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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0jua | Problem:
Find the number of positive integers less than $1000000$ which are less than or equal to the sum of their proper divisors. If your answer is $X$ and the actual value is $Y$, your score will be $\max \left(0,20-80\left|1-\frac{X}{Y}\right|\right)$ rounded to the nearest integer.
Proposed by: Allen Liu | [
"Solution:\n\nAnswer: $247548$\n\nThe following code computes the answer:\n\n```\nN = 1000000\ns = [0] * N\nans = 0\nfor i in range(1, N):\n if i <= s[i]:\n ans += 1\n for j in range(i + i, N, i):\n s[j] += i\nprint(ans)\n```\n\nHere, $s[i]$ stores the sum of proper divisors of $i$. For each $i$... | United States | HMMT November | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Discrete Mathematics > Algorithms"
] | null | final answer only | 247548 | |
0as1 | Problem:
A tournament consists of matches between exactly three players, each, respectively, garnering 2 points, 1 point, and a zero score. The ones who obtained no score are eliminated and the rest are grouped into threes to engage again in matches, with possibly one or two players having a bye. If there are 999 playe... | [
"Solution:\n997, which is equal to the number of players which earned a zero score in each of the 3-person matches."
] | Philippines | 13th Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | final answer only | 997 | |
0i4p | Problem:
A lattice point in the plane is a point of the form $(n, m)$, where $n$ and $m$ are integers. Consider a set $S$ of lattice points. We construct the transform of $S$, denoted by $S'$, by the following rule: the pair $(n, m)$ is in $S'$ if and only if any of $(n, m-1)$, $(n, m+1)$, $(n-1, m)$, $(n+1, m)$, and ... | [
"Solution:\n\nTransforming it $k \\geq 1$ times yields the \"diamond\" of points $(n, m)$ such that $|n| + |m| \\leq k$. The diamond contains $(k+1)^2 + k^2$ lattice points (this can be seen by rotating the plane 45 degrees and noticing the lattice points in the transforms form two squares, one of which is containe... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 421 | |
0cts | Quadratic polynomials $f_1(x), \dots, f_{100}(x)$ have equal leading coefficients, equal coefficients of $x$, but distinct constant terms; each of these polynomials has two real roots. Let $x_i$ be some root of $f_i(x)$. Determine all possible values of the sum $f_2(x_1) + f_3(x_2) + \dots + f_{100}(x_{99}) + f_1(x_{10... | [
"Пусть $i$-й трёхчлен имеет вид $f_i(x) = a x^2 + b x + c_i$. Тогда\n\n$f_2(x_1) + f_3(x_2) + \\dots + f_{100}(x_{99}) + f_1(x_{100}) = (c_2 - c_1) + (c_3 - c_2) + \\dots + (c_{100} - c_{99}) + (c_1 - c_{100}) = 0$.\n\nЗначит, единственное возможное значение суммы — ноль."
] | Russia | Russian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English; Russian | proof and answer | 0 | |
00q2 | Determine all positive integers $n$ such that $f_n(x, y, z) = x^{2n} + y^{2n} + z^{2n} - xy - yz - zx$ divides $g_n(x, y, z) = (x-y)^{5n} + (y-z)^{5n} + (z-x)^{5n}$, as polynomials in $x, y, z$ with integer coefficients. | [
"Assume that $f_n(x, y, z)$ divides $g_n(x, y, z)$, that is\n$$\ng_n(x, y, z) = f_n(x, y, z)h_n(x, y, z),\n$$\nwhere $h_n$ is a polynomial in $x, y, z$ with integer coefficients.\nConsidering $x = 2, y = 1, z = 0$, it follows $2^{2n} - 1$ must divide $(-2)^{5n} + 2$. The last property is equivalent to $2^{2n} - 1|2... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | 1 | |
04ee | Let $n$ be a composite positive integer and let $d_1, d_2, \dots, d_m$ be all of its divisors. Prove that the following equation holds:
$$
\frac{2}{\log n^m} \sum_{k=1}^{m} \log d_k = 1.
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof only | null | |
05o2 | Problem:
Soient $\Gamma_{1}$ et $\Gamma_{2}$ deux cercles se coupant en $A$ et $B$ distincts. Notons $O$ le centre de $\Gamma_{1}$. Soit $C$ un point de $\Gamma_{1}$, soit $D, E$ les intersections respectives de $(AC)$ et de $(BC)$ avec $\Gamma_{2}$.
Montrer que $(OC)$ et $(DE)$ sont perpendiculaires. | [
"Solution:\n\n\n\nOn procède par chasse aux angles.\nOn présente ici une solution qui se veut très (trop) rigoureuse, il n'y avait pas besoin d'être aussi pointilleux ni de traiter tous ces cas pour recevoir la note maximale à cet exercice.\nComme on remarque que la figure change beaucoup s... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00s1 | What is the least positive integer $k$ such that, in every convex $101$-gon, the sum of any $k$ diagonals is greater than or equal to the sum of the remaining diagonals? | [
"Let $PQ=1$. Consider a convex $101$-gon such that one of its vertices is at $P$ and the remaining $100$ vertices are within $\\varepsilon$ of $Q$ where $\\varepsilon$ is an arbitrarily small positive real. Let $k+l$ equal the total number $\\frac{101\\cdot 98}{2} = 4949$ of diagonals. When $k \\le 4851$, the sum o... | Balkan Mathematical Olympiad | BMO 2017 | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 4900 | |
0dnc | Problem:
Одредити све функције $f: \mathbb{R} \rightarrow \mathbb{R}$ такве да за све $x, y \in \mathbb{R}$ важи
$$
f(x f(y)-y f(x))=f(x y)-x y .
$$ | [
"Solution:\n\nЗамена $y=0$ даје $f(x f(0))=f(0)$. Ако је $f(0) \\neq 0$, израз $x f(0)$ узима све реалне вредности, па је $f$ константна функција, а она не задовољава услове. Према томе, $f(0)=0$.\n\nСтављањем $y=x$ добијамо $f(0)=f\\left(x^{2}\\right)-x^{2}$, тј. $f\\left(x^{2}\\right)=x^{2}$. Дакле, $f(x)=x$ за с... | Serbia | 8. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = x and f(x) = |x| | |
0eod | The value of $2 - (0 - (1 - 5))$ is
(A) 3
(B) 1
(C) 0
(D) -1
(E) -2 | [
"**E**\n\n$2 - (0 - (1 - 5)) = 2 - (0 - (-4)) = 2 - (0 + 4) = 2 - 4 = -2$"
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | E | |
075w | A positive integer $a$ is called a *double number* if it has an even number of digits (in base 10) and its base 10 representation has the form $a = a_1a_2\cdots a_k a_1a_2\cdots a_k$ with $0 \le a_i \le 9$ for $1 \le i \le k$, and $a_1 \ne 0$. For example, 283283 is a double number. Determine whether or not there are i... | [
"The answer is affirmative. Let $k \\ge 0$ be such that $k \\equiv 15 \\pmod{42}$ and $b = 5(10^k + 1)/7 + 1$ (which is an integer). Then $c = \\frac{5}{7}(b+1)$ is an integer and $b^2 - 1 = (10^k + 1)c$ is a double number."
] | India | Indija TS 2013 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | Yes, there are infinitely many. | |
05b9 | Points $E$ and $F$ are chosen respectively on the sides $CA$ and $AB$ of triangle $ABC$. Lines $BE$ and $CF$ intersect at $P$. Let $Q$ be a point such that $PBQC$ is a parallelogram and $R$ a point such that $AERF$ is a parallelogram. Prove that $PR \parallel AQ$. | [
"Let $S$ be a point such that $PESF$ is a parallelogram (Fig. 15). We first show that $PR \\parallel AS$. For this note that $\\angle PER = \\angle SFA$ since $PE \\parallel FS$ and $ER \\parallel AF$. Additionally $PE = FS$ and $ER = AF$, thus triangles $PER$ and $SFA$ are congruent. From this we deduce that $PR \... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof only | null | |
0e5b | Peter has $111$ red and $111$ blue marbles. Every day, Peter's uncle lets him exchange either $11$ red marbles for $7$ blue marbles or $20$ blue marbles for $28$ red marbles.
a. Can Peter increase the total number of marbles he has by $20$ after several exchanges?
b. Can Peter increase the total number of marbles he ... | [
"a. Peter can increase the number of marbles by $20$. The first three times he should exchange $20$ blue marbles for $28$ red ones. He will then have $111 - 60 = 51$ blue marbles and $111 + 3 \\cdot 28 = 111 + 84 = 195$ red ones. Then, he should exchange $11$ red marbles for $7$ blue ones. He will end up with $184$... | Slovenia | National Math Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a: Yes; b: No; c: No | |
05je | Problem:
Soit $ABCD$ un quadrilatère inscriptible, $L = (AC) \cap (BD)$, $J$ et $K$ les pieds des perpendiculaires à $(AD)$ et $(BC)$ passant par $L$ et $I$ le milieu de $[C, D]$. Montrer que $IJ = IK$.
 | [
"Solution:\n\nIl est facile de voir en utilisant le théorème de l'angle inscrit que $BLC$ et $ALD$ sont semblables, et que $CKL$ et $DJL$ sont semblables.\n\nNotons $M$ et $N$ les milieux respectifs de $[LD]$ et $[LC]$. Alors $L N I M$ est un parallélogramme. On en déduit que $NL = IM$. Or, $CKL$ est un triangle re... | France | null | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05us | Problem:
Pour tout entier $k \geqslant 0$, on note $a_{k}$ le premier chiffre du nombre $2^{k}$, écrit en base 10. Par exemple, $a_{5}=3$ est le premier chiffre de $2^{5}=32$.
Soit $n \geqslant 1$ un entier. Démontrer que, parmi les chiffres de 1 à 9, il y en a un qui est égal à au plus $n / 17$ des $n$ chiffres $a_{... | [
"Solution:\n\nSoit $\\mathrm{c}(k)$ le nombre de fois où le chiffre $k$ figure parmi les chiffres $a_{0}, a_{1}, a_{2}, \\ldots, a_{n-1}$. Lorsque $a_{i} \\geqslant 2$, on sait que $i \\geqslant 1$ et que $a_{i-1}=\\left\\lfloor a_{i} / 2\\right\\rfloor$.\n\nAinsi, pour tout chiffre $k \\leqslant 5$, on a $a_{i}=k$... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Other"
] | null | proof only | null | |
0lcw | Find all sets of 2014 rational numbers not necessarily distinct such that: If an arbitrary number in the set is removed, one always can divide the remaining 2013 numbers into three sets such that each set has exactly 671 elements and the products of all elements in each set are equal. | [
"Suppose that the tuple $(a_1, a_2, \\dots, a_{2014})$ satisfies the given condition. We investigate two following cases.\n\n**Case 1.** If there exists a zero in this tuple, it is easy to check that there are at least four zeros. Otherwise, if the tuple has four zeros then it satisfies the given condition.\n\n**Ca... | Vietnam | VMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | Either the multiset contains at least four zeros; or all numbers have the same nonzero absolute value and the number of negative numbers k satisfies k not in {1, 2, 2012, 2013}. | |
0dhi | Let $a_1 = 1$, $a_{n+1} = a_n + [\sqrt{a_n}]$. Find all $n$ such that $a_n$ is a perfect square. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | All n of the form n = 2^{k+1} + k - 1 for k ≥ 0; at these indices a_n = 4^k. | |
08ez | Problem:
Il piccolo Boole ha disegnato una striscia di 8 caselle, in ognuna delle quali può essere scritto uno 0 o un 1. Inizialmente ogni casella contiene uno 0. Ad ogni mossa, Boole compie una delle seguenti operazioni:
a. Sostituisce ogni 0 con un 1 e ogni 1 con uno 0;
b. Sceglie tre caselle consecutive e solo in... | [
"Solution:\n\nOsserviamo che tutte le operazioni commutano tra di loro, e che applicare un'operazione due volte è come non applicarla. Quindi possiamo associare ad ogni operazione un bottone ON/OFF (che corrisponde ad aver fatto o non aver fatto data operazione) e a ogni successione di operazioni uno e un solo stat... | Italy | Italian Mathematical Olympiad - February Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Linear Algebra > Vectors"
] | null | MCQ | C | |
0jrr | Problem:
Two circles centered at $O_{1}$ and $O_{2}$ have radii $2$ and $3$ and are externally tangent at $P$. The common external tangent of the two circles intersects the line $O_{1} O_{2}$ at $Q$. What is the length of $PQ$? | [
"Solution:\n\nLet the common external tangent intersect the circles centered at $O_{1}$, $O_{2}$ at $X$, $Y$ respectively. Then $\\frac{O_{2} Q}{O_{1} Q} = \\frac{O Y}{O X} = \\frac{3}{2}$, so $\\frac{O_{1} O_{2}}{O_{1} Q} = \\frac{O_{2} Q - O_{1} Q}{O_{1} Q} = \\frac{1}{2}$. Since $O_{1} O_{2} = 2 + 3 = 5$, $O_{1}... | United States | HMMT November 2016 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | 12 | |
0fzv | Problem:
Für welche natürlichen Zahlen $n$ ist es möglich, ein $n \times n$ Feld lückenlos und überlappungsfrei mit T-Tetrominos und einer ungeraden Anzahl an Square-Tetrominos zu bedecken?
Bemerkung: Es ist erlaubt, keine T-Tetrominos zu verwenden. | [
"Solution:\n\nPuisque les deux pièces à disposition comportent chacune 4 cases, on obtient que le nombre total de cases est divisible par 4. Comme ce nombre vaut $n^{2}$, $4 \\mid n^{2}$, donc $2 \\mid n$. Posons $n = 2k$.\n\nEn utilisant la coloration standard à deux couleurs (également appelée coloration de l'éch... | Switzerland | SMO - Vorrunde | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | n = 4k + 2 for some integer k | |
07ha | Let $ABCD$ be a quadrilateral and $w$ is a circle inscribed in it. The circle $\omega$ is tangent to $BC$ and $AD$ at $E$ and $F$, respectively and $DE$ meets $\omega$ at $X$, for the second time. If $AB$, $CD$ are tangent to the circumcircle of $DXF$, prove that $AFXC$ is cyclic. | [
"Let $\\Omega$ be the circumcircle of $DXF$ and $l$ be the perpendicular bisector of $FX$. Note that lines $AB$, $CD$ are external common tangents to $\\Omega$, $w$, therefore they are reflection of each other with respect to $l$. Suppose that $AB$ is tangent to $\\Omega$ at $Y$, and $Z$ is the reflection of $A$ wi... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
05pf | Problem:
Soit $O$ le centre d'un polygone régulier à 18 côtés de sommets $A_{1}, \ldots, A_{18}$. Soit $B$ le point de $[OA_{1}]$ tel que $\widehat{BA A_{2} O}=20^{\circ}$ et $C$ le point de $[OA_{2}]$ tel que $\widehat{CA_{1} O}=10^{\circ}$. Montrer que $BCA_{2} A_{3}$ sont cocycliques. | [
"Solution:\n\n\n\nOn prolonge déjà les différentes droites existantes : $A_{1}, O, A_{10}$ sont alignés, $A_{2}, O, A_{11}$ aussi.\n\nOn applique ensuite le théorème de l'angle inscrit : tout angle au centre qui intercepte un segment de longueur $A_{1}A_{2}$ vaut $20^{\\circ}$, tout angle i... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cal | Problem:
Fie $ABC$ un triunghi ascuţitunghic şi fie $B'$ şi $C'$ picioarele înălţimilor sale din $B$, respectiv $C$. Fie $B_A'$ şi $B_C'$ simetricele lui $B'$ în raport cu dreptele $BC$, respectiv $AB$. Cercul $BB_A'B_C'$, centrat în $O_B$, intersectează a doua oară dreapta $AB$ în $X_B$. Punctele $C_A'$, $C_B'$, $O_C... | [] | Romania | Olimpiada Nationala de Matematica 2022 baraj 2 de selectie seniori | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02ta | Problem:
Sejam $a$ e $b$ números reais positivos quaisquer. Determine o valor da expressão
$$
\frac{\sqrt{\frac{a b}{2}}+\sqrt{8}}{\sqrt{\frac{a b+16}{8}+\sqrt{a b}}}
$$ | [
"Solution:\nSeja $x=\\sqrt{\\frac{a b}{2}}+\\sqrt{8}$. Então:\n$$\n\\begin{aligned}\nx^{2} & =\\frac{a b}{2}+4 \\sqrt{a b}+8 \\\\\n& =4\\left(\\frac{a b+16}{8}+\\sqrt{a b}\\right) \\\\\n& =4\\left(\\sqrt{\\frac{a b+16}{8}+\\sqrt{a b}}\\right)^{2}\n\\end{aligned}\n$$\nAssim, o valor da expressão procurada é\n$$\n\\b... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof and answer | 2 | |
0bvv | Determine the triples of positive integers $(x, y, z)$ such that $x^4 + y^4 = 2z^2$ and $x, y$ are co-prime. | [
"Let $(x, y, z)$ be a solution of the problem. Then, notice that $x, y$ are odd, hence $z$ is also odd, and co-prime with $xy$. The equation can be written successively $x^8 + 2x^4y^4 + y^8 = 4z^4$, or $(x^4 - y^4)^2 = 4z^4 - 4x^4y^4$, or $z^4 - (xy)^4 = \\left(\\frac{x^4-y^4}{2}\\right)^2$.\n\nWe prove that the eq... | Romania | The Danube Mathematical Competition | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (1, 1, 1) | |
0do3 | Problem:
Нека су $a$ и $b$ непарни природни бројеви већи од 1. Посматрајмо таблу $a \times b$ којој недостају поља $(2,1),(a-2, b)$ и $(a, b)$ (под пољем $(i, j)$ подразумевамо поље у пресеку врсте $i$ и колоне $j$ ). Претпоставимо да је оваква табла поплочана помоћу $2 \times 1$ домина и $2 \times 2$ квадрата (домине... | [
"Solution:\n\nУпишимо у поље $(i, j)$ број $(-1)^{i+j}(i+j)$. Збир уписаних бројева у читавој таблици је $\\sum_{i=1}^{a}(-1)^{i} \\sum_{j=1}^{b}(-1)^{j}(i+j)=\\sum_{i=1}^{a}(-1)^{i+1}\\left(i+\\frac{b+1}{2}\\right)=\\frac{a+b+2}{2}$, а ако се три наведена поља избаце, збир у остатку таблице је\n$$\n\\frac{a+b+2}{2... | Serbia | 12. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0hj6 | Problem:
Each vertex of a cube is labeled with an integer. Prove that there exist four coplanar vertices the sum of whose numbers is even. | [
"Solution:\n\nLabel the vertices of the cube as shown. The quadruples of vertices\n$$\n\\begin{aligned}\n& a \\text{ and } b \\text{ and } c \\text{ and } d \\\\\n& a \\text{ and } b \\text{ and } e \\text{ and } f \\\\\n& c \\text{ and } d \\text{ and } e \\text{ and } f\n\\end{aligned}\n$$\nare all coplanar. Assu... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof only | null | |
00to | Let $n$ be a positive integer. Determine, in terms of $n$, the greatest integer which divides every number of the form $p+1$, where $p \equiv 2 \pmod{3}$ is a prime number which does not divide $n$. | [
"Let $k$ be the greatest such integer. We will show that $k = 3$ when $n$ is odd and $k = 6$ when $n$ is even.\nWe will say that a number $p$ is nice if $p$ is a prime number of the form $2 \\pmod{3}$ which does not divide $N$.\nNote first that $3 \\mid p+1$ for every nice number $p$ and so $k$ is a multiple of $3$... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic",
"Number Theory > Other"
] | null | proof and answer | 3 if n is odd; 6 if n is even | |
0f4z | Problem:
$ABCD$ is a parallelogram and $AB \ne BC$. $M$ is chosen so that
(1) $\angle MAC = \angle DAC$ and $M$ is on the opposite side of $AC$ to $D$, and
(2) $\angle MBD = \angle CBD$ and $M$ is on the opposite side of $BD$ to $C$.
Find $AM/BM$ in terms of $k = AC / BD$. | [] | Soviet Union | 16th ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | k^2 | |
0e3s | The equilateral triangles $ABC$ and $DEF$ with the sides of length $1$ are divided into congruent equilateral triangles as shown in the figure. A circle is inscribed into each smaller triangle. What is the ratio of the shaded area of the triangle $ABC$ to the shaded area of the triangle $DEF$?
... | [
"The proportion of the area of the triangle covered by the inscribed circle is $\\frac{\\pi\\sqrt{3}}{9}$. For each of the nine congruent triangles in the triangle $ABC$ the shaded area represents $\\frac{\\pi\\sqrt{3}}{9}$ of the total area, so this is also the proportion of the shaded area in the triangle $ABC$. ... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles"
] | null | MCQ | C | |
0izq | Problem:
The positive integer $i$ is chosen at random such that the probability of a positive integer $k$ being chosen is $\frac{3}{2}$ times the probability of $k+1$ being chosen. What is the probability that the $i^{\text{th}}$ digit after the decimal point of the decimal expansion of $\frac{1}{7}$ is a 2? | [
"Solution:\n\nAnswer: $\\frac{108}{665}$\n\nFirst we note that the probability that $n$ is picked is $\\frac{1}{2} \\times \\left(\\frac{2}{3}\\right)^{n}$, because this is the sequence whose terms decrease by a factor of $\\frac{2}{3}$ each time and whose sum is 1 (recall that probabilities must sum to 1).\n\nNow ... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 108/665 | |
0kz8 | The graph of $y = e^{x+1} + e^{-x} - 2$ has an axis of symmetry. What is the reflection of the point $(-1, \frac{1}{2})$ over this axis?
(A) $(-1, -\frac{3}{2})$ (B) $(-1, 0)$ (C) $(-1, \frac{1}{2})$ (D) $(0, \frac{1}{2})$ (E) $(3, \frac{1}{2})$ | [
"Let $f(x) = e^{x+1} + e^{-x} - 2$. Because $f(x)$ approaches infinity as $|x|$ increases without bound, the only possible axis of symmetry is a vertical line. If the axis of symmetry has equation $x = c$, then $f(x) = f(2c-x)$ for every real $x$, which is equivalent to $e \\cdot e^x + e^{-x} = e^{2c+1}e^{-x} + e^{... | United States | AMC 12 A | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | MCQ | D | |
0lg0 | Problem:
Determine all the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f\left(x^{3}+y^{3}+x y\right)=x^{2} f(x)+y^{2} f(y)+f(x y)
$$
for all $x, y \in \mathbb{R}$. | [
"Solution:\nLet $P(x, y)$ be the assertion that $f\\left(x^{3}+y^{3}+x y\\right)=x^{2} f(x)+y^{2} f(y)+f(x y)$.\n\nFrom $P(1,0)$ we get that $f(0)=0$, hence from $P(x, 0)$ we get $f\\left(x^{3}\\right)=x^{2} f(x)$.\n\n$P(x,-x)$ yields $f(x)=-f(-x)$.\n\nFrom $P(x, y)-P(x,-y)$ we get\n$$\nf\\left(y^{3}\\right)+f(x y)... | Zhautykov Olympiad | XI International Zhautykov Olympiad in Sciences | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = c x for all real x, where c is any real constant | |
0csl | The numbers $1, 2, \ldots, 10000$ are put in some order into the squares of a checkered $100 \times 100$ board, one number per square. Pete marks the squares according to the following rules. At the beginning, he just marks $k$ squares by his own choice. By any subsequent move, he may mark any unmarked square containin... | [
"Докажем вначале следующее утверждение.\n**Лемма.** Для любых двух клеток $A$ и $B$ существует такая клетка $C$, закрасив которую, можно затем закрасить и $A$, и $B$ (возможно, $C$ совпадает с $A$ или с $B$.)\n\n**Доказательство.** Можно считать, что номер $a$ клетки $A$ меньше, чем номер $b$ клетки $B$. Пусть $D$ ... | Russia | XL Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | k = 1 | |
0h66 | Nastia has 5 yellow coins, among which all are authentic. She has also 5 blue coins, among which three are authentic and two are fake. All 8 authentic coins have the same weight, one of the fake coins is heavier than authentic by 1 gram, and the other is lighter by 1 gram. Can Nastia, using scales without weights and 3... | [
"Denote blue coins by $B_1, \\ldots, B_5$. Firstly, weigh 3 yellow and 3 blue coins $B_1, B_2, B_3$:\n$$\n1) \\text{Ж}_1 + \\text{Ж}_2 + \\text{Ж}_3 \\ ? \\ B_1 + B_2 + B_3.\n$$\n\n$$\n1a) \\text{Ж}_1 + \\text{Ж}_2 + \\text{Ж}_3 = B_1 + B_2 + B_3.\n$$\nFake coins are among $B_1, B_2, B_3$, or among $B_4$ and $B_5$.... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Algorithms"
] | English | proof only | null | |
0bd0 | Solve $2^{\sin^4 x - \cos^2 x} - 2^{\cos^4 x - \sin^2 x} = \cos 2x$. | [
"The equation becomes $2^{\\sin^4 x + \\sin^2 x} - 2^{\\cos^4 x + \\cos^2 x} = 2 \\cos 2x$. Because $\\cos^4 x + \\cos^2 x - \\sin^4 x - \\sin^2 x = (\\cos^2 x + \\sin^2 x)(\\cos^2 x - \\sin^2 x) + \\cos^2 x - \\sin^2 x$, we can write $2^{\\sin^4 x+\\sin^2 x} + \\sin^4 x + \\sin^2 x = 2^{\\cos^4 x+\\cos^2 x} + \\co... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | x = (2k+1)π/4, k ∈ ℤ | |
03j0 | Problem:
Alice and Bob are in a hardware store. The store sells coloured sleeves that fit over keys to distinguish them. The following conversation takes place:
Alice: Are you going to cover your keys?
Bob: I would like to, but there are only 7 colours and I have 8 keys.
Alice: Yes, but you could always distinguish a ... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof and answer | 1 color if there is 1 key; 2 colors if there are 2 keys or at least 6 keys; 3 colors if there are 3, 4, or 5 keys. | |
0fzo | Problem:
Sei $n$ eine natürliche Zahl und $A=\{P_{1}, P_{2}, \ldots, P_{n}\}$ eine Menge von $n$ Punkten in der Ebene, von denen keine drei auf einer Geraden liegen. Ein Weg durch $A$ besteht aus $n-1$ Strecken $P_{\sigma(i)} P_{\sigma(i+1)}$ für $i=1, \ldots, n-1$, wobei $\sigma$ eine Permutation von $\{1,2, \ldots, ... | [
"Solution:\n\nBetrachte die konvexe Hülle der Punkte in $A$ und wähle einen Punkt $P_{\\sigma(1)}$ davon aus. Nun betrachte die konvexe Hülle der restlichen $n-1$ Punkte und verbinde $P_{\\sigma(1)}$ mit einem Punkt $P_{\\sigma(2)}$ auf der neuen konvexen Hülle, ohne diese zu schneiden. Wir wiederholen nun dieses V... | Switzerland | IMO-Selektionsprüfung | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof only | null | |
0bmw | Problem:
Adottak az $A, B \in \mathcal{M}_{2}(\mathbb{R})$ mátrixok úgy, hogy $(A-B)^{2}=\mathrm{O}_{2}$.
a) Igazold, hogy $\operatorname{det}\left(A^{2}-B^{2}\right)=(\operatorname{det}(A)-\operatorname{det}(B))^{2}$.
b) Bizonyítsd be, hogy $\operatorname{det}(A B-B A)$ akkor és csak akkor egyenlő 0 -val, ha $\operato... | [
"Solution:\na) $\\operatorname{Din}(A-B)^{2}=O_{2}$ obţinem $\\operatorname{det}(A-B)=0$. (1 punct)\nDe asemenea, deducem $\\operatorname{Tr}(A-B)=0$, deci $\\operatorname{Tr}(A)=\\operatorname{Tr}(B)=: a$. (1 punct) Notăm $b=\\operatorname{det}(A)-\\operatorname{det}(B)$. Conform relaţiei lui Cayley, avem\n$$\n\\l... | Romania | Olimpiada Naţională de Matematică, Etapa Judeţeană şi a Municipiului Bucureşti | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof only | null | |
0034 | En el cuadrilátero convexo $ABCD$, sean $E$ y $F$ los puntos medios de los lados $AD$ y $BC$, respectivamente. Los segmentos $CE$ y $DF$ se cortan en $O$. Demostrar que si las rectas $AO$ y $BO$ dividen al lado $CD$ en tres partes iguales entonces $ABCD$ es un paralelogramo. | [] | Argentina | XVII OLIMPIADA MATEMÁTICA DEL CONO SUR | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | Español | proof only | null | |
05jw | Problem:
Déterminer la plus grande valeur possible et la plus petite valeur possible de
$$
\sqrt{4-a^{2}}+\sqrt{4-b^{2}}+\sqrt{4-c^{2}}
$$
lorsque $a, b, c$ sont des réels strictement positifs vérifiant $a^{2}+b^{2}+c^{2}=6$. | [
"Solution:\n\nTout d'abord, on note que si l'on veut que l'expression ait un sens, il faut $a, b, c \\in [0; 2]$.\n\nD'après l'inégalité de Cauchy-Schwarz, on a\n$$\n\\left(\\sqrt{4-a^{2}}+\\sqrt{4-b^{2}}+\\sqrt{4-c^{2}}\\right)^{2} \\leqslant 3\\left(4-a^{2}+4-b^{2}+4-c^{2}\\right)=18,\n$$\nc'est-à-dire\n$$\n\\sqr... | France | Olympiades Françaises de Mathématiques, Envoi Numéro 3 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | maximum = 3√2, minimum = 2+√2 | |
089m | Problem:
Abbiamo un quadrilatero i cui lati misurano, nell'ordine, $1, 7, 5, 5$. Quanto vale al massimo la sua area?
(A) 12
(B) $6 \sqrt{6}$
(C) 16
(D) 20
(E) Un siffatto quadrilatero non esiste. | [
"Solution:\n\nLa risposta è (C). Sia $ABCD$ un quadrilatero, e supponiamo che la lunghezza di $AB$ sia $1$, quella di $BC$ $7$, e che $CD = AD = 5$. Si considerino il triangolo $ABC$ e il triangolo $CDA$; siano $AH$ l'altezza di $ABC$ relativa a $BC$, $AK$ l'altezza di $CDA$ relativa a $CD$. Abbiamo $AH \\leq AB$ (... | Italy | Progetto Olimpiadi della Matematica | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | MCQ | C | |
033k | Problem:
Let $H$ be the orthocenter of $\triangle ABC$. The points $A_1 \neq A$, $B_1 \neq B$ and $C_1 \neq C$ lie respectively on the circumcircles of $\triangle BCH$, $\triangle CAH$ and $\triangle ABH$, and $A_1H = B_1H = C_1H$. Denote by $H_1$, $H_2$ and $H_3$ the orthocenters of $\triangle A_1BC$, $\triangle B_1C... | [
"$$\n\\sin \\angle C_1AH = \\frac{C_1H}{2R_1} = \\frac{A_1H}{2R_2} = \\sin \\angle A_1BH\n$$\nand analogously\n$$\n\\sin \\angle C_1AH = \\sin \\angle A_1BH = \\sin \\angle A_1CH = \\sin \\angle C_1BH = \\sin \\angle B_1AH.\n$$\nLet $\\angle C_1AH = \\angle A_1BH = \\angle B_1CH = \\varphi$. Since $\\angle AHB = \\... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinat... | null | proof only | null | |
01u7 | Some positive integers are written on cards, at least two different numbers on each card. The same number may be written on several cards. Two cards are called *adjacent* if the maximum number on one of them is equal to the minimum number on the other.
Prove that if there are no adjacent cards then all written numbers ... | [
"We construct two required groups $G_1$ and $G_2$ as follows: we choose the smallest number on each card and put it in the group $G_1$. Therefore, every card contains at least one number from $G_1$. All other numbers we put in the group $G_2$. Since there are no adjacent cards, the largest number in any card does n... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0iv3 | Problem:
Two jokers are added to a $52$ card deck and the entire stack of $54$ cards is shuffled randomly. What is the expected number of cards that will be between the two jokers? | [
"Solution:\n\nEach card has an equal likelihood of being either on top of the jokers, in between them, or below the jokers. Thus, on average, $1 / 3$ of them will land between the two jokers."
] | United States | $12^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 52/3 | |
00lf | Let $ABC$ be a triangle with $\angle AC > \angle AB$ and circumcenter $O$. The tangents to the circumcircle at $A$ and $B$ intersect at $T$. The perpendicular bisector of the side $BC$ intersects $AC$ at $S$.
a. Prove that the points $A$, $B$, $O$, $S$ and $T$ lie on a common circle.
b. Prove that the line $ST$ is pa... | [
"Since $\\angle AT$ and $\\angle BT$ are perpendicular to $\\angle AO$ and $\\angle BO$, the points $A$, $B$, $T$ and $O$ lie on a circle $k_1$ by Thales' theorem. By the central angle theorem we have $\\angle AOB = 2\\gamma$. Since $BCS$ is an isosceles triangle, we find $\\angle BCS = \\angle CBS = \\gamma$. Now ... | Austria | Regional Competition | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ... | English | proof only | null | |
095b | Problem:
Fie $a, b, c$ numere reale pozitive, $a \leq b, a \leq c$, astfel încât
$$
\frac{a^{2}+b^{2}+c^{2}}{a(b+c)}=\frac{3}{2}.
$$
Să se arate că $a=b=c$. | [
"Solution:\n\nFără a restrânge generalitatea, considerăm $a \\leq b \\leq c$.\n\n1) $a \\leq b \\Rightarrow a c \\leq b c$; $a c+a b \\leq b c+a b$; $a(b+c) \\leq b(c+a)$\n\n2) $b \\leq c \\Rightarrow a b \\leq a c$; $a b+b c \\leq a c+b c$; $b(a+c) \\leq c(b+a)$\n\n3) Reunim (1) şi (2): $a(b+c) \\leq b(c+a) \\leq ... | Moldova | 61-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0203 | Problem:
Find all quadruples $(a, b, c, d)$ of positive real numbers such that $a b c d = 1$, $a^{2012} + 2012 b = 2012 c + d^{2012}$ and $2012 a + b^{2012} = c^{2012} + 2012 d$. | [
"Solution:\nRewrite the last two equations into\n$$\na^{2012} - d^{2012} = 2012(c - b) \\text{ and } c^{2012} - b^{2012} = 2012(a - d)\n$$\nand observe that $a = d$ holds if and only if $c = b$ holds. In that case, the last two equations are satisfied, and condition $a b c d = 1$ leads to a set of valid quadruples ... | Benelux Mathematical Olympiad | 4th Benelux Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (t, 1/t, 1/t, t) for t > 0 | |
0jgg | Problem:
Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be real numbers whose sum is $20$. Determine with proof the smallest possible value of
$$
\sum_{1 \leq i<j \leq 5}\left\lfloor a_{i}+a_{j}\right\rfloor.
$$ | [
"Solution:\nAnswer: $72$\n\nWe claim that the minimum is $72$. This can be achieved by taking $a_{1}=a_{2}=a_{3}=a_{4}=0.4$ and $a_{5}=18.4$.\n\nTo prove that this is optimal, note that\n$$\n\\sum_{1 \\leq i<j \\leq 5}\\left\\lfloor a_{i}+a_{j}\\right\\rfloor = \\sum_{1 \\leq i<j \\leq 5} (a_{i}+a_{j}) - \\{a_{i}+a... | United States | HMMT 2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 72 | |
095c | Problem:
Să se arate că există o infinitate de numere naturale $x$, $y$, $z$, cu $z$ număr prim, care satisfac ecuaţia
$$
x^{2}+y^{3}=2^{z}
$$ | [
"Solution:\n1. Observăm că numerele $x=8$, $y=4$ şi $z=7$ reprezintă o soluţie a ecuaţiei.\n\n2. Vom folosi faptul că există o infinitate de numere de forma $6n+1$, unde $n$ este număr natural.\n\n3. Examinăm mulţimea $A$ a tuturor numerelor naturale, pentru care $6n+1$ este un număr prim.\n\n4. Pentru oricare $n \... | Moldova | 61-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
02e2 | $A$, $B$ are real numbers. Find a necessary and sufficient condition for $A x + B [x] = A y + B [y]$ to have no solutions except $x = y$. | [
"If $A = 0$, then we have $B \\lfloor x \\rfloor = B \\lfloor y \\rfloor$ which obviously has infinitely many solutions with $x \\neq y$. So assume $A \\neq 0$. Then we can write the equation as $\\frac{B}{A}(\\lfloor x \\rfloor - \\lfloor y \\rfloor) = y - x$. If $x \\neq y$, we can assume $x < y$. We cannot have ... | Brazil | VII OBM | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | A ≠ 0 and B/A ∉ (-2, 0) (equivalently, A ≠ 0 and B/A ≤ −2 or B/A ≥ 0). | |
0i22 | Problem:
5 married couples gather at a party. As they come in and greet each other, various people exchange handshakes - but, of course, people never shake hands with themselves or with their own respective spouses. At the end of the party, one woman goes around asking people how many hands they shook, and she gets ni... | [
"Solution:\n\nSuppose that there were $n$ couples, and the woman asked all $2n-1$ other attendees how many hands they shook and received $2n-1$ different answers. We will show that she herself shook $n-1$ hands; hence, in our particular case, the answer is $4$.\n\nWe work by induction. When $n=1$, there is one coup... | United States | Berkeley Math Circle Monthly Contest #7 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 4 | |
05ov | Problem:
Les points $D$ et $E$ divisent le côté $[A B]$ d'un triangle équilatéral en trois parties égales, de telle manière que $D$ est situé entre $A$ et $E$. Le point $F$ est situé sur $[B C]$ de sorte que $C F = A D$. Calculer la somme des angles $\widehat{C D F} + \widehat{C E F}$.
 | [
"Solution:\n\nOn a $B F = B D$ et $\\widehat{D B F} = 60^{\\circ}$, donc le triangle $D B F$ est équilatéral. On a donc $\\widehat{B A C} = \\widehat{B D F} = 60^{\\circ}$ donc $(D F) \\parallel (A C)$, donc $\\widehat{C D F} = \\widehat{A C D}$.\n\nD'autre part, $\\widehat{A C D} = \\widehat{B C E}$ par symétrie, ... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | 30° | |
0jme | Problem:
Let $ABC$ be an acute triangle with circumcenter $O$ such that $AB = 4$, $AC = 5$, and $BC = 6$. Let $D$ be the foot of the altitude from $A$ to $BC$, and $E$ be the intersection of $AO$ with $BC$. Suppose that $X$ is on $BC$ between $D$ and $E$ such that there is a point $Y$ on $AD$ satisfying $XY \parallel ... | [
"Solution:\n\nAnswer: $\\dfrac{96}{41}$\n\nLet $AX$ intersect the circumcircle of $\\triangle ABC$ again at $K$. Let $OY$ intersect $AK$ and $BC$ at $T$ and $L$, respectively. We have $\\angle LOA = \\angle OYX = \\angle TDX = \\angle LAK$, so $AL$ is tangent to the circumcircle. Furthermore, $OL \\perp AK$, so $\\... | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line... | null | proof and answer | 96/41 | |
0ffy | Problem:
Resolver la ecuación
$$
tan^2 2x + 2 \tan 2x \tan 3x - 1 = 0
$$ | [
"Solution:\n\nPrimera solución (de M. Ascensión López Chamorro)\nLa ecuación\n$$\ntan^2 2x + 2 \\tan 2x \\tan 3x - 1 = 0\n$$\nse puede escribir como\n$$\n\\tan 3x = \\frac{1 - \\tan^2 2x}{2 \\tan 2x} = \\cot 4x\n$$\nasí que\n$$\n3x + 4x = \\frac{\\pi}{2} + k\\pi,\\ k \\in \\mathbb{Z}\n$$\nes decir\n$$\nx = \\frac{\... | Spain | OME 21 | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | x = π/14 + k·π/7, for any integer k | |
0j2p | Problem:
Suppose that a polynomial of the form $p(x) = x^{2010} \pm x^{2009} \pm \cdots \pm x \pm 1$ has no real roots. What is the maximum possible number of coefficients of $-1$ in $p$? | [
"Solution:\nAnswer: $1005$\n\nLet $p(x)$ be a polynomial with the maximum number of minus signs. $p(x)$ cannot have more than $1005$ minus signs, otherwise $p(1) < 0$ and $p(2) \\geq 2^{2010} - 2^{2009} - \\ldots - 2 - 1 = 1$, which implies, by the Intermediate Value Theorem, that $p$ must have a root greater than ... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof and answer | 1005 | |
0cxs | Let $ABCDEFG$ be a regular heptagon. If $AC = m$ and $AD = n$, prove that $AB = \frac{mn}{m + n}$. | [] | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | English | proof only | null | |
0fdd | Problem:
Sea $T$ un triángulo de ángulos $\alpha, \beta$ y $\gamma$. ¿Para qué valores de $\alpha, \beta$ y $\gamma$ el triángulo $T$ se puede dividir en tres triángulos congruentes entre sí? | [
"Solution:\n\nSi $\\alpha=\\beta=\\gamma$, el triángulo es equilátero y siendo $O$ el centro de $T$, los triángulos que se obtienen uniendo $O$ con cualquiera par de vértices son congruentes. Veremos que sólo en este caso se puede obtener una división con tres triángulos congruentes, con un vértice común en el inte... | Spain | null | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | Equilateral triangles, and right triangles with angles 30, 60, and 90. | |
0fc0 | Problem:
¿Qué número es mayor $999!$ o $500^{999}$? Justifica la respuesta. | [
"Solution:\n\nPongamos $A = 999!$, $B = 500^{999}$, tenemos\n\n$$\n\\frac{A}{B} = \\frac{500 - 499}{500} \\cdot \\frac{500 - 498}{500} \\ldots \\frac{500 - 1}{500} \\cdot \\frac{500}{500} \\cdot \\frac{500 + 1}{500} \\ldots \\frac{500 + 498}{500} \\cdot \\frac{500 + 499}{500} =\n$$\n$$\n\\left(1 - \\frac{499}{500}\... | Spain | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 500^{999} | |
01pp | Find all pairs $(n; p)$ of natural numbers $n$ and prime numbers $p$ satisfying the equality $p(p - 1) = 2(n^3 + 1)$. | [
"Answer: $(n; p) = (20; 127)$.\nIt is easy to see that the given equality\n$$\np(p-1) = 2(n^3 + 1) \\qquad (1)\n$$\ndoes not hold for $p=2$ and positive integer $n$. So $p \\ge 3$ is an odd prime number. Then $(n+1)(n^2 - n + 1)$ is divisible by $p$.\n\n1. If $(n+1) \\nmid p$, then $n+1 = kp$ for some positive inte... | Belarus | BelarusMO 2013_s | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (20, 127) | |
06zs | Problem:
A palindrome is a positive integer which is unchanged if you reverse the order of its digits. For example, $23432$. If all palindromes are written in increasing order, what possible prime values can the difference between successive palindromes take? | [
"Solution:\nLet $x$ be a palindrome and $x'$ the next highest palindrome. If $x < 101$, then it is easy to see by inspection that $x' - x = 1, 2$ or $11$, so the only prime differences are $2$ and $11$.\n\nSo assume $x > 100$. If $x$ and $x'$ have the same final digit, then their difference is divisible by $10$ and... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 2 and 11 | |
0fk9 | Problem:
Dado un triángulo acutángulo $ABC$, determinar para qué puntos de su interior se verifican las siguientes desigualdades:
$$
1 \leq \frac{\angle APB}{\angle ACB} \leq 2, \quad 1 \leq \frac{\angle BPC}{\angle BAC} \leq 2 \quad \text{y} \quad 1 \leq \frac{\angle CPA}{\angle CBA} \leq 2
$$ | [
"Solution:\n\nSea $O$ el circuncentro del triángulo $ABC$. El valor del ángulo $ACB$, por estar inscrito en la circunferencia, es la mitad del ángulo $AOB$. De nuevo, para cualquier punto $P$ sobre el arco $AOB$ se tiene $\\angle APB = \\angle AOB = 2 \\angle ACB$. Por tanto, este arco separa el interior del triáng... | Spain | XLV Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | The circumcenter O is the only interior point that satisfies all three inequalities. | |
09rs | Problem:
Bepaal alle paren $(p, q)$ van priemgetallen waarvoor $p^{q+1} + q^{p+1}$ een kwadraat is. | [
"Solution:\n\nStel eerst dat $p$ en $q$ beide oneven zijn. Dan zijn in $p^{q+1} + q^{p+1}$ beide exponenten even, waaruit volgt dat beide termen congruent $1 \\bmod 4$ zijn. De som is dus congruent $2 \\bmod 4$, maar dat is nooit een kwadraat.\n\nStel nu dat $p$ en $q$ beide even zijn. Dan zijn ze beide gelijk aan ... | Netherlands | IMO-selectietoets II | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | (2, 2) | |
0ggb | 設 $p \ge 3$ 為質數, 而整數 $r$ 滿足 $0 \le r \le p-3$。設整數 $x_1, x_2, \dots, x_{p-1+r}$ 滿足
$$
\sum_{i=1}^{p-1+r} x_i^k \equiv r \pmod{p}
$$
對所有的 $1 \le k \le p-2$ 均成立。則 $x_1, x_2, \dots, x_{p-1+r}$ 除以 $p$ 的餘數可能為何？ | [] | Taiwan | 2022 數學奧林匹亞競賽第二階段培訓營, 國際競賽實作(二) | [
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | Chinese; English | proof and answer | Modulo p, the multiset of residues must consist of every nonzero residue appearing exactly once together with r additional copies of the residue 1; in particular, 0 does not occur. | |
03c6 | Any integer from the set $A = \{1, 2, \dots, 2015\}$ is colored either red or green. For given positive integers $a$ and $b$ we are allowed to change the color of any $a$ or $b$ consecutive integers from $A$. A pair $(a, b)$ is called *good* if it is possible to change the color of all numbers after finite number of mo... | [] | Bulgaria | First Team Selection Test for 56th IMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1, 3, 7, 9, 21, 63 | |
07b6 | The equation $P(x) = Q(y)$ is called **interesting** if both $P$ and $Q$ are polynomials with integer coefficients and degree at least one and the equation has infinitely many solutions in natural numbers. We say that equation $F(x) = G(y)$ **results** from equation $P(x) = Q(y)$, if polynomial $R$ with rational coeffi... | [
"a) First, note that for each $x \\in \\mathbb{N}$ there are at most a finite number of elements of $S$ with first coordinate $x$. Therefore, for each positive number $R$, there are at most a finite number of elements of $S$ with a first coordinate less than $R$. A same assertion is true for the second coordinate.\... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
09op | Let us denote the set of positive real numbers by $\mathbb{R}_{>0} = \{x \in \mathbb{R} \mid x > 0\}$. Find all functions $f: \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that for all $x, y \in \mathbb{R}_{>0}$, the following holds:
$$f(x f(x) + y) = x^2 + f(y).$$
(Bilegdemberel Bat-Amgalan) | [] | Mongolia | MMO2025 Round 3 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = x for all x > 0 | |
0ghl | 求所有函數 $f: \mathbb{R} \to \mathbb{R}$ 滿足
$$
f(xy + f(y)) f(x) = x^2 f(y) + f(xy)
$$
對於所有實數 $x, y$ 皆成立。
Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that
$$
f(xy + f(y)) f(x) = x^2 f(y) + f(xy)
$$
holds for all real numbers $x, y$. | [
"令 $P(a, b)$ 為 $x = a, y = b$ 代入的結果。\n$$\nP(0,0) : f(f(0))f(0) = f(0) \\Rightarrow f(f(0)) = 1 \\text{ 或 } f(0) = 0\n$$\n\n**Case 1.** $f(f(0)) = 1$\n$$\nP(f(0), 0) : f(f(0))^2 = f(0)^3 + f(0) \\Rightarrow f(0)^3 + f(0) = 1 \\cdots (1)\n$$\n$$\nP(0, f(0)) : f(1)f(0) = f(0) \\Rightarrow f(1) = 1\n$$\n$$\nP(-1, 0) : ... | Taiwan | 2023 數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | Chinese (Traditional) | proof and answer | f(x) = 0 for all x, or f(x) = x for all x | |
00ur | For positive integers $a, b, c$ (not necessarily distinct), suppose that $a + bc$, $b + ca$, and $c + ab$ are all perfect squares. Prove that
$$
a^2(b + c) + b^2(c + a) + c^2(a + b) + 2abc
$$
can be written as the sum of two square numbers. | [
"We denote $x^2 = a + bc$, $y^2 = b + ca$, $z^2 = c + ab$. We make use of the following well-known lemma:\n\n*Lemma.* A positive integer $n$ can be written as the sum of two squares if and only if for all primes $p \\equiv 3 \\pmod{4}$, $v_p(n)$ is even.\n\nNote that we can write the target expression as:\n$$\nS = ... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof only | null | |
0jxa | Problem:
Let $S=\{1,2, \ldots, n\}$ for some positive integer $n$, and let $A$ be an $n$-by-$n$ matrix having as entries only ones and zeros. Define an infinite sequence $\{x_{i}\}_{i \geq 0}$ to be strange if:
- $x_{i} \in S$ for all $i$,
- $a_{x_{k} x_{k+1}}=1$ for all $k$, where $a_{ij}$ denotes the element in the $... | [
"Solution:\nConsider the directed graph $G$ on $n$ labeled vertices whose adjacency matrix is $A$. Then, observe that a strange sequence is simply an infinite path on $G$. Since powers of the adjacency matrix count paths, if $A$ is nilpotent, there exists no infinite path. If $A$ is not nilpotent, a cycle must exis... | United States | HMIC | [
"Discrete Mathematics > Graph Theory",
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
036e | Problem:
Find all positive integers $t, x, y, z$ such that
$$
2^{t} = 3^{x} 5^{y} + 7^{z}
$$ | [] | Bulgaria | Bulgarian Mathematical Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | t=6, x=1, y=1, z=2 | |
0cq0 | Let $ABC$ be an acute-angled triangle. A circle passing through its vertex $B$ and its circumcenter $O$ intersects the sides $BC$ and $BA$ at points $P \neq B$ and $Q \neq B$ respectively. Prove that the orthocenter of triangle $POQ$ lies on line $AC$. (T. Emelyanova, L. Emelyanov)
Дан остроугольный треугольник $ABC$.... | [
"Обозначим $\\angle OBA = \\angle OAB = \\alpha$, $\\angle OBC = \\angle OCB = \\gamma$; тогда $\\angle ACB = \\angle AOB = 90^\\circ - \\alpha$. Поскольку четырёхугольник $BPOQ$ вписан, $\\angle OPQ = \\alpha$ и $\\angle OQP = \\gamma$. Пусть $OO_1$ — высота треугольника $OPQ$, а $H$ — точка пересечения прямых $OO... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English, Russian | proof only | null | |
0jio | Problem:
Given that $x$ and $y$ are nonzero real numbers such that $x + \frac{1}{y} = 10$ and $y + \frac{1}{x} = \frac{5}{12}$, find all possible values of $x$. | [
"Solution:\nLet $z = \\frac{1}{y}$. Then $x + z = 10$ and $\\frac{1}{x} + \\frac{1}{z} = \\frac{5}{12}$. Since $\\frac{1}{x} + \\frac{1}{z} = \\frac{x + z}{x z} = \\frac{10}{x z}$, we have $x z = 24$. Thus, $x(10 - x) = 24$, so $x^{2} - 10x + 24 = (x - 6)(x - 4) = 0$, whence $x = 6$ or $x = 4$.\n\nAlternate solutio... | United States | HMMT 2014 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 4, 6 | |
0h2e | Let $H$ be the point of intersection of the altitudes $AP$ and $CQ$ of an acute-angled triangle $ABC$. On the median $BM$ points $E$ and $F$ are chosen so that $\angle APE = \angle BAC$, $\angle CQF = \angle BCA$, where the point $E$ lies inside the triangle $APB$, and the point $F$ lies inside the triangle $CQB$. Prov... | [
"We will prove that the lines $AE$ and $CF$ divide the line segment $BH$ in the same ratio (it is easy to see that the points of intersection of the lines $AE$ and $CF$ with the line $BH$ will belong to this segment). Let $T$ be the point of intersection of the lines $AE$ and $BH$ (Fig. 52). Then\n\n$$\n\\frac{BT}{... | Ukraine | Problems of Ukrainian Authors | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > M... | English | proof only | null | |
05nc | Problem:
Soit $n \geqslant 2$. On trace un circuit sur un échiquier $n \times n$ qui passe exactement une fois par chaque case et revient à son point de départ (deux cases consécutives sur le circuit doivent avoir un côté commun).
Montrer qu'il existe deux cases voisines sur l'échiquier telles que, si on "coupe" le c... | [
"Solution:\n\nRemarquons tout d'abord que si l'échiquier est pavé en noir et blanc de la manière habituelle, le circuit doit alterner des cases noires et blanches et revenir à sa case de départ donc sa longueur doit être paire donc $n^{2}$ est pair. On écrit donc $n=2m$.\n\n\n\nOn considère... | France | OCympiades Françaises de Mathématiques - Envoi Numéro 4 - Combinatoire | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Other"
] | null | proof only | null | |
03ue | There are $n$ white and $n$ black balls placed randomly on the circumference of a circle. Starting from a certain white ball, number all white balls in a clockwise direction by $1, 2, \dots, n$. Likewise, number all black balls by $1, 2, \dots, n$ in an anti-clockwise direction starting from a certain black ball. Prove... | [
"**Proof** Choose a black ball and a white ball with the same number, and the number of balls between the two balls is minimum. We can suppose the number of the two balls is $1$.\nFirstly, we shall prove that the balls between the two balls have the same color.\n\nIn fact, if they are of different color, then the w... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
03d9 | Let $P$ and $Q$ be fixed polynomials with real coefficients, let the degree of $Q$ be $2021$, and let $a_1, a_2, \dots, a_{2022}, b_1, b_2, \dots, b_{2022}$ be real numbers, such that $a_1 a_2 \dots a_{2022} \neq 0$. If
$$
P(a_1Q(x) + b_1) + \dots + P(a_{2021}Q(x) + b_{2021}) = P(a_{2022}Q(x) + b_{2022}), \quad \forall... | [
"If $P$ is a constant, then $P \\equiv 0$. If there exist $a, b$, such that $P(a) < 0$ and $P(b) > 0$, due to continuity $P$ will have a real root between $a$ and $b$. If no such $a, b$ exist, then $P$ attains either only positive or only negative values. WLOG $P(x) > 0$ for all $x$. If $a_i \\neq a_{2022}$ for som... | Bulgaria | Bulgaria 2022 | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0jsw | Problem:
How many perfect squares divide $10^{10}$? | [
"Solution:\n\nAnswer: 36\nA perfect square $s$ divides $10^{10}$ if and only if $s = 2^{a} \\cdot 5^{b}$ where $a, b \\in \\{0, 2, 4, 6, 8, 10\\}$. There are 36 choices, giving 36 different $s$'s."
] | United States | HMMT November | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | final answer only | 36 | |
0ekw | Problem:
Tone, Luka in Tine so zbirali star papir in dobili denarno nagrado. Prvotno naj bi bila nagrada razdeljena v razmerju $7:6:5$. Kasneje so dogovor spremenili in razdelili nagrado v razmerju $6:5:4$. Obe razmerji sta zapisani v istem vrstem redu, kot so navedena imena.
a. Katera delitev je za Tineta ugodnejša?... | [
"Solution:\n\na.\nTine bi po prvotnem dogovoru dobil $\\frac{5}{7+6+5} = \\frac{5}{18} = \\frac{25}{90}$ celotne nagrade, po spremembi pa dobi $\\frac{4}{6+5+4} = \\frac{4}{15} = \\frac{24}{90}$ nagrade. Zato je za Tineta ugodnejša prva delitev.\n\nb.\nVemo, da so razdelili nagrado v razmerju $6:5:4$. Tako je Tonet... | Slovenia | 22. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a) The first division is better for Tine. b) Tone 648 euros, Luka 540 euros, Tine 432 euros. | |
05fu | Problem:
Déterminer tous les nombres irrationnels $x$ pour lesquels les deux nombres $x^{2}+x$ et $x^{3}+2 x^{2}$ sont des entiers. | [
"Solution:\n\nSoit $x$ un nombre irrationnel pour lequel les deux nombres $x^{2}+x$ et $x^{3}+2 x^{2}$ sont des entiers.\nOn pose $x^{2}+x=a$ et $x^{3}+2 x^{2}=b$, où $a$ et $b$ sont des entiers. Alors $b-a x=x^{2}=a-x$, et ainsi $x(a-1)=b-a$. Si $a-1 \\neq 0$, on aurait $x=\\frac{b-a}{a-1}$, et $x$ serait rationne... | France | null | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (-1 - sqrt(5))/2 and (-1 + sqrt(5))/2 | |
0a9k | Problem:
Show that for any integer $n \geq 2$ the sum of the fractions $\frac{1}{a b}$, where $a$ and $b$ are relatively prime positive integers such that $a < b \leq n$ and $a + b > n$, equals $\frac{1}{2}$. | [
"$$\n\\sum_{\\substack{0 < a < n / 2 \\\\ (a, n-a) = 1}} \\frac{1}{n(a-n)} = \\sum_{\\substack{0 < a < n \\\\ (a, n) = 1}} \\frac{1}{a n}\n$$\n(We denote the greatest common factor of $x$ and $y$ by $(x, y)$.) Since $(a, n) = (n-a, n)$, the terms in the sum on the right hand side can be grouped into pairs\n$$\n\\fr... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 25 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1/2 | |
0jrj | Problem:
Calvin has a bag containing 50 red balls, 50 blue balls, and 30 yellow balls. Given that after pulling out 65 balls at random (without replacement), he has pulled out 5 more red balls than blue balls, what is the probability that the next ball he pulls out is red? | [
"Solution:\n\nThe only information this gives us about the number of yellow balls left is that it is even. A bijection shows that the probability that there are $k$ yellow balls left is equal to the probability that there are $30-k$ yellow balls left (flip the colors of the red and blue balls, and then switch the 6... | United States | HMMT February 2015 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 9/26 | |
0eon | Find three prime factors of: $5^{2015} + 5^{2016} + 5^{2017} + 31^{2016}$
NB: Show all your working! | [
"Let us consider the expression:\n$$\n5^{2015} + 5^{2016} + 5^{2017} + 31^{2016}\n$$\nFirst, factor $5^{2015}$ from the first three terms:\n$$\n5^{2015} + 5^{2016} + 5^{2017} = 5^{2015}(1 + 5 + 5^2) = 5^{2015}(1 + 5 + 25) = 5^{2015} \\times 31\n$$\nSo the expression becomes:\n$$\n5^{2015} \\times 31 + 31^{2016}\n$$... | South Africa | South African Mathematics Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 2, 3, 31 | |
05zk | Problem:
Soit $ABC$ un triangle et $\Omega$ son cercle circonscrit. On note $A'$ le point diamétralement opposé à $A$ dans le cercle $\Omega$. Soit $I$ le centre du cercle inscrit au triangle $ABC$, $E$ et $F$ les points de contact du cercle inscrit avec les côtés $AC$ et $AB$ respectivement. Le cercle circonscrit au ... | [
"Solution:\n\n\n\nPuisque $F$ est le point de contact du cercle inscrit avec le côté $[AC]$, l'angle $\\widehat{IFA}$ est droit et le segment $[IA]$ est un diamètre du cercle circonscrit au triangle $AEF$. On en déduit que\n$$\n\\widehat{AXI} = 90^\\circ = \\widehat{AXA'}\n$$\nou on a utili... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kba | Problem:
Submit an integer $x$ as your answer to this problem. The number of points you receive will be $\max (0,8-|8 x-100|)$. (Non-integer answers will be given 0 points.) | [
"Solution:\n\nWe want to minimize $|8 x-100|$, so $x$ should equal either the floor or the ceiling of $\\frac{100}{8}=12.5$. Note that no other answers receive any points, while both $12$ and $13$ receive $4$ points."
] | United States | HMMT February 2020 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 12 or 13 | |
0el0 | Problem:
Delitelj naravnega števila $n$ je pravi delitelj, če je različen od 1 in $n$. Za koliko naravnih števil $n$ je število 17 največji pravi delitelj?
(A) 0
(B) 1
(C) 3
(D) 5
(E) 7 | [
"Solution:\n\nKer je 17 pravi delitelj števila $n$, je $n = 17k$ za neko naravno število $k \\geq 2$. Ker je tudi $k$ pravi delitelj števila $n$, mora veljati $k \\leq 17$. Hkrati mora biti $k$ praštevilo, saj če bi bilo $k = ab$ za neki naravni števili $a, b > 1$, tedaj bi bil $17a$ pravi delitelj števila $n$, ki ... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | E | |
002t | Dados dos enteros positivos $a$ y $b$, se denota por $(a \nabla b)$ el residuo que se obtiene al dividir $a$ por $b$. Este residuo es uno de los números $0, 1, \ldots, b-1$. Encuentre todas las parejas de números $(a, p)$ tales que $p$ es primo y se cumple que
$$
(a \nabla p) + (a \nabla 2p) + (a \nabla 3p) + (a \nabla... | [] | Argentina | XX Olimpiada Iberoamericana de Matemáticas | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Español | proof and answer | All pairs (a, p) with a = 3p for any prime p, together with (a, p) = (1, 3) and (17, 3). | |
0crg | Петя поставил на доску $50 \times 50$ несколько фишек, в каждую клетку — не больше одной. Докажите, что у Васи есть способ поставить на свободные поля этой же доски не более 99 новых фишек (возможно, ни одной) так, чтобы по-прежнему в каждой клетке стояло не больше одной фишки, и в каждой строке и каждом столбце этой д... | [
"Построим граф с вершинами $r_1, \\dots, r_{50}$, соответствующими строкам доски, и вершинами $c_1, \\dots, c_{50}$, соответствующим её столбцам. Вершины $r_i$ и $c_j$ соединим ребром, если клетка в пересечении соответствующих строки и столбца свободна. Тогда Васина цель переформулируется так: требуется отметить не... | Russia | XL Russian mathematical olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0iy0 | Problem:
Paul fills in a $7 \times 7$ grid with the numbers $1$ through $49$ in a random arrangement. He then erases his work and does the same thing again (to obtain two different random arrangements of the numbers in the grid). What is the expected number of pairs of numbers that occur in either the same row as each... | [
"Solution:\n\nEach of the $\\binom{49}{2}$ pairs of numbers has a probability of $\\frac{14 \\cdot \\binom{7}{2}}{\\binom{49}{2}} = 1/4$ of being in the same row or column in one of the arrangements, so the expected number that are in the same row or column in both arrangements is\n$$\n\\binom{49}{2} \\cdot (1/4)^2... | United States | $12^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 147/2 | |
0k1o | Problem:
Pascal has a triangle. In the $n$th row, there are $n+1$ numbers $a_{n, 0}, a_{n, 1}, a_{n, 2}, \ldots, a_{n, n}$ where $a_{n, 0}=a_{n, n}=1$. For all $1 \leq k \leq n-1$, $a_{n, k}=a_{n-1, k}-a_{n-1, k-1}$. What is the sum of the absolute values of all numbers in the 2018th row?
Proposed by: Michael Ren | [
"Solution:\n\nLet $s_{n}$ be the sum of the absolute values of numbers in the $n$th row. For odd $n$, we have that $a_{n, 1}, \\ldots, a_{n, n-1}$ alternate in sign as $-,+,-,+, \\ldots,+$, with the last term being $a_{n, n-1}=1$. For even $n$, we have that $a_{n, 1}, \\ldots, a_{n, n-2}$ alternate in sign as $-,+,... | United States | HMMT November 2018 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (2^{2018} + 2) / 3 | |
01oh | $N$ boys ($N \ge 3$), no two of them having the same height, are arranged along a circle. A boy in the given arrangement is said to be *tall* if he is taller than both of his neighbors.
Find all possible numbers of tall boys in the arrangement. | [
"Answer: any integer number from $1$ to $[N/2]$.\n\nConsider arbitrary arrangement of the boys along the circle. We put the signs \"+\" or \"-\" before any boy in accordance with the following rule: we move clockwise along the circle and put the sign \"+\" before the boy if he is taller than the previous boy and we... | Belarus | Belorusija 2012 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | all integers from 1 to floor(N/2) | |
08lz | Problem:
Find the maximum value of $z + x$, if $(x, y, z, t)$ satisfies the conditions:
$$
\left\{\begin{array}{l}
x^{2} + y^{2} = 4 \\
z^{2} + t^{2} = 9 \\
x t + y z \geq 6
\end{array}\right.
$$ | [
"Solution:\nFrom the conditions we have\n$$\n36 = (x^{2} + y^{2})(z^{2} + t^{2}) = (x t + y z)^{2} + (x z - y t)^{2} \\geq 36 + (x z - y t)^{2}\n$$\nand this implies $x z - y t = 0$.\nNow it is clear that\n$$\nx^{2} + z^{2} + y^{2} + t^{2} = (x + z)^{2} + (y - t)^{2} = 13\n$$\nand the maximum value of $z + x$ is $\... | JBMO | 2009 Shortlist JBMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | sqrt(13) | |
05f1 | Problem:
Soit $ABC$ un triangle acutangle. Soient $P$ et $Q$ deux points sur le segment $[BC]$. On note respectivement $O_1$, $O_2$, $O_3$ et $O_4$ les centres des cercles circonscrits des triangles $ABP$, $ABQ$, $ACP$ et $ACQ$. Montrer que les points $O_1$, $O_2$, $O_3$ et $O_4$ sont cocycliques si et seulement si $\... | [
"Solution:\n\n\n\nLa figure comporte beaucoup d'objets, on va donc essayer de la simplifier un peu. On remarque donc dans un premier temps qu'il y a beaucoup d'axes radicaux (les droites $(AB)$, $(AP)$, $(AQ)$ et $(AC)$ sur la figure) et donc beaucoup d'angles droits avec les centres des ce... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03ea | Find all pairs $(a, b)$ of co-prime naturals such that $a < b$ and $b$ divides
$$
(n + 2)a^{n+1002} - (n + 1)a^{n+1001} - n a^{n+1000}
$$
for every natural number $n$. | [
"Since $a$ and $b$ are co-prime, so are $b$ and $a^{n+1000}$, respectively, what is requested is equivalent to $b$ dividing $(n + 2)a^2 - (n + 1)a - n$ for each $n$.\n\nFrom $n = 1$ and $n = 2$ we get that necessarily $b$ divides $3a^2 - 2a - 1$ and $4a^2 - 3a - 2$. Hence $b$ divides\n$$\n4(3a^2 - 2a - 1) - 3(4a^2 ... | Bulgaria | 1 Autumn tournament | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | (3, 5) | |
07cb | $n \ge 50$ is a natural number. Prove that $n$ can be written as the sum of two natural numbers whose prime factors do not exceed $\sqrt{n}$. For example, $94$ can be written as $80 + 14$, none of which has a prime factor greater than $\sqrt{94}$. | [
"The notation $x \\sqsubseteq y$ is used to show that none of the prime factors of $x$ exceed $y$. Obviously, if $r \\le m$, $r \\sqsubseteq m$; and if $r \\sqsubseteq m$ and $s \\sqsubseteq m$, then $rs \\sqsubseteq m$.\n\nSuppose that $[\\sqrt{n}] = m$. Therefore, $n = m^2 + r$, where $0 \\le r \\le 2m$ ($n \\ge ... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0h0f | Find the least possible value $k$ for which there exist $2010$ distinct natural numbers that satisfy the following condition: the product of any $k$ numbers from the chosen set is divisible by the product of the rest $2010 - k$ numbers. | [
"From one hand, $k$ cannot be less than $1006$ (otherwise, the product of the $k$ smallest numbers from our set is less than the product of the $2010 - k$ numbers which are left). We construct an example for $k = 1006$ that will do.\n\nLet $p_1, p_2, \\dots, p_{2010}$ be $2010$ distinct prime numbers and $a_i = p_1... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010) | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 1006 | |
0aio | For a positive integer $n$, two players $A$ and $B$ play the following game: Given a pile of $s$ stones, the players take turns alternatively with $A$ going first. On each turn the player is allowed to take either one stone, or a prime number of stones, or a multiple of $n$ stones. The winner is the one who takes the l... | [
"Denote by $k$ the sought number and let $\\{s_1, s_2, \\dots, s_k\\}$ be the corresponding values for $s$. We call each $s_i$ a losing number and every other nonnegative integer a winning number.\n\n(I) **Clearly every multiple of $n$ is a winning number.**\nSuppose there are two different losing numbers $s_i > s_... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof and answer | n - 1 | |
0d8v | Let $ABCD$ be a convex quadrilateral. Ray $AD$ meets ray $BC$ at $P$. Let $O, O'$ be the circumcenters of triangles $PCD, PAB$, respectively, $H, H'$ be the orthocenters of triangles $PCD, PAB$, respectively. Prove that circumcircle of triangle $DOC$ is tangent to circumcircle of triangle $AO'B$ if and only if circumci... | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Spiral similarit... | English | proof only | null | |
0dii | Calculate the sum of digits of the number
$$
1 + 11 + 111 + \cdots + \underbrace{111\dots111}_{\text{2023 1's}}.
$$ | [
"The given number can be expressed as\n$$\n\\begin{aligned}\n\\sum_{i=1}^{2023} \\underbrace{11\\dots11}_{i \\text{ 1's}} &= \\sum_{i=1}^{2023} \\frac{10^i - 1}{9} = \\frac{\\sum_{i=1}^{2023} 10^i - 2023}{9} \\\\\n&= \\frac{\\underbrace{111\\dots1110}_{2023 \\text{ 1's}} - 2023}{9} = \\frac{\\underbrace{111\\dots11... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | final answer only | 8308 | |
03n9 | Problem:
Let five points on a circle be labelled $A$, $B$, $C$, $D$, and $E$ in clockwise order. Assume $A E = D E$ and let $P$ be the intersection of $A C$ and $B D$. Let $Q$ be the point on the line through $A$ and $B$ such that $A$ is between $B$ and $Q$ and $A Q = D P$. Similarly, let $R$ be the point on the line ... | [
"Solution:\n\nWe are given $A Q = D P$ and $A P = D R$. Additionally $\\angle Q A P = 180^{\\circ} - \\angle B A C = 180^{\\circ} - \\angle B D C = \\angle R D P$, and so triangles $A Q P$ and $D P R$ are congruent. Therefore $P Q = P R$. It follows that $P$ is on the perpendicular bisector of $Q R$.\n\nWe are also... | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null |
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