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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
001l | Sean $x_1$ y $x_2$ enteros positivos. Se dan $n$ progresiones aritméticas infinitas de enteros no negativos tales que entre cualesquiera $x$ enteros no negativos consecutivos hay al menos uno que pertenezca a alguna de las $n$ progresiones. Sean $d_1, d_2, \ldots, d_n$ las diferencias de las progresiones y $d = \min\{d... | [] | Argentina | XII Olimpíada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | español | proof and answer | n x | |
08f8 | Problem:
Un numero di tre cifre, diverse fra loro e non nulle (diciamo $a b c$), si dice petaloso se esiste un intero $n \geq 1$ tale che il numero $c b a \underbrace{00 \cdots 0}_{n \text{ zeri}}$ sia multiplo di $a b c$. Il più piccolo $n$ che rende vera questa divisibilità è detto fiore di $a b c$.
Esempio. Il num... | [
"Solution:\n\na.\nDobbiamo dimostrare che, per $n$ sufficientemente grande, il numero $a b c=2^{x} \\cdot 3^{y} \\cdot 5^{z}$ divide $c b a \\cdot 10^{n}=c b a \\cdot 2^{n} \\cdot 5^{n}$. Mostreremo che questa condizione è rispettata con $n=\\max \\{x, z\\}$. Sicuramente con questa scelta di $n$ si ha che $2^{x} \\... | Italy | Italian Mathematical Olympiad - February Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 9 | |
032p | Problem:
Find all real numbers $a$ such that the equation
$$
\log_{4 a x}(x-3 a)+\frac{1}{2} \log_{x-3 a} 4 a x=\frac{3}{2}
$$
has exactly two solutions. | [
"Solution:\nThe equation is defined if\n$$\n4 a x > 0,\\ 4 a x \\neq 1,\\ x-3 a > 0,\\ x-3 a \\neq 1\n$$\nSetting $t = \\log_{4 a x}(x-3 a)$ we get the equation $t + \\frac{1}{2 t} = \\frac{3}{2}$ with roots $t_1 = 1$ and $t_2 = \\frac{1}{2}$.\nIf $\\log_{4 a x}(x-3 a) = 1$, then $x_1 = \\frac{3 a}{1-4 a}$, $a \\ne... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (-∞, -1/2) ∪ (-1/2, 0) ∪ (0, 1/6) ∪ (1/6, 1/4) | |
081b | Problem:
Quante sono le terne $(a, b, c)$ di numeri reali che verificano il seguente sistema?
$$
\left\{\begin{array}{l}
a^{2}+b^{2}+c^{2}=1 \\
a^{3}+b^{3}+c^{3}=1
\end{array}\right.
$$
(A) Nessuna
(B) 1
(C) 3
(D) 6
(E) infinite. | [
"Solution:\n$a^{2}+b^{2}+c^{2}=1$ e quindi $|a| \\leq 1$. Ciò implica che $a^{3} \\leq |a|^{3} \\leq a^{2}$, dove il segno di uguaglianza vale se e solo se $a=0$ oppure $a=1$. Analogamente si ragiona per $b$ e $c$. Perciò se uno dei tre numeri fosse diverso da 0 e da 1 si avrebbe\n$$\n1=a^{3}+b^{3}+c^{3}<a^{2}+b^{2... | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | C | |
09kb | For real numbers $a$, $b$, $c$ satisfying $0 \le a \le b \le c$ and $a + b + c = 1$, prove that the inequality
$$
ab\sqrt{b-a} + bc\sqrt{c-b} + ca\sqrt{c-a} < \frac{1}{4}
$$
holds. | [
"For any $x \\le y$, we have $\\sqrt{y-x} \\le \\frac{y-x+1}{2}$. Thus it suffices to prove that\n$$\nab(b-a+1)+bc(c-b+1)+ca(c-a+1) = ab(2b+c)+bc(2c+a)+ca(2c+b) < \\frac{1}{2}.\n$$\nSince $(a+b+c)^3 = a^3 + b^3 + c^3 + 3(ab^2 + bc^2 + ac^2) + 3(a^2b + b^2c + a^2c) + 6abc$, this is equivalent to proving the followin... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
08qs | Problem:
Alice and Bob play the following game: Alice picks a set $A=\{1,2, \ldots, n\}$ for some natural number $n \geqslant 2$. Then starting with Bob, they alternatively choose one number from the set $A$, according to the following conditions: initially Bob chooses any number he wants, afterwards the number chosen... | [
"Solution:\n\nTo say that Alice has a winning strategy means that she can find a number $n$ to form the set $A$, so that she can respond appropriately to all choices of Bob and always get at the end a composite number for the sum of her choices. If such $n$ does not exist, this would mean that Bob has a winning str... | JBMO | JBMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Alice | |
0al0 | Problem:
Let $f: \mathbb{N} \to \mathbb{N}$ be a monotonically increasing transformation of the natural numbers, such that $f(f(n)) = n^2$ for each natural number $n$. What are the smallest and largest possible value of $f(2023)$? | [] | North Macedonia | Team Selection Test for IMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | minimum: 2024; maximum: 4092528 | |
03ks | Problem:
Prove that
$$
\frac{1}{1999} < \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{1997}{1998} < \frac{1}{44}
$$ | [
"Solution:\nLet $P = \\frac{1}{2} \\cdot \\frac{3}{4} \\cdot \\frac{5}{6} \\cdots \\frac{1997}{1998}$.\n\nThere are $999$ terms in the product, since the numerators run through the odd numbers from $1$ to $1997$ and the denominators through the even numbers from $2$ to $1998$.\n\nWe can write:\n$$\nP = \\prod_{k=1}... | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0ih3 | Problem:
Suppose $x$ is a fifth root of unity. Find, in radical form, all possible values of
$$
2x + \frac{1}{1+x} + \frac{x}{1+x^{2}} + \frac{x^{2}}{1+x^{3}} + \frac{x^{3}}{1+x^{4}}.
$$ | [
"Solution:\nNote that\n$$\n\\begin{gathered}\n\\frac{x}{1+x^{2}} + \\frac{x^{2}}{1+x^{3}} = \\frac{x^{6}}{1+x^{2}} + \\frac{x^{4}}{x^{2}+x^{5}} = \\frac{x^{4}+x^{6}}{1+x^{2}} = x^{4} = \\frac{1}{x}, \\text{ and } \\\\\n\\frac{1}{1+x} + \\frac{x^{3}}{1+x^{4}} = \\frac{x^{5}}{1+x} + \\frac{x^{4}}{x+x^{5}} = \\frac{x^... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | 4, -1 + sqrt(5), -1 - sqrt(5) | |
07of | Let $a_1, a_2, a_3, \dots, b_1, b_2, b_3, \dots$ and $c_1, c_2, c_3, \dots$ be three sequences containing the numbers $+1$ and $-1$ only. Prove that the following inequality holds:
$$
3 \sum_{i=1}^{2014} a_i b_i + 3 \sum_{i=1}^{2014} b_i c_i \le \sum_{i=1}^{2014} a_i^2 + \sum_{i=1}^{2014} b_i^2 + \sum_{i=1}^{2014} c_i^... | [
"First observe that for $u \\le 1$ and $v \\le 1$ we always have $(1-u)(1-v) \\ge 0$, hence $u + v \\le 1 + uv$. We apply this now to $u = a_i b_i$ and $v = b_i c_i$. As $b_i = \\pm 1$, we have $b_i^2 = 1$ and so $uv = a_i c_i$. This shows that\n$$\na_i b_i + b_i c_i \\le 1 + a_i c_i \\quad \\text{and so} \\quad \\... | Ireland | Irska 2014 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0ab5 | During three hours a driver with a car has driven $180\,km$. During the first hour he has driven $0.375$ from the whole distance and during the second hour he has driven $0.9$ from the distance that he had driven during the first hour. What distance has the driver driven during the third hour? | [
"The third hour the driver has driven $180 - 0.375 \\cdot 180 - 0.9 \\cdot 0.375 \\cdot 180 = 51.75\\,km = 51\\,km\\ 75m$."
] | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | final answer only | 51.75 km | |
0bgj | Problem:
Cercul exînscris corespunzător vârfului $A$ al triunghiului $A B C$ este tangent la latura $B C$ în punctul $A_{1}$. Definim analog punctele $B_{1}$ pe latura $C A$ şi $C_{1}$ pe latura $A B$. Presupunem că centrul cercului circumscris triunghiului $A_{1} B_{1} C_{1}$ aparţine cercului circumscris triunghiulu... | [] | Romania | Romania Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hbk | There are $n$ rays $OA_1, OA_2, ..., OA_n$ that start at point $O$ and are enumerated counterclockwise on a plane so that $\angle A_1OA_n < 180^\circ$. Determine the smallest $n$ such that there exists a pair of $60^\circ$ angles, a pair of $45^\circ$ angles and a pair of $30^\circ$ angles among angles $\angle A_iOA_j$... | [
"Firstly, we will show that $n = 5$ satisfies the conditions (Fig. 1).\n\nSuppose there exists a configuration that satisfies the conditions with 4 rays (Fig. 2). Then these rays form 6 different angles. Thus, if conditions were satisfied then all the pairs of angles would be among these 6 angles. But if the greate... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 5 | |
0chu | Let $a$ be a given natural number. We consider the sequence $(x_n)_{n \ge 1}$ defined by $x_n = \frac{1}{1+na}$, for every natural nonzero number $n$.
Prove that, for every natural number $k \ge 3$, there are non-zero natural numbers $n_1 < n_2 < \dots < n_k$ so that the numbers $x_{n_1}, x_{n_2}, \dots, x_{n_k}$ are t... | [
"We will prove the requirement by induction on $k$.\n\nWe notice that\n$$\n\\frac{1}{1+ma} + \\frac{1}{(1+ma)(1+2ma)} = \\frac{2}{1+2ma},\n$$\ntherefore there is an arithmetic progression consisting of three terms of the sequence: $x_m, x_{2m}, x_{2m^2a+3m}$.\n\nWe assume that there are $k$ non-zero natural numbers... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0039 | En la isla Babba utilizan un alfabeto de dos letras, $a$ y $b$, y toda secuencia (finita) de letras es una palabra. Para cada conjunto $P$ de seis palabras de 4 letras cada una, denotamos $N_P$ al conjunto de todas las palabras que no contienen ninguna de las palabras de $P$ como silaba (subpalabra).
Demostrar que si $... | [] | Argentina | Argentina 2006 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Other"
] | Español | proof and answer | Upper bound: 10. One valid choice is P = {aaaa, abab, baaa, baab, babb, bbbb}, for which N_P is finite and contains the length-10 word aaabbbabaa. | |
041l | Let $N^*$ be the set of all positive integers. Prove that there exists a unique function $f: N^* \to N^*$ satisfying $f(1) = f(2) = 1$ and $f(n) = f(f(n-1)) + f(n-f(n-1))$, $n = 3, 4, \dots$. For such $f$, find the value of $f(2^m)$ for integer $m \ge 2$. | [
"We show by induction that for any integer $n > 1$, $f(n)$ is uniquely determined by the value of $f(1)$, $f(2)$, $\\dots$, $f(n-1)$, and $\\frac{n}{2} \\le f(n) \\le n$.\n\nFor $n = 2$, $f(2) = 1$, the claim is true.\nAssume that for any $k$, $1 \\le k < n$ ($n \\ge 3$), $f(k)$ is uniquely determined, and $\\frac{... | China | China Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | f(2^m) = 2^{m-1} for integer m ≥ 2 | |
00mt | Let $ABC$ be a triangle and $I$ its incenter. The circumcircle of $ACI$ intersects the line $BC$ a second time in the point $X$ and the circumcircle of $BCI$ intersects the line $AC$ a second time in the point $Y$.
Prove that the segments $AY$ and $BX$ are of equal length. | [
"We shall show that $AB = BX$ holds. Since $AB = AY$ then follows by the same argument, this completes the proof (see Figure 3).\n\n\nFigure 3: Problem 10\n\nIn this solution, we use oriented angles between lines (modulo $180^\\circ$) with the notation $\\angle PQR$. As usual the angles of ... | Austria | AUT_ABooklet_2020 | [
"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"
] | English | proof only | null | |
0a0q | Determine the largest real number $M$ such that for each infinite sequence $x_0, x_1, x_2, \dots$ of real numbers satisfying
a. $x_0 = 1$ and $x_1 = 3$,
b. $x_0 + x_1 + \dots + x_{n-1} \ge 3x_n - x_{n+1}$ for all $n \ge 1$,
the inequality
$$
\frac{x_{n+1}}{x_n} > M,
$$
holds for all $n \ge 0$. | [
"The largest possible $M$ for which the given property holds is $M = 2$.\n\nWe first show that the given property holds for $M = 2$. To do this, we show by induction on $n$ the stronger statement that $x_{n+1} > 2x_n > x_n + x_{n-1} + \\dots + x_0$ for all $n \\ge 0$.\n\nFor $n = 0$, this is the statement $x_1 > 2x... | Netherlands | IMO Team Selection Test 1 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 2 | |
0ccs | Let $ABC$ and $DEF$ be two congruent equilateral triangles, centered at $O_1$ and $O_2$, respectively, such that segment $AB$ meets segments $DE$ and $DF$ at $M$ and $N$, respectively, and segment $AC$ meets segments $DF$ and $EF$ at $P$ and $Q$, respectively. The bisectors of angles $EMN$ and $DPQ$ meet at $I$, and th... | [
"Denote by $C_1$ and $C_2$ the circumcircles of the triangles $ABC$ and $DEF$, respectively.\nSince $\\angle PAM = \\angle PDM = 60^\\circ$, the quadrilateral $APMD$ is cyclic, so $\\angle APD = \\angle AMD = x^\\circ$.\n\nLooking at the quadrilateral $APIM$ we have: $\\angle PIM = 360^\\circ - \\angle MAP - \\angl... | Romania | THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - THIRD SELECTION TEST | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0cnj | On a checkered paper, numbers $x$, $y$ and $z$ are placed in 3 squares (one number in each), other squares are empty. It is permitted to perform operations of two types:
1) To choose 2 numbers $a$ and $b$ from 2 non-empty squares and to place the number $a + b$ in some empty square;
2) To choose 3 numbers $a$, $b$, a... | [
"Пусть исходные числа $x, y, z$. Получим сначала числа $x+y, y+z, z+x$. Затем из клеток $x, y+z, y$ получим $x(y+z) + y^2$; аналогично, получим числа $y(z+x) + z^2$ и $z(x+y) + x^2$. Сложив их, получаем требуемое: $x(y+z) + y^2 + y(z+x) + z^2 + z(x+y) + x^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = (x+y+z)^2$."
] | Russia | Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English; Russian | proof only | null | |
05vr | Problem:
Soient $m$ et $n$ deux entiers tels que $m > n \geqslant 3$. Morgane a disposé $m$ jetons en cercle, et s'apprête à les peindre en utilisant $n$ couleurs distinctes. Elle souhaite que, parmi $n+1$ jetons consécutifs, il y ait toujours au moins un jeton de chacune des $n$ couleurs. Si elle peut y parvenir, on ... | [
"Solution:\n\nTout d'abord, si $m \\geqslant n^{2}-n$, Morgane peut réaliser son souhait en procédant comme suit, montrant au passage que $m$ est $n$-coloriable.\n\nSoit $a$ le résidu de $m$ modulo $n$, c'est-à-dire le plus petit entier naturel tel que $a \\equiv m (\\bmod\\ n)$. Morgane commence par placer $m-a$ j... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | n^2 - n - 1 | |
09ac | Denote by $d(A)$ the sum of all the elements of $A$ (if $A = \emptyset$, $d(A) = 0$). Let $S = \{1, 2, \dots, 2013\}$ and
$$
T_r = \{T \mid T \subseteq S, d(T) \equiv r \pmod 7\},
$$
for $r = 1, 2, \dots, 6$. Find the number of elements of $T_r$ for each $r$. | [
"Consider $f(x) = (1+x)(1+x^2)\\dots(1+x^{2013}) = \\sum_n a_n x^n$. Then\n$$\n|T_r| = \\sum_k [x^{7k+r}] f(x) = \\sum_k a_{7k+r}.\n$$\nLet $\\epsilon = e^{\\frac{2\\pi i}{7}}$, i.e. $\\epsilon$ is a 7th root of unity. We use the following well known facts.\n$$\n1 + \\epsilon + \\epsilon^2 + \\dots + \\epsilon^6 = ... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | For r = 3: (2^{2013} + 5·2^{287})/7. For r in {1, 2, 4, 5, 6}: (2^{2013} − 2^{288})/7. | |
0hvd | Problem:
Determine whether there exists a polynomial $f(x, y)$ of two variables, with real coefficients, with the following property: A positive integer $m$ is a triangular number if and only if there do not exist positive integers $x$ and $y$ such that $f(x, y)=m$. | [
"Solution:\nThe answer is yes.\n\nThe difference between the $n$th and $(n+1)$st triangular numbers is $n+1$. Therefore, a positive integer $m$ is not a triangular number if and only if it has the form\n$$\n\\frac{n^{2}+n}{2}+y\n$$\nwhere $1 \\leq y \\leq n$. Define $x = n - y + 1$; then the conditions $1 \\leq y$ ... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | Yes. For example, f(x, y) = ((x + y)(x + y − 1))/2 + y works. | |
02v8 | Problem:
Sérgio escolhe dois números inteiros positivos $a$ e $b$. Ele escreve 4 números no seu caderno: $a$, $a+2$, $b$ e $b+2$. Em seguida, todos os 6 produtos de dois desses números são escritos na lousa. Seja $Q$ a quantidade de quadrados perfeitos escritos nela, determine o valor máximo de $Q$. | [
"Solution:\n\nInicialmente provaremos que o produto $a(a+2)$ não é um quadrado perfeito para qualquer escolha de $a$. Temos dois casos a considerar:\n\ni) Se $a$ é ímpar, então nenhum primo que divide $a$ poderá dividir $a+2$. Daí, $a$ e $a+2$ deverão ser, cada um, um quadrado perfeito. Isso claramente não tem solu... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2 | |
073w | Find all functions $f : (0, \infty) \to (0, \infty)$ such that
$$
f(f(x) + y) = x f(1 + x y),
$$
for all $x, y$ in $(0, \infty)$. | [
"We show that $f(x) = 1/x$ is the only solution. We do this in several steps.\n\nStep 1 We show that $f(x)$ is a non-increasing function. Suppose the contrary; say $0 < a < b$ implies that $f(a) < f(b)$, for some $a, b$. Then\n$$\nw = \\frac{b f(b) - a f(a)}{b - a}\n$$\nis in $(0, \\infty)$. It is easy to check tha... | India | Indija TS 2008 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 1/x | |
0cxh | Show that in any triangle $ABC$ with $\widehat{A}=90^{\circ}$ the following inequality holds:
$$
(AB-AC)^{2}\left(BC^{2}+4 AB \cdot AC\right)^{2} \leq 2 BC^{6} .
$$ | [
"The inequality is equivalent to\n$$\n\\left(\\frac{AB}{BC}-\\frac{AC}{BC}\\right)^{2}\\left(1+4 \\frac{AB}{BC} \\cdot \\frac{AC}{BC}\\right)^{2} \\leq 2,\n$$\nhence\n$$\n(\\cos B-\\sin B)^{2}(1+4 \\cos B \\sin B)^{2} \\leq 2,\n$$\nthat is\n$$\n(1-\\sin 2B)\\left(1+4 \\sin 2B+4 \\sin ^{2} 2B\\right) \\leq 2 .\n$$\n... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof only | null | |
0g1w | Problem:
Sei $p \geq 2$ eine Primzahl. Louis und Arnaud wählen abwechselnd einen Index $i \in \{0,1, \ldots, p-1\}$, der bisher noch nicht gewählt wurde, und eine Ziffer $a_{i} \in \{0,1, \ldots, 9\}$. Louis beginnt. Wenn alle Indizes ausgewählt wurden, berechnen sie die folgende Summe:
$$
a_{0}+a_{1} \cdot 10+\ldots+... | [
"Solution:\n\nWir sagen, dass ein Spieler den Zug $\\left(i, a_{i}\\right)$ spielt, wenn er in diesem Zug den Index $i$ und die Ziffer $a_{i}$ gewählt hat.\n\nWenn $p=2$ oder $p=5$, dann macht Louis als erstes den Zug $(0,0)$ und gewinnt, da die Summe ein Vielfaches von $10$ sein wird.\n\nNehme nun an, dass $p$ wed... | Switzerland | SMO - Finalrunde | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0fc1 | Problem:
Los puntos de una superficie esférica de radio $4$, se pintan con cuatro colores distintos. Prueba que existen dos puntos sobre la superficie que tienen el mismo color y que están a distancia $4 \sqrt{3}$ o bien a distancia $2 \sqrt{6}$. | [
"Solution:\n\nDado que se pintan los puntos de la esfera con cuatro colores distintos, para resolver el problema será suficiente con encontrar cinco puntos distintos que cumplan las condiciones del enunciado y aplicar el Principio del Palomar.\n\nEsto puede hacerse, por ejemplo, observando que si inscribimos un tri... | Spain | LIV Olimpiada matemática Española (Concurso Final) | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof only | null | |
03de | Let $x > y > 2022$ are natural numbers such that $xy + x + y$ is a perfect square. Is it possible for each natural number $z$ in the interval $[x + 3y + 1, 3x + y + 1]$ the numbers $x + y + z$ and $x^2 + xy + y^2$ to not be coprimes? | [
"We will prove that it's not possible. Let $k = \\sqrt{xy + x + y}$ and $z = x + y + 2k + 1$. We have\n$$\ny < \\sqrt{y^2 + 2y} \\le \\sqrt{xy + x + y} = k \\le \\sqrt{x^2 + 2x} < x + 1.\n$$\nHence $k \\in (y, x]$ and $z \\in (x + 3y + 1, 3x + y + 1]$. It's easy to check that $xy + yz + zx = (x + y + k)^2$ and $xy ... | Bulgaria | Bulgaria 2022 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | No | |
0954 | Problem:
Calculați
$$
\lim _{n \rightarrow \infty}\left(\sum_{k=0}^{n} C_{2 n}^{2 k} 61^{k}\right) /\left(\sum_{k=0}^{n-1} C_{2 n}^{2 k+1} 61^{k}\right).
$$ | [
"Solution:\nAvem $x_{n}=\\sum_{k=0}^{n} C_{2 n}^{2 k} 61^{k}=\\sum_{k=0}^{n} C_{2 n}^{2 k}(\\sqrt{61})^{2 k}$ și $y_{n}=\\sum_{k=0}^{n-1} C_{2 n}^{2 k+1} 61^{k}=\\frac{1}{\\sqrt{61}} \\sum_{k=0}^{n-1} C_{2 n}^{2 k+1}(\\sqrt{61})^{2 k+1}$, $\\forall n \\geq 1$, ceea ce implică $x_{n}+y_{n} \\sqrt{61}=(1+\\sqrt{61})^... | Moldova | A 61-a OLIMPIADA DE MATEMATICA A REPUBLICII MOLDOVA | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | sqrt(61) | |
06nm | Let $n$ be a positive integer. In each cell of an $n \times n$ grid, we place a coin with either the head or the tail facing upwards. Each time we can select one cell, and flip all the $2^n - 1$ coins that lie in the same row or the same column as this cell. Find all $n$ such that it is always possible to have all the ... | [
"Answer: 1 and all positive even numbers.\n\nIn the following, to operate on the $(i, j)$th cell (the cell in the $i$th row and the $j$th column) means to perform an operation by selecting the $(i, j)$th cell in the operation.\n\n* When $n = 1$, it is clear that the goal can always be met.\n\n* When $n$ is even, it... | Hong Kong | Hong Kong Team Selection Test 2 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | n = 1 and all positive even n | |
0dbh | In triangle $ABC$ point $M$ is the midpoint of side $AB$, and point $D$ is the foot of altitude $CD$. Prove that $\angle A = 2 \angle B$ if and only if $AC = 2MD$. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0izk | Problem:
From the point $(x, y)$, a legal move is a move to $\left(\frac{x}{3}+u, \frac{y}{3}+v\right)$, where $u$ and $v$ are real numbers such that $u^{2}+v^{2} \leq 1$. What is the area of the set of points that can be reached from $(0,0)$ in a finite number of legal moves? | [
"Solution:\n\nAnswer: $\\frac{9 \\pi}{4}$\n\nWe claim that the set of points is the disc with radius $\\frac{3}{2}$ centered at the origin, which clearly has area $\\frac{9 \\pi}{4}$.\n\nFirst, we show that the set is contained in this disc. This is because if we are currently at a distance of $r$ from the origin, ... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 9π/4 | |
03lq | Problem:
In a rectangular array of nonnegative real numbers with $m$ rows and $n$ columns, each row and each column contains at least one positive element. Moreover, if a row and a column intersect in a positive element, then the sums of their elements are the same. Prove that $m=n$. | [
"Solution:\n\nConsider first the case where all the rows have the same positive sum $s$; this covers the particular situation in which $m=1$. Then each column, sharing a positive element with some row, must also have the sum $s$. Then the sum of all the entries in the matrix is $m s = n s$, whence $m = n$.\n\nWe pr... | Canada | 38th Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
02av | Problem:
Interseç̧ão de triângulos - Os $3$ triângulos da figura se cortam em $12$ pontos diferentes. Qual é o número máximo de pontos de intersecção de $3$ triângulos?
 | [] | Brazil | null | [
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | 18 | |
0ery | Jess is standing in a queue of people. She is 18th from the front and 35th from the back.
How many people are in the queue? | [
"There are $17$ people in front of Jess and $34$ behind her. Including herself this makes $17 + 1 + 34 = 52$ people."
] | South Africa | South African Mathematics Olympiad Second Round | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 52 | |
01we | Four positive integers $x$, $y$, $z$ and $t$ satisfy the relations
$$
xy - zt = x + y = z + t.
$$
Is it possible that both $xy$ and $zt$ are perfect squares? | [
"2. See IMO-2018 Shortlist, Problem N5."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | No | |
05cu | Digits $A$, $B$, $C$ are given distinct values from $1$ to $9$ to make the value of the expression $2024 \cdot AB \cdot CC \cdot BA$ a perfect square. How many distinct values can the expression $A+B+C$ obtain? | [
"Denote the value of the given expression by $k$. We know that $2024 = 2^3 \\cdot 11 \\cdot 23$ and $CC = C \\cdot 11$. In a perfect square, all prime exponents are even. Thus one of the numbers $AB$, $BA$, $CC$ must be divisible by $23$. This cannot be $CC$, so let it be $AB$ without loss of generality (the other ... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 6 | |
05c1 | Find the least positive integer $n$ such that:
a. both $n$ and $n+1$ are divisible by the squares of two distinct prime numbers;
b. both $n$ and $n+3$ are divisible by the squares of two distinct prime numbers. | [
"Both the pair $(n, n + 1)$ and $(n, n + 3)$ must contain one odd number. This odd number must be divisible by the square of one of the following numbers:\n$$\n15 = 3 \\cdot 5,\\ 21 = 3 \\cdot 7,\\ 33 = 3 \\cdot 11,\\ 35 = 5 \\cdot 7,\\ 39 = 3 \\cdot 13,\\ 45 = 3^2 \\cdot 5, \\dots\n$$\nTheir squares are $225$, $44... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | a: 675; b: 2025 | |
09u4 | A regular hexagon is filled with small circles of the same size, as illustrated in the figure. The circles can be tangent, but they do not overlap. Exactly four circles fit next to each other along the side of the hexagon.
What is the maximum number of circles that fit in the hexagon in this way?
![](attached_image_1.p... | [
"B) 37"
] | Netherlands | Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | MCQ | B | |
06tx | Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $A B C$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $A B$ and $A C$ at $X$ and $Y$ respectively. Let $P$ be the foot of ... | [
"Let the perpendicular from $S$ to $X Y$ meet line $Q M$ at $S'$. Let $E$ be the foot of altitude from $B$ to side $A C$. Since $Q$ and $S$ lie on the perpendicular bisector of $A D$, using directed angles, we have\n$$\n\\begin{aligned}\n\\measuredangle S D Q & =\\measuredangle Q A S=\\measuredangle X A S-\\measure... | IMO | IMO 2016 Shortlisted Problems | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Tri... | English | proof only | null | |
0bfc | Let $a$ and $b$ be distinct positive real numbers such that $\lfloor n a \rfloor$ divides $\lfloor n b \rfloor$ for every positive integer $n$. Show that $a$ and $b$ are both integers. | [
"Since the $\\lfloor n b \\rfloor / \\lfloor n a \\rfloor$ form a sequence of positive integers converging to $b/a$, it follows that $b = m a$ for some integer $m \\ge 2$, and $\\lfloor n b \\rfloor = m \\lfloor n a \\rfloor$ for all $n$ large enough. Consequently, if $n$ is large enough, then $\\lfloor n m a \\rfl... | Romania | 64th NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0cvj | Let $L$ be the foot of the internal angle bisector of $\angle B$ in an acute-angled triangle $ABC$. The points $D$ and $E$ are the midpoints of the smaller arcs $AB$ and $BC$, respectively, in the circumcircle $\omega$ of $\triangle ABC$. Points $P$ and $Q$ are marked on the extensions of the segments $BD$ and $BE$ bey... | [
"Обозначим $\\angle BAS = 2\\alpha$, $\\angle ACB = 2\\gamma$. Не умаляя общности считаем, что $\\alpha \\ge \\gamma$. Поскольку точки $D$ и $E$ — середины дуг $AB$ и $AC$ окружности $\\omega$, имеем $\\angle ABD = \\angle ACB = \\frac{\\angle ACB}{2} = \\gamma$ и $\\angle CBE = \\alpha$.\n\n![](attached_image_1.pn... | Russia | Final round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English; Russian | proof only | null | |
01y3 | There are $n \ge 5$ cities in some country. Some of the cities are connected with each other by roads, and the next three conditions are satisfied:
1) there is at most one road between any two cities;
2) not all cities are connected with each other;
3) there are exactly $k \ge 1$ roads between any four cities.
Find all... | [
"Answer: $n = 5, k = 3$."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | n = 5, k = 3 | |
0bo0 | A non-constant function $f : (0, \infty) \to (0, \infty)$ has the property $f(x^y) = (f(x))^{f(y)}$, for every $x, y > 0$. Prove that $f(xy) = f(x)f(y)$ and $f(x+y) = f(x)+f(y)$, for every $x, y > 0$. | [
"Take $a > 0$ such that $f(a) \\ne 1$. Then $f(a^{xy}) = f(a)^{f(xy)}$ and\n$$\nf(a^{xy}) = f((a^x)^y) = f(a^x)^{f(y)} = (f(a)^{f(x)})^{f(y)} = f(a)^{f(x)f(y)},\n$$\nwhence $f(xy) = f(x)f(y)$.\nAlso $f(a^{x+y}) = f(a)^{f(x+y)}$ and\n$$\nf(a^{x+y}) = f(a^x a^y) = f(a^x)f(a^y) = f(a)^{f(x)}f(a)^{f(y)} = f(a)^{f(x)+f(... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof only | null | |
0lfi | In a garden, which is organized as a $2024 \times 2024$ board, we plant three types of flowers: roses, daisies, and orchids. We want to plant flowers such that the following conditions are satisfied:
(i) Each cell is planted with at most one type of flower. Some cells can be left blank and not planted.
(ii) For each ... | [
"Let $P$ be the number of planted cells. The estimation of $P$ will be based on the following two simple lemmas.\n\n**Lemma 1.** If a row (or column) contains at least two types of flowers then it has at most $6$ planted cells.\n\n*Proof.* Suppose to the contrary that a row contains at least $7$ planted cells of at... | Vietnam | Team selection tests | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 24216 | |
00g1 | Let $ABC$ be an equilateral triangle. Let $P$ be a point on the side $AC$ and $Q$ be a point on the side $AB$ so that both triangles $ABP$ and $ACQ$ are acute. Let $R$ be the orthocentre of triangle $ABP$ and $S$ be the orthocentre of triangle $ACQ$. Let $T$ be the point common to the segments $BP$ and $CQ$. Find all p... | [
"We are going to show that this can only happen when\n$$\n\\angle CBP = \\angle BCQ = 15^{\\circ}.\n$$\n\nLemma. If $\\angle CBP > \\angle BCQ$, then $RT > ST$.\n\nProof. Let $AD$, $BE$ and $CF$ be the altitudes of triangle $ABC$ concurrent at its centre $G$. Then $P$ lies on $CE$, $Q$ lies on $BF$, and thus $T$ li... | Asia Pacific Mathematics Olympiad (APMO) | XIV APMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | English | proof and answer | angle CBP = angle BCQ = 15° | |
0c0s | The triangle $ABC$ is inscribed in the circle $\mathcal{C}(O, 1)$. Let $G_1$, $G_2$ and $G_3$ be the centroids of the triangles $OBC$, $OAC$ and $OAB$ respectively. Prove that the triangle $ABC$ is equilateral if and only if $AG_1 + BG_2 + CG_3 = 4$. | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | proof only | null | |
026t | Problem:
Qual o menor valor da fração
$$
\frac{x^{4}+x^{2}+5}{\left(x^{2}+1\right)^{2}} ?
$$ | [
"Solution:\nTemos\n$$\n\\begin{aligned}\ny & =\\frac{x^{4}+x^{2}+5}{\\left(x^{2}+1\\right)^{2}} \\\\\n& =\\frac{\\left(x^{2}+1\\right)^{2}-\\left(x^{2}+1\\right)+5}{\\left(x^{2}+1\\right)^{2}} \\\\\n& =1-\\frac{1}{1+x^{2}}+\\frac{5}{\\left(1+x^{2}\\right)^{2}}\n\\end{aligned}\n$$\nSe $v=\\frac{1}{1+x^{2}}$, então o... | Brazil | NÍVEL 3 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 19/20 | |
07xa | There are real numbers $\alpha$, $\beta$, such that the cubic functions
$$
f(x) = x^3 - 3x^2 + \alpha x + \beta \quad \text{and}
$$
$$
g(x) = x^3 + (\alpha - 3)x - 6
$$
have exactly two distinct non-zero roots in common. Find $\alpha$ and $\beta$. | [
"The two distinct non-zero common roots of $f$ and $g$ are also roots of $g - f$. These are the two roots of the quadratic\n$$\ng(x) - f(x) = 3 \\left( x^2 - x - \\frac{\\beta}{3} - 2 \\right),\n$$\nhence we must have\n$$\n\\begin{aligned}\ng(x) &= (x - \\lambda) \\left( x^2 - x - \\frac{\\beta}{3} - 2 \\right) \\\... | Ireland | IRL_ABooklet_2024 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | alpha = -4, beta = 12 | |
052e | Find all positive integers which are exactly $2013$ times bigger than the sum of their digits. | [
"Note that the minimal value of a $k$-digit number is $10^{k-1}$ and the maximal value of the cross-sum multiplied by $2013$ is $9k \\cdot 2013$. Since $9 \\cdot 7 \\cdot 2013 = 126819 < 1000000$ we can consider only numbers with up to $6$ digits. Since then the cross-sum is at most $54$, it is enough to consider n... | Estonia | Open Contests | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 36234 | |
0dyg | Find all real numbers $x$ such that $\sqrt{1-x^2} + \sqrt{5x-x^2}$ is an integer. | [
"Let us find an estimate for the value of this expression. Evidently, $1 - x^2 \\le 1$ and $5x - x^2$ is bounded by $5x - x^2 = \\frac{25}{4} - (x - \\frac{5}{2})^2 \\le \\frac{25}{4}$, so\n$$\n\\sqrt{1-x^2} + \\sqrt{5x-x^2} \\le \\sqrt{1} + \\sqrt{\\frac{25}{4}} = 1 + \\frac{5}{2} = 3 + \\frac{1}{2}.\n$$\nOn the o... | Slovenia | Slovenija 2008 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 0, 9/41, 1 | |
030j | Problem:
No triângulo $ABC$, temos os ângulos $\angle ACB = 65^{\circ}$ e $\angle ABC = 70^{\circ}$. Sobre os prolongamentos do lado $BC$, marcam-se o ponto $P$ de tal modo que $BP = AB$ e que $B$ esteja entre $P$ e $C$; e o ponto $Q$ de modo que $CQ = AC$ e que $C$ esteja entre $B$ e $Q$. Se $O$ é o centro da circunf... | [
"Solution:\n\nSejam $\\angle BAO = x$ e $\\angle OAC = y$. De $BP = AB$ e $AC = CQ$, podemos concluir que $\\angle APB = \\angle PAB = \\beta$ e $\\angle CAQ = \\angle AQC = \\gamma$. Como $O$ é o centro da circunferência que passa por $A$, $P$ e $Q$, temos $OP = OA = OQ$ e daí os triângulos $AOP$ e $AOQ$ são isósc... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | ∠OAP = 57°30′, ∠OAQ = 55° | |
0jhh | Problem:
Let $F_{1}, F_{2}, F_{3} \ldots$ be the Fibonacci sequence, the sequence of positive integers with $F_{1}=F_{2}=1$ and $F_{n+2}=F_{n+1}+F_{n}$ for all $n \geq 1$. A Fibonacci number is by definition a number appearing in this sequence.
Let $P_{1}, P_{2}, P_{3}, \ldots$ be the sequence consisting of all the int... | [
"Solution:\nLet $\\Phi=\\frac{1+\\sqrt{5}}{2}$ and $\\varphi=\\frac{1-\\sqrt{5}}{2}$. Note for later use that $\\Phi \\varphi=-1, \\Phi-\\varphi=\\sqrt{5}, \\Phi=\\Phi^{2}-1$, and $\\varphi=\\varphi^{2}-1$. We use Binet's formula for the Fibonacci numbers: $F_{n}=\\frac{1}{\\sqrt{5}}\\left(\\Phi^{n}-\\varphi^{n}\\r... | United States | Bay Area Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0bur | Problem:
Să se rezolve în mulţimea numerelor reale ecuaţia:
$$
\sum_{k=1}^{2n} \left([x] + [(-1)^k x + k]\right) = 2n(n+2).
$$
Am notat cu $[x]$ partea întreagă a numărului real $x$. | [] | Romania | OLIMPIADA DE MATEMATICĂ ETAPA LOCALĂ | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | (2,3) | |
0buc | Problem:
Rezolvaţi în $M_{2}(\mathbb{R})$ ecuaţia: $X^{n}=\left(\begin{array}{ll}3 & 1 \\ 6 & 2\end{array}\right),\ n \in \mathbb{N}^{*}$ | [] | Romania | Olimpiada Națională de Matematică | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof and answer | Let A = [[3, 1], [6, 2]]. All real solutions are scalar multiples of A. Specifically:
- If n is odd: X = 5^{1/n - 1} A.
- If n is even: X = ± 5^{1/n - 1} A. | |
0bt8 | Let $ABC$ be an acute triangle with $AB \neq AC$. The incircle $\omega$ of the triangle touches the sides $BC$, $CA$ and $AB$ in $D$, $E$ and $F$, respectively. The perpendicular line erected from $C$ to $BC$ intersects $EF$ at $M$, and, similarly, the perpendicular line erected at $B$ to $BC$ intersects $EF$ at $N$. T... | [
"Let $\\{T\\} = EF \\cap BC$. Applying Menelaus' theorem to the triangle $ABC$ and the transversal line $E-F-T$ we obtain: $\\frac{TB}{TC} \\cdot \\frac{EC}{EA} \\cdot \\frac{FA}{FB} = 1$, i.e. $\\frac{TB}{TC} \\cdot \\frac{s-c}{s-a} \\cdot \\frac{s-a}{s-b} = 1$, or $\\frac{TB}{TC} = \\frac{s-b}{s-c}$, where the no... | Romania | 67th NMO Selection Tests for JBMO | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous... | English | proof only | null | |
079e | In the isosceles triangle $ABC$, we have $AB = AC$ and $BC > AB$. $D$ and $M$ are the midpoints of $BC$ and $AB$ respectively. $X$ is a point that $BX \perp AC$ and $XD \parallel AB$. $H$ is the intersection of $BX$ and $AD$. If $P$ be the intersection of $DX$ with the circumcircle of $AHX$ (not $X$), prove that the ta... | [
"Obviously $X, P, D, M$ are collinear. Since $HAPX$ is an inscribed quadrilateral, then $\\angle APM = \\angle AHX$, and since $HCDT$ is inscribed ($T$ is the foot of the altitude from $B$ on $AC$), then $\\angle AHX = \\angle AHT = \\angle ACD$. Hence, $\\angle APM = \\angle C = \\angle B$. Now if we draw ray $Ax$... | Iran | 27th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0d4l | Let $X$ be a set of rational numbers satisfying the following two conditions:
(a) The set $X$ contains at least two elements,
(b) For any $x, y$ in $X$, if $x \neq y$ then there exists $z$ in $X$ such that either $\left|\frac{x-z}{y-z}\right|=2$ or $\left|\frac{y-z}{x-z}\right|=2$.
Prove that $X$ contains infinitely ma... | [
"The condition (b) tells that if $x \\neq y$ are in $X$, then at least one of the four numbers\n$z_1 = x - (y - x), \\quad z_2 = x + \\frac{1}{3}(y - x), \\quad z_3 = x + \\frac{2}{3}(y - x), \\quad z_4 = x + 2(y - x)$,\n\nis in $X$.\n\nAssume that $X$ is finite and let $x_0$ be the smalles... | Saudi Arabia | SAMC | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English, Arabic | proof only | null | |
0i0f | Problem:
Evaluate $\sum_{n=2}^{17} \frac{n^{2}+n+1}{n^{4}+2 n^{3}-n^{2}-2 n}$. | [
"Solution:\n\nObserve that the denominator $n^{4}+2 n^{3}-n^{2}-2 n = n(n-1)(n+1)(n+2)$. Thus we can rewrite the fraction as\n$$\n\\frac{n^{2}-n+1}{n^{4}+2 n^{3}-n^{2}-2 n} = \\frac{a}{n-1} + \\frac{b}{n} + \\frac{c}{n+1} + \\frac{d}{n+2}\n$$\nfor some real numbers $a$, $b$, $c$, and $d$. This method is called part... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 592/969 | |
07ea | A subgraph of a $K_n$ is chosen such that the number of its edges is a multiple of $3$ and the degree of each vertex is an even number. Prove that we can assign an integer weight to each triangle of the $K_n$ such that for each edge of the chosen subgraph, the sum of the weight of the triangles that contain that edge e... | [
"The following lemma is needed to approach the solution:\n**Lemma.** If some integer weight is assigned to any edge of a complete graph with more than $4$ vertices, such that for each vertex, the sum of the weight of the edges connected to this vertex is an even number, and the sum of the weight of all edges is a m... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0cfy | The sides $BC$, $CA$, $AB$ of the equilateral triangle $ABC$ contain the points $D$, $E$, respectively $F$, different from the sides' midpoints, so that $BD = CE = AF$. Denote $M$ the midpoint of $EF$ and $N$ the intersection of the lines $BC$ and $AM$. Prove that the segments $ND$ and $BC$ have the same midpoint.
Ghe... | [] | Romania | 74th NMO Shortlisted Problems | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
0h67 | Solve the equation:
$$
\frac{1}{\sqrt{2014}-\sqrt{x}} + \frac{1}{\sqrt{2015}-\sqrt{x+1}} + \frac{1}{\sqrt{2016}-\sqrt{x+2}} =
\frac{1}{\sqrt{2015-x}-\sqrt{1}} + \frac{1}{\sqrt{2016-x}-\sqrt{2}} + \frac{1}{\sqrt{2017-x}-\sqrt{3}}
$$ | [
"Multiply both numerator and denominator by the conjugate:\n$$\n\\frac{\\sqrt{2014+\\sqrt{x}}}{2014-x} + \\frac{\\sqrt{2015+\\sqrt{x+1}}}{2015-(x+1)} + \\frac{\\sqrt{2016+\\sqrt{x+2}}}{2016-(x+2)} = \\frac{\\sqrt{2015-x+\\sqrt{1}}}{(2015-x)-1} + \\frac{\\sqrt{2016-x+\\sqrt{2}}}{(2016-x)-2} + \\frac{\\sqrt{2017-x+\\... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | 1 | |
05j9 | Problem:
Chaque nombre rationnel strictement positif est colorié soit en rouge, soit en noir, de telle sorte que:
- les nombres $x$ et $x+1$ sont de couleurs différentes ;
- les nombres $x$ et $\frac{1}{x}$ sont de la même couleur;
- le nombre $1$ est colorié en rouge.
Quelle est la couleur de $\frac{2012}{2013}$ ?
(O... | [
"Solution:\n\nOn remarque que $\\frac{2012}{2013}$ s'écrit:\n$$\n\\frac{2012}{2013}=\\frac{1}{1+\\frac{1}{2012}}\n$$\nAinsi, d'après les règles de l'énoncé, il a la couleur inverse de $2012$. Or, ce dernier s'obtient en ajoutant $2011$ fois $1$ à $1$ ; il a donc la couleur inverse de $1$ (puisqu'on inverse $2011$ f... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Functional equations"
] | null | proof and answer | red | |
0hl7 | Problem:
Frank and Joe are playing ping pong. For each game, there is a $30\%$ chance that Frank wins and a $70\%$ chance Joe wins. During a match, they play games until someone wins a total of 21 games. What is the expected value of number of games played per match? | [
"Solution:\n\nThe expected value of the ratio of Frank's to Joe's score is $3:7$, so Frank is expected to win $9$ games for each of Frank's $21$. Thus the expected number of games in a match is $30$."
] | United States | null | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 30 | |
02j9 | Problem:
Encontre os algarismos que estão faltando sobre cada um dos traços:
a) $\frac{126}{8\_} = \frac{21}{\_}$;
b) $\frac{\_\_8}{33\_} = \frac{4}{5}$ | [
"Solution:\n\na) Observe que $126 \\div 6 = 21$, logo, o numerador $126$ foi dividido por $6$ para obter o numerador $21$ da outra fração. Logo, o denominador $8\\_$ também é divisível por $6$. O único número da forma $8\\_$ que é divisível por $6$ é $84$, e $84 \\div 6 = 14$. Podemos então completar as frações:\n\... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | final answer only | a) 84 and 14; b) 268 and 335 | |
0h2t | In a triangle $ABC$ $M$ is the midpoint of the side $BC$, and on the side $AB$ a point $N$ is chosen so that $NB = 2AN$. If $\angle CAB = \angle CMN$, find $\frac{AC}{BC}$? | [
"Let $D$ be a point on the half-line $CA$ such that $CA = AD$ (fig. 37).\n\n\n\nThen $DM$ and $BA$ are medians of the triangle $BCD$, and so the point of their intersection divides both of them in the ratio $2:1$ from the vertex. This means that they intersect at the point $N$. Hence $\\ang... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 1/2 | |
06u1 | Let $a$ be a positive integer which is not a square number. Denote by $A$ the set of all positive integers $k$ such that
$$
k = \frac{x^{2} - a}{x^{2} - y^{2}}
$$
for some integers $x$ and $y$ with $x > \sqrt{a}$. Denote by $B$ the set of all positive integers $k$ such that (1) is satisfied for some integers $x$ and $y... | [
"We first prove the following preliminary result.\n- Claim. For fixed $k$, let $x, y$ be integers satisfying (1). Then the numbers $x_{1}, y_{1}$ defined by\n$$\nx_{1} = \\frac{1}{2}\\left(x - y + \\frac{(x - y)^{2} - 4a}{x + y}\\right), \\quad y_{1} = \\frac{1}{2}\\left(x - y - \\frac{(x - y)^{2} - 4a}{x + y}\\rig... | IMO | IMO 2016 Shortlisted Problems | [
"Number Theory > Diophantine Equations > Pell's equations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0cch | Consider a table with $n$ lines and $m$ columns ($n, m \in \mathbb{N}, n, m \ge 2$) consisting of $n \cdot m$ squares $1 \times 1$, which we will call *cells*. We call *snake* a sequence of cells with the following properties: the first cell is located on the first (top) line of the table, the last cell is located on t... | [
"Consider a certain snake. For each line $i \\in \\{1, 2, \\dots, n\\}$ of the table, we denote by $a_i$ and $b_i$ the numbers of the columns corresponding to the first and last cell, respectively, that the snake has on line $i$. We observe that $a_{i+1} = b_i$ for any $i \\in \\{1, 2, \\dots, n-1\\}$, therefore th... | Romania | THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - FOURTH SELECTION TEST | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | n(m^2+3m-1)/(3m) | |
01jm | For integers $a$ and $n > 0$, we say that $a$ is *n-expressible* if it can be written as a sum of distinct positive divisors of $n$. The natural number $n$ is *good* if $a$ being *n-expressible* implies $a-1$ being *n-expressible* for all $a \ge 1$.
Determine for which $n$, both $n!$ and $n^n$ are good. | [
"For both versions of the problem, we begin with a lemma which classifies the good numbers in a more compact way.\n\nNote that the largest integer that can be written as a sum of distinct divisors of $n$ is the sum of all its positive divisors, denoted by $\\sigma(n)$. Hence, a positive integer $n$ is good iff all ... | Baltic Way | Baltic Way 2023 Shortlist | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | n! is good for all positive integers n. The number n^n is good exactly when n is even or n equals 1. Hence both n! and n^n are good precisely for n even or n = 1. | |
0c9e | Problem:
Fie $\triangle ABC$ cu $m(\angle A) > 90^\circ$. Considerăm $BE$ bisectoarea $\angle ABC$, $E \in AC$ și $CF$ bisectoarea $\angle ACB$, $F \in AB$, $BE \cap CF = \{I\}$. Știind că $IE = IF$, arătați că $\triangle ABC$ este isoscel. | [
"Solution:\n\nFie $IN \\perp AB$ și $IM \\perp AC$.\n\nDeoarece $I$ este centrul cercului înscris $\\triangle ABC \\Rightarrow AI$ este bisectoarea $\\angle BAC \\Rightarrow IN = IM$.\n\n$[IF] \\equiv [IE]$ și $[IN] \\equiv [IM] \\stackrel{I.C.}{\\Rightarrow} \\triangle IFN \\equiv \\triangle IEM \\Rightarrow \\ang... | Romania | Olimpiada Națională GAZETA MATEMATICĂ Etapa a III-a | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
00bf | A $27 \times 27$ square board is given. Carla paints some of the cells of the board in blue, so that at least one cell remains unpainted and the following two conditions hold simultaneously:
* In every $2 \times 2$ sub-board, the number of blue cells is even.
* In every $3 \times 3$ sub-board, the number of blue cells... | [
"We will first show that in a valid coloring of the board there cannot be a $3 \\times 3$ sub-board with all its cells painted in blue.\n\nConsider an $n \\times n$ sub-board $T$ with all cells painted in blue, for the maximum $n$, and assume $n \\ge 3$. Since the board has unpainted cells, we may find an $(n+1) \\... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 405 | |
03e5 | Let $G$ be a complete bipartite graph with partition sets $A$ and $B$ of sizes $km$ and $kn$, respectively. The edges of $G$ are colored in $k$ colors. Prove that there exists a monochromatic connected component with at least $m+n$ vertices (which means that there exists a color and a set of vertices, such that between... | [
"There are at least $kmn$ edges colored in the most used color. We delete the remaining edges and prove that there exists a connected component with at least $m+n$ vertices in the remaining graph. The idea is to consider all the connected components. If each one has less than $m+n$ vertices, then the graph is too f... | Bulgaria | 1 Autumn tournament | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Combinatorial optimization... | English | proof only | null | |
0efh | Problem:
Prvo število zaporedja šestih števil je enako $4$, zadnje pa $47$. Vsako število od vključno tretjega naprej je enako vsoti prejšnjih dveh števil. Naj bo $S$ vsota vseh šestih števil zaporedja. Tedaj $S$ leži na intervalu med
(A) $51$ in $90$
(B) $91$ in $100$
(C) $101$ in $110$
(D) $111$ in $120$
(E) $121$ i... | [
"Solution:\n\nOznačimo drugo število v zaporedju z $a$. Tedaj je zaporedje števil enako $4, a, a+4, 2a+4, 3a+8, 5a+12$. Zadnje število je enako $5a+12=47$, od koder izračunamo $a=7$. Vsota vseh šestih števil je enaka $S=4+a+(a+4)+(2a+4)+(3a+8)+(5a+12)=12a+32=12 \\cdot 7+32=116$. Pravilen odgovor je (D)."
] | Slovenia | Slovenian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
0fs9 | Problem:
Let $n$ be a positive integer. Consider the following game: Initially, $k$ stones are distributed among the $n^{2}$ squares of an $n \times n$ chessboard. A move consists of choosing a square containing at least as many stones as the number of its adjacent squares (two squares are adjacent if they share a com... | [
"Solution:\n\nClearly, the first requirement imposes an upper bound on $k$ (if there is an initial configuration with $k$ stones where no moves are possible, then taking away stones from this configuration will not suddenly make a move possible) and the second requirement imposes a lower bound on $k$ (similarly, if... | Switzerland | null | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | (a) All positive integers k with k ≤ 3n^2 − 4n. (b) All integers k with k ≥ 2n^2 − 2n. | |
07qo | A complex number $z$ is called a root of unity if $z^k = 1$, for some positive integer $k$.
Given a positive integer $n$, let $S_n$ be the set of all roots of unity of the form
$$
\frac{a+bi}{\sqrt{n}}
$$
where $i = \sqrt{-1}$ and $a$ and $b$ are positive integers.
(i) Prove that $S_n$ has less than $\sqrt{n}$ distinct... | [
"(i) For a root of unity $z = p + qi$, with $p$ and $q$ real and $z^k = 1$, we have $|z|^k = |z^k| = 1$, i.e. $p^2+q^2=1$. So, if $(a+bi)/\\sqrt{n} \\in S_n$, for positive integers $a$ and $b$, we have $a^2+b^2=n$.\nSince $a$ and $b$ are positive integers, $a < \\sqrt{n}$ and, once $a$ is known, $b$ is also known, ... | Ireland | Irish Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof only | null | |
0hlu | Problem:
Suppose that $k$ and $n$ are integers with $0 \leq k \leq n-3$. Prove that
$$
\binom{n}{k},\ \binom{n}{k+1},\ \binom{n}{k+2},\ \binom{n}{k+3}
$$
cannot form an arithmetic progression. | [
"Solution:\nSuppose otherwise. The common difference $d$ satisfies $\\binom{n}{k+1} - \\binom{n}{k} = d = \\binom{n}{k+2} - \\binom{n}{k+1}$, giving $2\\binom{n}{k+1} = \\binom{n}{k} + \\binom{n}{k+2}$. Expanding the binomial coefficients in terms of factorials gives\n$$\n2 \\frac{n!}{(k+1)! (n-k-1)!} = \\frac{n!}{... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
0ivo | Problem:
Define the sequence $\{x_{i}\}_{i \geq 0}$ by $x_{0}=2009$ and $x_{n}=-\frac{2009}{n} \sum_{k=0}^{n-1} x_{k}$ for all $n \geq 1$. Compute the value of $\sum_{n=0}^{2009} 2^{n} x_{n}$. | [
"Solution:\n\nWe have\n$$\n-\\frac{n x_{n}}{2009}=x_{n-1}+x_{n-2}+\\ldots+x_{0}=x_{n-1}+\\frac{(n-1) x_{n-1}}{2009}\n$$\n, which yields the recursion $x_{n}=\\frac{n-2010}{n} x_{n-1}$. Unwinding this recursion, we find $x_{n}=(-1)^{n} \\cdot 2009 \\cdot \\binom{2008}{n}$. Thus\n$$\n\\begin{aligned}\n\\sum_{n=0}^{20... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | final answer only | -2009 | |
0gk1 | Let $f: \mathbb{R}^+ \to \mathbb{R}^+$ be a function satisfying
$$
(f(xy))^2 = f(x^2)f(y^2)
$$
for all $x, y \in \mathbb{R}^+$ with $x^2y^3 > 2008$.
Prove that $(f(xy))^2 = f(x^2)f(y^2)$ for all $x, y \in \mathbb{R}^+$. | [
"Define $\\lambda(x, y) = \\frac{2008}{x^2y^3}$ for all $x, y > 0$.\nWe can see that $\\left(\\frac{x}{\\lambda(x, y)}\\right)^2 (y \\cdot \\lambda(x, y))^3 = 2008$. Thus,\n$$ \\left(\\frac{x}{z}\\right)^2 (yz)^3 > 2008 \\text{ for every } z > \\lambda(x, y). $$\nIt follows that $(f(xy))^2 = f(x^2/z^2)f(y^2z^2)$ fo... | Thailand | Thai Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof only | null | |
06gf | A finite sequence of integers $a_0, a_1, \dots, a_n$ is quadratic if for each $i = 1, 2, \dots, n$, $|a_i - a_{i-1}| = i^2$.
a. Show that for any two integers $b$ and $c$, $b < c$, there exists a natural number $n$ and a quadratic sequence with $a_0 = b$ and $a_n = c$.
b. Find the smallest natural number $n$ for whic... | [
"a.\nObserve that $f(k) = k^2 - (k+1)^2 - (k+2)^2 + (k+3)^2 = 4$ for any integer $k$.\n\n* If $c = b + 4m$ for some $m \\in \\mathbb{Z}$, then $c = b + \\sum_{k=1}^{m} f(4k - 3)$.\n* If $c = b + 4m + 1$ for some $m \\in \\mathbb{Z}$, then $c = b + 1^2 + \\sum_{k=1}^{m} f(4k - 2)$.\n* If $c = b + 4m + 2$ for some $m... | Hong Kong | CHKMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | null | proof and answer | 19 | |
08fn | Problem:
In una partita di palla Riemanniana si affrontano due squadre; in ogni momento, ciascuna schiera in campo $k>1$ giocatrici. Alla fine di ogni azione viene assegnato un punto a una delle due squadre; inoltre, ciascuna squadra può effettuare un numero arbitrario di sostituzioni prima che abbia inizio l'azione s... | [
"Solution:\n\nLa risposta è $\\mathbf{(B)}$. Costruiamo una tabella con una riga per ogni azione giocata e una colonna per ogni giocatrice delle Geodetiche. Scriviamo +1 in una casella se la giocatrice corrispondente alla riga era in campo durante l'azione corrispondente alla colonna e l'azione è stata vinta dalle ... | Italy | Gara di Febbraio | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | MCQ | B | |
03qz | In a convex quadrilateral $ABCD$, the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$. The point $P$ lies inside $ABCD$ and satisfies
$$\angle PBC = \angle DBA \text{ and } \angle PDC = \angle BDA.$$
Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP = CP$. | [
"(i) Necessity.\n\nAssume that $ABCD$ is a cyclic quadrilateral. Let the circle $\\Gamma$ be the circumcircle of quadrilateral $ABCD$. Extend $BP$ and $DP$ beyond $P$ to meet the circle $\\Gamma$ at $X$ and $Y$ respectively.\n\nSince $DB$ does not bisect $\\angle ABC$ and $P$ lies inside $ABCD$, it follows from $\\... | China | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry ... | English | proof only | null | |
0f8o | Problem:
A triangle with perimeter $1$ has side lengths $a$, $b$, $c$. Show that $a^2 + b^2 + c^2 + 4abc < \dfrac{1}{2}$. | [] | Soviet Union | 23rd ASU | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
01z0 | An arbitrary point $D$ is marked inside the triangle $ABC$. The lines $AD$, $BD$ and $CD$ intersect the sides $BC$, $CA$ and $AB$ at the points $K$, $L$ and $M$ respectively. Let $A_1$, $B_1$ and $C_1$ be the midpoints of the segments $BC$, $CA$ and $AB$, and $X$, $Y$ be the midpoints of the segments $ML$ and $KM$ resp... | [
"Note that $XA_1$ is the Gauss-Newton line of a complete quadrangle formed by the lines $BL$, $BA$ and $CM$, $CA$, hence it passes through the midpoint of the segment $AD$. Similarly $YB_1$ passes through the midpoint of the segment $BD$. Therefore $F$ is the center of mass of points $A$, $B$, $C$ and $D$ with unit... | Belarus | SELECTION and TRAINING SESSION | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
051h | In a cyclic quadrilateral $ABCD$ we have $|AD| > |BC|$ and the vertices $C$ and $D$ lie on the shorter arc $AB$ of the circumcircle. Rays $AD$ and $BC$ intersect at point $K$, diagonals $AC$ and $BD$ intersect at point $P$. Line $KP$ intersects the side $AB$ at point $L$. Prove that $\angle ALK$ is acute. | [
"From the properties of cyclic quadrilaterals we get $\\angle KAB = \\angle KC'D$ and $\\angle KBA = \\angle KDC$. Let $A', B', K'$ be the feet of the altitudes of the triangle $ABK$ drawn from the vertices $A, B, K$, respectively, and let $H$ be the orthocenter of the triangle $ABK$ (Fig. 18). The points $A, B, A'... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / C... | null | proof only | null | |
0gc2 | 給定一個有 $n$ 條邊的連通圖, 其中任兩點之間至多只有一條邊。對於此圖中的任兩個環 $C$ 和 $C'$, 定義其**外環**為
$$
C \star C' = \{x \mid x \in (C - C') \cup (C' - C)\}.
$$
(1) 令 $r$ 為最大的正整數, 使得我們能夠從這張圖中選出 $r$ 個環 $C_1, C_2, \dots, C_r$, 且對於所有 $1 \le k \le r$ 與 $1 \le i, j_1, j_2, \dots, j_k \le r$, 我們有
$$
C_i \neq C_{j_1} \star C_{j_2} \star \dots \star C_{j_k}.
$... | [
"考慮一種選出滿足條件的 $s$ 邊的方法, 並將這 $s$ 邊塗為紅色, 其他 $n-s$ 邊塗為藍色。若紅邊不是連通的, 則加入連接不同連通塊的邊必不會產生新的圈, 違反 $s$ 的最大性, 所以這些紅邊是連通且無圈的, 因此這 $s$ 邊會構成一棵樹。此時考慮藍邊 $b_1, b_2, \\dots, b_{n-s}$, 則由定義, 每條藍邊都會與某些紅邊構成恰好一個圈, 將其分別記為 $D_1, D_2, \\dots, D_{n-s}$。\n\n我們首先證明沒有 $D_i = D_{j_1} \\star D_{j_2} \\star \\dots \\star D_{j_k}$, 這是因為每條藍邊都唯一屬於一個 ... | Taiwan | 二〇一八數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
0g2x | Problem:
Sei $ABC$ ein Dreieck mit $\angle CAB = 2 \cdot \angle ABC$. Nehme an, dass ein Punkt $D$ im Inneren des Dreiecks $ABC$ existiert, sodass $AD = BD$ und $CD = AC$. Zeige, dass $\angle ACB = 3 \cdot \angle DCB$. | [
"Solution:\n\n\n\nWir wollen zunächst etwas mehr über die geometrische Situation dieser Aufgabe herausfinden: Damit $AD = BD$ gelten kann, muss $D$ auf der Mittelsenkrechten von $AB$ liegen.\n\nAus $CD = AC$ folgt, dass $D$ auf einem Kreis mit Mittelpunkt $C$ und Radius $AC$ liegt. Damit is... | Switzerland | SMO 2019 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0foh | Dado un conjunto $X$ y una función $f : X \to X$, denotamos, para cada $x \in X$, $f^{-1}(x) = f(x)$ y, para cada $j \ge 1$, $f^{j+1}(x) = f(f^j(x))$. Decimos que $a \in X$, es un punto fijo de $f$ si $f(a) = a$. Para cada número real $x$, definimos $\pi(x)$ como la cantidad de primos positivos menores o iguales que $x... | [] | Spain | XXIX Olimpiada Iberoamericana de Matemáticas | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | Spanish | proof only | null | |
016y | Let $x_1$ and $x_2$ be real numbers and $0 < p < 1$. Define $x_n = p x_{n-1} + (1-p) x_{n-2}$ for $n = 3, 4, \dots$. Show that the sequence $(x_n)$ converges and determine $\lim_{n \to \infty} x_n$. | [
"Set $q = 1-p$. Let $n \\ge 4$. Then\n$$\n\\begin{aligned}\nx_{n+1} - x_{n-1} &= p x_n + q x_{n-1} - p x_{n-2} - q x_{n-3} \\\\\n&= p^2 x_{n-1} + p q x_{n-2} + q x_{n-1} - p x_{n-2} - q x_{n-3} \\\\\n&= p^2 x_{n-1} + p(q-1) x_{n-2} + q(x_{n-1} - x_{n-3}) \\\\\n&= p^2 x_{n-1} - p^2 x_{n-2} + q(x_{n-1} - x_{n-3}) \\\... | Baltic Way | BALTIC WAY | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (x_2 + (1-p)x_1)/(2-p) | |
0jhb | Problem:
Let $ABCD$ be a quadrilateral such that $\angle ABC = \angle CDA = 90^{\circ}$, and $BC = 7$. Let $E$ and $F$ be on $BD$ such that $AE$ and $CF$ are perpendicular to $BD$. Suppose that $BE = 3$. Determine the product of the smallest and largest possible lengths of $DF$. | [
"Solution:\n\nBy inscribed angles, $\\angle CDB = \\angle CAB$, and $\\angle ABD = \\angle ACD$. By definition, $\\angle AEB = \\angle CDA = \\angle ABC = \\angle CFA$. Thus, $\\triangle ABE \\sim \\triangle ADC$ and $\\triangle CDF \\sim \\triangle CAB$. This shows that\n\n$$\n\\frac{BE}{AB} = \\frac{CD}{CA} \\tex... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 9 | |
07b0 | Simple polygon $A$ with perimeter $p$ is called a **rotund polygon**, if for each two points $x$ and $y$ on the perimeter of $A$ that have a distance of at most $1$ in the plane, their distance on $A$ (i.e. the smaller part of the perimeter of $A$ that lies between them) is at most $\frac{p}{4}$. We want to prove that ... | [
"Throughout the solution the following notations will be used:\n\n$C_1(a, b)$: The smaller part of the perimeter of the polygon which lies between $a$ and $b$.\n\n$C_2(a, b)$: The other part of the perimeter of the polygon joining $a$ and $b$.\n\n$d(a, b)$: The length of $C_1(a, b)$.\n\n$[a, b]$: A chord of the pol... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0hdh | Increasing geometric progression (sequence) of 5 natural numbers satisfies the following condition: the square of the sum of first and fourth elements is 100 times greater than the sum of first, fifth, and doubled third elements. Find the largest 3-digit number that can be a member of such progression.
(Bogdan Rublyov... | [
"Let $b_1 = b$, $b_2 = bq$, $b_3 = bq^2$, $b_4 = bq^3$, $b_5 = bq^4$ denote elements of the progression. Rewriting the problem by using our notation, we get:\n$$\n(b_2 + b_4)^2 = 100(b_1 + 2b_3 + b_5)\n$$\nSubstitute the terms:\n$$\n(bq + bq^3)^2 = 100(b + 2bq^2 + bq^4)\n$$\n$$\nb^2q^2(1 + q^2)^2 = 100b(1 + 2q^2 + ... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 625 | |
0c6b | Consider a set of integers $M$, having the properties:
i) $1$ belongs to $M$;
ii) if $x$ and $y$ are in $M$, then $2x + 3y$ is in $M$;
iii) if $x, y$ are integers such that $4x - 3y$ belongs to $M$, then $x \cdot y$ belongs to $M$.
Prove that $M$ contains the numbers $2$, $3$, $4$, $5$ and $2019$. | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
01yc | Integers from $1$ to $2022$ are written on the cards and placed in a row on the table. Each number is used once and there is exactly one number on each card. Mary plays the following game: on each move she takes any card from the table and puts it into her right pocket, then she takes the leftmost card and puts it into... | [
"Let Mary always choose the card with the maximal number among the two leftmost cards. Then on $i$-th move the difference between the sums of numbers in her pockets increases at least by $1$ which implies that this strategy allows to get at least $S = 2 + 4 + \\dots + 2022 = 1011 \\cdot 1012$. In particular, $S(P) ... | Belarus | SELECTION and TRAINING SESSION | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2^1011 | |
09q6 | Problem:
Zij $n \geq 2$ een positief geheel getal en $p$ een priemgetal zodat $n \mid p-1$ en $p \mid n^{3}-1$. Bewijs dat $4 p-3$ een kwadraat is. | [
"Solution:\n\nUit $n \\mid p-1$ volgt $n<p$. Omdat $p$ priem is, is $p$ een deler van één van de twee factoren van $n^{3}-1=(n-1)\\left(n^{2}+n+1\\right)$, maar $p$ is te groot om een deler van $n-1>0$ te zijn. Dus $p \\mid n^{2}+n+1$. Wegens $n \\mid p-1$ kunnen we $p$ schrijven als $k n+1$, met $k$ een positief g... | Netherlands | Dutch TST | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0cjw | Let $ABC$ be a triangle with $\angle ABC = 2 \cdot \angle ACB$. Let $X$ and $Y$ be the midpoints of arcs $AB$ and $BC$ (not containing $C$ and $A$, respectively) of the circumcircle of triangle $ABC$. Let $BL$ be the angle bisector of $\angle ABC$, with $L \in AC$. Given that $\angle XLY = 90^\circ$, determine the meas... | [
"From (1) and (2), we obtain that $\\triangle LBY \\equiv \\triangle LCY$ by the SSS criterion, and thus $\\angle BLY = \\angle CLY$, which implies that $LY$ is the angle bisector of $\\angle BLC$ (3).\nOn the other hand, $\\overline{AX} = \\overline{BX}$ implies that $\\angle ACX = \\angle XCB$, so $XC$ is the ang... | Romania | 75th NMO Selection Tests | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | ∠ABC = 90°, ∠ACB = 45°, ∠BAC = 45° | |
0bpa | Problem:
Fie $A \in \mathcal{M}_{5}(\mathbb{C})$ o matrice cu $\operatorname{tr}(A)=0$ şi cu proprietatea că $I_{5}-A$ este inversabilă. Să se arate că $A^{5} \neq I_{5}$. | [
"Solution:\n\nPresupunem prin reducere la absurd că $A^{5}=I_{5}$. Fie $\\lambda \\in \\mathbb{C}$ o valoare proprie a matricei $A$ (o soluţie a ecuaţiei $\\operatorname{det}\\left(x I_{5}-A\\right)=0$ ). Cum $\\lambda^{5}$ este o valoare proprie a matricei $A^{5}$, obţinem $\\lambda^{5}=1$.\n\nMatricea $I_{5}-A$ e... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0gik | A Facebook group has some members, some pairs of them are friends to each other. For any two members, if they are friends, then they have no common friend; on the other hand, if they are not friends, then they have exactly two common friends. Prove that there exist two members with the same number of friends.
某臉書社團有若干... | [
"以成員為點，朋友關係為邊作圖 $G = (V, E)$。對於任一點 $a$，令 $N(a)$ 為與 $a$ 相連的所有點所構成集合。由條件一知，對於有連線的任兩點 $a$ 與 $b$，$N(a) - \\{b\\}$ 與 $N(b) - \\{a\\}$ 沒有交集（否則交集的點會成為這兩個人的共同好友，矛盾）。此外，對於任何 $x \\in N(a) - \\{b\\}$，因為 $b$ 與 $x$ 不是朋友且 $a$ 是他們的兩個共同好友，故存在唯一的 $y \\in N(b) - \\{a\\}$ 與 $x$ 有連線。以上建立了一個 $N(a) - \\{b\\}$ 與 $N(b) - \\{a\\}$ 之間的一對一且映... | Taiwan | APMO Taiwan Preliminary Round 2 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | Chinese; English | proof only | null | |
0isl | Problem:
A triangle has altitudes of length $15$, $21$, and $35$. Find its area. | [
"Solution:\nAnswer: $245 \\sqrt{3}$\n\nIf $A$ is the area of the triangle, the sides are $\\frac{2A}{15}$, $\\frac{2A}{21}$, and $\\frac{2A}{35}$. So the triangle is similar to a $\\frac{1}{15}$, $\\frac{1}{21}$, $\\frac{1}{35}$ triangle, which is similar to a $3, 5, 7$ triangle. Let the sides be $3k$, $5k$, and $7... | United States | 1st Annual Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 245 sqrt(3) | |
05v6 | Problem:
Soit $ABC$ un triangle dans lequel $AB < AC$. Soit $\omega$ un cercle passant par $B$ et $C$ et on suppose que le point $A$ se trouve à l'intérieur du cercle $\omega$. Soient $X$ et $Y$ des points de $\omega$ tels que $\widehat{BX A} = \widehat{A Y C}$. On suppose que $X$ et $C$ sont situés de part et d'autre... | [
"Solution:\n\nTout d'abord, nous allons chercher à tracer une figure exacte, en espérant que les raisonnements mis en oeuvre pour le tracé nous éclairerons sur la dynamique de la figure.\n\n\n\nPour tracer la figure, on cherche à tirer profit de l'hypothèse que les points $X, B, C$ et $Y$ s... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.