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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
055r | Prove that every positive real number satisfies
$$
(x+1)(x+2)(x+5) \ge 36x.
$$ | [
"The given inequality is equivalent to $x^3 + 8x^2 - 19x + 10 \\ge 0$. Note that $x^3 + 8x^2 - 19x + 10 = (x-1)^2(x+10)$. As $(x-1)^2 \\ge 0$ and $x+10 > 0$ for positive $x$, this inequality holds indeed.",
"Let $f(x) = x^3 + 8x^2 - 19x + 10$. The given inequality is equivalent to $f(x) \\ge 0$. Note that $f'(x) ... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0eq6 | Let $A$, $B$, and $C$ be distinct points on a straight line with $AB = AC = 1$.
Square $ABDE$ and equilateral triangle $ACF$ are drawn on the same side of line $BC$.
Lines $EC$ and $BF$ cut in $G$.
What is the size of $\angle EGB$?
 | [
"We will compute the angle sizes of quadrilateral $DEGB$ and use the fact that the sum of the interior angles of a quadrilateral is equal to $360^\\circ$.\n\nSince $AFC$ is an equilateral triangle, $\\angle FAC = 60^\\circ$ and so $\\angle EAF = 90^\\circ - 60^\\circ = 30^\\circ$, which yields $\\angle BAE = 90^\\c... | South Africa | South African Mathematics Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 75° | |
0btw | Given a positive integer $n$, determine all functions $f$ from the first $n$ positive integers to the positive integers, satisfying the following two conditions:
$$
(1) \sum_{k=1}^{n} f(k) = 2n;
$$
$$
(2) \sum_{k \in K} f(k) = n \text{ for no subset } K \text{ of the first } n \text{ positive integers.}
$$ | [
"If $n$ is odd, the required functions are the constant function $f_0 \\equiv 2$ along with the $n$ functions $f_i: \\{1, \\dots, n\\} \\to \\mathbb{N}^*$,\n$$\nf_i(j) = \\begin{cases} n+1, & \\text{if } j = i, \\\\ 1, & \\text{if } j \\neq i, \\end{cases} \\quad i = 1, \\dots, n;\n$$\n\nnotice that $f_0 = f_1$ if ... | Romania | 67th NMO Selection Tests for BMO and IMO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | For odd n: either f(k) = 2 for all k, or there exists i such that f(i) = n + 1 and f(j) = 1 for all j ≠ i. For even n: the only solutions are those with one index i having f(i) = n + 1 and all other values equal to 1. | |
07lq | a) For each integer $a$ not divisible by $7$, let $N(a)$ denote the least among all positive integers $n$ such that $49$ divides $a^n - 1$. For each positive integer $N$, let $A_N := \{a \in \mathbb{Z} : 0 < a < 49 \text{ and } N(a) = N\}$. Show that
$$
\begin{align*}
|A_1| &= |A_2| = 1, & |A_3| &= |A_6| = 2, \\
|A_7| ... | [
"a) Unless otherwise specified, everywhere in part a), $\\equiv$ will denote congruence modulo $49$. Consider the six cases $a = 7k \\pm 1$, $7k \\pm 2$ and $7k \\pm 3$.\n\n1. If $a = 7k + 1$, with $k \\in \\{0, 1, ..., 6\\}$, then $a^n = 7kn + 1$, so $a^n \\equiv 1 \\pmod{7}$\n$k = 0$ or $n$ is a multiple of $7$. ... | Ireland | Irska | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | English | proof and answer | 840 | |
0iyt | Problem:
How many functions $f:\{1,2,3,4,5\} \rightarrow \{1,2,3,4,5\}$ satisfy $f(f(x))=f(x)$ for all $x \in \{1,2,3,4,5\}$? | [
"Solution:\nA fixed point of a function $f$ is an element $a$ such that $f(a)=a$. The condition is equivalent to the property that $f$ maps every number to a fixed point. Counting by the number of fixed points of $f$, the total number of such functions is\n\n$$\n\\begin{aligned}\n\\sum_{k=1}^{5}\\binom{5}{k} k^{5-k... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics"
] | null | proof and answer | 196 | |
0jsz | Problem:
For positive integers $n$, let $c_{n}$ be the smallest positive integer for which $n^{c_{n}}-1$ is divisible by 210, if such a positive integer exists, and $c_{n}=0$ otherwise. What is $c_{1}+c_{2}+\cdots+c_{210}$ ? | [
"Solution:\n\nIn order for $c_{n} \\neq 0$, we must have $\\operatorname{gcd}(n, 210)=1$, so we need only consider such $n$. The number $n^{c_{n}}-1$ is divisible by 210 iff it is divisible by each of $2,3,5$, and $7$, and we can consider the order of $n$ modulo each modulus separately; $c_{n}$ will simply be the L... | United States | HMMT February 2016 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | 329 | |
05rg | Problem:
Soit $a$ et $b$ deux réels tels que $a b \geqslant a^{3}+b^{3}$.
Démontrer que $a+b \leqslant 1$. | [
"Solution:\n\nPuisque $(a+b)^{2} = (a-b)^{2} + 4 a b \\geqslant 4 a b$, on en déduit que\n$$\n(a+b)^{3} = a^{3} + b^{3} + 3(a+b) a b \\leqslant (3a + 3b + 1) a b \\leqslant \\frac{(3a + 3b + 1)(a+b)^{2}}{4}\n$$\nSi $a+b > 1$, on peut diviser notre inégalité par $(a+b)^{2}/4$, ce qui signifie que $4(a+b) \\leqslant ... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0i8n | Problem:
A palindrome is a positive integer that reads the same backwards as forwards, such as $82328$. What is the smallest $5$-digit palindrome that is a multiple of $99$? | [
"Solution:\nWrite the number as $X Y Z Y X$. This is the same as $10000 X + 1000 Y + 100 Z + 10 Y + X = 99(101 X + 10 Y + Z) + 20 Y + 2 X + Z$. We thus want $20 Y + 2 X + Z$ to be a multiple of $99$, with $X$ as small as possible. This expression cannot be larger than $20 \\cdot 9 + 2 \\cdot 9 + 9 = 207$, and it is... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 54945 | |
0kyq | Problem:
Compute the largest prime factor of $3^{12}+3^{9}+3^{5}+1$. | [
"Solution:\n\nObserve\n\n$$\n\\left(3^{4}+1\\right)^{3}=3^{12}+3 \\cdot 3^{8}+3 \\cdot 3^{4}+1=3^{12}+3^{9}+3^{5}+1\n$$\n\nso the answer is the largest prime factor of $3^{4}+1=82$, which is 41."
] | United States | HMMT November 2024 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 41 | |
0cxt | Let $a > 0$ be a real number and let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function satisfying
$$
f\left(x_{1}\right) + f\left(x_{2}\right) \geq a f\left(x_{1} + x_{2}\right), \forall x_{1}, x_{2} \in \mathbb{R} .
$$
Prove that
$$
f\left(x_{1}\right) + f\left(x_{2}\right) + f\left(x_{3}\right) \geq \frac{3 a^{2}}... | [
"Using the inequality in the hypothesis we get successively:\n$$\n\\begin{aligned}\n& f\\left(x_{1}\\right) + f\\left(x_{2}\\right) + a f\\left(x_{3}\\right) \\geq a^{2} f\\left(x_{1} + x_{2} + x_{3}\\right) \\\\\n& f\\left(x_{1}\\right) + a f\\left(x_{2}\\right) + f\\left(x_{3}\\right) \\geq a^{2} f\\left(x_{1} + ... | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
005w | Sean $m$ y $n$ enteros tales que el polinomio $P(x) = x^3 + m x + n$ tiene la siguiente propiedad: si $x$ e $y$ son enteros y $107$ divide a $P(x) - P(y)$, entonces $107$ divide a $x - y$. Demuestre que $107$ divide a $m$. | [] | Argentina | XXIII Olimpíada Iberoamericana de Matemática | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | Spanish | proof only | null | |
0ehx | Problem:
Kateri od spodaj navedenih izrazov je enakovreden izrazu $(y-2)^{3}-y(y+5)(y-5)-29+6 y^{2}$?
(A) $12 y^{2}+37 y-37$
(B) $-13 y+37$
(C) $37 y-21$
(D) $37(y-1)$
(E) $12 y^{2}-37 y-37$ | [
"Solution:\n$(y-2)^{3}-y(y+5)(y-5)-29+6 y^{2}=y^{3}-6 y^{2}+12 y-8-y^{3}+25 y-29+6 y^{2}=37 y-37=37(y-1)$."
] | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | D | |
0h4b | Let $[x]$ denote the integer part of $x$ (i.e. the largest integer not exceeding $x$), $\{x\} = x - [x]$. Solve the equation $\{x\}^2 + 2\{x\} = 3x^2$. | [
"Оскільки $0 \\le \\{x\\} < 1$, то $x^2 < 1$, тобто $-1 < x < 1$.\n\nЯкщо $x \\in [0;1)$, то $x = \\{x\\}$, і знаходимо $x = 0$.\n\nДля $x \\in (-1;0)$ позначимо $u = \\{x\\}$, $0 < u < 1$, $[x] = -1$, $x = -1 + u$. Тоді з рівняння $2u^2 - 8u + 3 = 0$, з урахуванням нерівності $0 < u < 1$, одержуємо $u = 2 - \\frac... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | {0, 1 - sqrt(10)/2} | |
0i7d | For a pair of integers $a$ and $b$, with $0 < a < b < 1000$, the set $S \subseteq \{1, 2, \dots, 2003\}$ is called a *skipping set* for $(a, b)$ if for any pair of elements $s_1, s_2 \in S$, $|s_1 - s_2| \notin \{a, b\}$. Let $f(a, b)$ be the maximum size of a skipping set for $(a, b)$. Determine the maximum and minimu... | [
"The maximum and minimum values of $f$ are $1334$ and $668$, respectively.\n\na. First, we will show that the maximum value of $f$ is $1334$. The set $S = \\{1, 2, \\dots, 667\\} \\cup \\{1336, 1337, \\dots, 2002\\}$ is a skipping set for $(a, b) = (667, 668)$, so $f(667, 668) \\ge 1334$.\n\nNow we prove that for a... | United States | USA IMO 2003 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Maximum f is 1334; minimum f is 668. | |
01hx | A convex $n$-gon is triangulated with triangulation $P$ of $n-3$ non-intersecting diagonals. An *independent set* of vertices is a set of vertices of $P$ where no two vertices are connected by a side or by a diagonal. Let $i(P)$ be the number of ways to choose an independent set of vertices with respect to the triangul... | [
"1) Let $z_n$ be the number of ways to choose an independent set of vertices for a zig-zag triangulation of an $n$-gon. It is known that each triangulation has at least two vertices which are not the endpoints of any diagonal. Let $A$ be one of these vertices for the zig-zag triangulation (see figure above).\nIf th... | Baltic Way | Baltic Way 2021 Shortlist | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
0bdd | Denote $T$ the center of the face $CDHG$ of the cube $ABCDEFGH$ and consider the points $N \in (AD)$, $U \in (FG)$ so that $\frac{AN}{AD} = \frac{FU}{FG} = \frac{1}{3}$.
Let $P \in (CN)$, $R \in (CU)$ and $Q \in (BE)$ so that $\frac{NP}{NC} = \frac{UR}{UC} = \frac{BQ}{BE} = \frac{1}{4}$ and $S \in (EH)$ so that $\frac{... | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0jhw | Problem:
Define the sequence $\{x_{i}\}_{i \geq 0}$ by $x_{0}=x_{1}=x_{2}=1$ and $x_{k}=\frac{x_{k-1}+x_{k-2}+1}{x_{k-3}}$ for $k>2$. Find $x_{2013}$. | [
"Solution:\n\nWe have $x_{3}=\\frac{1+1+1}{1}=3$, $x_{4}=\\frac{3+1+1}{1}=5$, $x_{5}=\\frac{5+3+1}{1}=9$, $x_{6}=\\frac{9+5+1}{3}=5$. By the symmetry of our recurrence (or just further computation—it doesn't matter much), $x_{7}=3$ and $x_{8}=x_{9}=x_{10}=1$, so our sequence has period $8$. Thus $x_{2013}=x_{13}=x_... | United States | HMMT November | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 9 | |
0eh5 | Problem:
Poišči vsa praštevila $p$, $q$ in $r$, ki rešijo enačbo $r^{4} = p q + 4$. | [
"Solution:\n\nEnac̆bo preoblikujemo v $p q = r^{4} - 4$ in desno stran razcepimo po formuli za razliko kvadratov, da dobimo\n$$\np q = (r^{2} - 2)(r^{2} + 2)\n$$\nObe števili $r^{2} - 2$ in $r^{2} + 2$ sta večji od $1$, zato imamo le dve možnosti. Bodisi je $r^{2} - 2 = p$ in $r^{2} + 2 = q$ ali pa je $r^{2} - 2 = ... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"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 | p=7, q=11, r=3 and p=11, q=7, r=3 | |
0dtt | Find all 6-digit integers $n$ such that $n$ is a perfect square and that the number formed by the last 3 digits of $n$ is 1 more than the number formed by the first 3 digits of $n$. | [
"Let $n = y^2$ and $x$ be the number formed by the first 3 digits of $n$. Note that $y$ is a 3-digit number. Then\n$$\ny^2 = 1000x + x + 1 = 1001x + 1 \\quad \\Rightarrow \\quad (y-1)(y+1) = 7 \\times 11 \\times 13x.\n$$\n\nSince $y \\le 999$, not all of 7, 11, 13 can be factors of $y-1$. Thus we have 6 cases:\n\n*... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 183184, 328329, 528529, 715716 | |
04lh | Let $c$ and $d$ be positive divisors of a positive integer $n$. If $c > d$, prove that
$$
c > d + \frac{d^2}{n}.
$$
(Russia 2011) | [
"Consider $\\frac{n}{d}$ and $\\frac{n}{c}$. These are also positive divisors of $n$ which satisfy $\\frac{n}{d} > \\frac{n}{c}$.\nWe now have\n$$\n\\frac{n}{d} - \\frac{n}{c} \\ge 1,\n$$\nfrom which\n$$\nc - d \\ge \\frac{dc}{n} > \\frac{d^2}{n}\n$$\nfollows."
] | Croatia | Mathematical competitions in Croatia | [
"Number Theory > Divisibility / Factorization",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0hi5 | There are $n \ge 3$ children standing in a circle, each of them has two cards, one with the digit $0$ and the other with the digit $1$. At a certain moment, each child raises one of their cards at their discretion. Then every minute, each child whose card number is different from the numbers on both of their neighbors'... | [
"For even $n$, at the beginning, the kids flip every other sign, starting with $0$ and $1$. Then, every minute, they continue to flip the signs alternately, and this will go on forever.\n\nFor odd $n$, such a distribution is not possible. If two identical digits are next to each other, they will remain so forever. ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Yes, if the number of children is even (an alternating arrangement yields perpetual flipping); no, if the number is odd (the process eventually stops). | |
0j13 | Problem:
Dragoons take up $1 \times 1$ squares in the plane with sides parallel to the coordinate axes such that the interiors of the squares do not intersect. A dragoon can fire at another dragoon if the difference in the $x$-coordinates of their centers and the difference in the $y$-coordinates of their centers are ... | [
"Solution:\n\nAssign coordinates in such a way that the dragoon being fired on is centered at $(0,0)$. Any dragoon firing at it must have a center with $x$-coordinates and $y$-coordinates that are no smaller than $-6$ and no greater than $6$. That means that every dragoon firing at it must lie entirely in the regio... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 168 | |
0d5j | Let $BD$ and $CE$ be altitudes of an arbitrary scalene triangle $ABC$ with orthocenter $H$ and circumcenter $O$. Let $M$ and $N$ be the midpoints of sides $AB$, respectively $AC$, and $P$ the intersection point of lines $MN$ and $DE$. Prove that lines $AP$ and $OH$ are perpendicular. | [
"Because $\\angle ADH = \\angle HEA = 90^{\\circ}$, quadrilateral $AEHD$ is cyclic with $AH$ a diameter of its circumcircle $\\omega_{H}$.\n\nBecause $\\angle ANO = \\angle OMA = 90^{\\circ}$, quadrilateral $AMON$ is cyclic with $AO$ a diameter of its circumcircle $\\omega_{O}$.\n\nLet $\\omega_{9}$ be the nine poi... | Saudi Arabia | SAMC 2015 | [
"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"
] | English, Arabic | proof only | null | |
02x3 | Problem:
José arrancou algumas folhas consecutivas de um livro com páginas numeradas com inteiros consecutivos e escritos em ambos os lados de cada folha. A soma dos números das páginas arrancadas é $344$.
a) Determine a fatoração em números primos do número $344$.
b) Encontre a soma do primeiro e do último número d... | [
"Solution:\n\na) Como $44$ é múltiplo de $4$, então $344 = 4 \\cdot 86 = 2^{3} \\cdot 43$.\n\nb) Se $y$ é a quantidade de folhas arrancadas, sendo $2y$ a quantidade de páginas, e $x+1$ é o primeiro número que aparece nelas, temos\n$$\n\\begin{aligned}\n(x+1)+(x+2)+(x+3)+\\ldots+(x+2y) & = 344 \\\\\n2xy + (1+2+\\ldo... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) 344 = 2^3 * 43; b) 43; c) 16 | |
0ab1 | Find all injection functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that
$$
f(f(n)) \leq \frac{n+f(n)}{2}, \forall n \in \mathbb{N}.
$$ | [
"Let $f$ be the solution. Then\n$$\nf(f(n)) \\leq \\max\\{n, f(n)\\}, \\forall n \\in \\mathbb{N}. \\quad (*)\n$$\nLet $f^k(a) = \\underbrace{f(f(...f(a)))}_{k}$. Now there is $a$ such that $a > f(a)$. Then $f^2(a) < a$; by induction to $k$ it's easy to prove that\n$$\nf^k(a) < a, \\forall k > 0. \\quad (1)\n$$\nBe... | North Macedonia | Macedonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Diophantine Equations > Infinite descent / root flipping"
] | null | proof and answer | f(n) = n for all natural numbers n | |
02b4 | Problem:
Arqueólogos encontraram um colar de ouro feito de placas no formato de pentágonos regulares. Cada uma destas placas está conectada a outras duas placas, como ilustra a figura.
Figura 51.1
Quantas placas formam o colar? | [
"Solution:\n\nO ângulo interno de um pentágono regular mede $108^{\\circ}$. Assim, o ângulo interno do polígono determinado pelo colar mede $360^{\\circ} - 108^{\\circ} - 108^{\\circ} = 144^{\\circ}$. Devemos então encontrar $n$ tal que\n$$\n\\frac{180^{\\circ}(n-2)}{n} = 144^{\\circ}\n$$\nResolvendo esta equação, ... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 10 | |
0gyl | At the New year's party each boy gave each girl one candy and each girl gave each boy one cake. After that each boy ate two of given cakes and each girl ate three of given candies. It turned out that kids had eaten quarter of all given things. What is the maximal number of kids that could participate in this party? | [
"Let us denote the number of boys at the party by $x$ and the number of girls by $y$. Then from the problem statement we can get the following equality: $2xy = 4(3y + 2x)$, which is equivalent to the equation $(x-6)(y-4) = 24$. Now it is enough to consider all divisors of $24$ and to get all possible $x$ and $y$, a... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 35 | |
08sk | For a convex quadrilateral $ABCD$, the conditions $AB = BC = 2$, $CD = 2\sqrt{3}$, $DA = 2\sqrt{5}$ are satisfied. Also, if we let $M$, $N$ be the midpoint of $AC$ and $BD$, respectively, then $MN = \sqrt{2}$. Determine the area of the quadrilateral $ABCD$. Here, by $XY$ we are representing the length of the line segme... | [
"$[2 + 4\\sqrt{2}]$\n$$\n\\text{Choose a point } P \\text{ so that the quadrilateral } ABCP \\text{ becomes a parallelogram. Then, we have } AP = PC = 2. \\text{ Since } M \\text{ is the mid-point of } BC \\text{ and } N \\text{ is the mid-point of } BD, \\text{ we have } PD = 2MN = 2\\sqrt{2}. \\text{ Since } PC^2... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 2 + 4√2 | |
04bp | Determine all positive integers $n$ that satisfy the equality
$$
5^n + 2^{n+1} 3^n = 9^n + 4^n.
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | 2 | |
0b68 | Let $f : [a, b] \to \mathbb{R}$ be a continuous function with the property that for every $x \in [a, b)$ there exists $y \in (x, b]$ such that $f(x) \le f(y)$. Find $\max f(x)$. | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Precalculus > Functions",
"Precalculus > Limits"
] | English | proof and answer | f(b) | |
0cmx | In several cells of a $10 \times 10$ grid either a cross or a nought is placed. It appeared that there is no column or row occupied with $10$ signs of the same type, but if one places any sign into any empty cell, this condition would fail. Find the least possible number of cells occupied with signs. | [
"Let us fill the table according to the condition, and let cell $A$ be empty. Since placing any sign in it would create a row or column filled with identical signs, there must exist a line (row or column) containing $A$ in which all other cells are filled with crosses, and the same is true for noughts (these lines ... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | 98 | |
0fvj | Problem:
Bestimme alle positiven ganzzahligen Lösungen der Gleichung
$$
\operatorname{kgV}(a, b, c) = a + b + c
$$ | [
"Solution:\nWir können $a \\leq b \\leq c$ annehmen. Dann ist $a < c$, denn sonst wären alle drei Zahlen gleich $a$ und die Gleichung lautete $a = 3a$, ein Widerspruch. Nun gilt $c < a + b + c < 3c$ und da die linke Seite der Gleichung ein ganzzahliges Vielfaches von $c$ sein muss, folgt daraus $\\operatorname{kgV}... | Switzerland | Vorrundenprüfung | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All permutations of (t, 2t, 3t) for any positive integer t. | |
0dcd | All the cells in a $8 \times 8$ board are colored white. Omar and Asaad play the following game: in the beginning Omar colors $n$ cells red, then Asaad chooses 4 rows and 4 columns and colors them black. Omar wins if there is at least one red cell. Find the least possible value for $n$ such that Omar can always win reg... | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 25 | |
06ni | In a chess tournament, there were $n$ contestants and any two of them played at most one game against each other. Each contestant has played exactly $100$ games. For any two contestants $A$ and $B$ who had played against each other, there were exactly $50$ other participants who have played against both $A$ and $B$. On... | [
"Answer: $1326$\n\nPick any contestant $x$. Suppose he has played against $y_1, y_2, \\dots, y_{100}$ (call these Group Y contestants). Denote the contestants who have not played against $x$ by $z_1, z_2, \\dots, z_m$ (call these Group Z contestants).\n\n\n\nA Group Y contestant (who has pl... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 1326 | |
0kby | Problem:
Wendy is playing darts with a circular dartboard of radius $20$. Whenever she throws a dart, it lands uniformly at random on the dartboard. At the start of her game, there are $2020$ darts placed randomly on the board. Every turn, she takes the dart farthest from the center, and throws it at the board again. ... | [
"Solution:\n\nConsider an individual dart. There is a $\\frac{1}{4}$ probability it is already within $10$ units of the center. If not, for every throw there is a $\\frac{1}{4}$ probability it is not thrown again. Thus, if $E$ is the expected value of times it is thrown, we find $E = 1 + \\frac{3}{4} E \\Longrighta... | United States | HMMO 2020 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 6060 | |
04gz | Nikola is thinking of five numbers. The first number is $-2$, and the fifth number is $6$. The first four numbers are consecutive terms of an arithmetic progression, and the last three numbers are consecutive terms of a geometric progression. What numbers is Nikola thinking of? | [
"Let the five numbers be $a_1, a_2, a_3, a_4, a_5$.\n\nGiven:\n- $a_1 = -2$\n- $a_5 = 6$\n- $a_1, a_2, a_3, a_4$ are consecutive terms of an arithmetic progression (AP)\n- $a_3, a_4, a_5$ are consecutive terms of a geometric progression (GP)\n\nLet the common difference of the AP be $d$.\nThen:\n$$\na_2 = a_1 + d =... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | final answer only | -2, -2/3, 2/3, 2, 6 | |
08vr | Determine the smallest positive integer $n$ for which there are exactly 23 perfect squares greater than or equal to $n$ and less than or equal to $n + 2011$. | [
"Suppose we let $k^2$ (where $k$ is some positive integer) be the largest perfect square less than $n$. Then, the 23 perfect squares bigger than or equal to $n$ and less than or equal to $n+2011$ are $(k+1)^2, (k+2)^2, \\dots, (k+23)^2$. So, we must have $(k+24)^2 > n$.\n\nFrom $k^2 < n$ and $(k+24)^2 > n + 2011$, ... | Japan | Japan Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 901 | |
0052 | Se considera un tablero de $2007 \times 2007$. Se pintan algunas casillas del tablero. Se dice que el tablero es *charrúa* si ninguna fila está totalmente pintada y ninguna columna está totalmente pintada.
a) ¿Cuál es el máximo número $k$ de casillas pintadas que puede tener un tablero charrúa?
b) Para dicho número $k$... | [] | Argentina | XVIII Olimpiada Matemática del Cono Sur | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | Español | proof and answer | k = 2007*2006 and the number of boards is 2007! | |
048e | A triangle $ABC$ with $\angle BAC = 120^\circ$ is given. Let $D$ be the point in its interior such that $\angle DBC = 2\angle ABD$ and $\angle DCB = 2\angle ACD$. Determine $\angle BDC$. | [] | Croatia | CroatianCompetitions2011 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 140° | |
05ct | A TV show airs every 28 days. This century there was a year when the show aired on both January 1 and January 29. In how many years will the show air twice in January again? | [
"In a common year, there are $365 = 13 \\cdot 28 + 1$ days. Thus, after a common year, the dates of the airings shift 1 day earlier. After a leap year, the airings shift 2 days earlier. In every 4 years, there are 3 common years and 1 leap year, which means a total shift of $3 \\cdot 1 + 2 = 5$ days. Thus in $5 \\c... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 21 | |
0hqs | Problem:
Let $X$, $Y$, and $Z$ be the points on the sides $BC$, $CA$, and $AB$ of the triangle $ABC$, such that $\triangle XYZ \sim \triangle ABC$ ($\angle X = \angle A$, $\angle Y = \angle B$). Prove that the orthocenter of $\triangle XYZ$ coincides with the circumcenter of $\triangle ABC$. | [
"Solution:\n\nLet $x$, $y$, and $z$ be the points passing through $X$, $Y$, and $Z$, parallel to $YZ$, $ZX$, and $XY$. Let $P$, $Q$, and $R$ be the intersection point of the lines $y$ and $z$; $z$ and $x$; $x$ and $y$, respectively. Then we have $\\triangle PQR \\sim \\triangle XYZ$. The points $X$, $Y$, and $Z$ ar... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
066u | Let $A_1, A_2, \dots, A_{160}$ be set such that $|A_i|=i$, $i=1, 2, \dots, 160$. Using the elements of these sets we construct new sets $M_1, M_2, \dots, M_n$ with the following procedure: At the first step we choose some of the sets $A_1, A_2, \dots, A_{160}$ and we subtract from each of them the same number of elemen... | [
"We suppose that at the first step we select from all sets $k_1$ elements, at the second step we select from the remaining sets $k_2$ elements and similarly at the $n$-th step we select $k_n$ elements. After the depletion of the elements of all sets $A_1, A_2, \\dots, A_n$, then each $i = |A_i|$, $i = 1, 2, \\dots,... | Greece | Hellenic Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Other",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 8 | |
0g32 | Problem:
Soit $n$ un entier strictement positif. Maurice écrit sur une même ligne tous les $2^{n}-1$ sous-ensembles non-vides de l'ensemble $\{1,2, \ldots, n\}$. Ensuite, en-dessous de chaque sous-ensemble, il écrit le produit de ses éléments. Finalement, il écrit les inverses des nombres présents sur la deuxième lign... | [
"Solution:\n\nOn commence par tester les petites valeurs de $n$ et on constate que à chaque fois la valeur obtenue est $n$. On pose donc l'hypothèse\n$$\nH(n): \\text{ La somme obtenue avec l'ensemble } \\{1,2, \\ldots, n\\} \\text{ vaut } n\n$$\net on va essayer de prouver par induction que $H(n)$ est vérifiée pou... | Switzerland | Vorrunde 2019 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | n | |
07eg | Let $\mathbb{R}^{\ge 0}$ be the set of non-negative real numbers. Find all functions $f : \mathbb{R}^{\ge 0} \to \mathbb{R}^{\ge 0}$ such that for all $x, y \in \mathbb{R}^{\ge 0}$,
$$
f(x^3 + x f(xy)) = f(xy) + x^2 f(x + y).
$$ | [
"Let $P$ denote the assertion that:\n$$\nf(x^3 + x f(xy)) = f(xy) + x^2 f(x + y)\n$$\nBy $P(x, \\frac{y}{x})$ we have\n$$\nA(x, y) : \\quad f(x^3 + x f(y)) = f(y) + x^2 f\\left(x + \\frac{y}{x}\\right)\n$$\nIf there exists a pair $(x, y)$ such that $f(x^3 + x f(y)) = 0$. Then\n$$\nA(x, y) \\rightarrow f(y) + x^2 f\... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof and answer | f(x) = 0 for all x ≥ 0 | |
0a65 | Problem:
Find all non-zero real numbers $a$, $b$, $c$ such that the following polynomial has four (not necessarily distinct) positive real roots.
$$P(x) = a x^{4} - 8a x^{3} + b x^{2} - 32c x + 16c$$ | [
"Solution:\nAssume that $P(x)$ has 4 positive real roots $x_{1}$, $x_{2}$, $x_{3}$ and $x_{4}$. Using Viete's theorem, one can obtain the following equations\n\n$$\n\\begin{array}{c}\nx_{1} + x_{2} + x_{3} + x_{4} = \\frac{8a}{a} = 8 \\\\\nx_{1}x_{2} + x_{1}x_{3} + x_{1}x_{4} + x_{2}x_{3} + x_{2}x_{4} + x_{3}x_{4} ... | New Zealand | NZMO Round One | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | (a, 24a, a) for any non-zero real a | |
0e6b | A mole named Črt has 5 rooms in his lair. The rooms are numbered with numbers from 1 to 5. Črt has drilled tunnels between some of the rooms so that he can crawl from every room to any other room using some of the tunnels. No two tunnels intersect. Every tunnel starts in one room and ends in another room (different fro... | [
"We first notice that rooms $5$ and $1$ cannot be neighbouring. If they were, Črt could reach room $1$ from room $5$ by crawling through the connecting tunnel exactly three times—first from room $5$ to room $1$, then back to room $5$, and then again to room $1$. This would be in contradiction with the known facts a... | Slovenia | National Math Olympiad 2012 | [
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | Neighbouring pairs: (1,4), (2,3), (2,4), (3,4), (4,5). | |
06xh | Let $a_{1}, a_{2}, \ldots, a_{2023}$ be positive integers such that
- $a_{1}, a_{2}, \ldots, a_{2023}$ is a permutation of $1,2, \ldots, 2023$, and
- $\left|a_{1}-a_{2}\right|,\left|a_{2}-a_{3}\right|, \ldots,\left|a_{2022}-a_{2023}\right|$ is a permutation of $1,2, \ldots, 2022$.
Prove that $\max \left(a_{1}, a_{2023}... | [
"For the sake of clarity, we consider and prove the following generalisation of the original problem (which is the case $N=1012$):\nLet $N$ be a positive integer and $a_{1}, a_{2}, \\ldots, a_{2N-1}$ be positive integers such that\n- $a_{1}, a_{2}, \\ldots, a_{2N-1}$ is a permutation of $1,2, \\ldots, 2N-1$, and\n-... | IMO | International Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0066 | En el plano hay dibujadas $n$ rectas distintas. Cada una de ellas corta a exactamente otras 2007 de las rectas. Hallar todos los valores de $n$ para los cuales esto es posible. | [] | Argentina | Argentina 2008 | [
"Discrete Mathematics > Graph Theory",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | Spanish | proof and answer | 2008, 2010, 2016, 2230, 2676, 4014 | |
0a3d | Problem:
Vind alle functies $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ zo dat voor alle positieve gehele getallen $m$, $n$ en $a$ geldt dat
a) $f(f(m) f(n))=m n$,
b) $f(2024 a+1)=2024 a+1$. | [
"Solution:\n\nOplossing 1. Als we $a=0$ zouden mogen invullen, vinden we $f(1)=1$. Met $n=1$ krijgen we dan $f(f(m))=m$, dus $f$ is bijectief. Helaas mogen we $a=0$ niet invullen, maar het zet ons wel op het goede spoor.\nWe beginnen met invullen van $n=1$. Dit geeft dat $f(f(m) f(1))=m$, wat betekent dat $f$ bijec... | Netherlands | IMO-selectietoets II | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theo... | null | proof and answer | Exactly four multiplicative involutions:
1) The identity function f(n) = n for all positive integers n.
2) Swap 2 and 11 and fix all other primes: f(2) = 11, f(11) = 2, f(p) = p for all primes p not in {2, 11}; extend multiplicatively so f(p^k) = f(p)^k and f(n) = product over prime powers.
3) Swap 2 and 23 and fix all... | |
0brf | Find all real numbers $x \in (2, \infty)$ for which
$$
\cos(\pi \log_3 (x+6)) \cdot \cos(\pi \log_3 (x-2)) = 1.
$$ | [
"From the hypothesis, it is clear that\n$$\n\\cos(\\pi \\log_3 (x + 6)) = \\cos(\\pi \\log_3 (x - 2)) = \\pm 1,\n$$\nand thus, there exist $k, l \\in \\mathbb{Z}$, with the same parity, such that $\\pi \\log_3 (x + 6) = k\\pi$ and $\\pi \\log_3 (x - 2) = l\\pi$.\nWe get now the relations $x + 6 = 3^k$ and $x - 2 = ... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 3 | |
0hz7 | Problem:
An unfair coin has the property that when flipped four times, it has the same probability of turning up 2 heads and 2 tails (in any order) as 3 heads and 1 tail (in any order). What is the probability of getting a head in any one flip? | [
"Solution:\n\nLet $p$ be the probability of getting a head in one flip. There are $6$ ways to get $2$ heads and $2$ tails, each with probability $p^{2}(1-p)^{2}$, and $4$ ways to get $3$ heads and $1$ tail, each with probability $p^{3}(1-p)$. We are given that\n$$\n6 p^{2}(1-p)^{2} = 4 p^{3}(1-p).\n$$\nClearly $p$ ... | United States | Harvard-MIT Math Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 3/5 | |
0dbr | Consider a triangle $ABC$ and let $M$ be the midpoint of the side $BC$. Suppose $\angle MAC = \angle ABC$ and $\angle BAM = 105^\circ$. Find the measure of $\angle ABC$. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 30° | |
02zc | Problem:
A bandeira da figura é formada por um retângulo $40~\mathrm{cm} \times 20~\mathrm{cm}$ e possui duas cores (branca e cinza). A parte branca é composta por duas listras de mesma largura que se cruzam e são perpendiculares aos lados do retângulo. A parte cinza é a área do retângulo que não foi coberto pelas list... | [
"Solution:\nSeja $x$ a medida da largura das faixas. Assim, uma das faixas mede $x$ por $40~\\mathrm{cm}$, enquanto que a outra mede $x$ por $20~\\mathrm{cm}$. Se somarmos as áreas das duas faixas, estaremos contando sua intersecção duas vezes e, portanto, precisaremos descontá-la. Como a área das faixas deve ser a... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 30 − 10√5 cm | |
02rn | Let $f: (0, +\infty) \to (0, +\infty)$ be a twice differentiable function such that $f'(x) < 0$ for all $x > 0$. For each $x > 0$ consider the triangle determined by the line tangent to the graph of $f$ at $(x, f(x))$ and the two coordinate axes. It is known that the area of this triangle is constant and equal to $C$.
... | [] | Brazil | Brazilian Math Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | C/2 | |
0jjv | Problem:
A deck of $8056$ cards has $2014$ ranks numbered $1$-$2014$. Each rank has four suits—hearts, diamonds, clubs, and spades. Each card has a rank and a suit, and no two cards have the same rank and the same suit. How many subsets of the set of cards in this deck have cards from an odd number of distinct ranks? | [
"Solution:\n\nThe answer is $\\frac{1}{2}\\left(16^{2014}-14^{2014}\\right)$.\n\nThere are $\\binom{2014}{k}$ ways to pick $k$ ranks, and $15$ ways to pick the suits in each rank (because there are $16$ subsets of suits, and we must exclude the empty one). We therefore want to evaluate the sum\n$$\n\\binom{2014}{1}... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1/2*(16^2014 - 14^2014) | |
02wk | Problem:
O pai de João possui entre 200 e 300 livros em sua biblioteca. Um quinto destes livros está em inglês, um sétimo em francês, um quarto em italiano e o resto são livros em espanhol. Qual o total de livros em espanhol nessa biblioteca? | [
"Solution:\n\nA quantidade de livros é um múltiplo de $5$, $7$ e $4$. Como esses números não possuem fatores primos em comum, necessariamente deve ser um múltiplo de $5 \\cdot 7 \\cdot 4 = 140$. No intervalo dado, existe um único múltiplo de $140$, a saber, o número $280$. Portanto, as quantidades de livros que não... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | final answer only | 114 | |
0fmj | Sean $a$, $b$ y $n$ enteros positivos tales que $a > b$ y $ab - 1 = n^2$. Prueba que $a - b \ge \sqrt{4n - 3}$. Indica justificadamente cuando se alcanza la igualdad. | [] | Spain | Olimpiada Matemática Española | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Quadratic functions"
] | Spanish | proof only | null | |
07pu | Let $n \ge 3$ be an odd positive integer. We are given an $n \times n$ board which is divided into white unit squares by lines drawn parallel to the sides of the board. Let $N$ denote the number of ways of placing 2 identical red checkers and 2 identical blue checkers on the board such that
(a) in any such placement, n... | [
"Let $N_b = n^2$ and $k = (N_b - 1)/2$. There are $\\binom{N_b}{2}$ ways of placing 2 identical red checkers, and for any such placement there are $\\binom{N_b-2}{2}$ ways of next placing the 2 identical blue checkers. Therefore the total number of ways of placing 2 identical red checkers and 2 identical blue check... | Ireland | Ireland | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | ((n^2 - 1)/2)^4 - ((n^2 - 1)/2)^3 | |
0g5n | 令 $\triangle ABC$ 是一銳角三角形且 $CP$ 為邊 $AB$ 上的高, $H$ 為 $CP$ 上任一點。直線 $AH, BH$ 分別交 $BC, AC$ 於點 $M, N$.
(1) 證明: $\angle NPC = \angle MPC$
(2) 設 $O$ 是 $MN$ 與 $CP$ 的交點, 一條通過 $O$ 的任意的直線交四邊形 $CNHM$ 的邊於 $D, E$ 兩點。證明: $\angle EPC = \angle DPC$. | [
"(1) 記 $\\angle NPC = \\phi_1$, $\\angle MPC = \\phi_2$, 則\n$$\n\\frac{S_{\\triangle NPC}}{S_{\\triangle NPA}} = \\frac{CN}{AN} = \\frac{CP \\sin \\phi_1}{AP \\cos \\phi_2}.\n$$\n所以, $\\tan \\phi_1 = \\frac{CN}{AN} \\cdot \\frac{AP}{CP}$.\n同理可得, $\\tan \\phi_2 = \\frac{CM}{BM} \\cdot \\frac{BP}{CP}$.\n於是, $\\tan \\... | Taiwan | 二〇一一數學奧林匹亞競賽第一階段選訓營,模擬競賽(一) | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00rk | The plane is divided into unit squares by means of two sets of parallel lines. The unit squares are coloured in $1201$ colours so that no rectangle of perimeter $100$ contains two squares of the same colour. Show that no rectangle of size $1 \times 1201$ contains two squares of the same colour. | [
"Consider the set $D$ of all unit squares $(x, y)$ such that $|x| + |y| \\le 24$. Any translate of $D$ is called a *diamond*.\nSince any two unit squares that belong to the same diamond also belong to some rectangle of perimeter $100$, a diamond cannot contain two unit squares of the same colour. Since a diamond co... | Balkan Mathematical Olympiad | BMO 2016 Short List Final | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
04lm | In the rectangle with vertices $A(-1, 0)$, $B(1, 0)$, $C(1, 1)$, $D(-1, 1)$ a point $T$ is randomly chosen. Find the probability that $T$ is closer to $A$ than to the point $P(2, 1)$. | [] | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 3/4 | |
05j4 | Problem:
Soit $a, b, c$ des réels tels que $-1 \leqslant a x^{2}+b x+c \leqslant 1$ pour $x=-1, x=0$ et $x=1$. Prouver que
$$
-\frac{5}{4} \leqslant a x^{2}+b x+c \leqslant \frac{5}{4} \text{ pour tout réel } x \in[-1,1]
$$ | [
"Solution:\n\nPosons $P(x)=a x^{2}+b x+c$. Alors $P(-1)=a-b+c$, $P(0)=c$ et $P(1)=a+b+c$. Et, d'après l'énoncé, on a $|P(-1)| \\leqslant 1$, $|P(0)| \\leqslant 1$ et $|P(1)| \\leqslant 1$.\n\nOr, pour tout réel $x$, on vérifie directement que\n$$\nP(x)=\\frac{x(x+1)}{2} P(1)-\\frac{x(1-x)}{2} P(-1)+\\left(1-x^{2}\\... | France | Olympiades Françaises de Mathématiques, Envoi Numéro 3 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0fxq | Problem:
Sei $ABCD$ ein konvexes Viereck und seien $P$ und $Q$ Punkte innerhalb des Vierecks $ABCD$, so dass $PQDA$ und $Q P B C$ Sehnenvierecke sind. Nehme an, dass ein Punkt $E$ auf der Strecke $PQ$ existiert, so dass $\angle PAE = \angle QDE$ und $\angle PBE = \angle QCE$. Zeige, dass $ABCD$ ein Sehnenviereck ist. | [
"Solution:\n\nDie Gerade $PQ$ ist die Potenzlinie der beiden Kreise durch $PQDA$ und $Q P B C$. Wenn sich die Geraden $AD$ und $BC$ in einem Punkt $Z$ schneiden, so ist $ABCD$ genau dann ein Sehnenviereck, wenn $ZD \\cdot ZA = ZC \\cdot ZB$ gilt. Die Potenz muss also zu beiden Kreisen gleich groß sein. Dies ist abe... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01py | a) Three of six segments (four sides and two diagonals of an isosceles trapezoid) are painted red, and three others are painted green.
Prove that one can construct a triangle using the segments of the same color as its sides.
b) Is the previous statement true if these six segments are four sides and two diagonals of a... | [
"b) The statement may not hold.\n\na) See the solution of Problem D.4.\n\nb) Consider, for example, a trapezoid $ABCD$ such that its diagonal $AC$ is perpendicular to the bases $AD$ and $BC$ (see the Fig.). Let $CA = 1$. Let $AB, AC, CD$ be painted green and $BC, BD, AD$ be painted red. For these red segments the f... | Belarus | BelarusMO 2013_s | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof and answer | No | |
0933 | Problem:
Let $n \geqslant 3$ be an integer. A sequence $P_{1}, P_{2}, \ldots, P_{n}$ of distinct points in the plane is called good if no three of them are collinear, the polyline $P_{1} P_{2} \ldots P_{n}$ is non-self-intersecting and the triangle $P_{i} P_{i+1} P_{i+2}$ is oriented counterclockwise for every $i=1,2, ... | [
"Solution:\nFix $n$ points on a plane, no three of which are collinear. Let $\\mathcal{P}$ be their convex hull. Let the vertices of $\\mathcal{P}$ be $A_{1}, A_{2}, \\ldots, A_{m}$ (lying in this order on the boundary of $\\mathcal{P}$ counterclockwise). We denote $A_{m+1}=A_{1}$. Also, let $\\mathcal{I}$ be the s... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | n^2 - 4n + 6 | |
04tx | Let $ABC$ be an acute triangle with altitude $AD$. The bisectors of angles $BAD$, $CAD$, intersect side $BC$ at $E$, $F$, respectively. The circumcircle of triangle $AEF$ intersects sides $AB$, $AC$ at $G$, $H$, respectively. Prove that lines $EH$, $FG$, and $AD$ pass through a common point. (Patrik Bak) | [
"Let $K$ be the intersection of segments $FG$ and $AE$ and $L$ the intersection of $EH$ and $AF$ (Fig. 2). Inscribed angles give\n\n$$\n\\angle AGF = \\angle AEF = 90^\\circ - \\angle DAE = 90^\\circ - \\angle GAE,\n$$\n\nFig. 2\n\nthat is $\\angle AGF + \\angle GAE = 90^\\circ$, hence\n$$\... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0btk | Given positive integers $k$ and $m$, show that $m$ and $\binom{n}{k}$ are coprime for infinitely many integers $n \ge k$. | [
"Let $n = k + l m k!$, where $l$ is an arbitrary nonnegative integer, let $p$ be any prime factor of $m$, and let $p^h$ be the highest power of $p$ that divides $k!$ — that is, $p^h$ divides $k!$ but $p^{h+1}$ does not. Notice that $n \\equiv k \\pmod{p^{h+1}}$, to deduce that $n(n-1)\\cdots(n-k+1) \\equiv k! \\pmo... | Romania | 67th NMO Selection Tests for BMO and IMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
0jvi | Problem:
Determine the smallest positive integer $n \geq 3$ for which
$$
A \equiv 2^{10 n} \pmod{2^{170}}
$$
where $A$ denotes the result when the numbers $2^{10}, 2^{20}, \ldots, 2^{10 n}$ are written in decimal notation and concatenated (for example, if $n=2$ we have $A=10241048576$ ). | [
"Solution:\nNote that\n$$\n2^{10 n} = 1024^{n} = 1.024^{n} \\times 10^{3 n}\n$$\nSo $2^{10 n}$ has roughly $3 n + 1$ digits for relatively small $n$'s. (Actually we have that for $0 < x < 1$,\n$$\n(1 + x)^{2} = 1 + 2x + x^{2} < 1 + 3x\n$$\nTherefore, $1.024^{2} < 1.03^{2} < 1.09$, $1.09^{2} < 1.27$, $1.27^{2} < 1.8... | United States | HMMT February 2016 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 14 | |
0bs5 | Let $n$ be a positive integer, and let $S_1, \dots, S_n$ be a collection of finite non-empty sets such that
$$
\sum_{1 \le i < j \le n} \frac{|S_i \cap S_j|}{|S_i| |S_j|} < 1.
$$
Prove that there exist pairwise distinct elements $x_1, \dots, x_n$ such that $x_i$ is a member of $S_i$ for each index $i$. | [
"A *choice function* or simply a *choice* for the collection $S_1, \\dots, S_n$ is a function $c$ from the first $n$ positive integers to the union $S_1 \\cup \\dots \\cup S_n$ such that $c(i)$ is a member of $S_i$ for each $i$. We must show that an injective choice is always possible under the conditions in the st... | Romania | 67th NMO Selection Tests for BMO and IMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof only | null | |
0j4t | Problem:
Sarah and Hagar play a game of darts. Let $O_{0}$ be a circle of radius $1$. On the $n$th turn, the player whose turn it is throws a dart and hits a point $p_{n}$ randomly selected from the points of $O_{n-1}$. The player then draws the largest circle that is centered at $p_{n}$ and contained in $O_{n-1}$, an... | [
"Solution:\n\n$\\boxed{\\dfrac{6\\pi}{7}}$\n\nLet $f(r)$ be the average area colored red on a dartboard of radius $r$ if Sarah plays first. Then $f(r)$ is proportional to $r^{2}$. Let $f(r) = (\\pi x) r^{2}$ for some constant $x$. We want to find $f(1) = \\pi x$.\n\nIn the first throw, if Sarah's dart hits a point ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | 6π/7 | |
0i7g | Problem:
a) One Sunday, Zvezda wrote 14 numbers in a circle, so that each number is equal to the sum of its two neighbors. Prove that the sum of all 14 numbers is 0.
b) On the next Sunday, Zvezda wrote 21 numbers in a circle, and this time each number was equal to half the sum of its two neighbors. What is the sum of ... | [
"Solution:\na)\nDenoting the numbers $a_{1}, a_{2}, \\ldots, a_{14}$, and their sum as $S$ we have $a_{i} = a_{i-1} + a_{i+1}$ for $i = 1, \\ldots, 14$ (we take $a_{15} = a_{1}, a_{0} = a_{14}$). Summing all these equalities we get $S = 2S$ (since each $a_{i}$ appears exactly once on the left and exactly twice on t... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) 0; b) 63 | |
0jwv | Problem:
Points $A$, $B$, $C$, and $D$ lie on a line in that order such that $\frac{AB}{BC}=\frac{DA}{CD}$. If $AC=3$ and $BD=4$, find $AD$. | [
"Solution:\n\nLet $BC = x$, then the equation becomes $\\frac{3-x}{x}=\\frac{7-x}{4-x}$. This simplifies to a quadratic equation with solutions $x=1$ and $x=6$. Since $x<3$, we have $x=1$ and $AD=7-x=6$."
] | United States | HMMT November | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 6 | |
08ge | Problem:
Sia $N$ il massimo numero ottenibile come prodotto di interi positivi tutti distinti la cui somma è $2024$. Con quante cifre "0" termina la rappresentazione decimale di $N$?
(A) 9
(B) 10
(C) 12
(D) 13
(E) 14 | [
"Solution:\n\nLa risposta è (D). Chiamo $A$ l'insieme dei numeri che sommano a $2024$ e il cui prodotto è dato dal massimo possibile, cioè $N$. Chiamo \"buco\" un numero $n$ che non è in $A$, ma tale che $A$ contiene sia un numero maggiore di $n$, sia un numero minore di $n$.\n\nDimostro che $A$ non contiene $1$: s... | Italy | Olimpiadi di Matematica | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | D | |
0kyb | In a long line of people arranged left to right, the $1013$th person from the left is also the $1010$th person from the right. How many people are in the line?
(A) $2021$ (B) $2022$ (C) $2023$ (D) $2024$ (E) $2025$ | [
"There are $1012$ people to the left of the specified person and $1009$ people to the right of that person. There are therefore $1012 + 1 + 1009 = 2022$ people in the line."
] | United States | AMC 10 B | [
"Math Word Problems"
] | null | MCQ | B | |
07h6 | Let $ABC$ be an acute triangle and $M$ be the midpoint of $AB$. Let $K$ be a point such that $KB$ intersects the line $AC$, $\angle KMC = 90^\circ$ and $\angle KAC = 180^\circ - \angle ABC$. The tangent line to circumcircle of triangle $ABC$ at $A$ intersects line $CK$ at $E$. Prove that the reflection of line $BC$ wit... | [
"Note that $KA$ is parallel to the tangent line from $C$ to the circumcircle of $ABC$. Therefore, if we denote by $O$ the circumcenter of $ABC$, then $CO \\perp KA$. So, denoting by $F$ the intersection point of $KA$ and $CO$, we have $\\angle KFC = 90^\\circ$. On the other hand, $\\angle KMC = 90^\\circ$, therefor... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0leh | In a board of $2021 \times 2021$ grids, we pick $k$ unit squares such that every picked square shares at least one vertex with at most one other picked square. Determine the maximum of $k$. | [
"We say two squares are connected if they share at least one vertex. The condition states that every picked square is connected to at most one other picked square. Hence, the set of picked squares can be partitioned into many connected components, where each component contains at most two squares that are connected... | Vietnam | TST | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 2022^2/3 | |
02p7 | Let $a_1, a_2, \dots, a_{2011}$ be nonnegative reals with sum $\frac{2011}{2}$. Prove that
$$
\left| \prod_{cyc} (a_n - a_{n+1}) \right| = \left| (a_1 - a_2)(a_2 - a_3) \dots (a_{2011} - a_1) \right| \le \frac{3\sqrt{3}}{16}.
$$ | [
"In what follows, indices are taken modulo $2011$ and $E = \\left| \\prod_{cyc} (a_n - a_{n+1}) \\right|$.\n\n**Lemma.** If $E$ is maximum, for every $i \\in \\{1, 2, \\dots, 2011\\}$, one of the numbers $a_{i-1}, a_i, a_{i+1}$ is zero.\n\n*Proof.* Suppose, by means of contradiction, that $E$ is maximum and there e... | Brazil | Brazilian Math Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
06z6 | Problem:
Given positive real numbers $x$, $y$, $z$ each less than $\pi / 2$, show that $\pi / 2 + 2 \sin x \cos y + 2 \sin y \cos z > \sin 2x + \sin 2y + \sin 2z$. | [
"Solution:\n\nWe have $\\sin 2x + \\sin 2y + \\sin 2z - 2 \\sin x \\cos y - 2 \\sin y \\cos z = 2 \\sin x (\\cos x - \\cos y) + 2 \\sin y (\\cos y - \\cos z) + 2 \\sin z \\cos z$, so we wish to show that\n\n$\\sin x (\\cos x - \\cos y) + \\sin y (\\cos y - \\cos z) + \\sin z \\cos z < \\pi / 2$ (*).\n\n![](attached... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0l0e | Problem:
Suppose $a$ and $b$ are positive integers. Isabella and Vidur both fill up an $a \times b$ table. Isabella fills it up with numbers $1,2, \ldots, a b$, putting the numbers $1,2, \ldots, b$ in the first row, $b+1, b+2, \ldots, 2 b$ in the second row, and so on. Vidur fills it up like a multiplication table, pu... | [
"Solution:\n\nUsing the formula $1+2+\\cdots+n=\\frac{n(n+1)}{2}$, we get\n\n$$\n\\begin{aligned}\n\\frac{a b(a b+1)}{2}-\\frac{a(a+1)}{2} \\cdot \\frac{b(b+1)}{2} & =\\frac{a b(2(a b+1)-(a+1)(b+1))}{4} \\\\\n& =\\frac{a b(a b-a-b+1)}{4} \\\\\n& =\\frac{a b(a-1)(b-1)}{4} \\\\\n& =\\frac{a(a-1)}{2} \\cdot \\frac{b(b... | United States | HMMT February 2024 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functi... | null | proof and answer | 21 | |
08u2 | Let $n$ be a positive integer greater than or equal to 3. Determine all those $n$'s for which there exists an $n$-gon with all the sides having the same length and with all its interior angles having either $120^\circ$ or $240^\circ$. | [
"We will show that $n = 6$ and $n = 2k$, where $k$ is an integer greater than or equal to 5, satisfy the condition of the problem and that there are no other $n$ satisfying the requirement.\n\nFirst, let us show that $n = 6$ and $n = 2k$ with $k \\ge 5$ satisfy the condition of the problem.\n\n![](attached_image_1.... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | n = 6 or n is even and at least 10 | |
04bg | The numbers $2009$, $2012$ and $2015$ are written on the board. In each step Željko marks the numbers on the board by $a$, $b$ and $c$ in some order, and then replaces them with the numbers $3a - b$, $3b - c$ and $3c - a$. By repeating this procedure multiple times, can Željko achieve that at some point all the numbers... | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Linear Algebra > Linear transformations",
"Algebra > Linear Algebra > Matrices"
] | English | proof and answer | No | |
07xe | A non-negative integer $p$ is a *3-choice* if $p = k(k-1)(k-2)/6$ for some positive integer $k$. Let $p$ and $q$ be 3-choices with $p < q$. Show there is an integer $n$ such that $p \le n^2 < q$. | [
"The 3-choices are exactly the binomial coefficients $p = \\binom{k}{3}$ with $k \\ge 1$. Without loss of generality we can consider consecutive $p$ and $q$, so that $p = \\binom{k}{3}$ and $q = \\binom{k+1}{3}$ for $k \\ge 1$. For $k = 1, 2, 3, 4$ the required statement is true, because\n$$\n\\binom{1}{3} = 0 \\qu... | Ireland | IRL_ABooklet_2024 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0596 | A rectangle of integral side lengths is divided into $2022$ unit squares. At least one unit square is coloured black. There are equally many black squares in every row and also equally many black squares in every column. Find all possibilities of how many black unit squares there can be in total. | [
"**Answer:** $2022$.\n\nLet the rectangle be of size $a \\times b$. Let there be $k$ black unit squares in every row and $l$ black unit squares in every column; then $ak = bl$. Since $ab = 2022 = 2 \\cdot 3 \\cdot 337$ where the factors are primes, numbers $a$ and $b$ must be coprime. Thus $b \\mid k$, implying $ak... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 2022 | |
0gpn | Find the smallest value of $k$ for which $2012$ bags each containing finite number of colored balls, no matter how the contents of bags are arranged, can be distributed into $k$ boxes so that for each box at least one of the following two conditions is held:
i. all bags of a box contain a ball of the same color
ii. eac... | [
"The answer is $62$.\nLet us show that $k \\ge 62$. Suppose that there are $62$ bags each containing one ball colored $1$, $61$ bags each containing $1$ ball colored $2$, ..., $1$ bag containing $1$ ball colored $62$. Let us show that these $1953$ bags can not be distributed into $61$ boxes. A bag containing a ball... | Turkey | 17th Junior Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 62 | |
09lf | Let $p$ be a prime number of the form $p = 12k + 5$ and let $\{A_n\}_{n=1}^{\infty}$ be a sequence given by $A_n = (n^1 - 1)(n^2 - 1) \cdots (n^{6k+1} - 1)(n^{6k+2} - 1)$. Show that
$$
A_2^2 + A_8^2 \equiv 0 \pmod{p}.
$$ | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof only | null | |
09os | An olympiad committee held $20$ meetings, each attended by exactly $5$ members. No pair of members attended more than one meeting together. What is the minimum possible number of committee members? | [] | Mongolia | MMO2025 Round 3 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 21 | |
0494 | In the triangle $ABC$ with $|AB| = |AC|$ angle bisector of $\angle ABC$ meets $\overline{AC}$ in $D$. If $|BC| = |BD| + |AD|$, find all the angles. (Canada 1996) | [
"\n\nSince $BD$ is the bisector of the angle $\\angle CBA$, it follows that $|CD| : |AD| = |BC| : |AB|$, and because of that $\\frac{|CD|}{|CE|} = \\frac{|CD|}{|AD|} = \\frac{|BC|}{|AB|} = \\frac{|CB|}{|CA|}$.\n\nTriangles $ABC$ and $EDC$ have a common angle in vertex $C$ and equal ratios o... | Croatia | CroatianCompetitions2011 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 40°, 40°, 100° | |
0666 | Let $\triangle ABC$ be an acute-angled triangle with $AB < AC$, inscribed in to the circle $c(O, R)$. The extension of the altitude $AD$ intersects the circumcircle $c(O, R)$ at $E$ and the perpendicular bisector ($\mu$) of the side $AB$ meets $AD$ at $L$. The line $BL$ meets $AC$ at $M$ and the circumcircle $c(O, R)$ ... | [
"**Direct.** Let $MZ \\perp BC$. Then $MZ \\parallel AD$, since $AD \\perp BC$. Therefore we have: $Z\\hat{M}C = E\\hat{A}C$. From the quadrilateral $EANC$ we have: $E\\hat{A}C = E\\hat{N}C$ and hence we get\n$$\nZ\\hat{M}C = E\\hat{A}C = E\\hat{N}C = Z\\hat{N}C.\n$$\nTherefore the quadrilateral $MNCZ$ is cyclic an... | Greece | 28th Hellenic 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"
] | English | proof only | null | |
09y5 | Problem:
De rij positieve gehele getallen $a_{0}, a_{1}, a_{2}, \ldots$ is gedefinieerd door $a_{0}=3$ en
$$
a_{n+1}-a_{n}=n\left(a_{n}-1\right)
$$
voor alle $n \geq 0$. Bepaal alle gehele getallen $m \geq 2$ waarvoor geldt dat $\operatorname{ggd}\left(m, a_{n}\right)=1$ voor alle $n \geq 0$. | [
"Solution:\n\nEen directe formule voor de rij wordt gegeven door $a_{n}=2 \\cdot n!+1$ voor $n \\geq 0$. (We gebruiken de gangbare definitie $0!=1$, die voldoet aan $1!=1 \\cdot 0!$, zoals ook voor grotere $n$ geldt $n!=n \\cdot(n-1)!$.) We bewijzen de directe formule met inductie. Er geldt $a_{0}=3$ en dat is geli... | Netherlands | IMO-selectietoets | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | m = 2^i for i ≥ 1 | |
0km2 | Problem:
In triangle $A B C$, let $M$ be the midpoint of $B C$, $H$ be the orthocenter, and $O$ be the circumcenter. Let $N$ be the reflection of $M$ over $H$. Suppose that $O A = O N = 11$ and $O H = 7$. Compute $B C^{2}$. | [
"Solution:\n\nLet $\\omega$ be the circumcircle of $\\triangle A B C$. Note that because $O N = O A$, $N$ is on $\\omega$. Let $P$ be the reflection of $H$ over $M$. Then, $P$ is also on $\\omega$. If $Q$ is the midpoint of $N P$, note that because\n$$\nN H = H M = M P,\n$$\n$Q$ is also the midpoint of $H M$. Since... | United States | HMMT Spring 2021 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 288 | |
0ejy | Problem:
Dani sta enačbi $(y+3)^2 - x(x-4) = 7 + (y-x)(y+x)$ in $x + \frac{y-1}{2} = \frac{x}{2} - y$.
a) Reši sistem enačb in rešitev zapiši kot točko $A$ v koordinatnem sistemu.
b) Zapiši enačbo premice, ki poteka skozi točko $A$ in ima isto presečišče z osjo $x$ kot premica $3x + 2y - 15 = 0$ v eksplicitni, impli... | [
"Solution:\n\na.\nPoenostavimo prvo enačbo do zapisa npr. $4x + 6y + 2 = 0$. Poenostavimo drugo enačbo do zapisa npr. $x + 3y - 1 = 0$. Rešimo sistem enačb s katerokoli metodo. Rešitev sistema enačb zapišemo kot točko $A(-2, 1)$.\n\nb.\nIzračunamo presečišče premice $3x + 2y - 15 = 0$ z $x$ osjo $(5, 0)$. Izračunam... | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Odbirno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | Point A: (-2, 1). Line through A with the same x-intercept as 3x + 2y - 15 = 0:
- Explicit: y = -1/7 x + 5/7
- Implicit: x + 7y - 5 = 0
- Intercept: x/5 + y/(5/7) = 1 | |
02ij | Problem:
Na figura, os três círculos são concêntricos, e as áreas do menor círculo e do maior anel (em cinza) são iguais. O raio do menor círculo é $5~\mathrm{cm}$ e do maior $13~\mathrm{cm}$. Qual o raio do círculo intermediário?
(A) 12
(B) 11
(C) $10 \sqrt{65}$
(D) $5 \sqrt{3}$
(E) $12 \sqrt{2}$
![](attached_image... | [
"Solution:\n\nA área do menor círculo é $5^2 \\pi = 25\\pi~\\mathrm{cm}^2$ e do maior é $13^2 \\pi = 169\\pi~\\mathrm{cm}^2$. Seja $r$ o raio do círculo intermediário, então a área do maior anel é $169\\pi - \\pi r^2$. Logo,\n$$\n169\\pi - \\pi r^2 = 25\\pi \\Rightarrow r^2 = 169 - 25 = 144,\n$$\ndonde $r = 12~\\ma... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | A | |
04k5 | Gargamel caught $N$ Smurfs and distributed them into three bags. After moving Papa Smurf from the first bag to the second, Grouchy Smurf from the second bag to the third, and Smurfette from the third bag to the first one, the average Smurfs' height in the first bag decreased by 8 millimetres, while the average Smurfs' ... | [] | Croatia | Mathematical competitions in Croatia | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 21 | |
0esx | Let $n$ be a positive integer, and let $x_1, x_2, \dots, x_n$ be distinct positive integers with $x_1 = 1$. Construct an $n \times 3$ table where the entries of the $k$-th row are $x_k, 2x_k, 3x_k$ for $k = 1, 2, \dots, n$. Now follow a procedure where, in each step, two identical entries are removed from the table. Th... | [
"a.\nWe write $X$ as a shorthand for the set $\\{x_1, x_2, \\dots, x_n\\}$. Note that the number $1$ can only appear in the first row, so it remains at the end of the procedure. Next, let $x_k$ be the largest power of $2$ that occurs in $X$. This is possible since $x_1 = 1$ is a power of $2$. The number $2x_k$ in t... | South Africa | The South African Mathematical Olympiad Third Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0ao9 | Problem:
A fair die is thrown three times. What is the probability that the largest outcome of the three throws is a $3$?
(a) $1/36$
(b) $19/216$
(c) $25/36$
(d) $1/8$ | [] | Philippines | QUALIFYING STAGE | [
"Statistics > Probability > Counting Methods > Other"
] | null | MCQ | b | |
0gso | Find all pairs $(m, n)$ of positive integers such that both
$$
\frac{3n^2}{m} \quad \text{and} \quad \sqrt{n^2 + m}
$$
are integers. | [
"Answer: All pairs $(3n^2, n)$, where $n$ is a positive integer.\n\nLet $k = \\frac{3n^2}{m}$ and $\\sqrt{n^2+m} = \\frac{n\\sqrt{k+3}}{\\sqrt{k}}$ both should be integers. Since $k$ is integer, $\\frac{n\\sqrt{k+3}}{\\sqrt{k}} \\cdot k = n\\sqrt{k(k+3)}$ is also integer. Therefore, $k(k+3)$ is a perfect square. If... | Turkey | Junior Turkish Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (m, n) = (3n^2, n) for all positive integers n | |
01v3 | Let $q \in \mathbb{R}$. There are 10 distinct real numbers on the blackboard. Alex writes the following three lines of numbers:
1. In the first line Alex writes down every number of the form $a-b$, where $a, b$ are two (not necessarily distinct) numbers from the board;
2. In the second line Alex writes down every num... | [
"2. See IMO-2017 Shortlist, Problem A2."
] | Belarus | Selection and Training Session | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | q = 2 or q = -2 | |
0h3v | A real number sequence $a_1, a_2, \dots, a_n, \dots$ is such that $a_1 = 1$, $a_2 = 9$ and $a_{n+2} = 14a_{n+1} - a_n - 4$ for all positive integers $n$. Prove that for each positive integer $n$ the number $a_n$ is a square of integer. | [
"З умови задачі випливає, що всі члени даної послідовності є цілими числами, причому $a_3 = 14 \\cdot 9 - 1 - 4 = 121$. Доведемо індукцією за $n \\ge 2$, що $a_{n-1}a_{n+1} = (a_n + 2)^2$. Справді, для $n = 2$ маємо: $a_1a_3 = 1 \\cdot 121 = (9+2)^2 = (a_n + 2)^2$. Припустимо тепер, що $a_{n-1}a_{n+1} = (a_n + 2)^2... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
06pz | Prove that for any four positive real numbers $a$, $b$, $c$, $d$ the inequality
$$
\frac{(a-b)(a-c)}{a+b+c}+\frac{(b-c)(b-d)}{b+c+d}+\frac{(c-d)(c-a)}{c+d+a}+\frac{(d-a)(d-b)}{d+a+b} \geq 0
$$
holds. Determine all cases of equality. | [
"Solution 1. Denote the four terms by\n$$\nA=\\frac{(a-b)(a-c)}{a+b+c}, \\quad B=\\frac{(b-c)(b-d)}{b+c+d}, \\quad C=\\frac{(c-d)(c-a)}{c+d+a}, \\quad D=\\frac{(d-a)(d-b)}{d+a+b} .\n$$\nThe expression $2A$ splits into two summands as follows,\n$$\n2A=A' + A'' \\quad \\text{ where } \\quad A' = \\frac{(a-c)^2}{a+b+c... | IMO | 49th International Mathematical Olympiad Spain | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | Equality holds if and only if a = c and b = d. | |
0c60 | Let $P$ be a point in the interior of the triangle $ABC$, such that $BP = AC$. Let $M$ and $R$ be the midpoints of line segments $AP$ and $BC$, respectively, and let $BP \cap AC = \{E\}$. Prove that the angle bisector of $\angle BEA$ is perpendicular to $MR$. | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null |
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