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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
03tn | Suppose $f(x+2\pi) = f(x)$ for any $x \in \mathbb{R}$. Prove: there are $f_i(x)$ ($i = 1, 2, 3, 4$) such that
(1) $f_i(x)$ ($i = 1, 2, 3, 4$) is an even function, and $f_i(x + \pi) = f_i(x)$ for any $x \in \mathbb{R}$;
(2) $f(x) = f_1(x) + f_2(x)\cos x + f_3(x)\sin x + f_4(x)\sin 2x$ for any $x \in \mathbb{R}$. | [
"Let $g(x) = \\frac{f(x)+f(-x)}{2}$, and $h(x) = \\frac{f(x)-f(-x)}{2}$.\nThen $f(x) = g(x) + h(x)$, $g(x)$ is an even function, $h(x)$ is an odd function, and $g(x+2\\pi) = g(x)$, $h(x+2\\pi) = h(x)$ for any $x \\in \\mathbb{R}$.\n\nDefine\n$$\nf_1(x) = \\frac{g(x) + g(x + \\pi)}{2},\n$$\n$$\nf_2(x) = \\begin{case... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof only | null | |
0cn5 | A point $D$ is chosen on side $BC$ of an acute triangle $ABC$ so that $AB = AD$. The circumcircle of triangle $ABD$ intersects segment $AC$ at points $A$ and $K$. Line $DK$ intersects the line perpendicular to $AC$ and passing through $B$ at point $L$. Prove that $CL = BC$. (I. Bogdanov) | [
"Так как $AB = AD$, имеем $\\angle ADB = \\angle ABD$. Поскольку четырехугольник $ABDK$ вписан, $\\angle AKB = \\angle ADB$ и $\\angle ABD = 180^\\circ - \\angle AKD = \\angle LKA$. Таким образом, в треугольнике $BKL$ высота $KA$ является биссектрисой, а значит, и медианой; тогда точки $L$ и $B$ симметричны относит... | Russia | Russian mathematical olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
09az | The nine-point circle of scalene $\triangle ABC$ touches with incircle of $\triangle ABC$ at point $S$. $AD$, $BE$ and $CF$ are heights. If reflection of $DS$ over $AD$ is $\ell_1$, reflection of $ES$ over $BE$ is $\ell_2$ and reflection of $FS$ over $CF$ is $\ell_3$, then prove that $\ell_1 \parallel \ell_2 \parallel ... | [
"Let $\\omega$ be the nine-point circle and $l_1 \\cap \\omega = X$, $l_2 \\cap \\omega = Y$. Since $AFDC$ is inscribed in a circle, $\\angle FDA = \\angle FCA = 90^\\circ - \\angle BAC$. Also from $AEDB$ is inscribed in a circle, we have $\\angle EDA = \\angle EBA = 90^\\circ - \\angle BAC$. Therefore, $\\angle FD... | Mongolia | 46th Mongolian 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 > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0d8c | Given a polynomial $f(x) = x^{4} + a x^{3} + b x^{2} + c x$. It is known that each of the equations $f(x) = 1$ and $f(x) = 2$ has four real roots (not necessarily distinct). Prove that if the roots of the first equation satisfy the equality $x_{1} + x_{2} = x_{3} + x_{4}$, then the same equation holds for the roots of ... | [
"Consider the equation $f(x) = 1 \\Leftrightarrow x^{4} + a x^{3} + b x^{2} + c x = 1$, since it has four roots $x_{1}, x_{2}, x_{3}, x_{4}$ then we can write it as\n$$\n(x - x_{1})(x - x_{2})(x - x_{3})(x - x_{4}) = 0.\n$$\nNote that $x_{1} + x_{2} = x_{3} + x_{4} = -\\frac{a}{2}$ then\n$$\n\\left(x^{2} - (x_{1} +... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
054p | Find all solutions of the equation $a + b + c = 61$ in natural numbers that satisfy $\text{gcd}(a, b) = 2$, $\text{gcd}(b, c) = 3$, and $\text{gcd}(c, a) = 5$. | [
"As $\\text{gcd}(a, b) = 2$, $\\text{gcd}(b, c) = 3$ and $\\text{gcd}(c, a) = 5$, the number $a$ is divisible by both $2$ and $5$, the number $b$ is divisible by both $2$ and $3$, and the number $c$ is divisible by both $3$ and $5$. Hence $a$ is divisible by $10$, $b$ is divisible by $6$ and $c$ is divisible by $15... | Estonia | National Olympiad Final Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | English | proof and answer | (a, b, c) = (10, 6, 45), (10, 36, 15), (40, 6, 15) | |
0jer | Problem:
There exist unique nonnegative integers $A, B$ between $0$ and $9$, inclusive, such that
$$
(1001 \cdot A + 110 \cdot B)^2 = 57,108,249
$$
Find $10 \cdot A + B$. | [
"Solution:\nAnswer: $75$\nWe only need to bound for $AB00$; in other words, $AB^2 \\leq 5710$ but $(AB+1)^2 \\geq 5710$. A quick check gives $AB = 75$. (Lots of ways to get this...)"
] | United States | HMMT November 2013 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | 75 | |
0aa1 | Problem:
Determine all sequences of non-negative integers $a_{1}, \ldots, a_{2016}$ all less than or equal to $2016$ satisfying $i+j \mid i a_{i}+j a_{j}$ for all $i, j \in \{1,2, \ldots, 2016\}$. | [
"Solution:\nAll constant sequences of non-negative integers.\n\nThe condition rewrites to $i+j \\mid i\\left(a_{i}-a_{j}\\right)$. Since $2k-1$ and $k$ are coprime, we see that $2k-1 \\mid a_{k}-a_{k-1}$. Thus if $2k-1>2016$, then $a_{k}=a_{k-1}$ since $a_{k}$ and $a_{k-1}$ are non-negative and at most $2016$. All ... | Nordic Mathematical Olympiad | Nordic Mathematical Contest | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | All constant sequences with common value c, where c is an integer between 0 and 2016 inclusive. | |
0heg | a) For which positive integer $l$ there exists a pair of positive integers $(m, n)$, which consist of no more than $l$ digits, have different last digits and which satisfy the condition: the last $l$ digits of $m^2$ form the number $n$, and analogously, the last $l$ digits of $n^2$ form number $m$ (discarding any zeros... | [
"a) Consider the last digits of numbers satisfying the conditions of the problem. We follow the last digit of their squares, so we put the matching the last digit to each square:\n\n$1^2 \\rightarrow 1, 2^2 \\rightarrow 4, 3^2 \\rightarrow 9, 4^2 \\rightarrow 6, 5^2 \\rightarrow 5, 6^2 \\rightarrow 6, 7^2 \\rightar... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | a) No positive integer l. b) Every positive integer l. | |
0jvm | Problem:
Define $\phi^{!}(n)$ as the product of all positive integers less than or equal to $n$ and relatively prime to $n$. Compute the number of integers $2 \leq n \leq 50$ such that $n$ divides $\phi^{!}(n)+1$. | [
"Solution:\n\nNote that, if $k$ is relatively prime to $n$, there exists a unique $0 < k^{-1} < n$ such that $k k^{-1} \\equiv 1 \\pmod{n}$. Hence, if $k^{2} \\not\\equiv 1 \\pmod{n}$, we can pair $k$ with its inverse to get a product of $1$.\n\nIf $k^{2} \\equiv 1 \\pmod{n}$, then $(n-k)^{2} \\equiv 1 \\pmod{n}$ a... | United States | HMMT February 2016 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 30 | |
0l0h | Problem:
Paul is in the desert and has a pile of gypsum crystals. No matter how he divides the pile into two nonempty piles, at least one of the resulting piles has a number of crystals that, when written in base 10, has a sum of digits at least 7. Given that Paul's initial pile has at least two crystals, compute the s... | [
"Solution:\nDenote the digit sum of a positive integer $m$ as $s(m)$.\nLet the pile have $n$ gypsum crystals, so that $n$ can be written as $10^{i_{1}} + 10^{i_{2}} + \\cdots + 10^{i_{s(n)}}$.\n\nFirst, $s(n)$ cannot be $1$ (i.e. $n$ cannot be a power of $10$), since otherwise we could split the gypsum pile into tw... | United States | HMMT November 2024 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 49 | |
000g | Dado un cuadrilátero $ABCD$ se construyen triángulos isósceles $ABK$, $BCL$, $CDM$ y $DAN$, cuyas bases son los lados $AB$, $BC$, $CD$ y $DA$, y tales que $K$, $L$, $M$ y $N$ son puntos distintos y no hay tres de ellos alineados. La perpendicular a la recta $KL$ trazada por $B$ corta a la perpendicular a la recta $LM$ ... | [] | Argentina | XI Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | español | proof only | null | |
0j4u | Problem:
The ordered pairs $(2011,2),(2010,3),(2009,4), \ldots,(1008,1005),(1007,1006)$ are written from left to right on a blackboard. Every minute, Elizabeth selects a pair of adjacent pairs $\left(x_{i}, y_{i}\right)$ and $\left(x_{j}, y_{j}\right)$, with $\left(x_{i}, y_{i}\right)$ left of $\left(x_{j}, y_{j}\righ... | [
"Solution:\n\nAnswer: 504510\n\nFirst, note that none of the numbers will ever be $0$. Let $\\star$ denote the replacement operation. For each pair on the board $\\left(x_{i}, y_{i}\\right)$ define its primary form to be $\\left(x_{i}, y_{i}\\right)$ and its secondary form to be $\\left[x_{i} y_{i}, \\frac{x_{i}}{y... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 504510 | |
038p | Find all positive integers $x$ and $y$ such that $xy^2 + 2y$ divides $2x^2y + xy^2 + 8x$. | [
"Since $xy^2 + 2y$ divides\n$$(2x + y)(xy^2 + 2y) - y(2x^2y + xy^2 + 8x) = 2y^2 - 4xy,$$\nwe conclude that $xy + 2$ divides $2y - 4x$. We consider two cases.\n\nCase 1. Let $2y - 4x \\ge 0$. Then we have two possibilities:\n\n1.1) If $x \\ge 2$ then $xy + 2 > 2y - 4x$. Hence $2y - 4x = 0$, i.e. $x = a$ and $y = 2a$... | Bulgaria | Winter Mathematical Competition | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All solutions are (x, y) = (a, 2a) for any positive integer a, and (x, y) = (3, 1), (8, 1). | |
0add | Prove that $a^3 + b^3 + 3abc > c^3$, where $a$, $b$, $c$ are the sides of a triangle. | [
"Let $a$, $b$, $c$ be the sides of a triangle. From the triangle inequality $a + b > c$, we have\n$$\n\\begin{aligned}\na^3 + b^3 + 3abc &= (a+b)(a^2 - ab + b^2) + 3abc \\\\&> c \\cdot (a^2 - ab + b^2) + 3abc \\\\\n&= c \\cdot (a^2 - ab + b^2 + 3ab) \\\\\n&= c \\cdot (a^2 + 2ab + b^2) \\\\\n&= c \\cdot (a+b)^2 \\\\... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
0b5c | Problem:
Fie $a, b, c, d$ numere naturale nenule astfel încât numărul perechilor $(x, y), x, y \in (0,1)$, cu proprietatea că $a x + b y$ şi $c x + d y$ sunt întregi, este egal cu $2004$. Dacă $\operatorname{cmmdc}(a, c) = 6$, aflați $\operatorname{cmmdc}(b, d)$. | [] | Romania | Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | gcd(b,d) ∈ {1, 7, 49} | |
04zf | Let $n$ be a positive integer. Find the largest integer $N$ for which there exists a set of $n$ weights such that it is possible to determine the mass of all bodies with masses of $1$, $2$, $\ldots$, $N$ using a balance scale (i.e. to determine whether a body with unknown mass has a mass $1$, $2$, $\ldots$, $N$, and wh... | [
"The possibility to determine mass $m$ means the possibility to place the weights on the two scalepans so that the difference of total masses on the two scalepans is exactly $m$.\n\nEvery weight can be placed on either of the two pans or on neither of the pans. For $n$ weights this makes $3^n$ different placements.... | Estonia | Estonija 2010 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | (3^n - 1)/2 | |
01s6 | Find all functions $f, g: [0; 1] \to [0; 1]$, satisfying the following conditions:
$$
\begin{cases}
\{f(x)\} \sin^2 x + \{x\} \cos f(x) \cos x = f(x), \\
f(f(x)) = f(x),
\end{cases}
$$
for all real $x$.
(Here $\{y\}$ stands for the fractional part of $y$.) | [
"Let $f(x)$ satisfies the problem condition:\n$$\n\\left\\{ \n\\begin{array}{l} \n\\{f(x)\\} \\sin^2 x + \\{x\\} \\cos f(x) \\cos x = f(x), \\\\\n f(f(x)) = f(x), \n\\end{array} \n\\right. \n\\qquad (1) \n\\qquad (2)\n$$\nfor all real $x$.\nReplacing $x$ by $f(x)$ in (1) we obtain:\n$$\n\\{f(f(x))\\} \\sin^2 f(x) +... | Belarus | FINAL ROUND | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x for all x in [0,1] | |
0417 | In an acute triangle $ABC$, $AB > AC$, the bisector of angle $BAC$ and side $BC$ intersect at point $D$, two points $E$ and $F$ are in sides $AB$ and $AC$, respectively, such that $B, C, F, E$ are concyclic. Prove that the circumcenter of triangle $DEF$ coincides with the innercenter of triangle $ABC$ if and only if $B... | [
"Let $I$ be the innercenter of $\\triangle ABC$.\n\n(Sufficiency) Suppose $BC = BE + CF$. Let $K$ be the point on $BC$ such that $BK = BE$, thus $CK = CF$. Since $BI$ bisects $\\angle ABC$, $CI$ bisects $\\angle ACB$, $\\triangle BIK$ and $\\triangle BIE$ are reflection with respect to $BI$, $\\triangle CIK$ and $\... | China | China 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 | proof only | null | |
0bqp | Problem:
Determinați numerele $\overline{ab}$ pentru care $\overline{ba} + \overline{ab}$ și $\overline{ba} - \overline{ab}$ sunt pătrate perfecte. | [] | Romania | Olimpiada de Matematică - Etapa Locală | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 56 | |
0f9n | Problem:
If every altitude of a tetrahedron is at least $1$, show that the shortest distance between each pair of opposite edges is more than $2$. | [] | Soviet Union | 24th ASU | [
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
0aq8 | Problem:
If $a = 2^{-1}$ and $b = \frac{2}{3}$, what is the value of $\left(a^{-1} + b^{-1}\right)^{-2}$? | [] | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 4/49 | |
0h4d | Let $a$ and $b$ be positive integers such that
$$
\frac{a^4-1}{b+1} + \frac{b^4-1}{a+1}
$$
is integer. Prove that $a^{2010}b^{2012} - 1$ is divisible by $a+1$. | [
"Якщо виконується умова задачі, і принаймні одне з чисел $a, b$ дорівнює 1, то твердження є очевидним. Нехай $a > 1$ і $b > 1$. Позначимо $\\frac{a^4-1}{b+1} = \\frac{x}{y}$, $\\frac{b^4-1}{a+1} = \\frac{z}{t}$, де $x, y, z, t$ — натуральні числа, причому $(x;y) = (z;t) = 1$. За умовою задачі, сума $\\frac{x}{y} + ... | Ukraine | Ukrainian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0dnh | Problem:
Симетрале унутрашњих углова код темена $A$ и $B$ троугла $ABC$ секу наспрамне странице у тачкама $D$ и $E$, редом. Ромб је уписан у четвороугао $ABDE$ тако да се на свакој страници четвороугла налази тачно једно теме ромба. Ако је $\angle BAC=\alpha$ и $\angle ABC=\beta$, доказати да је барем један угао ромба... | [
"Solution:\n\nНека је $KLMN$ ромб са $K \\in AB$, $L \\in BD$ и $N \\in EA$. Посматрајмо трапез $PQRS$ са $P, Q \\in AB$, $PQ \\parallel RS$ и $\\angle PQR=\\angle QPS=\\angle KNM=\\varphi$ такав да су $K, L, M, N$ на $PQ, QR, RS, SP$ редом. Претпоставимо да је $\\varphi>\\alpha, \\beta$. Тада $R$ и $S$ леже ван $\... | Serbia | 8. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof only | null | |
097o | Problem:
Să se afle toate perechile de numere naturale prime $p$, $q$, care satisfac ecuaţia
$$
3 p^{4}+5 q^{4}+15=13 p^{2} q^{2}
$$ | [
"Solution:\nNumerele $p$ şi $q$ nu pot fi concomitent impare. Pentru verificare, se foloseşte congruenţa modulo $4$. Partea stângă, $3 p^{4}+5 q^{4}+15 \\equiv 3 \\pmod{4}$, iar partea dreaptă, $13 p^{2} q^{2} \\equiv 1 \\pmod{4}$.\nNumerele $p$ şi $q$ nu pot fi concomitent pare, adică nu pot fi egale fiecare cu $2... | Moldova | Olimpiada Republicană la Matematică, Ziua a doua | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (2, 3) | |
019o | Polynomials $P(x)$ and $Q(x)$ of rational coefficients are sums of three squares of polynomials of rational coefficients. Show that polynomial $P(x) \cdot Q(x)$ is a sum of four squares of polynomials of rational coefficients. | [
"Thesis follows from equality:\n$$\n(a^2 + b^2 + c^2)(x^2 + y^2 + z^2) = (a^2x^2 + b^2y^2 + c^2z^2) + (a^2y^2 + b^2x^2) + (b^2z^2 + c^2y^2) + (a^2z^2 + c^2x^2) = \\\\ = (ax + by + cz)^2 + (ay - bx)^2 + (az - cx)^2 + (bz - cy)^2.\n$$"
] | Baltic Way | Baltic Way 2013 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
09mm | Let $\mathbb{R}_{>0} = \{x \in \mathbb{R} \mid x > 0\}$ denote the set of positive real numbers. Find all functions $f: \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ satisfying
$$
f(x)f(y + f(x)) = f(1 + xy)
$$
for all $x, y \in \mathbb{R}_{>0}$.
(Otgonbayar Uuye) | [
"Answer: $f = 1$ and $f = 1/x$.\nIt is easy to check that these are solutions, so we prove there are no other solutions.\n\n*f = 1 if f is not injective:* Suppose $a > b > 0$ and $f(a) = f(b)$. Then we have\n$$\nf(1 + a x) = f(a) f(x + f(a)) = f(b) f(x + f(b)) = f(1 + b x)\n$$\nfor any $x \\in \\mathbb{R}_{>0}$. Le... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = 1 and f(x) = 1/x for all x > 0 | |
0ies | Problem:
Suppose $S$ tiles $\mathbb{N}$. Show that $S$ is symmetric; that is, if $-S=\{-s_{n}, \ldots,-s_{0}\}$, show that $S \sim -S$. | [
"Solution:\nAssume without loss of generality that the minimum element of $S$ is $0$. By the previous problem, $S$ tiles the set $\\{1,2, \\ldots, k\\}$ for some positive integer $k$. Then let $P(x)$ be the polynomial $\\sum_{i=0}^{n} x^{s_{i}}$. To say that the set $\\{1,2, \\ldots, k\\}$, or equivalently the set ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof only | null | |
09hv | Let $p \ge 3$ be a prime number. There are $n = 2p$ integers on the blackboard. A student chooses one or more numbers and writes down their sum modulo $n$. Suppose that the remainders $1, 2, \dots, p-1, p+1, \dots, n-1$ appear the same number of times when the student writes down all $2^n - 1$ possible sums modulo $n$.... | [
"Let $a_1, a_2, \\dots, a_n$ denote the numbers on the blackboard and let $X = \\{1, 2, \\dots, n\\}$ be the index set.\nThe remainder corresponding to an index subset $A \\subseteq X$ is the remainder of the sum $\\sum_{k \\in A} a_k$ modulo $n$ and we denote it by $\\sigma(A)$. Here we assume $\\sigma(\\emptyset)... | Mongolia | Round 3 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Other"
] | null | proof only | null | |
09iv | Let $n \ge 2$. There are $2n$ people standing in a circle, each holding a different painting in an art conference. Each person has their own fixed preference order of the paintings, which can be different from person to person. A trade of paintings between two adjacent people can happen if and only if both people are g... | [
"Answer: $2n^2 - n$.\nConsider the position of the painting for each person in their preference order, and let $S$ denote the sum of these positions. The maximum of $S$ is $2n \\cdot 2n = 4n^2$ and the minimum of $S$ is $2n \\cdot 1 = 2n$. For each trade, $S$ is reduced at least by 2. Hence the maximum number of tr... | Mongolia | Mongolian Mathematical Olympiad Round 3 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 2n^2 - n | |
01kr | Points $B$ and $C$ are marked on the half-hyperbola $y = 1/x$ which lies in the first quadrant of the Cartesian plane. The abscissa of $C$ is greater than the abscissa of $B$. Let $A$ be the intersection point of the other half-hyperbola and the line passing through the origin and $B$.
Prove that the angle $BAC$ is eq... | [
"Let $\\ell$ be tangent to the hyperbola at point $B$, $F$ be the intersection of $x$-axis and $BC$, $E$ be the intersection point of the $x$-axis and $\\ell$,\n\n\n\n$D$ be the intersection point of $\\ell$ and the line through $A$ parallel to $y$-axis (see the Fig.). Since $A$ and $B$ are... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0c5x | Let $n$ be a positive integer with $n \ge 2$, and let $A, B \in \mathcal{M}_n(\mathbb{C})$ be two matrices with complex entries having the property that there exists an idempotent matrix $C \in \mathcal{M}_n(\mathbb{C})$, such that $C^* = AB - BA$. Prove that $(AB - BA)^2 = O_n$. (A matrix $C$ is called *idempotent* if... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | English | proof only | null | |
0eu7 | Find all pair $(p, q)$ of primes such that $p^p + q^q + 1$ is divisible by $pq$. | [
"Clearly, $p \\ne q$. So we may assume that $p < q$ without loss of generality. Assume that $p = 2$. Then since\n$$\nq^q + 5 \\equiv 5 \\equiv 0 \\pmod{q},\n$$\nthe only possible prime for $q$ is $5$. Furthermore, $(p, q) = (2, 5)$ satisfies the above condition. Now we assume that both $p$ and $q$ are odd primes. S... | South Korea | 20th Korean Mathematical Olympiad Final Round | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequaliti... | null | proof and answer | (2, 5) and (5, 2) | |
033r | Problem:
The edges of a graph with $2n$ vertices, $n \geq 4$, are colored in blue and red such that there is no blue triangle and there is no red complete subgraph with $n$ vertices. Find the least possible number of the blue edges. | [
"Solution:\n\nWe call a graph, satisfying the given condition, $n$-purple. Let $f(n)$ be the smallest possible number of blue edges in an $n$-purple graph.\n\nSuppose that $f(n) < n + 5$ for $n \\geq 5$. If any vertex of an $n$-purple graph $G$ with $f(n)$ blue edges is a head of at least two blue edges, then the t... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | f(4) = 10; for n ≥ 5, f(n) = n + 5 | |
0gou | Let $A$ and $B$ be sets with $2011^2$ and $2010$ elements, respectively. Show that there is a function $f: A \times A \to B$ satisfying the condition $f(x,y) = f(y,x)$ for all $(x,y) \in A \times A$ such that for every function $g: A \to B$ there exists $(a_1, a_2) \in A \times A$ with $g(a_1) = f(a_1, a_2) = g(a_2)$ a... | [
"Let $p = 2011$, $A = \\{(i,j) : 0 \\le i, j < p\\}$ and $B = \\{k : 0 \\le k < p-1\\}$. We define $f$ as follows:\n$$\nf((i_1, j_1), (i_2, j_2)) = \\begin{cases} \\dfrac{j_1 - j_2}{i_1 - i_2} \\pmod{p} & \\text{if } i_1 \\neq i_2 \\text{ and } \\dfrac{j_1 - j_2}{i_1 - i_2} \\neq -1 \\pmod{p} \\\\ 0 & \\text{otherw... | Turkey | Team Selection Test for IMO 2011 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof only | null | |
0167 | Let $\delta(n)$ denote the number of positive divisors of positive integer $n$. Prove that there exist infinitely many positive integers that can not be represented in the form
$$
\left( \frac{2\sqrt{n}}{\delta(n)} \right)^2
$$
for positive integers $n$. | [
"$$\n\\left(\\frac{2\\sqrt{n}}{\\delta(n)}\\right)^2 = k^2.\n$$\nTaking the square root, one gets $\\frac{2\\sqrt{n}}{\\delta(n)} = k$, leading to\n$$\n\\sqrt{n} = \\frac{k \\cdot \\delta(n)}{2}. \\qquad (4)\n$$\nHence $\\sqrt{n}$ is rational, meaning that $n$ is a perfect square.\n\nLet $n = s^2$ where $s > 0$. Su... | Baltic Way | Baltic Way SHL | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
06og | Prove the inequality
$$
\sum_{i<j} \frac{a_{i} a_{j}}{a_{i}+a_{j}} \leq \frac{n}{2\left(a_{1}+a_{2}+\cdots+a_{n}\right)} \sum_{i<j} a_{i} a_{j}
$$
for positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$. | [
"Let $S=\\sum_{i} a_{i}$. Denote by $L$ and $R$ the expressions on the left and right hand side of the proposed inequality. We transform $L$ and $R$ using the identity\n$$\n\\sum_{i<j}\\left(a_{i}+a_{j}\\right)=(n-1) \\sum_{i} a_{i} . \\tag{1}\n$$\nAnd thus:\n$$\nL=\\sum_{i<j} \\frac{a_{i} a_{j}}{a_{i}+a_{j}}=\\sum... | IMO | IMO 2006 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
013b | Problem:
An $m \times n$ table is given, in each cell of which a number $+1$ or $-1$ is written. It is known that initially exactly one $-1$ is in the table, all the other numbers being $+1$. During a move, it is allowed to choose any cell containing $-1$, replace this $-1$ by $0$, and simultaneously multiply all the ... | [
"Solution:\n\nAnswer: Those $(m, n)$ for which at least one of $m, n$ is odd.\n\nLet us erase a unit segment which is the common side of any two cells in which two zeroes appear. If the final table consists of zeroes only, all the unit segments (except those which belong to the boundary of the table) are erased. We... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | All pairs where at least one of m or n is odd. | |
08nq | Problem:
Let $ABC$ be an equilateral triangle, and $P$ a point on the circumcircle of the triangle $ABC$ and distinct from $A$, $B$ and $C$. If the lines through $P$ and parallel to $BC$, $CA$, $AB$ intersect the lines $CA$, $AB$, $BC$ at $M$, $N$ and $Q$ respectively, prove that $M$, $N$ and $Q$ are collinear. | [
"Solution:\n\nWithout any loss of generality, let $P$ be in the minor arc of the chord $AC$ as in Figure 1. Since $\\angle PNA = \\angle NPM = 60^{\\circ}$ and $\\angle NAM = \\angle PMA = 120^{\\circ}$, it follows that the points $A$, $M$, $P$ and $N$ are concyclic. This yields\n$$\n\\angle NMP = \\angle NAP\n$$\n... | JBMO | JBMO Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0h3l | Prove that for any positive integer $n$ the number $5^n (5^n + 1) - 6^n (3^n + 2^n)$ is divisible by $91$. | [
"Позначимо значення даного виразу через $a_n$. Маємо:\n$$\n\\begin{aligned}\na_n &= 25^n + 5^n - 18^n - 12^n = (25^n - 18^n) - (12^n - 5^n) = \\\\\n&= (25 - 18)(25^{n-1} + 25^{n-2} \\cdot 18 + \\dots + 25 \\cdot 18^{n-2} + 18^{n-1}) - \\\\\n&\\quad -(12 - 5)(12^{n-1} + 12^{n-2} \\cdot 5 + \\dots + 12 \\cdot 5^{n-2}... | Ukraine | Ukrainian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
085p | Problem:
Un mercante ha 6 barili di capacità 15, 16, 18, 19, 20 e 31 litri. Cinque di essi sono pieni di vino e solo uno di essi è pieno di birra. Il mercante tiene per sé il barile di birra e vende tutti i barili di vino a due persone diverse, senza frazionarne il contenuto. Se uno dei due acquirenti ha comprato una ... | [
"Solution:\n\nLa risposta è (D). La somma delle capienze in litri di tutti i barili è $119$, numero che diviso per $3$ dà resto $2$. Poiché la somma dei litri di vino venduti deve essere multipla di $3$, occorre togliere dalla lista delle capacità l'unico valore che diviso per $3$ dà resto $2$, cioè $20$. Si verifi... | Italy | Progetto Olimpiadi di Matematica 2007 GARA di SECONDO LIVELLO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | MCQ | D | |
0ky1 | Problem:
For any positive integers $a$ and $b$ with $b>1$, let $s_{b}(a)$ be the sum of the digits of $a$ when it is written in base $b$. Suppose $n$ is a positive integer such that
$$
\sum_{i=1}^{\left\lfloor\log _{23} n\right\rfloor} s_{20}\left(\left\lfloor\frac{n}{23^{i}}\right\rfloor\right)=103 \text{ and } \sum_{... | [
"Solution:\nFirst we will prove that\n$$\ns_{a}(n)=n-(a-1)\\left(\\sum_{i=1}^{\\infty}\\left\\lfloor\\frac{n}{a^{i}}\\right\\rfloor\\right) .\n$$\nIf $n=\\left(n_{k} n_{k-1} \\cdots n_{1} n_{0}\\right)_{a}$, then the digit $n_{i}$ contributes $n_{i}$ to the left side of the sum, while it contributes\n$$\nn_{i}\\lef... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | null | proof and answer | 81 | |
0icx | Problem:
$S$ is a set of complex numbers such that if $u, v \in S$, then $u v \in S$ and $u^{2}+v^{2} \in S$. Suppose that the number $N$ of elements of $S$ with absolute value at most $1$ is finite. What is the largest possible value of $N$? | [
"Solution:\nFirst, if $S$ contained some $u \\neq 0$ with absolute value $<1$, then (by the first condition) every power of $u$ would be in $S$, and $S$ would contain infinitely many different numbers of absolute value $<1$. This is a contradiction.\n\nNow suppose $S$ contains some number $u$ of absolute value $1$ ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof and answer | 13 | |
08cc | Problem:
Dimostrare che esistono infiniti interi positivi $m$ tali che il numero di fattori primi distinti dispari di $m(m+3)$ è un multiplo di 3. | [
"Solution:\n\nChiamiamo $f(m)=m(m+3)$, e $d(m)$ il numero di fattori primi dispari distinti di $f(m)$. Si ha che $f(m) f(m+1)=f\\left(m^{2}+4 m\\right)$; inoltre, $f(m)$ e $f(m+1)$ hanno 2 come unico fattore primo in comune. Infatti, supponiamo che $p$ sia un primo dispari che divide sia $f(m)$, sia $f(m+1)$. Allor... | Italy | Olimpiade Italiana di Matematica | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
02lg | Problem:
Os bombons misturados - Marta e Carmem ganharam, cada uma, muitos bombons. Elas misturaram os bombons e agora não sabem mais qual o número de bombons que cada uma ganhou. Vamos ajudá-las a descobrir os números sabendo que:
- juntas ganharam 200 bombons;
- cada número é múltiplo de 8;
- Marta se lembra que gan... | [
"Solution:\n\nSejam $x$ o número de bombons que Marta ganhou e $y$ o que Carmem ganhou. Temos $x + y = 200$. Como $x < 100$ então $y \\geq 100$. Por outro lado, $x > \\frac{4}{5} y$ e $y \\geq 100$, concluímos que $x > \\frac{4}{5} \\times 100 = 80$. Logo, $x$ é um inteiro compreendido entre 80 e 100 e múltiplo de ... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Marta 96, Carmem 104 | |
0ei3 | Problem:
V nekem večkotniku je vsota notranjih kotov enaka $4140^{\circ}$. Kolikšno je število diagonal tega večkotnika?
(A) 250
(B) 275
(C) 205
(D) 725
(E) 257 | [
"Solution:\n\nUporabimo obrazec za vsoto notranjih kotov v večkotniku $S_{n} = (n-2) \\cdot 180^{\\circ}$ in izračunamo število stranic (oglišč) iskanega večkotnika. Dobimo $n = 25$. Uporabimo še obrazec za število diagonal $d_{n} = \\frac{n(n-3)}{2}$ in dobimo $d_{25} = 275$."
] | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | MCQ | B | |
0374 | Problem:
Let $k$ be a circle with diameter $AB$ and let $C \in k$ be an arbitrary point. The excircles of $\triangle ABC$ tangent to the sides $AC$ and $BC$ are tangent to the line $AB$ at points $M$ and $N$, respectively. Denote by $O_{1}$ and $O_{2}$ the circumcenters of $\triangle AMC$ and $\triangle BNC$. Prove tha... | [
"Solution:\nSet $AB = c$, $BC = a$, $AC = b$, $p = \\frac{a + b + c}{2}$ and let $O$, $P$ and $Q$ be the midpoints of $AB$, $AM$ and $BN$, respectively.\nSince $OO_{1} \\perp AC$ and $O_{1}P \\perp AM$, we have $\\triangle O_{1}OP \\sim \\triangle ABC$, implying that $\\frac{OO_{1}}{c} = \\frac{OP}{a}$. Further, it... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | null | proof only | null | |
0l6f | Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$ intersects line $BC$ at two distinct points $X$ and $Y$. Prove that $C$ is the midpoint of $XY$. | [] | United States | USAMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
036i | Problem:
Let $a$ be an integer. Find all real numbers $x, y, z$ such that
$$
a(\cos 2 x + \cos 2 y + \cos 2 z) + 2(1-a)(\cos x + \cos y + \cos z) + 6 = 9a
$$ | [] | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Let u = cos x, v = cos y, w = cos z.
- If a = 1: each of u, v, w equals ±1, i.e., x, y, z are each 0 or π modulo 2π (independently).
- If a ≠ 1: u = v = w = −1, i.e., x ≡ y ≡ z ≡ π (mod 2π). | |
0fck | Problem:
¿Existirán 16 números naturales distintos y menores de 100 tales que al colocarlos en las casillas de un tablero $4 \times 4$ el producto de los situados en cada fila sea el mismo y, a su vez, coincida con el de los colocados en cada columna y en las dos diagonales principales?
Si la respuesta es afirmativa, ... | [
"Solution:\n\nVeamos que sí existen.\nConsideremos estos dos conjuntos: $A = (1, 2, 3, 5)$ y $B = \\{7, 11, 13, 17\\}$.\n\nLos 16 productos que se obtienen al multiplicar un número de $A$ por uno de $B$ son todos distintos y menores de 100.\n\nLos colocaremos de forma que, en cada fila, en cada columna y cada dia... | Spain | Fase Local | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Other"
] | null | proof and answer | Yes. One example is the grid with rows:
17, 55, 14, 39
51, 22, 35, 13
85, 7, 33, 26
34, 21, 11, 65
All rows, columns, and main diagonals have common product 510510. | |
0bhv | In the right parallelepiped $ABCD'A'B'C'D'$, with $AB = 12\sqrt{3}$ cm and $AA' = 18$ cm, we consider the points $P \in [AA']$ and $N \in [A'B']$ such that $A'N = 3B'N$. Determine the length of the line segment $[AP]$ such that for any position of the point $M \in [BC]$, the triangle $MNP$ is right angled at $N$.
Dami... | [
"We have $BC \\perp (ABB')$, therefore $BC \\perp PN$. Since $PN \\perp NM$, it follows that $PN \\perp (NBC)$, hence $PN \\perp NB$, that is, triangle $NBP$ is right angled at $N$. Let $AP = x$; we obtain $BP^2 = x^2 + 432$, $PN^2 = (18-x)^2 + 243$ and $BN^2 = 351$. But $BP^2 = PN^2 + BN^2$, and hence $x = 13, 5$ ... | Romania | 65th Romanian Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 13.5 cm | |
04dw | Which number has more divisors in the set of all positive integers, $2013^2$ or $20480$? | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 2013^2 | |
0e9x | Problem:
Poišči najmanjše naravno število, ki ga lahko zapišemo v obliki $3 a^{2} - a b^{2} - 2 b - 4$, kjer sta $a$ in $b$ naravni števili. | [
"Solution:\n\nOdgovor je $2$. Če vzamemo $a = 4$ in $b = 3$, dobimo $3 a^{2} - a b^{2} - 2 b - 4 = 2$. Dovolj je torej pokazati, da enačba $3 a^{2} - a b^{2} - 2 b - 4 = 1$ nima rešitev v naravnih številih.\n\nEnačbo preuredimo v $3 a^{2} - a b^{2} = 2 b + 5$. Ker je desna stran liha, mora biti tudi leva stran liha... | Slovenia | 58. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2 | |
0ct9 | Each cell of a $100 \times 100$ table is colored either in black or in white, so that all columns contain equal numbers of black cells, but no two rows contain equal numbers of black cells. Determine the greatest possible number of pairs of neighboring cells whose colors are distinct. (Two cells are neighboring if they... | [
"Set $2n = 100$. By a counting argument, each column contains $n$ black cells, while the rows contain $0, 1, 2, \\dots, n-1, n+1, n+2, \\dots, 2n$ black cells in some order.\nA row with $i$ black cells contains at most $\\min\\{2i, 2(2n-i)\\}$ dichromatic pairs, so there are at most $2 \\cdot (2 \\cdot 0 + 2 \\cdot... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | 14751 | |
0djl | In an oral exam, there are 10 examiners and 1024 contestants. Each contestant will be asked by each examiner and receive a result of either “pass” or “fail”. It is given that for any two contestants, there is some examiner who will rate one passed and the other failed. Two contestants are said to be “separative” if the... | [
"Number the examiners from $1$ to $10$ and consider the results of contestants as binary sequences of the form $x_1x_2\\ldots x_{10}$, in which $x_i = 0$ or $1$ if the $i$th examiner gave pass or fail response for the corresponding candidate. We can directly construct $24$ binary strings that satisfy the “separativ... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof only | null | |
0gxg | The knight stands in the left lower corner of a $2009 \times 2009$ chessboard and it is known that the first column and the last row are colored in red. The knight moves on the chessboard according to the chess rules but it can't get on the cell which is already colored in red. After every move of the knight we color i... | [
"As it is shown on the fig.15, we can color in red the first four columns and the first four rows of the chessboard. Now consider our chessboard without these four columns and rows. As we can see on the same figure, the fifth move of the knight brings it to the left lower corner of the new reduced $2005 \\times 200... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Yes | |
0b6j | Let $x$ and $y$ be two elements of a ring $R$, such that $xyx \ne 0$ and $(xy)^n = 0$, where $n \ge 2$ is an integer. Show that there exists an element $z$ in $R$, such that $xz \ne 0$ and $xyxz = 0$. | [
"Let $z = (yx)^{n-1}$. Then\n\n$xyxz = x y (yx)^{n-1} z = x (y x)^{n} z = x \\cdot 0 \\cdot z = 0$.\n\nNow, suppose $xz = 0$. Then $x (yx)^{n-1} = 0$. But $xyx \\ne 0$, and $(xy)^n = 0$ implies $(xy)^{n-1} x y = 0$. If $x (yx)^{n-1} = 0$, then $xyx = 0$ (since $n \\ge 2$), which contradicts the hypothesis. Therefor... | Romania | Shortlisted Problems for the Romanian NMO | [
"Algebra > Abstract Algebra > Ring Theory"
] | English | proof only | null | |
0ham | Let $1 < d_1 < d_2 < \ldots < d_{l-1} < d_l < n$ denote divisors of composite natural number $n$, that is not a square of prime number, $l \ge 2$. For which $n$ are there natural numbers $a$, $b$ and $N$, that satisfy the conditions: $d_1 + d_2 = N^a$ and $d_{l-1} + d_l = N^b$? | [
"Notice, that from conditions of the task, it follows, that $d_1 \\cdot d_l = d_2 \\cdot d_{l-1} = n$. Using this equality and first equality from conditions, it is possible to rewrite the second equality from the task:\n$$\nN^b = d_{l-1} + d_l = \\frac{n}{d_2} + \\frac{n}{d_l} = \\frac{n(d_1+d_2)}{d_1d_2} = \\frac... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | Exactly the following n: (i) n has exactly two proper divisors greater than one, i.e., n is either the product of two primes or the cube of a prime (then taking both exponents equal to one works); and (ii) for any pair of twin primes p and p plus two and any integer b at least one, n = 2 · p · (p + 2)^(b − 1) with N = ... | |
0hc7 | Find the largest fraction out of 1010 fractions given below:
$$
\begin{array}{ccccccc}
\frac{1}{2019}, & \frac{1+2018}{2019+2}, & \frac{1+2018+3}{2019+2+2017}, & \frac{1+2018+3+2016}{2019+2+2017+4}, & \\
& \frac{1+2018+3+2016+5}{2019+2+2017+4+2015}, & \dots, & & \\
& & \frac{1+2018+3+2016+...+1009}{2019+2+2017+4+...+10... | [
"Let us split all fractions into two natural groups: those on the odd position and those on the even.\n$$\n\\begin{align*}\na_1 &= \\frac{1}{2019}, \\\\\na_2 &= \\frac{1+2018+3}{2019+2+2017} = \\frac{1+2021}{2019+2019}, \\\\\na_3 &= \\frac{1+2018+3+2016+5}{2019+2+2017+4+2015} = \\frac{1+2021+2}{2019+2019+2}, \\dots... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 2019/2021 | |
04xz | Given triangle $ABC$ and its circumcircle. Point $P$ is midpoint of arc $BAC$. Circle with diameter $CP$ cuts angle bisector of $\angle BAC$ at points $K, L$ (point $K$ lies closer to $A$ than $L$). Point $M$ is symmetric to $L$ with respect to line $BC$. Prove that circumcircle of triangle $BKM$ bisects segment $BC$. | [
"Let $D$ be the midpoint of arc $BC$, $N$ the midpoint of $BC$, $X$ the orthogonal projection of $P$ onto $AC$. Then points $P, X, K, L, N, C$ are concyclic (Fig. 7).\n\n\nFig. 7\n\nWe have $\\angle PNX = \\angle PCA = \\angle PDA$ therefore $XN \\parallel KL$. It follows that $LN = KX$, an... | Czech-Polish-Slovak Mathematical Match | Cesko-Slovacko-Poljsko 2013 | [
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0fcl | Problem:
¿Cuál es el número máximo de vértices de un polígono regular de 21 lados que podemos elegir para que, al trazar los segmentos que los unen entre sí, no haya dos con la misma longitud? | [
"Solution:\n\n\nPor la simetría de la figura, sólo hay 10 distancias distintas.\n\nComo mucho, podremos elegir 5 vértices. Pues, entre cinco puntos no alineados se pueden trazar $C_{5,2} = 10$ segmentos.\n\nNos faltará constatar si con $5$, y con qué $5$, vértices se puede.\n\nLa figura de ... | Spain | Fase Local | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 5 | |
08f3 | Problem:
Maddalena scrive su un foglio tutte le potenze di $2$ da $1$ a $2^{100}$ (estremi inclusi). Quanti dei numeri che ha scritto iniziano per $1$?
(A) 28 o meno
(B) 29
(C) 30
(D) 31
(E) 32 o più. | [
"Solution:\n\nLa risposta è (D). Iniziamo con un'osservazione più generale. Dato un intero positivo $n \\geq 2$, consideriamo la più piccola potenza $2^{a}$ di $2$ che in base $10$ si scriva con almeno $n$ cifre. Siccome $2^{a-1}$ si scrive con al massimo $n-1$ cifre abbiamo $2^{a-1} \\leq 10^{n-1}-1$, e quindi $10... | Italy | Italian Mathematical Olympiad - February Round | [
"Algebra > Intermediate Algebra > Exponential functions",
"Number Theory > Other"
] | null | MCQ | D | |
0e0r | Find all prime numbers $p$ such that $p^2 + 7^3$ is a perfect cube. | [
"Let $p^2 + 7^3 = n^3$. Then $p^2 = n^3 - 7^3 = (n-7)(n^2 + 7n + 49)$. Obviously, $n > 7$ and $n - 7 < n^2 + 7n + 49$. So there is only one possible case: $n - 7 = 1$ and $n^2 + 7n + 49 = p^2$. We conclude that $n = 8$ and $p^2 = 169$, so $p = 13$. The only prime number with the required property is $p = 13$."
] | Slovenia | National Math Olympiad | [
"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 | 13 | |
01i5 | Let $AD$ be the $A$-altitude of an acute-angled triangle $ABC$. The internal bisector of angle $DAC$ intersects $BC$ at $K$. Let $L$ be the projection of $K$ onto $AC$. Let $M$ be the intersection point of $BL$ and $AD$. Let $P$ be the intersection point of $MC$ and $DL$. Prove that $PK \perp AB$. | [
"**Solution.** Since $\\angle IFK = 90^\\circ$, then $IK$ is the diameter of the circumcircle of $CFI$, hence also $\\angle ICK = 90^\\circ$. Similarly is $IL$ the diameter of the circumcircle of $BGI$ and $\\angle IBL = 90^\\circ$. Therefore are the lines $CK$ and $GL$ parallel, also $BL$ and $FK$ are parallel.\nL... | Baltic Way | Baltic Way 2021 Shortlist | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
00lb | On the occasion of the 47th Mathematical Olympiad 2016 the numbers $47$ and $2016$ are written on the blackboard. Alice and Bob play the following game. Alice begins and in turns they choose two numbers $a$ and $b$ with $a > b$ written on the blackboard, whose difference $a - b$ is not yet written on the blackboard and... | [
"We consider the set $B$ of the numbers on the blackboard at the end of the game. It is clear that $B \\subseteq \\{1, \\dots, 2016\\}$. Let $m = \\min B$ and $n \\in B$. We claim that $m \\mid n$. Otherwise, write $n = qm + r$ with $0 < r < m$. By induction on $k$, we have $n - k m \\in B$ for $0 \\le k \\le q$ (b... | Austria | Regional Competition | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
03aw | Do there exist 6-digit numbers of the form $n^k$, $k \ge 3$, such that difference between the number, formed by the last three digits, and the number, formed by the first three digits, is equal to 4? | [
"If $a = \\overline{a_1a_2a_3a_4a_5a_6}$ satisfies the second condition, then $10^5 < a < 10^6$ and $\\overline{a_4a_5a_6} = \\overline{a_1a_2a_3} + 4$. It follows that\n\n$$\na = 1000\\overline{a_1a_2a_3} + \\overline{a_4a_5a_6} = 1001\\overline{a_1a_2a_3} + 4.\n$$\n\nSo, we are looking for 3-digit numbers $A = \\... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis,... | English | proof and answer | No, there do not exist such six-digit numbers. | |
0g92 | 求證：從任何包含 2047 個正整數的集合中，必能取出 1024 個正整數，使得這 1024 個正整數的總和可被 1024 整除。 | [
"假設 $n = 2^k$, 其中 $k$ 為正整數。我們以下將用數學歸納法證明，從任何包含 $2n - 1$ 個正整數的集合中，總能取出 $n$ 個整數，使得這 $n$ 個整數的總和可被 $n$ 整除。當 $k = 10$ 時，即為題目所求。\n\n1. 當 $k=1$ 時, $n=2$, $2n - 1 = 3$。在任何三個正整數中, 一定有兩個數的奇偶性相同, 其和可被 2 整除。\n\n2. 假設 $n = 2^k$ 時成立。考慮 $n = 2^{k+1}$。因為 $2n - 1 = 2^{k+2} - 1 > 2^{k+1} - 1$，由歸納假設知, 存在 $2^k$ 個正整數, 使得其和可被 $2^k$ 整除。再... | Taiwan | 二〇一五數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0kv8 | Problem:
Let $a_{0}, a_{1}, a_{2}, \ldots$ be an infinite sequence where each term is independently and uniformly random in the set $\{1,2,3,4\}$. Define an infinite sequence $b_{0}, b_{1}, b_{2}, \ldots$ recursively by $b_{0}=1$ and $b_{i+1}=a_{i}^{b_{i}}$. Compute the expected value of the smallest positive integer $... | [
"Solution:\nDo casework on what $a_{0}$ is.\n\nIf $a_{0}=1$ then $k=1$.\n\nIf $a_{0}=4$ then $k=2$.\n\nIf $a_{0}=3$ then\n- if $a_{1}=1$, then $k=2$\n- if $a_{1}=2$ or $4$, then $k=3$\n- if $a_{1}=3$, then you make no progress.\nSo in expectation it requires $E=(2+3+(E+1)+3)/4 \\Longrightarrow E=3$.\n\nIf $a_{0}=2$... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | 35/16 | |
021i | Problem:
O triângulo $ABC$ é isósceles de base $BC$ e $\angle BAC = 48^\circ$. Os pontos $D$ e $E$ estão sobre os lados $AB$ e $AC$, respectivamente, tais que $\angle DAC = 9^\circ$ e $\angle EBC = 33^\circ$. Determine a medida do ângulo $CDE$. | [] | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | null | |
0ceh | For $a$ and $b$ real numbers, define
$$
E(a,b) = \sqrt{\left(\frac{1}{2} - a\right)^2 + \left(\frac{1}{2} - b\right)^2} + \sqrt{\left(\frac{1}{2} + a\right)^2 + \left(\frac{1}{2} + b\right)^2}.
$$
a) Prove that $E(a, b) \ge \sqrt{2}$, for every real numbers $a, b$, and equality takes place if and only if $a = b \in [-... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | sqrt(10) | |
0k1p | Problem:
Compute the value of
$$
\frac{\cos 30.5^{\circ}+\cos 31.5^{\circ}+\ldots+\cos 44.5^{\circ}}{\sin 30.5^{\circ}+\sin 31.5^{\circ}+\ldots+\sin 44.5^{\circ}}
$$ | [
"Solution:\nConsider a 360-sided regular polygon with side length $1$, rotated so that its sides are at half-degree inclinations (that is, its sides all have inclinations of $0.5^{\\circ}, 1.5^{\\circ}, 2.5^{\\circ}$, and so on). Go to the bottom point on this polygon and then move clockwise, numbering the sides $1... | United States | HMMT February 2018 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | final answer only | 2 - sqrt(2) - sqrt(3) + sqrt(6) | |
01q2 | $2n$ girls and $2n$ boys take part in a dancing party. It is known that Bob has a dance with every girl, and Ann has a dance with every boy. Moreover, for any two girls the number of the boys who have a dance with exactly one of these two girls is equal to $n$.
Prove that
a) any girl, except for Ann, has a dance with... | [
"We use the solution of Problem A.8.\n\na) Let Ann get number $1$. Each vector $S_i$, $i \\neq 1$ differs from $S_1$ at exactly $n$ positions and all entries of $S_1$ are equal to $1$. Therefore, $S_i$ have exactly $n$ entries equal to $1$, which proves the statement.\n\nb) The same proof as in a). We consider the ... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Vectors"
] | English | proof only | null | |
0aud | Problem:
Let $f(x) = a^{x} - 1$. Find the largest value of $a > 1$ so that if $0 \leq x \leq 3$, then $0 \leq f(x) \leq 3$. | [] | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 4^{1/3} | |
0ilw | Problem:
Determine the largest integer $n$ such that $7^{2048}-1$ is divisible by $2^{n}$. | [
"Solution:\nAnswer: 14. We have\n$$\n7^{2048}-1=(7-1)(7+1)\\left(7^{2}+1\\right)\\left(7^{4}+1\\right) \\cdots\\left(7^{1024}+1\\right)\n$$\nIn the expansion, the eleven terms other than $7+1$ are divisible by $2$ exactly once, as can be checked easily with modulo $4$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 14 | |
03jd | Problem:
Let $u_{1}, u_{2}, u_{3}, \ldots$ be a sequence of integers satisfying the recurrence relation $u_{n+2} = u_{n+1}^{2} - u_{n}$. Suppose $u_{1} = 39$ and $u_{2} = 45$. Prove that $1986$ divides infinitely many terms of the sequence. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0buh | Problem:
Demonstrați că numerele naturale, care împărțite în $\mathbb{N}$ la 102 dau restul 78, sunt divizibile cu 3. | [
"Solution:\n\nFie $n \\in \\mathbb{N}, n: 102 = c$ rest 78.\n\nAplicând teorema împărțirii cu rest obținem $n = 102c + 78 = 3 \\cdot (34c + 26)$, de unde numărul $n$ este divizibil cu 3."
] | Romania | Olimpiada de Matematică - Etapa Locală | [
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
0gly | Let $ABC$ be a triangle with $\angle A = 60^\circ$, orthocenter $H$ and centroid $G$. Let $A'$ be the point on the circumcircle of $\triangle ABC$ diametrically opposite to $A$. Let $M$ be the midpoint of $BC$. The line through $M$ parallel to $AA'$ intersects $BH, CH$ at $K, L$, respectively. $P$ and $Q$ are distinct ... | [] | Thailand | T3MO 2017 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Rotation",
"Ge... | English | proof only | null | |
0fi3 | Problem:
Sean $a$, $b$ y $c$ tres números reales. Se consideran las funciones
$$
f(x) = a x^{2} + b x + c \quad \text{y} \quad g(x) = c x^{2} + b x + a
$$
Sabiendo que
$$
|f(-1)| \leq 1, \quad |f(0)| \leq 1, \quad y \quad |f(1)| \leq 1
$$
probar que si $-1 \leq x \leq 1$, entonces $|f(x)| \leq 5/4$ y $|g(x)| \leq 2$. | [
"Solution:\nPodemos conseguir coeficientes $A$, $B$, $C$, tales que se tenga idénticamente\n$$\nf(x) = A x(x+1) + B x(x-1) + C\\left(x^{2} - 1\\right)\n$$\nParticularizando para $x = 1, -1, 0$ y resolviendo el sistema queda\n$$\nf(x) = \\frac{f(1)}{2} x(x+1) + \\frac{f(-1)}{2} x(x-1) + f(0)\\left(1 - x^{2}\\right)\... | Spain | Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0gzh | Olesya and Andrew have thrown a dice once. Find the probability that the number of points that was thrown out by Olesya is strictly greater than the number of points that was thrown out by Andrew. Justify the answer. | [
"There are $6^2 = 36$ different combinations. Let us calculate the number of cases that satisfy the statement.\n\nOlesya was thrown out $1$, that is impossible.\n\nOlesya was thrown out $2$, the only possible case is $1$.\n\nOlesya was thrown out $3$, possible cases are $1$ and $2$.\n\nOlesya was thrown out $4$, po... | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Statistics > Probability > Counting Methods > Other"
] | English | proof and answer | 5/12 | |
0ffz | Problem:
Consideremos tres números naturales $a, b, c$ tales que la razón
$$
\frac{a+b+c}{a b c}
$$
sea el inverso de un número $k$ entero y positivo. Se pide demostrar:
a) $a^{3}+b^{3}+c^{3}$ no es primo.
b) Para cada $k \in \mathbb{N}$ existen ternas de naturales $a, b, c$ que cumplen las condiciones. | [
"Solution:\n\nTenemos la identidad\n$$\na^{3}+b^{3}+c^{3}=\\left((a+b+c)^{2}-3(a b+b c+a c)\\right)(a+b+c)+3 a b c\n$$\npero como\n$$\n\\frac{a+b+c}{a b c}=\\frac{1}{k} \\Longrightarrow a b c=k(a+b+c)\n$$\nla igualdad anterior se puede escribir\n$$\na^{3}+b^{3}+c^{3}=(a+b+c)\\left[(a+b+c)^{2}-3(a b+b c+a c)+3 k\\ri... | Spain | OME 21 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
05yk | Problem:
Soient $A, B, C, D$ et $E$ cinq points dans cet ordre sur un cercle tels que $A E = D E$. Soit $P$ le point d'intersection de $(A C)$ et $(B D)$. Soit $Q$ le point de la demi-droite $[B A)$ tel que $A Q = D P$. Soit $R$ le point de la demi-droite $[C D)$ tel que $D R = A P$. Montrer que les droites $(P E)$ et ... | [
"Solution:\nOn se place dans le cas de la figure suivante:\n\nDans cet énoncé on a de nombreuses égalités de longueur $(A Q = D P, D R = A P, A E = D E)$. Comme sur la figure, il ne faut pas hésiter à colorier les longueurs égales pour voir des potentiels triangles isométriques, et avancer ... | France | Préparation Olympique Française de Mathématiques - ENVOI 5 : Pot-POURRI | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
05c2 | The coefficients of the polynomial
$$
P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0
$$
are real numbers such that $a_i = a_{n-i}$ for every $i = 0, 1, \dots, n$, and $a_n \neq 0$. Let $x_1, x_2, \dots, x_k$ be all the real roots of the polynomial $P(x)$ without repetitions.
a. Prove that
$$
|x_1| + |x_2| + \d... | [
"a. Let $c$ be any root of $P(x)$. As $P(0) = a_0 = a_n \\neq 0$, we must have $c \\neq 0$. Notice that\n$$\n\\begin{aligned}\nP\\left(\\frac{1}{c}\\right) &= a_n \\left(\\frac{1}{c}\\right)^n + a_{n-1} \\left(\\frac{1}{c}\\right)^{n-1} + \\dots + a_1 \\left(\\frac{1}{c}\\right) + a_0 \\\\\n&= \\left(\\frac{1}{c}\\... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | a. The sum of the absolute values of the distinct real roots is at least the number of such roots. b. No; the inequality need not be strict. For example, (x−1)²(x+1) has real roots 1 and −1 with sum of absolute values equal to 2. | |
029b | Problem:
Usando a fatoração da diferença de quadrados, $x^{2}-y^{2}=(x-y)(x+y)$, podemos escrever
$$
x^{2^{n}}-y^{2^{n}}=\left(x^{2^{n-1}}+y^{2^{n-1}}\right)\left(x^{2^{n-1}}-y^{2^{n-1}}\right) \text{.}
$$
a) Explique por que $3^{2^{2018}}-2^{2^{2018}}$ pode ser escrito como produto de 2018 inteiros maiores que 1 e di... | [
"Solution:\n\na) Usando a fatoração da diferença de quadrados, temos:\n$$\n\\begin{aligned}\nx^{2^{n}}-y^{2^{n}} & =\\left(x^{2^{n-1}}+y^{2^{n-1}}\\right)\\left(x^{2^{n-1}}-y^{2^{n-1}}\\right) \\\\\nx^{2^{n-1}}-y^{2^{n-1}} & =\\left(x^{2^{n-2}}+y^{2^{n-2}}\\right)\\left(x^{2^{n-2}}-y^{2^{n-2}}\\right) \\\\\nx^{2^{n... | Brazil | null | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 3^{2^{2017}} - 2^{2^{2017}} | |
07gm | Let us denote by $\Omega(n)$ and $\omega(n)$ the largest and the smallest prime divisors of $n$, respectively. Alireza and Amin decided to play the following game. Starting with Alireza, he chooses 1400 polynomials with integer coefficients. Then, Amin chooses 700 polynomials among them and denotes their sets by $A$ an... | [
"Let us denote by $A_1, \\dots, A_N$ all the 700-element subsets of $\\{1, 2, \\dots, 1400\\}$. Let $N = \\binom{1400}{700}$. Now, choose 1400 polynomials $P_1, \\dots, P_{1400}$ of the following form:\n$$\nP_i(x) = 1 + 10^{10} \\prod_{1 \\le j \\le N,\\ j \\in A_i} (x - j).\n$$\nWe now claim that if Alireza choose... | Iran | 38th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Other"
] | null | proof and answer | Alireza | |
0igs | Problem:
Let $k$ be an integer such that $k \mid a$ and $k \mid b$. Prove that if an $m \times n$ rectangle is $(a, b)$-tileable, then $2k \mid m$ or $2k \mid n$. | [
"Solution:\n\nWe prove the following lemma.\n\nLemma. Let $k$ be a positive integer such that $k \\mid a$ and $k \\mid b$. Then an $m \\times n$ rectangle is $(a, b)$-tileable if and only if an $m' \\times n'$ rectangle is $\\left(\\frac{a}{k}, \\frac{b}{k}\\right)$-tileable for $\\left\\lfloor\\frac{m}{k}\\right\\... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0ial | Problem:
In triangle $ABC$, $\angle ABC = 50^{\circ}$ and $\angle ACB = 70^{\circ}$. Let $D$ be the midpoint of side $BC$. A circle is tangent to $BC$ at $B$ and is also tangent to segment $AD$; this circle intersects $AB$ again at $P$. Another circle is tangent to $BC$ at $C$ and is also tangent to segment $AD$; this... | [
"Solution:\n\nSuppose the circles are tangent to $AD$ at $E, F$, respectively; then, by equal tangents, $DE = DB = DC = DF \\Rightarrow E = F$ (as shown). So, by the Power of a Point Theorem, $AP \\cdot AB = AE^{2} = AF^{2} = AQ \\cdot AC \\Rightarrow AP / AQ = AC / AB \\Rightarrow \\triangle APQ \\sim \\triangle A... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 70 | |
0jkv | Problem:
Find all ordered pairs $(a, b)$ of complex numbers with $a^{2}+b^{2} \neq 0$, $a+\frac{10 b}{a^{2}+b^{2}}=5$, and $b+\frac{10 a}{a^{2}+b^{2}}=4$. | [
"Solution:\n\nAnswer: $(1,2),(4,2),\\left(\\frac{5}{2}, 2 \\pm \\frac{3}{2} i\\right)$\n\nFirst, it is easy to see that $a b \\neq 0$. Thus, we can write\n\n$$\n\\frac{5-a}{b}=\\frac{4-b}{a}=\\frac{10}{a^{2}+b^{2}}.\n$$\n\nThen, we have\n$$\n\\frac{10}{a^{2}+b^{2}}=\\frac{4 a-a b}{a^{2}}=\\frac{5 b-a b}{b^{2}}=\\fr... | United States | HMMT 2014 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | [(1, 2), (4, 2), (5/2, 2 + 3i/2), (5/2, 2 - 3i/2)] | |
031u | Problem:
Is it true that for any permutation $a_{1}, a_{2}, \ldots, a_{2002}$ of $1,2, \ldots, 2002$ there are positive integers $m$ and $n$ of the same parity such that $1 \leq m<n \leq 2002$ and $a_{m}+a_{n}=2 a_{\frac{m+n}{2}}$? | [
"Solution:\n\nThe answer is no. We shall prove by induction that for any $k \\geq 3$ there is a permutation $a_{1}, a_{2}, \\ldots, a_{k}$ of $1,2, \\ldots, k$ such that\n$$\na_{m}+a_{n} \\neq 2 a_{\\frac{m+n}{2}} \\text{ for any } 1 \\leq m<n \\leq k \\text{ of the same parity. }\n$$\nFor $k=3$ and $k=4$ take the ... | Bulgaria | Team selection test for 44. IMO | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | No | |
08v3 | How many positive integers $n$ less than or equal to $1000$ are there for which the sum of digits of $5n$ is the same as the sum of digits of $n$? | [
"$1000$ does not satisfy the requirement, so let us consider only positive integers $n \\le 999$. One can represent such an integer in the form $100a + 10b + c$ where $a, b, c$ are 1-digit non-negative integers. Since the number obtained by multiplying a 1-digit integer by $5$ is at most $45$ and its one's digit is... | Japan | Japan Junior Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 91 | |
00ya | Problem:
The Wonder Island is inhabited by Hedgehogs. Each Hedgehog consists of three segments of unit length having a common endpoint, with all three angles between them equal to $120^{\circ}$ (see Figure 3). Given that all Hedgehogs are lying flat on the island and no two of them touch each other, prove that there i... | [
"Solution:\n\nIt suffices to prove that if the distance between the centres of two Hedgehogs is less than $0.2$, then these Hedgehogs intersect. To show this, consider two Hedgehogs with their centres at points $O$ and $M$, respectively, such that $|OM| < 0.2$. Let $A$, $B$ and $C$ be the endpoints of the needles o... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0euk | Find all pairs of positive integers $m$, $n$ satisfying the equation $3^m - 7^n = 2$. | [
"Since $3^m - 2 \\equiv 0 \\pmod{7}$, one may easily show that $m \\equiv 2 \\pmod{6}$. If $m = 2$ then $n = 1$, which is a solution of the equation.\n\nAssume that $m = 2s \\ge 4$. Note that $2 + 7^n$ is divisible by $27$ in this case and\n$$\n7^9 \\equiv 1 \\pmod{27}, \\quad 7^1 \\equiv -2 \\pmod{27}.\n$$\nHence ... | South Korea | 22nd Korean Mathematical Olympiad Final Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof and answer | (m, n) = (2, 1) | |
0cq2 | A convex $2011$-gon is drawn on a blackboard. Pete draws its diagonals one by one so that each diagonal $d$ should intersect (by an interior point) not more than one diagonal drawn before $d$. Find the maximal possible number of diagonals that Pete can draw according to these rules. (S. Berlov)
На доске нарисован выпу... | [
"**Первое решение.** Покажем, что в выпуклом $n$-угольнике максимальное количество диагоналей, которое можно провести указанным способом, равно $2n - 6$; при $n = 2011$ тогда получится указанный ответ. Пусть $A_1A_2...A_n$ — данный многоугольник. Тогда Петя может провести последовательно диагонали $A_2A_4, A_3A_5, ... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F"
] | English, Russian | proof and answer | 4016 | |
05e7 | Problem:
For a sequence $a_{1} < a_{2} < \cdots < a_{n}$ of integers, a pair $(a_{i}, a_{j})$ with $1 \leq i < j \leq n$ is called interesting if there exists a pair $(a_{k}, a_{l})$ of integers with $1 \leq k < l \leq n$ such that
$$
\frac{a_{l} - a_{k}}{a_{j} - a_{i}} = 2
$$
For each $n \geq 3$, find the largest pos... | [
"Solution:\n\nAnswer. $\\frac{1}{2}(n-1)(n-2)+1$.\n\nConsider the numbers $a_{i} = 2^{i}$ for $2 \\leq i \\leq n$ and $a_{1} = 0$, and choose any pair $(i, j)$ with $1 \\leq i < j \\leq n$.\nIf $i = 1$, the pair $(1, j)$ is interesting for all $2 \\leq j \\leq n-1$, as $\\frac{a_{j+1} - a_{1}}{a_{j} - a_{1}} = \\fr... | European Girls' Mathematical Olympiad (EGMO) | EGMO 2024 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (n-1)(n-2)/2 + 1 | |
0iej | Problem:
Let $S=\{s_{0}, \ldots, s_{n}\}$ be a finite set of integers, and define $S+k=\{s_{0}+k, \ldots, s_{n}+k\}$. We say that $S$ and $T$ are equivalent, written $S \sim T$, if $T=S+k$ for some $k$. Given a (possibly infinite) set of integers $A$, we say that $S$ tiles $A$ if $A$ can be partitioned into subsets equ... | [
"Solution:\nThis can be done by brute force. Alternatively, note that if $P(x)$ and $Q(x)$ are polynomials with coefficients either $0$ or $1$ with $P(x) Q(x)=x+x^{2}+\\cdots+x^{12}$, then the set consisting of the exponents of nonzero terms in $P$ tiles $A$. Either way, we find that $S$ is one of the following: $\... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | {1}, {1,2}, {1,3}, {1,4}, {1,7}, {1,2,3}, {1,3,5}, {1,5,9}, {1,2,3,4}, {1,2,7,8}, {1,4,7,10}, {1,2,3,4,5,6}, {1,2,3,7,8,9}, {1,2,5,6,9,10}, {1,3,5,7,9,11}, {1,2,3,4,5,6,7,8,9,10,11,12} | |
0dbv | Quadrilateral $ABCD$ is circumscribed about a circle $\Lambda$ and $K$, $L$, $M$, $N$ are points of tangency of sides $AB$, $BC$, $CD$, $DA$ with $\Lambda$ respectively. Let $S$ be the intersection point of lines $KM$ and $LN$. If quadrilateral $SKBL$ is cyclic then show that $SNDM$ is also cyclic. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformat... | English | proof only | null | |
06ug | Let $R$ and $S$ be distinct points on circle $\Omega$, and let $t$ denote the tangent line to $\Omega$ at $R$. Point $R'$ is the reflection of $R$ with respect to $S$. A point $I$ is chosen on the smaller arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $ISR'$ intersects $t$ at two different points. D... | [
"In the circles $\\Omega$ and $\\Gamma$ we have $\\angle JRS = \\angle JIS = \\angle AR'S$. On the other hand, since $RA$ is tangent to $\\Omega$, we get $\\angle SJR = \\angle SRA$. So the triangles $ARR'$ and $SJR$ are similar, and\n$$\n\\frac{R'R}{RJ} = \\frac{AR'}{SR} = \\frac{AR'}{SR'}.\n$$\nThe last relation,... | IMO | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ko6 | Problem:
Let $ABC$ be a triangle with $\angle A = 60^\circ$. Line $\ell$ intersects segments $AB$ and $AC$ and splits triangle $ABC$ into an equilateral triangle and a quadrilateral. Let $X$ and $Y$ be on $\ell$ such that lines $BX$ and $CY$ are perpendicular to $\ell$. Given that $AB = 20$ and $AC = 22$, compute $XY$... | [
"Solution:\n\nLet the intersection points of $\\ell$ with $AB$ and $AC$ be $B'$ and $C'$. Note that $AB' + AC' = 2B'C'$, $BB' = 2XB'$, and $CC' = 2YC'$. Adding gives us\n$$\nAB + AC = AB' + AC' + BB' + CC' = 2\\left(B'C' + XB' + YC'\\right) = 2XY\n$$\nThus, $XY = \\frac{20 + 22}{2} = 21$."
] | United States | HMMT February 2022 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 21 | |
00a7 | Find all $n \in \mathbb{N}$ that can be represented in the form $n = [a,b] + [b,c] + [c,a]$ with $a, b, c \in \mathbb{N}$. Here $[u,v]$ denotes the least common multiple of $u$ and $v$. | [
"All $n \\in \\mathbb{N}$ are representable except the powers of $2$. Set $f(a,b,c) = [a,b] + [b,c] + [c,a]$. Take an arbitrary $k \\in \\mathbb{N}$ and let $a = k$, $b = c = 1$ to obtain $f(k,1,1) = 2k + 1$. Hence all odd $n$, $n \\ge 3$, are representable. If $n$ is representable then so is $2n$\n\nbecause $[2u, ... | Argentina | Argentine National Olympiad 2015 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All natural numbers that are not powers of 2 | |
05se | Problem:
Soit $x$ et $y$ deux nombres réels. On pose
$$
M=\max \{x y+1, x y-x-y+3,-2 x y+x+y+2\} .
$$
Démontrer que $M \geqslant 2$, et déterminer les cas d'égalité. | [
"Solution:\n\nParmi les trois nombres $x y+1$, $x y-x-y+3$ et $-2 x y+x+y+2$, on note $K$ le plus petit, $L$ le deuxième plus petit et $M$ le plus grand. Alors $K \\leqslant L \\leqslant M$ donc\n$$\n3 M \\geqslant K+L+M=(x y+1)+(x y-x-y+3)+(-2 x y+x+y+2)=6,\n$$\nce qui signifie que $M \\geqslant 2$.\n\nEn outre, s... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | M ≥ 2, with equality if and only if x = y = 1 | |
0gc3 | 試求所有正整數對 $(m, n)$, 滿足 $m$ 和 $(n + 1)$ 互質, 且
$$
\sum_{k=1}^{n} \frac{m^{k+1}}{k+1} \binom{n}{k}
$$
為整數。 | [
"There exists no such $(m, n)$.\nSuppose $(m, n)$ is a solution. By the fact that $(n + 1) \\binom{n}{k} = (k + 1) \\binom{n+1}{k+1}$,\nwe have\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} \\frac{m^{k+1}}{k+1} \\binom{n}{k} &= \\sum_{k=1}^{n} \\frac{m^{k+1}}{n+1} \\binom{n+1}{k+1} = \\frac{1}{n+1} \\sum_{l=0}^{n+1} m^l \... | Taiwan | 二〇一八數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"A... | null | proof and answer | There exists no such (m, n). | |
0hvm | Problem:
At a certain school, there are 6 subjects offered, and a student can take any combination of them. It is noticed that for any two subjects, there are fewer than 5 students taking both of them and fewer than 5 students taking neither. Determine the maximum possible number of students at the school. | [
"Solution:\n\nLet us count in two ways the number of ordered pairs $\\left(s,\\left\\{c_{1}, c_{2}\\right\\}\\right)$, where $s$ is a student, $\\left\\{c_{1}, c_{2}\\right\\}$ is an unordered pair of two distinct courses, and $s$ is either taking both $c_{1}$ and $c_{2}$ or neither $c_{1}$ nor $c_{2}$.\n\nFirst, w... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 20 |
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