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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0lcj | Let be given a square $ABCD$ and $2009$ points inside the square such that no three points from these $2013$ points are collinear (including $4$ points $A$, $B$, $C$, $D$). We connect some points inside the square (as well as the vertices $A$, $B$, $C$, $D$) to partite the square into triangles. Each connecting segment... | [] | Vietnam | Vietnamese Mathematical Competitions | [
"Discrete Mathematics > Graph Theory > Menger's theorem / max-flow, min-cut",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
062g | Problem:
Es sei $ABCD$ ein gleichschenkliges Trapez mit $AB \parallel CD$ und $\overline{BC} = \overline{AD}$. Die Parallele zu $AD$ durch $B$ treffe die Senkrechte zu $AD$ durch $D$ im Punkt $X$. Ferner treffe die durch $A$ gezogene Parallele zu $BD$ die Senkrechte zu $BD$ durch $D$ im Punkt $Y$. Man beweise, dass di... | [
"Solution:\n\nErste Lösung. Es sei $M$ der Mittelpunkt der Strecke $AB$. Die durch $M$ gezogene Parallele zu $DY$ treffe $BD$ in $G$ und die durch $A$ gezogene Parallele zu $BD$ in $H$. Aus $\\overline{AM} = \\overline{MB}$ schließt man leicht $\\overline{HM} = \\overline{MG}$, und da $DY \\perp DB$ vorausgesetzt i... | Germany | IMO-Auswahlklausur | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
02pm | Problem:
Para um inteiro positivo $n$ considere a função
$$
f(n)=\frac{4 n+\sqrt{4 n^{2}-1}}{\sqrt{2 n+1}+\sqrt{2 n-1}}
$$
Calcule o valor de
$$
f(1)+f(2)+f(3)+\cdots+f(40)
$$ | [
"Solution:\n\nSeja $a=\\sqrt{2 n+1}$ e $b=\\sqrt{2 n-1}$. Então $a b=\\sqrt{4 n^{2}-1}$, $a^{2}+b^{2}=4 n$ e $a^{2}-b^{2}=2$. Portanto,\n$$\nf(n)=\\frac{a^{2}+b^{2}+a b}{a+b}\n$$\nComo $a-b \\neq 0$, podemos escrever\n$f(n)=\\frac{a^{2}+b^{2}+a b}{a+b} \\cdot \\frac{a-b}{a-b}=\\frac{a^{3}-b^{3}}{a^{2}-b^{2}}=\\frac... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 364 | |
02e8 | $p(x_1, x_2, \dots, x_n)$ is a polynomial with integer coefficients. For each positive integer $r$, $k(r)$ is the number of $n$-tuples $(a_1, a_2, \dots, a_n)$ such that $0 \le a_i \le r - 1$ and $p(a_1, a_2, \dots, a_n)$ is prime to $r$. Show that if $u$ and $v$ are coprime then $k(u \cdot v) = k(u) \cdot k(v)$, and i... | [
"First observe that if $a_i \\equiv b_i \\pmod r$, $i = 1, 2, \\dots, n$ then $p(a_1, a_2, \\dots, a_n) \\equiv p(b_1, b_2, \\dots, b_n) \\pmod r$. If $u$ and $v$ are coprime then given $b_1, b_2, \\dots, b_n$ and $c_1, c_2, \\dots, c_n$ with $0 \\le b_i < u$ and $0 \\le c_i < v$ by the chinese remainder theorem th... | Brazil | IX OBM | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof only | null | |
06n5 | Given a $24 \times 24$ square grid, initially all its unit squares are coloured white. A *move* consists of choosing a row, or a column, and changing the colours of all its unit squares, from white to black, and from black to white. Is it possible that after finitely many moves, the square grid contains exactly 574 bla... | [
"No, it is not possible.\nSuppose it is possible that after finitely many moves, the square grid contains exactly 574 black unit squares. Then there are exactly $24 \\times 24 - 574 = 2$ white unit squares. No matter where the white unit squares are, we can always find a $2 \\times 2$ square that contains exactly 1... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No | |
03sg | Suppose four solid iron balls are placed in a cylinder with the radius of $1$ cm, such that every two of the four balls are tangent to each other, and the two balls in the lower layer are tangent to the cylinder base. Now put water into the cylinder. Then, to just submerge all the balls, we need a volume of ______ cm³ ... | [
"Let points $O_1$, $O_2$, $O_3$, $O_4$ be the centers of the four solid iron balls respectively, with $O_1$, $O_2$ belonging to the two balls in the lower layer, and $A$, $B$, $C$, $D$ be the projective points of $O_1$, $O_2$, $O_3$, $O_4$ on the base of the cylinder. $ABCD$ constitute a square with the side of $\\... | China | China Mathematical Competition | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes"
] | English | proof and answer | (1/3 + sqrt(2)/2) * pi | |
07pq | Let $p(x)$ and $q(x)$ be non-constant polynomial functions with integer coefficients. It is known that the polynomial
$$
p(x)q(x) - 2015
$$
has at least 33 different integer roots. Prove that neither $p(x)$ nor $q(x)$ can be a polynomial of degree less than three. | [
"Let $a_1, a_2, \\dots, a_{33}$ be different integer roots of $f(x) = p(x)q(x) - 2015$. Hence, $p(a_i)q(a_i) = 2015$ for all $i = 1, 2, 3, \\dots, 33$. It follows that all integers $p(a_i)$, $i = 1, 2, 3, \\dots, 33$ are divisors of $2015$. Because $2015 = 5 \\cdot 13 \\cdot 31$, this number has $16$ distinct integ... | Ireland | Ireland | [
"Algebra > Algebraic Expressions > Polynomials",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof only | null | |
0c7f | Find all positive integers $n \ge 4$ for which the following property holds true: *any distinct and nonzero complex numbers $a, b, c$, satisfying
$$
(a - b)^n + (b - c)^n + (c - a)^n = 0,
$$
*are the complex coordinates of the vertices of an equilateral triangle.* | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | English | proof and answer | n = 4, 5, 7 | |
0gdv | 設 $\mathbb{N}$ 為全體正整數所成之集合，並令
$$
A = \{2^a + a^2 \mid a \in \mathbb{N}\}.
$$
考慮一個定義在 $\mathbb{N}$ 上的函數序列 $\{h_i\}_{i \in \mathbb{N}}$ 如下：
$$
h_1(n) = n + \log_2 n; \quad h_{i+1}(n) = n + \log_2 h_i(n), \forall i \in \mathbb{N}.
$$
證明：存在一個單射函數 $g : \mathbb{R}^+ \to \mathbb{R}^+$ 滿足下列性質：
(a) 對每個正整數 $i$ 及 $n$, $n + \lfloo... | [
"The statement holds when $g$ is the inverse function of $2^x + x^2 + 1$ in $\\mathbb{R}^+$. The detail will be stated as follows.\n\nNote that $\\mathbb{N} \\setminus A = \\{1, 2, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16, 18, \\dots\\} = \\{b_n : n \\in \\mathbb{N}\\}$.\nAfter a rearrangement, we can assume $b_n ... | Taiwan | 2020 Taiwan IMO 2J | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Intermediate Algebra > Exponential functions",
... | null | proof only | null | |
0kwo | Suppose $a$, $b$, and $c$ are three complex numbers with product $1$. Assume that none of $a$, $b$, and $c$ are real or have absolute value $1$. Define
$$
p = (a + b + c) + \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \quad \text{and} \quad q = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}.
$$
Given that both $p$ a... | [
"Let us denote $a = \\frac{y}{x}$, $b = \\frac{z}{y}$, $c = \\frac{x}{z}$, where $x, y, z$ are nonzero complex numbers. Then\n$$\np+3 = 3 + \\sum_{\\text{cyc}} \\left(\\frac{x}{y} + \\frac{y}{x}\\right) = 3 + \\frac{x^2(y+z) + y^2(z+x) + z^2(x+y)}{xyz} \\\\\n= \\frac{(x+y+z)(xy+yz+zx)}{xyz}.\n$$\n$$\nq-3 = -3 + \\s... | United States | USA TSTST | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | (-3, 3) | |
08fi | Problem:
Lucia vuole scrivere tre interi positivi $a, b, c$ in modo che ognuno di essi sia un divisore di $30$ e che i massimi comuni divisori fra due termini consecutivi (cioè $\operatorname{MCD}(a, b)$ e $\operatorname{MCD}(b, c)$) siano numeri primi. In quanti modi può farlo?
(A) 69
(B) 72
(C) 105
(D) $2^{7}$
(E) ... | [
"Solution:\n\nLa risposta è (C). Distinguiamo diversi casi, a seconda del numero di fattori primi di $b$:\n\na. Se $b$ ha 0 fattori primi, allora $b=1$, e qualunque sia il valore di $a$ si ha $(a, b)=1$, che non è un numero primo.\n\nb. Se $b$ ha esattamente un fattore primo, cioè è esso stesso primo, allora la con... | Italy | Italian Mathematical Olympiad - February Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | MCQ | C | |
0hps | Problem:
Mr. Fat moves around on the lattice points according to the following rules: From point $(x, y)$ he may move to any of the points $(y, x)$, $(3x, -2y)$, $(-2x, 3y)$, $(x+1, y+4)$ and $(x-1, y-4)$. Show that if he starts at $(0,1)$ he can never get to $(0,0)$. | [
"Solution:\n\nObserve that for each of Mr. Fat's moves, the value of $x + y \\pmod{5}$ is invariant. Therefore, Mr. Fat can never reach $(0,0)$ from $(0,1)$."
] | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0in9 | Problem:
On the Cartesian grid, Johnny wants to travel from $(0,0)$ to $(5,1)$, and he wants to pass through all twelve points in the set $S=\{(i, j) \mid 0 \leq i \leq 1, 0 \leq j \leq 5, i, j \in \mathbb{Z}\}$. Each step, Johnny may go from one point in $S$ to another point in $S$ by a line segment connecting the tw... | [
"Solution:\n\nAnswer: 252. Observe that Johnny needs to pass through the points $(0,0), (1,0), (2,0), \\ldots, (5,0)$ in that order, and he needs to pass through $(0,1), (1,1), (2,1), \\ldots, (5,1)$ in that order, or else he will intersect his own path. Then, the problem is equivalent to interlacing those two sequ... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 252 | |
04bc | Determine the number of pairs of integers $(x, y)$ that satisfy the equality:
$$
(x + y + 2012)^2 = x^2 + y^2 + 2012^2 .
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 30 | |
0951 | Problem:
Să se determine toate numerele de patru cifre, ce sunt divizibile cu 11 și au suma cifrelor cu 1 mai mică decât produsul lor. | [
"Solution:\nNumărul $\\overline{a b c d}$ este divizibil cu $11$ dacă și numai dacă $[a+c-(b+d)]$ este divizibil cu $11$. Deoarece $b+d \\leq 18$ și $a+c \\leq 18$, atunci avem 2 cazuri posibile: \n1) $a+c-(b+d)=0$ și \n2) $a+c-(b+d)= \\pm 11$. \nDin condiția problemei avem $a+b+c+d+1=a b c d$, ceea ce implică $a b... | Moldova | A 61-a OLIMPIAD DE MATEMATICA A REPUBLICII MOLDOVA | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1133, 3113, 1331, 3311 | |
0460 | Let $m$ be a positive integer and $A$ be a finite set. Let $A_1, A_2, \dots, A_m$ be subsets of $A$ (not necessarily distinct). It is known that for any nonempty set $I \subseteq \{1, 2, \dots, m\}$,
$$
\left| \bigcup_{i \in I} A_i \right| \geq |I| + 1.
$$
Prove: the elements of $A$ can be coloured black or white, suc... | [
"We give three solutions as follows.\n\n**Solution 1**\n\nConstruct a bipartite graph $G$ whose two parts are $X = \\{A_1, A_2, \\dots, A_m\\}$ and $Y = A$: for $1 \\le i \\le m$ and $a \\in A$, $A_i \\in X$ and $a \\in Y$ are adjacent if and only if $a \\in A_i$. Since for each nonempty $I \\subseteq \\{1, 2, \\do... | China | China National Team Selection Test | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0j18 | Problem:
Let $S_{0}=0$ and let $S_{k}$ equal $a_{1}+2 a_{2}+\ldots+k a_{k}$ for $k \geq 1$. Define $a_{i}$ to be 1 if $S_{i-1}<i$ and -1 if $S_{i-1} \geq i$. What is the largest $k \leq 2010$ such that $S_{k}=0$ ? | [
"Solution:\n\nAnswer: 1092\n\nSuppose that $S_{N}=0$ for some $N \\geq 0$. Then $a_{N+1}=1$ because $N+1 \\geq S_{N}$. The following table lists the values of $a_{k}$ and $S_{k}$ for a few $k \\geq N$ :\n\n| $k$ | $a_{k}$ | $S_{k}$ |\n| :--- | ------: | :------ |\n| $N$ | | 0 |\n| $N+1$ | 1 ... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 1092 | |
09j1 | Let $a$ and $b$ be integers with $|a| \ge 2$. Prove that the sequence $a^1 + b, a^2 + b, \dots, a^n + b, \dots$ has 2022 consecutive members consisting of composite numbers. | [
"It is clear if $a$ and $b$ are not relatively prime, thus we assume that $a$ and $b$ are relatively prime.\nFor $n \\ge 1$, let $a_n = a^n + b$. First we choose $l \\ge 0$ such that $|a|^l > |b|$. Then for any $n \\ge l+1$, we have $|a_n| < |a|^n + |a|^l \\le 2|a|^n - |a|^l < |a_{n+1}|$.\nIn particular, for any $n... | Mongolia | Mongolian Mathematical Olympiad Round 3 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
01xw | The circles $\omega_1$ and $\omega_2$ centered at $O_1$ and $O_2$, respectively, intersect at the point $X$. The line $XO_1$ intersects $\omega_1$ at the points $X$ and $A$. The line $XO_2$ intersects $\omega_2$ at the points $X$ and $B$. Let $M$ be the midpoint of the segment $AB$. The ray $MO_1$ intersects $\omega_1$... | [] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof only | null | |
08yy | Determine the number of tuples of integers $(a_1, a_2, \dots, a_{17})$ which satisfy $2 \le a_i \le 20$ for all $i = 1, \dots, 17$ and
$$
a_1^{a_2} \cdots a_{16}^{a_{17}} \equiv a_2^{a_3} \cdots a_{17}^{a_{17}} \equiv 1 \pmod{17}.
$$
Here the exponential is calculated in order from upper right two numbers. | [
"$\\boxed{2042 \\cdot 19^{14}}$\n\nFirst we show the following lemma.\n\n**Lemma.** Let $a, b, c$ be integers satisfying $a \\neq 0 \\pmod{17}$, $b \\ge 1$, $c \\ge 4$. Then $a^{b^c} \\equiv 1 \\pmod{17}$ holds if and only if $a \\equiv 1 \\pmod{17}$ or $b$ is even.\n\n**Proof.** If $a \\equiv 1 \\pmod{17}$, obviou... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | 2042 * 19^14 | |
0eos | Some balls were distributed into $2015$ boxes which were arranged in a row as indicated below. Any four consecutive boxes always had a total of $30$ balls. How many balls were there in the $2015$th box?
 | [
"If the number of balls in the boxes are denoted $x_1, x_2, \\dots$, then\n\n$$\nx_n + x_{n+1} + x_{n+2} + x_{n+3} = 30 = x_{n+1} + x_{n+2} + x_{n+3} + x_{n+4}.\n$$\n\nThus $x_n = x_{n+4}$ for all $n$, which means that the numbers recur in cycles of length four. Since $2015 = 4 \\times 503 + 3$, it follows that $x_... | South Africa | South African Mathematics Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 7 | |
0ak7 | Let $H$ be the orthocenter of triangle $ABC$. Let $M$ and $N$ be the midpoints of the sides $AB$ and $AC$ respectively. Assume that $H$ lies inside the quadrilateral $BMNC$ and that the circumcircles of triangles $BMH$ and $CNH$ are tangent to each other. The line through $H$ parallel to $BC$ intersects the circumcircl... | [] | North Macedonia | Asian-Pacific Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Transformations > Homot... | English | proof only | null | |
03ct | Let $\triangle ABC$ be a triangle with an incenter $I$. The line $CI$ intersects for a second time the circumcircle of $\triangle ABC$ at $L$, where $CI = 2 \cdot IL$. Points $M$ and $N$ lie on the segment $AB$, such that $\angle AIM = \angle BIN = 90^\circ$. Prove that $AB = 2 \cdot MN$. | [
"We adopt the standard notation for $\\triangle ABC$. Denote by $P$ and $Q$ the midpoints of $AC$ and $BC$, respectively. Let $J$ be the center of the excircle, tangent to the segment $AB$. It is well known that $I$ is the midpoint of $CJ$, therefore $PI$ is a midsegment for $\\triangle AJC$ and $PI \\parallel AJ$.... | Bulgaria | Bulgaria 2022 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | null | proof only | null | |
0isg | Problem:
A Vandal and a Moderator are editing a Wikipedia article. The article originally is error-free. Each day, the Vandal introduces one new error into the Wikipedia article. At the end of the day, the moderator checks the article and has a $2/3$ chance of catching each individual error still in the article. After... | [
"Solution:\n\nConsider the error that was introduced on day 1. The probability that the Moderator misses this error on all three checks is $1/3^3$, so the probability that this error gets removed is $1 - \\frac{1}{3^3}$. Similarly, the probability that the moderator misses the other two errors are $1 - \\frac{1}{3^... | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 416/729 | |
0ai3 | Which of the following claims are true, and which of them are false? If a fact is true you should prove it, if it isn't, find a counterexample.
a) Let $a, b, c$ be real numbers such that $a^{2013} + b^{2013} + c^{2013} = 0$. Then $a^{2014} + b^{2014} + c^{2014} = 0$.
b) Let $a, b, c$ be real numbers such that $a^{201... | [
"Firstly, we know that for every real number $x$, $x^2 \\ge 0$ holds. The key idea in this problem is that the expression $a^{2014} + b^{2014} + c^{2014}$ is a sum of squares (which are nonnegative numbers). Thus $a^{2014} + b^{2014} + c^{2014} = 0 \\Leftrightarrow a = b = c = 0$.\n\na) No: It is sufficient to find... | North Macedonia | European Mathematical Cup | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | a) False. For example, take a = 1, b = 2^(1/2013), c = -3^(1/2013), then a^2013 + b^2013 + c^2013 = 1 + 2 − 3 = 0 but a^2014 + b^2014 + c^2014 > 0.
b) True. If a^2014 + b^2014 + c^2014 = 0 then a = b = c = 0, hence a^2015 + b^2015 + c^2015 = 0.
c) False. For example, a = 1, b = 0, c = −1 gives a^2013 + b^2013 + c^2013 ... | |
02h5 | Determine the smallest real number $C$ such that the inequality
$$
C(x_1^{2005} + x_2^{2005} + x_3^{2005} + x_4^{2005} + x_5^{2005}) \geq x_1 x_2 x_3 x_4 x_5 (x_1^{125} + x_2^{125} + x_3^{125} + x_4^{125} + x_5^{125})^{16}
$$
holds for all positive real numbers $x_1, x_2, x_3, x_4, x_5$. | [
"We have\n$$\n5 (x_1^{2005} + x_2^{2005} + x_3^{2005} + x_4^{2005} + x_5^{2005}) \\geq (x_1^5 + x_2^5 + x_3^5 + x_4^5 + x_5^5) (x_1^{2000} + x_2^{2000} + x_3^{2000} + x_4^{2000} + x_5^{2000})\n$$\nby Chebyshev. Also,\n$$\nx_1^5 + x_2^5 + x_3^5 + x_4^5 + x_5^5 \\geq 5x_1x_2x_3x_4x_5\n$$\nby AM-GM and\n$$\n\\frac{x_1... | Brazil | Brazil | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | 5^{15} | |
0hz4 | Problem:
One of the receipts for a math tournament showed that 72 identical trophies were purchased for $\$-99.9-$, where the first and last digits were illegible. How much did each trophy cost? | [
"Solution:\n\nThe price must be divisible by $8$ and $9$. Thus the last $3$ digits must be divisible by $8$, so the price ends with $992$, and the first digit must be $7$ to make the total divisible by $9$. $\\$799.92 / 72 = \\$11.11$."
] | United States | Harvard-MIT Math Tournament | [
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | $11.11 | |
0d1p | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfy
$$
\begin{aligned}
& f\left(\frac{\sqrt{3}}{3} x\right)=\sqrt{3} f(x)-\frac{2 \sqrt{3}}{3} x \\
& f(x) f(y)=f(x y)+f\left(\frac{x}{y}\right)
\end{aligned}
$$
for all $x, y \in \mathbb{R}$, with $y \neq 0$. | [
"Let $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ be a function which satisfies the two functional equations.\nBecause $f\\left(\\frac{\\sqrt{3}}{3} x\\right)-\\sqrt{3} f(1)=-\\frac{2 \\sqrt{3}}{3} \\neq 0$, either $f(1) \\neq 0$ or $f\\left(\\frac{\\sqrt{3}}{3} x\\right) \\neq 0$.\nFix $x_{0} \\in \\mathbb{R}$ with $... | Saudi Arabia | Selection tests for the Gulf Mathematical Olympiad 2013 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x + 1/x for x ≠ 0, and f(0) = 0 | |
0l10 | A right pyramid has regular octagon $ABCDEFGH$ with side length $1$ as its base and apex $V$. Segments $\overline{AV}$ and $\overline{DV}$ are perpendicular. What is the square of the height of the pyramid?
(A) $1$ \quad (B) $\frac{1+\sqrt{2}}{2}$ \quad (C) $\sqrt{2}$ \quad (D) $\frac{3}{2}$ \quad (E) $\frac{2+\sqrt{2... | [
"**Answer (B):** Let $O$ be the center of the octagon, and let $r = AO$. As can be seen from the figure below, $AD = 1 + \\sqrt{2}$.\n\n\n\nBecause $\\triangle AVD$ is an isosceles right triangle,\n$$\nAV = \\frac{\\sqrt{2}}{2} \\cdot AD = \\frac{2+\\sqrt{2}}{2}.\n$$\nApplying the Law of Co... | United States | 2024 AMC 12 B | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Linear Algebra > Vectors"
] | null | MCQ | B | |
00v0 | Prove that for every positive integer $k$ there exists an integer $n$ and distinct primes $p_1, p_2, \dots, p_k$ such that, if $A(n)$ denotes the number of integers in $\{1, 2, \dots, n\}$ which are relatively prime to $p_1p_2\dots p_k$, then
$$
\left| n \left(1 - \frac{1}{p_1}\right) \left(1 - \frac{1}{p_2}\right) \cd... | [
"If $k=1$, choose $p_1=3$, $n=2$. If $k=2$, choose $p_1=3$, $p_2=7$, $n=5$. Assume $k \\ge 3$. Let $p_1, p_2, \\dots, p_k$ be primes congruent to $3$ modulo $4$. By the Chinese Remainder Theorem choose $n \\equiv \\frac{p_i+1}{4} \\pmod{p_i}$ for every $i$, that is, choose an integer $n$ such that $p_1 p_2 \\cdots ... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induc... | English | proof only | null | |
00x0 | Problem:
Find the number of solutions of the equation $a e^{x} = x^{3}$. | [
"Solution:\nStudying the graphs of the functions $a e^{x}$ and $x^{3}$ it is easy to see that the equation always has one solution if $a < 0$ and can have $0$, $1$ or $2$ solutions if $a > 0$. Moreover, in the case $a > 0$ the number of solutions can only decrease as $a$ increases and we have exactly one positive v... | Baltic Way | Baltic Way | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | Number of real solutions: one solution for a ≤ 0 and for a = 27/e^3; two solutions for 0 < a < 27/e^3; no solutions for a > 27/e^3. | |
0bq9 | Problem:
Rezolvaţi, în mulţimea numerelor reale, ecuaţia $\frac{1}{\{x\}} = \frac{1}{x} + \frac{1}{[x]}$, unde $[x]$ reprezintă partea întreagă a numărului real $x$, iar $\{x\}$ reprezintă partea fracţionară a numărului real $x$. | [] | Romania | Olimpiada Națională de Matematică - Etapa Locală | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (1+sqrt(5))/2 | |
0hqr | Problem:
Suppose you have three children and 40 pieces of candy. How many ways are there to distribute the candy such that each child gets more than one but fewer than 20 pieces? | [
"Solution:\nWe can use the \"stars and bars\" method, since this is the equivalent of giving 40 pieces of candy to three children, such that each child gets at least two pieces. This is the same as giving $40-6=34$ pieces to the three children with no restrictions (since we can pretend six pieces are already given ... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 171 | |
0f4x | Problem:
A non-negative real is written at each vertex of a cube. The sum of the eight numbers is $1$. Two players choose faces of the cube alternately. A player cannot choose a face already chosen or the one opposite, so the first player plays twice, the second player plays once. Can the first player arrange that the... | [] | Soviet Union | 16th ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Yes | |
043o | The maximum of $f(x) = 2 \sin^2 x - \tan^2 x$ is ______. | [
"$$\n\\begin{aligned}\nf(x) &= 2(1 - \\cos^2 x) - \\frac{1 - \\cos^2 x}{\\cos^2 x} \\\\\n&= 3 - \\left( 2 \\cos^2 x + \\frac{1}{\\cos^2 x} \\right) \\\\\n&\\le 3 - 2\\sqrt{2 \\cos^2 x \\cdot \\frac{1}{\\cos^2 x}} = 3 - 2\\sqrt{2},\n\\end{aligned}\n$$\nWhen $2 \\cos^2 x = \\frac{1}{\\cos^2 x}$ (e.g., take $x = \\arc... | China | China Mathematical Competition | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | final answer only | 3 - 2\sqrt{2} | |
0i9u | Problem:
A room is built in the shape of the region between two semicircles with the same center and parallel diameters. The farthest distance between two points with a clear line of sight is $12$ m. What is the area (in $\mathrm{m}^{2}$) of the room? | [
"Solution:\n$18 \\pi$\nThe maximal distance is as shown in the figure (next page). Call the radii $R$ and $r$, $R > r$. Then $R^{2} - r^{2} = 6^{2}$ by the Pythagorean theorem, so the area is $(\\pi / 2) \\cdot (R^{2} - r^{2}) = 18 \\pi$.\n\n"
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 18π | |
0kac | Problem:
Given two distinct points $A$, $B$ and line $\ell$ that is not perpendicular to $A B$, what is the maximum possible number of points $P$ on $\ell$ such that $A B P$ is an isosceles triangle? | [
"Solution:\n\nIn an isosceles triangle, one vertex lies on the perpendicular bisector of the opposite side. Thus, either $P$ is the intersection of $A B$ and $\\ell$, or $P$ lies on the circle centered at $A$ with radius $A B$, or $P$ lies on the circle centered at $B$ with radius $A B$. Each circle-line intersecti... | United States | HMMT February 2019 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 5 | |
02fq | $A$ is a set of $n$ non-negative integers. We say it has property $P$ if the set $\{x + y \mid x, y \in A\}$ has $\frac{n(n+1)}{2}$ elements. We call the largest element of $A$ minus the smallest element the *diameter* of $A$. Let $f(n)$ be the smallest diameter of any set $A$ with property $P$. Show that $\frac{n^2}{4... | [
"Let $a_1 < a_2 < \\dots < a_n$ be the elements of $A$. The diameter of $A$ is $d(A) = a_n - a_1$. The minimum and maximum of $A + A = \\{x + y \\mid x, y \\in A\\}$ are $a_1 + a_1$ and $a_n + a_n$. Since $A + A$ has $\\frac{n(n+1)}{2}$ elements, $2a_n - 2a_1 + 1 \\ge \\frac{n(n+1)}{2} \\iff d(A) \\ge \\frac{n^2}{4... | Brazil | XIX OBM | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof only | null | |
0bkd | Let $ABCD$ be a square and consider the points $K \in (AB)$, $L \in (BC)$, and $M \in (CD)$ such that $KLM$ is a right isosceles triangle, with the right angle at $L$. Prove that the lines $AL$ and $DK$ are perpendicular to each other.
Bogdan Enescu
 | [
"It is not difficult to observe that $\\triangle KLB \\equiv \\triangle LMC$, hence $KB = LC$. Because $AB = BC$, it follows that $AK = BL$. But then, $\\triangle AKD \\equiv \\triangle BLA$, and since $AK \\perp BL$ and $AD \\perp BA$, we deduce that $AL \\perp KD$, as well."
] | Romania | 65th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0e7o | At most how many prime numbers can be contained in a non-constant geometric sequence of positive real numbers? | [
"The answer is $2$. As an example of such a sequence we can take $a_n = 2\\left(\\frac{3}{2}\\right)^{n-1}$, which contains two primes, $2$ and $3$.\n\nAssume that a geometric sequence $a_n = a q^{n-1}$, where $q \\neq 1$, contains three primes. Assume also that these primes are $a_k$, $a_m$ and $a_n$, where $k < m... | Slovenia | National Math Olympiad 2013 - Final Round | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | proof and answer | 2 | |
01rm | Given two hyperbolae $H_1$ and $H_2$ with the equations $y = 1/x$ and $y = -1/x$, respectively. A straight line meets $H_1$ at points $A$ and $B$, and meets $H_2$ at points $C$ and $D$. Let $O$ be the origin of coordinates.
Prove that the areas of the triangles $OAC$ and $OBD$ are equal. (S. Mazanik) | [
"Without loss of generality we can assume that the positions of all hyperbolae, lines, and points look like in the figure (otherwise we can rotate the plane by the angle which is a multiple of $90^\\circ$, and rename the points).\n\nLet $A(a; 1/a)$, $B(b; 1/b)$, $C(c; -1/c)$, $D(d; -1/d)$. ... | Belarus | FINAL ROUND | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof only | null | |
07l0 | Given an $n$-tuple of numbers $(x_1, x_2, \dots, x_n)$ where each $x_i = +1$ or $-1$, form a new $n$-tuple
$$
(x_1x_2, x_2x_3, x_3x_4, \dots, x_nx_1),
$$
and continue to repeat this operation. Show that if $n = 2^k$ for some integer $k \ge 1$, then after a certain number of repetitions of the operation, we obtain the $... | [
"Use induction on $k$. Result clear for $k=1$. Assume it is true for some $k > 1$ and now consider an arbitrary $n$-tuple $(x_1, x_2, \\dots, x_n)$ of length $n = 2^{k-1}$. Since $x_i^2 = 1$ for all $i$, the second iteration\n$$\n(x_1 x_2^2 x_3, x_2 x_3^2 x_4, \\dots, x_{n-1} x_n^2 x_1, x_n x_1^2 x_2)\n$$\ncan be w... | Ireland | Irska | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof only | null | |
02sg | Problem:
Considere a lista de números $a_{1}, a_{2}, \ldots$, onde
$$
a_{n}=\underbrace{111111 \ldots 1}_{3^{n} \text{ algarismos }}
$$
ou seja, $a_{1}=\underbrace{111}_{\text{três uns }}$, $a_{2}=\underbrace{111111111}_{\text{nove uns }}$, $a_{3}=\underbrace{111 \ldots 1}_{\text{vinte e sete uns }}$, e assim por diant... | [
"Solution:\na) Lembremos que um número é múltiplo de $3$ se, e somente se, a soma dos seus algarismos é um múltiplo de $3$. De modo similar, um número é múltiplo de $9$ se, e somente se, a soma de seus algarismos é um múltiplo de $9$. Como $a_{1}=111$, vemos que a soma dos seus algarismos é igual a $3$. Usando entã... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
04jo | On the board $8 \times 8$ tromino-tiles of the shape $\square$ have to be placed in such a way that each tile covers exactly three cells of the board and the tiles cannot overlap.
What is the least possible number of tromino-tiles that one can place on the board so that no additional tromino-tile can be placed afterwar... | [
"Let us divide the board into 16 $2 \\times 2$ squares as in the picture.\n\nIn each of those squares at least two cells have to be covered, otherwise we could place a tromino-tile on three uncovered cells. Hence, at least 32 cells have to be covered, and we need at least 11 tromino-tiles t... | Croatia | Croatia Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 11 | |
099a | Төрөл 1: Дурын хоосон биш $B_j$, $1 \le j \le 5$ хайрцгийг сонгож түүнээс 1 мөнгө авах ба $B_{j+1}$ хайрцагт 2 мөнгө нэмж хийнэ.
Төрөл 2: Дурын хоосон биш $B_k$, $1 \le k \le 4$ хайрцгийг сонгож түүнээс 1 мөнгө авч $B_{k+1}$ ба $B_{k+2}$ хайрцагт буй зүйлсийн байрыг сольж хийнэ. $B_1, B_2, B_3, B_4, B_5$, хайрцгууд хоо... | [
"Хэрэв бодлогын нөхцөлд зөвшөөрөгдсөн үйлдлүүдийг дараалсан тус бүр $a_1, \\dots, a_n$ ширхэг мөнгөтэй хайрцгууд дээр хийгээд харгалзан $a_1', \\dots, a_n'$ ширхэг мөнгөтэй хайрцгууд гаргаж авч болдог бол $(a_1, \\dots, a_n) \\to (a_1', \\dots, a_n')$ гэж тэмдэглэе. $A = 2010^{2010^{2010}}$ гэе. Бидний зорилго\n$$\... | Mongolia | International Mathematical Olympiad 51 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | Mongolian | proof and answer | Yes | |
08dm | Problem:
La calcolatrice di Pierino ha un display, che inizialmente mostra il numero $0$, e due tasti: il tasto $+1$, che aggiunge $1$ al numero scritto sul display, e il tasto $\times 3$, che moltiplica il numero scritto sul display per $3$. Se si preme il tasto $+1$ per due volte consecutive, la calcolatrice esplode... | [
"Solution:\n\nLa risposta è (A). I numeri che si possono ottenere con un'opportuna sequenza di tasti $+1$ e $\\times 3$ senza due $+1$ consecutivi sono tutti e soli quelli che in base $3$ si scrivono con solo cifre $0$ e $1$. Infatti, dato un tale numero, per ottenerlo sulla calcolatrice basta scorrere da sinistra ... | Italy | Gara di Febbraio | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | A | |
06af | Determine all positive integers $a$, $b$, $c$ for which there exist positive integers $x$, $y$, $z$, such that: $ab + 1 = x!$, $bc + 1 = y!$, $ca + 1 = z!$, where $n!$ denotes the product $1 \cdot 2 \cdot 3 \ldots \cdot n$. | [
"We can easily see that $x, y, z \\ge 2$.\nIf $x, y, z \\ge 3$, then observe that $3$ doesn't divide $a, b, c$. Indeed, if for example, $3 \\mid a$, then $3$ doesn't divide $ab + 1$, but $3$ divides $x!$, contradiction.\nThis means that two of the numbers are congruent mod $3$. Due to the symmetry,\nsuppose that $a... | Greece | Selection examinations | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | All solutions are the cyclic permutations of (1, 1, n! − 1) with n ≥ 2. | |
0c85 | Given an integer $n \ge 3$, determine the least value the sum $\sum_{i=1}^{n} (1/x_i - x_i)$ may achieve, as the $x_i$ run through the positive real numbers subject to $\sum_{i=1}^{n} \frac{1}{x_i + n - 1} = 1$. Also, determine the $x_i$ at which this minimum is achieved. | [
"The required minimum is $0$ and is achieved if and only if the $x_i$ are all equal to $1$. Let $x_1, \\dots, x_n$ be positive real numbers satisfying the condition in the statement. Let $y_i = x_i/(x_i + n - 1)$, $i = 1, 2, \\dots, n$, and notice that the $y_i$ are positive real numbers that add up to $1$. Express... | Romania | SELECTION TESTS FOR THE 2019 BMO AND IMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | The minimum value is 0, achieved exactly when all variables are equal to 1. | |
0kix | Problem:
Find (with proof) the units digit of the product of any 5 consecutive positive integers (consecutive means all in a row, like $5,6,7,8,9$). | [
"Solution:\nFor any 5 numbers in a row, one of them must be a multiple of $5$. Also, at least two of them are even. Thus, the product will be even and a multiple of $5$, so it has units digit $0$."
] | United States | Berkeley Math Circle: Monthly Contest 3 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 0 | |
08wj | Given two triangles $PAB$ and $PCD$ such that $PA = PB$, $PC = PD$, $P$, $A$, $C$ and $B$, $P$, $D$ are collinear in this order respectively. The circle $S_1$ passing through $A$, $C$ intersects with the circle $S_2$ passing through $B$, $D$ at distinct points $X$, $Y$. Prove that the circumcenter of the triangle $PXY$... | [] | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
08gw | Problem:
Consider a regular $2n+1$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $SE$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$... | [
"Solution:\n\nAnswer: $n=1$ is clearly a solution, we can just color each side of the equilateral triangle in a different color, and the conditions are satisfied. We prove there is no larger $n$ that fulfills the requirements.\n\nLemma 1. Given a regular $2n+1$-gon in the plane, and a sequence of $n+1$ consecutive ... | JBMO | null | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1 | |
04df | Determine all real solutions of the equations
$$
\begin{align*}
x^2 - y &= z^2 \\
y^2 - z &= x^2 \\
z^2 - x &= y^2.
\end{align*}
$$ | [
"By summing the given equations we get\n$$\nx + y + z = 0.\n$$\nIt follows that $z = -x - y$, and from the first equation we get\n$$\n\\begin{aligned}\nx^2 - y &= (-x - y)^2 \\\\\nx^2 - y &= x^2 + 2xy + y^2 \\\\\n2xy + y^2 + y &= 0 \\\\\ny(2x + y + 1) &= 0,\n\\end{aligned}\n$$\nso $y = 0$ or $2x + y + 1 = 0$.\n\nIf... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | [(0, 0, 0), (1, 0, -1), (0, -1, 1), (-1, 1, 0)] | |
019z | There are $n$ rooms in a sauna, each has unlimited capacity. At one time a room may be attended by people of the same gender (males or females). What's more, males want to share a room only with males that they don't know and females want to share a room only with females that they know. What's the biggest number $k$ s... | [
"First we'll show it by induction that it is possible for $n-1$ pairs to visit the sauna at the same time. Base of induction is clear.\n\nAssume that $n-2$ pairs may be placed in $n-1$ rooms. Take additional pair. Let $k$ be the number of pairs that they know and $m$ be the number of rooms taken by males.\n\nIf $m ... | Baltic Way | Baltic Way 2013 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n - 1 | |
0ivy | Problem:
If real numbers $a, b, c, d$ satisfy
$$
\frac{a+b}{c+d} = \frac{b+c}{a+d} \neq -1,
$$
prove that $a = c$. | [
"Solution:\nAssume for the sake of contradiction that $a \\neq c$. Cross-multiplying,\n$$\n\\begin{aligned}\n(a+b)(a+d) &= (c+b)(c+d) \\\\\na^2 + a b + a d + b d &= c^2 + b c + c d + b d \\\\\na^2 - c^2 + a b - b c + a d - c d &= 0 \\\\\n(a-c)(a+c) + (a-c) b + (a-c) d &= 0 \\\\\na+c+b+d &= 0 \\\\\nb+c &= -a-d.\n\\e... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof only | null | |
0goo | In an acute triangle $ABC$, let $D$ be point on the side $[BC]$ different than the vertices. Let $M_1, M_2, M_3, M_4, M_5$ be the midpoints of the line segments $[AD], [AB], [AC], [BD], [CD]$, respectively; $O_1, O_2, O_3, O_4$ be the circumcenters of the triangles $ABD, ACD, M_1M_2M_4, M_1M_3M_5$, respectively; $S$ an... | [
"As $\\angle O_1M_1A = \\angle O_1M_2A = 90^\\circ$, we have that $O_1M_1AM_2$ is a cyclic quadrilateral and the point $S$ is its circumcenter. Hence $S$ is the circumcenter of the triangle $AM_1M_2$.\n\nNext we observe that the triangles $AM_1M_2$ and $M_4M_2M_1$ are congruent since $AM_1 = M_4M_2 = \\frac{AD}{2}$... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous ... | English | proof only | null | |
03vc | Outside a convex quadrilateral $ABCD$ we construct equilateral triangles $ABQ$, $BCR$, $CDS$ and $DAP$. Denoting by $x$ the sum of the diagonals of $ABCD$, and by $y$ the sum of line segments joining the midpoints of opposite sides of $PQRS$, we find the maximum value of $\frac{y}{x}$. | [
"If $ABCD$ is a square, then $\\frac{y}{x} = \\frac{1+\\sqrt{3}}{2}$.\n\nNow we prove that $\\frac{y}{x} \\le \\frac{1+\\sqrt{3}}{2}$.\n\nDenote by $P_1, Q_1, R_1, S_1$ the midpoints of $DA$, $AB$, $BC$, $CD$, and by $E$, $F$, $G$, $H$ the midpoints of $SP$, $PQ$, $QR$, $RS$. Then $P_1Q_1R_1S_1$ is a parallelogram.... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous... | English | proof and answer | (1+sqrt(3))/2 | |
04xx | Let $a$, $b$ be integers with $b$ not a perfect square. Show that $x^2 + a x + b$ can be a perfect square only for finitely many integers $x$. | [
"Let us examine the diophantine equation $x^2 + a x + b = y^2$ with unknown integers $x$ and $y$. It can be transformed into the form $(2x + 2y + a)(2x - 2y + a) = a^2 - 4b$. Since we assume $b$ is not a perfect square, $a^2 - 4b \\neq 0$. There are only finitely many ways to write $a^2 - 4b$ as a product of two in... | Czech-Polish-Slovak Mathematical Match | Cesko-Slovacko-Poljsko 2013 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
0l7n | Let $n$ be a positive integer, and let $a_0, a_1, \dots, a_n$ be nonnegative integers such that $a_0 \ge a_1 \ge \dots \ge a_n$. Prove that
$$
\sum_{i=0}^{n} i \binom{a_i}{2} \le \frac{1}{2} \binom{a_0 + a_1 + \dots + a_n}{2}.
$$ | [] | United States | 16th United States of America Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0017 | Pablo estaba copiando el siguiente problema:
Considere todas las sucesiones de $2004$ números reales $(x_0, x_1, x_2, ..., x_{2001})$, tales que
$$
\begin{array}{l}
x_0 = 1, \\
0 \le x_1 \le 2x_0, \\
0 \le x_2 \le 2x_1, \\
\vdots \\
0 \le x_{2003} \le 2x_{2002}.
\end{array}
$$
Entre todas estas sucesiones, determine aq... | [] | Argentina | XVIII Olimpíada Iberoamericana de Matemática | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | español | proof only | null | |
02ka | Problem:
A função $f$ é dada pela tabela a seguir.
| $\boldsymbol{x}$ | 1 | 2 | 3 | 4 | 5 |
| :---: | :--- | :--- | :--- | :--- | :--- |
| $f(x)$ | 4 | 1 | 3 | 5 | 2 |
Por exemplo, $f(2)=1$ e $f(4)=5$. Quanto vale $\underbrace{f(f(f(f(\ldots \ldots . . f}_{2004 \text{ veces }}(4) \ldots .)))}$?
A) 1
B) 2
C) 3
D) 4
... | [
"Solution:\n\nDa tabela temos:\n\n$$\n\\begin{gathered}\nf(4)=5 \\, , \\, f(f(4))=f(5)=2 \\, , \\, f(f(f(4)))=f(f(5))=f(2)=1 \\, , \\\\\nf(f(f(f(4))))=f(f(f(5)))=f(f(2))=f(1)=4 \\\\\n\\text{Logo, } \\underbrace{f(f(f(f(4))))}_{\\text{4 vezes}}=4.\n\\end{gathered}\n$$\n\nComo $2004$ é múltiplo de $4$, segue que $\\u... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | MCQ | D | |
0f7u | Problem:
Your opponent has chosen a $1 \times 4$ rectangle on a $7 \times 7$ board. At each move you are allowed to ask whether a particular square of the board belongs to his rectangle. How many questions do you need to be certain of identifying the rectangle? How many questions are needed for a $2 \times 2$ rectangl... | [] | Soviet Union | 21st ASU | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 1×4: 7 questions. 2×2: 9 questions. | |
0eqq | In the school library there are $1\,024$ books of which $64$ are detective stories. Twenty-four of these detective stories are about Sherlock Holmes. What is the probability that a book you selected randomly is a detective story which is not about Sherlock Holmes?
(A) $\frac{1}{16}$ (B) $\frac{3}{8}$ (C) $\frac{3}{128... | [
"There are $64 - 24 = 40$ detective stories which are not about Sherlock Holmes. The probability of choosing one of these at random out of $1\\,024$ books is $\\frac{40}{1\\,024} = \\frac{5}{128}$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Statistics > Probability > Counting Methods > Other",
"Math Word Problems"
] | English | MCQ | D | |
00ts | A *super-integer* triangle is defined to be a triangle whose lengths of all sides and at least one height are positive integers. We will deem certain positive integer numbers to be *good* with the condition that if the lengths of two sides of a super-integer triangle are two (not necessarily different) good numbers, th... | [
"Evidently, all right-angle triangles with integer sides are super-integer triangles. We will use the following notation $(a, b, c\\{h\\})$ to denote a super-integer triangle whose sides are $a$, $b$ and $c$ and the height of integer length is $h$. The height will be written in curly brackets next to the correspond... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Other"
] | null | proof only | null | |
07mh | Tangents at points $A$ and $B$ on a circle, centre $O$, meet at $P$. The point $A$ is joined to the mid-point $M$ of $PB$ meeting the circle again at $C$. The line through $C$ and the mid-point $D$ of $PO$ meets the circle again at $E$. Prove $AE \parallel PO$. | [
"Let $AM$ meet $PO$ at $F$ and let $PO$ extended meet the circle at $G$ and $H$ with $G$ between $P$ and $O$. Let $AB$ meet $PO$ at $Q$.\n\n\n\nFirst note that $F$ is the centroid of $\\triangle ABP$ and so $|FQ| = \\frac{1}{3}|PQ|$. Using that $D$ is the mid-point of $PO$ and denoting the ... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | null | proof only | null | |
0epi | At a certain school, break starts at 11:45 and ends at 12:12. How long is break, in minutes? | [
"Since $11{:}45$ is $45$ minutes after $11{:}00$ and $12{:}12$ is $60 + 12 = 72$ minutes after $11{:}00$, the length of break is $72 - 45 = 27$ minutes.\n\nAlternatively, there are $15$ minutes from $11{:}45$ to $12{:}00$ and $12$ minutes from $12{:}00$ to $12{:}12$, so the length of break is $15 + 12 = 27$ minutes... | South Africa | South African Mathematics Olympiad | [
"Math Word Problems"
] | English | final answer only | 27 | |
05rx | Problem:
Dans un tournoi auxquels participent $n$ joueurs, numérotés de 1 à $n$, chaque paire de joueurs se rencontre exactement une fois. Cette rencontre se termine par la victoire d'un des deux joueurs et la défaite de l'autre joueur. On note $v_{k}$ le nombre de victoires du joueur $k$ au cours du tournoi, et $d_{k... | [
"Solution:\n\nPuisque chaque joueur a disputé $n-1$ parties et que l'ensemble des joueurs a totalisé $n(n-1)/2$ victoires, on sait que\n$$\n\\sum_{k=1}^{n} v_{k} = n(n-1)/2\n$$\net que $v_{k} + d_{k} = n-1$ pour tout $k \\leqslant n$. On en déduit que\n$$\n\\sum_{k=1}^{n} d_{k}^{2} = \\sum_{k=1}^{n} (n-1-v_{k})^{2}... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0kbd | Problem:
Prove that $\sqrt{n+1}+\sqrt{n}$ is irrational for every positive integer $n$. | [
"Solution:\nAssume for contradiction that it was rational, and let $q$ denote its value. Squaring, we find that\n$$\nq^{2} = (n+1) + 2 \\sqrt{n} \\cdot \\sqrt{n+1} + n\n$$\nso\n$$\n\\frac{q^{2} - (2n+1)}{2} = \\sqrt{n(n+1)}\n$$\nThe left-hand side is also rational, so we conclude the quantity $n(n+1)$ is the square... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0byh | Let $a$ and $n$ be two fixed positive integers.
a) Prove that there exist $n$ positive integers $a_1, a_2, \dots, a_n$ such that
$$
1 + \frac{1}{a} = \left(1 + \frac{1}{a_1}\right) \left(1 + \frac{1}{a_2}\right) \dots \left(1 + \frac{1}{a_n}\right).
$$
b) Prove that $1 + \frac{1}{a}$ has only finitely many possible re... | [
"a) If $a_1 < a_2 < \\dots < a_n$ are consecutive positive integers, then we have\n$$\n\\left(1 + \\frac{1}{a_1}\\right) \\left(1 + \\frac{1}{a_2}\\right) \\dots \\left(1 + \\frac{1}{a_n}\\right) = \\frac{a_n + 1}{a_1}.\n$$\nChoosing $a_k = an + k - 1, \\forall k = \\overline{1, n}$, we get\n$$\n\\left(1 + \\frac{1... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
03jv | Problem:
Let $ABCD$ be a convex quadrilateral inscribed in a circle, and let diagonals $AC$ and $BD$ meet at $X$. The perpendiculars from $X$ meet the sides $AB$, $BC$, $CD$, $DA$ at $A'$, $B'$, $C'$, $D'$ respectively. Prove that
$$
|A'B'| + |C'D'| = |A'D'| + |B'C'|
$$
($|A'B'|$ is the length of line segment $A'B'$, e... | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
054a | Let $m$ be an integer, $m \ge 2$. Each student in a school is practising $m$ hobbies the most. Among any $m$ students there exist two students who have a common hobby. Find the smallest number of students for which there must exist a hobby which is practised by at least 3 students. | [
"If the number of students is $m^2 - 1 = (m-1)(m+1)$, let us split the students to $m-1$ groups, each with $m+1$ students. Assume that each student has a unique common hobby with every other student in the same group and none of the students is practising any other hobby. Then each student has exactly $m$ hobbies b... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | m^2 | |
0hzg | Problem:
Find
$$
\int_{-4 \pi \sqrt{2}}^{4 \pi \sqrt{2}}\left(\frac{\sin x}{1+x^{4}}+1\right) d x
$$ | [
"Solution:\nThe function $\\frac{\\sin x}{1+x^{4}}$ is odd, so its integral over this interval is $0$. Thus we get the same answer if we just integrate $d x$, namely, $8 \\pi \\sqrt{2}$."
] | United States | Harvard-MIT Math Tournament | [
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | final answer only | 8*pi*sqrt(2) | |
0dqr | In the acute-angled non-isosceles triangle $ABC$, $O$ is its circumcentre, $H$ is its orthocentre and $AB > AC$. Let $Q$ be a point on $AC$ such that the extension of $HQ$ meets the extension of $BC$ at the point $P$. Suppose $BD = DP$, where $D$ is the foot of the perpendicular from $A$ onto $BC$. Prove that $\angle O... | [
"Drop perpendiculars $OM$ and $QX$ onto $BC$, and $QY$ from $Q$ onto $AD$. First $2DM = DM + BD - BM = BD - (BM - DM) = PD - (CM - DM) = PD - CD = PC$. It is a well-known fact that $2OM = AH$.\n\n\n\nNext $\\angle CPQ = \\angle DBH = \\angle HAQ$ so that the triangles $CPQ$ and $HAQ$ are si... | Singapore | Singapore Mathematical Olympiad (SMO) 2011 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Co... | null | proof only | null | |
0eu9 | Let $f: \mathbb{N} \to \mathbb{N}$ be a function satisfying
$$
k f(n) \le f(kn) \le k f(n) + k - 1
$$
for every $k, n \in \mathbb{N}$, where $\mathbb{N}$ is the set of all positive integers.
(1) Show that, for every $a, b \in \mathbb{N}$,
$$
f(a) + f(b) \le f(a + b) \le f(a) + f(b) + 1.
$$
(2) Show that, if $f$ satisfi... | [
"(1) Let $a, b$ be positive integers. Since\n$$\n\\begin{aligned}\n(a+b)(f(a)+f(b)) &= (a+b)f(a) + (a+b)f(b) \\\\\n&\\le f(a(a+b)) + f(b(a+b)) \\\\\n&\\le af(a+b) + a-1 + bf(a+b) + b-1 \\\\\n&< (a+b)(f(a+b)+1),\n\\end{aligned}\n$$\nwe have $f(a) + f(b) < f(a + b) + 1$. Because $f$ is an integer-valued function, $f(... | South Korea | 20th Korean Mathematical Olympiad Final Round | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0f6w | Problem:
Two points $A$ and $B$ are inside a convex 12-gon. Show that if the sum of the distances from $A$ to each vertex is $a$ and the sum of the distances from $B$ to each vertex is $b$, then $|a - b| < 10 |AB|$. | [] | Soviet Union | 20th ASU | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
07pc | In the triangle $ABC$, the length of the altitude from $A$ to $BC$ is equal to $1$. $D$ is the midpoint of $AC$. What are the possible lengths of $BD$? | [
"**Solution 1.** Place $A$ at $(0, 1)$, $B$ at $(b, 0)$ and $C$ at $(c, 0)$ with $b \\neq c$. Then the coordinates of $D$ are $(c/2, 1/2)$. The length of $BD$ is thus $\\sqrt{(b-c/2)^2 + 1/4}$. This has a minimum value of $1/2$ when $b=c/2$ and $b \\neq 0$. There is no upper bound.\n\n**Solution 2.** Let the positi... | Ireland | Ireland | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | [1/2, ∞) | |
074b | Let $v_1, v_2, \dots, v_n$ be $n (\ge 2)$ unit vectors in the plane. Prove that there exist $\lambda_1, \lambda_2, \dots, \lambda_n$, each equal to $+1$ or $-1$, such that
$$
|\lambda_1 v_1 + \lambda_2 v_2 + \dots + \lambda_n v_n| \le \sqrt{2}.
$$
(Here $|v|$ denotes the length of the vector $v$.) | [
"We prove the result for a more general class of vectors having magnitude not exceeding $1$. We use induction on $n$. If $n = 2$, parallelogram law gives\n$$\n|v_1 + v_2|^2 + |v_1 - v_2|^2 = 2(|v_1|^2 + |v_2|^2) \\le 4.\n$$\nHence either $|v_1 + v_2| \\le \\sqrt{2}$ or $|v_1 - v_2| \\le \\sqrt{2}$, giving the resul... | India | Indija TS 2009 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Linear Algebra > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0jdd | In triangle $ABC$, points $P, Q, R$ lie on sides $BC, CA, AB$, respectively. Let $\omega_A, \omega_B, \omega_C$ denote the circumcircles of triangles $AQR, BRP, CPQ$, respectively. Given the fact that segment $AP$ intersects $\omega_A, \omega_B, \omega_C$ again at $X, Y, Z$ respectively, prove that $YX/XZ = BP/PC$.
(T... | [
"Assume that $\\omega_B$ and $\\omega_C$ intersect again at second point $S$ other than $P$. If not, the degenerate case where $\\omega_B$ and $\\omega_C$ are tangent at $P$ can be dealt similarly. Because $BPSR$ and $CPSQ$ are cyclic, we have $\\angle RSP = 180^\\circ - \\angle PBR$ and $\\angle PSQ = 180^\\circ -... | United States | USAMO | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
053j | Each point at the sides of an equilateral triangle is coloured either red or blue. Is it sure that there exists a right triangle whose all vertices are of the same colour? | [
"Let the equilateral triangle be $XYZ$. We show that there exists a point on some side that has the same colour as its projection to another side. For that, take points $P$, $Q$ and $R$ on sides $XY$, $YZ$ and $ZX$, respectively, in such a way that $XP : XY = YQ : YZ = ZR : ZX = 1 : 3$ (Fig. 17).\n\nThen $PQ \\perp... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Yes | |
02db | Given 5 points of a sphere of radius $r$, show that two of the points are a distance less than or equal to $r\sqrt{2}$ apart. | [
"Suppose the result is false so that we can find 5 points with the distance between any two greater than $r\\sqrt{2}$. Then the angle subtended by any two at the center of the sphere is greater than $90^\\circ$. Take one of the points to be at the north pole. Then the other four must all be south of the equator. Tw... | Brazil | II OBM | [
"Geometry > Solid Geometry > 3D Shapes",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
0kg1 | Let $\mathbb{N}^2$ denote the set of ordered pairs of positive integers. A finite subset $S$ of $\mathbb{N}^2$ is *stable* if whenever $(x, y)$ is in $S$, then so are all points $(x', y')$ of $\mathbb{N}^2$ with both $x' \le x$ and $y' \le y$.
Prove that if $S$ is a stable set, then among all stable subsets of $S$ (inc... | [
"Suppose $|S| \\ge 2$. For any $p \\in S$, let $R(p)$ denote the stable rectangle with upper-right corner $p$. We say such $p$ is *pivotal* if $p + (1, 1) \\notin S$ and $|R(p)|$ is even.\n\n\n**Claim** — If $|S| \\ge 2$, then a pivotal $p$ always exists.\n*Proof.* Consider the top row of $... | United States | USA TSTST | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0hsb | Problem:
The sum of the squares of five real numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ equals $1$. Prove that the least of the numbers $(a_{i}-a_{j})^{2}$, where $i, j=1,2,3,4,5$ and $i \neq j$, does not exceed $1/10$. | [
"Solution:\n\nAssume w.l.o.g. that $a_{1} \\leq a_{2} \\leq a_{3} \\leq a_{4} \\leq a_{5}$. If $m$ is the least value of $|a_{i}-a_{j}|$, $i \\neq j$, then $a_{i+1}-a_{i} \\geq m$ for $i=1,2, \\ldots, 4$, and consequently $a_{i}-a_{j} \\geq (i-j) m$ for any $i, j \\in \\{1, \\ldots, 5\\}$, $i>j$. Then it follows th... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0ka0 | Problem:
Compute the number of ordered pairs of integers $(x, y)$ such that $x^{2}+y^{2}<2019$ and
$$
x^{2}+\min (x, y)=y^{2}+\max (x, y)
$$ | [
"Solution:\n\nWe have\n$$\nx^{2}-y^{2} = \\max(x, y) - \\min(x, y) = |x-y|\n$$\nNow if $x \\neq y$, we can divide by $x-y$ to obtain $x+y= \\pm 1$. Thus $x=y$ or $x+y= \\pm 1$.\n\nIf $x=y$, we see that $2019 > x^{2} + y^{2} = 2x^{2}$, so we see that $-31 \\leq x \\leq 31$. There are 63 ordered pairs in this case.\n... | United States | HMMT November 2019 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 127 | |
0cg8 | Let $I \subset \mathbb{R}$ be an open interval and consider $f : I \to \mathbb{R}$ a function that is twice differentiable on $I$, such that $f(x) \cdot f''(x) = 0$, for any $x \in I$. Show that $f''$ is the zero function.
Sorin Rădulescu and Mihai Piticari | [
"Consider the set $A = \\{x \\in I \\mid f''(x) \\ne 0\\}$. If, by way of contradiction, $A \\ne \\emptyset$, as $f'' = (f')'$ has the intermediate point value on $I$, the set $A$ cannot be a singleton.\n\nSo, let $a, b \\in A$, $a < b$. We get $f(a) = f(b) = 0$. The function $g : I \\to \\mathbb{R}$, $g(x) = f(x)f... | Romania | 74th Romanian Mathematical Olympiad | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications",
"Precalculus > Functions"
] | English | proof only | null | |
07e7 | Ten distinct positive integers are given. Hessam calculates all the gcd's and the lcm's of these numbers and gives them to Masoud. Is it possible for Masoud to use these 90 received numbers and figure out the original 10 numbers? | [
"No.\n\nLet $p$ and $q$ be two distinct prime numbers. We define two sets as follows,\n$$\n\\begin{aligned}\nA &= \\{a_1, a_2, c_3, c_4, \\dots, c_{10}\\}, \\quad B = \\{b_1, b_2, c_3, c_4, \\dots, c_{10}\\} \\\\\na_1 &= pq^{11}, \\quad a_2 = p^2q^{12}, \\quad b_1 = p^2q^{11}, \\quad b_2 = pq^{12}, \\quad c_i = p^i... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | No | |
01uv | For every integer $n \ge 2$ prove the inequality
$$
\frac{1}{2!} + \frac{2}{3!} + \dots + \frac{2^{n-2}}{n!} \le \frac{3}{2},
$$
where $k! = 1 \cdot 2 \cdot \dots \cdot k$. | [
"Transform the expression on the left side\n$$\n\\begin{aligned}\n\\frac{1}{2!} + \\frac{2}{3!} + \\dots + \\frac{2^{n-2}}{n!} &= \\frac{1}{2!} \\left( 1 + \\frac{2}{3} + \\frac{2^2}{3 \\cdot 4} + \\dots + \\frac{2^{n-2}}{3 \\cdot 4 \\dots n} \\right) \n\\\\\n&\\le \\frac{1}{2} \\left( 1 + \\frac{2}{3} + \\frac{2^2... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0kdn | Problem:
Find the smallest real constant $\alpha$ such that for all positive integers $n$ and real numbers $0 = y_{0} < y_{1} < \cdots < y_{n}$, the following inequality holds:
$$
\alpha \sum_{k=1}^{n} \frac{(k+1)^{3 / 2}}{\sqrt{y_{k}^{2}-y_{k-1}^{2}}} \geq \sum_{k=1}^{n} \frac{k^{2}+3 k+3}{y_{k}} \text{.}
$$ | [
"Solution:\nWe first prove the following lemma:\nLemma. For positive reals $a, b, c, d$, the inequality\n$$\n\\frac{a^{3 / 2}}{c^{1 / 2}} + \\frac{b^{3 / 2}}{d^{1 / 2}} \\geq \\frac{(a+b)^{3 / 2}}{(c+d)^{1 / 2}}\n$$\nholds.\nProof. Apply Hölder's inequality in the form\n$$\n\\left(\\frac{a^{3 / 2}}{c^{1 / 2}} + \\f... | United States | HMMT February 2020 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 16 sqrt(2) / 9 | |
0dbp | Let $S$ be a given set of real numbers such that:
i) $1 \in S$,
ii) for any $a, b \in S$ (not necessarily different), then $a-b \in S$,
iii) for $a \in S$, $a \neq 0$ then $\frac{1}{a} \in S$.
Prove that for any $a, b \in S$ then $a b \in S$. | [
"If $a=0 \\in S$ or $a=1 \\in S$ then for any $b \\in S$ we have $a b \\in S$ which is obvious. So we can suppose that $a, b \\notin \\{0 ; 1\\}$.\nFrom $1 \\in S$, $a \\in S$ we have $1-a \\in S \\rightarrow (1-a)-1 = -a \\in S$, then\n$$\nb-(-a) = a+b \\in S.\n$$\nWe also have $a-1 \\in S$ so $\\frac{1}{a}, \\fra... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Algebra > Prealgebra / Basic Algebra > Other"
] | English | proof only | null | |
0goa | A teacher wants to divide the 2010 questions she asked in the exams during the school year into three folders of 670 questions and give each folder to a student who solved all 670 questions in that folder. Determine the minimum number of students in the class that makes this possible for all possible situations in whic... | [
"If there are four students $S_i$, $1 \\le i \\le 4$, and $S_1$ and $S_2$ solved the same half of the questions, and $S_3$ and $S_4$ solved the other half; then we cannot partition 2010 questions into three sets of 670 questions so that each set can be assigned to a student who solved all of those questions. Now we... | Turkey | Team Selection Test for IMO 2010 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 5 | |
05cm | A gardener Andres wants to plant one currant bush to each cell of his garden of shape $24 \times 2024$. He wants to plant as many blackcurrant bushes as possible under the following conditions: There must be at least one redcurrant and at least one whitecurrant bush, and for any cell with a blackcurrant bush, cells tha... | [
"Firstly, we show that $46632$ blackcurrant bushes is possible. Let's alternate redcurrant and whitecurrant bushes on the falling diagonal starting from the top left corner; when we reach the bottom edge of the garden, skip one column and proceed similarly along a rising diagonal starting from the next column, and ... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 46632 | |
0gpj | Find all triples $(x, y, p)$ satisfying
$$
x^2 - 3xy + p^2y^2 = 12p
$$
where $x, y$ are integers and $p$ is a prime number. | [
"Since $x^2 + p^2y^2 \\equiv 0 \\pmod{3}$, if $p \\ne 3$ then $x^2 + y^2 \\equiv 0 \\pmod{3}$, and consequently $x \\equiv y \\equiv 0 \\pmod{3}$. Then $9 \\mid x^2 - 3xy + py^2$ but $9 \\nmid 12$, a contradiction. Thus, $p = 3$ and $x^2 - 3xy + 9y^2 = 36$. Therefore, $3 \\mid x$. $x = 3k$, and we get a second orde... | Turkey | 17th Junior Turkish Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | [(-6, 0, 3), (6, 0, 3), (0, 2, 3), (6, 2, 3), (0, -2, 3), (-6, -2, 3)] | |
033t | Problem:
In a volleyball tournament for the Euro-African cup the European teams are 9 more than the African teams. Every two teams met exactly once and the European teams gained 9 times more points than the African teams (the winner takes 1 point and the loser takes 0 point). What are the maximum possible points gaine... | [
"Solution:\n\nDenote by $x$ the number of African teams. Then the number of European teams equals $x+9$. The African teams played each other $\\frac{(x-1)x}{2}$ games and therefore the points won by them are $\\frac{(x-1)x}{2} + k$, where $k$ is the number of wins over European teams.\n\nFurther, the points won by ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 11 | |
08xz | A sequence $\{a_n \mid n = 1, 2, \dots\}$ of positive integers is called an ascending sequence if $a_n < a_{n+1}$ and $a_{2n} = 2a_n$ are satisfied for every positive integer $n$.
(1) Suppose $\{a_n\}$ is an ascending sequence. For any prime $p$ greater than $a_1$, show that a multiple of $p$ appears among the terms o... | [
"(1) Since $\\{a_n\\}$ is ascending, $a_{n+1} - a_n$ takes a positive integral value for every $n$. So, let $s$ be the minimum value for $a_{n+1} - a_n$ where $n \\ge 1$. Then, $s$ is a positive integer.\nLet $m$ be one positive integer for which $a_{m+1} - a_m = s$, and let $k$ be an integer satisfying $2^k > p$. ... | Japan | Japan 2015 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0k0n | Problem:
The area of square $ABCD$ is $196~\mathrm{cm}^2$. Point $E$ is inside the square, at the same distances from points $D$ and $C$, and such that $\angle DEC = 150^\circ$. What is the perimeter of $\triangle ABE$ equal to? Prove your answer is correct. | [
"Solution:\n\nSince the area of square $ABCD$ is $196 = 14^2~(\\mathrm{cm}^2)$, then the side of square $ABCD$ is $14~\\mathrm{cm}$.\n\nWe claim that $\\triangle ABE$ is equilateral. To prove this, we make a reverse construction, starting from an equilateral $\\triangle ABE'$, building up to square $ABCD$, and even... | United States | 19th Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 42 cm | |
02wr | Problem:
Os pontos $D$ e $E$ dividem o lado $AB$ do triângulo equilátero $ABC$ em três partes iguais, $D$ entre $A$ e $E$. O ponto $F$ está sobre o lado $BC$ de modo que $CF = AD$. Encontre a soma dos ângulos $\angle CDF + \angle CEF$.
 | [
"Solution:\n\nSejam $\\angle CDF = \\alpha$ e $\\angle CEF = \\beta$. Como $BF = BD = \\frac{2}{3} \\cdot AB$ e $\\angle ABC = 60^\\circ$, segue que $\\triangle BDF$ é equilátero. Além disso, como $E$ é ponto médio de $BD$, temos que $EF$ é altura e bissetriz relativa ao vértice $F$. Assim, $\\angle BFE = 30^\\circ... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 30° | |
0if2 | Problem:
A plane curve is parameterized by $x(t) = \int_{t}^{\infty} \frac{\cos u}{u} du$ and $y(t) = \int_{t}^{\infty} \frac{\sin u}{u} du$ for $1 \leq t \leq 2$. What is the length of the curve? | [
"Solution: $\\ln 2$\nBy the Second Fundamental Theorem of Calculus, $\\frac{d x}{d t} = -\\frac{\\cos t}{t}$ and $\\frac{d y}{d t} = -\\frac{\\sin t}{t}$. Therefore, the length of the curve is\n$$\n\\int_{1}^{2} \\sqrt{\\left(\\frac{d x}{d t}\\right)^{2} + \\left(\\frac{d y}{d t}\\right)^{2}} d t = \\int_{1}^{2} \\... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Trigonometric functions"
] | null | proof and answer | ln 2 | |
0h94 | Two players – Andriy and Olesya play the following game. On a table there is a rounded cake, which is cut by one of them into $2n$, $n>1$ different in weight sectors (pieces). Weight of every piece is known by each player. After that they choose pieces according the following rules. At first Olesya chooses 1 piece, the... | [
"Let's consider, which values $n$ can take.\n\nIf $n = 2k+1 > 1$, let's consider the following distribution of weights. Pieces $n-1$, $n$ and $n+1$ have weight 1, and other ones – 0. Then after Olesya's first step Andriy can not take non-zero pieces. Until Andriy takes by his turn piece $n-2$ or $n+2$, Olesya just ... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Exactly for odd n greater than one; for even n, Olesya cannot guarantee a win. | |
08gx | Problem:
We have two piles with $2000$ and $2017$ coins respectively. Ann and Bob take alternate turns making the following moves: The player whose turn is to move picks a pile with at least two coins, removes from that pile $t$ coins for some $2 \leqslant t \leqslant 4$, and adds to the other pile $1$ coin. The playe... | [
"Solution:\n\nDenote the number of coins in the two piles by $X$ and $Y$. We say that the pair $(X, Y)$ is losing if the player who begins the game loses and that the pair $(X, Y)$ is winning otherwise. We shall prove that $(X, Y)$ is losing if $X-Y \\equiv 0,1,7 \\pmod{8}$, and winning if $X-Y \\equiv 2,3,4,5,6 \\... | JBMO | null | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Bob has a winning strategy. | |
0axh | Problem:
Determine the number of ordered pairs of integers $(p, q)$ for which $p^{2} + q^{2} < 10$ and $-2^{p} \leq q \leq 2^{p}$. | [] | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Other"
] | null | final answer only | 17 | |
04ob | Let $ABCD$ be a square, and let $k$ be the circle centred at $B$ passing through $A$, $C$ and the point $T$ inside the square. Tangent on $k$ at $T$ intersects the segments $\overline{CD}$ and $\overline{DA}$ at $E$ and $F$, respectively. Let $G$ and $H$ be the intersections of the lines $BE$ and $BF$ with the segment ... | [
"Note that the lines $FA$ and $FT$ are tangent to the circle $k$, hence $|FA| = |FT|$, and the triangles $ABF$ and $TBF$ are congruent. Analogously, $|EC| = |ET|$, and the triangles $CBE$ and $TBE$ are congruent.\n\n\n\nLet us denote $\\alpha = \\angle FBA = \\angle FBT$ and $\\beta = \\ang... | Croatia | Croatian Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ... | English | proof only | null | |
0kqb | Problem:
Compute the nearest integer to
$$
100 \sum_{n=1}^{\infty} 3^{n} \sin^{3}\left(\frac{\pi}{3^{n}}\right)
$$ | [
"Solution:\nNote that we have\n$$\n\\sin 3x = 3 \\sin x - 4 \\sin^{3} x \\Longrightarrow \\sin^{3} x = \\frac{1}{4}(3 \\sin x - \\sin 3x)\n$$\nwhich implies that\n$$\n\\frac{\\sin^{3} x}{3x} = \\frac{1}{4}\\left(\\frac{\\sin x}{x} - \\frac{\\sin 3x}{3x}\\right)\n$$\nSubstituting $x = \\frac{\\pi}{3^{n}}$ and simpli... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | final answer only | 236 |
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