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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
024c | Problem:
Existe algum número inteiro $N$ tal que valha
$$
2008 \times N = 222\ldots2 ?
$$ | [
"Solution:\nSolução 1: Se existir esse número $N$, então\n$$\nN = \\frac{222\\ldots2}{2008} = \\frac{2 \\times 111\\ldots1}{2 \\times 1004} = \\frac{111\\ldots1}{1004}\n$$\nLogo, $N$ não é inteiro, por ser o quociente do número ímpar $111\\ldots1$ pelo número par $1004$. Portanto, não existe tal $N$.\n\n\nSolução 2... | Brazil | Nível 2 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
02b3 | Problem:
Para qual número natural $k$ a expressão $\frac{k^{2}}{1,001^{k}}$ atinge seu maior valor? | [
"Solution:\n\nEstamos procurando o valor de $k$ para o qual é máximo o termo da sequência:\n$$\n\\frac{1^{2}}{1,001}, \\frac{2^{2}}{1,001^{2}}, \\frac{3^{2}}{1,001^{3}}, \\ldots, \\frac{k^{2}}{1,001^{k}}, \\ldots\n$$\nConsidere as seguintes inequações equivalentes:\n$$\n\\frac{(k+1)^{2}}{1,001^{k+1}}<\\frac{k^{2}}{... | Brazil | Nível 3 | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | k = 2001 | |
0l9v | Consider a set $S$ of 2006 distinct numbers. A subset $T$ of $S$ is called stubborn if for every $u, v$ (not necessarily distinct) in $T$, the number $u + v$ does not belong to $T$. Prove that
i) if $S$ is the set of 2006 first positive integers, then the number of elements of every stubborn subset $T$ of $S$ does not... | [
"i) Let $A$ be a stubborn subset of $S = \\{1, 2, \\dots, 2006\\}$ consisting of $x$ elements $a_1 < a_2 < \\dots < a_x$. Consider the set $B := \\{a_2 - a_1, a_3 - a_1, \\dots, a_x - a_1\\}$. It is a subset of $S$ and consists of $x-1$ elements. As $A$ is a stubborn subset of $S$, $A \\cap B = \\emptyset$. It impl... | Vietnam | Vijetnam 2006 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0ast | Problem:
For what real values of $k$ does the system of equations
$$
\left\{\begin{array}{l}
x - k y = 0 \\
x^{2} + y = -1
\end{array}\right.
$$
have real solutions? | [
"Solution:\n\n$-\\frac{1}{2} \\leq k \\leq \\frac{1}{2}$"
] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | [-1/2, 1/2] | |
0hno | Problem:
Count the number of sequences $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ of integers such that $a_{i} \leq 1$ for all $i$ and all partial sums $\left(a_{1}, a_{1}+a_{2}, a_{1}+a_{2}+a_{3}, \ldots, a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right)$ are non-negative. | [
"Solution:\n\n$C($ length $+1 ) = C(6) = 132$."
] | United States | null | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 132 | |
01wb | The polynomial $f(x) = x^2 + bx + 8$, $b \neq 0$, has two different real roots $x_1$ and $x_2$. The numbers $x_1 + 1/x_2$ and $x_2 + 1/x_1$ are the roots of the monic polynomial $g(x)$ of degree 2.
Find all possible values of $g(1)$ if $g(1) = f(1)$. | [
"Answer: $-8$.\n\nVieta's formulas for $f(x)$ imply $x_1 + x_2 = -b$ and $x_1 \\cdot x_2 = 8$. Let $g(x) = x^2 + cx + d$, then from Vieta's formulas for $g(x)$:\n$$\n\\begin{aligned}\n-c &= x_1 + \\frac{1}{x_2} + x_2 + \\frac{1}{x_1} = -b + \\frac{x_1 + x_2}{x_1 \\cdot x_2} = -b - \\frac{b}{8} \\\\ \nd &= \\left(x_... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | -8 | |
0478 | Find the smallest real number $\lambda$ with the following property: For any positive integers $a$, $b$, $n$ where $a + b$ is not divisible by $n$, there exists a positive integer $k < n$ such that
$$
\left\{ \frac{ak}{n} \right\} + \left\{ \frac{bk}{n} \right\} \le \lambda,
$$
where $\{x\} = x - \lfloor x \rfloor$ den... | [
"*Proof.* The minimal $\\lambda$ is $\\frac{2}{3}$.\n\nFirst, we show $\\lambda \\ge \\frac{2}{3}$. Consider $n = 3$ with $a = b = 1$. For $k = 1, 2$, we have:\n$$\n\\left\\{ \\frac{k}{3} \\right\\} + \\left\\{ \\frac{k}{3} \\right\\} = \\frac{2k}{3} \\ge \\frac{2}{3},\n$$\nthus $\\lambda$ cannot be smaller than $\... | China | 2024 CGMO | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | 2/3 | |
0d4e | Let $ABC$ be a triangle and $I$ its incenter. The line $AI$ intersects the side $BC$ at $D$ and the perpendicular bisector of $BC$ at $E$. Let $J$ be the incenter of triangle $CDE$. Prove that triangle $CIJ$ is isosceles. | [
"It is well known that the bisector of the angle at vertex $A$ and the perpendicular bisector of side $BC$ intersect on the circumcircle of triangle $ABC$. We give here two solutions.\n\nSolution 1. We have\n$$\n\\begin{aligned}\n\\measuredangle JDI & = 180^\\circ - \\measuredangle EDJ = 180^\\circ - \\frac{1}{2} \... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English, Arabic | proof only | null | |
0g82 | 令 $x$ 和 $y$ 是正整數。試證若對所有的正整數 $n$, $2^n y + 1$ 整除 $x 2^n - 1$, 則 $x = 1$。 | [
"首先我們證明對於所有的正整數 $y$, 存在無限多個質數 $p \\equiv 3 \\pmod 4$ 使得 $p$ 整除某些形如 $2^n y + 1$ 的數。\n顯然只要考慮 $y$ 是奇數的例子。令\n$$\n2y + 1 = p_1^{e_1} \\cdots p_r^{e_r}\n$$\n是 $2y+1$ 的質因數分解。假設存在有限多個質數 $p_{r+1}, \\cdots, p_{r+s} \\equiv 3 \\pmod 4$ 整除某些形如 $2^n y + 1$ 的數, 但不整除 $2y + 1$。\n我們想要找到一個 $n$ 使得 $p_i^{e_i} \\mid 2^n y + 1, 1 \\le i... | Taiwan | 二〇一三數學奧林匹亞競賽第二階段選訓營 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | x = 1 | |
0gtd | Let $a, b, c, d$ be positive integers such that only finite number of prime numbers divide at least one element of the set
$$
\{a \cdot b^n + c \cdot d^n : n = 1, 2, 3, \dots\}.
$$
Prove that $b = d$. | [
"Let $x_n = a \\cdot b^n + c \\cdot d^n$, $\\gcd(a, c) = e$ and $\\gcd(b, d) = f$. Then $a = e a_1$, $c = e c_1$, $b = f b_1$, $d = f d_1$, $\\gcd(a_1, c_1) = 1$ and $\\gcd(b_1, d_1) = 1$. Let $P$ be the set of all primes such that each element of $P$ divides at least one element of the sequence $(y_n)_{n \\ge 1}$,... | Turkey | Turkish Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
0bb9 | Let $a$, $b$, $c$ be three positive real numbers. Prove that the function $f: \mathbb{R} \to \mathbb{R}$, $f(x) = \frac{a^x}{b^x + c^x} + \frac{b^x}{a^x + c^x} + \frac{c^x}{a^x + b^x}$ is increasing on $[0, \infty)$ and decreasing on $(-\infty, 0]$. | [
"We use straightforward computation: if $x \\le y$ are real numbers, then\n\n$$\n\\begin{align*} \nf(y) - f(x) &= \\sum_{\\text{cyc}} \\frac{a^y(b^x + c^x) - a^x(b^y + c^y)}{(b^x + c^x)(b^y + c^y)} \\\n&= \\sum_{\\text{cyc}} (a^y b^x - a^x b^y) \\left( \\frac{1}{(b^x + c^x)(b^y + c^y)} - \\frac{1}{(a^x + c^x)(a^y +... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof only | null | |
0gvt | In the space, $2005$ points were marked, none four of which lie in the same plane. Planes were drawn through all the triples of marked points. Prove that all the marked points can be colored into two colors in such a way that for any two points of the same color the number of the drawn planes separating them is odd. (W... | [
"**Доведення леми.** Розглянемо три можливі випадки взаємного розташування п’яти точок.\n1. Пряма $XY$ паралельна площині $PQR$. Тоді серед трьох прямих, що проходять через точки $P, Q, R$ паралельно $XY$, тільки одна лежить між двома іншими. Відповідна площина розділяє точки, а дві інші площини цього не роблять.\n... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
06fb | Let $p$ be a prime number and $f(x)$ be a polynomial with rational coefficients and is of degree $p$. Furthermore (i) $f(m)$ is an integer for any integer $m$ and (ii) $p$ divides $f(p+1) - f(1)$. Find all such polynomials. | [
"$f(x)$ can be any polynomial of the form\n$$\np a_p \\binom{x}{p} + a_{p-1} \\binom{x}{p-1} + a_{p-2} \\binom{x}{p-2} + \\dots + a_1 \\binom{x}{1} + a_0,\n$$\nwhere $a_0, a_1, \\dots, a_p$ are arbitrary integers with $a_p \\neq 0$.\n\nNote that we can always write\n$$\nf(x) = c_p \\binom{x}{p} + c_{p-1} \\binom{x}... | Hong Kong | Year 2008 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All polynomials of the form f(x) = p a_p \binom{x}{p} + a_{p-1} \binom{x}{p-1} + a_{p-2} \binom{x}{p-2} + \dots + a_1 \binom{x}{1} + a_0, where a_0, a_1, \dots, a_p are integers with a_p \neq 0. | |
0amj | Problem:
If $x \neq y$ and $\frac{x}{y} + x = \frac{y}{x} + y$, find the sum $\frac{1}{x} + \frac{1}{y}$. | [] | Philippines | Area Stage | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | -1 | |
01o3 | A parabola $y = a x^2$, $a > 0$, and the hyperbola $y = 1/x$ meet at point $T$. The common tangent of these curves touches the hyperbola at point $Q$ and touches the parabola at point $P$.
a) Prove that some two medians of the triangle $PQT$ are perpendicular.
b) Determine the area of the triangle $PQT$.
(V. Karamzin) | [
"b) Answer: $S = 27/4$.\n\nWe find the coordinates $(x_T; y_T)$ of the point of intersection of the parabola $y = a x^2$ and the hyperbola $y = 1/x$. We have $a x_T^2 = 1/x_T$, so $x_T = 1/\\sqrt{a} = 1/t$ and $y_T = 1/x_T = t$. The equation of the tangent to the parabola at point $P(x_P; y... | Belarus | Belorusija 2012 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof and answer | 27/4 | |
05vo | Problem:
Trouver tous les couples d'entiers naturels non nuls $(x, n)$ tels que
$$
3 \cdot 2^{x} + 4 = n^{2}
$$ | [
"Solution:\nPour $x=0$, l'équation devient $n^{2}=7$ qui n'a pas de solution. Supposons $x>0$, on factorise : $3 \\cdot 2^{x} = (n-2)(n+2)$. Le membre de gauche est pair donc le membre de droite est pair. Les deux facteurs du membre de droite sont de même parité donc ils sont tous les deux pairs et on a deux cas :\... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | [(2, 4), (5, 10), (6, 14)] | |
0khc | Each of 6 balls is randomly and independently painted either black or white with equal probability. What is the probability that every ball is different in color from more than half of the other 5 balls?
(A) $\frac{1}{64}$ (B) $\frac{1}{6}$ (C) $\frac{1}{4}$ (D) $\frac{5}{16}$ (E) $\frac{1}{2}$ | [] | United States | AMC 10 A | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | MCQ | D | |
0jv2 | Problem:
Let $ABCD$ be a rectangle, and let $E$ and $F$ be points on segment $AB$ such that $AE = EF = FB$. If $CE$ intersects the line $AD$ at $P$, and $PF$ intersects $BC$ at $Q$, determine the ratio of $BQ$ to $CQ$. | [
"Solution:\n\nAnswer: $\\frac{1}{3}$\n\nBecause $\\triangle PAE \\sim \\triangle PDC$ and $AE : DC = 1 : 3$, we have that $PA : PD = 1 : 3 \\Longrightarrow PA : AB = PA : BC = 1 : 2$. Also, by similar triangles $\\triangle PAF \\sim \\triangle QBF$, since $AF : BF = 2 : 1$, $PA : BQ = 2 : 1$. Then $BQ = \\frac{1}{2... | United States | HMMT November | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 1/3 | |
0dux | Problem:
Naj bo $M$ razpolovišče osnovnice $AB$ trapeza $ABCD$. V notranjosti daljice $AC$ leži taka točka $E$, da se premici $BC$ in $ME$ sekata v točki $F$, premici $FD$ in $AB$ se sekata v točki $G$ ter premici $DE$ in $AB$ v točki $H$. Dokaži, da je $M$ razpolovišče daljice $GH$. | [
"Solution:\n\nOznačimo z $N$ presečišče premic $CD$ in $EF$.\n\n\n\nKer je $DC \\parallel GB$ in so točke $M$, $N$ in $F$ kolinearne, je $$\\frac{|GM|}{|MB|} = \\frac{|DN|}{|NC|}.$$ Ker je $AH \\parallel DC$ in so točke $M$, $E$ ter $N$ kolinearne, je $$\\frac{|AM|}{|MH|} = \\frac{|NC|}{|DN... | Slovenia | 46. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | null | proof only | null | |
028u | Problem:
Um quadrilátero - O quadrilátero $ABCD$ da figura é um paralelogramo?
 | [
"Solution:\n\nPara que $ABCD$ seja um paralelogramo, seus lados devem ser dois a dois paralelos, isto é: $AB \\parallel CD$ e $AD \\parallel BC$.\n\nComo\n$$\n\\widehat{DAB} + \\widehat{ABC} = 180^\\circ\n$$\nentão as retas $AD$ e $BC$ são paralelas. Além disso, temos dois ângulos alternos internos de $45^\\circ$ e... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof only | null | |
0ds9 | In $\triangle ABC$, $AB = AC$, $D$ is a point on the side $BC$ and $E$ is a point on the segment $AD$. Given that $\angle BED = \angle BAC = 2\angle CED$, prove that $BD = 2CD$. | [
"Let $\\angle CED = x$ and $\\angle ABE = y$. Then\n$$\n\\angle BAC = \\angle BED = 2x \\Rightarrow \\angle BAE = 2x - y \\text{ and } \\angle EAC = y.\n$$\nLet $F$ be the point on $BE$ so that $\\angle AFE = \\pi$. (Note that $x = \\angle CED = y + \\angle ECA$ implying $x > y$. Thus $F$ is in fact in the interior... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
00dn | Prove that there are infinitely many positive integers $n$ such that the equation
$$
x^2 + y^{11} - z^{2022!} = n
$$
has no solution $(x, y, z)$ over the integers. | [
"We claim that if $n = 23k + 20$ with $k$ a non-negative integer then the equation has no solution over the integers. Let's assume that there is one and get a contradiction. Indeed, by Fermat's little theorem,\n$$\nx^2 + y^{11} \\equiv n + z^{2022!} \\equiv 20 + (0 \\text{ or } 1) \\equiv 20 \\text{ or } 21 \\pmod{... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0jzt | Problem:
Determine all primes $p$ such that there exists an integer $x$ satisfying $x^{2010} + x^{2009} + \cdots + 1 \equiv p^{2010} \pmod{p^{2011}}$. | [
"Solution:\nThe answer is all $p \\equiv 1 \\pmod{2011}$.\n\nFirst, note that if $x$ satisfies the problem condition, then we in particular have $\\Phi_{2011}(x) \\equiv 0 \\pmod{p}$ which implies $p \\equiv 1 \\pmod{2011}$.\n\nConversely, suppose $p \\equiv 1 \\pmod{2011}$ and fix an $a \\pmod{p}$ with order $2011... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof and answer | All primes p with p ≡ 1 (mod 2011). | |
0cl3 | Let $n \ge 3$ be a natural number. John draws a regular $n$-gon and connects every pair of vertices. On each segment, John writes a nonzero natural number such that in any triangle formed by three vertices, one of the numbers on its sides equals the sum of the other two.
Determine the smallest number of distinct value... | [
"We will show that the number sought is $n-1$ for $n \\neq 4$ and $2$ if $n=4$.\n\nFor $n = 3$ and $n = 4$, it can be easily verified that the answer is $2$. Suppose now that $n \\ge 5$. Denote by $a_{XY}$ the number written on segment $XY$.\n\nIf we label the vertices of the regular polygon as $A_1, A_2, \\dots, A... | Romania | 75th NMO Selection Tests | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n − 1 for n ≠ 4; when n = 4, the minimum is 2 | |
0alw | Problem:
If $x$ satisfies $\frac{\log_{2} x}{\log_{2} 2x - \log_{8} 2} = 3$, the value of $1 + x + x^{2} + x^{3} + x^{4} + \cdots$ is
(a) 1
(b) 2
(c) $\frac{1}{2}$
(d) the value does not exist | [] | Philippines | QUALIFYING STAGE | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | b | |
0k6z | Problem:
There are 100 people in a room with ages $1, 2, \ldots, 100$. A pair of people is called cute if each of them is at least seven years older than half the age of the other person in the pair. At most how many pairwise disjoint cute pairs can be formed in this room? | [
"Solution:\n\nFor a cute pair $(a, b)$ we would have\n$$\na \\geq \\frac{b}{2} + 7, \\quad b \\geq \\frac{a}{2} + 7\n$$\nSolving the system, we get that $a$ and $b$ must both be at least $14$. However, $14$ could only be paired with itself or a smaller number; therefore, only people with age $15$ or above can be pa... | United States | HMMT February 2019 | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 43 | |
0fwl | Problem:
Finde alle Tripel $(a, b, c)$ natürlicher Zahlen, sodass gilt:
$$
a \mid b c - 1, \quad b \mid c a - 1, \quad c \mid a b - 1
$$ | [
"Solution:\n\nOffenbar sind $a, b, c$ paarweise teilerfremd, denn zum Beispiel ist jeder gemeinsame Teiler von $a$ und $b$ wegen $a \\mid b c - 1$ auch ein Teiler von $1$. Da $b c - 1$ durch $a$ teilbar ist, muss auch $a b + b c + c a - 1$ durch $a$ teilbar sein, und aus Symmetriegründen gilt dasselbe für $b$ und $... | Switzerland | IMO Selektion 2008 | [
"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"
] | null | proof and answer | All permutations of (2, 3, 5) and of (1, 1, n) with n any positive integer. | |
02dv | Each day 289 students are divided into 17 groups of 17. No two students are ever in the same group more than once. What is the largest number of days that this can be done? | [
"Each day student $X$ is with 16 different students. But there are only 288 students available, so it can be done for at most $288/16 = 18$ days.\n\nWe now show explicitly how it can be done for 18 days. Label the students as $(m, n)$, where $0 \\le m, n \\le 16$. Now for $0 \\le d, g \\le 16$, take group $g$ on da... | Brazil | VI OBM | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | 18 | |
07lj | Let $(a_n)$ be the "iterated Fibonacci sequence": $a_n = f(f(n))$, where $f(1) = f(2) = 1$, and $f(n+2) = f(n+1) + f(n)$. Prove that $a_n$ is a multiple of 144 whenever it is a multiple of 14. | [
"This is actually true for the full Fibonacci sequence, not just the iterated one. Since the Fibonacci numbers depend only on the previous two values. Fibonacci modulo $n$ must be preperiodic with period at most $n^2$. In fact it is periodic for all of the values that we examine.\n\nCalculation shows that $f(n)$ mo... | Ireland | Irska | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof only | null | |
05zp | Problem:
Déterminer toutes les fonctions $f$ de $\mathbb{R}$ dans $\mathbb{R}$ telles que pour tous $x, y$ réels :
$$
f(x) f(y) f(x-y) = x^{2} f(y) - y^{2} f(x)
$$ | [
"Solution:\nSoit $f$ une solution de l'équation.\nPour $x = y = 0$, on a :\n$$\nf(0)^3 = 0\n$$\nAinsi, $f(0) = 0$. On exclut la fonction $f \\equiv 0$ qui est effectivement solution du problème.\nOn considère maintenant un nombre $x$ tel que $f(x) = 0$. On a alors pour un $y$ quelconque :\n$$\n0 = x^{2} f(y) - 0\n$... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 2 : AlgèBre | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) ≡ 0 for all x; or f(x) = x for all x; or f(x) = -x for all x. | |
0jjs | Problem:
Find the integer closest to
$$
\frac{1}{\sqrt[4]{5^{4}+1}-\sqrt[4]{5^{4}-1}}
$$ | [
"Solution:\nLet $x = (5^{4} + 1)^{1/4}$ and $y = (5^{4} - 1)^{1/4}$. Note that $x$ and $y$ are both approximately $5$. We have\n$$\n\\begin{aligned}\n\\frac{1}{x - y} &= \\frac{(x + y)(x^{2} + y^{2})}{(x - y)(x + y)(x^{2} + y^{2})} = \\frac{(x + y)(x^{2} + y^{2})}{x^{4} - y^{4}} \\\\\n&= \\frac{(x + y)(x^{2} + y^{2... | United States | HMMT 2014 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 250 | |
06ay | (A) Prove that for all real numbers $\kappa, \lambda, \mu$ holds the following inequality:
$$ (\kappa + \lambda + \mu)^2 \ge 3(\kappa\lambda + \lambda\mu + \mu\kappa). $$
When is equality valid?
(B) Let $x, y, z$ be positive real numbers and let $\alpha, \beta$ be real numbers such that
$$
\alpha(x + y + z) = \beta(xy... | [
"(A) It is enough to prove\n$$\n\\begin{aligned}\n& \\kappa^2 + \\lambda^2 + \\mu^2 + 2\\kappa\\lambda + 2\\lambda\\mu + 2\\mu\\kappa \\ge 3\\kappa\\lambda + 3\\lambda\\mu + 3\\mu\\kappa \\\\\n\\Leftrightarrow & \\kappa^2 + \\lambda^2 + \\mu^2 - \\kappa\\lambda - \\lambda\\mu - \\mu\\kappa \\ge 0 \\\\\n\\Leftrighta... | Greece | Hellenic Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0kxu | Problem:
Call a number feared if it contains the digits $13$ as a contiguous substring and fearless otherwise. (For example, $132$ is feared, while $123$ is fearless.) Compute the smallest positive integer $n$ such that there exists a positive integer $a<100$ such that $n$ and $n+10a$ are fearless while $n+a, n+2a, \l... | [
"Solution:\n\nFirst of all, note that we cannot have $n, n+a, \\ldots, n+10a$ be less than $1000$, since we cannot have fearless numbers have $13$ as their last two digits since $a<100$, and $129,130,131, \\ldots, 139$ doesn't work as $139$ is feared.\n\nThus, we must utilize numbers of the form $13xy$, where $1,3,... | United States | HMMT November | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 1287 | |
04h0 | For a positive integer $n$ denote by $s(n)$ the sum of all positive divisors of $n$ and by $d(n)$ the number of positive divisors of $n$. Determine all positive integers $n$ such that
$$
s(n) = n + d(n) + 1.
$$
(Nikola Adžaga) | [
"Observe that $n = 1$ is not a solution, so $d(n) \\ge 2$.\nIt is impossible that $d(n) = 2$, since then $n$ would be prime and given equation would reduce to $1 + n = n + 2 + 1$, so $d(n) \\ge 3$.\nLet $1 = D_1 < D_2 < \\cdots < D_d = n$ be the divisors of $n$. The given equation can then be written as\n$$\n1 + \\... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | English | proof and answer | 9 | |
00ym | Problem:
Find all pairs of positive integers $(a, b)$ such that $2^{a} + 3^{b}$ is the square of an integer. | [
"Solution:\nConsidering the equality $2^{a} + 3^{b} = n^{2}$ modulo $3$, it is easy to see that $a$ must be even. Obviously $n$ is odd so we may take $a = 2x$, $n = 2y + 1$ and write the equality as $4^{x} + 3^{b} = (2y + 1)^{2} = 4y^{2} + 4y + 1$. Hence $3^{b} \\equiv 1 \\pmod{4}$ which implies $b = 2z$ for some p... | Baltic Way | Baltic Way | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (4, 2) | |
0cgn | Let $a$ and $b$ be two real numbers in the interval $(0, 1)$, so that $a$ is a rational number and
$$
\{na\} \ge \{nb\}, \text{ for every natural number } n.
$$
Prove that $a = b$. (We denote by $\{x\}$ the fractional part of the real number $x$.) | [
"Let $a = \\frac{p}{q}$, where $p$ and $q$ are non-zero natural numbers, relatively prime, with $p < q$.\nThen $0 = \\{qa\\} \\ge \\{qb\\} \\ge 0$, thus $qb$ is a natural number. It follows that $b = \\frac{s}{q}$, where $s$ is a nonzero natural number.\n\nConsidering $n = q - 1$ in the relation from the hypothesis... | Romania | 74th Romanian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
0gz6 | For positive numbers $a$, $b$, $c$, such that $abc = 1$, prove the inequality:
$$
\frac{a^2 + b^2}{c^2 + a + b} + \frac{b^2 + c^2}{a^2 + b + c} + \frac{c^2 + a^2}{b^2 + c + a} \le 2.
$$ | [
"$abc = 1 \\Rightarrow \\frac{1}{3}(a+b+c)$. Then:\n$$\n\\begin{aligned}\nc^2 + a + b &\\le c^2 + (a+b)\\frac{1}{3}(a+b+c) \\\\\n&= \\frac{1}{3}(3c^2 + a^2 + b^2 + 2ab + ac + bc) \\\\\n&\\le \\frac{1}{6}(8c^2 + 5a^2 + 5b^2).\n\\end{aligned}\n$$\nSo\n$$\n\\frac{a^2 + b^2}{c^2 + a + b} + \\dots \\ge \\frac{a^2 + b^2}... | Ukraine | The Problems of Ukrainian Authors | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0cyf | The shape of a military base is an equilateral triangle of side $10$ kilometers. Security constraints make cellular phone communication possible only within $2.5$ kilometers. Each of $17$ soldiers patrols the base randomly and tries to contact all others. Prove that at each moment at least two soldiers can communicate. | [
"Divide any side of the equilateral triangle into four equal segments as follows. Then construct through the interior points to the sides parallel lines to the equilateral triangle.\n\n\n\nIn this way we get a net of the military base consisting in $16$ equilateral triangles of side $2.5$ k... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
0doz | The numbers $2$, $3$, $5$, $\ldots$, $2003$ are written on the board (here listed all prime numbers from the segment $[2; 2007]$). By operation of *simplification* we call a changing of two numbers $a$, $b$ by maximal prime number not exceeding $\sqrt{a^2 - ab + b^2}$. At the beginning student removes the number $q$, $... | [
"Note that $a < \\sqrt{a^2 - ab + b^2} < b$ if $a < b$. That's why simplification of two successive prime numbers is equivalent to removing the maximal (from that two) number. If between prime numbers $a < b$ there is exactly one prime number $c$, then operation of simplification to pair $(a, b)$ gives $a$ or $c$. ... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Maximum: 1999, independent of the initial removal. Minimum: 2 if the removed prime q ≥ 17; otherwise 3 for q ∈ {3, 5, 7, 11, 13}. | |
05er | Problem:
Les cases d'une grille à 10 lignes et 10 colonnes sont coloriées en blanc et en noir. Un coloriage de ces cases est dit homogène s'il contient un carré $3 \times 3$ monochrome, et inhomogène sinon. Montrer qu'il existe plus de coloriages inhomogènes que de coloriages homogènes. | [
"Solution:\n\nIl y a $2^{100}$ coloriages possibles. Si $c$ est un carré $3 \\times 3$, le nombre de coloriages tels que $c$ soit monochrome est égal à $2 \\times 2^{91} = 2^{92}$. Or, il y a 64 carrés $3 \\times 3$, donc il y a au plus $64 \\times 2^{92} = 2^{98}$ coloriages homogènes."
] | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0in0 | Problem:
Triangle $A B C$ has $A B=4$, $B C=6$, and $A C=5$. Let $O$ denote the circumcenter of $A B C$. The circle $\Gamma$ is tangent to and surrounds the circumcircles of triangles $A O B$, $B O C$, and $A O C$. Determine the diameter of $\Gamma$. | [
"Solution:\n\nDenote by $\\omega$, $\\Gamma_{1}$, $\\Gamma_{2}$, and $\\Gamma_{3}$ the circumcenters of triangles $A B C$, $B O C$, $C O A$, and $A O B$, respectively. An inversion about $\\omega$ interchanges $\\Gamma_{1}$ and line $B C$, $\\Gamma_{2}$ and line $C A$, and $\\Gamma_{3}$ and line $A B$. This inversi... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | 256√7/17 | |
07og | In a triangle $ABC$, $D$ is the foot of the altitude from $A$ to $BC$, and $M$ is the midpoint of the line segment $BC$. The three angles $\angle BAD$, $\angle DAM$ and $\angle MAC$ have equal measure.
Find the measures of the angles of the triangle $ABC$. | [
"**Solution 1.** Since $\\angle BAD = \\angle DAM$ and $\\angle ADB = \\angle ADM = 90^\\circ$, the triangles $ADB$ and $ADM$ which share the side $AD$, are congruent. Therefore $|BD| = |DM|$, and so $2|MD| = |BM| = |MC|$.\nSince $\\angle DAM = \\angle MAC$, $AM$ bisects $\\angle DAC$ and we have\n$$\n\\cos(\\angle... | Ireland | Irska 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 90°, 60°, 30° | |
081c | Problem:
Un esagono equiangolo ha quattro lati consecutivi lunghi nell'ordine $5$, $3$, $6$ e $7$. Determinare le lunghezze degli altri due lati. | [
"Solution:\n\nSiano $a, b, c, d, e, f$ i lati dell'esagono con $a=5$, $b=3$, $c=6$, $d=7$. Si prolunghino i lati $a, c, e$ fino ad incontrarsi nei punti $B, C, A$ (vedi figura). Poiché gli angoli interni dell'esagono sono tutti di $120^{\\circ}$, il triangolo $ABC$ e i tre triangolini determinati da ciascuno dei la... | Italy | Gara Nazionale di Matematica | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 1 and 8 | |
05w5 | Problem:
Soit $n \geqslant 2$ un entier. Une grenouille se déplace dans une grille de $3n \times 3n$ cases, sautant chaque fois d'une case vers la case située juste au-dessus ou juste à droite. Elle souhaite aller du coin en bas à gauche, marqué d'un $S$, au coin en haut à droite, marqué d'un $F$. Malheureusement, cer... | [
"Solution:\n\nColorions notre grille en trois couleurs :\n- en noir, les cases piégées ;\n- en rose, les cases non piégées auxquelles la grenouille peut accéder depuis la case $S$ ;\n- en vert, les cases non piégées depuis lesquelles la grenouille peut accéder à la case $F$.\n\nNotons que notre grille contient éven... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0l9m | In plane, let be given a fixed circle $(O)$ with radius $R$, two fixed points $A$, $B$ on the circle $(O)$ such that $A$, $B$, $O$ are not collinear.
Let $C$ be a point on $(O)$ distinct from $A$ and $B$. Construct the circle $(O_1)$ passing through $A$, touching the line $BC$ at $C$, construct the circle $(O_2)$ passi... | [
"1) From the construction of the circles $(O_1)$ and $(O_2)$, we have:\n$$\n\\angle CAD = \\angle DCB \\ (\\text{subtend the arc } CD \\text{ of } (O_1)) \\quad (1)\n$$\n$$\n\\angle ACD = \\angle CBD \\ (\\text{subtend the arc } CD \\text{ of } (O_2)) \\quad (2)\n$$\nTherefore $\\triangle DAC \\sim \\triangle DCB$ ... | Vietnam | Vietnam Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0dlh | On the extensions of the sides $CA$ and $AB$ of triangle $ABC$ beyond points $A$ and $B$, points $E$ and $F$ are chosen such that $AE = BC$ and $BF = AC$. The $A$-excircle is tangent to segment $BF$ at point $N$. Let $M$ be the midpoint of segment $EF$. Prove that $MN$ is parallel to the internal angle bisector of angl... | [] | Saudi Arabia | Saudi Booklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
07gi | Suppose that $A$-excircle of triangle $ABC$ is tangent to $AC$ and $AB$ at $E$ and $F$, respectively. Denote by $I_a$ the center of preceding circle. Point $D$ is the reflection of $A$ around the line $I_aB$. Lines $DI_a$ and $EF$ meet at $K$. Prove that the circumcenter of $DKE$, midpoint of $BC$ and $I_a$ are colline... | [
"Let $M$ be the midpoint of $BC$ and $E'$ be the reflection of $E$ with respect to $MI_a$. Also, let $\\omega$ be the $A$-excircle. It is clear that $E'$ lies on $\\omega$ and the problem is equivalent to proving that $DE'KE$ is cyclic. To do this we will show that $\\angle E'DK = \\angle E'EK$. Note that $EF \\per... | Iran | 38th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homot... | null | proof only | null | |
0bxc | Let $ABCD$ be a convex quadrangle and let $P$ and $Q$ be variable points inside this quadrangle so that $\angle APB = \angle CPD = \angle AQB = \angle CQD$. Prove that the lines $PQ$ obtained in this way all pass through a fixed point, or they are all parallel. | [
"By the condition in the statement, the points $A$, $B$, $P$, and $Q$ lie on some circle $\\omega_1$, and the points $C$, $D$, $P$, and $Q$ lie on some circle $\\omega_2$. The line $PQ$ is the radical axis of these two circles, and we proceed to prove that $M$ lies on this radical axis.\nLet $MA$ meet $\\omega_1$ a... | Romania | THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"... | English | proof only | null | |
0kt3 | Problem:
Alice and Bob are playing in the forest. They have six sticks of length $1,2,3,4,5,6$ inches. Somehow, they have managed to arrange these sticks, such that they form the sides of an equiangular hexagon. Compute the sum of all possible values of the area of this hexagon. | [
"Solution:\n\nLet the side lengths, in counterclockwise order, be $a, b, c, d, e, f$. Place the hexagon on the coordinate plane with edge $a$ parallel to the $x$-axis and the intersection between edge $a$ and edge $f$ at the origin (oriented so that edge $b$ lies in the first quadrant). If you travel along all six ... | United States | HMMT November 2022 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 33√3 | |
09ly | In a hexagonal grid with unit side lengths, vertices are connected by paths of unit length. Two vertices are considered adjacent if they are connected directly by such a path.
Determine the minimum number of cities that can be placed on the vertices such that no two cities are placed on adjacent vertices, and no more c... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 7 | |
0a4e | Problem:
We zeggen dat een rij $a_1, \dots, a_n$ van reële getallen afnemend stijgend is als voor alle $1 < i < n$ geldt dat $0 < a_{i+1} - a_i < a_i - a_{i-1}$. Vind voor elk positief geheel getal $m$ het kleinste positieve gehele getal $k$ waarvoor er een afnemend stijgende rij bestaat van lengte $k$ zo dat 1 op zij... | [
"Solution:\n\nOplossing I. We bewijzen eerst dat $k \\ge 2m$. We definiëren $b_i = a_{i+1} - a_i$. Dan is $b_1, b_2, \\dots, b_{k-1}$ een dalende rij positieve reële getallen. En elk van de manieren om 1 te schrijven is in deze schrijfwijze een som van opvolgende elementen in deze rij $b_j + b_{j+1} + \\dots + b_{j... | Netherlands | IMO-selectietoets III | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | k = 2m | |
0iyw | Problem:
An icosidodecahedron is a convex polyhedron with 20 triangular faces and 12 pentagonal faces. How many vertices does it have? | [
"Solution:\nSince every edge is shared by exactly two faces, there are $$(20 \\cdot 3 + 12 \\cdot 5) / 2 = 60$$ edges. Using Euler's formula $v - e + f = 2$, we see that there are 30 vertices."
] | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > 3D Shapes",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F"
] | null | final answer only | 30 | |
0aci | a) Calculate $(-3ab + a^2 - 1) - (-4a^2 + 5ab - 7) + 2(a - 2b)(a + b)$
b) If $A = 7x^2$ and $B = -2y$ calculate $(A+B)^2$ and $(A+B)(A-B)$. | [
"a)\n$$\n(-3ab + a^2 - 1) - (-4a^2 + 5ab - 7) + 2(a - 2b)(a + b) = -3ab + a^2 - 1 + 4a^2 - 5ab + 7 + 2a^2 + 2ab - 4ab - 4b^2 = 7a^2 - 10ab - 4b^2 + 6\n$$\n\nb)\n$$(A+B)^2 = A^2 + 2AB + B^2 = (7x^2)^2 + 2 \\cdot 7x^2 \\cdot (-2y) + (-2y)^2 = 49x^4 - 28x^2y + 4y^2$$\n$$\n(A+B)(A-B) = (7x^2)^2 - (-2y)^2 = 49x^4 - 4y^2... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | a) 7a^2 - 10ab - 4b^2 + 6; b) (A+B)^2 = 49x^4 - 28x^2y + 4y^2 and (A+B)(A-B) = 49x^4 - 4y^2 | |
0gr8 | Let $ABCD$ be a cyclic quadrilateral whose sides $BC$ and $AD$ are not parallel. Let $E$ be a point inside the circumcircle of $ABCD$ which is on the opposite side of the line $AB$ with respect to the point $C$. The lines $DE$ and $AB$ meet at $F$. Let $G$ be a point inside $ABCD$ and also on the line which is tangent ... | [
"Let the lines $GE$ and $AB$ meet at a point $M$. Let $\\angle GAD = \\angle BAE = a$, $\\angle GAB = b$. By angle chasing, we have $\\angle GED = \\angle MEF = \\angle MAE = a$. Therefore, the points $A$, $E$, $G$, $D$ are concyclic. We also get $DGA = \\angle DEA = c$. Let $\\angle ABE = d$. We obtain that $\\ang... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
09gy | Гурван овоонд нийт $2N$ ширхэг чулуу байв. Нэг удаагийн үйлдлээр тэгш тооны чулуутай овоог сонгон авч, уг овоонд байгаа чулууны хагасыг өөр овоонд шилжүүлж болно. Энэ үйлдлийн тусламжтайгаар аль нэг овоог $N$ чулуутай болгож чадахыг батал. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
0gmt | Find all non-negative triplets $(m, n, k)$ of integers satisfying $5^m + 7^n = k^3$. | [] | Turkey | XIII. National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (m, n, k) = (0, 1, 2) | |
0hg1 | We say that the triple $a, b, c$ from the segment $[-1; 1]$ is *worthy*, if these numbers satisfy the inequality $1 + 2abc > a^2 + b^2 + c^2$. Prove that if the triples $a, b, c$ and $x, y, z$ are worthy, then $ax, by, cz$ is worthy as well.
(Voloshyn Denys) | [
"We are given the following inequalities:\n$$\n1 + 2abc \\ge a^2 + b^2 + c^2 \\text{ and } 1 + 2xyz \\ge x^2 + y^2 + z^2, \\quad (1)\n$$\nand we need to show that the following inequality holds:\n$$\n1 + 2abcxyz \\ge (ax)^2 + (by)^2 + (cz)^2. \\quad (2)\n$$\nFirst observation: we can assume that all numbers are non... | Ukraine | Problems from Ukrainian Authors | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0a9i | Problem:
When $a_{0}, a_{1}, \ldots, a_{1000}$ denote digits, can the sum of the 1001-digit numbers $a_{0} a_{1} \ldots a_{1000}$ and $a_{1000} a_{999} \ldots a_{0}$ have odd digits only? | [
"Solution:\n\nThe answer is no. The following diagram illustrates the calculation of the sum digit by digit.\n\n| $a_{0}$ | $a_{1}$ | $\\ldots$ | $a_{i}$ | $\\ldots$ | $a_{500}$ | $\\ldots$ | $a_{1000-i}$ | $\\ldots$ | $a_{999}$ | $a_{1000}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---:... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 25 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | No | |
0g51 | Problem:
Trouver toutes les valeurs entières que l'expression
$$
\frac{p q+p^{p}+q^{q}}{p+q}
$$
peut prendre, où $p$ et $q$ sont des nombres premiers. | [
"Solution:\n\nRéponse : Le seul entier relatif atteignable est $3$.\n\nSi $p$ et $q$ sont impairs, le numérateur est impair mais le dénominateur est pair. Puisqu'un nombre pair ne divise jamais un nombre impair, l'expression n'est pas un entier naturel. Ainsi, nous pouvons supposer qu'au moins un des deux nombres p... | Switzerland | Deuxième tour 2023 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 3 | |
00dv | Consider a $99 \times 99$ table with its columns and rows labelled $1$ to $99$ from left to right and bottom to top respectively. Lucía writes the numbers from $1$ to $3160$ in increasing order by steps, in the following way:
| | | | | | | | | | |
|--------|----|----|----|----|----|--... | [
"First, let us observe that every step contains four more numbers than the previous one. In fact, the number of cells filled in step $k$ is $4k + 1$. After $31$ steps, there are $1+5+7+...+(4 \\cdot 31+1) = 2016$ numbers in the table. We can compute that sum by multiplying by $4$ the sum of the numbers from $1$ to ... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Discrete Mathematics > Algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | Step 32; row 6; column 18 | |
0al1 | In a city of gnomes there are $1000$ identical towers, each of which has $1000$ stories, with exactly one gnome living on each story. Every citizen proudly wears a hat colored in one of $1000$ possible colors and any two residents of the same tower wear differently colored hats. A pair of gnomes are friends if they wea... | [
"Let us analyze the problem. There are $1000$ towers, each with $1000$ stories, so there are $1000 \\times 1000 = 10^6$ gnomes. Each gnome wears a hat of one of $1000$ colors, and in each tower, all $1000$ colors are present (since any two residents of the same tower wear differently colored hats).\n\nA pair of gno... | North Macedonia | Macedonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | 999000 | |
0eep | Problem:
Naj bo
$$
f(x) = \ln x \quad \text{in} \quad g(x) = \frac{1}{3} x^{3} - x^{2} - 6x + 4
$$
ter $p$ tangenta na graf funkcije $f$ v točki $S(a, b)$, $q$ pa tangenta na graf funkcije $g$ v točki $T(a, c)$. Tangenta $p$ je pravokotna na tangento $q$. Izračunaj vrednost $a$. | [
"Solution:\n\nOdvod funkcije $f$: $f'(x) = \\frac{1}{x}$\n\nOdvod funkcije $g$: $g'(x) = x^{2} - 2x - 6$\n\nUpoštevanje $k_1 = f'(a) = \\frac{1}{a}$ in $k_2 = g'(a) = a^{2} - 2a - 6$\n\nUpoštevanje $k_1 k_2 = -1$ oziroma $\\frac{a^{2} - 2a - 6}{a} = -1$\n\nPreoblikovanje enačbe v $a^{2} - a - 6 = 0$\n\nZapis rešite... | Slovenia | 16. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 3 | |
02yk | Problem:
Em um certo país, existem exatamente $2019$ cidades e entre quaisquer duas delas existe exatamente um voo direto operado por alguma companhia aérea, isto é, dadas as cidades $A$ e $B$ ou existe um voo de $A$ para $B$ ou um voo de $B$ para $A$. Encontre o menor número de companhias aéreas que operam no país, s... | [
"Solution:\n\nA resposta é $2019$. Como existem $1009$ pares disjuntos de cidades, cada companhia aérea pode operar em no máximo $1009$ pares. Existem exatamente $2019 \\cdot 2018/2$ voos diretos e, portanto, o número de companhias aéreas é pelo menos $\\frac{2019 \\cdot 2018}{2 \\cdot 1009} = 2019$. Resta exibirmo... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | null | proof and answer | 2019 | |
0cqe | Ten pairwise distinct positive real numbers are given. Prove that among them there are either three numbers whose product is greater than the product of some other two, or three numbers whose product is greater than the product of some other four.
Даны десять положительных чисел, любые два из которых различны. Докажит... | [
"Первое решение. Возьмем любые 5 из данных чисел: $a, b, c, d, e$. Если $abc > de$, то утверждение задачи верно. Если же $de \\ge abc$, возьмем еще два числа $f$ и $g$. Пусть скажем $f > g$. Тогда $def > abcg$; значит, и в этом случае утверждение задачи верно.\n\nВторое решение. Упорядочим данные числа по убыванию:... | Russia | Russian mathematical olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof only | null | |
05ja | Problem:
Prouver que, pour tous entiers strictement positifs $a$, $b$, $c$, $d$, on a
$$
(2a-1)(2b-1)(2c-1)(2d-1) \geqslant 2abcd-1
$$
et déterminer les cas d'égalité. | [
"Solution:\nSi $x$ et $y$ sont des entiers strictement positifs, on a $x \\geqslant 1$ et $y \\geqslant 1$, et ainsi $(x-1)(y-1) \\geqslant 0$. Cela conduit à $xy - x - y + 1 \\geqslant 0$, et donc à $(2x-1)(2y-1) \\geqslant 2xy-1$. Notons que l'égalité a lieu si et seulement si $x=1$ ou $y=1$.\n\nSi $a$, $b$, $c$,... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Equality holds if and only if at least three of a, b, c, d are equal to 1. | |
0ig5 | Problem:
Let $X Y Z$ be a triangle with $\angle X = 60^{\circ}$ and $\angle Y = 45^{\circ}$. A circle with center $P$ passes through points $A$ and $B$ on side $X Y$, $C$ and $D$ on side $Y Z$, and $E$ and $F$ on side $Z X$. Suppose $A B = C D = E F$. Find $\angle X P Y$ in degrees. | [
"Solution:\n\nSince $P A B$, $P C D$, and $P E F$ are all isosceles triangles with equal legs and equal bases, they are congruent. It follows that the heights of each are the same, so that $P$ is equidistant from the sides of $X Y Z$. Therefore, $P$ is the incenter and therefore lies on the angle bisectors of $X Y ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 255/2 degrees | |
0ila | Problem:
Let $n \geq 3$ be a positive integer. Prove that given any $n$ angles $0 < \theta_{1}, \theta_{2}, \ldots, \theta_{n} < 180^{\circ}$, such that their sum is $180(n-2)$ degrees, there exists a convex $n$-gon having exactly those angles, in that order. | [
"Solution:\n\nWe induct on $n$. The statement holds trivially for $n=3$, as all triangles are convex.\n\nNow, suppose that the statement is true for $n-1$, where $n \\geq 4$. Let $\\theta_{1}, \\theta_{2}, \\ldots, \\theta_{n}$ be $n$ angles less than $180^{\\circ}$ whose sum equals $180(n-2)$ degrees. The statemen... | United States | Harvard-MIT Mathematics Tournament, Team Round A | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jr1 | Problem:
Let $\triangle ABC$ be a right triangle with right angle $C$. Let $I$ be the incenter of $ABC$, and let $M$ lie on $AC$ and $N$ on $BC$, respectively, such that $M$, $I$, $N$ are collinear and $\overline{MN}$ is parallel to $AB$. If $AB = 36$ and the perimeter of $CMN$ is $48$, find the area of $ABC$. | [
"Solution:\n\nAnswer: $252$\n\nNote that $\\angle MIA = \\angle BAI = \\angle CAI$, so $MI = MA$. Similarly, $NI = NB$. As a result, $CM + MN + NC = CM + MI + NI + NC = CM + MA + NB + NC = AC + BC = 48$.\n\nFurthermore, $AC^2 + BC^2 = 36^2$. As a result, we have $AC^2 + 2AC \\cdot BC + BC^2 = 48^2$, so $2AC \\cdot ... | United States | HMMT November 2015 | [
"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 and answer | 252 | |
0170 | Let $AD$, $BE$ and $CF$ be the angle bisectors of triangle $ABC$. Assume
$$
\frac{1}{AE} + \frac{1}{AF} = \left( \frac{1}{\sqrt{AB}} + \frac{1}{\sqrt{AC}} \right)^2
$$
Prove that $AE + AF = BC$. | [
"Let $AB = c$, $BC = a$, $CA = b$. By the well-known property of angle bisectors,\n$$\nAE = \\frac{c}{a+c} \\cdot b, \\quad AF = \\frac{b}{a+b} \\cdot c.\n$$\nThe equality of the problem now implies\n$$\n\\frac{1}{AE} + \\frac{1}{AF} = \\frac{a+c}{bc} + \\frac{a+b}{bc} = \\left(\\frac{1}{\\sqrt{c}} + \\frac{1}{\\sq... | Baltic Way | BALTIC WAY | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0gkk | Let the incircle of a scalene triangle $ABC$ touch the sides $BC$, $CA$, $AB$ at the points $A'$, $B'$, $C'$, respectively. Let $D$ be the midpoint of the arc $B'C'$ that does not contain $A'$ ($D$ lies on the opposite side of $A'$). If the common tangent of the incircles of triangles $A'CB'$ and $A'C'B$ (opposite to t... | [
"Since $D$ is the midpoint of the arc $B'C'$ and $AC$ touches the incircle at $B'$, we see that\n$$\n\\therefore \\angle AB'D = \\angle BA'D = \\angle DA'C' = \\angle DB'C'\n$$\nThus, $B'D$ bisects $\\angle AB'C'$ and similarly, $DC'$ bisects $\\angle B'C'A$. Hence $D$ is the incenter of $\\triangle AB'C'$. By the ... | Thailand | Thai 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"
] | English | proof only | null | |
0313 | Problem:
Cut $2003$ rectangles from an acute $\triangle ABC$ such that any of them has a side parallel to $AB$ and the sum of their areas is maximal. | [
"Solution:\n1. It is not difficult to see that the rectangles must be placed one over another so that any of them has two vertices on the sides $AC$ and $BC$, and the first one has a base on the side $AB$.\nNow we shall prove by induction that the sum of the areas of $n$ such rectangles is maximal when the side $AC... | Bulgaria | Team selection test for 44. IMO | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | The maximum total area is 2003/2004 of the area of triangle ABC, achieved by stacking the rectangles so that the contact points on one side divide it into 2004 equal parts. | |
0jfm | Problem:
The digits $1,2,3,4,5,6$ are randomly chosen (without replacement) to form the three-digit numbers $M=\overline{A B C}$ and $N=\overline{D E F}$. For example, we could have $M=413$ and $N=256$. Find the expected value of $M \cdot N$. | [
"Solution:\n\nAnswer: 143745 By linearity of expectation and symmetry,\n$$\n\\mathbb{E}[M N]=\\mathbb{E}[(100 A+10 B+C)(100 D+10 E+F)]=111^{2} \\cdot \\mathbb{E}[A D]\n$$\nSince\n$$\n\\mathbb{E}[A D]=\\frac{(1+2+3+4+5+6)^{2}-\\left(1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}\\right)}{6 \\cdot 5}=\\frac{350}{30}\n$$\nour an... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 143745 | |
0fiu | Problem:
Dos circunferencias $C_{1}$ y $C_{2}$ de radios $r_{1}$ y $r_{2}$ se cortan en los puntos $A$ y $B$. Por $B$ se traza una recta variable que corta de nuevo a $C_{1}$ y $C_{2}$ en dos puntos que llamaremos $P_{r}$ y $Q_{r}$, respectivamente.
Demuestra la siguiente propiedad: Existe un punto $M$, que depende só... | [
"Solution:\n\n\n\nSea $O$ el punto medio del segmento $M_{1} M_{2}$. Vamos a demostrar que todas las mediatrices de los segmentos $P_{r} Q_{r}$ pasan por el simétrico de $B$ respecto de $O$.\n\nSean $\\varepsilon=\\{\\widehat{P_{r} B M}\\}_{1}$, $\\gamma=\\{\\widehat{M_{1} B M}\\}_{2}$. Ent... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
087j | Problem:
È dato un triangolo acutangolo isoscele $ABC$ di base $AC$. All'interno di tale triangolo sono dati un punto $M$, dalla parte di $C$ rispetto all'asse di $AC$ e tale che $C\widehat{M}A = 2C\widehat{B}A$, e un punto $N$ all'interno del segmento $AM$ tale che $B\widehat{N}M = C\widehat{B}A$.
- Dimostrare che $C... | [
"Solution:\n\nSia $P$ il punto di intersezione tra il prolungamento di $CM$ e $BN$ e sia $R$ il punto di intersezione tra il prolungamento di $AM$ e $BC$; indichiamo inoltre con $\\beta$ l'angolo $C\\widehat{B}A$.\n\nDimostrazione della prima parte\n\nDiamo due dimostrazioni di questo punto.\n\nPrimo argomento\n\nP... | Italy | UNIONE MATEMATICA ITALIANA SCUOLA NORMALE SUPERIORE DI PISA Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nin... | null | proof only | null | |
0c0j | Let $\triangle ABC$ be a triangle and let $\omega$ be its $A$-excircle (excircle opposite vertex $A$). Let $D, E, F$ be the points where $\omega$ touches the lines $BC, CA, AB$, respectively. The circle $AEF$ crosses the line $BC$ at $P$ and $Q$. Let $M$ be the midpoint of the line segment $AD$. Prove that the circles ... | [
"For convenience, let $\\Omega$ and $\\gamma$ denote the circles $AEFPQ$ and $MPQ$, respectively. Letting the line $AD$ cross $\\omega$ again at $T$, we will show that $\\gamma$ and $\\omega$ are tangent at $T$.\n\nWe first prove that the points $M, P, Q, T$ are concyclic, so $T$ lies on $\\gamma$. Let $A'$ be the ... | Romania | 69th NMO Selection Tests for BMO and IMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | null | proof only | null | |
02tu | Problem:
É dado um tabuleiro $8 \times 8$.
a) Qual o número mínimo de casinhas que devemos marcar nesse tabuleiro, de modo que cada um de seus subtabuleiros $3 \times 3$ possua pelo menos uma casinha marcada?
b) Qual o número mínimo de casinhas que devemos marcar nesse tabuleiro, de modo que cada um de seus subtabulei... | [
"Solution:\n\na)\nConsidere a figura a seguir.\n\n\n\nCada um dos quatro subtabuleiros $3 \\times 3$ assinalados na figura deve ter pelo menos uma casa marcada. Além disso, com as quatro casas marcadas na figura acima, temos a propriedade desejada. Portanto, o mínimo é $4$.\n\nb)\nConsidere... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | a) 4; b) 16 | |
0ejt | Problem:
Poišči vsa cela števila $n$, ki jih lahko zapišemo v obliki $n=\frac{m+2021}{2021-m}$, kjer je $m$ celo število. | [
"Solution:\n\nEnakost pomnožimo z $2021-m$, da dobimo $2021 n - m n = m + 2021$. Nato jo preuredimo do $2021(n-1) = m(n+1)$ in izrazimo $m = \\frac{2021(n-1)}{n+1} = 2021 - \\frac{2 \\cdot 2021}{n+1}$. Torej lahko v predpisani obliki zapišemo vsa tista cela števila $n$, za katera je $\\frac{2 \\cdot 2021}{n+1}$ cel... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | -4043, -2022, -95, -87, -48, -44, -3, -2, 0, 1, 42, 46, 85, 93, 2020, 4041 | |
0j9d | Problem:
What is the sum of all of the distinct prime factors of $25^{3}-27^{2}$? | [
"Solution:\nWe note that $25^{3}-27^{2}=5^{6}-3^{6}=(5^{3}-3^{3})(5^{3}+3^{3})=(5-3)(5^{2}+5 \\cdot 3+3^{2})(5+3)(5^{2}-5 \\cdot 3+3^{2})=2 \\cdot 7^{2} \\cdot 2^{3} \\cdot 19$, so the sum of the distinct prime factors is $2+7+19=28$."
] | United States | HMMT November 2012 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 28 | |
00r5 | Define the sequence $(a_n)_{n \ge 0}$ by $a_0 = 0$, $a_1 = 1$, $a_2 = 2$, $a_3 = 6$, and
$$
a_{n+4} = 2a_{n+3} + a_{n+2} - 2a_{n+1} - a_n, \quad n \ge 0.
$$
Prove that $n^2$ divides $a_n$ for infinitely many positive integers $n$. | [
"From the recursive relation it follows that $a_4 = 12$, $a_5 = 25$, $a_6 = 48$, hence we have $\\frac{a_1}{1} = 1$, $\\frac{a_2}{2} = 1$, $\\frac{a_3}{3} = 2$, $\\frac{a_4}{4} = 3$, $\\frac{a_5}{5} = 5$, $\\frac{a_6}{6} = 8$, that is $\\frac{a_n}{n} = F_n$, for all $n = 1, 2, 3, 4, 5, 6$, where $(F_n)_{n \\ge 1}$ ... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0e15 | Let $\triangle ABC$ be an acute triangle such that $|AB| > |AC|$. Let $D$ be a point different from $C$ on the segment $BC$, such that $|AC| = |AD|$. Let $H$ denote the orthocentre of the triangle $ABC$, and let $A_1, B_1$ be the feet of the altitudes from $A$ and $B$, respectively. The line $DH$ intersects the line $A... | [
"The triangle $CAD$ is isosceles since $|AC| = |AD|$. The line $AA_1$ is the altitude in this isosceles triangle, so $\\angle HDA = \\angle ACH$. In the quadrilateral $HA_1CB_1$ we have $\\angle CA_1H = \\frac{\\pi}{2} = \\angle CB_1H$, so this quadrilateral is cyclic and $\\angle B_1A_1H = \\angle B_1CH$.\n\n![](a... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08xp | There are two blackboards $A$ and $B$, and on each of these blackboards certain number of distinct integers greater than or equal to $2$ and less than or equal to $20$ are written in such a way that any pair of numbers one of which is chosen from $A$ and the other from $B$ are relatively prime. Determine the maximum po... | [
"Denote by $A$ and $B$ the set of numbers written on the blackboard $A$ and $B$, respectively. Denote also by $|X|$ the number of elements belonging to the set $X$. If we let\n$$\nA = \\{2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20\\}, \\quad B = \\{7, 11, 13, 17, 19\\},\n$$\nthen we see that arbitrary pair of numbe... | Japan | Japan Mathematical Olympiad Initial Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 65 | |
0438 | Suppose that geometric sequence $\{a_n\}$ satisfies $a_1 - a_2 = 3$, $a_1 - a_3 = 2$. Then the common ratio of $\{a_n\}$ is ________. | [
"Let the common ratio of $\\{a_n\\}$ be $q$. Then\n$$\n\\begin{aligned}\na_1(1-q) &= a_1 - a_2 = 3, \\\\\na_1(1-q^2) &= a_1 - a_3 = 2,\n\\end{aligned}\n$$\nand thus $1 + q = \\frac{a_1(1 - q^2)}{a_1(1 - q)} = \\frac{2}{3}$. Therefore, $q = -\\frac{1}{3}$.\n$\\square$"
] | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | final answer only | -1/3 | |
0br8 | Prove that a regular quadrilateral pyramid has two opposite lateral faces perpendicular if and only if the angle of two consecutive lateral faces has measure $120^\circ$. | [
"Let $ABCD$ be the base of the pyramid, $V$ its apex, $VO$ its altitude and $M$, $N$ the midpoints of the edges $AD$, respectively $BC$. Denote $a$ the length of the edge $AB$. Faces $VAD$ and $VBC$ are perpendicular if and only if the triangle $VMN$ is right and isosceles, with sides $VM = VN = \\frac{a\\sqrt{2}}{... | Romania | 67th Romanian Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
00lr | Anthony writes down in order all positive integers which are divisible by $2$. Bertha writes down in order all positive integers which are divisible by $3$. Claire writes down in order all positive integers which are divisible by $4$. Orderly Dora writes all numbers written by the other three. Thereby she puts them in ... | [
"Dora can ignore Claire's numbers, since Anthony has already written them all. Considering the numbers up to $3000$ we see that Anthony has already written down $1500$ of them and Bertha has written $1000$ of them, $500$ of which have been written twice, which are ignored by Dora in their second occurrence. Hence D... | Austria | 48th Austrian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | final answer only | 3026 | |
0kf7 | Problem:
Let $S$ be a set of $a+b+3$ points on a sphere, where $a, b$ are nonnegative integers and no four points of $S$ are coplanar (that is, no four points lie on a plane). Determine how many planes pass through three points of $S$ and separate the remaining points into $a$ points on one side of the plane and $b$ po... | [
"Solution:\nLet $f(a, b)$ be the number of oriented planes through three of the given $a+b+3$ points, such that exactly $a$ points lie above the plane (a plane divides 3D-space into two regions; an oriented plane can be thought of as a plane and a choice of one side \"above\" the plane). We then recover $g$ as\n$$\... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls... | null | proof and answer | Number of planes equals 2(a+1)(b+1) if a ≠ b, and (a+1)^2 if a = b. | |
0grg | The Retired Linguist (R.L.) in his first move writes a word consisting of $n$ distinct letters to his notebook. Thereafter in each move he writes a new word to the notebook which is obtained by reversing the longest sub-word (starting from the first letter) of the last added word provided the word is not written to the... | [
"Let the first written word be $(1,2,\\dots,n)$. By induction on $n$ we prove that the number of written words is $n!$.\n\nIf $n = 2$ then R.L. starts with $(1, 2)$ and then reverses it and writes $(2, 1)$.\n\nSuppose that for $n = k-1$ the words\n$$\nW_1^{k-1}, W_2^{k-1}, \\dots, W_{(k-1)!}^{k-1}\n$$\nare written ... | Turkey | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | n! | |
0009 | Una pista para carreras de automóviles consiste de 6 circunferencias $\Gamma_1, \Gamma_2, \Gamma_3, \Gamma_4, \Gamma_5$ y $\Gamma_6$, tangentes interiores en un punto $P$. Las longitudes de las circunferencias son 1 km, 2 km, 4 km, 8 km, 16 km y 32 km, respectivamente. Dos pilotos, $A$ y $B$, viajan a lo largo de la pi... | [] | Argentina | XI Olimpiada Matemática Rioplatense | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Other"
] | español | proof and answer | No | |
0940 | Problem:
Find all pairs $(n, p)$ of positive integers such that $p$ is prime and
$$
1+2+\cdots+n=3 \cdot\left(1^{2}+2^{2}+\cdots+p^{2}\right)
$$ | [
"Solution:\nThe equation can be rewritten as\n$$\nn(n+1)=p(p+1)(2p+1)\n$$\nWe conclude that $p$ divides $n$ or $n+1$, so we divide the solution in two cases.\n\nIn the first case, if $n=kp$ for some integer $k>0$, then $k(kp+1)=(p+1)(2p+1)$. Firstly, after observing the equation modulo $p$ we can deduce that $p \\m... | Middle European Mathematical Olympiad (MEMO) | 15th Middle European Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequali... | null | proof and answer | (5, 2) | |
0dm0 | Let $ABC$ be a non-isosceles triangle with $M$ as the midpoint of $BC$. Take a point $D \neq M$ on the segment $BC$ and the circumcircles of triangles $ABD$, $ACD$ intersect $AC$, $AB$ again at $E$, $F$ respectively. Line $BE$, $CF$ meet at $K$ and $AK$, $EF$ meet at $I$. Denote $O$ as the circumcenter of triangle $AEF... | [] | Saudi Arabia | Saudi Booklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry... | null | proof only | null | |
00dp | Let $I$ be the incenter of triangle $ABC$. The incircle of $ABC$ is tangent to side $BC$ at $D$. Let $P$ and $Q$ be points on rays $IB$ and $IC$ respectively such that $\angle IAP = \angle CAD$ and $\angle IAQ = \angle BAD$. Prove that $AP = AQ$. | [
"Let's assume the incircle is tangent to $AB$ at $F$, then we have that $\\angle FAD = \\angle IAQ$ and $\\angle AFD = \\frac{180^\\circ - \\angle B}{2} = \\angle AIQ$ where the first equality is from the statement and the other are well-known identities. It follows that $\\triangle AFD$ and $\\triangle AIQ$ are si... | Argentina | XXIX Rioplatense 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 > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03ps | Suppose the sum of distances from any point $P$ in a convex quadrilateral $ABCD$ to lines $AB$, $BC$, $CD$ and $DA$ is constant. Prove that $ABCD$ is a parallelogram. (posed by Xiong Bin) | [
"Let $d(P, l)$ denote the distance from point $P$ to line $l$. We first prove the following lemma.\n\n**Lemma** Let $\\angle SAT = \\alpha$ be a given angle and $P$ a moving point in $\\angle SAT$. If the sum of distances from $P$ to lines $AS$ and $AT$ is a constant number $m$, then the trace of $P$ is a segment o... | China | China Western Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof only | null | |
0hpr | Problem:
Find the number of triangulations of a general convex $7$-gon into $5$ triangles by $4$ diagonals that do not intersect in their interiors. | [
"Solution:\nDefine the Catalan numbers by $C(n) = \\frac{1}{n+1} \\binom{2n}{n}$. The current solution is the $C($ number of triangles $) = C(5) = 42$."
] | United States | null | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | final answer only | 42 | |
07zt | Problem:
Andrea torna dalla pesca con una borsa piena di pesci. Giunto a casa, dà al più grande dei suoi due gatti i tre pesci più grossi: così facendo il peso della borsa si riduce del $38\%$. A questo punto dà all'altro gatto i tre pesci più piccoli: così facendo il peso della borsa si riduce nuovamente del $38\%$ (... | [] | Italy | Italian Mathematical Olympiad - Febbraio Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 10 | |
0kz2 | Problem:
Let $ABCD$ be a rectangle such that $AB = 20$ and $AD = 24$. Point $P$ lies inside $ABCD$ such that triangles $PAC$ and $PBD$ have areas $20$ and $24$, respectively. Compute all possible areas of triangle $PAB$. | [
"Solution:\n\nThere are four possible locations of $P$ as shown in the diagram. Let $O$ be the center. Then, $[PAO] = 10$ and $[PBO] = 12$. Thus, $[PAB] = [AOB] \\pm [PAO] \\pm [PBO] = 120 \\pm 10 \\pm 12$, giving the four values $98, 118, 122$, and $142$."
] | United States | HMMT February 2024 Guts Round | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | [98, 118, 122, 142] | |
0guw | Find all pairs of positive integers $(a, b)$ such that
$$
\frac{10^{a} - 3^{b} + 1}{2^{a}}
$$
is a perfect square. | [
"Answer: $(a, b) = (1, 1)$ and $(1, 2)$.\nAssume that\n$$\n\\frac{10^{a} - 3^b + 1}{2^a} = m^2\n$$\nholds for some integer $m$. Let us rewrite the equation as\n$$\n3^b = 10^{a} - m^2 2^a + 1.\n$$\nIf $a=1$ then clearly the only solutions are $(a, b) = (1, 1)$ and $(1, 2)$. If $a \\ge 2$ then $a$ is even. Then, sinc... | Turkey | Team Selection Test for IMO 2024 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (1, 1), (1, 2) | |
0k3v | Problem:
Suppose that $2002$ numbers, each equal to $1$ or $-1$, are written around a circle. For every two adjacent numbers, their product is taken; it turns out that the sum of all $2002$ such products is negative. Prove that the sum of the original numbers has absolute value less than or equal to $1000$. (The absol... | [
"Solution:\n\nSuppose that $n$ of the original numbers are $+1$, so that the remaining $2002-n$ are $-1$, and their sum is\n$$\n1 \\cdot n + (-1) \\cdot (2002-n) = 2n - 2002.\n$$\nAlso suppose that $m$ of the products equal $-1$; then the remaining $2002-m$ equal $1$, and the sum of the products is\n$$\n1 \\cdot (2... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0fmm | For all positive integer $n$ we consider the number $a_n = 4^{6^n} + 1943$. Prove that $a_n$ is divisible by $2013$ for all $n \ge 1$, and find all values of $n$ for which $a_n - 207$ is the cube of a positive integer. | [
"To prove the first part, we begin observing that $2013 = 3 \\cdot 11 \\cdot 61 = 33 \\cdot 61$. Since $9^5 = 81 \\cdot 81 \\cdot 9 = (61+20)(61+20) \\cdot 9$ and $20 \\cdot 20 \\cdot 9 = 3600 = 61 \\cdot 59 + 1$, then we have $9^5 \\equiv 1 \\pmod{61}$. Since $4^6 = 4096 = 61 \\cdot 67 + 9 \\equiv 9 \\pmod{61}$ an... | Spain | International Mathematical Arhimede Contest | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Spanish | proof and answer | n = 1 | |
005y | Dos amigos $A$ y $B$ deben resolver el siguiente acertijo. Cada uno de ellos recibe un número del conjunto $\{1, 2, ..., 250\}$ pero no ve el número que recibió el otro. El objetivo es que cada amigo descubra el número del otro. El procedimiento que deben seguir es anunciar, por turnos, números enteros positivos no nec... | [] | Argentina | XIX Olimpiada de Matemática de Países del Cono Sur | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | Spanish | proof only | null | |
0b3g | Problem:
Bryce plays a game in which he flips a fair coin repeatedly. In each flip, he obtains two tokens if the coin lands on heads, and loses one token if the coin lands on tails. At the start, Bryce has nine tokens. If after nine flips, he also ends up with nine tokens, what is the probability that Bryce always had... | [] | Philippines | 24th Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | MCQ | a | |
0j6k | Problem:
In-Young generates a string of $B$ zeroes and ones using the following method:
- First, she flips a fair coin. If it lands heads, her first digit will be a $0$, and if it lands tails, her first digit will be a $1$.
- For each subsequent bit, she flips an unfair coin, which lands heads with probability $A$. If... | [
"Solution:\n\nAnswer: $2$\n\nSince each digit is dependent on the previous, and the first digit is random, we note that the probability that In-Young obtains a particular string is the same probability as that she obtains the inverse string (i.e., that where the positions of the $0$s and $1$s are swapped). Conseque... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 2 |
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