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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
06p8 | Let $n$ be a positive integer, and let $x$ and $y$ be positive real numbers such that $x^{n}+y^{n}=1$. Prove that
$$
\left(\sum_{k=1}^{n} \frac{1+x^{2 k}}{1+x^{4 k}}\right)\left(\sum_{k=1}^{n} \frac{1+y^{2 k}}{1+y^{4 k}}\right)<\frac{1}{(1-x)(1-y)} .
$$ | [
"For each real $t \\in (0,1)$,\n$$\n\\frac{1+t^{2}}{1+t^{4}}=\\frac{1}{t}-\\frac{(1-t)\\left(1-t^{3}\\right)}{t\\left(1+t^{4}\\right)}<\\frac{1}{t}\n$$\nSubstituting $t=x^{k}$ and $t=y^{k}$,\n$$\n0<\\sum_{k=1}^{n} \\frac{1+x^{2 k}}{1+x^{4 k}}<\\sum_{k=1}^{n} \\frac{1}{x^{k}}=\\frac{1-x^{n}}{x^{n}(1-x)} \\quad \\tex... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0ib5 | Problem:
There exists a polynomial $P$ of degree $5$ with the following property: if $z$ is a complex number such that $z^{5} + 2004 z = 1$, then $P\left(z^{2}\right) = 0$. Calculate the quotient $P(1) / P(-1)$. | [
"Solution:\nLet $z_{1}, \\ldots, z_{5}$ be the roots of $Q(z) = z^{5} + 2004 z - 1$. We can check these are distinct (by using the fact that there's one in a small neighborhood of each root of $z^{5} + 2004 z$, or by noting that $Q(z)$ is relatively prime to its derivative). And certainly none of the roots of $Q$ i... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | -2010012/2010013 | |
0k79 | Problem:
Let $f(n)$ be the number of distinct digits of $n$ when written in base 10. Compute the sum of $f(n)$ as $n$ ranges over all positive 2019-digit integers. | [
"Solution:\n\n$$\nf(n) = f_{0}(n) + \\cdots + f_{9}(n)\n$$\nwhere $f_{d}(n) = 1$ if $n$ contains the digit $d$ and $0$ otherwise. The sum of $f_{d}(n)$ over all 2019-digit positive integers $n$ is just the number of 2019-digit positive integers that contain the digit $d$. For $1 \\leq d \\leq 9$,\n$$\n\\sum_{n} f_{... | United States | HMMT November 2019 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 9(10^{2019} - 9^{2019}) | |
02s6 | Problem:
Arnaldo, Bernaldo, Cernaldo, Dernaldo e Ernaldo são estudantes de distintas partes do Brasil que foram escolhidos para representar o seu país nas olimpíadas internacionais. Depois de várias semanas de treino, algumas amizades foram formadas. Perguntamos, então, a cada um deles quantos amigos tinham feito no g... | [
"Solution:\n\nDigamos que Ernaldo tem $x$ amigos dentro do grupo. Como Dernaldo tem 4 amigos, e o grupo tem 5 integrantes, então todos são amigos de Dernaldo. Tiremos Dernaldo do grupo. Assim, Arnaldo, Bernaldo, Cernaldo e Ernaldo ficam agora, respectivamente, com $0,1,2$ e $x-1$ amigos dentro do subgrupo. Como Arn... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Logic"
] | null | proof and answer | 2 | |
0cn8 | Given integers $x, y \in [2, 100]$, prove that there exists a positive integer $n$ such that the number $x^2 + y^2$ is composite.
(S. Berlov, A. Belov) | [
"11.8. См. решение задачи 10.8."
] | Russia | Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English; Russian | proof only | null | |
0jky | Problem:
How many ways can you remove one tile from a $2014 \times 2014$ grid such that the resulting figure can be tiled by $1 \times 3$ and $3 \times 1$ rectangles? | [
"Solution:\nNumber the rows and columns of the grid from $0$ to $2013$, thereby assigning an ordered pair to each tile. We claim that a tile $(i, j)$ may be selected if and only if $i \\equiv j \\equiv 0 \\pmod{3}$; call such a square good.\n\nFirst, let us show that this condition is sufficient. Observe that any s... | United States | HMMT November 2014 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 451584 | |
04jv | In a country there are $N$ cities and among some of them there are (two-way) flight connections. Each flight connects exactly two cities. No city is not directly connected to every other city. For any two cities $A$ and $B$ there is exactly one way to fly from $A$ to $B$ using at most two flights.
Prove that $N - 1$ is... | [
"Let $a$ be an arbitrary city and let it be connected to cities $x_1, \\dots, x_k$. We denote by $B_i$ the set of all cities connected to $x_i$ that are different from $a$, for $i = 1, \\dots, k$. Sets $B_1, \\dots, B_k$ are pairwise disjoint because there is exactly one way to fly from $x_i$ to $x_j$ in at most tw... | Croatia | Croatian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
0abm | In the triangle $ABC$, $\angle BAC = 120^\circ$. On the bisector of the angle $\angle BAC$, a point $D$ is chosen such that $\overline{AD} = \overline{AB} + \overline{AC}$. Prove that $\triangle BCD$ is equilateral. | [
"Let $M$ be a point on $AD$, such that $\\overline{AM} = \\overline{AC}$. $\\angle MAC = 60^\\circ$ implies that $\\triangle ACM$ is an equilateral triangle. Because $\\angle CAB = \\angle CMD = 120^\\circ$, $\\overline{MC} = \\overline{AC}$ and $\\overline{AB} = \\overline{MD}$ we have that $\\triangle CAB \\cong ... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
07y0 | Find all pairs $(n, p)$ of positive integers such that:
* $p$ is the smallest positive integer that is not a factor of $n$, and;
* $n$ is a factor of $3p^2 + 1$. | [
"A complete list of solutions is\n\n| $n$ | 1 | 2 | 4 | 13 | 14 | 28 |\n|-----|---|---|---|----|----|----|\n| $p$ | 2 | 3 | 3 | 2 | 3 | 3 |\n| $3p^2 + 1$ | 13 | 28 | 28 | 13 | 28 | 28 |\n\nSolution:\n\n**Solution 1.** Any factor of $n$ is a factor of $3p^2 + 1$. Hence, $3$ cannot be a factor of $n$. Hence, $p$, ... | Ireland | IRL_ABooklet_2025 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | Pairs (n, p): (1, 2), (13, 2), (2, 3), (4, 3), (14, 3), (28, 3) | |
0k5q | Problem:
For breakfast, Mihir always eats a bowl of Lucky Charms cereal, which consists of oat pieces and marshmallow pieces. He defines the luckiness of a bowl of cereal to be the ratio of the number of marshmallow pieces to the total number of pieces. One day, Mihir notices that his breakfast cereal has exactly $90$... | [
"Solution:\n\nLet $x$ be the number of marshmallows to add. We are given that\n$$\n2 \\cdot \\frac{9}{99} = \\frac{9 + x}{99 + x}\n$$\nRearranging this gives\n$$\n2(99 + x) = 11(9 + x)\n$$\nThus $9x = 99$ and $x = 11$."
] | United States | HMMT November 2019 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 11 | |
073p | Let $\alpha$ be a real number larger than $1$. Define a sequence $(s_n)_{n \ge 1}$ as follows: $s_1 = 1$, $s_2 = \alpha$; if $s_1, s_2, s_3, \dots, s_{2^n}$ are defined for some $n \ge 1$, then $s_{2^n+1}, s_{2^n+2}, \dots, s_{2^{n+1}}$ are defined by $s_j = \alpha s_{j-2^n}$, for $2^n+1 \le j \le 2^{n+1}$. (Thus the f... | [
"We prove this in four steps. We use $s_n = s(n)$ and $c_n = c(n)$,\n\n(A) If $b(n)$ denotes the number of $1$'s in the binary representation of a non-negative integer $n$, we have $s(n) = \\alpha^{b(n-1)}$, $n \\ge 1$.\n**Proof:** We use induction. Note that $s(1) = 1 = \\alpha^{b(0)}$. Suppose $s(r) = \\alpha^{b(... | India | Indija TS 2008 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
02wo | Problem:
Em torneios de xadrez, geralmente, cada vitória vale 1 ponto, cada empate, 0,5 ponto e cada derrota, zero ponto. No "Campeonato Magistral" de xadrez participaram apenas Mestres Internacionais - MI's e Grandes Mestres - GM's. O número de GM’s foi dez vezes o número de MI's. Cada enxadrista jogou apenas uma vez... | [
"Solution:\n\na) Sejam $i$ e $g$, os números de MI's e GM's, respectivamente. Denote as somas de pontuações de cada categoria por $I$ e $G$ e o total de jogadores por $n$. Do enunciado, $g=10i$, $G=4,5I$ e $n=i+g=i+10i=11i$. Agora, o total de jogos (e de pontos disputados) foi\n$$\n\\begin{aligned}\n\\frac{n(n-1)}{... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) 1 b) 10 c) 55 | |
0ev8 | Given a triangle $ABC$ with $AC = BC > AB$, let $E$ and $F$ be the midpoints of $AC$ and $AB$, respectively. The perpendicular bisector $\ell$ of $AC$ meets $AB$ at $K$ and the line parallel to $KC$ and passing through the point $B$ intersects $AC$ at a point $L$. For a point $P$ on the line segment $BF$, let $H$ be th... | [
"Let $W'$ be the circumcenter of the triangle $ACP$, and let $L'$ be the intersection point of the line $FW'$ and the side $AC$. Note that the points $A, F, C$ and the foot $T$ of the perpendicular from $A$ to $PC$ are on the circle with a diameter $AC$, and thus $\\angle PAT - \\angle PCF$. Since $AC = BC$ and $F$... | South Korea | Korean Mathematical Olympiad Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03i5 | Problem:
Find all pairs $a, b$ of positive integers satisfying the equation $2 a^{2} = 3 b^{3}$. | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All solutions are a = 18 t^3 and b = 6 t^2 for any positive integer t. | |
01mg | All cells of a $7 \times 8$ board are colored red, blue, and green. In any row of the board the number of red cells is no smaller than the number of blue cells and no smaller than the number of green ones. In any column of the board the number of blue cells is no smaller than the number of red cells and no smaller than... | [] | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | smallest = 8, greatest = 8 | |
05bc | Find the greatest possible value of the expression
$$
(((a * b) * c) * d) * e,
$$
if each star is replaced with one of the operations $+$, $-$, $\cdot$ and the numbers $a, b, c, d, e$ are $-2, -1, 0, 1, 2$ in some order. Different stars can correspond to different operations. | [
"We will show that a value greater than $8$ is impossible. For this, let us assume the opposite. In this case, we can simplify the expression by removing the $0$ and the operation to the left of it. The remaining operation consists of the numbers $-2, -1, 1, 2$ and three operations and has a value of at least $9$. ... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 8 | |
0iyr | Problem:
The squares of a $3 \times 3$ grid are filled with positive integers such that $1$ is the label of the upper-leftmost square, $2009$ is the label of the lower-rightmost square, and the label of each square divides the one directly to the right of it and the one directly below it. How many such labelings are p... | [
"Solution:\n\nWe factor $2009$ as $7^{2} \\cdot 41$ and place the $41$'s and the $7$'s in the squares separately. The number of ways to fill the grid with $1$'s and $41$'s so that the divisibility property is satisfied is equal to the number of nondecreasing sequences $a_{1}, a_{2}, a_{3}$ where each $a_{i} \\in \\... | United States | $12^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2448 | |
08yb | Let $S = \{1, 2, \dots, 6\}$. Let $\mathcal{S}$ be the set of all subsets of $S$. How many mappings $F: \mathcal{S} \to \mathcal{S}$ are there that satisfy the following condition:
$$
F(F(A) \cup B) = A \cap F(B) \text{ for any pair of subsets of } S.
$$ | [
"Let us fix $C \\subset S$ and consider the situation where $F(\\emptyset) = C$. If we let $A = \\emptyset$ in the given identity, we get $F(C \\cup B) = \\emptyset$ for any set $B$ in $\\mathfrak{S}$. So, if $D \\supset C$ then we have $F(D) = \\emptyset$. Therefore, if we let $A = C$, we get $F(B) = C \\cap F(B)$... | Japan | 2019 Japan Mathematical Olympiad First Stage | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof and answer | 499 | |
0jci | Problem:
For integer $n, m \geq 1$, let $A(n, m)$ denote the number of functions $f:\{1,2, \ldots, n\} \rightarrow \{1,2, \ldots, m\}$ such that $f(j)-f(i) \leq j-i$ for all $1 \leq i < j \leq n$, and let $B(n, m)$ denote the number of functions $g:\{0,1, \ldots, 2n+m\} \rightarrow \{0,1, \ldots, m\}$ such that $g(0)=... | [
"Solution:\n\nWe first note that the condition for $f$ is equivalent to $i - f(i) \\leq j - f(j)$ for all $1 \\leq i < j \\leq n$. Letting $f'(x) = x - f(x)$, we see this is equivalent to saying that $f'$ is decreasing. Thus, we only need that $f'(x) \\leq f'(x+1)$; in other words, we only require the statement to ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
00kf | Let $a$, $b$, $c$ and $d$ be real numbers with $a < b < c < d$.
Sort the numbers $x = a b + c d$, $y = b c + a d$ and $z = c a + b d$ in ascending order and prove the correctness of your result. | [
"By the rearrangement inequality we have (writing $(\\cdot, \\cdot)$ for the scalar product of two vectors)\n$$\n\\begin{aligned}\ny &= b c + a d = \\langle (b, a), (c, d) \\rangle < \\langle (a, b), (c, d) \\rangle = a c + b d = z \\\\\n&= \\langle (a, d), (c, b) \\rangle < \\langle (a, d), (b, c) \\rangle = a b +... | Austria | Austria 2014 | [
"Algebra > Equations and Inequalities",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof and answer | y < z < x | |
00yq | Problem:
Assume we have $95$ boxes and $19$ balls distributed in these boxes in an arbitrary manner. We take six new balls at a time and place them in six of the boxes, one ball in each of the six. Can we, by repeating this process a suitable number of times, achieve a situation in which each of the $95$ boxes contain... | [
"Solution:\n\nSince $6 \\cdot 16 = 96$, we can put $16$ times $6$ balls in the boxes so that the number of balls in one of the boxes increases by two, while in all other boxes it increases by one. Repeating this procedure, we can either diminish the difference between the number of balls in the box which has most b... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0dcy | Among numbers $a+b$, $a-b$, $a b$, $\frac{a}{b}$ exactly two are positive and exactly two are negative. Find the sign of $b$. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | b is negative | |
04y6 | Let $\omega$ be the circumcircle of an acute-angled triangle $ABC$. Point $D$ lies on the arc $BC$ of $\omega$ not containing point $A$. Point $E$ lies in the interior of the triangle $ABC$, does not lie on the line $AD$, and satisfies $\angle DBE = \angle ACB$ and $\angle DCE = \angle ABC$. Let $F$ be a point on the l... | [
"Denote $\\alpha = \\angle CAB$, $\\beta = \\angle ABC$, and $\\gamma = \\angle BCA$. Let $K$ and $L$ be the second intersections of lines $BE$ and $CE$ with $\\omega$, respectively, different from $B$ and $C$. Observe that\n$$\n\\angle BAK = \\angle BAD + \\angle DAK = \\angle BAD + \\angle DBE = \\angle BAD + \\g... | Czech-Polish-Slovak Mathematical Match | null | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0enk | The *liars guessing game* is a game played between two players *A* and *B*. The rules of the game depend on two positive integers *k* and *n* which are known to both players.
At the start of the game *A* chooses integers $x$ and $N$ with $1 \le x \le N$. Player *A* keeps $x$ secret, and truthfully tells $N$ to player ... | [
"The game can be reformulated in an equivalent one: The player $A$ chooses an element $x$ from the set $S$ (with $|S| = N$) and the player $B$ asks the sequence of questions. The $j$-th question consists of $B$ choosing a set $D_j \\subseteq S$ and player $A$ selecting a set $P_j \\in \\{Q_j, Q_j^C\\}$. The player ... | South Africa | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
004t | Sea $ABC$ un triángulo con todos sus ángulos agudos, de alturas $AD$, $BE$ y $CF$ (con $D$ en $BC$, $E$ en $AC$ y $F$ en $AB$). Sea $M$ el punto medio del segmento $BC$. La circunferencia circunscrita al triángulo $AEF$ corta a la recta $AM$ en $A$ y en $X$. La recta $AM$ corta a la recta $CF$ en $Y$. Sea $Z$ el punto ... | [] | Argentina | XVIII Olimpiada Matemática del Cono Sur | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Español | proof only | null | |
0kk6 | Problem:
Caroline starts with the number $1$, and every second she flips a fair coin; if it lands heads, she adds $1$ to her number, and if it lands tails she multiplies her number by $2$. Compute the expected number of seconds it takes for her number to become a multiple of $2021$. | [
"Solution:\n\nConsider this as a Markov chain on $\\mathbb{Z} / 2021 \\mathbb{Z}$. This Markov chain is aperiodic (since $0$ can go to $0$) and any number can be reached from any other number (by adding $1$), so it has a unique stationary distribution $\\pi$, which is uniform (since the uniform distribution is stat... | United States | HMMT Spring 2021 Guts Round | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 4040 | |
003s | Sean $H$ y $O$, respectivamente, el ortocentro y circuncentro del triángulo acutángulo $ABC$ ($AB \neq AC$); sea $\Gamma$ la circunferencia circunscripta al triángulo $ABC$. El prolongamiento de la mediana $AM$, relativa al lado $BC$, corta a $\Gamma$ en el punto $N$; la circunferencia de diámetro $AM$ corta a $\Gamma$... | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"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 > Miscellaneous > Angle chasing"
] | Español | proof only | null | |
0i89 | Problem:
As shown, $U$ and $C$ are points on the sides of triangle $M N H$ such that $M U = s$, $U N = 6$, $N C = 20$, $C H = s$, $H M = 25$. If triangle $U N C$ and quadrilateral $M U C H$ have equal areas, what is $s$?
 | [
"Solution:\nUsing brackets to denote areas, we have $[M C H] = [U N C] + [M U C H] = 2[U N C]$. On the other hand, triangles with equal altitudes have their areas in the same ratio as their bases, so\n$$\n2 = \\frac{[M N H]}{[U N C]} = \\frac{[M N H]}{[M N C]} \\cdot \\frac{[M N C]}{[U N C]} = \\frac{N H}{N C} \\cd... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 4 | |
05pi | Problem:
Soit $ABC$ un triangle et $(I)$ le centre de son cercle inscrit. La droite $(AI)$ recoupe $[BC]$ en $D$. La médiatrice de $[AD]$ recoupe $(BI)$ en $M$ et $(CI)$ en $N$.
Montrer que $A$, $M$, $N$ et $I$ sont cocycliques.
 | [
"Solution:\n\nOn commence par rappeler le théorème du Pôle Sud :\n\nLemme 1 (Théorème du Pôle Sud). Soit $ABC$ un triangle, $\\Gamma$ son cercle circonscrit et $I$ le centre de son cercle inscrit. Soit $S$ le second point d'intersection de la bissectrice de $\\widehat{BAC}$ avec $\\Gamma$. Alors $SB = SC$. Le point... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0c83 | Consider a cyclic quadrilateral $ABCD$ and let $M$ and $N$ be the midpoints of the diagonals $AC$ and $BD$, respectively. If $\angle AMB = \angle AMD$, prove that $\angle ANB = \angle BNC$. | [
"(Cristian Mangra)\n\n**Lemma:** Let $ABCD$ be an isosceles trapezoid ($AB \\parallel CD$) inscribed in a circle $C$ of center $O$, $M$ is the intersection point of the diagonals $AC$ and $BD$, and $T$ is the intersection point of lines $AD$ and $BC$. If the line through $M$ parallel to $AB$ meets the circle $C$ at... | Romania | The DANUBE Mathematical Competition | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Tria... | English | proof only | null | |
0jid | Given a set $S$ of $n$ variables, a binary operation $\times$ on $S$ is called *simple* if it satisfies $(x \times y) \times z = x \times (y \times z)$ for all $x, y, z \in S$ and $x \times y \in \{x, y\}$ for all $x, y \in S$. Given a simple operation $\times$ on $S$, any string of elements in $S$ can be reduced to a ... | [
"**Solution 1.** The answer is $n!^2$.\nWe use $s_1 \\sim s_2$ to denote that strings $s_1$ and $s_2$ are equivalent. If $x_1x_2 \\cdots x_k$ is any string in $S$, we can see that\n$$\nx_1x_2\\cdots x_kx_1x_2\\cdots x_k = (x_1\\cdots x_k)(x_1\\cdots x_k) \\sim x_1\\cdots x_k\n$$\nbecause $xx \\in \\{x\\}$. But also... | United States | IMO Team Selection Team Selection Test | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Algebra > Abstract Algebra > Other",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Other"
] | null | proof and answer | n!^2 | |
081r | Problem:
Determinare il numero dei parallelepipedi retti con base quadrata che hanno tutti gli spigoli di lunghezza intera e volume uguale a $270000$. | [
"Solution:\n\nSiano $x$ il lato della base e $y$ l'altezza di un parallelepipedo che soddisfa le condizioni date. Allora $x$ e $y$ sono interi positivi e\n$$\nx^2 y = 270000 = 3^3 \\cdot 2^4 \\cdot 5^4.\n$$\nPer la fattorizzazione unica degli interi, si ha $x = 3^a 2^b 5^c$, $y = 3^d 2^e 5^f$, dove gli esponenti so... | Italy | Progetto Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 18 | |
0hf4 | Prove that for natural numbers $a \ge b \ge c \ge d$ the inequality:
$$
ab+bc+cd-b^{2}-c^{2}-d^{2} \ge a-d.
$$ | [
"We have:\n$$\nab+bc+cd-b^{2}-c^{2}-d^{2} = b(a-b)+c(b-c)-d(c-d) \\ge \\\\\n\\ge (a-b)+(b-c)+(c-d)=a-d.\n$$"
] | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0b6y | Let $S$ be the sum of all invertible elements of a finite ring.
Prove that $S^2 = S$ or $S^2 = 0$. | [
"If $1 + 1 \\ne 0$, then $x \\ne -x$, for every invertible $x$, whence $S = 0$ (we can group the invertible elements into pairs $(x, -x)$).\n\nIf this is not the case, notice that $xS = S$, for every invertible $x$. Adding all these relations, $S^2 = kS$, where $k$ is the number of invertible elements. Then $S^2 = ... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Abstract Algebra > Group Theory"
] | English | proof only | null | |
01rf | Let $H$ be an intersection point of the altitudes $AA_1$, $BB_1$, $CC_1$ of an acute-angled triangle $ABC$. Let $M$ and $N$ be the midpoints of the segments $BC$ and $AH$, respectively.
Prove that $MN$ is the perpendicular bisector of the segment $B_1C_1$. | [
"Note that the segments $B_1M$ and $C_1M$ are medians in the right-angled triangles $B_1BC$ ($\\angle B_1 = 90^\\circ$) and $C_1CB$ ($\\angle C_1 = 90^\\circ$) respectively. So, $B_1M = 0.5CB$ and $C_1M = 0.5CB$, therefore $B_1M = C_1M$, i.e. $M$ is equidistant from the ends of the segment $B_1C_1$.\n\nSimilarly, w... | Belarus | Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0cup | A prestidigitator and his assistant have a deck of cards; back sides of all the cards look identical, the front side of each card is painted in one of $2017$ colors (there are $1000000$ cards of each color in the deck). They want to perform the following trick. The prestidigitator goes out of the room. The spectators p... | [
"Answer: $n = 2018$.\n\nSet $k = 2017$. If $n = k + 1$, the assistant merely encodes the color of the last card by choosing which of the others remains open.\n\nAssume that $k \\ge n$. A strategy of the prestidigitator and the assistant may be expressed as a set of instructions of the form $(a, i) \\to (b, j)$ mean... | Russia | XLIII Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | 2018 | |
00gx | Larry and Rob are two robots travelling in one car from Argovia to Zillis. Both robots have control over the steering and steer according to the following algorithm: Larry makes a $90^{\circ}$ left turn after every $\ell$ kilometer driving from start; Rob makes a $90^{\circ}$ right turn after every $r$ kilometer drivin... | [
"Let Zillis be $d$ kilometers away from Argovia, where $d$ is a positive real number. For simplicity, we will position Argovia at $(0,0)$ and Zillis at $(d, 0)$, so that the car starts out facing east. We will investigate how the car moves around in the period of travelling the first $\\ell r$ kilometers, the secon... | Asia Pacific Mathematics Olympiad (APMO) | XXI Asian Pacific Mathematics Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | ℓ ≡ r ≡ 1 (mod 4) or ℓ ≡ r ≡ 3 (mod 4) | |
00bn | For $n \in \mathbb{N}$ let $D_2(n)$ (respectively $D_3(n)$) denote the number of divisors of $n$ that are perfect squares (respectively perfect cubes). Prove that there is an $n$ such that $D_2(n) = 999D_3(n)$. | [
"For $a \\in \\mathbb{N}$ denote $u(a) = \\left[ \\frac{a}{2} \\right]$, $v(a) = \\left[ \\frac{a}{3} \\right]$. If $n = p_1^{a_1} \\cdots p_k^{a_k}$ is the prime factorization of $n \\in \\mathbb{N}$, it is straightforward that $D_2(n) = (u(a_1)+1)\\cdots(u(a_k)+1)$, $D_3(n) = (v(a_1)+1)\\cdots(v(a_k)+1)$.\n\nDefi... | Argentina | Argentina_2018 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0gq1 | Alice chooses an odd number $k = 2l+1$, places $k$ plates around a circle and distributes 2013 beads into some of these plates as she wishes. After that Bob marks one of the plates. Determine the maximal possible value of $N$ such that no matter how Bob marks a plate Alice can choose exactly $l$ unmarked non-neighbouri... | [
"For each $k$ we label plates clockwise $1, \\ldots, k$. Alice can take $k = 9$ and put $671$ beads in each of the plates numbered $1$, $4$ and $7$. It can be readily checked out that for any choice of Bob Alice can choose two nonempty plates containing $1342$ beads in total.\n\nNow let us show that Alice can not g... | Turkey | 18th Junior Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 1342 | |
0d1u | Find the maximum and the minimum values of
$$
S = (1 - x_{1})(1 - y_{1}) + (1 - x_{2})(1 - y_{2})
$$
for real numbers $x_{1}, x_{2}, y_{1}, y_{2}$ with $x_{1}^{2} + x_{2}^{2} = y_{1}^{2} + y_{2}^{2} = 2013$. | [
"We present for this problem two solutions, one using trigonometric functions and the other using classical inequalities.\n\nFirst solution. The condition $x_{1}^{2} + x_{2}^{2} = y_{1}^{2} + y_{2}^{2} = c^{2}$, here $c^{2} = 2013$, is equivalent to saying that there exist $\\alpha, \\beta \\in \\mathbb{R}$ such th... | Saudi Arabia | Selection tests for the International Mathematical Olympiad 2013 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | maximum = 2015 + 2 sqrt(4026), minimum = -2012 | |
04hn | Prove that there exist infinitely many positive integers $n$ such that the largest prime divisor of $n^4 + n^2 + 1$ is equal to the largest prime divisor of $(n+1)^4 + (n+1)^2 + 1$. (IMO Shortlist 2013) | [
"Let $p_n$ be the largest prime divisor of $n^4 + n^2 + 1$ and let $q_n$ be the largest prime divisor of $n^2 + n + 1$. Then $p_n = q_{n^2}$, and from\n$$\nn^4 + n^2 + 1 = (n^2 + 1)^2 - n^2 = (n^2 - n + 1)(n^2 + n + 1) = ((n-1)^2 + (n-1) + 1)(n^2 + n + 1)\n$$\nit follows that $p_n = \\max\\{q_n, q_{n-1}\\}$ for $n ... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
03l7 | Problem:
Call a positive integer $n$ practical if every positive integer less than or equal to $n$ can be written as the sum of distinct divisors of $n$.
For example, the divisors of $6$ are $\{1, 2\}$, $\{3\}$, and $\{6\}$. Since
$$
1=\{1\}, \quad 2=\{2\}, \quad 3=\{3\}, \quad 4=\{1\}+\{3\}, \quad 5=\{2\}+\{3\}, \qu... | [
"Solution:\n\nLet $p$ and $q$ be practical. For any $k \\leq p q$, we can write\n$$\nk = a q + b \\text{ with } 0 \\leq a \\leq p,\\ 0 \\leq b < q.\n$$\nSince $p$ and $q$ are practical, we can write\n$$\na = c_1 + \\ldots + c_m, \\quad b = d_1 + \\ldots + d_n\n$$\nwhere the $c_i$'s are distinct divisors of $p$ and ... | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0fph | Let $ABC$ be a triangle. $D$ is the foot of the internal bisector of the angle $A$. The perpendicular from $D$ to the tangent $AT$ ($T$ belongs to $BC$) to the circumscribed circle of $ABC$ intersects the altitude $AH_a$ at the point $I$ ($H_a$ belongs to $BC$). If $P$ is the midpoint of $AB$ and $O$ is the circumcente... | [
"Let $Q$ be the midpoint of $AC$ and $N$ the intersection of $AD$ and $PQ$. Then $N$ is the midpoint of $AD$. As $DE$ is perpendicular to $AT$, being $E$ the intersection point of $DI$ and $AT$, and as $OA$ is perpendicular to $AT$, we get that $DE$ is parallel to $OA$, and so the angles $OAN$ and $ADE$ are equal. ... | Spain | MEDITERRANEAN MATHEMATICAL COMPETITION | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > ... | Spanish | proof only | null | |
01mz | Find all triples $(x; n; p)$ of positive integers $x$, $n$, and primes $p$, such that
$$
x^3 + 3x + 14 = 2 \cdot p^n.
$$ | [
"Answer: $(x; n; p) = (1; 2; 3)$, $(x; n; p) = (3; 2; 5)$.\n\nIt is easy to see that $x^3 + 3x + 14 = (x+2)(x^2 - 2x + 7)$, so the initial equality can be rewritten as\n$$\n(x+2)(x^2-2x+7)=2 \\cdot p^n \\quad (1)\n$$\nIt is evident that $x^2-2x+7 > x+2$ for all $x \\in \\mathbb{N}$. So for the case $x+2 = 2 \\cdot ... | Belarus | 62nd Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (1, 2, 3) and (3, 2, 5) | |
02ov | Problem:
Sejam
$$
R=3 \times 9+4 \times 10+5 \times 11+\cdots+2003 \times 2009
$$
$$
S=1 \times 11+2 \times 12+3 \times 13+\cdots+2001 \times 2011
$$
(a) Qual é o maior número: $R$ ou $S$?
(b) Calcule a diferença entre o maior e o menor. | [
"Solution:\n(a) Cada parcela de $S$ é da forma $n \\times (n+10) = n^{2} + 10n$ e cada parcela de $R$ é da forma $(n+2) \\times (n+8) = n^{2} + 10n + 16$ com $n = \\{1, 2, \\ldots, 2001\\}$ em ambos os casos. Assim, para todo $n$, cada parcela de $R$ é maior que a correspondente em $S$, o que torna $R > S$.\n\n(b) ... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | R is larger; the difference is 32016 | |
0eh7 | Problem:
Poišči vsa praštevila $p, q$ in $r$, za katera ima polinom $f(x)=x^{3}-p x^{2}+q x-r^{2}$ same racionalne ničle. | [
"Solution:\n\nKer ima polinom $f$ celoštevilske koeficiente in vodilni koeficient enak 1, so vse njegove racionalne ničle v resnici celoštevilske. Če je $x$ negativno število, potem je tudi $f(x)=x^{3}-p x^{2}+q x-r^{2}$ negativno število, saj so vsi členi v tem primeru negativni. Torej polinom $f$ nima negativnih ... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | p = 11, q = 19, r = 3 | |
011m | Problem:
In a $40 \times 50$ array of control buttons, each button has two states: $\mathrm{ON}$ and OFF. By touching a button, its state and the states of all buttons in the same row and in the same column are switched. Prove that the array of control buttons may be altered from the all-OFF state to the all-ON state ... | [
"Solution:\n\nAnswer: $2000$.\n\nAltering the state from all-OFF to all-ON requires that the state of each button is changed an odd number of times. This is achieved by touching each button once. We prove that the desired result cannot be achieved if some button is never touched. In order to turn this button ON, th... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 2000 | |
0hfv | Let $ABCD$ be a cyclic quadrilateral. Suppose that there exists a line $l \parallel BD$, which is tangent to the inscribed circles of triangles $ABC$ and $CDA$. Prove that the line $l$ contains the incenter of one of $\triangle ABCD$ and $\triangle DAB$.
(Fedir Yudin) | [
"Suppose it's false, wlog $BC < CD$. Let $CK$ and $CL$ be the bisectors of the correspondent angles in triangles $\\triangle ABC$ and $\\triangle CDA$, and $B'$ and $D'$ be the points symmetric to the points $B$ and $D$ with respect to the lines $CK$ and $CL$ correspondingly (fig. 20). Then lines $KB$ and $KB'$ are... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"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 c... | English | proof only | null | |
0eay | How many two-digit numbers are there, such that the predecessor and the successor are a prime and a perfect square, not necessarily in this order?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5 | [
"Perfect squares that are also predecessors or successors of two-digit numbers are $9$, $16$, $25$, $36$, $49$, $64$, $81$ and $100$. Two-digit numbers that have a perfect square as their successors or predecessors are therefore $10$, $15$, $17$, $24$, $26$, $35$, $37$, $48$, $50$, $63$, $65$, $80$, $82$ and $99$. ... | Slovenia | National Math Olympiad in Slovenia | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | E | |
06nc | $ABCD$ is a trapezium with $AD \parallel BC$, $AD > BC$ and $AB = CD$. $P$ is a point on the plane whose distance from $A$, $B$, $C$, $D$ are $1$, $2$, $3$, $4$ respectively. Find $AD : BC$. | [
"Answer: $3 : 1$\n\nSuppose the line through $P$ perpendicular to $BC$ and $AD$ meets $AD$ at $E$ and $BC$ at $F$. Then we have\n$$\n4^2 - 1^2 = (DE^2 + PE^2) - (AE^2 + PE^2) = DE^2 - AE^2 = (DE + AE)(DE - AE).\n$$\nIf we draw another line perpendicular to $BC$ and $AD$ such that its distance from $D$ is equal to $... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3 : 1 | |
080a | Problem:
Un parallelepipedo retto ha spigoli di lunghezza $a, b, c$, con $a < b < c$. Se la lunghezza di uno degli spigoli viene aumentata di una quantità $q$, allora il volume del parallelepipedo aumenta. In quale dei seguenti casi si ha il massimo incremento di volume?
(A) Quando viene aumentato $a$
(B) quando viene... | [
"Solution:\n\nlati non incrementati. Tale prodotto è chiaramente massimo quando i due lati non incrementati sono $b$ e $c$ (i due più lunghi), cioè quando viene incrementato $a$."
] | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Solid Geometry > Volume"
] | null | MCQ | A | |
01xc | Cells of an $11 \times 11$ table are colored with $n$ colors (each cell is colored with exactly one color). For each color the total amount of the cells of this color is not less than $7$ and not greater than $13$.
Prove that there exists at least one row or column which contains cells of at least four different colors... | [] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0ihd | Problem:
Eight coins are arranged in a circle heads up. A move consists of flipping over two adjacent coins. How many different sequences of six moves leave the coins alternating heads up and tails up? | [
"Solution:\n\nImagine we flip over two adjacent coins by pushing a button halfway between them. Then the outcome depends only on the parities of the number of times that each button is pushed. To flip any coin, we must push the two buttons adjacent to that coin a total of an odd number of times. To flip every other... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 7680 | |
0401 | Prove that, among any $n$ vertexes of a regular $2n-1$ polygon ($n \ge 3$), there are three points, which are the vertexes of an isosceles triangle. | [
"Let the vertices of the regular $(2n-1)$-gon be $A_0, A_1, \\ldots, A_{2n-2}$, labeled in order around the circle.\n\nSuppose we select any $n$ vertices from these $2n-1$ vertices. For each vertex $A_i$, consider the set of possible isosceles triangles with $A_i$ as a vertex and the other two vertices also among t... | China | China Western Invitational Mathematical Competition | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
03u8 | Let $x_1, \dots, x_n$ ($n \ge 2$) be real numbers such that
$$
A = \left| \sum_{i=1}^{n} x_i \right| \neq 0
$$
and
$$
B = \max_{1 \le i < j \le n} |x_i - x_j| \neq 0.
$$
Prove that for every $n$ vectors $\alpha_1, \dots, \alpha_n$ on the plane, there exists a permutation $(k_1, k_2, \dots, k_n)$ of $(1, 2, \dots, n)$ s... | [
"**Proof** Let $|\\alpha_k| = \\max_{1 \\le i \\le n} |\\alpha_i|$. It is sufficient to prove that\n$$\n\\max_{(k_1, \\dots, k_n) \\in S_n} \\left| \\sum_{i=1}^{n} x_{k_i} \\alpha_i \\right| \\ge \\frac{AB}{2A+B} |\\alpha_k|,\n$$\nwhere $S_n$ is the set of all permutations of $(1, 2, \\dots, n)$.\n\nWithout loss of... | China | China National Team Selection Test | [
"Algebra > Linear Algebra > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
010h | Problem:
Let $P$ be a polynomial with integer coefficients. Suppose that for $n=1,2,3, \ldots, 1998$ the number $P(n)$ is a three-digit positive integer. Prove that the polynomial $P$ has no integer roots. | [
"Solution:\n\nLet $m$ be an arbitrary integer and define $n \\in \\{1,2, \\ldots, 1998\\}$ to be such that $m \\equiv n \\pmod{1998}$. Then $P(m) \\equiv P(n) \\pmod{1998}$. Since $P(n)$ as a three-digit number cannot be divisible by $1998$, then $P(m)$ cannot be equal to $0$. Hence $P$ has no integer roots."
] | Baltic Way | Baltic Way 1998 | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
022j | Problem:
Um professor enfurecido - Para castigar os alunos de sua turma por indisciplina, o professor Zerus decidiu descontar da nota mensal de cada aluno uma percentagem igual à nota da prova, isto é: quem tirou $60$, terá um desconto de $60\%$ na nota, quem tirou $20$, um desconto de $20\%$ da nota, e assim por dian... | [
"Solution:\n\nQuem teve $x$ como nota mensal vai ter um desconto de $x\\%$ sobre essa nota, ou seja vai perder\n$$\nx\\% \\text{ de } x = \\frac{x}{100} \\times x = \\frac{x^{2}}{100}\n$$\nLogo, depois do castigo, a nota fica sendo $x - \\frac{x^{2}}{100}$, onde $x$ era a nota inicial.\n\nConsideremos a função \"no... | Brazil | Nível 3 | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a) The highest final grade occurs for those with an original grade of fifty, and the final grade is twenty five. b) The lowest final grade is zero, occurring for original grades zero and one hundred. c) Yes; students with original grades equidistant from fifty (for example, whose grades add to one hundred) receive the ... | |
0d8r | A polynomial $P(x)$ with integer coefficients is called cube-represented if it can be represented as the sum of several cubes of polynomials with integer coefficients.
Examples: $3x + 3x^{2}$ is cube-represented because $3x + 3x^{2} = (x+1)^{3} + (-x)^{3} + (-1)^{3}$.
1. Is $3x^{2}$ a cube-represented polynomial?
2.... | [
"A polynomial with integer coefficients is called \"nice\" if the coefficients in the terms whose degree is not a multiple of $3$, are multiples of $3$, and their sum is a multiple of $6$. Clearly, the sum of nice polynomials is a nice polynomial, and the sum of cube-represented polynomials is a cube-represented po... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof and answer | 1) No; 2) 227711232 | |
0asq | Problem:
On a cyclic quadrilateral $A B C D$, there is a point $P$ on side $A D$ such that the triangle $C D P$ and the quadrilateral $A B C P$ have equal perimeters and equal areas. Prove that two sides of $A B C D$ have equal lengths. | [
"Solution:\n\nWe denote by $(X Y Z)$ and $(W X Y Z)$ the areas of $\\triangle X Y Z$ and quadrilateral $W X Y Z$, respectively. We use the labels depicted in the following figure.\n\n\n\nWith equal perimeters, we get\n$$\na+b+z+x=c+y+z\n$$\nor\n$$\na+b+x=c+y.\n$$\nWith equal areas, we get\n... | Philippines | Philippines Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0gzl | Andriy has cards which are the same from one side, and numbers $1, 3, 5, \ldots, 2009$ are written on the other side. Lesya has the same cards but with numbers $2, 4, 6, \ldots, 2010$. Lesya places cards with numbers down (so we can not see them at the beginning), after that Andriy places his cards on the Lesya's (so, ... | [
"In order to get at least $502$ points Andriy can place his cards as follows, starting from the left: $2009, 2007, 2005, \\ldots, 3, 1$. We prove by induction, that he gets $502$ points in this case. Lesya has $1005$ options to place her cards and we present all of them in the following table:\n\n| 2 | 4 | 6 ... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010) | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 502 | |
0jod | Problem:
Bassanio has three red coins, four yellow coins, and five blue coins. At any point, he may give Shylock any two coins of different colors in exchange for one coin of the other color; for example, he may give Shylock one red coin and one blue coin, and receive one yellow coin in return. Bassanio wishes to end ... | [
"Solution:\n\nLet $r, y, b$ denote the numbers of red, yellow, and blue coins respectively. Note that each of the three possible exchanges do not change the parities of $y - r$, $b - y$, or $b - r$, and eventually one of these differences becomes zero. Since $b - r$ is the only one of these differences that is orig... | United States | HMMT November 2015 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 7 yellow coins | |
096l | Problem:
Fie funcția continuă $f:(0 ;+\infty) \rightarrow \mathbb{R}$, pentru care $(x+1) f(x)+\frac{1}{x^{3}} f\left(\frac{1}{x}\right)=\frac{2 x+1}{x\left(x^{2}+1\right)}$.
Determinaţi primitivele $F:(0 ;+\infty) \rightarrow \mathbb{R}$ ale funcției $f$. | [
"Solution:\n\nNotăm prin $t=\\frac{1}{x}$ și obținem $t^{3} f(t)+\\frac{t+1}{t} f\\left(\\frac{1}{t}\\right)=\\frac{(2+t) t^{2}}{t^{2}+1}$.\nPentru a determina funcția $f$, rezolvăm sistemul $\\left\\{\\begin{array}{c}(x+1) f(x)+\\frac{1}{x^{3}} f\\left(\\frac{1}{x}\\right)=\\frac{2 x+1}{x\\left(x^{2}+1\\right)} \\... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | F(x) = ln(x / sqrt(x^2 + 1)) + C, for x > 0 | |
06vs | We say that a set $S$ of integers is rootiful if, for any positive integer $n$ and any $a_{0}, a_{1}, \ldots, a_{n} \in S$, all integer roots of the polynomial $a_{0}+a_{1} x+\cdots+a_{n} x^{n}$ are also in $S$. Find all rootiful sets of integers that contain all numbers of the form $2^{a}-2^{b}$ for positive integers ... | [
"First, note that $0=2^{1}-2^{1} \\in S$ and $2=2^{2}-2^{1} \\in S$. Now, $-1 \\in S$, since it is a root of $2 x+2$, and $1 \\in S$, since it is a root of $2 x^{2}-x-1$. Also, if $n \\in S$ then $-n$ is a root of $x+n$, so it suffices to prove that all positive integers must be in $S$.\n\nNow, we claim that any po... | IMO | IMO 2019 Shortlisted Problems | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algeb... | English | proof and answer | all integers | |
0duu | Problem:
Imamo $n \geq 3$ listov, ki jih oštevilčimo od 1 do $n$. Liste nato razdelimo na 2 kupa in ugotavljamo, ali sta v vsaj 1 kupu lista, označena s številoma, katerih vsota je popolni kvadrat. Dokaži, da
a) Če je $n \geq 15$, taka lista obstajata ne glede na to, kako liste razdelimo.
b) Če je $n \leq 14$, taka ... | [
"Solution:\n\nRešimo najprej prvi del naloge. Dovolj je dokazati trditev za $n=15$. Če namreč taka lista obstajata za $n=15$, obstajata tudi za $n \\geq 16$, saj lahko liste s številkami od 16 dalje izločimo in sklepamo kot prej. Če pri tem en kup v celoti izpraznimo, ostaneta na drugem kupu denimo števili 1 in 3.\... | Slovenia | 46. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0g1h | Problem:
Au camp SMO, il y a au moins quatre Romands. Deux Romands sont soit mutuellement amis, soit mutuellement ennemis. Dans chaque groupe de quatre Romands, au moins un des Romands est ami avec les trois autres. Existe-t-il toujours un Romand qui est ami avec tous les autres? | [
"Solution:\n\nOn numérote les personnes $P_{1}, P_{2}, \\ldots, P_{n}$. Considérons la première personne $P_{1}$. Si $P_{1}$ est ami avec tout le monde, l'exercice est terminé. Autrement on peut supposer sans perte de généralité que $P_{1}$ et $P_{2}$ ne sont pas amis. Alors, dans le groupe $P_{1}, P_{2}, P_{3}, P_... | Switzerland | SMO Finalrunde | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Yes | |
0dpq | Triangle $ABC$ is inscribed into circle $\omega$. On sides $AB$, $BC$, $CA$ there are points $K$, $L$, $M$, respectively, such that $CM \cdot CL = AM \cdot BL$. Ray $LK$ intersects line $AC$ at point $P$. The common chord of $\omega$ and the circumscribed circle of $KMP$ intersects segment $AM$ at point $S$. Prove that... | [
"\n\nSuppose that $AC \\le BC$. Let $D$ be a point on the side $AB$ such that $DM \\parallel BC$. Then\n$$\n\\frac{DB}{DA} = \\frac{CM}{AM} = \\frac{BL}{CL},\n$$\ni. e. $DL \\parallel AC$. On the tangent line to $\\omega$ at point $C$ let's choose a point $T$, such that $KT \\parallel BC$. ... | Silk Road Mathematics Competition | XIX Silk Road Mathematical Competition | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > A... | English | proof only | null | |
0co5 | Each cell of a square grid $100 \times 100$ contains a positive integer. A grid rectangle is called *good* if the sum of numbers in all its cells is divisible by $17$. By one move it is allowed to color all the cells of some good rectangle. It is not allowed to color a certain cell twice. Find the maximal $d$ such that... | [
"$9744 = 100^2 - 16^2$ cells.\n\n**Lemma.** Let a strip $1 \\times k$ be filled with positive integers. Then it is possible to color several non-overlapping good rectangles in it, covering at least $k - 16$ cells.\n\n**Proof.** Induction on $k$. For $k \\le 16$ nothing needs to be colored. Suppose $k \\ge 17$. Let ... | Russia | Final round | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Modular Arithmetic"
] | English; Russian | proof and answer | 9744 | |
05vf | Problem:
Aline et Théo jouent au jeu suivant. Tout d'abord, Théo choisit des nombres premiers $p_{1}, p_{2}, \ldots$ Puis Aline choisit des entiers naturels non nuls $n_{1}, n_{2}, \ldots$ deux à deux distincts. Théo choisit ensuite deux entiers $k \geqslant 1$ et $\ell \geqslant 1$. Il gagne la partie s'il existe une... | [
"Solution:\n\na) C'est Aline qui dispose d'une stratégie gagnante. Une telle stratégie consiste, par exemple, à choisir\n$$\nn_{i}=\\left(p_{1} \\times p_{2} \\times \\cdots \\times p_{i}\\right)^{i}\n$$\npour tout $i$. En effet, tout entier $p_{k}^{\\ell}$ divise chacun des entiers $\\left(p_{1} \\times p_{2} \\ti... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) Aline. b) Théo. | |
03kk | Problem:
Find all real solutions to the following system of equations. Carefully justify your answer.
$$
\left\{\begin{array}{l}
\frac{4 x^{2}}{1+4 x^{2}}=y \\
\frac{4 y^{2}}{1+4 y^{2}}=z \\
\frac{4 z^{2}}{1+4 z^{2}}=x
\end{array}\right.
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (0, 0, 0) and (1/2, 1/2, 1/2) | |
06bq | Find all positive integers which cannot be written as the sum of at least two consecutive positive integers. | [
"The answer is any power of $2$.\nSuppose $n = a + (a + 1) + \\dots + (a + k)$ for some positive integers $a$ and $k$. Then we have\n$$\n2n = (2a + k)(k + 1).\n$$\nLet $n = 2^r m$ where $m$ is odd. Note that $2a + k$ and $k + 1$ have different parities. Therefore, $2^{r+1}$ must divide one of $2a + k$ and $k + 1$, ... | Hong Kong | IMO HK TST | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | powers of 2 | |
0iu4 | Problem:
Five guys join five girls for a night of bridge. Bridge games are always played by a team of two guys against a team of two girls. The guys and girls want to make sure that every guy and girl play against each other an equal number of times. Given that at least one game is played, what is the least number of ... | [
"Solution:\n\nAnswer: 25\n\nSuppose that each guy plays each girl $t$ times. Since each guy plays against two girls in one game, the total number of games each guy plays is $\\frac{5 t}{2}$. Then the total number of games is $\\frac{25 t}{4}$, which is a multiple of $25$ and therefore at least $25$.\n\nTo check tha... | United States | 2nd Annual Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 25 | |
0dmq | Problem:
У оштроуглом $\triangle ABC$ тачка $M$ је средиште странице $BC$, а тачке $D, E$ и $F$ су подножја висина из темена $A, B$ и $C$, редом. Нека је $H$ ортоцентар $\triangle ABC$, $S$ средиште дужи $AH$, а $G$ пресек дужи $FE$ и $AH$. Ако је $N$ тачка пресека тежишне дужи $AM$ и описане кружнице $\triangle BCH$,... | [
"Solution:\n\nНека је $A'$ тачка таква да је $ABA'C$ паралелограм. Тада важи $\\varangle BA'C = \\varangle BAC = 180^{\\circ} - \\varangle BHC = 180^{\\circ} - \\varangle BNC$, па су тачке $A', B, C, H, N$ на истом кругу, тј. кругу над пречником $HA'$. Одавде је $\\varangle ANH = 90^{\\circ}$, дакле $N$ је на описа... | Serbia | Serbian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous ... | null | proof only | null | |
0l55 | Let $x_1, x_2, x_3, \ldots$ be a sequence of rational numbers defined by $x_1 = \frac{25}{11}$ and
$$
x_{k+1} = \frac{1}{3} \left( x_k + \frac{1}{x_k} - 1 \right)
$$
for all $k \ge 1$. Then $x_{2025}$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find the remainder when $m... | [] | United States | AIME II | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | final answer only | null | |
0epw | The dimensions of a box and a book are, respectively, $36\ \text{cm} \times 30\ \text{cm} \times 20\ \text{cm}$ and $20\ \text{cm} \times 15\ \text{cm} \times 6\ \text{cm}$, as shown in the diagram. What is the maximum number of books that can be packed into the box with no books sticking out?
![](attached_image_1.png... | [
"We must decide whether the books are to be packed flat, on their spines, or upright (or possibly a combination). In this case it is easy, because if they are packed upright, then their height is $20\\ \\text{cm}$, the same as the depth of the box. The width of each book is $15\\ \\text{cm}$, which goes twice into ... | South Africa | South African Mathematics Olympiad | [
"Geometry > Solid Geometry > 3D Shapes"
] | English | proof and answer | 12 | |
04sc | A natural number $n$ is given. Square with side of length $n$ is divided into $n^2$ unit squares. For the distance between two squares we consider the distance from centre to centre. Find the number of pairs of squares whose distance is $5$. | [] | Czech Republic | Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof and answer | For n ≥ 5: 6n^2 − 38n + 48; for n ≤ 4: 0 | |
009f | There are $2k$ chips placed in one row. A move consists in switching two adjacent chips. Several moves must be made until each chip occupies at some point the first and the last position. Which is the smallest amount of moves to achieve that? | [] | Argentina | XXIX Olimpíada Matemática Argentina National Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | k(2k-1) | |
09p9 | Let point $D$ lie on side $AB$ and point $E$ lie on side $BC$ of triangle $ABC$, such that quadrilateral $ADEC$ is cyclic. Let $X$, $Y$ be two distinct points on arc $DE$ of the circumcircle of triangle $BDE$, not containing $B$, such that $BX = BY$. Prove that line $XY$ is parallel to line $AC$.
(Khulan Tumenbayar) | [] | Mongolia | MMO2025 Round 3 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
02yj | Problem:
Todo lado de um triângulo equilátero é dividido em $n$ partes iguais. Linhas paralelas aos lados do triângulo são desenhadas através desses pontos dividindo o triângulo em $n^{2}$ triângulos menores. Dizemos que uma sequência de triângulos distintos é uma cadeia se dois triângulos sucessivos compartilham um l... | [
"Solution:\n\nA resposta é $n^{2}-n+1$ e uma cadeia com essa quantidade de triângulos é mostrada a seguir. Para mostrar que esse número é o máximo, pinte os triângulos de forma alternada com duas cores como também indica a figura abaixo. Existem $n$ triângulos brancos a mais que cinzas. Como os triângulos da cadeia... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n^2 - n + 1 | |
0161 | Let $AB$ be the diameter of the circle $\Gamma$ with centre $O$ and let $C$ and $D$ be points on $\Gamma$, on different sides on $AB$ and such that $AD$ and $CB$ intersect at $R$. The circumscribed circles of the triangles $AOC$ and $BOD$ meet also at $Q$. $CD$ and $AB$ meet at $P$. Show that $Q, P$ and $R$ are colline... | [
"\nThere is no loss in generality if we assume that $R$ and $D$ are on the same side of $AB$. We first show that $A, Q, B$ and $R$ are concyclic. To this end, we note that $\\angle AQB = \\angle AQO + \\angle OQR = \\angle ACO + 180^\\circ - \\angle ODB = \\angle CAO + 90^\\circ + \\angle B... | Baltic Way | Baltic Way SHL | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08h7 | Problem:
Consider triangle $A B C$ such that $A B \leq A C$. Point $D$ on the arc $B C$ of the circumcircle of $A B C$ not containing point $A$ and point $E$ on side $B C$ are such that
$$
\Varangle B A D=\Varangle C A E<\frac{1}{2} \Varangle B A C .
$$
Let $S$ be the midpoint of segment $A D$. If $\Varangle A D E=\Va... | [
"Solution:\n\nLet the tangent to the circumcircle of $\\triangle A B C$ at point $A$ intersect line $B C$ at $T$. Since $A B \\leq A C$ we get that $B$ lies between $T$ and $C$. Since $\\Varangle B A T=\\Varangle A C B$ and $\\Varangle A B T=\\Varangle 180^{\\circ}-\\Varangle A B C$ we get $\\Varangle E T A=\\Varan... | JBMO | null | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | null | proof only | null | |
0e3m | Problem:
Kovinsko kocko s površino $72~\mathrm{cm}^2$ pretopimo v pravilno štiristrano piramido enake prostornine, katere dolžina osnovnega roba je enak tretjini dolžine telesne diagonale kocke. Kolikšna je višina piramide? | [
"Solution:\n\nUporabimo obrazec za površino kocke $P = 6a^2$ in iz nje izračunamo dolžino roba $a = 2\\sqrt{3}$ $\\mathrm{cm}$. Nato izračunamo dolžino telesne diagonale kocke $D = a\\sqrt{3} = 6~\\mathrm{cm}$ ter prostornino kocke $V = a^3 = (2\\sqrt{3})^3 = 24\\sqrt{3}~\\mathrm{cm}^3$. Upoštevamo, da sta prostorn... | Slovenia | 10. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | 18√3 cm | |
06t3 | Let $ABCD$ be a convex quadrilateral with $\angle B = \angle D = 90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. The points $S$ and $T$ are chosen on the sides $AB$ and $AD$, respectively, in such a way that $H$ lies inside triangle $SCT$ and
$$
\angle SHC - \angle BSC = 90^{\circ}, \quad \ang... | [
"Let the line passing through $C$ and perpendicular to the line $SC$ intersect the line $AB$ at $Q$ (see Figure 1). Then\n$$\n\\angle SQC = 90^{\\circ} - \\angle BSC = 180^{\\circ} - \\angle SHC,\n$$\nwhich implies that the points $C$, $H$, $S$, and $Q$ lie on a common circle. Moreover, since $SQ$ is a diameter of ... | IMO | 55th International Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
... | null | proof only | null | |
01gx | Let $\omega$ be a semicircle with diameter $XY$. Let $M$ be the midpoint of $XY$. Let $A$ be an arbitrary point on $\omega$ such that $AX < AY$. Let $B$ and $C$ be points lying on the segments $XM$ and $YM$, respectively, such that $BM = CM$. The line through $C$ parallel to $AB$ intersects $\omega$ at $P$. The line th... | [
"Let $D$ be the point symmetric to $A$ with respect to $M$. Then $ABDC$ is a parallelogram as $M$ bisects both $BC$ and $AD$. It follows that $AB \\parallel CD$ and $AC \\parallel BD$. This means that $P$ lies on $CD$ and $Q$ lies on $BD$.\nWe shall prove that the tangent line at $A$ to $\\omega$ intersects $PQ$ at... | Baltic Way | Baltic Way 2020 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof only | null | |
0g3j | Problem:
We call a positive integer silly if the sum of its positive divisors is a square. Prove that there are infinitely many silly numbers. | [
"Solution:\n\nLet $\\sigma(n)$ denote the sum of all positive divisors of the integer $n$, and we will order the prime numbers $2=p_{1}<p_{2}<p_{3}<\\ldots$ An important observation is that $\\sigma(n)$ is multiplicative, in the sense that if $n_{1}, n_{2}$ are coprime, then $\\sigma\\left(n_{1} n_{2}\\right)=\\sig... | Switzerland | IMO Selection | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
00cz | Sea $k > 1$ un entero. Determinar el menor entero positivo $n$ tal que algunas casillas de un tablero de $n \times n$ se pueden pintar de negro de modo que en cada fila y en cada columna haya exactamente $k$ casillas negras, y además, las casillas negras no compartan ni un lado ni un vértice con otra casilla negra.
**... | [
"Observamos que todo subtablero de $2 \\times 2$ puede tener como máximo una casilla negra. Consideremos dos filas consecutivas del tablero. Entre las dos deben tener, en total, exactamente $2k$ casillas negras. Dividiendo las casillas de dos filas en cuadrados de $2 \\times 2$, comenzando desde la izquierda, vemos... | Argentina | Nacional OMA | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Spanish | proof and answer | 4k | |
00i1 | Let $n$ be a positive integer and $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers. Prove that
$$
\sum_{i=1}^{n} \frac{1}{2^{i}}\left(\frac{2}{1+a_{i}}\right)^{2^{i}} \geq \frac{2}{1+a_{1} a_{2} \ldots a_{n}}-\frac{1}{2^{n}} .
$$ | [
"We first prove the following lemma:\nLemma 1. For $k$ positive integer and $x, y>0$,\n$$\n\\left(\\frac{2}{1+x}\\right)^{2^{k}}+\\left(\\frac{2}{1+y}\\right)^{2^{k}} \\geq 2\\left(\\frac{2}{1+x y}\\right)^{2^{k-1}} .\n$$\nThe proof goes by induction. For $k=1$, we have\n$$\n\\left(\\frac{2}{1+x}\\right)^{2}+\\left... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
09lu | Determine all positive integer values of $n$, which are perfect squares, such that when dividing $2024$ by $n$ the remainder is $188 - \frac{5}{4}n$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 144 | |
06s6 | Let $x$ and $y$ be positive integers. If $x^{2^{n}}-1$ is divisible by $2^{n} y+1$ for every positive integer $n$, prove that $x=1$. | [
"First we prove the following fact: For every positive integer $y$ there exist infinitely many primes $p \\equiv 3 \\pmod{4}$ such that $p$ divides some number of the form $2^{n} y+1$.\n\nClearly it is enough to consider the case $y$ odd. Let\n$$\n2y+1 = p_{1}^{e_{1}} \\cdots p_{r}^{e_{r}}\n$$\nbe the prime factori... | IMO | 53rd International Mathematical Olympiad Shortlisted Problems with Solutions | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof only | null | |
00dj | Let $ABC$ be a triangle with $AB < AC$. On the angle bisector of $\angle BAC$, two points $X$ and $Y$ are marked such that $X$ is between $A$ and $Y$, and $BX$ is parallel to $CY$. Let $Z$ be the reflection of $X$ with respect to $BC$. Let $P$ be the point of intersection of lines $YZ$ and $BC$. If lines $BY$ and $CX$ ... | [
"Let $P_1$ be the point where the circumcircle of $ACY$ intersects $BC$ for the second time. Let $D$ be the foot of the bisector of $\\angle BAC$. Clearly, $D$ is interior to the circumcircle of $ACY$ and by power of a point, we have that\n$$DP_1 \\cdot DC = DA \\cdot DY.$$ \nSince $BX$ and $CY$ are parallel, $\\fr... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle center... | English | proof only | null | |
03b5 | Let $n \ge 3$ be a positive integer and let
$$
S = \{k \in \mathbb{N} : (k, n) = (k + 1, n), 1 \le k \le n - 1\}.
$$
Find the remainder of $\prod_{k \in S} k$ divided by $n$. | [
"It is easy to see that $(k, n) = (k + 1, n) = 1$. Then for any $k \\in S$ there exists a unique $k_1 \\in \\{1, 2, \\dots, n\\}$ such that $(k_1, n) = 1$ and $k k_1 \\equiv 1 \\pmod{n}$. Since $k_1 + 1 \\equiv k_1(k + 1) \\pmod{n}$, then $(k_1 + 1, n) = 1$, i.e. $k_1 \\in S$. Moreover, $(k-1)(k_1+1) \\equiv k-k_1 ... | Bulgaria | Bulgaria | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 1 | |
0lcd | Let be given a fixed circle $(O, R)$ and a fixed point $A$ on this circle. Suppose that $B, C$ are two arbitrary points on $(O)$ such that $\angle BAC = \alpha$ with $\alpha$ is a constant.
On rays $BA, CA$ respectively the points $E, F$ be taken in such that $BE = CF = BC$.
a) Let $r$ be the radius of $(AEF)$. Prove t... | [] | Vietnam | Vietnamese Mathematical Competitions | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous... | null | proof only | null | |
00vz | Let $a$, $b$ and $k$ be positive integers. Suppose that for any integer $n \ge 2025$ there exists a positive integer $x_n > n$ such that $x_n \mid n^2 + a$ and $x_n + k \mid n^2 + b$. Prove that $k = b - a$. | [
"**Solution 1.** Let $n \\ge 2025$ and $y_n = \\frac{n^2 + a}{x_n}$. Also let $z_n = \\frac{n^2 + b}{x_n + k}$. Consider $c_n = y_n - z_n$. We have\n$$\n|c_n| = \\left| \\frac{n^2}{x_n} - \\frac{n^2}{x_n + k} + \\frac{a}{x_n} - \\frac{b}{x_n + k} \\right| = \\left| k \\cdot \\frac{n^2}{x_n(x_n + k)} + \\frac{a}{x_n... | Balkan Mathematical Olympiad | 42nd Balkan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalitie... | English | proof only | null | |
05nk | Problem:
Soit $ABC$ un triangle, $D, E, F$ les points de tangence du cercle inscrit sur les côtés $(BC)$, $(AC)$ et $(AB)$, $M$ le milieu de $[BC]$, $I$ le centre du cercle inscrit de $ABC$. $G$ et $H$ sont définis comme les symétriques de $E$ et $F$ par rapport à $I$. On note $Q$ l'intersection entre les droites $(BC... | [
"Solution:\n\n\n\nSoit $\\omega$ le cercle inscrit de $ABC$.\nOn notera pour tout point $X$ du plan $p(X)$ la polaire de $X$ par rapport à $\\omega$. On sait que pour tout point $X$ du plan, $p(X)$ est une droite perpendiculaire à $(IX)$. Nous allons donc montrer que $p(Q)$ et $(IM)$ sont p... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry >... | null | proof only | null | |
01kn | From the digits $1, 2, 3, 4, 5, 6, 7, 8$, find the smallest possible value of $N$ such that, for any two distinct digits from this set, there exists a number among the $N$ numbers (each number is a four-digit number formed from these digits) which contains both of them. | [
"Answer: $N = 6$.\n\nLet some digit, say $1$, appear exactly in $k$ numbers from $N$ given numbers. Hence, $1$ forms at most $3$ distinct pairs with the remaining $3$ digits of any of these $k$ numbers. Since the total number of all distinct pairs formed by $1$ and the other $7$ numbers ($2,3,\\ldots,8$) is equal t... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 6 | |
08jf | Problem:
Let $m$ and $n$ be the arbitrary digits of the decimal system and $a, b, c$ be the positive distinct integers of the form $2^{m} \cdot 5^{n}$. Find the number of the equations $a x^{2}-2 b x+c=0$, if it is known that each equation has a single real solution. | [] | JBMO | The first selection test for JBMO 2003 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 2400 | |
08hp | Problem:
The real numbers $\alpha, \beta, \gamma$ satisfy the relations $\sin \alpha+\sin \beta+\sin \gamma=0$ and $\cos \alpha+\cos \beta+\cos \gamma=0$. Find all positive integers $n \geq 0$ for which $\sin (n \alpha+\pi / 4)+\sin (n \beta+\pi / 4)+\sin (n \gamma+\pi / 4)=0$. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | All n ≥ 0 with n not divisible by 3 (i.e., n ≡ 1 or 2 mod 3). | |
0jmj | Problem:
Find all real numbers $k$ such that $r^{4}+k r^{3}+r^{2}+4 k r+16=0$ is true for exactly one real number $r$. | [
"Solution:\nAnswer: $\\pm \\frac{9}{4}$ (OR $\\frac{9}{4},-\\frac{9}{4}$ OR $\\left.-\\frac{9}{4}, \\frac{9}{4}\\right)$ OR $\\pm 2 \\frac{1}{4}$ OR $\\pm 2.25$\n\nAny real quartic has an even number of real roots with multiplicity, so there exists real $r$ such that $x^{4}+k x^{3}+x^{2}+4 k x+16$ either takes the ... | United States | HMMT 2014 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 9/4 and -9/4 | |
0ec7 | Let $P$ be the midpoint of the side $AB$ of the triangle $ABC$. The mirror image of the ray $PC$ by reflection across the line $AB$ intersects the circumcircle of the triangle $ABC$ in the point $D$. Let $E$ be the other intersection of the line $CP$ with the circumcircle of the triangle $ABC$. Prove that $|AE| = |BD|$... | [
"Let $\\mathcal{K}$ be the circumcircle of the triangle $ABC$. Since the ray $PD$ is the mirror image of the ray $PC$ by reflection across the line $AB$ we have $\\angle DPB = \\angle BPC = \\angle APE$. Let's look at the reflection across the bisector of the side $AB$. The point $A$ is reflected to the point $B$. ... | Slovenia | National Math Olympiad 2015 – Final Round | [
"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 | |
0bye | Consider the sequence of rational numbers defined by $x_1 = 4/3$, and $x_{n+1} = x_n^2/(x_n^2 - x_n + 1)$, $n \ge 1$. Show that the numerator of the lowest term expression of each sum $\sum_{k=1}^n x_k$ is a perfect square. | [
"It is easily seen that the $x_n$ are all rational numbers greater than $1$. Rewrite the recurrence formula in the form $x_n = 1/(x_{n+1} - 1) - 1/(x_n - 1)$, $n \\ge 1$, to get\n$$\n\\sum_{k=1}^{n} x_k = \\frac{1}{x_{n+1} - 1} - \\frac{1}{x_1 - 1} = \\frac{x_n^2 - x_n + 1}{x_n - 1} - 3 = \\frac{(x_n - 2)^2}{x_n - ... | Romania | THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof only | null | |
0bfq | Prove that for every $x \in K \setminus L$ there exists $a, b \in L$ so that $x^{2013} + a x + b = 0$.
**8.** The function $f : \mathbb{R} \to [0, \infty)$ is continuous and $\lim_{x \to \infty} f(x) = \infty$. Prove that, for every $n \in \mathbb{N}^*$ and every $y \in \mathbb{R}$, there exists $t = t(n, y)$ so that ... | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Calculus > Integral Calculus > Applications",
"Precalculus > Limits"
] | null | proof only | null |
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