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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0gct | 設 $a_1 \ge a_2 \ge \cdots \ge a_{107} > 0$, 滿足 $\sum_{k=1}^{107} a_k \ge M$, 且 $0 < b_1 \le b_2 \le \cdots \le b_{107}$, 滿足 $\sum_{k=1}^{107} b_k \le N$.
試證：對任意 $m \in \{1, 2, \dots, 107\}$, 數列
$$
\frac{a_1}{b_1}, \frac{a_2}{b_2}, \dots, \frac{a_m}{b_m}
$$
之算術平均數都不小於 $\frac{M}{N}$. | [
"原題欲證明：對每一 $m \\in \\{1, 2, \\dots, n\\}$，\n$$\n\\frac{1}{m} \\sum_{k=1}^{m} \\frac{a_k}{b_k} \\geq \\frac{M}{N}, \\text{ 其中 } n = 107.\n$$\n\n設 $\\sum_{k=1}^{n} a_k = a \\ge M$ 且 $\\sum_{k=1}^{n} b_k = b \\le N$，並定義\n$$\nx_k = \\frac{a_k}{a}, \\quad y_k = \\frac{b_k}{b}, \\quad k = 1, 2, \\dots, n.\n$$\n\n則有 $\\su... | Taiwan | 二〇一九數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0j7e | Find a real number $t$ such that for any set of 120 points $P_1, \dots, P_{120}$ on the boundary of a unit square, there exists a point $Q$ on this boundary with $|P_1Q| + \dots + |P_{120}Q| = t$. | [
"The answer is $t = 30 + 30\\sqrt{5}$.\n\nWe work in the Cartesian plane, and we let $\\mathcal{B}$ denote the boundary of the unit square with corners $(\\pm \\frac{1}{2}, \\pm \\frac{1}{2})$. We complete the solution in three steps.\n\n**Step 1:**\nLet $A_1 = (-\\frac{1}{2}, 0)$ and $A_2 = (\\frac{1}{2}, 0)$, and... | United States | Team Selection Test | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 30 + 30√5 | |
03y8 | Let $a_1, a_2, a_3, b_1, b_2, b_3$ be pairwise distinct positive integers such that
$$
(n+1)a_1^2 + n a_2^2 + (n-1)a_3^2 \mid (n+1)b_1^2 + n b_2^2 + (n-1)b_3^2
$$
holds for all positive integers $n$. Prove that there exists a positive integer $k$ such that $b_i = k a_i$ for all $i = 1, 2, 3$. | [
"Suppose that $r$ is any positive integer. Since there are infinitely many primes, there is a prime $p$, such that\n$$\np > (a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2). \\quad ①\n$$\nBecause $p$ is prime and ①, we have $(p, a_1^2 + a_2^2 + a_3^2) = 1$. $p$ is coprime to $p-1$; from the Chinese remainder theorem,... | China | Chinese Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
05z8 | Problem:
Trouver tous les entiers $a$, $b$, $c \in \mathbb{N}$ tels que $1517^{a} + 15^{b} = 1532^{c}$. | [
"Solution:\nOn commence par traiter les petites valeurs de $c$.\n\nSi $c = 0$, $1517^{a} + 15^{b} = 1$ n'a pas de solution.\n\nSi $c = 1$, on doit trouver $a, b$ tels que $1517^{a} + 15^{b} = 1532$. Si $a \\geq 2$ ou si $a = 0$ il n'y a pas de solution. Si $a = 1$, alors $15^{b} = 1532 - 1517 = 15$ donc $b = 1$ : c... | France | Préparation Olympique Française de Mathématiques - Envoi 3: Arithmétique | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Fac... | null | proof and answer | (1, 1, 1) | |
0jhs | In a table with $n$ rows and $2n$ columns where $n$ is a fixed positive integer, we write either zero or one into each cell so that each row has $n$ zeros and $n$ ones. For $1 \le k \le n$ and $1 \le i \le n$, we define $a_{k,i}$ so that the $i$th zero in the $k$th row is the $a_{k,i}^{th}$ column. Let $\mathcal{F}$ be... | [
"We first give a bijection between tables $C \\in \\mathcal{F}$ and partitions of a fixed regular hexagon of side length $n$ into parallelograms given by two unit equilateral triangles glued together. Call such a partition a *well-partitioned* hexagon. For a well-partitioned hexagon, align one of its edges parallel... | United States | TST | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0dtp | Find all integer solutions of the equation
$$
y^2 + 2y = x^4 + 20x^3 + 104x^2 + 40x + 2003.
$$ | [
"Suppose that there are integer solutions. By completing squares, the equation becomes\n$$\n(y+1)^2 = (x^2 + 10x + 2)^2 + 2000.\n$$\nLet $a = y + 1$, $b = x^2 + 10x + 2$, $u = a - b$, $v = a + b$. Then $uv = 2000$ and $b = \\frac{v-u}{2}$.\nThe equation $x^2 + 10x + 2 - b = x^2 + 10x + 2 + \\frac{u-v}{2} = 0$ in th... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (7, 128), (7, -130), (-17, -128), (-17, -130) | |
0ins | Let $n$ be a positive integer. Consider
$$
S = \{(x, y, z) : x, y, z \in \{0, 1, \dots, n\}, x + y + z > 0\}.
$$
as a set of $(n+1)^3 - 1$ points in three-dimensional space. Determine the smallest possible number of planes, the union of which contains $S$ but does not include $(0, 0, 0)$.
(This problem was suggested by... | [
"We establish the following key lemma.\n\n**Lemma 1.** Consider a nonzero polynomial $P(x_1, \\dots, x_k)$ in $k$ variables. Suppose that $P$ vanishes at all points $(x_1, \\dots, x_k)$ such that $x_1, \\dots, x_k \\in \\{0, 1, \\dots, n\\}$ and $x_1 + \\dots + x_k > 0$, while $P(0, 0, \\dots, 0) \\ne 0$. Then $\\d... | United States | IMO | [
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Linear Algebra > Determinants"
] | null | proof and answer | 3n | |
0k3r | Problem:
Let $a, b, c, n$ be positive real numbers such that $\frac{a+b}{a}=3$, $\frac{b+c}{b}=4$, and $\frac{c+a}{c}=n$. Find $n$. | [
"Solution:\nAnswer: $\\frac{7}{6}$\nWe have\n$$\n1=\\frac{b}{a} \\cdot \\frac{c}{b} \\cdot \\frac{a}{c}=(3-1)(4-1)(n-1) .\n$$\nSolving for $n$ yields $n=\\frac{7}{6}$."
] | United States | HMMT November 2018 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 7/6 | |
0kri | Problem:
Let $a$ be any positive integer. Show that there is always a Fibonacci number divisible by $a$. | [
"Solution:\n\nTake the pair of values $(F_{i}, F_{i+1})$ modulo $a$. There can be no more than $a^{2}$ unique such pairs, so if we take these Fibonacci pairs up to $(F_{a^{2}}, F_{a^{2}+1})$, there must be two pairs which coincide modulo $a$, say they are $(F_{a}, F_{a+1})$ and $(F_{b}, F_{b+1})$ with $a < b$. But ... | United States | Berkeley Math Circle | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
04ke | The unit squares of an $N \times N$ board are coloured black and white so that squares that share a side have different colours, and so that at least one corner square is coloured black. In each step we choose a $2 \times 2$ square and change the colour of all four unit squares inside that square, so that white unit sq... | [
"We claim that the sought numbers are all multiples of $3$.\nNote that for $N = 3$ it is possible to achieve that black unit squares become white, and vice versa, by choosing each of the four $2 \\times 2$ squares exactly twice.\n\n\nMoreover, for a $3 \\times 3$ board whose corner and cent... | Croatia | Mathematical competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | All multiples of 3 | |
0b4v | Problem:
Let $S$ be the sum of all positive integers less than $10^{6}$ which can be expressed as $m! + n!$, where $m$ and $n$ are nonnegative integers. Determine the last three digits of $S$. | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Other"
] | null | proof and answer | 130 | |
0b4q | Problem:
Silverio is very happy for the 25th year of the PMO. In his jubilation, he ends up writing a finite sequence of $A$'s and $G$'s on a nearby blackboard. He then performs the following operation: if he finds at least one occurrence of the string $"AG"$, he chooses one at random and replaces it with $"GAAA"$. He... | [
"Solution:\n\nWe assign the weight $4^{k}$ to each $G$ in the sequence, where $k$ is the number of $A$'s to the right of this $G$. In each operation, if $4^{k}$ is the weight of the $G$ in the $\"AG\"$ being replaced, then each of the three $G$'s in $\"GAAA\"$ have a weight of $4^{k-1}$. So the sum of the weights d... | Philippines | 25th Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
0442 | In $\triangle ABC$, $\tan A$ and $\tan B$ are the two roots of equation $x^2 - 10x + 6 = 0$. Then the value of $\cos C$ is ______. | [
"By the condition, we know that $\\tan A + \\tan B = 10$, $\\tan A \\tan B = 6$. Thus,\n$$\n\\begin{aligned}\n\\tan C &= \\tan(\\pi - A - B) \\\\\n&= -\\tan(A + B) \\\\\n&= -\\frac{\\tan A + \\tan B}{1 - \\tan A \\tan B} = 2.\n\\end{aligned}\n$$\nTherefore, $C$ is an acute angle, and thus $\\cos C = \\frac{1}{\\sqr... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | sqrt(5)/5 | |
0l3s | Problem:
Let $ABC$ be an acute triangle. Let $D$, $E$, and $F$ be the feet of altitudes from $A$, $B$, and $C$ to sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively, and let $Q$ be the foot of altitude from $A$ to line $EF$. Given that $AQ = 20$, $BC = 15$, and $AD = 24$, compute the perimeter o... | [
"Solution:\n\n\n\nNote that $A$ is the excenter of $\\triangle DEF$ and $AQ$ is the length of the exradius. Let $T$ be the tangency point of the $A$-excircle to line $DF$. We have $AQ = AT = 20$. It is well known that the length of $DT$ is the semiperimeter of $DEF$. Note that $\\triangle A... | United States | HMMT February 2024 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 8*sqrt(11) | |
0aca | On training in a football club there were 225 children and 105 balls. The children were split in few equal groups. The coaches gave to each group equal number of balls. How many groups were formed and how many balls did every group get? How many solutions does the problem have? | [
"The common divisors of $225$ and $105$ are $1$, $3$, $5$ and $15$. Each of these numbers can represent the number of groups that can be formed from the children in order for each group to get equal number of balls. So the problem has $4$ solutions:\n\n- $1$ group with $225$ children and $105$ balls\n- $3$ groups w... | North Macedonia | Macedonian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | Four solutions: 1 group with 225 children and 105 balls; 3 groups with 75 children and 35 balls each; 5 groups with 45 children and 21 balls each; 15 groups with 15 children and 7 balls each. Number of solutions: 4. | |
0g8g | 三角形 $ABC$ 中, 令點 $D$ 與 $E$ 分別是角 $A$ 和角 $B$ 的角平分線與對邊的交點。將一菱形內接於四邊形 $AEDB$ 中, 且菱形的頂點分別位於 $AEDB$ 不同的邊上。設 $\phi$ 為此菱形非鈍角的內角。證明 $\phi \le \max\{\angle BAC, \angle ABC\}$。
In a triangle $ABC$, let $D$ and $E$ be the feet of the angle bisectors of angles $A$ and $B$, respectively. A rhombus is inscribed into the quadrilateral... | [
"令點 $K, L, M, N$ 分別為菱形在 $AE, ED, DB, BA$ 邊上的頂點。定義 $d(X, YZ)$ 表點 $X$ 到直線 $YZ$ 的距離。因為 $D$ 與 $E$ 為角平分線與對邊的交點，故有 $d(D, AB) = d(D, AC)$，$d(E, AB) = d(E, BC)$ 以及 $d(D, BC) = d(E, AC) = 0$，由此得\n$$\nd(D, AC) + d(D, BC) = d(D, AB),\n$$\n$$\nd(E, AC) + d(E, BC) = d(E, AB).\n$$\n因為 $L$ 在線段 $DE$ 上，而等式 $d(X, AC) + d(X, BC) = d(... | Taiwan | 二〇一四年國際數學奧林匹亞競賽第二階段選訓營 獨立研究(三) | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry... | null | proof only | null | |
0irz | Problem:
Suppose that $a, b, c, d$ are real numbers satisfying $a \geq b \geq c \geq d \geq 0$, $a^{2} + d^{2} = 1$, $b^{2} + c^{2} = 1$, and $a c + b d = 1/3$. Find the value of $a b - c d$. | [
"Solution:\nAnswer: $\\frac{2 \\sqrt{2}}{3}$\nWe have\n$$\n(a b - c d)^2 = (a^2 + d^2)(b^2 + c^2) - (a c + b d)^2 = (1)(1) - \\left(\\frac{1}{3}\\right)^2 = \\frac{8}{9}\n$$\nSince $a \\geq b \\geq c \\geq d \\geq 0$, $a b - c d \\geq 0$, so $a b - c d = \\frac{2 \\sqrt{2}}{3}$."
] | United States | 11th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | 2*sqrt(2)/3 | |
08uh | Suppose 5 line segments are given on a plane satisfying the following property:
Of the 10 possibilities for choosing 3 line segments from the given 5, in 9 cases, one can form an acute triangle with the chosen 3 line segments.
Prove that in the remaining 10-th possibility for the choice of 3 line segments, there is a t... | [
"When 3 line segments with their lengths $x, y, z$ ($x \\le y \\le z$) are given, the necessary and sufficient condition for these line segments to form a triangle is $x + y > z$, and if the triangle they form is acute, then the condition $x^2 + y^2 > z^2$ is also satisfied.\nLet us denote by $a, b, c, d, e$ ($a \\... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
06tk | Let $A B C$ be an acute triangle with $A B > A C$, and let $\Gamma$ be its circumcircle. Let $H$, $M$, and $F$ be the orthocenter of the triangle, the midpoint of $B C$, and the foot of the altitude from $A$, respectively. Let $Q$ and $K$ be the two points on $\Gamma$ that satisfy $\angle A Q H = 90^{\circ}$ and $\angl... | [
"Consider any point $T$ such that $T K$ is tangent to the circle $K Q H$ at $K$ with $Q$ and $T$ lying on different sides of $K H$ (see Figure 1). Then $\\angle H K T = \\angle H Q K$ and we are to prove that $\\angle M K T = \\angle C F K$. Thus it remains to show that $\\angle H Q K = \\angle C F K + \\angle H K ... | IMO | 56th International Mathematical Olympiad Shortlisted Problems | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadril... | English | proof only | null | |
0lfs | Problem:
Some squares of a $n \times n$ table $(n>2)$ are black, the rest are white. In every white square we write the number of all the black squares having at least one common vertex with it. Find the maximum possible sum of all these numbers.
The answer is $3 n^{2}-5 n+2$. | [
"Solution:\n\nThe sum attains this value when all squares in even rows are black and the rest are white. It remains to prove that this is the maximum value.\n\nThe sum in question is the number of pairs of differently coloured squares sharing at least one vertex. There are two kinds of such pairs: sharing a side an... | Zhautykov Olympiad | XVI International Zhautykov Olympiad in Mathematics | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3 n^2 - 5 n + 2 | |
0gfy | 找出所有符合後列條件的整數 $n \ge 2$：所有總和不被 $n$ 整除、兩兩相異的 $n$ 個整數，都可以被重新排列為 $a_1, a_2, \dots, a_n$，使得 $n$ 整除 $1 \cdot a_1 + 2 \cdot a_2 + \dots + n \cdot a_n$。 | [
"If $n = 2^k a$, where $a \\ge 3$ is odd and $k$ is a positive integer, we can consider a set containing the number $2^k + 1$ and $n - 1$ numbers congruent to $1$ modulo $n$. The sum of these numbers is congruent to $2^k$ modulo $n$ and therefore is not divisible by $n$; for any permutation $(a_1, a_2, \\dots, a_n)... | Taiwan | 2022 數學奧林匹亞競賽第三階段選訓營, 國際競賽實作(一) | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | Chinese; English | proof and answer | All odd integers and all powers of two. | |
0h0b | Find the least possible natural number $n$, such that each number from the set $1, 2, \dots, 10$ can be expressed as a digit or as a sum of consecutive digits of $n$. | [
"It is obvious, that we can not find number with given property and three digits. Suppose, that it has 4 digits: $a b c d$, then we can construct 10 different sums of consecutive digits, more precisely: $a$, $b$, $c$, $d$, $a+b$, $b+c$, $c+d$, $a+b+c$, $b+c+d$, $a+b+c+d$. To satisfy the conditions of the problem th... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010) | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 11134 | |
063o | Problem:
Man bestimme alle positiven ganzen Zahlen $m$ mit folgender Eigenschaft:
Die Folge $a_{0}, a_{1}, a_{2}, \ldots$ mit $a_{0}=\frac{2m+1}{2}$ und $a_{k+1}=a_{k}\left\lfloor a_{k}\right\rfloor$ für $k=0,1,2, \ldots$ enthält wenigstens eine ganze Zahl.
Hinweis: $\lfloor x\rfloor$ bezeichnet den größten ganzen Te... | [
"Solution:\n\nEs gilt $a_{0}=m+\\frac{1}{2}$ und $a_{1}=a_{0}\\left\\lfloor a_{0}\\right\\rfloor=\\left(m+\\frac{1}{2}\\right) \\cdot m=m^{2}+\\frac{m}{2}$. Dieser Ausdruck ist für gerades $m$ offensichtlich ganz, so dass hier die gesuchte Eigenschaft der Folge vorliegt.\n\nWeiterhin gilt bei $m=1$, dass $a_{0}=\\f... | Germany | 2. Auswahlklausur | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | All positive integers except 1 | |
086s | Problem:
All'interno di un cerchio di raggio $1$ si tracciano $3$ archi di circonferenza, anch'essi di raggio $1$, centrando nei vertici di un triangolo equilatero inscritto nella circonferenza. Quanto vale l'area della zona ombreggiata?
(A) $\frac{\sqrt{3}}{4} \pi$
(B) $\pi-\frac{3 \sqrt{3}}{4}$
(C) $\pi-\frac{3 \sq... | [
"Solution:\n\nLa risposta è (D). Gli archi tracciati all'interno del cerchio hanno lo stesso raggio del cerchio stesso. Di conseguenza, per equiscomposizione del cerchio, l'area della parte ombreggiata in figura 1 è uguale a quella della parte ombreggiata in figura 2, ovvero è pari all'area dell'esagono inscritto i... | Italy | Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | D | |
02vf | Problem:
Existem 2017 cadeiras não ocupadas em uma fila. A cada minuto, uma pessoa chega e se senta em uma delas que esteja vazia e, no mesmo instante, caso esteja ocupada, uma pessoa em uma cadeira vizinha se levanta e vai embora. Qual o número máximo de pessoas que podem estar simultaneamente sentadas na fileira de ... | [
"Solution:\n\nNão é possível todas as cadeiras estarem simultaneamente ocupadas, pois o último a sentar inevitavelmente sentaria ao lado de uma cadeira ocupada e forçaria, de acordo com a regra do enunciado, alguém a ir embora. Nosso objetivo agora é mostrar uma sequência de movimentos onde é possível 2016 pessoas ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2016 | |
01wo | Points $M$ and $N$ are the midpoints of the sides $BC$ and $AD$, respectively, of a convex quadrilateral $ABCD$. Is it possible that
$$
AB + CD > \max(AM + DM, BN + CN)
$$ | [
"**Answer: no.**\nSince $(\\angle ABC + \\angle BCD) + (\\angle BAD + \\angle CDA) = 360^\\circ$, one of these summands is not less than $180^\\circ$. Without loss of generality, assume that $\\angle ABC + \\angle BCD \\ge 180^\\circ$. Denote the reflection of the triangle $MCD$ about $M$ by $MBD_1$. The inequality... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof and answer | No | |
015k | Find all integer solutions to the equation
$$
(1 + (x - 1)^2)^{x^2+1} + (4 - (x - 2)^2)^{(x-1)^2} = 2.
$$ | [
"The equation may be rewritten as\n$$\n(x^2 - 2x + 2)^{x^2+1} + (x(4-x))^{(x-1)^2} = 2,\n$$\nfrom which we see that if $x$ is even, the second term of the left-hand side will be divisible by 4. So is the first term, provided the exponent be greater than 1, which happens iff $x \\neq 0$. Thus the only even solution ... | Baltic Way | Baltic Way SHL | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | x = 0, 1 | |
0eox | Five squares (labelled $A$, $B$, $C$, $D$ and $E$) are drawn along a straight line, each touching the next square at a vertex, as shown in the diagram. Square $B$ has an area of $20$. Square $C$ has a side length of $4$. Square $D$ has double the area of square $B$. What is the area of square $E$?
![](attached_image_1... | [
"Denote the side lengths of squares $A$, $B$, $C$, $D$, $E$ by $a$, $b$, $c$, $d$, $e$ respectively. The two right-angled triangles enclosed by squares $C$, $D$, $E$ and the line are congruent to each other and have sides $c$, $d$, $e$, so by Pythagoras' theorem $d^2 = c^2 + e^2$. We are given that $c^2 = 4^2 = 16$... | South Africa | South African Mathematics Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | final answer only | 24 | |
04p4 | Find all pairs $(m, n)$ of integers such that $m^2 = n^5 + n^4 + 1$ and $m - 7n$ divides $m - 4n$. (Nikola Adžaga, Petar Bakić) | [
"Note that $n^5 + n^4 + 1 = (n^3 - n + 1)(n^2 + n + 1)$, and that\n$$\n\\begin{align*}\nd &= \\gcd(n^3 - n + 1, n^2 + n + 1) \\\\\n&= \\gcd(n^2 + n + 1, -n^2 - 2n + 1) \\\\\n&= \\gcd(n^2 + n + 1, n - 2) \\\\\n&= \\gcd(n - 2, 7),\n\\end{align*}\n$$\nhence we have two cases:\n\n1) $d = 7$\n\nThis implies $7 \\mid m$ ... | Croatia | Croatian Mathematical Society Competitions | [
"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"
] | English | proof and answer | (-1, 0), (1, 0) | |
0k2p | Problem:
Let $a$, $b$, $c$ be positive integers. All the roots of each of the quadratics
$$
a x^{2}+b x+c,\quad a x^{2}+b x-c,\quad a x^{2}-b x+c,\quad a x^{2}-b x-c
$$
are integers. Over all triples $(a, b, c)$, find the triple with the third smallest value of $a+b+c$. | [
"Solution:\nThe quadratic formula yields that the answers to these four quadratics are $\\frac{ \\pm b \\pm \\sqrt{b^{2} \\pm 4 a c}}{2 a}$. Given that all eight of these expressions are integers, we can add or subtract appropriate pairs to get that $\\frac{b}{a}$ and $\\frac{\\sqrt{b^{2} \\pm 4 a c}}{a}$ are integ... | United States | HMMT February 2018 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (1, 10, 24) | |
038b | Problem:
In an acute $\triangle ABC$ the altitudes $AA_1$ ($A_1 \in BC$) and $BB_1$ ($B_1 \in AC$) are drawn, $I$ is the incenter and the line $CI$ meets $AB$ at $L$. It is known that $I$ lies on the circumcircle of $\triangle A_1B_1C$.
a) Prove that $L$ is the center of excircle of $\triangle A_1B_1C$ tangent to the... | [
"Solution:\n\na) Since $\\Varangle CB_1A_1 = \\Varangle CIA_1$ and the quadrilateral $ABA_1B_1$ is cyclic, we have $\\Varangle CIA_1 = \\Varangle CB_1A_1 = \\Varangle ABC$. This shows that the quadrilateral $LBA_1I$ is cyclic. Then\n$$\n\\begin{aligned}\n\\Varangle LA_1B &= \\Varangle LIB = \\Varangle ICB + \\Varan... | Bulgaria | Spring Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof and answer | 60 degrees | |
0b4b | Problem:
Seven people are seated together around a circular table. Each one will toss a fair coin. If the coin shows a head, then the person will stand. Otherwise, the person will remain seated. The probability that after all of the tosses, no two adjacent people are both standing, can be written in the form $p / q$, ... | [
"Solution:\n\nLet $n = 7$ (number of people). Each person independently stands with probability $1/2$ (heads), sits with probability $1/2$ (tails).\n\nWe want the probability that no two adjacent people are both standing.\n\nLet $a_n$ be the number of ways to seat $n$ people around a table so that no two adjacent p... | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 157 | |
0cak | Problem:
Fie $ABC$ un triunghi scalen ascuţitunghic şi fie $\omega$ cercul său Euler. Tangenta $t_{A}$ a lui $\omega$, prin piciorul înălţimii din $A$ a triunghiului $ABC$, intersectează a doua oară cercul de diametru $AB$ în punctul $K_{A}$. Dreapta determinată de picioarele înălţimilor din $A$ şi $C$ ale triunghiulu... | [] | Romania | Olimpiada Nationala de Matematica 2022 baraj 2 de selectie seniori | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > ... | null | proof only | null | |
0k7h | Problem:
Let $ABC$ be a triangle with $AB = 13$, $BC = 14$, $CA = 15$. Let $I_{A}$, $I_{B}$, $I_{C}$ be the $A$, $B$, $C$ excenters of this triangle, and let $O$ be the circumcenter of the triangle. Let $\gamma_{A}$, $\gamma_{B}$, $\gamma_{C}$ be the corresponding excircles and $\omega$ be the circumcircle. $X$ is one ... | [
"Solution:\nLet $r_{A}$, $r_{B}$, $r_{C}$ be the exradii. Using $OX = R$, $XI_{A} = r_{A}$, $OI_{A} = \\sqrt{R(R + 2r_{A})}$ (Euler's theorem for excircles), and the Law of Cosines, we obtain\n$$\n\\cos \\angle O X I_{A} = \\frac{R^{2} + r_{A}^{2} - R(R + 2r_{A})}{2 R r_{A}} = \\frac{r_{A}}{2R} - 1.\n$$\nTherefore ... | United States | HMMT February 2019 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | -49/65 | |
03lc | Problem:
Find all real positive solutions (if any) to
$$
\begin{gathered}
x^{3}+y^{3}+z^{3}=x+y+z, \text{ and } \\
x^{2}+y^{2}+z^{2}=x y z .
\end{gathered}
$$ | [
"Solution:\nLet $f(x, y, z)=\\left(x^{3}-x\\right)+\\left(y^{3}-y\\right)+\\left(z^{3}-z\\right)$. The first equation above is equivalent to $f(x, y, z)=0$. If $x, y, z \\geq 1$, then $f(x, y, z) \\geq 0$ with equality only if $x=y=z=1$. But if $x=y=z=1$, then the second equation is not satisfied. So in any solutio... | Canada | 2003 CMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | No positive real solutions exist. | |
0jkk | Problem:
Let $ABCDEF$ be a regular hexagon. Let $P$ be the circle inscribed in $\triangle BDF$. Find the ratio of the area of circle $P$ to the area of rectangle $ABDE$. | [
"Solution:\n\n$\\boxed{\\frac{\\pi \\sqrt{3}}{12}}$ Let the side length of the hexagon be $s$. The length of $BD$ is $s \\sqrt{3}$, so the area of rectangle $ABDE$ is $s^{2} \\sqrt{3}$. Equilateral triangle $BDF$ has side length $s \\sqrt{3}$. The inradius of an equilateral triangle is $\\sqrt{3} / 6$ times the len... | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | π√3/12 | |
0c1b | Given the square $ABCD$, let $E$ be a point on the side $[AB]$, and $F$ be the foot of the perpendicular line from $B$ on the line $DE$. Let $L$ be a point on the line $DE$ so that $F$ is situated between $E$ and $L$, and $FL = BF$. If $N$ is the symmetric of $A$ with respect to the line $DE$ and $P$ is the symmetric o... | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Quadrilater... | null | proof only | null | |
0ez6 | Problem:
Given a circle, diameter $AB$ and a point $C$ on $AB$, show how to construct two points $X$ and $Y$ on the circle such that (1) $Y$ is the reflection of $X$ in the line $AB$, (2) $YC$ is perpendicular to $XA$. | [
"Solution:\n$AB$ is a diameter, so $BX$ is perpendicular to $AX$ and hence parallel to $YC$. $YX$ is perpendicular to $BC$, so $YC = YB$. Hence $X$ and $Y$ lie on the perpendicular bisector of $BC$."
] | Soviet Union | 4th ASU | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
01m7 | Points $K$ and $L$ are marked on the side $AB$ of a triangle $ABC$ so that $AK = KL = LB$. Points $M$ and $N$ are the midpoints of the sides $AC$ and $BC$, respectively. $X$ is the intersection point of the segments $AN$ and $CK$, $Y$ is the intersection point of the segments $BM$ and $CL$. Find the length of $XY$ if $... | [] | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 9 | |
0kyd | Problem:
Six consecutive positive integers are written on slips of paper. The slips are then handed out to Ethan, Jacob, and Karthik, such that each of them receives two slips. The product of Ethan's numbers is $20$, and the product of Jacob's numbers is $24$. Compute the product of Karthik's numbers. | [
"Solution:\n\nEach person's numbers differ by at most $5$, so Alice must have $4$ and $5$. Bob could have $4$ and $6$ or $3$ and $8$. Since Alice already has $4$, Bob cannot have $4$ and $6$. So, Bob has $3$ and $8$. Then the six numbers must be $3$ through $8$, so Charlie has $6$ and $7$, multiplying to $42$."
] | United States | HMMT November | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 42 | |
0bdx | Let $k \ge 2$ be an integer and $0 < a_1 < a_2 < \dots < a_k$ be real numbers. Compute $\lim_{n \to \infty} \{ \sqrt[n]{a_1 n + a_2^n + \dots + a_k^n} \}$, where $\{x\}$ denotes the fractional part of $x$. | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Precalculus > Limits"
] | null | proof and answer | If a_k ≥ 1, the limit is {a_k}; if a_k < 1, the limit is 0. | |
0eaw | During the course of a crime investigation 4 suspects have been arrested. Each of them gave a statement.
Žan: "Of all the suspects only I am innocent."
Alen: "Of all the suspects only I am guilty."
Zala: "We are all innocent."
Beno: "At least 2 of the suspects are guilty."
Further investigation has shown that at leas... | [
"Alen cannot be innocent since that would imply that he is telling the truth and would be, according to his own statement, guilty. Hence, Alen is guilty. From here we conclude that Zala was lying and must, therefore, also be guilty. Now, we see that Beno was telling the truth and is innocent. Finally, Žan must have... | Slovenia | National Math Olympiad in Slovenia | [
"Discrete Mathematics > Logic"
] | null | MCQ | C | |
0kbv | Problem:
In a single-elimination tournament consisting of $2^{9}=512$ teams, there is a strict ordering on the skill levels of the teams, but Joy does not know that ordering. The teams are randomly put into a bracket and they play out the tournament, with the better team always beating the worse team. Joy is then give... | [
"Solution:\n\nThe best team must win the tournament. The second-best team has to be one of the 9 teams that the first best team beat; call these teams marginal. The third best team must have lost to either the best or the second-best team, so it must either be marginal or have lost to a marginal team. Since there i... | United States | HMMO 2020 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 45 | |
02iy | Problem:
O quociente de $50^{50}$ por $25^{25}$ é igual a:
A) $25^{25}$
B) $10^{25}$
C) $100^{25}$
D) $2^{25}$
E) $2 \times 25^{25}$ | [
"Solution:\n\nSolução 1:\n$$\n\\frac{50^{50}}{25^{25}} = \\frac{\\left(2 \\times 5^{2}\\right)^{50}}{\\left(5^{2}\\right)^{25}} = \\frac{2^{50} \\times 5^{100}}{5^{50}} = 2^{50} \\times 5^{50} = \\left(2^{2} \\times 5^{2}\\right)^{25} = 100^{25}\n$$\n\nSolução 2:\n$$\n\\frac{50^{50}}{25^{25}} = \\frac{(2 \\times 25... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | C | |
04xk | Points $A$, $B$, $C$, $D$ lie on a circle in this order, where $AB$ and $CD$ are not parallel. The length of the arc $\widehat{AB}$ that contains points $C$, $D$ is twice as large as the length of the arc $\widehat{CD}$ that does not contain points $A$, $B$. Point $E$ is chosen such that $AC = AE$ and $BD = BE$ and $E$... | [
"We use the following lemma\n\nLEMMA. Given two circles $\\Gamma_1$, $\\Gamma_2$ with the centre $S_2$ of $\\Gamma_2$ lying on the circle $\\Gamma_1$. The circles intersect in two points $K$ and $L$. Let $M$ be the point on the circle $\\Gamma_1$ (different from $K$ and $L$) and the line $KM$ meets $\\Gamma_2$ agai... | Czech-Polish-Slovak Mathematical Match | 11-th Czech-Slovak-Polish Match, 2011 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06fk | For each permutation $x_1, x_2, \dots, x_{10}$ of $1, 2, \dots, 10$, compute
$$
|2x_1 - 3x_2| + |2x_2 - 3x_3| + |2x_3 - 3x_4| + \dots + |2x_{10} - 3x_1|.
$$
Let $S$ be the maximum possible value of this sum. Find the number of permutations attaining $S$. | [
"The answer is 28800.\nDefine\n$$\n\\begin{align*}\nX &= \\{2x_j : 1 \\le j \\le 10\\} = \\{2, 4, 6, \\dots, 20\\}, \\\\\nY &= \\{3x_j : 1 \\le j \\le 10\\} = \\{3, 6, 9, \\dots, 30\\}.\n\\end{align*}\n$$\nAfter removing the absolute value signs of the given expression, we get a sum of 10 terms from $X \\cup Y$ min... | Hong Kong | Year 2008 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 28800 | |
0cew | Let $p$ be an odd prime. Does there exist a permutation $a_1, a_2, \dots, a_p$ of $1, 2, \dots, p$ satisfying
$$
(i-j)a_k + (j-k)a_i + (k-i)a_j \neq 0,
$$
for all pairwise distinct $i, j, k$? | [
"The answer is in the affirmative. To define the desired permutation, let $a$ be a quadratic non-residue modulo $p$, let $ia_i \\equiv a \\pmod p$, $i = 1, 2, \\dots, p-1$, and let $a_p = p$. Clearly, the $a_i$ form a permutation of $1, 2, \\dots, p$; moreover, $a_{p-i} = p-a_i$, $i = 1, 2, \\dots, p-1$, and, since... | Romania | Nineteenth IMAR Mathematical Competition | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof only | null | |
0crm | На доске написано выражение $\frac{a}{b} \cdot \frac{c}{d} \cdot \frac{e}{f}$, где $a, b, c, d, e, f$ — натуральные числа. Если число $a$ увеличить на 1, то значение этого выражения увеличится на 3. Если в исходном выражении увеличить число $c$ на 1, то его значение увеличится на 4; если же в исходном выражении увеличи... | [
"Ответ. 60.\n\n**Первое решение.** Пусть значение исходного выражения равно $A$. Тогда в результате первой операции произведение примет значение $\\frac{a+1}{a} \\cdot A = A + 3$, откуда $A = 3a$. Значит, $A$ — натуральное число. Кроме того, из этого равенства следует, что оно делится на 3.\nАналогично доказывается... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 60 | |
09q2 | Problem:
Zij $ABC$ een scherphoekige driehoek met de eigenschap $\angle BAC=45^{\circ}$. Zij $D$ het voetpunt van de loodlijn vanuit $C$ op $AB$. Zij $P$ een inwendig punt van het lijnstuk $CD$. Bewijs dat de lijnen $AP$ en $BC$ loodrecht op elkaar staan dan en slechts dan als $|AP|=|BC|$. | [
"Solution:\n\nZij $E$ het snijpunt van $AP$ en $BC$. Merk op dat $\\angle DCA=90^{\\circ}-\\angle CAD=90^{\\circ}-\\angle CAB=45^{\\circ}$, dus $\\triangle ACD$ is gelijkbenig: $|AD|=|CD|$.\n\nStel nu dat $|AP|=|BC|$. Omdat $\\angle ADP=90^{\\circ}=\\angle CDB$, geldt $\\triangle ADP \\cong \\triangle CDB$ wegens (... | Netherlands | Dutch TST | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jxo | Problem:
A string of digits is defined to be similar to another string of digits if it can be obtained by reversing some contiguous substring of the original string. For example, the strings $101$ and $110$ are similar, but the strings $3443$ and $4334$ are not. (Note that a string is always similar to itself.) Consid... | [
"Solution:\n\nWe first count the number of substrings that one could pick to reverse to yield a new substring. If we insert two dividers into the sequence of $50$ digits, each arrangement of $2$ dividers among the $52$ total objects specifies a substring that is contained between the two dividers, for a total of $\... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 1126 | |
06ov | Circles $\omega_{1}$ and $\omega_{2}$ with centres $O_{1}$ and $O_{2}$ are externally tangent at point $D$ and internally tangent to a circle $\omega$ at points $E$ and $F$, respectively. Line $t$ is the common tangent of $\omega_{1}$ and $\omega_{2}$ at $D$. Let $AB$ be the diameter of $\omega$ perpendicular to $t$, s... | [
"Point $E$ is the centre of a homothety $h$ which takes circle $\\omega_{1}$ to circle $\\omega$. The radii $O_{1} D$ and $O B$ of these circles are parallel as both are perpendicular to line $t$. Also, $O_{1} D$ and $O B$ are on the same side of line $E O$, hence $h$ takes $O_{1} D$ to $O B$. Consequently, points ... | IMO | IMO 2006 Shortlisted Problems | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Pappus theorem",
"Geometry > Plane Geometry > Triangles > T... | English | proof only | null | |
04vm | Ten boys and ten girls met at a party. Assume that every girl likes exactly $k$ boys and every boy likes exactly $k$ girls. Is it always possible to find a couple where both the partners like each other? Solve the problem for:
a) $k = 5$,
b) $k = 6$. | [
"a) For $k=5$, it may happen that there are no such couples, with one counterexample given as follows. Split the boys into two disjoint quintuples $A, B$ and the girls into two disjoint quintuples $C, D$. Consider the configuration where every boy from $A$ likes all the girls in $C$, every boy in $B$ likes all the ... | Czech Republic | First Round of the 73rd Czech and Slovak Mathematical Olympiad (take-home part) | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | a) No. b) Yes. | |
0g5u | 已知圓 $O_1, O_2$ 交於 $M, N$ 兩點。靠近 $M$ 的公切線分別與 $O_1, O_2$ 切於點 $A, B$。點 $C, D$ 分別為 $A, B$ 關於 $M$ 的對稱點，$ riangle DCM$ 的外接圓與 $O_1, O_2$ 分別交於不同於 $M$ 的點 $E, F$。證明 $ riangle MEF$ 和 $ riangle NEF$ 的外接圓半徑相等。 | [
"取 $N'$ 使得四邊形 $NEN'F'$ 為平行四邊形，延長 $AD$ 交圓 $O_1$ 於 $E'$，延長 $BC$ 交圓 $O_2$ 於 $F'$。\n由 $M, C, F, E, D$ 五點共圓得 $\\angle MFC = \\angle MDC = \\angle MBA = \\angle MFB$，\n故 $B, C, F$ 三點共線。即 $F = F'$。同理 $E = E'$。\n連結 $NM$ 並延長交 $AB$ 於點 $L$，因為 $LA^2 = LM \\cdot LN$, $LB^2 = LM \\cdot LN$，\n故 $LA = LB$。\n又 $MB = ND$，則 $LM \\par... | Taiwan | 二〇一一數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0j75 | Problem:
Let $p$ be a prime positive integer. Define a mod-$p$ recurrence of degree $n$ to be a sequence $\{a_k\}_{k \geq 0}$ of numbers modulo $p$ satisfying a relation of the form $a_{i+n} = c_{n-1} a_{i+n-1} + \ldots + c_1 a_{i+1} + c_0 a_i$ for all $i \geq 0$, where $c_0, c_1, \ldots, c_{n-1}$ are integers and $c_... | [
"Solution:\n\nAnswer: $1 - n \\frac{p-1}{p+1} + \\frac{p^{2}(p^{2n}-1)}{(p+1)^{2}}$\n\nIn the solution all polynomials are taken modulo $p$. Call a polynomial nice if it is monic with nonzero constant coefficient. We can associate each recurrence relation with a polynomial: associate\n$$\nc_{n} a_{i+n} + c_{n-1} a_... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | 1 - n * (p - 1) / (p + 1) + p^2 * (p^(2n) - 1) / (p + 1)^2 | |
01sk | Find all positive integers $n$ ($n < 60$) such that the set $M = \{n, n+1, \dots, 60\}$ can be partitioned into disjoint subsets so that in each subset one of the numbers is equal to the sum of all other numbers of this subset.
(V. Kaskevich) | [
"A number $a$ from a subset of the desired partition is called major if $a$ is equal to the sum of all other numbers of this subset. All numbers in each subset must be distinct, since one of them is major we see that there exist at least three numbers in each subset. Let $k$ be the number of the subsets of the desi... | Belarus | FINAL ROUND | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / B... | English | proof and answer | 1, 4, 5 | |
025v | Problem:
No tabuleiro abaixo, é permitido mover qualquer objeto de seu quadrado para qualquer quadrado adjacente vazio acima, abaixo, ao lado ou em diagonal.
a) Mostre como trocar a posição de todos os chapéus com todos os troféus em apenas cinco movimentos. Argumente porque não é possível t... | [
"Solution:\n\na) Abaixo, mostramos uma sequência de cinco movimentos para trocar os chapéus com os troféus (há outra!).\n\n\nO argumento para mostrar que não é possível trocá-los com menos do que 5 movimentos é o seguinte: no primeiro movimento, precisamos mover um troféu ou um chapéu para ... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | a) 5; b) 7; c) 2001 | |
03yk | It is known that $f(x) = ax^3 + bx^2 + cx + d$ ($a \neq 0$), and $|f'(x)| \le 1$ for $0 \le x \le 1$. Please find the maximum value of $a$. | [
"$f'(x) = 3a x^2 + 2b x + c$. We have\n$$\n\\begin{cases}\nf'(0) = c, \\\\\nf'\\left(\\frac{1}{2}\\right) = \\frac{3}{4}a + b + c, \\\\\nf'(1) = 3a + 2b + c.\n\\end{cases}\n$$\nThen\n$$\n3a = 2f'(0) + 2f'(1) - 4f'\\left(\\frac{1}{2}\\right).\n$$\nWe get\n$$\n\\begin{aligned}\n3 | a | &= \\left| 2f'(0) + 2f'(1) - 4f... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 8/3 | |
01rc | A sequence $(a_n)$, $n \in \mathbb{N}$ is defined as
$$
a_1 = 1,\ a_2 = 2,\ a_3 = 3 \text{ and } a_n = \frac{a_{n-1} a_{n-2} + 7}{a_{n-3}}, \text{ for } n \ge 4.
$$
Prove that all terms of this sequence are integers. | [
"It is evident that all terms of the sequence are positive. Find $a_4$. Since $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, we have\n$$\na_4 = \\frac{a_3 a_2 + 7}{a_1} = 13. \\quad (1)\n$$\nBy condition,\n$$\n\\begin{aligned}\na_{n+1} &= \\frac{a_n a_{n-1} + 7}{a_{n-2}} = [7 = a_n a_{n-3} - a_{n-1} a_{n-2}] = \\\\\n&= \\frac{a_... | Belarus | Final Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0ilp | Problem:
$\alpha_{1}, \alpha_{2}, \alpha_{3}$, and $\alpha_{4}$ are the complex roots of the equation $x^{4}+2 x^{3}+2=0$. Determine the unordered set
$$
\left\{\alpha_{1} \alpha_{2}+\alpha_{3} \alpha_{4}, \alpha_{1} \alpha_{3}+\alpha_{2} \alpha_{4}, \alpha_{1} \alpha_{4}+\alpha_{2} \alpha_{3}\right\}
$$ | [
"Solution:\n\n$\\{1 \\pm \\sqrt{5},\\ -2\\}$. Same as Algebra $\\# 9$."
] | United States | $10^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | {-2, 1 - sqrt(5), 1 + sqrt(5)} | |
0ald | Problem:
In rectangle $ABCD$, $E$ and $F$ are chosen on $\overline{AB}$ and $\overline{CD}$, respectively, so that $AEFD$ is a square. If $\frac{AB}{BE} = \frac{BE}{BC}$, determine the value of $\frac{AB}{BC}$. | [
"Solution:\n\nLet $x$ be $BE$ and $y$ be $AE$. Note that $AEFD$ is a square so $AE = BC = y$. Also, $AB = BE + AE$ so $AB = x + y$. Since $\\frac{AB}{BE} = \\frac{BE}{BC}$ then $\\frac{x + y}{x} = \\frac{x}{y}$. Thus, we have $xy + y^2 = x^2$ which yields $x^2 - x y - y^2 = 0$. Solving for $x$ using the quadratic f... | Philippines | Area Stage | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (3+\sqrt{5})/2 | |
0a6u | Problem:
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(f(x) f(y)+y)=f(x) y+f(y-x+1)
$$
for all $x, y \in \mathbb{R}$. | [
"Solution:\nFirst note that $f$ satisfies the functional equation if and only if $-f$ does as well. We can therefore assume that $f(0) \\geq 0$. We consider two cases, depending on whether $f$ is injective or not.\n\n1. First assume that $f$ is not injective. Then there exist $a, t \\in \\mathbb{R}, t \\neq 0$ such... | Nordic Mathematical Olympiad | Nordic Mathematical Contest | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x)=0; f(x)=1-x; f(x)=x-1 | |
0hy1 | Problem:
How many positive integers less than $1998$ are relatively prime to $1547$? (Two integers are relatively prime if they have no common factors besides $1$.) | [
"Solution:\n\nAnswer: $1487$. The factorization of $1547$ is $7 \\cdot 13 \\cdot 17$, so we wish to find the number of positive integers less than $1998$ that are not divisible by $7$, $13$, or $17$. By the Principle of Inclusion-Exclusion, we first subtract the numbers that are divisible by one of $7$, $13$, and $... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 1487 | |
0kta | Problem:
Random sequences $a_{1}, a_{2}, \ldots$ and $b_{1}, b_{2}, \ldots$ are chosen so that every element in each sequence is chosen independently and uniformly from the set $\{0,1,2,3, \ldots, 100\}$. Compute the expected value of the smallest nonnegative integer $s$ such that there exist positive integers $m$ and... | [
"Solution:\n\nLet's first solve the problem, ignoring the possibility that the $a_{i}$ and $b_{i}$ can be zero. Call a positive integer $s$ an $A$-sum if $s=\\sum_{i=1}^{m} a_{i}$ for some nonnegative integer $m$ (in particular, 0 is always an $A$-sum). Define the term $B$-sum similarly. Let $E$ be the expected val... | United States | HMMT February 2022 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 2550 | |
08e7 | Problem:
Sia $S$ l'insieme degli interi maggiori o uguali a 2. Una funzione $f: S \rightarrow S$ si dice primordiale se verifica le seguenti proprietà:
- è surgettiva (cioè per ogni $s \in S$ esiste almeno un $n \in S$ tale che $f(n)=s$ ),
- è crescente sui primi (cioè se $p_{1}<p_{2}$ sono numeri primi, allora $f\lef... | [] | Italy | XXXVI Olimpiade Italiana di Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | minimum 216, maximum 584 | |
06ro | Let $p$ be an odd prime number. For every integer $a$, define the number
$$
S_{a} = \frac{a}{1} + \frac{a^{2}}{2} + \cdots + \frac{a^{p-1}}{p-1}.
$$
Let $m$ and $n$ be integers such that
$$
S_{3} + S_{4} - 3 S_{2} = \frac{m}{n}
$$
Prove that $p$ divides $m$. | [
"For rational numbers $p_{1} / q_{1}$ and $p_{2} / q_{2}$ with the denominators $q_{1}, q_{2}$ not divisible by $p$, we write $p_{1} / q_{1} \\equiv p_{2} / q_{2} (\\bmod p)$ if the numerator $p_{1} q_{2} - p_{2} q_{1}$ of their difference is divisible by $p$.\n\nWe start with finding an explicit formula for the re... | IMO | 52nd International Mathematical Olympiad 2011 Shortlist | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
06b5 | Let $n \ge 2$ be an integer. We consider two finite subsets $A, B$ of the set of integers, such that the subset $A$ contains at most $n$ elements and let $\Gamma$ be a subset of the set $\{(\alpha, \beta): \alpha \in A, \beta \in B\}$. Achilles writes on a blackboard all possible differences $\alpha - \beta$, with $(\a... | [
"For the set $X$, we use the symbol $|X|$ for the number of its elements. The number of all possible differences $\\alpha - \\beta$, with $(\\alpha, \\beta) \\in \\Gamma$, is at most equal to the number of the elements of the set $\\Gamma$, that is\n$$\nd \\leq |\\Gamma|. \\qquad (*)\n$$\nLet $A = \\{a_1, a_2, ...,... | Greece | Hellenic Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0kgu | Problem:
Real numbers $x, y, z$ satisfy
$$
x + x y + x y z = 1, \quad y + y z + x y z = 2, \quad z + x z + x y z = 4
$$
The largest possible value of $x y z$ is $\frac{a + b \sqrt{c}}{d}$, where $a, b, c, d$ are integers, $d$ is positive, $c$ is square-free, and $\operatorname{gcd}(a, b, d) = 1$. Find $1000 a + 100 b ... | [
"Solution:\n\nLet $p = x y z$ and $q = (x+1)(y+1)(z+1)$. Then, we get\n$$\np q = [x(1+y)] \\cdot [y(1+z)] \\cdot [z(1+x)] = (1-p)(2-p)(4-p)\n$$\nAdditionally, note that\n$$\nq - p = x y + y z + z x + x + y + z + 1 = (x + x y) + (y + y z) + (z + x z) + 1 = 8 - 3p\n$$\nTherefore, we have $q = 8 - 2p$. Substituting th... | United States | HMMT November 2021 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 5272 | |
099j | In the right triangle $ABC$, $\angle A = 90^\circ$. The midpoint of side $BC$ is denoted by $M$ and the point $D$ lies on $AC$ such that $AD = AM$. Let $P$ be a different from $C$ and intersection of the circumcircles of triangles $AMC$ and $BDC$. Prove that $CP$ is the bisector of angle $BCA$.
(Iran, 2008) | [
"Since quadrilateral $APMC$ is cyclic, we have $\\angle PMB = \\angle PAD$.\n\nAlso, $BPDC$ is cyclic $\\Rightarrow \\angle PBM = \\angle PDA$.\n\n\n\nBy the given condition: $AD = AM = BM$. Hence, $\\triangle PMB = \\triangle PAD$. Showing that $PM = PA$, in other words $PM = PA$. This imp... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
07bc | Let $ABC$ be an acute-angled triangle. Point $Z$ on the altitude of vertex $A$ and points $X$ and $Y$ on the extensions of the altitudes of vertices $B$ and $C$ are selected such that,
$$
\angle AYB = \angle BZC = \angle CXA = 90^\circ.
$$
Prove that $X$, $Y$ and $Z$ are collinear if and only if the length of the tange... | [
"We denote the feet of altitudes of vertices $A$, $B$ and $C$ by $D$, $E$ and $F$, respectively. Firstly, we prove that if $X$, $Y$ and $Z$ are collinear then the length of the tangent from $A$ to the nine-point circle of triangle $ABC$ (which we denote by $\\omega$) is equal to the sum of the lengths of tangents f... | Iran | Iranian 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 > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadril... | null | proof only | null | |
0dmj | Problem:
Нека су $\alpha$ и $\beta$ углови неједнакокраког троугла $A B C$ код темена $A$ и $B$, редом. Нека симетрале ових углова секу наспрамне странице троугла у $D$ и $E$, редом. Доказати да оштар угао између правих $D E$ и $A B$ није већи од $\frac{|\alpha-\beta|}{3}$. | [
"Solution:\n\nКао и обично, означимо $\\varangle A C B=\\gamma$ и $B C=a, C A=b, A B=c$, при чему је без смањења општости $a>b$ и $\\alpha>\\beta$. Нека је $F$ тачка пресека правих $D E$ и $A B$, а $\\varphi$ угао између ових правих. Из односа $\\frac{B D}{D C}=\\frac{c}{b}$ и $\\frac{C E}{E A}=\\frac{a}{c}$ лако н... | Serbia | Српска математичка олимпијада | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
028l | Problem:
Área de triângulo - Se $AC = 1{,}5\ \mathrm{cm}$ e $AD = 4\ \mathrm{cm}$, qual é a relação entre as áreas dos triângulos $\triangle ABC$ e $\triangle DBC$?
 | [
"Solution:\n\nOs triângulos $\\triangle ABC$ e $\\triangle DBC$ têm bases $AC$ e $CD$ respectivamente, e a mesma altura $h$ em relação a essas bases.\n\n\n\nAssim temos:\n$$\n\\text{área } \\triangle ABC = \\frac{AC \\times h}{2} \\quad \\text{e área } \\triangle DBC = \\frac{CD \\times h}{... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Triangles"
] | null | final answer only | 3/5 | |
0adg | Prove that if $2a_m = a_n$, then $a_{2m-n}$ is a perfect square, where $a_n = 1+2+...+n$, for every $n \in \mathbb{N}$. | [
"We have $a_n = \\frac{n(n+1)}{2}$, $n \\in \\mathbb{N}$. Since $2a_m = a_n$, it follows that\n$$\n2 \\frac{m(m+1)}{2} = \\frac{n(n+1)}{2}, \\quad 2m(m+1) = n(n+1).\n$$\nNow\n$$\n\\begin{align*}\na_{2m-n} &= \\frac{1}{2}(2m-n)(2m-n+1) = \\frac{1}{2}(4m^2 - 4mn + n^2 + 2m - n) = \\\\\n&= \\frac{1}{2}\\left(2m^2 + 2m... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
051k | On an $m \times m$ board, at the midpoints of the unit squares there are some ants. At the time $0$ each ant starts moving with speed $1$ parallel to some edge of the board until it meets an ant moving in the opposite direction or until it reaches the edge of the board. When two ants moving in the opposite direction me... | [
"Let the lower left corner of the board be the origin. Divide the units of time and space by $2$; then the squares are of dimensions $2 \\times 2$, the coordinates of the midpoints of the squares are odd positive integers, and the speed of the ants is still $1$.\n\nWe prove by induction that at integer time moments... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 3m/2 - 1 | |
02kk | Problem:
Pedro montou um quadrado com quatro das cinco peças abaixo. Qual é a peça que ele não usou?
(a)
(d)
(b)
(c)
(e)
 | [
"Solution:\n\nSolução 1 - Contando o total de quadrados nas peças.\nPara que seja possível montar o quadrado, o número total de quadradinhos deve ser um quadrado perfeito (Um número é um quadrado perfeito se ele é igual ao quadrado de um número inteiro. Por exemplo, $1, 9$ e $16$ são quadrados perfeitos pois $1=1^{... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | b | |
0a4m | Problem:
Let $X$ be the intersection of the diagonals $AC$ and $BD$ of convex quadrilateral $ABCD$. Let $P$ be the intersection of lines $AB$ and $CD$, and let $Q$ be the intersection of lines $PX$ and $AD$. Suppose that $\angle ABX = \angle XCD = 90^{\circ}$. Prove that $QP$ is the angle bisector of $\angle BQC$. | [
"Solution:\n\nFirst note that quadrilateral $ABCD$ is cyclic because $\\angle ABD = \\angle ACD = 90^{\\circ}$. Also, since $AP \\perp DX$ and $DP \\perp AX$, we see that $X$ is the orthocentre of triangle $APD$. Hence $PX \\perp AD$. Therefore quadrilaterals $ABXQ$ and $QXCD$ are cyclic (opposite angles are supple... | New Zealand | New Zealand Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
05nu | Problem:
Trouver tous les entiers $m \geqslant 1$ et $n \geqslant 1$ tels que $\frac{5^{m}+2^{n+1}}{5^{m}-2^{n+1}}$ soit le carré d'un entier. | [
"Solution:\n\nLa démonstration qui suit est valable pour $m, n \\in \\mathbb{N}$.\n\nDéjà, $5^{m}-2^{n+1}$ doit diviser $5^{m}+2^{n+1}$, donc divise $5^{m}+2^{n+1}-\\left(5^{m}-2^{n+1}\\right)=2^{n+2}$, par conséquent c'est une puissance de 2. Or, $5^{m}-2^{n+1}$ est impair, donc $5^{m}-2^{n+1}=1$.\n\nÉcrivons $5^{... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | m = 1, n = 1 | |
0038 | a) Para cada $k \ge 3$, hallar un entero positivo $n$ que se pueda representar como suma de exactamente $k$ divisores positivos de $n$ distintos entre sí.
b) Supongamos que $n$ se puede expresar como suma de exactamente $k$ divisores positivos de $n$ distintos entre sí, para algún $k \ge 3$. Sea $p$ el menor divisor pr... | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Español | proof only | null | |
0jlw | Prove that there is a constant $c > 0$ with the following property: If $a, b, n$ are positive integers such that $\gcd(a + i, b + j) > 1$ for all $i, j \in \{0, 1, \dots, n\}$, then
$$
\min\{a, b\} > c^n \cdot n^{\frac{n}{2}}.
$$ | [
"(by Titu Andreescu and Gabriel Dospinescu). Let $a, b, n$ be positive integers as in the statement of the problem. Let $P_n$ be the set of prime numbers not exceeding $n$. We will need the following\n\n**Lemma 1.** There is a positive integer $n_0$ such that for all $n \\ge n_0$ we have\n$$\n\\sum_{p \\in P_n} \\l... | United States | USAMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
063s | Problem:
Die Menge der positiven ganzen Zahlen sei mit $\mathbb{N}$ bezeichnet. Man bestimme alle Funktionen $f: \mathbb{N} \rightarrow \mathbb{N}$ mit der folgenden Eigenschaft: Für alle positiven ganzen Zahlen $m$ und $n$ ist die Zahl $f(m)+f(n)-m n$ von 0 verschieden und ist ein Teiler der Zahl $m f(m)+n f(n)$. | [
"Solution:\n\nAntwort: Es gibt genau eine Funktion, die die beschriebene Bedingung erfüllt, nämlich $f(k)=k^{2}$ für alle $k$.\n\nZum Beweis sei $f$ wie verlangt.\n\nSchritt 1: Einsetzen von $m=n=1$ liefert $2 f(1)-1 \\mid 2 f(1)$, also auch $2 f(1)-1 \\mid 2 f(1)-(2 f(1)-1)=1$ und damit $2 f(1)-1=1$, also $f(1)=1$... | Germany | 1. Auswahlklausur | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | f(k) = k^2 for all k in N | |
02bw | Problem:
a) Uma calculadora do país de Cincolândia tem apenas os algarismos de 0 a 9 e dois botões $\square$ e $\triangle$. O botão $\square$ eleva ao quadrado o número que está no visor da calculadora. O botão $\triangle$ subtrai 5 do número que está no visor da calculadora. Mônica digita o número 7 e depois aperta $\... | [
"Solution:\na) Mônica começa digitando o número 7. Daí,\n$$\n7 \\xrightarrow{\\square} 7^{2}=49 \\xrightarrow{\\triangle} 49-5=44.\n$$\nLogo, o resultado final que aparece na calculadora é o número 44.\n\nb) Se um número natural $x$ deixa resto 4 quando dividido por 5, isso quer dizer que $x$ é da forma\n$$\nx=5q+4... | Brazil | null | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) 44; b) remainder 1 when divided by 5; c) no | |
0bjd | Let $O$ be the circumcenter of the acute triangle $ABC$. An arbitrary diameter intersects side $[AB]$ in $D$ and side $[AC]$ in $E$. If $F$ is the midpoint of $[BE]$ and $G$ is the midpoint of $[CD]$, show that $\angle FOG = \angle BAC$.
 | [
"We shall make use of the following:\n\n**Lemma.** Let $ABC$ be a triangle and $MN$ be a chord of its circumcircle which intersects side $[AB]$ in $D$ and side $[AC]$ in $E$. Then $\\frac{DM}{DN} : \\frac{EM}{EN} = \\frac{BM}{BN} : \\frac{CM}{CN}$.\n\n*Proof of the lemma.* Using the law of sines in triangles $BDM$ ... | Romania | 65th NMO Selection Tests for JBMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angl... | null | proof only | null | |
06l1 | Let $n \ge 3$ be an integer. Prove that there exists a set of $2n$ positive integers satisfying the following property: For every $m = 2, 3, \dots, n$ the set $S$ can be partitioned into two subsets with equal sums of elements, with one of the subsets of cardinality $m$. | [] | Hong Kong | HKG TST | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0b0m | Problem:
A committee of three is to be selected from a pool of candidates consisting of five men and four women. If all the candidates are equally likely to be chosen, what is the probability that the committee will have an odd number of female members? | [
"Solution:\n\nWe either have exactly one or three female members. Therefore, the required probability is\n$$\n\\frac{\\binom{4}{1}\\binom{5}{2} + \\binom{4}{3}}{\\binom{9}{3}} = \\frac{44}{84} = \\frac{11}{21}\n$$"
] | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 11/21 | |
05bq | A mother has 7 apples, 6 pears, and 5 oranges. She wants to divide them among 2 children so that each gets the same number of fruits. In how many different ways can this be done?
*Remark:* We consider the distributions of fruit to be different if a child receives a different number of some types of fruit. | [
"According to the conditions, each child must receive 9 fruits. It suffices to find how many possibilities there are to give the first child 9 fruits, because the second child receives all the remaining fruits.\nThe first child can be given 0 to 6 pears and 0 to 5 oranges. Disregarding the condition that he must re... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 36 | |
05ti | Problem:
Soit $S$ un ensemble d'entiers relatifs. On dit que $S$ est beau s'il contient tous les entiers de la forme $2^{a}-2^{b}$, où $a$ et $b$ sont des entiers naturels non nuls. On dit également que $S$ est fort si, pour tout polynôme $P(X)$ non constant et à coefficients dans $S$, les racines entières de $P(X)$ a... | [
"Solution:\n\nL'ensemble $\\mathbb{Z}$ est clairement beau et fort. Nous allons démontrer que c'est le seul. Pour ce faire, considérons un ensemble $S$ beau et fort : nous allons en fait prouver, par récurrence forte sur $n$, que les entiers $n$ et $-n$ appartiennent nécessairement à $S$.\n\nTout d'abord, puisque $... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | the set of all integers | |
0ihh | Problem:
Let $x$ and $y$ be two distinct roots of unity. Prove that $x+y$ is also a root of unity if and only if $\frac{y}{x}$ is a cube root of unity. | [
"Solution:\n\nThis is easiest to see geometrically. The vectors corresponding to $x$, $y$, and $-x-y$ sum to $0$, so they form a triangle. In order for them all to be roots of unity, they must all have length one, so the triangle must be equilateral. Therefore the angle between $x$ and $y$ is $\\pm \\frac{2 \\pi}{3... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof only | null | |
0aa9 | Problem:
The quadrilateral $ABCD$ satisfies $\angle ACD = 2 \angle CAB$, $\angle ACB = 2 \angle CAD$ and $CB = CD$.
Show that $\angle CAB = \angle CAD$. | [
"Solution:\nLet the angle bisectors from angle $C$ in triangle $ACB$ and $ACD$ intersect $AB$ and $AD$ in points $E$ and $F$ respectively. From $\\angle ACE = \\angle CAD$ it follows that $CE$ and $AD$ are parallel. Similarly $CF$ and $AB$ are parallel. Hence $AECF$ is a parallelogram. From this it follows that $\\... | Nordic Mathematical Olympiad | Nordic Mathematical Contest | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0jku | Problem:
Flat Albert and his buddy Mike are watching the game on Sunday afternoon. Albert is drinking lemonade from a two-dimensional cup which is an isosceles triangle whose height and base measure $9$ cm and $6$ cm; the opening of the cup corresponds to the base, which points upwards. Every minute after the game beg... | [
"Solution:\n\nLet $A_{0} = \\frac{1}{2} (6)(9) = 27$ denote the area of Albert's cup; since area varies as the square of length, at time $n$ Mike adds\n\n$$\nA\\left(1-\\left(1-\\frac{1}{9 n^{2}}\\right)^{2}\\right)\n$$\n\nwhence in all, he adds\n\n$$\nA_{0} \\sum_{n=1}^{\\infty}\\left(\\frac{2}{9 n^{2}}-\\frac{1}{... | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 26 | |
04v0 | Let $ABCD$ be a convex quadrilateral with $AB = BC = CD$ and $P$ its intersection of diagonals. Denote by $O_1, O_2$ the circumcenters of triangles $ABP, CDP$, respectively. Prove that $O_1BCO_2$ is a parallelogram. | [] | Czech Republic | Final Round | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ixo | Problem:
Let $ABC$ be a right triangle with hypotenuse $AC$. Let $B'$ be the reflection of point $B$ across $AC$, and let $C'$ be the reflection of $C$ across $AB'$. Find the ratio of $[BCB']$ to $[BC'B']$. | [
"Solution:\n\nSince $C$, $B'$, and $C'$ are collinear, it is evident that $[BCB'] = \\frac{1}{2}[BCC']$. It immediately follows that $[BCB'] = [BC'B']$. Thus, the ratio is $1$."
] | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 1 | |
02lx | Carlitos has several pieces formed by four unit squares, shaped as an L:
He assembles bigger figures with these pieces, making them share one or more sides of the little squares. In the following example, the figure in the left was assembled by two pieces sharing a unit side. The figures are n... | [
"a) For example,\n\n(of course, there are other possibilities)\n\nb) For example,\n\nwhich consists of a rectangle with dimensions $4 \\times 1001$ minus a square of side $2$.\n\nc) No, it's not possible, because $2009$ is odd. Each piece has perimeter $10$, which i... | Brazil | XXXI Brazilian Math Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | a) One can arrange the pieces to obtain a figure with perimeter fourteen (for example, as in the provided sample drawing). b) Construct a rectangle of four by one thousand one using the pieces and remove a two by two corner notch; this yields perimeter two thousand ten. c) No; an odd perimeter is impossible because eve... | |
0k3s | Problem:
Find the number of eight-digit positive integers that are multiples of $9$ and have all distinct digits. | [
"Solution:\n\nNote that $0+1+\\cdots+9=45$. Consider the two unused digits, which must then add up to $9$. If it's $0$ and $9$, there are $8 \\cdot 7!$ ways to finish; otherwise, each of the other four pairs give $7 \\cdot 7!$ ways to finish, since $0$ cannot be the first digit. This gives a total of $36 \\cdot 7! ... | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics",
"Number Theory > Other"
] | null | proof and answer | 181440 | |
0ghx | 泓江的兩岸各有 $n$ 座城市。江上有若干條雙向渡輪航班，每一條都連接左岸的一座城市與右岸的一座城市。我們稱一座城市是**便利的**，若且唯若該城市有通往對岸所有城市的航班。我們稱泓江是**暢通的**，若且唯若我們可以找到 $n$ 條航班，使得其兩端點的城市恰包含全部 $2n$ 座城市。
已知泓江目前不是暢通的，且只要增設任何一條新航班，泓江便是暢通的。試求便利城市數量的所有可能值。
There are $n$ cities on each side of Hung river, with two-way ferry routes between some pairs of cities across the river. A c... | [
"Graph theoretic statement: in a balanced bipartite graph $G(V_1, V_2, E)$ with $n$ vertices on both parties, if there is no perfect matching but adding any other edges would lead to one, determine the number of vertices with degree $n$. We will show that this number is $n-1$.\n\n(1) Since $G(V_1, V_2, E)$ has no p... | Taiwan | 2023 數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | Chinese (Traditional) | proof and answer | n-1 | |
010f | Problem:
A triple of positive integers $(a, b, c)$ is called quasi-Pythagorean if there exists a triangle with lengths of the sides $a, b, c$ and the angle opposite to the side $c$ equal to $120^{\circ}$. Prove that if $(a, b, c)$ is a quasi-Pythagorean triple then $c$ has a prime divisor greater than 5. | [
"Solution:\n\nBy the cosine law, a triple of positive integers $(a, b, c)$ is quasi-Pythagorean if and only if\n$$\nc^{2} = a^{2} + a b + b^{2}\n$$\nIf a triple $(a, b, c)$ with a common divisor $d > 1$ satisfies (1), then so does the reduced triple $\\left(\\frac{a}{d}, \\frac{b}{d}, \\frac{c}{d}\\right)$. Hence i... | Baltic Way | Baltic Way 1998 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequali... | null | proof only | null | |
0cvs | Let $ABCD$ be a triangular pyramid. A sphere $\omega_A$ is tangent to the face $BCD$, and to the planes of the other faces outside the faces. Similarly, a sphere $\omega_B$ is tangent to the face $ACD$, and to the planes of the other faces outside the faces. Let $\omega_A$ meet the plane $ACD$ at $K$, and let $\omega_B... | [
"Отметим точки $K_1$ и $L_1$ касания вписанной сферы $\\omega$ тетраэдра с гранями $ACD$ и $BCD$ соответственно, а также точки $K_2$ и $L_2$ касания сфер $\\omega_B$ и $\\omega_A$ с этими гранями. Сферы $\\omega$ и $\\omega_A$ гомотетичны с центром в точке $A$, поэтому точка $K_1$ лежит на отрезке $AK$. Аналогично,... | Russia | Final round | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Ci... | English; Russian | proof only | null | |
078t | Let $P(x) \in \mathbb{Q}[x]$ be a polynomial with rational coefficients and degree $d \ge 2$. Prove there is no infinite sequence $a_0, a_1, \dots$ of rational numbers such that $P(a_i) = a_{i-1} + i$ for all $i \ge 1$. | [
"FTSOC, assume that there is such a sequence.\nWe proceed with the solution in two steps. In the first step, we show that all $a_i$ must be of the form $\\frac{k_i}{n}$ where $k_i \\in \\mathbb{Z}$ and $n$ is a fixed natural dependent only on $P$ and $a_1$.\nFirst, we write the polynomial $P$ as $\\frac{Q}{N}$ wher... | India | IMO TST | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof only | null | |
01hk | Determine all sequences $(a_1, a_2, ...)$ of positive integers satisfying
$$ a_{n+1}^2 = 1 + (n + 2021)a_n $$
for all $n \ge 1$. | [
"Clearly for $C = 1$ we have the solution $(a_n)_{n=1}^{\\infty} = (n + 2019)_{n=1}^{\\infty}$. Let's prove that this is the only value for $C$ that works.\nAssume $(a_n)_{n=1}^{\\infty}$ is a solution and let $(b_n)_{n=1}^{\\infty} = (a_n - n)_{n=1}^{\\infty}$. We claim that for $n > |C| + 2021^2$:\n(i) If $b_n < ... | Baltic Way | Baltic Way 2021 Shortlist | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | a_n = n + 2019 | |
0kap | Problem:
Let $S$ be a finite set of nonzero real numbers, and let $f: S \rightarrow S$ be a function with the following property: for each $x \in S$, either
$$
f(f(x))=x+f(x) \quad \text{or} \quad f(f(x))=\frac{x+f(x)}{2}
$$
Prove that $f(x)=x$ for all $x \in S$. | [
"Solution:\nWe will use the notation $f^{n}(x)$ to denote $f(f(\\cdots f(x) \\cdots))$, where we iterate the function $n$ times. Suppose, to the contrary, that $f(x) \\neq x$ for some $x \\in S$. This implies that $f(f(x)) \\neq f(x)$ as well, since $f(f(x))$ is either the sum or the average of $x$ and $f(x)$ and t... | United States | Bay Area Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0343 | Problem:
Find the least number of colors with the following property: the integers $1,2, \ldots, 2004$ can be colored such that there are no integers $a < b < c$ of the same color for which $a$ divides $b$ and $b$ divides $c$. | [
"Solution:\nDenote by $f(n)$ the least number of colors such that the integers $1,2, \\ldots, n$ can be colored in the required way. We shall prove that $f(n) = \\lfloor (k+1)/2 \\rfloor$, where $2^{k-1} \\leq n < 2^{k}$.\n\nObserve that in the sequence $1, 2, 2^{2}, \\ldots, 2^{k-1}$ we have no three numbers of th... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 6 | |
0jhn | Problem:
Marty and three other people took a math test. Everyone got a non-negative integer score. The average score was $20$. Marty was told the average score and concluded that everyone else scored below average. What was the minimum possible score Marty could have gotten in order to definitively reach this conclusi... | [
"Solution:\n\n$61$\n\nSuppose for the sake of contradiction Marty obtained a score of $60$ or lower. Since the mean is $20$, the total score of the $4$ test takers must be $80$. Then there exists the possibility of $2$ students getting $0$, and the last student getting a score of $20$ or higher. If so, Marty could ... | United States | HMMT November 2013 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 61 |
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