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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
04fh | Find all positive integers $a$ and $b$ such that
$$
(a^2 + b) \mid (a^2b + a) \quad \text{and} \quad (b^2 - a) \mid (ab^2 + b).
$$ | [
"From the fact that $(a^2 + b) \\mid (a^2b + a)$ we conclude that\n$$\n\\frac{a^2b + a}{a^2 + b} = \\frac{b(a^2 + b) + a - b^2}{a^2 + b} = b - \\frac{b^2 - a}{a^2 + b}\n$$\nis an integer.\nHence $a^2 + b \\le b^2 - a$ or $b^2 - a \\le 0$. In the latter case we would have $a < a^2 + b \\le a - b^2 < a$, which is not... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All pairs (n, n+1) with n a positive integer | |
0hes | Let $\Gamma$ be a half circle with diameter $AB$. Let $C$ be the point on the diameter, and let $D$ and $E$ be the points on the half circle so that $E$ lies between $B$ and $D$. It turns out that $\angle ACD = \angle ECB$. Let $F$ be the point of intersection of the lines tangent to $\Gamma$ at the points $D$ and $E$.... | [
"\nFig. 18\n\nLet $O$ be the center of the half circle $\\Gamma$, then by the properties of the tangent lines $\\angle OEF = 90^\\circ = \\angle FDO$ (Fig. 18). Thus, $O, E, F, D$ are cyclic.\n\nLet $\\angle ACD = \\angle ECB = \\theta$. By the Law of Sines for $\\triangle COD$ we obtain th... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05x3 | Problem:
Soit $G$ un graphe à $n$ sommets. Une arête $e$ de $G$ est un cul-de-sac s'il est possible de partitionner $G$ en deux ensembles $A$ et $B$ tels que :
- Il y a au plus $2023$ arêtes de $G$ ayant une extrémité dans $A$ et une extrémité dans $B$.
- L'arête $e$ a l'une de ses extrémités dans $A$ et l'autre dans ... | [
"Solution:\n\nOn le démontre par récurrence forte sur $n$. Le cas $n=1$ est immédiat. S'il n'existe aucun cul-de-sac dans $G$, la propriété est triviale. Sinon, soit $e$ un cul-de-sac de $G$. On obtient une partition des sommets de $G$ en deux ensembles $A$, $B$ telle qu'il y a au plus $2023$ arêtes entre $A$ et $B... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
04h9 | Each of hundred dots in a $10 \times 10$ grid is coloured red or blue. A dot is called *corner-dot* if it is in the corner of the grid. A dot is called *edge-dot* if it is at the end of some row or column, but it is not a corner-dot. Two dots are *neighbouring* if they are in the same row or column, and there are no ot... | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 37 | |
023b | Problem:
Duas caixas contêm juntas 65 bolas de vários tamanhos. Cada bola é branca, preta, vermelha ou amarela. Cada vez que pegamos cinco bolas da mesma cor, pelo menos duas são do mesmo tamanho.
a. Qual é o número máximo de tipos de bolas que existem nas caixas? Duas bolas são consideradas de tipos distintos quando... | [
"Solution:\n\n(a) Não podem existir cinco bolas da mesma cor e tamanhos diferentes porque cada vez que pegamos cinco bolas da mesma cor, duas devem ser do mesmo tamanho. Assim, existem no máximo quatro tamanhos para cada cor. Logo, existem no máximo $4 \\times 4 = 16$ tipos de bolas.\n\n(b) As duas caixas possuem j... | Brazil | Nível 3 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | a) 16. b) There exist at least three balls in the same box that are of the same type. | |
0078 | Pablo tiene cierta cantidad de rectángulos cuyas áreas suman $3$ y cuyos lados son todos menores o iguales que $1$. Demostrar que con estos rectángulos es posible cubrir un cuadrado de lado $1$ de modo que los lados de los rectángulos sean paralelos a los lados del cuadrado.
**Nota:** Los rectángulos se pueden superpo... | [] | Argentina | XX Olimpiada Matemática del Cono Sur | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Spanish | proof only | null | |
0dge | Given $\triangle ABC$ where $AB < AC$, $M$ is the midpoint of $BC$. The circle $O$ passes through $A$ and is tangent to $BC$ at $B$, intersecting the lines $AM$, $AC$ at $D$, $E$ respectively. Let $CF \parallel BE$, intersecting $BD$ extended at $F$. Let the lines $BC$ and $EF$ intersect at $G$. Show that $AG = DG$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
047c | Let $m > 1$ be an integer. It is known that there exists a prime number in the interval $[2m - \sqrt{m} + 1, 2m]$. Prove that among any $m$ pairwise distinct positive integers $a_1, a_2, \dots, a_m$, there exist two numbers $a_i$ and $a_j$ ($1 \le i, j \le m$) such that
$$
\frac{a_i}{(a_i, a_j)} \ge m,
$$
where $(a_i, ... | [
"*Proof.* Without loss of generality, assume that the greatest common divisor of $a_1, a_2, \\dots, a_m$ is $1$, otherwise divide all $a_1, a_2, \\dots, a_m$ by their greatest common divisor, which does not change the conclusion.\nLet $p$ be a prime number in the interval $[2m - \\sqrt{m} + 1, 2m]$. If $p$ divides ... | China | The 65th IMO China National Team Selection Test | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0ajw | Determine all the sets of six consecutive positive integers such that the product of some two of them, added to the product of some other two of them, is equal to the product of the remaining two numbers. | [
"Exactly two of the six numbers are multiples of $3$ and these two need to be multiplied together, otherwise two of the three terms of the equality are multiples of $3$ but the third one is not.\nLet $n$ and $n+3$ denote these multiples of $3$. Two of the four remaining numbers give remainder $1$ when divided by $3... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | [1,2,3,4,5,6], [2,3,4,5,6,7], [6,7,8,9,10,11] | |
0d6w | Let $n \geq 1$ be a fixed positive integer. We consider all the sets $S$ which consist of sub-sequences of the sequence $0,1,2, \ldots, n$ satisfying the following conditions:
i) If $\left(a_{i}\right)_{i=0}^{k}$ belongs to $S$, then $a_{0}=0, a_{k}=n$ and $a_{i+1}-a_{i} \leq 2$ for all $0 \leq i \leq k-1$.
ii) If $\le... | [
"Firstly, denote by $A_{n}$ the set of all sub-sequences $\\left(a_{i}\\right)_{i=0}^{k}$ of $(0,1,2, \\ldots, n)$ such that:\n- $a_{0}=0, a_{1}=1, a_{k}=n$,\n- $a_{i+1}-a_{i} \\leq 2$ for all $0 \\leq i \\leq k-1$.\nObviously, $A_{n}$ is one of the sets $S$ satisfying Conditions i)-ii).\nIt is also clear that $|A_... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | F_n (the nth Fibonacci number with F_1 = 1, F_2 = 1) | |
0ifd | Problem:
In triangle $ABC$, $AC = 3 AB$. Let $AD$ bisect angle $A$ with $D$ lying on $BC$, and let $E$ be the foot of the perpendicular from $C$ to $AD$. Find $[ABD]/[CDE]$. (Here, $[XYZ]$ denotes the area of triangle $XYZ$.) | [
"Solution:\n\n$1/3$\n\nBy the Angle Bisector Theorem, $DC/DB = AC/AB = 3$. We will show that $AD = DE$. Let $CE$ intersect $AB$ at $F$. Then since $AE$ bisects angle $A$, $AF = AC = 3 AB$, and $EF = EC$. Let $G$ be the midpoint of $BF$. Then $BG = GF$, so $GE \\parallel BC$. But then since $B$ is the midpoint of $A... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 1/3 | |
089f | Problem:
Sia $ABC$ un triangolo isoscele con base $BC$, sia $D$ il punto medio di $AC$. Sapendo che $BCD$ è a sua volta isoscele con base $CD$ e che $BC=2$, quanto misura l'area di $ABC$?
(A) 2
(B) $\sqrt{5}$
(C) $\sqrt{6}$
(D) $\sqrt{7}$
(E) 3. | [
"Solution:\n\nLa risposta è $\\mathbf{( D )}$. Tracciamo l'altezza $BH$ di $BCD$, che sarà anche una mediana poiché il triangolo è isoscele. Allora $CD=2CH$, e $BA=CA=2CD=4CH$, perciò $HA=3CH$ e quindi $BH=\\sqrt{7}CH=\\frac{\\sqrt{7}}{2}CD$ per il teorema di Pitagora. Inoltre $BCD \\sim ABC$ poiché sono entrambi i... | Italy | Italian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | D | |
0izx | Problem:
An ant starts at the origin, facing in the positive $x$-direction. Each second, it moves 1 unit forward, then turns counterclockwise by $\sin^{-1}\left(\frac{3}{5}\right)$ degrees. What is the least upper bound on the distance between the ant and the origin? (The least upper bound is the smallest real number ... | [
"Solution:\n\nAnswer: $\\sqrt{10}$\n\nWe claim that the points the ant visits lie on a circle of radius $\\frac{\\sqrt{10}}{2}$. We show this by saying that the ant stays a constant distance $\\frac{\\sqrt{10}}{2}$ from the point $\\left(\\frac{1}{2}, \\frac{3}{2}\\right)$.\n\nSuppose the ant moves on a plane $P$. ... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geom... | null | proof and answer | sqrt(10) | |
0iz4 | Problem:
If $A = 10^{9} - 987654321$ and $B = \frac{123456789 + 1}{10}$, what is the value of $\sqrt{A B}$? | [
"Solution:\n\nAnswer: $12345679$\n\nBoth $A$ and $B$ equal $12345679$, so $\\sqrt{A B} = 12345679$ as well."
] | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 12345679 | |
0a28 | Find all functions $f: \mathbb{Z}_{>0} \to \mathbb{Z}_{>0}$ such that, for all positive integers $m$, $n$, and $a$, the following holds:
1.
$f(f(m)f(n)) = mn;$
2.
$f(2024a + 1) = 2024a + 1.$ | [
"We first show that $f(f(m)) = m$ for all $m$.\nWe start by substituting $n = 1$. This yields $f(f(m)f(1)) = m$. Then on the one hand $f$ is injective, since if $f(m) = f(n)$ then it immediately follows that $m = f(f(m)f(1)) = f(f(n)f(1)) = n$. On the other hand, $f$ is surjective, since for every $m$ there is an $... | Netherlands | IMO Team Selection Test 2 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | Exactly four functions:
1) The identity function.
2) The function that swaps two and eleven and fixes all other primes.
3) The function that swaps two and twenty-three and fixes all other primes.
4) The function that swaps eleven and twenty-three and fixes all other primes.
In each case the function is multiplicative a... | |
0lbw | Let $A$ be the set of positive integers that can be written in the form $2x^2 + 3y^2$ where $x, y$ are integers with $x^2 + y^2 \neq 0$. Prove that:
a) All numbers from $A$ are not perfect square.
b) The product of an odd number of numbers from $A$ is not a perfect square. | [
"a) Without loss of generality, we can consider this problem for nonnegative integers. We know that a perfect square gives the remainder $0$ or $1$ when divided by $3$, so all numbers from $A$ when divided by $3$ give the remainder $3$ or $2$.\n\nAssume that there exists a pair $(x_0, y_0)$ such that $2x_0^2 + 3y_0... | Vietnam | Vietnamese Mathematical Competitions | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Algebraic Number Theory > Quadratic forms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0d48 | Let $ABCDE$ be a cyclic pentagon such that the diagonals $AC$ and $AD$ intersect $BE$ at $P$ and $Q$, respectively, with $BP \cdot QE = PQ^{2}$. Prove that $BC \cdot DE = CD \cdot PQ$. | [
"Because triangles $PBC$ and $PAE$ are similar, and triangles $QED$ and $QAB$ are similar, we have\n$$\n\\frac{BC}{AE} = \\frac{PB}{PA}, \\quad \\text{and} \\quad \\frac{DE}{BA} = \\frac{EQ}{AQ}.\n$$\n\nTherefore\n$$\n\\begin{equation*}\n\\frac{BC \\cdot DE}{AE \\cdot BA} = \\frac{PB \\cdot... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | English, Arabic | proof only | null | |
0fsg | Problem:
Bestimme alle reellen Lösungen $(x, y, z)$ des Systems
$$
\frac{4 x^{2}}{1+4 x^{2}}=y, \quad \frac{4 y^{2}}{1+4 y^{2}}=z, \quad \frac{4 z^{2}}{1+4 z^{2}}=x
$$ | [
"Solution:\n\nOffensichtlich sind die Tripel $(x, y, z)=(0,0,0)$ und $(x, y, z)=\\left(\\frac{1}{2}, \\frac{1}{2}, \\frac{1}{2}\\right)$ Lösungen. Wir zeigen, dass es die einzigen sind. Setze $f(t)=\\frac{4 t^{2}}{1+4 t^{2}}$. Es gilt $f(t) \\geq 0$ für alle $t \\in \\mathbb{R}$, also sind $x, y, z$ nichtnegativ. A... | Switzerland | IMO - Selektion | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (0, 0, 0) and (1/2, 1/2, 1/2) | |
02d6 | Show that the number of positive integer solutions to
$$
x_1 + 2^3 x_2 + 3^3 x_3 + \cdots + 10^3 x_{10} = 3025 \quad (*)
$$
equals the number of non-negative integer solutions to the equation
$$
y_1 + 2^3 y_2 + 3^3 y_3 + \cdots + 10^3 y_{10} = 0
$$
Hence show that (*) has a unique solution in positive integers and find... | [
"We have $1^3 + 2^3 + \\cdots + 10^3 = 3025$. Now $x_i$ is a positive integer solution to $(*)$ iff $y_i = x_i - 1$ are all non-negative and satisfy $(y_1+1)+2^3(y_2+1)+3^3(y_3+1)+\\cdots+10^3(y_{10}+1) = 3025$ and hence $y_1+2^3y_2+3^3y_3+\\cdots+10^3y_{10} = 0$. But that clearly has the unique solution $y_i = 0$,... | Brazil | I OBM | [
"Number Theory > Diophantine Equations",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | x1 = x2 = ... = x10 = 1 | |
09wh | Lieneke is making bracelets with beads. Each bracelet has six beads: two white, two grey, and two black beads. Some bracelets look different on first sight, but are actually not different: by turning or flipping the first one over, it looks the same as the other one. For example, the following three bracelets are the s... | [
"B) 11"
] | Netherlands | First Round | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | MCQ | B) 11 | |
00c9 | We say a sequence $a_1, a_2, a_3, \dots$ of positive integers is *alagoana* if, for every positive integer $n$, the following two conditions hold simultaneously:
* $a_{n!} = a_1 \cdot a_2 \cdot \dots \cdot a_n$.
* $a_n$ is the $n$th power of a positive integer.
Determine all the sequences that are *alagoanas*.
(Note th... | [
"In order to do this, we will show that if the sequence is *alagoana*, then $a_n$ cannot have any prime factor.\nConsider a prime $p$. For every positive integer $n$, let $\\alpha(n)$ be the exponent of $p$ in the factorization of $a_n$. We will prove that $\\alpha(n) = 0$ for every $n$.\nBy the second condition on... | Argentina | 29° Olimpiada Matemática del Cono Sur | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | The unique sequence is a_n = 1 for all n. | |
06u4 | Find all polynomials $P(x)$ of odd degree $d$ and with integer coefficients satisfying the following property: for each positive integer $n$, there exist $n$ positive integers $x_{1}, x_{2}, \ldots, x_{n}$ such that $\frac{1}{2}<\frac{P\left(x_{i}\right)}{P\left(x_{j}\right)}<2$ and $\frac{P\left(x_{i}\right)}{P\left(x... | [
"Let $P(x)=a_{d} x^{d}+a_{d-1} x^{d-1}+\\cdots+a_{0}$. Consider the substitution $y=d a_{d} x+a_{d-1}$. By defining $Q(y)=P(x)$, we find that $Q$ is a polynomial with rational coefficients without the term $y^{d-1}$. Let $Q(y)=b_{d} y^{d}+b_{d-2} y^{d-2}+b_{d-3} y^{d-3}+\\cdots+b_{0}$ and $B=\\max _{0 \\leqslant i ... | IMO | IMO 2016 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other"
] | English | proof and answer | All such polynomials are P(x) = a (r x + s)^d with integers a, r, s and d odd. | |
032n | Problem:
Prove that if $a, b, c \geq 1$ and $a+b+c=9$, then
$$
\sqrt{a b+b c+c a} \leq \sqrt{a}+\sqrt{b}+\sqrt{c}
$$ | [
"Solution:\nSet $a=\\frac{9 x^{2}}{x^{2}+y^{2}+z^{2}}$, $b=\\frac{9 y^{2}}{x^{2}+y^{2}+z^{2}}$, $c=\\frac{9 z^{2}}{x^{2}+y^{2}+z^{2}}$, where $x, y, z>0$ and $x+y+z=1$. We have to prove that\n$$\nx^{2}+y^{2}+z^{2} \\geq 9\\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\\right)\n$$\nSince $a \\geq 1$, then\n$$\n9 x^{2}=a\... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
02bq | Problem:
Qual dos seguintes triângulos não pode existir?
a) triângulo agudo isósceles
b) triângulo retângulo isósceles
c) triângulo retângulo obtusângulo
d) triângulo retângulo escaleno
e) triângulo escaleno obtusângulo | [
"Solution:\n\nA soma dos três ângulos internos de um triângulo é $180^\\circ$. Logo, se um deles mede $90^\\circ$, a soma dos outros dois é $90^\\circ$, e por isso não podem ser maiores do que $90^\\circ$. Portanto, não existem triângulos retângulos obtusângulos.\n\nOs seguintes exemplos de comprimentos de lados mo... | Brazil | Lista 4 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | MCQ | c | |
0fzh | Problem:
Deux cercles $k_{1}$, $k_{2}$ ayant pour centres $M_{1}$ resp. $M_{2}$ se coupent aux points $A$ et $B$. La tangente à $k_{1}$ par $A$ coupe $k_{2}$ une nouvelle fois au point $P$. De plus, la droite $M_{1} B$ coupe aussi une nouvelle fois $k_{2}$ au point $Q$. Supposons que $Q$ se trouve en dehors de $k_{1}$... | [
"Solution:\n\n$1^{\\text{ère}}$ solution : Posons $\\alpha = \\angle B M_{1} M_{2}$. Comme les triangles $\\triangle M_{1} A M_{2}$ et $\\triangle M_{1} B M_{2}$\n\n\n\nsont égaux, $\\angle A M_{1} M_{2} = \\alpha$. Par le théorème de l'angle tangent,\n$$\n\\angle B A P = \\frac{1}{2} \\ang... | Switzerland | Vorrundenprüfung | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02u8 | Problem:
Os números que são inteiros positivos elevados ao quadrado são chamados quadrados perfeitos, por exemplo, 16 é um quadrado perfeito pois é igual a $4^{2}$. Um fato curioso é que números que são quadrados perfeitos deixam apenas restos 0 ou 1 na divisão por 4. Com isso podemos provar, por exemplo, que 2014 não... | [
"Solution:\n\na) Desenvolvamos o produto notável relativo ao quadrado de um número ímpar $(2k+1)^{2}$:\n$$\n\\begin{aligned}\n(2k+1)^{2} & = 4k^{2} + 4k + 1 \\\\\n& = 4k(k+1) + 1\n\\end{aligned}\n$$\nComo pelo menos um dentre $k$ e $k+1$ é par, segue que $k(k+1)$ é par e que $4k(k+1)$ é múltiplo de $4 \\cdot 2 = 8$... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) Odd perfect squares leave remainder one upon division by eight. b) No, such a sequence of forty five odd integers does not exist. | |
0am7 | Problem:
Let $P$ be the product of all prime numbers less than $90$. Find the largest integer $N$ so that for each $n \in \{2, 3, 4, \ldots, N\}$, the number $P + n$ has a prime factor less than $90$. | [] | Philippines | 18th PMO Area Stage | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 96 | |
0dbo | Let $ABC$ be an acute, non-isosceles triangle. Take two points $D, E$ inside this triangle such that
$$
\begin{aligned}
& \angle DAB = \angle DCB, \quad \angle DAC = \angle DBC; \\
& \angle EAB = \angle EBC, \quad \angle EAC = \angle ECB.
\end{aligned}
$$
Prove that triangle $ADE$ is right. | [
"First, we will show that $D$ is the orthocenter of triangle $ABC$. Denote $A' = AD \\cap BC$, $B' = BD \\cap CA$, $C' = CD \\cap AB$. Since $\\angle DAB = \\angle DCB$, we have $ACA'C'$ is cyclic. Similarly, $ABA'B'$ is also cyclic. So we have\n$$\n\\angle DA'B = \\angle DB'A, \\quad \\angle DA'C = \\angle DC'A \\... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
003u | Diremos que un entero positivo es ganador si se puede escribir como suma de un cuadrado perfecto más un cubo perfecto. Por ejemplo, $33$ es ganador porque $33 = 5^2 + 2^3$.
Gabriel elige dos enteros positivos, $r$ y $s$, y Germán debe hallar $2005$ enteros positivos $n$ tales que para cada $n$, los números $r + n$ y $... | [] | Argentina | Argentina 2006 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Other"
] | Español | proof only | null | |
0b8e | Let $VABCD$ be a regular pyramid, having the square $ABCD$ as basis. Suppose that on the line $AC$ lies a point $M$ such that $VM = MB$ and $(VMB) \perp (VAB)$. Prove that $4AM = 3AC$. | [
"Since $MV = MB = MD$ and $MO \\perp (VBD)$, it follows that $O$ is the circumcenter of the triangle $VBD$. Furthermore, triangle $VBD$ is isosceles and right-angled, implying that the lateral faces of the pyramid are equilateral triangles.\nLet $P$ be the midpoint of the edge $VB$. The angle of the planes $(VAB)$ ... | Romania | Romanian Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
01uk | Given a convex hexagon $H$ with obtuse inner angles and parallel opposite sides.
a) Prove that there exists a pair of the opposite sides of $H$ which possesses the following property: there exists a straight line that is perpendicular to these sides and intersects each of them.
b) Is it true that there exist two pair... | [
"b) it is not true.\n\na) Let $ABCDDEF$ be a hexagon with all obtuse inner angles and parallel opposite sides ($AB \\parallel DE$, $BC \\parallel EF$, $CD \\parallel FA$, see Fig. 1). Consider the greatest side of this hexagon (one of such sides if there are more than one). Let it be the side $AB$. Since the hexago... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | a) Yes. b) No. | |
01pz | $2n$ girls and $2n$ boys take part in a dancing party. It is known that for any two girls the number of the boys who have a dance with exactly one of these two girls is equal to $n$.
Prove that for any two boys the number of the girls who have a dance with exactly one of these two boys is equal to $n$ too. | [
"Number girls and boys by numbers from $1$ to $2n$. For each $i$-th girl we correspond a vector $S_i$ with $j$-th entry equal to $1$ if the $i$-th girl has a dance with the $j$-th boy, and equal to $-1$ otherwise. Then the condition is equivalent to $S_i \\cdot S_j^T = 0$ for every $i \\neq j$, because $S_i$ and $S... | Belarus | BelarusMO 2013_s | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Vectors",
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
00gi | In a triangle $A B C$, points $M$ and $N$ are on sides $A B$ and $A C$, respectively, such that $M B = B C = C N$. Let $R$ and $r$ denote the circumradius and the inradius of the triangle $A B C$, respectively. Express the ratio $M N / B C$ in terms of $R$ and $r$. | [
"Let $\\omega$, $O$ and $I$ be the circumcircle, the circumcenter and the incenter of $A B C$, respectively. Let $D$ be the point of intersection of the line $B I$ and the circle $\\omega$ such that $D \\neq B$. Then $D$ is the midpoint of the arc $A C$. Hence $O D \\perp C N$ and $O D = R$.\n\nWe first show that t... | Asia Pacific Mathematics Olympiad (APMO) | XVII APMO - March, 2005 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geom... | null | proof and answer | sqrt(1 - 2r/R) | |
0irb | Problem:
Let $n > 4$ be a positive integer. Determine the number of ways to walk from $(0,0)$ to $(n, 2)$ using only up and right unit steps such that the path does not meet the lines $y = x$ or $y = x - n + 2$ except at the start and at the end. | [
"Solution:\n\nIt is easy to see that the first two steps and the last two steps must all be right steps. So we need to compute the number of walks from $(2,0)$ to $(n-2,2)$ that do not pass through $(2,2)$ and $(n-2,0)$. There are $\\binom{n-2}{2}$ paths from $(2,0)$ to $(n-2,2)$, and exactly two of them are invali... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 1/2*(n^2 - 5n + 2) | |
0i5d | Problem:
Two circles have radii $13$ and $30$, and their centers are $41$ units apart. The line through the centers of the two circles intersects the smaller circle at two points; let $A$ be the one outside the larger circle. Suppose $B$ is a point on the smaller circle and $C$ a point on the larger circle such that $... | [
"Solution:\n\n$12 \\sqrt{13}$\n\nCall the large circle's center $O_1$. Scale the small circle by a factor of $2$ about $A$; we obtain a new circle whose center $O_2$ is at a distance of $41 - 13 = 28$ from $O_1$, and whose radius is $26$. Also, the dilation sends $B$ to $C$, which thus lies on circles $O_1$ and $O_... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 12 \sqrt{13} | |
0lce | On the plane, let $ABC$ be the acute and non-isosceles triangle. Let $B', C'$ be the symmetric points of $B, C$ with respect to the lines $CA, AB$, respectively. Assume that $BC', B'C$ meet at $A_0$. Let $A_1$ be the circumcenter of triangle $AB'C'$ and $A_2$ be the center of the circle passing through the projections ... | [
"First, we will prove two following lemmas:\n\n**Lemma 1.**\n\nLet $ABC$ be the acute and non-isosceles triangle. The points $M,N$ are the feet of the perpendicular line from $A$ to the exterior angle bisector $B,C$ of triangle $ABC$. Let $P$ be the tangent point of line $BC$ and the excircle $(I)$ with respect to ... | Vietnam | Vietnamese Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"G... | null | proof only | null | |
00i6 | Let $n \geq 3$ be an integer. There are $n$ cells on a circle, and each cell is assigned either $0$ or $1$. There is a rooster on one of these cells, and it repeats the following operations:
- If the rooster is on a cell assigned $0$, it changes the assigned number to $1$ and moves to the next cell counterclockwise.
- ... | [
"Reformulate the problem as a $n$-string of numbers in $\\{0,1\\}$ and a position at which the action described in the problem is performed, and add $1$ or $2$ modulo $n$ to the position according to the action. Say that a lap is complete for each time the position resets to $0$ or $1$. We will prove that the state... | Asia Pacific Mathematics Olympiad (APMO) | APMO 2025 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
0bwh | Let $f: [0, 1] \to \mathbb{R}$ be a continuous function such that $f((1-t)x + ty) \le \max\{f(x), f(y)\}$, for every $t, x, y \in [0, 1]$, and let $m = \inf_{0 \le x \le 1} f(x)$. Show that:
a) the set $\{x \mid 0 \le x \le 1, f(x) = m\}$ is a closed interval;
b) if $f(0) = 0$, $\int_{0}^{1} f(x) \, dx = 0$ and $g: [... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
0dir | Find the least value of $k > 0$ such that
$$
|x - ky| + |y - kz| + |z - kx| \geq 7
$$
for all real numbers $x, y, z$ satisfying $x^2 + y^2 + z^2 = 21$. | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Linear Algebra > Vectors",
"Algebra > Linear Algebra > Matrices"
] | English | proof and answer | 2 | |
06sh | Let $r$ be a positive integer, and let $a_{0}, a_{1}, \ldots$ be an infinite sequence of real numbers. Assume that for all nonnegative integers $m$ and $s$ there exists a positive integer $n \in [m+1, m+r]$ such that
$$
a_{m}+a_{m+1}+\cdots+a_{m+s}=a_{n}+a_{n+1}+\cdots+a_{n+s}
$$
Prove that the sequence is periodic, i.... | [
"For every indices $m \\leqslant n$ we will denote $S(m, n)=a_{m}+a_{m+1}+\\cdots+a_{n-1}$; thus $S(n, n)=0$. Let us start with the following lemma.\n\nLemma. Let $b_{0}, b_{1}, \\ldots$ be an infinite sequence. Assume that for every nonnegative integer $m$ there exists a nonnegative integer $n \\in [m+1, m+r]$ suc... | IMO | International Mathematical Olympiad Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
07y9 | Problem:
Dimostrare che, comunque si prendano 18 numeri interi positivi consecutivi minori o uguali a 2005, ce ne è almeno uno divisibile per la somma delle sue cifre. | [
"Solution:\n\nTra i 18 numeri considerati vi sono due multipli consecutivi di 9. Per il criterio di divisibilità per 9 la somma delle cifre di questi numeri è multipla di 9. Considerando che la somma massima delle cifre di un numero minore di 2005 è 28 (nel caso di 1999), analizziamo due casi.\n\nSe la somma delle ... | Italy | null | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0bsz | Show that every non-empty subset of the set
$$
Y = \{113! + 2, 113! + 3, \dots, 113! + 15\}
$$
has the sum of its elements a composite number. | [
"A non-empty subset of $Y$ has the sum of its elements $S = 113! + s$, where $2 \\leq s \\leq 2 + 3 + \\dots + 15$, whence $2 \\leq s \\leq 119$.\n\nIf $2 \\leq s \\leq 113$, then $S = 113! + s = Ms > s$.\n\nIf $s \\in \\{114, 116, 118\\}$, then $S = M2$.\n\nIf $s = 115$, then $S = M5$.\n\nIf $s = 117$, then $S = M... | Romania | 67th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof only | null | |
040c | Let $f(x) = (x+a)(x+b)$ where $a$, $b$ are given positive real numbers, $n \ge 2$ be a given integer. For non-negative real numbers $x_1, x_2, \dots, x_n$ that satisfy $x_1 + x_2 + \dots + x_n = 1$, find the maximum of $F = \sum_{1 \le i < j \le n} \min\{f(x_i), f(x_j)\}$. | [
"As\n$$\n\\begin{align*}\n\\min\\{f(x_i), f(x_j)\\} &= \\min\\{(x_i + a)(x_i + b), (x_j + a)(x_j + b)\\} \\\\\n&\\le \\sqrt{(x_i + a)(x_i + b)(x_j + a)(x_j + b)} \\\\\n&\\le \\frac{1}{2}((x_i + a)(x_j + b) + (x_i + b)(x_j + a)) \\\\\n&= x_i x_j + \\frac{1}{2}(x_i + x_j)(a + b) + ab,\n\\end{align*}\n$$\nso\n$$\n\\be... | China | China Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | (n-1)/2 * (1/n + a + b + n a b) | |
00wf | Problem:
Prove that, for any real $a_{1}, a_{2}, \ldots, a_{n}$,
$$
\sum_{i, j=1}^{n} \frac{a_{i} a_{j}}{i+j-1} \geq 0
$$ | [
"Solution:\nConsider the polynomial $P(x) = a_{1} + a_{2} x + \\cdots + a_{n} x^{n-1}$. Then $P^{2}(x) = \\sum_{k, l=1}^{n} a_{k} a_{l} x^{k+l-2}$ and\n$$\n\\int_{0}^{1} P^{2}(x) d x = \\sum_{k, l=1}^{n} \\frac{a_{k} a_{l}}{k+l-1}.\n$$"
] | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0l4c | Problem:
On each cell of a $200 \times 200$ grid, we place a car, which faces in one of the four cardinal directions. In a move, one chooses a car that does not have a car immediately in front of it, and slides it one cell forward. If a move would cause a car to exit the grid, the car is removed instead. The cars are p... | [
"Solution:\nLet $n=100$. The answer is $\\frac{1}{2} n\\left(12 n^{2}+3 n-1\\right)=6014950$.\nA construction for an $8 \\times 8$ grid instead (so $n=4$ ):\n\n\n\nLabel the rows and columns from 1 to $2 n$, and let $(r, c)$ denote the cell at row $r$, column $c$. The cars can be cleared in... | United States | HMMT February 2024 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 6014950 | |
0bh6 | A sequence $(a_n)_n$ of positive integers is such that the sequence $\frac{a_n}{n}$ is bounded and $a_n - a_m$ is divisible by $n - m$ for every positive integers $m, n$. Prove that $(a_n)_n$ is an arithmetic sequence. | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | null | proof only | null | |
00f6 | Let $G$ be the centroid of triangle $A B C$ and $M$ be the midpoint of $B C$. Let $X$ be on $A B$ and $Y$ on $A C$ such that the points $X, Y$, and $G$ are collinear and $X Y$ and $B C$ are parallel. Suppose that $X C$ and $G B$ intersect at $Q$ and $Y B$ and $G C$ intersect at $P$. Show that triangle $M P Q$ is simila... | [
"Let $R$ be the midpoint of $A C$; so $B R$ is a median and contains the centroid $G$.\n\n\nIt is well known that $\\frac{A G}{A M}=\\frac{2}{3}$; thus the ratio of the similarity between $A X Y$ and $A B C$ is $\\frac{2}{3}$. Hence $G X=\\frac{1}{2} X Y=\\frac{1}{3} B C$.\nNow look at the ... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1991 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distanc... | null | proof only | null | |
0ig0 | Consider the sequence $a_1, a_2, \dots$ defined by
$$
a_n = 2^n + 3^n + 6^n - 1.
$$
for all positive integers $n$. Determine all positive integers that are relatively prime to every term of the sequence. | [
"The answer is that $1$ is the only such number. It suffices to show that every prime $p$ divides $a_n$ for some positive integer $n$. Note that both $p=2$ and $p=3$ divide $a_2 = 2^2 + 3^2 + 6^2 - 1 = 48$.\n\nNow we assume that $p \\ge 5$. By **Fermat's Little Theorem**, we have $2^{p-1} \\equiv 3^{p-1} \\equiv 6^... | United States | IMO | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 1 | |
0b5a | Problem:
Un grup este format din $n$ turişti. Oricum am alege 3 turişti, există 2 dintre ei care nu se cunosc. Pentru orice partiție a turiştilor în două autobuze putem găsi doi turişti care se cunosc şi sunt în acelaşi autobuz. Demonstrați că în grupul de turişti există un turist care are cel mult $\frac{2}{5} n$ cun... | [] | Romania | 53. Bulgarian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0idd | Problem:
a and b are positive integers. When written in binary, $a$ has 2004 1's, and $b$ has 2005 1's (not necessarily consecutive). What is the smallest number of 1's $a+b$ could possibly have? | [
"Solution:\n\nConsider the following addition:\n$$\n\\begin{array}{r}\n111\\cdots 100\\cdots 01 \\\\\n+ \\\\\n11\\cdots 11\n\\end{array}\n$$\nBy making the blocks of 1's and 0's appropriately long, we can ensure that the addends respectively contain 2004 and 2005 1's. (To be precise, we get $a = 2^{4008} - 2^{2005}... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 1 | |
0dbj | An acute-angled triangle $ABC$ is inscribed in a circle $\omega$. The tangents to the circle $\omega$ at $A$ and $C$ intersect at $F$. The perpendicular bisector of segment $AB$ intersects the line $BC$ at $E$. Prove that the lines $FE$ and $AB$ are parallel. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians"
] | English | proof only | null | |
09xi | A circle of radius $1$ and a square are given, such that the circle is tangent to one side of the square and also two of the vertices of the square lie on the circle. What is the length of a side of the square?
 | [
"$\\dfrac{8}{5}$"
] | Netherlands | Dutch Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | final answer only | 8/5 | |
00mk | Alice and Bob determine a number with 2018 digits in the decimal system by choosing digits from left to right. Alice starts and then they each choose a digit in turn. They have to observe the rule that each digit must differ from the previously chosen digit modulo 3.
Since Bob will make the last move, he bets that he ... | [
"It is well-known that every number is congruent to its sum of digits in the decimal system modulo 3. It is therefore sufficient to consider the digits modulo 3. In particular, it is enough to only consider digits in $\\{1, 2, 3\\}$.\n\nIn the fourth move from the end, Alice makes sure that the sum of the digits is... | Austria | 49th Austrian Mathematical Olympiad, National Competition (Final Round, part 1) | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0kmj | Problem:
In triangle $A B C$, let $M$ be the midpoint of $B C$ and $D$ be a point on segment $A M$. Distinct points $Y$ and $Z$ are chosen on rays $\overrightarrow{C A}$ and $\overrightarrow{B A}$, respectively, such that $\angle D Y C = \angle D C B$ and $\angle D B C = \angle D Z B$. Prove that the circumcircle of $... | [
"Solution:\n\nWe first note that the circumcircles of $D B Z$ and $Y D C$ are tangent to $B C$ from our angle criteria. By power of a point, we obtain that $M$ lies on the radical axis of the two circles and clearly $D$ does as well. Therefore, we find that $A$ lies on the radical axis so $A Y \\cdot A C = A B \\cd... | United States | HMMT Spring 2021 Team Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00cl | Llamaremos números *similares* a los números enteros positivos que tienen exactamente los mismos dígitos. Por ejemplo, $1241$, $2114$, $4211$ son números similares, pero $1424$ no es similar a los anteriores.
Decidir si existen tres números similares de $300$ dígitos cada uno, con sus dígitos distintos de $0$, y tales ... | [] | Argentina | Nacional OMA 2019 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Spanish | proof and answer | Yes. Example: let A = (10^300 − 1) / 7, B = 3·(10^300 − 1) / 7, and C = 4·(10^300 − 1) / 7. Each is a 300-digit number with digits formed by repeating the block 142857, 428571, and 571428 fifty times, respectively; they are similar and satisfy A + B = C. | |
07lr | Suppose that each point of the plane is coloured either black or white. Show that there is a set of three points of the same colour which form the vertices of an equilateral triangle. | [
"Assume, if possible, that the result is false so every equilateral triangle in the plane has two vertices of different colours. Consider a regular hexagon $ABCDEF$ with centre $S$.\n\n\n\nAssume $S$ is coloured black. One of the vertices of the triangle $BDF$ must be black; assume it is $B... | Ireland | Irska | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
00xk | Problem:
$\overline{a_{1} a_{2} a_{3}}$ and $\overline{a_{3} a_{2} a_{1}}$ are two three-digit decimal numbers, with $a_{1}, a_{3}$ being different non-zero digits. The squares of these numbers are five-digit numbers $\overline{b_{1} b_{2} b_{3} b_{4} b_{5}}$ and $\overline{b_{5} b_{4} b_{3} b_{2} b_{1}}$ respectively... | [
"Solution:\n\nAssume $a_{1} > a_{3} > 0$. As the square of $\\overline{a_{1} a_{2} a_{3}}$ must be a five-digit number we have $a_{1} \\leq 3$. Now a straightforward case study shows that $\\overline{a_{1} a_{2} a_{3}}$ can be 301, 311, 201, 211 or 221."
] | Baltic Way | Baltic Way 1993 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 301, 311, 201, 211, 221 | |
01h4 | Let $A$ and $B$ be sets of positive integers with $|A| \ge 2$ and $|B| \ge 2$. Let $S$ be a set consisting of $|A| + |B| - 1$ numbers of the form $ab$ where $a \in A$ and $b \in B$. Prove that there exist pairwise distinct $x, y, z \in S$ such that $x$ is a divisor of $yz$. | [
"We use induction on $k = |A| + |B| - 1$.\nFor $k = 3$ we have $|A| = |B| = 2$. Let $A = \\{x, y\\}$, $B = \\{z, t\\}$. Then $S$ consists of three numbers from the set $\\{xz, yz, xt, yt\\}$. Relabelling the elements of $A$ and $B$ if necessary, we can assume without loss of generality that the missing number is $y... | Baltic Way | Baltic Way 2020 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
0koo | Problem:
A regular dodecagon $P_{1} P_{2} \cdots P_{12}$ is inscribed in a unit circle with center $O$. Let $X$ be the intersection of $P_{1} P_{5}$ and $O P_{2}$, and let $Y$ be the intersection of $P_{1} P_{5}$ and $O P_{4}$. Let $A$ be the area of the region bounded by $X Y, X P_{2}, Y P_{4}$, and minor arc $\wideha... | [
"Solution:\nThe area of sector $O P_{2} P_{4}$ is one sixth the area of the circle because its angle is $60^{\\circ}$. The desired area is just that of the sector subtracted by the area of equilateral triangle $O X Y$.\n\nNote that the altitude of this triangle is the distance from $O$ to $P_{1} P_{5}$, which is $\... | United States | HMMT February | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | 45 | |
085l | Problem:
Lorenza si trova su una pista avente la forma di un poligono regolare con 2007 lati, i cui vertici sono numerati da 1 a 2007 in senso antiorario. Lorenza, partendo dal vertice 6, salta ogni volta 4 vertici e cade sul quinto più avanti (ad esempio, dal 20 salta al 25), ma salta indietro di 2 vertici quando cad... | [
"Solution:\n\nLa risposta è 405. Lorenza parte dal vertice 6; già dopo 2 salti arriva sul vertice $16=2^{4}$, perciò torna al 14. Lorenza a questo punto salterà alternativamente su un vertice dispari (che non può essere potenza di 2) e su uno pari; in particolare tra un vertice pari e il successivo impiegherà due s... | Italy | Olimpiadi di Matematica | [
"Number Theory > Other",
"Discrete Mathematics > Other"
] | null | proof and answer | 405 | |
0ile | Problem:
A 15-inch-long stick has four marks on it, dividing it into five segments of length $1$, $2$, $3$, $4$, and $5$ inches (although not necessarily in that order) to make a "ruler." Here is an example.
Using this ruler, you could measure $8$ inches (between the marks $B$ and $D$) and ... | [
"Solution:\n\nIn order to measure $14$ inches, one mark must be $1$ inch from an end of the ruler. Likewise, in order to measure $13$ inches, there must be another mark that is $2$ inches from an end of the ruler. Without loss of generality, suppose the leftmost mark is $1$ inch from the end, and the rightmost mark... | United States | 9th Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
003n | Sea $n$ un número natural. La sucesión finita $\alpha$ de enteros positivos tiene, entre sus términos, exactamente $n$ números distintos ($\alpha$ puede tener números repetidos). Además, si a uno cualquiera de sus términos se le resta $1$, se obtiene una sucesión que tiene, entre sus términos, al menos $n$ números posi... | [] | Argentina | XVII OLIMPIADA MATEMÁTICA DEL CONO SUR | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | Español | proof and answer | ceil((3/4)·n^2 + n) = {3m^2 + 2m if n=2m; 3m^2 + 5m + 2 if n=2m+1} | |
0gci | 設凸四邊形 $ABCD$ 有內切圓, 其圓心為 $I$。令點 $I_a, I_b, I_c, I_d$ 分別為 $\triangle DAB, \triangle ABC, \triangle BCD$ 與 $\triangle CDA$ 的內心。設 $\triangle AI_bI_d$ 與 $\triangle CI_bI_d$ 的兩外接圓的兩條外公切線交於點 $X$, $\triangle BI_aI_c$ 與 $\triangle DI_aI_c$ 的兩外接圓的兩條外公切線交於點 $Y$。
試證: $\angle XIY$ 為直角。 | [
"Denote by $\\omega_a, \\omega_b, \\omega_c$, and $\\omega_d$ the circles $AI_bI_d, BI_aI_c, CI_bI_d$, and $DI_aI_c$, let their centers be $O_a, O_b, O_c$, and $O_d$, and let their radii be $r_a, r_b, r_c$, and $r_d$, respectively.\n\n**Claim 1.** $I_bI_d \\perp AC$ and $I_aI_c \\perp BD$.\n | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle ... | null | proof only | null | |
06or | To each side $a$ of a convex polygon we assign the maximum area of a triangle contained in the polygon and having $a$ as one of its sides. Show that the sum of the areas assigned to all sides of the polygon is not less than twice the area of the polygon.
(Serbia) | [
"Lemma. Every convex $(2n)$-gon, of area $S$, has a side and a vertex that jointly span a triangle of area not less than $S / n$.\n\nProof. By main diagonals of the $(2n)$-gon we shall mean those which partition the $(2n)$-gon into two polygons with equally many sides. For any side $b$ of the $(2n)$-gon denote by $... | IMO | IMO 2006 Shortlisted Problems | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof only | null | |
0dru | In the triangle $ABC$, $\angle A = 90^\circ$, the bisector of $\angle B$ meets the altitude $AD$ at the point $E$, and the bisector of $\angle CAD$ meets the side $CD$ at $F$. The line through $F$ perpendicular to $BC$ intersects $AC$ at $G$. Prove that $B, E, G$ are collinear. | [
"\n\nFirst $\\triangle ABC \\sim \\triangle DAC$ so that $AC/BA = DC/AD$. Also $\\triangle DAC \\sim \\triangle DBA$. It follows that $\\triangle AFC \\sim \\triangle BEA$ so that $FC/AC = EA/BA$. Thus $FC/EA = AC/BA = DC/AD$. This shows that $EF$ is parallel to $AC$. Hence $AEFG$ is a para... | Singapore | Singapur | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof only | null | |
0e9q | Problem:
Na vrhu $75~\mathrm{m}$ dolgega klanca je oddajnik, ki ga od vznožja vidimo pod kotom $10^\circ$ (glej sliko). Kako visok je oddajnik (njegova višina je na sliki označena s $h$), če je naklon klanca $20^\circ$? Rezultat zaokroži na dve mesti natančno.
 | [] | Slovenia | 14. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 15.04 | |
093j | Problem:
There are $n$ boys and $n$ girls in a school class, where $n$ is a positive integer. The heights of all the children in this class are distinct. Every girl determines the number of boys that are taller than her, subtracts the number of girls that are taller than her, and writes the result on a piece of paper.... | [
"Solution:\n\nWe prove the statement by induction. The case $n=1$ is easy (either both children write down $0$, or both write down $1$).\n\nFor the induction step, suppose that the children are standing in a row in order of height (the tallest first), and consider a boy and a girl standing next to each other (such ... | Middle European Mathematical Olympiad (MEMO) | MEMO Team Competition | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
05bp | The product $abc$ of positive integers $a$, $b$, and $c$ is divisible by $3$, and the equations $a = \frac{b^2}{2} = \frac{c}{4}$ hold. Find the smallest possible sum of the numbers $a$, $b$, and $c$ under these conditions. | [
"The given equations are equivalent to $2a = b^2 = \\frac{c}{2}$. So $b^2$ is divisible by $2$, therefore $b$ is also divisible by $2$. Since the product $abc$ is divisible by $3$, one of the numbers $a$, $b$ and $c$ must be divisible by $3$. If $a$ is divisible by $3$, then $b^2$ is also divisible by $3$, therefor... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 96 | |
0jup | Problem:
Allen and Brian are playing a game in which they roll a 6-sided die until one of them wins. Allen wins if two consecutive rolls are equal and at most 3. Brian wins if two consecutive rolls add up to 7 and the latter is at most 3. What is the probability that Allen wins? | [
"Solution:\n\nNote that at any point in the game after the first roll, the probability that Allen wins depends only on the most recent roll, and not on any rolls before that one. So we may define $p$ as the probability that Allen wins at any point in the game, given that the last roll was a $1, 2$, or $3$, and $q$ ... | United States | HMMT November 2016 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 5/12 | |
0hr5 | Problem:
Let $x$, $y$, $z$, and $u$ be real numbers satisfying the equation
$$
\frac{x-y}{x+y}+\frac{y-z}{y+z}+\frac{z-u}{z+u}+\frac{u-x}{u+x}=0
$$ | [
"Suppose that $x$, $y$, and $z$ are rational (i.e. each is the quotient of two integers) and distinct. Prove that $u$ is rational as well.\n\nLet us begin by combining the fractions in pairs.\n$$\n\\begin{aligned}\n0 & =\\left(\\frac{x-y}{x+y}+\\frac{y-z}{y+z}\\right)+\\left(\\frac{z-u}{z+u}+\\frac{u-x}{u+x}\\right... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0i2i | There are 51 senators in a senate. The senate needs to be divided into $n$ committees so that each senator is on one committee. Each senator hates exactly three other senators. (If senator $A$ hates senator $B$, then senator $B$ does *not* necessarily hate senator $A$.) Find the smallest $n$ such that it is always poss... | [
"The smallest such number is $7$. Assume that there are $7$ senators $A_1, \\dots, A_7$ such that each $A_i$ hates $A_{i+1}$, $A_{i+2}$, and $A_{i+3}$ (where indices are taken modulo $7$). In this situation, for any different $A_i, A_j$, either $A_i$ hates $A_j$ or vice versa. The senators $A_1, \\dots, A_7$ must b... | United States | USA IMO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 7 | |
0j58 | Problem:
Let $f:[0,1] \rightarrow [0,1]$ be a continuous function such that $f(f(x)) = 1$ for all $x \in [0,1]$. Determine the set of possible values of $\int_{0}^{1} f(x) \, dx$. | [
"Solution:\nAnswer: $\\left(\\frac{3}{4}, 1\\right]$\n\nSince the maximum value of $f$ is $1$, $\\int_{0}^{1} f(x) \\, dx \\leq 1$.\n\nBy our condition $f(f(x)) = 1$, $f$ is $1$ at any point within the range of $f$. Clearly, $1$ is in the range of $f$, so $f(1) = 1$. Now $f(x)$ is continuous on a closed interval so... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | (3/4, 1] | |
0fms | Let $ABC$ be a triangle with $\angle ABC = 120^\circ$ and triangle bisectors $(AA_1)$, $(BB_1)$, $(CC_1)$, respectively. $B_1F \perp A_1C_1$, where $F \in (A_1C_1)$. Let $R$, $I$ and $S$ be the centers of circles which are inscribed in triangles $C_1B_1F$, $C_1B_1A_1$, $A_1B_1F$, and $B_1S \cap A_1C_1 = \{Q\}$. Show th... | [
"First, we will show that $\\angle C_1B_1A_1 = 90^\\circ$. Let $K \\in BC$ so that $B \\in (KA_1)$, then $\\angle ABK = 60^\\circ$. Point $C_1$ is on the bisector of $\\angle ACB$ and this implies that $d(C_1, BC) = d(C_1, AC)$ or $C_1F_1 = C_1F_3$, where $F_1$ is the projection of $C_1$ on $BC$ and $F_3$ is the pr... | Spain | International Mathematical Arhimede Contest | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | Spanish | proof only | null | |
0ku8 | Problem:
Let $x_{0} = x_{101} = 0$. The numbers $x_{1}, x_{2}, \ldots, x_{100}$ are chosen at random from the interval $[0,1]$ uniformly and independently. Compute the probability that $2 x_{i} \geq x_{i-1} + x_{i+1}$ for all $i = 1, 2, \ldots, 100$. | [
"Solution:\n\nWe solve for general $n$ where $n = 100$ in the problem. Notice that the points $(i, A_{i})$ must form a convex hull, so there is some unique maximal element $A_{i}$. Consider the $i-1$ points $A_{1}, \\ldots, A_{i-1}$ left of $i$, and the $i$ slopes formed between these points of segments $\\overline... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | binom(200, 99) / (100 * (100!)^2) | |
00ou | Let $n > 2$ be a positive integer. Consider all numbers $S$ of the form
$$
S = a_1 a_2 + a_2 a_3 + \dots + a_{k-1} a_k,
$$
with $k > 1$, and $a_i$ being positive integers such that $a_1 + a_2 + \dots + a_k = n$. Determine all numbers that can be represented in the given form. | [
"Let $\\lfloor x \\rfloor$ be the largest integer less than or equal to $x$ and $\\lceil x \\rceil$ the smallest integer greater than or equal to $x$. The smallest number $S$ that can be represented in the given form is $n-1$, while the largest number is $\\lfloor \\frac{n^2}{4} \\rfloor$.\nSince $a_2 a_3 \\ge a_3$... | Balkan Mathematical Olympiad | BMO 2010 Shortlist | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | All integers S with n - 1 <= S <= floor(n^2/4). | |
0gj0 | 設 $\triangle ABC$ 為銳角三角形，且其外接圓為 $\omega$，圓心 $O$。令點 $D \neq B$ 和 $E \neq C$ 位於 $\omega$ 上，使得 $BD \perp AC$ 和 $CE \perp AB$ 且 $CO$ 與 $AB$ 相交於 $X$，$BO$ 與 $AC$ 相交於 $Y$。證明 $\triangle BXD$ 和 $\triangle CYE$ 的外接圓在直線 $AO$ 上有共同交點。
Let $\triangle ABC$ be an acute-angled triangle with circumcircle $\omega$ and circumcentre $O$. ... | [
"Note that $AO = OC$ implies the lines $AO, XO$ are reflections of each other about the line parallel to $AC$ through $O$, which is the perpendicular bisector of $BD$. Call this line $l$. Let $P \\neq X$ be the second intersection of circle $\\odot BXD$ with line $XO$, and let $Z$ be the intersection of circle $\\o... | Taiwan | IMO 1J, Mock Exam 2 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous ... | Chinese; English | proof only | null | |
0az4 | Problem:
The letters of the word MATHEMATICS are rearranged to form distinct strings of the same 11 letters. What proportion of these strings do not contain the string MATH? | [
"Solution:\n\nBy treating the string MATH as a single character, we find that there are exactly eight distinct \"letters\" to rearrange. Hence, there are $8!$ such words out of a total of $\\frac{11!}{2!2!2!}$. This makes for a proportion of $\\frac{8! \\cdot 8}{11!} = \\frac{4}{495}$, and so $\\frac{491}{495}$ of ... | Philippines | 20th Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 491/495 | |
0d8z | Let $(O)$ be a circle, and $BC$ be a chord of $(O)$ such that $BC$ is not a diameter. Let $A$ be a point on the larger arc $BC$ of $(O)$, $E$ and $F$ be two feet of the perpendiculars from $B$ and $C$ to $AC$, $AB$ respectively.
1. Prove that two tangents of $(AEF)$ at $E$ and $F$ intersect to each other at a fixed po... | [
"Denote $M$ as the midpoint of $BC$. Since $\\angle AEH = \\angle AFH = 90^\\circ$ then $A, H, E, F$ belong to the same circle and the center of this circle is midpoint $N$ of $AH$. It is easy to check that $NE = NH$ and $ME = MB$, so\n$$\n\\angle MEN = \\angle MEB + \\angle NEB = \\angle MBE + \\angle NHE = \\angl... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadril... | English | proof only | null | |
0h9x | Point $P$ was chosen inside the triangle $ABC$ so that $BC = AP$ and $\angle APC = 180^\circ - \angle ABC$. On side $AB$, there exists point $K$ such that $AK = KB + PC$. Prove $\angle AKC = 90^\circ$. | [
"We extend the ray $AB$ further after $B$ to find the point $T$ such that $BT = PC$ (Fig. 1). Then, $\\Delta TBC = \\Delta CPA$ due to two equal length sides and the same angle between them. Hence, $TC = CA$. Analogously,\n$$\nAK = KB + PC = KB + BT = KT.\n$$\nTherefore, in the isosceles triangle $ATC$ segment $KC$... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0j47 | Problem:
Determine the remainder when $1+2+\cdots+2014$ is divided by $2012$. | [
"Solution:\nWe wish to find the value of $1+2+\\cdots+2014$ modulo $2012$. We have\n$$\n1+2+\\cdots+2014=\\frac{1}{2}(2014)(2015)=1007 \\cdot 2015 \\equiv 1007 \\cdot 3=3021 \\equiv 1009 \\pmod{2012}\n$$"
] | United States | Harvard-MIT November Tournament | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 1009 | |
0ipn | Problem:
Let $P(x)$ be a polynomial with degree $2008$ and leading coefficient $1$ such that
$$
P(0)=2007,\ P(1)=2006,\ P(2)=2005,\ \ldots,\ P(2007)=0.
$$
Determine the value of $P(2008)$. You may use factorials in your answer. | [
"Solution:\nAnswer: $2008! - 1$\n\nConsider the polynomial $Q(x) = P(x) + x - 2007$. The given conditions imply that $Q(x)$ is also a polynomial with degree $2008$ and leading coefficient $1$. It follows that\n$$\nQ(x) = x(x-1)(x-2)\\cdots(x-2007).\n$$\nThus,\n$$\nP(x) = x(x-1)(x-2)\\cdots(x-2007) - x + 2007.\n$$\n... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 2008! - 1 | |
001d | Expresar $\frac{1}{2}$ como suma de fracciones todas distintas y todas de la forma $\frac{1}{n^2}$ con $n$ un número natural. | [] | Argentina | XIX Olimpíada Matemática Argentina | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | español | final answer only | 1/2 = ∑_{k=1}^{∞} (1/2^{2k} + 1/3^{2k} + 1/5^{2k}) = (1/4 + 1/16 + 1/64 + ⋯) + (1/9 + 1/81 + 1/729 + ⋯) + (1/25 + 1/625 + 1/15625 + ⋯). | |
0bwu | Fix an integer $n \ge 3$ and let $a_0 = n$. Does there exist a permutation $a_1, a_2, \dots, a_{n-1}$ of the first $n-1$ positive integers such that $\sum_{j=0}^{k-1} a_j$ is divisible by $a_k$ for all indices $k < n$? | [
"The answer is in the affirmative. If $n$ is odd, set $a_1 = 1$ and $a_2 = 2$, and if $n > 3$ define the other $a_k$ recursively by $a_k a_{k-1} = a_0 + a_1 + \\dots + a_{k-1}$, $k = 3, \\dots, n$. It is easily seen that $a_{k+1} = a_{k-1} + 1$, $k = 3, \\dots, n-1$, so $a_{2k} = k+1$ and $a_{2k+1} = k + (n+1)/2$, ... | Romania | THE Fourteenth IMAR MATHEMATICAL COMPETITION | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | Yes | |
0br2 | Solve the equation
$$
\sqrt{2^{x+1}-1} + \sqrt{2^x-1} + 2^x = 2 \cdot (3 \cdot 4^x - 2^{x+1} \cdot \sqrt{2^{2x+1}-1}).
$$ | [] | Romania | 67th NMO Shortlisted Problems | [
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | x = 0 and x = log2(5) | |
0htn | Problem:
One marks 16 points on a circle. What is the maximum number of acute triangles with vertices in these points? | [
"Solution:\n\nConsider the set of all angles $M_{1} M_{2} M_{3}$, where $M_{1}$, $M_{2}$ and $M_{3}$ is an arbitrary triple of selected points. There are $\\frac{16 \\cdot 15 \\cdot 14}{2} = 1680$ different angles in this set. Suppose $n$ of them are not acute. We shall prove $n \\geq 392$.\n\nFor each integer $m$ ... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 168 | |
0c40 | Problem:
Arătaţi că $2^{-x}+2^{-1 / x} \leq 1$, oricare ar fi numărul real $x>0$. | [
"Solution:\n\nFie $f:(0, \\infty) \\rightarrow \\mathbb{R}$, $f(x)=2^{-x}+2^{-1 / x}$. Cum $f(x)=f(1 / x)$, este suficient să arătăm că $f(x) \\leq 1$, oricare ar fi $x \\in(0,1]$.\n\nCum $f$ este derivabilă şi $\\lim_{x \\rightarrow 0} f(x)=1=f(1)$, pentru a demonstra inegalitatea din enunţ, este suficient să arăt... | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
0h7j | Grasshopper is sitting at point $O$ of coordinate line. He makes $2016$ jumps in positive direction. Firstly his jump has length $1$, every next jump is $k \in \mathbb{N}$ times longer than the previous one. There are holes in every point with coordinate $2016l$, $l \in \mathbb{N}$. Determine $k$ for which the grasshop... | [
"$$\na_1 = 1, \\quad a_n = 1 + k + k^2 + \\dots + k^{n-1}, \\quad n = 2, \\ldots, 2016.\n$$\nWe have to find such $k$, for which $a_n$ is not divisible by $2016$.\n\nLet us suggest that $(k, 2016) = d > 1$. As we see, every coordinate after the jump has residue $1$ modulo $d$ thus cannot be divisible by $2016$. So ... | Ukraine | UkraineMO | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | All natural numbers k with gcd(k, 2016) > 1 | |
07s9 | Suppose $a, b, c > 0$ and $\sqrt{a-b} + \sqrt{a-c} > \sqrt{b+c}$. Prove that $a > \frac{3}{4}(b+c)$. | [
"Squaring the given inequality and rearranging gives\n$$\n\\sqrt{\\frac{(a-b)(a-c)}{b+c-a}} > b+c-a.\n$$\nIf $a > b + c$, the above inequality is always true and we also have\n$$\na > b + c > \\frac{3}{4}(b + c).\n$$\n\nIf $a \\le b + c$, then both sides of the previous inequality are non-negative, so squaring give... | Ireland | Irish | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
01t4 | Any diagonal connecting two opposite vertices of a convex hexagon divides this hexagon into two quadrilaterals. Six quadrilaterals can be obtained in this way.
Find the greatest number of these quadrilaterals that can occur circumscribed quadrilaterals.
(S. Mazanik, I. Voronovich) | [
"Answer: 3.\n\nWe show that at most three circumscribed quadrilaterals can be obtained. Suppose, contrary to our claim, that there are at least four circumscribed quadrilaterals. Then some two of them are obtained when one diagonal (say, $AD$) is constructed.\n\n\n\nFig. 1\n\n![](attached_i... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 3 | |
01nu | Find all functions $f, f: \mathbb{Q} \to \mathbb{Q}$, such that
$$
f(x + f(y + f(z))) = y + f(x + z)
$$
for all $x, y, z \in \mathbb{Q}$. | [
"(Solution of S. Dabryneuski, A. Tanana, A. Zhuk.) Set $x = 0$ and $z = 0$ in\n$$\nf(x + f(y + f(z))) = y + f(x + z), \\quad (*)\n$$\n\nthen $f(f(y + f(0))) = y + f(0)$, or\n$$\nf(f(x)) = x, \\quad \\forall x \\in \\mathbb{Q}, \\quad (1)\n$$\nhence $f$ is bijective. Let $f(0) = a$, then $f(a) = 0$. Further, the rig... | Belarus | Belorusija 2012 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x for all rational x; and f(x) = -x + a for any fixed rational a | |
08f1 | Problem:
Determinare per quali interi positivi $n$ esiste un intero positivo $A$ tale che
- $A$ è multiplo di 2022,
- l'espressione decimale di $A$ contiene solo cifre 0 e 7,
- l'espressione decimale di $A$ contiene esattamente $n$ volte la cifra 7. | [
"Solution:\n\nI valori di $n$ richiesti sono tutti e soli i multipli di 3.\n\nCondizione necessaria Sia $A$ un multiplo di 2022 la cui espressione decimale contiene esattamente $n$ volte la cifra 7, ed eventualmente altre cifre 0. Poiché 2022 è multiplo di 3, anche $A$ deve essere multiplo di 3, e quindi (per il cr... | Italy | XXXVIII Olimpiade Italiana di Matematica | [
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | All positive integers divisible by 3 | |
08s8 | $ABC$ is a triangle and $D$, $E$, $F$ are midpoints of $BC$, $CA$, $AB$ respectively. If $AD = 3$, $BE = 4$ and $CF = 5$, what is the area of $ABC$? | [
"Let $|PQR|$ denote the area of triangle $PQR$. $AD$, $BE$, $CF$ cross at a point $G$ and $AG : GD = BG : GE = CG : GF = 2 : 1$. Take a point $C'$ on line $GC$ so that $G$ is the midpoint of $CC'$. From $C'G = GC$ and $CG : GF = 2 : 1$, it follows that $C'F = GF$. Since $AF = BF$, $C'F = GF$ and $\\angle AFC' = \\a... | Japan | Japanese Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | 8 | |
05sg | Problem:
Une grille de dimensions $20 \times 20$ est divisée en 400 cases unité de dimensions $1 \times 1$. Clara colorie chaque case en blanc ou en noir, puis Isabelle découpe la grille en rectangles dont les côtés sont contenus dans la grille. Chacun de ces rectangles doit contenir au plus 2 cases noires, et elle do... | [
"Solution:\n\nNous allons démontrer pour tout entier $n \\geqslant 3$ que, si Clara et Isabelle jouent sur une grille de dimensions $n \\times n$, Isabelle donnera $n$ chocolats à Clara.\n\nTout d'abord, pour limiter le nombre de chocolats qu'elle donnera à Clara, Isabelle peut procéder comme suit. Elle commence pa... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 20 | |
0564 | Parents have $n$ children, where $n$ is a given natural number. Find all possibilities for how many children in this family can have both a brother and a sister. | [
"If all the children have the same gender, then nobody can have a brother as well as a sister. In that case the number of children that match the condition is 0, regardless of $n$.\n\nIf there are children of either gender, but for at least one gender there is exactly one child of that gender, then this child does ... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Logic"
] | English | proof and answer | For up to two children: 0. For three children: 0 or 2. For four or more children: 0, n−1, or n. | |
0hvy | Problem:
Determine whether there exists a $2011 \times 2011$ matrix with the following properties:
- Every cell is filled with an integer from $1$ to $4021$.
- For every integer $i$ ($1 \leq i \leq 2011$), the $i$th row and the $i$th column together contain every integer from $1$ to $4021$. | [
"Solution:\n\nAnswer: no. Fix an integer $k$ from $1$ to $4021$. Let us say that an index $i$ \"hits\" a cell containing the number $k$ if the cell is in either the $i$th row or the $i$th column. The conditions stipulate that each index hits exactly one instance of $k$, so the total number of hits is $4021$. On the... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | no | |
0how | Problem:
A building has the plan of a $5 \times 5$ grid of rooms, each of which has a door in each of its four walls: thus there are 20 doors leading to the outside. The doors are to be opened and closed so that every room has exactly 3 open doors leading from it. Determine the minimum and maximum number of doors to th... | [
"Solution:\nIf we consider the building as possessing 26 \"spaces,\" namely the outside and the 25 rooms, then each door belongs to two spaces; hence the total of the numbers of open doors to each space must be even. Each room has three open doors, an odd number; hence the number of open doors to the outside must a... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | minimum 5, maximum 19 | |
0b3n | Problem:
How many five-digit numbers containing each of the digits $1,2,3,4,5$ exactly once are divisible by $24$?
(a) 8
(b) 10
(c) 12
(d) 20 | [] | Philippines | 24th Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization"
] | null | MCQ | b | |
0kmk | Problem:
Prove that there exists infinitely many integers $n$ such that $n^{4}+2020$ has a prime divisor larger than $2 n$. | [
"Solution:\n\nNote that if $p$ is a prime that divides $x^{4}+2020$ for some $x$, then we can find $n$ such that $p \\mid n^{4}+2020$ and $n < p/2$. Indeed, just take $n$ to be the remainder when $p-x$ or $x$ is divided by $p$, whichever is smaller.\n\nSuppose $S$ is a set of good integers satisfying the property; ... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
0csr | К натуральному числу $N$ прибавили наибольший его делитель, меньший $N$, и получили степень десятки. Найдите все такие $N$.
(Н. Атаханов) | [
"**Ответ.** 75.\nПусть $m$ — наибольший делитель числа $N$, меньший, чем $N$. Тогда $n = mp$, где $p$ — наименьший простой делитель числа $N$. Имеем $N + m = 10^k$, то есть $m(p + 1) = 10^k$. Число в правой части не делится на 3, поэтому $p > 2$. Отсюда следует, что $N$ — нечётное число, а тогда и $m$ нечётно. Знач... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 75 | |
0l7p | Let $S$ be a set of positive integers with the following properties:
* $\{1, 2, \dots, 2025\} \subseteq S$.
* If $a, b \in S$ and $\gcd(a, b) = 1$, then $ab \in S$.
* If for some $s \in S$, $s+1$ is composite, then all positive divisors of $s+1$ are in $S$.
Prove that $S$ contains all positive integers. | [] | United States | 16th United States of America Junior Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.