pairs involve segments from two sides of... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Pla... | English | proof only | null | |
00bi | There are 8 weights, all of different colors, and a two-plate scale. Ana and Beto know that the weights are of $1$, $2$, $3$, $4$, $5$, $6$, $7$ and $8$ grams, but only Ana knows which color corresponds to each weight.
An *operation* consists in putting weights on each side of the scale so that it stays balanced. Ana ... | [
"Let us see that the minimum number of operations that Ana has to make is $2$.\n\nIn the first operation, Ana balances five weights in one side with two in the other. The weight of five weights is at least $1+2+3+4+5=15$, and the weight of two weights is at most $7+8=15$. Then, the only possibility to achieve balan... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Logic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 2 | |
0c15 | Find the smallest possible sum of the digits of a non-null multiple of $99$. | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 18 | |
0864 | Problem:
Determinare il più grande numero di due cifre tale che:
a) sia un numero primo;
b) scambiando di posto le due cifre resti un numero primo;
c) il prodotto delle due cifre sia un numero primo. | [
"Solution:\n\nLa risposta è $71$. Per la condizione c) il prodotto delle due cifre deve essere un numero primo. Perché questo avvenga è necessario che una delle due cifre sia $1$, altrimenti il numero che ne deriva moltiplicando è composto.\n\nSupponendo che $1$ sia la cifra delle decine, la condizione a) restringe... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 71 | |
0eum | Let $I$, $O$, $H$ be the incenter, circumcenter, orthocenter of an acute triangle $ABC$, respectively, and the inscribed circle of $\triangle ABC$ touch side $BC$ at $D$. Suppose that $\angle B < \angle C$, and $AO$ is parallel to $HD$. Show that the four points $E$, $F$, $I$, $O$ lie on a common circle, where $E$ is t... | [
"Let $E' (\\neq A)$ be the intersecting point of line $AH$ and the circumscribed circle of $\\triangle ABC$, and $D'$ be the intersecting point of line $OE'$ and side $BC$. We have\n$$\n\\angle OEA' = \\angle D'E'A = \\angle D'HE'\n$$\nsince points $H$ and $E'$ locate symmetrically about side $BC$. And it also hold... | South Korea | 23rd Korean Mathematical Olympiad Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tang... | English | proof only | null | |
04ls | There are $3000$ points marked on a circle, and there is a grasshopper standing on one of these points. With every jump, the grasshopper jumps over one or two marked points (in the clockwise direction) and lands on the next marked point. Determine the smallest possible number of jumps made by the grasshopper if it land... | [] | Croatia | Mathematical competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 3001 | |
0ex5 | Problem:
a. The convex hexagon $ABCDEF$ has all angles equal. Prove that $AB - DE = EF - BC = CD - FA$.
b. Given six lengths $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$, $a_{5}$, $a_{6}$ satisfying $a_{1} - a_{4} = a_{5} - a_{2} = a_{3} - a_{6}$, show that you can construct a hexagon with sides $a_{1}$, $a_{2}$, $a_{3}$, $a_{... | [
"Solution:\n\na. Extend $AB$, $CD$, $EF$. We get an equilateral triangle with sides $AF + AB + BC$, $BC + CD + DE$, $ED + EF + FA$. Hence $AB - DE = CD - FA = EF - BC$, as required.\n\nb. Take an equilateral triangle with sides $s$, $t$, $u$ of lengths $a_{2} + a_{3} + a_{4}$, $a_{4} + a_{5} + a_{6}$, and $a_{6} + ... | Soviet Union | 4th ASU | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Translation"
] | null | proof only | null | |
09oo | Let $ABCD$ be a cyclic quadrilateral with obtuse angle at $C$. Let $H$ be such that $BCDH$ is a parallelogram. Let line $AH$ meet $CD$ at $M$ and $BC$ at $N$. A line through $H$ parallel to $AC$ meets $CD$ at $K$ and $BC$ at $L$. If the points $K, L, M, N$ lie on a common circle, prove that $\angle ABC = 90^\circ$.
(K... | [] | Mongolia | MMO2025 Round 2 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0bs9 | Let $m, n$ be positive integers and $x, y, z \in [0, 1]$ be real numbers. Prove that
$$
0 \le x^{m+n} + y^{m+n} + z^{m+n} - x^m y^n - y^m z^n - z^m x^n \le 1
$$
and find when equality holds. | [
"Without loss of generality, we may assume that $x$ is the largest of $x, y, z$; then we have:\n$x^{m+n} + y^{m+n} + z^{m+n} - x^m y^n - y^m z^n - z^m x^n = (x^m - z^m)(x^n - y^n) + (y^m - z^m)(y^n - z^n) \\ge 0$.\nThe minimum is $0$ and is obtained if $x = \\max\\{y, z\\}$ and $y = z$, so if $x = y = z$.\n\nThe ma... | Romania | 67th NMO Selection Tests for JBMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | The expression lies between zero and one for all inputs. The minimum zero occurs exactly when all three numbers are equal. The maximum one occurs exactly for the six triples obtained by permuting either one coordinate equal to one with the other two zero, or two coordinates equal to one with the remaining one zero: (1,... | |
0b1r | Problem:
What is the sum of all four-digit numbers that each use the digits $2, 4, 6$, and $8$ exactly once? | [
"Solution:\nEvery digit appears in each decimal place exactly $6$ times. Therefore, the sum is $6 \\cdot (2+4+6+8) \\cdot 1111 = 133,320$."
] | Philippines | 22nd Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 133,320 | |
0ifx | Problem:
Let $a_{1}, a_{2}, \ldots, a_{2005}$ be real numbers such that
$$
\begin{array}{ccccccccccc}
a_{1} \cdot 1 & + & a_{2} \cdot 2 & + & a_{3} \cdot 3 & + & \cdots & + & a_{2005} \cdot 2005 & = & 0 \\
a_{1} \cdot 1^{2} & + & a_{2} \cdot 2^{2} & + & a_{3} \cdot 3^{2} & + & \cdots & + & a_{2005} \cdot 2005^{2} & = &... | [
"Solution:\n$1 / 2004!$\n\nThe polynomial $p(x) = x(x-2)(x-3) \\cdots (x-2005) / 2004!$ has zero constant term, has the numbers $2, 3, \\ldots, 2005$ as roots, and satisfies $p(1) = 1$. Multiplying the $n$th equation by the coefficient of $x^{n}$ in the polynomial $p(x)$ and summing over all $n$ gives\n$$\na_{1} p(... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1/2004! | |
01h2 | Let $A$, $B$, $C$, $P$ and $Q$ be five pairwise different points in the plane. Suppose that $A$, $B$ and $C$ are not collinear and that
$$
\frac{AP}{BP} = \frac{AQ}{BQ} = \frac{21}{20},
$$
$$
\frac{BP}{CP} = \frac{BQ}{CQ} = \frac{20}{19}.
$$
Prove that the line $PQ$ contains the circumcentre of the triangle $ABC$. | [
"We choose a coordinate system such that the circumcentre $O$ of $\\triangle ABC$ lies in the origin and the circumradius of $\\triangle ABC$ is equal to $1$. Let us write $A = (a_1, a_2)$, $B = (b_1, b_2)$ and $C = (c_1, c_2)$. The following statements about a point $X = (x_1, x_2)$ are equivalent:\n$$\n\\frac{AX}... | Baltic Way | Baltic Way 2020 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0bh1 | Find all strictly increasing sequences $(a_n)_n$ of positive integers with the following two properties:
a) $a_1^3 + a_2^3 + \dots + a_n^3 = (a_1 + a_2 + \dots + a_n)^2$, for every integer $n \ge 2014$;
b) for every integer $k \ge 2015$, the number $a_1 + a_2 + \dots + a_{2014}$ cannot be written as a sum of $k$ cons... | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | The unique sequence is a_n = n for all n ≥ 1. | |
01s2 | The incircle of the triangle $ABC$ touches the sides $AC$ and $BC$ at points $P$ and $Q$, respectively, $N$ and $M$ are the midpoints of $AC$ and $BC$, respectively. Let $X = AM \cap BP$, $Y = BN \cap AQ$.
Given $C, X, Y$ are collinear prove that $CX$ is the angle bisector of the angle $ACB$. | [
"(Solution by V. Vityaz, A. Gaponenko.) Let without loss of generality $CX \\ge CY$. Then $AP \\le AN, BM \\le BQ$ (see the Fig.) By the Menelaus theorem, we have: for the triangle $AQM$ and the line $XC$\n\n\n\n$$\n\\frac{AY}{YQ} \\cdot \\frac{QC}{CM} \\cdot \\frac{MX}{XA} = 1, \\quad (1)\... | Belarus | SELECTION and TRAINING SESSION | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null | |
0hr7 | Problem:
Prove that for each $n \geq 1$, there is a number $N$ having $n$ digits, each of which is either $1$ or $2$, such that $N$ is divisible by $2^{n}$. | [
"Solution:\n\nWe proceed by induction. For $n=1$, the number $N=2$ works. Suppose $N$ works for a given $n$. Consider the two $(n+1)$-digit numbers\n$$\nN_{1}=10^{n}+N \\quad \\text{ and } \\quad N_{2}=2 \\cdot 10^{n}+N .\n$$\nformed by attaching a single $1$ or $2$ to the left-hand side of $N$. Note that $N_{1}$ a... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
00tv | Find all functions $f: (0, \infty) \to (0, \infty)$ such that
$$
f(yf(x)^3 + x) = x^3 f(y) + f(x)
$$
for all $x, y > 0$. | [
"Setting $y = \\frac{t}{f(x)^3}$ we get\n$$\nf(x+t) = x^3 f\\left(\\frac{t}{f(x)^3}\\right) + f(x) \\quad (1)\n$$\nfor every $x, t > 0$.\nFrom (1) it is immediate that $f$ is increasing.\n\n**Claim.** $f(1) = 1$\n\n**Proof of Claim.** Let $c = f(1)$. If $c < 1$, taking $x = 1$ and $y = \\frac{1}{1-c^3}$ we have $y ... | Balkan Mathematical Olympiad | BMO 2022 shortlist | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | f(x) = x for all x > 0 | |
0c3y | Problem:
Demonstraţi că, dacă $a, b, c \geq 0$ şi $a+b+c=3$, atunci
$$
\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a} \geq \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}
$$ | [
"Solution:\nEliminând numitorii obţinem inegalitatea echivalentă $a^{2} c+b^{2} a+c^{2} b+\\sum a^{2}+\\sum a b+\\sum a \\geq 3+2 \\sum a+\\sum a b$, adică $a^{2} c+b^{2} a+c^{2} b+\\sum a^{2} \\geq 6$.\n\nAvem $a^{2} c \\geq 2 a c-c$ şi analoagele.\n\nEste deci suficient să arătăm că $\\sum a^{2}+2 \\sum a b-\\sum... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
092o | Problem:
Determine all triples $(a, b, c)$ of real numbers satisfying the system of equations
$$
\begin{aligned}
& a^{2}+a b+c=0 \\
& b^{2}+b c+a=0 \\
& c^{2}+c a+b=0
\end{aligned}
$$ | [
"Solution:\nIf one of the numbers $a, b$ and $c$ is equal to zero, it is easy to see that the other two numbers also have to be equal to zero, which gives us the solution $(0,0,0)$.\nNow assume that $a, b, c \\neq 0$.\nIf all three numbers are positive, then the left-hand side of each equation is positive, while th... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | {(0,0,0), (-1/2,-1/2,-1/2)} | |
0a56 | Problem:
Josie and Ross are playing a $20 \times 20$ chessboard game. Initially the chessboard is empty. The two players alternately take turns, with Josie going first. On Josie's turn, she selects any two different empty cells, and places one white stone in each of them. On Ross' turn, he chooses any one white stone ... | [
"Solution:\n\nRoss can't stop Josie winning — Josie has a strategy in which she can ensure that there will be 8 white stones in a row. We will give an explicit example of such a strategy.\n\nTo simplify notation, we define a $k$-strip to be a $1 \\times 8$ rectangle, in which the first $k$ cells are filled with whi... | New Zealand | NZMO Round One | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | No. Ross cannot stop Josie; Josie has a winning strategy to force eight consecutive white stones. | |
0f60 | Problem:
A $3 \times 3$ chess board is colored alternately black and white in the usual way with the center square white. Each white square is given the number $1$ or the number $-1$. A move consists of simultaneously changing each number to the product of the adjacent numbers. So the four corner squares are each chan... | [] | Soviet Union | 18th ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0e0j | In a triangle $ABC$ we have $|AB| = 2|AC|$. Let $D$ and $E$ be the two points on the segments $AB$ and $BC$, such that $\angle BAE = \angle ACD$. The segments $AE$ and $CD$ intersect at $F$, and $CFE$ is an equilateral triangle. Find the angles of the triangle $ABC$. | [
"Since $CEF$ is an equilateral triangle we have $\\angle EFC = 60^\\circ$. This implies that $\\angle CFA = 120^\\circ$, so $\\angle FAC = 180^\\circ - \\angle CFA - \\angle ACF = 60^\\circ - \\angle ACF$ and $\\angle BAC = \\angle BAE + \\angle FAC = \\angle BAE + 60^\\circ - \\angle ACD = 60^\\circ$. We have $|AB... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | ∠A = 60°, ∠B = 30°, ∠C = 90° | |
0h74 | Find all $x$, $y$, $z$ which satisfy:
$$
\begin{cases} x^2 + x y + x z = y, \\ y^2 + y z + y x = z, \\ z^2 + z x + z y = x. \end{cases}
$$ | [
"Summing up the equalities we get\n$$\nx^2 + y^2 + z^2 + 2 x y + 2 x z + 2 y z = x + y + z \\text{ or } (x + y + z)^2 = x + y + z.\n$$\nNow we face two cases.\n\nCase 1. $x + y + z = 1$. Then from the first equality we get\n$$\nx^2 + x y + x z = x(x + y + z) = x = y,\n$$\nAnalogously, $x = y = z$, so the only solut... | Ukraine | UkraineMO | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | x = y = z = 0 or x = y = z = 1/3 | |
02pw | Problem:
Uma formiga se movimenta uma unidade por segundo sobre os pontos $0$, $1$ e $2$ da figura a seguir, começando do ponto $0$.
a) Quais são os possíveis percursos da formiga até $3$ segundos?
b) Quantos possíveis percursos pode fazer a formiga até $10$ segundos? | [
"Solution:\n\n(a) Até três segundos temos dois possíveis percursos: $0-1-0-1$ ou $0-1-2-1$.\n\n(b) Observemos que quando a formiga está nos pontos $0$ e $2$ ela somente tem uma possibilidade para caminhar no segundo seguinte, que é ir para $1$. Quando está em $1$ ela tem duas possibilidades no segundo seguinte, que... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) 0-1-0-1 and 0-1-2-1. b) 32 | |
0f2m | Problem:
a and b are positive real numbers. $x_i$ are real numbers lying between $a$ and $b$. Show that
$$(x_1 + x_2 + \ldots + x_n)\left(\frac{1}{x_1} + \frac{1}{x_2} + \ldots + \frac{1}{x_n}\right) \leq \frac{n^2 (a + b)^2}{4ab}.$$ | [] | Soviet Union | ASU | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
019k | Find all polynomials $P$ with non-negative integer coefficients such that for all primes $p$ and positive integers $n$ there exist a prime $q$ and a positive integer $m$ such that $P(p^n) = q^m$. | [
"Notice that among the constant polynomials the only solutions are $P(t) = q^m$ where $q$ is a prime and $m$ a positive integer. Assume that\n$$\nP(t) = a_k t^k + \\cdots + a_0,\n$$\nwhere $a_k \\neq 0$ and $a_0, a_1, \\ldots, a_k$ are non-negative integers, is a polynomial that fulfills the conditions.\n\nFirst co... | Baltic Way | Baltic Way 2013 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | All solutions are P(t) = t^m with m a positive integer, and constant polynomials P(t) = q^m where q is prime and m is a positive integer. | |
04jb | Let $a = \sqrt[2015]{2015}$ and $(a_n)$ be the sequence such that $a_1 = a$ and $a_{n+1} = a^{a_n}$ for $n \ge 1$. Does there exist a positive integer $n$ such that $a_n \ge 2015$? | [
"Let $a = \\sqrt[2015]{2015}$. Note that $a > 1$ since $2015 > 1$.\n\nWe have $a_1 = a$ and $a_{n+1} = a^{a_n}$ for $n \\ge 1$.\n\nLet us compute the first few terms:\n\n$a_1 = a$\n\n$a_2 = a^{a_1} = a^a$\n\n$a_3 = a^{a_2} = a^{a^a}$\n\nand so on.\n\nLet us try to estimate $a_n$ and see if it can ever reach $2015$.... | Croatia | Croatia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | No | |
05nq | Problem:
Prouver que, pour tout entier $n \geq 2$, on a :
$$
\sum_{k=2}^{n} \frac{1}{\sqrt[k]{(2 k)!}} \geq \frac{n-1}{2 n+2}
$$
N.B. Si $a>0$, on note $\sqrt[k]{a}$ l'unique nombre réel $b>0$ tel que $b^{k}=a$. | [
"Solution:\n\nOn raisonne par récurrence sur $n \\geq 2$.\n\nPour $n=2$, on a bien $\\frac{1}{\\sqrt{24}} > \\frac{1}{6}$.\n\nSupposons que l'inégalité désirée soit vraie pour la valeur $n-1 \\geq 2$. Pour la valeur $n$, le membre de droite augmente de\n$$\n\\frac{n-1}{2 n+2} - \\frac{n-2}{2 n} = \\frac{1}{n(n+1)}.... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0e6u | Let $ABC$ be a right-angled triangle with the right angle at $C$. Let the points $K$, $L$ and $M$ lie on the sides $CA$, $AB$ and $BC$, respectively, so that the angle $\angle MLK$ is a right angle and $|KC| = |KL|$. Prove that the bisectors of the angles $\angle AKL$ and $\angle LMB$ are parallel. | [
"Let $K'$ and $L'$ be the intersection points of the side $AB$ with the bisectors of the angles $\\angle AKL$ and $\\angle LMB$, respectively. The triangles $KMC$ and $KML$ coincide in two sides and the angle that lies opposite the longer among the two sides, hence the triangles are congruent.\n![](attached_image_1... | Slovenia | National Math Olympiad 2012 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02wc | Problem:
Juquinha marca pontos sobre uma circunferência e desenha triângulos ligando 3 destes pontos. Os comprimentos dos arcos de 2 pontos consecutivos são iguais.
a) Marcando 4 pontos na circunferência, quantos triângulos ele conseguirá desenhar?
b) Marcando 5 pontos na circunferência, quantos triângulos equiláter... | [
"Solution:\n\na) $4$. Temos que escolher $3$ pontos de um total de $4$. Em situações como esta, ao escolhermos um triângulo qualquer, por exemplo o triângulo $ABC$, todas as permutações dos pontos $A$, $B$ e $C$ continuam sendo o triângulo $ABC$. Dessa forma, ao utilizarmos o princípio fundamental da contagem, deve... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | a) 4; b) 0; c) 12 | |
00xu | Problem:
A convex quadrangle $A B C D$ is inscribed in a circle with the centre $O$. The angles $\angle A O B, \angle B O C, \angle C O D$ and $\angle D O A$, taken in some order, are of the same size as the angles of quadrangle $A B C D$. Prove that $A B C D$ is a square. | [
"\nFigure 7\n\n\nFigure 8\n\nSolution:\n\nAs the quadrangle $A B C D$ is inscribed in a circle, we have $\\angle A B C + \\angle C D A = \\angle B C D + \\angle D A B = 180^{\\circ}$. It suffices to show that if each of these angles is equal to $90^{\\circ}$, then e... | Baltic Way | Baltic Way 1993 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jc1 | Problem:
For a permutation $\sigma$ of $1,2, \ldots, 7$, a transposition is a swapping of two elements. (For instance, we could apply a transposition to the permutation $3,7,1,4,5,6,2$ and get $3,7,6,4,5,1,2$ by swapping the $1$ and the $6$.)
Let $f(\sigma)$ be the minimum number of transpositions necessary to turn $\... | [
"Solution:\n\nTo solve this problem, we use the idea of a cycle in a permutation. If $\\sigma$ is a permutation, we say that $(a_{1} a_{2} \\cdots a_{k})$ is a cycle if $\\sigma(a_{i})=a_{i+1}$ for $1 \\leq i \\leq k-1$ and $\\sigma(a_{k})=a_{1}$. Any permutation can be decomposed into disjoint cycles; for instance... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof and answer | 22212 | |
0cvy | Five points are marked in the plane. It is permitted to choose some of them and change their positions; the distances between the chosen points should be preserved. Prove that it is possible to perform such change so as to obtain a configuration of five points possessing an axis of symmetry. | [
"Let us denote the given points by $A$, $B$, $C$, $D$, and $E$. Choose among them two points that are the farthest apart; let these be $A$ and $B$. We will show that it is possible to move them as required.\n\nDraw the perpendicular bisector $\\alpha$ to the segment $CD$. If $E$ lies on $\\alpha$, then it is suffic... | Russia | Final round | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English; Russian | proof only | null | |
0ee9 | Problem:
Kocko z robom dolžine $a$ razrežemo na 8 malih skladnih kock. Površino ene take male kocke označimo s $P$. Koliko je $P$?
(A) $\frac{a^{2}}{2}$
(B) $\frac{3 a^{2}}{4}$
(C) $\frac{a^{2}}{8}$
(D) $\frac{3 a^{2}}{2}$
(E) $a^{8}$ | [
"Solution:\n\nOznačimo z $a_{1}$ dolžino roba male kocke. Upoštevamo, da je prostornina male kocke $a_{1}^{3} = \\frac{a^{3}}{8}$ in dobimo $a_{1} = \\frac{a}{2}$. Torej velja $P = 6 a_{1}^{2} = 6\\left(\\frac{a}{2}\\right)^{2} = \\frac{3 a^{2}}{2}$."
] | Slovenia | 16. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area"
] | null | MCQ | D | |
04yx | A regular $2010$-gon is divided into pieces of triangular shape. Find the least possible number of pieces. | [
"All the interior angles of the $2010$-gon can be built from the inner angles of the triangular pieces. As the sum of the inner angles of the $2010$-gon is $2008 \\cdot 180^{\\circ}$ and that of every triangle is $180^{\\circ}$, there must be at least $2008$ triangles. On the other hand, each convex $2010$-gon can ... | Estonia | Estonija 2010 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 2008 | |
04m0 | In an organization there are three committees. Each person belongs to exactly one committee. For any two persons belonging to different committees, in the third committee there are exactly $10$ people that both persons know and exactly $10$ people that both persons do not know. All acquaintances are mutual. How many pe... | [
"Let $A$, $B$, $C$ be committees and let them have exactly $a$, $b$, $c$ persons, respectively. To each person assign a point in the plane so that no three points are collinear. We connect the points corresponding to persons that know each other by a blue segment, and for those that do not know each other by a red ... | Croatia | Croatian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 120 | |
0iix | Problem:
Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of real numbers defined by $a_{0}=21$, $a_{1}=35$, and $a_{n+2}=4 a_{n+1}-4 a_{n}+n^{2}$ for $n \geq 2$. Compute the remainder obtained when $a_{2006}$ is divided by $100$. | [
"Solution:\nNo pattern is evident in the first few terms, so we look for a formula for $a_{n}$. If we write $a_{n}=A n^{2}+B n+C+b_{n}$ and put $b_{n+2}=4 b_{n+1}-4 b_{n}$. Rewriting the original recurrence, we find\n$$\n\\begin{aligned}\n& A n^{2}+(4 A+B) n+(4 A+2 B+C)+b_{n+2} \\\\\n& =4\\left(A n^{2}+(2 A+B) n+(A... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 0 | |
062r | Problem:
Auf einem Tisch liegen nebeneinander 2009 Karten in einer Reihe. Zunächst ist bei allen Karten die Oberseite weiß und die Unterseite schwarz. Die Karten seien von $1$ bis $2009$ nummeriert.
Zwei Spieler $A$ und $B$ führen abwechselnd einen Spielzug aus, wobei $A$ beginnt. Jeder Spielzug besteht darin, dass d... | [
"Solution:\n\na) Wenn das Spiel nicht endet, dann muss wegen der höchstens $2^{2008}$ verschiedenen möglichen Spielzustände eine sich periodisch wiederholende Folge von Zuständen existieren. Dabei sei $k$ die kleinste Nummer der Karten, die innerhalb dieser Periode umgedreht werden. Um die Karte mit der Nummer $k$ ... | Germany | IMO-Auswahlklausur | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | The game necessarily ends; Player B has a winning strategy. | |
08js | Problem:
Se consideră un poligon convex având $n$ vârfuri, $n \geq 4$. Descompunem arbitrar poligonul in triunghiuri ale căror vârfuri sunt printre vârfurile poligonului, astfel încât orice două triunghiuri să nu aibă puncte interioare comune. Colorăm cu negru triunghiurile ce au două laturi care sunt şi laturi ale po... | [] | JBMO | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F"
] | null | proof only | null | |
0h5v | Determine the maximum positive integer that is divisible by $7$, and all digits of which are odd, and their sum equals $2015$. | [
"It is obvious that the greater the number is, the more digits it has. Since every digit equals at least $1$, the maximum amount of digits is $2015$. Such number is unique: $\\underline{11\\dots1}$, since every digit is $1$ and their sum is $2015$.\n\nHowever, $111\\dots1$ (with $2015$ digits) is not divisible by $... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | A = (10^2013 − 1)/9 + 2·10^2008, i.e., the 2013-digit number with four ones, then a three, followed by 2008 ones | |
02u0 | Problem:
No desenho abaixo, $ABCD$ é um retângulo e $E$ é o pé da perpendicular traçada de $A$ até a diagonal $BD$. As distâncias do ponto $E$ aos lados $DC$, $BC$ e $AB$ são $n$, $1$ e $x$, respectivamente. Seja ainda $d$ o comprimento da diagonal $BD$.
a) Verifique que $DE = x^{2} \sqrt{1 ... | [
"Solution:\n\na) Seja $y = DG$. Pelo Teorema de Pitágoras no triângulo $\\triangle BEF$, temos $BE^{2} = x^{2} + 1$. Pelas relações métricas do triângulo retângulo $\\triangle AEB$, temos\n$$\n\\begin{aligned}\nx^{2} & = y \\cdot 1 \\\\\n& = y \\\\\nAE^{2} & = y(y+1) \\\\\n& = x^{2}(x^{2}+1)\n\\end{aligned}\n$$\nAg... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
05bo | A rectangular grid is divided by two perpendicular straight lines into four smaller rectangles with integral side lengths. It is possible to remove one among these four rectangles in such a way that the remaining figure can be exactly covered by rectangles of size $2 \times 3$ and $3 \times 2$. Prove that it is possibl... | [
"Call a figure *coverable* if it can be exactly covered by rectangles of size $2 \\times 3$ and $3 \\times 2$. Let the original rectangle be of size $(a+d) \\times (b+c)$ and let the figure remaining after cutting out the upper left corner of size $a \\times b$ be *coverable* (Fig. 52). This would be impossible if ... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
06pu | For a prime $p$ and a positive integer $n$, denote by $\nu_{p}(n)$ the exponent of $p$ in the prime factorization of $n!$. Given a positive integer $d$ and a finite set $\{p_{1}, \ldots, p_{k}\}$ of primes. Show that there are infinitely many positive integers $n$ such that $d \mid \nu_{p_{i}}(n)$ for all $1 \leq i \le... | [
"For arbitrary prime $p$ and positive integer $n$, denote by $\\operatorname{ord}_{p}(n)$ the exponent of $p$ in $n$. Thus,\n$$\n\\nu_{p}(n)=\\operatorname{ord}_{p}(n!)=\\sum_{i=1}^{n} \\operatorname{ord}_{p}(i)\n$$\nLemma. Let $p$ be a prime number, $q$ be a positive integer, $k$ and $r$ be positive integers such ... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0kgr | Problem:
Jude repeatedly flips a coin. If he has already flipped $n$ heads, the coin lands heads with probability $\frac{1}{n+2}$ and tails with probability $\frac{n+1}{n+2}$. If Jude continues flipping forever, let $p$ be the probability that he flips 3 heads in a row at some point. Compute $\lfloor 180 p\rfloor$. | [
"Solution:\n\nLet $p_{n}$ be the probability that the $n$\\text{th} head is flipped after a tail and Jude has yet to flip 3 heads consecutively to this point. For example, $p_{2}=\\frac{2}{3}$, as it is impossible for 3 heads to be flipped consecutively and the second head comes after a tail exactly when the first ... | United States | HMMT Spring 2021 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 47 | |
04gn | In a triangle $ABC$ bisector of the angle $\angle ACB$ meets $\overline{AB}$ at $D$. If $|CB| = |CD|$, $|AD| = 4$ and $|DB| = 3$, determine $|AC|$. (Hong Kong) | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 8 | |
01yu | The polynomial $P(x, y)$ of two variables with integer coefficients satisfies the following two conditions: 1) for every integer $a$ there exists exactly one integer $y$ such that $P(a, y) = 0$; and 2) for every integer $b$ there exists exactly one integer $x$ such that $P(x, b) = 0$.
a) Prove that if the degree of $P(... | [
"**a)** Let us prove the following\n**Lemma.** If for each integer $x$ the value of the polynomial $Q(x) = bx^2 + cx + d$ with integer coefficients is a square of an integer, then there exist integers $m$ and $n$ such that $b = m^2$, $c = 2mn$, $d = n^2$ and, in particular, $Q(x) = (mx + n)^2$.\n\n**Proof of the le... | Belarus | Belarus2022 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | Yes. For example, P(x, y) = (3x − 2y)(3x − 4y + 1)(3x − 4y − 1). | |
008w | For each natural number $n$, let $a_n$ be the greatest perfect square number lower than or equal to $n$ and $b_n$ the smallest perfect square number greater than $n$. For instance, $a_9 = 3^2$, $b_9 = 4^2$, $a_{20} = 4^2$ and $b_{20} = 5^2$.
Calculate the sum of the 600 terms
$$
\frac{1}{a_1b_1} + \frac{1}{a_2b_2} + \d... | [] | Argentina | XXIX Olimpíada Matemática Argentina National Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | 599/600 | |
009c | An integer is written in each square of a $100 \times 100$ checkerboard. An allowed operation consists in choosing four squares in one of the following layouts:
or any rotation of them, and add $1$ to the numbers written on each square. The goal is to obtain, by me... | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 3 | |
0jow | Problem:
Let $ABC$ be a triangle whose incircle has center $I$ and is tangent to $\overline{BC}$, $\overline{CA}$, $\overline{AB}$, at $D, E, F$. Denote by $X$ the midpoint of major arc $\widehat{BAC}$ of the circumcircle of $ABC$. Suppose $P$ is a point on line $XI$ such that $\overline{DP} \perp \overline{EF}$.
Give... | [
"Solution:\n\n$\\boxed{\\dfrac{4 \\sqrt{5}}{5}}$\n\nLet $H$ be the orthocenter of triangle $DEF$. We claim that $P$ is the midpoint of $\\overline{DH}$. Indeed, consider an inversion at the incircle of $ABC$, denoting the inverse of a point with an asterisk. It maps $ABC$ to the nine-point circle of $\\triangle DEF... | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
... | null | final answer only | 4√5/5 | |
0gif | Let $\mathbb{R}_{>0}$ be the set of positive real numbers. Determine all functions $f : \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that
$$
x(f(x) + f(y)) \geq (f(f(x)) + y)f(y)
$$
for every $x, y \in \mathbb{R}_{>0}$.
令 $\mathbb{R}_{>0}$ 為所有正實數所成的集合。試找出所有函數 $f: \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ 滿足：
$$
x(f(x) + f(y))... | [
"All functions $f(x) = \\frac{c}{x}$ for some $c > 0$.\n\nLet $f : \\mathbb{R}_{>0} \\to \\mathbb{R}_{>0}$ be a function that satisfies the inequality of the problem statement. We will write $f^k(x) = f(f(\\cdots f(x)\\cdots))$ for the composition of $f$ with itself $k$ times, with the convention that $f^0(x) = x$.... | Taiwan | IMO 1J, Mock Exam 1 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | Chinese; English | proof and answer | f(x) = c/x for some c > 0 | |
03ap | Consider a unit cell $2 \times 2010$ table. Ivan puts horizontal domino $[\ ]$, which covers exactly $2$ of the cells, then Peter puts vertical domino $[\ ]$, which covers exactly $2$ of the cells then again Ivan puts horizontal domino and so on. The player who has no move loses the game. Determine which of the two pla... | [
"We describe winning strategy for Ivan. He divides the table into $502$ tables $2 \\times 4$ and a table $2 \\times 2$. Then he puts his first domino in the table $2 \\times 2$ (no matter in which row). Peter is forced to put his domino in a table $2 \\times 4$. Now Ivan puts a domino in any place in the same table... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Ivan | |
05hm | Problem:
Soit $ABC$ un triangle acutangle non isocèle en $A$. Soit $M$ le milieu du segment $[BC]$, $H$ l'orthocentre du triangle $ABC$, $O_{1}$ le milieu du segment $[AH]$ et $O_{2}$ le centre du cercle circonscrit au triangle $CBH$. Montrer que le quadrilatère $O_{1}AMO_{2}$ est un parallélogramme.
![](attached_ima... | [
"Solution:\n\nSoit $H_{A}$ le pied de la hauteur issue du sommet $A$, $H_{B}$ le pied de la hauteur issue du sommet $B$ et $O$ le centre du cercle circonscrit au triangle $ABC$.\n\nOn remarque déjà que $O_{2}$ est sur la médiatrice de $[BC]$ donc les droites $(MO_{2})$ et $(AH)$ sont parallèles car elles sont perpe... | France | ENVOI 1 : GÉOMÉTRIE Corrigé | [
"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 > Miscellaneous > Distance... | null | proof only | null | |
05g8 | Problem:
Soit $x$ un réel strictement positif tel que $x^{5}-x^{3}+x \geq 3$. Prouver que $x^{6} \geq 5$. | [
"Solution:\nLa clé du problème est la factorisation $x^{6}+1=\\left(x^{2}+1\\right)\\left(x^{4}-x^{2}+1\\right)$, valable pour tout réel $x$.\nSoit $x>0$ tel que $x^{5}-x^{3}+x \\geq 3$. D'après l'identité ci-dessus, on a alors\n$$\nx^{6}+1=\\left(x^{2}+1\\right)\\left(x^{4}-x^{2}+1\\right)=\\frac{x^{2}+1}{x}\\left... | France | null | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0kgo | Forty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, “I can’t tell who has the larger number.” Then Bob says, “I know who has the larger number.” Alice says, “You do? Is your number p... | [
"Based on Alice's first statement, Bob can deduce that her number is not $1$ or $40$. Bob says that he knows who has the larger number, which implies that his number must be $1$, $2$, $39$, or $40$. The number $2$ is the only prime among them, which tells Alice that Bob's number is $2$. Alice then says that her num... | United States | Fall 2021 AMC 10 B | [
"Discrete Mathematics > Logic",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | A |
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