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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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
09no | (1) Find all pairs of natural numbers $(a, b)$ such that $a^3 = b^3 + 61$.
(2) Find all pairs of natural numbers $(a, b)$ such that $a^5 = b^5 + 61$. | [] | Mongolia | MMO2025 Round 2 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | (1) (a, b) = (5, 4). (2) No solutions in natural numbers. | |
0hdw | A convex quadrilateral $ABCD$ is given, and $\angle CBD = 90^\circ$, $\angle BCD = \angle CAD$ and $AD = 2BC$. Prove that $CA = CD$. | [
"We denote by $C_1$ the point symmetric to $C$ with regards to (w.r.t) point $B$ (Fig. 23). Then $\\angle BCD = \\angle CAD = \\angle DC_1C$, hence, $ADCC_1$ is inscribed. Since $AD = 2BC = CC_1$,\n\n\n\nFig. 23\n\nthen $\\angle CDC_1 = \\angle DC_1A$, since they are based on the arcs of th... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07aa | Let $ABCD$ be a parallelogram. Consider circles $\omega_1$ and $\omega_2$ such that $\omega_1$ is tangent to segments $AB$, $AD$ and $\omega_2$ is tangent to segments $BC$, $CD$. Suppose that there exists a circle tangent to lines $AD$, $DC$ and externally tangent to $\omega_1$, $\omega_2$. Prove that there exists a ci... | [
"Suppose $\\omega_1$ is tangent to $AB$, $AD$ at $P_1$, $Q_1$ respectively and $\\omega_2$ is tangent to $CD$, $BC$ at $P_2$, $Q_2$ respectively.\n\n**Lemma.** Let a circle $\\omega$ be tangent to the half-lines $BA$, $BC$ at $P$, $Q$ respectively. $\\omega$ is tangent to $\\omega_1$ if and only if $\\sqrt{BP} = \\... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
001x | En el pizarraón están escritos los $2003$ números enteros desde $1$ hasta $2003$. Lucas debe borrar $90$ números. A continuación, Mauro debe elegir $37$ de los números que permanecen escritos. Si los $37$ números que elige Mauro forman una progresión aritmética, gana Mauro. Si no, gana Lucas.
Decidir si Lucas puede ele... | [] | Argentina | XX OLIIMPÍADA MATEMÁTICA ARGENTINA | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | español | proof and answer | Yes, Lucas can ensure victory by deleting all numbers in one residue class modulo thirty seven that has fifty four elements and one number from each of the other thirty six classes, placed to break any run of thirty seven equally spaced terms. | |
003p | a) Explique por qué no existe ningún heptágono tal que todos sus lados tengan la misma longitud y sus ángulos midan, exactamente en este orden, $120^\circ$, $150^\circ$, $120^\circ$, $120^\circ$, $120^\circ$, $150^\circ$ y $120^\circ$.
b) Justifique por qué existe algún heptágono tal que todos sus lados tienen la mism... | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | Español | proof only | null | |
0jwe | Problem:
Find the sum of all positive integers whose largest proper divisor is $55$. (A proper divisor of $n$ is a divisor that is strictly less than $n$.) | [
"Solution:\n\nThe largest proper divisor of an integer $n$ is $\\frac{n}{p}$, where $p$ is the smallest prime divisor of $n$. So $n = 55p$ for some prime $p$. Since $55 = 5 \\cdot 11$, we must have $p \\leq 5$, so $p = 2, 3, 5$ gives all solutions. The sum of these solutions is $55(2 + 3 + 5) = 550$."
] | United States | HMMT November 2017 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 550 | |
0cx9 | In triangle $A B C$ the circumcircle has radius $R$ and center $O$ and the incircle has radius $r$ and center $I \neq O$. Let $G$ denote the centroid of triangle $A B C$. Prove that $I G \perp B C$ if and only if $A B=A C$ or $A B+A C=3 B C$. | [
"Let $\\alpha = B C$, $\\beta = C A$, $\\gamma = A B$. Without loss of generality we may assume that $\\beta \\geq \\gamma$. Let $M$ be the midpoint of $B C$, and let $A'$, $I'$, and $G'$ be the orthogonal projections of $A$, $I$ and $G$ on $M B$.\n\n\n\nWe have\n$$\nM I' = M B - B I' = \\f... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0711 | Problem:
Show that there are arbitrarily large numbers $n$ such that: (1) all its digits are 2 or more; and (2) the product of any four of its digits divides $n$. | [
"Solution:\n$3232 = 16 \\times 202$ and $10000 = 16 \\times 625$. So any number with $3232$ as its last $4$ digits is divisible by $16$. So consider $N = 22223232$. Its sum of digits is $18$, so it is divisible by $9$. Hence it is divisible by $9 \\cdot 16 = 144$. But any four digits have at most four $2$'s and at ... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
0fbn | Problem:
Estudiar la función real
$$
f(x)=\left(1+\frac{1}{x}\right)^{x}
$$
definida para $x \in \mathbb{R} - [-1,0]$. Representación gráfica. | [
"Solution:\n\nSi analizamos su comportamiento tenemos que\n$$\n\\begin{gathered}\n\\lim _{x \\longrightarrow-\\infty}\\left(1+\\frac{1}{x}\\right)^{x}=e ; \\quad \\lim _{\\substack{x \\rightarrow-1 \\\\ x<-1}}\\left(1+\\frac{1}{x}\\right)^{x}=\\infty \\\\\n\\lim _{\\substack{x \\rightarrow 0 \\\\ x>0}}\\left(1+\\fr... | Spain | OME 12 | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications",
"Precalculus > Limits"
] | null | proof only | null | |
00pz | Find all positive integers $n$ such that there exist non-constant polynomials with integer coefficients $f_1(x), \dots, f_n(x)$ (not necessarily distinct) and $g(x)$ such that
$$
1 + \prod_{k=1}^{n} (f_k^2(x) - 1) = (x^2 + 2013)^2 g^2(x).
$$ | [
"Consider the complex number $\\omega = (a \\pm i\\sqrt{b})^2 - 1$, where $a, b \\in \\mathbb{Z}$, $b \\ge 0$. It is easy to see that $|\\omega| = \\sqrt{(a^2 - b - 1)^2 + 4a^2b} > 1$ or $|\\omega| = 0$, where the latter is possible only when $a = 0$ or $b = 0$.\nSet in the condition $x = i\\sqrt{2013}$ and conside... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | all odd positive integers | |
0h8u | Two players – Andriy and Olesya play the following game. On a table, there is a rounded cake, which is cut by one of them into $4n$ different in weight sectors (pieces). Weight of every part is known by each player. After that they choose pieces for themselves upon following rules. At first Andriy chooses 1 piece, then... | [
"a) Let's show how Olesya can cut the cake to win. Let $M$ denote the weight of the cake. She cuts it into 3 neighboring pieces weighting $\\frac{1}{3}M$ (heavy), and all the rest have zero weight (light). Let's number the pieces clockwise 1; 2; ...; $4n$. Piece number $4n$ is neighboring to 1. Let pieces $4n-3$, $... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | a) Yes. Olesya can cut and play so she surely gets more than half. b) Yes. Andriy can cut and play so he surely gets more than half even if he cannot take the heaviest piece on his first move. | |
0fw5 | Problem:
Finde alle Funktionen $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$, sodass für alle $x, y>0$ gilt
$$
f\left(x^{y}\right)=f(x)^{f(y)}
$$ | [
"Solution:\nDie konstante Funktion $f(x)=1$ für alle $x>0$ ist offensichtlich eine Lösung der Gleichung. Wir nehmen nun an, dass $f$ nicht konstant gleich $1$ ist und wählen ein $a>0$ mit $f(a) \\neq 1$. Nach den Potenzgesetzen gilt für alle $x, y>0$\n$$\nf(a)^{f(x y)}=f\\left(a^{x y}\\right)=f\\left(\\left(a^{x}\\... | Switzerland | IMO Selektion | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 1 for all x > 0, and f(x) = x for all x > 0 | |
0b86 | A nonconstant polynomial $f$ with integral coefficients has the property that, for each prime $p$, there exist a prime $q$ and a positive integer $m$ such that $f(p) = q^m$. Prove that $f = X^n$ for some positive integer $n$. | [
"We claim that for every prime $p$, $f(p) = p^m$, where $m$ is a positive integer which (possibly) depends on $p$. Assume the claim for the time being. Since the degree $n$ of $f$ is positive, the claim forces $f(p) = p^n$ for sufficiently large primes $p$. Consequently, the polynomials $f$ and $X^n$ agree infinite... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
04jw | Circles $k_1$ and $k_2$ meet at points $M$ and $N$. The line $l$ meets the circle $k_1$ in points $A$ and $C$, and the circle $k_2$ in points $B$ and $D$ so that the points $A, B, C$ and $D$ are on the line $l$ in that order. Let $X$ be a point on the line $MN$ such that the point $M$ is between points $X$ and $N$. Let... | [
"Let $Y$ be the second intersection of the circle $k_1$ with the line $AX$ and let $Z$ be the second intersection of the circle $k_2$ with the line $DX$.\nSince lines $AY$, $DZ$ and $MN$ pass through the point $X$, by the converse of the radical centre theorem, quadrilateral $AYZD$ is cyclic. Hence $\\angle YZX = \... | Croatia | Croatian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a1v | Problem:
Zij $\triangle A B C$ een gelijkbenige driehoek met $|A B|=|A C|$. Gegeven is een punt $P$ in $\triangle A B C$ ongelijk aan het middelpunt van de omgeschreven cirkel. Zij $\omega$ de cirkel door $C$ met middelpunt $P$. Gegeven is dat de cirkel $\omega$ de lijnstukken $B C$ en $A C$ een tweede keer snijdt in ... | [
"Solution:\n\nOplossing I. Omdat $|F P|=|P E|$ zijn de hoeken op deze koorden van $\\Gamma$ ook gelijk: $\\angle F A P=\\angle P A E=\\angle P A C$. Dan vinden we met behulp van koordenvierhoek $A E P F$ dat $\\angle P F A=180^{\\circ}-\\angle P E A=\\angle P E C$. Dus met hoekensom volgt\n$$\n\\begin{aligned}\n\\a... | Netherlands | IMO-selectietoets III | [
"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"
] | null | proof only | null | |
0d9y | Let $n$ be an even positive integer. We fill in a number on each cell of a rectangle table of $n$ columns and multiple rows as following:
i. Each row is assigned to some positive integer $a$ and its cells are filled by $0$ or $a$ (in any order);
ii. The sum of all numbers in each row is $n$.
Note that we cannot add any... | [
"Denote $\\sigma(n), \\tau(n)$ as the sum of divisors and number of divisors of $n$, respectively.\nConsider a row with assigned number is $a$, and suppose that there are $b$ cells on that row are filled by $a$ then $n = a b$ or $a$ is a divisor of $n$. By condition ii), the number of rows of the table is exactly t... | Saudi Arabia | Team selection tests for IMO 2018 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)"
] | English | proof only | null | |
0hjp | Problem:
Two players play a game with pennies, which are circles of radius $1$, on an $m \times n$ rectangular table. Each player takes turns putting a penny on the table so that it touches no other penny. The first player who is unable to do so loses. The table starts with no pennies on it. Assuming that there is an ... | [
"Solution:\n\nIf $m<2$ or $n<2$, then it is not possible to fit even one penny on the table, so the first player loses immediately. On the other hand, if $m \\geq 2$ and $n \\geq 2$, we will show that the first player can win. Let $O$ denote the center of the table, and let the first player place his first penny so... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | m >= 2 and n >= 2 | |
09pg | Let $n \ge 3$ be a given integer. Find the least number of digits in the number formed with only digits $1$ and $2$ such that the number is divisible by $10^n - 7$. | [
"Answer: $2n + 1$\nLet $N = \\overline{a_m \\dots a_2 a_1}$ be a multiple of $d$ such that $a_i \\in \\{1, 2\\}$ for every $1 \\le i \\le m$, where $d$ denotes $10^n - 7$. Since $N \\ge 2d$ it is clear that $m \\ge n + 1$. Suppose that $2n \\ge m$. By setting $S = 7 \\times \\overline{a_m \\dots a_{n+1}} + \\overli... | Mongolia | MMO2025 Round 4 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Other"
] | English | proof and answer | 2n + 1 | |
0he6 | Petryk and Vasyl' are playing a game with the numbers written on the board. In a single one move – Petryk goes first – the player chooses two co-prime numbers out of the ones written on the board, erases them, and writes down their sum instead. The one who can't make the move loses. Who will win if both players play co... | [
"a) Let us show that Vasyl' can achieve the situation that after each his move, the board contains some odd number $n$ and $2019-n$ numbers $1$. Then, after $1009$ pairs of moves, only number $2019$ remains on the board, and Petryk will not be able to make a move and will lose. As his first move, Petryk wipes two n... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) Vasyl wins. b) Vasyl wins. | |
04ph | Find all pairs $(m, n)$ of positive integers for which there exists a prime number $p$ such that
$$
9^m + 3^m - 2 = 2p^n.
$$ | [] | Croatia | Croatian Mathematical Society Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (1,1) | |
07ix | Let $x$, $y$, $z$ be real numbers. Prove that:
$$
(x+y+z)^2 + \frac{(x+y)(y+z)}{1+|x-z|} + \frac{(y+z)(z+x)}{1+|y-x|} + \frac{(z+x)(x+y)}{1+|z-y|} \ge xy+yz+zx
$$ | [
"After the change of variables $(a, b, c) = (x + y, y + z, z + x)$ it suffices to prove that\n$$\na^2 + b^2 + c^2 + \\frac{2ab}{1 + |a - b|} + \\frac{2bc}{1 + |b - c|} + \\frac{2ac}{1 + |c - a|} \\geq 0\n$$\nfor all real numbers $a$, $b$, $c$. Let $(A, B, C) = (|b - c|, |c - a|, |a - b|)$, since $(ab) \\cdot (bc) \... | Iran | 41th Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
04cg | Determine the smallest integer $N$ larger than $1000$ such that exactly half of the numbers from $1$ to $N$ have at least one digit $1$ in their decimal notation. | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Other"
] | English | proof and answer | 1456 | |
0h3o | Let $S(a)$ denote the sum of decimal digits of a positive integer $a$. A positive integer $n$ is such that $S(n) = 503$, $S(121n) = 2012$. Find all possible values of $S(11n)$. | [
"Зауважимо, що $121n$ — це сума одного числа $100n$, двох чисел $10n$ і одного числа $n$. Якщо додавати ці числа в стовпчик, то в кожному розряді суми, починаючи з другого і закінчуючи передостаннім, буде сума двох цифр попереднього розряду, однієї цифри цього розряду і однієї цифри наступного розряду числа $n$. Як... | Ukraine | Ukrainian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 1006 | |
0jy8 | Problem:
Let $v_{1}, v_{2}, \ldots, v_{m}$ be vectors in $\mathbb{R}^{n}$, such that each has a strictly positive first coordinate. Consider the following process. Start with the zero vector $w=(0,0, \ldots, 0) \in \mathbb{R}^{n}$. Every round, choose an $i$ such that $1 \leq i \leq m$ and $w \cdot v_{i} \leq 0$, and ... | [
"Solution:\n\nLet $w_{0}=0$, let $w_{r}$ be the vector $w$ after $r$ rounds. Also, let $e_{1}$ denote the vector $(1,0, \\ldots, 0)$. Note that\n$$\nw_{r+1}^{2}=\\left(w_{r}+v_{i}\\right)^{2}=w_{r}^{2}+v_{i}^{2}+2 w_{r} \\cdot v_{i} \\leq w_{r}^{2}+v_{i}^{2}\n$$\nas we have by the condition that $w_{r} \\cdot v_{i}... | United States | HMIC | [
"Algebra > Linear Algebra > Vectors",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0d38 | Let $ABC$ be a triangle with incenter $I$, and let $D, E, F$ be the midpoints of sides $BC, CA, AB$, respectively. Lines $BI$ and $DE$ meet at $P$, and lines $CI$ and $DF$ meet at $Q$. Line $PQ$ meets sides $AB$ and $AC$ at $T$ and $S$, respectively. Prove that $AS = AT$. | [
"Because sides of triangles $ABC$ and $DEF$ are parallel (homothetic triangles), we have\n$$\n\\begin{aligned}\n\\angle BPD & = 180^\\circ - \\angle EDF - \\angle FDB - \\frac{1}{2} \\angle CBA \\\\\n& = 180^\\circ - \\angle BAC - \\angle ACB - \\frac{1}{2} \\angle CBA \\\\\n& = \\frac{1}{2} \\angle CBA = \\angle D... | Saudi Arabia | Selection tests for the Balkan Mathematical Olympiad 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0itw | Problem:
Evaluate the sum:
$$
11^{2}-1^{2}+12^{2}-2^{2}+13^{2}-3^{2}+\ldots+20^{2}-10^{2}
$$ | [
"Solution:\nThis sum can be written as\n$$\n\\sum_{a=1}^{10} \\left( (a+10)^2 - a^2 \\right)\n$$\nExpanding $(a+10)^2 - a^2$ gives:\n$$\n(a+10)^2 - a^2 = a^2 + 20a + 100 - a^2 = 20a + 100\n$$\nSo the sum becomes:\n$$\n\\sum_{a=1}^{10} (20a + 100) = 20 \\sum_{a=1}^{10} a + 100 \\times 10\n$$\nWe know $\\sum_{a=1}^{1... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 2100 | |
0d88 | Find the smallest prime $q$ such that
$$
q = a_{1}^{2} + b_{1}^{2} = a_{2}^{2} + 2 b_{2}^{2} = a_{3}^{2} + 3 b_{3}^{2} = \ldots = a_{10}^{2} + 10 b_{10}^{2}
$$
where $a_{i}, b_{i}$ $(i=1,2, \ldots, 10)$ are positive integers. | [
"Since $q = a_{1}^{2} + b_{1}^{2}$ then $q$ must have the form $4k+1$; otherwise, $a_{1}^{2} + b_{1}^{2}$ is divisible by $q = 4k+3$ which implies that $q \\mid a_{1}, q \\mid b_{1}$, a contradiction.\n\nNote that $q = a_{2}^{2} + 2 b_{2}^{2}$ so in modulo $8$, number $q$ is congruent to either $1$ or $3$ modulo $8... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | English | proof and answer | 1009 | |
04nc | Determine all positive integers which are squares of integers, and which have exactly two non-zero digits in the decimal representation, one of which is equal to $3$. | [] | Croatia | Croatia_2018 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | All numbers of the form 36 × 100^k for nonnegative integers k. | |
0ilu | Problem:
Let $ABCD$ be a cyclic quadrilateral, and let $P$ be the intersection of its two diagonals. Points $R$, $S$, $T$, and $U$ are feet of the perpendiculars from $P$ to sides $AB$, $BC$, $CD$, and $AD$, respectively. Show that quadrilateral $RSTU$ is bicentric if and only if $AC \perp BD$. (Note that a quadrilater... | [
"Solution:\nFirst we show that $RSTU$ is always inscriptible. Note that in addition to $ABCD$, we have cyclic quadrilaterals $ARPU$ and $BSPR$. Thus,\n$$\n\\angle PRU = \\angle PAU = \\angle CAD = \\angle CBD = \\angle SBP = \\angle SRP\n$$\nand it follows that $P$ lies on the bisector of $\\angle SRU$. Analogously... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01t5 | A number $\alpha$ is a root of the equation $x^3 - 12x + 8 = 0$.
Prove that the number $2 - \frac{4}{\alpha}$ is also the root of this equation. | [
"Set $x = 2 - \\frac{4}{\\alpha}$ in the left hand side of the given equation:\n$$\n\\begin{aligned}\n& \\left(2 - \\frac{4}{\\alpha}\\right)^3 - 12\\left(2 - \\frac{4}{\\alpha}\\right) + 8 = 8\\left(\\left(1 - \\frac{2}{\\alpha}\\right)^3 - 3\\left(1 - \\frac{2}{\\alpha}\\right) + 1\\right) = \\\\\n& = 8\\left(1 -... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0d7q | Let $m, n$ be odd integers such that $n^{2}-1$ is divisible by $m^{2}+1-n^{2}$. Prove that $\left|m^{2}+1-n^{2}\right|$ is a perfect square. | [
"By assumption, there is an integer $t$ such that $n^{2}-1 = t\\left(m^{2}+1-n^{2}\\right)$. Put $k = t+1$. It is clear that $k \\neq 0$. We have $k\\left(n^{2}-1\\right) = t m^{2}$. Note that $(k, t) = 1$ we get there is an integer $l$ such that $n^{2}-1 = l t$ (which gives $l = m^{2}+1-n^{2}$) and $m^{2} = l k = ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
01va | Given $n \ge 2$. We call a group of people *n-compact* if for every person of group one can find *n* people (different from that person) which are acquainted with each other.
Find the maximum possible $N$ such that every *n*-compact group of $N$ people contains a subgroup of $n + 1$ people acquainted with each other. | [
"3. See III Silk Road Math. Competition 2004, Problem 4."
] | Belarus | Selection and Training Session | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2n | |
004f | Problem:
Halle todos los dígitos $a, b, c$, distintos de cero, tales que
$$
\frac{\overline{abc} + a + b + c}{ab + bc + ca}
$$
es un número entero.
Notación: $\overline{abc}$ indica el número cuyos dígitos son $a, b, c$, en ese orden. | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Spanish | proof and answer | null | |
0irt | Problem:
How many numbers less than $1,000,000$ are the product of exactly 2 distinct primes? You will receive $\left\lfloor 25-50 \cdot\left|\frac{N}{A}-1\right|\right\rfloor$ points, if you submit $N$ and the correct answer is $A$. | [
"Solution:\n\nAnswer: 209867 While it is difficult to compute this answer without writing a program or using a calculator, it can be approximated using the fact that the number of primes less than a positive integer $n$ is about $\\frac{n}{\\log n}$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | final answer only | 209867 | |
0222 | Problem:
Uma expressão - A expressão $\frac{a^{-2}}{a^{5}} \times \frac{4 a}{\left(2^{-1} a\right)^{-3}}$ onde $a \neq 0$, é igual a:
(a) $\frac{a^{3}}{2}$
(b) $\frac{2}{a^{3}}$
(c) $\frac{1}{2 a^{3}}$
(d) $\frac{a^{5}}{2}$
(e) $\frac{2}{a^{5}}$ | [
"Solution:\n\nTemos:\n$$\n\\begin{aligned}\n\\frac{a^{-2}}{a^{5}} \\times \\frac{4 a}{\\left(2^{-1} a\\right)^{-3}} & = a^{-2-5} \\times \\frac{2^{2} a}{2^{3} a^{-3}} \\\\\n& = a^{-7} \\times \\frac{a^{1-(-3)}}{2} \\\\\n& = a^{-7} \\times \\frac{a^{4}}{2} \\\\\n& = \\frac{a^{-7+4}}{2} = \\frac{a^{-3}}{2} = \\frac{1... | Brazil | Lista 4 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | MCQ | c | |
0hvp | Problem:
Aerith writes $50$ consecutive positive integers in a circle on a whiteboard. Each minute after, she simultaneously replaces each number $x$ with $2020a - x + 2020b$, where $a$ and $b$ were the numbers next to $x$. Can she choose her initial numbers such that she will never write down a negative number? | [
"Solution:\n\nLet $R_{0}$ be the initial sum of every other number that Aerith wrote down, and let $S_{0}$ be the sum of the remaining numbers. Let $R_{t}$ and $S_{t}$ be the corresponding sums after $t$ minutes. Because $D = R_{0} + S_{0}$ is a sum of $25$ even and $25$ odd numbers, it is itself odd, so $R_{0} - S... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
06z1 | Problem:
The positive integers $a$, $b$, $c$, $d$, $p$, $q$ satisfy $ad - bc = 1$ and $a / b > p / q > c / d$. Show that $q \geq b + d$ and that if $q = b + d$, then $p = a + c$. | [
"Solution:\n\n$p / q > c / d$ implies $pd > cq$ and hence $pd \\geq cq + 1$, so $p / q \\geq c / d + 1 / (qd)$. Similarly, $a / b > p / q$ implies $a / b \\geq p / q + 1 / (bq)$. So $a / b - c / d \\geq 1 / (qd) + 1 / (qb) = (b + d) / (qbd)$. But $a / b - c / d = 1 / bd$. Hence $q \\geq b + d$.\n\nNow assume $q = b... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0dpz | In an infinite sequence $\{\alpha\}$, $\{\alpha^2\}$, $\{\alpha^3\}$, ... there are only finitely many distinct values. Show that $\alpha$ is an integer. ($\{x\}$ denotes the fractional part of $x$, i.e. $\{x\} = x - [x]$, where $[x]$ is the greatest integer not greater than $x$.) (Golovanov A.S.) | [
"Step 1. We show that there is a positive integer $l$ such that $\\alpha^l$ is rational. Say the sequence is of length $k-1$. For any positive integer $n$ the sequence $\\{\\alpha^{nk}\\}$, $\\{\\alpha^{nk+1}\\}$, ..., $\\{\\alpha^{nk+k-1}\\}$ contains two equal elements. Hence, there are infinitely many pairs $i, ... | Silk Road Mathematics Competition | XXI SILK ROAD MATHEMATICAL COMPETITION | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Other"
] | English | proof only | null | |
00tx | Alice is drawing a shape on a piece of paper. She starts by placing her pencil at the origin, and then draws line segments of length $1$, alternating between vertical and horizontal segments. Eventually, her pencil returns to the origin, forming a closed, non-self-intersecting shape. Show that the area of this shape is... | [
"Colour the horizontal segments in every other line of the grid alternately red and blue as shown below:\n\n\n\nLet there be $r$ red segments on the perimeter and $s$ red segments in the interior of the shape. By considering the possibilities starting from a red segment, we see that every f... | Balkan Mathematical Olympiad | BMO 2022 shortlist | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
0emj | Find all triangular numbers which are also perfect squares. | [
"Suppose that $n^2 = T_m = \\frac{m(m+1)}{2}$. Multiplying by $8$,\n$$\n8n^2 = 4m^2 + 4m = (2m + 1)^2 - 1 \\Leftrightarrow (2m + 1)^2 - 8n^2 = 1.\n$$\nLetting $A = 2m+1$ and $B = 2n$, this gives Pell's equation, $A^2 - 2B^2 = 1$. The first solution is $A = 3, B = 2$, and the general solution is given by $a_k + \\sq... | South Africa | South-Afrika 2011-2013 | [
"Number Theory > Diophantine Equations > Pell's equations"
] | null | proof and answer | All square triangular numbers are given by x^2 = (1/32) ((3 + 2√2)^k − (3 − 2√2)^k)^2 for k ∈ ℕ; equivalently, they correspond to solutions of A^2 − 2B^2 = 1 with A = 2m + 1 and B = 2n. | |
09j8 | Find all positive integer solutions to the equation
$$
4z^2(2z^2 + 1) = 4xy - 2x - y.
$$ | [
"Answer: There is no solution.\nOur equation is equivalent to\n$$\n(4z^2 + 1)^2 = (4x - 1)(2y - 1).\n$$\nSince $4x - 1 \\equiv 3 \\pmod{4}$, there exists a prime divisor $p \\ge 3$ of $4x - 1$ such that $p \\equiv 3 \\pmod{4}$. Then $p \\mid 4z^2 + 1$, thus $4z^2 \\equiv -1 \\pmod{p}$. By Fermat's theorem, we have\... | Mongolia | Mongolian Mathematical Olympiad Round 3 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers... | null | proof and answer | No positive integer solutions | |
07ry | Let $n$ be a positive integer and define
$$
f(j) = \binom{2j}{j} \cdot \binom{2n-2j}{n-j}, \quad j = 0, 1, \dots, n.
$$
Prove that
(a) $f(j) \ge f(j+1)$, if $0 \le j < n/2$;
(b) $f$ is strictly convex, i.e.,
$$
f(j+1) + f(j-1) > 2f(j), \quad j = 1, 2, \dots, n-1.
$$ | [
"Let $u_j = \\binom{2j}{j}$, $j = 0, 1, 2, \\dots$. For fixed $n, j$ we use the abbreviation $m = n-j$ to get $f(j) = u_j u_{n-j} = u_j u_m$ and $f(j+1) = u_{j+1} u_{n-j-1} = u_{j+1} u_{m-1}$.\n\na.\nSince\n$$\nu_{j+1} = \\frac{2(2j+1)}{j+1} u_j \\quad \\text{and} \\quad u_m = \\frac{2(2m-1)}{m} u_{m-1}, \\text{ we... | Ireland | Irish | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
007c | One of the numbers $1$, $2$, $3$ is written in each cell of a rectangular table with $4$ rows and $n$ columns. For every three different columns there is a row that intersects them at cells with different numbers. Find the maximum $n$ for which there exists such a table. | [
"The maximum $n$ is $9$. An example with $n = 9$ is the table to the right.\n\nSuppose that there is such a table $T$ with $n \\ge 10$ columns. Let $3$ be the least represented number in row $4$. Then $1$ and $2$ combined occur at least $7$ times in row $4$. So we can select $7$ columns whose intersections with row... | Argentina | Mathematical Olympiad Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 9 | |
0i1h | Problem:
What is the last digit of $17^{103} + 5$? | [
"Solution:\n\nLet $\\langle N \\rangle$ be the last digit of $N$. $\\left\\langle 17^{2} \\right\\rangle = 9$, $\\left\\langle 17^{3} \\right\\rangle = 3$, $\\left\\langle 17^{4} \\right\\rangle = 1$, and $\\left\\langle 17^{5} \\right\\rangle = 7$. Since this pattern keeps on repeating itself, $\\left\\langle 17^{... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | final answer only | 8 | |
02ig | Problem:
Uma certa máquina é capaz de produzir $8$ réguas em cada minuto. Quantas réguas esta máquina consegue produzir em $15$ minutos?
A) $104$
B) $110$
C) $112$
D) $128$
E) $120$ | [
"Solution:\n\nSe a máquina produz $8$ réguas em $1$ minuto, em $15$ minutos ela produzirá $8 \\times 15 = 120$ réguas."
] | Brazil | Brazilian Mathematical Olympiad | [
"Math Word Problems"
] | null | MCQ | E | |
0cyw | In triangle $A B C$, let $I_{a}, I_{b}, I_{c}$ be the centers of the excircles tangent to sides $B C, C A, A B$, respectively. Let $P$ and $Q$ be the tangency points of the excircle of center $I_{a}$ with lines $A B$ and $A C$. Line $P Q$ intersects $I_{a} B$ and $I_{a} C$ at $D$ and $E$. Let $A_{1}$ be the intersectio... | [
"We shall prove that $A_{1}$ is the orthocenter of triangle $I_{a} B C$. Indeed, we have\n$$\n\\begin{aligned}\n\\widehat{P E I_{a}} & =180^{\\circ}-\\widehat{P Q A}-\\widehat{E C Q} \\\\\n& =180^{\\circ}-\\left(90^{\\circ}-\\frac{1}{2} \\widehat{A}\\right)-\\left(90^{\\circ}-\\frac{1}{2} \\widehat{C}\\right) \\\\\... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
00mp | Let $ABC$ be a triangle and $P$ a point inside the triangle such that the centers $M_B$ and $M_A$ of the circumcircles $k_B$ and $k_A$ of triangles $ACP$ and $BCP$, respectively, lie outside the triangle $ABC$. In addition, we assume that the three points $A$, $P$ and $M_A$ are collinear as well as the three points $B$... | [
"We put $\\varphi := \\angle CBP$, cf. Figure 1. Then we get for the corresponding central angle $\\angle CM_A P =$\n\nFigure 1: Problem 4\n$2\\varphi$. Since $M_A C M_B P$ is a deltoid having $M_A M_B$ as its axis of symmetry, we deduce $\\angle C M_A M_B = \\varphi = \\angle C B M_B$. The... | Austria | 49th Austrian Mathematical Olympiad, National Competition (Final Round, part 2) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0i3l | Find all pairs of nonnegative integers $(m, n)$ such that
$$
(m + n - 5)^2 = 9mn.
$$ | [
"**First Solution.** The equation is symmetric in $m$ and $n$. The solutions are the unordered pairs\n$$\n\\{5F_{2k}^2, 5F_{2k+2}^2\\}, \\quad \\{L_{2k-1}^2, L_{2k+1}^2\\},\n$$\nwhere $k$ is a nonnegative integer and $\\{F_j\\}$, $\\{L_j\\}$ are the **Fibonacci** and **Lucas sequences**, respectively — that is, the... | United States | USA IMO | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common div... | English | proof and answer | All solutions are given (up to order) by {5 F_{2k}^2, 5 F_{2k+2}^2} and {L_{2k-1}^2, L_{2k+1}^2} for nonnegative integers k, where F denotes the Fibonacci numbers and L denotes the Lucas numbers (extended with L_{-1} = −1 and L_0 = 2). | |
0hf3 | In the triangle $ABC$, $H$ is a midpoint of the altitude $AD$ and $O$ is a centre of the circumscribed circle. A line perpendicular to the line $HO$ passing through $H$ intersects $AB$ and $AC$ at $P$ and $Q$ respectively. Prove that the midpoints of $BP$, $CQ$ and $O$ are collinear. | [
"Define $M_C$ and $M_B$ as midpoints of $AB$ and $AC$ respectively. Then $H$ lies on the line $M_B M_C$ due to its initial position. Note that projections of $O$ on the lines $AB$, $AC$ and $PQ$ are points $M_C$, $M_B$ and $H$ which lie on the Simson line (Fig 33). From this can be concluded that $O$ lies on the ci... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0g3l | Problem:
Let $n$ be a positive integer. Call a sequence of positive integers $a_{1}, a_{2}, \ldots, a_{n}$ tame if it satisfies
$$
1 \cdot a_{1} \leq 2 \cdot a_{2} \leq \ldots \leq n \cdot a_{n}
$$
Determine the number of tame permutations of $1,2, \ldots, n$. | [
"Solution:\nWe prove that the number of possibilities is the $n^{\\text{th}}$ Fibonacci number $F_{n}$ by induction $\\left(F_{0}=1, F_{1}=1\\right)$. For the base case, observe that there is 1 way to do it for $n=1$ and 2 ways for $n=2$ (all the arrangements work in both cases).\n\nNow suppose this holds true up t... | Switzerland | IMO Selection | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | F_n | |
06wr | Determine all integers $n \geqslant 2$ with the following property: every $n$ pairwise distinct integers whose sum is not divisible by $n$ can be arranged in some order $a_{1}, a_{2}, \ldots, a_{n}$ so that $n$ divides $1 \cdot a_{1}+2 \cdot a_{2}+\cdots+n \cdot a_{n}$. | [
"Answer: All odd integers and all powers of $2$.\n\nIf $n=2^{k} a$, where $a \\geqslant 3$ is odd and $k$ is a positive integer, we can consider a set containing the number $2^{k}+1$ and $n-1$ numbers congruent to $1$ modulo $n$. The sum of these numbers is congruent to $2^{k}$ modulo $n$ and therefore is not divis... | IMO | IMO 2021 Shortlisted Problems | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | All odd integers and all powers of 2 | |
0k6b | Problem:
Four players stand at distinct vertices of a square. They each independently choose a vertex of the square (which might be the vertex they are standing on). Then, they each, at the same time, begin running in a straight line to their chosen vertex at $10$ mph, stopping when they reach the vertex. If at any ti... | [
"Solution:\n\nObserve that no two players can choose the same vertex, and no two players can choose each other's vertices. Thus, if two players choose their own vertices, then the remaining two also must choose their own vertices (because they can't choose each other's vertices), thus all $4$ players must choose th... | United States | HMMT November 2019 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 11 | |
09zi | Kjell has a large piece of graph paper of $100 \times 100$ squares.
How many squares can Kjell colour at most without there being three coloured squares in a row, all directly next to each other or all directly above each other? | [] | Netherlands | Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 6667 | |
0l9c | Let $f: N \times Z \to N$ be a given function satisfying simultaneously the following conditions:
i/ $f(0, 0) = 5^{2003}$, $f(0, n) = 0$ for every integer $n \neq 0$;
ii/ $f(m, n) = f(m - 1, n) - 2\left[\frac{f(m - 1, n)}{2}\right] + \left[\frac{f(m - 1, n - 1)}{2}\right] + \left[\frac{f(m - 1, n + 1)}{2}\right]$ for e... | [] | Vietnam | Vietnamese Team Selection Contest for the 44th IMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
04vk | Suppose that we have three natural numbers $a$, $b$ and $c$ such that one of the values
$$
gcd(a, b) \cdot \text{lcm}(b, c), \gcd(b, c) \cdot \text{lcm}(c, a), \gcd(c, a) \cdot \text{lcm}(a, b),
$$
is equal to the product of the other two. Prove that one of the numbers $a$, $b$ and $c$ is a multiple of a different one. | [
"Without loss of generality, we can assume that\n$$\nP = \\gcd(a, b) \\cdot \\text{lcm}(b, c) = \\gcd(b, c) \\cdot \\text{lcm}(c, a) \\cdot \\gcd(c, a) \\cdot \\text{lcm}(a, b).\n$$\nUsing the well-known relation $\\gcd(x, y) \\cdot \\text{lcm}(x, y) = xy$, we get:\n$$\n(abc)^2 = (ab) \\cdot (bc) \\cdot (ca) = \\gc... | Czech Republic | Final Round of the 73rd Czech and Slovak Mathematical Olympiad (March 17–20, 2024) | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof only | null | |
0d3o | Find all ordered triples $(a, b, c)$ of positive integers which satisfy
$$
5^{a} + 3^{b} - 2^{c} = 32
$$ | [
"By considering this equation modulo $3$, we get\n$$\n5^{a} + 3^{b} - 2^{c} \\equiv 2^{a} - 2^{c} \\equiv 2 \\pmod{3}\n$$\nThis occurs only when $a$ is even and $c$ is odd.\n\nIf $c = 1$, the equation becomes $5^{a} + 3^{b} = 34$, which is equivalent to $a = 2$ and $b = 1$.\n\nIf $c \\neq 1$, then $c \\geq 3$, sinc... | Saudi Arabia | SAMC | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | English, Arabic | proof and answer | (2, 2, 1) | |
0h1o | Numbers $a_1, a_2, a_3, a_4, a_5$ and $b_1, b_2, b_3, b_4, b_5$ are permutations of $1, 2, 3, 4, 5$. Prove that among five numbers $a_1b_1, a_2b_2, a_3b_3, a_4b_4, a_5b_5$ at least two have the same remainders modulo $5$. | [
"WLOG, $a_5 = 5$. If $b_5 \\neq 5$, then there will be two numbers that are divisible by $5$. Let $b_5 \\neq 5$ and suppose that the statement of the problem is false. Then the numbers $a_1b_1, a_2b_2, a_3b_3, a_4b_4$ have remainders $1, 2, 3$ and $4$ modulo $5$ (in some order). Then from one hand the product has r... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof only | null | |
03ih | Problem:
If $a679b$ is a five digit number (in base 10) which is divisible by $72$, determine $a$ and $b$. | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a=3, b=2 | |
063x | Problem:
Es sei $A B C D E$ ein konvexes Fünfeck mit den Eigenschaften
$$
\overline{A B}=\overline{B C}=\overline{C D}, \angle B A E=\angle D C B \text{ und } \angle E D C=\angle C B A .
$$
Man beweise, dass die Lotgerade von $E$ auf $B C$ durch den Schnittpunkt von $A C$ und $B D$ verläuft. | [
"Solution:\n\nWegen der Voraussetzungen sind $\\triangle A B C$ und $\\triangle B C D$ gleichschenklig. Daher geht die Mittelsenkrechte zu $A C$ durch $B$ und die Mittelsenkrechte zu $B D$ durch $C$. Beide schneiden sich im Punkt $I$ (siehe Figur).\n\nWegen $B D \\perp C I$ und $A C \\perp B I$ schneiden sich $A C$... | Germany | 2. Auswahlklausur | [
"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"
] | null | proof only | null | |
04ix | Determine all values that the expression
$$ \frac{1 + \cos^2 x}{\sin^2 x} + \frac{1 + \sin^2 x}{\cos^2 x}, $$
can attain, where $x$ is a real number. | [] | Croatia | Croatia Mathematical Competitions | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | [6, ∞) | |
08ur | On the left scale of a balance 4 weights weighing $22$, $24$, $26$, $28$ grams each are placed and on the right scale 4 weights weighing $23$, $25$, $27$, $29$ grams each are placed. This balance gets tilted toward the side having heavier total weight, and settles in equilibrium position when the total weights of both ... | [
"Suppose we keep on removing weights from the scale on the tilted side, and instead of stopping when the balance gets in the equilibrium position with some weights still left on the scales, remove another weight from the left-side scale and keep on removing weights from the tilted side again until there are no more... | Japan | Japan Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 480 | |
0fgn | Problem:
Demostrar que para todo número natural $n>1$ se cumple
$$
1 \cdot \sqrt{\left(\begin{array}{l}
n \\
1
\end{array}\right)}+2 \cdot \sqrt{\left(\begin{array}{l}
n \\
2
\end{array}\right)}+\cdots+n \cdot \sqrt{\left(\begin{array}{l}
n \\
n
\end{array}\right)}<\sqrt{2^{n-1} n^{3}}
$$ | [
"Solution:\nPongamos\n$$\nu=(1,2, \\ldots, n) \\quad \\mathrm{y} \\quad v=\\left(\\left(\\begin{array}{c}\nn \\\\\n1\n\\end{array}\\right),\\left(\\begin{array}{l}\nn \\\\\n2\n\\end{array}\\right), \\ldots,\\left(\\begin{array}{l}\nn \\\\\nn\n\\end{array}\\right)\\right)\n$$\nSabemos que $|u \\cdot v| \\leq\\|u\\| ... | Spain | OME 23 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
0asa | Problem:
How many pairs of integers solve the system $|x y| + |x - y| = 2$ if $-10 \leq x, y \leq 10$? | [
"Solution:\n\n(ans. $4: (2, 0), (0, 2), (-2, 0), (0, -2)$.\nThe cases are (A) $|x y| = 0$ and $|x - y| = 2 \\Rightarrow (x, y) = (0, \\pm 2), (\\pm 2, 0)$; (B) $|x y| = 2$ and $|x - y| = 0 \\Rightarrow$ no solution; (C) $|x y| = |x - y| = 1 \\Rightarrow$ no solution.)"
] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | 4 | |
02dk | $S$ is a $(k+1) \times (k+1)$ array of lattice points. How many squares have their vertices in $S$? | [
"The key is to consider how many squares have their vertices on the perimeter of a given $(n+1) \\times (n+1)$ array whose sides are parallel to the sides of the array.\n\n\n\nThe diagram shows that there are $n$ such squares. There are $(k+1-n)^2$ such arrays. So the total number of square... | Brazil | IV OBM | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | k(k+1)^2(k+2)/12 | |
0ied | Problem:
Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$. How many positive integers less than $2005$ can be expressed in the form $\lfloor x\lfloor x\rfloor\rfloor$ for some positive real $x$? | [
"Solution:\n\nLet $\\{x\\} = x - \\lfloor x\\rfloor$ be the fractional part of $x$. Note that\n$$\n\\lfloor x\\lfloor x\\rfloor\\rfloor = \\lfloor (\\lfloor x\\rfloor + \\{x\\})\\lfloor x\\rfloor \\rfloor = \\lfloor x\\rfloor^{2} + \\lfloor \\{x\\}\\lfloor x\\rfloor \\rfloor.\n$$\nBecause $\\{x\\}$ may take on any ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 990 | |
04j1 | In a country between every two cities there is a direct bus or a direct train line (all lines are two-way and they don't pass through any other city). Prove that all cities in that country can be arranged in two disjoint sets so that all cities in one set can be visited using only train so that no city is visited twice... | [
"Let $G$ be the set of all cities in the country. We call a pair $(A, Z)$, where $A$ and $Z$ are disjoint subsets of $G$ good if all cities in the set $A$ can be visited using only bus such that no city is visited twice and all cities in the set $Z$ can be visited using only train such that no city is visited twice... | Croatia | Croatia Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
02u3 | Problem:
Encontre as soluções da equação
$$
\sqrt{x+\sqrt{4 x+\sqrt{16 x+\sqrt{\ldots+\sqrt{4^{n} x+3}}}}}=1+\sqrt{x}
$$ | [
"Solution:\nElevemos a equação dada ao quadrado $n$ vezes, obtendo\n$$\n\\begin{aligned}\nx+\\sqrt{4 x+\\sqrt{16 x+\\sqrt{\\ldots+\\sqrt{4^{n} x+3}}}} & =1+2 \\sqrt{x}+x \\\\\n\\sqrt{4 x+\\sqrt{16 x+\\sqrt{\\ldots+\\sqrt{4^{n} x+3}}}} & =1+2 \\sqrt{x} \\\\\n4 x+\\sqrt{16 x+\\sqrt{\\ldots+\\sqrt{4^{n} x+3}}} & =1+4 ... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | x = 2^{-2n} | |
07xo | Three circles of radius $1$ are packed without overlapping in a square of side $a$. What is the smallest value of $a$? | [
"We start with three circles of radius $1$ that are tangent to each other externally and consider all rectangles that contain these circles such that each side is tangent to at least one of the circles. A square that contains the three circles and has sides parallel to one such rectangle has side length not smaller... | Ireland | IRL_ABooklet_2025 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / ... | null | proof and answer | 2 + (sqrt(6) + sqrt(2))/2 | |
001z | Las diagonales $AC$ y $BD$ de un cuadrilátero convexo $ABCD$ se cortan en $E$ y $\frac{CE}{AC} = \frac{3}{7}$, $\frac{DE}{BD} = \frac{4}{9}$. Sean $P$ y $Q$ los puntos que dividen el segmento $BE$ en tres partes iguales, con $P$ entre $B$ y $Q$, y sea $R$ el punto medio del segmento $AE$. Calcular $\frac{\text{area}(AP... | [] | Argentina | XX Olimpiada Matemática Argentina | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | español | proof and answer | 10/63 | |
0hb5 | The inscribed circle of an acute triangle **ABC** touches sides **AB** and **BC** in points $C_1$ and $A_1$ respectively. Let **M** be the midpoint of the side **AC**, and **N** be the midpoint of an arc **ABC** of the circumcircle of **ABC**. Also, let **P** be the projection of a point **M** to the segment $A_1C_1$. ... | [
"Let's denote the second intersection point of the line **BI** with the circumcircle of $\\triangle ABC$ as **W** (fig. 35).\nClearly, **W** is a midpoint of the smaller arc **AC**, and points **M**, **W**, and **N** lie on the same line.\n\nHence, it is enough to prove that triangles $PMN$ and $IWN$ are similar (w... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | English | proof only | null | |
00de | Let $L$ be the 2022-digit number made only by ones, that is $L = \underbrace{111\ldots11}_{2022 \text{ digits}}$. Find the sum of the digits of $9L^2 + 2L$. | [
"Notice that $9L^2 + 2L = (9L + 2)L$.\nThe number $9L$ is the 2022-digit number made only by nines. By adding 2 we get $9L + 2 = \\underbrace{1000\\ldots01}_{2022 \\text{ digits}} = 10^{2022} + 1$. Now it is clear that\n$$\n(9L + 2)L = 10^{2022}L + L = \\underbrace{111\\ldots11}_{2022 \\text{ digits}}\\underbrace{0... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 4044 | |
05qf | Problem:
Soit $ABC$ un triangle. On note $L, M, N$ les milieux de $[BC]$, $[CA]$ et $[AB]$. Notons $(d)$ la tangente en $A$ au cercle circonscrit à $ABC$. La droite $(LM)$ coupe $(d)$ en $P$, et la droite $(LN)$ coupe $(d)$ en $Q$. Montrer que $(CP)$ et $(BQ)$ sont parallèles. | [
"Solution:\n\n\nOn a $(AQ, AB) = (CA, CB) = (LQ, LB)$ donc $A, Q, B, L$ sont cocycliques, et de même $A, P, C, M$ sont cocycliques. On a donc $(CP, BC) = (CP, CL) = (AP, AL) = (AQ, AL) = (BQ, BL) = (BQ, BC)$, donc $(CP)$ et $(BQ)$ sont parallèles."
] | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
040e | Let real numbers $a, b, c$ and $d$ satisfy
$$
f(x) = a \cos x + b \cos 2x + c \cos 3x + d \cos 4x \le 1
$$
for any real number $x$. Find the values of $a, b, c$ and $d$ such that $a + b - c + d$ takes the maximum number. (posed by Li Shenghong) | [
"Since\n$$\n\\begin{aligned}\nf(0) &= a + b + c + d, \\\\\nf(\\pi) &= -a + b - c + d, \\\\\nf\\left(\\frac{\\pi}{3}\\right) &= \\frac{a}{2} - \\frac{b}{2} - c - \\frac{d}{2},\n\\end{aligned}\n$$\nthen\n$$\na + b - c + d = f(0) + \\frac{2}{3}f(\\pi) + \\frac{4}{3}f\\left(\\frac{\\pi}{3}\\right) \\le 3\n$$\nif and on... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | Maximum is 3; a=1, b=1/2, c=−1, d=1/2 | |
0def | Let $ABC$ be a non-isosceles triangle with incenter $I$ and let the circumcircle of the triangle $ABC$ has radius $R$. Let $AL$ be the external angle bisector of $\angle BAC$ with $L \in BC$. Let $K$ be the point on the perpendicular bisector of $BC$ such that $IL \perp IK$. Prove that $OK = 3R$. | [
"Denote $M$ as the midpoint of the arc $BC$ not containing $A$ of $(O)$, we need to prove $MK = 2OM$. Denote $BD, CE$ as the internal bisectors of $ABC$ then $D, E, L$ are collinear. Note that $L(AI, DB) = -1$ and take the orthogonal projection from $I$ to get $(Ix, Iy, IK, IM) = -1$ in which\n$$\nIM \\perp LA, IK ... | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plan... | null | proof only | null | |
0jtd | Problem:
Let $\Delta A_{1} B_{1} C$ be a triangle with $\angle A_{1} B_{1} C=90^{\circ}$ and $\frac{C A_{1}}{C B_{1}}=\sqrt{5}+2$. For any $i \geq 2$, define $A_{i}$ to be the point on the line $A_{1} C$ such that $A_{i} B_{i-1} \perp A_{1} C$ and define $B_{i}$ to be the point on the line $B_{1} C$ such that $A_{i} B_... | [
"Solution:\nWe claim that $\\Gamma_{2}$ is the incircle of $\\triangle B_{1} A_{2} C$. This is because $\\triangle B_{1} A_{2} C$ is similar to $A_{1} B_{1} C$ with dilation factor $\\sqrt{5}-2$, and by simple trigonometry, one can prove that $\\Gamma_{2}$ is similar to $\\Gamma_{1}$ with the same dilation factor. ... | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"... | null | final answer only | 4030 | |
0ep3 | $ABCD$ is a rectangle and $P$ is a point on $BC$. If the area of triangle $ABP$ is one third of the area of the rectangle, then the ratio $BP : PC$ is
(A) $5 : 2$
(B) $3 : 2$
(C) $2 : 1$
(D) $3 : 1$
(E) $9 : 4$ | [
"$\\frac{\\text{area } \\triangle ABP}{\\text{area } ABCD} = \\frac{1}{3}$ so $\\frac{\\frac{1}{2} \\text{BP} \\cdot \\text{AB}}{\\text{BC} \\cdot \\text{AB}} = \\frac{1}{3}$ so $\\frac{\\frac{1}{2} \\text{BP}}{\\text{BC}} = \\frac{1}{3}$ and therefore $\\frac{\\text{BP}}{\\text{BC}} = \\frac{2}{3}$\nand then $BP :... | South Africa | South African Mathematics Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | MCQ | C | |
0cb2 | Find all the sequences of equal ratios of the form $\frac{a_1}{a_2} = \frac{a_3}{a_4} = \frac{a_5}{a_6} = \frac{a_7}{a_8}$ fulfilling the conditions:
- the set $\{a_1, a_2, \dots, a_8\}$ is the set of the positive divisors of $24$;
- the common value of the ratios is an integer. | [
"The common value $r$ of the ratios can be only a divisor of $24$, different from $1$; these divisors are $2$, $3$, $4$, $6$, $8$, $12$ and $24$.\nIf $r = 2$, then $2 = \\frac{24}{12} = \\frac{8}{4} = \\frac{6}{3} = \\frac{2}{1}$;\nif $r = 3$, then $3 = \\frac{24}{8} = \\frac{12}{4} = \\frac{6}{2} = \\frac{3}{1}$;\... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 24/12 = 8/4 = 6/3 = 2/1; 24/8 = 12/4 = 6/2 = 3/1; 24/6 = 12/3 = 8/2 = 4/1 | |
06th | Let $ABC$ be an acute triangle with orthocenter $H$. Let $G$ be the point such that the quadrilateral $ABGH$ is a parallelogram. Let $I$ be the point on the line $GH$ such that $AC$ bisects $HI$. Suppose that the line $AC$ intersects the circumcircle of the triangle $GCI$ at $C$ and $J$. Prove that $IJ = AH$.
(Australi... | [
"Since $HG \\parallel AB$ and $BG \\parallel AH$, we have $BG \\perp BC$ and $CH \\perp GH$. Therefore, the quadrilateral $BGCH$ is cyclic. Since $H$ is the orthocenter of the triangle $ABC$, we have $\\angle HAC = 90^\\circ - \\angle ACB = \\angle CBH$. Using that $BGCH$ and $CGJI$ are cyclic quadrilaterals, we ge... | IMO | 56th International Mathematical Olympiad Shortlisted Problems | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03f8 | Find all natural numbers $n$ for which the number of positive divisors of $\text{LCM}(1, 2, \dots, n)$ is a power of 2. | [
"For each prime $p$, the numbers of the interval $n \\in [p^2, p^3)$ are not solutions, because the degree of $p$ in the decomposition of the LCM is 2 and so contributes by a factor of 3 to the number of divisors. Therefore this number is not a power of 2. From Bertrand's postulate for a prime $p$ there is a prime ... | Bulgaria | Bulgarian Spring Tournament | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 1, 2, 3, 8 | |
03q1 | Suppose $a$, $b$, $c$ and $d$ are positive real numbers satisfying $ab + cd = 1$ and $P_i(x_i, y_i)$ ($i = 1, 2, 3, 4$) are four points on the unit circle which has the origin as its center. Prove that:
$$(ay_1 + by_2 + cy_3 + dy_4)^2 + (ax_4 + bx_3 + cx_2 + dx_1)^2 \le 2\left(\frac{a^2+b^2}{ab}+\frac{c^2+d^2}{cd}\rig... | [
"**Proof I** Set $u = ay_1 + by_2$, $v = cy_3 + dy_4$, $u_1 = ax_4 + bx_3$ and $v_1 = cx_2 + dx_1$. Then\n$$\nu^2 \\le (ay_1 + by_2)^2 + (ax_1 - bx_2)^2 \\\\ = a^2 + b^2 + 2ab(y_1 y_2 - x_1 x_2),$$\nthat is\n$$x_1 x_2 - y_1 y_2 \\le \\frac{a^2 + b^2 - u^2}{2ab}. \\qquad ①$$\n$$v_1^2 \\le (cx_2 + dx_1)^2 + (cy_2 - d... | China | China Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
02a1 | Problem:
Três circunferências de raios $1~\mathrm{cm}$, $2~\mathrm{cm}$ e $3~\mathrm{cm}$ são duas a duas tangentes exteriormente, como na figura ao lado.
Determine o raio da circunferência tangente exteriormente às três circunferências.
 | [
"Solution:\n\nLigando os centros das três circunferências obtemos o triângulo $\\triangle ABC$ de lados $AB=3~\\mathrm{cm}$, $AC=4~\\mathrm{cm}$ e $BC=5~\\mathrm{cm}$. Como $3^{2}+4^{2}=5^{2}$, esse triângulo é retângulo, com hipotenusa $BC$.\n\n\n\nConstrua o retângulo $ABDC$, fazendo uma ... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 6 cm | |
0i27 | For a set $S$, let $|S|$ denote the number of elements in $S$. Let $A$ be a set of positive integers with $|A| = 2001$. Prove that there exists a set $B$ such that
(i) $B \subseteq A$;
(ii) $|B| \geq 668$;
(iii) for any $u, v \in B$ (not necessarily distinct), $u + v \notin B$. | [
"**First Solution.** (By Reid Barton) For a positive integer $n$, let $Z_n$ denote the set of residues modulo $n$. Let $\\phi(n)$ be the **Euler function** which is defined to be the number of integers between $1$ and $n$ relatively prime to $n$. Call a set $A$ of residues modulo $3^n$ **sum-free** if for any $a, b... | United States | USA IMO | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"... | English | proof only | null | |
01ep | Find all the triples of non-negative integers $(a, b, c)$ for which the number
$$
\frac{(a+b)^4}{c} + \frac{(b+c)^4}{a} + \frac{(c+a)^4}{b}
$$
is integer and $a + b + c$ is prime. | [
"Answer $(1, 1, 1)$, $(1, 2, 2)$, $(2, 3, 6)$.\nLet $p = a + b + c$, then $a + b = p - c$, $b + c = p - a$, $c + a = p - b$ and\n$$\n\\frac{(p-c)^4}{c} + \\frac{(p-a)^4}{a} + \\frac{(p-b)^4}{b}\n$$\nis a non-negative integer. By expanding brackets we obtain that the number $p^4\\left(\\frac{1}{a}+\\frac{1}{b}+\\fra... | Baltic Way | Baltic Way shortlist | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | (1, 1, 1), (1, 2, 2), (2, 3, 6) | |
00n2 | Let $ABCD$ be a square. The equilateral triangle $BCS$ is constructed on the exterior of the side $BC$. Let $N$ denote the midpoint of the line segment $AS$ and let $H$ be the midpoint of the side $CD$.
Prove: $\angle NHC = 60^{\circ}$. | [
"Let $P$ be the midpoint of $BS$, see Figure 1.\n\nSince triangles $\\triangle SNP$ and $\\triangle SAB$ are similar with factor $2$, the segment $NP$ is parallel to $AB$ and half the length of the segment $AB$. Therefore $NPCH$ is a parallelogram.\n\nAs $NP$ and $BC$ are orthogonal and $PC$ and $BS$ are orthogonal... | Austria | Austria2019 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
02ms | Problem:
Um código - Na expressão abaixo, cada letra corresponde a um algarismo, sendo que letras diferentes correspondem a algarismos diferentes. Determine esses algarismos.
$$
6 \times A O B M E P = 7 \times M E P A O B
$$ | [] | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | A=5, O=3, B=8, M=4, E=6, P=1 | |
0ax3 | Problem:
In how many ways can the integers
$$
-5,-4,-3,-2,-1,1,2,3,4,5
$$
be arranged in a circle such that the product of each pair of adjacent integers is negative? (Assume that arrangements which can be obtained by rotation are considered the same.) | [] | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 2880 | |
0e6k | How many polynomials of degree $5$ with the coefficients being $1$ or $-1$ have a root at $1$?
(A) $5$
(B) $10$
(C) $15$
(D) $20$
(E) $24$ | [
"The value of the polynomial at $1$ is equal to the sum of all the coefficients of the polynomial. Since this value must be $0$, exactly three coefficients must be $1$ and three must be $-1$. The number of such polynomials is thus $\\binom{6}{3} = 20$."
] | Slovenia | National Math Olympiad 2012 | [
"Algebra > Algebraic Expressions > Polynomials"
] | null | MCQ | D | |
091v | Problem:
We call a positive integer $n$ amazing if there exist positive integers $a, b, c$ such that the equality
$$
n = (b, c)(a, b c) + (c, a)(b, c a) + (a, b)(c, a b)
$$
holds. Prove that there exist $2011$ consecutive positive integers which are amazing. (By $(m, n)$ we denote the greatest common divisor of positiv... | [
"Solution:\nWe may choose such positive integers $x_{1}, x_{2}, \\ldots, x_{2011}$ that the numbers\n$$\ny_{1} = x_{1}^{2}(x_{1} + 2), \\quad y_{2} = x_{2}^{2}(x_{2} + 2), \\quad \\ldots, \\quad y_{2011} = x_{2011}^{2}(x_{2011} + 2)\n$$\nare pairwise coprime. For example, we may choose $x_{1} = 1$ and $x_{i} = y_{1... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof only | null | |
0g7u | 一個 $3n \times 3n$ 的棋盤, 行與列都依序編號為 1 到 $3n$。方格 $(x, y)$ 依據 $x + y$ 模 3 的餘數為 0, 1 或 2, 分別塗成顏色 A, B 或 C。在每一個方格中擺上一個籌碼, 每個籌碼的顏色為 A, B 或 C。三種顏色的籌碼各有 $3n^2$ 個。
假設我們可以安排下列籌碼的重新排列方式：每個籌碼移動的距離不超過 $d$, 顏色 A 的籌碼取代顏色 B 的籌碼、顏色 B 的籌碼取代顏色 C 的籌碼、且顏色 C 的籌碼取代顏色 A 的籌碼。試證：我們還有另一種籌碼重新排列的方式，使得每個籌碼移動的距離不超過 $d+2$，並且移動後每個方格內都有與該方格顏色相同的籌碼。 | [
"不失一般性，只需證明所有顏色 A 的籌碼（簡稱 A-籌碼）可以移動到不同的顏色 A 的方格（簡稱 A-方格）中，並且每個 A-籌碼從原來的方格所移動的距離不超過 $d+2$。這就是說，我們可以將 $3n^2$ 個 A-籌碼與 $3n^2$ 個 A-方格作完美配對，並且每一對之間的距離最多是 $d+2$。\n爲了找出這個的完美配對，我們建構一個雙色圖：所有的 A-方格為一群頂點，所有的 A-籌碼為另一群頂點。\n將原來的棋盤分割成 $3 \\times 1$ 的三方塊；每一個三方塊恰包含一個 A-方格。找一個把 A-籌碼送到 B-籌碼、B-籌碼送到 C-籌碼、C-籌碼送到 A-籌碼，且每一個移動的距離都不超過 $d$ 的排列... | Taiwan | 二〇一三數學奧林匹亞競賽第二階段選訓營 | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
05xh | Problem:
Soit $k, n \geqslant 1$ deux entiers fixés. Thanima possédait $2n$ bonbons de chaque couleur. Elle a donné deux bonbons de couleurs différentes à chacun des enfants de sa famille. Sachant que, quelle que soit la manière de choisir $k+1$ enfants, il y en a deux parmi eux qui ont reçu une couleur de bonbon en c... | [
"Solution:\n\nOn peut modéliser le problème par un graphe : chaque sommet de ce graphe est une couleur. Si Thanima a donné un bonbon d'une couleur $c_1$ et un bonbon d'une couleur $c_2$ à un enfant de sa famille, on ajoute une arête entre $c_1$ et $c_2$.\n\nComme Thanima n'a que $2n$ bonbons de chaque couleur, le d... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3kn | |
065w | Let $\triangle ABC$ be a scalene triangle with $AB < AC$, with circumcircle $c(O, R)$. The circle $c_1(A, AB)$ intersects the side $BC$ at $E$ and the circle $c$ at $F$. The line $EF$ meets for a second time the circle $c$ at point $D$ and the side $AC$ at point $M$. The line $AD$ intersects the side $BC$ at point $K$.... | [
"The angle $\\hat{F}_1$ is inscribed into the circle $c_1$ with corresponding central angle $\\hat{BAE}$. Hence:\n\nFigure 4\n---\n$$\n\\hat{F}_1 = \\frac{BA\\hat{E}}{2} = \\frac{\\hat{A}_1 + \\hat{A}_2}{2} \\qquad (1)\n$$\nFrom the cyclic quadrilateral *AFDB* we have:\n$$\n\\hat{F}_1 = \\h... | Greece | SELECTION EXAMINATION | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
07hp | The triangle $ABC$ is given. Point $T$ is the intersection of the $A$-symmedian and the circumcircle of $ABC$. Point $D \neq A$ lies on the line $AC$ in such a way that $BD = BA$. The line tangent to the circumcircle of $ADT$ at point $D$ intersects the circumcircle of $DCT$, for the second time at point $K$. Prove tha... | [
"According to the assumptions of the problem, we have $\\angle ADT = \\angle TKC$ and $\\angle TCK = \\angle TDK = \\angle TAD$. From these two equations, it follows that $\\triangle TDA \\sim \\triangle TKC$. Therefore, we can write:\n$$\n\\frac{KC}{AD} = \\frac{TC}{AT} \\qquad (1)\n$$\nOn the other hand, if we de... | Iran | 40th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0f2i | Problem:
Let $a_n$ be the nearest integer to $\sqrt{n}$. Find
$$
\frac{1}{a_1} + \frac{1}{a_2} + \ldots + \frac{1}{a_{1980}}.
$$ | [] | Soviet Union | ASU | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 88 | |
0edk | Problem:
Jure je na ravnini risal po 4 različne premice v različnih položajih in z $n$ označil število presečišč premic v posamezni legi. S katero množico so podane natanko vse možne vrednosti števila $n$?
(A) $\{0,2,3,4,5,6\}$
(B) $\{0,1,3,4,5,6\}$
(C) $\{0,1,3,4,6\}$
(D) $\{1,3,4,5,6\}$
(E) $\{0,1,2,3,4,6\}$ | [
"Solution:\n\nKot prikazujejo slike, se lahko štiri različne premice v ravnini sekajo v $0,1,3,4$, 5 ali 6 točkah.\n\n\n\nDenimo, da sta presečišči natanko 2. V vsakem od teh presečišč se potem sekata bodisi 2 premici bodisi 3 premice. Denimo, da se v enem presešišču, označimo ga z $A$, sek... | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | MCQ | B | |
03m0 | Problem:
Find all ordered pairs $(a, b)$ such that $a$ and $b$ are integers and $3^{a} + 7^{b}$ is a perfect square. | [
"Solution:\nIt is obvious that $a$ and $b$ must be non-negative.\nSuppose that $3^{a} + 7^{b} = n^{2}$. We can assume that $n$ is positive. We first work modulo $4$. Since $3^{a} + 7^{b} = n^{2}$, it follows that\n$$\nn^{2} \\equiv (-1)^{a} + (-1)^{b} \\pmod{4}\n$$\nSince no square can be congruent to $2$ modulo $4... | Canada | CANADIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | (1, 0) and (2, 1) | |
07qc | Given a set of 2016 distinct points in the plane, show that we can choose a "circle of evil" $C$ in the plane such that exactly 666 of these points lie strictly inside $C$, and none of them lies on $C$. | [
"Let $S$ denote the set of the 2016 given points. Consider the collection of perpendicular bisectors of pairs of distinct points in $S$. This is a finite collection of lines, so we can pick a point $P$ not on any one of these lines, and also not in $S$.\n\nFor $r \\ge 0$, let $N(r)$ be the number of points in $S$ w... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
09sk | Problem:
Zij $A B C$ een scherphoekige driehoek. Zij $H$ het voetpunt van de hoogtelijn vanuit $C$ op $A B$. Veronderstel dat $|A H| = 3|B H|$. Laat $M$ en $N$ de middens van respectievelijk $A B$ en $A C$ zijn. Zij $P$ een punt zodat $|N P| = |N C|$ en $|C P| = |C B|$ en zodat $B$ en $P$ aan verschillende kanten van ... | [
"Solution:\n\nEr is maar één configuratie. Omdat $N$ het midden van $A C$ is, is $|N C| = |N A|$. Gegeven was ook dat $|N P| = |N C|$, dus $N$ is het middelpunt van een cirkel door $A, C$ en $P$. Met de stelling van Thales volgt hieruit dat $\\angle A P C = 90^{\\circ}$. Omdat $|A H| = 3|B H|$ en $M$ het midden van... | Netherlands | IMO-selectietoets | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02oo | Prove that, for all convex pentagons $P_1P_2P_3P_4P_5$ with area $1$, there are indices $i$ and $j$ (assume $P_6 = P_1$ and $P_7 = P_2$) such that:
$$
\text{area } \triangle P_i P_{i+1} P_{i+2} \le \frac{5 - \sqrt{5}}{10} \le \text{area } \triangle P_j P_{j+1} P_{j+2}
$$ | [
"Let's prove that there exists a triangle $P_j P_{j+1} P_{j+2}$ with area less than or equal to $\\alpha = \\frac{5-\\sqrt{5}}{10}$. Suppose that all triangles $P_j P_{j+1} P_{j+2}$ have area greater than $\\alpha$.\n\n\n\nLet diagonals $P_1P_4$ and $P_3P_5$ meet at $Q$. Since $Q \\in P_3P_... | Brazil | Brazilian Math Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0g49 | Problem:
a school class of $n \geq 2$ children is taking several group pictures. For every group with at least one child, there is exactly one picture containing this specific group. The pictures are now hung up in different rooms in the school, such that every child appears in at most one photo per room.
a. Show tha... | [
"Solution:\n\na.\nOne way of doing this is to pair each photo with its complement and put the photo with all children in a room by itself. Indeed, this is the only way (see appendix).\n\nb.\nThe simplest way of doing this is to provide a set of $2^{n-1}$ photos, of which no two are disjoint. There are many ways of ... | Switzerland | Second round 2022 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2^{n-1} rooms are necessary and sufficient | |
0ejw | Problem:
Na tabelo velikosti $2 \times n$, kjer je $n$ naravno število, naključno na različni polji postavimo dva žetona. Kolikšna je verjetnost, da se da preostanek tabele prekriti z dominami velikosti $2 \times 1$? Domine se ne smejo prekrivati in ne smejo segati čez rob tabele. | [
"Solution:\n\nTabelo pobarvamo kot šahovnico. Tedaj je v tabeli enako število belih in črnih polj. Ker vsaka domina prekrije 1 belo in 1 črno polje, vse domine skupaj prekrijejo enako število belih in črnih polj. Da lahko preostanek tabele prekrijemo z dominami velikosti $2 \\times 1$, morata biti torej žetona post... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | n/(2n-1) |
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