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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
027q | Problem:
Pedrinho faz uma lista de todos os números de 5 algarismos distintos que se formam com os dígitos $1,2,3,4,5$. Nesta lista os números estão ordenados de forma crescente.
a) Qual o número que ocupa a posição 10 da lista?
b) Qual o número que ocupa a posição 85 da lista? | [
"Solution:\n\na) Começamos escrevendo os primeiros números da lista:\n12345, 12354, 12435, 12453, 12534, 12543, 13245, 13254, 13425, 13452.\nLogo, o décimo número é 13452.\n\nb) Para encontrar o número que ocupa a posição 85, percebemos que sempre que um número de 5 dígitos começa com o dígito 1, este número é meno... | Brazil | null | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | final answer only | a) 13452; b) 43125 | |
069i | Prove that there exist infinitely many integers $x, y, z$ for which the sum of the digits in the decimal representation of $4x^4 + y^4 - z^2 + 4xyz$ is at most 2. | [
"This is an easy problem with many solutions. We rewrite\n$$\n\\begin{aligned}\n4x^4 + y^4 - z^2 + 4xyz &= (4x^4 + y^4 + 4x^2y^2) - (4x^2y^2 + z^2 - 4xyz) \\\\\n&= (2x^2 + y^2)^2 - (2xy - z)^2 \\\\\n&= (2x^2 + y^2 - 2xy + z)(2x^2 + y^2 + 2xy - z)\n\\end{aligned}\n$$\nThe two factors $A = 2x^2 + y^2 - 2xy + z$ and $... | Greece | 21st Mediterranean Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
088b | Problem:
Sia $ABCD$ un quadrilatero convesso. Sia $P$ l'intersezione delle bisettrici esterne di $D\widehat{A}C$ e $D\widehat{B}C$.
Dimostrare che $A\widehat{P}D = B\widehat{P}C$ se e solo se $AD + AC = BC + BD$.
[Nota : Si ricorda che la bisettrice esterna ad un angolo è la retta passante per il vertice dell'angolo ... | [
"Solution:\n\nChiamiamo $r$ ed $s$ rispettivamente le bisettrici esterne di $D\\widehat{A}C$ e $D\\widehat{B}C$. Si costruiscano i punti $C'$ e $D'$ rispettivamente come simmetrico di $C$ rispetto a $s$ e come simmetrico di $D$ rispetto a $r$. Poiché $r$ è bisettrice esterna si ha che $C', B$ e $D$ sono allineati e... | Italy | Cesenatico | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
01z8 | The $2022 \times 2022$ board was cut into tetraminoes of two types: L-tetramino $L$ and Z-tetramino $Z$. Each tetramino consists of four unit squares, tetriminoes can be rotated and flipped.
Find the smallest number of $Z$-tetriminoes that could be obtained. | [
"Answer: $1$.\nLet's label the columns of the table from bottom to top with the numbers from $1$ to $2022$ and color all the cells of the rows with odd numbers in yellow, and with even numbers — in blue. Each $L$-tetromino consists of three squares of one color and one square of another. Let's assume that not a sin... | Belarus | Belarus2022 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 1 | |
0hby | Equilateral triangle $ABC$ is inscribed in circle $w$. Points $F$ and $E$ are chosen on sides $AB$ and $AC$, respectively, so that $\angle ABE + \angle ACF = 60^\circ$. Circumscribed circle of $\triangle AFE$ intersects circle $w$ at point $D$. Rays $DE$ and $DF$ intersect line $BC$ at points $X$ and $Y$, respectively.... | [
"Let us denote by $P$ intersection point of $CF$ and $BE$. Then, $\\angle CBE = 60^\\circ - \\angle ABE = \\angle FCA$. Analogously, $\\angle FCB = \\angle ABE$ (fig. 27). Then, $\\angle FKE = \\angle CKB = 180^\\circ - \\angle CBE - \\angle FCB = 120^\\circ$.\n\nThus, quadrilateral $AFPE$ is inscribed in circle $\... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0i4l | Problem:
Find the number of pentominoes (5-square polyominoes) that span a $3$-by-$3$ rectangle, where polyominoes that are flips or rotations of each other are considered the same polyomino. | [
"Solution:\nBy enumeration, the answer is $6$."
] | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 6 | |
0c07 | Problem:
Fie $A$ un inel finit şi $a, b \in A$ cu proprietatea $a(a b-1) b=0$. Arătaţi că $b(a b-1)=0$. | [
"Solution:\n\nEgalitatea din ipoteză este echivalentă cu $a b^{2}=b$, iar cea de demonstrat cu $b a b=b$.\n\nDacă elementul $b$ este idempotent (i.e., $b^{2}=b$), atunci $b a b=b a b^{2}=b \\cdot b=b^{2}=b$.\n\nDacă $b^{m}=b$, cu $m>2$, atunci $b a b=b a b^{m}=b a b^{2} b^{m-2}=b \\cdot b \\cdot b^{m-2}=b^{m}=b$.\n... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Algebra > Abstract Algebra > Ring Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0dmu | Problem:
Нека су $a_{0}$ и $a_{n}$ различити делиоци природног броја $m>1$, а низ природних бројева $a_{0}, a_{1}, a_{2}, \ldots, a_{n}$ такав да задовољава
$$
a_{i+1}=\left|a_{i} \pm a_{i-1}\right| \quad \text{ за } 0<i<n
$$
Ако је НЗД $\left(a_{0}, \ldots, a_{n}\right)=1$, доказати да у низу постоји члан који је мањ... | [
"Solution:\n\nПосматрајмо два најмања (различита) члана низа, $p$ и $q$. Ако је $\\min \\{p, q\\}= 1$, тврђење тривијално важи; зато надаље претпостављамо да је $p, q>1$.\n\nЛема 1. Постоје индекси $k$ и $l$ за које је $a_{k}=p, a_{l}=q$ и $|k-l| \\leq 2$.\n\nДоказ. Нека је $a_{k}=p$ и $a_{l}=q$ ($k<l$). Претпостав... | Serbia | Serbian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0c8w | Determine the continuous functions $f : \mathbb{R} \to \mathbb{R}$ having the property that, for all $x, y \in \mathbb{R}$, there exist $t \in (0, 1)$ such that
$$
f((1-t)x + ty) = (1-t)f(x) + tf(y).
$$ | [
"We shall prove that the answer is $f : \\mathbb{R} \\to \\mathbb{R}$, $f(x) = mx + n$, with $m, n \\in \\mathbb{R}$. It is trivial to see that this family of functions satisfy the property.\n\nFor the converse, let $f$ with the given property. Consider $a, b \\in \\mathbb{R}$, $a < b$, and define\n$$\nm = \\frac{f... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | All affine functions f(x) = m x + n for real m, n. | |
09oi | Let $\mathbb{N} = \{1, 2, 3, \dots\}$ be the set of natural numbers.
Consider those functions $f: \mathbb{N} \to \mathbb{N}$ which satisfy the conditions $f(1) = 1$, $f(f(n)) = 3f(n) - 2n$ and $\frac{n-1}{f(n)-n}$ is an integer for $n \ge 2$. Find all possible values of $f(2024)$.
(Nursoltan Khavalbolot) | [] | Mongolia | MMO2025 Round 2 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | 4047 | |
09yy | Let $ABC$ be an acute triangle, and let $D$ be the foot of the altitude from $A$. The circle with centre $A$ passing through $D$ intersects the circumcircle of triangle $ABC$ in $X$ and $Y$, in such a way that the order of the points on this circumcircle is: $A$, $X$, $B$, $C$, $Y$. Show that $\angle BXD = \angle CYD$. | [
"\n\nAs the radius $AD$ is perpendicular to $BC$, the line $BC$ is tangent to the circumcircle of $\\triangle DXY$. By the inscribed angle theorem (tangent case), we have $\\angle XDB = \\angle XYD$. Moreover, the quadrilateral $BCYX$ is cyclic, so $\\angle CBX + \\angle XYC = 180^\\circ$. ... | Netherlands | BxMO Team Selection Test | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
01s3 | Alice has $n^2$ sweets. These sweets are distributed among $n$ boxes ($n > 2$). Alice chooses some two of these boxes and if the total number of the sweets in these two boxes is even, then she redistributes the sweets so that the numbers of the sweets in these two boxes will be equal. Otherwise she chooses another pair... | [
"Answer: $n = 2^m$, where $m \\in \\mathbb{N}$, $m > 1$.\n\nThe goal of Alice is to obtain exactly $n$ sweets in any box. The number $a-n$ is said to be the distance if $a$ is a number of the sweets in the box. Note that the sum of the distances is equal to zero at any moment. Now we see that Alice's goal is to mak... | Belarus | FINAL ROUND | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | n = 2^m with m > 1 | |
0jqf | Problem:
For positive integers $x$, let $g(x)$ be the number of blocks of consecutive 1's in the binary expansion of $x$. For example, $g(19)=2$ because $19=10011_{2}$ has a block of one 1 at the beginning and a block of two 1's at the end, and $g(7)=1$ because $7=111_{2}$ only has a single block of three 1's. Compute... | [
"Solution:\n\nAnswer: 577\n\nWe prove that $g(1)+g(2)+\\cdots+g\\left(2^{n}\\right)=1+2^{n-2}(n+1)$ for all $n \\geq 1$, giving an answer of $1+2^{6} \\cdot 9=577$.\n\nFirst note that $g\\left(2^{n}\\right)=1$, and that we can view $0,1, \\ldots, 2^{n}-1$ as $n$-digit binary sequences by appending leading zeros as ... | United States | HMMT February 2015 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 577 | |
0i3z | Problem:
Find the number of positive integer solutions to $n^{x} + n^{y} = n^{z}$ with $n^{z} < 2001$. | [
"Solution:\nIf $n = 1$, the relation cannot hold, so assume otherwise. If $x > y$, the left hand side factors as $n^{y} (n^{x-y} + 1)$ so $n^{x-y} + 1$ is a power of $n$. But it leaves a remainder of $1$ when divided by $n$ and is greater than $1$, a contradiction. We reach a similar contradiction if $y > x$. So $y... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 9 | |
0cv6 | Determine whether for every three distinct positive integers $a$, $b$, and $c$ there exists a quadratic trinomial with integer coefficients and positive coefficient of $x^2$ such that at some integer points this trinomial takes the values $a^3$, $b^3$, and $c^3$.
Верно ли, что для любых трёх различных положительных це... | [
"Yes.\nThe polynomial $f(x) = (a+b+c)x^2 - (ab+bc+ca)x + abc = x^3 - (x-a)(x-b)(x-c)$ fits.",
"Да, верно.\nПокажем, что трёхчлен\n$$\nf(x) = (a+b+c)x^2 - (ab+bc+ca)x + abc = x^3 - (x-a)(x-b)(x-c)\n$$\nподходит. Ясно, что его коэффициенты целые и старший коэффициент $a+b+c$ положителен. Наконец, легко видеть, что ... | Russia | XLIII Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English; Russian | proof and answer | Yes | |
0dkz | Let $k$ be a positive integer. Determine all $k$-tuples $(n_1, n_2, ..., n_k)$ of positive integers such that
$$(n_1! - 1)(n_2! - 1) \cdots (n_k! - 1) - 16$$
is a perfect square of a positive integer. | [] | Saudi Arabia | Saudi Booklet | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Algebraic Number Theory > Unique factorization",
"Number Theory > Algebraic Number Theory > Quadratic fields"
] | null | proof and answer | All k-tuples consisting of exactly two entries equal to 3 and all remaining entries equal to 2. | |
0civ | Find the real numbers $a$, $b$, $c$, with sum $s$, such that
$$
(\sqrt{s} - \sqrt{a})(\sqrt{s} - \sqrt{b}) + (\sqrt{s} - \sqrt{b})(\sqrt{s} - \sqrt{c}) + (\sqrt{s} - \sqrt{c})(\sqrt{s} - \sqrt{a}) = \frac{s}{2}.
$$ | [] | Romania | 75th NMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | a = b = c = 0 (and s = 0) | |
06i8 | Let $\triangle ABC$ be a scalene triangle, and let $D$ and $E$ be points on sides $AB$ and $AC$ respectively such that the circumcircles of triangles $ACD$ and $ABE$ are tangent to $BC$. Let $F$ be the intersection point of $BC$ and $DE$. Prove that $AF$ is perpendicular to the Euler line of $ABC$.
(Recall that the Eul... | [
"Let $O$ and $H$ be the circumcentre and orthocentre of $\\triangle ABC$ respectively. Let $X$ and $Y$ be the points on the extension of $AB$ and $AC$ such that $HA = HX = HY$. The the centres of $(ABC)$ and $(AXY)$ are $O$ and $H$ respectively. It remains to show that $F$ lies on the radical axis of these circles.... | Hong Kong | CHKMO | [
"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 > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity",
... | null | proof only | null | |
08ou | Problem:
We have a $5 \times 5$ chessboard and a supply of $\{L\}$-shaped triominoes, i.e. $2 \times 2$ squares with one corner missing. Two players $A$ and $B$ play the following game: A positive integer $k \leq 25$ is chosen. Starting with $A$, the players take alternating turns marking squares of the chessboard unt... | [
"Solution:\n\nWe will show that player $A$ wins if $k=1,2$ or $3$, but player $B$ wins if $k=4$. Thus the smallest $k$ for which $B$ has a winning strategy exists and is equal to $4$.\n\nIf $k=1$, player $A$ marks the upper left corner of the square and then fills it as follows.\n\n\n\nIf $... | JBMO | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 4 | |
081g | Problem:
Sia $ABC$ un triangolo e sia $\gamma$ la circonferenza inscritta in $ABC$. La circonferenza $\gamma$ è tangente al lato $AB$ nel punto $T$. Sia $D$ il punto di $\gamma$ diametralmente opposto a $T$, e sia $S$ il punto di intersezione della retta passante per $C$ e $D$ con il lato $AB$.
Dimostrare che $AT = S... | [
"Solution:\n\nCon riferimento alla figura a fianco, tracciamo la retta $r$ passante per $D$ e parallela al lato $AB$. Siano $L$ ed $M$, rispettivamente, le intersezioni di $r$ con i lati $AC$ e $BC$. Denotiamo inoltre con $H$ e $K$, rispettivamente, i punti di tangenza di $\\gamma$ con i lati $AC$ e $BC$. Poiché su... | Italy | Gara Nazionale di Matematica | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0k4r | Problem:
A triple of integers $(a, b, c)$ satisfies $a + b c = 2017$ and $b + c a = 8$. Find all possible values of $c$. | [
"Solution:\nAdd and subtract the two equations to find\n$$\n\\begin{aligned}\n& (b + a)(c + 1) = 8 + 2017 \\\\\n& (b - a)(c - 1) = 2017 - 8\n\\end{aligned}\n$$\nWe see that $c$ is even and then that every integer $c$ with $c + 1 \\mid 2025$, $c - 1 \\mid 2009$ works. We factor and solve.\nThe full solutions are $(2... | United States | HMMT February 2018 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | {-6, 0, 2, 8} | |
08gn | Problem:
Determinare il numero di coppie $(m, n)$ di numeri interi di modulo minore di $10000$ che risolvono l'equazione $n^{2}+2024 n+2024 m=2024$.
(A) 4
(B) 9
(C) 10
(D) 18
(E) 19 | [
"Solution:\n\nLa risposta è (B). Innanzitutto notiamo che $2024=2^{3} \\cdot 11 \\cdot 23$ deve dividere $n^{2}$, quindi possiamo scrivere $n=\\left(2^{2} \\cdot 11 \\cdot 23\\right) \\cdot k=1012 k$, con $k$ intero di modulo minore o uguale a $9$, così che $|n|=|1012 k|<10000$. Sostituendo nell'equazione abbiamo\n... | Italy | Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | B | |
09mc | Find all non-negative integer solutions $(k, m, n)$ of the equation
$$
211 \cdot 9^k + 125^m = 2024^n.
$$ | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (1, 1, 1) | |
0dop | Determine all prime numbers $p$ for which there are integers $m$ and $n$ such that $p = m^2 + n^2$ and $p \mid m^3 + n^3 - 4$. | [
"Considering all integers $|m|, |n| \\le 3$ we can get the solutions $2 = 1^2 + 1^2$, $5 = 1^2 + 2^2$ and $13 = (-3)^2 + (-2)^2$. Now, let's prove that there is no other prime numbers satisfying the statement. We have\n$$\np = m^2 + n^2 \\Rightarrow p = (m + n)^2 - 2mn \\Rightarrow mn = \\frac{(m + n)^2 - p}{2},\n$... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | 2, 5, 13 | |
0l82 | Find the number of ordered pairs $(x, y)$, where both $x$ and $y$ are integers between $-100$ and $100$, inclusive, such that $12x^2 - xy - 6y^2 = 0$. | [
"The given equation can be written as $(3x + 2y)(4x - 3y) = 0$. Thus the graph of this equation consists of two lines intersecting at the origin. Lattice points (points with integer coordinates) on this graph that make the first factor equal to $0$ are of the form $(2k, -3k)$ for some integer $k$. In order for the ... | United States | 2025 AIME I | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 117 | |
0dgj | Let $1 \le r \le n$. We consider all $r$-element subsets of $(1, 2, \dots, n)$. Each of them has a minimum. Prove that the average of these minima is $\frac{n+1}{r+1}$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
0dfq | Prove that for any 2 positive integers $m$ and $n$ with $m > n$ holds the following inequality
$$
\operatorname{lcm}(m, n) + \operatorname{lcm}(m + 1, n + 1) > \frac{2mn}{\sqrt{m-n}}.
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0bbn | Find all positive integers $m$ such that
$$
\sqrt{m} = \sqrt{m+2011}.
$$ | [
"Rewrite the equation $\\sqrt{m} - [\\sqrt{m}] = \\sqrt{m+2011} - [\\sqrt{m+2011}]$ to get $\\sqrt{m+2011} - \\sqrt{m} = [\\sqrt{m+2011}] - [\\sqrt{m}] = p \\in \\mathbb{N}$. Squaring $\\sqrt{m+2011} = p + \\sqrt{m}$ yields $2011 = p^2 + 2p\\sqrt{m} \\in \\mathbb{N}$, hence $m = k^2, k \\in \\mathbb{N}^*$. The rela... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 1010025 | |
0jw1 | Problem:
The diagram below is an example of a rectangle tiled by squares:
Each square has been labeled with its side length. The squares fill the rectangle without overlapping.
In a similar way, a rectangle can be tiled by nine squares whose side lengths are $2$, $5$, $7$, $9$, $16$, $25$,... | [
"Solution:\n\nTo tile a rectangle, the areas of the squares must add to match the area of the rectangle. The total area of the 9 squares is:\n$$\n2^{2}+5^{2}+7^{2}+9^{2}+16^{2}+25^{2}+28^{2}+33^{2}+36^{2}=4209.\n$$\nWhen we factor $4209$ we obtain $4209=3 \\times 23 \\times 61$.\n\nTo fit the largest square, the re... | United States | BAMO-8 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | null | |
0crl | Петя и Вася играют в игру на клетчатой доске $n \times n$ (где $n > 1$). Изначально вся доска белая, за исключением угловой клетки — она чёрная, и в ней стоит ладья. Игроки ходят по очереди. Каждым ходом игрок передвигает ладью по горизонтали или вертикали, при этом все клетки, через которые ладья перемещается (включая... | [
"Одна из выигрышных стратегий для Пети состоит в том, чтобы каждым своим ходом делать самый длинный из возможных вертикальных ходов (например, первым ходом он пойдёт по вертикали в другой угол доски). Покажем, что, действуя согласно ей, он выиграет.\n\nНазовём белую клетку *достижимой*, если из текущего положения л... | Russia | XL Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Petya | |
0b9s | Let $n$ be a positive integer and $a_1 \le a_2 \le \cdots \le a_n$ be positive real numbers. Show that
$$
\left( \sum_{k=1}^{n} a_k^2 \right) \left( \sum_{k=1}^{n} k a_k \right) \le \left( \sum_{k=1}^{n} a_k \right) \left( \sum_{k=1}^{n} k a_k^2 \right).
$$ | [
"The hypothesis leads to $(i-j)(a_i - a_j)a_i a_j \\ge 0$ for every $i$ and $j$, hence\n$$\n\\begin{align*}\n0 &\\le \\sum_{i,j=1}^{n} (i-j)(a_i - a_j)a_i a_j \\\\\n&= \\sum_{i,j=1}^{n} (i a_i^2 a_j - i a_i a_j^2 - j a_i^2 a_j + j a_i a_j^2) \\\\\n&= \\left(\\sum_{i=1}^{n} i a_i^2\\right) \\sum_{j=1}^{n} a_j - \\le... | Romania | 2011 CLOCK-TOWER SCHOOL SENIORS COMPETITION | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0b6t | Let $VABC$ be a regular tetrahedron of side length $a$ and points $M$, $N$ on the edges $VB$, $VC$, such that $VM = \frac{a}{3}$ and $VN = \frac{a}{6}$. If $Q$ is the projection of $V$ onto the plane $(AMN)$ and $O$ is the projection of $V$ onto the plane $(ABC)$, compute the lengths of the segments $VO$, $VQ$, and a t... | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | VO = a*sqrt(2/3), VQ = a*sqrt(42)/42, and cos(angle between VQ and VO) = 5/(3*sqrt(7)) | |
0c6j | Find the maximum value of the expression
$$
E(a, b) = \frac{a + b}{(4a^2 + 3)(4b^2 + 3)}
$$
when $a, b \in \mathbb{R}$. | [
"We will show that the maximum value is $\\frac{1}{16}$, obtained when $a = b = \\frac{1}{2}$.\n\nThe inequality $E(a, b) \\le \\frac{1}{16}$ is equivalent to $16(a+b) \\le (4a^2+3)(4b^2+3)$, which we rewrite as $(4ab-1)^2+4(a+b-1)^2+2(2a-1)^2+2(2b-1)^2 \\ge 0$, obviously true.",
"We will show that the maximum va... | Romania | 70th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 1/16 | |
0aeh | За целите броеви $a$ и $b$ важи $a = a^2 + b^2 - 8b - 2ab + 16$. Покажи дека $a$ е полн квадрат. | [
"Имаме\n$$\n9a = a^2 + b^2 + 8a - 8b - 2ab + 16 = (a - b + 4)^2,\n$$\nпа затоа $9a$ е полн квадрат, а оттука и $a$ е полн квадрат."
] | North Macedonia | ЈММО 2009 година | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Other"
] | Macedonian, English | proof only | null | |
00re | Find all monotonic functions $f : \mathbb{R} \to \mathbb{R}$ satisfying the condition: for every real number $x$ and every natural number $n$
$$
\left| \sum_{i=1}^{n} i (f(x + i + 1) - f^2(x + i)) \right| < C
$$
where $C > 0$ is independent of $x$ and $f^2(x) = f(f(x))$. | [
"From the condition of the problem we get $\\left| \\sum_{i=1}^{n-1} i (f(x + i + 1) - f^2(x + i)) \\right| < C$. Then\n$$ \\left| n (f(x + n + 1) - f^2(x + n)) \\right| = \\left| \\sum_{i=1}^{n} i (f(x + i + 1) - f^2(x + i)) - \\sum_{i=1}^{n-1} i (f(x + i + 1) - f^2(x + i)) \\right| < 2C $$\nimplying $|f(x + n + 1... | Balkan Mathematical Olympiad | BMO 2016 Short List Final | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = x + 1 | |
0blg | Let $M$ be the midpoint of the altitude $[VO]$ of the regular pyramid $VABCD$ and $N$ be the midpoint of the segment $[BM]$. Let $P \in (AO)$ be such that $AP = 3 \cdot PO$. Prove that if we denote $d(a, b)$ the distance between the straight lines $a, b$, then
$$
\frac{d(PN, VD)}{d(PN, AB)} = \frac{5}{3}.
$$
Devian Aug... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
06ye | Let $ABC$ be a triangle with $AB < AC < BC$, incentre $I$ and incircle $\omega$. Let $X$ be the point in the interior of side $BC$ such that the line through $X$ parallel to $AC$ is tangent to $\omega$. Similarly, let $Y$ be the point in the interior of side $BC$ such that the line through $Y$ parallel to $AB$ is tange... | [
"\nWe have $\\angle APB = \\angle ACB$ in the circumcircle and $\\angle ACB = \\angle A'XC$ because $A'X \\parallel AC$. Hence, $\\angle APB = \\angle A'XC$, and so quadrilateral $BPA'X$ is cyclic. Similarly, it follows that $CYA'P$ is cyclic.\nNow we are ready to transform $\\angle KIL + \... | IMO | IMO2024 Shortlisted Problems | [
"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 > Transformations > Homot... | English | proof only | null | |
0erl | What is the remainder when $2^{2016}$ is divided by $13$? | [
"(We do this by inspection, by trying out some values until we can see the pattern.) Draw up a list of the remainders left by the powers of $2$ after division by $13$:\n\n| $n$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |\n|-----|---|---|---|---|---|---|----|----|----|----|----|----|----|\n| $2^n \\bmod... | South Africa | South African Mathematics Olympiad Second Round | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | final answer only | 1 | |
093s | Problem:
Determine all real numbers $A$ such that every sequence of non-zero real numbers $x_{1}, x_{2}, \ldots$ satisfying
$$
x_{n+1}=A-\frac{1}{x_{n}}
$$
for every integer $n \geqslant 1$, has only finitely many negative terms. | [
"Solution:\nLet us assume that $A \\geqslant 2$ holds and there is some $n \\geqslant 1$ with $x_{n}<0$. Then $x_{n+1}>A \\geqslant 2$.\nWe claim that $x_{n+k}>1$ for all $k \\geq 1$. This is easily proven by induction: we already did this for $k=1$, and the induction step follows from\n$$\nx_{n+k+1}=A-\\frac{1}{x_... | Middle European Mathematical Olympiad (MEMO) | 15th Middle European Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | A ≥ 2 | |
01s5 | Prove that $(a+b+c)^5 \ge 81(a^2+b^2+c^2)abc$ for any positive real numbers $a$, $b$, $c$. | [
"Write $\\sigma_1 = a + b + c$, $\\sigma_2 = ab + bc + ca$, $\\sigma_3 = abc$. Then the inequality we have to prove becomes\n$$\n\\sigma_1^5 \\ge 81(\\sigma_1^2 - 2\\sigma_2)\\sigma_3 \\iff \\sigma_1^5 + 162\\sigma_2\\sigma_3 \\ge 81\\sigma_1^2\\sigma_3. \\quad (1)\n$$\n\nNow using the AM-GM inequality for three su... | Belarus | SELECTION and TRAINING SESSION | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0iuw | Let $ABC$ be an acute triangle. Point $D$ lies on side $BC$. Let $O_B$ and $O_C$ be the circumcenters of triangles $ABD$ and $ACD$, respectively. Suppose that points $B$, $C$, $O_B$, $O_C$ lie on a circle centered at point $X$. Let $H$ be the orthocenter of triangle $ABC$. Prove that $\angle DAX = \angle DAH$. | [
"**Solution 1.** First, we claim that, if $I$ is the incenter of triangle $ABC$, then $AD$ trisects $\\angle HAI$ with $\\angle IAH = 3\\angle IAD$. Extend rays $AI$ and $AH$ to meet side $BC$ at $I_A$ and $H_A$, respectively. We want to show that $D$ lies on segment $I_A H_A$ with $\\angle I_A A H_A = 3\\angle I_A... | United States | Team Selection Test 2009 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Analyt... | null | proof only | null | |
080u | Problem:
Sia $A$ l'area del sottoinsieme del piano costituito dai punti $(x, y)$ che verificano le due relazioni $x^{2}+y^{2} \leq 100, \pi x+\sqrt{17} y \leq 0$. Allora:
(A) $A<100$
(B) $100 \leq A<150$
(C) $150 \leq A<200$
(D) $200 \leq A<250$
(E) $A \geq 250$. | [
"Solution:\n\nLa risposta è $\\mathbf{( C )}$. La relazione $x^{2}+y^{2} \\leq 100$ rappresenta un cerchio con centro nell'origine e raggio $10$. La relazione $\\pi x+\\sqrt{17} y \\leq 0$ rappresenta un semipiano delimitato da una retta passante per l'origine, la quale pertanto divide il cerchio in due parti ugual... | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | C | |
0dmh | Problem:
Нека је низ $\left(a_{n}\right)_{n \geqslant 1}$ дефинисан са $a_{1}=3, a_{2}=11$ и $a_{n}=4 a_{n-1}-a_{n-2}$, за $n \geqslant 3$. Доказати да је сваки члан овог низа облика $a^{2}+2 b^{2}$ за неке природне $a$ и $b$.
(Ђорђе Баралић) | [
"Solution:\n\nИмамо $a_{1}=1+2 \\cdot 1^{2}, a_{2}=3^{2}+2 \\cdot 1^{2}, a_{3}=3^{2}+2 \\cdot 4^{2}, a_{4}=11^{2}+2 \\cdot 4^{2}$, итд. Доказаћемо индукцијом по $n$ да важи\n$$\na_{2 n-1}=a_{n-1}^{2}+2\\left(\\frac{a_{n}-a_{n-1}}{2}\\right)^{2} \\quad \\text { и } \\quad a_{2 n}=a_{n}^{2}+2\\left(\\frac{a_{n}-a_{n-... | Serbia | СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | null | proof only | null | |
01k0 | Find the value of the expression
$$(a+b+c) \left( \frac{1}{a+b-5c} + \frac{1}{b+c-5a} + \frac{1}{c+a-5b} \right),$$
if real $a$, $b$, $c$ satisfy the equality $$(a+b+c) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) = \frac{27}{2}$$ (all denominators are supposed to be different from zero). | [
"Set $a+b+c = x$ and $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = y$. Let $k = \\frac{x}{6}$, then\n$$\na+b-5c=6(k-c), \\quad b+c-5a=6(k-a), \\quad c+a-5b=6(k-b).\n$$\nIt follows that\n$$\n(a+b+c) \\left( \\frac{1}{a+b-5c} + \\frac{1}{b+c-5a} + \\frac{1}{c+a-5b} \\right) = \\frac{x}{6} \\left( \\frac{1}{k-a} + \\f... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 9/5 | |
00ln | a. Determine the maximum $M$ of $x + y + z$ where $x$, $y$ and $z$ are positive real numbers with
$$
16xyz = (x + y)^2(x + z)^2.
$$
b. Prove the existence of infinitely many triples $(x, y, z)$ of positive rational numbers that satisfy $16xyz = (x + y)^2(x + z)^2$ and $x + y + z = M$. | [
"**a.** The given equation and the AM-GM-inequality imply\n$$\n4\\sqrt{xyz} = (x + y)(x + z) = x(x + y + z) + yz \\ge 2\\sqrt{xyz(x + y + z)}.\n$$\nTherefore, $2 \\ge \\sqrt{x + y + z}$ which gives $4 \\ge x + y + z$. Since we will explicitly give infinitely many triples with $x + y + z = 4$ in the second part, $M ... | Austria | 48th Austrian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 4 | |
0i9f | Problem:
Simplify: $2 \sqrt{1.5+\sqrt{2}}-(1.5+\sqrt{2})$. | [
"Solution:\nThe given expression equals $\\sqrt{6+4 \\sqrt{2}}-(1.5+\\sqrt{2}) = \\sqrt{6+2 \\sqrt{8}}-(1.5+\\sqrt{2})$.\n\nBut on inspection, we see that $(\\sqrt{2}+\\sqrt{4})^{2} = 6+2 \\sqrt{8}$, so the answer is $(\\sqrt{2}+\\sqrt{4})-(1.5+\\sqrt{2}) = 2-3/2 = 1/2$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Other",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | proof and answer | 1/2 | |
01pk | A code lock has 10 keys labeled by digits from 0 to 9. The lock is opened by entering a code of 4 keys that are not necessarily different. It is opened as soon as the correct keys are pressed consequently, no matter what keys have been pressed earlier. For example, if the code happens to be $2002$, the sequence $456778... | [
"Answer: $10003$.\nWe need to find the shortest sequence of decimal digits that includes every 4-digit sequence as a (consecutive) subsequence. Since there are $10^4$ possible codes, and an $n$-digit sequence has at most $n-3$ distinct 4-digit subsequences, the answer must be at least $10^4 + 3 = 10003$.\n\nWe will... | Belarus | BelarusMO 2013_s | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 10003 | |
0jkp | Problem:
Find the smallest positive integer $n$ such that, if there are initially $n+1$ townspeople and $n$ goons, then the probability the townspeople win is less than $1\%$. | [
"Solution:\nAnswer: $6$\n\nBy a similar inductive argument, the probability for a given $n$ is\n$$\np_{n} = \\frac{n!}{(2n+1)!!}.\n$$\nClearly this is decreasing in $n$. It is easy to see that\n$$\np_{5} = \\frac{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5}{3 \\cdot 5 \\cdot 7 \\cdot 9 \\cdot 11} = \\frac{8}{693} > 0.01\... | United States | HMMT November 2014 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 6 | |
08q0 | Problem:
Let $S$ be a set of 100 positive integer numbers having the following property:
"Among every four numbers of $S$, there is a number which divides each of the other three or there is a number which is equal to the sum of the other three."
Prove that the set $S$ contains a number which divides all other 99 numb... | [
"Solution:\n\nLet $a < b$ be the two smallest numbers of $S$ and let $d$ be the largest number of $S$. Consider any two other numbers $x < y$ of $S$. For the quadruples $(a, b, x, d)$ and $(a, b, y, d)$ we cannot get both of $d = a + b + x$ and $d = a + b + y$, since $a + b + x < a + b + y$. From here, we get $a \\... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
09od | Let $\phi(n)$ denote the number of positive integers less than or equal to a given integer $n$ that are relatively prime to $n$. For example, $\phi(6) = |\{1, 5\}| = 2$ and $\phi(1) = |\{1\}| = 1$.
Let $a, b, c, d$ be nonnegative integers such that $\phi(2^a(2b+1)) = 2^c(2d+1)$. If $b \ge 1$ then show that
(1) $a \le c... | [
"Consider the prime factorization $2b+1 = (2b_1+1)^{\\alpha_1} \\dots (2b_l+1)^{\\alpha_l}$, where $\\alpha_i$ are the exponents. Then\n$$\n\\varphi(2b+1) = 2^l(2b_1+1)^{\\alpha_1-1} \\dots (2b_l+1)^{\\alpha_l-1}b_1 \\dots b_l.\n$$\nIt is clear that $l \\ge 1$ since $b \\ge 1$.\n(i) The case $a=0$ is clear. So assu... | Mongolia | MMO2025 Round 4 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
09di | Ангид 48 сурагчтай ба тэдгээрээс хотын сурагчид нь орон нутгаас ирсэн сурагчдаас олон байсан байг. Ямар ч орон нутгаас ирсэн сурагчид үнэн хариулдаг ба хотын сурагчид үнэн эсвэл худал хариулдаг байв. Ангийн шинэ байшинд нүүхдээ сурагчдын хаанаас ирсний нь мэдэхийн тулд дурын сурагчаас "тэр сурагч хаанаас ирсэн бэ?" гэс... | [] | Mongolia | ММО-48 | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | Mongolian | proof and answer | 94 | |
08u1 | How many sextuples $(a, b, c, d, e, f)$ of positive integers are there for which conditions $a > b > c > d > e > f$ and $a+f = b+e = c+d = 22$ are satisfied? | [
"To each $n$ chosen from the set of integers $1, 2, \\ldots, 10$, there is a pair $(m, n)$ of integers satisfying the conditions $m > n$ and $m + n = 22$. Therefore, there are exactly $10$ such pairs $(m, n)$. Finding sextuples $(a, b, c, d, e, f)$ satisfying the conditions of this problem is equivalent to finding ... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 120 | |
097i | Problem:
Determinați funcțiile continue $f:\left[\frac{1}{e^{2}} ; e^{2}\right] \rightarrow \mathbb{R}$, pentru care
$$
\int_{-2}^{2} \frac{1}{\sqrt{1+e^{x}}} f\left(e^{-x}\right) d x-\int_{-2}^{2} f^{2}\left(e^{x}\right) d x=\frac{1}{2}
$$ | [
"Solution:\nObținem\n$$\n\\int_{-2}^{2} \\frac{1}{\\sqrt{1+e^{x}}} f\\left(e^{-x}\\right) d x=\\left|\\begin{array}{c}\nx=-t \\\\\nd x=-d t \\\\\nx=-2 \\Rightarrow t=2 \\\\\nx=2 \\Rightarrow t=-2\n\\end{array}\\right|=\\int_{-2}^{2} \\frac{1}{\\sqrt{1+e^{-t}}} f\\left(e^{t}\\right) d t\n$$\nAtunci\n$$\n\\begin{arra... | Moldova | Olimpiada Republicană la Matematică | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Functions"
] | null | proof and answer | f(t) = (1/2) * sqrt(t / (1 + t)) for t in [1/e^2, e^2] | |
0ftc | Problem:
$m$ sei eine beliebige natürliche Zahl. Bestimme in Abhängigkeit von $m$ die kleinste natürliche Zahl $k$, für die gilt: Ist $\{m, m+1, \ldots, k\}=A \cup B$ eine beliebige Zerlegung in zwei Mengen $A$ und $B$, dann enthält $A$ oder $B$ drei Elemente $a, b, c$ (die nicht notwendigerweise verschieden sein müss... | [
"Solution:\n\nAntwort: $k = m^{m^{m+2}}$\n\nWir zeigen zuerst, dass man $\\{m, m+1, \\ldots, m^{m^{m+2}}-1\\}$ in zwei Mengen zerlegen kann, sodass es keine drei Elemente $a, b, c$ gibt, wie in der Aufgabe gefordert. Setze\n$$\n\\begin{aligned}\n& A = A_1 \\cup A_2 = \\{m, \\ldots, m^{m}-1\\} \\cup \\{m^{m^{m+1}}, ... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | k = m^{m^{m+2}} | |
0is1 | Problem:
Find the real solution(s) to the equation $(x+y)^2 = (x+1)(y-1)$. | [
"Solution:\n\nAnswer: $(-1,1)$\n\nSet $p = x + 1$ and $q = y - 1$, then we get $(p + q)^2 = p q$, which simplifies to $p^2 + p q + q^2 = 0$. Then we have $\\left(p + \\frac{q}{2}\\right)^2 + \\frac{3 q^2}{4} = 0$, and so $p = q = 0$. Thus $(x, y) = (-1, 1)$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | (-1, 1) | |
01eq | Find all quadruples $(x, y, z, t)$ of positive integers that satisfy the system of equations
$$
\begin{cases} xyz = t! \\ (x+1)(y+1)(z+1) = (t+1)! \end{cases}
$$ | [
"Answer: $t = 3$ and $(x, y, z)$ is any permutation of $(1, 2, 3)$.\nSince the equations are symmetrical with respect to variables $x$, $y$ and $z$, we can assume that $x \\le y \\le z$. Dividing the second equation by the first one we obtain the equality\n$$\nt + 1 = \\frac{(t + 1)!}{t!} = \\left(1 + \\frac{1}{x}\... | Baltic Way | Baltic Way shortlist | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | t = 3 and (x, y, z) is any permutation of (1, 2, 3) | |
05lb | Problem:
Soit $n \geqslant 1$ un entier.
On considère un ensemble de $4n+5$ points du plan, trois jamais alignés, et chacun coloré soit en rouge soit en bleu.
Prouver qu'il existe $n$ triangles dont les sommets sont tous d'une même couleur (la même pour tous les triangles), et dont les intérieurs respectifs sont deux ... | [
"Solution:\n\nLemme. Soit $n \\geqslant 3$ et $m \\geqslant 1$ des entiers.\nOn considère un $n$-gone convexe $P$ et $m$ de ses points intérieurs, trois des $n+m$ points jamais alignés. Alors, on peut trianguler $P$ en $n+2m-2$ triangles dont les sommets sont tous parmi les $n+m$ points, aucun de ces triangles ne c... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0hv9 | Problem:
Let $ABC$ be an acute triangle, and $P$ a point on the interior of side $BC$. Let $I$ be the incenter of triangle $ABC$, and denote by $D$ the foot of the altitude from $I$ to $BC$. Line $BI$ meets the internal angle bisector of $\angle APC$ at $X$, while line $CI$ meets the internal angle bisector of $\angle... | [
"Solution:\n\nFirst, notice that $\\angle YPX$ is a right angle. Thus, we claim that in fact $D$ and $P$ lie on a circle with diameter $\\overline{XY}$; in light of this it suffices to prove that\n$$\nDX^{2} + DY^{2} = PX^{2} + PY^{2}\n$$\nLet $K$ and $L$ be the feet of the altitudes from $X$ and $Y$ to $\\overline... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0326 | Problem:
Let $ABCD$ be a parallelogram and let $\angle BAD$ be acute. Denote by $E$ and $F$ the feet of the perpendiculars from the vertex $C$ to the lines $AB$ and $AD$, respectively. A circle through $D$ and $F$ is tangent to the diagonal $AC$ at a point $Q$ and a circle through $B$ and $E$ is tangent to the segment ... | [
"Solution:\nLet $DH \\perp AC$ ($H \\in AC$). Then $\\triangle AHD \\sim \\triangle AFC$ and $\\triangle CHD \\sim \\triangle AEC$. Hence\n$$\n\\begin{aligned}\nAC^{2} &= AH \\cdot AC + HC \\cdot AC \\\\\n&= AF \\cdot AD + AE \\cdot CD \\\\\n&= AQ^{2} + AE \\cdot AB = AQ^{2} + AP^{2}\n\\end{aligned}\n$$\nSetting $Q... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 5/3 | |
0hbg | Find all functions $f:(0, +\infty) \rightarrow (0, +\infty)$ for which for any positive numbers $x, y$ the following equality is true:
$$
f(f(x)+y) = f(f(x))+2yf(x)-f(y)+2y^2+1. \qquad (\text{Ihor Voronovych})
$$ | [
"From the task of the problem when substituting $y = f(y)$ we have that\n$$\n\\begin{aligned}\nf(f(x)+f(y)) &= f(f(x))+2f(y)f(x)-f(f(y))+2f^2(y)+1 \\Rightarrow \\\\\nf(f(x)+f(y))-f(f(x))-2f(y)f(x)-f(f(y)) &= -2f(f(y))+2f^2(y)+1.\n\\end{aligned}\n$$\nTherefore, due to the symmetry of the left part relative to the va... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(x) = x^2 + 1 for all x > 0 | |
07ib | The game of **Hive** is played on a regular hexagonal grid (as shown in the figure) by 3 players. The grid consists of $k$ layers (where $k$ is a natural number) surrounding a regular hexagon, with each layer constructed around the previous layer. The figure below shows a grid with 2 layers.
The players, *Ali*, *Shaya... | [
"The first two players can play in a way that the third one always loses. In the first two moves the first and the second player fill the second layer like below.\n\n\n\nFrom then, they rotate every piece that the third person places on the board, $120^\\circ$ and $240^\\circ$ respectively ... | Iran | 40th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
09xq | In triangle $ABC$ we have $\angle ACB = 90^\circ$. The point $M$ is the midpoint of $AB$. The line through $M$ parallel to $BC$ intersects $AC$ in $D$. The midpoint of line segment $CD$ is $E$. The lines $BD$ and $CM$ are perpendicular.
Be aware: the figure is not drawn to scale.
a. Prove tha... | [
"a. We first prove the similarity $\\triangle CMD \\sim \\triangle ABC$. Since $BC$ and $MD$ are parallel, we find that $\\angle ADM = \\angle ACB = 90^\\circ$ and also $\\angle AMD = \\angle ABC$. It follows that $\\triangle ABC \\sim \\triangle AMD$. Because $|AB| = 2|AM|$ we also have that $|AC| = 2|AD|$ and thu... | Netherlands | Dutch Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
01ks | For all admissible $a$, $b$, $c$ find all possible values of the expression
$$
\frac{(a+b-c)^2}{(a-c)(b-c)} + \frac{(b+c-a)^2}{(b-a)(c-a)} + \frac{(c+a-b)^2}{(c-b)(a-b)}
$$ | [
"Set $A = a + b + c$. Then\n$$\n\\begin{aligned}\nM &= \\frac{(a+b-c)^2}{(a-c)(b-c)} + \\frac{(b+c-a)^2}{(b-a)(c-a)} + \\frac{(c+a-b)^2}{(c-b)(a-b)} = \\\\\n&= \\frac{(A-2c)^2}{(a-c)(b-c)} + \\frac{(A-2a)^2}{(b-a)(c-a)} + \\frac{(A-2b)^2}{(c-b)(a-b)} = \\\\\n&= \\frac{(A-2c)^2(b-a) + (A-2a)^2(c-b) + (A-2b)^2(a-c)}{... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | 4 | |
0cqd | A straight stick of length $2$ m is cut into $N$ pieces, the length of each (measured in centimeters) is an integer number. Find the least value of $N$ such that it is always possible to make a rectangular contour using all the obtained pieces. (It is not allowed to break the pieces.)
Прямую палку длиной $2$ метра рас... | [
"Answer: $N = 102$.\n\nFirst solution. Suppose $N \\leq 101$. Cut the stick into $N-1$ pieces of length $1$ cm and one piece of length $(201 - N)$ cm. From this set, it is impossible to form a rectangle, since each side of the rectangle is less than the semiperimeter, and therefore the piece of length $201 - N \\ge... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English, Russian | proof and answer | 102 | |
01uh | Point $M$ is marked inside a convex quadrilateral $ABCD$. It appears that $AM = BM$, $CM = DM$, and $\angle AMB = \angle CMD = 60^\circ$. Let $K$, $L$, and $N$ be the midpoints of the segments $BC$, $AM$, and $DM$, respectively. Find the value of the angle $LKN$.
(S. Mazanik) | [
"Answer: $60^\\circ$.\n\nLet $E$ and $F$ be the midpoints of the segments $BM$ and $CM$, respectively. Since $\\angle AMB = \\angle CMD = 60^\\circ$, we have\n$$\n\\begin{aligned}\n\\angle BMC &= 360^\\circ - \\angle AMB - \\angle CMD - \\angle LMN = 360^\\circ - 60^\\circ - 60^\\circ - \\angle LMN = \\\\\n&= 240^\... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 60° | |
02tg | Problem:
Pedro acabou de se mudar para sua nova casa e ganhou um novo quarto. A figura a seguir mostra uma vista superior simplificada de seu novo quarto que possui $2~\mathrm{m}$ de largura por $2{,}5~\mathrm{m}$ de comprimento.
A porta indicada na figura tem $50~\mathrm{cm}$ de compriment... | [
"Solution:\n\nSeja $L$ o comprimento de cada porta da janela. Considerando que, quando as duas portas se abrem elas encostam nas paredes dos lados, temos então: $4 \\cdot L = 2$, ou seja, $L = 0,5~\\mathrm{m}$.\n\nChamemos de $A$ a área que Pedro tem para colocar seus móveis. Para determiná-la, basta considerar a á... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | (80 - 5π)/16 | |
094h | Problem:
Let $\mathbb{Z}$ denote the set of all integers and $\mathbb{Z}_{>0}$ denote the set of all positive integers.
a. A function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ is called $\mathbb{Z}$-good if it satisfies $f\left(a^{2}+b\right)=f\left(b^{2}+a\right)$ for all $a, b \in \mathbb{Z}$. Determine the largest pos... | [
"Solution:\n\na.\nNote that\n$$\nf\\left(a^{2}+b\\right)=f\\left(b^{2}+a\\right)=f\\left((-b)^{2}+a\\right)=f\\left(a^{2}-b\\right)\n$$\nIn particular, by setting $a \\in\\{0,1\\}$ we get $f(b)=f(-b)$ and $f(1+b)=f(1-b)$. This then yields\n$$\nf(2+b)=f(1+(1+b))=f(1-(1+b))=f(-b)=f(b)\n$$\nhence by induction the func... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a = 2; b = 1077 | |
0480 | There are $2025$ people and $66$ given colors. Each person has $66$ balls, one of each color, with a total weight of $1$ for all $66$ balls.
Find the smallest real number $C$ such that, no matter how the balls are weighted, one can always select exactly one ball from each person so that for every color, the total weigh... | [
"Let's generalize the problem by replacing $2025$ with $n$ and $66$ with $m$. For any positive integers $m \\le n$, define:\n$$\nf_m(n) = \\min_{\\substack{a_1+a_2+\\dots+a_m=n \\\\ a_1, a_2, \\dots, a_m \\text{ are positive integers}}} \\left( \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_m} \\right).\n$$... | China | China-TST-2025A | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 248/517 | |
0fqx | Problem:
Ana y Bernardo juegan al siguiente juego. Se empieza con una bolsa que contiene $n \geq 1$ piedras. En turnos sucesivos y empezando por Ana, cada jugador puede hacer los siguientes movimientos: si el número de piedras en la bolsa es par, el jugador puede coger una sola piedra o la mitad de las piedras. Si el n... | [
"Solution:\nAna tiene una estrategia ganadora para $n=1,3$ y para todos los números pares mayores que 3. En primer lugar, observamos que en los casos $n=1,2,3$ todos los movimientos están determinados y Ana gana cuando $n=1$ o $n=3$. Si el número inicial de piedras es par y mayor que 4, Ana puede coger una sola pie... | Spain | FASE LOCAL DE LA OLIMPIADA MATEMÁTICA ESPAÑOLA. | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Ana wins for starting sizes equal to one, three, or any even number at least four. | |
0l1m | Problem:
Let $ABC$ be a triangle such that $AB = 3$, $AC = 4$, and $\angle BAC = 75^{\circ}$. Square $BCDE$ is constructed outside triangle $ABC$. Compute $AD^{2} + AE^{2}$. | [
"Solution:\n\n\n\nConstruct point $X$ such that $\\triangle CBD \\sim \\triangle CXA$. Then, $\\triangle CBX \\cong \\triangle CAD$. Thus, $AD = BX$. We have $AB = 3$, $AX = 4\\sqrt{2}$, and $\\angle BAX = 120^{\\circ}$, so law of cosine gives\n\n$$\nAD^{2} = BX^{2} = 3^{2} + (4\\sqrt{2})^{... | United States | HMMT November 2024 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions a... | null | final answer only | 75 + 24√2 | |
01xe | The fence consists of less than $40$ equally spaced boards. On some of them sits a sparrow, in total $10$ sparrows.
a) Prove that among the pairwise distances between the sparrows there are two equal.
b) Prove that this statement will remain true even if one of the sparrows flies away. | [] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0htg | Problem:
Does there exist a function $f$ from the positive integers to itself, such that for any positive integers $a$ and $b$, we have $\operatorname{gcd}(a, b)=1$ if and only if $\operatorname{gcd}(f(a), f(b))>1$ holds? | [
"Solution:\n\nThe answer is no. Assume that $f$ satisfies the hypothesis. Let $k$ denote the number of distinct primes dividing $f(1)$. For every integer $e$, the number $f\\left(2^{e}\\right)$ shares some prime factor with $f(1)$. So among $f(2), f(4), \\ldots, f\\left(2^{k+1}\\right)$ two of them have the same sh... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | No | |
093b | Problem:
A group of pirates had an argument and now each of them holds some other two at gunpoint. All the pirates are called one by one in some order. If the called pirate is still alive, he shoots both pirates he is aiming at (some of whom might already be dead). All shots are immediately lethal. After all the pirate... | [
"Solution:\nCall a pirate mortal if someone is aiming at him in the beginning. Since some order of shooting results in 28 pirates dead, there are at least 28 mortal pirates.\n\nFor the sake of contradiction, suppose that some order of shooting results in at most 9 dead pirates. Then at least 19 mortal pirates survi... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0hhk | Given $n \ge 3$ pairwise distinct real numbers. Prove that there are either 3 numbers with a positive sum or 2 numbers with a negative sum. | [
"If there are no positive numbers in the given set, then there is at most one number equal to zero, and all other numbers are negative, which means that there are $n-1$ negative numbers. Therefore, as the required set of 2 numbers, we can take any two numbers.\n\nOtherwise, if there is at least one positive number ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Logic"
] | English | proof only | null | |
05xo | Problem:
Déterminer toutes les fonctions $f: \mathbb{N}^{\star} \rightarrow \mathbb{N}^{\star}$ telles que :
(i) Les entiers $f(1), f(2), \ldots$ sont premiers entre dans leur ensemble.
(ii) Il existe $N \geqslant 1$ tel que pour tout $n \geqslant N, f(n) \neq 1$ et pour tous $a, b \in \mathbb{N}^{\star}$,
$$
f(a)^{... | [
"Solution:\n\nSoit $f$ une solution. Comme les valeurs de $f$ sont premières entre elles dans leur ensemble, si $f$ est constante, elle égale $1$, ce qui est exclu.\n\nAvec $a=1$, on voit que $f(1)^{n} \\mid f(b+1)-f(b)$ pour tout $b \\geqslant 1$ et tout $n$ assez grand. Comme $f$ est non constante, il existe $b$ ... | France | Préparation Olympique Française de Mathématiques - Envoi 3: Arithmétique | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Number Theor... | null | proof and answer | f(n) = n for all positive integers n | |
01hc | Prove that there are infinitely many mutually coprime positive integers $a$, $b$ and $c$ such that
$$
\left\lfloor \frac{a^2}{2020} \right\rfloor + \left\lfloor \frac{b^2}{2020} \right\rfloor = \left\lfloor \frac{c^2}{2020} \right\rfloor.
$$ | [
"At first we note that if $b$ is divisible by $2020$ and $a^2 + b^2 = c^2$ then also\n$$\n\\lfloor \\frac{a^2}{2020} \\rfloor + \\lfloor \\frac{b^2}{2020} \\rfloor = \\lfloor \\frac{c^2}{2020} \\rfloor.\n$$\nConsider the equality $(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2$ and take $n = 2020$ and $m > 2020$ to be any... | Baltic Way | Baltic Way 2020 | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
05kj | Problem:
Soit $ABCD$ un quadrilatère inscriptible. On note $K$ le point d'intersection des diagonales. Soient $M$ et $N$ les milieux de $[A, C]$ et $[B, D]$. Les cercles circonscrits à $ADM$ et $BCM$ se recoupent en un point $L$. Montrer que $K, L, M, N$ sont cocycliques. | [
"Solution:\n\nNotons $E$ le centre radical des trois cercles $ABCD$, $ADM$, $BCM$. Les triplets $(E, A, D)$, $(E, L, M)$ et $(E, B, C)$ sont alignés.\n\nSoit $F$ le point où les cercles $ADK$ et $BCK$ se recoupent. On a $(FA, FD) = (KA, KD) = (KC, KB) = (FC, FB)$ et $(AD, AF) = (KD, KF) = (... | France | null | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a3n | Problem:
Zij $\triangle A B C$ een scherphoekige driehoek met $|A B|<|A C|$. Gegeven zijn punten $D$ en $E$ op respectievelijk de lijnstukken $A B$ en $A C$. Zij $P$ een punt zodanig dat $|P B|=|P D|$ en $|P C|=|P E|$. Zij $X$ een punt op de boog $A C$ van de omgeschreven cirkel van $\triangle A B C$ die niet $B$ beva... | [
"Solution:\n\nWe bekijken de configuratie zoals in het plaatje waarbij $Y$ op de boog $A E$ ligt. Wanneer $Y$ op de boog $A Z$ ligt, klapt er een hoekje om maar dit geval gaat verder analoog. Zij $Z$ het tweede snijpunt van de omgeschreven cirkels van $\\triangle A B C$ en $\\triangle A D E$. Dan vinden we met geri... | Netherlands | Maarttoets | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneo... | null | proof only | null | |
0jww | Problem:
Mrs. Toad has a class of 2017 students, with unhappiness levels $1,2, \ldots, 2017$ respectively. Today in class, there is a group project and Mrs. Toad wants to split the class in exactly 15 groups. The unhappiness level of a group is the average unhappiness of its members, and the unhappiness of the class i... | [
"Solution:\n\nOne can show that the optimal configuration is $\\{1\\}, \\{2\\}, \\ldots, \\{14\\}, \\{15, \\ldots, 2017\\}$. This would give us an answer of $1+2+\\cdots+14+\\frac{15+2017}{2}=105+1016=1121$."
] | United States | February 2017 | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | 1121 | |
0574 | Find all positive integers $n$ such that one can choose $n-3$ non-intersecting diagonals of a regular $n$-gon that divide the $n$-gon into triangles in such a way that every chosen diagonal is a side of minimal length in some triangle. | [
"Let $\\triangle$ be the triangle containing the center $O$ of the $n$-gon (colored in Fig. 23; if the center lies on a diagonal then choose either of the triangles having this diagonal as a side). Let $d$ be any side of $\\triangle$. If $d$ is a diagonal of the $n$-gon then $d$ separates $\\triangle$ from a neighb... | Estonia | Final Round of National Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | n = 3 · 2^k for natural k | |
02wm | Problem:
O rei Luis estava desconfiado de alguns de seus cortesãos. Ele fez uma lista completa de cada um dos seus cortesãos e disse a cada um deles para espionar um outro cortesão. O primeiro da lista foi espionar o cortesão que estava espionando o segundo da lista, o segundo da lista foi espionar o cortesão que esta... | [
"Solution:\n\nSeja $n$ o número de cortesãos da lista e suponha que $n$ é par. Coloque-os sentados ao redor de uma mesa circular de modo que cada um esteja espionando o seu vizinho da esquerda.\n\n\n\nO cortesão $1$ espia o cortesão $X$ que espia o cortesão $2$, o cortesão $2$ espia o corte... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
08vj | Let $OAB$ be a sector of a circle with its center $O$ as in the figure below. A point $Q$ is chosen on the chord $AB$. Let $P$ be the point of intersection of the arc of the circular sector and the line $OQ$. If $AQ = 5$, $BQ = 6$, $OQ = PQ$, determine the value of the radius of this sector. Here for a line segment $XY... | [
"Let $r$ be the radius of the circle. Consider the full circle obtained by extending the arc of the circular sector in question. Let $R$ be the point of intersection, other than $P$, of the line $PO$ and the full circle. By the well-known theorem on the power of a point with respect to a circle, we have $AQ \\cdot ... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof and answer | 2*sqrt(10) | |
0c7y | We consider the isosceles rectangular triangle $ABC$, $m(\hat{A}) = 90^\circ$, and the point $D \in (AB)$ such that $AD = \frac{1}{3}AB$. In the halfplane determined by the line $AB$ and the point $C$, one takes the point $E$ for which $m(\angle BDE) = 60^\circ$ and $m(\angle DBE) = 75^\circ$. The lines $BC$ and $DE$ i... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
08we | Let $p$ be a prime. Determine all positive integers $n$ for which the following condition is satisfied for all the integers $x$:
Condition: If $x^n - 1$ is divisible by $p$, then it is also divisible by $p^2$. | [
"If, for a pair of integers $a, b$ and a positive integer $m$, $a - b$ is divisible by $m$, we write $a - b \\equiv 0 \\pmod{m}$.\nWe will show that the numbers $n$ we seek are those of the form $kp$, where $k$ is a positive integer.\nLet us first show that if $n$ satisfies the condition of the problem, then $n$ mu... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | All positive multiples of p; that is, n = k p for some positive integer k. | |
024r | Problem:
Brincando com números ímpares - Beatriz adora números ímpares. Quantos números entre $0$ e $1000$ ela pode escrever usando apenas algarismos ímpares? | [
"Solution:\n\nComo cada algarismo é ímpar, temos:\n- cinco possibilidades de números com um algarismo: $1$, $3$, $5$, $7$ e $9$;\n- para números com dois algarismos, temos cinco possibilidades na casa das unidades e cinco na casa das dezenas, totalizando $5 \\times 5 = 25$ possibilidades;\n- para números com três a... | Brazil | Nível 2 | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 155 | |
0hbm | There is an infinite sequence of letters $a$ and $b$. In this sequence, one can perform a substitution $abb \rightarrow baa$. It is known that regardless of the order of these substitutions, one can only make a finite number of them. Prove that, in that case, substitutions $aabb \rightarrow bbaa$ can also be made only ... | [
"Since there can only be made a finite number of substitutions, there is a number $N$, starting from which substitutions are not possible. That means that there are no pairs of subsequent letters $b$. Because, otherwise, on the left from them there is letter $a$, and substitution is possible. But then the second ty... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
06wl | Version 1. Let $n$ be a fixed positive integer, and let $S$ be the set of points $(x, y)$ on the Cartesian plane such that both coordinates $x$ and $y$ are nonnegative integers smaller than $2n$ (thus $|S| = 4n^2$). Assume that $\mathcal{F}$ is a set consisting of $n^2$ quadrilaterals such that all their vertices lie i... | [
"Answer for both Versions: The largest possible sum of areas is $\\Sigma(n) := \\frac{1}{3} n^2 (2n+1)(2n-1)$.\n\nCommon remarks. Throughout all solutions, the area of a polygon $P$ will be denoted by $[P]$.\nWe say that a polygon is legal if all its vertices belong to $S$. Let $O = \\left(n - \\frac{1}{2}, n - \\f... | IMO | IMO 2021 Shortlisted Problems | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | n^2(2n+1)(2n-1)/3 | |
0dfd | Let the sequence $a_1, a_2, \dots, a_{20}$ is the permutation of integers $1, 2, \dots, 20$. Find the maximum possible value of
$$
\min\{|a_2 - a_1|, |a_3 - a_2|, \dots, |a_{20} - a_{19}|, |a_1 - a_{20}|\}.
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 9 | |
0efj | Problem:
Katero število je rešitev enačbe $2^{x} \cdot 5^{x}=0,01 \cdot\left(10^{x-2}\right)^{4}$ ?
(A) $\frac{2}{3}$
(B) $\frac{10}{3}$
(C) $\frac{1}{3}$
(D) $-2$
(E) $-\frac{3}{2}$ | [
"Solution:\nUpoštevamo pravila za računanje s potencami in preoblikujemo enačbo v obliko $10^{x}=10^{4 x-10}$. Rešitev enačbe je $x=\\frac{10}{3}$. Pravilen odgovor je (B)."
] | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | MCQ | (B) | |
0fqa | Problem:
Sea $n$ un número natural. Probar que si la última cifra de $7^{n}$ es 3, la penúltima es 4. | [
"Solution:\n\nLos restos potenciales de $7$ respecto del módulo $10$ coinciden con la última cifra, o cifra de las unidades, de las potencias de $7$ y siendo\n$$\n\\begin{aligned}\n& 7^{0} \\equiv 1 \\quad(\\text{mód. } 10) \\\\\n& 7^{1} \\equiv 7 \\quad(\\text{mód. } 10) \\\\\n& 7^{2} \\equiv 9 \\quad(\\text{mód. ... | Spain | null | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
09id | 70 pairwise distinct positive integers are written on the blackboard, each not exceeding 100. Show that it is possible to choose pairwise distinct integers $a, b, c, d, e$ from the blackboard satisfying $a + b + c = d + e$. | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0dh5 | Let $AD$ be the altitude of the triangle $ABC$. Let $J$, $K$ be the incenters of the triangles $ABD$, $ACD$ respectively. Let $JK$ intersect $AB$, $AC$ at $E$, $F$ respectively. Prove that $AE = AF$ if and only if $AB = AC$ or $\angle A = 90^\circ$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Advanced Configur... | English | proof only | null | |
0heo | Prove the following inequality for positive $a, b, c$:
$$
\frac{a^3 + b^3 + c^3}{abc} + 6 \ge 9 \cdot \frac{a^2 + b^2 + c^2}{ab + bc + ca}.
$$ | [
"Subtract $9$ from both sides and use the well-known identity\n$$\na^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca).\n$$\n\n$$\n\\frac{a^3 + b^3 + c^3}{abc} - 3 \\ge 9 \\cdot \\frac{a^2 + b^2 + c^2}{ab + bc + ca} - 9 \\Leftrightarrow \\\\\n\\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc}{abc} - 3 \\ge 9 \\cdot \\fr... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0gpz | Some pairs of cities of a country consisting of $100$ cities are connected by $2013$ round trip flights operated by $n$ air companies. There are at least two cities such that one is not reachable from the other one by one or two flights. Given that for any pair of cities there is an air company connecting these two cit... | [
"The answer is $n = 2015$. The solution will be given in terms of graph theory: vertices are cities, edges corresponding to flights will be identically colored if they belong to the same air company. Let all edges of a tree connecting all $100$ vertices be identically colored and the remaining edges be differently ... | Turkey | 21st Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | 1915 | |
080l | Problem:
Quante sono le progressioni aritmetiche costituite da quattro numeri interi $a, b, c, d$ con $1 \leq a < b < c < d \leq 100$? | [
"Solution:\nLa risposta è 1617. Se $a, b, c, d$ è una progressione aritmetica di quattro termini interi, anche la ragione $r$ è un intero $\\geq 1$ e si ha $d = a + 3r$. D'altra parte, il termine iniziale e il termine finale di una progressione aritmetica di quattro termini individuano completamente anche gli altri... | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | null | proof and answer | 1617 | |
096w | Problem:
Din mulţimea $\{1,2,3, \ldots, n\}$ se aleg 9 numere, distincte două câte două, care se scriu în celulele unui tabel $3 \times 3$ astfel, încât produsele numerelor din fiecare linie, coloană şi din diagonale să fie egale. Să se determine cea mai mică valoare a lui $n$, pentru care un asemenea tabel există. | [
"Solution:\n\nFie $k$ produsul menţionat în enunţ, iar $x$ numărul din celula centrală a tabelului. Numărul $x$ figurează în 4 produse - în linia a 2-a, în coloana 2 şi în ambele diagonale. Produsul tuturor numerelor din linia 2, coloana 2 şi din ambele diagonale este $k^{4}$. În acest produs numărul $x$ participă ... | Moldova | A 63-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 36 | |
0c3x | Problem:
Dacă $a, b, c > 0$, arătaţi că
$$
\frac{a}{\sqrt{(a+2b)^3}} + \frac{b}{\sqrt{(b+2c)^3}} + \frac{c}{\sqrt{(c+2a)^3}} \geq \frac{1}{\sqrt{a+b+c}}.
$$ | [
"Solution:\nDin inegalitatea lui Hölder avem:\n$$\n\\sum_{\\text{cycl}} \\frac{a}{\\sqrt{(a+2b)^3}} \\cdot \\sum_{\\text{cycl}} a \\sqrt{a+2b} \\cdot \\sum_{\\text{cycl}} a \\sqrt{a+2b}.\n$$\n$\\sum_{\\text{cycl}} a \\sqrt{a+2b} \\geq (a+b+c)^4$, iar din inegalitatea Cauchy-Buniakowsky-Schwarz rezultă\n$$\n\\begin{... | Romania | Al patrulea test de selecţie pentru OBMJ | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0ek1 | Problem:
Naj bo $x$ takšno realno število, za katerega velja $\cos \left(60^\circ-x\right) \neq 0$ in $\sin \left(120^\circ-x\right) \neq 0$. Brez uporabe žepnega računala izračunaj natančno vrednost izraza:
$$
\frac{\sqrt{3}+4 \sin x \cos x}{\cos \left(60^\circ-x\right) \cdot \sin \left(120^\circ-x\right)}
$$ | [
"Solution:\n\nNajprej dvakrat uporabimo adicijska izreka in izraza kar se da poenostavimo:\n\n$\\cos \\left(60^\\circ-x\\right) = \\cos 60^\\circ \\cdot \\cos x + \\sin 60^\\circ \\cdot \\sin x = \\frac{1}{2} \\cdot \\cos x + \\frac{\\sqrt{3}}{2} \\cdot \\sin x$\n\nin\n\n$\\sin \\left(120^\\circ-x\\right) = \\sin 1... | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 4 | |
0h1r | The trapezoid $ABCD$ with parallel sides $BC = a$ and $AD = 2a$ is drawn on the plane. Using only ruler construct triangle with the area equals to the area of trapezoid. | [
"We first construct the point $O$ of the intersection of diagonals, and the point $T$ of intersection of sides $AB$ and $CD$. It is well known that $OT$ contains midpoints $P$ and $E$ of parallel sides (fig. 11). We have $AE = ED = BC = a$. Hence $ABCE$ and $BCDE$ are parallelograms, $M$ and $N$ are the points of i... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | English | proof and answer | null | |
07n7 | Find with proof all solutions in nonnegative integers $a, b, c, d$ of the equation
$$
11^a 5^b - 3^c 2^d = 1.
$$ | [
"We consider four cases: $d=0$, $d=1$, $d=2$ and $d > 2$.\n\n*Case 1:* $d=0$. In this case $3^c 2^d$ is odd, so there are no solutions.\n\n*Case 2:* $d=1$. We have $11^{a} 5^{b} - 2 \\cdot 3^{c} = 1$, so $c > 0$, and, reading mod 4, we see that $a$ is odd. Next reading mod 3, we get that $a+b$ is even. Hence $a = 2... | Ireland | Ireland | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | [[1, 1, 3, 1], [0, 1, 0, 2], [0, 2, 1, 3]] |
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