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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
07o0 | Let $ABC$ be an isosceles triangle with $|AB| = |AC|$. The points $D, E$ and $F$ are on the sides $BC, CA$ and $AB$, respectively, such that $\angle FDE = \angle ABC$ and $FE$ is not parallel to $BC$. Prove that $BC$ is tangent to the circumcircle of $\triangle DEF$ if and only if $D$ is the midpoint of $BC$. | [
"Assume first that $BC$ is tangent to the circumcircle of $\\triangle DEF$. Then $\\angle EDC = \\angle DFE$. By assumption $\\angle FDE = \\angle ABC = \\angle ACB$, thus the two triangles $DEC$ and $DEF$ are similar. Hence, $\\frac{|BD|}{|FD|} = \\frac{|DE|}{|FE|}$. Similarly we obtain that $\\triangle FBD$ and $... | Ireland | Ireland | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
06su | Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that
$$
n^{2}+4 f(n)=f(f(n))^{2}
$$
for all $n \in \mathbb{Z}$. | [
"Part I. Let us first check that each of the functions above really satisfies the given functional equation. If $f(n)=n+1$ for all $n$, then we have\n$$\nn^{2}+4 f(n)=n^{2}+4 n+4=(n+2)^{2}=f(n+1)^{2}=f(f(n))^{2} .\n$$\nIf $f(n)=n+1$ for $n>-a$ and $f(n)=-n+1$ otherwise, then we have the same identity for $n>-a$ and... | IMO | 55th International Mathematical Olympiad Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadrat... | null | proof and answer | All functions f from the integers to the integers satisfying the equation are exactly the following:
1) f(n) = n + 1 for all integers n.
2) For any integer a ≥ 1, the piecewise function given by f(n) = 1 − n for all n ≤ −a and f(n) = n + 1 for all n ≥ −a + 1.
3) The sign-symmetric piecewise function with f(n) = n + 1 f... | |
04vb | In a convex quadrilateral $ABCD$, $|AB| = |BC| = |CD|$. Let furthermore, for the intersection $P$ of its diagonals, $|\angle APD| < 90^\circ$. Let $R$ and $S$ be reflections of $A$ and $D$ with respect to $BD$ and $AC$, respectively. Prove that the lines $BC$ and $RS$ are parallel. | [
"$$\n|AB| = |BC| = |CD| = |BR| = |CS|. \\qquad (1)\n$$\nBy construction, the points $R, S$ are obviously different and the midpoint $X$ of the segment $AR$ lies on its perpendicular bisector $BD$ and the midpoint $Y$ of the segment $DS$ lies on its perpendicular bisector $AC$. In the following paragraph we prove th... | Czech Republic | 72nd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
0l3o | Problem:
Positive integers $a$, $b$, and $c$ have the property that $a^{b}$, $b^{c}$, and $c^{a}$ end in $4$, $2$, and $9$, respectively. Compute the minimum possible value of $a+b+c$. | [
"Solution:\n\nThis minimum is attained when $(a, b, c) = (2, 2, 13)$. To show that we cannot do better, observe that $a$ must be even, so $c$ ends in $3$ or $7$. If $c \\geq 13$, since $a$ and $b$ are even, it's clear $(2, 2, 13)$ is optimal. Otherwise, $c = 3$ or $c = 7$, in which case $b^{c}$ can end in $2$ only ... | United States | HMMT February 2024 Guts Round | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 17 | |
061q | Problem:
Sei $N$ eine natürliche Zahl und $x_{1}, x_{2}, \ldots, x_{n}$ weitere natürliche Zahlen kleiner als $N$ und so, dass das kleinste gemeinsame Vielfache von beliebigen zwei dieser $n$ Zahlen größer als $N$ ist.
Man beweise, dass die Summe der Kehrwerte dieser $n$ Zahlen stets kleiner $2$ ist; also
$$
\frac{1}{... | [
"Solution:\n\nDa das kgV von $x_{i}$ und $x_{j}$ größer als $N$ ist, gibt es unter den Zahlen $1,2, \\ldots, N$ keine zwei, die sowohl Vielfache von $x_{i}$, als auch von $x_{j}$ sind.\nUnter den Vielfachen der natürlichen Zahl $x$ gibt es zwei so, dass $k x \\leq N < (k+1) x$, woraus $k \\leq \\frac{N}{x} < k+1$ f... | Germany | Auswahlwettbewerb zur IMO | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0hp7 | Problem:
The sequence of numbers $1, 2, 3, \ldots, 100$ is written on the blackboard. Between each two consecutive numbers a square box is drawn. Player $A$ starts the game and the players $A$ and $B$ alternate the moves. In each turn a player chooses an empty box and places "+" or "." sign in it. After all the boxes ... | [
"Solution:\n\n$A$ has the winning strategy. She should first place the sign \"+\" between the numbers $1$ and $2$. After that she should group the square boxes into pairs: each pair consisting of two boxes adjacent to the same odd number. Then the player $A$ should make sure that there is at least one sign \".\" in... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Player A. First place plus between one and two to isolate one as a separate summand. Then partition the remaining boxes into pairs around each odd integer and, whenever Player B plays in a box of a pair, immediately play in the other box of that same pair with a multiplication sign. This guarantees at least one multipl... | |
0gc0 | 令 $n$ 為大於 $1$ 的正整數。一個 $n \times n \times n$ 的大立方體由 $n^3$ 個邊長為 $1$ 的小立方體所構成。每個小立方體被塗上一種顏色。從這個大正方體中，我們沿著 $x$ 軸的方向截取出 $n$ 個不同的 $1 \times n \times n$ 的長方體 $R_1, R_2, \cdots, R_n$。令 $C_i$ 為 $R_i$ 中所有小立方體的顏色所成集合，並令 $\mathcal{X} = \{C_1, C_2, \cdots, C_n\}$。以同樣的方式，令 $\mathcal{Y}$ 為沿著 $y$ 軸的方向截取 $n \times 1 \times n$ 長方體下造出來的集... | [
"The maximal number is $\\frac{n(n+1)(2n+1)}{6}$.\n\nCall a $n \\times n \\times 1$ box an *x-box*, an *y-box*, or a *z-box* according to its orientation.\nLet $N$ be the number of colors in a valid configuration. We start with the upper bound for $N$.\n\nLet $D_1, D_2$ and $D_3$ be the sets of colors which appear ... | Taiwan | 2018 數學奧林匹亞競賽第二階段選訓營, 獨立研究(三) | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | n(n+1)(2n+1)/6 | |
0bqt | Let $a, b \in \mathbb{R}$ and the function $f_{a,b}: \mathbb{R} \to \mathbb{R}$, $f_{a,b}(x) = a[x] + b\{x\}$. Prove that this function is bijective if and only if $a^2 = b^2 \ne 0$. | [] | Romania | 67th NMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof only | null | |
08hq | Problem:
For every positive integer $n \geq 1$ we define the polynomial $P(X) = X^{2n} - X^{2n-1} + \ldots - X + 1$. Find the remainder of the division of the polynomial $P\left(X^{2n+1}\right)$ by the polynomial $P(X)$. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2n+1 | |
037e | Problem:
Find the area of the triangle determined by the straight line with equation $x - y + 1 = 0$ and the tangent lines to the graph of the parabola $y = x^{2} - 4x + 5$ at its common points with the line. | [
"Solution:\nThe common points of the graphs of the line and the parabola are $A(1, 2)$ and $B(4, 5)$. The equations of the tangents to the graph of the parabola at $A$ and $B$ are $y = -2x + 11$ and $y = 4x - 11$, respectively. The intersecting point of the two tangents is the point $C\\left(\\frac{5}{2}, -1\\right... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 27/4 | |
05vg | Problem:
Le gouvernement de Bosnie-Herzégovine a décidé de mettre en place un nouveau système de plaques d'immatriculations. Chaque plaque d'immatriculation devra contenir 8 chiffres, chacun pouvant valoir $0,1,2,3,4,5,6,7,8$ ou $9$. En outre, deux plaques d'immatriculation distinctes devront toujours avoir au moins d... | [
"Solution:\n\nNous allons montrer que le nombre maximal de plaques disponibles est égal à $10^{7}$.\n\nTout d'abord, on dit que deux plaques font partie de la même famille si leurs sept premiers chiffres sont différents. Chaque famille contient dix plaques, donc il y a $10^{7}$ familles distinctes. Or, le gouvernem... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 10000000 | |
0epv | When simplified, the fraction $\frac{2}{0 + \frac{1}{5 + \frac{2}{0 + \frac{1}{5}}}}$ equals | [
"$$\n\\frac{2}{0 + \\frac{1}{5 + \\frac{2}{0 + \\frac{1}{5}}}} = \\frac{2}{0 + \\frac{1}{5 + \\frac{2}{5 + \\frac{1}{5}}}} = \\frac{2}{0 + \\frac{1}{5 + 10}} = \\frac{2}{\\frac{1}{15}} = 2 \\times 15 = 30\n$$"
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | final answer only | 30 | |
06jf | Let $k$ be an integer. If the equation $kx^2 + (4k-2)x + (4k-7) = 0$ has an integral root, find the sum of all possible values of $k$. | [
"Let $m$ be an integral root of the equation. Then we have $km^2 + (4k - 2)m + (4k - 7) = 0$, which can be rewritten as $(m^2 + 4m + 4)k = 2m + 7$, or $(m + 2)^2k = 2m + 7$. Hence $m + 2$ divides $2m + 7$, and so $m + 2$ divides $2m + 7 - 2(m + 2) = 3$ as well. Thus $m$ can only be $-5, -3, -1$ or $1$, which we plu... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 6 | |
0jn6 | Problem:
Let $R$ be the rectangle in the Cartesian plane with vertices at $(0,0)$, $(2,0)$, $(2,1)$, and $(0,1)$. $R$ can be divided into two unit squares, as shown.
The resulting figure has 7 segments of unit length, connecting neighboring lattice points (those lying on or inside $R$). Comput... | [
"Solution:\nAnswer: 4\n\nJust count them directly. If the first step is to the right, there are 2 paths. If the first step is downwards (so the next step must be to the right), there are again 2 paths. This gives a total of 4 paths."
] | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Other"
] | null | final answer only | 4 | |
0hid | The country has $n \ge 3$ airports, some pairs of which are connected by bidirectional flights. Every day, the government closes the airport from which the largest number of flights is flying. What is the maximum number of days this can continue?
Fig. 17 | [
"We will use the obvious interpretation in the language of graphs, and for convenience, we will move on to graph complements. From now on, every day the vertex of strictly lowest degree will be removed.\n\nIt is clear that when 2 vertices remain, nothing else will happen. For $n = 3$ this bound is achieved by conne... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | n - 3 for n ≥ 4; 1 for n = 3 | |
0b5n | Let $n$ be a positive integer. Determine the least number of equilateral triangles of side $1$ which can cover an equilateral triangle of side $n + \frac{1}{2n}$. | [
"The ratio of the areas of the equilateral triangle $\\Delta$ of side $n + \\frac{1}{2n}$ and that of the equilateral triangle $\\Delta_1$ of side $1$ is the square of the ratio of the lengths of their sides, i.e.\n$$(n + \\frac{1}{2n})^2 > n^2 + 1,$$\nhence at least $n^2 + 2$ triangles $\\Delta_1$ are needed.\n\nF... | Romania | Local Mathematical Competitions | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | n^2 + 2 | |
0l6h | Problem:
Let $\lfloor z\rfloor$ denote the greatest integer less than or equal to $z$. Compute
$$
\sum_{j = -1000}^{1000} \left\lfloor \frac{2025}{j + 0.5} \right\rfloor.
$$ | [
"Solution:\nThe key idea is to pair up the terms $\\left\\lfloor \\frac{2025}{x} \\right\\rfloor$ and $\\left\\lfloor \\frac{2025}{x} \\right\\rfloor$. There are 1000 such pairs and one lone term, $\\left\\lfloor \\frac{2025}{1000.5} \\right\\rfloor = 2$. Thus,\n$$\n\\sum_{j = -1000}^{1000} \\left\\lfloor \\frac{20... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | final answer only | -984 | |
04b7 | Prove that for all real numbers $a, b, c$ the following inequality holds
$$
\frac{1}{3}(a+b+c)^2 \leq a^2 + b^2 + c^2 + 2(a-b+1).
$$ | [
"Transforming the right hand side of the inequality we get\n$$\n\\begin{aligned}\na^2 + b^2 + c^2 + 2(a - b + 1) &= a^2 + 2a + 1 + b^2 - 2b + 1 + c^2 \\\\\n&= (a + 1)^2 + (b - 1)^2 + c^2.\n\\end{aligned}\n$$\nFrom the A–K inequality it follows\n$$\n\\sqrt{\\frac{(a+1)^2 + (b-1)^2 + c^2}{3}} \\ge \\frac{(a+1) + (b-1... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
087z | Problem:
$x$ e $y$ sono due interi positivi tali che $x^2 - y^2$ è positivo, multiplo di $2011$ e ha esattamente $2011$ divisori positivi. Quante sono le coppie ordinate $(x, y)$ che verificano tali condizioni? Nota: $2011$ è un numero primo
(A) 2010
(B) 2011
(C) 1005
(D) 0
(E) Ne esistono infinite. | [
"Solution:\n\nLa risposta è (C). Innanzitutto notiamo che un numero ha esattamente $2011$ divisori positivi se e solo se è della forma $p^{2010}$ con $p$ primo. Dobbiamo quindi risolvere $x^2 - y^2 = 2011^{2010}$.\n\nImponendo che $x + y = p^{\\alpha}$ e che $x - y = p^{\\beta}$, con $\\alpha + \\beta = 2010$, si o... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | C | |
0iuk | Problem:
Let $f(x) = x^{4} + a x^{3} + b x^{2} + c x + d$ be a polynomial whose roots are all negative integers. If $a + b + c + d = 2009$, find $d$. | [
"Solution:\n\nCall the roots $-x_{1}$, $-x_{2}$, $-x_{3}$, and $-x_{4}$. Then $f(x)$ must factor as $(x + x_{1})(x + x_{2})(x + x_{3})(x + x_{4})$.\n\nIf we evaluate $f$ at $1$, we get $(1 + x_{1})(1 + x_{2})(1 + x_{3})(1 + x_{4}) = a + b + c + d + 1 = 2009 + 1 = 2010$.\n\n$2010 = 2 \\cdot 3 \\cdot 5 \\cdot 67$.\n\... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 528 | |
0e2p | Let $ABC$ be an acute triangle. A line parallel to $BC$ intersects the sides $AB$ and $AC$ at $D$ and $E$. The circumcircle of the triangle $ADE$ intersects the segment $CD$ at $F$, $F \neq D$. Prove that the triangles $AFE$ and $CBD$ are similar. | [
"The lines $DE$ and $BC$ are parallel, so $\\angle DCB = \\angle CDE$. The inscribed angles over the chord $EF$ in the cyclic quadrilateral $ADFE$ are equal, $\\angle FDE = \\angle FAE$. This implies\n$$\n\\angle DCB = \\angle CDE = \\angle FDE = \\angle FAE.\n$$\nThe lines $DE$ and $BC$ are parallel, so $\\angle A... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0gtr | Let $ABC$ be a triangle, $\omega$ be its circumcircle and $I$ be its incentre. Let the line $BI$ meet $AC$ at $E$ and $\omega$ at $M$ for the second time. The line $CI$ meet $AB$ at $F$ and $\omega$ at $N$ for the second time. Let the circumcircles of $BFI$ and $CEI$ meet at $K$ for the second time. Prove that the line... | [
"\nLet $AK \\cap MN = L$. Since $K$ is the Miquel point of the quadrilateral $AFIE$ we get that $ABKE$ and $AFKC$ are conicyclic. Therefore, we have $\\angle BK_A = \\angle BEA = B/2 + C$ and since $\\angle BNM = B/2 + A$ we get that quadrilateral $BNKL$ is cyclic. Similarly the quadrilater... | Turkey | Team Selection Test for EGMO 2024 | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line,... | English | proof only | null | |
0bg4 | Problem:
In a mathematical competition some competitors are friends; friendship is always mutual, that is to say that when $A$ is a friend of $B$, then also $B$ is a friend of $A$. We say that $n \geq 3$ different competitors $A_{1}, A_{2}, \ldots, A_{n}$ form a weakly-friendly cycle if $A_{i}$ is not a friend of $A_{... | [] | Romania | 30th Balkan Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
03y6 | We operate on piles of cards placed at $n+1$ positions $A_1, A_2, \dots, A_n$ ($n \ge 3$) and $O$. In one operation, we can do either of the following:
(1) If there are at least three cards at $A_i$, we may take three cards from $A_i$ and place one at each of $A_{i-1}, A_{i+1}$ and $O$ (assume that $A_0 = A_n, A_{n+1} ... | [
"**Proof** One only needs to consider the case with the total number of cards equal $n^2 + 3n + 1$. We take the following strategy. If there are at least three cards at some $A_i$, then use operation (1) at this position. Such operations can be done for only finitely many steps. Then we have no more than two cards ... | China | Chinese Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
0d9d | Let $ABC$ be an acute triangle inscribed in circle $(O)$, with orthocenter $H$. Median $AM$ of triangle $ABC$ intersects circle $(O)$ at $A$ and $N$. $AH$ intersects $(O)$ at $A$ and $K$. Three lines $KN$, $BC$ and the line through $H$ and perpendicular to $AN$ intersect each other and form triangle $XYZ$. Prove that t... | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Advanced Configurations ... | English | proof only | null | |
08b0 | Problem:
Sia $n$ un intero positivo e siano $1 = d_{1} < d_{2} < d_{3} < \ldots < d_{k} = n$ i suoi divisori positivi, ordinati per grandezza. Si sa che $k \geq 4$ e che $d_{3}^{2} + d_{4}^{2} = 2n + 1$.
a. Trovare tutti i possibili valori di $k$.
b. Trovare tutti i possibili valori di $n$. | [
"Solution:\n\na.\nPer prima cosa, osserviamo $d_{4}$ non può essere minore o uguale a $\\sqrt{n}$, perché altrimenti $d_{3}^{2} + d_{4}^{2}$ sarebbe minore di $(\\sqrt{n} - 1)^{2} + \\sqrt{n}^{2} < 2n + 1$, mentre sappiamo che vale $2n + 1$. Allo stesso modo, se $d_{3}$ fosse maggiore o uguale a $\\sqrt{n}$ si avre... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | k = 6; n ∈ {12, 20} | |
0lf6 | Let $a, b, c, \alpha, \beta$ be the given integers and the sequence $(u_n)$ is defined by $u_1 = \alpha, u_2 = \beta, u_{n+2} = a u_{n+1} + b u_n + c$ for all $n \ge 1$.
a) Prove that if $a = 3, b = -2, c = -1$ then there are infinitely many pairs of integers $(\alpha; \beta)$ so that $u_{2023} = 2^{2022}$.
b) Prove ... | [
"a) For $a = 3, b = -2, c = -1$, we have $u_{n+2} = 3u_{n+1} - 2u_n - 1, \\forall n \\ge 1$. By induction, one can prove\n$$\nu_n = 2\\alpha - \\beta + (\\beta - \\alpha - 1) \\cdot 2^{n-1} + n$$\nfor all $n \\ge 1$. Then $u_{2023} = 2\\alpha - \\beta + (\\beta - \\alpha - 1)2^{2022} + 2023$.\nFor any $t \\in \\mat... | Vietnam | Vietnamese Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
08uy | Let $x$ be a 2-digit positive integer and $y$ be a 1-digit positive integer. Suppose that the ten's digit of $x$, the one's digit of $x$ and $y$ are all distinct. Determine the maximum possible value the product $xy$ can take. | [
"Let $a$ and $b$ be the ten's digit and the one's digit of $x$, respectively. Since we are concerned with the maximum value of $xy$ and since $a, b, y$ are distinct by assumption, it is clear that it suffices to consider the situation where $a, b, y$ are chosen from $7, 8, 9$. If the numbers for $a$ and $b$ are cho... | Japan | Japan Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 783 | |
00v5 | Let $ABC$ be a scalene acute triangle $ABC$, $D$ be the orthogonal projection of $A$ on $BC$, $M$ and $N$ are the midpoints of $AB$ and $AC$ respectively. Let $P, Q$ are points on the minor arcs $\widehat{AB}$ and $\widehat{AC}$ of circumcircle of $\triangle ABC$ respectively such that $PQ \parallel BC$. Show that the ... | [
"\nConsider $i$ = the inversion of pole $A$ and $k = \\frac{AB \\cdot AC}{2}$ followed by the reflection with respect to angle bisector of $\\angle BAC$. Denote $X' = i(X)$ for any $X$ in the plane.\nNotice that $B' = N$, $C' = M$, the midpoints of $AC$ and $AB$ respectively and that Euler'... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal... | English | proof only | null | |
0gf9 | 坐標平面上兩坐標皆為整數的點稱為格子點。以下考慮的三角形，其頂點皆為格子點。一個合法的動作，係指將這樣的三角形的其中一個頂點，沿著與其對邊平行的方向移到另一個格子點的操作。證明若兩個三角形有相同的面積，則必存在一系列合法的動作，將其中一個變成另外一個。 | [] | Taiwan | 2022 數學奧林匹亞競賽第二階段培訓營, 獨立研究 (二) | [
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Linear Algebra > Determinants",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Chinese; English | proof only | null | |
0fdu | Problem:
Halla todas las ternas de números enteros positivos $a \leq b \leq c$ primitivas (es decir, que no tengan ningún factor primo común) tales que cada uno de ellos divide a la suma de los otros dos. | [
"Solution:\nSupongamos que $a = b$. Como $a$ y $b$ no tienen factores en común, debe ser $a = b = 1$. Como $c$ divide a $a + b = 2$, esto da lugar a las ternas $(1, 1, 1)$ y $(1, 1, 2)$.\n\nSupongamos ahora que $a < b$. Como $c$ divide a $a + b < c + c = 2c$, debe ser $a + b = c$. Pero entonces, como $b$ divide a $... | Spain | XLVII Olimpiada Matemática Española Primera Fase | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (1,1,1), (1,1,2), (1,2,3) | |
0j4k | Problem:
For positive integers $m, n$, let $\operatorname{gcd}(m, n)$ denote the largest positive integer that is a factor of both $m$ and $n$. Compute
$$
\sum_{n=1}^{91} \operatorname{gcd}(n, 91) .
$$ | [
"Solution:\n\nAnswer: 325\n\nSince $91 = 7 \\times 13$, we see that the possible values of $\\operatorname{gcd}(n, 91)$ are $1, 7, 13, 91$.\n\nFor $1 \\leq n \\leq 91$, there is only one value of $n$ such that $\\operatorname{gcd}(n, 91) = 91$.\n\nThen, we see that there are 12 values of $n$ for which $\\operatorna... | United States | Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | final answer only | 325 | |
0hii | An acute-angled triangle is given $ABC$ with the circumcircle $\omega$. The points $F$ on $AC$, $E$ on $AB$ and $P, Q$ on $\omega$ are selected so that $\angle AFB = \angle AEC = \angle APE = \angle AQF = 90^\circ$. Prove that the lines $BC, EF$ and $PQ$ intersect at one point.
**Fig. 12** | [
"Let $A'$ be the point diametrically opposite to $A$ in $\\omega$, and let $R$ be the projection of $A$ onto $EF$ (fig. 12). Since $\\angle BAA' = 90^\\circ - \\angle ACB = 90^\\circ - \\angle AEF$, the line $AR$ passes through $A'$. Lines $PE$ and $QF$ also pass through $A'$. Since the quadrilaterals $AREP$ and $A... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0j5x | Problem:
Determine the set of all real numbers $p$ for which the polynomial $Q(x) = x^{3} + p x^{2} - p x - 1$ has three distinct real roots. | [
"Solution:\n\nAnswer: $p > 1$ and $p < -3$\n\nFirst, we note that\n$$\nx^{3} + p x^{2} - p x - 1 = (x - 1)\\left(x^{2} + (p + 1)x + 1\\right)\n$$\nHence, $x^{2} + (p + 1)x + 1$ has two distinct roots. Consequently, the discriminant of this equation must be positive, so $(p + 1)^{2} - 4 > 0$, so either $p > 1$ or $p... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | p > 1 or p < -3 | |
0f3k | Problem:
$S$ is a set of integers. Its smallest element is $1$ and its largest element is $100$. Every element of $S$ except $1$ is the sum of two distinct members of the set or double a member of the set. What is the smallest possible number of integers in $S$? | [
"Solution:\n\nLet $\\{n\\} = \\{M\\}(\\{n\\}) + \\{m\\}(\\{n\\})$, where $\\{M\\}(\\{n\\}) \\geq \\{m\\}(\\{n\\})$. Put $\\{M\\}^{1}(\\{n\\}) = \\{M\\}(\\{n\\})$, $\\{M\\}^{2}(\\{n\\}) = \\{M\\}(\\{M\\}(\\{n\\}))$ etc. Then $\\{M\\}(100) \\geq 50$, $\\{M\\}^{2}(100) \\geq 25$, $\\{M\\}^{3}(100) \\geq 13$, $\\{M\\}^... | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 9 | |
0k23 | Problem:
John has a 1 liter bottle of pure orange juice. He pours half of the contents of the bottle into a vat, fills the bottle with water, and mixes thoroughly. He then repeats this process 9 more times. Afterwards, he pours the remaining contents of the bottle into the vat. What fraction of the liquid in the vat i... | [
"Solution:\n\nAll the liquid was poured out eventually. $5$ liters of water was poured in, and he started with $1$ liter of orange juice, so the fraction is $\\frac{5}{1+5}=\\frac{5}{6}$."
] | United States | HMMT February 2018 | [
"Math Word Problems"
] | null | proof and answer | 5/6 | |
0385 | Problem:
Let $m$ be a positive integer and $u_{m} = \underbrace{11 \ldots 1}_{m}$. Prove that there is no positive integer multiple of $u_{m}$ such that the sum of its digits is less than $m$. | [
"Solution:\n\nAssume the contrary. Then there exists a positive integer multiple of $u_{m}$ such that the sum of its digits is less than $m$ and let $t$ be the smallest number with this property. Since $t > 10^{m}$, the number $t$ can be written as $t = 10^{m} a + b$, where $0 < b < 10^{m}$.\n\nWe have $t = 10^{m} ... | Bulgaria | Spring Mathematical Competition | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0ken | Problem:
Determine, with proof, whether or not there exist distinct positive integers $a_{1}, a_{2}, \ldots, a_{n}$ such that
$$
\frac{1}{a_{1}} + \frac{1}{a_{2}} + \cdots + \frac{1}{a_{n}} = 2019.
$$ | [
"Solution:\n\nYes, the decomposition exists.\nRecall that the harmonic series diverges. We first take the largest partial sum of the harmonic series that is smaller than $2019$, subtract it from $2019$ to get a \"remainder\" $r$. We then use the greedy algorithm to pick the rest of the unit fractions: pick the larg... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Other"
] | null | proof and answer | Yes, such distinct positive integers exist. | |
05xr | Problem:
Une ligne maritime Est-Ouest voit partir chaque matin 10 bateaux à des moments tous distincts, 5 bateaux partent du côté Ouest et 5 du côté Est. On suppose qu'ils naviguent tous à la même vitesse et que dès que deux bateaux se rencontrent ils se retournent et repartent chacun de leur côté, toujours à la même ... | [
"Solution:\n\nLe nombre de croisements ne change pas si on dit que les bateaux continuent tout droit au lieu de faire demi-tour lors d'une rencontre. Chaque bateau qui part d'un côté croise tous les autres bateaux qui partent de l'autre côté. Il y a donc $5 \\times 5=25$ rencontres."
] | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 25 | |
0g12 | Problem:
Finde alle Funktionen $f: \mathbb{R} \rightarrow \mathbb{R}$, sodass für alle $x, y \in \mathbb{R}$ gilt:
$$
(f(x)+y)(f(x-y)+1)=f(f(x f(x+1))-y f(y-1))
$$ | [
"Solution:\nSei $x=0, y=1$ :\n$$\n(f(0)+1)(f(-1)+1)=f(0)\n$$\nSetzen wir $y=-f(x)$, folgt sofort, dass ein $a \\in \\mathbb{R}$ existiert mit $f(a)=0$. Nun setzen wir $x=a, y=0$ :\n$$\n0=f(f(a f(a+1)))\n$$\nUnd nun $x=a, y=a+1$ :\n$$\n(a+1)(f(-1)+1)=f(f(a f(a+1)))=0\n$$\nWäre $a=-1$ würde die erste Gleichung $f(0)+... | Switzerland | IMO-Selektion | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = x | |
09n8 | Let $N$ be an arbitrary integer. Prove that the equation
$$
(m+n)^2 + mn + 1 = N(m+n)
$$
has infinitely many integer $(m, n)$ solutions.
(Otgonbayar Uuye) | [
"Let $s = m + n$ and $p = mn$. Then the equation becomes:\n$$\ns^2 + p + 1 = N s\n$$\nSo,\n$$\np = N s - s^2 - 1\n$$\nRecall that $m$ and $n$ are roots of the quadratic equation $x^2 - s x + p = 0$.\n\nThe discriminant is:\n$$\nD = s^2 - 4p = s^2 - 4(N s - s^2 - 1) = s^2 - 4N s + 4s^2 + 4\n$$\n$$\n= 5s^2 - 4N s + 4... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof only | null | |
04cm | Determine and sketch in the Gaussian plane the set of all $z$ that satisfy
$$
|z - 1| - |z + 1| = \sqrt{3}.
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0l5b | Problem:
The Cantor set is defined as the set of real numbers $x$ such that $0 \leq x < 1$ and the digit $1$ does not appear in the base-$3$ expansion of $x$. Two numbers are uniformly and independently selected at random from the Cantor set. Compute the expected value of their absolute difference.
(Formally, one can ... | [
"Solution:\nLet $d$ be the expected value of the absolute difference. Observe that the Cantor set is made up of two smaller copies of itself, each scaled down by a factor of $3$. There is a $\\frac{1}{2}$ chance that the two selected numbers are in the same copy, in which case the expected value of their absolute d... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 2/5 | |
0dun | Problem:
Dana je funkcija $f(x) = \frac{x-2}{x^{2} + x - 2}$. Za katere vrednosti $x$ bo graf funkcije $f(x)$ ležal nad premico z enačbo $y = 1$? | [
"Solution:\n\nZapis pogoja $\\frac{x-2}{x^{2} + x - 2} > 1$\n\n"
] | Slovenia | 2. matematično tekmovanje dijakov srednjih tehniških in strokovnih šol | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (-2,0) ∪ (0,1) | |
0kql | Problem:
Let $a$ and $b$ be integers. Show that $29$ divides $3a + 2b$ if and only if $29$ divides $11a + 17b$. | [
"Solution:\n\nWe give two solutions to this problem:\n\nFirst, observe that $11a + 17b \\equiv 23(3a + 2b) \\pmod{29}$, so if $3a + 2b \\equiv 0 \\pmod{29}$, then $11a + 17b \\equiv 0 \\pmod{29}$. Similarly, if $11a + 17b \\equiv 0 \\pmod{29}$, then since $23$ is coprime to $29$, we can invert $23 \\pmod{29}$ and f... | United States | Berkeley Math Circle | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
06ox | Let $ABCD$ be a convex quadrilateral. A circle passing through the points $A$ and $D$ and a circle passing through the points $B$ and $C$ are externally tangent at a point $P$ inside the quadrilateral. Suppose that
$$
\angle PAB + \angle PDC \leq 90^{\circ} \quad \text{ and } \quad \angle PBA + \angle PCD \leq 90^{\cir... | [
"We start with a preliminary observation. Let $T$ be a point inside the quadrilateral $ABCD$. Then:\n$$\n\\begin{align*}\n& \\text{Circles } (BCT) \\text{ and } (DAT) \\text{ are tangent at } T \\\\\n& \\text{if and only if} \\quad \\angle ADT + \\angle BCT = \\angle ATB. \\tag{1}\n\\end{align*}\n$$\nIndeed, if the... | IMO | IMO 2006 Shortlisted Problems | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Op... | English | proof only | null | |
07pi | The point $P$ is on the side $BC$ of triangle $ABC$. The incircles of $\triangle ABC$, $\triangle ABP$ and $\triangle ACP$ are denoted by $K$, $K_1$ and $K_2$, respectively.
Prove that the angle at which the radical axis of $K$ and $K_1$ meets the radical axis of $K$ and $K_2$ does not depend on the position of $P$. Fi... | [
"Let $I$ be the incentre of $K$. The incentres of $K$ and $K_1$ both lie on the bisector of $\\angle ABC$. The incentres of $K$ and $K_2$ both lie on the bisector of $\\angle ACB$. Note that $\\angle BIC = 180^\\circ - \\frac{1}{2}(\\angle ABC + \\angle ACB) = 90^\\circ + \\frac{1}{2}\\angle BAC$.\n\n![](attached_i... | Ireland | Ireland | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 40° (acute angle; the supplementary angle is 140°) | |
058o | A set of 8 dominoes is given, each consisting of two unit squares:
Is it possible to completely cover a rectangular grid of size $4 \times 4$ with these dominoes in such a way that all rows and columns of the grid contain the same number of pips? | [
"Figure 10 shows one possibility.\n\n"
] | Estonia | Estonian Math Competitions | [
"Math Word Problems"
] | English | proof and answer | Yes | |
09gj | Let $a$, $b$ and $c$ be positive real numbers such that $a + b + c = 3$. Prove that
$$
\frac{a + b}{2ab + 1} + \frac{b + c}{2bc + 1} + \frac{c + a}{2ca + 1} \ge 2.
$$ | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
00n4 | Let $p$, $q$, $r$ and $s$ be prime numbers satisfying
$$
5 < p < q < r < s < p + 10.
$$
Prove that the sum of these four prime numbers is divisible by 60. | [
"The four prime numbers have to fulfill $p > 5$ and $s < p + 10$ and hence they must be among the five consecutive odd numbers $p$, $p + 2$, $p + 4$, $p + 6$ and $p + 8$.\nAs we have to choose 4 out of the five numbers $p$, $p + 2$, $p + 4$, $p + 6$, $p + 8$, we have to omit exactly one of these numbers. If we omit... | Austria | Austria2019 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0gev | 令 $m, n$ 為正整數。有一張 $m \times n$ 的方格紙，每一格 $(r, c)$ 寫有一個實數 $a(r, c)$。考慮列集 $R \subseteq \{1, 2, \dots, m\}$ 與行集 $C \subseteq \{1, 2, \dots, n\}$。我們稱滿足以下兩個條件的 $(R, C)$ 為好組合：
1. 對於每個 $r' \in \{1, 2, \dots, m\}$，存在 $r \in R$ 使得 $a(r, c) \ge a(r', c)$ 對於所有 $c \in C$ 皆成立；
2. 對於每個 $c' \in \{1, 2, \dots, n\}$，存在 $c \in C$ 使得 $a(r... | [
"以 $(R', C') \\le (R, C)$ 表示 $R' \\subseteq R$ 且 $C' \\subseteq C$，並以 $(R', C') < (R, C)$ 表示至少其中一個等號不成立。此外，我們以 $(r, c) \\in (R, C)$ 表示 $r \\in R$ 且 $c \\in C$。\n考慮兩個好組合 $(R_1, C_1)$ 跟 $(R_2, C_2)$，其中 $|R_1| > |R_2|$。我們將證明，我們可以找到 $(R', C') \\le (R_1, C_1)$ 滿足 $|R'| \\le |R_2|$。注意到這直接得證原命題。\n\n步驟一：首先我們構造映射 $\\rho : R... | Taiwan | 2021 數學奧林匹亞競賽第二階段選訓營, 獨立研究(二) | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0aqf | Problem:
Determine the smallest positive integer $n$ such that $n$ is divisible by $20$, $n^{2}$ is a perfect cube, and $n^{3}$ is a perfect square. | [] | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1000000 | |
0273 | Problem:
Os elementos do conjunto $\{1,2,3,4,5,6,7,8,9,10,11\}$ podem ser separados nos conjuntos $\{3,9,10,11\}$ e $\{1,2,4,5,6,7,8\}$ de modo que cada um deles possua soma dos elementos igual a 33.
a. Exiba um modo de separar os elementos do conjunto $\{1,2,3,4,5,6,7,8\}$ em três conjuntos tais que as somas dos ele... | [
"Solution:\n\na. A soma dos números de $1$ até $8$ é igual a $36$, então cada um dos três conjuntos deve ter soma $12$. Basta separar nos conjuntos: $\\{4,8\\}$, $\\{5,7\\}$ e $\\{1,2,3,6\\}$.\n\nb. A soma dos números de $1$ até $10$ é igual a $55$. Para separar em dois conjuntos de mesma soma, cada um teria elemen... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | N = 2n − 1 | |
0ixi | Problem:
At the start of this problem, six frogs are sitting with one at each of the six vertices of a regular hexagon. Every minute, we choose a frog to jump over another frog using one of the two rules illustrated below. If a frog at point $F$ jumps over a frog at point $P$, the frog will land at point $F'$ such tha... | [
"Solution:\n\na.\nAssign coordinate axes making a $120^{\\circ}$ angle so that the center of the hexagon is at $(0,0)$, the rightmost frog is at $(1,0)$, and the top left frog is at $(0,1)$. Then, the remaining four frogs are at $(1,1)$, $(-1,0)$, $(0,-1)$, and $(-1,-1)$. At each jump, if two frogs' coordinates dif... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof and answer | a. No; b. No | |
0a17 | A green, a blue and a red dragon all do not like one of the three vegetables leek, spinach and carrot; each a different one. They also all have a favourite vegetable out of these three, again each a different one. They all make two statements.
* The green dragon says: "My favourite vegetable is leek; the red dragon doe... | [
"B) green: leek, blue: carrot, red: spinach."
] | Netherlands | Junior Mathematical Olympiad | [
"Discrete Mathematics > Logic"
] | English | MCQ | B | |
0a7b | Problem:
It is possible to perform three operations $f$, $g$, and $h$ for positive integers: $f(n) = 10n$, $g(n) = 10n + 4$, and $h(2n) = n$; in other words, one may write $0$ or $4$ at the end of the number and one may divide an even number by $2$. Prove: every positive integer can be constructed starting from $4$ an... | [
"Solution:\n\nAll odd numbers $n$ are of the form $h(2n)$. All we need is to show that every even number can be obtained from $4$ by using the operations $f$, $g$, and $h$. To this end, we show that a suitably chosen sequence of inverse operations $F = f^{-1}$, $G = g^{-1}$, and $H = h^{-1}$ produces a smaller even... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 4 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
02f9 | The edges of a cube are labeled from $1$ to $12$ in an arbitrary manner. Show that it is not possible to get the sum of the edges at each vertex the same. Show that we can get eight vertices with the same sum if one of the labels is changed to $13$. | [
"Each edge contributes to the total sum twice, one for each of its vertices. So if each vertex has sum $v$, the sum of all numbers is $8v = 2(1+2+\\cdots+12) = 4 \\cdot 39 \\Rightarrow v = \\frac{39}{2}$, which can't be possible.\n\nThe following diagram shows a solution for sums equal to $13$.\n\n![](attached_imag... | Brazil | XVI OBM | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
091i | Problem:
For a nonnegative integer $n$, define $a_{n}$ to be the positive integer with decimal representation
$$
1 \underbrace{0 \ldots 0}_{n} 2 \underbrace{0 \ldots 0}_{n} 2 \underbrace{0 \ldots 0}_{n} 1
$$
Prove that $a_{n} / 3$ is always the sum of two positive perfect cubes but never the sum of two perfect squares... | [
"Solution:\n\nFirst we prove that $a_{n} / 3$ is never the sum of two perfect squares. Note that perfect squares give only remainders $0$ and $1$ when divided by four; therefore, integers expressible as the sum of two squares give only remainders $0$, $1$, and $2$. On the other hand, the number $a_{n} / 3$ gives re... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
08az | Problem:
Sia $ABCD$ un quadrilatero convesso tale che $AB = AC = AD$ e $BC < CD$. La bisettrice dell'angolo $\widehat{BAD}$ interseca internamente $CD$ in $M$ e il prolungamento di $BC$ in $N$. Dimostrare che
a. il quadrilatero $ABCM$ è inscrittibile in una circonferenza;
b. i triangoli $ANB$ e $ABM$ sono simili. | [
"Solution:\n\nSia $\\Gamma$ la circonferenza centrata in $A$ e passante per $B, C, D$. Sia inoltre $P$ un qualsiasi punto sull'arco $BD$ che non contiene $C$. Poiché l'angolo alla circonferenza $\\widehat{BPD}$ insiste sullo stesso arco dell'angolo al centro $\\widehat{BAD}$, vale $\\widehat{BAD} = 2 \\cdot \\wideh... | Italy | Progetto Olimpiadi della Matematica | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fyi | Problem:
Sei $P$ eine endliche Menge von Primzahlen und sei $\ell(P)$ die grösstmögliche Anzahl aufeinanderfolgender natürlicher Zahlen, sodass jede dieser Zahlen durch mindestens eine Primzahl aus $P$ teilbar ist. Beweise die Ungleichung $\ell(P) \geq |P|$ und zeige, dass genau dann Gleichheit gilt, wenn das kleinste... | [
"Solution:\n\nSei $|P|=n$ und $P=\\{p_{1}, \\ldots, p_{n}\\}$. Wir nehmen zuerst an, dass $p_{k}>n$ gilt für alle $k$, und zeigen, dass keine $n+1$ aufeinanderfolgende Zahlen existieren können, die alle durch ein Element von $P$ teilbar sind. Nach dem Schubfachprinzip müssten dann nämlich zwei dieser Zahlen durch d... | Switzerland | IMO Selektion | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
02nu | Problem:
26. Festa de aniversário - A festa de aniversário de André tem menos do que 120 convidados. Para o jantar, ele pode dividir os convidados em mesas completas de seis pessoas ou em mesas completas de sete pessoas. Em ambos os casos, são necessárias mais do que 10 mesas e todos os convidados ficam em alguma mesa... | [
"Solution:\n\n26. Festa de aniversário - Como podemos repartir o total de convidados em mesas de 6 ou 7, o número de convidados é um múltiplo de 6 e de 7. Como o menor múltiplo comum de 6 e 7 é 42, podemos ter $42, 84, 126, \\ldots$ convidados. Como são menos do que 120 convidados, só podemos ter 42 ou 84 convidado... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | 84 | |
0e9k | Problem:
Naj bo $x > 0$. Reši enačbo $\left(\frac{2}{5}\right)^{\log^2 x + 1} = \left(\frac{25}{4}\right)^{2 - \log x^3}$. | [] | Slovenia | 14. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 10 and 100000 | |
0hpq | Problem:
A circle is inscribed in $\triangle ABC$ that touches side $BC$ at $D$, side $AC$ at $E$, and side $AB$ at $F$. Show that $\triangle DEF$ must be acute. | [
"Solution:\n\nSince $AE$ and $AF$ are tangents from the same point to the same circle, we must have $AE = AF$. Thus, $\\triangle AEF$ is isosceles, so\n$$\n\\angle AEF = \\angle AFE = \\frac{180^\\circ - \\angle A}{2} = 90^\\circ - \\frac{1}{2} \\angle A.\n$$\nNow, since $AE$ is a tangent to the circle, $\\angle ED... | United States | Berkeley Math Circle: Monthly Contest 8 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cgk | Consider two sets $A$ and $B$ of real numbers that have the following properties:
a. $0 \in A$;
b. if $1+x \in A$, then $\sqrt{1+x+x^2} \in B$;
c. if $\sqrt{x^2 - x + 1} \in B$, then $2+x \in A$.
Prove that $\sqrt{3}$, $\sqrt{13}$, $\sqrt{31}$ are elements of the set $B$ and $2024 \in A$. | [
"Since $1 + (-1) = 0 \\in A$, according to (b) we obtain $1 \\in B$ and, since $1 = \\sqrt{0^2 - 0 + 1} \\in B$, we infer from (c) that $2 \\in A$, from which $\\sqrt{3} \\in B$.\n\nSince $\\sqrt{2^2 - 2 + 1} = \\sqrt{3} \\in B$, it follows from (c) that $2 + 2 = 4 \\in A$ and, from (b), we infer $\\sqrt{13} \\in B... | Romania | 74th Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0228 | Problem:
Número curioso - O número $81$ tem a seguinte propriedade: ele é divisível pela soma de seus algarismos $8+1=9$. Quantos números de dois algarismos cumprem esta propriedade? | [
"Solution:\n\nSeja $ab$ um tal número. Por hipótese, $ab = 10a + b$ é divisível por $a + b$. Logo, a diferença $(10a + b) - (a + b) = 9a$, também é divisível por $a + b$. Além disso, sabemos que $10a + b$ é divisível por $a + b$ se, e somente se, $(10a + b) - (a + b) = 9a$ é divisível por $a + b$ (prove isso).\n\nA... | Brazil | Nível 2 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 23 | |
0jh5 | Problem:
How many functions $f: \mathbb{Z} \rightarrow \mathbb{R}$ satisfy the following three properties?
(a) $f(1)=1$;
(b) For all $m, n \in \mathbb{Z}$, $f(m)^2 - f(n)^2 = f(m+n) f(m-n)$;
(c) For all $n \in \mathbb{Z}$, $f(n) = f(n+2013)$. | [
"Solution:\n\nBy plugging $m=n=0$ into (b) we easily get $f(0)=0$. For any $u \\in \\mathbb{Z}$, we have\n$$\n\\begin{aligned}\nf(u+1)^2 - f(u-1)^2 & = f(2u) f(2) \\\\\nf(u+1)^2 - f(u)^2 & = f(2u+1) f(1) = f(2u+1) \\\\\nf(u)^2 - f(u-1)^2 & = f(2u-1) f(1) = f(2u-1)\n\\end{aligned}\n$$\nwhence\n$$\nf(2u) f(2) = f(2u+... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 1006 | |
0g6b | 給定平面上一直線 $L$。假設 $A$ 與 $B$ 為直線 $L$ 同側的兩相異點。試證：直線 $AB$ 與 $L$ 不垂直的充要條件為：$L$ 上有唯一的一點 $R$，使得對於直線 $L$ 上任意一點 $P$，$\angle APB \le \angle ARB$。 | [
"首先，若 $AB$ 平行於 $L$，作 $AB$ 中垂線交 $L$ 於 $R$，並作 $\\triangle ABP$ 的外接圓 $C$。則，對於 $L$ 上任一異於 $R$ 的點 $P$，令 $PB$ 交圓 $C$ 於 $S$，則 $\\angle APB < \\angle ASB = \\angle ARB$。故存在唯一的一點 $R$ 滿足題設。\n\n接著，若 $AB$ 不平行於 $L$，令直線 $AB$ 交直線 $L$ 於 $K$。易知平面上存在唯一的兩圓 $C$ 和 $C'$，使得此兩圓皆過點 $A$ 和 $B$，且與 $L$ 相切。令此兩圓分別切 $L$ 於 $R$ 和 $R'$。\n\n現在，不失一般性假設... | Taiwan | 二〇一二數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02sp | Problem:
Uma folha de papel é retangular, com base igual a $20~\mathrm{cm}$ e altura $10~\mathrm{cm}$. Esta folha é dobrada nas linhas pontilhadas conforme a figura abaixo, e no final recortada por uma tesoura na linha indicada, a qual é paralela à base e está na metade da altura do triângulo.
![](attached_image_1.png... | [
"Solution:\n\na) Vamos marcar a linha cortada pela tesoura em cinza, e fazer o processo inverso, que corresponde a abrir a folha depois de cortada:\n\nLogo, a folha foi dividida em três pedaços.\n\nb) Como se pode observar, os dois quadrados recortados nos cantos superior esquerdo e inferio... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | a) 3 parts; b) 150 cm^2 | |
047q | For a finite non-empty set $A$ of real numbers, let $\max(A)$ denote its maximum element, and define:
$$
P(A) = \sum_{\substack{B \subseteq A \\ |B| \text{ odd}}} m(B), \quad Q(A) = \sum_{\substack{\emptyset \neq B \subseteq A \\ |B| \text{ even}}} m(B),
$$
where $m(B)$ is the median of finite non-empty set $B$: if $B ... | [
"Without loss of generality, let the elements of $A$ in increasing order be $x_0, x_1, \\dots, x_{2024}$. We first compute $P(A)$. For each $x_m$, we count how many odd-sized subsets have $x_m$ as their median. A subset $B$ of odd size has $x_m$ as its median if and only if\n$$\n|B \\cap \\{x_0, \\dots, x_{m-1}\\}|... | China | 2025 International Mathematical Olympiad China National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | ((2024 choose 990) - (2024 choose 989) + 1)/2 | |
0kko | Problem:
Let $N$ be a positive integer. Brothers Michael and Kylo each select a positive integer less than or equal to $N$, independently and uniformly at random. Let $p_{N}$ denote the probability that the product of these two integers has a units digit of $0$. The maximum possible value of $p_{N}$ over all possible ... | [
"Solution:\n\nFor $k \\in \\{2, 5, 10\\}$, let $q_{k} = \\frac{\\lfloor N / k \\rfloor}{N}$ be the probability that an integer chosen uniformly at random from $[N]$ is a multiple of $k$. Clearly, $q_{k} \\leq \\frac{1}{k}$, with equality iff $k$ divides $N$.\n\nThe product of $p_{1}, p_{2} \\in [N]$ can be a multip... | United States | HMMT Spring 2021 | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 2800 | |
05vw | Problem:
Soient $m$ et $n$ deux entiers tels que $m > n \geqslant 3$. Morgane a disposé $m$ jetons en cercle, et s'apprête à les peindre en utilisant $n$ couleurs distinctes. Elle souhaite que, parmi $n+1$ jetons consécutifs, il y ait toujours au moins un jeton de chacune des $n$ couleurs. Si elle peut y parvenir, on ... | [
"Solution:\n\nTout d'abord, si $m \\geqslant n^{2}-n$, Morgane peut réaliser son souhait en procédant comme suit, montrant au passage que $m$ est $n$-coloriable.\n\nSoit $a$ le résidu de $m$ modulo $n$, c'est-à-dire le plus petit entier naturel tel que $a \\equiv m \\pmod{n}$. Morgane commence par placer $m-a$ jeto... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | n^2 - n - 1 | |
0hru | Problem:
Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers whose sum is equal to $1$. If
$$
\begin{aligned}
S= & \frac{a_{1}^{2}}{2 a_{1}}+\frac{a_{1} a_{2}}{a_{1}+a_{2}}+\frac{a_{1} a_{3}}{a_{1}+a_{3}}+\cdots+\frac{a_{1} a_{n}}{a_{1}+a_{n}} \\
& +\frac{a_{2} a_{1}}{a_{2}+a_{1}}+\frac{a_{2}^{2}}{2 a_{2}}+\fra... | [
"Solution:\n\nIt is easy to show that $\\left(\\frac{a+b}{2}\\right)^{2} \\geq a b$. Now applying this inequality to $a_{1}, a_{2}$ we get $\\frac{a_{1} a_{2}}{a_{1}+a_{2}} \\leq \\frac{\\left[\\left(a_{1}+a_{2}\\right) / 2\\right]^{2}}{a_{1}+a_{2}} = \\frac{a_{1}+a_{2}}{4}$. We get similar inequalities for all ter... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0lga | Problem:
Prove that there are at least $100!$ ways to partition the number $100!$ into summands from the set $\{1!, 2!, 3!, \ldots, 99!\}$. (Partitions differing in the order of summands are considered the same; any summand can be taken multiple times. We remind that $n! = 1 \cdot 2 \cdot \ldots \cdot n$.) | [
"Solution:\n\nLet us prove by induction on $n \\geqslant 4$ that there are at least $n!$ ways to partition the number $n!$ into summands from $\\{1!, 2!, \\ldots, (n-1)!\\}$.\n\nFor $n=4$, if we use only the summands $1!$, $2!$ there are $13$ ways to partition $4!$ as $2!$ can be used from $0$ to $12$ times. If $3!... | Zhautykov Olympiad | XV International Zhautykov Olympiad in Mathematics | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
077r | Let $g: \mathbb{N} \to \mathbb{N}$ be a bijective function and suppose that $f: \mathbb{N} \to \mathbb{N}$ is a function such that:
• For all naturals $x$, $f^{2023}(x) = x$
• For all naturals $x, y$ such that $x \mid y$, we have $f(x) \mid g(y)$.
Prove that $f(x) = x$. | [
"Claim 0 $f$ is bijective.\n*Proof.* First, it is easy to see that $f$ is surjective, since for any given $x \\in \\mathbb{N}$, $f^{n-1}(x)$ exists and $f(f^{n-1}(x)) = x$. Now, assume $f(x) = f(y)$. Choose $m = x^{2023}$ and $n = y^{2023}$. Thus, $f^m(x) = x$ and $f^n(y) = y$, and note that\n$$\n\\begin{align*}\nf... | India | IMO TST Day 1 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Algebr... | null | proof only | null | |
041f | Let $n$ be an integer, $n \ge 2$, and $x_1, x_2, \dots, x_n \in [0, 1]$. Prove that
$$
\sum_{1 \le k < l \le n} kx_k x_l \le \frac{n-1}{3} \sum_{k=1}^{n} kx_k.
$$ | [
"As $x_1, x_2, \\dots, x_n \\in [0, 1]$, $x_i x_j \\le x_i$, so we have\n$$\n3 \\sum_{1 \\le k < l \\le n} kx_k x_l = \\sum_{1 \\le k < l \\le n} 3k x_k x_l \\le \\sum_{1 \\le k < l \\le n} (k x_k + 2k x_l).\n$$\nFor $1 \\le k \\le n$, the coefficient of $x_k$ in the last sum is\n$$\n2[1 + 2 + \\dots + (k-1)] + k(n... | China | China Western Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0eot | $\triangle ABC$ is right-angled at $A$, and $\triangle ABP$ is equilateral with $AB = 2$. The length of $AC$ is
(A) $\sqrt{6}$
(B) $\sqrt{8}$
(C) $\sqrt{10}$
(D) $\sqrt{12}$
(E) $4$ | [
"$BAP = 60^\\circ$ while $BAC = 90^\\circ$, so $PAC = 30^\\circ$. With $BPA = 60^\\circ$ that means $C = 30^\\circ$, and so $\\angle PAC$ is isosceles. Then $BC$ has length $4$, and so by Pythagoras the length of $AC$ is $\\sqrt{4^2 - 2^2}$"
] | South Africa | South African Mathematics Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | MCQ | D | |
06mi | Let $a$ and $b$ be the two roots of the equation $x^{\frac{4}{3}} - 2022x^{\frac{2}{3}} + 2023 = 0$. If $p = a + 3a^{\frac{1}{3}}b^{\frac{2}{3}}$ and $q = b + 3a^{\frac{2}{3}}b^{\frac{1}{3}}$, find the value of $(p+q)^{\frac{1}{3}} + (p-q)^{\frac{1}{3}}$. | [
"Answer: 4044\nLet $a = u^3$ and $b = v^3$. Then we have $u^4 - 2022u^2 + 2023 = 0$ and $v^4 - 2022v^2 + 2023 = 0$, so $u^2$ and $v^2$ are the two roots of $t^2 - 2022t + 2023 = 0$. In particular we have $u^2 + v^2 = 2022$. Note also that $p+q = u^3 + 3uv^2 + v^3 + 3u^2v = (u+v)^3$ and similarly $p-q = (u-v)^3$. It... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 4044 | |
0d3c | A perfect number is an integer that equals half the sum of its positive divisors. For example, because $2 \cdot 28 = 1 + 2 + 4 + 7 + 14 + 28$, $28$ is a perfect number.
a. A square-free integer is an integer not divisible by a square of any prime number. Find all square-free integers that are perfect numbers.
b. Prov... | [
"Recall that if\n$$\nn = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_k^{\\alpha_k}\n$$\nwhere $p_1 < p_2 < \\cdots < p_k$ are prime numbers and $\\alpha_1, \\alpha_2, \\ldots, \\alpha_k$ are positive integers, then the sum of the positive divisors of $n$ is\n$$\n\\sigma(n) = \\prod_{i=1}^{k} \\left( \\sum_{j=0}^{\\al... | Saudi Arabia | SAMC | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English, Arabic | proof and answer | 6; no perfect square is a perfect number | |
08tr | Suppose there are $6$ red, $3$ blue and $3$ yellow balls. When you place all of these balls on a straight line, how many distinguishable ways of lining them up are there, in which no adjacent balls are of the same color? Assume that balls of the same color are non-distinguishable. | [
"Since at least one non-red ball has to be placed between any pair of red balls and since there are $6$ red balls and $6$ non-red balls, it is easy to see that a yellow and a blue ball can be placed next to each other at most once.\n\nIn the case where no $2$ non-red balls are placed next to each other, red balls a... | Japan | Japan Mathematical Olympiad First Round | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 100 | |
0f24 | Problem:
Given a sequence $a_1$, $a_2$, ..., $a_n$ of positive integers. Let $S$ be the set of all sums of one or more members of the sequence. Show that $S$ can be divided into $n$ subsets such that the smallest member of each subset is at least half the largest member. | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0ckh | Consider an arbitrary triangle $ABC$ with incenter $I$, and let $I_a$, $I_b$, and $I_c$ be the excenters of triangle $ABC$ opposite to vertices $A$, $B$, and $C$, respectively, tangent to sides $BC$, $CA$, and $AB$. Let $E$, $F$, and $G$ be the points of tangency of the incircle with the sides $BC$, $CA$, and $AB$, res... | [
"Denote by $J_a$, $J_b$, and $J_c$ the centers of the circumcircles of triangles $IEI_a$, $IFI_b$, and $IGI_c$, respectively. We prove that $J_a$, $J_b$, and $J_c$ are collinear.\n\nLet $L_a$, $L_b$, and $L_c$ be the points diametrically opposite to $I$ on the circumcircles of triangles $IEI_a$, $IFI_b$, and $IGI_c... | Romania | 75th NMO Selection Tests | [
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler ... | English | proof only | null | |
04ou | Find all pairs $(m, n)$ of integers such that $mn + 5m + 2n = 121$. (Nikola Adžaga) | [] | Croatia | Croatian Mathematical Society Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (-1, 126), (129, -4), (-3, -136), (-133, -6) | |
0hny | Problem:
A position of the hands of a (12-hour, analog) clock is called valid if it occurs in the course of a day. For example, the position with both hands on the 12 is valid; the position with both hands on the 6 is not. A position of the hands is called bivalid if it is valid and, in addition, the position formed by... | [
"Solution:\nLet $h$ and $m$ denote the respective angles of the hour and minute hands, measured clockwise in degrees from 12 o'clock $(0 \\leq h, m < 360)$. Since the minute hand moves twelve times as fast as the hour hand, we have\n$$\nm = 12h - 360a\n$$\nwith $a$ an integer, for any valid time. Conversely, it is ... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Math Word Problems"
] | null | proof and answer | 143 | |
0chn | For a positive integer $m$ denote $S(m)$ the sum of its natural divisors, and if $n$ and $p$ are positive integers, denote $Q(n, p)$ the sum of the quotients of the division of $n$ by the natural divisors of $p$ (for instance, $Q(18, 10) = 18 + 9 + 3 + 1 = 31$).
Let $a$ and $b$ be two positive integers.
a) Prove that, ... | [
"a) If $d_1, d_2, \\dots, d_p$ are the positive divisors of a positive integer $n$, then $\\{d_1, d_2, \\dots, d_p\\} = \\{\\frac{n}{d_1}, \\frac{n}{d_2}, \\dots, \\frac{n}{d_p}\\}$.\nLet $b_1, b_2, \\dots, b_q$ be the positive divisors of $b$. Then\n$$\nQ(a, b) \\le \\frac{a}{b_1} + \\dots + \\frac{a}{b_q} = \\fra... | Romania | 74th Romanian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | a) a = b. b) No; for example a = 2 and b = 5. | |
02fr | Let $R$ be the set of real numbers. Show that there are no functions $f, g: R \to R$ such that $g(f(x)) = x^3$ and $f(g(x)) = x^2$ for all $x$. Let $S$ be the set of all real numbers greater than $1$. Show that there are functions $f, g: S \to S$ satisfying the condition above. | [
"First of all, since $g(f(x)) = x^3$ and $x^3$ is injective, $f$ is injective. Indeed,\n$f(x) = f(y) \\implies g(f(x)) = g(f(y)) \\iff x^3 = y^3 \\iff x = y$. Plugging\n$f(x)$ in $f(g(x)) = x^2$, we obtain $f(g(f(x))) = f(x)^2 \\iff f(x^3) = f(x)^2$. If\n$f, g: R \\to R$, consider $x = -1, 0, 1$ (i.e., the roots of... | Brazil | XIX OBM | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof only | null | |
0ch2 | Let $k \in \mathbb{N}^*$. We say that the ring $(A, +, \cdot)$ has the property $CP(k)$, if for every $a, b \in A$ there is a $c \in A$, such that $a^k = b^k + c^k$.
a) Give an example of a finite ring $(A, +, \cdot)$, which does not have the property $CP(k)$ for any positive integer $k$, with $k \ge 2$.
b) Let $n \in ... | [
"For any $k \\in \\mathbb{N}^*$ we denote $P_k(A) = \\{a^k \\mid a \\in A\\}$. The condition $CP(k)$ is then equivalent with\n$$\nx - y \\in P_k(A), \\quad \\text{for any } x, y \\in P_k(A),\n$$\nmeaning that $P_k(A)$ is a subgroup of the additive group $(A, +)$.\n\na) For $A = \\mathbb{Z}_4$, we have that $P_{2k}(... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Abstract Algebra > Group Theory",
"Number Theory > Other"
] | English | proof only | null | |
08ub | There are four spade cards with numbers $1$, $2$, $3$, $4$, six heart cards with numbers $1$, $2$, $3$, $\ldots$, $6$ and eight diamond cards with numbers $1$, $2$, $3$, $\ldots$, $8$. Suppose you choose three cards, one from each group. How many possible choices are there if the total of the numbers on the chosen card... | [
"28 ways\n\nLet $a$, $b$, $c$ be the number on the spade, heart and diamond card chosen, respectively. If $a + c$ is not a multiple of $7$, let $k$ be the remainder obtained when $a + c$ is divided by $7$. Then $k$ satisfies $1 \\le k \\le 6$, and $a + b + c$ becomes a multiple of $7$ when and only when $b = 7 - k$... | Japan | Japan Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics",
"Number Theory > Modular Arithmetic"
] | null | final answer only | 28 | |
003y | Determine todas las parejas $(a, b)$ de enteros positivos tales que $2a+1$ y $2b-1$ sean primos relativos y $a+b$ divida a $4ab+1$. | [] | Argentina | 21º OLIMPIADA IBEROAMERICANA DE MATEMÁTICA | [
"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"
] | Español | proof and answer | All pairs with b = a − 1 and a ≥ 2. | |
01dj | Starting with three points $A$, $B$, $C$ in general position and the circumcircle $k$ of $\triangle ABC$, a step consists of drawing a line $l$ and obtaining all points of intersection of $l$ with the lines already drawn and with $k$, where $l$ is
(1) the line through two distinct points that had been obtained before o... | [
"The answer is yes. It suffices to describe how to obtain the point $E \\neq A$ on $k$ with $AE \\perp BC$. We can then obtain the point $F \\neq B$ on $k$ with $BF \\perp AC$ analogously, and the orthocentre of $\\triangle ABC$ is where $AE$ and $BF$ intersect.\nLet the bisector of $\\angle BAC$ intersect $k$ in $... | Baltic Way | Baltic Way 2016 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0j2b | Problem:
Circle $O$ has chord $AB$. A circle is tangent to $O$ at $T$ and tangent to $AB$ at $X$ such that $AX = 2XB$. What is $\frac{AT}{BT}$? | [
"Solution:\n\nLet $TX$ meet circle $O$ again at $Y$. Since the homothety centered at $T$ takes $X$ to $Y$ and also takes $AB$ to the tangent line of circle $O$ passing through $Y$, we have $Y$ is the midpoint of arc $AB$. This means that $\\angle ATY = \\angle YTB$. By the Angle Bisector Theorem, $\\frac{AT}{BT} = ... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | final answer only | 2 | |
0jcy | Problem:
64 people are in a single elimination rock-paper-scissors tournament, which consists of a 6-round knockout bracket. Each person has a different rock-paper-scissors skill level, and in any game, the person with the higher skill level will always win. For how many players $P$ is it possible that $P$ wins the fi... | [
"Solution:\n\nAnswer: 49\n\nNote that a sub-bracket, that is, a subset of games of the tournament that themselves constitute a bracket, is always won by the person with the highest skill level. Therefore, a person wins her first four rounds if and only if she has the highest skill level among the people in her 16-p... | United States | HMMT November 2012 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 49 | |
07by | For each nonnegative integer $n$, polynomial $K_n(x_1, x_2, ..., x_n)$ is defined recursively as follows,
$$
\begin{array}{l}
K_0 = 1 \\
K_1(x_1) = x_1 \\
K_n(x_1, \dots, x_n) = x_n K_{n-1}(x_1, \dots, x_{n-1}) + (x_n^2 + x_{n-1}^2) K_{n-2}(x_1, \dots, x_{n-2}).
\end{array}
$$
Prove that $K_n(x_1, x_2, ..., x_{n-1}, x_... | [
"Consider a $1 \\times n$ table with cells labelled by variables $x_1, x_2, \\dots, x_n$ from left to right.\n\n| $x_1$ | $x_2$ | ... | $x_n$ |\n\nWe assign a polynomial in terms of variables $x_1, x_2, \\dots, x_n$ to each tiling of this table by $1 \\times 1$ and $1 \\times 2$ tiles, as follows. We define the wei... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
054x | Juku conjectured the following in his mathematics circle: whenever the product of two coprime integers $x$ and $y$ is divisible by the product of some two coprime integers $a$ and $b$, at least one of $x$ and $y$ is divisible by $a$ or $b$. Does his proposition hold? | [
"Let $x = 20$, $y = 21$, $a = 14$, $b = 15$. Then $x$ and $y$ are coprime, as they are consecutive, similarly $a$ and $b$ are coprime. The product $xy = 420$ is divisible by $ab = 210$ but neither of $20$ and $21$ is divisible by $14$ or $15$."
] | Estonia | Open Contests | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | No; counterexample: x = 20, y = 21, a = 14, b = 15. | |
0db0 | Let $ABC$ be a triangle, let $D$ be the touch point of the side $BC$ and the incircle of the triangle $ABC$, and let $J_b$ and $J_c$ be the incentres of the triangles $ABD$ and $ACD$, respectively. Prove that the circumcentre of the triangle $A J_b J_c$ lies on the bisector of the angle $BAC$. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Isogon... | English | proof only | null | |
0016 | En el pizarrón había un cuadrilátero $ABCD$ en el que se marcaron los puntos $P$, $Q$, $R$, $S$ en los lados $AB$, $BC$, $CD$, $DA$, respectivamente, tales que
$$
\frac{AP}{PB} = \frac{BQ}{QC} = \frac{CR}{RD} = \frac{DS}{SA} = \frac{1}{2}.
$$
Se borró toda la figura, excepto los cuatro puntos $P$, $Q$, $R$, $S$.
Descri... | [] | Argentina | XIX Olimpíada Matemática Argentina | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | español | proof only | null | |
0k0a | Problem:
Let $ABC$ be a triangle with circumradius $R=17$ and inradius $r=7$. Find the maximum possible value of $\sin \frac{A}{2}$. | [
"Solution:\n\nLetting $I$ and $O$ denote the incenter and circumcenter of triangle $ABC$ we have by the triangle inequality that\n$$\nAO \\leq AI + OI \\Longrightarrow R \\leq \\frac{r}{\\sin \\frac{A}{2}} + \\sqrt{R(R-2r)}\n$$\nand by plugging in our values for $r$ and $R$ we get\n$$\n\\sin \\frac{A}{2} \\leq \\fr... | United States | February 2017 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | (17 + sqrt(51)) / 34 | |
04by | Let $\triangle ABC$ be an acute triangle, and let $M$ and $N$ be the feet of the altitudes from $A$ and $B$, respectively. If $|AN| = |NM|$, prove that the incentre of $\triangle ABC$ lies on the altitude $\overline{BN}$. | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
07au | Find all polynomials with integer coefficients $P$ such that the set $P(\mathbb{Z}) = \{P(a) : a \in \mathbb{Z}\}$ contains an infinite geometric progression. | [
"Suppose that the image of integer numbers under the polynomial $P(x) = a_nx^n + a_{n-1}x^{n-1} + \\dots + a_1x + a_0$ contains an infinite geometric progression with common ratio $a \\in \\mathbb{Z} - \\{0\\}$. For each $b \\in \\mathbb{Z}$ we have\n$$\nP(ax + b) = a_n a^n x^n + (n a_n a^{n-1} b + a_{n-1} a^{n-1})... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | All such polynomials are of the form P(x) = s (q x − p)^n, where p, q, s are integers with gcd(p, q) = 1. | |
0eln | For a positive integer $n$, $S(n)$ denotes the sum of its digits and $U(n)$ its unit digit. Determine all positive integers $n$ with the property that
$$
n = S(n) + U(n)^2.
$$ | [
"Write $n$ as $a_0 + 10a_1 + 100a_2 + \\dots$, where $a_0, a_1, a_2, \\dots$ are the digits of $n$. Then the stated equation is equivalent to\n$$\na_0 + 10a_1 + 100a_2 + \\dots = a_0 + a_1 + a_2 + \\dots + a_0^2\n$$\nor\n$$\n9a_1 + 99a_2 + 999a_3 + \\dots = a_0^2.\n$$\nThe right hand side is at most $9^2 = 81 < 99$... | South Africa | The South African Mathematical Olympiad Third Round | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 13, 46, 99 | |
02be | Problem:
Augusto tem um arame com $10~\mathrm{m}$ de comprimento. Ele realiza um corte em um ponto do arame obtendo assim dois arames. Um com comprimento $x$ e outro com comprimento $10-x$ como mostra a figura abaixo:
Augusto usa os dois pedaços do arame para fazer dois quadrados.
a) Qual é o... | [
"Solution:\nNessa questão todos os comprimentos são dados em metros e as áreas em metros quadrados.\n\na) Um pedaço de corda tem comprimento $x$ e outro pedaço de corda tem comprimento $10-x$. Como um quadrado tem quatro lados de tamanhos iguais, um quadrado terá lado de comprimento igual a $\\frac{x}{4}$ e outro q... | Brazil | null | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | a) Lados: x/4 e (10 − x)/4. Áreas: x^2/16 e (10 − x)^2/16.
b) Corte no meio: comprimentos 5 e 5.
c) Dez pedaços iguais: cada um com comprimento 1. |
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