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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
09vo | In an acute triangle $ABC$, the centre of the incircle is $I$, and $|AC| + |AI| = |BC|$. Prove that $\angle BAC = 2\angle ABC$. | [
"Let $D$ be a point on $BC$ such that $|CD| = |AC|$. Because $|BC| = |AC| + |AI|$, the point $D$ lies on the interior of side $BC$, and we have $|BD| = |AI|$. Because triangle $ACD$ is isosceles, the angle bisector $CI$ is also the perpendicular bisector of $AD$, hence $A$ is the reflection of $D$ in $CI$. Hence, w... | Netherlands | IMO Team Selection Test 1, June 2020 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
0c1o | The sequence $(a_n)_{n \ge 1}$ has the properties that $a_n > 1$ and $a_{n+1}^2 \ge a_n a_{n+2}$, for $n \ge 1$. Show that the sequence $(x_n)_{n \ge 1}$, defined by $x_n = \log_{a_n} a_{n+1}$, for $n \ge 1$, has a finite limit. Find it. | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 1 | |
03sw | Eight persons join a party.
(1) If there exist three persons who know each other in any group of five, prove that we can find that four persons know each other.
(2) If there exist three persons in a group of six who know each other in a cyclical manner, can we find four persons who know each other in a cyclical manne... | [
"(1) By means of graph theory, use $8$ vertices to denote $8$ persons. If two persons know each other, we connect them with an edge. With the given condition, there will be a triangle in every induced subgraph with five vertices, while every triangle in the graph belongs to different $\\binom{8-3}{2} = \\binom{5}{2... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Part (1): Yes, there exists a group of four mutual acquaintances. Part (2): No. | |
0ikd | Assign to each side $b$ of a convex polygon $\mathcal{P}$ the maximum area of a triangle that has $b$ as a side and is contained in $\mathcal{P}$. Show that the sum of the areas assigned to the sides of $\mathcal{P}$ is at least twice the area of $\mathcal{P}$. | [
"Define the *weight* of a side $XY$ to be the area assigned to it, and define an *antipoint* of a side of a polygon to be one of the points in the polygon farthest from that side (and consequently forming the triangle with greatest area).\n\n**Lemma** For any side $XY$, $Z$ is an antipoint if and only if the line $... | United States | IMO | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0ahr | We say that a rectangle is inscribed in a triangle if two of the rectangle's neighbouring vertices lie on one side of the triangle, and the other two lie on the remaining two sides of the triangle. Assume that the lengths of the sides of the triangle $ABC$ are known. What is the smallest possible length of the diagonal... | [
"Let the quadrilateral $EFGH$ be inscribed in the triangle $ABC$ so that $E$ and $F$ lie on $BC$ and $G$ lies on $AC$ and $H$ lies on $AB$. Let us denote the side-lengths of the triangle $ABC$ by $a$, $b$ and $c$ and let $h$ denote the length of the height drawn from $A$ to $BC$. We put $\\overline{AH} = x$, $\\ove... | North Macedonia | Team Selection Test for IMO | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | Minimum diagonal length is 2P / sqrt(a^2 + 4P^2 / a^2), attained when the rectangle has two neighboring vertices on the longest side a; here P is the triangle’s area (obtainable from a, b, c via Heron’s formula). | |
017r | Prove the inequality for positive real numbers $x_1$, $x_2$, $\dots$, $x_n$:
$$
\frac{x_1}{x_2+x_3} + \frac{x_2}{x_3+x_4} + \dots + \frac{x_{n-2}}{x_{n-1}+x_n} + \frac{x_{n-1}}{x_n+x_1} + \frac{x_n}{x_1+x_2}
\ge \frac{x_2}{x_1+x_2} + \frac{x_3}{x_2+x_3} + \dots + \frac{x_n}{x_{n-1}+x_n} + \frac{x_1}{x_n+x_1}
$$ | [
"Observe that the product of $n$ fractions $\\frac{x_k+x_{k+1}}{x_{k+1}+x_{k+2}}$ is equal to $1$ (we assume that $x_{n+1} = x_1$, etc.). Then by the Cauchy inequality we conclude that\n$$\n\\sum_{k=1}^{n} \\frac{x_k + x_{k+1}}{x_{k+1} + x_{k+2}} \\ge n = \\sum_{k=1}^{n} \\frac{x_{k+1} + x_{k+2}}{x_{k+1} + x_{k+2}}... | Baltic Way | BALTIC WAY | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0alr | Problem:
There are $5$ shmacks in $2$ shicks, $3$ shicks in $5$ shures, and $2$ shures in $9$ shneids. How many shmacks are there in $6$ shneids?
(a) $5$
(b) $8$
(c) $2$
(d) $1$ | [] | Philippines | Qualifying Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | c | |
0iu9 | Problem:
Suppose that instead there are 6 rooms with 4 doors. In each room, 1 door leads to the next room in the sequence (or, for the last room, Bowser's level), while the other 3 doors lead to the first room. Now what is the expected number of doors through which Mario will pass before he reaches Bowser's level? | [
"Solution:\nAnswer: 5460 This problem works in the same general way as the last problem, but it can be more succintly solved using the general formula, which is provided below in the solution to the next problem."
] | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 5460 | |
090x | Sea $S$ un subconjunto finito de los números enteros. Definimos $d_2(S)$ y $d_3(S)$ de la siguiente manera:
* $d_2(S)$ es el número de elementos $a \in S$ para los que existen $x, y \in \mathbb{Z}$ tales que $x^2 - y^2 = a$.
* $d_3(S)$ es el número de elementos $a \in S$ para los que existen $x, y \in \mathbb{Z}$ tales... | [
"En primer lugar, observamos que un número se puede escribir como diferencia de cuadrados si y solo si no es de la forma $4k + 2$. Para ver esto, escribimos $x^2 - y^2 = \\alpha \\cdot \\beta$, donde $\\alpha$ y $\\beta$ tienen la misma paridad y simplemente ponemos $x = \\frac{\\alpha+\\beta}{2}$, $y = \\frac{\\be... | Mexico | LVI Olimpiada Matemática Española (Concurso Final) | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Spanish | proof only | null | |
06dq | Let $\triangle ABC$ be a triangle. $M$ is the midpoint of $AC$, $D$ is a point on $AB$. $BM$ and $CD$ meet at $O$, with $AB = CO$. $E$ is a point on $AC$ such that $DE // BM$. Prove that $AB \perp BC$ if and only if $ADOM$ is a cyclic quadrilateral. | [
"Using the parallel lines, we have $\\triangle ADE \\sim \\triangle ABM$ and $\\triangle CMO \\sim \\triangle CED$. This implies\n$$\n\\frac{AB}{BD} = \\frac{AM}{ME} = \\frac{CM}{ME} = \\frac{CO}{OD}.\n$$\nAs $AB = CO$, we obtain $BD = OD$.\nNow,\n$$\n\\begin{align*}\n& \\Leftrightarrow \\quad \\angle BAM = \\angle... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
0c8i | Let $k$ and $n$ be positive integers, and let $G$ be a group of order $n$. Prove that the following two statements are equivalent:
(a) The numbers $k$ and $n$ are relatively prime.
(b) For every subgroup $H$ of $G$, the set $\{x: x \in G \text{ and } x^k \in H\}$ is contained in $H$. | [
"We show that (a) implies (b). Since $k$ and $n$ are relatively prime, $kp + nq = 1$ for some integers $p$ and $q$. Let $H$ be a subgroup of $G$, and let $x$ be a member of $G$ such that $x^k \\in H$. Since $x^n = e$, the unit of $G$, it follows that\n$$\nx = x^{kp + nq} = (x^k)^p \\cdot (x^n)^q = (x^k)^p \\in H.\n... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Group Theory"
] | English | proof only | null | |
0l8c | Let triangle $ABC$ have orthocenter $H$. Let $B_1, C_1, B_2$ and $C_2$ be collinear points which lie on lines $AB, AC, BH$, and $CH$, respectively. Let $\omega_B$ and $\omega_C$ be the circumcircles of triangles $BB_1B_2$ and $CC_1C_2$, respectively. Prove that the radical axis of $\omega_B$ and $\omega_C$ intersects t... | [] | United States | USA TST Selection Test for 67th IMO and 15th EGMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0c73 | $$
A_n = \{ (x, y) \in \mathbb{N} \times \mathbb{N} \mid \sqrt{x^2 + y + n} + \sqrt{y^2 + x + n} \in \mathbb{N} \}.
$$
Prove that for all $n \ge 1$, $A_n$ is a finite, nonempty set. | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
00iy | Let $x$, $y$ be positive real numbers such that
$$
x + y + xy = 3.
$$
Prove that
$$
x + y \ge 2.
$$
When does equality hold? | [
"We have\n$$\n\\begin{aligned}\nx + y + xy &= 3 \\\\\n\\Leftrightarrow xy + x + y + 1 &= 4 \\\\\n\\Leftrightarrow (x + 1)(y + 1) &= 4.\n\\end{aligned}\n$$\nTherefore we can rewrite the inequality in the following way:\n$$\n\\begin{aligned}\n&x+y \\ge 2 \\\\\n\\Leftrightarrow &x+1+y+1 \\ge 4 \\\\\n\\Leftrightarrow &... | Austria | AustriaMO2011 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds at x = y = 1 | |
0hs3 | Problem:
Given a circle $k$, let $AB$ be its diameter. An arbitrary line $l$ intersects the circle $k$ at the points $P$ and $Q$. If $A_{1}$ and $B_{1}$ are the feet of perpendiculars from $A$ and $B$ to $PQ$, prove that $A_{1}P = B_{1}Q$. | [
"Solution:\n\nLet $A_{2}$ and $B_{2}$ be the intersections of the lines $AA_{1}$ and $BB_{1}$ with the circle $k$. Furthermore, let $s$ be the line through the center of $k$ perpendicular to $l$. Then $AB_{2}BA_{2}$ is a rectangle (because $AA_{2} \\parallel BB_{2}$ and $\\angle AB_{2}B = 90^{\\circ}$ since $AB$ is... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
061w | Problem:
Eine unendliche Folge $a_{0}, a_{1}, a_{2}, \ldots$ reeller Zahlen erfüllt die Bedingung $a_{n}=\left|a_{n+1}-a_{n+2}\right|$ für alle $n \geq 0$, wobei $a_{0}$ und $a_{1}$ verschiedene positive Zahlen sind.
Kann diese Folge beschränkt sein? Die Antwort ist zu begründen. | [
"Solution:\n\nZunächst beweisen wir, dass zwei aufeinander folgende Glieder niemals gleich sein können. Aus $a_{n}=a_{n+1}=c$ folgte nämlich sofort $a_{n-1}=0$ und $a_{n-2}=a_{n-3}=c \\quad(n>2)$. Schließlich müsste $a_{0}=a_{1}$ oder $a_{0}=0$ bzw. $a_{1}=0$ sein, was ausgeschlossen ist. Daher gilt auch $a_{n}>0$ ... | Germany | Auswahlwettbewerb zur IMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | The sequence is unbounded. | |
09nk | A $4 \times 4$ grid consists of 25 vertex points formed by its horizontal and vertical lines. Coloring a cell colors its four corner vertex points. In how many distinct ways can cells be colored so that every vertex point is colored at least once?
(Batzorig Undrakh) | [] | Mongolia | MMO2025 Round 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1215 | |
0074 | Dada una sucesión $S$ de $1001$ números reales positivos no necesariamente distintos, y dado un conjunto $A$ de números enteros positivos distintos, la operación permitida es: satisface un $k \in A$, seleccionar $k$ números de $S$, calcular el promedio de los $k$ números (media aritmética) y reemplazar cada uno de los ... | [] | Argentina | XX Olimpiada Matemática del Cono Sur | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Spanish | proof and answer | 13 | |
08i6 | Problem:
Calculate the sum
$$
\frac{2^{4}+2^{2}+1}{2^{7}-2}+\frac{3^{4}+3^{2}+1}{3^{7}-3}+\ldots+\frac{2003^{4}+2003^{2}+1}{2003^{7}-2003}+\frac{1}{2 \cdot 2003 \cdot 2004}
$$ | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1/4 | |
04hj | Let $I$ be the incentre of the acute triangle $ABC$. Rays $AI$ and $BI$ intersect the circumcircle $k$ of the triangle $ABC$ in points $D$ and $E$ respectively. Segments $DE$ and $CA$ intersect in point $F$, line through point $E$ parallel to the line $FI$ intersects circle $k$ also in point $G$, and lines $FI$ and $DG... | [
"Let us denote $\\angle BAC = \\alpha$, $\\angle CBA = \\beta$, $\\angle ACB = \\gamma$, and let $J$ be the intersection of the segments $\\overline{DE}$ and $\\overline{CB}$.\n\nInscribed angles $\\angle BED$, $\\angle BCD$ and $\\angle BAD$ over the chord $BD$ are equal, so $\\angle BED =... | Croatia | Mathematica competitions in Croatia | [
"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 | |
0g7b | 設四邊形 $ABCD$ 有內切圓, 其圓心為 $I$。令對角線 $AC$, $BD$ 相交於 $E$ 點。
若 $AD$, $BC$, $EI$ 三線段的中點共線, 試證 $AB = CD$。 | [
"設 $AB$ 與 $CD$ 交於 $F$ 點，並設線段 $EF$, $AD$, $BC$ 的中點分別為 $X$, $Y$, $Z$。\n根據 Gauss 定理，$X$, $Y$, $Z$ 三點共線（牛頓線）。原本包含 $X$, $Y$, $Z$ 的直線通過 $F$ 點，經過以 $E$ 點為中心放大為 $2$ 倍的動作會使該直線通過 $I$ 點。因此 $YZ \\parallel FI$。\n設 $AC$, $BD$ 的中點分別為 $U$, $V$。因為 $FI$ 是 $\\angle DFA$ 的角平分線，所以\n$YZ$ 也是 $\\angle UYV$ 的角平分線。但 $YZ$ 平分線段 $UV$ ($UYVZ$ 是平... | Taiwan | 二〇一三數學奧林匹亞競賽第三階段選訓營, 獨立研究 (二) | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | null | proof only | null | |
0d2g | Find the area of the set of points of the plane whose coordinates $(x, y)$ satisfy
$$
x^{2}+y^{2} \leq 4|x|+4|y|
$$ | [
"Notice that, if $(x, y)$ is a solution of the inequality, then $(-x, y)$, $(x,-y)$, and $(-x,-y)$ are all solutions of the inequality. Therefore, we can assume $x, y \\geq 0$ and deduce the other solutions by symmetries with respect to $x$-axis and $y$-axis.\n\nAssume $x, y \\geq 0$. The inequality is equivalent t... | Saudi Arabia | Selection tests for the Balkan Mathematical Olympiad 2013 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 32 + 16π | |
0f2w | Problem:
Let $p(x) = x^2 + x + 1$. Show that for every positive integer $n$, the numbers $n$, $p(n)$, $p(p(n))$, $p(p(p(n)))$, ... are relatively prime. | [] | Soviet Union | ASU | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0692 | A set $S$ of integers is Balearic, if there are two (not necessarily distinct) elements $s, s' \in S$ whose sum $s+s'$ is a power of two; otherwise it is called a non-Balearic set. Find an integer $n$ such that $\{1, 2, \dots, n\}$ contains a 99-element non-Balearic set, whereas all the 100-element subsets are Balearic... | [] | Greece | 20th Mediterranean Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 203 | |
0jkb | Problem:
Let $x$ be a complex number such that $x + x^{-1}$ is a root of the polynomial $p(t) = t^{3} + t^{2} - 2 t - 1$. Find all possible values of $x^{7} + x^{-7}$. | [
"Solution:\nSince $x + x^{-1}$ is a root,\n$$\n\\begin{aligned}\n0 & = \\left(x + x^{-1}\\right)^{3} + \\left(x + x^{-1}\\right)^{2} - 2\\left(x + x^{-1}\\right) - 1 \\\\\n& = x^{3} + x^{-3} + 3x + 3x^{-1} + x^{2} + 2 + x^{-2} - 2x - 2x^{-1} - 1 \\\\\n& = x^{3} + x^{-3} + x^{2} + x^{-2} + x + x^{-1} + 1 \\\\\n& = x... | United States | HMMT November 2014 | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2 | |
09fc | Let $ABCDEF$ be a circumscribed hexagon. Let $AB \cap CD = K$, $CD \cap EF = L$, $DE \cap AF = M$ and $AF \cap BC = N$. Prove that the lines $KM$, $LN$ and $BE$ are concurrent. | [
"Let the circle $\\omega$ inscribed in a hexagon $ABCDEF$ tangents the sides $AB$, $BC$, $CD$, $DE$, $EF$, $FA$ at the points $A_4$, $A_3$, $A_2$, $A_1$, $A_6$, $A_5$ respectively.\n\n\n\n(1) Let $A_6A_2$ be the polar line of the point $L$ with respect to $\\omega$ and let $A_3A_5$ be the p... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0awg | Problem:
A triangle $ABC$ is to be constructed so that $A$ is at $(3,2)$, $B$ is on the line $y = x$, and $C$ is on the $x$-axis. Find the minimum possible perimeter of $\triangle ABC$. | [
"Solution:\n\nLet $D(2,3)$ be the reflection of $A$ with respect to $y = x$, and $E(3,-2)$ the reflection of $A$ with respect to the $x$-axis. Then if $B$ is on $y = x$ and $C$ is on the $x$-axis, then $|AB| = |DB|$ and $|AC| = |CE|$. Then the perimeter of $\\triangle ABC$ is\n$$\n|AB| + |BC| + |AC| = |DB| + |BC| +... | Philippines | Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loc... | null | proof and answer | sqrt(26) | |
02ah | Problem:
Inflação - Márcia está numa loja comprando um gravador que ela queria há muito tempo. Quando o caixa registra o preço ela exclama: "Não é possível, você registrou o número ao contrário, trocou a ordem de dois algarismos, lembro que na semana passada custava menos que 50 reais!" Responde o caixa: Sinto muito, ... | [
"Solution:\n\nO preço antigo era menor que 50 reais e sofreu um acréscimo de $20\\%$. Logo, o novo preço ainda é um número de 2 algarismos. Vamos representá-lo por $ab$, onde $a$ é o algarismo das dezenas e $b$ é o algarismo das unidades. Logo, o novo preço é $ba$, e temos:\n$$\n10b + a = 1,2(10a + b)\n$$\n$$\n10b ... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 54 | |
0f5v | Problem:
Find all pairs of digits $(b, c)$ such that the number $b \ldots b6c \ldots c4$, where there are $n$ $b$s and $n$ $c$s, is a square for all positive integers $n$. | [] | Soviet Union | 18th ASU | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (b, c) = (9, 0) | |
068m | Prove that the number $A = \frac{4n!}{n!\ 2n!}$, where $n$ is a positive integer, is an integer and has a factor of the form $2^{n+1}$. (Note: The number $n!$ for $n \in \mathbb{N}$, is defined by: $n! = 1 \cdot 2 \cdot \dots \cdot n$, and $0! = 1$.) | [
"We can write: $A = \\frac{4n!}{n!\\ 2n!} = \\binom{3n}{n} \\cdot 3n+1\\ 3n+2\\ \\dots\\ 4n-1 \\cdot 4n \\in \\mathbb{Z}$, since $\\binom{3n}{n} \\in \\mathbb{Z}$.\n\nNext we observe that in the prime factorization of $n!$ the exponent of $2$ is\n$$\n\\exp n = \\left\\lfloor \\frac{n}{2} \\right\\rfloor + \\left\\l... | Greece | SELECTION EXAMINATION | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
03kb | Problem:
Twenty-five men sit around a circular table. Every hour there is a vote, and each must respond yes or no. Each man behaves as follows: on the $n^{\text{th}}$ vote, if his response is the same as the response of at least one of the two people he sits between, then he will respond the same way on the $(n+1)^{th... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0icn | Alice and Bob play a game on a $6$ by $6$ grid. On his or her turn, a player chooses a rational number not yet appearing in the grid and writes it in an empty square of the grid. Alice goes first and then the players alternate. When all squares have numbers written in them, in each row, the square with the greatest num... | [
"**First Solution:** Bob can win as follows.\nAfter each of his moves, Bob can insure that the maximum number in each row is a square in $A \\cup B$, where $A$ and $B$ are the sets of squares marked with A's and B's in the following diagram, respectively.\n\n**Proof:** Bob pairs each square... | United States | USA IMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Bob has a winning strategy. | |
0ddr | Let $ABC$ be triangle with the symmedian point $L$ and circumradius $R$. Construct parallelograms $ADLE$, $BHLK$, $CILJ$ such that $D, H \in AB$; $K, I \in BC$; $J, E \in CA$. Suppose that $DE$, $HK$, $IJ$ pairwise intersect at $X$, $Y$, $Z$. Prove that inradius of $XYZ$ is $\frac{R}{2}$. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, ... | null | proof and answer | R/2 | |
093d | Problem:
Let $ABC$ be an acute-angled triangle with $AB < AC$, and let $D$ be the foot of its altitude from $A$. Points $B'$, $C'$ lie on the rays $AB$ and $AC$, respectively, so that points $B'$, $C'$, and $D$ are collinear and points $B$, $C$, $B'$, and $C'$ lie on one circle with center $O$. Prove that if $M$ is th... | [
"Solution:\n\nWithout loss of generality assume that $AB < AC$. Let $H'$, $H''$ be the points symmetric to $H$ with respect to $BC$ and with respect to $M$, respectively. It is well-known that $H'$, $H''$ both lie on the circumcircle of $ABC$; furthermore, $AH''$ is the diameter of this circumcircle. Since $DH \\pa... | Middle European Mathematical Olympiad (MEMO) | Middle European 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 > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneo... | null | proof only | null | |
04os | Let $k > 1$ be a positive integer. $k+2$ distinct positive integers are given, all less than $3k+1$. Prove that we can find two numbers among them whose difference is greater than $k$ and less than $2k$. (Mathematical Excalibur 2015) | [
"Denote by $S$ the set of $k + 2$ given numbers. Without loss of generality, we can assume that $S$ contains $1$. Indeed, if $1$ is not in $S$, we can subtract the smallest element of $S$ from all elements of $S$ and add $1$ to all of them, which preserves the differences between all elements of $S$.\n\nIf at least... | Croatia | Croatian Mathematical Society Competitions | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
07ax | Let $ABC$ be a triangle with circumcircle $\omega$ and circumcenter $O$. Denote by $M$ the midpoint of that arc $BC$ of $\omega$ which does not contain vertex $A$. Lines passing through $O$ parallel to $MB$ and $MC$ intersect sides $AB$ and $AC$ at points $K$ and $L$, respectively. If the perpendicular from vertex $A$ ... | [
"First, we prove that $KB = LC$.\n\n\nLet $F$ and $E$ be the feet of perpendiculars from $K$ and $L$ to $MB$ and $MC$, respectively.\nWe have\n$$\nOK = MB \\Rightarrow KF = \\text{The distance from } O \\text{ to } MB.\n$$\n$$\nOL = MC \\Rightarrow LE = \\text{The distance from } O \\text{ ... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
0kvt | Problem:
There are 17 people at a party, and each has a reputation that is either $1, 2, 3, 4$, or $5$. Some of them split into pairs under the condition that within each pair, the two people's reputations differ by at most $1$. Compute the largest value of $k$ such that no matter what the reputations of these people ... | [
"Solution:\n\nFirst, note that $k=8$ fails when there are $15, 0, 1, 0, 1$ people of reputation $1, 2, 3, 4, 5$, respectively. This is because the two people with reputation $3$ and $5$ cannot pair with anyone, and there can only be at maximum $\\left\\lfloor\\frac{15}{2}\\right\\rfloor = 7$ pairs of people with re... | United States | HMMT November | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 7 | |
0clx | Problem:
Prove that there are infinitely many positive integer numbers $n$ such that $2^{2^{n}+1}+1$ is divisible by $n$, but $2^{n}+1$ is not. | [
"Solution:\nThroughout the solution $n$ stands for a positive integer. By Euler's theorem, $(2^{3^{n}}+1)(2^{3^{n}}-1)=2^{2 \\cdot 3^{n}}-1 \\equiv 0 \\pmod{3^{n+1}}$. Since $2^{3^{n}}-1 \\equiv 1 \\pmod{3}$, it follows that $2^{3^{n}}+1$ is divisible by $3^{n+1}$.\n\nThe number $(2^{3^{n+1}}+1)/(2^{3^{n}}+1)=2^{2 ... | Romanian Master of Mathematics (RMM) | Romanian Master of Mathematics Competition | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
01wg | Find all possible values of digits $a$ and $b$ such that
$$
(ab)^3 = \overline{(a-3)(b-3)(b+2)(a+2)ab}.
$$
(As usual, by $\overline{xyz}$ we denote an integer number, which decimal representation consists of digits $x, y, \dots, z$ in that order.) | [
"$a = 7, b = 6$."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a = 7, b = 6 | |
0fme | Un conjunto $S$ de enteros positivos se llama **canalero** si para cualesquiera tres números $a, b, c \in S$, todos diferentes, se cumple que $a$ divide $bc$, $b$ divide $ca$ y $c$ divide $ab$.
a. Demostrar que para cualquier conjunto finito de enteros positivos $\{c_1, c_2, \dots, c_n\}$ existen infinitos enteros pos... | [
"**Solución por Daniel Lasaosa Medarde, Pamplona, España.**\n\na.\nSea $M$ el mínimo común múltiplo de $c_1, c_2, \\dots, c_n$, y sea $k = k'M$, donde $k'$ toma cualquier valor entero positivo. Nótese que, para cualesquiera $u, v, w \\in \\{1, 2, \\dots, n\\}$ distintos, se tiene que\n$$\n\\frac{(kc_u)(kc_v)}{kc_w}... | Spain | Olimpiada Iberoamericana de Matemáticas | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | Spanish | proof only | null | |
051c | For any positive integer $n$ let $a_n$ be the largest power of $2$ that divides $n$ (e.g. $a_{2011} = 1$, $a_{2012} = 4$). Prove that for any positive integers $i$ and $j$ with $i < j$, the sum $\frac{1}{a_i} + \frac{1}{a_{i+1}} + \dots + \frac{1}{a_j}$ is a fractional number. | [
"First prove that the largest power of $2$ among the numbers $a_i$, $a_{i+1}$, $\\dots$, $a_j$ is unique. Let $2^s$ be the largest of the numbers $a_i$, $a_{i+1}$, $\\dots$, $a_j$. If there were $k$ and $l$ with $i \\le k < l \\le j$ such that $a_k = a_l = 2^s$, then they must be of the form $k = 2^s u$ and $l = 2^... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
02oj | Consider a regular $n$-gon inscribed in the unit circle. Compute the sum of the areas of all triangles determined by the vertices of the $n$-gon. | [
"First consider a triangle $ABC$ and its circumcenter $O$. Then the area of $ABC$ is $\\frac{R^2}{2}(\\sin 2\\angle A + \\sin 2\\angle B + \\sin 2\\angle C)$. Notice that if $\\angle B > 90^\\circ$ then $\\sin 2\\angle B < 0$.\n\n\n\nSo the sum is equal to the sum of the areas of triangles ... | Brazil | Brazilian Math Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | n^2/4 * cot(pi/n) | |
00og | Initially, the numbers $1, 2, \ldots, 2024$ are written on a blackboard. Trixi and Nana play a game, taking alternate turns. Trixi plays first.
The player whose turn it is chooses two numbers $a$ and $b$, erases both, and writes their (possibly negative) difference $a - b$ on the blackboard. This is repeated until onl... | [
"We will prove that Nana has a winning strategy.\n\nThe only relevant property of all numbers in the game is their residue modulo $3$. Therefore, we will call all numbers $0$, $1$ or $2$ according to their residue, and we will also call $1$s and $2$s non-zeros.\n\nWe observe that each move either does not change th... | Austria | Austrian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Nana | |
01e5 | Grandfather has a finite number of empty dustbins in his attic. Each dustbin is a rectangular parallelepiped with integral side lengths. A dustbin can be thrown away into another iff the side lengths of these dustbins can be set to one-to-one correspondence in such a way that the side lengths of the first dustbin are l... | [
"Suppose grandfather has 6 dustbins with sizes $20 \\times 20 \\times 20$, $19 \\times 19 \\times 19$, $16 \\times 16 \\times 16$, $21 \\times 18 \\times 15$, $18 \\times 15 \\times 12$ and $17 \\times 14 \\times 11$. The first dustbin can contain the second one, the second can contain the third or the fifth, the f... | Baltic Way | Baltic Way shortlist | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | No | |
0i8d | Problem:
A grasshopper lives on a coordinate line. It starts off at $1$. It can jump either $1$ unit or $5$ units either to the right or to the left. However, the coordinate line has holes at all points with coordinates divisible by $4$ (e.g. there are holes at $-4, 0, 4, 8$ etc.), so the grasshopper can not jump to an... | [
"Solution:\nEach jump changes the parity of grasshopper's coordinate. After $2003$ jumps the grasshopper will be at an even point on the coordinate line, and therefore can not be at $3$."
] | United States | Berkeley Math Circle Monthly Contest 1 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
0g69 | 已知銳角 $\triangle ABC$ 不是等腰三角形, 點 $O$ 與 $I$ 分別為 $\triangle ABC$ 之外心與內心。$\triangle ABC$ 的內切圓分別與三邊 $BC$, $CA$, $AB$ 相切於點 $D$, $E$, $F$。若直線 $AI$ 與 $OD$ 相交於 $P$ 點, $BI$ 與 $OE$ 相交於 $Q$ 點, $CI$ 與 $OF$ 相交於 $R$ 點, 且 $M$ 為 $\triangle PQR$ 的外心。試證: $I$, $M$, $O$ 三點共線。 | [
"(i) 令 $R$, $r$ 分別為 $\\triangle ABC$ 的外接圓與內切圓半徑。先證明 $OP : PD = R : r$。\n\n證明如下。延長 $AP$ 交 $\\triangle ABC$ 外接圓於 $A'$。因為 $AI$ 平分 $\\angle BAC$，故 $A'$ 為弧 $BA'C$ 的中點，從而 $OA'$ 與 $BC$ 垂直。又 $BC$ 與內切圓相切於 $D$，故 $ID$ 垂直於 $BC$，因此 $ID$ 平行於 $OA'$。故 $\\triangle IPD \\sim \\triangle A'PO$，因此 $OP : PD = OA' : ID = R : r$。得證。\n\n接下... | Taiwan | 二〇一二數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0idp | Problem:
We have a polyhedron such that an ant can walk from one vertex to another, traveling only along edges, and traversing every edge exactly once. What is the smallest possible total number of vertices, edges, and faces of this polyhedron? | [
"Solution:\nThis is obtainable by construction. Consider two tetrahedrons glued along a face; this gives us 5 vertices, 9 edges, and 6 faces, for a total of 20, and one readily checks that the required Eulerian path exists.\n\nNow, to see that we cannot do better, first notice that the number $v$ of vertices is at ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 20 | |
03jp | Problem:
The integers $1, 2, \ldots, n$ are placed in order so that each value is either strictly bigger than all the preceding values or is strictly smaller than all preceding values. In how many ways can this be done? | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 2^{n-1} | |
0d7k | Find the number of permutations $\left(a_{1}, a_{2}, \ldots, a_{2016}\right)$ of the first $2016$ positive integers satisfying the following two conditions:
1. $a_{i+1}-a_{i} \leq 1$ for all $i=1,2,3, \ldots, 2015$.
2. There are exactly two indices $i<j$ with $1 \leq i<j \leq 2016$ such that $a_{i}=i$ and $a_{j}=j$. | [
"For each positive integer $n \\geq 1$, we denote $s_{n}$ as the number of permutations of the first $n$ positive integers that satisfy the condition\n$$\na_{i+1}-a_{i} \\leq 1, \\quad i=1,2,\\ldots,n-1.\n$$\nWe call these permutations \"nice\".\n\nFirst, we shall prove that $s_{n}=2^{n-1}$ for all $n \\geq 1$.\nLe... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 2^{2012} | |
0cbx | A triple of integers $(a, b, c)$ is called *artistic* if the number $\frac{ab+bc+ca}{a+b+c}$ is also an integer.
a) Determine the integers $n$ for which the triples $(n, n+1, n+3)$ are artistic.
b) If $(x, y, z)$ is an artistic triplet, prove that $\frac{x^4+y^4+z^4}{x+y+z}$ is an integer. | [
"a) The triple $(n, n+1, n+3)$, $n \\in \\mathbb{Z}$, is artistic iff $\\frac{3n^2+8n+3}{3n+4} = n + \\frac{4n+3}{3n+4}$ is an integer. It follows that $3n+4$ divides $4(3n+4)-3(4n+3) = 7$, whence we obtain $n \\in \\{-1, 1\\}$, therefore the solutions are $(-1, 0, 2)$ and $(1, 2, 4)$.\n\nb) Since $x + y + z$ and $... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) n = -1 or n = 1, yielding the triples (-1, 0, 2) and (1, 2, 4).
b) The quantity (x^4 + y^4 + z^4) / (x + y + z) is an integer for any artistic triple (x, y, z). | |
03up | (1) Can one divide the set $\{1, 2, \ldots, 96\}$ into $32$ subsets, each containing three elements, and the sums of the three elements in each subset are all equal?
(2) Can one divide the set $\{1, 2, \ldots, 99\}$ into $33$ subsets, each containing three elements, and the sums of the three elements in each subset ar... | [
"(1) No. As\n$$\n1+2+\\cdots+96=\\frac{96\\times(96+1)}{2}=48\\times97,\n$$\nand $32 \\nmid 48 \\times 97$.\n\n(2) Yes. The sum of the three elements in each set is\n$$\n\\frac{1+2+\\cdots+99}{33} = \\frac{99 \\times (99+1)}{33 \\times 2} = 150.\n$$\n\nWe can divide $1, 2, 3, \\ldots, 66$ into $33$ pairs, such that... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | (1) No. (2) Yes. | |
09bv | a_n = \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} C_{n-k}^k \cdot 6^k$ гэе. Хэрэв $p > 5$ анхны тоо бол $p \mid a_{p-2}$ гэж батал. | [
"$a_n = \\sum_{k=0}^{\\lfloor \\frac{n}{2} \\rfloor} C_{n-k}^k \\cdot 6^k$ гэе. Тэгвэл\n$$\na_n = 1 + \\sum_{k=1}^{\\lfloor \\frac{n}{2} \\rfloor} (C_{n-k-1}^k + C_{n-k-1}^{k-1}) \\cdot 6^k = \\sum_{k=0}^{\\lfloor \\frac{n}{2} \\rfloor} C_{n-k-1}^k \\cdot 6^k + \\sum_{k=1}^{\\lfloor \\frac{n}{2} \\rfloor} C_{n-k-1}... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | Mongolian | proof only | null | |
0ie6 | Problem:
On an infinite checkerboard, the union of any two distinct unit squares is called a (disconnected) domino. A domino is said to be of type $(a, b)$, with $a \leq b$ integers not both zero, if the centers of the two squares are separated by a distance of $a$ in one orthogonal direction and $b$ in the other. (Fo... | [
"Solution:\n\nWe must have $0 \\leq a < m$, $0 \\leq b < n$, $a \\leq b$, and $a$ and $b$ not both $0$. The number of pairs $(a, b)$ with $b < a < m$ is $m(m-1)/2$, so the answer is\n$$\nm n - \\frac{m(m-1)}{2} - 1 = m n - \\frac{m^2 - m + 2}{2}.\n$$"
] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | mn - m(m-1)/2 - 1 | |
0bjt | Given a triangle $A_0A_1A_2$, determine the locus of the centers of the equilateral triangles $X_0X_1X_2$ satisfying the condition that each of the lines $X_kX_{k+1}$ passes through $A_k$ (all indices are reduced modulo 3). | [
"From any point $X_0$ on $\\gamma_0$ draw lines $X_0A_2X_1$ and $X_0A_1X_2$, where $X_1$ lies on $\\gamma_1$ and $X_2$ lies on $\\gamma_2$. The points $X_1, A_0, X_2$ are collinear, and the triangle $X_0X_1X_2$ is an equilateral triangle satisfying the conditions in the statement.\n\n\n\nLe... | Romania | 65th NMO Selection Tests for BMO and IMO | [
"Geometry > Plane Geometry > Advanced Configurations > Napoleon and Fermat points",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | A pair of circles: the circumcircles through the three centers of the inner and the outer Napoleon triangles of the given triangle. | |
0hay | Find positive integers $a_1, a_2, \dots, a_{2019}$, which satisfy the equation
$$
a_1 + a_2 + \dots + a_{2019} = a_1 a_2 \dots a_{2019} = \sqrt[2018]{2019^{2019}}.
$$ | [
"From the Arithmetic mean - Geometric mean Inequality,\n$$\n\\frac{1}{2019}(a_1 + a_2 + \\dots + a_{2019}) \\ge \\sqrt[2019]{a_1 a_2 \\dots a_{2019}}, \\text{ or} \\\\\n(a_1 + a_2 + \\dots + a_{2019})^{2019} \\ge 2019^{2019} a_1 a_2 \\dots a_{2019}.\n$$\nFrom problem statement,\n$$\n(a_1 + a_2 + \\dots + a_{2019})^... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | No positive integers exist | |
0ld0 | a) Find all positive integers $k$ with the property $T(20)$, where a positive integer $k$ has the property $T(m)$ if for any positive integer $a$, there exists a positive integer $n$ such that
$$
1^k + 2^k + \dots + n^k \equiv a \pmod{m}.
$$
b) Find the smallest positive integer $k$ with the property $T(20^{15})$. | [
"Let $s_k(n) = 1^k + 2^k + \\dots + n^k$. We rewrite the property $T(m)$ as follows: a positive integer $k$ has the property $T(m)$ if $s_k(n)$ covers the complete residue system modulo $m$ when $n$ is a positive integer.\n\na.\nWe note that if $k > 1$ has the property $T(20)$ then so does $k + 4$. This fact follow... | Vietnam | IMO 2015 Team Selection Tests | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | a) All positive integers divisible by 4. b) 4. | |
0j57 | Problem:
Let $A=\{1,2, \ldots, 2011\}$. Find the number of functions $f$ from $A$ to $A$ that satisfy $f(n) \leq n$ for all $n$ in $A$ and attain exactly 2010 distinct values. | [
"Solution:\nAnswer: $2^{2011}-2012$\n\nLet $n$ be the element of $A$ not in the range of $f$. Let $m$ be the element of $A$ that is hit twice.\n\nWe now sum the total number of functions over $n, m$. Clearly $f(1)=1$, and by induction, for $x \\leq m$, $f(x)=x$. Also unless $n=2011$, $f(2011)=2011$ because $f$ can ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2^{2011}-2012 | |
0bbe | Find the sum of the elements of the set
$$
M = \left\{ \frac{n}{2} + \frac{m}{5} \mid m, n = 0, 1, 2, \dots, 100 \right\}.
$$ | [
"Consider the set $A = \\{2a + 5b \\mid a, b = 1, 2, \\dots, 100\\}$ and notice that $1 \\notin A$ and $3 \\notin A$.\n\nThe largest even number from $A$ is equal to $700$ and it is obtained for $a = b = 100$. The number $698$ is obtained for $a = 99, b = 100$. The largest odd number from $A$ is equal to $695$ and ... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 24395 | |
08z5 | A circle with radius $3$ is inscribed in a trapezoid $ABCD$ such that $AD \parallel BC$ and both angles $B$ and $C$ are acute. If $AB = 7$ and $CD = 8$, find the area of trapezoid $ABCD$.
 | [
"Let $E$, $F$, $G$ and $H$ be the points of contact of side $AB$, $BC$, $CD$ and $DA$ respectively to the inscribed circle of trapezoid $ABCD$. Since the lengths of tangents from $A$ to this inscribed circle are equal, $AE = AH$ holds. Similarly we have $BE = BF$, $CF = CG$ and $DG = DH$. Then we have $AD + BC = AH... | Japan | Japan 2022 | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 45 | |
0810 | Problem:
Tre ragazzi dicono, rispettivamente: "I nostri nomi sono Andrea, Bruno, Carlo"; "I nostri nomi sono Andrea, Carlo, Daniele"; "I nostri nomi sono Bruno, Daniele, Enrico".
Sapendo che ciascuno di loro ha detto un nome sbagliato e due giusti, come si chiamano i tre?
(A) Andrea, Carlo, Enrico
(B) Andrea, Bruno, ... | [] | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Logic"
] | null | MCQ | E | |
0ib8 | Problem:
A convex quadrilateral is drawn in the coordinate plane such that each of its vertices $(x, y)$ satisfies the equations $x^{2}+y^{2}=73$ and $x y=24$. What is the area of this quadrilateral? | [
"Solution:\nThe vertices all satisfy $(x+y)^{2}=x^{2}+y^{2}+2 x y=73+2 \\cdot 24=121$, so $x+y= \\pm 11$. Similarly, $(x-y)^{2}=x^{2}+y^{2}-2 x y=73-2 \\cdot 24=25$, so $x-y= \\pm 5$. Thus, there are four solutions: $(x, y)=(8,3),(3,8),(-3,-8),(-8,-3)$. All four of these solutions satisfy the original equations. Th... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | final answer only | 110 | |
05j3 | Problem:
Soit $n \geqslant 5$ un entier. Soient $a_{1}, \ldots, a_{n}$ des entiers dans $\{1, \ldots, 2n\}$ deux à deux distincts. Montrer qu'il existe des indices $i, j \in \{1, \ldots, n\}$ avec $i \neq j$ tels que
$$
\operatorname{PPCM}\left(a_{i}, a_{j}\right) \leqslant 6\left(E\left(\frac{n}{2}\right)+1\right)
$$
... | [
"Solution:\nSupposons, pour commencer, qu'il existe un indice $i$ tel que $a_{i} \\leqslant n$. S'il existe $j$ tel que $a_{j}=2 a_{i}$, on a alors\n$$\n\\operatorname{PPCM}\\left(a_{i}, a_{j}\\right)=a_{j} \\leqslant 2 n \\leqslant 6 \\cdot\\left(E\\left(\\frac{n}{2}\\right)+1\\right)\n$$\net on a trouvé un couple... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof only | null | |
0i49 | Problem:
A certain cafeteria serves ham and cheese sandwiches, ham and tomato sandwiches, and tomato and cheese sandwiches. It is common for one meal to include multiple types of sandwiches. On a certain day, it was found that 80 customers had meals which contained both ham and cheese; 90 had meals containing both ham ... | [
"Solution: 230. Everyone who ate just one sandwich is included in exactly one of the first three counts, while everyone who ate more than one sandwich is included in all four counts. Thus, to count each customer exactly once, we must add the first three figures and subtract the fourth twice: $80+90+100-2 \\cdot 20=... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 230 | |
09wd | A set $S$ consisting of $2019$ (distinct) positive integers has the following property: the product of any $100$ elements of $S$ is a divisor of the product of the other $1919$ elements. What is the maximum number of prime numbers that $S$ could contain? | [
"We start with the construction. Choose distinct primes $p_1, p_2, \\dots, p_{1819}$, and let $P = p_1p_2\\cdots p_{1819}$. Let\n$$\nS = \\{p_1, p_2, \\dots, p_{1819}, P, P \\cdot p_1, \\dots, P \\cdot p_{199}\\}.\n$$\nFor each $p_i$, there are $201$ numbers in $S$ that are divisible by $p_i$ (namely, $p_i$ and all... | Netherlands | BxMO Team Selection Test, March 2020 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1819 | |
0ik2 | Let $n$ be a given integer with $n$ greater than $7$, and let $\mathcal{P}$ be a convex polygon with $n$ sides. Any set of $n-3$ diagonals of $\mathcal{P}$ that do not intersect in the interior of the polygon determine a triangulation of $\mathcal{P}$ into $n-2$ triangles. A triangle in the triangulation of $\mathcal{P... | [
"The answer is\n$$\nn2^{n-9} \\binom{n-4}{4}\n$$\nDenote the vertices of $\\mathcal{P}$ counter-clockwise by $A_0, A_1, \\dots, A_{n-1}$. We will count first the number of triangulations of $\\mathcal{P}$ with two interior triangles positioned as in the following figure. We say that such a triangulation starts at $... | United States | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | n * 2^{n-9} * binom(n-4, 4) | |
0hcd | For which positive integers $n \ge 2$, there exist $n$ odd (not necessarily different) numbers $a_1, a_2, \dots, a_n$ such that
$a_1^2 + a_2^2 + \dots + a_n^2$ is a square of some positive integer? | [
"Clearly, square of an integer number can give a remainder of $0$, $1$ or $4$ modulo $8$. Therefore, only for $n$ of the form $8k + r$, where $r \\in \\{0, 1, 4\\}$, such numbers can exist. Let us show how they can be constructed.\n\n$$\nn = 4t,\\ a_1 = \\dots = a_{n-1} = 1,\\ a_n = (2t-1):\n$$\n$$\na_1^2 + a_2^2 +... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof and answer | Exactly those n with n ≡ 0, 1, or 4 (mod 8). | |
0hvo | Problem:
Let $R$ be the region consisting of all points inside or on the boundary of a given circle of radius $1$. Find, with proof, all positive real numbers $d$ such that it is possible to color each point of $R$ red, green or blue such that any two points of the same color are separated by a distance less than $d$. | [
"Solution:\n\nThe answer is all $d \\geq \\sqrt{3}$.\n\nIf $d \\geq \\sqrt{3}$, refer to the diagram at right. Color $120^{\\circ}$ sectors $OAB$, $OBC$, and $OCA$ red, green, and blue respectively, including their boundaries on the circle. For the boundaries between the sectors, color $OA$ red, $OB$ green, and $OC... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | d >= sqrt(3) | |
0560 | Every sound in a certain language can be either long or short. A sound is classified either as a vowel or as a consonant. Every word consists of exactly two sounds (without repetitions) and satisfies the following conditions.
1) Every word contains a short sound.
2) Words beginning with a vowel contain a long sound.
3... | [
"a) If the word consists of two vowels, then based on the rule 3) the second of them is short and based on the rule 2) the first of them is long. If the word begins with a vowel and ends with a consonant, then the length of the second sound is determined uniquely by its writing and the length of first sound must be... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Logic"
] | English | proof and answer | a) Yes; b) No | |
06wi | Consider a checkered $3m \times 3m$ square, where $m$ is an integer greater than $1$. A frog sits on the lower left corner cell $S$ and wants to get to the upper right corner cell $F$. The frog can hop from any cell to either the next cell to the right or the next cell upwards.
Some cells can be sticky, and the frog g... | [
"a.\nIn the following example the square is divided into $m$ stripes of size $3 \\times 3m$. It is easy to see that $X$ is a minimal blocking set. The first and the last stripe each contains $3m-1$ cells from the set $X$; every other stripe contains $3m-2$ cells, see Figure 1. The total number of cells in the set $... | IMO | IMO 2021 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
048n | Two players play the following game: after one of them tells number $n$, the other has to tell a number of the form $a \cdot b$ where $a, b$ are positive integers such that $a + b = n$. The game continues in the same way. If at some point one of the players has told $2011$, which are the possible numbers that the game ... | [] | Croatia | CroatianCompetitions2011 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | All integers greater than or equal to 5 | |
03hy | Problem:
Let $O$ be the centre of a circle and $A$ a fixed interior point of the circle different from $O$. Determine all points $P$ on the circumference of the circle such that the angle $OPA$ is a maximum.
 | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | The two points where the triangle with vertices at the center, the interior point, and the point on the circle is right-angled at the interior point; equivalently, the intersections of the circle with the line through the interior point perpendicular to the line from the center to that point. | |
0g7j | 給定圓內接四邊形 $ABCD$。直線 $L$ 為過外心 $O$ 的一直線，$P$ 為 $L$ 上一動點。設圓 $c_1$ 是通過 $P$ 在 $AB$, $BC$, $CA$ 三個垂足的圓；圓 $c_2$ 是通過 $P$ 在 $AB$, $BD$, $DA$ 三個垂足的圓。設圓 $c_1, c_2$ 交於 $P_1, Q$ 兩點，其中 $P_1$ 為 $P$ 在 $AB$ 的垂足。試問當 $P$ 點在 $L$ 上移動時，$Q$ 點的軌跡為何？ | [
"如圖，設 $P_1, P_2, P_3, P_4, P_5$ 分別代表 $P$ 點在 $AB$, $BC$, $AC$, $BD$, $AD$ 線段上的垂足。\n\n設 $M_1, M_2, M_3, M_4, M_5$ 分別是 $AB$, $BC$, $AC$, $BD$, $AD$ 各邊上的中點。考慮 $\\triangle ABC$。顯然外心 $O$ 是 $\\triangle M_1M_2M_3$ 的垂心。由斯坦納定理，直線 $L$ 對 $M_1M_2$, $M_2M_3$, $M_3M_1$ 做反射後必會交於 $\\triangle ABC$ 九點圓上一點 $K_... | Taiwan | 二0一三數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordin... | null | proof and answer | The locus is the fixed straight line K1K2. | |
0hit | Problem:
Consider an $8 \times 8$ chessboard, on which we place some bishops in the 64 squares. Two bishops are said to attack each other if they lie on a common diagonal.
a. Prove that we can place 14 bishops in such a way that no two attack each other.
b. Prove that we cannot do so with 15 bishops. | [
"Solution:\n\nFor the first part, here is one maximal arrangement, where the location of the bishops are indicated by the letter $B$.\n\n| $B$ | | | | | | | |\n| :---: | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| $B$ | | | | | | | $B$ |\n| $B$ | | | | | | | $B$ |\n| $B... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 14 | |
0eqp | One horse eats $40\%$ of a bale of hay and another horse eats $P\%$ of what is left. If both horses ate the same amount, the value of $P$ is
(A) $43$
(B) $66\frac{2}{3}$
(C) $50$
(D) $75$
(E) $80$ | [
"The second horse eats $P\\%$ of $60\\%$ and this is the same as $40\\%$: so $P\\%$ is $\\frac{2}{3}$, i.e. $P = 66\\frac{2}{3}$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | B | |
0c56 | Let $m$ be a positive integer. Find the number of the real solutions of the equation
$$
\left| \sum_{k=0}^{m} \binom{2m}{2k} x^k \right| = |x - 1|^m.
$$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | m | |
0905 | Let $ABC$ be an acute triangle with $AB < AC$, $O$ be its circumcenter, and $M$ be the midpoint of arc $BC$ of triangle $ABC$'s circumcircle which does not include $A$. There is a point $D$ on the extension of side $AB$ beyond $B$ satisfying $BD = BM$, and there is a point $E$ on side $AC$ (except for end points) satis... | [
"For distinct three points $P$, $Q$, $R$, the description $\\angle QPR = \\theta$ means that line $PQ$ rotated around $P$ by angle $\\theta$ counterclockwise coincides with line $PR$. Here $180^\\circ$ difference is ignored.\nBy the inscribed angle theorem, we have $\\angle XBD = \\angle XEC$, $\\angle XDB = \\angl... | Japan | Japan 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 > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
01lp | Let $A$ be the sum of all 10 pairwise products of the sides of a convex pentagon, $S$ be the area of the pentagon.
a) Prove that $S \le \frac{1}{5}A$.
b) Does there exist a constant $c < 1/5$ such that $S \le cA$?
(I. Voronovich) | [
"Answer: b) yes.\n\nWe use the following well-known\n\n**Lemma**. Let $a, b, c, d$ be the lengths of the sides of some quadrilateral, and $S$ be its area. Then $2S \\le ab + cd$ and $2S \\le ac + bd$.\n\nLet now $a, b, c, d, e$ be the lengths of the sides of the given pentagon, $f$ be the length of one of its diago... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof and answer | yes | |
00a2 | Call a natural number *acceptable* if it has at most 9 distinct prime divisors. There is given a pile of $100! = 1 \cdot 2 \cdot \dots \cdot 100$ stones. A legal move is to remove $k$ stones from the pile where $k$ is an *acceptable number*. Players $A$ and $B$ take turns in making legal moves; $A$ goes first. The one ... | [
"Let $P = 2 \\cdot 3 \\cdot 5 \\cdot \\dots \\cdot 29$ be the product of the first 10 primes $2, 3, 5, 7, 11, 13, 17, 19, 23, 29$. Observe that $P$ is the smallest unacceptable number. Apparently $P$ divides $100!$, and acceptable numbers are not divisible by $P$.\nLet $A$ remove $k_1$ stones on his first move. Bec... | Argentina | Argentine National Olympiad 2015 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | B | |
0kb4 | Let $\triangle ABC$ be an acute triangle with orthocenter $H$ and circumcircle $\Gamma$. A line through $H$ intersects segments $AB$ and $AC$ at $E$ and $F$, respectively. Let $K$ be the circumcenter of $\triangle AEF$, and suppose line $AK$ intersects $\Gamma$ again at a point $D$. Prove that line $HK$ and the line th... | [
"We present several solutions.\n\n**First solution (Andrew Gu)** We begin with the following two observations.\n\n**Claim** — Point $K$ lies on the radical axis of $(BEH)$ and $(CFH)$.\n*Proof.* Actually we claim $\\overline{KE}$ and $\\overline{KF}$ are tangents. Indeed,\n$$\n\\angle HEK = 90^\\circ - \\angle EAF ... | United States | USA TSTST | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Transformations > Spiral sim... | null | proof only | null | |
0bwf | Consider a 7-point configuration consisting of the vertices of a quadrangle (not necessarily convex) along with three other points lying in the interior or on the boundary of the quadrangle. Every pair of distinct points in the configuration are at least $1$ distance apart. Show that the diameter of the quadrangle is g... | [
"Let $C$ denote the $7$-point configuration and let $[C]$ denote its convex hull. The latter is either a triangle formed by three vertices of the quadrangle or the quadrangle itself. Since the diameter of $[C]$ is the longest distance determined by some pair of vertices, it is sufficient to show that this diameter ... | Romania | Eleventh STARS OF MATHEMATICS Competition | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
00lk | Gegeben sind die nichtnegativen reellen Zahlen $a$ und $b$ mit $a + b = 1$. Man beweise:
$$
\frac{1}{2} \le \frac{a^3 + b^3}{a^2 + b^2} \le 1
$$
Wann gilt Gleichheit in der linken Ungleichung, wann in der rechten? | [
"Durch Umformen der Angabe erhalten wir\n$$\n\\frac{a^3 + b^3}{a^2 + b^2} = (a + b)\\frac{a^2 - ab + b^2}{a^2 + b^2} = 1 - \\frac{ab}{a^2 + b^2}\n$$\nDaraus sieht man sofort die rechte Ungleichung mit Gleichheit für $ab = 0$, also $a = 0$, $b = 1$ und für $a = 1, b = 0$.\n\nDie linke Ungleichung ist äquivalent zu\n... | Austria | 48. Österreichische Mathematik-Olympiade Landeswettbewerb für Anfängerinnen und Anfänger | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | German | proof and answer | Left equality holds when a = b = 1/2. Right equality holds when {a, b} = {0, 1}. | |
0ksh | Problem:
If $f(n, k)$ is the number of ways to divide the set $\{1,2, \ldots, n\}$ into $k$ nonempty subsets and $m$ is a positive integer, find a formula for
$$
\sum_{k=1}^{n} f(n, k)\, m(m-1)(m-2) \cdots (m-k+1).
$$ | [
"Solution:\n\nWe claim that the sum is equal to $m^{n}$.\n\nWe note that $m^{n}$ counts the number of ways to color $n$ objects each with one of $m$ different colors, so it suffices to show that the left side counts the same thing.\n\nWe can consider cases based on how many different colors get used. If $k$ colors ... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | m^n | |
09f1 | Let $I_{\sigma} = \{ |\sigma_i - i| : i \in I \}$ be sets formed by every permutation $\sigma = (\sigma_1, \sigma_2, \dots, \sigma_{2014})$ of the set $I = \{1, 2, \dots, 2014\}$. Find all possible values of $|I_{\sigma}|$. | [
"If $\\sigma = \\{2014, 2013, \\dots, 1008, 1, 1007, 1006, \\dots, 2\\}$ then $|I_\\sigma| = 2013$.\nIf $\\sigma = \\{2013, 2012, \\dots, 1008, 1007, 1, 1006, 1005, \\dots, 2, 2014\\}$ then $|I_\\sigma| = 2012$.\nIf $\\sigma = \\{2k, 2k-1, \\dots, k+1, 1, k, k-1, \\dots, 2, 2k+1, 2k+2, \\dots, 2014\\}$ then $|I_\\s... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof and answer | All integers from 1 to 2013 | |
0kit | Problem:
Suppose $a$ and $b$ are positive integers for which $8 a^{a} b^{b} = 27 a^{b} b^{a}$. Find $a^{2} + b^{2}$. | [
"Solution:\nWe have\n$$\n8 a^{a} b^{b} = 27 a^{b} b^{a} \\Longleftrightarrow \\frac{a^{a} b^{b}}{a^{b} b^{a}} = \\frac{27}{8} \\Longleftrightarrow \\frac{a^{a-b}}{b^{a-b}} = \\frac{27}{8} \\Longleftrightarrow \\left(\\frac{a}{b}\\right)^{a-b} = \\frac{27}{8}.\n$$\nSince $27 = 3^{3}$ and $8 = 2^{3}$, there are only ... | United States | HMMT November 2021 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 117 | |
09zg | At a fish market there are 10 stalls, each selling the same 10 kinds of fish. Each fish was caught in either the North Sea or the Mediterranean Sea, and each stall has, for each kind of fish, only fish of one origin. A number, say $k$, of customers buy exactly one fish from each stall, in such a way that they obtain ex... | [
"The largest possible value of $k$ is $2^{10} - 10$. First note that there are $2^{10}$ possible combinations for the origins per kind of fish. We show that there are always at least 10 exceptions (combinations that cannot be obtained by a customer), and that there is a way to supply the stalls for which there are ... | Netherlands | BxMO Team Selection Test | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | 2^10 - 10 | |
0cnb | In Leonardland, each road has one-way movement, connects two cities and does not pass through another city. The Statistics Department calculated for each city $A$ the total number $f(A)$ of citizens in the cities, to which the roads from $A$ lead, and the total number $g(A)$ of citizens in the cities, from which the ro... | [
"Первое решение. Построим граф, вершины которого соответствуют жителям страны, причем две вершины соединены направлённым ребром в том и только том случае, когда их города соединены дорогой (направление на ребре будет такое же, как и на дороге между городами). Для каждой вершины $v$ обозначим через $f(v)$ разность к... | Russia | Euler olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English; Russian | proof only | null | |
04ip | A field of the shape of a circular sector needs to be fenced using a wire of length $d$. What is the maximal area of that field? (Ilko Brnetić) | [] | Croatia | Croatia Mathematical Competitions | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | d^2/16 | |
033b | Problem:
Prove that if $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n} \geq 0$ and $c_{k}=\prod_{i=1}^{k} b_{i}^{\frac{1}{k}}$, $1 \leq k \leq n$, then
$$
n c_{n}+\sum_{k=1}^{n} k\left(a_{k}-1\right) c_{k} \leq \sum_{k=1}^{n} a_{k}^{k} b_{k}
$$ | [
"Solution:\nThe Arithmetic mean - Geometric mean inequality (for any $k=2,3, \\ldots, n$) implies that\n$$\nk a_{k} c_{k}=k a_{k} b_{k}^{\\frac{1}{k}} \\underbrace{c_{k-1}^{\\frac{1}{k}} \\ldots c_{k-1}^{\\frac{1}{k}}}_{k-1 \\text{ times }} \\leq a_{k}^{k} b_{k}+(k-1) c_{k-1}\n$$\nSumming up these inequalities and ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0kek | Problem:
The classrooms at MIT are each identified with a positive integer (with no leading zeroes). One day, as President Reif walks down the Infinite Corridor, he notices that a digit zero on a room sign has fallen off. Let $N$ be the original number of the room, and let $M$ be the room number as shown on the sign.
... | [
"Solution:\n\nLet $A$ represent the portion of $N$ to the right of the deleted zero, and $B$ represent the rest of $N$. For example, if the unique zero in $N=12034$ is removed, then $A=34$ and $B=12000$. Then, $\\frac{M}{N}=\\frac{A+B / 10}{A+B}=1-\\frac{9}{10} \\frac{B}{N}$.\n\nThe maximum value for $B / N$ is 1, ... | United States | HMMO 2020 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 2031 | |
0da4 | Let $C$ be a point lies outside the circle $(O)$ and $CS, CT$ are tangent lines of $(O)$. Take two points $A, B$ on $(O)$ with $M$ is the midpoint of the minor $\operatorname{arc} AB$ such that $A, B, M$ differ from $S, T$. Suppose that $MS, MT$ cut line $AB$ at $E, F$. Take $X \in OS$ and $Y \in OT$ such that $EX, FY$... | [
"First, note that $OM \\perp AB$ then $OM \\parallel XE$. But $OMS$ is isosceles triangle implies that triangle $XES$ is also isosceles, or $XE = XS$. Similarly, $YE = YT$. Denote $\\left(\\omega_{1}\\right), \\left(\\omega_{2}\\right)$ as the circle of center $X$, radius $XS$ and center $Y$, radius $YT$. Since $CS... | Saudi Arabia | Team selection tests for GMO 2018 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0jry | Problem:
Let $z$ be a complex number such that $|z|=1$ and $|z-1.45|=1.05$. Compute the real part of $z$. | [
"Solution:\n\nFrom the problem, let $A$ denote the point $z$ on the unit circle, $B$ denote the point $1.45$ on the real axis, and $O$ the origin. Let $A H$ be the height of the triangle $O A H$ and $H$ lies on the segment $O B$. The real part of $z$ is $O H$. Now we have $O A = 1$, $O B = 1.45$, and $A B = 1.05$. ... | United States | HMMT February 2016 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 20/29 | |
069k | Determine all pairs $(\alpha, \beta)$ of prime numbers $\alpha, \beta$ for which the number $A = 3\alpha^2\beta + 16\alpha\beta^2$ is the square of an integer. | [
"We distinguish the cases:\n\n1. $\\alpha = \\beta$\nLet: $A = 3\\alpha^2\\alpha + 16\\alpha\\alpha^2 = 19\\alpha^3 = \\kappa^2, \\kappa \\in \\mathbb{Z}, \\alpha$ prime.\n\nThen $19|\\kappa^2 \\Rightarrow 19|\\kappa \\Rightarrow 19^2|\\kappa^2 \\Rightarrow \\kappa^2 = 19^2\\omega, \\omega \\in \\mathbb{Z}$, and he... | Greece | SELECTION EXAMINATION 2019 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (19, 19) and (2, 3) | |
01h9 | (a) Is there a positive integer $a$ such that $((a^2 - 2)^3 + 1)^a - 1$ is a perfect square?
(b) Is there a positive integer $a$ such that $((a^2 - 2)^3 + 1)^{a+1} - 1$ is a perfect square? | [
"Answer: (a) No; (b) No.\n\na. If $a$ is even then $((a^2 - 2)^3 + 1)^a$ is clearly a perfect square. An integer that differs from it by 1 cannot be a square of a positive integer.\nIf $a$ is odd then $a^2 \\equiv 1 \\pmod 4$, implying that $a^2 - 2 \\equiv -1 \\pmod 4$ and $(a^2 - 2)^3 \\equiv -1 \\pmod 4$. Then $... | Baltic Way | Baltic Way 2020 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (a) No; (b) No | |
0hf1 | Let $n$ be a positive integer. Equilateral triangle with the side length $n$ is divided into $n^2$ smaller equilateral triangles with the side length $1$ (Fig. 31 shows this division for $n=10$). Initially, one of these triangles is blue and the rest are yellow. The blue triangle is guaranteed not to have any common po... | [
"\nLet's consider one step and assume that some triangle $a$ and his neighbors changed color on this step. Pairs that include $a$ do not influence $X$, because if they had the same color - they will stay the same, and if they had different colors, after the recoloring they will also have di... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No | |
0309 | Problem:
Em um tabuleiro $7 \times 7$, dizemos que 4 casas assinaladas com $X$ formam um castelo se elas são vértices de um retângulo com lados paralelos aos do tabuleiro, como indicado na figura a seguir:
a) Marque 21 casas com $X$ no tabuleiro $7 \times 7$ sem que exista qualquer castelo e... | [
"Solution:\n\na) Um exemplo é o que está na seguinte figura a seguir.\n\n| $X$ | $X$ | | $X$ | | | |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| | $X$ | $X$ | | $X$ | | |\n| | | $X$ | $X$ | | $X$ | |\n| | | | $X$ | $X$ | | $X$ |\n| $X$ | | | | $X$ | $X$ | |\n| | ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 21 | |
00zj | Problem:
Let $ABCD$ be a cyclic convex quadrilateral and let $r_{a}, r_{b}, r_{c}, r_{d}$ be the radii of the circles inscribed in the triangles $BCD$, $ACD$, $ABD$, $ABC$ respectively. Prove that $r_{a} + r_{c} = r_{b} + r_{d}$. | [
"Solution:\n\nFor a triangle $MNK$ with in-radius $r$ and circumradius $R$, the equality\n$$\n\\cos \\angle M + \\cos \\angle N + \\cos \\angle K = 1 + \\frac{r}{R}\n$$\nholds; this follows from the cosine theorem and formulas for $r$ and $R$.\n\nWe have $\\angle ACB = \\angle ADB$, $\\angle BDC = \\angle BAC$, $\\... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscella... | null | proof only | null | |
03db | Let $m$ and $n$ be positive integers, while $p$ is a prime number. Find the maximal $s \in \mathbb{N}$ (as a function of $m, n$, and $p$) such that from an arbitrary group of $mnp$ positive integers, one can only choose $snp$ among them, satisfying the following property: The numbers can be split into $s$ disjoint subs... | [
"$s = m - 1$. Assume $s = m$ and consider a set of $mnp - 1$ positive integers, congruent $1 \\pmod{p}$, and $p$. Clearly this set does not fulfill the statement, thus $s \\le m - 1$.\n\n**Lemma.** Among every $np + p - 1$ positive integers, there exist $np$ with sum, divisible by $p$.\n\n*Proof.* We apply inductio... | Bulgaria | Bulgaria 2022 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | m - 1 | |
0aua | Problem:
What is the remainder when
$$
16^{15} - 8^{15} - 4^{15} - 2^{15} - 1^{15}
$$
is divided by $96$? | [] | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | final answer only | 31 | |
07v1 | Let $\mathbb{N}$ denote the strictly positive integers. A function $f : \mathbb{N} \to \mathbb{N}$ has the following properties which hold for all $n \in \mathbb{N}$:
(a) $f(n) < f(n + 1)$;
(b) $f(f(f(n))) = 4n$.
Find $f(2022)$. | [
"**Solution 1.** We first prove inductively that the following equations hold for all $i \\in \\mathbb{N}$:\n$$\n\\begin{aligned}\nf(1 \\cdot 4^{i-1}) &= 2 \\cdot 4^{i-1}, \\\\\nf(2 \\cdot 4^{i-1}) &= 3 \\cdot 4^{i-1}, \\\\\nf(3 \\cdot 4^{i-1}) &= 4 \\cdot 4^{i-1} = 4^i.\n\\end{aligned}\n$$\nThe base case follows e... | Ireland | IRL_ABooklet | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 3046 | |
0j5y | Problem:
Let $a$, $b$, and $c$ be complex numbers such that $|a|=|b|=|c|=|a+b+c|=1$. If $|a-b|=|a-c|$ and $b \neq c$, evaluate $|a+b||a+c|$. | [
"Solution:\n\nSince $|a|=1$, $a$ cannot be $0$. Let $u=\\frac{b}{a}$ and $v=\\frac{c}{a}$. Dividing the given equations by $|a|=1$ gives $|u|=|v|=|1+u+v|=1$ and $|1-u|=|1-v|$. The goal is to prove that $|1+u||1+v|=2$.\n\nBy squaring $|1-u|=|1-v|$, we get $(1-u) \\overline{(1-u)}=(1-v) \\overline{(1-v)}$, and thus $... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | proof and answer | 2 |
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