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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0ie2 | Problem:
Prove that for any two types of dominoes, there exists a rectangle that can be tiled by dominoes of either type. | [
"Solution:\nNote that a type $(a, b)$ domino tiles a $\\max\\{1, 2a\\} \\times 2b$ rectangle (see diagram for $a > 0$). Then both type $(a, b)$ and type $(a', b')$ dominoes tile a $\\left(\\max\\{1, 2a\\}, \\max\\{1, 2a'\\}\\right) \\times (2b \\cdot 2b')$ rectangle.\n\n"
] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Other"
] | null | proof only | null | |
08gi | Problem:
Lucio ha una scatola con tanti dadi a 14 facce di forma e dimensioni uguali ma con numerazioni differenti. La forma dei dadi è un cubo tronco, cioè un cubo a cui si tagliano i vertici in modo da formare un poliedro con 14 facce, 6 ottagoni regolari e 8 triangoli equilateri. Ogni dado riporta tutti i numeri da... | [
"Solution:\n\nLa risposta è $26880$. È immediato notare che la somma delle facce opposte deve essere necessariamente $15$. Partizioniamo quindi i numeri da $1$ a $14$ nelle coppie $\\{1,14\\},\\{2,13\\}, \\ldots,\\{7,8\\}$, in modo che la somma in ogni coppia sia $15$: ogni coppia di numeri deve essere assegnata a ... | Italy | Italian Mathematical Olympiad - February Round | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 26880 | |
07zr | Problem:
Ad una gara a punti su pista partecipano nove concorrenti. Ad ogni traguardo intermedio vengono assegnati 9 punti al primo, 8 al secondo, 7 al terzo e così via fino ad assegnare 1 punto all'ultimo. Prima dell'ultimo sprint (in cui il punteggio assegnato vale doppio) la classifica vede al comando Abdujaparov c... | [] | Italy | Italy Febbraio Contest | [
"Discrete Mathematics > Other"
] | null | proof and answer | 30 | |
0i8q | Let $n \neq 0$. For every sequence of integers
$$
a = a_0, a_1, a_2, \dots, a_n
$$
satisfying $0 \le a_i \le i$, for $i = 0, \dots, n$, define another sequence
$$
t(a) = t(a)_0, t(a)_1, t(a)_2, \dots, t(a)_n
$$
by setting $t(a)_i$ to be the number of terms in the sequence $a$ that precede the term $a_i$ and are differe... | [
"**First Solution.** Note first that the transformed sequence $t(a)$ also satisfies the inequalities $0 \\le t(a)_i \\le i$, for $i = 0, \\dots, n$. Call any integer sequence that satisfies these inequalities an *index bounded sequence*.\nWe prove now that $a_i \\le t(a)_i$, for $i = 0, \\dots, n$. Indeed, this is ... | United States | USA IMO 2003 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series"
] | English | proof only | null | |
05fy | Problem:
Montrer que si $a, b, c$ sont des nombres réels positifs vérifiant $a+b+c=1$ alors
$$
\frac{7+2 b}{1+a}+\frac{7+2 c}{1+b}+\frac{7+2 a}{1+c} \geqslant \frac{69}{4}
$$ | [
"Solution:\n\nComme $7+2 b=5+2(1+b)$, on écrit le membre de gauche sous la forme\n$$\n5\\left(\\frac{1}{1+a}+\\frac{1}{1+b}+\\frac{1}{1+c}\\right)+2\\left(\\frac{1+b}{1+a}+\\frac{1+c}{1+b}+\\frac{1+a}{1+c}\\right)\n$$\nEn utilisant l'inégalité $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z} \\geqslant \\frac{9}{x+y+z}$, on... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0bej | Consider $a \in (0,1)$ and $C$ the set of all increasing functions $f: [0,1] \to [0, \infty)$, such that $\int_0^1 f(x) dx = 1$. Determine:
a. $\max_{f \in C} \int_0^a f(x) dx$,
and
b. $\max_{f \in C} \int_0^a (f(x))^2 dx$. | [
"a.\n\nWe show that\n$$\n\\int_0^a f(x) dx \\le a.\n$$\nTo this end, write\n$$\n\\begin{aligned}\na - \\int_0^a f(x) dx &= a \\int_0^1 f(x) dx - \\int_0^a f(x) dx \\\\\n&= a \\int_a^1 f(x) dx - (1-a) \\int_0^a f(x) dx \\\\\n&\\ge a \\int_a^1 f(a) dx - (1-a) \\int_0^a f(a) dx \\\\\n&= 0.\n\\end{aligned}\n$$\nHere eq... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | a) The maximum is a. b) The maximum is a if a ≤ 1/2, and 1/(4(1−a)) if a > 1/2. | |
09ni | It takes $54$ seconds to descend the escalator while standing still. When Bayaraa walks down the escalator at a speed of $1$ m/s, it takes $36$ seconds. How many seconds will it take Bayaraa to descend if he walks at a speed of $2$ m/s? | [] | Mongolia | MMO2025 Round 2 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | final answer only | 27 | |
0jqz | Problem:
Consider a $7 \times 7$ grid of squares. Let $f:\{1,2,3,4,5,6,7\} \rightarrow \{1,2,3,4,5,6,7\}$ be a function; in other words, $f(1), f(2), \ldots, f(7)$ are each (not necessarily distinct) integers from $1$ to $7$. In the top row of the grid, the numbers from $1$ to $7$ are written in order; in every other ... | [
"Solution:\n\nConsider the directed graph with $1,2,3,4,5,6,7$ as vertices, and there is an edge from $i$ to $j$ if and only if $f(i)=j$. Since the bottom row is equivalent to the top one, we have $f^{6}(x)=x$. Therefore, the graph must decompose into cycles of length $6,3,2$, or $1$. Furthermore, since no other ro... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Other"
] | null | proof and answer | 1470 | |
0ctp | In the space, $2016$ distinct spheres are arranged. Some of these spheres are red, and the others are green. Each tangency point of two spheres of different colors is marked in blue. Find the greatest possible number of blue points.
(A. Kuznetsov)
В пространстве расположены $2016$ сфер, никакие две из них не совпадают... | [
"$1008^2 = 1016064$ points.\n\nOne may arrange the spheres so that there are $1008$ red and $1008$ green spheres, and each pair of a red and a green sphere is tangent at a separate blue point. Arrange the centers $R_i$ of equal red spheres along a circle centered at $O$, arrange the centers $G_i$ of green spheres a... | Russia | Russian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | 1016064 | |
05dt | Problem:
Let $ABC$ be a triangle such that $\angle CAB > \angle ABC$, and let $I$ be its incentre. Let $D$ be the point on segment $BC$ such that $\angle CAD = \angle ABC$. Let $\omega$ be the circle tangent to $AC$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumc... | [
"Solution:\n\nLet $S$ be the intersection point of $BC$ and the angle bisector of $\\angle BAD$, and let $T$ be the intersection point of $BC$ and the angle bisector of $\\angle BXC$. We will prove that both quadruples $A, I, B, S$ and $A, I, B, T$ are concyclic, which yields $S = T$.\n\nFirstly denote by $M$ the m... | European Girls' Mathematical Olympiad (EGMO) | EGMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Inver... | null | proof only | null | |
0dnx | Problem:
Одредити колико се највише краљица може поставити на таблу $2017 \times 2017$, при чему свака краљица сме да напада највише једну од преосталих.
(Бојан Башић и комисија) | [
"Solution:\n\nОзначимо $n=2017$. Претпоставимо да је постављено $m>n$ краљица. Ни у једној врсти нема више од две краљице, па се у бар $m-n$ врста налазе по две краљице, тако да има највише $m-2(m-n)=2 n-m$ краљица које су саме у својој врсти. Слично, највише $2 n-m$ краљица су саме у својој колони. С друге стране,... | Serbia | 11. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2689 | |
075d | Let $\mathbb{R}^+$ denote the set of all positive real numbers. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}$ satisfying
$$
f(x) + f(y) \le \frac{f(x+y)}{2}, \quad \frac{f(x)}{x} + \frac{f(y)}{y} \ge \frac{f(x+y)}{x+y},
$$
for all $x, y \in \mathbb{R}^+$. | [
"Put $x = y = t$ ($t > 0$). We get\n$$\n4f(t) \\le f(2t), \\quad f(2t) \\ge 4f(t),\n$$\nfor all $t > 0$. Hence $f(2t) = 4f(t)$, for all $t > 0$. By induction\n$$\nf(2^m t) = 2^{2m} f(t), \\text{ for all } t > 0.\n$$\nLet $g(x) = f(x)/x$, $x > 0$, $g(0) = 0$. We show that $g(nt) = ng(t)$ for all $n \\in \\mathbb{N}$... | India | Indija TS 2012 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = a x^2 for all x > 0, where a ≤ 0 | |
03ar | Any of two lines, parallel to the $x$-axis, has two common points with the graph of the function $f(x) = x^3 + a x^2 + b x + c$. Prove that the quadrilateral with vertices at these four points is a rhombus if and only if its area is equal to $6$. | [
"Lines $p$ and $q$ with the given properties exist if and only if $f(x)$ has local minima and maxima, and $p$ and $q$ pass through the point of maxima $D$ and the point of minima $B$ on the graph $f(x)$, respectively. Using transformations of the forms $g(x) = f(x) + a$ and $g(x) = f(x + a)$, we may move this graph... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Geometry > Plane Geometry > Transformations > Translation"
] | English | proof only | null | |
08k4 | Problem:
Let $E$ and $F$ be two distinct points inside of a parallelogram $A B C D$. Find the maximum number of triangles with the same area and having the vertices in three of the following five points: $A, B, C, D, E, F$. | [
"Solution:\nWe shall use the following two well known results:\n\nLemma 1. Let $A, B, C, D$ be four points lying in the same plane such that the line $A B$ does not intersect the segment $C D$ (in particular $A B C D$ is a convex quadrilateral). If $[A B C]=[A B D]$, then $A B \\parallel C D$.\n\nLemma 2. Let $X$ b... | JBMO | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 10 | |
07uf | Find all solutions in positive integers $a, b, c$ to the equation $a! = \frac{1}{5}b! + \frac{4}{5}c!$. | [
"There are trivial solutions when $a = b = c$, and one non-trivial solution when $a = 2, b = 3, c = 1$ as:\n$$\n2! = 2 = \\frac{1}{5} \\times 6 + \\frac{4}{5} \\times 1 = \\frac{1}{5} \\times 3! + \\frac{4}{5} \\times 1!\n$$\nWe will prove that there are no other solutions. If $b = c$ the equation becomes $a! = b!$... | Ireland | IRL_ABooklet | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | All solutions are a = b = c (any positive integer) and the nontrivial solution (a, b, c) = (2, 3, 1). | |
0jcr | Problem:
Let $ABC$ be a triangle with $AB < AC$. Let $M$ be the midpoint of $BC$. Line $l$ is drawn through $M$ so that it is perpendicular to $AM$, and intersects line $AB$ at point $X$ and line $AC$ at point $Y$. Prove that $\angle BAC = 90^{\circ}$ if and only if quadrilateral $XBYC$ is cyclic. | [
"Solution:\n\n\n\nFirst, note that $XBYC$ cyclic is equivalent to $\\measuredangle BXM = \\measuredangle ACB$. However, note that $\\measuredangle BXM = 90^{\\circ} - \\measuredangle BAM$, so $XBYC$ cyclic is in turn equivalent to $\\measuredangle BAM + \\measuredangle ACB = 90^{\\circ}$.\n... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
01ap | Let $n$ be a positive integer greater than $1$. The teacher writes $n+1$ positive integers on the blackboard, whereby the last of them, let it be $c$, is not divisible by $n$. Can Mary always denote the first $n$ integers written by the teacher by $a_1, \dots, a_n$ in such an order that the product $(a_1 - a_2) \cdot (... | [
"**Answer:** Yes.\n\nIf some two of the first $n$ integers are congruent modulo $n$ then Mary can choose them consecutively and obtain a product divisible by $n$. Hence we may assume in the rest that the first $n$ integers written by the teacher are pairwise incongruent modulo $n$. This means that these $n$ integer... | Baltic Way | Baltic Way 2013 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | Yes | |
0l0u | A group of 16 people will be partitioned into 4 indistinguishable 4-person committees. Each committee will have one chairperson and one secretary. The number of different ways to make these assignments can be written as $3^r M$, where $r$ and $M$ are positive integers and $M$ is not divisible by 3. What is $r$?
(A) 5 (... | [
"The 16 people can be partitioned into the 4 committees, each of size 4, in\n$$\n\\frac{16!}{(4!)^5}\n$$\nways; four of the $4!$ factors come from permuting the members of the committees and one $4!$ factor comes from permuting the committees. Then there are $4^4$ ways to choose the four chairpersons and $3^4$ ways... | United States | AMC 10 B | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | A | |
0fi9 | Problem:
Sea $p$ un número primo. Determinar todos los enteros $k \in \mathbb{Z}$ tales que $\sqrt{k^{2}-p k}$ es un entero positivo. | [
"Solution:\n\nSi ponemos $\\sqrt{k^{2}-k p}=n$ nos queda $k^{2}-p k-n^{2}=0$, de donde se deduce\n$$\nk=\\frac{p \\pm \\sqrt{p^{2}+4 n^{2}}}{2}\n$$\nEl radicando ha de ser cuadrado perfecto; llamémosle $a$. Se tiene\n$$\np^{2}+4 n^{2}=a^{2} \\quad \\text{ o bien } \\quad p^{2}=(a+2 n)(a-2 n)\n$$\nComo $p$ es primo ... | Spain | Olimpiada Matemática Española | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | If p = 2: k ∈ {0, 2}. If p ≠ 2: k ∈ { ((p+1)/2)^2, -((p-1)/2)^2, p, 0 }. | |
0cb8 | The elements of a set $A$ are 13 consecutive positive integers, the elements of a set $B$ are 12 consecutive positive integers, and the elements of the set $A \cup B$ are 15 consecutive integers.
a)
Find the number of elements of $A \setminus B$.
b) If, moreover, the sum of the elements of the set $A$ is equal to the s... | [
"a) The elements of the set $A \\setminus B$ are the elements of $A \\cup B$ which are not in $B$; there are $15 - 12 = 3$ such elements.\n\nb) The sum of the elements from $A \\setminus B$ equals the sum of the elements from $B \\setminus A$.\nLet $A \\cap B = \\{n, n+1, n+2, \\dots, n+9\\}, n \\in \\mathbb{N}$. T... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) 3
b) A = {24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36}, B = {27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38} | |
0d98 | Given a set $S$ of 200 points on the plane: 100 points are the vertices of a convex polygon $A$, and 100 other points are in the interior of the polygon. Moreover, there does not exist 3 collinear points. A triangulation is a way to partition the polygon $A$ into triangles by drawing the edges between some two points o... | [
"1) Suppose that we have $k$ triangles in some triangulation. By calculating the sum of all angles of these triangles, we have $180^\\circ \\cdot k$.\nThe sum of interior angles of $A$ is $180^\\circ \\cdot 98$.\nThe sum of angle around each point among 100 points is $360^\\circ \\cdot 100$. Hence, we have\n$$\n180... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combin... | English | proof only | null | |
0ejj | Problem:
Poišči vsa cela števila $a$, za katera je tudi $\log_{2}\left(a^{2}-4a-1\right)$ celo število. | [
"Solution:\n\nOznačimo $\\log_{2}\\left(a^{2}-4a-1\\right)=n$, kjer je $n$ celo število. Potem je $a^{2}-4a-1=2^{n}$ oziroma $a^{2}-4a-(1+2^{n})=0$. To je kvadratna enačba za $a$, ki ima rešitvi\n$$\na_{1}=2+\\sqrt{5+2^{n}}, \\quad a_{2}=2-\\sqrt{5+2^{n}}.\n$$\nEnačba ima celoštevilsko rešitev le v primeru, ko je $... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | a = -1 and a = 5 | |
024n | Problem:
Para fazer a separação em regiões da correspondência que deve ser entregue, um serviço postal indica sobre os envelopes um código postal com uma série de 5 grupos de bastões, que podem ser lidos por um leitor ótico. Os algarismos são codificados como a seguir:
| $0 \bullet\|I\|$ | $5 \% \cdot \\|$ |
| :---: ... | [
"Solution:\n\n47679 e 47779"
] | Brazil | null | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 47679, 47779 | |
0boa | Find all triples $(x, y, z)$ of natural numbers such that $21^x + 4^y = z^2$. | [
"Rewrite the equation as $(z - 2^y)(z + 2^y) = 21^x$. Denote by $d$ the largest common divisor of $z - 2^y$ and $z + 2^y$; then $d$ divides $(z + 2^y) - (z - 2^y)$, so $d \\mid 2^{y+1}$. Since $d$ is a divisor of $z + 2^y$ and $z + 2^y$ divides $21^x$, it follows that $d \\mid (2^{y+1}, 21^x)$, so $d = 1$.\nConsequ... | Romania | 66th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (1, 1, 5) | |
00mo | There are $n$ children in a room. Each child has at least one piece of candy. In Round 1, Round 2, etc., additional pieces of candy are distributed among the children according to the following rule:
*In Round $k$, each child whose number of pieces of candy is relatively prime to $k$ receives an additional piece.*
*Sho... | [
"We observe that a child that has $k - 1$ or $k + 1$ pieces of candy at the start of Round $k$ will receive an additional piece because of $\\text{gcd}(k, k \\pm 1) = 1$ and be in the same situation in the next round. Furthermore, in each round, the number of the round will increase by 1 and the number of pieces of... | Austria | 49th Austrian Mathematical Olympiad, National Competition (Final Round, part 2) | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0bi0 | Find, with proof, all positive integers $\overline{abc}$ satisfying
$$
b \cdot \overline{ac} = c \cdot \overline{ab} + 10.
$$ | [
"The given condition rewrites as $b(10a + c) = c(10a + b) + 10$, from which $a(b - c) = 1$, and hence $a = b - c = 1$. The numbers are\n110, 121, 132, 143, 154, 165, 176, 187, 198."
] | Romania | 65th Romanian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 110, 121, 132, 143, 154, 165, 176, 187, 198 | |
0epz | If today is Thursday, what day of the week will it be in 150 days from now?
(A) Sunday
(B) Monday
(C) Tuesday
(D) Wednesday
(E) Thursday | [
"Every multiple of $7$ represents a full week. Since today is Thursday, in one day's time it will be Friday. Thus each full week after today starts on a Friday and ends on a Thursday. $150$ days divided by $7$ equals $21$ full weeks with a remainder of $3$ days. The $147$th day from now will thus be a Thursday (end... | South Africa | South African Mathematics Olympiad | [
"Number Theory > Modular Arithmetic"
] | English | MCQ | A | |
0jd0 | Problem:
Consider triangle $ABC$ where $BC = 7$, $CA = 8$, and $AB = 9$. $D$ and $E$ are the midpoints of $BC$ and $CA$, respectively, and $AD$ and $BE$ meet at $G$. The reflection of $G$ across $D$ is $G'$, and $G'E$ meets $CG$ at $P$. Find the length $PG$. | [
"Solution:\n\nAnswer: $\\frac{\\sqrt{145}}{9}$\n\nObserve that since $G'$ is a reflection and $GD = \\frac{1}{2} AG$, we have $AG = GG'$ and therefore, $P$ is the centroid of triangle $ACG'$. Thus, extending $CG$ to hit $AB$ at $F$, $PG = \\frac{1}{3} CG = \\frac{2}{9} CF = \\frac{2}{9} \\sqrt{\\frac{2(8^2 + 7^2) -... | United States | HMMT November 2012 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Reflection",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | sqrt(145)/9 | |
0k56 | Problem:
20 players are playing in a Super Smash Bros. Melee tournament. They are ranked $1-20$, and player $n$ will always beat player $m$ if $n < m$. Out of all possible tournaments where each player plays 18 distinct other players exactly once, one is chosen uniformly at random. Find the expected number of pairs of... | [
"Solution:\n\nConsider instead the complement of the tournament: The 10 possible matches that are not played. In order for each player to play 18 games in the tournament, each must appear once in these 10 unplayed matches. Players $n$ and $n+1$ will win the same number of games if, in the matching, they are matched... | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof and answer | 4 | |
0awm | Problem:
The boat is sinking! Passengers must then be saved, but the rescuer must know their count. If the passengers group themselves into $7$, one group will only have $4$ passengers. If the passengers group themselves into $11$, one group will only have $7$ passengers. If the passengers group themselves into $13$, ... | [] | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | 634 | |
0l4z | Problem:
The circumference of a circle is divided into $45$ arcs, each of length $1$. Initially, there are $15$ snakes, each of length $1$, occupying every third arc. Every second, each snake independently moves either one arc left or one arc right, each with probability $\frac{1}{2}$. If two snakes ever touch, they me... | [
"Solution:\nWe solve the problem generally for $n$ snakes and $3n$ arcs. Without loss of generality, fix the two snakes $A$ and $B$ that will eventually form the ends of the last snake. Note that $A$ and $B$ must be consecutive in the initial configuration; assume $B$ lies immediately clockwise from $A$.\n\nLet $d$... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 28 | |
01vf | Three $n \times n$ squares form the figure $\Phi$ on the checkered plane as
shown on the picture. (Neighboring squares are touching along the segment of length $1$).
Find all $n > 1$ for which the figure $\Phi$ can be covered with tiles $1 \times 3$ and $3 \times 1$ without overlapping. | [
"Answer: $n = 3k$, $n = 3k + 1$.\nIt is clear that any rectangle with one of the sides divided by $3$ can be covered with tiles. Therefore, if $n = 3k$, each of the three squares and the entire figure can be tiled.\n\nWe prove that $\\Phi$ can be covered for any $n$ of the form $3k+1$. Put two horizontal tiles so t... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | n = 3k or n = 3k + 1 for integers k ≥ 1 | |
09o4 | Let quadrilateral $ABCD$ be inscribed in a circle. Let $E$ be the midpoint of arc $AB$ that does not contain $C$ or $D$, and let $F$ be the midpoint of arc $CD$ that does not contain $A$ or $B$. Let $AC \cap BF = K$, $BD \cap AF = L$, $AC \cap DE = M$, and $BD \cap CE = N$. Prove that $\angle DML = \angle CNK$.
(Khula... | [] | Mongolia | MMO2025 Round 3 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0e0n | Find all natural numbers $n$ for which there exists a prime $p$ such that $p^2 + 7^n$ is a perfect square. | [
"Let $p^2 + 7^n = m^2$. Then $7^n = m^2 - p^2 = (m - p)(m + p)$. We consider two possible cases.\n\nIf $m - p = 1$ and $m + p = 7^n$, then $2p = 7^n - 1$. Assume that $n \\ge 2$. Then $2p = 7^n - 1 = (7 - 1)(7^{n-1} + 7^{n-2} + \\dots + 7 + 1)$, so $p = 3(7^{n-1} + 7^{n-2} + \\dots + 7 + 1)$, but this is not possib... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1 | |
0dde | Let $ABC$ be a triangle inscribed in a fixed circle $(O)$ with $BC$ fixed and $A$ varying on $(O)$. Denote $H$ as the orthocenter of triangle $ABC$ and take $D$, $E$ on $AB$, $AC$ respectively such that $H$ is the midpoint of $DE$. Prove that when $A$ moves on $(O)$, the center of $(ADE)$ belongs to a fixed circle. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous ... | null | proof only | null | |
00m5 | Es sei $ABCDE$ ein regelmäßiges Fünfeck. Auf der Strecke zwischen dem Mittelpunkt $M$ des Fünfecks und dem Punkt $D$ wird ein Punkt $P \neq M$ gewählt. Der Umkreis von $ABP$ schneidet die Seite $AE$ in den Punkten $A$ und $Q$ und die Normale auf $CD$ durch $P$ in den Punkten $P$ und $R$.
Man zeige, dass $AR$ und $QR$ g... | [
"\n\nLösung 1. Es sei $S$ der Schnittpunkt von $RP$ und $AE$, vgl. Abbildung 1. Die Winkel im Dreieck $ABE$ sind bekannt als $\\angle BAE = 108^\\circ$ und $\\angle EBA = \\angle AEB = 36^\\circ$. Da $BE$ und $CD$ parallel sind, steht $RP$ auch auf $BE$ normal, und wir bezeichnen deren Schn... | Austria | 48. Österreichische Mathematik-Olympiade Bundeswettbewerb für Fortgeschrittene, Teil 1 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | German | proof only | null | |
08kb | Problem:
Government of Greece wants to change the building. So they will move in building with square shape $2005 \times 2005$. Only demand is that every room has exactly 2 doors and doors cannot lead outside the building. Is this possible? What about a building $2004 \times 2005$ ? | [
"Solution:\n\nThe key idea is to color the table $2005 \\times 2005$ in two colors, like chessboard. One door connects two adjacent squares, so they have different colors. We can count the number of doors in two ways. The number of doors is twice the number of white squares and it is also twice the number of black ... | JBMO | OJBM | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 2005 by 2005: impossible; 2004 by 2005: possible. | |
0ki1 | An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$, which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of the pillars at $A$, $B$, and $C$ are $12$, $9$, and $10$ m... | [
"Let $M$ be the midpoint of $\\overline{AC}$. The pillars at $A$ and $C$ are $12$ and $10$ meters high, respectively, so the height of the panel at $M$ is $11$ meters. Because $\\triangle BMA$ is a $30$-$60$-$90^\\circ$ triangle, $BM$ is half of the side length of the hexagon. Therefore $BM = \\frac{1}{4}BE$. It fo... | United States | AMC 10 A | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | D | |
014u | Problem:
Consider a subset $A$ of 84 elements of the set $\{1,2, \ldots, 169\}$ such that no two elements in the set add up to 169. Show that $A$ contains a perfect square. | [
"Solution:\n\nIf $169 \\in A$, we are done. If not, then\n$$\nA \\subset \\bigcup_{k=1}^{84}\\{k, 169-k\\}\n$$\nSince the sum of the numbers in each of the sets in the union is $169$, each set contains at most one element of $A$; on the other hand, as $A$ has 84 elements, each set in the union contains exactly one ... | Baltic Way | Baltic Way 2008 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0eu6 | Find all triplets $(x, y, z)$ of positive integers satisfying $1 + 4^x + 4^y = z^2$. | [
"Without loss of generality, we may assume that $x \\le y$. Suppose that $2x < y + 1$. Then\n$$\n(2^y)^2 < 1 + 4^x + 4^y < (1 + 2^y)^2,\n$$\nwhich implies that $1 + 4^x + 4^y$ is not a square of an integer. If $2x = y + 1$, then $1 + 4^x + 4^y = 1 + 2^{y+1} + 4^y = (1 + 2^y)^2$. Hence\n$$\n(x, y, z) = (x, 2x-1, 1 +... | South Korea | 20th Korean Mathematical Olympiad Final Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (x, y, z) = (t, 2t - 1, 1 + 2^{2t - 1}) or (2t - 1, t, 1 + 2^{2t - 1}) for any positive integer t | |
00lz | Find all pairs $(a, b)$ of non-negative integers such that
$$
2017^a = b^6 - 32b + 1.
$$ | [
"Answer: The two solutions are $(0, 0)$ and $(0, 2)$.\n\nSince $2017^a$ is always odd, $b$ must be even, so $b = 2c$, $c$ integer. Therefore, $2017^a = 64(c^6 - c) + 1$ and thus $2017^a \\equiv 1 \\pmod{64}$. But we find $2017 \\equiv 33 \\pmod{64}$ and $2017^2 \\equiv (1+32)^2 = 1+2\\cdot32+32^2 \\equiv 1 \\pmod{6... | Austria | 48th Austrian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | [(0, 0), (0, 2)] | |
0akc | Let $n$ be a positive integer and $C$ be nonnegative real number. Find the number of the sequences of real numbers $1, x_2, \dots, x_n, 1$, such that the absolute value of the difference of every two consecutive terms is equal to $C$. | [
"Let us suppose that the sequence $1, x_2, \\dots, x_n, 1$ satisfy the condition of the problem. Then\n$$\n|1 - x_2| = |x_2 - x_3| = \\dots = |x_n - 1| = C. \\quad (1)\n$$\nAlso\n$$\nx_n - 1 = (x_n - x_{n-1}) + (x_{n-1} - x_{n-2}) + \\dots + (x_2 - 1) \\quad (2)\n$$\nwhere the number of pairs of brackets of the rig... | North Macedonia | Macedonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | r(n, C) = 1 if C = 0; r(n, C) = 0 if n is odd and C > 0; r(n, C) = 2 * binomial(n - 1, n/2) if n is even and C > 0. | |
03dq | Consider a quadratic polynomial $f(x)$ with integer coefficients. If the values $f(0)$, $f(3)$, and $f(4)$ are pairwise different numbers from the set $\{2, 20, 202, 2022\}$, then determine all possible values for $f(1)$. | [
"$f(1) \\in \\{-80, -990\\}$.\nSince $f \\in \\mathbb{Z}[x]$, we have that $y - x \\mid f(y) - f(x)$, $\\forall x, y \\in \\mathbb{Z}$. Thus, $3 \\mid |f(3) - f(0)|$, and $4 \\mid |f(4) - f(0)|$. In the set $\\{2, 20, 202, 2022\\}$, only $2$ and $20$ are congruent modulo $3$, so $\\{f(0), f(3)\\} = \\{2, 20\\}$. Mo... | Bulgaria | Bulgaria 2022 | [
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | {-80, -990} | |
0hib | The positive integers $x$, $y$ satisfy the conditions:
$$
\{\sqrt{x^2 + 2y}\} > \frac{2}{3}, \{\sqrt{y^2 + 2x}\} > \frac{2}{3}.
$$
Prove that $x = y$.
Here, $\{a\} \in [0; 1)$ denotes the fractional part of the number $a$, that is, there exists an integer $n$ for which the equality $a = n + \{a\}$ holds. For example, $... | [
"Suppose that for some positive integers $x < y$ these inequalities are true: $\\{\\sqrt{x^2 + 2y}\\} > \\frac{2}{3}, \\{\\sqrt{y^2 + 2x}\\} > \\frac{2}{3}$. Note that $y^2 < y^2 + 2x < (y + 1)^2$, so we have\n$$\ny^2 + 2x > \\left(y + \\frac{2}{3}\\right)^2 \\Leftrightarrow 2x > \\frac{4}{3}y + \\frac{4}{9} \\Righ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0086 | 2010 cards are enumerated $1, 2, \dots, 2010$. All cards whose number has odd digit sum are chosen. Find the sum of the numbers on the chosen cards. | [
"Denote the digit sum of $a$ by $S(a)$. Add a card with $0$ and assume $S(0) = 0$. Among $0, 1, \\dots, 999$ there are $500$ numbers $a$ with $S(a)$ odd and $500$ with $S(a)$ even. Indeed $0, 1, \\dots, 999$ can be divided into $500$ pairs $(a, b)$ with sum $999$ of every pair. There is no carryover in the addition... | Argentina | Mathematical Olympiad Rioplatense | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 1011535 | |
0etw | Consider three circles $\Gamma_1, \Gamma_2$ and $\Gamma_3$, with centres $O_1, O_2$ and $O_3$, respectively, such that each pair of circles is externally tangent. Suppose we have another circle $\Gamma$ with centre $O$ on the line segment $O_1O_3$ such that $\Gamma_1, \Gamma_2$ and $\Gamma_3$ are each internally tangen... | [
"Let the points of tangency of $\\Gamma$ with $\\Gamma_1, \\Gamma_2$ and $\\Gamma_3$ be $A, B$ and $C$, respectively.\n\n\n\nThen $AOB$ is a straight line, and since it is a diameter, $\\angle ACB = 90^\\circ$. Let $E$ be the other intersection point of $AC$ with $\\Gamma_3$ and $F$ be the ... | South Africa | The South African Mathematical Olympiad Third Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0geo | 設 $ABCD$ 為凸四邊形，其中任兩邊皆不等長，且 $AC \perp BD$。設 $O_1, O_2$ 分別為三角形 $ABD$ 與 $CBD$ 的外心。證明：直線 $AO_2$、$CO_1$ 以及三角形 $ABC$ 的尤拉線、三角形 $ADC$ 的尤拉線四線共點。
（註：三角形的尤拉線為其外心、重心、垂心所在的直線。）
Let $ABCD$ be a convex quadrilateral with pairwise distinct side lengths such that $AC \perp BD$. Let $O_1, O_2$ be the circumcenters of $\triangle ABD, \... | [
"By symmetry it suffices to show that $AO_2$, $CO_1$ and the Euler line of $\\triangle ABC$ are concurrent. Note that $O_1O_2 \\parallel AC$. Let $\\ell_A, \\ell_C$ be the perpendicular bisector of $BC, AB$, respectively. Also denote by $M_A, M_C$ the midpoints of $BC, AB$, respectively. Let $G, O, H$ be the centro... | Taiwan | 2021 數學奧林匹亞競賽第二階段選訓營, 獨立研究 (一) | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
... | null | proof only | null | |
08w5 | Suppose a cube $ABCD$-$EFGH$ having a side length of $2012$ and a plane are placed in space, and the intersection of the plane and the cube forms a hexagon $IJKLMN$, where the points $I$, $J$, $K$, $L$, $M$, $N$ lie on the sides $AE$, $EF$, $FG$, $GC$, $CD$, $DA$, respectively, as shown in figure 3 below.
If $AI - GL =... | [
"Take three points $P$, $Q$, $R$ inside of the cube in such a way that the quadrilaterals $KLMP$, $MNIQ$ and $IJKR$ are parallelograms. Since $IJ$ and $ML$ are parallel, so are $PK$ and $RK$, and we see that the points $P$, $R$, $K$ lie on a same straight line. Therefore, we conclude that the hexagon $IJKLMN$ is pa... | Japan | Japan Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof and answer | 4\sqrt{61} | |
09je | A cube number is a cube of a positive integer. For which values of a positive integer $n \ge 3$, the sum of $n$ cube numbers can again be a cube number? | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All integers n greater than or equal to 3 | |
03ye | The lengths of the nine edges of regular triangular prism $ABC-A_1B_1C_1$ are equal, $P$ is the midpoint of $CC_1$, and the dihedral angle $B-A_1P-B_1 = \alpha$. Then $\sin \alpha = \underline{\hspace{2cm}}$. | [
"Let the line through segment $AB$ be $x$-axis with the origin $O$ being the midpoint of $AB$ and let the line through segment $OC$ be $y$-axis to establish a space rectangular coordinate system shown in Fig. 7.1. Assuming the length of an edge is 2, we have $B(1, 0, 0)$, $B_1(1, 0, 2)$, $A_1(-1, 0, 2)$, $P(0, \\sq... | China | China Mathematical Competition | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Linear Algebra > Vectors"
] | English | proof and answer | sqrt(10)/4 | |
046h | Let $a_1, a_2, \dots, a_{2023}$ be nonnegative real numbers such that $a_1 + a_2 + \dots + a_{2023} = 100$. Let $N$ denote the number of elements in the following set
$$
\{(i, j) \mid 1 \le i \le j \le 2023, a_i a_j \ge 1\}.
$$
Prove that $N \le 5050$, and determine the necessary and sufficient condition for $N = 5050$... | [
"**Proof.** Let $S$ be the number of pairs $(i, j)$ that satisfy the following conditions:\n$$\n1 \\le i < j \\le 2023, \\quad a_i a_j \\ge 1.\n$$\nLet $T$ be the number of elements among $a_1, a_2, \\dots, a_{2023}$ that are not less than $1$. Then\n$$\n100 = a_1 + a_2 + \\dots + a_{2023} \\ge T, \\qquad (3)\n$$\n... | China | Chinese Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | N ≤ 5050, with equality if and only if exactly 100 of the numbers are equal to one and the remaining 1923 are zero. | |
0kga | Problem:
Suppose you have 9 evenly spaced dots in a circle on a piece of paper. You want to draw a 9-pointed star by connecting dots around the circle without lifting your pencil, skipping the same number of dots each time.
Determine the number of different stars that can be drawn, if the regular nonagon does not count... | [
"Solution:\n\nThere are two different stars that can be drawn, by skipping 1 dot each time (as in the first star shown above) or skipping 3 dots each time (as in the third star shown above). Skipping 2 dots each time does not work because we end up drawing three separate equilateral triangl... | United States | Berkeley Math Circle: Monthly Contest 3 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 2 | |
0gbh | 試求最大的實數 $a$ 使得對所有 $n \ge 1$ 與所有實數 $x_0, x_1, \dots, x_n$ 滿足
$$
0 = x_0 < x_1 < x_2 < \dots < x_n,
$$
我們有
$$
\frac{1}{x_1 - x_0} + \frac{1}{x_2 - x_1} + \dots + \frac{1}{x_n - x_{n-1}} \ge a \left( \frac{2}{x_1} + \frac{3}{x_2} + \dots + \frac{n+1}{x_n} \right).
$$ | [
"The largest $a$ is $4/9$.\n\nWe first show that $a = 4/9$ is admissible. For each $2 \\le k \\le n$ by the Cauchy-Schwarz Inequality, we have\n$$\n(x_{k-1} + (x_k - x_{k-1})) \\left( \\frac{(k-1)^2}{x_{k-1}} + \\frac{3^2}{x_k - x_{k-1}} \\right)^2 \\ge (k-1+3)^2,\n$$\nwhich can be rewritten as\n$$\n\\frac{9}{x_k -... | Taiwan | 二〇一七數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 4/9 | |
0bhr | Find all natural $a$ such that there exist prime numbers $p, q, r$ so that
$$
a = \frac{p+q}{r} + \frac{q+r}{p} + \frac{r+p}{q}.
$$ | [
"We will show that the only solution is $a = 6$ (for $p = q = r$).\n\nIf exactly two of the three prime numbers are equal; e.g. $p = q \\neq r$, then\n$$\na = 2\\left(\\frac{p}{r} + \\frac{r}{p}\\right) + 2 \\in \\mathbb{N},\n$$\nso there exists $n \\in \\mathbb{N}$ so that\n$$\n\\frac{n}{2} = \\frac{p}{r} + \\frac... | Romania | THE 2014 DANUBE MATHEMATICAL COMPETITION | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 6 | |
0k1k | Problem:
Michael picks a random subset of the complex numbers $\{1, \omega, \omega^{2}, \ldots, \omega^{2017}\}$ where $\omega$ is a primitive $2018^{\text{th}}$ root of unity and all subsets are equally likely to be chosen. If the sum of the elements in his subset is $S$, what is the expected value of $|S|^{2}$? (The... | [
"Solution:\n\nConsider $a$ and $-a$ of the set of complex numbers. If $x$ is the sum of some subset of the other complex numbers, then expected magnitude squared of the sum including $a$ and $-a$ is\n$$\n\\begin{gathered}\n\\frac{(x+a)(\\overline{x+a})+x \\bar{x}+x \\bar{x}+(x-a)(\\overline{x-a})}{4} \\\\\nx \\bar{... | United States | HMMT February 2018 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | proof and answer | 1009/2 | |
0i4m | Problem:
For how many integers $a$ ($1 \leq a \leq 200$) is the number $a^{a}$ a square? | [
"Solution:\n\nIf $a$ is even, we have $a^{a} = \\left(a^{a / 2}\\right)^{2}$, which is always a square.\n\nIf $a$ is odd, $a^{a} = \\left(a^{(a-1) / 2}\\right)^{2} \\cdot a$, which is a square precisely when $a$ is a square.\n\nThus we have 100 even values of $a$ and 7 odd square values ($1^{2}, 3^{2}, \\ldots, 13^... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 107 | |
02f1 | In a chess tournament each player plays every other player once. A player gets 1 point for a win, $\frac{1}{2}$ point for a draw and 0 for a loss. Both men and women played in the tournament and each player scored the same total of points against women as against men. Show that the total number of players must be a squ... | [
"Let $m$ and $w$ be the number of men and women in the tournament. Each game assigns a total of 1 point to the players, the total of points assigned to games between two men and two women are $\\binom{m}{2}$ and $\\binom{w}{2}$. Since each player scored the same total of points against men as against women and the ... | Brazil | XIV OBM | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
0cfb | Determine the real numbers $a$, $b$, $c$ and $d$ for which the inequality
$$
|x+a| + |y+b| + |z+c| + |x+y+z+d| \geq |x| + |y| + |z| + |x+y+z|
$$
holds for any real numbers $x$, $y$ and $z$. | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | a = b = c = d = 0 | |
0gf2 | 試求滿足下列條件的最大正整數 $K$:
任給有限多個長度皆為 1 的閉區間 $A_1, A_2, \dots, A_N$ ($N$ 為任意正整數)。若其聯集為 $[0, 2021]$, 則我們必定可以在 $A_1, \dots, A_N$ 中找到 $K$ 個兩兩交集皆為空集合的區間。 | [
"最大的 $K$ 為 1011。\n\n首先證明 $K \\ge 1011$。令 $\\epsilon \\in (0, \\frac{1}{1009})$, 並考慮集合 $\\mathcal{A} = \\{0, 2 + \\epsilon, 4 + 2\\epsilon, \\dots, 2018 + 1009\\epsilon\\}$。顯然 $\\mathcal{A}$ 的每個點都必須存在一個 $A_i$ 包含之, 且這些 $A_i$ 兩兩互斥 (因為 $A_i$ 的長度皆為 1), 故這邊共選到 1010 個 $A_i$。又基於 $\\cup A_i = [0, 2021]$, 我們必須有一個 $A_j$ 是 $[2... | Taiwan | 2021 年台灣數學奧林匹亞考試試題 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1011 | |
09lt | A grandmother has 21 bags filled with candies. Her grandson Bob can ask his grandma for the sum of any two bags and always gets the exact number. Is it possible for Bob to determine exactly how many candies his grandma has with only 12 questions? | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Algebra > Linear Algebra > Vectors"
] | English | proof only | null | |
0cru | Плоскость $\alpha$ пересекает ребра $AB$, $BC$, $CD$ и $DA$ треугольной пирамиды $ABCD$ в точках $K$, $L$, $M$ и $N$ соответственно. Оказалось, что двугранные углы $\angle(KLA, KLM)$, $\angle(LMB, LMN)$, $\angle(MNC, MNK)$ и $\angle(NKD, NKL)$ равны. (Здесь через $\angle(PQR, PQS)$ обозначается двугранный угол при ребр... | [
"Обозначим через $A'$, $B'$, $C'$, $D'$ проекции вершин $A$, $B$, $C$, $D$ соответственно на плоскость $\\alpha$. Пусть $X$ — произвольная точка на продолжении отрезка $KL$ за точку $K$. Тогда имеем $\\angle(KXA, KXN) = \\angle(KLA, KLM)$ и $\\angle(KNA, KNX) = \\angle(NKD, NKL)$. По условию, эти углы равны; значит... | Russia | XL Russian mathematical olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04eo | Determine the product
$$
\left(1 - \frac{\cos 61^{\circ}}{\cos 1^{\circ}}\right) \left(1 - \frac{\cos 62^{\circ}}{\cos 2^{\circ}}\right) \dots \left(1 - \frac{\cos 119^{\circ}}{\cos 59^{\circ}}\right).
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 1 | |
04s5 | We are given an acute-angled triangle $ABC$. Denote by $k$ the circle with diameter $AB$. A circle touching the bisector of the angle $BAC$ at the point $A$ and passing through the point $C$ meets the circle $k$ in a point $P$, $P \neq A$. Similarly, a circle touching the bisector of the angle $ABC$ at the point $B$ an... | [
"Besides the Thales' circle $k$, denote as $l_A = AP$ and $l_B = BP$ the other two circles under consideration. Let us deal, for example, with the circle $l_B$ drawn in Fig. 4. In what follows the interior angles of $\\triangle ABC$ are denoted as usual.\n\n\nFig. 4\n\nLet us explain that i... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
... | English | proof only | null | |
04g1 | Students decided to play a game with $960$ tokens. First they distributed all of the tokens so that each student had the same number of tokens. Once they did that, their teacher arrived wanting to join the game. Each student gave him $4$ of his tokens, so that everyone had the same number of tokens and the game could s... | [
"Let $n$ be the number of students.\n\nInitially, the $960$ tokens are distributed equally, so each student gets $\\dfrac{960}{n}$ tokens.\n\nWhen the teacher arrives, each student gives him $4$ tokens, so the teacher receives $4n$ tokens in total. Now there are $n+1$ people (the $n$ students and the teacher), and ... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 15 | |
00sz | Let $\triangle ABC$ be a triangle. On the sides $BC$, $CA$, $AB$ of the triangle, construct outwardly three squares with centres $O_a$, $O_b$, $O_c$ respectively. Let $\omega$ be the circumcircle of $\triangle O_aO_bO_c$.
Given that $A$ lies on $\omega$, prove that the centre of $\omega$ lies on the perimeter of $\tri... | [
"Let the vertices of the squares be $AC_1C_2B$, $BA_1A_2C$, $CB_1B_2A$.\n\n**Lemma:** $BB_2 = CC_1$ and $BB_2 \\perp CC_1$.\n\n**Proof:** Notice that by rotating $\\triangle AC_1C$ by $90^\\circ$ we get $\\triangle ABB_2$ proving the lemma.\n\n**Claim:** $AO_a \\perp O_bO_c$\n\n**Proof:** Let $M$ be the midpoint of... | Balkan Mathematical Olympiad | BMO Short List | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chas... | English | proof only | null | |
02im | Problem:
Seja $v$ a soma das áreas das regiões pertencentes unicamente aos três discos pequenos (em cinza claro), e seja $w$ a área da região interior unicamente ao maior disco (em cinza escuro). Os diâmetros dos círculos são 6, 4, 4 e 2. Qual das igualdades abaixo é verdadeira?
(A) $3 v=\p... | [
"Solution:\n\n(C) Os raios dos três discos menores são $1$, $2$ e $2$; e do disco maior $3$.\n\nDenotemos por $b$ a área em branco, temos:\n$$\n\\begin{aligned}\n& v = 9\\pi - b \\\\\n& \\text{de raio } 3 \\\\\n& w = 9\\pi - b\n\\end{aligned}\n$$\nLogo, $v = w$.\n\nA área do disco de raio $r$ é $\\pi r^{2}$."
] | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | MCQ | C | |
0da9 | Find all positive integers $k$ such that there exists some permutation of $(1,2, \ldots, 1000)$ namely $\left(a_{1}, a_{2}, \ldots, a_{1000}\right)$ and satisfy $\left|a_{i}-i\right|= k$ for all $i=1,1000$. | [
"Note that $k=1$ is an answer since we can choose a permutation with $a_{2i-1}=2i$, $a_{2i}=2i-1$.\n\nNow we take $k>1$, denote $S_{0}, S_{1}, S_{2}, \\ldots, S_{k-1}$ as the subsets of $\\{1,2,3, \\ldots, 1000\\}$ and all elements of $S_{i}$ are congruent to $i$ modulo $k$. We have:\n- The number of elements of $S... | Saudi Arabia | Team selection tests for IMO 2018 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | all positive divisors of 500 | |
01wm | Let $O$ be the circumcenter and $H$ be the orthocenter of an acute-angled triangle $ABC$. Point $T$ is the midpoint of the segment $AO$. The perpendicular bisector of $AO$ intersects the line $BC$ at point $S$.
Prove that the circumcircle of the triangle $AST$ bisects the segment $OH$. | [
"Denote the foot of the altitude from $A$ by $E$ and the midpoint of $OH$ by $L$. Since $\\angle ATS = \\angle AES = 90^\\circ$, the quadrilateral $ATLE$ is cyclic. Thus it is enough to prove that the quadrilateral $ALES$ is cyclic. Let $K$ be the reflection of $H$ about $E$. It is well known that $K$ lies on the c... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0knc | A right rectangular prism whose surface area and volume are numerically equal has edge lengths $\log_2 x$, $\log_3 x$, and $\log_4 x$. What is $x$?
(A) $2\sqrt{6}$ (B) $6\sqrt{6}$ (C) 24 (D) 48 (E) 576 | [
"Solution:\nDenote the edge lengths by $a = \\log_2 x$, $b = \\log_3 x$, and $c = \\log_4 x$. The condition that the surface area numerically equals the volume is equivalent to $2(ab + ac + bc) = abc$. Dividing both sides by $2abc$ gives $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = \\frac{1}{2}$. By the Change of ... | United States | AMC 12 A | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area"
] | null | MCQ | E | |
012y | Problem:
Let $P$ be the intersection point of the diagonals $AC$ and $BD$ in a cyclic quadrilateral. A circle through $P$ touches the side $CD$ at the midpoint $M$ of this side and intersects the segments $BD$ and $AC$ at the points $Q$ and $R$, respectively. Let $S$ be a point on the segment $BD$ such that $BS = DQ$.... | [
"Solution:\n\nWith reference to the figure below we have $CR \\cdot CP = DQ \\cdot DP = CM^2 = DM^2$, which is equivalent to $RC = \\frac{DQ \\cdot DP}{CP}$. We also have $\\frac{AT}{BS} = \\frac{AP}{BP} = \\frac{AT}{DQ}$, so $AT = \\frac{AP \\cdot DQ}{BP}$. Since $ABCD$ is cyclic the result now comes from the fact... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0fh2 | Problem:
Demostrar que el número $1989$ y todas sus potencias enteras $1989^{n}$ se pueden escribir como suma de dos cuadrados de enteros positivos, y como mínimo, de dos formas diferentes. | [
"Solution:\nComprobamos que\n$$\n1989 = 9 \\cdot 221 = 9\\left(10^{2} + 11^{2}\\right) = 9\\left(14^{2} + 5^{2}\\right)\n$$\nPor lo tanto\n$$\n1989^{2n+1} = 1989^{2n} \\cdot 1989 = \\left((1989)^{n}\\right)^{2} \\cdot 9\\left(10^{2} + 11^{2}\\right) = \\left((1989)^{n}\\right)^{2} \\cdot 9\\left(14^{2} + 5^{2}\\rig... | Spain | OME 25 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0ebr | Problem:
Naj bo $a_{1}, a_{2}, a_{3}, \ldots$ zaporedje neničelnih realnih števil, za katerega velja $a_{n}^{2} = -a_{n+1} a_{n-1}$ za vsa naravna števila $n$, $n \geq 2$. Dokaži, da je zaporedje $a_{2}, a_{4}, a_{6}, \ldots$ geometrijsko. | [
"Solution:\n\n1. način. Za vsak $n \\geq 2$ lahko dano rekurzivno formulo preoblikujemo v $\\frac{a_{n+1}}{a_{n}} = -\\frac{a_{n}}{a_{n-1}}$. S pomočjo te formule izračunamo\n$$\n\\frac{a_{2n+2}}{a_{2n}} = \\frac{a_{2n+2}}{a_{2n+1}} \\cdot \\frac{a_{2n+1}}{a_{2n}} = \\left(-\\frac{a_{2n+1}}{a_{2n}}\\right) \\cdot \... | Slovenia | 59. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
02di | The angles of the triangle $ABC$ satisfy $\frac{\angle A}{\angle C} = \frac{\angle B}{\angle A} = 2$. The incenter is $O$. $K$, $L$ are the excenters of the excircles opposite $B$ and $A$ respectively. Show that triangles $ABC$ and $OKL$ are similar. | [
"$\\angle AOC = 180^\\circ - \\angle A/2 - \\angle C/2$. But $\\angle KAO = \\angle KCO = 90^\\circ$, so $\\angle AKC = \\angle A/2 + \\angle C/2$. So considering triangle $AKL$, $\\angle ALK = 90^\\circ - \\angle A/2 - \\angle C/2 = \\angle B/2$. Similarly, $\\angle BLC = \\angle B/2 + \\angle C/2$, so $\\angle BK... | Brazil | IV OBM | [
"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 | |
0gaa | 令 $\mathbb{Z}^+$ 代表所有正整數所成的集合。試求所有滿足下列條件的映成函數 $f: \mathbb{Z}^+ \times \mathbb{Z}^+ \to \mathbb{Z}^+$：對任意 $a, b, c \in \mathbb{Z}^+$，下列三條件均成立：
(i) $f(a, b) \le a + b$
(ii) $f(a, f(b, c)) = f(f(a, b), c)$
(iii) $\binom{f(a, b)}{a}$ 及 $\binom{f(a, b)}{b}$ 都是奇數 (其中 $\binom{n}{k}$ 為二項式係數 $C_k^n$)
Let $\mathbb{Z}^+$ denote ... | [
"設正整數 $n$ 的二進位表示為 $n = \\sum_{i=1}^{k} 2^{r_i}$；透過此二進位展開，我們可以得到 $\\mathbb{Z}^+$ 與 $\\mathbb{Z}^+ \\cup \\{0\\}$ 的有限非空子集之間的一對一對應：$S_n := \\{r_1, r_2, \\dots, r_k\\}$。於是定義 $f(a, b)$ 為滿足 $S_{f(a,b)} = S_a \\cup S_b$ 的唯一函數。\n\n由於 $f(a, a) = a$，所以 $f$ 為滿射函數。又易知 $f(a, b) \\le a + b$，等號成立的充要條件為 $S_a \\cap S_b = \\emptyset... | Taiwan | 二〇一六數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | The unique solution is the bitwise OR: for each pair of positive integers a and b, f(a, b) is the unique integer whose binary expansion has ones exactly in the positions where at least one of a or b has a one. Equivalently, if S_n denotes the set of positions of ones in the binary expansion of n, then S_{f(a,b)} = S_a ... | |
09sn | Problem:
Bepaal alle paren $(a, b)$ van gehele getallen met de volgende eigenschap: er is een gehele $d \geq 2$ zodat $a^{n}+b^{n}+1$ deelbaar is door $d$ voor alle positieve gehele getallen $n$. | [
"Solution:\n\nBekijk een paar $(a, b)$ dat voldoet, met bijbehorende $d$. Zij $p$ een priemdeler van $d$ (die bestaat omdat $d \\geq 2$). Omdat $d \\mid a^{n}+b^{n}+1$ voor alle $n$, geldt ook $p \\mid a^{n}+b^{n}+1$ voor alle $n$. Bekijk nu $n=p-1$. Dan geldt $a^{n} \\equiv 0 \\bmod p$ als $p \\mid a$ en $a^{n} \\... | Netherlands | IMO-selectietoets II | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All integer pairs (a, b) such that either a ≡ b ≡ 1 mod 3, or exactly one of a and b is even (i.e., a ≡ 0, b ≡ 1 mod 2 or a ≡ 1, b ≡ 0 mod 2). | |
0gm9 | Each of the $2001$ boys chooses an integer and makes a list by writing the names of some of the boys other than himself. For each boy, one computes the sum of the numbers which the boys whose names are included in this boy's list chose, as well as the sum of the numbers which the boys who included the name of this boy ... | [] | Turkey | TEAM SELECTION EXAMINATION FOR THE 42nd INTERNATIONAL MATH- EMATICAL OLYMPIAD. TURKEY. | [
"Discrete Mathematics > Other",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | No, they cannot all be positive. | |
00i3 | Line $\ell$ intersects sides $B C$ and $A D$ of cyclic quadrilateral $A B C D$ in its interior points $R$ and $S$ respectively, and intersects ray $D C$ beyond point $C$ at $Q$, and ray $B A$ beyond point $A$ at $P$. Circumcircles of the triangles $Q C R$ and $Q D S$ intersect at $N \neq Q$, while circumcircles of the ... | [
"We start with the following lemma.\nLemma 1. Points $M, N, P, Q$ are concyclic.\nPoint $M$ is the Miquel point of lines $A P=A B, P S=\\ell, A S=A D$, and $B R=B C$, and point $N$ is the Miquel point of lines $C Q=C D, R C=B C, Q R=\\ell$, and $D S=A D$. Both points $M$ and $N$ are on the circumcircle of the trian... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates... | English | proof only | null | |
065d | Let $ABCD$ be a convex quadrilateral inscribed in a circle $(O, R)$. With centers the vertices of the quadrilateral and radius $R$ we draw circles $C_A(A, R)$, $C_B(B, R)$, $C_C(C, R)$, $C_D(D, R)$. Circles $C_A$ and $C_B$ meet at $K$, circles $C_B$ and $C_C$ meet at $L$, circles $C_C$ and $C_D$ meet at $M$ and the cir... | [
"The line segment $AB$ connects the centers of the circles $C_A$ and $C_B$, and therefore it is the perpendicular bisector of the common chord $OK$. Since the circles $C_A$ and $C_B$ have the same radius, the quadrilateral $AOBK$ is rhombus. Thus point $K_1$ is the middle of $AB$.\n\n\n\nSi... | Greece | SELECTION EXAMINATION | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0b3z | Problem:
How many ending zeroes does the decimal expansion of $2022!$ have?
(a) 404
(b) 484
(c) 500
(d) 503 | [] | Philippines | 24th Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | d | |
0g5h | 對於任意正整數 $m$, 我們定義 $\phi(m)$ 為小於 $m$ 且與 $m$ 互質的正整數個數。
試問: 是否存在無窮正整數數列 $a_1, a_2, \dots, a_n, \dots$, 滿足:
$$
(i) \ a_1 = (2011)! = 1 \times 2 \times \dots \times 2011.
$$
$$ (ii) \ 對每個正整數 i, 有 $a_i = \phi(a_{i+1})$. $$ | [
"不存在這樣的數列。我們用反證法，假設存在這樣的無窮數列，我們將每個 $a_i$ 表示成 $a_i = 2^{r_i}b_i$，其中 $b_i$ 是奇數。則有由 $\\phi$-函數的性質有 $a_i = \\phi(a_{i+1}) = 2^{r_{i+1}-1}\\phi(b_{i+1})$。\n\n注意到 $b_{i+1}$ 不可能是 $1$, 否則我們會得到 $b_i = 1$, $b_{i-1} = 1, \\dots$, 最後得到 $b_1 = 1$ 與 $a_1 = (2011)!$ 顯然矛盾。此時我們有 $\\phi(b_{i+1})$ 是偶數。比較 $2^{r_{i+1}-1}\\phi(b_{i+1}) ... | Taiwan | 二〇一一數學奧林匹亞競賽第三階段選訓營 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0boh | Eight numbers, all zero, are written on a blackboard. A *move* consists in randomly selecting four of them, $a$, $b$, $c$, $d$, and replacing them by $a+3$, $b+3$, $c+2$ and $d+1$, respectively.
a) What is the smallest number of moves after which on the blackboard can appear eight consecutive numbers?
b) Is there a s... | [
"a) After each move, the sum of the numbers increases by $9$. Since the sum of the smallest $8$ consecutive numbers is $0 + 1 + 2 + \\dots + 7 = 28$, but the table below uses $1$ to $8$ (sum $36$), so let's check the table for correctness. The table shows the process:\n\n| Start | 0 | 0 | 0 | 0 | 0 | 0 | 0 ... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) 4; b) No | |
0hqy | Problem:
Let $x$, $y$, and $z$ be real numbers such that $x y z = 1$. Prove that
$$
x^{2} + y^{2} + z^{2} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z}
$$ | [
"Solution:\n\nReplacing the $1$ in the numerators of the fractions on the right by $x y z$, it suffices to prove that\n$$\nx^{2} + y^{2} + z^{2} \\geq y z + z x + x y\n$$\nwhich is true because\n$$\nx^{2} + y^{2} + z^{2} - y z - z x - x y = \\frac{(x - y)^{2} + (y - z)^{2} + (z - x)^{2}}{2} \\geq 0.\n$$"
] | United States | Berkeley Math Circle Monthly Contest 3 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
02ne | Problem:
Quinze minutos a mais - Dois carros partem, ao mesmo tempo, de uma cidade $A$ em direção a uma cidade $B$. Um deles viaja à velocidade constante de $60~\mathrm{km}/\mathrm{h}$ e o outro à velocidade constante de $70~\mathrm{km}/\mathrm{h}$. Se o carro mais rápido faz a viagem de $A$ a $B$ em $15$ minutos a me... | [
"Solution:\n\nSeja $d$ a distância entre as cidades $A$ e $B$.\n\nO tempo gasto pelo carro mais lento ($60~\\mathrm{km}/\\mathrm{h}$) é $t_1 = \\dfrac{d}{60}$ (em horas).\n\nO tempo gasto pelo carro mais rápido ($70~\\mathrm{km}/\\mathrm{h}$) é $t_2 = \\dfrac{d}{70}$ (em horas).\n\nSabemos que o carro mais rápido g... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 105 km | |
008p | Let $T$ be a non-isosceles triangle and $n \ge 4$ be an integer. Show that $T$ can be divided in $n$ triangles and one interior bisector can be traced in each one of them so that those $n$ bisectors are parallel. | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"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"
] | English | proof only | null | |
0ak5 | Find all positive integers $n$ such that $9^n - 7$ can be represented as a product of at least two consecutive positive integers. | [
"The product of three consecutive positive integers is divisible by $3$, and $9^n - 7 \\equiv 2 \\pmod{3}$, so we can conclude that $9^n - 7$ cannot be written as a product of three or more consecutive positive integers.\n\nLet $9^n - 7 = m(m+1)$ for some positive integer $m$. The last equation is equivalent with t... | North Macedonia | Macedonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | n = 1 | |
0i2e | Problem:
Calculate the sum of the coefficients of $P(x)$ if $(20 x^{27} + 2 x^{2} + 1) P(x) = 2001 x^{2001}$. | [
"Solution:\nThe sum of coefficients of $f(x)$ is the value of $f(1)$ for any polynomial $f$. Plugging in $1$ to the above equation, $P(1) = \\frac{2001}{23} = 87$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 87 | |
03ve | Let $n > 1$ be a given integer and $A$ be an infinite set of positive integers satisfying: for any prime $p \nmid n$, there exist infinitely many elements of $A$ not divisible by $p$. Prove that for any integer $m > 1$, $(m, n) = 1$, there exists a finite subset of $A$ whose sum of elements, say $S$, satisfies $S \equi... | [
"Suppose a prime $p$ satisfies $p^a \\mid m$. Then from the given conditions, there exists an infinite subset $A_1$ of $A$ such that $p$ is coprime to every element in $A_1$.\nBy the pigeonhole principle, there is an infinite subset $A_2$ of $A_1$, such that $x \\equiv a \\pmod{mn}$, for each element $x \\in A_2$, ... | China | China Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0dff | Does there exist a polynomial $P(x)$ of second degree with integer coefficients such that the leading coefficient is not divisible by $2022$ and all numbers $P(1), P(2), \dots, P(2022)$ give different residues mod $2022$? | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof and answer | Yes. For example, P(x) = 1011x^2 + 2x works. | |
08cu | Problem:
Sia $N$ un intero maggiore di $1$. Chiamiamo $x$ il più piccolo intero positivo con la seguente proprietà: esiste un intero positivo $y$ strettamente minore di $x-1$ tale che $x$ divide $N+y$. Dimostrare che $x$ è il doppio di un numero primo o una potenza di un numero primo.
Nota: si ricorda che $x$ è una p... | [
"Solution:\n\nSi ricorda che la scrittura $a \\mid b$ significa che $a$ divide $b$. Diciamo che una coppia di interi positivi $(x, y)$ è bella se valgono $x \\mid N+y$ e $0<y<x-1$.\n\nLa prima osservazione è che, dato un qualunque intero positivo $x'$, fra $x'$ interi consecutivi c'è un multiplo di $x'$, dunque è s... | Italy | XXXIV Olimpiade Italiana di Matematica | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0k96 | Problem:
Wendy eats sushi for lunch. She wants to eat six pieces of sushi arranged in a $2 \times 3$ rectangular grid, but sushi is sticky, and Wendy can only eat a piece if it is adjacent to (not counting diagonally) at most two other pieces. In how many orders can Wendy eat the six pieces of sushi, assuming that the... | [
"Solution:\n\nCall the sushi pieces $A, B, C$ in the top row and $D, E, F$ in the bottom row of the grid. Note that Wendy must first eat either $A, C, D$, or $F$. Due to the symmetry of the grid, all of these choices are equivalent. Without loss of generality, suppose Wendy eats piece $A$.\n\nNow, note that Wendy c... | United States | HMMT November 2019 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 360 | |
03gw | Problem:
A quadrilateral has one vertex on each side of a square of side-length $1$. Show that the lengths $a, b, c$ and $d$ of the sides of the quadrilateral satisfy the inequalities
$$
2 \leq a^{2}+b^{2}+c^{2}+d^{2} \leq 4
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
045d | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that, for any real numbers $x, y$, the following two multisets are equal
$$
\{f(xf(y) + 1), f(yf(x) - 1)\} = \{xf(f(y)) - 1, yf(f(x)) + 1\}.
$$
*Remark: $\{a, b\} = \{c, d\}$ are equal as multisets if $a = c$ and $b = d$, or $a = d$ and $b = c$.* | [
"**Solution:** All functions satisfying (*) are $f(x) = x$ and $f(x) = -x$. It is easy to verify that these two functions satisfy (*).\nIn what follows, all sets are multisets. Take $x = y = 0$ in (*) gives $\\{f(1), f(-1)\\} = \\{1, -1\\}$. First consider the case when $f(1) = 1$. We want to show that $f(x) = x$ f... | China | 2022 China Team Selection Test | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | f(x) = x and f(x) = -x | |
03nt | Problem:
Let $n \geq 2$ be an integer. Initially, the number $1$ is written $n$ times on a board. Every minute, Vishal picks two numbers written on the board, say $a$ and $b$, erases them, and writes either $a+b$ or $\min \{a^{2}, b^{2}\}$. After $n-1$ minutes there is one number left on the board. Let the largest pos... | [
"Solution:\n\nClearly $f(n)$ is a strictly increasing function, as we can form $f(n-1)$ with $n-1$ ones, and add the final one. However, we can do better; assume Vishal generates $f(n)$ on the board. After $n-2$ minutes, there are two numbers left, say they were formed by $x$ ones and $y$ ones, where $x+y=n$. Clear... | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof only | null | |
0jwp | Problem:
Let $ABC$ be a triangle in the plane with $AB = 13$, $BC = 14$, $AC = 15$. Let $M_n$ denote the smallest possible value of $\left(AP^n + BP^n + CP^n\right)^{\frac{1}{n}}$ over all points $P$ in the plane. Find $\lim_{n \rightarrow \infty} M_n$. | [
"Solution:\n\nLet $R$ denote the circumradius of triangle $ABC$. As $ABC$ is an acute triangle, it isn't hard to check that for any point $P$, we have either $AP \\geq R$, $BP \\geq R$, or $CP \\geq R$. Also, note that if we choose $P = O$ (the circumcenter) then $\\left(AP^n + BP^n + CP^n\\right) = 3 \\cdot R^n$. ... | United States | February 2017 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 65/8 | |
09hy | Let $d_1, d_2, \dots, d_n$ be nonnegative real numbers satisfying $1 \ge d_1 \ge \dots \ge d_n \ge 0$. Prove
$$
\frac{(1 + d_1 + d_2 + \dots + d_n)^2}{n + 1} \ge 2 \cdot \frac{d_1^2 + 2d_2^2 + \dots + n d_n^2}{n}.
$$ | [
"Setting $d_0 = 1$, we have $j d_j \\le d_0 + d_1 + \\dots + d_{j-1}$ for any $1 \\le j \\le n$. Hence\n$$\n\\sum_{j=1}^{n} j d_j^2 \\le \\sum_{j=1}^{n} \\sum_{i=0}^{j-1} d_i d_j = \\sum_{0 \\le i < j \\le n} d_i d_j \\le \\frac{n}{2(n+1)} \\left( \\sum_{k=0}^{n} d_k \\right)^2,\n$$\nwhere the last part follows fro... | Mongolia | Round 3 | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0cbp | The triangle $ABC$ is isosceles, with $AB = AC$ and $\angle ABC = 72^\circ$. The point $D$ is taken on the line $BC$, so that $C$ is on the segment $BD$ and $CD = AB$.
a) Prove that $AC$ is the bisector of the angle $\angle BAD$.
b) The point $E$ is taken on the parallel to $AB$ through $D$, on the same side of $BD$ ... | [
"a) In the triangle $ABC$, $\\angle BAC = 36^\\circ$. From the isosceles triangle $ACD$, with $\\angle ACD = 180^\\circ - 36^\\circ = 144^\\circ$, it follows that $\\angle CAD = \\angle ADC = 36^\\circ$, hence $AC$ is the bisector of the angle $\\angle BAD$.\n\nb) From $\\angle ABD = \\angle BAD = 72^\\circ$ it fol... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
04sg | Find all the pairs of prime numbers $p$, $q$ such that the value of the expression $p^2 + 5pq + 4q^2$ is a perfect square. | [] | Czech Republic | Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | [(13, 3), (5, 11), (7, 5)] | |
0gwn | Let's examine all the possible sequences $a_1, a_2, ..., a_{2008}$ of non-negative integers where $a_1 \le a_2 \le ... \le a_{2008}$ and $a_k \le (k-1)$ for all values of $k$ from $1$ to $2008$. Prove that the number of such sequences exceeds
a) $2^{2007}$;
b) $2^{2008}$. | [
"a.\nTo begin, we form $2^{2007}$ different sequences satisfying the given conditions. For all the sequences let $a_1 = 0$. Let each successive member of the sequence be \"by 0 greater\" or \"by 1 greater\" than the preceding one. Thus we obtain $2^{2007}$ different sequences. It's easy to show that $a_k \\le k-1$ ... | Ukraine | Ukrajina 2008 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | English | proof only | null | |
06x0 | Let $n$ be a positive integer. We start with $n$ piles of pebbles, each initially containing a single pebble. One can perform moves of the following form: choose two piles, take an equal number of pebbles from each pile and form a new pile out of these pebbles. For each positive integer $n$, find the smallest number of... | [
"We can combine two piles of $2^{k-1}$ pebbles to make one pile of $2^{k}$ pebbles. In particular, given $2^{k}$ piles of one pebble, we may combine them as follows:\n$$\n\\begin{array}{lll}\n2^{k} \\text{ piles of } 1 \\text{ pebble } & \\rightarrow & 2^{k-1} \\text{ piles of } 2 \\text{ pebbles } \\\\\n2^{k-1} \\... | IMO | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 1 if n is a power of 2; otherwise 2 | |
02fs | Let $F_n$ be the Fibonacci sequence $F_1 = F_2 = 1$, $F_{n+2} = F_{n+1} + F_n$. Put $V_n = \sqrt{F_n^2 + F_{n+2}^2}$. Show that $V_n$, $V_{n+1}$, $V_{n+2}$ are the sides of a triangle of area $1/2$. | [
"Let $A = (F_{n+4}, 0)$, $B = (0, F_{n+2})$ and $C = (F_{n+3}, F_n)$. Notice that\n$$\nAB = \\sqrt{F_{n+4}^2 + F_{n+2}^2} = V_{n+2}, \\quad BC = \\sqrt{F_{n+3}^2 + (F_{n+2} - F_n)^2} = \\sqrt{F_{n+3}^2 + F_{n+1}^2} = V_{n+1}\n$$\nand $CA = \\sqrt{(F_{n+4} - F_{n+3})^2 + F_n^2} = \\sqrt{F_{n+2}^2 + F_n^2} = V_n$. It... | Brazil | XIX OBM | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 1/2 |
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