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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0igd | Problem:
Doug and Ryan are competing in the 2005 Wiffle Ball Home Run Derby. In each round, each player takes a series of swings. Each swing results in either a home run or an out, and an out ends the series. When Doug swings, the probability that he will hit a home run is $1/3$. When Ryan swings, the probability that... | [
"Solution:\n\nDenote this probability by $p$. Doug hits more home runs if he hits a home run on his first try when Ryan does not, or if they both hit home runs on their first try and Doug hits more home runs thereafter. The probability of the first case occurring is $\\frac{1}{3} \\cdot \\frac{1}{2} = \\frac{1}{6}$... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 1/5 | |
08w2 | Let $n$ be a positive integer and suppose a $2n \times 2n$ square grid is given. Suppose we color exactly $2n^2$ square boxes of the grid in such a way that the following condition is satisfied.
Condition: If a box is colored then none of the boxes which share only a vertex with that box is not colored.
How many ways... | [
"$(2nC_n)^2$ ways\n\nLet us consider the $n \\times n$ grid of squares of side length $2$ obtained from the original grid by forming $n^2$ non-overlapping squares of side length $2$ by coalescing $4$ neighboring squares of the original grid into one large square. Let us call each of the squares of the new grid a *b... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | (2n choose n)^2 | |
08k9 | Problem:
Solve the equation
$$
8 x^{3}+8 x^{2} y+8 x y^{2}+8 y^{3}=15\left(x^{2}+y^{2}+x y+1\right)
$$
in the set of integers. | [
"Solution:\nWe transform the equation to the following one\n$$\n\\left(x^{2}+y^{2}\\right)(8 x+8 y-15)=15(x y+1)\n$$\nSince the right side is divisible by $3$, then $3 /\\left(x^{2}+y^{2}\\right)(8 x+8 y-15)$. But if $3 /\\left(x^{2}+y^{2}\\right)$, then $3 / x$ and $3 / y$, which will give $15(x y+1)$ and $3 /(x y... | JBMO | OJBM | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (1,2) and (2,1) | |
0cpb | In a group of people some pairs are friends. A group is called *k-indivisible* if for each decomposition of this group into $k$ subgroups, at least one subgroup contains a pair of friends. Suppose that $A$ is a finite 3-indivisible group of people having no subgroup of 4 persons such that each two of them are friends. ... | [
"Предположим противное. Рассмотрим граф $G$, в котором люди являются вершинами, а два человека соединены ребром, если они знакомы. Тогда граф $k$-разбиваем, если его вершины можно правильно окрасить в $k$ цветов (т. е. окрасить так, чтобы соединённые вершины имели разные цвета). Мы будем пользоваться следующей изве... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English, Russian | proof only | null | |
0i8j | Problem:
Let $ABC$ be an isosceles triangle with apex $A$. Let $I$ be the incenter. If $AI = 3$ and the distance from $I$ to $BC$ is $2$, then what is the length of $BC$? | [
"Solution:\n\nLet $X$ and $Y$ be the points where the incircle touches $AB$ and $BC$, respectively. Then $AXI$ and $AYB$ are similar right triangles. Since $I$ is the incenter, we have $IX = IY = 2$. Using the Pythagorean theorem on triangle $AXI$, we find $AX = \\sqrt{5}$. By similarity, $AY / AX = BY / IX$. Plugg... | United States | Harvard-MIT Mathematics Tournament | [
"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 chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | null | proof and answer | 4√5 | |
0a64 | Problem:
Let $n, m$ be positive integers. Let $A_{1}, A_{2}, A_{3}, \ldots, A_{m}$ be sets such that $A_{i} \subseteq \{1,2,3,\ldots, n\}$ and $|A_{i}| = 3$ for all $i$ (i.e. $A_{i}$ consists of three different positive integers each at most $n$). Suppose for all $i < j$ we have
$$|A_{i} \cap A_{j}| \leqslant 1$$
(i.e.... | [
"Solution:\nEach set $A_{i}$ has exactly three pairs of elements. But each unordered pair chosen from $\\{1,2,\\ldots, n\\}$ can be in at most one such set. Therefore\n$$\\binom{n}{2} \\geq 3m.$$ \nThis establishes the required upper bound on the size of $m$.\n\nLet $T$ be the set of all triples, i.e.\n$$T = \\{(a,... | New Zealand | NZMO Round One | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0k0i | Problem:
Let $A, B, C, D$ be points chosen on a circle, in that order. Line $BD$ is reflected over lines $AB$ and $DA$ to obtain lines $\ell_{1}$ and $\ell_{2}$ respectively. If lines $\ell_{1}, \ell_{2}$, and $AC$ meet at a common point and if $AB=4, BC=3, CD=2$, compute the length $DA$. | [
"Solution:\nLet the common point be $E$. Then since lines $BE$ and $BD$ are symmetric about line $BA$, $BA$ is an exterior bisector of $\\angle DBE$, and similarly $DA$ is also an exterior bisector of $\\angle BDE$. Therefore $A$ is the $E$-excenter of triangle $BDE$ and thus lies on the interior bisector of $\\ang... | United States | HMMT November 2017 | [
"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 and answer | sqrt(21) | |
0a13 | Problem:
We spelen een spelletje stoelendans met $n$ stoelen genummerd 1 tot en met $n$. Je hangt $n$ blaadjes, genummerd 1 tot en met $n$, op de stoelen zodanig dat het nummer op een blaadje niet overeenkomt met het nummer op de stoel.
Op elke stoel zit een speler. Wanneer je klapt, kijkt elke speler naar het nummer ... | [
"Solution:\n\nAls $m=n$, dan kan je op stoel $i$ het blaadje ophangen met nummer $i+1$. Iedereen schuift dan elke klap een stoel op en de eerste keer dat iemand weer op zijn eigen stoel zit is na $n$ klappen. En dan zit ook iedereen weer op zijn eigen stoel. Stel nu dat $m<n$.\nAangezien $m$ geen priemmacht is kunn... | Netherlands | Selectietoets | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0l1u | Problem:
Let $ABC$ be a scalene triangle and $M$ be the midpoint of $BC$. Let $X$ be the point such that $CX \parallel AB$ and $\angle AMX = 90^{\circ}$. Prove that $AM$ bisects $\angle BAX$. | [
"Solution:\n\nLet $Y$ be the intersection of lines $AB$ and $XM$. Since $BY \\parallel CX$, we have $\\angle YBM = \\angle XCM$. Furthermore, we have $BM = CM$, since $M$ is the midpoint of $BC$. Thus,\n$$\n\\triangle BMY \\cong \\triangle CMX\n$$\nThus, $MY = MX$. Combined with the conditi... | United States | HMMT February 2024 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
02zx | Problem:
Os vértices do quadrilátero $ABCD$ estão em uma circunferência. Cada uma de suas diagonais bissecta um ângulo e trisecta o ângulo oposto. Determine as medidas dos ângulos do quadrilátero.
Observação: Dizemos que uma semirreta $OM$ trisecta um ângulo $\angle AOB$ se a amplitude de u... | [
"Solution:\n\nSuponha que a diagonal $AC$ bissecta o ângulo $\\angle DCB$. A outra diagonal bissecta o ângulo $\\angle ABC$ ou $\\angle ADC$. Podemos supor sem perda de generalidade que ela bissecta o ângulo $\\angle ADC$. Como $\\angle ADB = \\angle BDC$ e $\\angle ACB = \\angle ACD$, os arcos $AD$, $AB$ e $BC$ co... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | Either 72°, 72°, 108°, 108° or 720°/7, 720°/7, 540°/7, 540°/7. | |
0g3x | Problem:
Let $n$ be a positive integer. Prove that the numbers
$$
1^{1}, 3^{3}, 5^{5}, \ldots,\left(2^{n}-1\right)^{2^{n}-1}
$$
all give different remainders when divided by $2^{n}$. | [
"Solution:\nLet's prove the statement by induction.\n\nThe base case $n=1$ is trivial.\n\nLet's prove the inductive step. Given that $1^{1}, 3^{3}, 5^{5}, \\ldots,\\left(2^{n}-1\\right)^{2^{n}-1}$ have different residues $(\\bmod\\ 2^{n})$, we want to show that $1^{1}, 3^{3}, 5^{5}, \\ldots,\\left(2^{n+1}-1\\right)... | Switzerland | Final round | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0imk | Problem:
Find the number of 7-tuples $\left(n_{1}, \ldots, n_{7}\right)$ of integers such that
$$
\sum_{i=1}^{7} n_{i}^{6}=96957
$$ | [
"Solution:\nAnswer: 2688. Consider the equation in modulo 9. All perfect 6th powers are either 0 or 1. Since 9 divides 96957, it must be that each $n_{i}$ is a multiple of 3. Writing $3 a_{i}=n_{i}$ and dividing both sides by $3^{6}$, we have $a_{1}^{6}+\\cdots+a_{7}^{6}=133$. Since sixth powers are nonnegative, $\... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2688 | |
0him | You are given $n \ge 2$ distinct positive integers. A pair of integers is called *elegant* if their sum is $2^k$ for some positive integer $k$. For each $n$ find the largest possible number of *elegant* pairs.
(Oleksii Masalitin) | [
"**Answer:** $n-1$\n\nWe prove that there exist at most $n-1$ *elegant* pairs by induction by $n$. The base case for $n=2$ is obvious, we prove the transition. Let the statement be proved for $n-1$, let us prove it for $n$ numbers. Let us denote by $a$ the largest of them, suppose that the number $a$ is contained i... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | n-1 | |
047u | Find all functions $f : \mathbb{Z} \to \mathbb{Z}$ satisfying:
(1) For any positive integer $M$, there exists an integer $k$ with $|f(k)| \ge M$;
(2) For any integers $m, n$,
$$2f(m)f(n) - f(m - n) - 1$$ is a perfect square. | [
"Let $\\alpha = 3 + 2\\sqrt{2}$. All functions satisfying the given conditions are of the form $f(n) = \\frac{1}{2}(\\alpha^{2n} + \\alpha^{-2n})$ for some fixed positive integer $t$.\n\nFirst, we verify that such $f(n)$ satisfies $(\\star)$. The left-hand side of $(\\star)$ becomes:\n$$\n2 \\cdot \\frac{\\alpha^{2... | China | 2025 International Mathematical Olympiad China National Team Selection Test | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Algebraic Number Theory > Quadratic fields",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | All such functions are f(n) = ((3+2√2)^{2tn} + (3−2√2)^{2tn}) / 2 for some fixed positive integer t. | |
07sz | Let $a$, $b$, $c$, $d$ be real numbers. Show there is a pair $(x, y)$ of real numbers such that $|x| \le 1$, $|y| \le 1$ and $|a + bx + cy + dxy - \frac{3}{1 + x^2 + y^2}| \ge 1$. | [
"Let\n$$\ng(x, y) = a + bx + cy + dxy - \\frac{3}{1 + x^2 + y^2}.\n$$\nThen taking the centre of the square $[-1, 1]^2$ and its four corners, we have\n$$\n\\begin{align*}\ng(0, 0) &= a - 3 \\\\\ng(1, 1) &= a + b + c + d - 1 \\\\\ng(-1, 1) &= a - b + c - d - 1 \\\\\ng(-1, -1) &= a - b - c + d - 1 \\\\\ng(1, -1) &= a... | Ireland | IRL_ABooklet_2020 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
02ty | Problem:
Em uma lousa são escritos os 2014 inteiros positivos de 1 até 2014. A operação permitida é escolher dois números $a$ e $b$, apagá-los e escrever em seus lugares os números $mdc(a, b)$ (Máximo Divisor Comum) e $mmc(a, b)$ (Mínimo Múltiplo Comum). Essa operação pode ser feita com quaisquer dois números que estã... | [
"Solution:\n\nA maior quantidade de números 1 que podemos deixar é 1007. Primeiro vamos mostrar como obtê-los. Para isso, basta tomar os pares de números consecutivos, $(1,2),(3,4),(5,6), \\ldots, (2013,2014)$ e realizar a operação em cada par. Sabendo que números consecutivos não têm fator comum, cada um dos máxim... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | 1007 | |
0ahu | We denote the set of all nonzero integers and the set of all nonnegative integers by $\mathbb{Z}^*$ and $\mathbb{N}_0$, respectively. Find all functions $f:\mathbb{Z}^* \to \mathbb{N}_0$ for which the following two conditions hold:
(1) for each $a,b \in \mathbb{Z}^*$ such that $a+b \in \mathbb{Z}^*$ it holds that $f(a... | [
"One trivial solution is the constant function $f \\equiv 0$. Let $f$ be a nontrivial function for which the conditions (1) and (2) hold. We will show that there exists a natural number $c$ and a prime number $p$ for which it holds that $f(a)=cv_p(a)$ for each $a \\in \\mathbb{Z}^*$, where $v_p(a):=\\text{the expon... | North Macedonia | Team Selection Test for IMO | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | Either the constant zero function, or for some fixed prime p and positive integer c, the function given by f(a) = c times the exponent of p in the prime factorization of a. | |
0fbh | Problem:
Designaremos por $Z_{(5)}$ un cierto subconjunto del conjunto $\mathbb{Q}$ de los números racionales. Un racional pertenece a $Z_{(5)}$ si y sólo si existen fracciones pertenecientes a este racional tales que $5$ no sea divisor de su denominador. (Por ejemplo, el número racional $13/10$ no pertenece a $Z_{(5)... | [
"Solution:\n\nSean $\\alpha \\in \\mathbb{Z}_{(5)}$, con $\\alpha=\\frac{p}{q}$ y $q \\neq 5$; y $\\beta \\in \\mathbb{Z}_{(5)}$ con $\\beta=\\frac{p'}{q'}$ y $q' \\neq 5$.\n\nEntonces:\n\n$\\alpha-\\beta=\\frac{p}{q}-\\frac{p'}{q'}=\\frac{p q'-q p'}{q q'}$ siendo $q q' \\neq 5$. Luego $\\mathbb{Z}_{(5)}$ es subgru... | Spain | OME 11 | [
"Algebra > Abstract Algebra > Group Theory",
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Abstract Algebra > Field Theory"
] | null | proof only | null | |
02ge | For a given integer $a_0 > 1$ one defines a sequence $(a_n)_{n \ge 0}$ as follows: for each $k \ge 0$, define $a_{k+1}$ as the smallest integer greater than $a_k$ such that $\text{gcd}(a_{k+1}, a_0 a_1 \cdots a_k) = 1$. Determine all values $a_0$ for which all terms $a_k$ are primes or powers of primes. | [
"We show first that if (1) $a_n$ is prime, (2) all terms $a_i$ for $i < n$ are primes or prime powers, (3) given any prime $p < a_n$, we have $p \\mid a_i$ for some $i < n$, then $a_i$ is prime for all $i > n$. For suppose $a_n < k < q$, where $q$ is the next larger prime than $a_n$. Then $k$ lies between two conse... | Brazil | XXIII OBM | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > ... | English | proof and answer | a0 ∈ {2, 3, 4, 7, 8} | |
0dxx | Problem:
Stranica $AD$ trapeza $ABCD$ je pravokotna na osnovnici $AB$ in $CD$. Dolžine stranic tega trapeza so: $|AB|=2~\mathrm{cm}$, $|CD|=8~\mathrm{cm}$ in $|AD|=10~\mathrm{cm}$. Izračunaj oddaljenost presečišča diagonal $S$ od osnovnice $AB$. | [
"Solution:\n\nNarišemo ustrezno skico trapeza, kjer je $a=2$, $c=8$ in $d=10$. Na skici označimo presečišče diagonal z $S$, razdaljo presečišča od stranice $a$ z $x$ in razdaljo od stranice $d$ z $y$.\n\nIz podobnih trikotnikov $\\triangle AES \\approx \\triangle ADC$ in $\\triangle FBS \\approx \\triangle ABD$ izr... | Slovenia | 7. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2 cm | |
0lg5 | Problem:
Points $N$, $K$, $L$ lie on the sides $AB$, $BC$, $CA$ of a triangle $ABC$ respectively so that $AL = BK$ and $CN$ is the bisector of the angle $C$. The segments $AK$ and $BL$ meet at the point $P$. Let $I$ and $J$ be the incentres of the triangles $APL$ and $BPK$ respectively. The lines $CN$ and $IJ$ meet at... | [
"Solution:\n\nThe case $CA = CB$ is trivial. If $CA \\neq CB$, we may suppose, without loss of generality, that $CN$ meets the segment $PK$.\n\nLet the circumcircles $\\omega_{1}$ and $\\omega_{2}$ of the triangles $APL$ and $BPK$ respectively meet again at point $T$. Then\n$$\n\\angle LAT = \\angle TPB = \\angle T... | Zhautykov Olympiad | IZhO | [
"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 | |
066r | Let $A = abcd = a \cdot 10^3 + b \cdot 10^2 + c \cdot 10 + d$ be a four digit positive integer such that: $a \ge 7$ and $a > b > c > d > 0$. We consider the positive integer $B = \overline{dcba} = d \cdot 10^3 + c \cdot 10^2 + b \cdot 10 + a$. If all digits of $A + B$ are odd, determine all possible values of $A$. | [
"$$\nA+B = (a+d) \\cdot 10^3 + (b+c) \\cdot 10^2 + (b+c) \\cdot 10 + (a+d).\n$$\nAll digits of $A + B$ are odd. However, in order to find the digits of $A + B$ we must know if the integers $a+d$ and $b+c$ are less than $10$. Hence we have the cases:\n\n(α) Let $a+d \\ge 10$ and $b+c \\ge 10$. Then, because $a > b >... | Greece | Hellenic Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 8721, 8631, 8541, 7632, 7542, 8521, 8431, 7432, 8321 | |
06r1 | Let $a_{1}, \ldots, a_{r}$ be positive real numbers. For $n>r$, we inductively define
$$
a_{n}=\max _{1 \leq k \leq n-1}\left(a_{k}+a_{n-k}\right) \tag{1}
$$
Prove that there exist positive integers $\ell \leq r$ and $N$ such that $a_{n}=a_{n-\ell}+a_{\ell}$ for all $n \geq N$. | [
"First, from the problem conditions we have that each $a_{n}(n>r)$ can be expressed as $a_{n}=a_{j_{1}}+a_{j_{2}}$ with $j_{1}, j_{2}<n, j_{1}+j_{2}=n$. If, say, $j_{1}>r$ then we can proceed in the same way with $a_{j_{1}}$, and so on. Finally, we represent $a_{n}$ in a form\n$$\na_{n}=a_{i_{1}}+\\cdots+a_{i_{k}} ... | IMO | 51st IMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
095e | Problem:
Fie funcția $f: \mathbb{R} \rightarrow \mathbb{R}$, $f(x) = (x - a_{1})^{2} + (x - a_{2})^{2} + \ldots + (x - a_{2018})^{2}$, unde $a_{1}, a_{2}, \ldots, a_{2018}$ sunt numere reale date.
a) Să se arate că $f(x) \geq f\left(\frac{a_{1} + a_{2} + \ldots + a_{2018}}{2018}\right)$, oricare ar fi valorile număru... | [
"Solution:\n\na) Putem scrie $f(x) = 2018 x^{2} - 2(a_{1} + a_{2} + \\ldots + a_{2018}) x + a_{1}^{2} + a_{2}^{2} + \\ldots + a_{2018}^{2}$. Funcția $f$ este de gradul al doilea. Întrucât $2018 > 0$, funcția are un minim egal cu $f\\left(-\\frac{b}{2a}\\right) = f\\left(\\frac{a_{1} + a_{2} + \\ldots + a_{2018}}{20... | Moldova | A 62-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0fax | Problem:
A country contains $n$ cities and some towns. There is at most one road between each pair of towns and at most one road between each town and each city, but all the towns and cities are connected, directly or indirectly. We call a route between a city and a town a gold route if there is no other route between... | [
"Solution:\n\nLet the cities be $C_1$, $C_2$, ..., $C_n$. For each town $T$ take the shortest path from $T$ to a city. If there are shortest paths to more than one city, then take one to the city with the smallest index. We assign $T$ to that city.\n\nNow suppose $T$ is assigned to $C_i$. Let $G$ be a gold route fr... | Soviet Union | 1st CIS | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
08zo | For positive real numbers $x, y$, the positive real number $x \star y$ is defined as $x \star y = \frac{x}{xy+1}$.
Calculate the following expression:
$$
(((\cdots (((100 \star 99) \star 98) \star 97) \star \cdots) \star 3) \star 2) \star 1.
$$ | [
"$\\frac{100}{495001}$\n\nFor any positive real numbers $x, y$, and $z$, the following equation holds:\n$$\n(x \\star y) \\star z = \\frac{x \\star y}{(x \\star y)z + 1} = \\frac{\\frac{x}{xy+1}}{\\frac{x}{xy+1}z + 1} = \\frac{x}{xy + xz + 1} = \\frac{x}{x(y+z) + 1} = x \\star (y+z).\n$$\nUsing this repeatedly, we ... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 100/495001 | |
07rq | Find all real-valued functions $f$ satisfying
$$
f(2x + f(y)) + f(f(y)) = 4x + 8y
$$
for all real numbers $x$ and $y$. | [
"Letting $y = 0$, we see that\n$$\nf(2x + f(0)) = 4x - f(0). \\quad (1)\n$$\nReplacing $x$ by $(x - f(0))/2$ above, we get\n$$\nf(x) = 2x + c, \\quad (2)\n$$\nwhere $c = -2f(0) - f(0)$. Thus,\n$$\nf(f(x)) = 2(2x + c) + c = 4x + 3c. \\quad (3)\n$$\nUsing (2) and (3) in the original functional equation, we get\n$$\n2... | Ireland | Irish | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 2x | |
0eeb | Find all real numbers $x$ and $y$ that solve the system of equations
$$
\log_3 x^2 + \log_2 y^3 = 1,
$$
$$
\log_9 x^4 + \log_4 y^9 = 2.
$$ | [
"Denote $a = \\log_3 x^2$ and $b = \\log_2 y^3$. Then\n$$\n\\log_9 x^4 = \\frac{\\log_3 x^4}{\\log_3 9} = \\frac{2 \\log_3 x^2}{2} = a \\quad \\text{and} \\quad \\log_4 y^9 = \\frac{\\log_2 y^9}{\\log_2 4} = \\frac{3 \\log_2 y^3}{2} = \\frac{3}{2}b.\n$$\nInserting this into the initial equations we get $a+b=1$ and ... | Slovenia | Slovenija 2016 | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | x = ±1/√3, y = 4^(1/3) | |
0diw | Consider function $f : \mathbb{R} \to \mathbb{R}$ such that $x^2 f(x - y^2) \geq (x + 2y + 1) f(x)$ for all $x, y \in \mathbb{R}$. Calculate $f(2022)$. | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | 0 | |
0121 | Problem:
From a sequence of integers $(a, b, c, d)$ each of the sequences
$$(c, d, a, b),(b, a, d, c),(a+n c, b+n d, c, d),(a+n b, b, c+n d, d),$$
for arbitrary integer $n$ can be obtained by one step. Is it possible to obtain $(3,4,5,7)$ from $(1,2,3,4)$ through a sequence of such steps? | [
"Solution:\nAnswer: no.\nUnder all transformations $(a, b, c, d) \\rightarrow (a', b', c', d')$ allowed in the problem we have $|a d - b c| = |a' d' - b' c'|$, but $|1 \\cdot 4 - 2 \\cdot 3| = 2 \\neq 1 = |3 \\cdot 7 - 4 \\cdot 5|$."
] | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Linear Algebra > Determinants"
] | null | proof and answer | no | |
0kwl | Problem:
Philena and Nathan are playing a game. First, Nathan secretly chooses an ordered pair $(x, y)$ of positive integers such that $x \leq 20$ and $y \leq 23$. (Philena knows that Nathan's pair must satisfy $x \leq 20$ and $y \leq 23$.) The game then proceeds in rounds; in every round, Philena chooses an ordered p... | [
"Solution:\n\nIt suffices to show the upper bound and lower bound.\n\nUpper bound. Loosen the restriction on $y$ to $y \\leq 24$. We'll reduce our remaining possibilities by binary search; first, query half the grid to end up with a $10 \\times 24$ rectangle, and then half of that to go down to $5 \\times 24$. Simi... | United States | HMMT February | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 9 | |
0bfj | Let $S$ be the set of rational numbers of the form
$$
\frac{(a_1^2 + a_1 - 1)(a_2^2 + a_2 - 1) \cdots (a_n^2 + a_n - 1)}{(b_1^2 + b_1 - 1)(b_2^2 + b_2 - 1) \cdots (b_n^2 + b_n - 1)},
$$
where $n, a_1, a_2, \dots, a_n, b_1, b_2, \dots, b_n$ run through the positive integers. Show that $S$ contains infinitely many primes... | [
"Clearly, $S$ is closed under multiplication and division: if $r$ and $s$ are members of $S$, so are $rs$ and $r/s$.\n\nIf $a$ is a positive integer, and $p \\neq 5$ is a prime factor of $a^2 + a - 1$, then $p \\equiv \\pm 1 \\pmod 5$. To prove this, notice that $(2a + 1)^2 \\equiv 5 \\pmod p$, so $5$ is a quadrati... | Romania | 64th NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
06lq | Let $S$ be a set of $2020$ distinct points in the plane. Let
$M = \{P : P$ is the midpoint of $XY$ for some distinct points $X, Y$ in $S\}$.
Find the least possible value of the number of points in $M$. | [
"There are at least $4037$ points in $M$.\nSince there are finitely many points in $S$, we can find a coordinate system such that all points in $S$ have distinct $x$-coordinates. Indeed, we just need to find a line which is not perpendicular to any line joining two of the points in $S$.\n\nNow, let the $n = 2020$ p... | Hong Kong | Year 2021 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | 4037 | |
032w | Problem:
The points $D$ and $E$ lie respectively on the perpendicular bisectors of the sides $AB$ and $BC$ of $\triangle ABC$. It is known that $D$ is an interior point for $\triangle ABC$, $E$ does not and $\Varangle ADB = \Varangle CEB$. If the line $AE$ meets the segment $CD$ at a point $O$, prove that the areas of... | [
"Solution:\n\nSince the isosceles triangles $ABD$ and $CBE$ are similar, we have $\\frac{AB}{DB} = \\frac{CB}{BE}$. Using that $\\Varangle ABC = \\Varangle DBE$, we get that $\\triangle ABC \\sim \\triangle DBE$ and $\\frac{DB}{AB} = 2 \\cos \\varphi$, where $\\varphi = \\Varangle ABD$. Hence $\\Varangle ACB = \\Va... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0g0k | Problem:
Seien $a$, $b$ und $c$ die Seiten eines Dreiecks, das heisst: $a + b > c$, $b + c > a$ und $c + a > b$. Zeige, dass gilt:
$$
\frac{a b + 1}{a^{2} + c a + 1} + \frac{b c + 1}{b^{2} + a b + 1} + \frac{c a + 1}{c^{2} + b c + 1} > \frac{3}{2}
$$ | [
"Solution:\n\nJeder der drei Summanden kann wie folgt umgeformt werden:\n$$\n\\frac{a c + 1}{c^{2} + b c + 1} = 1 - \\frac{c(b + c - a)}{c^{2} + b c + 1} > 1 - \\frac{c(b + c - a)}{c^{2} + b c} = \\frac{a}{b + c}\n$$\nDie Ungleichung gilt, da $a$, $b$ und $c$ die Seiten eines Dreiecks sind, also insbesondere positi... | Switzerland | SMO - Finalrunde | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof only | null | |
07bz | Find the maximum number of rectangles with sides equal to $1$ and $2$ and parallel to the coordinate axes such that each two have an area equal to $1$ in common. | [
"The answer is $5$. The picture below shows five $1 \\times 2$ rectangles mutually intersecting at rectangles with unit area.\n\nIt is enough to show that there are no six $1 \\times 2$ horizontal or vertical rectangles with such a property. Assume that there are $6$ such rectangles.\nNote ... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 5 | |
0dga | In each cell of a chessboard (sizes $8 \times 8$) is put a rock. At each step one can remove from the board one rock which beats an odd number of other rocks (for example in initial configuration top-left rock beats 2 rocks). Find the maximal possible number of rocks one can remove from the board. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | 62 | |
0ht1 | Problem:
Let $a$, $b$, $c$ be positive real numbers with $a + b + c = 1$. Prove that
$$
a^{4} + b^{4} + c^{4} \geq a b c .
$$ | [
"Solution:\nSince $a + b + c = 1$, we can multiply the right side by $a + b + c$ to get the equivalent inequality\n$$\na^{4} + b^{4} + c^{4} \\geq a^{2} b c + a b^{2} c + a b c^{2} .\n$$\nBy AM-GM,\n$$\n\\frac{2 a^{4} + b^{4} + c^{4}}{4} = \\frac{a^{4} + a^{4} + b^{4} + c^{4}}{4} \\geq \\sqrt[4]{a^{4} a^{4} b^{4} c... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0983 | Problem:
Fie numărul complex $z$, pentru care $z^{2020}|z|+ar{z} i=0$. Determinați mulțimea valorilor expresiei
$$
E=z^{2020}-z^{2019} i+\cdots+(-1)^{k} z^{2020-k} i^{k}+\cdots+z i+1
$$ | [
"Solution:\n$z^{2020}|z|+\\bar{z} i=0 \\Leftrightarrow z^{2021}|z|+z \\bar{z} i=0 \\Leftrightarrow z^{2021}|z|+|z|^{2} i=0 \\Leftrightarrow$\n$$\n\\Leftrightarrow z^{2021}|z|=-|z|^{2} i \\Rightarrow |z|^{2021}=|z| \\Leftrightarrow \\left[\\begin{array}{l}\n|z|=0 \\\\\n|z|=1\n\\end{array}\\right.\n$$\nPentru $|z|=0$... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | {0, 1, 2021} | |
09yi | Problem:
Gegeven is een driehoek $A B C$ met de eigenschap dat $|A B|+|A C|=3|B C|$. Zij $T$ het punt op lijnstuk $A C$ zodat $|A C|=4|A T|$. Laat $K$ en $L$ punten zijn op het inwendige van respectievelijk lijnstukken $A B$ en $A C$ zodat ten eerste $K L \| B C$ en ten tweede $K L$ raakt aan de ingeschreven cirkel va... | [
"Solution:\n\nDe vierhoek $K B C L$ raakt met alle zijden aan de ingeschreven cirkel van $\\triangle A B C$. Bij elk hoekpunt zitten twee gelijke raaklijnstukjes. Twee overstaande zijden van de vierhoek bestaan samen precies uit de vier verschillende raaklijnstukjes, dus geldt $|K B|+|L C|=|K L|+|B C|$.\n\nVanwege ... | Netherlands | Selectietoets | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscel... | null | proof and answer | 2/3 | |
0702 | Problem:
$n$ and $r$ are positive integers. Find the smallest $k$ for which we can construct $r$ subsets $A_1, A_2, \ldots, A_r$ of $\{0,1,2, \ldots, n-1\}$ each with $k$ elements such that each integer $0 \leq m < n$ can be written as a sum of one element from each of the $r$ subsets. | [
"Solution:\n\nWe can form at most $k^r$ distinct sums, so $k^r$ must be $\\geq n$.\n\nNow consider $A_1 = \\{0, 1, 2, \\ldots, k-1\\}$, $A_2 = \\{0, k, 2k, \\ldots, (k-1)k\\}$, $A_3 = \\{0, k^2, 2k^2, \\ldots, (k-1)k^2\\}, \\ldots, A_r = \\{0, k^{r-1}, 2k^{r-1}, \\ldots, (k-1)k^{r-1}\\}$.\n\nThen for any non-negati... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | ceil(n^(1/r)) | |
0ksb | Problem:
In a plane, equilateral triangle $A B C$, square $B C D E$, and regular dodecagon $D E F G H I J K L M N O$ each have side length $1$ and do not overlap. Find the area of the circumcircle of $\triangle A F N$. | [
"Solution:\n\nNote that $\\angle A C D = \\angle A C B + \\angle B C D = 60^{\\circ} + 90^{\\circ} = 150^{\\circ}$. In a dodecagon, each interior angle is $180^{\\circ} \\cdot \\frac{12-2}{12} = 150^{\\circ}$, meaning that $\\angle F E D = \\angle D O N = 150^{\\circ}$. Since $E F = F D = 1$ and $D O = O N = 1$ (ju... | United States | HMMT November | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | (2 + sqrt(3)) * pi | |
0agw | Let $a_1, a_2, \dots, a_n$ be real numbers such that $a_1 + a_2 + \dots + a_n = 0$. Prove that:
$$
\frac{n}{n-1} \min\{a_1, a_2, \dots, a_n\} \ge \min\{a_1, a_2\} + \min\{a_2, a_3\} + \dots + \min\{a_n, a_1\}.
$$ | [
"Firstly, we will prove one useful statement:\n**Lemma.** If $x_1, x_2, \\dots, x_{n-1} \\in \\mathbb{R}$ are such that $x_1 + x_2 + \\dots + x_{n-1} = 0$, then\n$$\nx_1 + |x_1 - x_2| + |x_2 - x_3| + \\dots + |x_{n-2} - x_{n-1}| + x_{n-1} \\ge 0\n$$\n(1)\n**Proof.** It is clear that if $x_1 + x_{n-1} \\ge 0$ the in... | North Macedonia | IMO Team Selection Test | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
0h1j | Prove that there exists infinitely many squares that can be represented as $2^n + 2^m$, where $n, m$ are distinct natural numbers. | [
"Let $n = m + 3$, then $N = 2^n + 2^m = 2^m(2^3 + 1) = 2^m \\cdot 9$, and we can take $m = 2s$."
] | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0e79 | Let $n > 5$. Prove that a square can be divided into $n$ quadrilaterals having the side ratio of $2 : 3$. The quadrilaterals need not be of the same size. | [
"We shall call a quadrilateral suitable if it has side ratio of $2 : 3$. Suppose that we are able to divide a square into $k$ suitable quadrilaterals for some $k \\in \\mathbb{N}$. Then we are also able to divide it into $k+3$ suitable quadrilaterals by additionally dividing one of the $k$ suitable quadrilaterals i... | Slovenia | Selection Examinations for the IMO 2013 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Quadrilaterals",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0gt3 | The circles $\omega_1$ and $\omega_2$ which do not intersect and which have different sizes are tangent to the line $\ell$ at $K$ and $L$, and are tangent to the circle $\Gamma$ at $M$ and $N$ respectively, such that all three circles lie on the same side of $\ell$. A circle which passes through $K$ and $L$ intersects ... | [
"Denote the tangent lines to $\\Gamma$ at $M$ and $N$ by $m$ and $n$, and let\n$$\n\\alpha := \\angle LKM = \\angle (KM, m), \\\\ \\beta := \\angle (m, MN) = \\angle (MN, n), \\quad \\gamma := \\angle (n, NL) = \\angle NLK.\n$$\nThe internal angles of $KMNL$ add up to $360^\\circ$, thus $\\alpha + \\beta + \\gamma ... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0bya | Fix an integer $n \ge 2$, let $Q_n$ be the graph consisting of all vertices and all edges of an $n$-cube, and let $T$ be a spanning tree in $Q_n$. Show that $Q_n$ has an edge whose adjunction to $T$ produces a simple cycle of length at least $2n$. | [
"For every vertex $v$ of $Q_n$, let $v'$ be the antipodal (opposite) vertex, consider the unique path in $T$ from $v$ to $v'$ and orient its first edge away from $v$. Since $T$ has fewer edges than vertices, some edge, say $xy$, has been assigned two orientations. The (combinatorial) distance between two antipodes ... | Romania | THE Fourteenth IMAR MATHEMATICAL COMPETITION | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0k4x | Problem:
Suppose a real number $x > 1$ satisfies
$$
\log_{2}\left(\log_{4} x\right) + \log_{4}\left(\log_{16} x\right) + \log_{16}\left(\log_{2} x\right) = 0
$$
Compute
$$
\log_{2}\left(\log_{16} x\right) + \log_{16}\left(\log_{4} x\right) + \log_{4}\left(\log_{2} x\right)
$$ | [
"Solution:\nLet $A$ and $B$ be these sums, respectively. Then\n$$\n\\begin{aligned}\nB - A &= \\log_{2}\\left(\\frac{\\log_{16} x}{\\log_{4} x}\\right) + \\log_{4}\\left(\\frac{\\log_{2} x}{\\log_{16} x}\\right) + \\log_{16}\\left(\\frac{\\log_{4} x}{\\log_{2} x}\\right) \\\\\n&= \\log_{2}\\left(\\log_{16} 4\\right... | United States | HMMT February 2018 | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | -1/4 | |
0kwv | Problem:
Suppose $E, I, L, V$ are (not necessarily distinct) nonzero digits in base ten for which
- the four-digit number $\underline{E}\,\underline{V}\,\underline{I}\,\underline{L}$ is divisible by $73$, and
- the four-digit number $\underline{V}\,\underline{I}\,\underline{L}\,\underline{E}$ is divisible by $74$.
Co... | [
"Solution:\n\nLet $\\underline{E} = 2k$ and $\\underline{V}\\,\\underline{I}\\,\\underline{L} = n$. Then $n \\equiv -2000k \\pmod{73}$ and $n \\equiv -k/5 \\pmod{37}$, so $n \\equiv 1650k \\pmod{2701}$. We can now exhaustively list the possible cases for $k$:\n- if $k=1$, then $n \\equiv 1650$ which is not possible... | United States | HMMT February | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization"
] | null | final answer only | 9954 | |
03v8 | Every positive integer is colored by blue or red. Prove that there is a sequence $\{a_n\}$ which has infinite terms, and $a_1 < a_2 < \dots$ are positive integers, such that $a_1$, $\frac{a_1+a_2}{2}$, $a_2$, $\frac{a_2+a_3}{2}$, $a_3$, $\dots$ is a positive integer sequence with the same color. | [
"We need three lemmas. Firstly, define $N^*$ as the set of all the positive integers.\n\nLemma 1 If there is an arithmetic progression having infinite positive integer terms with the same color, then the conclusion holds.\n\nProof: Let $c_1 < c_2 < \\dots < c_n < \\dots$ be a red arithmetic progression of positive ... | China | China National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
014f | Problem:
There are 2006 points marked on the surface of a sphere. Prove that the surface can be cut into 2006 congruent pieces so that each piece contains exactly one of these points inside it. | [
"Solution:\n\nChoose a North Pole and a South Pole so that no two points are on the same parallel and no point coincides with either pole. Draw parallels through each point. Divide each of these parallels into 2006 equal arcs so that no point is the endpoint of any arc. In the sequel, \"to connect two points\" mean... | Baltic Way | Baltic Way | [
"Geometry > Non-Euclidean Geometry > Spherical Geometry",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0803 | Problem:
a. Determinare tutte le coppie $(x, k)$ di interi positivi che soddisfano l'equazione
$$
3^{k}-1=x^{3}.
$$
b. Dimostrare che se $n$ è un intero maggiore di $1$ e diverso da $3$ non esistono coppie $(x, k)$ di interi positivi che soddisfano l'equazione
$$
3^{k}-1=x^{n}.
$$ | [
"Solution:\n\na. Riscriviamo l'equazione nella forma\n$$\n3^{k}=x^{3}+1=(x+1)\\left(x^{2}-x+1\\right)\n$$\nPoiché gli unici divisori di una potenza di $3$ sono a loro volta potenze di $3$, ogni soluzione deve soddisfare il sistema\n$$\n\\left\\{\\begin{array}{l}\nx+1=3^{a} \\\\\nx^{2}-x+1=3^{b}\n\\end{array}\\right... | Italy | XV Gara Nazionale di Matematica | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials ... | null | proof and answer | a) (x, k) = (2, 2). b) For n > 1 with n ≠ 3, there are no solutions. | |
042a | Let $n$ be an integer greater than $1$. Denote the first $n$ primes in increasing order by $p_1, p_2, \dots, p_n$ (i.e., $p_1 = 2, p_2 = 3, \dots$). Let $A = p_1^{p_1} p_2^{p_2} \cdots p_n^{p_n}$. Find all positive integers $x$ such that $\frac{A}{x}$ is even and has exactly $x$ distinct positive divisors. | [
"By $2x \\mid A$, note that $A = 4 \\cdot p_2^{p_2} \\cdots p_n^{p_n}$. We may suppose that $x = 2^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_n^{\\alpha_n}$, where $0 \\le \\alpha_1 \\le 1$, $0 \\le \\alpha_i \\le p_i$ ($i = 2, 3, \\dots, n$). Then, we have\n$$\n\\frac{A}{x} = 2^{2-\\alpha_1} p_2^{p_2-\\alpha_2} \\cdots... | China | China Southeastern Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | the product of the first n primes, p1 p2 ... pn | |
0g6u | 令 $p$ 為一奇質數。對每一個正整數 $a$, 定義 $S_a$ 如下:
$$
S_a = \frac{a}{1} + \frac{a^2}{2} + \cdots + \frac{a^{p-1}}{p-1}.
$$
令 $m$ 與 $n$ 為正整數使得
$$
S_3 + S_4 - 3S_2 = \frac{m}{n}.
$$
試證: $p$ 能整除 $m$. | [
"對於有理數 $p_1/q_1$ 與 $p_2/q_2$ 其中 $p \\nmid q_1, p \\nmid q_2$。若 $p|(p_1q_2 - p_2q_1)$, 記為 $p_1/q_1 \\equiv p_2/q_2 \\pmod p$。\n由 $S_a$ 模 $p$ 開始。觀察: $p|(\\binom{p}{k})$, $k = 1, \\cdots, p-1$ 且\n$$\n\\begin{aligned}\n\\frac{1}{p}\\binom{p}{k} &= \\frac{(p-1)(p-2)\\cdots(p-k+1)}{k!} \\\\\n&\\equiv \\frac{(-1)\\cdot(-2... | Taiwan | 二〇一二數學奧林匹亞競賽第三階段選訓營 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
0adq | Дали може во правоаголен триаголник во кој должините на страните се природни броеви, должините на катетите да се непарни броеви? (Одговорот да се образложи.) | [
"Нека претпоставиме дека должините на катетите се непарни броеви, т.е. $a = 2k + 1$ и $b = 2n + 1$. Тогаш од Питагорина теорема добиваме\n$$\nc^2 = a^2 + b^2 = (2k+1)^2 + (2n+1)^2 = 4(k^2 + n^2 + k + n) + 2\n$$\nПоследното не е можно бидејќи $c$ е природен број а квадратот на природен број е делив со четири или дав... | North Macedonia | Републички натпревар по математика за основно образование | [
"Geometry > Plane Geometry > Triangles",
"Number Theory > Modular Arithmetic"
] | Macedonian, English | proof and answer | No | |
0cqn | A Magician performs the following trick. He lays out 36 cards in a $6 \times 6$ square (6 columns of 6 cards) and asks the Spectator to select (mentally) a card and remember the column containing it. Then the Magician collects the cards, and again lays out the cards in a $6 \times 6$ square. After that the Magician ask... | [
"Пусть Фокусник после первого действия не тасует карты, а собирает их, не нарушая порядок в столбцах, и складывает в колоду один столбец за другим. Второй раз он выкладывает карты построчно, т.е. бывшие столбцы становятся строками (это нетрудно сделать, выкладывая ту же колоду по горизонтали). После ответа Зрителя,... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English; Russian | proof only | null | |
0bg2 | The closure (interior and boundary) of a convex quadrangle is covered by four closed discs centered at each vertex of the quadrangle each. Show that three of these discs cover the closure of the triangle determined by their centers. | [
"Amongst the four discs, choose one, say $\\Delta_0$, containing the point $O$ where the diagonals of the quadrangle cross one another. Let $A_0$ be the center of $\\Delta_0$, label the other three centers in circular order, $A_1, A_2, A_3$, so that the opposite angles $A_0OA_1$ and $A_2OA_3$ be not obtuse, and let... | Romania | The Tenth IMAR Mathematical Competition | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0ftp | Problem:
Gegeben sei ein spitzwinkliges Dreieck $\triangle A B C$ mit Höhen $\overline{A U}$, $\overline{B V}$, $\overline{C W}$ und Höhenschnittpunkt $H$. $X$ liege auf $\overline{A U}$, $Y$ auf $\overline{B V}$ und $Z$ auf $\overline{C W}$. $X$, $Y$ und $Z$ sind alle von $H$ verschieden. Zeige
a. Wenn $X$, $Y$, $Z$... | [
"Solution:\n\n(a) Da das Sehnenviereck $X Y Z H$ insbesondere konvex ist, können wir oBdA annehmen, dass $Z$ auf $\\overline{C H}$ und $X$ auf $\\overline{H U}$ liegt. Wir betrachten zuerst den Fall, dass $Y$ auf $\\overline{B H}$ liegt. Eine kurze Winkeljagd zeigt, dass $\\angle X Y Z=\\beta$ und $\\angle Y Z X=\\... | Switzerland | IMO Selektion | [
"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 | |
0kce | Problem:
Four friends, Anna, Bob, Celia, and David, exchanged some money. For any two of these friends, exactly one gave some money to the other. For example, Celia could have given money to David but then David would not have given money to Celia. In the end, each person broke even (meaning that no one made or lost a... | [
"Solution:\n\na. The answer to (a) is yes. For example, the transactions could be as follows:\n\n\n\nb. The answer to (b) is no. Let's reason by contradiction: assume that it is possible for the four friends to exchange the amounts $\\$ 20, \\$ 30, \\$ 40, \\$ 50, \\$ 60$, and $\\$ 70$. Wit... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a: yes; b: no | |
02v1 | Problem:
João possui uma folha de papel quadriculado com 10 quadrados de comprimento e 7 quadrados de largura. Ele escolheu 30 triângulos com vértices nas interseções das linhas desse papel quadriculado. Explique por que obrigatoriamente existem pelo menos dois triângulos escolhidos com vértices em comum. | [
"Solution:\n\nO papel quadriculado possui 11 linhas horizontais e 8 linhas verticais. Consequentemente, existem $11 \\cdot 8 = 88$ pontos de interseções entre elas. Se não existirem dois triângulos com vértices em comum, precisaremos de pelo menos $3 \\cdot 30 = 90$ vértices distintos. Como $90 > 88$, esse absurdo ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
06s9 | Prove that in any set of $2000$ distinct real numbers there exist two pairs $a > b$ and $c > d$ with $a \neq c$ or $b \neq d$, such that
$$
\left|\frac{a-b}{c-d}-1\right|<\frac{1}{100000}
$$ | [
"For any set $S$ of $n=2000$ distinct real numbers, let $D_{1} \\leqslant D_{2} \\leqslant \\cdots \\leqslant D_{m}$ be the distances between them, displayed with their multiplicities. Here $m = n(n-1)/2$. By rescaling the numbers, we may assume that the smallest distance $D_{1}$ between two elements of $S$ is $D_{... | IMO | International Mathematical Olympiad Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0c0f | Let $n$ be a positive integer, and let $W$ be the set of all words of length $3n$ containing each of the letters $a, b, c$ exactly $n$ times. Prove that for any word $w$ in $W$ there exists a word $w'$ in $W$ that cannot be obtained from $w$ by less than $3n^2/2$ successive transpositions of adjacent letters.
IMO 2017 ... | [
"For convenience, a transposition of two adjacent letters in a word will be referred to as a *swap*. Notice that swapping identical letters does not change a word, so assume that no swaps are performed.\n\nDefine the *distance* $\\text{dist}(w, w')$ of two words $w$ and $w'$ in $W$ to be the minimal number of swaps... | Romania | 69th NMO Selection Tests for BMO and IMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
09sv | Problem:
Vind alle functies $f: \mathbb{R} \rightarrow \mathbb{R}$ die voldoen aan
$$
f(x y-1)+f(x) f(y)=2 x y-1
$$
voor alle $x, y \in \mathbb{R}$. | [
"Solution:\n\nAls $f$ constant is, staat links altijd hetzelfde, terwijl rechts kan variëren, tegenspraak. Dus $f$ is niet constant.\n\nVul nu $x=0$ in:\n$$\nf(-1)+f(0) f(y)=-1,\n$$\ndus $f(0) f(y)$ neemt voor alle $y \\in \\mathbb{R}$ dezelfde waarde aan. Omdat $f$ niet constant is, geeft dit $f(0)=0$. We kunnen d... | Netherlands | IMO-selectietoets | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = x for all real x; f(x) = -x^2 for all real x | |
084k | Problem:
a, b, c sono tre numeri reali positivi tali che $a+b+c=1$. Quale delle seguenti condizioni è equivalente a imporre che $a, b, c$ siano le misure dei lati di un triangolo non degenere?
(A) $0<|b-a|<\frac{1}{2},\ 0<|c-b|<\frac{1}{2},\ 0<|c-a|<\frac{1}{2}$
(B) $a<\frac{1}{2},\ b<\frac{1}{2},\ c<\frac{1}{2}$
(... | [
"Solution:\n\nLa risposta è (B). Infatti, dalla disuguaglianza triangolare abbiamo\n$$\na < b + c \\Longleftrightarrow a + a < a + b + c = 1 \\Longleftrightarrow a < \\frac{1}{2}\n$$\ne le altre due relazioni simmetriche che si ottengono scambiando $a, b$ e $c$ nelle disequazioni qui sopra. Quindi, le seguenti tre ... | Italy | Progetto Olimpiadi di Matematica 2005 GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | B | |
0kwf | Problem:
There are six empty slots corresponding to the digits of a six-digit number. Claire and William take turns rolling a standard six-sided die, with Claire going first. They alternate with each roll until they have each rolled three times. After a player rolls, they place the number from their die roll into a re... | [
"Solution:\n\nA number being divisible by $6$ is equivalent to the following two conditions:\n- the sum of the digits is divisible by $3$\n- the last digit is even\n\nRegardless of Claire and William's strategies, the first condition is satisfied with probability $\\frac{1}{3}$. So Claire simply plays to maximize t... | United States | HMMT November 2023 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 43/192 | |
0978 | Problem:
Fie șirul $\left(a_{n}\right)_{n=0}^{\infty}$, definit prin relațiile $a_{0}=1$, $a_{1}=\frac{1+\sqrt{3}}{2 \sqrt{2}}$ și $a_{n}=2 a_{1} a_{n-1}-a_{n-2}$, $\forall n \geq 2$. Să se calculeze valoarea $a_{2020}$ și să se determine $\lim _{n \rightarrow \infty} \frac{a_{n}}{n^{2}}$. | [
"Solution:\n\nAvem $a_{2}=2 a_{1}^{2}-a_{0}=2\\left(\\frac{1+\\sqrt{3}}{2 \\sqrt{2}}\\right)^{2}-1=\\frac{\\sqrt{3}}{2}$, $a_{3}=2 a_{1} a_{2}-a_{1}=2 \\cdot \\frac{\\sqrt{3}}{2} \\cdot\\left(\\frac{1+\\sqrt{3}}{2 \\sqrt{2}}\\right)-\\frac{1+\\sqrt{3}}{2 \\sqrt{2}}=\\frac{\\sqrt{2}}{2}$, $a_{4}=2 a_{1} a_{3}-a_{2}=... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials"
] | null | proof and answer | a_2020 = 1/2; limit = 0 | |
06m6 | Suppose $a$, $b$ and $c$ are nonzero real numbers satisfying $abc = 2$. Prove that among the three numbers $2a - \frac{1}{b}$, $2b - \frac{1}{c}$ and $2c - \frac{1}{a}$, at most two of them are greater than $2$. | [
"Suppose on the contrary that all three numbers $2a - \\frac{1}{b}$, $2b - \\frac{1}{c}$ and $2c - \\frac{1}{a}$ are greater than $2$. Since $abc = 2$, at least one of $a$, $b$, $c$ is positive. WLOG assume $a > 0$. Then $2c - \\frac{1}{a} > 2$ implies $c > 1 + \\frac{1}{2a} > 0$, and $2b - \\frac{1}{c} > 2$ implie... | Hong Kong | CHKMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
02rc | Consider all matrices of order $4n$ with $4n$ entries equal to $1$, $4n$ entries equal to $-1$ and all remaining entries equal to $0$. Find, in terms of $n$, the maximum of its determinant. | [] | Brazil | Brazilian Math Olympiad | [
"Algebra > Linear Algebra > Determinants",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 4^n | |
0a90 | Problem:
The real numbers $x$, $y$ and $z$ are not all equal and they satisfy
$$
x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} = k
$$
Determine all possible values of $k$. | [
"Solution:\nLet $(x, y, z)$ be a solution of the system of equations. Since\n$$\nx = k - \\frac{1}{y} = \\frac{k y - 1}{y} \\quad \\text{and} \\quad z = \\frac{1}{k - y}\n$$\nthe equation\n$$\n\\frac{1}{k - y} + \\frac{y}{k y - 1} = k\n$$\ncan be simplified into\n$$\n\\left(1 - k^{2}\\right)\\left(y^{2} - k y + 1\\... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 20 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | k = 1 or k = -1 | |
0b57 | Problem:
O mulţime finită de drepte, oricare trei neconcurente, partiţionează planul într-un număr de regiuni. Două regiuni se numesc "vecine" dacă frontierele lor au în comun mai mult decât un segment nedegenerat.
În fiecare regiune trebuie scris un număr întreg astfel încât:
(i) produsul numerelor scrise în oricare... | [] | Romania | BMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof only | null | |
085r | Problem:
Uno studente universitario ha superato un certo numero di esami, riportando la media di 23. Dopo aver superato un altro esame, la sua media scende a 22,25. Sapendo che il voto di ciascun esame è un numero intero compreso fra 18 e 30 inclusi, che voto ha riportato lo studente all'ultimo esame?
(A) 18
(B) 19
(... | [
"Solution:\n\nLa risposta è (C). Sia $n$ il numero degli esami superati dallo studente con la media di 23. Siano poi $S$ la somma dei voti riportati dallo studente prima dell'ultimo esame e $x$ il voto riportato nell'ultimo esame. Abbiamo:\n$$\n\\left\\{\\begin{array}{l}\nS=23 n \\\\\nS+x=22.25(n+1)\n\\end{array}\\... | Italy | Olimpiadi di Matematica | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | C | |
0frv | Halla todas las cuaternas $(a, b, c, d)$ de números enteros positivos que cumplen que
$$
a^2 + b^2 = c^2 + d^2
$$
y de manera que $ac + bd$ es divisor de $a^2 + b^2$. | [
"Usando la identidad\n$$\n(ac + bd)^2 + (ad - bc)^2 = (a^2 + b^2)(c^2 + d^2) = (a^2 + b^2)^2,\n$$\nobservamos que si $k$ es el valor (entero positivo) de $\\frac{a^2+b^2}{ac+bd}$, entonces\n$$\n(ad - bc)^2 = (k^2 - 1)(ac + bd)^2,\n$$\ny por tanto\n$$\n\\frac{ad - bc}{ac + bd} = \\pm\\sqrt{k^2 - 1}.\n$$\n\nEs conoci... | Spain | LIX Olimpiada Matemática Española | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | Spanish | proof and answer | All quadruples of the form (a, b, a, b) with positive integers a, b. | |
0k9p | Problem:
To play the lottery game Sum Thing, you choose five distinct numbers from $1$ to $50$, then the lottery master chooses five distinct numbers from $1$ to $50$. If there exist a nonempty subset of your five numbers and a nonempty subset of the lottery master's five numbers such that both subsets have the same su... | [
"Solution:\nYes, there is such a set. One such set is $\\{4,8,16,32,42\\}$.\nWith this ticket, the subsums are all multiples of $4$ from $4$ to $60$, and all integers congruent to $2 \\pmod{4}$ from $42$ to $102$. So we must show that any five numbers chosen by the lottery master will have a subsum equal to one of ... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | {4, 8, 16, 32, 42} | |
0kyr | Problem:
Let $S=\{1,2,3, \ldots, 64\}$. Compute the number of ways to partition $S$ into 16 arithmetic sequences such that each arithmetic sequence has length 4 and common difference 1, 4, or 16. | [
"Solution:\nThe key observation is the following:\nClaim 1. No partition can contain all three common differences.\nProof. Indeed, suppose the sequences $x, x+16, x+32, x+48$ and $y, y+4, y+8, y+12$ are both present for some $x$ and $y$ in $S$. Without loss of generality, assume $y \\leq 26$; otherwise, we can take... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | proof and answer | 203 | |
07v8 | Suppose $a$, $b$, $c$ are real numbers such that $a + b + c = 1$. Prove that
$$
a^3 + b^3 + c^3 + 3(1-a)(1-b)(1-c) = 1.
$$ | [
"$$\n\\begin{align*} \na^3 + b^3 + c^3 - 3abc &= (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \\\n&= a^2 + b^2 + c^2 - ab - bc - ca \\\n&= (a+b+c)^2 - 3(ab+bc+ca) \\\n&= 1 - 3(ab+bc+ca), \n\\end{align*}\n$$\nand\n$$\n\\begin{align*} \n(1-a)(1-b)(1-c) &= 1 - (a+b+c) + (ab+bc+ca) - abc \\\n&= ab + bc + ca - abc. \n\\end{align*}\n$$... | Ireland | IRL_ABooklet | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
06xa | Let $A A^{\prime} B C C^{\prime} B^{\prime}$ be a convex cyclic hexagon such that $A C$ is tangent to the incircle of the triangle $A^{\prime} B^{\prime} C^{\prime}$, and $A^{\prime} C^{\prime}$ is tangent to the incircle of the triangle $A B C$. Let the lines $A B$ and $A^{\prime} B^{\prime}$ meet at $X$ and let the l... | [
"Denote by $\\omega$ and $\\omega^{\\prime}$ the incircles of $\\triangle A B C$ and $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ and let $I$ and $I^{\\prime}$ be the centres of these circles. Let $N$ and $N^{\\prime}$ be the second intersections of $B I$ and $B^{\\prime} I^{\\prime}$ with $\\Omega$, the circum... | IMO | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cycl... | English | proof only | null | |
04r8 | Each of $n$ Robin Hoods ($n \ge 3$) robbed some coins. Together they have earned $100n$ coins. They have decided to cut the loot in a following way: in one step one Robin can take his two coins and give to some other two Hoods, one coin each. Find all positive integers $n \ge 3$ for which they can split the loot in equ... | [
"Let $z_i$ denote through the process the number of coins of the $i$-th Robin Hood.\n\nLet $n = 3$. After any step, $z_1 - z_2$ modulo $3$ does not change. That means for $z_1 = 101$, $z_2 = 100$, and $z_3 = 99$ (out of many choices), $z_1$ will never be the same as $z_2$. Thus $n = 3$ is not a solution.\n\nNow we ... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | all integers n ≥ 4 | |
095o | Problem:
Demonstrați, că ecuația
$$
[x]+[3x]+[9x]+[27x]+[81x]=19480
$$
nu are soluții în $\mathbb{R}$ (prin $[\alpha]$ s-a notat partea întreagă a numărului real $\alpha$). | [
"Solution:\nLema. Pentru orice $k \\in \\mathbb{N}$ este adevărată inegalitatea:\n$$\nk \\cdot [x] \\leq k \\cdot x \\leq k \\cdot [x] + (k-1)\n$$\nDemonstrație. Notăm\n$$\n[x] = n, \\quad \\{x\\} = \\alpha,\n$$\natunci\n$$\nx = n + \\alpha, \\quad n \\leq x, \\quad 0 \\leq \\alpha < 1, \\quad n \\in \\mathbb{N}\n$... | Moldova | A 62 - A OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0hw3 | Problem:
Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be a function such that $f(0)=2$ and for all integers $x$,
$$
f(x+1)+f(x-1)=f(x) f(1) .
$$
Prove that for all integers $x$ and $y$,
$$
f(x+y)+f(x-y)=f(x) f(y) .
$$ | [
"Solution:\nLet us first prove the result for nonnegative $y$ by strong induction on $y$. If $y=0$, (2) becomes\n$$\nf(x)+f(x)=f(x) f(0),\n$$\nwhich is true since $f(0)=2$, and if $y=1$, (2) is the same as (1). Let us assume (2) for $y=z-1$ and $y=z$ and try to prove it for $y=z+1$. We have\n$$\n\\begin{gathered}\n... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
07kg | Find the number of zeros in which the decimal expansion of the integer $2007!$ ends. Also find its last non-zero digit. | [
"For each positive integer $k$, we let $F(k)$ denote the number of positive integers less than or equal to $2007$ which are divisible by $k$. Clearly, $F(k) = \\lfloor \\frac{2007}{k} \\rfloor$, the integer part of the rational number $\\frac{2007}{k}$.\nBecause $5^5 > 2007$, the exponent of $5$ in the decimal expa... | Ireland | Irish Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Other"
] | null | proof and answer | 500 trailing zeros; last non-zero digit is 2 | |
0exb | Problem:
A tourist arrives in Moscow by train and wanders randomly through the streets on foot. After supper he decides to return to the station along sections of street that he has traversed an odd number of times. Prove that this is always possible. [In other words, given a path over a graph from $A$ to $B$, find a p... | [
"Solution:\nDisregard all edges except those used in the path from $A$ to $B$, and for each of those let the multiplicity be the number of times it was traversed. Let the degree of a vertex be the sum of the multiplicities of its edges. The key is to notice that the degree of every vertex except $A$ and $B$ must be... | Soviet Union | 5th ASU | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
01d3 | Does there exist a hexagon with side lengths $1$, $2$, $3$, $4$, $5$, $6$ (not necessarily in this order) that can be tiled with
a) $31$
b) $32$
equilateral triangles with side length $1$? | [
"a) Yes, for example, see Figure 1.\n\nFigure 1:\n\nb) No, the number of triangles cannot be an even number. Denote the number of triangles by $x$. Then in total there are $3x$ sides. Some sides touch each other: let there be $n$ such places. Others form the perimeter of the hexagon, whose ... | Baltic Way | Baltic Way 2016 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) Yes. b) No. | |
07xk | We say that a pair $(a, b)$ of positive integers is a *link* if no proper factor of $a + b$ exceeds $\min\{a, b\}$.
Show that for any positive integers $u$ and $v$ there is a positive integer $n$ and a chain $c_0, c_1, \dots, c_n$ of positive integers such that
* $c_0 = u$
* $c_n = v$
* Each pair $(c_{k-1}, c_k)$ for $... | [
"For every positive integer $r$, the pair $(r, r+1)$ is a link. This follows because the sum $2r + 1$ of the pair is an odd number, which implies that any proper factor is at least $3$. Therefore, any proper factor of $2r+1$ is at most $(2r+1)/3$. As $2r+1 \\le 2r+1 + (r-1) = 3r$ we deduce that any proper factor of... | Ireland | IRL_ABooklet_2025 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
04c8 | Calculate the sum
$$
\sum_{n=1}^{2012} \operatorname{tg} n \operatorname{tg} (n + 1).
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | cot(1)·(tan(2013) − tan(1)) − 2012 | |
0d89 | Prove that there are infinitely many positive integer $n$ such that $n!$ is divisible by $n^{3}-1$. | [
"We have $n^{3}-1=(n-1)\\left(n^{2}+n+1\\right)$. We shall prove that there are infinitely many positive integers $n$ such that $n^{2}+n+1$ can be represented as the product of some factors less than $n-1$ and then deduce that $n^{3}-1$ divides $n!$. In fact, we first choose $n=m^{2}$ for some positive integer $m$,... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
07tx | Suppose $a$, $b$, $c$ are the side lengths of a triangle $ABC$ with area $S$, and semi-perimeter $s$. Prove that
$$
s < \sqrt{a^2 + b^2 - 4S} + \sqrt{b^2 + c^2 - 4S} + \sqrt{c^2 + a^2 - 4S}.
$$ | [
"Since $2S = ab \\sin C$, we can write\n$$\n\\begin{aligned}\na^2 + b^2 - 4S &= a^2 + b^2 - 2ab \\sin C = a^2 + (b - a \\sin C)^2 - a^2 \\sin^2 C \\\\\n&= a^2 \\cos^2 C + (b - a \\sin C)^2 \\geq a^2 \\cos^2 C\n\\end{aligned}\n$$\nand similarly\n$$\n\\begin{aligned}\na^2 + b^2 - 4S &= b^2 + (a - b \\sin C)^2 - b^2 \... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
044e | Given positive integers $n$, $r$ and distinct prime numbers $p_1$, $p_2$, $\dots$, $p_r$. Initially, there are $(n+1)^r$ numbers on the blackboard: $p_1^{i_1} p_2^{i_2} \cdots p_r^{i_r}$ ($0 \le i_1, i_2, \dots, i_r \le n$). Alice and Bob take turns (Alice goes first) to make the following moves, until only one number ... | [
"The least $M$ is $(p_1 \\cdots p_r)^{\\lfloor \\frac{n}{2} \\rfloor}$.\n\nDenote $N = (p_1 \\cdots p_r)^n$, $M = (p_1 \\cdots p_r)^{\\lfloor \\frac{n}{2} \\rfloor}$. For each divisor $a$ of $N$, call $\\left(a, \\frac{N}{a}\\right)$ a “pair”. Evidently, $\\gcd(a, \\frac{N}{a})$ is a divisor of $M$, while $\\text{l... | China | China National Team Selection Test | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | (p_1 \cdots p_r)^{\lfloor \frac{n}{2} \rfloor} | |
01kj | Is it possible to mark $6$ distinct points ($1$ red, $2$ blue, and $3$ green points) on a plane so that the sum of the distances between the red point and the blue points is $8$, the sum of the distances between the red point and the green points is $6$, and the sum of the distances between the blue points and the gree... | [
"Suppose, contrary to our claim, that there exist points satisfying the problem condition. We label the red point, two blue points, and three green points $R$, $B_1$, $B_2$, and $G_1$, $G_2$, $G_3$ respectively. By triangle inequality,\n$$\nB_1G_1 + RG_1 \\ge RB_1, \\quad B_2G_1 + RG_1 \\ge RB_2, \\quad B_1G_2 + RG... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | No | |
0d44 | Show that it is possible to write a $n \times n$ array of non-negative numbers (not necessarily distinct) such that the sums of entries on each row and each column are pairwise distinct perfect squares. | [
"First solution suggested by the student Salman Saleh. We will construct an $n \\times n$ array by induction on $n \\geq 2$.\nFor $n=2$, consider the following example:\n| 0 | 9 |\n| :---: | :---: |\n| 0 | 16 |\nAssume there exists an $n \\times n$ array\n| $a_{1,1}$ | $a_{1,2}$ | $\\ldots$ | $a_{1, n}$ |\n| :---: ... | Saudi Arabia | SAMC | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Other",
"Discrete Mathematics > Other"
] | English, Arabic | proof only | null | |
0j0y | In acute triangle $ABC$, denote by $h_a$, $h_b$, $h_c$ the lengths of the altitudes to bases $BC$, $CA$, $AB$, respectively. Point $P$ lies inside the triangle. Prove that
$$
\frac{PA}{h_b + h_c} + \frac{PB}{h_c + h_a} + \frac{PC}{h_a + h_b} \ge 1.
$$ | [
"**Solution 1.** We begin with a key lemma, for which we provide two proofs.\n**Lemma 1.** Let $a = BC$, $b = CA$, $c = AB$, and let $p_a$, $p_b$, $p_c$ denote the distances from $P$ to sides $BC$, $CA$, $AB$, respectively. We have\n$$\n2a \\cdot AP \\geq (b+c)(p_b+p_c). \\qquad (1)\n$$\n\n*First proof of Lemma 1.*... | United States | Team Selection Test 2010 | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellane... | null | proof only | null | |
03k1 | Problem:
Let $n$ be a fixed positive integer. Find the sum of all positive integers with the following property: In base $2$, it has exactly $2n$ digits consisting of $n$ $1$'s and $n$ $0$'s. (The first digit cannot be $0$.) | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | C(2n-1, n-1) * [ 2^(2n-1) + ((n-1)/(2n-1)) * (2^(2n-1) - 1 ) ] | |
0ijd | Problem:
$A B C$ is an acute triangle with incircle $\omega$. $\omega$ is tangent to sides $\overline{B C}$, $\overline{C A}$, and $\overline{A B}$ at $D$, $E$, and $F$ respectively. $P$ is a point on the altitude from $A$ such that $\Gamma$, the circle with diameter $\overline{A P}$, is tangent to $\omega$. $\Gamma$ ... | [
"Solution:\n\nBy the Law of Sines we have $\\sin \\angle A = \\frac{X Y}{A P} = \\frac{4}{5}$. Let $I$, $T$, and $Q$ denote the center of $\\omega$, the point of tangency between $\\omega$ and $\\Gamma$, and the center of $\\Gamma$ respectively. Since we are told $A B C$ is acute, we can compute $\\tan \\frac{\\ang... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety"... | null | final answer only | 675/4 | |
0dj2 | Let $n \le 100$ be a positive integer. There are 101 numbers written in a row:
$$
0 \cdot n \text{ mod } 101, 1 \cdot n \text{ mod } 101, \dots, 100 \cdot n \text{ mod } 101.
$$
How many pairs of neighbouring numbers are there in this row such that the one on the left is bigger than the one on the right? | [
"The answer is $n-1$.\n\nIndeed, the main claim is that $an \\bmod 101$ is larger than $(a+1)n \\bmod 101$ if and only if there is a number divisible by $101$ between $an$ and $(a+1)n$. Denote by $f(x)$ the remainder of $x$ when divided by $101$. Note that $f(an) \\ne f((a+1)n)$, otherwise $101 \\mid n$, a contradi... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | n-1 | |
0hxr | Problem:
In the fourth annual Swirled Series, the Oakland Alphas are playing the San Francisco Gammas. The first game is played in San Francisco and succeeding games alternate in location. San Francisco has a $50\%$ chance of winning their home games, while Oakland has a probability of $60\%$ of winning at home. Norma... | [
"Solution:\n\nLet $F(x)$ be the probability that the Gammas will win the series if they are ahead by $x$ games and are about to play in San Francisco, and let $A(x)$ be the probability that the Gammas will win the series if they are ahead by $x$ games and are about to play in Oakland. Then we have\n$$\n\\begin{gath... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 34/73 | |
0kjj | A quadratic polynomial $p(x)$ with real coefficients and leading coefficient $1$ is called *disrespectful* if the equation $p(p(x)) = 0$ is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial $\tilde{p}(x)$ for which the sum of the roots is maxim... | [] | United States | AMC 12 A | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | A | |
0em7 | Consider all the subsets of $\{1, 2, \ldots, N\}$ that do not contain two consecutive numbers. For each subset, calculate the product of the members of the set. What is the sum of the squares of these products?
Note: the product of the elements of an empty set is one. | [
"Let $P(N)$ be the sum of squares of the product of these subsets. For $N = 0$ there is only the empty subset, so $P(0) = 1$. For $N = 1$, we add the subset $\\{1\\}$, so $P(1) = 2$. For $N = 2$, we add an extra set $\\{2\\}$, so $P(2) = 6$. Evaluating $P(3) = 24$, one can guess that $P(N) = (N+1)!$. We show this b... | South Africa | South-Afrika 2011-2013 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | (N + 1)! | |
0445 | For a positive integer $n$, let $\varphi(n)$ represent the number of positive integers not exceeding $n$ and relatively prime to $n$. Find all functions $f:\mathbb{N}_+ \to \mathbb{N}_+$ satisfying that for any positive integers $m, n$ with $m \ge n$,
$$f(m\varphi(n^3)) = f(m)\varphi(n^3).$$ | [
"The desired functions have the form\n$$\nf(n) = \\begin{cases} c, & n = 1, \\\\ dn, & n \\ge 2, \\end{cases}\n$$\nwhere $c, d$ are positive integers. It is straightforward to check that the problem conditions are met.\nIn the governing equation, taking $n = 2$ leads to $f(4m) = 4f(m)$ for $m \\ge 2$. By iterations... | China | China National Team Selection Test | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All functions with f(1)=c for any positive integer c and f(n)=d n for all n≥2, where d is a fixed positive integer. | |
0k8i | Problem:
Let $\mathbb{N}$ be the set of positive integers, and let $f: \mathbb{N} \rightarrow \mathbb{N}$ be a function satisfying
- $f(1)=1$;
- for $n \in \mathbb{N}$, $f(2n)=2f(n)$ and $f(2n+1)=2f(n)-1$.
Determine the sum of all positive integer solutions to $f(x)=19$ that do not exceed $2019$. | [
"Solution:\nFor $n=2^{a_{0}}+2^{a_{1}}+\\cdots+2^{a_{k}}$ where $a_{0}>a_{1}>\\cdots>a_{k}$, we can show that $f(n)=2^{a_{0}}-2^{a_{1}}-\\cdots-2^{a_{k}}=2^{a_{0}+1}-n$ by induction: the base case $f(1)=1$ clearly holds; for the inductive step, when $n$ is even\nwe note that $f(n)=2 f\\left(\\frac{n}{2}\\right)=2\\... | United States | HMMT February 2019 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 1889 | |
0720 | Problem:
Suppose the $n^{2}$ numbers $1,2,3, \ldots, n^{2}$ are arranged to form an $n$ by $n$ array consisting of $n$ rows and $n$ columns such that the numbers in each row (from left to right) and each column (from top to bottom) are in increasing order. Denote by $a_{j k}$ the number in $j$-th row and $k$-th column... | [
"Solution:\n\nSince $a_{j j}$ has to exceed all the numbers in the top left $j \\times j$ submatrix (excluding itself), and since there are $j^{2}-1$ entries, we must have $a_{j j} \\geq j^{2}$. Similarly, $a_{j j}$ must not exceed each of the numbers in the bottom right $(n-j+1) \\times (n-j+1)$ submatrix (other t... | India | INMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0k4n | Problem:
At Easter-Egg Academy, each student has two eyes, each of which can be eggshell, cream, or cornsilk. It is known that $30\%$ of the students have at least one eggshell eye, $40\%$ of the students have at least one cream eye, and $50\%$ of the students have at least one cornsilk eye. What percentage of the stud... | [
"Solution:\nAnswer: $80\\%$\n\nFor the purposes of this solution, we abbreviate \"eggshell\" by \"egg\", and \"cornsilk\" by \"corn\". We know that there are only six combinations of eye color possible: egg-cream, egg-corn, egg-egg, cream-corn, cream-cream, corn-corn. If we let the proportions for each of these be ... | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 80% |
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