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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0hmi | Problem:
Show that there are infinitely many prime numbers whose last digit is not $1$. | [
"Solution:\nAssume there are only finitely many such primes $p_{1}, \\ldots, p_{k}$. Consider the number\n$$\nN = 10 p_{1} p_{2} \\ldots p_{k} - 1\n$$\nSince $N$ has last digit $9$, there must be a prime $p$ dividing $N$ which does not have last digit $1$ (otherwise $N$ must have last digit $1$). But by constructio... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
019t | In a one man game there are $n$ boxes numbered $1, 2, 3, \dots, n$. In the beginning there are $21k$ balls in box $k$, $k = 1, 2, 3, \dots, n$. In each turn you take two balls from box $k$, put one of the balls in box $k-1$, and throw the last ball away, $1 < k \le n$. For which $n$ is it possible to get an equal numbe... | [
"It is possible for $n = 1, 2, 3, 6$.\nA ball in box $i$, $i = 1, 2, \\ldots, n$, is given the value $2^{n-i}$. In the beginning, the sum of the values of all the balls is\n$$\nS_n = 21(1 \\cdot 2^{n-1} + 2 \\cdot 2^{n-2} + \\dots + n \\cdot 2^{n-n})\n$$\nWhen two balls are taken from box $i$, $i = 2, 3, \\ldots, n... | Baltic Way | Baltic Way 2013 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | n = 1, 2, 3, 6 | |
07kq | How many sequences $a_1, a_2, \dots, a_{2008}$ are there such that each of the numbers $1, 2, \dots, 2008$ occurs once in the sequence, and $i \in \{a_1, a_2, \dots, a_i\}$ for each $i$ such that $2 \le i \le 2008$? | [
"There are several different solutions:\n\n**Solution 1:** The function $f: \\{1, \\dots, 2008\\} \\to \\{1, \\dots, 2008\\}$ given by $f(i) = a_i$ is a bijection. Let us consider its inverse $f^{-1}: \\{1, \\dots, 2008\\} \\to \\{1, \\dots, 2008\\}$. The condition that $i \\in \\{a_1, \\dots, a_i\\}$ for $i \\ge 2... | Ireland | Irish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 2^{2007} | |
0l12 | Problem:
Consider a quarter-circle with center $O$, arc $\widehat{A B}$, and radius $2$. Draw a semicircle with diameter $\overline{O A}$ lying inside the quarter-circle. Points $P$ and $Q$ lie on the semicircle and segment $\overline{O B}$, respectively, such that line $P Q$ is tangent to the semicircle. As $P$ and $... | [
"Solution:\n\n\n\nNote that we can bound the area of $\\triangle B Q P$ by\n$$\n\\begin{aligned}\n{[B Q P]} & = \\frac{1}{2} B Q \\cdot Q P \\sin \\angle B Q P \\\\\n& \\leq \\frac{1}{2} B Q \\cdot Q P \\\\\n& = \\frac{1}{2} B Q (2 - B Q) \\\\\n& \\leq \\frac{1}{2} .\n\\end{aligned}\n$$"
] | United States | HMMT November 2024 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 1/2 | |
0fqi | Problem:
Demuestra que para todo $n \geq 2$ podemos encontrar $n$ números reales $x_{1}, x_{2}, \cdots, x_{n} \neq 1$ de manera que los productos
$$
x_{1} \cdot x_{2} \cdot \ldots \cdot x_{n} \quad \text{ y } \quad \frac{1}{1-x_{1}} \cdot \frac{1}{1-x_{2}} \cdot \ldots \cdot \frac{1}{1-x_{n}}
$$
son iguales. | [
"Solution:\nDado $x \\neq 1$, notemos que la ecuación\n$$\nx=\\frac{1}{1-x} \\Longleftrightarrow x^{2}-x+1=0\n$$\nno tiene soluciones reales. Sin embargo, dados $x, y \\neq 1$,\n$$\nx \\cdot y=\\frac{1}{1-x} \\cdot \\frac{1}{1-y}\n$$\ntiene una solución sencilla ya que\n$$\nx \\cdot y=\\frac{1}{1-x} \\cdot \\frac{1... | Spain | OME fase local | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0b83 | Consider the plane partitioned in unit squares. The interior of each square is coloured in either *red* or *black* (the sides of the squares are not considered to be coloured). Prove that given any positive integer $\alpha$ there exists an equilateral triangle of integer area $A \ge \alpha$, having monochromatic vertic... | [
"Like usually in such situations, there is some doubt on the colouring of the separating lines. One idea would be for a unit square to be considered to be made of its interior, together with its left and lower sides, but less its north-west and south-east corners. This establishes a partition of the plane, given by... | Romania | Local Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0g9p | 令 $n \ge 4$. $M$ 是平面上 $n$ 個點所成集合, 其中任三點不共線。在遊戲開始前, 在平面上做 $n$ 條線段, 使得每條線段的兩端點都屬於 $M$, 且 $M$ 裡的每個點都恰為兩條線段的端點。
我們接著考慮以下操作: 選擇兩個有交點 (含端點) 的線段 $AB$ 與 $CD$, 將這兩條線段擦掉, 並畫上 $AC$ 與 $BD$.
試證：我們不可能執行 $\frac{n^3}{4}$ 或更多次操作。 | [
"（為方便說明起見，以下線段都不包含其兩端點。）\n對於平面上的任一條直線，我們說它是”紅”的，若且唯若它包含 $M$ 中的兩點。基於 $M$ 中任三點不共線，每條紅線唯一決定 $M$ 中的兩個點。此外，紅線的數量顯然為 $C_2^n < n^2/2$。此外，對於每個線段，令它的”相交數”為有多少條紅線與之相交。對於一組線段，其”相交數”則定義為各線段相交數的總和。我們將證明\n1. 遊戲開始時，全部線段的相交數小於 $n^3/2$.\n2. 每經過一次操作，全部線段的相交數將至少減 2.\n基於全部線段相交數永遠不能為負值（否則無法進行操作），以上兩點即證明原命題。\n\n**Pf of 1.** 基於每條線段的相交數小於... | Taiwan | 二〇一五數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof only | null | |
0ky6 | There are 2022 equally spaced points on a circular track $\gamma$ of circumference 2022. The points are labeled $A_1, A_2, \dots, A_{2022}$ in some order, each label used once. Initially, Bunbun the Bunny begins at $A_1$. She hops along $\gamma$ from $A_1$ to $A_2$, then from $A_2$ to $A_3$, until she reaches $A_{2022}... | [
"Replacing 2022 with $2n$, the answer is $2n^2 - 2n + 2$.\n\n\n\n**Construction** The construction for $n=5$ shown on the left half of the figure easily generalizes for all $n$.\n\n**First proof of bound** Let $d_i$ be the shorter distance from $A_{2i-1}$ to $A_{2i+1}$.\n\n**Claim** — The d... | United States | USA TST | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 2042222 | |
06h3 | $S = \{(x, y) : x, y$ are integers, $0 \le x \le 5$ and $0 \le y \le 5\}$ is a set of lattice points on the rectangular coordinate plane. How many triangles can be formed by taking any three elements of $S$? | [
"There are 6768 triangles.\nSince there are 36 points in $S$, there are $\\binom{36}{3} = 7140$ triples of points. We have to remove those triples which contain collinear points.\n\nFirstly, there are 6 rows, and there are $\\binom{6}{3} = 20$ ways to choose a triple from each row. Therefore, there are $6 \\times 2... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | final answer only | 6768 | |
0klj | Problem:
In a regular $n$-gon, all the diagonals are drawn, forming smaller regular $n$-gons inside. If the outer regular $n$-gon has side length $1$, show that the $k$th largest regular $n$-gon formed has side length
$$
\frac{\cos (k \pi / n)}{\cos (\pi / n)}
$$
(where the original regular $n$-gon is the 1st largest)... | [
"Solution:\n\nDraw a diagonal of the small $n$-gon to form the red triangle $\\triangle ABC$ as shown in the diagram, where $A$ is a vertex of the large $n$-gon and $B$ and $C$ are the vertices of the small $n$-gon closest to $A$ on the two diagonals from $A$. Also draw the diagonal $AD$ of the large $n$-gon such t... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | cos(kπ/n) / cos(π/n) | |
0f74 | Problem:
The incircle of a triangle has radius $1$. It also lies inside a square and touches each side of the square. Show that the area inside both the square and the triangle is at least $3.4$. Is it at least $3.5$? | [] | Soviet Union | 20th ASU | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | The overlap area is at least 3.4, but not necessarily at least 3.5. In fact, the minimum can approach 6√2 − 5 ≈ 3.485. | |
0fsy | Problem:
Bestimme alle Funktionen $f: \mathbb{R} \rightarrow \mathbb{R}$, sodass für alle reellen Zahlen $x$ und $y$ gilt
$$
f(f(x)+y)=f\left(x^{2}-y\right)+4 y f(x)
$$ | [
"Solution:\nSetzt man in der Gleichung $y=-f(x)$ ein, dann folgt\n$$\nf(0)=f\\left(x^{2}+f(x)\\right)-4 f(x)^{2}\n$$\nFür $y=x^{2}$ erhält man\n$$\nf\\left(f(x)+x^{2}\\right)=f(0)+4 x^{2} f(x)\n$$\nKombination von (1) und (2) ergibt nun $4 f(x)^{2}=f\\left(f(x)+x^{2}\\right)-f(0)=4 x^{2} f(x)$, also $f(x)\\left(f(x... | Switzerland | IMO - Selektion | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) ≡ 0 for all real x, or f(x) = x^2 for all real x | |
09rx | Problem:
Zij $\triangle ABC$ een driehoek. Zij $M$ het midden van $BC$ en zij $D$ een punt op het inwendige van zijde $AB$. Het snijpunt van $AM$ en $CD$ noemen we $E$. Veronderstel dat $|AD|=|DE|$. Bewijs dat $|AB|=|CE|$. | [
"Solution:\n\nEr is maar één configuratie. Zij $N$ het midden van $AC$. Dan is $MN$ een middenparallel. Wegens Z-hoeken is $\\angle EAD=\\angle MAB=\\angle AMN$, maar ook geldt $\\angle EAD=\\angle DEA=\\angle MEC$. Dus $\\angle AMN=\\angle CEM$. Als nu $S$ het snijpunt van $MN$ en $CE$ is, dan is $\\triangle SEM$ ... | Netherlands | IMO-selectietoets | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0iiq | Problem:
Let $O$ be the intersection of the diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$. Let $S_{1}, S_{2}, S_{3}$, and $S_{4}$ denote the areas of the triangles $ABO$, $BCO$, $CDO$, and $DAO$.
a) Prove that $S_{1} \cdot S_{3} = S_{2} \cdot S_{4}$.
b) Does there exist a quadrilateral $ABCD$ such that ... | [
"Solution:\n\na. Let $M$ and $N$ be feet of perpendiculars from $B$ and $D$ to $AC$. Then $S_{1} = AO \\cdot BM / 2$, $S_{2} = CO \\cdot BM / 2$, $S_{3} = CO \\cdot DN / 2$, and $S_{4} = AO \\cdot DN / 2$. Now the desired statement follows immediately from the previous four relations.\n\nb. We will prove that the a... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | no | |
0h2m | In a parallelogram $ABCD$ $\angle ABC = 105^\circ$. It is known that inside this parallelogram there is a point $M$, such that the triangle $BMC$ is equilateral and $\angle CMD = 135^\circ$. Let $K$ be a midpoint of the side $AB$. Find $\angle BKC$.
 | [
"Drop the perpendicular $CL$ to the line $DM$. Then in the triangle $MCL$ we know: $\\angle MLC = 90^\\circ$, $\\angle LMC = 45^\\circ$. Therefore, $\\angle LCM = 45^\\circ$ and $CL = \\frac{1}{\\sqrt{2}} CM$ (fig. 36).\n\nSince $\\angle LCD = 60^\\circ$, from the right triangle $LCD$ we find that $CD = 2CL = \\sqr... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 45° | |
024p | Problem:
Cubo - Pedro quer pintar uma caixa de formato cúbico de tal maneira que as faces que tenham uma aresta em comum sejam pintadas em cores diferentes. Calcule o número mínimo de cores que serão necessárias para pintar a caixa dessa maneira. | [
"Solution:\n\nUm cubo tem seis faces distintas, duas a duas opostas, sendo que as faces opostas não têm aresta em comum. Temos três pares de faces opostas, logo três cores são suficientes, bastando pintar as faces opostas de uma mesma cor. Por outro lado, é claro que duas cores somente não bastam."
] | Brazil | Nível 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | 3 | |
025w | Problem:
Na figura abaixo, os segmentos $AB$ e $CD$ são paralelos. Se $\angle CIJ = 10\beta$, $\angle AGJ = 10\alpha$, $\angle CEJ = 6\alpha$ e $\angle JFG = 6\beta$, determine o valor do ângulo $\angle IJK$.
 | [
"Solution:\n\n\nComo $\\angle CEJ$ e $\\angle JFA$ são ângulos colaterais internos, $6\\alpha + 6\\beta = 180^\\circ$, ou seja, $\\alpha + \\beta = 30^\\circ$. Pelo ponto $J$, considere o segmento $ML$ paralelo a $AB$. Temos\n$$\n\\begin{aligned}\nx & = 180^\\circ - \\angle IJG \\\\\n & = ... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 120° | |
01tv | We call two quadratic trinomials $P(x) = x^2 + a x + b$ and $Q(x) = x^2 + c x + d$ *friendly* if each of them has distinct real roots and $x_1 + x_3$, $x_2 + x_4$ are the roots of the quadratic trinomial $x^2 + (a + c)x + b + d$, where $x_1 < x_2$ are the roots of $P(x)$, $x_3 < x_4$ are the roots of $Q(x)$.
Let $M$ be... | [
"Let the trinomials $P(x) = x^2 + a x + b$ and $Q(x) = x^2 + c x + d$ with the roots $x_1 < x_2$ and $x_3 < x_4$, respectively, be friendly. By condition, $x_1 + x_3$ and $x_2 + x_4$ are the roots of the trinomial $x^2 + (a + c)x + b + d$. By Vieta's theorem, we have\n$$\nb + d = (x_1 + x_3)(x_2 + x_4) = x_1 x_2 + ... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
0e2u | Let $ABCD$ be a rectangle with $|AB| > |BC|$. The bisector of the diagonal $AC$ meets the side $CD$ at $E$. The circle with the centre at $E$ and the radius $AE$ meets the segment $AB$ again at $F$. Let $G$ be the orthogonal projection of the point $C$ to the line $EF$. Show that $G$ lies on the diagonal $BD$. | [
"Denote $\\angle FAE = \\alpha$. Since the point $E$ lies on the bisector of the segment $AC$, its distances to the points $A$ and $C$ are the same. Hence, $E$ is the centre of the circle containing the points $A$, $C$ and $F$ and we have $|AE| = |CE| = |FE|$. So, $\\angle EFA = \\angle FAE = \\alpha$. Since $AB$ a... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
083w | Problem:
Quanto vale la somma
$$
\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\cdots+\frac{1}{\sqrt{100}+\sqrt{99}} ?
$$
(A) 9
(B) $\sqrt{101}-1$
(C) $2 \sqrt{\frac{101}{2}}$
(D) 10
(E) nessuna delle precedenti. | [] | Italy | Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO TRIENNIO | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | MCQ | A | |
0gqk | Find all functions $f : \mathbf{R} \to \mathbf{R}$ satisfying
$$
f(f(y) + x^2 + 1) + 2x = y + (f(x+1))^2
$$
for all real numbers $x$ and $y$. | [
"We will show that $f(x) = x$ for all $x$. Substituting $0$ for $x$ in $*$ we obtain\n$$\n(1) \\qquad f(f(y) + 1) = y + (f(1))^2.\n$$\nThen substituting $0$ for $x$ and $f(y) + 1$ for $y$ in $*$ and using (1) we get\n$$\nf(y + (f(1))^2 + 1) = f(y) + 1 + (f(1))^2\n$$\nand now substituting $x^2 + y$ for $y$ gives\n$$... | Turkey | Team Selection Test | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x | |
0acx | Let $H$ be the orthocenter of the triangle $\triangle ABC$ and $A_1$, $B_1$, $C_1$ be the feet of the altitudes from $A$, $B$, $C$ respectively.
Find $\frac{\overline{AH}}{HA_1} \cdot \frac{\overline{BH}}{HB_1} \cdot \frac{\overline{CH}}{HC_1}$ if $\frac{\overline{AH}}{HA_1} + \frac{\overline{BH}}{HB_1} + \frac{\overli... | [
"We have\n$$\n\\frac{\\overline{AH}}{HA_1} + 1 = \\frac{\\overline{AH}}{HA_1} + \\frac{\\overline{HA_1}}{HA_1} = \\frac{\\overline{AA_1}}{HA_1} = \\frac{P_{\\triangle ABC}}{P_{\\triangle HBC}}.\n$$\nSimilarly, $\\frac{\\overline{BH}}{HB_1} + 1 = \\frac{P_{\\triangle HBC}}{P_{\\triangle HCA}}$ and $\\frac{\\overline... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | 2010 | |
0fer | Problem:
Encontrar, razonadamente, dos números enteros positivos $a$ y $b$, tales que
$$
\begin{aligned}
& b^{2} \text{ sea múltiplo de } a, \\
& a^{3} \text{ sea múltiplo de } b^{2}, \\
& b^{4} \text{ sea múltiplo de } a^{3}, \\
& a^{5} \text{ sea múltiplo de } b^{4}, \\
& \text{pero } b^{6} \text{ no sea múltiplo de... | [
"Solution:\n\nEscribamos\n$$\na = p_{1}^{\\alpha_{1}} \\cdots p_{r}^{\\alpha_{r}}, \\quad b = p_{1}^{\\beta_{1}} \\cdots p_{r}^{\\beta_{r}}\n$$\ndonde $\\alpha_{i}, \\beta_{i} \\geq 0$, $p_{i}$ es primo para cada $i$, y $p_{i} \\neq p_{j}$ si $i \\neq j$.\nLas condiciones del problema son entonces equivalentes a\n\... | Spain | TANDA I | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a = 16, b = 8 | |
057j | Let $n$ and $m$ be positive integers. On one turn, an $n$-$m$-knight can move either horizontally by $n$ squares and vertically by $m$ squares or vertically by $n$ squares and horizontally by $m$ squares. (For instance, the usual chess knight, all possible target squares of one move of which are depicted by bullets in ... | [
"Consider three cases:\n\n*Exactly one of the numbers $n$, $m$ is odd.* Color the squares like on a chessboard. Every move changes the color of the square where the knight is, whence after an odd number of moves, the knight is on a square of the opposite color. Thus the knight cannot be on the initial square after ... | Estonia | Open Contests | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | No | |
03zh | We call a point sequence $(A_0, A_1, \dots, A_n)$ interesting, if the abscissa and ordinate are equal for each $A_i$, and the slopes of segment $OA_0, OA_1, \dots, OA_n$ strictly increase ($O$ is the origin), and the area of each $\triangle OA_iA_{i+1}$ ($0 \le i \le n-1$) is $\frac{1}{2}$.
For a point sequence $(A_0,... | [
"We see that by the condition of the problem, an expansion of an interesting sequence is still interesting.\n\nFirst, we construct the interesting sequence $(C_0, C_1, \\dots, C_k)$ containing all points of sequences $(A_0, A_1, \\dots, A_n)$ and $(B_0, B_1, \\dots, B_m)$, and $C_0 = A_0 = B_0$, $C_k = A_n = B_m$.\... | China | China National Team Selection Test | [
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof only | null | |
09h9 | Let $ABCD$ be a cyclic quadrilateral with circumcenter $\omega$, and $E$ be the intersection of the diagonals $AC$ and $BD$. A line passing through $E$ intersects lines $AB$, $BC$ at $P, Q$, respectively. Let $R$ ($R \neq D$) be the intersection point of $\omega$ and a circle that passes through $D, E$ and tangents the... | [
"Let $\\omega_1$ be the circle that passes through $D, E$ and tangents the line $PQ$ at $E$. Since $PQ$ is tangent to the circle $\\omega_1$ at $E$ we have $\\angle EDR = \\angle QER$. So it follows from $\\angle EDR = \\angle BAR$ that $\\angle PAR = \\angle PER$, i.e, $P, A, E, R$ are cyclic. Thus, since $\\angle... | Mongolia | Mongolian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0fv6 | Problem:
Entscheide, ob es eine ganze Zahl $n>1$ gibt mit folgenden Eigenschaften:
a. $n$ ist keine Primzahl.
b. Für alle ganzen Zahlen $a$ ist $a^{n}-a$ durch $n$ teilbar. | [
"Solution:\n\nJa, zum Beispiel $n=561$.\n\nEine solche Zahl $n$ ist quadratfrei: nehme an, $m^{2}>1$ sei ein Teiler von $n$ und setze $a=m$. Dann ist $a^{n}-a$ durch $m$, aber nicht durch $m^{2}$ teilbar, also auch nicht durch $n$. Es muss somit $n=p_{1} \\cdots p_{r}$ gelten mit verschiedenen Primzahlen $p_{i}$. I... | Switzerland | SMO Finalrunde | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | Yes; n = 561 | |
01ia | Find all integer triples $(a, b, c)$ satisfying the equation:
$$
5a^2 + 9b^2 = 13c^2
$$ | [
"Observe that $(a, b, c) = (0, 0, 0)$ is a solution. Assume that the equation has a solution $(a_0, b_0, c_0) \\neq (0, 0, 0)$. Let $d = \\gcd(a_0, b_0, c_0) > 0$. Let $(a, b, c) = (a_0/d, b_0/d, c_0/d)$. Then $\\gcd(a, b, c) = 1$. From $5a_0^2 + 9b_0^2 = 13c_0^2$ it follows that:\n$$\n5a^2 + 9b^2 = 5 \\left(\\frac... | Baltic Way | Baltic Way 2021 Shortlist | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (g... | null | proof and answer | (0, 0, 0) | |
0dcj | There are $2016$ digits written around a circle. If you write down these digits going around the circle clockwise and starting with a certain digit, you get a number divisible by $81$. Prove that if you write down the digits going around the circle clockwise, but starting with any other digit, you will again get a numb... | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof only | null | |
0258 | Problem:
Triângulos - Quantos triângulos existem que tenham um perímetro de 15 unidades e lados medindo números inteiros? | [
"Solution:\n\nVamos supor que $a$, $b$ e $c$ sejam os comprimentos dos lados do triângulo. Não há perda de generalidade em supor que $a \\leq b \\leq c$, de modo que $a+b+c \\leq 3c$. Como cada lado de um triângulo é menor do que a soma dos outros dois, temos que $c < a+b$ e, portanto, obtemos $2c < a+b+c \\leq 3c$... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 7 | |
0gf3 | 令 $n$ 為一正奇數。平面上的整點集 $C = \{(i,j): i,j = 0,1,\dots,2n-1\}$ 構成一個 $2n \times 2n$ 的陣列，每個點上各有一隻天竺鼠，各自面向 $x$ 軸正向、$x$ 軸負向、$y$ 軸正向或 $y$ 軸負向中的其中一個方向。傑夫想要保留其中 $n^2+1$ 隻天竺鼠，並將其餘天竺鼠移除。
接著天竺鼠們作以下運動：在每一回合，被保留的每一隻天竺鼠同時往牠所面向的方向前進一單位長，並保持其面向；但如果一隻天竺鼠要前進的點 $(i,j) \notin C$，則牠改為前進到 $(p,q) \in C$，其中 $p \equiv i \pmod{2n}$ 而 $q \equiv j \p... | [
"先將天竺鼠的面向以箭頭表示，並用西洋棋盤的方式黑 (B) 白 (W) 塗色。這樣我們便有八種箭頭：$B \\to$, $B \\leftarrow$, $B \\uparrow$, $B \\downarrow$, $W \\to$, $W \\leftarrow$, $W \\uparrow$, $W \\downarrow$。\n\n注意到如果兩隻天竺鼠一黑一白，且處於相同或是垂直的方向，則牠們永遠不會停在同一個點上。從而我們得到十二種不會對撞的天竺鼠組合：\n$$\n\\begin{align*}\n& (B \\uparrow, W \\uparrow), (B \\downarrow, W \\downarrow... | Taiwan | 2021 年台灣數學奧林匹亞考試試題 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
088c | Problem:
Determinare tutte le soluzioni $(p, n)$ dell'equazione
$$
n^{3}=p^{2}-p-1
$$
dove $p$ è un numero primo e $n$ è un numero intero. | [
"Solution:\n\nLe soluzioni dell'equazione sono $(p, n)=(2,1)$ e $(p, n)=(37,11)$.\nRiscriviamo l'equazione nella forma\n$$\np(p-1)=(n+1)\\left(n^{2}-n+1\\right) .\n$$\nOsserviamo innanzitutto che per ogni intero $n$ il valore di $n^{2}-n+1$ è positivo, quindi tutti i fattori dell'equazione scritta devono essere pos... | Italy | Cesenatico | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (p, n) = (2, 1) and (37, 11) | |
0kwt | Problem:
Let $ABCDE$ be a convex pentagon such that
$$
\begin{aligned}
& AB + BC + CD + DE + EA = 64 \text{ and } \\
& AC + CE + EB + BD + DA = 72.
\end{aligned}
$$
Compute the perimeter of the convex pentagon whose vertices are the midpoints of the sides of $ABCDE$. | [
"Solution:\n\n\nBy the midsegment theorem on triangles $ABC$, $BCD$, $CDE$, $DEA$, and $EAB$, the side lengths of the said pentagon are $AC/2$, $BD/2$, $CE/2$, $DA/2$, and $EB/2$. Thus, the answer is\n$$\n\\frac{AC + BD + CE + DA + EB}{2} = \\frac{72}{2} = 36\n$$"
] | United States | HMMT November 2023 | [
"Geometry > Plane Geometry > Triangles"
] | null | final answer only | 36 | |
0dmz | Problem:
Да ли постоје природни бројеви $a, b$ и $c$, већи од 2011, такви да у децималном запису важи једнакост
$$
(a+\sqrt{b})^{c}=\ldots 2010,2011 \ldots ?
$$ | [
"Solution:\n\nПоказаћемо да такви бројеви $a, b$ и $c$ постоје. Број $x=(a+\\sqrt{b})^{c}+(a-\\sqrt{b})^{c}$ је цео. Довољно је одабрати $a, b, c$ тако да $x$ буде дељиво са $10^{4}$ и $7989,7989> (a-\\sqrt{b})^{c}>7989,7988$.\n\nЗа непарно $c$, број $x=2 a^{c}+2\\binom{c}{2} a^{c-2}+\\cdots+2\\binom{c}{c-1} a$ је ... | Serbia | СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0a7w | Problem:
Determine all real numbers $x$, such that
$$
x^{n}+x^{-n}
$$
is an integer for all integers $n$. | [
"Solution:\nSet $f_{n}(x)=x^{n}+x^{-n}$. $f_{n}(0)$ is not defined for any $n$, so we must have $x \\neq 0$. Since $f_{0}(x)=2$ for all $x \\neq 0$, we have to find out those $x \\neq 0$ for which $f_{n}(x)$ is an integer for every $n>0$. We note that\n$$\nx^{n}+x^{-n}=(x+x^{-1})(x^{n-1}+x^{1-n})-(x^{n-2}+x^{2-n})\... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 10 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | All real nonzero x such that x + 1/x is an integer; equivalently, x = (m ± sqrt(m^2 − 4)) / 2 for some integer m with |m| ≥ 2. | |
009z | Find all pairs $a, b \in \mathbb{N}, a \neq b$, such that $a+b$ and $a \cdot b+1$ are powers of $2$. | [
"If $a=1$ or $b=1$ we obtain the solutions $(1, 2^n-1)$ and $(2^n-1, 1)$, with $n>1$.\n\nLet $a,b \\ge 2$ and $a<b$. Note that then $a+b < ab+1$ due to the identity $(ab+1)-(a+b) = (a-1)(b-1)$. Let $a+b = 2^n$, $n \\ge 2$; in fact then $n \\ge 3$ as $n=2$ forces $a=b=2$. Then $a = 2^{n-1}-c$, $b = 2^{n-1}+c$ with $... | Argentina | Argentine National Olympiad 2015 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All pairs are (1, 2^n − 1) and (2^n − 1, 1) for n > 1, and (2^n − 1, 2^n + 1) and (2^n + 1, 2^n − 1) for n > 1. | |
0fmx | Sea $A = \{1, 2, 3, \dots, n\}$ con $n > 5$. Demostrar que existe un conjunto finito $B$ de enteros positivos distintos tal que $A \subseteq B$ y tiene la propiedad
$$
\prod_{x \in B} x = \sum_{x \in B} x^2,
$$ | [
"**Solución oficial.** Vamos a demostrar el resultado tomando $A = \\{1, 2, ..., n\\}$, para $n > 5$ (ya que de ser cierto, también será cierto para cualquier subconjunto de éste).\nDado un conjunto $X$ de números naturales, denotemos por $P(X) = \\prod_{x \\in B} x$ y $S(X) = \\sum_{x \\in B} x^2$. Con esta notaci... | Spain | Olimpiada Iberoamericana de Matemáticas | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | Spanish | proof only | null | |
0g75 | 決定所有的整數 $m \ge 2$ 使得所有的 $n, \frac{m}{3} \le n \le \frac{m}{2}$, 整除二項係數 $\binom{n}{m-2n}$。 | [
"解：滿足題設的整數是所有質數。\n\n首先我們檢查所有的質數都會滿足題目的條件。也就是說, 如果 $p$ 是質數則對於所有的 $n$, $1 \\le n \\le \\frac{p}{2}$, 整除二項係數 $\\binom{n}{p-2n}$。當 $p=2$ 的時候是對的。對於奇質數 $p$ 取 $n \\in [1, \\frac{p}{2}]$ 且考慮\n$$\n(p-2n) \\cdot \\binom{n}{p-2n} = n \\cdot \\binom{n-1}{p-2n-1}.\n$$\n因為 $p \\ge 2n$ 且 $p$ 是奇數, 上式每一項因數皆非零。若 $d = \\gcd(p-2n, n)$... | Taiwan | 二〇一三數學奧林匹亞競賽第三階段選訓營, 獨立研究 (三) | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All prime numbers | |
05hn | Problem:
Soit $k \geq 3$ un entier. On définit la suite $\left(a_{n}\right)_{n \geq k}$ par $a_{k}=2k$, et
$$
a_{n}= \begin{cases}a_{n-1}+1 & \text{ si } \operatorname{pgcd}\left(a_{n-1}, n\right)=1 \\ 2n & \text{ sinon. }\end{cases}
$$
Montrer que la suite $\left(a_{n+1}-a_{n}\right)_{n \geq k}$ a une infinité de te... | [
"Solution:\n\nPartons d'un entier $n$ tel que $a_{n}=2n$. Montrons par récurrence que si $p$ est le plus petit facteur premier de $n-1$, alors pour tout $i \\in\\{0, \\ldots, p-2\\}, a_{n+i}=2n+i$. En effet, c'est vrai pour $i=0$, et si c'est vrai pour un certain $i<p-2$, alors\n$$\n\\operatorname{pgcd}(n+i+1,2n+i)... | France | Envoi 1 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
02n6 | Problem:
Os vértices de um cubo são numerados de 1 a 8, de tal modo que uma das faces tem os vértices $\{1,2,6,7\}$ e as outras cinco têm os vértices $\{1,4,6,8\}$, $\{1,2,5,8\}$, $\{2,3,5,7\}$, $\{3,4,6,7\}$ e $\{3,4,5,8\}$. Qual é o número do vértice que está mais distante do vértice de número 6?
(a) 1
(b) 3
(c) 4
... | [
"Solution:\n\nDesenhando o cubo e numerando seus vértices de acordo com o enunciado da questão, obtemos uma figura em que podemos ver que o vértice 5, por ser diametralmente oposto, é o mais distante do vértice 6.\n\n\n\nO vértice 6 está nas faces $\\{1,2,6,7\\}$, $\\{1,4,6,8\\}$ e $\\{3,4,... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes"
] | null | MCQ | d | |
07bm | $b_1 < b_2 < \dots$ is the sequence of all natural numbers that can be written as the sum of squares of two natural numbers. Prove that for infinitely many natural numbers $n$, $b_{n+1} - b_n = 2015$. | [
"We prove that for any odd integer $m > 0$, there are infinitely many positive integers $n$, such that $b_{n+1} - b_n = m$. For sake of this reason we will use the following lemma.\n\n**Lemma 1.** Let $a$ be a positive integer which is not a perfect square. There exist infinitely many primes $p \\equiv 3 \\pmod{4}$... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Other"
] | null | proof only | null | |
0ico | Problem:
Let $f(x) = \cos (\cos (\cos (\cos (\cos (\cos (\cos (\cos x)))))))$, and suppose that the number $a$ satisfies the equation $a = \cos a$. Express $f'(a)$ as a polynomial in $a$. | [
"Solution: $a^{8} - 4 a^{6} + 6 a^{4} - 4 a^{2} + 1$\nThis is an exercise using the chain rule. Define $f_{0}(x) = x$ and $f_{n}(x) = \\cos f_{n-1}(x)$ for $n \\geq 0$. We will show by induction that $f_{n}(a) = a$ and $f_{n}'(a) = (-\\sin a)^{n}$ for all $n$. The case $n = 0$ is clear. Then $f_{n}(a) = \\cos f_{n-... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a^8 - 4 a^6 + 6 a^4 - 4 a^2 + 1 | |
029d | Problem:
Seja $ABC$ um triângulo acutângulo com $AB \neq AC$. Considere todos os retângulos com dois vértices sobre o lado $BC$, um sobre o lado $AB$ e um sobre o lado $AC$. Chamaremos de centro do retângulo o ponto de encontro das diagonais. Na figura a seguir, o centro do retângulo $DEFG$ é o ponto $P$.
![](attache... | [
"Solution:\n\n\n\nComo as coordenadas do ponto médio de um segmento são as médias aritméticas das coordenadas de seus extremos, obtemos:\n$$\n\\begin{aligned}\nM & =\\left(0, \\frac{a}{2}\\right) \\\\\nN & =\\left(\\frac{b+c}{2}, 0\\right)\n\\end{aligned}\n$$\nPara determinar a equação da r... | Brazil | NÍVEL 3 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
09jo | The numbers $1$, $2$, $3$, $\ldots$, $11$, and $12$ are to be divided into several groups. If the maximum number in each group is equal to the sum of the other numbers in that group, how many groups can the numbers be divided into? | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 4 | |
0ihu | Problem:
Somewhere in the universe, $n$ students are taking a 10-question math competition. Their collective performance is called laughable if, for some pair of questions, there exist 57 students such that either all of them answered both questions correctly or none of them answered both questions correctly. Compute ... | [
"Solution:\n\nLet $c_{i, j}$ denote the number of students correctly answering questions $i$ and $j$ ($1 \\leq i < j \\leq 10$), and let $w_{i, j}$ denote the number of students getting both questions wrong. An individual student answers $k$ questions correctly and $10-k$ questions incorrectly. This student answers... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 253 | |
066p | Show that
$$
\left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right) \ge 16,
$$
for all positive real numbers $a$ and $b$ such that $ab \ge 1$. | [] | Greece | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0fav | Problem:
1992 vectors are given in the plane. Two players pick unpicked vectors alternately. The winner is the one whose vectors sum to a vector with larger magnitude (or they draw if the magnitudes are the same). Can the first player always avoid losing? | [
"Solution:\n\nSuppose the vectors sum to $s$. Take the $x$-axis along $s$ (or in any direction if $\\mathbf{s} = 0$). At each move the first player picks the vector with biggest $x$-coordinate. Each player makes $996$ moves and the $x$-coordinate the first player picks on any move is larger than the $x$-coordinate ... | Soviet Union | 1st CIS | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Yes | |
0gwe | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(x^3 + y^3) = x^2 f(x) + y f(y^2)
$$ | [
"Відповідь: $f(x) = kx$, $k \\in \\mathbb{R}$ — довільне.\n\nЗ вихідного рівняння дістаємо, що $f(y^3) = y f(y^2)$, $f(x^3) = x^2 f(x)$. А тому при всіх $x, y \\in \\mathbb{R}$\n$$\nf(x^3 + y^3) = f(x^3) + f(y^3).\n$$\nОтже, ми встановили, що функція $f$ є адитивною: при всіх $u, v \\in \\mathbb{R}$\n$$\nf(u+v) = f... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All functions of the form f(x) = k x for arbitrary real constant k. | |
0eg7 | Problem:
Naj bo $M$ točka na stranici $BC$ pravokotnega trikotnika $ABC$ s pravim kotom pri $A$. Razpolovišče daljice $AM$ označimo s $P$. Naj bosta $P_{B}$ in $P_{C}$ zrcalni sliki točke $P$ preko stranic $AC$ in $AB$, $B'$ in $C'$ pa zrcalni sliki točk $B$ in $C$ preko točke $P$. Denimo, da so točke $B'$, $C'$, $P_{... | [
"Solution:\n\n\n\nKer sta $P_{B}$ in $P_{C}$ zrcalni sliki točke $P$ preko stranic trikotnika, velja $\\Varangle PAC = \\Varangle CAP_{B}$ in $\\Varangle BAP = \\Varangle P_{C}AB$. Torej je\n$$\n\\Varangle P_{C}AP_{B} = \\Varangle P_{C}AB + \\Varangle BAP + \\Varangle PAC + \\Varangle CAP_{... | Slovenia | Slovenian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0bkz | Find all positive primes $p$ such that the system of equations $x^2(y+z) + y^2(z+x) + z^2(x+y) = 3xyz = p^2$ has positive integer solutions. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | no primes | |
0h43 | Positive integers are written on $2012$ cards, one number on each side. On the first card, numbers $1$ and $2$ are written, on the second, $3$ and $4$, ..., on the $2012$th, $4023$ and $4024$. All cards are arranged in line along a table. Two players play the following game. During each move a player chooses $5$ cards ... | [
"Доведемо, що виграє перший гравець. Для цього йому достатньо розбити всі можливі способи розташування чисел на пари, узявши до однієї пари два розташування, які відрізняються одне від одного лише зміною перших п'яти чисел розташувань. Тоді виграшна стратегія першого гравця полягає в тому, щоб завжди перевертати ли... | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | First player | |
05qw | Problem:
Déterminer tous les nombres réels $a$ tels qu'il existe une suite infinie de nombres réels strictement positifs $x_{0}, x_{1}, x_{2}, x_{3}, \ldots$ vérifiant pour tout $n$ l'égalité
$$
x_{n+2}=\sqrt{a x_{n+1}-x_{n}} .
$$ | [
"Solution:\n\nSi $a>1$, on peut prendre la suite constante égale à $a-1$.\n\nSupposons $a \\leqslant 1$. Comme la racine carrée est bien définie, on a nécessairement $a x_{n+1}-x_{n}>0$, donc déjà $a \\geqslant 0$, et de plus $x_{n+1} \\geqslant a x_{n+1}>x_{n}$. Par conséquent, la suite est strictement croissante.... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | a > 1 | |
0j9j | Problem:
Let $P$ be a polynomial such that $P(x) = P(0) + P(1) x + P(2) x^{2}$ and $P(-1) = 1$. Compute $P(3)$. | [
"Solution:\nPlugging in $x = -1, 1, 2$ results in the trio of equations\n$1 = P(-1) = P(0) - P(1) + P(2)$,\n$P(1) = P(0) + P(1) + P(2) \\Rightarrow P(1) + P(2) = 0$,\nand $P(2) = P(0) + 2 P(1) + 4 P(2)$.\nSolving these as a system of equations in $P(0), P(1), P(2)$ gives $P(0) = -1, P(1) = -1, P(2) = 1$.\nConsequen... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | final answer only | 5 | |
0ifk | Problem:
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a continuous function with $\int_{0}^{1} f(x) f^{\prime}(x) d x=0$ and $\int_{0}^{1} f(x)^{2} f^{\prime}(x) d x=18$. What is $\int_{0}^{1} f(x)^{4} f^{\prime}(x) d x$? | [
"Solution:\n\n$0=\\int_{0}^{1} f(x) f^{\\prime}(x) d x=\\int_{f(0)}^{f(1)} u d u=\\frac{1}{2}\\left(f(1)^{2}-f(0)^{2}\\right)$, and\n\n$18=\\int_{0}^{1} f(x)^{2} f^{\\prime}(x) d x=\\int_{f(0)}^{f(1)} u^{2} d u=\\frac{1}{3}\\left(f(1)^{3}-f(0)^{3}\\right)$.\n\nThe first equation implies $f(0)= \\pm f(1)$. The secon... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | proof and answer | 486/5 | |
0f0n | Problem:
$OA$ and $OB$ are tangent to a circle at $A$ and $B$. The line parallel to $OB$ through $A$ meets the circle again at $C$. The line $OC$ meets the circle again at $E$. The ray $AE$ meets the line $OB$ at $K$. Prove that $K$ is the midpoint of $OB$. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0h3r | Let $n \geq 1$ be integer. Sequence $a_1, a_2, \dots, a_{2n}$ is called lucky if
1) $a_1, \dots, a_n$ is a permutation of $\{1, 2, \dots, n\}$;
2) $a_k = a_{n+k}$ for all $k = 1, 2, \dots, n$;
3) there exist indices $1 \le i_1 < i_2 < \dots < i_n \le 2n$ such that $a_{i_k} = k$ for all $k = 1, 2, \dots, n$.
For eac... | [
"Через $m$, $0 \\le m \\le n$, позначимо кількість чисел множини $\\{1, 2, \\dots, n\\}$, які знаходяться серед індексів $i_1 < i_2 < \\dots < i_n$. Неважко довести, що такими $m$ індексами визначаються $n - m$ чисел із множини $\\{n+1, n+2, \\dots, 2n\\}$, які увійдуть до даного набору індексів $i_1 < i_2 < \\dots... | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 2^n - n | |
02a2 | Problem:
Cálculo de segmentos - As medidas do retângulo $ABCD$ são $1200~\mathrm{m}$ por $150~\mathrm{m}$. Além disso, $P$ está no prolongamento do lado $BC$ e dista $350~\mathrm{m}$ de $C$. Determine $AP$, $PQ$, $PD$, $CQ$ e $DP$.
 | [
"Solution:\n\nO triângulo $\\triangle ABP$ é retângulo com catetos $AB = 1200$ e $BP = 150 + 350 = 500$. Pelo Teorema de Pitágoras, temos:\n$$\nAP^2 = 1200^2 + 500^2 = (144 + 25) \\times 10^4 = 169 \\times 10^4 = (13 \\times 10^2)^2\n$$\nLogo, $AP = 13 \\times 10^2 = 1300~\\mathrm{m}$.\n\nAnalogamente, considerando... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | AP = 1300 m, PQ = 910 m, CQ = 840 m, DP = 1250 m, PD = 1250 m | |
08yt | There are three distinct positive integers written on the blackboard. When real numbers $a$, $b$, $c$ are on the blackboard, consider the procedure that replaces those three numbers with $\frac{b+c}{2}$, $\frac{c+a}{2}$, $\frac{a+b}{2}$ at the same time. After this procedure is done $2021$ times, all three numbers on t... | [
"$3 \\cdot 2^{2021} + 3$\n\nFor integers $n = 0, \\dots, 2021$, let $a \\le b \\le c$ be three numbers on the blackboard after the procedure is done $n$ times and let $p_n = c - b$, $q_n = b - a$. Here $n = 0$ means the initial status. When the procedure is done once, three numbers $a \\le b \\le c$ are replaced wi... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 3 * 2^2021 + 3 | |
0g18 | Problem:
Soient $a, c \in \mathbb{N}$ et $b \in \mathbb{Z}$. Prouver qu'il existe $x \in \mathbb{N}$ tel que
$$
a^{x}+x \equiv b \quad \bmod c
$$ | [
"Solution:\n\nSupposons dans un premier temps que $c=p^{n}$ est une puissance d'un nombre premier. Nous allons construire une suite $\\left(x_{1}, \\ldots, x_{n}\\right)$ telle que pour tout $1 \\leq k \\leq n$ on ait $a^{x_{k}}+x_{k} \\equiv b$ $\\bmod p^{k}$ et $x_{k} \\equiv x_{k-1} \\bmod (p-1) p^{k-1}$. Rappel... | Switzerland | IMO-Selektion | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Div... | null | proof only | null | |
00ex | Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive real numbers, and let
$$
S = x_{1} + x_{2} + \cdots + x_{n} .
$$
Prove that
$$
\left(1 + x_{1}\right)\left(1 + x_{2}\right) \cdots \left(1 + x_{n}\right) \leq 1 + S + \frac{S^{2}}{2!} + \frac{S^{3}}{3!} + \cdots + \frac{S^{n}}{n!} .
$$ | [
"Let $\\sigma_{k}$ be the $k$th symmetric polynomial, namely\n$$\n\\sigma_{k} = \\sum_{\\substack{|S| = k \\\\ S \\subseteq \\{1,2, \\ldots, n\\}}} \\prod_{i \\in S} x_{i},\n$$\nand more explicitly\n$$\n\\sigma_{1} = S, \\quad \\sigma_{2} = x_{1} x_{2} + x_{1} x_{3} + \\cdots + x_{n-1} x_{n}, \\quad \\text{and so o... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1989 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
03ws | Let $n \ge 3$ be a given integer, and $a_1, a_2, \dots, a_n$ be real numbers satisfying $\min_{1 \le i < j \le n} |a_i - a_j| = 1$. Find the minimum value of $\sum_{k=1}^n |a_k|^3$. (Posed by Zhu Huawei) | [
"Without loss of generality, we may assume that $a_1 < a_2 < \\dots < a_n$, and note also that\n$$\n|a_k| + |a_{n-k+1}| \\ge |a_{n-k+1} - a_k| \\ge n + 1 - 2k\n$$\nfor $1 \\le k \\le n$. So\n$$\n\\begin{align*}\n\\sum_{k=1}^{n} |a_k|^3 &= \\frac{1}{2} \\sum_{k=1}^{n} (|a_k|^3 + |a_{n+1-k}|^3) \\\\\n&= \\frac{1}{2} ... | China | Chinese Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | Minimum equals (n^2 - 1)^2 / 32 for odd n, and n^2 (n^2 - 2) / 32 for even n; attained by a_i = i - (n + 1) / 2. | |
00x5 | Problem:
Prove that for any positive $x_{1}, x_{2}, \ldots, x_{n}$ and $y_{1}, y_{2}, \ldots, y_{n}$ the inequality
$$
\sum_{i=1}^{n} \frac{1}{x_{i} y_{i}} \geq \frac{4 n^{2}}{\sum_{i=1}^{n}\left(x_{i}+y_{i}\right)^{2}}
$$
holds. | [
"Solution:\nSince $\\left(x_{i}+y_{i}\\right)^{2} \\geq 4 x_{i} y_{i}$, it is sufficient to prove that\n\n$$\n\\left(\\sum_{i=1}^{n} \\frac{1}{x_{i} y_{i}}\\right)\\left(\\sum_{i=1}^{n} x_{i} y_{i}\\right) \\geq n^{2}\n$$\n\nThis can easily be done by induction using the fact that $a+\\frac{1}{a} \\geq 2$ for any $... | Baltic Way | Baltic Way 1992 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0jh2 | Let $ABC$ be an acute triangle. Circle $\omega_1$, with diameter $AC$, intersects side $BC$ at $F$ (other than $C$). Circle $\omega_2$, with diameter $BC$, intersects side $AC$ at $E$ (other than $C$). Ray $AF$ intersects $\omega_2$ at $K$ and $M$ with $AK < AM$. Ray $BE$ intersects $\omega_1$ at $L$ and $N$ with $BL <... | [
"Let $D$ be the foot of the perpendicular from $C$ to $AB$ and $H$ be the orthocenter of $\\triangle ABC$. Note first that $\\omega_1$ and $\\omega_2$ both intersect $AB$ at $D$. By Power of a Point, $LH \\cdot HN = CH \\cdot HD = KH \\cdot HM$, implying that $KLMN$ is a cyclic quadrilateral. Noting that $AC$ and $... | United States | IMO Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis ... | null | proof only | null | |
09ra | Problem:
Vind alle viertallen $(a, b, c, d)$ van reële getallen waarvoor geldt
$$
\begin{aligned}
& a b + c + d = 3, \\
& b c + d + a = 5, \\
& c d + a + b = 2, \\
& d a + b + c = 6.
\end{aligned}
$$ | [
"Solution:\nDe eerste twee vergelijkingen van elkaar aftrekken geeft\n$$\n2 = 5 - 3 = (b c + d + a) - (a b + c + d) = b(c - a) + a - c = (b - 1)(c - a).\n$$\nDe laatste twee vergelijkingen van elkaar aftrekken geeft\n$$\n4 = 6 - 2 = (d a + b + c) - (c d + a + b) = d(a - c) + c - a = (1 - d)(c - a)\n$$\nWe zien dat ... | Netherlands | MO-selectietoets | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (2, 0, 0, 3) | |
05sk | Problem:
Soit $\left(x_{n}\right)_{n \in \mathbb{N}}$ une suite réelle telle que $x_{0}=0$ et $x_{1}=1$ et pour tout $n \geqslant 0$, $x_{n+2}=3 x_{n+1}-2 x_{n}$. On pose également $y_{n}=x_{n}^{2}+2^{n+2}$ pour tout entier naturel $n$. Montrer que pour tout entier $n>0$, $y_{n}$ est le carré d'un entier impair. | [
"Solution:\n\nOn commence par tester l'énoncé pour des petites valeurs de $n$. On calcule donc les premières valeurs des suites $\\left(x_{n}\\right)$ et $\\left(y_{n}\\right)$. On trouve par exemple $x_{2}=3$, $x_{3}=7$, $x_{4}=15$, $\\ldots$ On peut donc conjecturer que $x_{n}=2^{n}-1$. On s'empresse de le démont... | France | Envoi 5: Pot Pourri | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0bc5 | The matrix $A \in M_k(\mathbb{R})$ ($k \ge 2$) has the property: for every positive integer $n$, $\det(n^2 I_k - A^2) \ge n^k (\det(nI_k - A) + \det(nI_k + A) - 1)$. Prove that $\operatorname{tr}(A) = 0$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Algebra > Linear Algebra > Determinants",
"Algebra > Linear Algebra > Matrices",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof only | null | |
0iyo | Problem:
Daniel wrote all the positive integers from $1$ to $n$ inclusive on a piece of paper. After careful observation, he realized that the sum of all the digits that he wrote was exactly $10,\!000$. Find $n$. | [
"Solution:\n\nLet $S(n)$ denote the sum of the digits of $n$, and let $f(x) = \\sum_{n=0}^{x} S(n)$. (We may add $n=0$ because $S(0)=0$.) Observe that:\n\n$$\nf(99) = \\sum_{a=0}^{9} \\left( \\sum_{b=0}^{9} (a+b) \\right) = 10 \\sum_{b=0}^{9} b + 10 \\sum_{a=0}^{9} a = 900\n$$\n\nIf $a$ is an integer between $1$ an... | United States | Harvard-MIT November Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | null | final answer only | 799 | |
00ii | We are given the set $M_n = \{0, 1, 2, ..., n\}$ of all non-negative integers less than or equal to $n$. We call a subset $S$ of $M_n$ *outstanding* if it is not empty and a $k$-element subset of $S$ exists for all $k \in S$. Determine the number of outstanding subsets of $M_n$. | [
"If $k$ is the largest element of an outstanding subset $S$, it follows that $S$ must contain $k$ elements. This is possible if it either contains all elements not greater than $k$ or all but one. We see that each outstanding subset of $M_n$ corresponds to an ordered pair $(a, b)$ of integers with $n \\ge a \\ge b ... | Austria | Austria 2010 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | binom(n+2, 2) | |
04tz | Let $ABC$ be an acute triangle with orthocenter $H$. The bisector of angle $BHC$ intersects side $BC$ at $D$. Denote by $E$, $F$ the reflections of $D$ about $AB$, $AC$, respectively. Prove that the circumcircle of triangle $AEF$ passes through the midpoint of arc $BAC$.
(Patrik Bak) | [
"Clearly, the directed angle $EAF$ is $\\widehat{EAF} = 2\\alpha$. Let $H_2$, $H_3$ be the reflections of the orthocenter $H$ about the sides $AC$, $AB$, respectively. It is well-known that $H_2$ and $H_3$ lie on the circumcircle $k$ of triangle $ABC$.\n\n\nFig. 1\n\nAs $DH$ is the bisector... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0952 | Problem:
Într-un trapez dreptunghic $ABCD$, cu unghiurile drepte în vârfurile $A$ şi $B$, se află două cercuri. Unul din ele este tangent la laturile laterale și la baza mare $AD$, iar celălalt este tangent la laturile laterale, la baza mică $BC$ şi la primul cerc.
1) Dreapta ce trece prin centrele cercurilor interse... | [
"Solution:\n\n1) Deoarece cercurile sunt tangente între ele, construim tangenta lor comună $LZ$. Notăm: $m(\\angle D) = \\alpha$, $R$ – raza cercului mare cu centrul în $O_1$, $r$ – raza cercului mic cu centrul în $O_2$, $M$ – piciorul perpendicularei dusă din $O_1$ pe $AD$, $N$ – piciorul perpendicularei dusă din ... | Moldova | A 61-a OLIMPIAD DE MATEMATICA A REPUBLICII MOLDOVA | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 116/7 | |
03fh | A three-digit natural number $n$ is initially written on the board. Two players, $A$ and $B$, take turns, with $A$ going first. The turner reduces the number on the board by some divisor of his own (i.e., other than $1$ and the number itself). For example, if at some point the number on the board is $6$, it can be redu... | [
"If there is a prime number on the board, the player loses by definition. If there is an even number on the board that is not a power of $2$, then the player can always reduce it by its odd divisor, leaving an odd number on the board. If the number on the board is odd and is reduced by its (odd) divisor $a$, i.e. a... | Bulgaria | Bulgarian Spring Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | All even three-digit numbers except 128 and 512 (448 total). | |
08h0 | Problem:
Find all pairs of positive integers $(x, y)$ such that $2^{x} + 3^{y}$ is a perfect square. | [
"Solution:\nIn order for the expression $2^{x} + 3^{y}$ to be a perfect square, a positive integer $t$ such that $2^{x} + 3^{y} = t^{2}$ should exist.\n\nCase 1. If $x$ is even, then there exists a positive integer $z$ such that $x = 2z$. Then\n$$\n(t - 2^{z})(t + 2^{z}) = 3^{y}\n$$\nSince $t + 2^{z} - (t - 2^{z}) ... | JBMO | null | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (4, 2) | |
0czr | Find all positive integers $n$ for which the equation
$$
x^{3}+y^{3}=n!+4
$$
has solutions in integers. | [
"For any integer $a$ we have $a^{3} \\equiv 0,1,8 \\pmod{9}$, hence for any integers $x$ and $y$, we have\n$$\nx^{3}+y^{3} \\equiv 0,1,2,7,8 \\quad (\\bmod 9).\n$$\nFor $n \\geq 6$, we have $n!+4 \\equiv 4 \\pmod{9}$, that is there are no solutions in this case.\n\nIf $n=1$, then the equation becomes $x^{3}+y^{3}=5... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | n = 4 and n = 5 | |
0k5o | Problem:
Each person in Cambridge drinks a (possibly different) 12 ounce mixture of water and apple juice, where each drink has a positive amount of both liquids. Marc McGovern, the mayor of Cambridge, drinks $\frac{1}{6}$ of the total amount of water drunk and $\frac{1}{8}$ of the total amount of apple juice drunk. H... | [
"Solution:\n\nThe total amount of liquid drunk must be more than 6 times the amount that Marc drinks but less than 8 times the amount that Marc drinks. Therefore, the number of people in Cambridge is more than 6 but less than 8, so it must be 7."
] | United States | HMMT November 2019 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 7 | |
0cwc | Several cardboard rectangles, possibly of different sizes, are put onto a rectangular table; the sides of rectangles are parallel to those of the table. The rectangles may overlap, but there are no two rectangles with four common vertices. May it happen that each point which is a vertex of some rectangle is in fact a v... | [
"\nРис. 4\n\nНа рис. 4 показано, как можно положить три пары прямоугольников так, чтобы для каждой пары все точки $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$ были вершинами ровно по разу. Одинаковыми точками отмечены вершины одного из прямоугольников пары.\n\n**Замечание.** Существует много друг... | Russia | Regional round | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English; Russian | proof and answer | Yes; such configurations exist. | |
0jqm | Problem:
Let $X=\{1,2, \ldots, 100\}$. How many functions $f: X \rightarrow X$ satisfy $f(b)<f(a)+(b-a)$ for all $1 \leq a<b \leq 100$? | [
"Solution:\nThe answer is $\\binom{199}{100}$. We claim that the functions are precisely those of the form $f(n)=n+a_{n}$, where\n\n$$\n-99 \\leq a_{100}<a_{99}<\\cdots<a_{1} \\leq 99\n$$\n\nis an arbitrary sequence. The answer follows from this.\n\nTo see that all functions are of this form, we rewrite the given a... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | C(199,100) | |
050i | On a square board with $m$ rows and $n$ columns, where $m \le n$, some squares are colored black in such a way that no two rows are alike. Find the biggest integer $k$ such that for every possible coloring to start with one can always color $k$ columns entirely red in such a way that no two rows are still alike.
**Ans... | [
"Prove that if $m \\le n$, one of the columns can always be colored red. Then, when excluding this column, we can continue the process until the number of columns is smaller than the number of rows, i.e. $n-m+1$ times. Suppose we cannot color a single column red such that no two rows still appear alike. Then, for e... | Estonia | IMO Team Selection Contest | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Graph Theo... | English | proof and answer | n - m + 1 | |
08u9 | Suppose there are $2010$ airports. Each airport has a number of direct flights to some of the other airports and the following conditions (1), (2) are known to be satisfied:
(1) For any pair of airports, say $A$ and $B$, one can go from $A$ to $B$, by making connections of several direct flights.
(2) If any one of the ... | [
"Let us say that the airport $B$ is accessible from the airport $A$ if one can reach $B$ starting from $A$ by making connections of direct flights.\n\nFirst, we will show that the answer we seek is no more than $1004^2$.\n\nIn the sequel until we say otherwise, we will assume that we are in the situation where one ... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | 1008016 | |
00hs | Let $n$ and $k$ be positive integers. Cathy is playing the following game. There are $n$ marbles and $k$ boxes, with the marbles labelled $1$ to $n$. Initially, all marbles are placed inside one box. Each turn, Cathy chooses a box and then moves the marbles with the smallest label, say $i$, to either any empty box or t... | [
"We claim Cathy can win if and only if $n \\leq 2^{k-1}$.\nFirst, note that each non-empty box always contains a consecutive sequence of labeled marbles. This is true since Cathy is always either removing from or placing in the lowest marble in a box. As a consequence, every move made is reversible.\n\nNext, we pro... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | All pairs (n, k) with n ≤ 2^{k-1} | |
0c7m | Given a positive integer $n$, a triangular array $(a_{i,j})$ of zeroes and ones, where $i$ and $j$ run through the positive integers such that $i + j \le n + 1$, is called a *binary anti-Pascal n-triangle* if $a_{i,j} + a_{i,j+1} + a_{i+1,j} \equiv 1 \pmod 2$ for all possible values $i$ and $j$ may take on. Determine t... | [
"In what follows, part of the generic configurations referred to may not exist for the first few values of $n$; in this case, simply consider the corresponding induced\n\nWe now show by induction on $n$ that a binary anti-Pascal $n$-triangle contains at least $\\lfloor n(n+1)/6 \\rfloor$ ones, of which at least $n-... | Romania | Stars of Mathematics Competition | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | ⌊n(n+1)/6⌋ | |
0e7x | Problem:
Največ koliko praštevil lahko vsebuje nekonstantno geometrijsko zaporedje pozitivnih realnih števil? | [
"Solution:\n\nOdgovor je $2$. Primer takega zaporedja je na primer zaporedje $a_{n} = 2 \\left(\\frac{3}{2}\\right)^{n-1}$, ki vsebuje praštevili $2$ in $3$.\n\nDenimo, da geometrijsko zaporedje $a_{n} = a q^{n-1}$, kjer je $q \\neq 1$, vsebuje tri praštevila. Denimo, da so to $a_{k}$, $a_{m}$ in $a_{n}$, kjer je $... | Slovenia | 57. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | proof and answer | 2 | |
07et | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for any two real numbers $x$ and $y$,
$$
f(2xy)^2 + f(f(x)^2 - y^2)^2 = f(x^2 + y^2)^2.
$$ | [
"Obviously, the constant function $f(x) = 0$ is an answer. Let $f$ be a non-constant function satisfying the problem.\nDefine $g$ to be $g(x) = f(x)^2$ for all $x$. Since $f(x)^2 \\ge 0$, $g$ would always be non-negative.\nLet $P$ denote the assertion that\n$$\ng(2xy) + g(g(x) - y^2) = g(x^2 + y^2)\n$$\nFor every $... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | All solutions are: (1) the constant zero function; and (2) all functions satisfying f(x)^2 = x^2 for all real x (i.e., for each real x, f(x) equals either x or negative x). | |
0kv9 | Problem:
There is a $6 \times 6$ grid of lights. There is a switch at the top of each column and on the left of each row. A light will only turn on if the switches corresponding to both its column and its row are in the "on" position. Compute the number of different configurations of lights. | [
"Solution:\n\nTake any configuration of switches such that there exists at least one row and one column which are switched on. There are $\\left(2^{6}-1\\right)^{2}=3969$ such configurations.\n\nWe prove that any two such configurations $A$ and $B$ lead to a different set of lights. Without loss of generality assum... | United States | HMMT November 2023 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 3970 | |
00df | Let $ABCDEFGHI$ be a regular 9-gon with its vertices labelled anticlockwise, and let $ABJKLM$ be a regular hexagon with its vertices also labelled anticlockwise. Prove that the angles $\angle HMG$ and $\angle KEL$ are equal. | [
"First we recall that since $ABCDEFGHI$ is a regular 9-gon all of its sides are equal and each of its angles is equal to $\\frac{7 \\cdot 180^\\circ}{9} = 140^\\circ$; likewise, all sides of $ABJKLM$ are equal and each of its angles is equal to $120^\\circ$.\n\nBy symmetry, $\\angle KEL = \\angle LGK$, so it suffic... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
08bl | Problem:
Sia $ABCD$ un rettangolo con $AB > BC$ e sia $\omega$ la sua circonferenza circoscritta. Siano $E$ e $F$ rispettivamente le intersezioni (distinte da $A$) della bisettrice dell'angolo $\widehat{BAD}$ con il lato $CD$ e la circonferenza $\omega$. La perpendicolare a $DF$ passante per $E$ interseca la corda $DF... | [
"Solution:\n\n(a) Per dimostrare (a) ci basta mostrare che il triangolo $BDF$ è isoscele su base $BD$. Siccome $AF$ è bisettrice dell'angolo $\\widehat{BAD}$, abbiamo $\\widehat{FAD} = \\widehat{FAB}$. Inoltre, $\\widehat{FBD} = \\widehat{FAD}$ poiché insistono sullo stesso arco di circonferenza e, per il medesimo ... | Italy | Progetto Olimpiadi della Matematica | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02hb | Let $P$ be a convex $2006$-gon. The $1003$ diagonals connecting opposite vertices and the $1003$ lines connecting the midpoints of opposite sides are concurrent, that is, all $2006$ lines have a common point. Prove that the opposite sides of $P$ are parallel and congruent. | [
"Let $A_1A_2\\ldots A_{2006}$ be a polygon such that $A_1A_{1004}$, $A_2A_{1005}$, $\\ldots$, $A_{1003}A_{2006}$ and $M_1M_{1004}$, $M_2M_{1005}$, $\\ldots$, $M_{1003}M_{2006}$ are concurrent at $O$, where $M_1, M_2, \\ldots, M_{2006}$ are the midpoints of $A_1A_2, A_2A_3, \\ldots, A_{2006}A_1$, respectively.\n\nNo... | Brazil | Brazilian Math Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0jen | Problem:
Two fair six-sided dice are rolled. What is the probability that their sum is at least $10$? | [
"Solution:\n\nThere are $3$, $2$, $1$ outcomes with sum $10$, $11$, $12$, so the probability is\n$$\n\\frac{3+2+1}{6^2} = \\frac{6}{36} = \\frac{1}{6}.\n$$"
] | United States | HMMT November 2013 | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 1/6 | |
0jzd | Problem:
Let $ABC$ be an acute triangle. The altitudes $BE$ and $CF$ intersect at the orthocenter $H$, and point $O$ denotes the circumcenter. Point $P$ is chosen so that $\angle APH = \angle OPE = 90^{\circ}$, and point $Q$ is chosen so that $\angle AQH = \angle OQF = 90^{\circ}$. Lines $EP$ and $FQ$ meet at point $T... | [
"Solution:\n\nObserve that $T$ is the radical center of the circles with diameter $OE$, $OF$, $AH$. So $T$ lies on the radical axis of $(OE), (OF)$ which is the altitude from $O$ to $EF$, hence passing through $A$.\n\nSo $A T O$ are collinear, done."
] | United States | February 2017 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00fe | Find the total number of different integer values the function
$$
f(x) = [x] + [2x] + \left[\frac{5x}{3}\right] + [3x] + [4x]
$$
takes for real numbers $x$ with $0 \leq x \leq 100$.
Note: $[t]$ is the largest integer that does not exceed $t$. | [
"Note that, since $[x+n]=[x]+n$ for any integer $n$,\n$$\nf(x+3) = [x+3] + [2(x+3)] + \\left[\\frac{5(x+3)}{3}\\right] + [3(x+3)] + [4(x+3)] = f(x) + 35,\n$$\none only needs to investigate the interval $[0,3)$.\nThe numbers in this interval at which at least one of the real numbers $x, 2x, \\frac{5x}{3}, 3x, 4x$ is... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1993 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 734 | |
0krs | Problem:
Let triangle $ABC$ be such that $AB = AC = 22$ and $BC = 11$. Point $D$ is chosen in the interior of the triangle such that $AD = 19$ and $\angle ABD + \angle ACD = 90^\circ$. The value of $BD^2 + CD^2$ can be expressed as $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Compute $100a ... | [
"Solution:\n\nRotate triangle $ABD$ about $A$ so that $B$ coincides with $C$. Let $D$ map to $D'$ under this. Note that $CDD'$ is a right triangle with right angle at $C$. Also, note that $ADD'$ is similar to $ABC$. Thus, we have $DD' = \\frac{AD}{2} = \\frac{19}{2}$. Finally, note that\n$$\nBD^2 + CD^2 = CD'^2 + C... | United States | HMMT February 2022 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 36104 | |
02ri | What is the smallest positive integer $n$ such that there exists a positive integer $k$ such that the last 2012 digits of $n^k$ are all equal to 1? | [
"*Answer: 71.*\n\nSince $\\underbrace{11\\dots1}_{2012 \\text{ ones}} = \\frac{10^{2012}-1}{9}$, the problem is equivalent to finding the smallest $n$ such that, for some $k$,\n$$\n\\begin{align*}\nn^k &\\equiv \\frac{10^{2012} - 1}{9} \\pmod{10^{2012}} \\\\\n9n^k &\\equiv -1 \\pmod{10^{2012}} \\\\\n&\\iff n^{-k} \... | Brazil | Brazilian Math Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 71 | |
0baz | Let $A$, $B \in \mathcal{M}_2(\mathbb{C})$ be two non-zero matrices with $AB + BA = O_2$ and $\det(A + B) = 0$. Prove that $\text{tr}(A) = \text{tr}(B) = 0$. | [
"So, subtracting we get $\\text{tr}(A)B = \\text{tr}(B)A$.\nIf $\\text{tr}(A) = 0$, then $\\text{tr}(B) = 0$ for else $A = O_2$, a contradiction.\nHence $A = \\lambda B$, so $AB + BA = O_2$ leads to $\\lambda B^2 = O_2$. Therefore $\\lambda = 0$, which yields $\\text{tr}(A) = \\text{tr}(B) = 0$."
] | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof only | null | |
03vx | Suppose that the minimum of $f(x) = \cos 2x - 2a(1 + \cos x)$ is $-\frac{1}{2}$. Then $a = \underline{\hspace{2cm}}$. | [
"$$\n\\begin{aligned}\nf(x) &= 2\\cos^2 x - 1 - 2a - 2a \\cos x \\\\\n&= 2\\left(\\cos x - \\frac{a}{2}\\right)^2 - \\frac{1}{2}a^2 - 2a - 1.\n\\end{aligned}\n$$\nFor $a > 2$, $f(x)$ takes the minimum value of $1 - 4a$ when $\\cos x = 1$; for $a < -2$, $f(x)$ takes the minimum $1$ when $\\cos x = -1$; for $-2 \\le ... | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | a = -2 + √3 | |
08ev | Problem:
Nello studiare il polinomio $p(x) = x^{2} + 2x - 6$, Enrica ha scoperto due numeri reali distinti $\alpha$ e $\beta$ tali che $p(\alpha) = \beta$ e $p(\beta) = \alpha$. Quanto vale $\alpha + \beta$?
(A) -3
(B) $-2 \sqrt{2}$
(C) 0
(D) 2
(E) 6 | [
"Solution:\n\nLa risposta è $\\mathbf{(A)}$. Le condizioni imposte si traducono nel sistema\n$$\n\\left\\{\\begin{array}{l}\n\\alpha^{2} + 2\\alpha - 6 = \\beta \\\\\n\\beta^{2} + 2\\beta - 6 = \\alpha\n\\end{array}\\right.\n$$\nSottraendo le due equazioni otteniamo $\\alpha^{2} + 2\\alpha - (\\beta^{2} + 2\\beta) ... | Italy | Italian Mathematical Olympiad - February Round | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | A | |
0iy6 | Problem:
Let $F_{n}$ be the Fibonacci sequence, that is, $F_{0}=0$, $F_{1}=1$, and $F_{n+2}=F_{n+1}+F_{n}$. Compute $\sum_{n=0}^{\infty} F_{n} / 10^{n}$. | [
"Solution:\n\nWrite $F(x)=\\sum_{n=0}^{\\infty} F_{n} x^{n}$. Then the Fibonacci recursion tells us that $F(x)-x F(x)-x^{2} F(x)=x$, so $F(x)=x /\\left(1-x-x^{2}\\right)$. Plugging in $x=1 / 10$ gives the answer."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 10/89 | |
0l0t | Problem:
Kelvin the frog currently sits at $(0,0)$ in the coordinate plane. If Kelvin is at $(x, y)$, either he can walk to any of $(x, y+1)$, $(x+1, y)$, or $(x+1, y+1)$, or he can jump to any of $(x, y+2)$, $(x+2, y)$ or $(x+1, y+1)$. Walking and jumping from $(x, y)$ to $(x+1, y+1)$ are considered distinct actions. ... | [
"Solution:\nObserve there are $\\binom{14}{6} = 3003$ up-right paths from $(0,0)$ to $(6,8)$, each of which are $14$ steps long. Any two of these steps can be combined into one: $UU$, $RR$, and $RU$ as jumps, and $UR$ as walking from $(x, y)$ to $(x+1, y+1)$. The number of ways to combine steps is the number of way... | United States | HMMT February 2024 Guts Round | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 1831830 | |
01qw | Construct a tetramino by attaching two $2 \times 1$ dominoes along their sides such that the midpoint of the longer side of one domino is a corner of other domino. This construction yields two kinds of the tetraminoes with opposite orientations. Let us call them S- and Z-tetraminoes, respectively.
![](attached_image_1.... | [
"Consider the following arrangement of numbers in the cells of the lattice (see Fig.1).\n\nIt is evident that the sum of the numbers in any S-tetramino is always zero, so the sum of the numbers in all cells of the polygon $P$ from the problem condition equals zero.\n\nFig. 2\n![](attached_i... | Belarus | SELECTION and TRAINING SESSION | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0i9h | Problem:
All the sequences consisting of five letters from the set $\{T, U, R, N, I, P\}$ (with repetitions allowed) are arranged in alphabetical order in a dictionary. Two sequences are called "anagrams" of each other if one can be obtained by rearranging the letters of the other. How many pairs of anagrams are there... | [
"Solution:\n\nConvert each letter to a digit in base $6$: $I \\mapsto 0$, $N \\mapsto 1$, $P \\mapsto 2$, $R \\mapsto 3$, $T \\mapsto 4$, $U \\mapsto 5$. Then the dictionary simply consists of all base-$6$ integers from $00000_{6}$ to $55555_{6}$ in numerical order. If one number can be obtained from another by a r... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 0 | |
0chz | Let $f : [0, \infty) \to \mathbb{R}$ be a differentiable function, with continuous derivative, such that $f(0) = 0$, and $0 \le f'(x) \le 1$, for any $x > 0$. Prove that
$$
\int_{0}^{a} f(t)^{2n+1} dt \le (n+1) \cdot \left( \int_{0}^{a} f(t)^{n} dt \right)^{2},
$$
for any $a > 0$ and $n \in \mathbb{N}^*$. | [
"Because $f'(x) \\ge 0$ for any $x \\ge 0$, the function $f$ is monotonically increasing, so that $f(x) \\ge f(0) = 0$, for any $x \\ge 0$.\n\nConsider an arbitrary fixed $n \\in \\mathbb{N}^*$ and the function $F : [0, \\infty) \\to \\mathbb{R}$ defined for any $x \\ge 0$ by\n$$\nF(x) = (n + 1) \\cdot \\left( \\in... | Romania | 74th Romanian Mathematical Olympiad | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Integral Calculus > Applications",
"Calculus > Differential Calculus > Applications"
] | English | proof only | null | |
0bgx | a) Prove that, for every positive integers $a, b, k$ such that $a < b$,
$$
\frac{a}{b} \le \frac{a+k}{b+k}.
$$
b) Prove that
$$
\frac{1}{100} + \frac{4}{101} + \frac{7}{102} + \dots + \frac{148}{149} \ge 25.
$$ | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null |
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