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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0fik | Problem:
El baricentro del triángulo $ABC$ es $G$. Denotemos por $g_{a}, g_{b}, g_{c}$ las distancias desde $G$ a los lados $a, b, c$, respectivamente. Sea $r$ el radio de la circunferencia inscrita.
a) Probar que
$$
g_{a} \geq \frac{2r}{3}, \quad g_{b} \geq \frac{2r}{3}, \quad g_{c} \geq \frac{2r}{3}
$$
b) Probar q... | [
"Solution:\n\nPrimera solución\n\ni) Es sabido que uniendo $G$ con cada vértice, se forman tres triángulos: $BGC$ de base $a$ y altura $g_{a}$, $AGC$ de base $b$ y altura $g_{b}$ y $AGB$ de base $c$ y altura $g_{c}$. Los tres tienen la misma área que es un tercio del área total del triángulo.\nPor tanto, llamando $... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Ineq... | null | proof only | null | |
0bmt | a) Prove that if $p$, $q$, $\sqrt{2p-q}$ and $\sqrt{2p+q}$ are integers, then $q$ is even.
b) Find out how many positive integers $p$ have the property that both $\sqrt{2p-4030}$ and $\sqrt{2p+4030}$ are integers. | [
"a) We know that $2p - q = k^2$, $2p + q = r^2$, hence $r^2 - k^2 = 2q$, with $k$, $r$ positive integers. Then $(r - k)(r + k) = 2q$, and the conclusion follows from the fact that $r - k$ and $r + k$ have the same parity.\n\nb) Answer: four numbers.\nWith the above notations, $(r - k)(r + k) = 2 \\cdot 4030 = 2^2 \... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 4 | |
0kqd | Problem:
Let $a \neq b$ be positive real numbers and $m, n$ be positive integers. An $m+n$-gon $P$ has the property that $m$ sides have length $a$ and $n$ sides have length $b$. Further suppose that $P$ can be inscribed in a circle of radius $a+b$. Compute the number of ordered pairs $(m, n)$, with $m, n \leq 100$, for... | [
"Solution:\nLetting $x=\\frac{a}{a+b}$, we have to solve\n$$\nm \\arcsin \\frac{x}{2} + n \\arcsin \\frac{1-x}{2} = \\pi\n$$\nThis is convex in $x$, so if it is to have a solution, we must find that the LHS exceeds $\\pi$ at one of the endpoints. Thus $\\max (m, n) \\geq 7$. If $\\min (m, n) \\leq 5$ we can find a ... | United States | HMMT February | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | 940 | |
06pc | Let $n > 1$ be an integer. Find all sequences $a_{1}, a_{2}, \ldots, a_{n^{2}+n}$ satisfying the following conditions:
(a) $a_{i} \in \{0,1\}$ for all $1 \leq i \leq n^{2}+n$;
(b) $a_{i+1}+a_{i+2}+\ldots+a_{i+n} < a_{i+n+1}+a_{i+n+2}+\ldots+a_{i+2n}$ for all $0 \leq i \leq n^{2}-n$. | [
"Consider a sequence ($a_{i}$) satisfying the conditions. For arbitrary integers $0 \\leq k \\leq l \\leq n^{2}+n$ denote $S(k, l] = a_{k+1} + \\cdots + a_{l}$. (If $k = l$ then $S(k, l] = 0$.) Then condition (b) can be rewritten as $S(i, i+n] < S(i+n, i+2n]$ for all $0 \\leq i \\leq n^{2}-n$. Notice that for $0 \\... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | Partition the indices into n+1 consecutive blocks of length n. For block v (0 ≤ v ≤ n), the first n−v entries are 0 and the last v entries are 1. Equivalently, writing any index as u + v n with 1 ≤ u ≤ n and 0 ≤ v ≤ n, the sequence is given by a_{u+vn} = 0 if u + v ≤ n and a_{u+vn} = 1 if u + v ≥ n + 1. | |
0i2q | Problem:
Boris was given a Connect Four game set for his birthday, but his color-blindness makes it hard to play the game. Still, he enjoys the shapes he can make by dropping checkers into the set. If the number of shapes possible modulo (horizontal) flips about the vertical axis of symmetry is expressed as $9(1+2+\cd... | [
"Solution:\n\nThere are $9^{7}$ total shapes possible, since each of the 7 columns can contain anywhere from 0 to 8 checkers. The number of shapes symmetric with respect to a horizontal flip is the number of shapes of the leftmost four columns, since the configuration of these four columns uniquely determines the c... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 729 | |
098r | Problem:
Fie numărul complex $z=\frac{1}{2}-\frac{\sqrt{3}}{4}+\frac{1}{4} i$. Arătați că valoarea raportului $\sqrt{3} \cdot \frac{\operatorname{Re}\left(z^{2024}\right)}{\operatorname{Im}\left(z^{2024}\right)}$ este un număr rațional. | [
"Solution:\n\n$z=\\frac{1}{2}\\left(1-\\frac{\\sqrt{3}}{2}+\\frac{1}{2} i\\right)=\\frac{1}{2}\\left(1-\\cos \\frac{\\pi}{6}+i \\sin \\frac{\\pi}{6}\\right)=\\frac{1}{2}\\left(2 \\sin ^{2} \\frac{\\pi}{12}+2 i \\sin \\frac{\\pi}{12} \\cos \\frac{\\pi}{12}\\right)=$\n$$\n\\begin{gathered}\n=\\sin \\frac{\\pi}{12}\\l... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
09o0 | Let $ABCD$ be a convex quadrilateral with $\angle ABC = \angle ADC = 90^\circ$, and diagonals intersecting at $P$. A line through $P$ parallel to $AD$ intersects $BC$ at $K$, and a line through $P$ parallel to $AB$ intersects $DC$ at $L$. Let $\omega$ be the circumcircle of triangle $KLC$. Let $AC$ intersect $\omega$ a... | [] | Mongolia | MMO2025 Round 2 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0gmq | Prove that
$$
\sqrt{a^4 + c^4} + \sqrt{a^4 + d^4} + \sqrt{b^4 + c^4} + \sqrt{b^4 + d^4} \ge 2\sqrt{2}(ad + bc)
$$
for all real numbers $a$, $b$, $c$, $d$. | [] | Turkey | XIII. National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
07x8 | A sequence of real numbers $(x_n)_{n=0}^{\infty}$ is *concave* if
$$
x_n \ge \frac{x_{n-1} + x_{n+1}}{2}, \quad \text{for all } n \in \mathbb{N}.
$$
Given two sequences of real numbers $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=0}^{\infty}$, we define a new sequence $(c_n)_{n=0}^{\infty}$ by:
$$
c_n = \max\{a_i + b_{n-i} \mi... | [
"We claim that concavity of $(a_n)$ implies that\n$$\na_i + a_j \\le 2a_{(i+j)/2}, \\quad \\text{for all non-negative } i, j \\in \\mathbb{Z}, \\qquad (31)\n$$\nwhere $a_r$ is defined for $r$ halfway between two integers as follows:\n$$\na_{n-1/2} := \\frac{a_{n-1} + a_n}{2} \\quad \\text{for all } n \\in \\mathbb{... | Ireland | IRL_ABooklet_2024 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof only | null | |
075c | Problem:
Let $ABCD$ be a quadrilateral inscribed in a circle. Suppose $AB = \sqrt{2+\sqrt{2}}$ and $AB$ subtends $135^{\circ}$ at the centre of the circle. Find the maximum possible area of $ABCD$.
 | [
"Solution:\nLet $O$ be the centre of the circle in which $ABCD$ is inscribed and let $R$ be its radius. Using cosine rule in triangle $AOB$, we have\n$$\n2+\\sqrt{2}=2R^{2}\\left(1-\\cos 135^{\\circ}\\right)=R^{2}(2+\\sqrt{2})\n$$\nHence $R=1$.\n\nConsider quadrilateral $ABCD$ as in the second figure above. Join $A... | India | INMO | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Jensen/smoothing",
"Geomet... | null | proof and answer | (5+3√3)/(4√2) | |
06ub | Let $n$ be a positive integer. Define a chameleon to be any sequence of $3n$ letters, with exactly $n$ occurrences of each of the letters $a$, $b$, and $c$. Define a swap to be the transposition of two adjacent letters in a chameleon. Prove that for any chameleon $X$, there exists a chameleon $Y$ such that $X$ cannot b... | [
"To start, notice that the swap of two identical letters does not change a chameleon, so we may assume there are no such swaps.\nFor any two chameleons $X$ and $Y$, define their distance $d(X, Y)$ to be the minimal number of swaps needed to transform $X$ into $Y$ (or vice versa). Clearly, $d(X, Y)+d(Y, Z) \\geqslan... | IMO | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0h72 | Let $P(n)$ and $Q(n)$ be two polynomials (not constant) with positive integer or zero coefficients. For $n \ge 1$ we determine the sequence $x_n = 2016^{P(n)} + Q(n)$. Prove that there are infinitely many prime numbers $p$ for which there is a positive integer $m$ not divisible by the square of any prime, and for which... | [
"Suppose that it is not so. Consider the sequence $(x_n)$ of all members, the indices of which do not contain in their decomposition on factors the squares of prime numbers. Then in this subsequence there is only a finite number of prime divisors $p_1, p_2, \\dots, p_k$ other than $2$, $3$, $7$ (prime factors of th... | Ukraine | UkraineMO | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
046y | Let $p$ be an odd prime number, and $a$, $b$, $m$, $r$ be positive integers, such that $p \nmid ab$ and $ab > m^2$. Prove that there exists at most one pair of positive integers $(x, y)$ satisfying the following conditions: $x$ and $y$ are coprime, and $ax^2 + by^2 = mp^r$. | [
"**Proof:** By contradiction, suppose there exist two different pairs of positive integer solutions $(x_1, y_1)$ and $(x_2, y_2)$. Since $x_1$ and $y_1$ are coprime, it follows that $p \\nmid x_1y_1$. Similarly, $p \\nmid x_2y_2$. Given\n$$\nax_1^2 \\equiv -by_1^2 \\pmod{p^r}, \\quad ax_2^2 \\equiv -by_2^2 \\pmod{p... | China | 22nd Chinese Girls' Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0fmn | Una configuración de $4027$ puntos del plano, de los cuales $2013$ son rojos y $2014$ azules, y no hay tres de ellos que sean colineales, se llama colombiana. Trazando algunas rectas, el plano queda dividido en varias regiones. Una colección de rectas es buena para una configuración colombiana si se cumplen las dos sig... | [
"Consideremos un polígono regular de $4027$ lados con vértices $P_1, P_2, \\dots, P_{4027}$ numerados en el sentido de las agujas del reloj, y tales que $P_i$ es rojo si $i$ es par, y $P_i$ es azul si $i$ es impar. Claramente, los vértices de este polígono regular forman una configuración colombiana. Sea una colecc... | Spain | Olimpiada Internacional de Matemáticas | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | Spanish | proof and answer | 2013 | |
064o | Determine the natural numbers $v$ for which $2007 + 4v$ is perfect square. | [
"Let $2007 + 4v = \\kappa^2$, $\\kappa \\in \\mathbb{N}$. Then\n$$\n\\kappa^2 - 4v = 2007 \\Leftrightarrow \\kappa^2 - 2^{2v} = 2007 \\Leftrightarrow (\\kappa - 2^v)(\\kappa + 2^v) = 1 \\cdot 3 \\cdot 3 \\cdot 223.\n$$\nSince $\\kappa - 2^v < \\kappa + 2^v$, last equality is equivalent to the systems\n$$\n\\begin{c... | Greece | 24th Hellenic Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | none | |
0jtk | Problem:
The numbers $1$-$10$ are written in a circle randomly. Find the expected number of numbers which are at least $2$ larger than an adjacent number.
Proposed by: Shyam Narayanan | [
"Solution:\nFor $1 \\leq i \\leq 10$, let $X_{i}$ be the random variable that is $1$ if the $i$ in the circle is at least $2$ larger than one of its neighbors, and $0$ otherwise. The random variable representing number of numbers that are at least $2$ larger than one of their neighbors is then just $X_{1}+X_{2}+\\c... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 17/3 | |
00pe | Let $A_0, A_1, \dots, A_5$ be a circular labelling of six distinct points on a circle centred at the point $O$, and let $B_i$ be the midpoint of the segment $A_iA_{i+1}$, $i = 0, 1, \dots, 5$ (indices are reduced modulo $6$). Assume that no opposite sides of the hexagon $A_0A_1\cdots A_5$ are parallel. A line through t... | [
"Let $\\gamma$ be the circle through the $A_i$, centred at $O$, and let tangents to $\\gamma$ at $A_i$ and $A_{i+1}$ meet at $B'_i$. Notice that the line $B'_iB'_{i+3}$ is the image of the circle $B_iOB_{i+3}$ under the inversion of pole $O$ and power $r^2$, where $r$ is the radius of $\\gamma$. By Brianchon's theo... | Balkan Mathematical Olympiad | shortlistBMO 2011 | [
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06ht | Find the greatest positive integer $k$ for which the following holds: For all positive real numbers $a$, $b$ and $c$ with $abc = 1$, we have the inequality
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{k}{a+b+c+1} \ge \frac{k}{4} + 3.
$$ | [
"The greatest $k$ is $13$.\n\nFirstly, consider $a = b = \\frac{2}{3}$ and $c = \\frac{9}{4}$. They satisfy $abc = 1$. The inequality becomes\n$$\n\\frac{3}{2} + \\frac{3}{2} + \\frac{4}{9} + \\frac{12k}{55} \\ge \\frac{k}{4} + 3,\n$$\ni.e. $\\frac{4}{9} \\ge \\frac{7k}{220}$. As $k$ is a positive integer, this imp... | Hong Kong | IMO HK TST | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 13 | |
0a24 | For a positive integer $n$, we define $\alpha(n)$ as the average of all positive divisors of $n$, and $\beta(n)$ as the average of all positive integers $k \le n$ such that $\gcd(k, n) = 1$.
Find all positive integers $n$ for which $\alpha(n) = \beta(n)$. | [
"**Answer:** $n = 1$ and $n = 6$.\n\nWe first note that $n = 1$ satisfies.\n\nWe now prove that $\\beta(n) = \\frac{n}{2}$ for $n \\ge 2$. Indeed, $\\gcd(k, n) = \\gcd(k - n, n) = \\gcd(n - k, n)$, so $\\gcd(k, n) = 1$ if and only if $\\gcd(n - k, n) = 1$. This yields a partition of the positive integers $1 \\le k ... | Netherlands | IMO Team Selection Test 1 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)"
] | English | proof and answer | 1 and 6 | |
04sl | For positive real numbers $a$, $b$, $c$ the following holds:
$$ab + bc + ca = 16, \quad a \ge 3.$$
Find the smallest possible value of the expression $2a + b + c$. (Michal Rolínek) | [] | Czech Republic | Czech and Slovak Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 10 | |
073x | Find all positive integers $a, b$ such that $(a-b)^{a+b} = a^a$. | [
"Let $d = \\gcd(a, b)$. Write $a = dx$ and $b = dy$. Then the equation reduces to $d^b(x - y)^{a+b} = x^a$. Since $\\gcd(x, x-y) = 1$, it follows that $x - y = \\pm 1$ and $d^b = x^a$. Using $b = dx$ and $a = dy$, this further reduces to $dy = x^x$ or\n$$\nd^{x \\pm 1} = x^x\n$$\nNote that $d = 1 = x$ does not lead... | India | Indija TS 2009 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (a, b) = (8, 4) | |
0ijs | Problem:
Vernonia High School has 85 seniors, each of whom plays on at least one of the school's three varsity sports teams: football, baseball, and lacrosse. It so happens that 74 are on the football team; 26 are on the baseball team; 17 are on both the football and lacrosse teams; 18 are on both the baseball and foo... | [
"Solution:\n\nSuppose that $n$ seniors play all three sports and that $2n$ are on the lacrosse team. Then, by the principle of inclusion-exclusion,\n$$\n85 = (74 + 26 + 2n) - (17 + 18 + 13) + n = 100 + 2n - 48 + n = 52 + 3n.\n$$\nIt is easily seen that $n = 11$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 11 | |
0dli | Let $n > 2$ be an integer and let $S$ be a set of $k$ points in the plane whose coordinates both lie in $\{1, 2, \dots, n\}$. Find the minimal number of $k$ for which one can choose four points from $S$ that are the vertices of a nondegenerate parallelogram. | [
"If $k = 2n - 1$, we can choose all points with $x$-coordinate, $y$-coordinate are $1$. Then, clearly there is no parallelogram. Thus, any $k \\le 2n-1$ does not satisfy the condition.\n\nNext, we will prove that $k = 2n$ satisfies. Indeed, let $x_1, x_2, ..., x_n$ be the number of points whose $x$-coordinate are $... | Saudi Arabia | Saudi Booklet | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2n | |
0adc | Find all three digit numbers equal to the sum of the factorials of their digits. | [
"$100a + 10b + c = a! + b! + c!$\n$$\na! + b! + c! < 1000 \\Rightarrow \\max(a, b, c) \\le 6 \\\\\n(7! > 1000,\\ 6! < 1000)\n$$\nBecause $a! + b! + c! > 99$, one of the digits $a$, $b$, or $c$ has to be equal to $5$ or $6$ (since $4! + 4! + 4! < 99$).\nIf one of the digits is equal to $6$, then $a! + b! + c! > 6! =... | North Macedonia | Macedonian Mathematical Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 145 | |
0b1w | Problem:
In convex pentagon $A B C D E$, $A B = B C$, $C D = D E$, $\angle A B C = 100^{\circ}$, $\angle C D E = 80^{\circ}$, and $B D^{2} = \frac{100}{\sin 100^{\circ}}$. Find the area of the pentagon. | [
"Solution:\n\nLet $A B = B C = p$, $A C = r$, $C D = D E = q$, $C E = s$, and $\\theta = \\angle A C E$. Then $r = 2 p \\cos 40^{\\circ}$, $s = 2 q \\cos 50^{\\circ}$, and\n$$\n\\begin{aligned}\n\\frac{100}{\\sin 100^{\\circ}} & = B D^{2} = p^{2} + q^{2} - 2 p q \\cos \\left(90^{\\circ} + \\theta\\right) \\\\\n& = ... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 50 | |
04nn | Find the last two digits of the number $(1!)^2 + (2!)^2 + \dots + (2018!)^2$. | [
"For $n \\geq 10$, $n!$ is divisible by $10$, so $(n!)^2$ is divisible by $100$. Therefore, only the terms $(1!)^2, (2!)^2, \\dots, (9!)^2$ contribute to the last two digits.\n\nCompute each term:\n\n$(1!)^2 = 1^2 = 1$\n\n$(2!)^2 = 2^2 = 4$\n\n$(3!)^2 = 6^2 = 36$\n\n$(4!)^2 = 24^2 = 576$\n\n$(5!)^2 = 120^2 = 14400$... | Croatia | Croatia_2018 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | final answer only | 17 | |
00y7 | Problem:
Find the smallest number $a$ such that a square of side $a$ can contain five disks of radius $1$ so that no two of the disks have a common interior point. | [
"Solution:\nLet $PQRS$ be a square which has the property described in the problem. Clearly, $a > 2$. Let $P'Q'R'S'$ be the square inside $PQRS$ whose sides are at distance $1$ from the sides of $PQRS$, and, consequently, are of length $a - 2$. Since all the five disks are inside $PQRS$, their centres are inside $P... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 2 + 2√2 | |
05a5 | Let $p$ be a fixed prime number. Juku and Miku play the following game. One of the players chooses a natural number $a$ such that $a > 1$ and $a$ is not divisible by $p$, his opponent chooses any natural number $n$ such that $n > 1$. Miku wins if the natural number written as $n$ ones in the positional numeral system w... | [
"*Answer*: (a) Miku; (b) Juku.\n\nThe positional representation with radix $a$ consisting of $n$ ones denotes the sum $a^{n-1} + a^{n-2} + \\dots + a + 1$ which equals $\\frac{a^n-1}{a-1}$.\n\na. Let $a \\equiv 1 \\pmod{p}$. If $p > 2$ or $a \\equiv 1 \\pmod{4}$ then, by the lifting-the-exponent lemma, the exponent... | Estonia | Estonian Math Competitions | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | a) Miku; b) Juku | |
0kw2 | Problem:
Let $s(n)$ denote the sum of the digits (in base ten) of a positive integer $n$. Compute the number of positive integers $n$ at most $10^{4}$ that satisfy
$$
s(11 n)=2 s(n)
$$ | [
"Solution:\nNote $2 s(n)=s(10 n)+s(n)=s(11 n)$, so there cannot be any carries when adding $n$ and $10 n$. This is equivalent to saying no two consecutive digits of $n$ sum to greater than $9$.\n\nWe change the problem to nonnegative integers less than $10^{4}$ (as both $0$ and $10^{4}$ satisfy the condition) so th... | United States | HMMT November 2023 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2530 | |
0d4x | Let $ABC$ be a triangle, $I$ its incenter, and $D$ a point on the $\operatorname{arc}\overparen{BC}$ of the circumcircle of $ABC$ not containing $A$. The bisector of the angle $\angle ADB$ intersects the segment $AB$ at $E$. The bisector of the angle $\angle CDA$ intersects the segment $AC$ at $F$. Prove that the point... | [
"First solution. Let $P$ be the intersection point of the bisector of the angle $\\angle BAC$ with the segment $EF$. Because both $I$ and $P$ are on the bisector of $\\angle BAC$, the problem is equivalent to showing that $AP = AI$.\nBecause $DE$ is the bisector of $\\angle ADB$ we have from the bisector theorem\n$... | Saudi Arabia | SAMC 2015 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurre... | English, Arabic | proof only | null | |
0dte | Let $ABC$ be an isosceles right-angled triangle of area $1$. Find the length of the shortest segment that divides the triangle into $2$ parts of equal area. | [
"There are $2$ cases as shown in the figures below:\n\nFig.1\n\nFig.2\n\nIn Fig. 1,\n$$\nXY^2 = AX^2 + AY^2 = (AX - AY)^2 + 2AX \\cdot AY = (AX - AY)^2 + 2.\n$$\nThus $XY$ is minimum when $AX = AY$ and so $XY^2 = 2$.\n\nIn Fig. 2, let $B$ be the origin and $BA$ be t... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | sqrt(2*sqrt(2) - 2) | |
0ft2 | Problem:
Zeige, dass jede 1000-elementige Teilmenge $M \subset \{0,1, \ldots, 2001\}$ eine Zahl enthält, die eine Zweierpotenz ist, oder zwei verschiedene Zahlen, deren Summe eine Zweierpotenz ist. | [
"Solution:\nWir können annehmen, dass $M$ keine Zweierpotenz enthält (sonst ist alles klar). Wir ignorieren die Null und zeigen: Ist $M \\subset \\{1, \\ldots, 2001\\}$ eine Menge mit mindestens 999 Elementen, die keine Zweierpotenz enthält, dann gibt es zwei verschiedene Elemente in $M$, deren Summe eine Zweierpot... | Switzerland | IMO - Selektion | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0jdm | Problem:
Rosencrantz and Guildenstern each start with $\$ 2013$ and are flipping a fair coin. When the coin comes up heads Rosencrantz pays Guildenstern $\$ 1$ and when the coin comes up tails Guildenstern pays Rosencrantz $\$ 1$. Let $f(n)$ be the number of dollars Rosencrantz is ahead of his starting amount after $n... | [
"Solution:\n\nAnswer: $\\frac{-1}{2}+\\frac{(1007)\\binom{2013}{1006}}{2^{2012}}$\n\nWe want to calculate $\\Gamma=\\sum_{i=0}^{\\infty} i \\cdot P(\\max$ profit $=i)$, where we consider the maximum profit Rosencrantz has at any point over the first 2013 coin flips. By summation by parts this is equal to $\\sum_{a=... | United States | HMMT | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation"
] | null | proof and answer | (-1)/2 + (1007)*binom(2013,1006)/2^2012 | |
0b7e | Let $n$ be an integer, $n \ge 2$. For each $k = 1, 2, \dots, n$, let $a_k$ be the number of multiples of $k$ in the set $\{1, 2, \dots, n\}$, and let $x_k = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{a_k}$. Show that $\frac{x_1 + x_2 + \dots + x_n}{n} \le \frac{1}{1^2} + \frac{1}{2^2} + \dots + \frac{1}... | [
"The number $a_k$ of the multiples of $k$ in the set $\\{1, 2, \\dots, n\\}$ is $\\lfloor \\frac{n}{k} \\rfloor$. Let us count the appearances of the term $\\frac{1}{i}$ in the sum $x_1 + x_2 + \\dots + x_n$. The term $\\frac{1}{i}$ belongs in the sum $x_k = \\frac{1}{1} + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\f... | Romania | NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
0eq8 | If $\frac{n+3}{13}$ is a positive integer, then the remainder when $n$ is divided by $13$ is | [
"10 Suppose $\\frac{n+3}{13} = k$; then $n = 13k - 3$ or $n = 13(k-1) + 10$. This shows that $n$ leaves remainder $10$ on division by $13$."
] | South Africa | South African Mathematics Olympiad | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 10 | |
08mf | Problem:
Let $a, b, c$ be positive real numbers such that $a b c(a+b+c)=3$. Prove the inequality
$$
(a+b)(b+c)(c+a) \geq 8
$$
and determine all cases when equality holds. | [
"Solution:\nWe have\n$A=(a+b)(b+c)(c+a)=\\left(a b+a c+b^{2}+b c\\right)(c+a)=(b(a+b+c)+a c)(c+a)$,\nso by the given condition\n$$\nA=\\left(\\frac{3}{a c}+a c\\right)(c+a)=\\left(\\frac{1}{a c}+\\frac{1}{a c}+\\frac{1}{a c}+a c\\right)(c+a)\n$$\nApplying the AM-GM inequality for four and two terms respectively, we... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | The minimum value is 8, achieved only when a = b = c = 1. | |
0hao | In triangle $ABC$ there are points $M_1, M_2, M_3$ – midpoints of sides $BC$, $AC$, $AB$ correspondingly. Point $K$ is symmetrical to $M_2$ with respect to line $BC$, $AH$ is the altitude of triangle $ABC$. Prove that line $KM_3$ splits the segment $HM_1$ in half. | [
"From right-angled triangles $ABH$ and $ACH$ we get that $M_3H = \\frac{1}{2}AB$, $M_2H = \\frac{1}{2}AC$ (Fig. 42).\n\nAlso $M_1M_2 = \\frac{1}{2}AB$, $M_1M_3 = \\frac{1}{2}AC$ as centerlines. From symmetry $M_2M_1 = M_1K$ and $M_2H = HK$. So $M_3H = M_1K$ and $M_3M_1 = HK$, so quadrilateral $HM_3M_1K$ is a parall... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
06ta | Let $2 \mathbb{Z} + 1$ denote the set of odd integers. Find all functions $f: \mathbb{Z} \rightarrow 2 \mathbb{Z} + 1$ satisfying
$$
f(x + f(x) + y) + f(x - f(x) - y) = f(x + y) + f(x - y)
$$
for every $x, y \in \mathbb{Z}$. | [
"Throughout the solution, all functions are assumed to map integers to integers.\nFor any function $g$ and any nonzero integer $t$, define\n$$\n\\Delta_{t} g(x) = g(x + t) - g(x)\n$$\nFor any nonzero integers $a$ and $b$, notice that $\\Delta_{a} \\Delta_{b} g = \\Delta_{b} \\Delta_{a} g$. Moreover, if $\\Delta_{a}... | IMO | 56th International Mathematical Olympiad Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof and answer | All solutions are as follows: choose an odd integer d, an integer k, and odd integers ℓ_0, ℓ_1, …, ℓ_{d−1}. For every integer x written uniquely as x = m d + i with m ∈ ℤ and i ∈ {0, 1, …, d−1}, define
f(m d + i) = d(2 k m + ℓ_i).
Equivalently, Δ_d f(x) = f(x + d) − f(x) = 2 d k is constant, and for each residue clas... | |
06i5 | Given a triangle $ABC$ with $AB = BC = 1$ and $CA = \sqrt{2}$, $P$ is a point inside the triangle $ABC$ such that $\angle PAB = \angle PBC = \angle PCA$. Find $BP$. | [
"We have $BP = \\frac{1}{\\sqrt{5}}$.\nNote that $\\angle CBA = 90^\\circ$ and $AB = BC$. Since $\\angle CBP = \\angle BAP$, the line $BC$ is tangent to ($ABP$). As $\\angle CBA = 90^\\circ$, the centre of this circle must lie on $AB$. Thus, the centre is the midpoint $M$ of $AB$. This implies $\\angle APB = 90^\\c... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 1/sqrt(5) | |
04a5 | Let $z$ be a root of the polynomial $z^2 - 2z \cos \frac{\pi}{n} + 1$. Determine all the possible values of $z^n$. | [] | Croatia | Hrvatska 2011 | [
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | -1 | |
0hj9 | Problem:
Determine whether there exists a natural number having exactly $10$ divisors (including itself and $1$), each ending in a different digit. | [
"Solution:\n\nThe answer is no.\nSuppose that such a number $n$ exists. Since $n$ has a divisor ending in $5$, $n$ is divisible by $5$. Then for each divisor $d$ of $n$ not divisible by $5$, $5d$ also divides $n$. This contradicts the assumption that $n$ has eight divisors ending in $1,2,3,4,6,7,8$, or $9$ and only... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | no | |
01yg | Let $n$ be a positive integer. On the segment $[0, n]$ of the real line there are marked $m$ pairwise distinct segments with integer endpoints. It is known that it's impossible to choose a set of these segments such that the sum of their lengths is $n$ and their union is $[0, n]$. (We say that the two segments are dist... | [
"Answer: $\\frac{n(n-1)}{2}$.\n\nEach set of segments that satisfies the condition of the problem is called *good*. Consider a set consisting of all segments of the form $[a, b]$, where $a, b \\in \\mathbb{N}$ and $a < b \\le n$. This collection contains $n(n-1)/2$ segments and it is good since even the union of al... | Belarus | Belarus2022 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | n(n-1)/2 | |
03t3 | Let $M = \{1, 2, \dots, 19\}$ and $A = \{a_1, a_2, \dots, a_k\} \subseteq M$. Find the minimum value of $k$ so that there exist $a_i, a_j \in A$ such that if $b \in M$, then $a_i = b$ or $a_i \pm a_j = b$. | [
"By the definition of $A$, we have $k(k+1) \\ge 19$, implying $k \\ge 4$.\n\nIf $k=4$, we have $k(k+1)=20$. We may assume that $a_1 < a_2 < a_3 < a_4$. Then, $a_4 \\ge 10$.\n(1) If $a_4 = 10$, then $a_3 = 9$, and we have $a_2 = 8$ or $7$. If $a_2 = 8$, then $20, 10-9=1, 9-8=1$, impossible. If $a_2 = 7$, then $a_1 =... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 5 | |
0hkr | Problem:
Determine, with proof, the value of
$$
1^{2}-2^{2}+3^{2}-4^{2}+5^{2}-\cdots+97^{2}-98^{2}+99^{2} .
$$ | [
"Solution:\nObserve that $3^{2}-2^{2}=(3-2)(3+2)=3+2$, $5^{2}-4^{2}=(5-4)(5+4)=5+4$, and so on. Thus this sum, call it $S$, is actually equal to\n$$\nS=1+(2+3)+(4+5)+\\cdots+(97+98)+99 .\n$$\nWe can also write it in reverse as\n$$\nS=99+98+97+96+95+\\cdots+3+2+1 .\n$$\nAdding these two, we get\n$$\n2 S=\\underbrace... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 4950 | |
0fog | Para cada entero positivo $n$, el Banco de Ciudad del Cabo produce monedas de valor $\frac{1}{n}$. Dada una colección finita de tales monedas (no necesariamente de distintos valores) cuyo valor total no supera $99 + \frac{1}{2}$, demostrar que es posible separar esta colección en 100 o menos montones, de modo que el va... | [
"**Solución por Daniel Lasaosa Medarde, Pamplona, España.** Antes de empezar a particionar el conjunto de monedas, vayamos a la ventanilla del Banco de Ciudad del Cabo y realicemos los siguientes cambios:\n\n* Para cada entero $n = 2m$ par, para el que haya al menos dos monedas de valor $\\frac{1}{n}$, tomemos dos ... | Spain | LV Olimpiada Internacional de Matemáticas | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof only | null | |
011a | Problem:
Let $a_{1}, a_{2}, \ldots, a_{n}$ be an arithmetic progression of integers such that $i$ divides $a_{i}$ for $i=1,2, \ldots, n-1$ and $n$ does not divide $a_{n}$. Prove that $n$ is a power of a prime. | [
"Solution:\nAssume $a_{i} = k + d i$ for $i = 1, 2, \\ldots, n$. Then $k$ is a multiple of every $i \\in \\{1, 2, \\ldots, n-1\\}$ but not a multiple of $n$. If $n = a b$ with $a, b > 1$ and $\\gcd(a, b) = 1$, then $k$ is divisible by both $a$ and $b$, but not by $n$, which is a contradiction. Hence, $n$ has only o... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0g59 | 令 $n$ 是大於 1 的正整數且 $p$ 為一質數使得 $n$ 能整除 $p-1$ 且 $p$ 能整除 $n^3-1$。
試證: $4p-3$ 是一完全平方數。 | [
"因 $n|(p-1)$, $n < p$。由於 $p|(n^3-1) = (n-1)(n^2+n+1)$, $p|(n^2+n+1)$。\n令 $p-1 = n\\ell$。又\n$$\nn^2 + n + 1 = n^2 + n + p - n\\ell = n(n + 1 - \\ell) + p\n$$\n即得 $p|n(n+1-\\ell)$。因 $n < p$, $p|(n+1-\\ell)$。又因 $\\ell > 0$, $n+1-\\ell \\le n$。今\n$p|(n+1-\\ell)$ 且 $n < p$, 得 $\\ell \\ge n+1$。若 $\\ell > n+1$, 則 $p = 1+n... | Taiwan | 二〇一一數學奧林匹亞競賽第一階段選訓營,獨立研究(三) | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0gi2 | Find all polynomials $P$ with real coefficients satisfying that there exist infinitely many pairs $(m, n)$ of coprime positive integers such that $P\left(\frac{m}{n}\right) = \frac{1}{n}$. | [
"Easy to prove that $P$ must have rational coefficients.\nIf $P$ has a degree $d \\ge 2$, consider $P(x) = \\frac{p}{q}x^d + \\frac{1}{M}Q(x)$, where $p, q \\in \\mathbb{Z}$, $Q(x)$ is a polynomial of degree $< d$ with integer coefficients. Suppose $P\\left(\\frac{m}{n}\\right) = \\frac{1}{n}$, $m, n$ coprime.\nWe ... | Taiwan | 2023 數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Chinese (Traditional) | proof and answer | All such polynomials are linear with rational coefficients: P(x) = a x + b, where a and b are rational, a·b ≤ 0, and writing a = p/q and b = r/s in lowest terms with positive denominators, the numerators satisfy gcd(p, r) = 1. | |
08p5 | Problem:
Let $x$, $y$, $z$ be positive real numbers such that $x + y + z = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$.
a) Prove the inequality
$$
x + y + z \geq \sqrt{\frac{x y + 1}{2}} + \sqrt{\frac{y z + 1}{2}} + \sqrt{\frac{z x + 1}{2}}
$$
b) When does the equality hold? | [
"Solution:\n\na. We rewrite the inequality as\n$$\n(\\sqrt{x y + 1} + \\sqrt{y z + 1} + \\sqrt{z x + 1})^2 \\leq 2 \\cdot (x + y + z)^2\n$$\nand note that, from CBS,\n$$\n\\text{LHS} \\leq \\left(\\frac{x y + 1}{x} + \\frac{y z + 1}{y} + \\frac{z x + 1}{z}\\right)(x + y + z)\n$$\nBut\n$$\n\\frac{x y + 1}{x} + \\fra... | JBMO | Junior Balkan Mathematics Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | Equality holds if and only if x = y = z = 1. | |
04sx | Let $ABC$ be a triangle with the shortest side $BC$. Let $X$, $Y$, $K$, $L$ be points on sides $AB$, $AC$ and on rays opposite to rays $BC$, $CB$ respectively such that $BX = BK = BC = CY = CL$. Line $KX$ intersects line $LY$ in a point $M$. Prove that centroid of triangle $KLM$ coincides with incentre of triangle $ABC... | [
"Since $ABC$ is an external angle of the isosceles triangle $XKB$ with the apex $B$ (see the picture), a line $KX$ is parallel to a bisectrix of the angle $ABC$.\n\n\n\nFig. 4\n\nThe ratio $LB : LK = 2 : 3$ yields, that the bisectrix of $ABC$ meets the centroid of the triangle $KLM$. If we ... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0bg0 | Problem:
Let $S$ be the set of positive real numbers. Find all functions $f: S^{3} \rightarrow S$ such that, for all positive real numbers $x, y, z$ and $k$, the following three conditions are satisfied:
(a) $x f(x, y, z)=z f(z, y, x)$,
(b) $f\left(x, y k, k^{2} z\right)=k f(x, y, z)$,
(c) $f(1, k, k+1)=k+1$.
Problem:... | [] | Romania | 30th Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | f(x,y,z) = (y/(2x)) * (1 + sqrt(1 + 4xz/y^2)) | |
0b38 | Problem:
Let $a$, $b$, $c$, and $d$ be real numbers such that $a \geq b \geq c \geq d$ and
$$
\begin{aligned}
a+b+c+d & = 13 \\
a^{2}+b^{2}+c^{2}+d^{2} & = 43
\end{aligned}
$$
Show that $ab \geq 3 + cd$. | [
"Solution:\n\nSince $(a-d)(b-c) \\geq 0$ and $(a-b)(c-d) \\geq 0$, then\n$$\nab + cd \\geq ac + bd \\geq ad + bc\n$$\nFrom Equations (4) and (5), we have\n$$\n\\begin{gathered}\n(ab + cd) + (ac + bd) + (ad + bc) = \\frac{1}{2}\\left[(a+b+c+d)^{2} - \\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)\\right] = \\\\\n\\frac{1}{2}... | Philippines | 23rd Philippine Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0efz | Problem:
Za točko $P$ znotraj trikotnika $ABC$ pravimo, da je izjemna, če velja naslednji pogoj:
Če so $A'$, $B'$ in $C'$ zaporedoma presečišča premic $AP$, $BP$ in $CP$ s stranicami $BC$, $AC$ in $AB$, lahko daljice $AA'$, $BB'$ in $CC'$ vzporedno premaknemo tako, da oblikujemo trikotnik, katerega dolžine stranic so ... | [
"Solution:\n\n\n\nOznačimo $\\vec{a}=\\overrightarrow{AB}$ in $\\vec{b}=\\overrightarrow{AC}$.\n\na.\nČe je $P$ težišče, tedaj je\n$$\n\\begin{aligned}\n& \\overrightarrow{AA'}=\\overrightarrow{AB}+\\frac{1}{2} \\overrightarrow{BC}=\\vec{a}+\\frac{1}{2}(-\\vec{a}+\\vec{b})=\\frac{1}{2} \\ve... | Slovenia | Slovenian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0ec5 | Let $ABCD$ be a quadrilateral which satisfies $\angle BAC = \angle ACB = 20^\circ$, $\angle DCA = 30^\circ$ and $\angle CAD = 40^\circ$. Determine the size of the angle $\angle CBD$. | [
"\n\n**Solution 1.** Notice that the triangle $ABC$ is isosceles with apex at vertex $B$. Let $E$ be such a point on the line $CD$ that $\\angle CAE = 30^\\circ$, so that the triangle $ACE$ is also isosceles with apex at vertex $E$. Since $\\angle CAD = 40^\\circ$ the point $E$ lies between... | Slovenia | National Math Olympiad 2015 – Final Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 80° | |
06ir | $\triangle ABC$ has area $1$ and satisfies $AB > AC$. The internal bisector of $\angle A$ meets $BC$ at $D$, and $\triangle DBC'$ is similar to $\triangle ABC$ where $C'$ is the image of $C$ upon reflection across $AD$. When the perimeter of $\triangle ABC$ is minimised, find the length of $AB$. | [
"As $AD$ is the internal bisector of $\\angle A$, point $C'$ lies on $AB$. Since $\\triangle DBC'$ is similar to $\\triangle ABC$, we have $\\angle BC'D = \\angle ACB$. Together with $\\angle AC'D = \\angle ACB$, we have $\\angle BC'D = \\angle AC'D$ and so each of $\\angle BC'D$, $\\angle AC'D$ and $\\angle ACB$ i... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 2 | |
0k2s | Problem:
Allen and Yang want to share the numbers $1,2,3,4,5,6,7,8,9,10$. How many ways are there to split all ten numbers among Allen and Yang so that each person gets at least one number, and either Allen's numbers or Yang's numbers sum to an even number? | [
"Solution:\n\nSince the sum of all of the numbers is odd, exactly one of Allen's sum and Yang's sum must be odd. Therefore any way of splitting the numbers up where each person receives at least one number is valid, so the answer is $2^{10}-2=1022$."
] | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 1022 | |
0d1a | Let $n \ge 3$ be a positive integer. Suppose we have a circle with $n$ positions labeled $1, 2, \dots, n$ in clockwise order. $n$ counters, each with one side white and the other side black, are placed on the circle with one counter in each of the numbered positions. Initially all counters have the white side facing up... | [
"The operation is invertible; its inverse is to take consecutive counters $Y, Z, X$ with $X$ black, move $X$ to the front, and flip $Y$. We will work with this operation instead, and show that from any initial position with at least one black counter, we can reach the configuration with one black counter on positio... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0c88 | Show that for infinitely many positive integers $n$ there exist pairwise distinct positive integers $a_1, a_2, \dots, a_n$ such that $a_1^2 a_2^2 \cdots a_n^2 - 4(a_1^2 + a_2^2 + \cdots + a_n^2)$ is the square of an integer. | [
"Letting $f_n(a_1, a_2, \\dots, a_n) = a_1^2 a_2^2 \\cdots a_n^2 - 4(a_1^2 + a_2^2 + \\cdots + a_n^2)$, the conclusion is a straightforward consequence of the following two facts:\n(1) For infinitely many positive integers $n$ there exist positive integers $a_1, a_2, \\dots, a_n$ such that $f_n(a_1, a_2, \\dots, a_... | Romania | SELECTION TESTS FOR THE 2019 BMO AND IMO | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0a1j | How many pairs of positive integers $a$ and $b$ with $a < b$ are there such that
$$
\frac{1}{a} + \frac{1}{b} = \frac{4}{15} ?
$$
A) 1 B) 2 C) 3 D) 4 E) 5 | [
"C) 3"
] | Netherlands | Dutch Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | C | |
093e | Problem:
Let $ABC$ be a triangle. The internal bisector of $\angle ABC$ intersects the side $AC$ at $L$ and the circumcircle of triangle $ABC$ again at $W \neq B$. Let $K$ be the perpendicular projection of $L$ onto $AW$. The circumcircle of triangle $BLC$ intersects line $CK$ again at $P \neq C$. Lines $BP$ and $AW$ m... | [
"Solution:\nNote that\n$$\n\\angle BPC = \\angle BLC = \\angle BAC + \\frac{\\angle CBA}{2} = \\angle BAW\n$$\nTherefore, $ABPK$ is cyclic.\nLet $KL$ intersect the circumcircle of triangle $BLC$ at a point $S \\neq L$. Then\n$$\n\\angle KSC = \\angle LSC = \\angle LBC = \\angle WBC = \\angle WAC = \\angle KAC.\n$$\... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line,... | null | proof only | null | |
0j97 | Problem:
Let $f$ be the function such that
$$
f(x)= \begin{cases}2x & \text{ if } x \leq \frac{1}{2} \\ 2-2x & \text{ if } x > \frac{1}{2}\end{cases}
$$
What is the total length of the graph of $\underbrace{f(f(\ldots f}_{2012\ f\text{'s}}(x) \ldots))$ from $x=0$ to $x=1$? | [
"Solution:\nAnswer: $\\sqrt{4^{2012}+1}$\n\nWhen there are $n$ copies of $f$, the graph consists of $2^{n}$ segments, each of which goes $1/2^{n}$ units to the right, and alternately $1$ unit up or down. So, the length is\n$$\n2^{n} \\sqrt{1+\\frac{1}{2^{2n}}} = \\sqrt{4^{n}+1}\n$$\nTaking $n=2012$, the answer is\n... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | sqrt(4^{2012}+1) | |
0dkr | Let $p \ge 5$ be a prime number and $a$ be an integer with $1 < a < p-1$. Let $a_1, a_2, \dots, a_n$ be all possible remainders that powers of $a$ (that is, $1, a, a^2, a^3, \dots$) can leave when divided by $p$. If the remainders $a_1 \cdot a_2 \cdot \dots \cdot a_n$ form an arithmetic sequence modulo $p$ in some orde... | [] | Saudi Arabia | Saudi Booklet | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof only | null | |
0je1 | Problem:
A bug is on one exterior vertex of solid $S$, a $3 \times 3 \times 3$ cube that has its center $1 \times 1 \times 1$ cube removed, and wishes to travel to the opposite exterior vertex. Let $O$ denote the outer surface of $S$ (formed by the surface of the $3 \times 3 \times 3$ cube). Let $L(S)$ denote the lengt... | [
"Solution:\n$\\boxed{\\frac{\\sqrt{29}}{3 \\sqrt{5}}}$ OR $\\frac{\\sqrt{145}}{15}$\n\nBy $\\left(^*\\right)$, the shortest route in $O$ has length $2 \\sqrt{1.5^2+3^2} = 3 \\sqrt{5}$.\n\nBy $\\left(^{**}\\right)$, the shortest route overall (in $S$) has length $2 \\sqrt{1.5^2+1^2+2^2} = \\sqrt{3^2+2^2+4^2} = \\sqr... | United States | HMMT November 2013 | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | sqrt(145)/15 | |
0ei6 | Problem:
Klara ima 6 enakih kock. Na mejne ploskve vsake kocke je zapisala črke $B, A, B, I, C$ in $A$, na vsako mejno ploskev po črko. Nato je hkrati vrgla vseh 6 kock. Kolikšna je verjetnost, da lahko iz črk, ki so po metu na zgornih ploskvah kock, sestavi besedo $B A B I C A$ ?
(A) $\frac{6!}{(2!)^{2}}\left(\frac{... | [
"Solution:\n\nVerjetnosti, da na eni kocki pade črka $A, B, C$ oz. $I$ so zaporedoma enake $\\frac{1}{3}, \\frac{1}{3}, \\frac{1}{6}$ oz. $\\frac{1}{6}$. Mislimo si, da Klara kocke vrže tako, da so urejene v vrsto. Tedaj je verjetnost, da pade beseda $B A B I C A$ enaka $\\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\fr... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | MCQ | A | |
0ks9 | Find the number of ordered pairs of integers $(a, b)$ such that the sequence
$3, 4, 5, a, b, 30, 40, 50$
is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression. | [
"Neither $a$ nor $b$ can be $6$ or $20$ because either of those numbers would make an arithmetic progression with either the least three numbers or the greatest three numbers. Therefore $a$ and $b$ must be chosen from the remaining $22$ values between $7$ and $29$ not including $20$. Setting $(a, b) = (7, 9)$ resul... | United States | 2022 AIME I | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 228 | |
05sh | Problem:
Martin et Théo jouent à un jeu : Martin écrit un nombre entier au tableau. Théo a ensuite le droit d'effacer le nombre et de lui ajouter $2$, ou d'effacer le nombre et de lui enlever $3$, ceci autant de fois qu'il veut. Théo gagne s'il arrive à obtenir $2020$ après un nombre fini d'étapes, sinon Martin gagne.... | [
"Solution:\n\nOn commence par tester l'énoncé pour des entiers proches de $2020$. Par exemple, on remarque que si Martin écrit les entiers $2018$, $2016$, ... Théo peut écrire $2020$ au tableau en rajoutant au nombre écrit par Martin un nombre suffisamment grand de $2$. On remarque aussi que si Martin écrit $2019$,... | France | Envoi 5: Pot Pourri | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Théo | |
0dft | Let the triangle $ABC$ is given and $D$, $E$, $F$ are on sides $BC$, $AC$, $AB$, respectively, such that
$$
\frac{BD}{CD} = \frac{CE}{AE} = \frac{AF}{BF}.
$$
Show that if the circumcircle of $ABC$ and $DEF$ coincide, then $ABC$ is equilateral. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
... | English | proof only | null | |
020a | Problem:
Let $k \geq 1$ be an integer.
We consider $4k$ chips, $2k$ of which are red and $2k$ of which are blue. A sequence of those $4k$ chips can be transformed into another sequence by a so-called move, consisting of interchanging a number (possibly one) of consecutive red chips with an equal number of consecutive b... | [
"Solution:\nThe answer is $n = k$.\n\nWe will first show that $n \\geq k$. Let us count the number $C$ of times a red chip is directly to the right of a blue chip. In the final position this number equals $0$. In the position $b r b r b r \\cdots b r$ this number equals $2k$. We claim that any move reduces this num... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | k | |
0c3h | Let $a, b, c \in (0, \infty)$, not all equal. Solve the equation
$$
(a^x + b^x + c^x) (a^{-x} + b^{-x} + c^{-x}) = 3 + \frac{a^2 + b^2}{ab} + \frac{b^2 + c^2}{bc} + \frac{c^2 + a^2}{ca}.
$$ | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | x = 1 or x = -1 | |
042l | Let $n$ be a given integer which is greater than $1$. Find the greatest constant $\lambda(n)$ such that for any non-zero complex $z_1, z_2, \dots, z_n$, we have
$$
\sum_{k=1}^{n} |z_k|^2 \geq \lambda(n) \min_{1 \leq k \leq n} \{|z_{k+1} - z_k|^2\},
$$
where $z_{n+1} = z_1$. | [
"Let $\\lambda_0(n) = \\begin{cases} \\frac{n}{4}, & 2 \\mid n, \\\\ \\frac{n}{4 \\cos^2 \\frac{\\pi}{2n}}, & \\text{otherwise.} \\end{cases}$. We prove $\\lambda_0(n)$ is the greatest constant.\n\nIf there exists $k$ ($1 \\leq k \\leq n$) such that $|z_{k+1} - z_k| = 0$, the inequality holds obviously. So without ... | China | China Team Selection Test | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | lambda(n) = { n/4 if n is even; n / (4 cos^2(pi/(2n))) if n is odd } | |
02l5 | Problem:
A árvore de Emília cresce de acordo com a seguinte regra: após 2 semanas do aparecimento de um galho, esse mesmo galho produz um novo galho a cada semana, e o galho original continua a crescer. A árvore tem 5 galhos depois de 5 semanas, como mostra a figura. Quantos galhos, incluindo o galho principal, a árvor... | [
"Solution:\nDenotemos por $f_{n}$ o número de galhos da árvore depois de $n$ semanas. Como depois de duas semanas aparece um galho então $f_{2}=1$. Na seguinte semana este galho produz um novo galho, logo $f_{3}=2$. Pela regra, o número de galhos na $n+1$-ésima semana é igual ao número de galhos que existiam na $n$... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 21 | |
0cw2 | Determine the smallest positive integer $n$ for which there exist integers $a_1, a_2, \dots, a_n$ such that the polynomial
$$x^2 - 2(a_1 + a_2 + \dots + a_n)^2 x + (a_1^4 + a_2^4 + \dots + a_n^4 + 1)$$
has at least one integer root. (P. Kozlov) | [
"For $n = 6$, one can take $a_1 = a_2 = a_3 = a_4 = 1$ and $a_5 = a_6 = -1$; then the quadratic from the problem becomes $x^2 - 8x + 7$ and has two integer roots: $1$ and $7$. It remains to show that this is the smallest possible value of $n$.\n\nSuppose the numbers $a_1, a_2, \\dots, a_n$ satisfy the problem's con... | Russia | Final round | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials"
] | English; Russian | proof and answer | 6 | |
07tg | In how many ways can $2021$ be expressed as a sum of four positive integers none of which is a multiple of $3$? Sums with different orders count as distinct. | [
"Let the four integers be $n_1, n_2, n_3, n_4$ and write them as\n$$\nn_j = 3x_j - r_j \\quad x_j \\ge 1, r_j = 1 \\text{ or } 2.\n$$\nWithout any further constraint, there are $16$ possible combinations of values $1$ and $2$ for the $r_j$. However, as $2021 \\equiv 2 \\pmod 3$, we must have\n$$\nr_1 + r_2 + r_3 + ... | Ireland | IRL_ABooklet | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Other"
] | null | proof and answer | 254924324 | |
0fb2 | Problem:
An $m \times n$ rectangle is divided into $mn$ unit squares by lines parallel to its sides. A gnomon is the figure of three unit squares formed by deleting one unit square from a $2 \times 2$ square. For what $m, n$ can we divide the rectangle into gnomons so that no two gnomons form a rectangle and no vertex... | [
"Solution:\n\nSuppose an $m \\times n$ rectangle could be tiled as described. We will establish a contradiction by counting gnomon vertices.\n\nA gnomon cannot touch a side of the rectangle along a length $1$, because then the gnomon that fitted under the overhang would form a rectangle with the first. So each gnom... | Soviet Union | 1st CIS | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | None | |
0kx8 | Problem:
The pairwise greatest common divisors of five positive integers are
$$
2, 3, 4, 5, 6, 7, 8, p, q, r
$$
in some order, for some positive integers $p, q, r$. Compute the minimum possible value of $p+q+r$. | [
"Solution:\n\nTo see that $9$ can be achieved, take the set $\\{6, 12, 40, 56, 105\\}$, which gives\n\n$$\n\\{p, q, r\\} = \\{2, 3, 4\\}\n$$\n\nNow we show it's impossible to get lower.\n\nNotice that if $m$ of the five numbers are even, then exactly $\\binom{m}{2}$ of the gcd's will be even. Since we're shown four... | United States | HMMT November 2023 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 9 | |
0bti | The unit squares of a $n \times n$ board, $n \ge 2$, are colored either black or white so that any black square has at least three white neighbors (a neighbor is a unit square with a common side). What is the maximum number of black unit squares? | [
"The answer is $\\frac{n^2-1}{2}$ if $n$ is odd and $\\frac{n^2-4}{2}$ if $n$ is even.\n\nNotice that:\n\n* the unit squares from the corners are white (because they only have two neighbors, not three);\n* the other squares from the border of the board have only three neighbors, so there can't be two consecutive bl... | Romania | 67th NMO Selection Tests for JBMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | If n is odd: (n^2 − 1)/2; if n is even: (n^2 − 4)/2. | |
03e1 | In every cell of a board $101 \times 101$ is written a positive integer. For any choice of 101 cells from different rows and columns, their sum is divisible by 101. Show that the number of ways to choose a cell from each row of the board, so that the total sum of the numbers in the chosen cells is divisible by 101, is ... | [
"(Vlad Spataru) Index the rows and columns from $0$ to $100$. We shall only work modulo $101$ in what follows. Observe that we may let $(0,0) = 0$ by adding some constant to all the terms of the table. Next, because of the condition in the statement,\n$$\n(u, v) + (i, j) = (u, j) + (i, v),\n$$\nfor any $u, v, i, j$... | Bulgaria | Autumn tournament | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Linear Algebra > Matrices"
] | English | proof only | null | |
04ws | The sum of four real numbers is $9$, the sum of their squares is $21$. Prove that these numbers can be signed $a$, $b$, $c$, $d$ so that inequality $ab - cd \geq 2$ holds. | [
"Let the numbers be $p$, $q$, $r$, $s$. Up to a permutation we may assume that $p \\geq q \\geq r \\geq s$. We first consider the case where $p+q \\geq 5$. Then\n$$\np^2 + q^2 + 2pq \\geq 25 = 4 + (p^2 + q^2 + r^2 + s^2) \\geq 4 + p^2 + q^2 + 2rs,\n$$\nwhich is equivalent to $pq - rs \\geq 2$.\n\nAssume now that $p... | Czech-Polish-Slovak Mathematical Match | Cesko-Slovacko-Poljsko 2006 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
08x4 | Determine all real-valued functions $f$ defined on the set of all integers and satisfying the following identity for an arbitrary pair $m, n$ of integers:
$$
f(m) + f(n) = f(mn) + f(m + n + mn).
$$ | [
"Let $f(1) = a$. By letting $n = 1$, we get from the given equation that $f(m) + f(1) = f(m) + f(2m + 1)$ is valid for any integer $m$. Thus we see that $f(d) = a$ must hold for any odd integer $d$.\n\nAn arbitrary non-zero integer can be written in the form $2^k d$ where $d$ is an odd integer and $k$ is a non-nega... | Japan | Japan 2013 Final Round | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All functions of the form: f(n) = a for n odd, and f(n) = b for n even (including zero), where a and b are arbitrary real constants. | |
0ayu | Problem:
How many distinct numbers are there in the sequence $\left\lfloor\frac{1^{2}}{2018}\right\rfloor, \left\lfloor\frac{2^{2}}{2018}\right\rfloor, \ldots, \left\lfloor\frac{2018^{2}}{2018}\right\rfloor$? | [] | Philippines | 21st PMO Area Stage | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 1514 | |
006q | Germán escribió números en las casillas de un tablero de $11 \times 11$ de modo que en cada fila la suma de los $11$ números es igual a $3$, en cada columna la suma de los $11$ números es igual a $3$, y en cada cuadrado de $3 \times 3$ la suma de los $9$ números es igual a $1$.
Dar un ejemplo de un tablero como el de ... | [] | Argentina | Argentina 2009 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | final answer only | null | |
0du7 | Problem:
a. Dokaži, da za vsako naravno število $n$ velja
$$
\sqrt{n+1}-\sqrt{n}<\frac{1}{2 \sqrt{n}}<\sqrt{n}-\sqrt{n-1}
$$
b. Dokaži, da je celi del izraza
$$
1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{m^{2}-1}}+\frac{1}{\sqrt{m^{2}}}
$$
kjer je $m$ naravno število, enak $2 m-2$ ali $2 m-1$. | [
"Solution:\n\na. Ko neenakost $\\sqrt{n+1}-\\sqrt{n}<\\frac{1}{2 \\sqrt{n}}$ pomnožimo s $\\sqrt{n+1}+\\sqrt{n}$, dobimo $1< \\frac{\\sqrt{n+1}+\\sqrt{n}}{2 \\sqrt{n}}$, kar očitno drži. Ko pa neenakost $\\frac{1}{2 \\sqrt{n}}<\\sqrt{n}-\\sqrt{n-1}$ pomnožimo s $\\sqrt{n}+\\sqrt{n-1}$, dobimo $\\frac{\\sqrt{n}+\\sq... | Slovenia | 45. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | [S] ∈ {2 m - 2, 2 m - 1} | |
04b8 | Each digit of a positive integer $n$ (except the first one) is larger than the digit next left to it. Determine the sum of all digits of the number $9n$. (Russia 1999) | [
"Let $n = \\overline{a_{m-1}a_{m-2}\\dots a_1a_0}$. We know that $a_0 > a_1 > \\dots > a_{m-1}$.\n\nSince $9n = 10n - n$, from\n$$\n\\begin{array}{c c c c c c c c c c c c c c c}\n& a_{m-1} & a_{m-2} & a_{m-3} & \\dots & a_1 & a_0 & 0 \\\\\n- & \\multicolumn{1}{c}{a_{m-1}} & a_{m-2} & \\dots & a_2 & a_1 & a_0 \\\\\n... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | 9 | |
03y2 | Let $a_1, a_2, \dots, a_n, b_1, b_2, \dots, b_n$ be non-negative numbers satisfying the following conditions simultaneously:
$$
(1) \quad \sum_{i=1}^{n} (a_i + b_i) = 1;
$$
$$
(2) \quad \sum_{i=1}^{n} i(a_i - b_i) = 0;
$$
$$
(3) \quad \sum_{i=1}^{n} i^2 (a_i + b_i) = 10.
$$
Prove that $\max\{a_k, b_k\} \le \frac{10}{10... | [
"For any $1 \\le k \\le n$, it follows from the given conditions and Cauchy's Inequality that\n$$\n\\begin{align*}\n(ka_k)^2 &\\le \\left(\\sum_{i=1}^n i a_i\\right)^2 = \\left(\\sum_{i=1}^n i b_i\\right)^2 \\\\\n&\\le \\left(\\sum_{i=1}^n i^2 b_i\\right) \\left(\\sum_{i=1}^n b_i\\right) \\\\\n&= (10 - \\sum_{i=1}^... | China | China Western Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0i4t | Problem:
A certain cafeteria serves ham and cheese sandwiches, ham and tomato sandwiches, and tomato and cheese sandwiches. It is common for one meal to include multiple types of sandwiches. On a certain day, it was found that 80 customers had meals which contained both ham and cheese; 90 had meals containing both ham ... | [
"Solution:\n230. Everyone who ate just one sandwich is included in exactly one of the first three counts, while everyone who ate more than one sandwich is included in all four counts. Thus, to count each customer exactly once, we must add the first three figures and subtract the fourth twice: $80+90+100-2 \\cdot 20... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 230 | |
0g42 | Problem:
Let $n \geq 2$ be an integer. Prove that if
$$
\frac{n^{2}+4^{n}+7^{n}}{n}
$$
is an integer, then it is divisible by $11$. | [
"Solution:\nThe \"Lifting the Exponent\" lemma tells us that\n$$\nv_{11}\\left(4^{n}+7^{n}\\right)=v_{11}(4+7)+v_{11}(n)=1+v_{11}(n)\n$$\nand we therefore have $11 \\mid\\left(4^{n}+7^{n}\\right) / n$.\n\nNow let $p$ denote the smallest prime factor of $n$. By assumption we have $p \\mid n \\mid 4^{n}+7^{n}$, with ... | Switzerland | IMO Selection | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Mul... | null | proof only | null | |
0fa0 | Problem:
The figure above is cut along the lines into polygons (which need not be convex). No polygon contains a $2 \times 2$ square. What is the smallest possible number of polygons? | [
"Solution:\n\nWe can clearly cut the polygon into $12$ strips width $1$, so the smallest number is $\\leq 12$.\n\nThere are $84$ unit squares in the figure. Each cut along the edge of a unit square not already cut (and not on the boundary) increases the number of pieces by at most $1$. So it is sufficient to show t... | Soviet Union | 25th ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 12 | |
080t | Problem:
Sia $ABCD$ un quadrilatero convesso; poniamo $D\hat{A}B=\alpha$; $A\hat{D}B=\beta$; $A\hat{C}B=\gamma$; $D\hat{B}C=\delta$; $D\hat{B}A=\epsilon$. Sapendo che $\alpha<90^\circ$, $\beta+\gamma=90^\circ$, $\delta+2\epsilon=180^\circ$, dimostrare che
$$
(DB+BC)^2=AD^2+AC^2.
$$ | [
"Solution:\n\nSia $P$ il simmetrico di $C$ rispetto alla retta $AB$. Dalla congruenza dei triangoli $ABC$ e $ABP$ si ricava che\n1. $A\\hat{B}P=A\\hat{B}C=\\delta+\\epsilon$;\n2. $A\\hat{P}B=A\\hat{C}B=\\gamma$;\n3. $AP=AC$ e $BP=BC$.\n\nUsando la prima informazione abbiamo che\n$$\nD\\hat{B}P=D\\hat{B}A+A\\hat{B}P... | Italy | Cesenatico | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0f69 | Problem:
The incircle of the triangle $ABC$ has center $I$ and touches $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. The segments $AI$, $BI$, $CI$ intersect the circle at $D'$, $E'$, $F'$ respectively. Show that $DD'$, $EE'$, $FF'$ are collinear. | [] | Soviet Union | 18th ASU | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05ds | Problem:
The $n$ contestants of an EGMO are named $C_{1}, \ldots, C_{n}$. After the competition they queue in front of the restaurant according to the following rules.
- The Jury chooses the initial order of the contestants in the queue.
- Every minute, the Jury chooses an integer $i$ with $1 \leq i \leq n$.
- If cont... | [
"Solution:\n\nThe maximal number of euros is $2^{n}-n-1$.\n\nTo begin with, we show that it is possible for the Jury to collect this number of euros. We argue by induction. Let us assume that the Jury can collect $M_{n}$ euros in a configuration with $n$ contestants. Then we show that the Jury can collect at least ... | European Girls' Mathematical Olympiad (EGMO) | EGMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2^n - n - 1 | |
0gzm | A rectangle with the lengths of sides $2010$ and $11$ is partitioned into unit squares. The external slice of squares is colored with yellow, the next slice (all squares that share a vertex with external slice) are colored with blue. A slice of squares that touches the previous slice is colored with yellow and so on. F... | [
"Do allocate external slices of squares of each color step by step:\n$2010 \\times 11$ - yellow; $2008 \\times 9$ - blue; $2006 \\times 7$ - yellow; $2004 \\times 5$ - blue; $2002 \\times 3$ - yellow;\n$2000 \\times 1$ - blue.\n\nNow calculate the number of squares in each slice as difference between numbers of squ... | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Discrete Mathematics > Other"
] | English | proof and answer | Yellow: 12066; Blue: 10044 | |
0hus | Problem:
Show that $\sin 10^{\circ}$ is irrational. | [
"Solution:\nOne can show the triple-angle identity\n$$\n\\sin (3 \\theta) = 3 \\sin \\theta - 4 \\sin^{3} \\theta.\n$$\nThus letting $x = 2 \\sin (10^{\\circ})$ we derive\n$$\nx^{3} - 3x + 1 = 0.\n$$\nThis polynomial has no rational roots (by, say, Rational Root Theorem), hence $x$ is irrational, whence $\\sin 10^{... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | null | proof only | null | |
0815 | Problem:
Sia $p(x)$ un polinomio a coefficienti interi tale che $p(0)=0$ e $0 \leq p(1) \leq 10^{7}$. Sapendo che esistono due interi positivi $a, b$ tali che $p(a)=1999$ e $p(b)=2001$, si determinino i valori possibili di $p(1)$.
(Nota: si osservi che 1999 è un numero primo) | [
"Solution:\n\nI valori possibili di $p(1)$ sono $1, 1999, 3.996.001, 7.992.001$.\n\nPer giustificare questi risultati, incominciamo a dimostrare che\n- $a$ può essere solo $1$ o $1999$;\n- $b$ è un divisore positivo di $2001$;\n- $b-a$ vale $\\pm 1$ oppure $\\pm 2$.\n\nPer dimostrare queste affermazioni, osserviamo... | Italy | Cesenatico | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1, 1999, 3996001, 7992001 | |
0jp8 | Problem:
Find the number of pairs of union/intersection operations $\left(\square_{1}, \square_{2}\right) \in \{\cup, \cap\}^{2}$ satisfying the following condition: for any sets $S, T$, function $f: S \rightarrow T$, and subsets $X, Y, Z$ of $S$, we have equality of sets
$$
f(X) \square_{1}\left(f(Y) \square_{2} f(Z)\... | [
"Solution:\nAnswer: 1 If and only if $\\square_{1}=\\square_{2}=\\cup$. See http://math.stackexchange.com/questions/359693/overview-of-1"
] | United States | HMMT February | [
"Discrete Mathematics > Other"
] | null | final answer only | 1 | |
0d39 | Ayman wants to color the cells of a $50 \times 50$ chessboard into black and white so that each $2 \times 3$ or $3 \times 2$ rectangle contains an even number of white cells. Determine the number of ways Ayman can color the chessboard. | [
"Let us associate a $1$ to each cell with a black color and a $0$ to each cell with a white color. The condition is equivalent to the sum of numbers in each $2 \\times 3$ rectangle and in each $3 \\times 2$ rectangle being even. Let $a_{i, j}$ be this number at the cell in the $i^{\\text{th}}$ row and $j^{\\text{th... | Saudi Arabia | Selection tests for the Balkan Mathematical Olympiad 2013 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 32 | |
0ii0 | Problem:
Triangle $ABC$ has side lengths $AB = 65$, $BC = 33$, and $AC = 56$. Find the radius of the circle tangent to sides $AC$ and $BC$ and to the circumcircle of triangle $ABC$. | [
"Solution:\n\nLet $\\Gamma$ be the circumcircle of triangle $ABC$, and let $E$ be the center of the circle tangent to $\\Gamma$ and the sides $AC$ and $BC$. Notice that $\\angle C = 90^{\\circ}$ because $33^{2} + 56^{2} = 65^{2}$. Let $D$ be the second intersection of line $CE$ with $\\Gamma$, so that $D$ is the mi... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 24 | |
0h2c | Triangle **ABC** is inscribed into the circle. Tangents at points **A** and **B** meet at point **T**. The line, that passes through the point **T** and is parallel to **AC**, intersects **BC** at point **D**. Prove that $AD = CD$. | [
"Since triangle $ATB$ is isosceles, $\\angle TBA$ and $\\angle BCA$ share common arc, then:\n\n$$\n\\angle TBA = \\angle TAB = \\angle BCA = \\angle BDT.\n$$\n\nThis implies, that $BTAD$ is cyclic (fig. 9). Hence $\\angle BTA = \\angle ADC$ and triangles $ABT$ and $ACD$ are similar. Triangle $ATB$ is isosceles, thu... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0l8d | Let $n$ be a positive integer. Ana and Banana play a game. Banana thinks of a function $f: \mathbb{Z} \to \mathbb{Z}$ and a prime number $p$. He tells Ana that $f$ is nonconstant, $p < 100$, and $f(x+p) = f(x)$ for all integers $x$. Ana's goal is to determine the value of $p$. She writes down $n$ integers $x_1, \dots, ... | [
"The answer is $n = 83 + 89 - 1 = 171$.\nIn general, if Ana has to decide between periods from the set $\\mathcal{P} := \\{p_1 > p_2 > \\dots > p_r\\}$ of pairwise distinct relatively prime positive integers for $r \\ge 3$, the answer is $p_2 + p_3 - 1$.\n\n### Bound\nSuppose for the sake of contradiction that Ana ... | United States | USA TST 2025 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 171 | |
0cly | Problem:
Given a positive integer number $n \geq 3$, colour each cell of an $n \times n$ square array one of $\left[(n+2)^{2} / 3\right]$ colours, each colour being used at least once. Prove that the cells of some $1 \times 3$ or $3 \times 1$ rectangular subarray have pairwise distinct colours. | [
"Solution:\nFor more convenience, say that a subarray of the $n \\times n$ square array bears a colour if at least two of its cells share that colour.\n\nWe shall prove that the number of $1 \\times 3$ and $3 \\times 1$ rectangular subarrays, which is $2 n(n-2)$, exceeds the number of such subarrays, each of which ... | Romanian Master of Mathematics (RMM) | Romanian Master of Mathematics Competition | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
096b | Problem:
Calculați limita
$$
\lim_{n \rightarrow \infty} \left( \left(1+\frac{1}{2}\right) \cdot \left(1+\frac{1}{2+3}\right) \cdot \ldots \cdot \left(1+\frac{1}{2+3+\ldots+n}\right) \right).
$$ | [
"Solution:\nFie $P_{n}=\\left(1+\\frac{1}{2}\\right) \\cdot\\left(1+\\frac{1}{2+3}\\right) \\cdots \\left(1+\\frac{1}{2+3+\\ldots+n}\\right)$.\n$$\n\\begin{aligned}\n& S_{n}=1+2+\\ldots+n=\\frac{n(n+1)}{2}, \\forall n \\in \\mathbb{N}^{*} \\Rightarrow \\\\\n& S_{n}-1=2+3+\\ldots+n=\\frac{n(n+1)}{2}-1=\\frac{n^{2}+n... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 3 |
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