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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
01bv | Let $p$ be a prime number. Find
$$
1! \cdot 2^2 + 2! \cdot 3^2 + 3! \cdot 4^2 + \dots + (p-3)! \cdot (p-2)^2 \quad \mod p.
$$ | [
"It can be shown by induction that $1! \\cdot 2^2 + 2! \\cdot 3^2 + 3! \\cdot 4^2 + \\dots + (p-3)! \\cdot (p-2)^2 = (p-1)! - 2$ which is equal to $-3 \\mod p$ by Wilson theorem."
] | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | -3 mod p | |
0epx | If we place a $3$ at both ends of a number, its value is increased by $3372$.
Find the original number. | [
"Placing a $3$ at both ends of the number increases it by $3372$, a four-digit number. Hence the original number is a two-digit number. Suppose the original number is $x$. Then the number formed by placing a $3$ at both ends is equal to $3003 + 10x$. This is $3372$ more than $x$, that is,\n$$\n3003 + 10x - x = 3372... | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 41 | |
0i0n | Problem:
Martha is a chicken who lives in a coop whose finitely many residents have a definite "pecking order": for every pair of distinct chickens, exactly one pecks the other. (However, it is not necessarily true that if $X$ pecks $Y$ and $Y$ pecks $Z$, then $X$ pecks $Z$.) A chicken $X$ is called a "leader" of the ... | [
"Solution:\n\nSuppose that no one pecks more chickens than $M$, yet $M$ is not a leader; we will seek a contradiction. Let $X_{1}, X_{2}, \\ldots, X_{n}$ be the chickens pecked by $M$. Since $M$ is not a leader, there is a chicken $N$ who is neither pecked by $M$ nor by any of the $X_{i}$. But it is given that of a... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0k54 | Problem:
Let $A$, $B$, $C$ be points in that order along a line, such that $AB = 20$ and $BC = 18$. Let $\omega$ be a circle of nonzero radius centered at $B$, and let $\ell_1$ and $\ell_2$ be tangents to $\omega$ through $A$ and $C$, respectively. Let $K$ be the intersection of $\ell_1$ and $\ell_2$. Let $X$ lie on se... | [
"Solution:\nNote that $B$ is the $K$-excenter of $KXY$, so $XB$ is the angle bisector of $\\angle AKY$. As $AB$ and $XY$ are parallel, $\\angle XAB + 2\\angle AXB = 180^\\circ$, so $\\angle XBA = 180^\\circ - \\angle AXB - \\angle XAB$. This means that $AXB$ is isosceles with $AX = AB = 20$. Similarly, $YC = BC = 1... | United States | HMMT November 2018 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 35 | |
0f6q | Problem:
The quadratic $p(x) = ax^2 + bx + c$ has $a > 100$. What is the maximum possible number of integer values $x$ such that $p(x) < 50$? | [] | Soviet Union | 19th ASU | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | No maximum; it can be made arbitrarily large. | |
0fib | Problem:
Para dar una vuelta completa en un coche a un circuito circular, la cantidad exacta de gasolina está distribuida en depósitos fijos situados en $n$ puntos distintos cualesquiera del circuito. Inicialmente el depósito del coche está vacío. Demostrar que cualquiera que sea la distribución del combustible en los... | [
"Solution:\n\nSean $c_{1}, c_{2}, \\ldots, c_{n}$ las cantidades de combustible en cada uno de los $n$ depósitos y sean $d_{1}, d_{2}, \\ldots, d_{n}$ las distancias a recorrer desde cada depósito hasta el siguiente.\nHagamos el gráfico del consumo comenzando en un punto de aprovisionamiento cualquiera. Notemos que... | Spain | Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
0hcn | Show that for positive numbers $x, y, z, t$ the following inequality holds:
$$
\frac{x^8+1}{x^4} + \frac{y^8+1}{y^4} + \frac{z^8+1}{z^4} + \frac{t^8+1}{t^4} \ge 2 \cdot \left( \frac{x}{y} + \frac{y}{z} + \frac{z}{t} + \frac{t}{x} \right).
$$ | [
"The inequality can be rewritten as follows:\n$$\n\\begin{aligned}\n& (x^4 + \\frac{1}{z^4} + z^4 + \\frac{1}{y^4}) + (y^4 + \\frac{1}{t^4} + t^4 + \\frac{1}{z^4}) + (z^4 + \\frac{1}{x^4} + x^4 + \\frac{1}{t^4}) + (t^4 + \\frac{1}{y^4} + y^4 + \\frac{1}{x^4}) \\\\ & \\ge 4 \\cdot \\left( \\frac{x}{y} + \\frac{y}{z}... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
084p | Problem:
Al variare del parametro reale $a$, qual è il numero massimo di soluzioni per l'equazione $||x-1|-4|+x=a$?
(A) 1
(B) 2
(C) 3
(D) 4
(E) può averne infinite. | [
"Solution:\n\nLa risposta è (E). Si ha infatti che, per ogni $x \\leq -3$, $x-1 \\leq 0$ e quindi $|x-1| = 1-x$, inoltre $|x-1|-4 = -x-3 \\geq 0$, e l'equazione si può semplificare così:\n$$\n||x-1|-4|+x = |-x+1-4|+x = -x-3+x = -3\n$$\nPertanto se $a = -3$ l'equazione ha infinite soluzioni."
] | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Equations and Inequalities"
] | null | MCQ | E | |
043w | Given positive integers $k$ and $n$ ($n \ge 2$), find the minimum constant $c$ satisfying this assertion: if $G$ is a simple $kn$-regular graph (the degree of each vertex is $kn$) with $m$ vertices, then each vertex can be coloured one of $n$ colours, such that the number of “mono edges” is at most $cm$. Here, a mono e... | [
"$c = \\frac{k(kn - n + 2)}{2(kn + 1)}$.\n\nFirst, we prove $c \\ge \\frac{k(kn - n + 2)}{2(kn + 1)}$. Consider a complete graph $K_{kn+1}$ with any $n$-colouring of the vertices. Suppose that the numbers of vertices coloured by 1, 2, ..., $n$ are $a_1, a_2, ..., a_n$, respectively. Then\n$$\n\\begin{aligned}\na_1^... | China | China National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Cau... | null | proof and answer | k(kn - n + 2)/(2(kn + 1)) | |
06cl | Suppose $p$ and $q$ are positive integers such that $\frac{p}{q} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{1335}$. Show that 2003 is a factor of $p$. | [
"$$\n\\begin{align*}\n& 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\dots - \\frac{1}{1334} + \\frac{1}{1335} \\\\\n&= \\left(1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\dots + \\frac{1}{1334} + \\frac{1}{1335}\\right) - 2\\left(\\frac{1}{2} + \\frac{1}{4} + \\dots + \\frac{1}{1334}\\right) \\\\\n&= \\... | Hong Kong | Test 1 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0cc9 | Determine the largest positive integer $k$ for which exists a positive integer $n$ such that:
$$
sin(n + 1) < \sin(n + 2) < \sin(n + 3) < \dots < \sin(n + k).
$$ | [
"We claim the largest number is $k = 5$.\n\nThe difference of two consecutive terms is:\n$$\n\\sin(n+i+1)-\\sin(n+i) = 2 \\sin \\frac{1}{2} \\cos \\frac{2n+2i+1}{2} > 0 \\Leftrightarrow \\cos \\frac{2n+2i+1}{2} > 0. \\ (\\star)\n$$\n\nSuppose that for $k = 6$ there exists a positive integer $n$ such that $\\sin(n +... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | 5 | |
0isu | Let $n$ be a positive integer not divisible by the cube of a prime. Given an integer-coefficient polynomial $f(x)$, define its *signature modulo* $n$ to be the (ordered) sequence $f(1), \dots, f(n)$ modulo $n$. Of the $n^n$ such $n$-term sequences of integers modulo $n$, how many are the signature of some polynomial $f... | [
"Let $p_1 \\cdots p_r q_1^2 \\cdots q_s$ be the prime factorization of $n$. Then the number of achievable signatures is $p_1^{p_1} \\cdots p_r^{p_r} q_1^{3q_1} \\cdots q_s^{3q_s}$.\n\nWe begin by reducing to the case in which $n$ is a prime or the square of a prime. Assume for now that our claim holds in these case... | United States | Team Selection Test | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Linear... | null | proof and answer | If n = p1 ··· pr · q1^2 ··· qs^2, the number of achievable signatures is p1^{p1} ··· pr^{pr} · q1^{3 q1} ··· qs^{3 q1}. | |
0idj | Problem:
Find all real solutions to $x^{4} + (2 - x)^{4} = 34$. | [
"Solution:\n$1 \\pm \\sqrt{2}$\nLet $y = 2 - x$, so $x + y = 2$ and $x^{4} + y^{4} = 34$. We know\n$$\n(x + y)^{4} = x^{4} + 4x^{3}y + 6x^{2}y^{2} + 4x y^{3} + y^{4} = x^{4} + y^{4} + 2 x y (2 x^{2} + 2 y^{2} + 3 x y)\n$$\nMoreover, $x^{2} + y^{2} = (x + y)^{2} - 2 x y$, so the preceding equation becomes $2^{4} = 3... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | x = 1 ± sqrt(2) | |
0295 | Problem:
Pedrinho tem várias peças de madeira na forma de um triângulo retângulo de catetos $1~\mathrm{cm}$ e $2~\mathrm{cm}$. Com 4 dessas peças, ele consegue montar um quadrado de lado $2~\mathrm{cm}$, como na figura abaixo.
Brincando com mais peças, ele conseguiu montar um quadrado usand... | [
"Solution:\n\nCada peça tem área $\\frac{1 \\times 2}{2}~\\mathrm{cm}^2 = 1~\\mathrm{cm}^2$. Logo, se usarmos 20 das peças, teremos um quadrado de área $20~\\mathrm{cm}^2$. Portanto, seu lado mede $\\sqrt{20} = 2\\sqrt{5}~\\mathrm{cm}$. Por outro lado, pelo Teorema de Pitágoras, vale que a hipotenusa de cada triâng... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2√5 cm | |
076o | Let $\triangle ABC$ be a right-angled triangle with $\angle B = 90^\circ$. Let $D$ be a point on $AC$ such that the in-radii of the triangles $ABD$ and $CBD$ are equal. If this common value is $r'$ and if $r$ is the in-radius of triangle $ABC$, prove that
$$
\frac{1}{r'} = \frac{1}{r} + \frac{1}{BD}.
$$ | [
"Let $E$ and $F$ be the incentres of triangles $ABD$ and $CBD$ respectively. Let the incircles of triangles $ABD$ and $CBD$ touch $AC$ in $P$ and $Q$ respectively. If $\\angle BDA = \\theta$, we see that\n\n$$\nr' = PD \\tan(\\theta/2) = QD \\cot(\\theta/2).\n$$\nHence\n$$\nPQ = PD + QD = r... | India | IND_National | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | null | proof only | null | |
07mp | Let $ABC$ be a triangle of area $1$ (square units), and let $P$ denote the midpoint of the side $BC$. Consider two points $M$ and $N$ interior to the sides $AB$ and $AC$ respectively, such that $|AM| = 2|MB|$ and $|CN| = 2|AN|$. The lines $AP$ and $MN$ intersect at a point $D$. Find the area of the triangle $ADN$. | [
"*First solution.* Let $Q$ be the point inside $AC$ such that $MQ$ is parallel to $BC$. Then, because $\\triangle ABC$ and $\\triangle AMQ$ are homothetic, $|AN| = |NQ| = |QC|$ and, if $(ABC)$ denotes the area of the triangle $ABC$ etc., we also have\n$$\n\\frac{(AMQ)}{(ABC)} = \\frac{|AM|^2}{|AB|^2} = \\frac{4}{9}... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 2/27 | |
03gz | Problem:
Describe a construction of a quadrilateral $ABCD$ given:
(i) the lengths of all four sides;
(ii) that $AB$ and $CD$ are parallel;
(iii) that $BC$ and $DA$ do not intersect. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
06so | Let $\mathbb{Z}_{>0}$ be the set of positive integers. Find all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that
$$
m^{2}+f(n) \mid m f(m)+n
$$
for all positive integers $m$ and $n$. | [
"Answer. $f(n)=n$.\n\nSolution 1. Setting $m=n=2$ tells us that $4+f(2) \\mid 2 f(2)+2$. Since $2 f(2)+2<2(4+f(2))$, we must have $2 f(2)+2=4+f(2)$, so $f(2)=2$. Plugging in $m=2$ then tells us that $4+f(n) \\mid 4+n$, which implies that $f(n) \\leqslant n$ for all $n$.\nSetting $m=n$ gives $n^{2}+f(n) \\mid n f(n)... | IMO | International Mathematical Olympiad Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | f(n) = n for all positive integers n | |
0c4t | A circle passing through the vertices $B$ and $C$ of the triangle $ABC$ meets again $AB$ and $AC$ at $C_1$ and $B_1$ respectively. Let $A'$ be the midpoint of $B_1C_1$ and $AA' \cap BC = \{A''\}$. Define similarly $B''$ and $C''$. Prove that the triangles $ABC$ and $A''B''C''$ have the same centroid. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous... | English | proof only | null | |
03zg | Let $AA_0$, $BB_0$ and $CC_0$ be angular bisectors of $\triangle ABC$. Let $A_0A_1 \parallel BB_0$ and $A_0A_2 \parallel CC_0$, where $A_1$ and $A_2$ lie on $AC$ and $AB$, respectively, and let line $A_1A_2$ intersect $BC$ at $A_3$. The points $B_3$ and $C_3$ are obtained similarly. Prove that points $A_3$, $B_3$, $C_3... | [
"By the Menelaus Inverse Theorem, we need only to show that\n$$\n\\frac{AB_3}{B_3C} \\cdot \\frac{CA_3}{A_3B} \\cdot \\frac{BC_3}{C_3A} = 1. \\qquad \\textcircled{1}\n$$\nSince line $A_1A_2A_3$ intersects $\\triangle ABC$, by Menelaus' Theorem, we have $\\frac{CA_3}{A_3B} \\cdot \\frac{BA_2}{A_2A} \\cdot \\frac{AA_... | China | China Southeastern Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0eh3 | Problem:
Poišči vsa realna števila $x$, ki rešijo enačbo
$$
\sqrt[3]{x+1}+\frac{6-6 \sqrt[3]{x+1}}{\sqrt{x+1}-\sqrt[6]{x+1}}=1
$$ | [
"Solution:\nV enačbo vpeljemo novo spremenljivko $z=\\sqrt[6]{x+1}$, da dobimo $z^{2}+\\frac{6-6 z^{2}}{z^{3}-z}=1$, in ulomek na levi strani preoblikujemo $z^{2}+\\frac{-6\\left(z^{2}-1\\right)}{z\\left(z^{2}-1\\right)}=1$. Od tod sklepamo, da $z$ ne sme biti enak $0$, $1$ ali $-1$, saj sicer ulomek ne bi bil defi... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | 63 | |
05i9 | Problem:
Soient $n$ et $k$ deux entiers strictement positifs. On considère une assemblée de $k$ personnes telle que, pour tout groupe de $n$ personnes, il y en ait une $(n+1)$-ième qui les connaisse toutes (si $A$ connaît $B$ alors $B$ connaît $A$ ).
1) Si $k=2 n+1$, prouver qu'il existe une personne qui connaît tout... | [
"Solution:\n\n1) On commence par construire, par récurrence, un groupe de $n+1$ personnes qui se connaissent deux à deux : il est clair que l'on peut trouver deux personnes qui se connaissent. Supposons que pour $p \\in \\{2, \\ldots, n\\}$ fixé, on ait réussi à trouver un groupe de $p$ personnes qui se connaissent... | France | Olympiades Françaises de Mathématiques - Test de Sélection | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0co8 | Nonzero real numbers $a$, $b$, and $c$ are chosen so that $a x^2 + b x + c > c x$ for each real $x$. Prove that $c x^2 - b x + a > c x - b$ for each real $x$. (M. Murashkin)
Ненулевые числа $a$, $b$, $c$ таковы, что $a x^2 + b x + c > c x$ при любом $x$. Докажите, что $c x^2 - b x + a > c x - b$ при любом $x$. (М. Мур... | [
"Так как для всех $x$ верно неравенство $P(x) = a x^2 + (b - c) x + c > 0$, то дискриминант трёхчлена $P(x)$ отрицателен: $D = (b - c)^2 - 4 a c = b^2 + c^2 - 2 b c - 4 a c < 0$. Кроме того, $c = P(0) > 0$. Значит, у трехчлена $Q(x) = c x^2 - (b + c) x + (a + b)$ положителен старший коэффициент, а его дискриминант ... | Russia | Regional round | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials"
] | English; Russian | proof only | null | |
07l7 | Find a polynomial $x^3 + a x^2 + b x + c$ for which $\max_{x \in [-1, 1]} |x^3 + a x^2 + b x + c|$ is minimised. You may assume that such a polynomial exists. | [
"Define $M(a, b, c) = \\max_{x \\in [-1, 1]} |x^3 + a x^2 + b x + c|$. Observe that\n$$\n\\max_{x \\in [-1, 0]} |x^3 + a x^2 + b x + c| = \\max_{x \\in [0, 1]} |-x^3 + a x^2 - b x + c|.\n$$\nSince $\\max(|\\alpha + \\beta|, |\\alpha - \\beta|) = |\\alpha| + |\\beta|$, then\n$$\nM(a, b, c) = \\max_{x \\in [0, 1]} (|... | Ireland | Irska | [
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials"
] | English | proof and answer | x^3 - 3/4 x | |
0733 | Consider the quadratic polynomial $p(x) = x^2 + ax + b$, where $a, b$ are in the interval $[-2, 2]$. Find the range of the real roots of $p(x) = 0$, as $a$ and $b$ vary over $[-2, 2]$. | [
"Suppose $a, b \\in [-2, 2]$ and $\\alpha$ is a real root of $x^2 + ax + b = 0$. Then\n$$\n\\alpha = \\frac{-a \\pm \\sqrt{a^2 - 4b}}{2},\n$$\nwhich shows that\n$$\n\\alpha \\leq \\frac{2 + \\sqrt{4 + 8}}{2} = 1 + \\sqrt{3}.\n$$\n(Take $a = -2$, $b = -2$.) Similarly, we see that $\\alpha \\ge -1 - \\sqrt{3}$, by ta... | India | Indija TS 2007 | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | [-1 - sqrt(3), 1 + sqrt(3)] | |
0jtx | Problem:
Evaluate $\frac{2016!^{2}}{2015!2017!}$. Here $n!$ denotes $1 \times 2 \times \cdots \times n$. | [
"Solution:\n$\\frac{2016!^{2}}{2015!2017!} = \\frac{2016!}{2015!} \\cdot \\frac{2016!}{2017!} = \\frac{2016}{1} \\cdot \\frac{1}{2017} = \\frac{2016}{2017}$"
] | United States | HMMT November | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 2016/2017 | |
0atl | Problem:
What is the largest number of $7~\mathrm{m} \times 9~\mathrm{m} \times 11~\mathrm{m}$ boxes that can fit inside a box of size $17~\mathrm{m} \times 37~\mathrm{m} \times 27~\mathrm{m}$? | [] | Philippines | Philippine Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 18 | |
08af | Problem:
Sia $ABC$ un triangolo tale che, detto $H$ il piede dell'altezza condotta da $C$, si ha $AH = 3 \cdot HB$. Siano inoltre:
- $M$ il punto medio di $AB$;
- $N$ il punto medio di $AC$;
- $P$ il punto dal lato opposto di $B$ rispetto alla retta $AC$ tale che $NP = NC$ e $PC = CB$.
Dimostrare che $\widehat{APM} =... | [] | Italy | Olimpiade Italiana di Matematica | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
055z | A circle with diameter $AB$ intersects side $BC$ of rhombus $ABCD$ at point $K$. A circle with diameter $AD$ intersects side $CD$ of rhombus $ABCD$ at point $L$. Find the angles of rhombus $ABCD$ if $\angle AKL = \angle ABC$. | [
"Let $\\angle ABC = \\angle ADC = \\alpha$; then $\\angle AKL = \\alpha$ (Fig. 28).\n\nAccording to Thales' theorem $AK$ is perpendicular to $BC$ and $AL$ is perpendicular to $CD$. But $\\angle ABK = \\alpha = \\angle ADL$ and $AB = AD$, so $ABK$ and $ADL$ are equal. Therefore, $AK = AL$ from which $\\angle ALK = \... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 60 degrees and 120 degrees | |
0909 | Determine the number of pairs $(p, a)$ of a prime number $p$ and an integer $a$ such that $p \ge 3$, $1 \le a \le 2024$, and the following condition is satisfied:
$a < p^4$ holds and $ap^4 + 2p^3 + 2p^2 + 1$ is a square number. | [
"When the condition is satisfied, there exists a non-negative integer $n$ such that $ap^4 + 2p^3 + 2p^2 + 1 = n^2$ and then\n$$\n(a - p^2 - 2p - 1)p^4 = n^2 - (p^3 + p^2 + 1)^2 = (n + p^3 + p^2 + 1)(n - p^3 - p^2 - 1) \\quad (*)\n$$\nholds. If we assume that both $n + p^3 + p^2 + 1$ and $n - p^3 - p^2 - 1$ are divi... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 16 | |
0kdm | Problem:
For some positive real $\alpha$, the set $S$ of positive real numbers $x$ with $\{x\} > \alpha x$ consists of the union of several intervals, with total length $20.2$. The value of $\alpha$ can be expressed as $\frac{a}{b}$, where $a, b$ are relatively prime positive integers. Compute $100 a + b$. (Here, $\{x... | [
"Solution:\n\nIf we note that $x = \\{x\\} + \\lfloor x \\rfloor$, then we can rewrite our given inequality as $\\{x\\} > \\frac{\\alpha}{1-\\alpha} \\lfloor x \\rfloor$. However, since $\\{x\\} < 1$, we know that we must have $\\frac{\\alpha}{1-\\alpha} \\lfloor x \\rfloor < \\{x\\} < 1$, so each interval is of th... | United States | HMMO 2020 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 4633 | |
065c | If $a$ is an even positive integer and $A = a^n + a^{n-1} + \dots + a + 1$, $n \in \mathbb{N}^*$, is a perfect square, prove that $a$ is a multiple of $8$. | [
"Since $a$ is an even positive integer, it follows that $A$ is odd. Therefore $A$ will be a perfect square of an odd integer, that is\n$$\nA = (2\\kappa + 1)^2 = 4\\kappa^2 + 4\\kappa + 1 = 4\\kappa(\\kappa + 1) + 1,\n$$\nwhere $\\kappa$ is a positive integer. Since one of the two integers $\\kappa$ and $\\kappa + ... | Greece | SELECTION EXAMINATION | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0b12 | Problem:
Suppose a real number $x > 1$ satisfies
$$
\log_{\sqrt[3]{3}}\left(\log_{3} x\right) + \log_{3}\left(\log_{27} x\right) + \log_{27}\left(\log_{\sqrt[3]{3}} x\right) = 1
$$
Compute $\log_{3}\left(\log_{3} x\right)$. | [
"Solution:\n\nLet $y = \\log_{3} x$. Then $x = 3^{y}$.\n\nWe compute the inner logarithms:\n\n- $\\log_{3} x = y$\n- $\\log_{27} x = \\frac{\\log_{3} x}{\\log_{3} 27} = \\frac{y}{3}$\n- $\\log_{\\sqrt[3]{3}} x = \\frac{\\log_{3} x}{\\log_{3} \\sqrt[3]{3}} = \\frac{y}{1/3} = 3y$\n\nNow substitute into the equation:\... | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 5/13 | |
06o3 | A test question reads 'write down three consecutive positive integers not exceeding $2023$ in ascending order, such that one of them is a multiple of $6$, another one is a multiple of $7$ and the remaining one is a multiple of $8$. How many different correct answers are there for this question?' | [
"Let $n$, $n + 1$ and $n + 2$ be the three consecutive positive integers. Then $1 \\leq n \\leq 2021$. Since multiples of $6$ and $8$ must be even, we must have $n + 1$ being divisible by $7$ whereas $n$ and $n + 2$ are divisible by $6$ and $8$ in either order. For each of these two cases, there are exactly two cho... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | 24 | |
00j2 | A sequence $\langle a_n \rangle$ of positive integers is given, such that $a_1 = 1$ and $a_{n+1}$ is the smallest positive integer such that
$$lcm(a_1, a_2, \dots, a_n, a_{n+1}) > lcm(a_1, a_2, \dots, a_n).$$
Which numbers are contained in the sequence? | [
"The first few elements of the sequence are given by\n$1, 2, 3, 4, 5, 7, 8, 9, 11, \\ldots$\nand we note that the first two positive integers not contained in the sequence are $6$ and $10$.\nThese would not have made the lcm larger when it was \"their turn\", and they would certainly\nnot do so at a later date. We ... | Austria | AustriaMO2011 | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | Exactly the number one and all powers of primes, listed in ascending order. | |
06el | Let $ABCD$ be a cyclic quadrilateral. Show that the orthocentres of the triangles $\triangle ABC$, $\triangle BCD$, $\triangle CDA$ and $\triangle DAB$ are the vertices of a quadrilateral congruent to $ABCD$ and show that the centroids of the same triangles are the vertices of a cyclic quadrilateral. | [
"Let $O$ be the centre of $(ABCD)$. Let $H_A, H_B, H_C, H_D$ and $G_A, G_B, G_C, G_D$ be the orthocentres and centroids of $\\triangle DAB$, $\\triangle ABC$, $\\triangle BCD$ and $\\triangle CDA$ respectively. Let $M$ be the midpoint of $AB$. Recall that $O$ lies on $H_A G_A$, etc., with $OG_A : G_A H_A = 1 : 2$. ... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof only | null | |
00s9 | Find all pairs $(x, y)$ of positive integers such that
$$
x^3 + y^3 = x^2 + 42xy + y^2.
$$ | [
"Let $d = (x, y)$ be the greatest common divisor of positive integers $x$ and $y$.\nSo, $x = ad$, $y = bd$, where $d \\in \\mathbb{N}$, $(a, b) = 1$, $a, b \\in \\mathbb{N}$. We have\n$$\n\\begin{aligned}\nx^3 + y^3 &= x^2 + 42xy + y^2 \\\\\n&\\Leftrightarrow d^3(a^3 + b^3) = d^2(a^2 + 42ab + b^2) \\\\\n&\\Leftrigh... | Balkan Mathematical Olympiad | BMO 2017 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials ... | English | proof and answer | [(1, 7), (7, 1), (22, 22)] | |
05p2 | Problem:
Soit $ABC$ un triangle dont tous les angles sont aigus.
Les hauteurs $[AA_{1}]$, $[BB_{1}]$ et $[CC_{1}]$ se coupent au point $H$. Soit $A_{2}$ le symétrique de $A$ par rapport à $[B_{1}C_{1}]$, et soit $O$ le centre du cercle circonscrit à $ABC$.
a) Prouver que les points $O, A_{2}, B_{1}, C$ sont cocycliqu... | [
"Solution:\n\n\n\na) $A, H, B_{1}, C_{1}$ sont cocycliques sur le cercle de diamètre $[AH]$, donc $\\widehat{AB_{1}C_{1}} = \\widehat{AHC_{1}} = 90^{\\circ} - \\widehat{C_{1}AH}$. Or, $(AA_{2}) \\perp (B_{1}C_{1})$ donc $\\widehat{A_{2}AB_{1}} = \\widehat{C_{1}AH}$, i.e. les demi-droites $[... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
... | null | proof only | null | |
0fxr | Problem:
Sei $ABC$ ein Dreieck mit $AB \neq AC$ und Inkreismittelpunkt $I$. Der Inkreis berühre $BC$ bei $D$. Der Mittelpunkt von $BC$ sei $M$. Zeige, dass die Gerade $IM$ die Strecke $AD$ halbiert. | [
"Solution:\n\nSeien $U$, $V$ und $W$ die Schnittpunkte von $MI$ mit $AD$, $AC$ bzw. $AB$. Menelaos am Dreieck $ABD$ liefert\n$$\n\\frac{DU}{UA} \\frac{AW}{WB} \\frac{BM}{MD} = -1 \\quad \\Leftrightarrow \\quad \\frac{DU}{UA} = \\frac{MD}{BM} \\frac{BW}{AW}\n$$\nWir wenden nun Menelaos am Dreieck $ABC$ an und verwen... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Pla... | null | proof only | null | |
0gp5 | Let $K(M, N)$ be the collection of the midpoints of the line segments with one endpoint belongs to $M$ and the other endpoint belongs to $N$ where $M$ and $N$ are regular convex polygonal regions in the plane. Determine all pairs $(M, N)$ for which $K(M, N)$ is a regular convex polygonal region. | [
"We will show that $K = K(M, N)$ is a regular polygon if and only if either $N$ is homothetic to $M$, or $N$ is obtained from $M$ by a $180/m$ degree rotation about the center of $M$ followed by a translation where $m$ is the number of edges of $M$.\n\nIn the following when we say an edge $XY$ of a convex polygon i... | Turkey | 19th Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
... | English | proof only | null | |
0kh8 | For $n$ a positive integer, let $f(n)$ be the quotient obtained when the sum of all positive divisors of $n$ is divided by $n$. For example, $f(14) = (1 + 2 + 7 + 14) \div 14 = \frac{12}{7}$. What is $f(768) - f(384)$?
(A) $\frac{1}{768}$ (B) $\frac{1}{192}$ (C) 1 (D) $\frac{4}{3}$ (E) $\frac{8}{3}$ | [
"Suppose all the positive divisors of $n$ are $d_1 < d_2 < \\dots < d_k$. Then\n$$\nf(n) = \\frac{d_1 + d_2 + \\dots + d_k}{n} = \\frac{d_1}{n} + \\frac{d_2}{n} + \\dots + \\frac{d_k}{n} = \\frac{1}{d_k} + \\frac{1}{d_{k-1}} + \\dots + \\frac{1}{d_1}.\n$$\nThus $f(n)$ is the sum of the reciprocals of the positive d... | United States | AMC 12 B | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | B | |
0af9 | Да се провери точноста на равенството
$$
\operatorname{tg} \alpha + 2 \operatorname{tg} 2\alpha + 2^2 \operatorname{tg} 2^2 \alpha + \dots + 2^n \operatorname{tg} 2^n \alpha = \operatorname{ctg} \alpha - 2^{\eta+1} \operatorname{ctg} 2^{\eta+1} \alpha.
$$ | [
"Не е тешко да се провери точноста на следното равенство\n$$\n\\operatorname{ctg} \\alpha - \\operatorname{tg} \\alpha = 2 \\operatorname{ctg} 2\\alpha. \\qquad (1)\n$$\nНавистина\n$$\n\\operatorname{ctg} \\alpha - \\operatorname{tg} \\alpha = \\frac{\\cos \\alpha}{\\sin \\alpha} - \\frac{\\sin \\alpha}{\\cos \\alp... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | Macedonian, English | proof only | null | |
09k7 | Let $n$, $m$, $k$ be integers such that $\frac{mk+n}{n^2+1}$ and $\frac{nk+m}{m^2+1}$ are integers. Show that $k^2 - 1$ is divisible by the greatest common divisor of $n^2+1$ and $m^2+1$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof only | null | |
0ec8 | Problem:
Dan je kvadrat $A B C D$ in taki točki $E$ in $F$ izven kvadrata, da sta trikotnika $B E C$ in $C F D$ enakostranična. Dokaži, da je tudi trikotnik $A E F$ enakostraničen.
 | [
"Solution:\n\n1. način. Ker je $A B C D$ kvadrat in sta trikotnika $B E C$ in $C F D$ enakostranična, je $|E B|=|B A|=|A D|=|D F|$. Torej sta trikotnika $E B A$ in $A D F$ enakokraka z vrhoma pri $B$ in $D$ in imata enako dolge krake. Hkrati velja $\\Varangle E B A=\\Varangle A D F=90^{\\circ}+60^{\\circ}=150^{\\ci... | Slovenia | 59. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05sr | Problem:
On considère une grille de taille $2019 \times 2019$. Sur cette grille sont posés des cailloux. Une configuration est dite belle s'il n'existe pas de parallélogramme formé par quatre cailloux $A B C D$, tels que $A, B, C$, et $D$ ne soient pas tous alignés.
Quel est le plus grand nombre de cailloux qu'il est ... | [
"Solution:\n\nDans ce problème, on cherche le plus grand entier satisfaisant une certaine propriété. Supposons que l'on veuille montrer que le plus grand entier recherché est l'entier $c$. Pour montrer que $c$ est bien le plus grand entier, on va d'une part montrer que si un entier $n$ satisfait la propriété, alors... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | 4037 | |
01j0 | Determine if there exists a triangle that can be cut into 101 congruent triangles. | [
"Answer: Yes, there is.\nChoose an arbitrary positive integer $m$ and draw a height in the right triangle with ratio of legs $1 : m$. This height cuts the triangle in two similar triangles with similarity coefficient $m$. The largest of them can further be cut into $m^2$ smaller equal triangles by splitting all sid... | Baltic Way | Baltic Way 2023 Shortlist | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | Yes | |
0j2u | Problem:
Into how many regions can a circle be cut by 10 parabolas? | [
"Solution:\nAnswer: $201$\n\nWe will consider the general case of $n$ parabolas, for which the answer is $2 n^{2} + 1$.\n\nWe will start with some rough intuition, then fill in the details afterwards. The intuition is that, if we make the parabolas steep enough, we can basically treat them as two parallel lines. Fu... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | 201 | |
0h4t | Find all positive integers $m, n$ satisfying the equation $2^m = 7n + 4$. | [
"The powers of $2$ give remainders $2$, $4$ and $1$ in division by $7$ with period $3$, in particular, $2^k \\equiv 4 \\pmod{7}$ if and only if $k \\equiv 2 \\pmod{3}$. But this means that $m^2 \\equiv 2 \\pmod{3}$, which is impossible. This contradiction completes the proof."
] | Ukraine | Ukrainian National Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | no positive integer solutions | |
0a7v | Problem:
Show that there exists an integer divisible by $1996$ such that the sum of its decimal digits is $1996$. | [
"Solution:\n\nThe sum of the digits of $1996$ is $25$ and the sum of the digits of $2 \\cdot 1996 = 3992$ is $23$. Because $1996 = 78 \\cdot 25 + 46$, the number obtained by writing $78$ $1996$'s and two $3992$ in succession satisfies the condition of the problem.\n\nAs $3 \\cdot 1996 = 5988$, the sum of the digits... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 10 | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0gei | C. 令 $n$ 與 $k$ 為滿足 $k \le 2n^2$ 的正整數。小李與小晴用一張 $2n \times 2n$ 的方格紙玩一個遊戲。首先, 小李在紙上的每一個方格中寫上一個至多為 1 的非負實數, 使得整張方格紙上的數字總和為 $k$。接著, 小晴照著方格紙上的格線, 將方格紙切成數片, 使得每一片都是由若干個完整的方格所組成, 且每一片上的數字總和至多為 1。小晴所切出的每一片的形狀沒有限制。
令遊戲最後得到的片數為 $M$。小李的目標是極大化 $M$, 而小晴的目標是極小化 $M$。
試求當小李與小晴都以最佳策略進行下, 遊戲結束時的 $M$ 值。
C. Let $n$ and $k$ be positive int... | [
"答案是 $2k-1$。讓我們依序考慮兩人的策略。\n\n- 對於小李, 他可以在 $2k-1$ 的格子裡寫 $\\frac{1}{2}+\\epsilon$, 一個格子裡寫 $\\frac{1}{2}-(2k-1)\\epsilon$, 其餘填 0。基於每一片至多只能有一個寫 $\\frac{1}{2}+\\epsilon$ 的格子, 小李可以用此策略保證 $M \\ge 2k-1$。\n\n- 對於小晴, 他先選擇一個一筆劃通過所有方格的路徑, 假設此路徑上的數字依序為 $a_1, a_2, \\dots, a_{4n^2}$。小晴的策略是, 先從 $a_1$ 開始, 找到最大的正整數 $M$ 使得 $\\sum_{i=... | Taiwan | 2021 數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2k-1 | |
0jkx | Problem:
An up-right path from $(a, b) \in \mathbb{R}^{2}$ to $(c, d) \in \mathbb{R}^{2}$ is a finite sequence $\left(x_{1}, y_{1}\right), \ldots,\left(x_{k}, y_{k}\right)$ of points in $\mathbb{R}^{2}$ such that $(a, b)=\left(x_{1}, y_{1}\right),(c, d)=\left(x_{k}, y_{k}\right)$, and for each $1 \leq i<k$ we have tha... | [
"Solution:\n\nAnswer: $0.2937156494680644 \\ldots$ Note that any up-right path must pass through exactly one point of the form $(n,-n)$ (i.e. a point on the upper-left to lower-right diagonal), and the number of such paths is $\\binom{800}{400-n}^{2}$ because there are $\\binom{800}{400-n}$ up-right paths from $(-4... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | final answer only | 0.2937156494680644 | |
0isp | Problem:
Let $S$ be a set of $5$ points in the $2$-dimensional lattice. Show that we can always choose a pair of points in $S$ whose midpoint is also a lattice point. | [
"Solution:\n\nConsider the parities of the coordinates. There are four possibilities: $(\\text{odd}, \\text{odd})$, $(\\text{odd}, \\text{even})$, $(\\text{even}, \\text{odd})$, $(\\text{even}, \\text{even})$. By the pigeonhole principle, two of the points must have the same parity in both coordinates (i.e., they a... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0grv | For integers $m \ge 3$, $n$ and $x_1, x_2, \dots, x_m$ if $x_{i+1} - x_i \equiv x_i - x_{i-1} \pmod n$ for every $2 \le i \le m-1$, we say that the $m$-tuple $(x_1, \dots, x_m)$ is an *arithmetic sequence in (mod n)*. Let $p \ge 5$ be a prime number and $1 < a < p-1$ be an integer. Let $\{a_1, a_2, \dots, a_k\}$ be the... | [
"Let $(b_1, b_2, \\dots, b_k)$ be a permutation of $\\{a_1, a_2, \\dots, a_k\\}$ which is an arithmetic sequence in (mod $p$). Then, for some integers $c$ and $d$ we have $b_i \\equiv c + i d \\pmod p$ for every $i = 1, 2, \\dots, k$.\n\nIt is easy to see that $k$ is the order of $a$ in (mod $p$). Then, we have $a^... | Turkey | Team Selection Test | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
02qs | Problem:
A figura mostra três circunferências de raios $1$, $2$ e $3$, tangentes duas a duas nos pontos destacados. Qual é o comprimento do segmento $AB$?
A) $1$
B) $\sqrt{2}$
C) $\frac{1+\sqrt{5}}{2}$
D) $\frac{3}{2}$
E) $\sqrt{3}$ | [
"Solution:\nLembramos primeiro que se duas circunferências são tangentes, então, a reta que passa por seus centros passa também pelo ponto de tangência. No nosso caso, chamando de $P$, $Q$ e $R$ os centros das circunferências (como na figura), isso mostra que $PR = 3$, $PQ = 4$ e $QR = 5$. Como $3^{2} + 4^{2} = 5^{... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | B | |
00oc | Let $n \ge 3$ be an integer. A *circle dance* is a dance that is performed according to the following rule: On the floor, $n$ points are marked at equal distances along a large circle. At each of these points is a sheet of paper with an arrow pointing either clockwise or counterclockwise. One of the points is labeled „... | [
"a) By the pigeon-hole principle, there exists at least one point that is visited infinitely often. If there is another point that is visited only finitely many times, then there are also two neighboring points where one point is visited infinitely many times and the other one finitely many times. But this is not p... | Austria | Austrian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | n | |
0k64 | Problem:
The sequence of integers $\{a_{i}\}_{i=0}^{\infty}$ satisfies $a_{0}=3$, $a_{1}=4$, and
$$
a_{n+2}=a_{n+1} a_{n}+\left\lceil\sqrt{a_{n+1}^{2}-1} \sqrt{a_{n}^{2}-1}\right\rceil
$$
for $n \geq 0$. Evaluate the sum
$$
\sum_{n=0}^{\infty}\left(\frac{a_{n+3}}{a_{n+2}}-\frac{a_{n+2}}{a_{n}}+\frac{a_{n+1}}{a_{n+3}}-... | [
"Solution:\n\nThe key idea is to note that $a_{n+1} a_{n}+\\sqrt{a_{n+1}^{2}-1} \\sqrt{a_{n}^{2}-1}$ is the larger zero of the quadratic\n$$\nf_{n}(x)=x^{2}-\\left(2 a_{n+1} a_{n}\\right) x+a_{n}^{2}+a_{n+1}^{2}-1 .\n$$\nSince $a_{n+2}$ is the smallest integer greater than or equal to this root, it follows that $a_... | United States | HMMT February 2019 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
... | null | proof and answer | 14/69 | |
05e5 | Problem:
Let $ABC$ be a triangle with $AC > AB$, and denote its circumcircle by $\Omega$ and incentre by $I$. Let its incircle meet sides $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. Let $X$ and $Y$ be two points on minor arcs $\widehat{DF}$ and $\widehat{DE}$ of the incircle, respectively, such that $\angle BXD = ... | [
"Solution:\n\nBy the alternate segment theorem we have that:\n$$\n180^\\circ = \\angle DCY + \\angle CYD + \\angle YDC = \\angle DCY + \\angle DXB + \\angle YXD = \\angle DCY + \\angle YXB\n$$\nTherefore opposite angle of $BXYC$ are supplementary and so $CYXB$ cyclic.\nOne can apply power of a point at $K$ :\n$$\nK... | European Girls' Mathematical Olympiad (EGMO) | EGMO 2024 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
... | null | proof only | null | |
0bp2 | Problem:
Să se determine funcţiile derivabile $f: \mathbb{R} \rightarrow \mathbb{R}$ care verifică simultan condiţiile:
i) $f'(x)=0$, pentru orice $x \in \mathbb{Z}$;
ii) pentru $x \in \mathbb{R}$, dacă $f'(x)=0$, atunci $f(x)=0$. | [
"Solution:\nFuncţia identic nulă verifică condiţiile din enunţ.\n\nPresupunem că există o funcţie derivabilă $f: \\mathbb{R} \\rightarrow \\mathbb{R}$, neidentic nulă, care satisface (i) şi (ii).\nAtunci $f$ este continuă pe $\\mathbb{R}$, cu $f(x)=f'(x)=0, \\forall x \\in \\mathbb{Z}$.\n\nExistă $x_{0} \\in \\math... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | null | proof and answer | The identically zero function f(x) = 0 for all real x. | |
09w3 | Find all functions $f: \mathbb{Z} \to \mathbb{Z}$ satisfying
$$
f(-f(x) - f(y)) = 1 - x - y
$$
for all $x, y \in \mathbb{Z}$. | [
"Substituting $x = y = 1$ yields $f(-2f(1)) = -1$.\n\nSubstituting $x = n$ and $y = 1$ yields $f(-f(n) - f(1)) = -n$.\n\nSubstituting $x = -f(n) - f(1)$ and $y = -2f(1)$ then yields\n$$\nf(-f(-f(n) - f(1)) - f(-2f(1))) = 1 - (-f(n) - f(1)) - (-2f(1))\n$$\nin which the left hand side expands as $f(-(-n) - (-1)) = f(... | Netherlands | IMO Team Selection Test 3, June 2020 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = x - 1 | |
08r1 | Problem:
Let $ABC$ be an acute triangle with circumcenter $O$. Let $D$ be the foot of the altitude from $A$ to $BC$ and let $M$ be the midpoint of $OD$. The points $O_{b}$ and $O_{c}$ are the circumcenters of triangles $AOC$ and $AOB$, respectively. If $AO = AD$, prove that the points $A$, $O_{b}$, $M$ and $O_{c}$ are ... | [
"Solution:\n\nNote that $AB = AC$ cannot hold since $AO = AD$ would imply that $O$ is the midpoint of $BC$, which is not possible for an acute triangle. So we may assume without loss of generality that $AB < AC$.\nLet $M_{b}$ and $M_{c}$ be the midpoints of $AC$ and $AB$, respectively. Sinc... | JBMO | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line,... | null | proof only | null | |
0jl4 | Problem:
Let $b$ and $c$ be real numbers, and define the polynomial $P(x) = x^{2} + b x + c$. Suppose that $P(P(1)) = P(P(2)) = 0$, and that $P(1) \neq P(2)$. Find $P(0)$. | [
"Solution:\n\nSince $P(P(1)) = P(P(2)) = 0$, but $P(1) \\neq P(2)$, it follows that $P(1) = 1 + b + c$ and $P(2) = 4 + 2b + c$ are the distinct roots of the polynomial $P(x)$. Thus, $P(x)$ factors:\n$$\n\\begin{aligned}\nP(x) &= x^{2} + b x + c \\\\\n&= (x - (1 + b + c))(x - (4 + 2b + c)) \\\\\n&= x^{2} - (5 + 3b +... | United States | HMMT 2014 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | -3/2 | |
05f4 | Problem:
Soit $ABC$ un triangle. Soit $t$ la tangente en $A$ au cercle circonscrit du triangle $ABC$. On note $A'$ le point de $t$ de telle sorte que la droite $(BC)$ coupe le segment $[AA']$ en son milieu. On note $C'$ le symétrique de $C$ par rapport à $t$. Montrer que les points $B$, $A$, $C'$ et $A'$ sont cocycliq... | [
"Solution:\n\n\n\nOn note $X$ le point d'intersection des droites $(AA')$ et $(BC)$. On note de plus $C^*$ le symétrique de $C$ par rapport au point $X$. On va montrer que les points $B$, $A$, $C'$, et $A'$ sont cocycliques sur un cercle qui contient de plus le point $C^*$. On commence par ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geom... | null | proof only | null | |
0coj | A quadrilateral $ABCD$ is inscribed into a circle with diameter $AC$. Let $K$ and $M$ be respectively the projections of $A$ and $C$ onto the line $BD$. Point $P$ is chosen on $AC$ so that $PK \parallel BC$. Prove that $\angle KPM = 90^\circ$. (T. Emelyanova)
Четырёхугольник $ABCD$ вписан в окружность с диаметром $AC$... | [
"First solution. Let $E$ be the intersection point of diagonals $AC$ and $BD$. Assume for definiteness that point $K$ lies on segment $BE$. Let the line passing through $K$ and parallel to $PM$ intersect $AC$ at point $N$ (see Fig. 8). Then $\\triangle NKE \\sim \\triangle PME$ (since their sides are parallel), hen... | Russia | Regional round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
0b85 | Let $n$ be a positive integer number and let $a_1, a_2, \dots, a_n$ be $n$ positive real numbers. Prove that $f : [0, \infty) \to \mathbb{R}$, defined by
$$
f(x) = \frac{a_1+x}{a_2+x} + \frac{a_2+x}{a_3+x} + \dots + \frac{a_{n-1}+x}{a_n+x} + \frac{a_n+x}{a_1+x},
$$
is a decreasing function. | [
"Set $a_{n+1} = a_1$ and let $0 \\le x \\le y$. Since\n$$\nf(y) - f(x) = (y - x) \\sum_{i=1}^{n} \\frac{a_{i+1} - a_i}{(a_{i+1} + x)(a_{i+1} + y)}\n$$\nshowing $f$ is decreasing amounts to showing\n$$\n\\sum_{i=1}^{n} \\frac{a_{i+1}}{(a_{i+1} + x)(a_{i+1} + y)} \\le \\sum_{i=1}^{n} \\frac{a_i}{(a_{i+1} + x)(a_{i+1}... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof only | null | |
0662 | Let $A$ be a finite set of positive reals, let $B = \{\frac{x}{y}: x, y \in A\}$ and let $C = \{xy: x, y \in A\}$. If we denote $cardX = |X|$, show that: $|A| \cdot |B| \le |C|^2$. | [] | Greece | Mediterranean Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof only | null | |
08x2 | Suppose two $20 \times 13$ rectangular grids consisting of $260$ small squares are given.
We insert into each square box of the two grids, numbers $1, 2, \ldots, 260$ in the following way:
* For the first grid, we start inserting numbers $1, 2, \ldots, 13$ into the boxes on the top row from left to right. Continue to ... | [
"The number inserted into the box located on the $i$-th row from the top and $j$-th column from the left is given by $13(i-1)+j$ for the case of the first grid, and by $20(13-j)+i$ for the case of the second grid. If the same number goes into the boxes located at the same position in the two grids, we must have $13... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 87, 174 | |
0is2 | Problem:
Let $p$ be a tropical polynomial:
$$
p(x)=a_{n} \odot x^{n} \oplus a_{n-1} \odot x^{n-1} \oplus \cdots \oplus a_{1} \odot x \oplus a_{0}, \quad a_{n} \neq \infty
$$
Prove that we can find $r_{1}, r_{2}, \ldots, r_{n} \in \mathbb{R} \cup\{\infty\}$ so that
$$
p(x)=a_{n} \odot\left(x \oplus r_{1}\right) \odot\le... | [
"Solution:\nAgain, we have\n$$\np(x)=\\min _{0 \\leq k \\leq n}\\left\\{a_{k}+k x\\right\\} .\n$$\nSo the graph of $y=p(x)$ can be drawn as follows: first, draw all the lines $y=a_{k}+k x$, $k=0,1, \\ldots, n$, then trace out the lowest broken line, which then is the graph of $y=p(x)$.\nSo $p(x)$ is piecewise linea... | United States | 11th Annual Harvard-MIT Mathematics Tournament - Team Round: B Division | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0dfi | 2000 consecutive integers (not necessarily positive) are written on the board. A student takes several turns. On each turn, he partitions the 2000 integers into 1000 pairs, and substitutes each pair by the difference and the sum of that pair (note that the difference does not need to be positive as the student may choo... | [
"Note that $(a-b)^2 + (a+b)^2 = 2(a^2 + b^2)$, so the sum of the squares of the numbers written on the board doubles on each turn. Note that\n$$\nn^2 + (n+1)^2 + \\dots + (n+1999)^2 = 2000n^2 + 1999 \\cdot 2000n + \\frac{1999 \\cdot 2000 \\cdot 3999}{6},\n$$\nwhich is congruent to $8$ modulo $16$. Obviously, when w... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
044c | Complex numbers $z_1, z_2, \dots, z_{100}$ satisfy $z_1 = 3 + 2i$, $z_{n+1} = \overline{z_n} \cdot i^n$, ($n = 1, 2, \dots, 99$), with $i$ as the imaginary unit. Then the value of $z_{99} + z_{100}$ is ______. | [
"By the given conditions, we have\n$$\nz_{n+2} = \\overline{z_{n+1}} \\cdot i^{n+1} = \\overline{z_n} \\cdot i^n \\cdot i^{n+1} = z_n i \\quad (n = 1, 2, \\dots, 98).\n$$\nAnd since $z_1 = 3 + 2i$, $z_{99} = z_1 i = -2 + 3i$. Therefore,\n$$\n\\begin{aligned}\nz_{99} + z_{100} &= z_{99} + \\overline{z_{99}} \\cdot i... | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | -5 + 5i | |
04ie | Determine all triples $(p, m, n)$ of positive integers such that $p$ is a prime number and
$$
p^m - n^3 = 8.
$$ | [
"By moving $n^3$, we get a sum of cubes on the right-hand side:\n$$\np^m = n^3 + 8 = (n+2)(n^2 - 2n + 4).\n$$\nSince $p$ is prime, each of the factors on the right-hand side must be a power of $p$:\n$$\nn + 2 = p^{\\alpha}, \\quad n^2 - 2n + 4 = p^{\\beta},\n$$\nwhere $\\alpha$ and $\\beta$ are obviously positive i... | Croatia | Croatia Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (2, 4, 2) and (3, 2, 1) | |
0k2w | Problem:
For how many positive integers $n \leq 100$ is it true that $10 n$ has exactly three times as many positive divisors as $n$ has? | [
"Solution:\n\nAnswer: 28\nLet $n=2^{a} 5^{b} c$, where $2,5 \\nmid c$. Then, the ratio of the number of divisors of $10 n$ to the number of divisors of $n$ is $\\frac{a+2}{a+1} \\frac{b+2}{b+1}=3$. Solving for $b$, we find that $b=\\frac{1-a}{2 a+1}$. This forces $(a, b)=(0,1),(1,0)$. Therefore, the answers are of ... | United States | HMMT November 2018 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 28 | |
09aq | Find all the positive integer $N$ such that there exists $M$ satisfying the following two conditions:
a. $M$'s first few digits coincide with $N$.
b. Let $S$ be a number obtained from $M$ by transferring the digits that express $N$ from the beginning of $M$ to the end of $M$. Then $S \cdot N = M$.
(For example, $M = 4... | [
"Let $N = a_1a_2\\ldots a_c$. Let us define a sequence of pair of integers by the following way. $a_0 = N$ and $b_0 = 0$.\n$$\n\\begin{cases} a_{n+1} \\equiv a_n \\cdot N + b_n \\pmod{10^c} \\\\ b_{n+1} = \\frac{a_n \\cdot N + b_n - a_{n+1}}{10^c} \\end{cases}\n$$\nLet us show that $b_n < N$, for $n \\in \\mathbb{N... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | null | proof and answer | all positive integers | |
0feh | Problem:
Se pide encontrar todos los números enteros positivos $n$ tales que $3^{n}+5^{n}$ es múltiplo de $3^{n-1}+5^{n-1}$. | [
"Solution:\nPara un tal $n$, puesto que\n$$\n3\\left(3^{n-1}+5^{n-1}\\right)<3^{n}+5^{n}<5\\left(3^{n-1}+5^{n-1}\\right)\n$$\nse verifica\n$$\n3^{n}+5^{n}=4\\left(3^{n-1}+5^{n-1}\\right)\n$$\nla cual se reduce a\n$$\n5^{n-1}=3^{n-1}\n$$\nque implica $n=1$.\n\nPues $n=1$ es solución (porque 8 es múltiplo de 2), se c... | Spain | null | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1 | |
06l6 | We choose $100$ points in the coordinate plane. Let $N$ be the number of triples $(A, B, C)$ of distinct chosen points such that $A$ and $B$ have the same $y$-coordinate, and $B$ and $C$ have the same $x$-coordinate. Find the greatest possible value that $N$ can attain by considering all possible ways to choose the poi... | [] | Hong Kong | HKG TST | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 8100 | |
04md | Determine all pairs $(m, n)$ of integers such that
$$
n^2 - 6n = m^2 + m - 10.
$$ | [
"We are given the equation\n$$\nn^2 - 6n = m^2 + m - 10.\n$$\nLet's rewrite it as\n$$\nn^2 - 6n - m^2 - m + 10 = 0.\n$$\nGroup terms:\n$$\nn^2 - m^2 - 6n - m + 10 = 0.\n$$\nRecall that $n^2 - m^2 = (n - m)(n + m)$, so\n$$\n(n - m)(n + m) - 6n - m + 10 = 0.\n$$\nLet us solve for $n$ in terms of $m$.\n\nAlternatively... | Croatia | Croatia_2018 | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | (1, 2), (1, 4), (-2, 2), (-2, 4) | |
0e43 | Prove that for all real $x$ and $y$ the inequality
$$
|x + y| + |x + 1| + |y + 1| \geq 2
$$
holds. For what $x$ does there exist $y$ such that $|x+y|+|x+1|+|y+1| = 2$? | [
"For all real $a$ we have $|a| \\ge a$ and $|a| \\ge -a$. So,\n$$\n|x + y| \\ge -(x + y), \\quad |x + 1| \\ge x + 1, \\quad |y + 1| \\ge y + 1, \\quad (1)\n$$\n\n$|x + y| + |x + 1| + |y + 1| \\geq -(x + y) + (x + 1) + (y + 1) = 2.$\n$$\nAssume that the equality holds. Then the equality case occurs in all three ineq... | Slovenia | National Math Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x in [-1, 1] | |
057l | Find the value of the expression
$$
\frac{1}{\left(\frac{1}{2019}\right)^2 + 1} + \frac{1}{\left(\frac{2}{2018}\right)^2 + 1} + \frac{1}{\left(\frac{3}{2017}\right)^2 + 1} + \dots + \frac{1}{\left(\frac{2018}{2}\right)^2 + 1} + \frac{1}{\left(\frac{2019}{1}\right)^2 + 1}
$$ | [
"Group the summands into pairs: the first one together with the last one, the second one together with the second last one, etc. Adding the members of each pair gives us\n$$\n\\frac{1}{\\left(\\frac{i}{j}\\right)^2 + 1} + \\frac{1}{\\left(\\frac{j}{i}\\right)^2 + 1} = \\frac{\\frac{i^2}{j^2} + 1 + \\frac{j^2}{i^2} ... | Estonia | Open Contests | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 2019/2 | |
0cr6 | На доске написано уравнение $x^3 + *x^2 + *x + * = 0$. Петя и Вася по очереди заменяют звёздочки на рациональные числа: вначале Петя заменяет любую из звёздочек, потом Вася — любую из двух оставшихся, а затем Петя — оставшуюся звёздочку. Верно ли, что при любых действиях Васи Петя сможет получить уравнение, у которого ... | [
"**Ответ.** Да.\n\n**Первое решение.** Пусть Петя первым ходом сделает свободный член уравнения нулём. Тогда полученное уравнение точно будет иметь корень $0$; значит, Пете достаточно добиться того, чтобы другим корнем было число $t = 2014$. Это всегда можно сделать: если после хода Васи получится уравнение $x^3 + ... | Russia | XL Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | Yes | |
0dt4 | Let $n \ge 2$ be a positive integer. For any integer $a$, let $Q_a(x)$ denote the polynomial $x^n + ax$. Let $p$ be a prime number and $S_a$ be the set
$$
S_a = \{b \mid 0 \le b \le p-1, \exists c \in \mathbb{Z}, Q_a(c) \equiv b \pmod{p}\}.
$$
Show that the expression $\frac{1}{p-1} \sum_{a=1}^{p-1} |S_a|$ is an intege... | [
"All congruences are taken modulo $p$. We count pairs $(a, b)$ taken modulo $p$ where $a \\neq 0$ and the equation $x^n + ax \\equiv b$ has a solution in $x$. Then $\\sum_{a=1}^{p-1} |S_a|$ is the total number of such pairs.\n\nFirst, consider all pairs $(a, b)$ where $a \\neq 0$. Defined the relation $(a, b) \\sim... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof only | null | |
0itd | Problem:
Show that for positive integers $n_{1}$, $n_{2}$ and $d$,
$$
f\left(n_{1} n_{2}, d\right) \leq f\left(n_{1}, d\right)+n_{1}\left(f\left(n_{2}, d\right)-1\right)
$$ | [
"Solution:\nGiven a multiset of $f\\left(n_{1}, d\\right)+n_{1}\\left(f\\left(n_{2}, d\\right)-1\\right)$ lattice points, we may select $l=f\\left(n_{2}, d\\right)$ pairwise disjoint submultisets $S_{1}, S_{2}, \\ldots, S_{l}$, each consisting of $n_{1}$ points, whose centroid is a lattice point.\n\nLet $\\varphi$ ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof only | null | |
033w | Problem:
Find all values of $a$ such that the equation
$$
(a^{2}-a-9) x^{2}-6 x-a=0
$$
has two distinct positive roots. | [
"Solution:\nThe equation has two distinct positive roots if and only if\n$$\n\\left\\lvert\\,\n\\begin{aligned}\n& D=a^{3}-a^{2}-9 a+9>0 \\\\\n& x_{1}+x_{2}=\\frac{6}{a^{2}-a-9}>0 \\\\\n& x_{1} x_{2}=\\frac{-a}{a^{2}-a-9}>0\n\\end{aligned}\n\\right.\n$$\nThe first inequality is satisfied for $a \\in(-3,1) \\cup(3,+... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a ∈ (-3, (1 − √37)/2) | |
03et | The sides and diagonals of a regular $n$-gon are colored in $k \ge 3$ colors. For each color $i$, between every two vertices of the polygon there exists a path consisting only of segments of color $i$. Prove that there exist three vertices of the polygon $A$, $B$, and $C$ such that the line segments $AB$, $BC$, and $AC... | [
"We need to prove that in a complete graph with $n$ vertices and $k$ colors, where the induced graph on each color is connected, there exists a multicolor triangle. Denote the colors by $1$, $2$, $3$, $\\ldots$, $k$ and recolor all edges that are in any of the colors $4$, $5$, $\\ldots$, $k$ into color $3$. The new... | Bulgaria | Bulgarian Winter Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0i82 | Problem:
Let $r, s, t$ be the solutions to the equation $x^{3}+a x^{2}+b x+c=0$. What is the value of $(r s)^{2}+(s t)^{2}+(r t)^{2}$ in terms of $a, b$, and $c$? | [
"Solution:\n$b^{2}-2 a c$\n\nWe have $(x-r)(x-s)(x-t)=x^{3}+a x^{2}+b x+c$, so\n$$\na=-(r+s+t), \\quad b=r s+s t+r t, \\quad c=-r s t .\n$$\nSo we have\n$$\n(r s)^{2}+(s t)^{2}+(r t)^{2}=(r s+s t+r t)^{2}-2 r s t(r+s+t)=b^{2}-2 a c .\n$$"
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | final answer only | b^2 - 2 a c | |
0l7v | Alex divides a disk into four quadrants with two perpendicular diameters intersecting at the center of the disk. He draws 25 more line segments through the disk, drawing each segment by selecting two points at random on the perimeter of the disk in different quadrants and connecting those two points. Find the expected ... | [
"Because the endpoints of the segments are chosen at random from a uniform distribution, the probability that two segments share an endpoint is 0, so that possibility can be ignored. For the same reason, it can be assumed that no more than two of the segments intersect at the same point. Before any of the 27 line s... | United States | 2025 AIME I | [
"Discrete Mathematics > Combinatorics > Expected values",
"Geometry > Plane Geometry > Circles"
] | null | proof and answer | 204 | |
0ike | Problem:
Octagon $A B C D E F G H$ is equiangular. Given that $A B=1$, $B C=2$, $C D=3$, $D E=4$, and $E F=F G=2$, compute the perimeter of the octagon. | [
"Solution:\n\nExtend sides $A B$, $C D$, $E F$, $G H$ to form a rectangle: let $X$ be the intersection of lines $G H$ and $A B$; $Y$ that of $A B$ and $C D$; $Z$ that of $C D$ and $E F$; and $W$ that of $E F$ and $G H$.\n\nAs $B C=2$, we have $B Y=Y C=\\sqrt{2}$. As $D E=4$, we have $D Z=Z E=2\\sqrt{2}$. As $F G=2$... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 20 + sqrt(2) | |
0a6x | Problem:
Let $ABCD$ be a parallelogram in the plane. We draw two circles of radius $R$, one through the points $A$ and $B$, the other through $B$ and $C$. Let $E$ be the other point of intersection of the circles. We assume that $E$ is not a vertex of the parallelogram. Show that the circle passing through $A$, $D$, a... | [
"Solution:\n\n(See Figure 1.) Let $F$ and $G$ be the centers of the two circles of radius $R$ passing through $A$ and $B$; and $B$ and $C$, respectively. Let $O$ be the point for which the rectangle $ABGO$ is a parallelogram. Then $\\angle OAD = \\angle GBC$, and the triangles $OAD$ and $GBC$ are congruent (sas). S... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 1 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Transformations > Translation"
] | null | proof only | null | |
05st | Problem:
Trouver tous les entiers $n \geqslant 1$ tel que pour tout nombre premier $p < n$, $n - \left\lfloor \frac{n}{p} \right\rfloor p$ n'est pas divisible par un carré différent de 1. | [
"Solution:\n\nAnalysons le problème : on cherche les entiers $n$ vérifiant une certaine propriété. On peut déjà commencer par chercher les petites valeurs de $n$ qui conviennent : en testant les entiers $n$ entre 1 et 20 on trouve $\\{1,2,3,5,7,13\\}$.\n\nUne des premières choses à faire devant ce problème est de r... | France | Envoi 5: Pot Pourri | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > ... | null | proof and answer | {1, 2, 3, 5, 7, 13} | |
0fzd | Problem:
Trouver toutes les fonctions $f: \mathbb{N} \rightarrow \mathbb{N}$ telles que pour tous $m, n \in \mathbb{N}$, on ait
$$
m^{2}+f(n) \mid m f(m)+n
$$ | [
"Solution:\n\nAvec $m=n$, on obtient $m^{2}+f(m) \\mid m f(m)+m$.\n$$\n\\Rightarrow m^{2}+f(m)\\left|m f(m)+m-m\\left(m^{2}+f(m)\\right) \\Rightarrow m^{2}+f(m)\\right| m^{3}-m\n$$\nDonc, avec $m=2$, on a $4+f(2) \\mid 6$ et comme $f(2)>0$, on a $f(2)=2$.\nSubstituons $m=2$ au départ, on a $4+f(n) \\mid 4+n$ et don... | Switzerland | IMO-Selektionsprüfung | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | f(n) = n for all natural numbers n | |
03tv | Nine balls of the same size and color, numbered $1, 2, \dots, 9$, were put into a packet. Now $A$ draws a ball from the packet, noted that it is of number $a$, and puts back it. Then $B$ also draws a ball from the packet and noted that it is of number $b$. Then the probability for the inequality $a - 2b + 10 > 0$ to ho... | [
"Since each has equally $9$ different possible results for $A$ and $B$ to draw a ball from the packet independently, the total number of possible events is $9^2 = 81$.\n\nFrom $a - 2b + 10 > 0$ we get $2b < a + 10$.\n\nWe find that when $b = 1, 2, 3, 4, 5$, $a$ can take any value in $1, 2, \\dots, 9$ to make the in... | China | China Mathematical Competition | [
"Statistics > Probability > Counting Methods > Permutations"
] | English | proof and answer | 61/81 | |
097v | Problem:
Determinați toate valorile parametrului real $m$ pentru care ecuația $m(m+2) \cdot x^{2}-(m-2) \cdot x(x^{2}+1)-2(x^{2}+1)^{2}=0$ are două soluții reale distincte. | [
"Solution:\n\nObservăm că $x=0$ nu este soluție a ecuației. Împărțind ambii membri ai ecuației la $x^{2}$, obținem ecuația echivalentă\n$$\n2 \\cdot\\left(\\frac{x^{2}+1}{x}\\right)^{2}+(m-2) \\cdot \\frac{x^{2}+1}{x}-m(m+2)=0\n$$\nNotând $\\frac{x^{2}+1}{x}=t$, obținem ecuația $2 t^{2}+(m-2) \\cdot t-m(m+2)=0$.\nV... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (-6, -2) ∪ {2} | |
0gnl | Let $N$ denote the set of nonnegative integers, and $Z$ denote the set of all integers. If a function $f: N \times Z \rightarrow Z$ satisfies the conditions that
$$
i.\ f(0, 0) = 1,\ f(0, 1) = 1,
$$
$$
ii.\ \text{for all}\ k \notin \{0, 1\},\ f(0, k) = 0\ \text{and}
$$
$$
iii.\ \text{for all}\ n \ge 1\ \text{and}\ k,\ ... | [
"It can be shown by induction on $n \\ge 0$ that $f(n, k) = 0$ if $k < 0$ or $n^2+n+1 < k$, $f(n, n^2+n+1-k) = f(n, k)$ for all $k$, and $\\sum_{k=0}^{n^2+n+1} f(n, k) = 2^{n+1}$. For example,\nthe induction step for the second claim can be verified as follows:\n$$\n\\begin{align*}\n& f(n + 1, (n + 1)^2 + (n + 1) +... | Turkey | 16th Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 2^{2008} | |
02jh | Problem:
Quais dos números abaixo são negativos?
$10 - 3 \sqrt{11}$;
$3 \sqrt{11} - 10$;
$18 - 5 \sqrt{13}$;
$51 - 10 \sqrt{26}$;
$10 \sqrt{26} - 51$. | [
"Solution:\n\nComo $100 > 99$ então $\\underbrace{\\sqrt{100}}_{10} > \\underbrace{\\sqrt{99}}_{3 \\sqrt{11}}$. Logo, $10 - 3 \\sqrt{11} > 0$ e $3 \\sqrt{11} - 10 < 0$.\n\nAnalogamente: $2601 > 2600 \\Rightarrow \\underbrace{\\sqrt{2601}}_{51} > \\underbrace{\\sqrt{2600}}_{10 \\sqrt{26}}$.\nAssim, $51 - 10 \\sqrt{2... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | 3\sqrt{11} - 10; 10\sqrt{26} - 51; 18 - 5\sqrt{13} | |
0i8i | Problem:
Compute the radius of the inscribed circle of a triangle with sides $15$, $16$, and $17$. | [
"Solution:\n\nHeron's formula gives that the area is $\\sqrt{24 \\cdot 9 \\cdot 8 \\cdot 7} = 24 \\sqrt{21}$. Then, using the result that the area of a triangle equals the inradius times half the perimeter, we see that the radius is $\\sqrt{21}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles"
] | null | final answer only | sqrt(21) | |
0e6p | Among the functions below, which takes the value $0$ exactly twice?
(A) $f(x) = \sin x - 1$
(B) $f(x) = |x^2 - 1| - 2$
(C) $f(x) = e^x - 1$
(D) $f(x) = |2x - 1|$
(E) $f(x) = x - 1$ | [
"The graph of the function given in (A) intersects the $x$ axis infinitely many times, the graph of the function in (B) intersects the $x$ axis in $(\\sqrt{3}, 0)$ and $(-\\sqrt{3}, 0)$, the graph of the function in (C) intersects the $x$ axis only in $(0, 0)$, the graph of the function in (D) intersects the $x$ ax... | Slovenia | National Math Olympiad 2012 | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | B | |
0088 | Bibi wrote a natural number $N$. The sum of all natural numbers less than $N$ is a 3-digit number with equal digits. Find $N$. | [
"A 3-digit number with equal digits is divisible by $111$ and hence by $37$ because $111 = 37 \\cdot 3$. By hypothesis such a number is $1 + 2 + \\cdots + (N-1) = \\frac{1}{2}N(N-1)$, so $N(N-1)$ is divisible by $37$. Hence $N = 37k$ or $N = 37k+1$ for some integer $k \\ge 1$. If $k \\ge 2$ then $N \\ge 74$, $\\fra... | Argentina | Mathematical Olympiad Rioplatense | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 37 | |
06fh | Let $a$, $b$ and $c$ be the sides of a triangle and $p = a + b + c$. Show that
a. $\frac{a}{b+c-a} + \frac{b}{c+a-b} + \frac{c}{a+b-c} \ge 3$;
b. $a^k + b^k > \frac{c^k}{2^{k-1}}$ for $k = 2, 3, 4, \dots$;
c. $a^k + b^k + c^k \ge \frac{p^k}{3^{k-1}}$ for $k = 2, 3, 4, \dots$. | [
"a. Let $x = \\frac{b+c-a}{2}$, $y = \\frac{c+a-b}{2}$ and $z = \\frac{a+b-c}{2}$. Then we have\n$$\n\\begin{align*}\n\\frac{a}{b+c-a} + \\frac{b}{c+a-b} + \\frac{c}{a+b-c} &\\ge 3 \\\\\n\\frac{y+z}{2x} + \\frac{z+x}{2y} + \\frac{x+y}{2z} &\\ge 3 \\\\\n\\frac{y}{x} + \\frac{z}{x} + \\frac{z}{y} + \\frac{x}{y} + \\f... | Hong Kong | Year 2008 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
03cp | A cyclic hexagon $ABCDEF$ is given with the following property:
$$
AB \cdot CD \cdot EF = BC \cdot DE \cdot AF.
$$
Let $B_1$ be the symmetric point of $B$ with respect to line $AC$, $D_1$ be the symmetric point of $D$ with respect to line $CE$, and $F_1$ be the symmetric point of $F$ with respect to line $EA$. Prove ... | [
"It is well known fact that for cyclic hexagon the equation $|AB| \\cdot |CD| \\cdot |EF| = |BC| \\cdot |DE| \\cdot |AF|$ holds iff the diagonals $AD$, $BE$ and $CF$ are concurrent (follows from the trigonometric form of sine theorem for $\\triangle ACE$). Let $P$ be the point of intersection of $AD$, $BE$ and $CF$... | Bulgaria | 68. National Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06yq | Problem:
Find the roots $r_{1}, r_{2}, r_{3}, r_{4}$ of the equation $4x^{4} - a x^{3} + b x^{2} - c x + 5 = 0$, given that they are positive reals satisfying $r_{1}/2 + r_{2}/4 + r_{3}/5 + r_{4}/8 = 1$. | [
"Solution:\n\nWe have $r_{1} r_{2} r_{3} r_{4} = 5/4$ and hence $(r_{1}/2)(r_{2}/4)(r_{3}/5)(r_{4}/8) = 1/4^{4}$. But AM/GM gives that $(r_{1}/2)(r_{2}/4)(r_{3}/5)(r_{4}/8) \\leq \\left( (r_{1}/2 + r_{2}/4 + r_{3}/5 + r_{4}/8)/4 \\right)^{4} = 1/4^{4}$ with equality iff $r_{1}/2 = r_{2}/4 = r_{3}/5 = r_{4}/8$. Henc... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | r1 = 1/2, r2 = 1, r3 = 5/4, r4 = 2 | |
0dgk | Problem:
Consider the positive numbers $x_1, x_2, \dots, x_n$ such that $\sum_{i=1}^n x_i = \sum_{i=1}^n \frac{1}{x_i}$. Prove that $\sum_{i=1}^n \frac{1}{n-1+x_i} \le 1$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0dat | Find all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$ such that
$$
n^{3}-n^{2} \leq f(n) \cdot (f(f(n)))^{2} \leq n^{3}+n^{2}
$$
for every $n$ is positive integers. | [
"Firstly, we prove that $f$ is injective. Let two natural numbers $a, b$ such that $a+1 \\leq b$ and $f(a)=f(b)$. By the given condition, we have\n$$\nf(a)[f(f(a))]^{2} \\leq a^{3}+a^{2}<a^{3}+2a^{2}+a=(a+1)^{3}-(a+1)^{2}.\n$$\nSince $a, b \\in \\mathbb{Z}^{+}$ and $a+1 \\leq b$, we have\n$$\n(a+1)^{3}-(a+1)^{2} \\... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | All functions obtained by choosing any collection of disjoint adjacent pairs of positive integers and swapping each chosen pair while fixing all other integers. Equivalently, f is an involution whose cycles are either fixed points or two-cycles of the form k and k plus one; for each chosen pair set f(k) = k plus one an... |
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