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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0bpr | The incircle of the triangle $ABC$ has center $I$, touches $BC$, $CA$ and $AB$ at $D$, $E$, respectively $F$ and meets $AI$, $BI$, $CI$ in $M$, $N$, respectively $P$. Prove that if the triangles $DEF$ and $MNP$ have the same centroid, then the triangle $ABC$ is equilateral. | [] | Romania | 67th NMO Shortlisted Problems | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Com... | English | proof only | null | |
04no | Let $z$ be a complex number such that $\arg z \in [\frac{\pi}{2}, \pi]$ and $z^6 + z^3 + 1 = 0$. Determine the modulus and the argument of $z$. The argument of the complex number $z = |z|(\cos\varphi + i\sin\varphi)$ is the number $\arg z = \varphi$. | [] | Croatia | Croatia_2018 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | English | proof and answer | |z| = 1, arg z = 8π/9 | |
0cu1 | Consider nine 9-digit numbers, each contains exactly the digits $1, 2, 3, 4, 5, 6, 7, 8, 9$ (in some order). The sum of these nine numbers ends with $k$ zeroes. Determine the maximal possible value of $k$.
Из цифр $1, 2, 3, 4, 5, 6, 7, 8, 9$ составлены девять (не обязательно различных) девятизначных чисел; каждая из ц... | [
"$k = 8$.\n\nПокажем, что сумма не может оканчиваться на 9 нулей. Каждое из составленных чисел делится на $9$, поскольку сумма его цифр делится на $9$. Поэтому их сумма также делится на $9$. Наименьшее натуральное число, делящееся на $9$ и оканчивающееся на девять нулей, равно $9 \\cdot 10^9$, так что сумма наших ч... | Russia | Russian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English; Russian | proof and answer | 8 | |
05c4 | Find all positive integers $n$ for which all $10^n$ non-negative integers consisting of $n$ digits can be ordered in such a way that all the following conditions are met:
(1) the first number consists of zeros only;
(2) every two numbers that are consecutive in this order differ at exactly one position and the digits a... | [
"Assume that for some $n$, a suitable ordering exists. From condition 2 we see that the sums of digits of any two consecutive numbers differ by exactly 1, and thus have opposite parities. As the first number has an even sum of digits, the $10^n$-th number must have an odd sum of digits. As the latter sum is $9n$, t... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All odd positive integers n | |
0a98 | Problem:
Assume that $n \geq 3$ people with different names sit around a round table. We call any unordered pair of them, say $M$ and $N$, dominating, if
(i) $M$ and $N$ do not sit on adjacent seats, and
(ii) on one (or both) of the arcs connecting $M$ and $N$ along the table edge, all people have names that come alph... | [
"Solution:\n\nWe will show by induction that the number of dominating pairs (hence also the minimal number of dominating pairs) is $n-3$ for $n \\geq 3$.\n\nIf $n=3$, all pairs of people sit on adjacent seats, so there are no dominating pairs.\n\nAssume that the number of dominating pairs is $n-3$ for some $n>3$. I... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 22 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | n-3 | |
04lz | Prove that it is possible to colour each positive integer with one of three colours so that the following conditions are satisfied:
i) For each $n \in \mathbb{N}_0$, all positive integers $x$ such that $2^n \le x < 2^{n+1}$ have the same colour.
ii) There are no positive integers $x$, $y$ and $z$ of the same colour (ex... | [
"Let $c_n$ denote the colour of positive integers $x$ such that $2^n \\le x < 2^{n+1}$. We will determine colours $c_n$ inductively. First, let us choose $c_0$, $c_1$ and $c_2$ to be three different colours. Next, for each $n \\ge 3$ let $c_n$ be the colour different from $c_{\\lfloor n/2 \\rfloor}$ and $c_{\\lfloo... | Croatia | Croatian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0kg8 | Problem:
The graphs of the equations
$$
\begin{aligned}
y & = -x + 8 \\
173y & = -289x + 2021
\end{aligned}
$$
on the Cartesian plane intersect at $(a, b)$. Find $a + b$. | [
"Solution:\nFrom the first equation, it is known that $(a, b)$ lies on the line $x + y = 8$, therefore $a + b = 8$."
] | United States | HMMT November 2021 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 8 | |
0akq | In a circle is inscribed regular $2018$-gon. The numbers $1,2,\ldots,2018$ are put in the vertices of the $2018$-gon, in each vertex only one number, such that the sum of every two consecutive numbers (of that configuration) is equal to the sum of their diametral opposite numbers. Find the number of all such configurat... | [
"Let us consider a configuration satisfying the conditions of the problem. Let $A,B$ be two consecutive numbers in that configuration and let their diametral opposite numbers be $a,b$, respectively. Then holds $A+B=a+b$ i.e. $A-a=b-B$. Since $A,B$ are arbitrary, we get that every difference of the numbers which are... | North Macedonia | Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 2 * 1008! | |
098n | Problem:
Rezolvați ecuația
$$
\cos 2x - \sin 2x + 2 \cos x + 1 = 0
$$ | [
"Solution:\nAvem ecuațiile echivalente:\n$$\n\\begin{gathered}\n\\cos 2x - \\sin 2x + 2 \\cos x + 1 = 0 \\\\\n1 - 2 \\sin^2 x - 2 \\sin x \\cos x + 2 \\cos x + 1 = 0 \\\\\n\\sin^2 x + \\sin x \\cos x - \\cos x - 1 = 0 \\\\\n(\\sin^2 x - \\sin x) + (\\sin x \\cos x - \\cos x) + (\\sin x - 1) = 0 \\\\\n(\\sin x - 1) ... | Moldova | Moldova National Olympiad | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | x = −π/2 + 2kπ, x = π/2 + 2kπ, or x = π + 2kπ for k ∈ ℤ | |
05oh | Problem:
Trouver tous les entiers $m$, $n \geq 0$ tels que $n^{3}-3 n^{2}+n+2=5^{m}$. | [
"Solution:\n\nOn peut factoriser $n^{3}-3 n^{2}+n+2=(n-2)(n^{2}-n-1)$. Pour que l'équation soit vraie, il faut que les deux facteurs soient, au signe près, des puissances de $5$.\n\nSi $n-2$ est négatif, comme $n$ est positif, pour que $-(n-2)$ soit une puissance de $5$, il faut prendre $n=1$, et dans ce cas $n^{3}... | France | Envoi 1: Arithmétique | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (m, n) = (0, 1) and (1, 3) | |
04yp | The sequence $(a_n)$ is defined by $a_1 = 1$ and $a_n = n \cdot (a_1 + \dots + a_{n-1})$ for all $n > 1$. Find all indices $n$ for which $a_n$ is divisible by $1 \cdot 2 \cdot \dots \cdot n$. (Grade 12.) | [
"For each $n \\ge 2$ denote $S_n = a_1 + \\dots + a_{n-1}$. Then $a_n = S_n \\cdot n$ and for all $n > 2$ we have $S_n = S_{n-1} + a_{n-1} = S_{n-1} + S_{n-1} \\cdot (n-1) = S_{n-1} \\cdot n$. Hence $S_n = S_{n-1} \\cdot n = S_{n-2} \\cdot (n-1)n = \\dots = S_2 \\cdot 3 \\cdot \\dots \\cdot n = \\frac{n!}{2}$ becau... | Estonia | Estonija 2010 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | n = 1 or n is even | |
0dbu | For $n \geq 3$, it is given an $2n \times 2n$ board with black and white squares. It is known that all border squares are black and no $2 \times 2$ subboard has all four squares of the same color. Prove that there exists a $2 \times 2$ subboard painted like a chessboard, i.e. with two opposite black corners and two opp... | [
"Assume for the sake of contradiction that there are no $2 \\times 2$ square painted like a chessboard. Then all $2 \\times 2$ subboards are of these paintings as follow (and their rotations).\n\n\n\nWe will count the length of black-white border in two ways.\n\n1. First way. Observe that f... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0dsz | Simon plays a game on an $n \times n$ grid of cells. Initially, each cell is filled with an integer. Every minute, Simon picks a cell satisfying the following:
(a) The magnitude of the integer in the chosen cell is less than $n^n$
(b) The sum of all the integers in the neighboring cells (sharing one side with the chose... | [] | Singapore | Singapore International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
04wl | Let a convex quadrilateral $ABCD$ be inscribed in a circle with center $O$ and circumscribed to a circle with center $I$, and let its diagonals $AC$ and $BD$ meet at a point $P$. Prove that the points $O$, $I$ and $P$ are collinear. | [
"Assume that the lines $AI$, $BI$, $CI$, $DI$ meet the circumcircle of the quadrilateral $ABCD$ at $E$, $F$, $G$, $H$, respectively. Since the lines $AI$, $BI$, $CI$, $DI$ are the bisectors of the respective angles of the quadrilateral $ABCD$, the lines $EG$ and $FH$ are the diameters of the circumcircle of $ABCD$.... | Czech-Polish-Slovak Mathematical Match | Czech-Polish-Slovak Match | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
08fv | Problem:
Il soggiorno della casa di Filippo ha pianta rettangolare. Filippo ha notato che se attacca l'aspirapolvere alla presa vicino alla porta d'ingresso riesce a pulire tutto il pavimento: questo vuol dire che tutti i punti del pavimento sono a distanza minore di $5~\mathrm{m}$ dal punto della parete dove si trova... | [
"Solution:\n\nLa risposta è 25. Consideriamo il rettangolo che forma la pianta del soggiorno, e orientiamolo in modo che la presa $P$ si trovi sul lato orizzontale $AB$, suddividendolo in due segmenti di lunghezza $a$ e $b$. Chiamiamo inoltre $c$ la lunghezza dei due lati verticali $BC$ e $DA$, come in figura. In q... | Italy | Olimpiadi di Matematica - Febbraio | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals"
] | null | proof and answer | 25 | |
0bbh | Let $A, B \in \mathcal{M}_2(\mathbb{C})$ such that $A^2 + B^2 = 2AB$. Prove $AB = BA$ and $\text{tr } A = \text{tr } B$. | [
"a) We firstly prove that $(AB - BA)^2 = 0$. Define the quadratic function $g$ by $f(x) = \\det(A^2 + B^2 + x(AB - BA)) = \\det(A^2 + B^2) + mx + x^2 \\det(AB - BA)$. As $f(-i) = \\det((A + iB)(A - iB))$, $f(i) = \\det((A - iB)(A + iB))$, so $f(-i) = f(i)$, we get $m = 0$. Moreover, from $f(0) = \\det(A^2 + B^2) = ... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof only | null | |
0hsv | Problem:
Let $M_{n}$ be the number of integers $N$ such that
(a) $0 \leq N < 10^{n}$;
(b) $N$ is divisible by $4$;
(c) The sum of the digits of $N$ is also divisible by $4$.
Prove that $M_{n} \neq 10^{n} / 16$ for all positive integers $n$. | [
"Solution:\nSince $10^{n} / 16$ is not an integer for $n=1,2,3$, we may assume that $n \\geq 4$. Let\n$$\n\\tilde{M}_{n} = M_{n} - \\frac{10^{n}}{16}\n$$\nWe note that numbers whose hundreds digit is between $2$ and $9$ inclusive make no total contribution to $\\tilde{M}_{n}$ for $n \\geq 3$. This is because, given... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0k1g | Problem:
Find the smallest positive integer $n$ for which
$$
1!2!\cdots(n-1)! > n!^{2}
$$ | [
"Solution:\nAnswer: 8\nDividing both sides by $n!^{2}$, we obtain\n$$\n\\begin{aligned}\n\\frac{1!2!\\ldots(n-3)!(n-2)!(n-1)!}{[n(n-1)!][n(n-1)(n-2)!]} & > 1 \\\\\n\\frac{1!2!\\ldots(n-3)!}{n^{2}(n-1)} & > 1 \\\\\n1!2!\\ldots(n-3)! & > n^{2}(n-1)\n\\end{aligned}\n$$\nFactorials are small at first, so we can rule ou... | United States | HMMT November 2018 | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 8 | |
02ua | Problem:
Um erro que muitos alunos cometem é pensar que dois quadriláteros são congruentes se tiverem os seus respectivos lados iguais. Isso não é verdade. Nesse problema, veremos que quadriláteros podem ter lados correspondentes iguais, mas áreas distintas.
a) Mostre que a maior área possível para um quadrilátero qu... | [
"Solution:\n\na) Existem dois modos de montar o quadrilátero com pares de lados iguais: ou eles ficam juntos ou ficam separados. Nos dois casos, o quadrilátero pode ser dividido em dois triângulos que serão congruentes pelo caso (L.L.L.). Veja a figura abaixo.\n\n\n\n![](attached_image_2.pn... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 12 | |
087q | Problem:
Per quanti interi relativi $n$ si ha che $\frac{3 n}{n+5}$ è intero e divisibile per $4$?
(A) 1
(B) 2
(C) 4
(D) 8
(E) più di 8. | [
"Solution:\n\nLa risposta è **(C)**. Sostituendo $m = n + 5$, l'espressione data diventa $\\frac{3m - 15}{m}$, cioè $3 - \\frac{15}{m}$. Affinché sia intera, quindi, $m$ deve essere un divisore di $15$, per cui le possibilità sono solo $1, 3, 5, 15$ e i loro opposti. Di queste, le uniche per cui l'espressione è un ... | Italy | Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | C | |
00yx | Problem:
Prove that if both coordinates of every vertex of a convex pentagon are integers, then the area of this pentagon is not less than $\frac{5}{2}$. | [
"Solution:\n\nThere are two vertices $A_{1}$ and $A_{2}$ of the pentagon that have their first coordinates of the same parity, and their second coordinates of the same parity. Therefore the midpoint $M$ of $A_{1}A_{2}$ has integer coordinates. There are two possibilities:\n\n(i) The considered vertices are not cons... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
06o7 | $ABCD$ is a square of side length $1$. $BC$ is extended to $E$ and $DC$ is extended to $F$ such that $BE = DF = 3$. The circumcircle of $\triangle AEF$ meets the extensions of $CB$ and $CD$ at $G$ and $H$ respectively. Find $GH$. | [
"Answer: $\\frac{3\\sqrt{2}}{2}$\n\nNote that $AG = AH$ by symmetry. Hence they subtend equal angles on the circumference of the circle, i.e. we have $\\angle AFG = \\angle AFH$. Suppose $FA$ meets $GE$ at $I$.\n\nBy similar triangles, we have $BI : IC = AB : CF = 1 : 2$. Hence $IC = \\frac{2}{3}$. Let $GC = x$. Us... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3√2/2 | |
009p | Let $x \ge 5$, $y \ge 6$, $z \ge 7$ and $x^2 + y^2 + z^2 \ge 125$. Find the minimum of $x+y+z$. | [
"The minimum of $x+y+z$ is $19$. The value $19$ is attained for $x=5$, $y=6$, $z=8$.\n\nConversely, we prove $x+y+z \\ge 19$ for all admissible $x$, $y$, $z$. One may assume $x<6$, $y<7$, $z<8$. Indeed, if one of the inequalities $x \\ge 6$, $y \\ge 7$, $z \\ge 8$ holds, then $x+y+z \\ge (5+6+7)+1=19$.\n\nSet $u=x-... | Argentina | NATIONAL XXX OMA | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 19 | |
0bdu | Let $f: \mathbb{R} \to \mathbb{R}$ be an arbitrary function and $g: \mathbb{R} \to \mathbb{R}$ be a quadratic function such that: for any reals $m$ and $n$, the equation $f(x) = mx + n$ has solutions if and only if the equation $g(x) = mx + n$ has solutions. Prove $f = g$.
Vasile Pop | [
"Remark that if the graph of $h: \\mathbb{R} \\to \\mathbb{R}$, $h(x) = mx + n$ is tangent to the graph of $g: \\mathbb{R} \\to \\mathbb{R}$, $g(x) = ax^2 + bx + c$, then the equation $ax^2 + bx + c = mx + n$, $a, b, c, m, n \\in \\mathbb{R}$, $a \\neq 0$ has zero discriminant, that is the function $k: \\mathbb{R} ... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0c24 | For real $a < b$, and $f : (a, b) \to \mathbb{R}$ a function such that the functions defined by $g : (a, b) \to \mathbb{R}$, $g(x) = (x-a)f(x)$, and $h : (a, b) \to \mathbb{R}$, $h(x) = (x-b)f(x)$ are non-decreasing, prove that $f$ is continuous.
Vladimir Cerbu and Sorin Rădulescu | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Precalculus > Functions",
"Precalculus > Limits"
] | null | proof only | null | |
0026 | Arnaldo elige un número $a$, $a \ge 0$, y Bernaldo elige un número $b$, $b \ge 0$. Ambos le dicen en secreto su número elegido a Cernaldo, quien escribe en una pizarra los números $5$, $8$ y $15$, siendo uno de ellos la suma $a + b$.
Cernaldo toca una campana y Arnaldo y Bernaldo, individualmente, escriben en papelitos... | [] | Argentina | 15ª Olimpiada Matemática del Cono Sur | [
"Discrete Mathematics > Logic"
] | español | proof and answer | 10 | |
0cme | Problem:
Initially, a non-constant polynomial $S(x)$ with real coefficients is written down on a board. Whenever the board contains a polynomial $P(x)$, not necessarily alone, one can write down on the board any polynomial of the form $P(C+x)$ or $C+P(x)$, where $C$ is a real constant. Moreover, if the board contains ... | [
"Solution:\n\nThe required polynomials are all polynomials of an even degree $d \\geq 2$, and all polynomials of odd degree $d \\geq 3$ with negative leading coefficient.\n\nPart I. We begin by showing that any (non-constant) polynomial $S(x)$ not listed above is not $(A, B)$-nice for some pair $(A, B)$ with either... | Romanian Master of Mathematics (RMM) | Romanian Master of Mathematics Competition | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof and answer | All non-constant polynomials of even degree at least two, and all odd-degree polynomials of degree at least three with negative leading coefficient. | |
03k9 | Problem:
Determine a triangle for which the three sides and an altitude are four consecutive integers and for which this altitude partitions the triangle into two right triangles with integer sides. Show that there is only one such triangle. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Number Theory > Diophantine Equations > Pythagorean triples"
] | null | proof and answer | The unique triangle has side lengths 13, 14, and 15, with the altitude of length 12 dropped to the side of length 14. | |
01c6 | A doubly infinite sequence $a_n$, for $n \in \mathbb{Z}$, has each $a_n$ equal to either $0$ or $1$. Prove that there exist numbers $p$ and $q > 1$ such that $a_{p+k} = a_{p+q+k}$ for $k = 0, 1, \dots, q-1$. | [
"(a) The run $010$ must extend to $00100$, since $1010$ and $0101$ are excluded.\n\n(b) The run $000$ must extend to $10001$, since $0000$ is excluded.\n\n(c) The run $000$ must extend to $1100011$. For $000$ extends to $10001$ by (b). If this were continued to $100010$, it would extend further to $10001000$ by (a)... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Other"
] | null | proof only | null | |
0jv3 | Problem:
Let $P_{1}, P_{2}, \ldots, P_{6}$ be points in the complex plane, which are also roots of the equation $x^{6}+6 x^{3}-216=0$. Given that $P_{1} P_{2} P_{3} P_{4} P_{5} P_{6}$ is a convex hexagon, determine the area of this hexagon.
Proposed by: Eshaan Nichani | [
"Solution:\n\nAnswer: $9 \\sqrt{3}$\n\nFactor $x^{6}+6 x^{3}-216=\\left(x^{3}-12\\right)\\left(x^{3}+18\\right)$. This gives us 6 points equally spaced in terms of their angles from the origin, alternating in magnitude between $\\sqrt[3]{12}$ and $\\sqrt[3]{18}$. This means our hexagon is composed of 6 triangles, e... | United States | HMMT November 2016 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Al... | null | proof and answer | 9 sqrt(3) | |
086g | Problem:
In un trapezio isoscele $ABCD$ di base maggiore $AB$, le diagonali vengono divise dal loro punto di incontro $O$ in parti proporzionali ai numeri 1 e 3. Sapendo che l'area del triangolo $BOC$ è 15, quanto misura l'area dell'intero trapezio?
(A) 60
(B) 75
(C) 80
(D) 90
(E) 105. | [
"Solution:\n\nLa risposta è $\\mathbf{(C)}$. Evidentemente il triangolo $AOD$ è uguale al triangolo $BOC$, quindi ha anch'esso area 15.\n\nI triangoli $ODC$ e $OCB$ hanno la stessa altezza $CH$, e poiché la base $OD$ di $ODC$ è $1/3$ della base $OB$ di $BOC$, l'area di $ODC$ è $1/3$ dell'area di $BOC$, cioè $15/3 =... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | C | |
07g2 | Let $S$ be a set with $n$ elements and $P(S)$ be the set of all subsets of $S$. We want to partition $P(S)$ to $m$ parts such that if $A$, $B$ and $A \cup B$ are in the same part then $A = B$. Find the minimum value of $m$ so that such a partition exists. | [
"The answer is $m = n + 1$. To give an example, for all integers $0 \\le i \\le n$ define\n$$\nA_i = \\{E \\mid E \\subset S, |E| = i\\}.\n$$\nThis partition satisfies the condition of problem, because if $A$, $B$ and $A \\cup B$ are in the same partition it means that they have the same number of elements but $A, ... | Iran | 38th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | m = n + 1 | |
0i8u | Problem:
What is the area of the region bounded by the curves $y = x^{2003}$ and $y = x^{1 / 2003}$ and lying above the $x$-axis? | [
"Solution:\nThe two curves intersect at $(0, 0)$ and $(1, 1)$, so the desired area is\n\n$$\n\\int_{0}^{1} \\left( x^{1 / 2003} - x^{2003} \\right) dx = \\left[ \\frac{x^{2004 / 2003}}{2004 / 2003} - \\frac{x^{2004}}{2004} \\right]_{0}^{1} = \\frac{1001}{1002}\n$$"
] | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Integral Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | final answer only | 1001/1002 | |
0ar4 | Problem:
Denote by $a$, $b$ and $c$ the sides of a triangle, opposite the angles $\alpha$, $\beta$ and $\gamma$, respectively. If $\alpha$ is sixty degrees, show that
$$
a^{2} = \frac{a^{3} + b^{3} + c^{3}}{a + b + c}.
$$ | [
"Solution:\n\n$a^{2} = b^{2} + c^{2} - 2 b c \\left( \\frac{1}{2} \\right)$ by cosine law, and $b^{3} + c^{3} = (b + c) (b^{2} - b c + c^{2}) = (b + c) a^{2}$. Add $a^{3}$ to both sides and move terms to get the desired equation. That is,\n$$\na^{3} + b^{3} + c^{3} = (b + c) a^{2} + a^{3} = (a + b + c) a^{2}\n$$\na... | Philippines | 13th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
09y3 | Problem:
Zij $\Gamma$ de omgeschreven cirkel van een driehoek $A B C$ en zij $D$ een punt op lijnstuk $B C$. De cirkel door $B$ en $D$ die raakt aan $\Gamma$ en de cirkel door $C$ en $D$ die raakt aan $\Gamma$ snijden in een punt $E \neq D$. De lijn $D E$ snijdt $\Gamma$ in twee punten $X$ en $Y$. Bewijs dat $|E X|=|E ... | [
"Solution:\nWe bekijken de configuratie zoals in de figuur, waarbij $E$ minstens zo dicht bij $B$ ligt als bij $C$. De configuratie waarbij dit andersom is, gaat analoog. Zij $O$ het middelpunt van $\\Gamma$. De hoek tussen de lijn $B C$ en de gemeenschappelijke raaklijn in $B$ is volgens de raaklijnomtrekshoekstel... | Netherlands | IMO-selectietoets II | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point... | null | proof only | null | |
0an6 | Problem:
If $f$ is a function such that $f(a+b) = \frac{1}{f(a)} + \frac{1}{f(b)}$, find all possible values of $f(2011)$. | [
"Solution:\nConsider $f(0) = f(0+0) = \\frac{1}{f(0)} + \\frac{1}{f(0)}$ which gives $[f(0)]^2 = 2$. Thus, $f(0) = \\pm \\sqrt{2}$.\n\nLet $x = f(2011)$.\n\nIf $f(0) = \\sqrt{2}$ then $x = f(2011) = f(2011+0) = \\frac{1}{f(2011)} + \\frac{1}{f(0)} = \\frac{1}{x} + \\frac{1}{\\sqrt{2}}$. So, $x = \\frac{\\sqrt{2} + ... | Philippines | Area Stage | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | {-√2, √2} | |
04p3 | Prove that for non-negative real numbers $a$ and $b$ such that $a + b \le 2$ the inequality
$$
\frac{1}{1+a^2} + \frac{1}{1+b^2} \le \frac{2}{1+ab}
$$
holds. When is the equality attained? (Austria 2018) | [] | Croatia | Croatian Mathematical Society Competitions | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds if and only if a = b (which, under the given constraint, means a = b in the interval from zero to one). | |
0e2s | Let $O$ be the circumcentre of the acute triangle $ABC$ and denote the circumcircle by $\kappa$. The bisector of the inner angle at $A$ meets $\kappa$ again at $D$. The bisector of the inner angle at $B$ meets $\kappa$ again at $E$. Let $I$ denote the incentre of the triangle $ABC$. How much does the angle $\angle ACB$... | [
"The triangle $ABC$ is acute, so the points $I$ and $O$ lie on the same side of the line $ED$. The condition that the points $D$, $E$, $I$ and $O$ lie on the same circle therefore implies that $\\angle DOE = \\angle DIE$.\n\nLet us denote the angles of the triangle by $\\alpha$, $\\beta$ and $\\gamma$ and let us ex... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | pi/3 | |
08m7 | Problem:
Let $A B C D E$ be a convex pentagon such that $A B + C D = B C + D E$ and let $k$ be a semicircle with center on side $A E$ that touches the sides $A B, B C, C D$ and $D E$ of the pentagon, respectively, at points $P, Q, R$ and $S$ (different from the vertices of the pentagon). Prove that $P S \| A E$. | [
"Solution:\n\nLet $O$ be center of $k$. We deduce that $B P = B Q$, $C Q = C R$, $D R = D S$, since those are tangents to the circle $k$. Using the condition $A B + C D = B C + D E$, we derive:\n$$\nA P + B P + C R + D R = B Q + C Q + D S + E S\n$$\nFrom here we have $A P = E S$.\nThus,\n$$\n\\triangle A P O \\cong... | JBMO | 2009 Shortlist JBMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
07m2 | Let $ABC$ be a triangle. Let $AD$ be the angle bisector of the angle $\angle BAC$ and let $BE$ be the angle bisector of the angle $\angle ABC$, with $D$ interior to the side $BC$ and $E$ to the side $AC$. Let $M$ be a point interior to the side $BC$ such that $|CM| = |AE|$, and let $N$ be a point interior to the side $... | [
"Let $a$, $b$ and $c$ be the lengths of the sides of the triangle $ABC$. Using that $AD$ is an angle bisector, that the two angles $\\angle ADB$ and $\\angle CDA$ are complementary (hence have equal sines) and the sine rule for the triangles $ABD$ and $ADC$ one obtains\n$$\n\\frac{|BD|}{|DC|} = \\frac{c}{b} \\quad ... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cuu | Initially, a positive integer $N$ is written on the board. At each moment, Misha may choose a number $a > 1$ on the board, remove it, and write down all its positive divisors except $a$ instead. After some time it happened that the board contains $N^2$ numbers. Determine all values of $N$ for which this can happen. (M.... | [
"Answer: $N = 1$.\n\nRecall that\n$$\n\\frac{1}{2^2} + \\frac{1}{3^2} + \\dots + \\frac{1}{n^2} < 1.\n$$\nfor all $n > 1$. Using this inequality, show by induction on $N$ that the board may get at most $N^2$ numbers, with equality achieved only for $N = 1$.\n\nLemma. For any natural $n > 1$, the inequality\n$$\n\\f... | Russia | XLIII Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English; Russian | proof and answer | 1 | |
045j | Fix an integer $n \ge 2$. Find all $n$-tuples $(a_1, a_2, \dots, a_n)$ of integers satisfying the following two conditions:
(1) $a_1$ is odd, $1 < a_1 \le a_2 \le \dots \le a_n$, and $M = \frac{1}{2^n}(a_1 - 1)a_2 \dots a_n$ is an integer; and
(2) there exist $M$ different $n$-tuples $(c_{i,1}, c_{i,2}, \dots, c_{i,n})... | [
"The $n$-tuples we seek are the ones satisfying the following condition:\nif there are exactly $r$ odd numbers in $a_2, \\dots, a_n$, then $2^r \\mid a_1 - 1$. $(*)$\n\nWe first verify the necessity of $(*)$. For this, we drop the condition $a_n \\ge a_{n-1} \\ge \\dots \\ge a_1$ and assume that $a_1, \\dots, a_r$ ... | China | 2022 China Team Selection Test for IMO | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Exactly those tuples for which, if r is the number of odd numbers among a2,…,an, then 2^r divides a1 − 1. | |
09el | Find the number of obtuse triangles with integer sides and perimeter equals to $50$. | [
"Let $a$, $b$, $c$ be sides of the triangle. Then $a + b + c = 50$. Without losing generality we may assume that $a \\ge b \\ge c$.\n\nNow $50 = a + b + c \\le 3a$ and consequently we get $a \\ge 17$. On the other hand $50 = a + b + c > 2a$ by triangle inequality and we came to conclusion $a < 25$.\n\nIt is well kn... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Discrete Mathematics > Combinatorics"
] | English | proof and answer | 30 | |
0g31 | Problem:
Sei $k_{1}$ ein Kreis und $l$ eine Gerade, die $k_{1}$ in zwei verschiedenen Punkten $A$ und $B$ schneidet. Sei $k_{2}$ ein weiterer Kreis ausserhalb von $k_{1}$, der $k_{1}$ in $C$ und $l$ in $D$ berührt. Sei $T$ der zweite Schnittpunkt von $k_{1}$ und der Geraden $CD$. Zeige, dass $AT = TB$ gilt. | [
"Solution:\n\nUm $AT = BT$ zu zeigen reicht es, die Winkelgleichung $\\angle ABT = \\angle TAB$ zu beweisen.\n\nSei $t$ die Tangente an $k_{1}$ in $C$ und $S$ der Schnittpunkt von $t$ und $l$. Dann ist $t$ auch die Tangente an $k_{2}$ in $C$, da die Kreise sich berühren. Sei $\\beta = \\angle SCB$ und $\\alpha = \\... | Switzerland | Vorrunde 2019 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ckn | For given positive integer $k$, consider the function $g_k : \mathbb{Z} \to \mathbb{Z}$, $g_k(x) = x^k$. Determine the set $M_k$ of positive integers $n$ with the property that there exist injective functions $f_1, f_2, \dots, f_n : \mathbb{Z} \to \mathbb{Z}$ such that $g_k = f_1 \cdot f_2 \cdot \dots \cdot f_n$. | [
"For $x = 2$, we get $2^k = |f_1(2)| \\cdot |f_2(2)| \\cdot \\dots \\cdot |f_n(2)|$. Using injectivity, we deduce that $|f_i(2)| \\ge 2$ for any $i = \\overline{1, n}$. Thus, $2^k = |f_1(2)| \\cdot |f_2(2)| \\cdot \\dots \\cdot |f_n(2)| \\ge 2^n$, so $k \\ge n$. Since $n$ and $k$ have the same parity, it follows th... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | M_k = {k, k-2, k-4, ..., k - 2\lfloor (k-1)/2 \rfloor} | |
0gl3 | Determine the largest real number $k$ such that the inequality
$$
(k + \frac{a}{b}) (k + \frac{b}{c}) (k + \frac{c}{a}) \le \left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\right) \left(\frac{b}{a} + \frac{c}{b} + \frac{a}{c}\right)
$$
holds for all positive real numbers $a$, $b$, and $c$. | [
"By setting $a = b = c$, it follows that $k \\leq \\sqrt[3]{9} - 1$. We claim that $k = \\sqrt[3]{9} - 1$ is the largest possible number so that the inequality holds. Let\n$$\nA = \\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a}, \\quad B = \\frac{b}{a} + \\frac{c}{b} + \\frac{a}{c}.\n$$\nBy AM-GM inequality, we get that... | Thailand | Tajland 2014 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | sqrt[3]{9} - 1 | |
0kak | Problem:
Let $a$ and $b$ be positive whole numbers such that $\frac{4.5}{11} < \frac{a}{b} < \frac{5}{11}$. Find the fraction $\frac{a}{b}$ for which the sum $a+b$ is as small as possible. Justify your answer. | [
"Solution:\nBy multiplying numerators and denominators by $7$, we can rewrite the inequalities as follows:\n$$\n\\frac{7 \\cdot 4.5}{7 \\cdot 11} < \\frac{a}{b} < \\frac{7 \\cdot 5}{7 \\cdot 11} \\Rightarrow \\frac{31.5}{77} < \\frac{a}{b} < \\frac{35}{77}\n$$\nWe now see that the fraction $\\frac{a}{b} = \\frac{33... | United States | Bay Area Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 3/7 | |
07tr | 2021 points are given, no three of them are collinear. Divide these points into 20 groups with different numbers of points in each group. Count the number of triangles with vertices in different groups. In order to get the maximum number of such triangles, how should you divide those point? | [
"For any division of the 2021 points into 20 groups $M_1, M_2, \\dots, M_{20}$ of different size we let $n_k$ be the number of points in the group $M_k$ and we suppose that $n_1 < n_2 < n_3 < \\dots < n_{20}$. The number $g = n_{20} - n_1 - 19$ is then equal to the number of integers between $n_1$ and $n_{20}$ whic... | Ireland | IRL_ABooklet | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Group sizes are the integers 91, 92, 93, 94, 95, 96, 97, 98, 99, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111 (i.e., 91 through 111 except 100). | |
04ud | Do there exist positive integers $n$, $k$ such that
$$
\frac{n}{11^k - n}
$$
is a square of an integer? | [
"Such numbers don't exist. For the sake of contradiction, assume that there exist positive integers $n$, $k$, $a$ such that\n$$\n\\frac{n}{11^k - n} = a^2\n$$\nwhich rewrites as\n$$\nn(a^2 + 1) = a^2 \\cdot 11^k.\n$$\nFrom $\\text{GCD}(a^2, a^2+1) = 1$ we deduce $a^2+1 \\mid 11^k$ and hence $a^2+1 = 11^t$ for $1 \\... | Czech Republic | 67th Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof and answer | No | |
00j1 | We call a set of three numbers "arithmetic" if one of its elements is the arithmetic mean of the other two. Similarly, we call a set of three numbers "harmonic" if one of its elements is the harmonic mean of the other two. How many three element subsets of the set
$$
\{z \mid -2011 < z < 2011\}
$$
of integers are both ... | [
"Choosing an arithmetic set $\\{u, v, w\\}$, we can assume that $u < v < w$ holds, and we can therefore write $u = a - d$, $v = a$ and $w = a + d$ with $d > 0$. We now wish this set to also be harmonic. If some number $q$ is the harmonic mean of numbers $p$ and $r$, we have $\\frac{1}{p} + \\frac{1}{r} = \\frac{2}{... | Austria | AustriaMO2011 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 1004 | |
09sz | Problem:
Bepaal alle paren $(m, n)$ van positieve gehele getallen waarvoor
$$
(m+n)^3 \mid 2 n\left(3 m^2+n^2\right)+8
$$ | [
"Solution:\nStel dat het quotiënt van $2 n\\left(3 m^2+n^2\\right)+8$ en $(m+n)^3$ niet gelijk aan 1 is. Dan is het minstens 2, dus geldt\n$$\n(m+n)^3 \\leq n\\left(3 m^2+n^2\\right)+4\n$$\noftewel\n$$\nm^3+3 m^2 n+3 m n^2+n^3 \\leq 3 m^2 n+n^3+4\n$$\noftewel\n$$\nm^3+3 m n^2 \\leq 4\n$$\nHieruit volgt meteen $m<2$... | Netherlands | Selectietoets | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (m, n) = (1, 1) and (m, n) = (n+2, n) for all positive integers n | |
0314 | Problem:
Let $D$ be a point on the side $AC$ of $\triangle ABC$ with $AC = BC$, and $E$ be a point on the segment $BD$. Prove that $\Varangle EDC = 2 \Varangle CED$ if $BD = 2AD = 4BE$. | [
"Solution:\n1. Note that\n$$\n\\Varangle EDC = 2 \\Varangle CED \\Longleftrightarrow DI = EI,\n$$\nwhere $DI$ ($I \\in CE$) is the bisector of $\\Varangle CDE$. Setting $BD = 2AD = 4BE = 4x$ and $AC = BC = y$ we get\n$$\n\\begin{aligned}\nDI & = \\frac{\\sqrt{ED \\cdot CD \\left((ED + CD)^2 - CE^2\\right)}}{ED + CD... | Bulgaria | Team selection test for 20. BMO | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0ce2 | The triangle $ABC$ has circumcircle $\Gamma$ with centre $O$, $AO \cap BC = \{X\}$, $AO \cap \Gamma = \{Y\}$, and $\frac{AX}{XY} = k$, $k > 1$. The tangent to $\Gamma$ at $Y$ meets $AB$ and $AC$ in $M$, respectively $N$. Find $k$ so that the quadrilateral $OMNC$ is cyclic.
Mihaela Berindeanu | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | k = 2 | |
0hn2 | Problem:
At a market, a buyer and a seller each have four exotic coins. You are allowed to label each of the eight coins with any positive integer value in cents. The labeling is called $n$-efficient if for any integer $k$, $1 \leq k \leq n$, it is possible for the buyer and the seller to give each other some of their... | [
"Solution:\n\nThe answer is $240$.\n\nTo see that $n > 240$ is impossible, note that there are $2^{8} = 256$ ways for the transaction to happen, since each coin either changes hands or does not change hands. However, the $2^{4} = 16$ ways in which the buyer keeps all four of his coins clearly cannot allow the buyer... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 240 | |
0bjn | Let $(A, +, \cdot)$ be a unit ring with the property: for all $x \in A$,
$$
x + x^2 + x^3 = x^4 + x^5 + x^6.
$$
a) Let $x \in A$ and let $n \ge 2$ be an integer such that $x^n = 0$. Prove that $x = 0$.
b) Prove that $x^4 = x$, for all $x \in A$. | [
"a) From the given equality we derive that\n$$\nx^{n-1} = x^n(x^4 + x^3 + x^2 - x - 1) = 0\n$$\nand, step by step, that $x^{n-2} = x^{n-3} = \\dots = x = 0$.\n\nb) Rewrite the given equation as\n$$\nx(x^3 - 1)(x^2 + x + 1) = 0.\n$$\nIt follows that\n$$\n(x^4 - x)^2 = x^2(x-1)(x^3 - 1)(x^2 + x + 1) = 0,\n$$\nhence $... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Ring Theory"
] | null | proof only | null | |
0k55 | Problem:
How many ordered sequences of 36 digits have the property that summing the digits to get a number and taking the last digit of the sum results in a digit which is not in our original sequence? (Digits range from 0 to 9.) | [
"Solution:\n\nWe will solve this problem for 36 replaced by $n$. We use $[n]$ to denote $\\{1,2, \\ldots, n\\}$ and $\\sigma_{s}$ to denote the last digit of the sum of the digits of $s$.\nLet $D$ be the set of all sequences of $n$ digits and let $S_{i}$ be the set of digit sequences $s$ such that $s_{i}=\\sigma_{s... | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 9^36 + 4 | |
0jhj | Problem:
Find all composite positive integers $n$ such that all the divisors of $n$ can be written in the form $a^{r}+1$, where $a$ and $r$ are integers with $a \geq 0$ and $r \geq 2$. | [
"Solution:\nThe only such number is $n=10$. It is easy to see that $n=10$ indeed satisfies the conditions. Call $n$ \"good\" if every divisor of $n$ has the form $a^{r}+1$, $a \\geq 0$, $r \\geq 2$ (a good $n$ may be prime or composite).\n\nFirst, it is easy to check that $4$ is not good, and so $4$ does not divide... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 10 | |
07yx | Problem:
Siano $a$, $b$, $c$ tre numeri positivi dispari distinti e minori di $100$. Quanto può essere, al massimo, il loro massimo comune divisore?
(A) $7$
(B) $11$
(C) $19$
(D) $25$
(E) Nessuna delle risposte precedenti | [] | Italy | Progetto Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | MCQ | C | |
066o | If $x, y, z > 0$ with $x^2y^2 + y^2z^2 + z^2x^2 = 6x^2y^2z^2$, prove that:
$$
\sqrt{\frac{x}{x+yz}} + \sqrt{\frac{y}{y+zx}} + \sqrt{\frac{z}{z+xy}} \ge \sqrt{3}.
$$ | [] | Greece | Mediterranean Mathematical Competition | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
05ga | Problem:
Trouver le plus petit entier positif qui ne s'écrit pas sous la forme $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$ pour $a, b, c, d \in \mathbb{N}$. | [
"Solution:\n\nSoit $E$ l'ensemble des entiers strictement positifs s'écrivant sous cette forme. Commençons par remarquer que\n$$\n\\frac{2^{a}-2^{b}}{2^{c}-2^{d}}=2^{b-d} \\frac{2^{a-b}-1}{2^{c-d}-1}\n$$\nAinsi, si $x>0$ s'écrit $x=\\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$, alors $b-d=v_{2}(x)$ et si on appelle $y$ l'entie... | France | Envoi 1 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | 11 | |
0gg0 | Determine all positive integers $d$ for which there exists a $k \ge 3$ such that you can put the numbers $d, 2d, 3d, \dots, kd$ in a sequence in such a way that the sum of every pair of neighbouring numbers is a square. | [
"For $d = 1$, we take $k = 15$ and the sequence\n\n8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9.\n\nTwo neighbouring numbers in this sequence always add up to 9, 16, or 25. For square $d > 1$, we also take $k = 15$ and the same sequence as above, except that we multiply all numbers by $d$. Two neighbouring num... | Taiwan | 2022 數學奧林匹亞競賽第二階段培訓營, 國際競賽實作(一) | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Chinese; English | proof and answer | All positive squares | |
0f1x | Problem:
$a_1$ and $a_2$ are positive integers less than $1000$. Define $a_n = \min\{|a_i - a_j| : 0 < i < j < n\}$. Show that $a_{21} = 0$. | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0chv | Consider the functions $f, g : \mathbb{R} \to \mathbb{R}$, such that $g(x) = 2f(x) + f(x^2)$, for all $x \in \mathbb{R}$.
a) Prove that if $f$ is locally bounded in the origin and $g$ is continuous in the origin, then $f$ is continuous in the origin.
b) Give an example of a function $f$, discontinuous in the origin, ... | [
"a) Let $\\varepsilon > 0$, be arbitrary. By the hypothesis there are $\\delta_1, M > 0$ such that $|f(x)| < M$, for all $x \\in (-\\delta_1, \\delta_1)$. As $g$ is continuous at $0$, we get a $\\delta_2 > 0$, depending on $\\varepsilon$, such that $|g(x) - g(0)| < \\frac{\\varepsilon}{2}$ for any $x \\in (-\\delta... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | null | |
0a0h | Tim has exactly six meters of iron wire. He cuts the iron wire into a number of pieces, in such a way that each piece is an integer number of meters long. From each of those pieces, he makes a circle. Then he stacks those circles perfectly vertically balancing on top of each other. A possible front view is shown in the... | [
"1"
] | Netherlands | Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | final answer only | 1 | |
0cx1 | Find all triples $(x, y, z)$ of positive integers such that
$$
x + y + z = 2010 \text{ and } x^2 + y^2 + z^2 - x y - y z - z x = 3.
$$ | [] | Saudi Arabia | SAMC | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All permutations of (669, 670, 671) | |
0kkj | Problem:
Among all polynomials $P(x)$ with integer coefficients for which $P(-10)=145$ and $P(9)=164$, compute the smallest possible value of $|P(0)|$. | [
"Solution:\n\nSince $a-b \\mid P(a)-P(b)$ for any integer polynomial $P$ and integers $a$ and $b$, we require that $10 \\mid P(0)-P(-10)$ and $9 \\mid P(0)-P(9)$. So, we are looking for an integer $a$ near $0$ for which\n$$\na \\equiv 5 \\bmod 10, \\quad a \\equiv 2 \\bmod 9\n$$\nThe smallest such positive integer ... | United States | HMMT Spring 2021 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 25 | |
05vl | Problem:
Pour tout entier $n \geqslant 1$, on note $f_{n}$ la somme de tous les restes obtenus en divisant $n$ par les nombres $1,2, \ldots, n$. Par exemple, si on divise $5$ par $1,2,3,4$ et $5$, les restes que l'on obtient sont $0,1,2,1$ et $0$, de sorte que $f_{5}=0+1+2+1+0=4$.
Trouver tous les entiers $n \geqslan... | [
"Solution:\n\nPour tous les entiers $a$ et $b$ tels que $1 \\leqslant a \\leqslant b$, on note $r_{a}(b)$ le reste obtenu en divisant $b$ par $a$. Puisque\n$$\nf_{n}-f_{n-1}=\\sum_{i=1}^{n} r_{i}(n)-\\sum_{i=1}^{n-1} r_{i}(n-1)=\\sum_{i=2}^{n-1} r_{i}(n)-\\sum_{i=2}^{n-1} r_{i}(n-1)=\\sum_{i=2}^{n-1}\\left(r_{i}(n)... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All prime numbers n ≥ 2 | |
0g83 | 證明對於所有的質數 $p > 100$ 和每一個整數 $r$, 存在兩個整數 $a$ 和 $b$ 使得 $p$ 整除 $a^2 + b^5 - r$。 | [
"在這整個解答當中，所有的同餘關係皆為模 $p$。\n\n固定 $p$, 令 $\\mathcal{P} = \\{0, 1, \\dots, p-1\\}$ 為模 $p$ 的完全剩餘類。對於所有的 $r \\in \\mathcal{P}$, 令 $S_r = \\{(a, b) \\in \\mathcal{P} \\times \\mathcal{P} : a^2 + b^5 \\equiv r\\}$, 且令 $s_r = |S_r|$。我們的目標是證明對於所有的 $r \\in \\mathcal{P}$, $s_r > 0$。\n\n我們將用已知的事實對於所有同餘類 $r \\in \\mathcal{P}$ 和... | Taiwan | 二〇一三數學奧林匹亞競賽第三階段選訓營, 模擬競賽(二) | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
07pw | Two circles $C_1$ and $C_2$, with centres at $D$ and $E$ respectively, touch at $B$. The circle having $DE$ as diameter intersects the circle $C_1$ at $H$ and the circle $C_2$ at $K$. The points $H$ and $K$ both lie on the same side of the line $DE$. $HK$ extended in both directions meets the circle $C_1$ at $L$ and me... | [
"Extend the line $DE$ in both directions to meet the circles again at $A$ and $C$. Let $F$ be the centre of the circle with diameter $DE$. From $D$, $F$ and $E$ draw perpendiculars $DR$, $FP$ and $ES$ on $LM$. Then $DR$, $PF$ and $SE$ are all parallel. Also, since $|DF| = |FE|$ we have $|RP| = |PS|$. Also $|RL| = |... | Ireland | Ireland | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0ih5 | Problem:
Let $ABCD$ be a convex quadrilateral inscribed in a circle with shortest side $AB$. The ratio $[BCD]/[ABD]$ is an integer (where $[XYZ]$ denotes the area of triangle $XYZ$.) If the lengths of $AB$, $BC$, $CD$, and $DA$ are distinct integers no greater than $10$, find the largest possible value of $AB$. | [
"Solution:\nNote that\n$$\n\\frac{[BCD]}{[ABD]} = \\frac{\\frac{1}{2} BC \\cdot CD \\cdot \\sin C}{\\frac{1}{2} DA \\cdot AB \\cdot \\sin A} = \\frac{BC \\cdot CD}{DA \\cdot AB}\n$$\nsince $\\angle A$ and $\\angle C$ are supplementary. If $AB \\geq 6$, it is easy to check that no assignment of lengths to the four s... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 5 | |
0b00 | Problem:
If the sum of the first $22$ terms of an arithmetic progression is $1045$ and the sum of the next $22$ terms is $2013$, find the first term. | [] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 53/2 | |
0lfu | Problem:
Do there exist functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
a) $f$ is a surjective function; and
b) $f(f(x)) = (x-1) f(x) + 2$ for all real $x$? | [
"Solution:\n\nWe have $f(f(f(x))) = f(f(f(x))) = f((x-1) f(x) + 2)$ and also $f(f(f(x))) = (f(x)-1) f(f(x)) + 2 = (f(x)-1)((x-1) f(x) + 2) + 2 = (f(x)-1)(x-1) f(x) + 2(f(x)-1) + 2 = f(x)((f(x)-1)(x-1) + 2)$, so\n$$\nf((x-1) f(x) + 2) = f(x)((x-1) f(x) + 2 - (x-1))\n$$\nLet $f(a) = 0$; then for $x = a$ we get $f(0) ... | Zhautykov Olympiad | International Zhautykov Olympiad in Sciences | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | No | |
0gty | Let $ABC$ be a triangle and let $P$ be a point in the interior of this triangle. Let $\omega_A$ be the circle that is tangent to the circumcircle of $BPC$ at $P$ internally and tangent to the circumcircle of $ABC$ at $A_1$ internally. Let $\Gamma_A$ be the circle that is tangent to the circumcircle of $BPC$ at $P$ exte... | [
"\nFrom the radical axis theorem on the circles $(ABC)$, $(BCP)$ and $\\omega_A$, the tangent lines to $\\omega_A$ passing through $P$ and $A_1$ intersect on the line $BC$, let us say at point $D$. From the radical axis theorem on the circles $(ABC)$, $(BCP)$ and $\\Gamma_A$, we see that th... | Turkey | 31st Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null | |
0h3c | Positive integers $A$ and $B$ have the following decimal forms: $A = \overline{abcabc}$ and $B = \overline{d00d}$, where $a, b, c, d$ are decimal digits, $a \neq 0, d \neq 0$. Find all possible values of $a, b, c, d$ such that $A+B$ is a full square. | [
"Оскільки\n$$\nA + B = \\overline{abcabc} + \\overline{d00d} = 1001 (\\overline{abc} + d) \\le 1001 (999 + 9) = 1001 \\cdot 1008,\n$$\nто для виконання умови задачі необхідно й достатньо, щоб $\\overline{abc} + d = 1001$. Звідси $a = 9, b = 9$ і $c + d = 11$.\n\n*Відповідь:* $a = 9, b = 9, c = 11 - d, a, d \\in \\{... | Ukraine | Ukrainian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a = 9, b = 9, d ∈ {2,3,4,5,6,7,8,9}, c = 11 − d | |
0h9d | It is given that the bookshelf can fit $9$ of the same thick books, but the $10$th one will not fit anymore. Similarly, it can hold $15$ of the same thin books, but $16$th will not fit anymore. Is it possible for that shelf to hold simultaneously:
a) $6$ thick and $5$ thin books?
b) $7$ thick and $5$ thin books? | [
"Let us re-write the statement of the problem as follows. Let us denote the length of the shelf by $S$, the width of the thick book by $x$ and the width of the thin book by $y$. Then, we have the conditions:\n$$\n9x \\leq S < 10x \\text{ and } 15y \\leq S < 16y \\Leftrightarrow \\frac{1}{10}S < x \\leq \\frac{1}{9}... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | a) Yes. b) No. | |
0av6 | Problem:
Define $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ by $f(x, y) = (2x - y, x + 2y)$. Let $f^0(x, y) = (x, y)$ and, for each $n \in \mathbb{N}$, $f^n(x, y) = f\left(f^{n-1}(x, y)\right)$. Determine the distance between $f^{2016}\left(\frac{4}{5}, \frac{3}{5}\right)$ and the origin. | [] | Philippines | 19th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Linear Algebra > Linear transformations"
] | null | proof and answer | 5^{1008} | |
02ha | Let $ABC$ be an acute triangle and $F$ its Fermat point, that is, the interior point of $ABC$ such that $\angle AFB = \angle BFC = \angle CFA = 120^\circ$. For each one of triangles $ABF$, $BCF$ and $CAF$, draw its Euler line, that is, the line connecting its circumcenter and its centroid.
Prove that these three lines ... | [
"First we'll prove the following well-known\n**Lemma.** Let $ABP$, $BCM$ and $CAN$ be the equilateral triangles constructed externally to triangle $ABC$. The lines $AM$, $BN$ and $CP$ concur on the Fermat point of $ABC$.\n*Proof.* Let $F$ be the intersection point of $BN$ and $CP$. Since triangles $BAN$ and $PAC$ a... | Brazil | Brazil | [
"Geometry > Plane Geometry > Advanced Configurations > Napoleon and Fermat points",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > ... | English | proof only | null | |
01i3 | Let $\omega_1$ and $\omega_3$ be two circles, touching externally in a common point $P$. Let further $\omega_2$ and $\omega_4$ be two circles touching externally in $P$. Suppose that for $i \in \{1, 2, 3, 4\}$ $\omega_i$ intersect $\omega_{(i \pmod 4)+1}$ again in $A_i$. Let $\ell_1$ be the common tangent of $\omega_1$... | [
"**Solution.** In figure 15, we have\n$$\n\\angle CNK = \\angle ANK = \\angle AOK = 2\\angle ABK = 2\\angle NBK.\n$$\nHence triangle $BNK$ is isosceles, so $NK = NB = NC$, and therefore $\\angle BKC$ is right.\nIf we let $KC$ intersect $\\omega$ at $K'$, we see that $K', O$ and $B$ are collinear, as $\\angle BKK'$ ... | Baltic Way | Baltic Way 2021 Shortlist | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kzc | One side of an equilateral triangle of height $24$ lies on line $l$. A circle of radius $12$ is tangent to $l$ and is externally tangent to the triangle. The area of the region exterior to the triangle and the circle and bounded by the triangle, the circle, and line $l$ can be written as $a\sqrt{b} - c\pi$, where $a$, ... | [
"The given situation is shown in the figure below, where $D$ is the center of the circle, $E$ is the point of tangency between the circle and the triangle, $F$ is the intersection of line $DE$ with line $l$, and $G$ is the projection of $D$ onto $l$.\n\n\nBecause $\\angle BAC = 60^\\circ$ a... | United States | AMC 10 A | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | D | |
05cq | Mama snail and her child want to visit a neighbour who lives at distance $75$ cm. Every hour, they have planned to use $45$ minutes to move and $15$ minutes to rest. On the $n$-th hour, they move $\frac{1}{n^2+1}$ metres forward, but instead of resting, the child pulls them backwards by $\frac{1}{n+1}$ of this hour's d... | [
"Combining both parts of the $n$-th hour, the total distance travelled forward is $\\frac{1}{n^2+1} - \\frac{1}{n+1} \\cdot \\frac{1}{n^2+1} = \\frac{n}{(n+1)(n^2+1)}$ metres. On the first hour, this means $\\frac{1}{4}$ metres. Notice that $\\frac{n}{(n+1)(n^2+1)} < \\frac{n}{(n+1)n} = \\frac{1}{n} - \\frac{1}{n+1... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | They will never reach the neighbour. | |
05pa | Problem:
Soit $ABC$ un triangle, $\Gamma$ son cercle circonscrit. Soit $\omega_{A}$ le cercle inscrit intérieurement à $(AB),(AC)$ et à $\Gamma$. On note $T_{A}$ le point de tangence de $\Gamma$ avec $\omega_{A}$. On définit de même $T_{B}$ et $T_{C}$.
Montrer que $\left(AT_{A}\right),\left(BT_{B}\right)$ et $\left(CT_... | [
"Solution:\n\nQuand on voit beaucoup de cercles tangents, on doit penser aux homothéties qui les échangent. En l'occurrence, on a trois homothéties positives, de centre respectifs $T_{A}, T_{B}, T_{C}$ et qui envoient respectivement $\\omega_{A}, \\omega_{B}$ et $\\omega_{C}$ sur $\\Gamma$.... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
0ibg | Problem:
On a spherical planet with diameter $10,000~\mathrm{km}$, powerful explosives are placed at the north and south poles. The explosives are designed to vaporize all matter within $5,000~\mathrm{km}$ of ground zero and leave anything beyond $5,000~\mathrm{km}$ untouched. After the explosives are set off, what is ... | [
"Solution:\n$100,000,000 \\pi$\n\nThe explosives have the same radius as the planet, so the surface area of the \"cap\" removed is the same as the new surface area revealed in the resulting \"dimple.\" Thus the area is preserved by the explosion and remains $\\pi \\cdot (10,000)^2$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > 3D Shapes"
] | null | final answer only | 100000000π | |
00k9 | We call a set of squares with sides parallel to the coordinate axes and vertices with integer coordinates friendly if any two of them have exactly two points in common. We consider friendly sets in which each of the squares has sides of length $n$. Determine the largest possible number of squares in such a friendly set... | [
"No two such vertices can lie on the same horizontal or vertical line, as the squares with these vertices would otherwise have a line segment in common, and not just two points.\nWe see that the highest possible number of possible vertices of other squares in the interior of the chosen square is equal to the number... | Austria | Austria 2014 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | n | |
0f1f | Problem:
Given real numbers $a_i$, $b_i$ and positive reals $c_i$, $d_i$, let $e_{ij} = (a_i + b_j) / (c_i + d_j)$. Let $M_i = \max_{0 \leq j \leq n} e_{ij}$, $m_j = \min_{1 \leq i \leq n} e_{ij}$. Show that we can find an $e_{ij}$ with $1 \leq i, j \leq n$ such that $e_{ij} = M_i = m_j$. | [] | Soviet Union | ASU | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0552 | The bisector of the exterior angle at vertex $C$ of the triangle $ABC$ intersects the bisector of the interior angle at vertex $B$ in point $K$. Consider the diameter of the circumcircle of the triangle $BCK$ whose one endpoint is $K$. Prove that $A$ lies on this diameter. | [
"\nFig. 9\n\nFig. 10\n\nLet $B'$ and $C'$ be respectively the second intersection points of the lines $AB$ and $AC$ with the circumcircle of the triangle $BCK$ (Fig. 9). Notice that $\\angle BKC = 180^\\circ - \\angle CBK - \\angle BCK = 180^\\circ - \\frac{\\angle ... | Estonia | Open Contests | [
"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"
] | English | proof only | null | |
0izh | Problem:
Let $ABC$ be a triangle with $AB = 8$, $BC = 15$, and $AC = 17$. Point $X$ is chosen at random on line segment $AB$. Point $Y$ is chosen at random on line segment $BC$. Point $Z$ is chosen at random on line segment $CA$. What is the expected area of triangle $XYZ$? | [
"Solution:\n\nLet $\\mathbb{E}(X)$ denote the expected value of $X$, and let $[S]$ denote the area of $S$. Then\n$$\n\\begin{aligned}\n\\mathbb{E}([\\triangle XYZ]) &= \\mathbb{E}([\\triangle ABC] - [\\triangle XYB] - [\\triangle ZYC] - [\\triangle XBZ]) \\\\\n&= [\\triangle ABC] - \\mathbb{E}([\\triangle XYB]) - \... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 15 | |
04lg | Find all positive integers $b$ such that $11 \cdot 22 \cdot 33 = 13310$ holds in base $b$. | [] | Croatia | Mathematical competitions in Croatia | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 6 | |
0k7i | Problem:
Consider the eighth-sphere $\{(x, y, z) \mid x, y, z \geq 0, x^{2}+y^{2}+z^{2}=1\}$. What is the area of its projection onto the plane $x+y+z=1$? | [
"Solution:\nConsider the three flat faces of the eighth-ball. Each of these is a quarter-circle of radius $1$, so each has area $\\frac{\\pi}{4}$. Furthermore, the projections of these faces cover the desired area without overlap. To find the projection factor one can find the cosine of the angle $\\theta$ between ... | United States | HMMT February 2019 | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof and answer | π√3/4 | |
079x | Problem:
In triangle $ABC$ we have $\angle BAC = 60^\circ$. The perpendicular line to $AB$ at $B$ intersects the bisector of $\angle BAC$ at $D$ and the perpendicular line to $BC$ at $C$ meets the bisector of $\angle ABC$ at $E$. Prove that $\angle BED \leq 30^\circ$. | [
"Solution:\n\nDenote by $I$ the intersection point of $AD$ and $BE$, so $I$ is the incenter of triangle $ABC$. Suppose that $\\alpha = \\frac{\\angle BAC}{2}$. We have\n$$\n\\angle IBD = 90^\\circ - \\angle IBA = 90^\\circ - \\frac{\\angle CBA}{2} = 90^\\circ - 30^\\circ = 60^\\circ,\n$$\n$$\n\\angle CEB = 90^\\cir... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triang... | English | proof only | null | |
0j7i | Problem:
Toward the end of a game of Fish, the $2$ through $7$ of spades, inclusive, remain in the hands of three distinguishable players: $\mathrm{DBR}$, $\mathrm{RB}$, and $DB$, such that each player has at least one card. If it is known that $DBR$ either has more than one card or has an even-numbered spade, or both... | [
"Solution:\n\nAnswer: $450$\n\nFirst, we count the number of distributions where each player has at least $1$ card. The possible distributions are:\n\n- Case 1: $4 / 1 / 1$ : There are $3$ choices for who gets $4$ cards, $6$ choices for the card that one of the single-card players holds, and $5$ choices for the car... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 450 | |
0eya | Problem:
$ABCD$ is a unit square. One vertex of a rhombus lies on side $AB$, another on side $BC$, and a third on side $AD$. Find the area of the set of all possible locations for the fourth vertex of the rhombus. | [
"Solution:\n\nAnswer: $2\\,1/3$\n\nLet the square be $ABCD$. Let the vertices of the rhombus be $P$ on $AB$, $Q$ on $AD$, and $R$ on $BC$. We require the locus of the fourth vertex $S$ of the rhombus. Suppose $P$ is a distance $x$ from $B$. We may take $x \\leq 1/2$, since the locus for $x > 1/2$ is just the reflec... | Soviet Union | 1st ASU | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Transformations > Translation"
] | null | proof and answer | 2 1/3 | |
0brp | The vertices of a prism are colored using two colors, so that each lateral edge has its vertices differently colored. Consider all the segments that join vertices of the prism and are not lateral edges. Prove that the number of such segments with endpoints differently colored is equal to the number of such segments wit... | [
"Denote $a$ the number of the vertices of the upper base which have the first color and $b = n - a$ the number of the vertices of the upper base which have the second color. Then the lower base has $b$ points with the first color and $a$ points with the second color.\nThe number of segments with endpoints different... | Romania | 67th Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof only | null | |
0d74 | Let $ABC$ be a triangle and $I$ its incenter. The point $D$ is on segment $BC$ and the circle $\omega$ is tangent to the circumcircle of triangle $ABC$ but is also tangent to $DC$, $DA$ at $E$, $F$, respectively. Prove that $E$, $F$ and $I$ are collinear. | [
"Denote $\\omega$ the circumcircle of $\\triangle ABC$ and $\\gamma$ the circle tangent to $\\omega$, $DA$, $DC$. Let $\\omega$ touch $\\gamma$ at $K$ and $M$ be the midpoint of $\\operatorname{arc} BC$ on $\\omega$ not containing $K$. One has the dilation with center $K$ sending $\\gamma$ to $\\omega$ and $BC$ to ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geome... | English | proof only | null | |
0fpp | Para pertenecer a un club cada nuevo socio debe pagar como cuota de inscripción a cada miembro del club la misma cantidad que él tuvo que pagar en total cuando ingresó más un euro. Si el primer socio pagó un euro, ¿cuanto deberá pagar en total el n-ésimo socio? | [
"Sea $a_n$ la cuota total del socio $n$-ésimo y sea $s_n = a_1 + \\dots + a_n$. El $n$-ésimo ($n \\ge 2$) socio tiene que pagar en total $(a_1 + 1) + (a_2 + 1) + \\dots + (a_{n-1} + 1) = s_{n-1} + n-1$ euros, luego $a_n = s_{n-1} + n-1$ y\n$$\ns_n = s_{n-1} + a_n = s_{n-1} + s_{n-1} + (n-1) = 2s_{n-1} + n - 1.\n$$\... | Spain | LII Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | Spanish | proof and answer | a_1 = 1; for n >= 2: a_n = 3*2^{n-2} - 1 | |
0ldi | Does there exist a polynomial $P(x)$ with integer coefficients such that $P(1+\sqrt[3]{2}) = 1+\sqrt[3]{2}$ and $P(1+\sqrt{5}) = 2+3\sqrt{5}$? | [
"Suppose that there exists such polynomial $P(x)$. Let $Q(x) = P(1+x) - 1$ then $Q(x) \\in \\mathbb{Z}[x]$. We have $Q(\\sqrt[3]{2}) = \\sqrt[3]{2}$ and $Q(\\sqrt{5}) = 1 + 3\\sqrt{5}$.\n\nTherefore, $Q(x) - x$ has an irrational root $\\sqrt[3]{2}$. Since $x^3 - 2$ is irreducible over $\\mathbb{Z}[x]$ and has the s... | Vietnam | Vietnamese Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | null | proof and answer | No | |
0kj4 | Regular polygons with $5$, $6$, $7$, and $8$ sides are inscribed in the same circle. No two of the polygons share a vertex, and no three of their sides intersect at a common point. At how many points inside the circle do two of their sides intersect?
(A) $52$ (B) $56$ (C) $60$ (D) $64$ (E) $68$ | [
"**Answer (E):** Consider a regular $m$-gon and a regular $n$-gon, with $m \\leq n$, inscribed in the same circle with no shared vertices. If $A$ and $B$ are adjacent vertices of the $m$-gon, then minor arc $AB$ contains at least one vertex of the $n$-gon. Thus side $AB$ intersects exactly two sides of the $n$-gon ... | United States | Fall 2021 AMC 10 B | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | MCQ | E | |
06xl | Let $n \geqslant 2$ be a positive integer. Paul has a $1 \times n^{2}$ rectangular strip consisting of $n^{2}$ unit squares, where the $i^{\text{th}}$ square is labelled with $i$ for all $1 \leqslant i \leqslant n^{2}$. He wishes to cut the strip into several pieces, where each piece consists of a number of consecutive... | [
"Answer: The minimum number of pieces is $2n-1$.\n\nSolution 1. For the entirety of the solution, we shall view the labels as taking values in $\\mathbb{Z} / n \\mathbb{Z}$, as only their values modulo $n$ play a role.\nHere are two possible constructions consisting of $2n-1$ pieces.\n1. Cut into pieces of sizes $n... | IMO | International Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2n-1 | |
09lv | Let $A$ be the sum of the squares of three consecutive positive integers, and let $B$ be the sum of the squares of four consecutive positive integers. Determine the number of pairs $(A, B)$ that satisfy the equation $3A - B = 2025$. (Batzorig Undrakh) | [
"Let the three consecutive positive integers be $n$, $n+1$, $n+2$.\nThen\n$$\nA = n^2 + (n+1)^2 + (n+2)^2 = n^2 + n^2 + 2n + 1 + n^2 + 4n + 4 = 3n^2 + 6n + 5.\n$$\n\nLet the four consecutive positive integers be $m$, $m+1$, $m+2$, $m+3$.\nThen\n$$\nB = m^2 + (m+1)^2 + (m+2)^2 + (m+3)^2 = m^2 + m^2 + 2m + 1 + m^2 + ... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 2 | |
00fh | Let $P_{1}, P_{2}, \ldots, P_{1993}=P_{0}$ be distinct points in the $xy$-plane with the following properties:
(i) both coordinates of $P_{i}$ are integers, for $i=1,2, \ldots, 1993$;
(ii) there is no point other than $P_{i}$ and $P_{i+1}$ on the line segment joining $P_{i}$ with $P_{i+1}$ whose coordinates are both in... | [
"Call a point $(x, y) \\in \\mathbb{Z}^{2}$ even or odd according to the parity of $x+y$. Since there are an odd number of points, there are two points $P_{i}=(a, b)$ and $P_{i+1}=(c, d), 0 \\leq i \\leq 1992$ with the same parity. This implies that $a+b+c+d$ is even. We claim that the midpoint of $P_{i} P_{i+1}$ i... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1993 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0795 | Let $p(x)$ be a polynomial of degree $2$ such that $|p(x)| \le 1$ holds for $x \in \{-1, 0, 1\}$. Show that for every $x \in [-1, 1]$
$$
|p(x)| \le \frac{5}{4}
$$ | [
"Notice that\n$$\np(x) = a x^2 + b x + c = \\frac{x(1+x)}{2} p(+1) - \\frac{x(1-x)}{2} p(-1) + (1-x^2) p(0)\n$$\nIf $0 \\le x \\le +1$, we see that\n$$\n|a x^2 + b x + c| \\le + \\frac{x(1+x)}{2} + \\frac{x(1-x)}{2} + (1-x^2) = \\frac{5}{4} - \\left(\\frac{1}{2} - x\\right)^2 \\le \\frac{5}{4}\n$$\nIf $-1 \\le x \\... | Iran | 27th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null |
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