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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0j2x | Points $P$ and $Q$ lie inside acute triangle $ABC$ such that $\angle PAB = \angle QAC$ and $\angle PBA = \angle QBC$. Point $D$ lies on side $BC$. Prove that
$$
\angle DPC + \angle APB = 180^{\circ} \quad \text{if and only if} \quad \angle DQB + \angle AQC = 180^{\circ}.
$$ | [
"\nSolution:\n\n**Solution 1.** Extend $BP$ through $P$ to $P_1$ so that $\\angle PAP_1 = \\angle PCD = C - z$. By (1), $\\angle DPC = 180^{\\circ} - \\angle APB = \\angle APP_1$. Hence triangles $PAP_1$ and $PCD$ are similar. In particular, there is a spiral similarity $S_1$ centered at $P... | United States | Team Selection Test 2010 | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > M... | null | proof only | null | |
0ir6 | Problem:
How many diagonals does a regular undecagon (11-sided polygon) have? | [
"Solution:\nThere are 8 diagonals coming from the first vertex, 8 more from the next, 7 from the next, 6 from the next, 5 from the next, etc., and 1 from the last, for $8+8+7+6+5+4+3+2+1=44$ total.\n\nThird method: Each vertex has 8 diagonals touching it. There are 11 vertices. Since each diagonal touches two verti... | United States | 1st Annual Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 44 | |
02b0 | Problem:
Conjunto de Cantor - Desenhe um segmento de reta de comprimento $1$, e denote-o por $C_{1}$. Remova o terço central (sem remover os extremos). Denote por $C_{2}$ o que sobrou. Agora, remova o terço central (sem os extremos) de cada segmento de reta de $C_{2}$. Denote por $C_{3}$ o que sobrou. Podemos continua... | [
"Solution:\n\na. De acordo com a definição do Conjunto de Cantor temos os seguintes desenhos:\n\n\n\nb. $1/3$ é um extremo de $C_{2}$, logo pertence ao conjunto de Cantor.\n\n$3/81 = 1/27$ e $1/27$ é um extremo de $C_{4}$, logo $3/81$ pertence ao conjunto de Cantor.\n\n$4/9$ está entre $1/3... | Brazil | Nível 3 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a) C1: [0, 1]. C2: [0, 1/3] and [2/3, 1]. C3: [0, 1/9], [2/9, 1/3], [2/3, 7/9], [8/9, 1].
b) 1/3 ∈ Cantor set; 3/81 = 1/27 ∈ Cantor set; 4/9 ∉ Cantor set; 4/81 ∉ Cantor set.
c) Lengths: |C3| = 4/9, |C4| = 8/27, |C5| = 16/81. In general, |Cn| = (2/3)^(n−1). | |
03hu | Problem:
Given four weights in geometric progression and an equal arm balance, show how to find the heaviest weight using the balance only twice. | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
01ac | Trapezoid $ABCD$ of bases $AB$ and $CD$ is such that the circumcircle of the triangle $BCD$ intersects line $AD$ in a point $E$ which is distinct from $A$ and $D$. Prove that the circumcircle of the triangle $ABE$ is tangent to the line $BC$. | [
"\nIf point $E$ lies on the segment $AD$ it is sufficient to prove $\\angle CBE = \\angle BAE$. It is true, since both angles are equal to $180^\\circ - \\angle ADC$. If point $D$ lies on the segment $AE$ we have $\\angle CBE = \\angle CDE = \\angle BAE$, which proves the thesis. In the end... | Baltic Way | Baltic Way 2013 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0f0e | Problem:
A $7$-gon is inscribed in a circle. The center of the circle lies inside the $7$-gon. $A$, $B$, $C$ are adjacent vertices of the $7$-gon. Show that the sum of the angles at $A$, $B$, $C$ is less than $450$ degrees. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0ke7 | Problem:
Nine fair coins are flipped independently and placed in the cells of a $3$ by $3$ square grid. Let $p$ be the probability that no row has all its coins showing heads and no column has all its coins showing tails. If $p=\frac{a}{b}$ for relatively prime positive integers $a$ and $b$, compute $100 a+b$. | [
"Solution:\n\nConsider the probability of the complement. It is impossible for some row to have all heads and some column to have tails, since every row intersects every column. Let $q$ be the probability that some row has all heads. By symmetry, $q$ is also the probability that some column has all tails. We can th... | United States | HMMO | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 8956 | |
07wn | Triangle $ABC$ is obtuse with $\angle ABC > 90^\circ$. Let $\ell$ be the external bisector of $\angle BCA$, and let $AB$ (extended) meet $\ell$ at $D$. The circumcircle $\Gamma$ of triangle $BCD$ has centre $O$. Let $E$ be the point where the line through $A$ perpendicular to $DO$ meets $\ell$. Prove that $BE$ is tange... | [
"Let $F$ be the second intersection point of $DO$ and $\\Gamma$ so that $DF$ is a diameter of $\\Gamma$ and $BDCF$ is a cyclic quadrilateral. Then $\\angle DCF = 90^\\circ$ and $CF$ is the internal bisector of $\\angle BCA$ as it is perpendicular to the external angle bisector $\\ell$.\nLet $G$ be the intersection ... | Ireland | IRL_ABooklet_2024 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hd9 | Do there exist pairwise distinct positive integers $a$, $b$ and $c$, such that $\left\{ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \right\} = 0$? The fractions are not necessarily irreducible.
Here $\{x\}$ denotes the difference between $x$ and the greatest integer that does not exceed $x$, for example, $\left\{ \frac{7}... | [
"A source for an example is the following equation: $\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{6} = 1$. Clearly, irreducible fractions would not work, thus we will choose $\\left\\{ \\frac{2}{12} \\right\\} = \\left\\{ \\frac{1}{6} \\right\\} = \\frac{1}{6}$. Then $\\left\\{ \\frac{12}{9} \\right\\} = \\left\\{ \\fra... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | Yes; for example a=2, b=12, c=9. | |
07zb | Problem:
Quale fra questi numeri non è il prodotto di un quadrato perfetto per un cubo perfetto?
(A) $900$
(B) $961$
(C) $968$
(D) $972$
(E) $980$ | [] | Italy | Progetto Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | E | |
0cnk | A convex quadrilateral $ABCD$ is given. It appears that on diagonal $AC$ there exists a point that lies on perpendicular bisectors of the segments $AB$ and $CD$, and on diagonal $BD$ there exists a point that lies on perpendicular bisectors of the segments $DA$ and $BC$.
Prove that $ABCD$ is a rectangle. (N. Agakhanov) | [
"Пусть $M$ — указанная точка на диагонали $AC$. Тогда, по свойству серединных перпендикуляров, $MA = MB$ и $MC = MD$. Поэтому $BD \\le BM + MD = AM + MC = AC$. Аналогично рассуждая для другой пары серединных перпендикуляров, получаем, что $AC \\le BD$. Значит, $BD = AC$, и все перавечства обратились в равенства.\nЭ... | Russia | Russian mathematical olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English; Russian | proof only | null | |
0clp | Determine the smallest positive integer $k$ for which there exists a set $A$ of 10 points in the plane, no three of which are collinear, having the following property: it is possible to color the 45 segments with endpoints in $A$ using $k$ colors so that any two segments of the same color intersect either in their inte... | [
"**Lemma.** Given 5 points in the plane, no three collinear, it is impossible to color the 10 segments determined by them with only 2 colors so that any two segments of the same color intersect.\n\nFirst, we show that any 4 points determine a monochromatic triangle. The 4 points form either a convex or concave quad... | Romania | 75th NMO Selection Tests | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English | proof and answer | 6 | |
0040 | Nacho debe escribir en un tablero de $26 \times 26$ los $26^2=676$ números enteros del 1 al 676, uno en cada casilla. Luego, Julián elige dos casillas vecinas (con un lado o un vértice común). El objetivo de Julián es que la suma de los números de esas dos casillas sea múltiplo de 4. Determinar si Nacho puede distribui... | [] | Argentina | Argentina 2006 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Español | proof and answer | Nacho cannot; Julian will always be able to choose two neighboring squares whose sum is a multiple of four. | |
0a3s | Let $n \ge 2$ be an integer, and let $z_1, \dots, z_n$ be positive integers satisfying:
* $z_j \le j$ for $j = 1, \dots, n$;
* $z_1 + \dots + z_n$ is even.
Prove that there exist $s_1, \dots, s_n \in \{-1, 1\}$ such that:
$$
s_1z_1 + s_2z_2 + \dots + s_nz_n = 0.
$$ | [
"We prove this with two-step induction to $n$. For $n=2$, it follows from the conditions that $z_1 = z_2 = 1$, and so then $z_1 - z_2 = 0$.\n\nFor $n=3$, it follows from the conditions that $(z_1, z_2, z_3)$ equals $(1, 1, 2)$, $(1, 2, 1)$, or $(1, 2, 3)$. In these cases, we choose $(s_1, s_2, s_3)$ to be equal to ... | Netherlands | BxMO/EGMO Team Selection Test | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
02zp | Problem:
Na figura abaixo, os segmentos $A M$, $A K$, $B_{2} C_{2}$ e $B_{1} C_{1}$ são tangentes à circunferência de centro $O$. Se $\angle B_{1} O B_{2} = 30^{\circ}$, determine o valor do ângulo $\angle B_{1} D B_{2}$.
 | [
"Solution:\n\nPelo Teorema do Ângulo Externo, temos\n$$\n\\begin{aligned}\n\\angle B_{1} O B_{2} &= \\angle A B_{1} O - \\angle A B_{2} O \\\\\n&= \\frac{\\angle A B_{1} C_{1} - \\angle A B_{2} C_{2}}{2} \\\\\n&= \\frac{\\angle B_{1} D B_{2}}{2}\n\\end{aligned}\n$$\nPortanto, $\\angle B_{1} D B_{2} = 2 \\cdot 30 = ... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 60° | |
0g8z | 已知 $x, y$ 為滿足 $x + y = 1$ 的正實數, $n$ 為大於或等於 $2$ 的整數。試證:
$$
\frac{x^n}{x+y^3} + \frac{y^n}{x^3+y} \geq \frac{2^{4-n}}{5}.
$$
Let $x, y$ be positive real numbers with $x + y = 1$, and $n$ be a positive integer with $n \ge 2$. Prove that
$$
\frac{x^n}{x+y^3} + \frac{y^n}{x^3+y} \geq \frac{2^{4-n}}{5}.
$$ | [
"因為函數 $z = t^m$ ($m$ 為大於 $0$ 的常數) 在開區間 $(0, \\infty)$ 上是增函數, 則有\n$$\n(x - y)(x^{n+3} - y^{n+3}) \\geq 0 \\\\\n(x - y)(x^{n-1} - y^{n-1}) \\geq 0.\n$$\n$$\n\\begin{align*}\n& \\text{故 } 2\\left(\\frac{x^{n+1}}{x+y^3} + \\frac{y^{n+1}}{x^3+y}\\right) - \\left(\\frac{x^n}{x+y^3} + \\frac{y^n}{x^3+y}\\right) \\\\\n&= \... | Taiwan | 二〇一五數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0fq5 | Problem:
En el triángulo $ABC$ los puntos medios respectivos de los lados $BC$, $CA$ y $AB$ son $D$, $F$, $E$. Sean: $M$ el punto donde la bisectriz interior del ángulo $\angle ADB$ corta al lado $AB$, y $N$ el punto donde la bisectriz interior del ángulo $\angle ADC$ corta al lado $AC$. Sean además $O$ el punto de in... | [
"Solution:\n\nAplicando el teorema la bisectriz a los triángulos $ADB$ y $ADC$ se obtiene\n$$\n\\frac{MB}{AM} = \\frac{BD}{AD} \\quad \\text{y} \\quad \\frac{NC}{AN} = \\frac{DC}{AD}\n$$\n\n\n\nComo $BD = DC$, $\\frac{MB}{AM} = \\frac{NC}{AN}$ y entonces $MN$ es paralelo a $BC$, de donde $\... | Spain | LIII Olimpiada matemática Española | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
094d | Problem:
Find all integers $n \geq 3$ for which it is possible to draw $n$ chords of one circle such that their $2 n$ endpoints are pairwise distinct and each chord intersects precisely $k$ other chords for:
(a) $k=n-2$,
(b) $k=n-3$. | [
"Solution:\n\n(a) Every chord avoids precisely one other chord, hence the avoiding chords form pairs and $n$ must be even. On the other hand, for any even $n \\geq 2$ the construction is simple (see the left figure).\n\n\n\n(b) For $n=3,4,5,8$ we can draw the chords as in the middle figure.... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | (a) All even n with n at least four.
(b) All integers n with n at least three. | |
06gd | Among a group of $451$ persons, every pair of persons together know exactly one other person (knowing is mutual). Denote by $n$ the difference between the number of persons known by a member of the group who knows the largest number of persons among the group, and the number of persons known by another member who knows... | [
"The only possible (and hence the maximum) $n$ is $448$.\n\nIndeed, by the friendship theorem, there must be one person who knows everyone else. We give a proof as follows.\n\nWLOG, assume $A$ and $B$ do not know each other. If $X$ knows $A$, then $X$ and $B$ know a person $Y$ in common. Note that the function mapp... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof and answer | Maximum 448; the only possible value is 448. | |
08ht | Problem:
Let $\left(F_{n}\right)_{n \in \mathbb{N}^{*}}$ be the Fibonacci sequence so that: $F_{1}=1$, $F_{2}=1$, $F_{n+1}=F_{n}+F_{n-1}$ for every positive integer $n \geq 2$. Show that $F_{n}<3^{n / 2}$ and calculate the limit
$$
\lim_{n \rightarrow \infty}\left(\frac{F_{1}}{2}+\frac{F_{2}}{2^{2}}+\ldots+\frac{F_{n}... | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof and answer | 2 | |
0amy | Problem:
Find the remainder if $(2001)^{2012}$ is divided by $10^{6}$. | [] | Philippines | AREA STAGE | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 24001 | |
016d | Let $a$, $b$ and $n$ be positive integers with $\text{GCD}(a, b) = 2009$. Prove that
$$
\text{GCD}(n^a - 1, n^b - 1) = n^{2009} - 1.
$$ | [
"A more general assertion is that\n$$\n\\text{GCD}(n^a - 1, n^b - 1) = n^{\\text{GCD}(a,b)} - 1.\n$$\nWrite $a = dp$, $b = dq$, and $m = n^d$, where $d = \\text{GCD}(a,b)$ and $p$ and $q$ are relatively prime. The statement is then that\n$$\n\\text{GCD}(m^p - 1, m^q - 1) = m - 1.\n$$\nClearly $m-1$ is a divisor of ... | Baltic Way | Baltic Way SHL | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
05pq | Problem:
Trouver tous les entiers strictement positifs $a, b$ et $c$ tels que $(a^{3}+b)(b^{3}+a)=2^{c}$. | [
"Solution:\nSoit $(a, b, c)$ un triplet solution.\n\nSi $a=b$, on a $a^{2}(a^{2}+1)^{2}=2^{c}$, ce qui implique $a=b=1$ et $c=2$.\n\nSi $a \\neq b$, supposons par symétrie que $a>b$. On voit aisément que $a$ et $b$ sont impairs et premiers entre eux. Il existe deux entiers $x, y \\geqslant 1$ tels que $2^{x}=a^{3}+... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (a,b,c) = (1,1,2), (3,5,12), (5,3,12) | |
08us | Consider an infinite sequence $a_1, a_2, a_3, \dots$ of positive integers. Suppose that the following 2 conditions are satisfied for an arbitrary positive integer $n$:
* $a_n$ is a multiple of $n$.
* $|a_n - a_{n+1}| \le 5$.
Find the maximum possible value that $a_1$ can take. | [
"Let $a_1, a_2, a_3, \\dots$ be an infinite sequence of positive integers satisfying the conditions of the problem. From $a_{n+1} \\le a_n + 5$ ($n = 1, 2, \\dots$) it follows that for every $n$ $a_n \\le 5 (n-1) + a_1$ is satisfied. In particular, for all sufficiently large $n$ $a_n < 6n$ is satisfied (in fact, $n... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 85 | |
027u | Problem:
Seja $ABC$ um triângulo inscrito na circunferência abaixo. Sejam também $I$ o incentro do triângulo $ABC$ e $D$ o ponto onde a reta $AI$ corta a circunferência. Mostre que $|\overline{DB}| = |\overline{DC}| = |\overline{DI}|$.
 | [
"Solution:\nComo $I$ é o incentro do triângulo $ABC$, então os segmentos $AI$, $BI$ e $CI$ são bissetrizes dos ângulos $\\hat{A}$, $\\hat{B}$ e $\\hat{C}$. Sejam então $\\alpha = \\angle BAI = \\angle CAI$ e $\\beta = \\angle ABI = \\angle CBI$.\n\n\n\nComo o ângulo $\\angle BID$ é ângulo e... | Brazil | null | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jlz | Problem:
Let $ABC$ be a triangle with $AB = 5$, $AC = 4$, $BC = 6$. The angle bisector of $C$ intersects side $AB$ at $X$. Points $M$ and $N$ are drawn on sides $BC$ and $AC$, respectively, such that $\overline{XM} \parallel \overline{AC}$ and $\overline{XN} \parallel \overline{BC}$. Compute the length $MN$. | [
"Solution:\n\nBy Stewart's Theorem on the angle bisector,\n$$\nCX^2 = AC \\cdot BC \\left(1 - \\frac{AB}{AC + BC}^2\\right)\n$$\nThus,\n$$\nCX^2 = 4 \\cdot 6 \\left(1 - \\frac{5}{10}^2\\right) = 18\n$$\nSince $\\overline{XM} \\parallel \\overline{AC}$ and $\\overline{XN} \\parallel \\overline{BC}$, we produce equal... | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3*sqrt(14)/5 | |
05tw | Problem:
On pose $n=2021$. Démontrer que le nombre $3^{2 n+1}+2^{n+2}$ est divisible par 35. | [
"Solution:\n\nNous allons démontrer que le nombre $S_{n}=3^{2 n+1}+2^{n+2}$ est divisible simultanément par 5 et par 7 lorsque $n=2021$.\nOn vérifie immédiatement que\n$$\nS_{n} \\equiv 9^{n} \\times 3+2^{n} \\times 4 \\equiv 2^{n} \\times 3+2^{n} \\times 4 \\equiv 2^{n} \\times 7 \\equiv 0 \\quad(\\bmod 7),\n$$\nc... | France | Préparation Olympique Française de Mathématiques | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof only | null | |
0awi | Problem:
Define $f(x) = x^{2} - x - 4$ and $g(x) = \log_{2}(x - 2)$. Find the domain of $g \circ f$. | [
"Solution:\n\nNote first that we need $f(x) > 2$ so that $g \\circ f$ is defined. Hence, we have $x^{2} - x - 4 > 2$, or $x^{2} - x - 6 > 0$. This factors into $(x + 2)(x - 3) > 0$, and quick interval testing shows that the intervals $(-\\infty, -2)$ and $(3, +\\infty)$ are the ones that work. (A table of signs may... | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x < -2 or x > 3 | |
0k10 | Problem:
When the number
$$
N = 1^{1} \times 2^{2} \times 3^{3} \times \cdots \times 9^{9}
$$
is written as a decimal number, how many zeros does it end in? | [
"Solution:\nThe number $N$ ends with five zeros. Indeed, $10^{5}$ divides $N$ since $4^{4} \\times 5^{5} = 800,000$ divides $N$. But $10^{6}$ does not divide $N$, since the only terms in the product which are divisible by $5$ are $5^{5}$, and hence $5^{6}$ does not divide $N$. So the answer is $5$."
] | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 5 | |
0itz | Problem:
How many ways can the integers from $-7$ to $7$ be arranged in a sequence such that the absolute value of the numbers in the sequence is nondecreasing? | [
"Solution:\n\nEach of the pairs $a, -a$ must occur in increasing order of $a$ for $a = 1, \\ldots, 7$, but $a$ can either occur before or after $-a$, for a total of $2^{7} = 128$ possible sequences."
] | United States | $12^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 128 | |
00vy | Given a positive integer $k$, a pair of positive integers $(a, b)$ is said to be $k$-nice if $a \mid b^k + 1$ and $b \mid a^k + 1$. For which positive integers $k$ are there infinitely many $k$-nice pairs? | [
"First we will solve the case $k = 1$. Notice that a pair $(a, b)$ is 1-nice if $a \\mid b + 1$ and $b \\mid a + 1$. Assuming without loss of generality that $a \\le b$ and setting $d = b - a$, we observe that since $b \\mid a + 1$ then $b \\le a + 1$, i.e. $d \\le 1$. Since $a$ divides $(b - a) + 1 = d + 1$, we co... | Balkan Mathematical Olympiad | 42nd Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All positive integers k greater than or equal to 2 | |
0cp6 | For a positive integer $n$, denote by $S_n$ the sum of the $n$ least primes: $S_1 = 2$, $S_2 = 2+3=5$, $S_3 = 2+3+5=10$, and so on. Determine whether two consecutive terms of the sequence $S_1, S_2, S_3, \ldots$ can be perfect squares. (V. Sharich)
Для каждого натурального $n$ обозначим через $S_n$ сумму первых $n$ пр... | [
"**Ответ.** Не могут.\n\nОбозначим $n$-е простое число через $p_n$. Предположим, что нашлось $m > 1$, для которого $S_{m-1} = k^2$, $S_m = l^2$, где $k$ и $l$ — натуральные числа. Числа $S_2 = 5$, $S_3 = 10$ квадратами не являются, так что $m > 4$. Заметим, что $p_m = S_m - S_{m-1} = (l-k)(l+k)$; ввиду простоты $p_... | Russia | Regional round | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English; Russian | proof only | null | |
0g2l | Problem:
Sei $ABC$ ein Dreieck und seien $D$, $E$ und $F$ die Höhenfusspunkte der Höhen von $A$, $B$ respektive $C$. Sei $H$ der Höhenschnittpunkt von Dreieck $ABC$. Die Strecken $EF$ und $AD$ schneiden sich in $G$. Sei $K$ der Punkt auf dem Umkreis von Dreieck $ABC$, der diametral gegenüber von $A$ liegt. Die Gerade ... | [
"Solution:\n\nWir wollen $GM \\parallel HK$ zeigen. Wenn wir annehmen, dass das stimmt und das Ganze von $A$ aus betrachten, können wir das Dreieck $AMG$ zentrisch strecken, sodass es auf $AKH$ landet. Dies bedeutet, dass wir $\\frac{AG}{AH} = \\frac{AM}{AK}$ hätten. Die Umkehrung stimmt nun aber auch: Wenn $\\frac... | Switzerland | Selektion | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous ... | null | proof only | null | |
05my | Problem:
Soit $n \geqslant 1$. On a un nombre fini de bouteilles, chacune contenant une quantité d'eau inférieure à 1 litre, telles que la quantité totale d'eau est de $\frac{n}{2}$ litres. On dispose également de $n$ seaux vides.
Montrer qu'il est possible de vider chaque bouteille dans un seau (on peut vider plusie... | [
"Solution:\n\nOn raisonne par récurrence sur $n$ :\n\nPour $n = 1$, il suffit de tout vider dans l'unique seau.\n\nSupposons le résultat vrai au rang $n$ : si on arrive à vider dans un seau un certain nombre de bouteilles représentant au total au moins $\\frac{1}{2}$ litre et au plus 1 litre, alors il restera $\\fr... | France | OCympiades Françaises de Mathématiques - Envoi Numéro 4 - Combinatoire | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
06gx | For any positive integer $n$, define $f(n)$ as the sum of the squares of the digits of $n$ (in base ten). For example, $f(2012) = 2^2 + 0^2 + 1^2 + 2^2 = 9$. Find the greatest integer $n$ such that there exists a positive integer $d$ with $f(\underbrace{f(\cdots f(n)\cdots)}_{d \text{ times}}) = n$. | [
"The greatest $n$ is $145$.\n\nOne can easily check that\n$$\n145 \\mapsto 42 \\mapsto 20 \\mapsto 4 \\mapsto 16 \\mapsto 37 \\mapsto 58 \\mapsto 89 \\mapsto 145,\n$$\nso that $f^8(145) = 145$. It remains to check $n \\le 145$.\n\nWe first prove that $n$ has at most $3$ digits. Suppose on the contrary that $n = \\o... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 145 | |
0bk5 | Determine all positive integers $a$ for which there exist exactly 2014 positive integers $b$ such that $2 \le \frac{a}{b} \le 5$. | [
"Rewrite $2 \\le \\frac{a}{b} \\le 5$ as $2a \\le 10b \\le 5a$. It follows that the sequence $2a, 2a+1, \\dots, 5a$ contains 2014 multiples of 10, hence it contains at least 2013 and at most 2015 groups of 10 consecutive numbers. We deduce that $2013 \\cdot 10 \\le 5a - 2a < 2015 \\cdot 10$, which leads to $a \\in ... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 6710, 6712, 6713 | |
00r3 | Let scalene triangle $ABC$ have orthocentre $H$ and circumcircle $\Gamma$. $AH$ meets $\Gamma$ at $D$ distinct from $A$. $BH$ and $CH$ meet $CA$ and $AB$ at $E$ and $F$ respectively, and $EF$ meets $BC$ at $P$. The tangents to $\Gamma$ at $B$ and $C$ meet at $T$. Show that $AP$ and $DT$ are concurrent on the circumcirc... | [
"Let $Y$ be the point on $\\Gamma$ diametrically opposite $A$ and $X$ the second point of intersection of $YH$ with $\\Gamma$. We show that $X$ lies on lines $AP$, $DT$ and circle $AFE$, completing the proof.\n\nTo see that $X$ lies on circle $AFE$, note that as $AY$ is a diameter of $\\Gamma$, $\\angle HXA = \\fra... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis ... | null | proof only | null | |
01fk | Let $M(n)$ be the minimum number of cells that must be coloured in the $6 \times n$ square grid to ensure that every $2 \times 3$ rectangle (in any orientation) contains a coloured cell. Is it true that $M(n) = p_n + k_n^3$ for all $n \ge 2$ where $p_n$ is a prime and $k_n$ is a nonnegative integer? | [
"No, it is not true. At first we note that $M(n)$ is a nondecreasing function and $M(n) \\ge n$ for all $n \\ge 2$ what follows from the fact that $6 \\times n$ rectangle can be cut into $n$ rectangles of size $2 \\times 3$, each of whom should contain a coloured cell.\n\nNext we show that $M(n+1) \\le M(n)+2$ for ... | Baltic Way | Baltic Way 2019 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0kor | For any finite set $X$, let $|X|$ denote the number of elements in $X$. Define
$$
S_n = \sum |A \cap B|,
$$
where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\{1, 2, 3, \dots, n\}$ with $|A| = |B|$. For example, $S_2 = 4$ because the sum is taken over the pairs of subsets
$$
(... | [] | United States | 2022 AIME I | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 245 | |
0j4s | Problem:
The ordered pairs $(2011,2),(2010,3),(2009,4), \ldots,(1008,1005),(1007,1006)$ are written from left to right on a blackboard. Every minute, Elizabeth selects a pair of adjacent pairs $\left(x_{i}, y_{i}\right)$ and $\left(x_{j}, y_{j}\right)$, with $\left(x_{i}, y_{i}\right)$ left of $\left(x_{j}, y_{j}\righ... | [
"Solution:\n\nAnswer: 504510\n\nFirst, note that none of the numbers will ever be $0$. Let $\\star$ denote the replacement operation. For each pair on the board $\\left(x_{i}, y_{i}\\right)$ define its primary form to be $\\left(x_{i}, y_{i}\\right)$ and its secondary form to be $\\left[x_{i} y_{i}, \\frac{x_{i}}{y... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 504510 | |
09iu | Suppose $a \ge 0$ and $b \ge 0$. If the equation $x^4 - a x^3 + b x^2 - a x + 1 = 0$ has four pairwise distinct real solutions, prove that $2a + b > 14$. | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0kvs | Problem:
Two distinct similar rhombi share a diagonal. The smaller rhombus has area $1$, and the larger rhombus has area $9$. Compute the side length of the larger rhombus. | [
"Solution:\n\nLet $d$ be the length of the smaller diagonal of the smaller rhombus. Since the ratio of the areas is $9:1$, the ratio of the lengths is $3:1$. This means that the smaller diagonal of the larger rhombus (which is also the longer diagonal of the smaller rhombus) has length $3d$.\n\n![](attached_image_1... | United States | HMMT November 2023 | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(15) | |
04n2 | Let $ABCD$ be a trapezium with bases $\overline{AB}$ and $\overline{CD}$, such that $ABC$ is an acute-angled triangle. Let $O$ be the centre of the circle circumscribed to the triangle $ABC$, and let $E$ be the intersection of lines $OB$ and $CD$. If $\angle DBC = \angle CEB + 10^\circ$, find the measure of the angle b... | [] | Croatia | Croatia_2018 | [
"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 and answer | 20° | |
02fu | $f$ is a map on the plane such that two points a distance $1$ apart are always taken to two points a distance $1$ apart. Show that for any $d$, $f$ takes two points a distance $d$ apart to two points a distance $d$ apart. | [
"Observe first that the images of equilateral triangles of side $1$ are equilateral triangles of side $1$. Consider two equilateral triangles $ABC$ and $A'B'C$ with side $1$ and a common side. Notice that $AA' = \\sqrt{3}$ and $f(A)f(A') = 0$ or $\\sqrt{3}$. If $f(A) = f(A')$, then consider $B$ such that $AB = 1$ a... | Brazil | XIX OBM | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0e39 | Find the smallest three-digit number such that the following holds: if the order of the digits of this number is reversed and the number obtained by this is added to the original number the resulting number consists of only odd digits. | [
"Write the three-digit number as $abc$. The number we obtain by reversing the order of the digits is $cba$. The sum $x$ of these two numbers is\n$$\nx = \\overline{abc} + \\overline{cba} = (a + c)10^2 + (b + b)10 + (c + a).\n$$\nSince all the digits of $x$ are odd and $2b$ is even, we conclude that $c + a$ is equal... | Slovenia | National Math Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 209 | |
0bsy | Let $n$ be a positive integer, and let $a_1, \dots, a_n$ be pairwise distinct positive integers. Show that
$$
\sum_{k=1}^{n} \frac{1}{[a_1, \dots, a_k]} < 4,
$$
where $[a_1, \dots, a_k]$ is the least common multiple of the integers $a_1, \dots, a_k$. | [
"Since the number of positive divisors of a positive integer $m$ does not exceed $2\\sqrt{m}$, and $a_1, \\dots, a_k$ are pairwise distinct positive divisors of $[a_1, \\dots, a_k]$, it follows that $[a_1, \\dots, a_k] \\ge k^2/4$. Consequently,\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} \\frac{1}{[a_1, \\dots, a_k]} &... | Romania | 67th NMO Selection Tests for BMO and IMO | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof only | null | |
0d1y | $ABCD$ is a cyclic quadrilateral such that $AB = BC = CA$. Diagonals $AC$ and $BD$ intersect at $E$. Given that $BE = 19$ and $ED = 6$, find the possible values of $AD$. | [
"Applying Ptolemy relation to the cyclic quadrilateral $ABCD$, we get\n$$\nAB \\cdot CD + BC \\cdot DA = AC \\cdot BD\n$$\nwhich simplifies to\n$$\nCD + DA = 25\n$$\n\nLet $a = AB = BC = CA$ and $x = AE$.\nWe have, from similarity of triangles $AED$ and $BEC$, that\n$$\n\\frac{DA}{a} = \\fr... | Saudi Arabia | Selection tests for the Balkan Mathematical Olympiad 2013 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 10 or 15 | |
0b9g | Let $m$ be a positive integer and $p$ a prime. Suppose $A$ is a unitary ring having exactly $m$ invertible elements and such that $\underbrace{1+1+\cdots+1}_{p \text{ times}} = 0$.
Prove that $A$ has non-zero nilpotent elements if and only if $p$ divides $m$. | [
"Let $a \\in A$, $a \\neq 0$, be a nilpotent element and $k \\in \\mathbb{N}^*$, such that $a^{p^k} = 0$. From $(a+1)^{p^k} = a^{p^k} + 1 = 1$, we have that $a+1$ is invertible and the order of $a+1$ in the group of units $U(A)$ is a power of $p$. Because the order of $a+1$ divides the order $m$ of $U(A)$, we deduc... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
07ww | There are real numbers $\alpha, \beta, \gamma$ such that the cubic functions
$$
f(x) = x^3 - 3x^2 + \alpha x + \beta \quad \text{and} \quad g(x) = x^3 + \gamma x - 6
$$
have exactly two distinct non-zero roots in common. Prove that those two roots satisfy the equation $3(\alpha - \gamma - 9)x^2 + (\alpha - \gamma - 9)(... | [
"Suppose the common roots are $a, b$ and the third root of $g$ is $c$. Using Vieta's formulas, or comparing coefficients in\n$$\ng(x) = x^{3} - \\gamma x - 6 = (x - a)(x - b)(x - c) = x^{3} - (a+b+c)x^{2} + (ab+bc+ca)x - abc\n$$\ngive $a + b + c = 0$ and $abc = 6$. Hence, $-c = a + b$ and so\n$$\nab(a + b) = -6. \\... | Ireland | IRL_ABooklet_2024 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
03x5 | On a $10 \times 10$ chessboard, some $4n$ unit square fields are chosen to form a region $R$. This region $R$ can be tiled by $n$ $2 \times 2$ squares. If $R$ can also be tiled by a combination of $n$ pieces of the following types of shapes (with rotations allowed).
D... | [
"The answer is $n = 4$. We call those two kinds of nonsquare tiles *ducks*.\nFirst, the left-hand-side figure and the middle figure below show that $n = 4$ works.\n\n\n\n\nSecond, we show that $n$ must be even. We mark the (infinite) chessbo... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 4 | |
0h2w | In the decimal form of the number
$$
34! = 295232799039*041408476186096435*0000000
$$
two digits are replaced by stars. Find these digits. | [
"Нехай перша пропущена цифра — $a$, друга — $b$. Число $34!$ ділиться без остачі на $9$ та на $11$. Оскільки сума записаних цифр від ділення на $9$ дає остачу $1$, то сума $a$ і $b$ від ділення на $9$ має давати остачу $8$, звідки $(a, b)$ — одна з таких пар цифр\n$$(0,8), (1,7), (2,6), (3,5), (4,4), (5,3), (6,2), ... | Ukraine | Ukrainian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | first digit = 6, second digit = 2 | |
01rx | The medians $AM$ and $BN$ of a triangle $ABC$ are the diameters of the circles $\omega_1$ and $\omega_2$.
If $\omega_1$ touches the altitude $CH$ prove that $\omega_2$ also touches $CH$. | [
"The statement easily follows from the following fact: if points $A_1$, $B_1$ are chosen on the sides $AC$, $BC$ of the triangle $ABC$, respectively, then the radical axis of the circles with the diameters $AA_1$ and $BB_1$ contains the orthocenter of the triangle $ABC$.\n\nIn case when $AA_1$ and $BB_1$ are median... | Belarus | SELECTION and TRAINING SESSION | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0hd5 | Determine all possible pairs of positive integers $(m, n)$ that satisfy the following:
$$
m^{10} + 23mn + n! = 2020,
$$
where $n!$ denotes the product of all positive integers from $1$ to $n$. | [
"When $m \\ge 3$, $m^{10} \\ge 3^{10} > 3^7 = 2187 > 2020$, so the given equality doesn't hold. Thus, there are two possible values for $m$: $m=1$ or $m=2$.\n\nIf $m=1$, the condition to be satisfied is $23n + n! = 2019$. Since $7! = 5040 > 2019$, $n \\le 6$. For $n \\le 6$, $23n + n! \\le 138 + 720 = 850 < 2019$, ... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (2, 6) | |
07fc | Let $O$ be the circumcenter of the triangle $\triangle ABC$. Let $D, E$ on $AC, AB$ and $P, Q, R, S$ be points such that $P, C$ and $R, C$ lie on the different sides of $AB$, $Q, B$ and $S, B$ lie on the different sides of $AC$ and $R, S$ lie on circumcircle of $\triangle DAP$, $\triangle EAQ$. We have $\triangle BCE \... | [
"First we prove that it suffices to show that $ASCBR$ is cyclic. Let $\\omega$ be the circumcircle of $\\triangle ABC$. The fact $\\triangle PAE \\sim \\triangle BDC$ implies that $\\angle PAB = \\angle C$. Therefore, $PA$ is tangent to $\\omega$. Similarly, $\\triangle QAD \\sim \\triangle EBC$ implies $\\angle QA... | Iran | 37th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0ju5 | Problem:
The Lucas numbers are defined by $L_{0} = 2$, $L_{1} = 1$, and $L_{n+2} = L_{n+1} + L_{n}$ for every $n \geq 0$. There are $N$ integers $1 \leq n \leq 2016$ such that $L_{n}$ contains the digit $1$. Estimate $N$.
An estimate of $E$ earns $\lfloor 20 - 2|N - E| \rfloor$ or $0$ points, whichever is greater. | [
"Solution:\n\nlucas_ones $n = \\text{length} \\circ \\text{filter} (\\text{elem}~'1')~\\$~\\text{take}~(n+1)~\\text{lucas_strs}$\n\nwhere\n\n$\\text{lucas} = 2 : 1 : \\text{zipWith}~(+)~\\text{lucas}~(\\text{tail}~\\text{lucas})$\n\n$\\text{lucas_strs} = \\text{map}~\\text{show}~\\text{lucas}$\n\n$\\text{main} = \\... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Algorithms"
] | null | final answer only | 1975 | |
0kmn | Problem:
In $\triangle ABC$, suppose the incircle has center $I$ and is tangent to $BC$ at $D$, and the $A$ excircle has center $I_{a}$ and is tangent to $BC$ at $D'$. Show that $ID'$ and $I_{a}D$ intersect on the altitude from $A$ to $BC$. | [
"Solution:\n\nThe intersection point is the midpoint of the altitude.\n\nLet $E$ be the point on the incircle diametrically opposite from $D$. Then the homothety centered at point $A$ which takes the incircle to the $A$-excircle takes $E$ to $D'$, so $A$, $D'$, and $E$ are collinear. Since $DE$ is parallel to the a... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0fwu | Problem:
Zeige, dass jede Zahl der Form
$$
2^{5^{2^{5^{*}}}} + 4^{5^{4^{5^{*}}}}
$$
durch $2008$ teilbar ist, wobei die Exponententürme beliebige, voneinander unabhängige Höhen $\geq 3$ haben. | [
"Solution:\n\nEs gilt $2008 = 8 \\cdot 251$ und offensichtlich ist jeder der beiden Summanden in der Aufgabenstellung durch $8$ teilbar. Es genügt daher zu zeigen, dass die Summe auch durch $251$ teilbar ist. Modulo $251$ gilt $2^{8} = 256 \\equiv 5$ und daher $2^{25} \\equiv 2 \\cdot 5^{3} = 250 \\equiv -1$. Ausse... | Switzerland | SMO Finalrunde | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
00c4 | There were three candidates $A$, $B$, $C$ in the elections for a provincial governor. In the first round $A$ won $44\%$ of the number of votes given for $B$ and $C$ together and $C$ had fewest votes. No candidate had the majority necessary for a first-round win, so there was a second round for $A$ and $B$. The voters i... | [
"The journalist is right for $p \\ge 73$ and wrong for $p \\le 72$.\n\nLet $a, b, c$ denote the number of votes for $A, B, C$ in the first round, and let $N = a + b + c$ be the total number of voters in this round. By hypothesis $a = \\frac{44}{100}(b + c) = \\frac{11}{25}(N - a)$, hence $a = \\frac{11}{36}N$; also... | Argentina | Argentina_2018 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | p ≥ 73 | |
0hoa | Problem:
$$
\frac{a}{b} \quad \text{ and } \quad \frac{c}{d}
$$
are called approximately equal if $a, b, c, d$ are positive integers and
$$
\frac{a}{b}-\frac{c}{d}=\frac{1}{b d}
$$
Prove that given two approximately equal fractions, we can multiply the four numerators and denominators by the same positive integer and ... | [
"Solution:\n\nGiven the approximately equal fractions\n$$\n\\frac{a}{b} \\text{ and } \\frac{c}{d}\n$$\nwe multiply the four terms by $a+b+c+d$ to get\n$$\n\\frac{a^{2}+a b+a c+a d}{a b+b^{2}+b c+b d} \\quad \\text{ and } \\quad \\frac{a c+b c+c^{2}+c d}{a d+b d+c d+d^{2}}\n$$\nWe then adjust each member by 1 by ch... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
08mg | Problem:
The real positive numbers $x, y, z$ satisfy the relations $x \leq 2$, $y \leq 3$, $x+y+z=11$. Prove that $\sqrt{x y z} \leq 6$. | [
"Solution:\n\nFor $x=2$, $y=3$ and $z=6$ the equality holds.\nAfter the substitutions $x=2-u$, $y=3-v$ with $u \\in [0,2)$, $v \\in [0,3)$, we obtain that $z=6+u+v$ and the required inequality becomes\n$$\n(2-u)(3-v)(6+u+v) \\leqslant 36\n$$\nWe shall need the following lemma.\n\nLemma. If real numbers $a$ and $b$ ... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
03ze | Given integer $n \geqslant 4$. Find the maximum of
$$ \frac{\sum_{i=1}^{n} a_i (a_i + b_i)}{\sum_{i=1}^{n} b_i (a_i + b_i)} $$ for non-negative real numbers $a_1, a_2, \dots, a_n, b_1, b_2, \dots, b_n$ satisfying
$$ a_1 + a_2 + \dots + a_n = b_1 + b_2 + \dots + b_n > 0. $$ | [
"The maximum is $n - 1$. By homogeneity, we can assume without loss of generality that $\\sum_{i=1}^{n} a_i = \\sum_{i=1}^{n} b_i = 1$.\n\nFirst, it is clear that if $a_1 = 1$, $a_2 = a_3 = \\dots = a_n = 0$ and $b_1 = 0$, $b_2 = b_3 = \\dots = b_n = \\frac{1}{n-1}$, then $\\sum_{i=1}^{n} a_i (a_i + b_i) = 1$, $\\s... | China | China Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | n - 1 | |
07hi | Let $m$, $n$ and $a_1, a_2, \dots, a_m$ be arbitrary positive integers. Ali and Mohammad play the following game: At each step, Ali chooses $b_1, b_2, \dots, b_m \in \mathbb{N}$, then Mohammad chooses a positive integer $s$ and obtains a new sequence $\{c_i = a_i + b_{i+s}\}_{i=1}^m$, where
$$
b_{m+1} = b_1,\ b_{m+2} =... | [
"We claim that the second player has a winning strategy for $(m, n) = (p^{\\alpha}, p^{\\beta})$ where $p$ is a prime number and $\\alpha, \\beta \\in \\mathbb{Z}^{\\ge 0}$. We call a pair $(m, n)$ a good pair if the second player has a winning strategy.\n\n**Lemma 1.** If $(m, n)$ is a good pair and $d \\mid m$ th... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Linear Algebra > Linear transformations",
"Algebra > Linear Algebra > Vectors... | English | proof and answer | Exactly those pairs where both numbers are powers of the same prime; that is, m = p^alpha and n = p^beta for some prime p and nonnegative integers alpha and beta. | |
04h3 | There are $2014$ coins each of which is black on one side and white on the other side. A $2014 \times 1$ board is given. One coin is placed on each square. In each move we can remove one coin with black side up and, at the same time, flip the coins on the neighbouring squares (if they were not already removed).
Determi... | [
"We will prove the following claim by mathematical induction on the number of coins: We can remove all the coins from some starting arrangement if and only if there is an odd number of coins with black side up.\n\nIf there is only one coin on the board, we can remove it if and only if it is turned black side up.\n\... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Exactly those starting arrangements with an odd number of coins initially showing black. | |
04hu | Prove that there does not exist a positive integer $n$ such that $7^n - 1$ is divisible by $6^n - 1$. | [
"Assume that such $n$ exists.\nNote that $5$ divides $6^n - 1 = (6-1)(6^{n-1} + \\dots + 1)$, so $5$ also has to divide $7^n - 1$.\nThe powers of $7$ when divided by $5$ give remainders $2, 4, 3, 1, \\dots$ and these remainders repeat periodically. Therefore, $7^n - 1$ will be divisible by $5$ if and only if $n$ is... | Croatia | Croatia Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof only | null | |
0gf6 | 問題四. 設 $I$ 為三角形 $ABC$ 的內心，$D$ 為 $I$ 關於邊 $BC$ 的垂足。設 $D'$ 為 $D$ 關於 $I$ 的對稱點並且滿足 $\overline{AD'} = \overline{ID'}$。以 $D'$ 為圓心，作圓 $\Gamma$ 過 $A$, $I$ 並交 $AB$, $AC$ 於 $X$, $Y$。設 $Z$ 為 $\Gamma$ 上一點滿足 $AZ$ 垂直 $BC$。
證明：$AD$, $D'Z$, $XY$ 共點。 | [
"解. 注意到 $AI$ 平分 $\\angle XAY$, 所以 $D'I$ 垂直 $XY$, 因此 $\\triangle AXY$ 與 $\\triangle ABC$ 位似。令 $E$ 為 $I$ 關於 $CA$ 的垂足, $I'$ 為 $\\triangle AXY$ 的內心, 注意到 $AD$ 與 $XY$ 交於 $I'$ 關於 $XY$ 的垂足 $T$, 所以只需證明 $D'$, $T$, $Z$ 共線。由 $\\overline{IY} = \\overline{II'}$, $\\overline{IE} = \\overline{ID'}$ 及\n$$\n\\angle YIE = \\frac{1}{2... | Taiwan | 2021 年台灣數學奧林匹亞考試試題 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles",
"Geometry > ... | null | proof only | null | |
08x6 | On the circumference of the circle with radius $1$, points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$ are placed in this order, and they satisfy
$$
\angle A_5A_2A_4 = \angle A_1A_3A_5 = \angle A_2A_4A_1 = \angle A_3A_5A_2 = 30^{\circ}.
$$
Let $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ be the points of intersections of $A_2A_4$ and $A_3A... | [
"$\\frac{\\sqrt{3}}{6}$\n\nFirst note that we have $\\angle A_2A_3A_4 = \\angle A_2A_3A_1 + \\angle A_1A_3A_5 + \\angle A_5A_3A_4$. By the theorem on subtended angles by arcs on the circumference on a circle, we have $\\angle A_2A_3A_1 = \\angle A_2A_4A_1 = 30^{\\circ}$, $\\angle A_5A_3A_4 = \\angle A_5A_2A_4 = 30^... | Japan | Japan 2013 Initial Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | sqrt(3)/6 | |
001h | Sea $ABC$ un triángulo inscripto en una circunferencia. Sea $M$ el punto medio del arco $AB$ que no contiene a $C$ y $N$ el punto medio del arco $AC$ que no contiene a $B$.
Sean $E$ y $F$ los puntos donde la recta $MN$ corta a los lados $AB$ y $AC$ respectivamente.
Demuestre que si $ME = EF = FN$, entonces el triángulo... | [] | Argentina | XII Olimpiada Matemática Rióplatense | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | español | proof only | null | |
0hmy | Problem:
Alice picks four numbers from the set $\{1,2,3,4,5,6\}$, tells Bob their product and asks him to guess their sum. Bob realizes he cannot even determine for sure whether the sum is odd or even. What is the product of the numbers Alice chose? | [
"Solution:\n\nLet $P$ be said product. Evidently there are two distinct sets of numbers $\\{x_{1}, x_{2}, x_{3}, x_{4}\\}$ and $\\{y_{1}, y_{2}, y_{3}, y_{4}\\}$ such that $x_{1} x_{2} x_{3} x_{4} = y_{1} y_{2} y_{3} y_{4} = P$, but $x_{1} + x_{2} + x_{3} + x_{4}$ and $y_{1} + y_{2} + y_{3} + y_{4}$ have different ... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 60 | |
004v | Jorge elige 6 números enteros positivos distintos y escribe uno en cada cara de un cubo. Arroja su cubo tres veces.
La primera vez su cubo mostró el número $5$ hacia arriba y además, la suma de los números de las caras laterales fue $20$.
La segunda vez su cubo mostró el número $7$ hacia arriba y además, la suma de los... | [] | Argentina | XIIIª OLIMPÍADA de MAYO | [
"Discrete Mathematics > Logic"
] | Español | proof and answer | The numbers are 2, 3, 4, 5, 6, 7. The opposite face pairs are 5 opposite 2, 7 opposite 3, and 4 opposite 6. Up to rotation, this is the unique configuration consistent with the data. | |
0f55 | Problem:
a, b, c are positive integers. If $a^b$ divides $b^a$ and $c^a$ divides $a^c$, show that $c^b$ divides $b^c$. | [] | Soviet Union | 17th ASU | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
00ld | Consider $2016$ points arranged on a circle. We are allowed to jump ahead by $2$ or $3$ points in clockwise direction.
What is the minimum number of jumps required to visit all points and return to the starting point? | [
"If the problem could be solved with $2016$ jumps, the total distance covered by these jumps would be strictly between $2 \\cdot 2016$ and $3 \\cdot 2016$ which makes a return to the original point impossible. Therefore, at least $2017$ jumps are required.\nThis is indeed possible, for example with the following se... | Austria | National Competition | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 2017 | |
0l77 | Problem:
There exists a unique triple $(a, b, c)$ of positive real numbers that satisfies the equations
$$2(a^{2} + 1) = 3(b^{2} + 1) = 4(c^{2} + 1) \quad \mathrm{and} \quad ab + bc + ca = 1.$$
Compute $a + b + c$. | [
"Solution:\nThe crux of this problem is to apply the trigonometric substitutions $a = \\cot \\alpha$, $b = \\cot \\beta$, and $c = \\cot \\gamma$, with $0 < \\alpha, \\beta, \\gamma < \\pi / 2$. Then, the given equations translate to\n$$\n\\frac{2}{\\sin^{2}\\alpha} = \\frac{3}{\\sin^{2}\\beta} = \\frac{4}{\\sin^{2... | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 9/\sqrt{23} | |
0eag | A positive integer $n$ is written on the blackboard. In each step we replace the number by the sum of any two positive integers whose product is equal to the number on the board. Determine the smallest number that can be obtained after a finite number of steps in terms of the initial number $n$. | [
"First, let us show the following: if the number on the blackboard has decreased after the change, then the new number is greater than or equal to $5$. We cannot obtain the number $1$ because the sum of two positive integers can never be equal to $1$. The number $2 = 1 + 1$ can only be obtained from $1$, the number... | Slovenia | National Math Olympiad in Slovenia | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | min(n, 5) | |
0l86 | An isosceles trapezoid has an inscribed circle tangent to each of its four sides. The radius of the circle is $3$, and the area of the trapezoid is $72$. Let the parallel sides of the trapezoid have lengths $r$ and $s$, with $r \neq s$. Find $r^2 + s^2$. | [
"Let $ABCD$ be the trapezoid with $\\overline{AB} \\parallel \\overline{CD}$ and $AB = s < r = CD$. Let $K, L, M$, and $N$ be the points at which the circle is tangent to $\\overline{AB}$, $\\overline{BC}$, $\\overline{CD}$, and $\\overline{DA}$, respectively. Because $KM$ is a diameter of the circle perpendicular ... | United States | 2025 AIME I | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 504 | |
0ago | The point $O$ is the centre of the circumcircle of triangle $\Delta ABC$.
The line $AO$ intersects the side $BC$ in point $N$, and the line $BO$ the side $AC$ in point $M$. Prove that, if $CM = CN$, then $AC = BC$. | [
"Then $\\angle ABC > \\angle BAC$ i.e. $\\angle ADC = \\angle ABC > \\angle BAC = \\angle BDC$.\nThis implies\n\nBut\n$$\n\\angle AOE = 2\\angle ACE \\text{ and } \\angle BOE = 2\\angle BCE.\n$$\nHence\n$\\angle AOE < \\angle BOE$ i.e. $\\overline{AE} < \\overline{BE}$\n($\\triangle ABO$ is isosceles).\nFrom Ceva's... | North Macedonia | Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06a8 | On a blackboard we write in a line $n$ numbers, $n \ge 40$, each being equal to $1$ or $-1$, such that:
(i) The sum of every $40$ successive numbers is equal to $0$.
(ii) The sum of every $42$ successive numbers is not equal to $0$.
We denote by $\Sigma_n$ the sum of the $n$ numbers on the blackboard. Find the greates... | [
"Let $a_1, a_2, \\dots, a_n$ be the numbers in a line on the blackboard. Since $a_1 + a_2 + \\dots + a_{40} = 0$, it follows that half of the numbers $a_1, a_2, \\dots, a_{40}$ are equal to $1$ and the other half are equal to $-1$. Since $a_1 + a_2 + \\dots + a_{40} + a_{41} + a_{42} \\neq 0$ it follows that $a_{41... | Greece | 39th Hellenic Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 20 | |
0e1y | Problem:
Ploščina romba meri $120~\mathrm{cm}^2$. Razlika dolžin njegovih diagonal je $14~\mathrm{cm}$. Kolikšna je dolžina stranice romba? | [
"Solution:\n\nUpoštevamo smiselno zvezo med diagonalama $e - f = 14$ ali $e = 14 + f$. Uporabimo obrazec za ploščino romba $\\frac{e \\cdot f}{2} = 120$. Vstavimo $e$ in dobimo kvadratno enačbo $f^2 + 14f - 240 = 0$. Enačbo rešimo. Dobimo rešitvi $f_1 = 10$ in $f_2 = -24$, ki pa ni ustrezna. Dolžino stranice $a$ iz... | Slovenia | 10. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | 13 | |
0ht5 | Problem:
A number bracelet in base $m$ is made by choosing two nonnegative integers less than $m$ (not both $0$) and continuing in a clockwise loop, each succeeding number being the mod $m$ sum of its two predecessors. The figure is closed up as soon as it starts to repeat. The figure to the right shows two number bra... | [
"Solution:\n\nThe elements of a number bracelet $N$ will be denoted $N_{0}, N_{1}, N_{2}, \\ldots$; $N_{0}$ and $N_{1}$ being the starting numbers. Let $F$ be the bracelet with starting numbers $F_{0}=0, F_{1}=1$. All congruences are modulo $m$ unless otherwise noted.\n\nSince there are only $m^{2}$ possible pairs ... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other"
] | null | proof only | null | |
0gwv | Sequence $\{f_n\}$ is defined as follows: $f_1 = 1$, $f_2 = 2$, $f_{n+2} = f_{n+1} + f_n$ for random natural $n$. What may be the greatest number of members of the sequence $\{f_n\}$ among the consecutive members of the increasing arithmetic progression? | [
"Let the first consecutive members be $f_1 = a$, $f_m = a + d$. Since $a > 0$, $d > 0$, $f_{m+1} > f_m$, $f_{m+2} = f_{m+1} + f_m > 2a + 2d > a + 2d$, the third member may only be $f_{m+1}$.\n\nSo $f_{m+1} = a + 2d$. It's clear that the next member of progression does not satisfy the condition because $f_{m+2} = 2a... | Ukraine | Ukrajina 2008 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 3 | |
07zd | Problem:
Fissiamo un punto $O$ nell'intersezione di due linee di un foglio a quadretti e indichiamo le quattro direzioni parallele alle linee come Nord, Sud, Est, Ovest (il Nord in alto). Muoviamoci, partendo da $O$ di un quadretto verso Est, poi due verso Nord, tre verso Ovest, quattro verso Sud, cinque verso Est e c... | [] | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | B | |
0jaq | Problem:
Suppose $x$ and $y$ are real numbers such that $-1 < x < y < 1$. Let $G$ be the sum of the geometric series whose first term is $x$ and whose ratio is $y$, and let $G'$ be the sum of the geometric series whose first term is $y$ and ratio is $x$. If $G = G'$, find $x + y$. | [
"Solution:\n\nWe note that $G = \\dfrac{x}{1 - y}$ and $G' = \\dfrac{y}{1 - x}$. Setting them equal gives\n$$\n\\frac{x}{1 - y} = \\frac{y}{1 - x}\n$$\nCross-multiplying, we get\n$$\nx(1 - x) = y(1 - y)\n$$\nExpanding both sides:\n$$\nx - x^2 = y - y^2\n$$\nBringing all terms to one side:\n$$\nx - x^2 - y + y^2 = 0... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1 | |
0bom | Let $S$ be a finite planar set no three points of which are collinear, and let $D(S, r) = \{\{x, y\} : x, y \in S, \text{dist}(x, y) = r\}$, where $r$ is a positive real number, and $\text{dist}(x, y)$ is the Euclidean distance between the points $x$ and $y$. Show that
$$
\sum_{r>0} |D(S, r)|^2 \le 3|S|^2(|S| - 1)/4.
$... | [
"Given a point $x$ in $S$ and a real number $r$, let $S(x, r) = \\{y : y \\in S, \\text{dist}(x, y) = r\\}$, and notice that the $S(x, r)$, $r \\ge 0$, partition $S$.\nThe number of non-degenerate isosceles triangles with vertices in $S$ and apex at $x$\nis $\\sum_{r>0} (|S(x,r)|)^2$, so the total number of non-deg... | Romania | 2015 Ninth STARS OF MATHEMATICS Competition | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English | proof only | null | |
0eyn | Problem:
A circle radius $100$ is drawn on squared paper with unit squares. It does not touch any of the grid lines or pass through any of the lattice points. What is the maximum number of squares can it pass through? | [
"Solution:\n\nTake compass directions aligned with the grid. Let $N$, $E$, $S$, $W$ be the most northerly, easterly, southerly and westerly points on the circle. The arc from $N$ to $E$ must cross $100$ north-south grid lines and $100$ east-west grid lines. Each time it crosses a grid line it changes square (and it... | Soviet Union | 2nd ASU | [
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | 800 | |
0a6a | Problem:
A rectangular sheet of paper is folded so that one corner lies on top of the corner diagonally opposite. The resulting shape is a pentagon whose area is $20\%$ one-sheet-thick, and $80\%$ two-sheets-thick. Determine the ratio of the two sides of the original sheet of paper. | [
"Solution:\n\nLet the original rectangle be $ABCD$. Let the fold line be $XY$ with point $X$ lying on side $AB$ and point $Y$ lying on side $CD$. Consider the fold such that point $D$ lands on point $B$. Let $P$ be the location of corner $A$ after the fold.\n\n\n\nThe two-sheets-thick area ... | New Zealand | NZMO Round One | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(7):3 | |
0kks | Problem:
Let acute triangle $A B C$ have circumcenter $O$, and let $M$ be the midpoint of $B C$. Let $P$ be the unique point such that $\angle B A P = \angle C A M$, $\angle C A P = \angle B A M$, and $\angle A P O = 90^\circ$. If $A O = 53$, $O M = 28$, and $A M = 75$, compute the perimeter of $\triangle B P C$. | [
"Solution:\n\nThe point $P$ has many well-known properties, including the property that $\\angle B A P = \\angle A C P$ and $\\angle C A P = \\angle B A P$. We prove this for completeness.\n\nInvert at $A$ with radius $\\sqrt{A B \\cdot A C}$ and reflect about the $A$-angle bisector. Let $P'$ be the image of $P$. T... | United States | HMMT Spring 2021 Guts Round | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-poi... | null | final answer only | 192 | |
0089 | A cube with edge $10$ is cut into $27$ parallelepipeds by three pairs of planes parallel to its faces. The edges of the interior parallelepiped have lengths $1$, $2$ and $3$. Find the sum of the volumes of the $8$ corner parallelepipeds. | [
"Imagine the eight corner parts yellow and the rest of the cube white. The two horizontal cuts produce three parallelepipeds. The middle one is white and has the same vertical dimension as the central piece. Assume the latter dimension to be $1$ and remove the middle part. A $10 \\times 10 \\times 9$ parallelepiped... | Argentina | Mathematical Olympiad Rioplatense | [
"Geometry > Solid Geometry > Volume"
] | English | proof and answer | 504 | |
0ekk | Problem:
Za padajočo linearno funkcijo $f(x)=k \cdot x+n$ velja $f(x+y)=f(x)+f(y)-3$ in $f(k)=2 f(1)$, za vsak $x, y \in \mathbb{R}$. Koliko je vrednost $f(-1)$ ?
(A) 4
(B) 0
(C) 2
(D) -2
(E) -4 | [
"Solution:\n\n$f(x+y)=k(x+y)+n=k x+k y+n$ in $f(x)+f(y)-3=k x+n+k y+n-3$. Iz tega dobimo $n=3$.\n\nIz $f(k)=2 f(1)$ dobimo enačbo $k \\cdot k+3=2 k+6$. Preoblikujemo jo v $k^{2}-2 k-3=0$.\n\nRešitev $3$ ne ustreza. Za rešitev $-1$ dobimo $f(x)=-x+3$.\n\nZato je $f(-1)=4$.\n\nPravilen je odgovor $A$."
] | Slovenia | 22. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | A | |
08ze | Find the smallest positive integer $n$ such that $10 + n$ and $10n$ are both perfect squares. | [
"Let $n$ be a positive integer such that $n + 10$ and $10n$ are both perfect squares. Then, there exists a positive integer $k$ such that $10n = k^2$. Since $k$ is a multiple of $2$ and $5$, we can write $k = 10\\ell$ for some positive integer $\\ell$, and we have $n = 10\\ell^2$. For $\\ell = 1$ or $\\ell = 2$, we... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 90 | |
05xb | Problem:
Soient $b_{1}, \ldots, b_{n}$ des réels positifs ou nuls tels que $b_{1}+b_{2}+\ldots+b_{n}=2$. Soient $a_{0}, a_{1}, \ldots, a_{n}$ des nombres réels tels que $a_{0}=a_{n}=0$. Supposons que pour tout $i$ compris entre $1$ et $n$ inclus,
$$
\left|a_{i}-a_{i-1}\right| \leqslant b_{i}
$$
Montrer que l'on a
$$
\l... | [
"Solution:\nPour tout indice $i$, on a deux majorations de $a_{i}$. En effet, on peut écrire\n$$\n\\begin{aligned}\na_{i}=a_{i}-a_{0}= & \\left(a_{i}-a_{i-1}\\right)+\\left(a_{i-1}-a_{i-2}\\right)+\\ldots+\\left(a_{1}-a_{0}\\right) \\\\\n& \\leqslant b_{i}+b_{i-1}+\\ldots+b_{1}=\\sum_{j=1}^{i} b_{j}\n\\end{aligned}... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOi 2 : Algèbre | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
091z | Problem:
Let $a$ and $b$ be positive integers. Prove that there exist positive integers $x$ and $y$ such that
$$
\binom{x+y}{2}=a x+b y
$$ | [
"Solution:\nDenoting $A=2a+1$ and $B=2b+1$, the equation can be translated into\n$$\n\\frac{B-(x+y)}{x}=\\frac{(x+y)-A}{y}\n$$\nFor $A=B$, any integers with $x+y=A$ satisfy the equation. Now suppose that $A<B$. Let $n$ be the integer in the interval $[A, B)$ which is divisible by $d=B-A$. Then $n \\neq A$ because $... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0jyi | Problem:
In unit square $ABCD$, points $E, F, G$ are chosen on side $BC, CD, DA$ respectively such that $AE$ is perpendicular to $EF$ and $EF$ is perpendicular to $FG$. Given that $GA=\frac{404}{1331}$, find all possible values of the length of $BE$. | [
"Solution:\n\nLet $BE = x$, then since triangles $ABE, ECF, FDG$ are all similar, we have $CE = 1 - x$, $CF = x(1 - x)$, $FD = 1 - x(1 - x)$, $DG = x - x^{2}(1 - x)$, $GA = 1 - x + x^{2}(1 - x) = (1 - x)(x^{2} + 1)$, therefore it remains to solve the equation\n$$\n(1 - x)(x^{2} + 1) = \\frac{404}{1331}\n$$\nWe firs... | United States | HMMT November | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | null | proof and answer | 9/11 | |
01fx | Let $ABCDEF$ be a convex hexagon in which $AB = AF$, $BC = CD$, $DE = EF$ and $\angle ABC = \angle EFA = 90^\circ$. Prove that $AD \perp CE$. | [
"Consider circle $\\omega$ with centre $A$ and radius $AB$. Note that $BC$ and $EF$ are tangent to $\\omega$ and so from the problem condition, the line $CE$ is a radical axis of $\\omega$ and $D$. Therefore $CE \\perp AD$."
] | Baltic Way | Baltic Way 2019 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | English | proof only | null | |
0fdv | Problem:
Halla todas las ternas $(x, y, z)$ de números reales que son soluciones del sistema de ecuaciones
$$
\left.\begin{array}{l}
3 \cdot 2^{y}-1=2^{x}+2^{-x} \\
3 \cdot 2^{z}-1=2^{y}+2^{-y} \\
3 \cdot 2^{x}-1=2^{z}+2^{-z}
\end{array}\right\}
$$ | [
"Solution:\nHaciendo la sustitución $2^{x}=a$, $2^{y}=b$, y $2^{z}=c$, se observa que $a, b, c>0$ y se obtiene\n$$\n\\left.\\begin{array}{l}\nb=\\frac{1}{3}\\left(a+1+\\frac{1}{a}\\right) \\\\\nc=\\frac{1}{3}\\left(b+1+\\frac{1}{b}\\right) \\\\\na=\\frac{1}{3}\\left(c+1+\\frac{1}{c}\\right)\n\\end{array}\\right\\}\... | Spain | XLVII Olimpiada Matemática Española Primera Fase | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (0, 0, 0) | |
0dlx | Find all functions $f : \mathbb{R}^+ \to \mathbb{R}$ such that for all $x, y > 0$ we have
$$
f(x^2 + x f(y) + y) = 2x + f(y).
$$ | [
"Let $r > s$ be two positive real numbers. Then, there would be a positive real number $t$ such that $t^2 + f(s)t + s - r = 0$. Now, put $(x, y) = (t, s)$ it follows that\n$$\nf(r) = f(t^2 + f(s)t + s) = 2t + f(s) > f(s).\n$$\nSo, $f$ is strictly increasing and hence injective. Plugging $(x, y) = (\\frac{f(z)}{2}, ... | Saudi Arabia | Saudi Booklet | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | All solutions are f(x) = sqrt(4x + C) for a constant C >= 0. | |
03bt | Consider a rectangular $n \times m$ table where $n \ge 2$ and $m \ge 2$ are positive integers. Each cell is colored in one of the four colors: white, green, red or blue. Call such a coloring *interesting* if any $2 \times 2$ square contains every color exactly once. Find the number of interesting colorings. | [
"Number the columns of the table by $1, 2, \\ldots, m$ and its rows by $1, 2, \\ldots, n$ and denote the cell in the $i$-th column and $j$-th row by $(i, j)$.\n\nConsider an interesting coloring $S$. We show first that either any row in $S$ contains only two colors or any column in $S$ contains only two colors.\n\n... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | proof and answer | 6*(2^m + 2^n) - 24 | |
0gky | Let $p(x) = x^{2013} + a_{2012}x^{2012} + a_{2011}x^{2011} + \dots + a_1x + a_0$ be a polynomial with real coefficients. Suppose that all the roots of $p(x)$ are
$$
-b_{1006}, -b_{1005}, \dots, -b_1, 0, b_1, \dots, b_{1005}, b_{1006}
$$
where $b_1, b_2, \dots, b_{1006}$ are positive real numbers with product equal to $... | [
"Since $0$ is a root of $p(x)$, we have $a_0 = 0$ and so\n$$\np(x) = x(x^{2012} + a_{2012}x^{2011} + a_{2011}x^{2010} + \\dots + a_2x + a_1).\n$$\nSince all the roots of $p(x)$ are $-b_{1006}, -b_{1005}, \\dots, -b_1, 0, b_1, \\dots, b_{1005}, b_{1006}$, we see that\n$$\np(x) = x(x + b_{1006})(x - b_{1006}) \\cdots... | Thailand | The 10th Thailand Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0byf | On the side $[CD]$ of the square $ABCD$ point $E$ is chosen such that $m(\angle ABE) = 60^{\circ}$. On the halfline $(BA$ point $F$ is chosen such that $[BE] = [BF]$. Let $M$ be the intersection between the lines $EF$ and $AD$.
a) Prove that $m(\angle BME) = 75^{\circ}$.
b) The bisector of angle $CBE$ intersects $CD$... | [
"a) We have $m(\\angle MEB) = m(\\angle CEB) = 60^{\\circ}$, hence $EB$ is an exterior angle bisector in the triangle $MDE$.\n\nBut $m(\\angle MDB) = m(\\angle BDE) = 45^{\\circ}$, hence $DB$ is an interior angle bisector in the triangle $MDE$.\n\nTherefore, $MB$ is the bisector of $AME$ and since $m(\\angle AME) =... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0glx | Let $n \ge 4$ be an integer. Define
$$
S = \{(x, y, z) \in \mathbb{Z}^3 \mid xyz = 0 \text{ and } 0 \le x, y, z < n\}.
$$
Let $S'$ be the projection of all points in $S$ onto a plane $P$. Given that $|S'| > n^2$, determine the least possible value of $|S'|$. | [] | Thailand | T3MO 2017 | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof and answer | n^2 + n - 1 | |
0baa | The incircle of a triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at points $D$, $E$, $F$, respectively. Let $X$ be a point on the incircle, different from the points $D$, $E$, $F$. The lines $XD$ and $EF$, $XE$ and $FD$, $XF$ and $DE$ meet at points $J$, $K$, $L$, respectively. Let further $M$, $N$, $P$ be points on... | [
"Let the lines $KL$ and $NP$, $LJ$ and $PM$, $JK$ and $MN$ meet at points $Q$, $R$, $S$, respectively. By Desargues' theorem on perspective triangles, the lines $JM$, $KN$ and $LP$ are concurrent if and only if the points $Q$, $R$ and $S$ are collinear.\n\n\n\nWithout loss of generality, we... | Romania | 62nd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
00p7 | Let $x$, $y$ and $z$ be positive real numbers such that $xyz = 3(x + y + z)$. Show that
$$
\frac{1}{x^2(y+1)} + \frac{1}{y^2(z+1)} + \frac{1}{z^2(x+1)} \geq \frac{3}{4(x+y+z)}
$$
and determine the cases of equality. | [
"The AM-GM inequality and the condition in the statement yield $x + y + z \\geq 9$, so\n$$\n\\frac{3}{4(x + y + z)} \\sum (x + 1) = \\frac{3}{4} \\left( 1 + \\frac{3}{x + y + z} \\right) \\leq 1.\n$$\nFinally, apply the Cauchy-Schwarz inequality and take into account the condition in statement to get\n$$\n\\left(\\... | Balkan Mathematical Olympiad | shortlistBMO 2011 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds if and only if x = y = z = 3. |
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